Untitled
CHAPTER 2
Protein Turnover
and Amino Acid Catabolism
3
Organisms derive energy from both stored and exogenous fuels. The catabolism of carbohydrates (Chapter 16) and fats (Chapter 22) have been discussed in previous chapters. In Chapter 23, the authors explain the role of proteins in energy metabolism. Although proteins are not stored as fuels per se as carbohydrates and fats are, supplies of proteins in excess of those needed to provide biosynthetic precursors are degraded for energy or are converted into fats or carbohydrates. Most of the amino groups of surplus amino acids are converted into urea through the urea cycle, whereas their carbon skeletons are transformed into acetyl CoA, pyruvate, or one of the citric acid cycle intermediates. The chapter begins with a discussion of the process of ubiquination, by which proteins are targeted for degradation. The N-terminal amino acid of a protein strongly determines its half-life. When tagged by ubiquitin, a large protease complex called the proteosome carries out the degradation of the protein. The proteosome then cleaves off the ubiquitin intact, so it can be recycled.
Once a protein is cleaved into individual amino acids, the amino acids are ei ther incorporated into newly synthesized proteins or degraded to specific compounds for entry into an energy transduction pathway. If entry into an energy transduction pathway is its fate, the nitrogen(s) must first be removed. The a-amino groups of most amino acids are transferred to a-ketoglutarate to form glutamate by transamination (catalyzed by aminotransferases), and the a-amino group of glutamate is then converted to ammonia by an oxidative deamination. The authors describe the reaction mechanism of aminotransferases and the role the coenzyme pyridoxal phosphate (PLP) plays in this enzyme and others. The urea cycle is introduced next, which carries out the condensation of ammonia, the a-amino group of aspartate, and CO2 to
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CHAPTER 23 form urea—a nontoxic excretory product of nitrogen in higher animals. The urea cycle is linked to the citric acid cycle (Chapter 17) due to its production of the citric acid cycle intermediate fumarate.
Because there are 20 amino acids, the catabolic pathways of their carbon skeletons are numerous and of varied types. The authors describe how the carbon atoms of each amino acid are funneled into one or more of seven primary products. Two of these, acetyl CoA and acetoacetyl CoA, can be converted to ketone bodies (Chapter 22), and the remaining five can be converted into glucose (Chapter 16) all of which can be oxidized in energy-generating pathways. The two groups of products lead to the glycogenic-ketogenic classification of the amino acids. The chapter ends emphasizing the importance of carrying out amino acid catabolism by examining the pathological consequences of defects or deficiencies in some of the enzymes involved in catabolism of amino acids and the synthesis of urea.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
INTRODUCTION
1. State the fate of exogenously supplied amino acids that are not used for protein synthesis.
Proteins Are Degraded to Amino Acids (Text Section 23.1) 2. Contrast the effect of different N-terminal amino acids on the half-lives of proteins in yeast.
Protein Turnover Is Tightly Regulated (Text Section 23.2) 3. Explain the functions of the three enzymes (E1, E2, and E3) that participate in the attachment of ubiquitin to proteins targeted for degradation.
4. List the advantages of poly-ubiquination of proteins targeted for degradation.
5. Describe the subunit structure and function of the 26S proteosome.
6. Explain how protein degradation plays a role in regulation of NF-kB.
The First Step in Amino Acid Degradation Is the Removal of Nitrogen
(Text Section 23.3) 7. Name the major organ of amino acid degradation in mammals.
8. Describe the reactions catalyzed by the aminotransferases (transaminases) and state the major function of these reactions.
9. Write the equations for the transamination reactions catalyzed by aspartate aminotrans-ferase and alanine aminotransferase.
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
409
10. Describe the reaction catalyzed by glutamate dehydrogenase, outline its regulation, and state its major function. Note that the participation of NAD+ in the reaction links nitrogen metabolism and energy generation.
11. Recognize the structures of pyridoxal phosphate (PLP) and pyridoxamine phosphate (PMP), indicate the reactive functional groups on each, and name the dietary precursor of PLP.
12. Describe the aminotransferase reaction mechanism and explain the involvement of a Schiff base and PLP. List the other kinds of reactions catalyzed by PLP and describe the common features of PLP catalysis. Define stereoelectronic control.
13. Explain how the a-amino groups of serine and threonine can be directly converted into NH +
4 .
Ammonium Ion Is Converted into Urea in Most Terrestrial Vertebrates
(Text Section 23.4) 14. Define the terms ureotelic, uricotelic, and ammonotelic.
15. Name the molecule that brings nitrogen and carbon into the urea cycle.
16. Name the enzymes and the other components of the urea cycle, note their intracellular locations, and indicate the molecular connection between this cycle and the citric acid cycle. Account for the ATP requirement of the cycle.
17. Distinguish the structural and functional properties of the two isozymes of carbamoylphosphate synthetase in mammals.
18. Explain how deficiencies in several different enzymes of the urea cycle give rise to hy-perammonemia.
Carbon Atoms of Degraded Amino Acids Emerge
as Major Metabolic Intermediates (Text Section 23.5) 19. State the strategy used by humans for catabolizing the carbon atom skeletons of amino acids and name the seven major metabolic products formed.
20. Describe the basis for the glycogenic-ketogenic designation of the amino acids and classify each amino acid accordingly. Appreciate the limitations of this classification.
21. List the amino acids that give rise to pyruvate, oxaloacetate, fumarate, succinyl CoA, acetylCoA, and acetoacetyl CoA respectively.
22. List the amino acids that give rise to a-ketoglutarate. Describe the role of tetrahydrofolate in one of the conversions.
Inborn Errors in Metabolism Can Disrupt Amino Acid Degradation
(Text Section 23.6) 23. Describe alcaptonuria. Relate Garrod’s hypothesis concerning this disorder.
24. Explain the biochemical bases of phenylketonuria and maple syrup urine disease. Describe some of the consequences of these diseases.
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CHAPTER 23
SELF-TEST
Introduction 1. Which of the following answers complete the sentence correctly? Surplus dietary amino acids may be converted into (a) proteins.
(b) fats.
(c) ketone bodies.
(d) glucose.
(e) a variety of biomolecules for which they are precursors.
Proteins Are Degraded to Amino Acids 2. Order the following events in the correct sequence in the digestion of dietary proteins.
(a) proteolysis by peptidases (b) proteolysis by pepsin in stomach (c) release of free amino acids into bloodstream (d) proteolysis in lumen of intestine by aminopeptidases (e) acidic denaturation of proteins in stomach.
(f) transport of free amino acids and di- and tri-peptides into intestinal cells Protein Turnover Is Tightly Regulated 3. Which of the following are features of the 26S proteosome?
(a) consists of a 20S catalytic subunit and a 19S regulatory subunit (b) releases only free amino acids as products (c) digests ubiquitin along with the protein to which it is attached (d) contains six ATPases The First Step in Amino Acid Degradation Is the Removal of Nitrogen 4. Which of the following compounds serves as an acceptor for the amino groups of many amino acids during catabolism?
(a) glutamine (c) a-ketoglutarate (b) asparagine (d) oxalate
5. Which of the following answers completes the sentence correctly? The removal of a-amino groups from amino acids for conversion to urea in animals may occur by (a) transamination.
(c) oxidative deamination.
(b) reductive deamination.
(d) transamidation.
6. Which of the following amino acids have their a-amino groups removed by dehy dratases?
(a) histidine (d) glutamine (b) tryptophan (e) threonine (c) serine (b) aspartate and a-ketoglutarate.
(c) alanine and oxaloacetate.
(d) alanine and a-ketoglutarate.
8. Explain the role of pyridoxal phosphate in aminotransferase reactions. Be sure to de scribe the Schiff base and the ketimine that are involved in the mechanism.
Ammonium Ion Is Converted into Urea in Most Terrestrial Vertebrates 9. Considering all forms of life, which of the following are major excretory forms of the a-amino groups of amino acids?
(a) urea (c) ammonia (b) uracil (d) uric acid
10. How many moles of ATP are required to condense two moles of nitrogen and one mole of CO2 into one mole of urea via the urea cycle? How many high-energy bonds are used in this process? Do both atoms of nitrogen enter the cycle as NH + 4 ?
11. What would be the net effect of linking the following two transamination reactions, and why would such coupling be useful when excess proteins were being catabolized for energy generation?
Alanine + a-ketoglutarate pyruvate + glutamate
Oxaloacetate + glutamate aspartate + a-ketoglutarate
12. Describe the role of ornithine in the urea cycle.
13. Explain how hyperammonemia, the increased concentration of NH +
4
in the serum, can arise from defects in more than one enzyme of the pathway that forms urea.
Carbon Atoms of Degradaed Amino Acids Emerge as Major Metabolic
Intermediates 14. Match the catabolic products in the right column with the amino acids in the left col umn from which they can be derived.
(a) alanine (1) succinyl
CoA
(b) aspartate (2) acetoacetate (c) glutamine (3) a-ketoglutarate (d) phenylalanine (4) oxaloacetate (e) leucine (5) pyruvate (f)
valine
(6) acetyl CoA (7) fumarate 15. Classify the following amino acids as glycogenic (G), ketogenic (K), or both (GK).
(a) leucine (e) histidine (b) alanine (f)
isoleucine (c) tyrosine (g) aspartate (d) serine (h) phenylalanine (a) histidine (b) phenylalanine (c) tyrosine (d) isoleucine (e) glutamine Inborn Errors in Metabolism Can Disrupt Amino Acid Degradation 18. Which of the following statements is true of the metabolic disease phenylketonuria?
(a) The disease is caused by an inability to synthesize phenylalanine.
(b) The disease can be caused by a deficiency in phenylalanine hydroxylase.
(c) The disease can be caused by a deficiency in tetrahydrobiopterin.
(d) The disease is treated with a high phenylalanine diet.
(e) The disease leads to a build-up of phenylalanine in the body.
ANSWERS TO SELF-TEST
1. All are correct. The amino acids that are needed for protein synthesis and as precursors for other biomolecules are used directly for those purposes. The carbon skeletons of any in excess can be converted into acetyl CoA or glucose, depending on the particular amino acid, and thus into products that are derivable from these two basic molecules.
2. The correct order is e, b, d, f, a, and c.
3. The correct answers are (a) and (d). (b) is incorrect because the proteosome also releases small peptides and (c) is incorrect because ubiquitin is spared degradation so it can be recycled.
4. c. The transamination of several different amino acids with a-ketoglutarate forms gluta mate and a-keto acids that can subsequently be catabolized.
5. a, c 6. c, e. Serine and threonine are deaminated by dehydratases that take advantage of the b-hydroxyl of these amino acids to carry out a dehydration followed by a rehydration to release NH + 4 .
7. d. A five-carbon amino acid will yield a five-carbon e-keto acid as a result of a transam ination; in this case, glutamate yields e-ketoglutarate. The other partner in the reaction, pyruvate, will yield alanine as a product.
8. Pyridoxal phosphate (PLP) acts as an amino carrier in transamination reactions. It is co valently bound to the a-amino group of a lysine residue in the enzyme by a Schiff base or aldimine bond, that is, by a carbon-nitrogen double bond between the a-amino group of the lysine and a carbon of PLP (see p. 641 of text). The enzyme catalyzes a displacement of the e-amino group of the lysine of the enzyme and forms an analogous aldimine bond with the a-amino group of an amino acid. Through a deprotonation and a reprotonation, the aldimine is isomerized to a ketimine in which the carbon-nitrogen bond is between the a-amino nitrogen and the a-carbon of the amino acid. The addition of H2O to the ketimine releases the carbon skeleton of the amino acid as an a-keto acid and leaves the coenzyme in the enzyme-bound pyridoxamine form, which now contains the a-amino group of the amino acid. After dissociation of the a-keto acid, a different a-keto acid binds to the enzyme to form a new ketimine, and the overall process reverses, resulting in the transfer of the enzyme-bound amino group to the keto acid to form a new amino acid and regenerate the pyridoxal phosphate.
9. a, c, d. Uracil is a pyrimidine component of RNA and is not a nitrogen excretory prod uct. Uric acid is a purine derivative excreted by uricotelic organisms.
10. Three moles of ATP are directly involved in the synthesis of urea. Two are converted to ADP and Pi by carbamoyl phosphate synthetase, and one is converted to AMP and PPi by argininosuccinate synthetase. Two ATP would be required to convert AMP back into ATP, so a total of four high-energy bonds are used. Only one molecule of nitrogen enters as NH + 4 ; the other enters in the a-amino group of aspartate, which can be formed by a transamination between oxaloacetate and glutamate.
11. The sum of the two reactions would be alanine + oxaloacetate aspartate + pyru vate, and the net effect would be that the a-amino group of alanine would appear as the a-amino group of aspartate, one of the two direct donors of nitrogen into the urea cycle—the other being carbamoyl phosphate. The a-amino groups of many other amino acids can be similarly collected on aspartate and fed into the cycle.
12. Ornithine serves as the carrier on which the urea molecule is constructed. The a-amino group of ornithine has a carbamoyl group added to it by ornithine transcarbamoylase to form citrulline. Subsequent steps of the cycle add aspartate to bring in the second nitrogen atom and also regenerate ornithine when arginase cleaves arginine to form urea.
13. A defect in carbamoyl phosphate synthesis would cause increased NH + 4 concentrations, as would a defect in any of the four reactions that condense carbamoyl phosphate with ornithine and regenerate the ornithine. Essentially, blocking a biochemical pathway at any of its steps may lead to increased concentrations of any of the precursors or members of the pathway.
14. (a) 5 (b) 4, 7; aspartate can be converted to fumarate via the urea cycle. (c) 3 (d) 2, 7 (e) 2, 6 (f) 1 15. (a) K (b) G (c) GK (d) G (e) G (f) GK (g) G (h) GK
16. Both give rise to methylmalonyl CoA, which can, in turn, be converted into succinyl CoA and ultimately into glucose.
17. b, c. Monoxygenases and dioxygenases are involved in the conversion of phenylalanine to tyrosine and the subsequent opening of the aromatic ring during tyrosine catabolism.
18. Answers (b), (c), and (e) are correct. Phenylketonuria is a metabolic disorder arising from an absence or deficiency in the enzyme phenylalanine hydroxylase or (more rarely) its cofactor tetrahydrobiopterin. It results in the build-up of phenylalanine in the body and is treated with a diet low in phenylalanine.
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PROBLEMS
1. Three enzymes are needed for the attachment of ubiquitin to proteins targeted for degra dation: E1, E2, and E3. For each enzyme state its name and give a brief description of its role in ubiquitin attachment.
2. Birds require arginine in their diet. Would you expect to find the production of urea in these animals? Explain.
3. Which would you expect to have a greater effect on the rate of urea biosynthesis, a de fect in fumarase activity or a defect in alanine aminotransferase?
4. Pyridoxal phosphate or related metabolites are required growth factors for Lactobacillus species (Morishita et al., J. Bacteriol. 148[1981]:64–71). For example, when amino acids such as alanine or glutamate are used as the sole source of nutrition, these bacilli do not grow nor do they generate metabolic energy unless pyridoxal phosphate or its metabolites are supplied. Explain these observations.
5. A male infant six weeks of age exhibited symptoms of pronounced hyperammone mia, which included vomiting, fever, irritability, and screaming episodes interspersed with periods of lethargy. The infant had low levels of blood urea, elevated serum transaminase, and generalized hyperaminoacidemia and aminoaciduria. The levels of citrulline, argininosuccinic acid, and arginine were relatively low in both blood and urine. Enzymatic assays of liver tissue established that the level of mitochondrial carbamoyl phosphate synthetase (CPS) was approximately 20% of normal; the enzyme was active only in the presence of relatively high concentrations of N-acetylglutamate.
All other urea cycle enzyme levels were relatively normal. The infant was treated with a supplement containing arginine, pyridoxine, and a-keto analogs of essential amino acids (those that cannot be synthesized in humans). As part of the therapy, dietary protein was also restricted.
(a) Why were blood levels of ammonia high in this infant?
(b) Explain why infants with CPS deficiency normally exhibit relatively low blood lev els of citrulline, argininosuccinic acid, arginine, and urea.
(c) Why is the administration of supplemental arginine recommended for this infant?
(d) Why were a-keto analogs of essential amino acids and pyridoxine administered?
(e) Why was dietary protein restricted?
(f) Why would you expect to see elevated concentrations of glutamate, glutamine, and alanine in the blood and urine of this infant?
6. Why is glutamate dehydrogenase a logical point for the control of ammonia production in cells?
7. During the process of glomerular filtration in the kidney, amino acids, as well as other metabolites, enter the lumen of the kidney tubule. Normally, a large portion of these amino acids are reabsorbed into the blood through the action of membrane-bound carrier systems that are specific for different classes of amino acids. Cystinuria is a disorder whose symptoms include urinary excretion with unusually high concentrations of cystine as well as excess amounts of ornithine, lysine, and arginine. Cystine is a dibasic amino acid, composed of two cysteine molecules joined by a disulfide linkage.
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
415
Patients with this disorder often have urinary tract stones, which are caused by the limited solubility of cystine. A related disorder found in other people is characterized by the appearance of ornithine, lysine, and arginine in the urine, although the levels of urinary cystine are normal.
(a) What is the most likely source of cystine in cells?
(b) What common structural feature of the four amino acids—cystine, ornithine, ly sine, and arginine—is recognized by the carrier in the kidney tubule membrane?
(c) How many carrier systems may exist for these molecules?
(d) Other amino acidurias are due to a deficiency in one or more of the enzymes in the catabolic pathway for an amino acid. This leads to higher concentrations of the amino acid in the blood and a corresponding increase in the concentrations in the glomerular filtrate. In this case, the capacity of the reabsorption system is surpassed, causing some amino acid to be lost in the urine. How could you distinguish between a defect in amino acid metabolism and a defect in a renal transport system?
8. Why is the catabolism of isoleucine said to be both glucogenic and ketogenic?
9. Brain cells take up tryptophan, which is then converted to 5-hydroxytryptophan by tryp tophan hydroxylase, an enzyme whose activity is similar to that of phenylalanine hydroxylase. Aromatic amino acid decarboxylase then catalyzes the formation of the potent neurotransmitter 5-hydroxytryptamine, also called serotonin. In the blood, tryptophan is bound to serum albumin, with an affinity such that about 10% of the tryptophan is freely diffusable. The rate of tryptophan uptake by brain cells depends on the concentration of free tryptophan. In these cells, tryptophan concentration is normally well below that of the KM for tryptophan hydroxylase. Aspirin and other drugs displace tryptophan from albumin, thereby increasing the concentration of free tryptophan.
(a) What cofactor is required for the activity of tryptophan hydroxylase?
(b) Dietary deficiencies in pyridoxin and related metabolites can induce a number of symptoms, including those that appear to be related to derangements in serotonin metabolism. What enzyme could be affected by a deficiency of vitamin B6?
(c) What effect does aspirin have on tryptophan metabolism in brain cells?
10. In many microorganisms, glutamate dehydrogenase (GDH) participates in the catabo lism of glutamate by generating ammonia and a-ketoglutarate, which undergoes oxidation in the citric acid cycle. However, when E. coli is grown with glutamate as the sole source of carbon, the synthesis of GDH protein is strongly repressed. Under these conditions, aspartase, an enzyme that catalyzes the removal of ammonia from aspartate to form fumarate, is required in order for the cell to grow in glutamate. Propose a cyclic pathway for the catabolism of glutamate that includes aspartate.
11. When E. coli is grown in glucose and ammonia, GDH synthesis is accelerated and the en zyme is active. Under these conditions, what role does GDH play in bacterial metabolism?
12. After an overnight fast, muscle tissue proteolysis generates free amino acids, many of which pass into the blood. Among the amino acids that are found in the blood are alanine, glutamate, and glutamine, all of which are rapidly taken up by the liver. What happens to these amino acids when they enter hepatic cells?
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13. Propionyl CoA and methylmalonyl CoA both inhibit N-acetylglutamate synthase activ ity in slices of liver tissue. What clinical symptom would you expect to see in patients suffering from methylmalonic aciduria as a result of this inhibition?
14. Dialysis of purified glutamate aminotransferase can be used to remove pyridoxal phos phate from the enzyme, but the dissociation of the cofactor from the enzyme is very slow.
Why would the addition of glutamate to the enzyme solution increase the rate of dissociation of the cofactor from the enzyme?
15. Early work by Esmond Snell on the enzymes that employ pyridoxal phosphate in cluded experiments in which free pyridoxal was heated with amino acids like glutamate. Snell found that the (-amino group of glutamate was transferred to pyridoxal, generating pyridoxamine. Why was this an important clue to the function of pyridoxal as a cofactor?
16. The enzyme serine dehydratase employs pyridoxal phosphate (PLP) in the direct deam ination of serine. In this reaction, the a-carbon of serine undergoes a two-electron oxidation through a-elimination. Show how PLP participates in the process by writing a mechanism for serine dehydration and deamination.
17. A small number of infants who have phenylketonuria have normal levels of phenylala nine hydroxylase activity, but on normal diets they continue to accumulate phenylalanine as well as other metabolites, including phenylpyruvate, phenyllactate, and phenylacetate. They also have high levels of quinonoid dihydrobiopterin.
(a) What is the probable enzyme deficiency in these infants? Rationalize the deficiency with the observed clinical symptoms.
(b) Write brief pathways for the formation of the phenylalanine metabolites found in these infants.
J
CH
OH
2
J
J
CKO
JCH J CJ COO: JCH J COO:
2
2
J
J
COO:
H
Phenylpyruvate
Phenyllactate
Phenylacetate
ANSWERS TO PROBLEMS
1. E1, ubiquitin-activating enzyme: adenylates ubiquitin on its terminal carboxylate with release of PPi. Transfers ubiquitin to a sulfhydryl on the enzyme with release of AMP, forming a thioester bond. E2, ubiquitin-conjugating enzyme: receives ubiquitin from E1.
Ubiquitin is also bound to a sulfhydryl of E2 via a thioester bond. E3, ubiquitin-protein ligase: transfers ubiquitin from E2 to the e-amino group of a target protein.
2. In the urea cycle, arginine is cleaved to yield urea and ornithine. The fact that birds require arginine in their diet indicates that they are unable to synthesize it for utilization in protein synthesis. As a result, they are also unable to synthesize urea to dispose of ammonia; instead, they synthesize uric acid. Birds do have carbamoyl phosphate synthetase activity; however, it is located in the cytosol, and it catalyzes the formation of carbamoyl phosphate, which is then utilized for pyrimidine synthesis.
3. Fumarase activity has an effect on the urea cycle because it is needed, along with malate dehydrogenase, for the regeneration of oxaloacetate, which in turn undergoes transamination to form aspartate. The amino group of aspartate contains one of the two nitrogen atoms that are used to synthesize urea. Alanine aminotransferase is one of a number of aminotransferases that can transfer amino groups from amino acids to a-ketoglutarate to generate glutamate. Subsequent deamination of glutamate provides ammonia for the urea cycle. If all the other aminotransferases in the cell are active, then alanine aminotransferase would not be particularly essential. Thus, a defect in fumarase activity would have the greater effect on the rate of urea biosynthesis.
4. In order to utilize amino acids as sources of oxidative energy or to generate glucose through gluconeogenesis, the bacilli must carry out transamination reactions to dispose of ammonia (or to use it for the biosynthesis of other nitrogen-containing compounds) as well as to generate a-keto acids that can be used in the citric acid cycle or other pathways. Pyridoxal phosphate is a required cofactor for the aminotransferase enzymes, and in bacteria in which it cannot be synthesized, it must be derived from the growth medium. Otherwise, amino acids cannot be metabolized. In this case, where an amino acid is the only source of carbon and of nitrogen, the bacilli will not be able to introduce the amino acid into any catabolic pathway. Pyridoxal phosphate also functions as a cofactor for a large number of other enzymes, including decarboxylases, racemases, aldolases, and deaminases. Therefore, deficiencies in pyridoxal phosphate would also adversely affect a large number of other pathways.
5. (a) Carbamoyl phosphate synthetase utilizes ammonia and bicarbonate to synthesize carbamoyl phosphate, which enters the urea cycle. A deficiency in the activity of this enzyme leads to an accumulation of ammonia in blood and in urine.
(b) Because carbamoyl phosphate condenses with ornithine to form citrulline, it is in effect a precursor of all the components of the urea cycle. When its synthesis is depressed, the rate of synthesis of other urea cycle components will be decreased.
(c) Arginine is needed as a component of most proteins; it must therefore be supplied to avoid arginine deficiency, which would result in a decrease in the rate of protein synthesis. In the urea cycle, arginine serves as a precursor of ornithine, which in turn serves as an acceptor of the relatively small number of carbamoyl groups that enter the urea cycle. Thus, supplemental arginine could accelerate the rate of urea synthesis in an attempt to drive the reaction catalyzed by the partially active CPS enzyme toward the net formation of carbamoyl phosphate. Finally, recent findings suggest that arginine is a feed-forward activator for the synthesis of N-acetyl glutamate, which activates CPS. Recall that the deficient enzyme appeared to be heavily dependent on levels of N-acetyl glutamate. Supplying arginine indirectly activates CPS, thereby increasing the rate of ammonia utilization.
(d) The administration of a-keto analogs of essential amino acids serves two functions.
First, these substrates undergo transamination with glutamate as the amino donor.
The corresponding increase in the synthesis of glutamate, through the action of glutamate dehydrogenase, utilizes more ammonia, removing at least some of it from the blood. Second, the transamination of the a-keto acids generates essential amino acids, which are used for protein synthesis. This is especially important in a situation in 418
CHAPTER 23 which dietary protein intake is restricted. Since the cofactor pyridoxyl phosphate is required for aminotransferases to function, adding its vitamin precursor (pyridoxine) to the diet will ensure sufficient quantities to process the added a-keto analogs into essential amino acids.
(e) When dietary protein is hydrolyzed to its component amino acids, subsequent cata bolic steps include transamination and deamination, which produce free ammonia.
The less protein consumed, the lower the production of ammonia from the breakdown of amino acids. A reduction in the blood ammonia level is the primary goal of the therapy for this infant.
(f) Increased levels of ammonia in the cells will drive the reaction catalyzed by glutamate dehydrogenase toward the net formation of glutamate and will also stimulate the synthesis of glutamine, which is in effect a carrier of two molecules of ammonia. With the high concentrations of glutamate, the equilibria for most transamination reactions would be shifted toward the net formation of other amino acids, such as alanine, a transaminated form of pyruvate. Concentrations of these and other amino acids would therefore be increased in both the blood and the urine.
6. The deamination of amino acids occurs through the action of transaminases as well as through the action of glutamate dehydrogenase (GDH). GDH is the only enzyme that catalyzes the oxidative deamination of an L amino acid. It deaminates glutamate, whose precursor, a-ketoglutarate, is the ultimate acceptor of amino groups from almost all the amino acids. In addition, while the reactions catalyzed by the transaminases are freely reversible, the GDH reaction is far from equilibrium; it is therefore a logical activity to control because large changes in its velocity can be achieved with small changes in the concentrations of allosteric effectors like ATP or NADH. Thus, GDH is the enzyme of choice for the control of ammonia synthesis.
7. (a) Disulfide linkages exist in many proteins, and when they are hydrolyzed by pro teases to yield free amino acids, cystine is often one of the products.
(b) Like cystine, the other three amino acids—arginine, ornithine, and lysine—have two basic groups. The carrier systems probably recognize and bind these groups for transport.
(c) From the disorders described, it is likely that there are two transport systems. One carries all four dibasic acids; when it is defective, reabsorption of all four species fails, allowing all four to spill over into the urine. Another system transports ornithine, arginine, and lysine but not cystine; its failure to function accounts for the appearance of these three amino acids but not cystine in the urine.
(d) A defect in the catabolic pathway for a particular amino acid causes elevation in the concentration of that single amino acid in blood and in urine as well, unless other amino acids carried by the same kidney transport system are lost to the urine. A defect in renal reabsorption means that all those amino acids that share the affected carrier system will be lost to the urine. Their concentration in blood will be lower than normal, while their concentration in urine will be higher.
8. The catabolic pathway for isoleucine leads to the formation of acetyl CoA and propionyl CoA. Acetyl CoA can be utilized for the net synthesis of fatty acids or ketone bodies, but it cannot be used for the net synthesis of glucose; thus, it is said to be ketogenic. In contrast to acetyl CoA, propionyl CoA is converted to succinyl CoA, which can be utilized through part of the citric acid cycle and the gluconeogenic pathway to give the net formation of glucose. The distinction between the two types of substrates is somewhat arbitrary, however. For example, succinyl CoA can also be converted via pyruvate and acetyl CoA to citrate, which, when transported to the cytosol, serves as the source of carbons for the synthesis of fatty acids. Thus, glucogenic substrates can, under certain conditions, be ketogenic; however, ketogenic substrates cannot be glucogenic, unless a cell has a functional glyoxylate pathway.
9. (a) Tetrahydrobiopterin is utilized as a reductant by many hydroxylase enzymes, in cluding phenylalanine hydroxylase and tryptophan hydroxylase.
(b) Pyridoxal phosphate, which is derived from dietary pyridoxine (vitamin B6), is a cofactor for a number of enzymatic reactions that occur at the a carbon of an amino acid including decarboxylations (see text, p. 642). In this case, the cofactor participates in the decarboxylation of 5-hydroxytryptophan by the enzyme aromatic amino acid decarboxylase to form serotonin. A deficiency of vitamin B6 would lead to a reduction in the rate of synthesis of serotonin.
(c) The higher the concentration of free tryptophan, the greater the rate of uptake of the amino acid by the brain cells. Because the normal concentration of tryptophan in these cells is below that of the K for tryptophan hydroxylase, an influx of more
M tryptophan into the cells provides more substrate for the enzyme. You would therefore expect an increase in the level of 5-hydroxytryptophan production.
10. A possible catabolic pathway for glutamate that includes aspartate is as follows:
Glutamate+OxaloacetateDAspartate+a-Ketoglutarate
NADH
Oxidation in the +H;
citric acid cycle NAD;
NH4 H O
2
MalateEFumarate
Glutamate undergoes transamination with oxaloacetate to generate a-ketoglutarate and aspartate. Oxidation of a-ketoglutarate is carried out in the citric acid cycle, whereas aspartate is cleaved to yield ammonia and fumarate. Fumarate is converted to malate, which is then oxidized to oxaloacetate in the citric acid cycle so that it can be regenerated to serve as an acceptor of the amino group from glutamate. It should be noted that glutamate will also be used as a source of amino groups by the aminotransferases and other enzymes involved in biosynthesis.
11. In cells grown in glucose and ammonia, GDH catalyzes the assimilation of ammonia by incorporating it into a-ketoglutarate to yield glutamate, which serves as a source of amino groups for other biosynthetic reactions.
12. In the liver, alanine, glutamate, and glutamine are utilized as sources of carbon for glu coneogenesis. Glutamine is deaminated to yield glutamate and ammonia. Glutamate undergoes oxidative deamination to form ammonia and a-ketoglutarate, a substrate for gluconeogenesis. Pyruvate is generated from the transamination of alanine, and carboxylation of the a-keto acid yields oxaloacetate, another source of carbon for gluconeogenesis. The ammonia generated by the conversion of the amino acids to their corresponding a-keto acids is used for the synthesis of urea. Glucose synthesized by liver can be returned through the blood to the muscle, where it serves as a source of energy.
Thus, muscle uses the amino acids as a means of contributing to the generation of glucose in liver as well as a means of transporting ammonia to the liver, where the synthesis of urea can be carried out.
420
CHAPTER 23
13. The activity of mitochondrial carbamoyl phosphate synthetase (CPS) depends on the availability of N-acetylglutamate, which is generated from glutamate and acetyl CoA. A reduction in the availability of the activating molecule will lead to a decrease in the activity of CPS, which utilizes ammonia and bicarbonate for the synthesis of carbamoyl phosphate. This, in turn, leads to an increase in the level of ammonia in blood and urine.
Over two-thirds of patients with methylmalonic aciduria are hyperammonemic.
14. In the native enzyme, the dissociation of the cofactor from the enzyme during dialysis is very slow because PLP is covalently bound to a lysine residue. The addition of glutamate, a substrate for the transamination reaction, leads to the formation of a Schiff base between glutamate and PLP, which means that the cofactor is no longer covalently attached to the enzyme molecule. Although PLP is bound to the enzyme by noncovalent forces, these forces are not as strong as a covalent bond, so during dialysis, the rate of dissociation of the cofactor from the enzyme is increased.
15. Snell’s observations of the formation of pyridoxamine by heating with a-amino acids suggested that pyridoxal is involved in transamination reactions. As an enzyme cofactor it could transfer an a-amino group from an amino acid to the a-keto group of an a-keto acid. It is now established that the action of pyridoxal phosphate in aminotransferase enzymes includes formation of pyridoxamine phosphate during the catalytic cycle (see text, p. 641) 16. In the accompanying figure, PLP in Schiff base linkage with a lysine residue in the enzyme forms a new Schiff base link with serine. A hydrogen atom is removed from the a-carbon, and then the hydroxyl group is removed by elimination from the b-carbanion, generating aminoacrylate attached to pyridoxal phosphate. Hydrolysis of the Schiff base to give aminoacrylate and PLP is followed by tautomerization to the imino form. This compound hydrolyzes spontaneously to form pyruvate and ammonia. PLP is once again linked covalently to the enzyme.
ENZ
HJO
J
J (CH ) ;
HJO
H CH JC:JCOO: :
H CKCJCOO: 2 4
H
2
OH
2
J
J
J
J
J
N
+
CH JCJCOO:
NJ
N
2
J
H
K
J
K
H
b-Elimination
K
HJC
NH
HJC
HJC
3
J
OH
J O:
J O:
J
Serine
J
J
PLP
Carbanion
H O
2
H O CH JCJCOO:
2
H CJCJCOO: H CKCJCOO:
3
3
2
K
Pyruvate
G
K
J
O
+NH
NH
2
2
+
+
Aminoacrylate
Ammonia
NH
Imino Form
PLP
3
17. (a) The enzyme that is deficient in these infants is dihydropteridine reductase, which converts quinonoid dihydrobiopterin to tetrahydrobiopterin, using NADH as a substrate. In the phenylalanine hydroxylase reaction, tetrahydrobiopterin, the reductant in the conversion of phenylalanine to tyrosine, is oxidized to quinonoid dihydrobiopterin. The reductase enzyme regenerates tetrahydrobiopterin so that it can be used for further use in tyrosine formation. Cells that are deficient in the reductase cannot carry out efficient conversion of phenylalanine to tyrosine because they cannot regenerate tetrahydrobiopterin.
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
421
(b) Phenylpyruvate can be generated from phenylalanine by transamination, and re duction of phenylpyruvate by NADH or NADPH generates phenyllactate, in a reaction similar to that catalyzed by lactate dehydrogenase. Phenylacetate can be generated by oxidative decarboxylation of phenylpyruvate, in a reaction reminiscent of the conversion of pyruvate to acetyl CoA, catalyzed by pyruvate dehydrogenase.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. (a) The energy from ATP hydrolysis could be used for maintaining the specificity of peptide-bond hydrolysis, unfolding the target protein, threading the target protein through the barrel, or other related functions.
(b) Small peptides would not need to be unfolded or threaded, and would not need to have any of their bonds protected (e.g., remember that the ubiquitin tags are not hydrolyzed).
2. (a) pyruvate, (b) oxaloacetate, (c) a-ketoglutarate, (d) a-ketoisocaproate, (e) phenylpyru vate, and (f) hydroxyphenylpyruvate
3. (a) Aspartate + a-ketoglutarate + GTP + ATP + 2 H2O + NADH + H+ 1⁄2 glucose + glutamate + CO +
2
ADP + GDP + NAC+ + 2Pi
The glucogenic route for aspartate involves transamination to oxaloacetate, con version of the latter to phosphoenolpyruvate, which is then converted to glucose (see the text, Chapter 20). PLP and NADH participate as coenzymes in the conversion of aspartate to glucose.
(b) Aspartate + CO + + +
2
NH4 3 ATP + NAD+ + 4 H2O oxaloacetate + urea + 2 ADP + 4 P +
i
AMP + NADH + H+
This equation represents the summation of the stoichiometry of urea synthesis, the hydrolysis of PPi, and the conversion of fumarate to oxaloacetate in the citric acid cycle.
4. Different sites could specialize in the hydrolysis of different categories of peptide bonds (e.g., those that join hydrophobic/hydrophilic, hydrophobic/hydrophobic, or hydrophilic/hydrophilic amino acids, etc.) and thereby optimize the overall kinetics of degrading diverse protein sequences.
5. By analogy with the 20S proteasome (Figure 23.7), the overall architecture could be conserved. Perhaps the six different AAA ATPase subunits from the 19S regulatory complex associate into a hexamer with pseudo six-fold symmetry. If these hexamers could be separated from the 19S complexes, they might be visulaized by electron microscopy or crystallography; suitable crosslinking experiments might then reveal which pairs of AAA ATPase subunits border each other.
6. thiamine pyrophosphate
7. It acts as an electron sink. See C. Walsh (1979), Enzymatic Reaction Mechanisms, p. 178 (W. H. Freeman).
8. CO + + +
2
NH4 3 ATP + NAD+ + 3H2O + glutamate urea + 2 ADP + 2 P +
+
i
AMP + PPi NADH + H+ + a-ketoglutarate
The answer in the text is for the conversion of fumarate to oxaloacetate rather than to aspartate. The equation above gives the correct stoichiometry for the synthesis of urea from NH +
4
and glutamate. It includes a non-energy-requiring transamination reaction.
Hence, the number of ~P spent remains at four. Note that aspartate does not appear in this equation, since it is resynthesized. Thus aspartate can be considered a nitrogencarrying cofactor in the synthesis of urea.
422
CHAPTER 23
9. The compound should inhibit ornithine transcarbamoylase because it appears to be a nonhydrolyzable analogue of an intermediate that should be formed between ornithine and carbamoyl phosphate. (The CH2 group in compound A will prevent the release of the phosphate.)
10. High concentrations of ammonia could increase the ratio of glutamate/a-ketoglutarate (glutamate dehydrogenase reaction) and increase the level of glutamate in the brain.
Ammonia also could increase the ratio of asparatate/oxaloacetate; the resulting lower level of oxaloacetate would decrease the availability of all citric acid cycle intermediates.
11. The mass spectrometric analysis strongly suggests that three enzymes—pyruvate dehy drogenase, a-ketoglutarate dehydrogenase, and the branched-chain a-keto dehydrogenase—are deficient. Most likely, the common E3 component of these enzymes is missing or defective. This proposal could be tested by purifying these three enzymes and assaying their capacity to catalyze the regeneration of lipoamide.
12. Benzoate, phenylacetate, and arginine would be given in a protein-restricted diet.
Nitrogen would emerge in hippurate, phenylacetylglutamine, and citrulline. See S. W.
Brusilow and A. L. Horwich in C. R. Scriver, A. L. Beaudet, W. S. Sly, and D. Valle (eds.)
(1989), The Metabolic Basis of Inherited Disease, 6th ed., pp. 629–663 (McGraw-Hill), for a detailed discussion of the therapeutic rationale.
This therapy is designed to facilitate nitrogen excretion in the absence of urea synthesis.
Note that much of the nitrogen would be excreted as glycine (hippurate is benzoylglycine), glutamine, and citrulline.
13. Aspartame, a dipeptide ester (aspartylphenylalanine methyl ester), is hydrolyzed to L-as partate and L-phenylalanine. High levels of phenylalanine are harmful in phenylketonurics.
14. N-acetylglutamate is synthesized from acetyl CoA and glutamate. Once again, acetyl CoA serves as an activated acetyl donor. This reaction is catalyzed by N-acetylglutamate synthase.
15. First serine forms a protonated Schiff base (external aldimine) with pyridoxal-5′-phos phate. Removal of the serine a-hydrogen leads to a quinonoid intermediate, which then can eliminate the b-OH to generate the Schiff base of aminoacrylate. The reaction is com H
HOCH COO:
2
H
H
HOCH COO:
OH
+
NH
2
3
COO:
(Lys)
H
N
H (Lys)
Ser
H
N
+
N
H
H
G
G O:
+ H
PO
O:
=O
3
=O PO
O:
3
=O PO
3
+
N
CH
3
N
CH
J
+
3
H
N
CH
J
3
J
H
H
Lys H CK COO: H C K COO:
2
2
(Lys)
+
+
+
H
N
NH
3
N
H
H
H
aminoacrylate
G
G O: O:
=O PO =O PO
3
3
+
N
CH3
+
N
CH3
J
J
H
H
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
423
pleted by the transfer of pyridoxal phosphate back to a Schiff base with a lysine of the enzyme (internal aldimine), with the concomitant release of aminoacrylate. (The aminoacrylate will react with water to give pyruvate and ammonia.)
(Note: Another family of serine dehydratases does not use pyridoxal phosphate, but rather an iron-sulfur cluster as the cofactor. These enzymes may use a mechanism similar to the dehydration of citrate catalyzed by aconitase. See Trends Biochem. Sci.
18[1993]:297–300.)
16. As in problem 15, an external aldimine (protonated Schiff base) is formed between serine and pyridoxal-5′-phosphate, and the serine a-hydrogen is removed. The a-hydrogen then can be reattached from either of two faces to give the Schiff base of either D-serine or L-serine (see below). The equilibrium constant for the reaction will be one.
L-approach
H;
H
HOCH COO:
HOCH COO:
2
2
H
N
H
N
D-approach
+
H
+ H
O: O:
G
=O PO =O PO
3
3
+
N
CH
N
CH
3
3
J
J
H
H
17. Protein-protein interactions often relate directly to biological function. Two or more as pects of these interactions may be relevant for degradation: (1) If an interaction domain becomes chemically damaged so that interaction with a partner protein is no longer feasible, then it would be appropriate to turn over (degrade) the protein. (2) If a partner protein is in short supply within the cell, then the interacting protein may not be needed and should be degraded. (Without the partner protein, the interaction domain could be exposed as a possible signal for degradation.)
18. (a) The initial surge originates from a need to supply glucose to the brain. When car bohydrates are depleted, mammals cannot resupply glucose from fatty acids.
Glycerol provides a small source of carbohydrate, but mammals also must break down amino acids to meet the short-term demands of the brain for glucose. The ammonia byproduct from amino acid degradation accounts for the initial surge of nitrogen excretion.
(b) Over time, the liver begins to metabolize acetyl-CoA (from fats) to ketone bodies, and the brain adapts to using ketone bodies. During this period, fats can provide many of the energy needs of the brain, and little nitrogen is excreted.
(c) When lipid stores have been depleted, the organism once again must metabolize amino acids to provide glucose for the brain, and nitrogen excretion again increases.
19. Isoleucine can give its amino group to a-ketoglutarate in a transamination reaction and then be oxidatively decarboxylated and dehydrogenated to form the corresponding (a,b)-unsaturated acyl-CoA derivative. Further reactions (see the figure on p. 424) then are identical to fatty acid oxidation until the carbon skeleton is split into acetylS-CoA and propionyl-S-CoA. The three subsequent steps for the conversion of the (odd-chain) propionyl-S-CoA to succinyl-S-CoA have been discussed for the oxidation of odd-chain fatty acids (see Chapter 22).
424
CHAPTER 23
NADH, CO
2
+
NH3
O
NAD;,
O a-KG Glu
CoASH
CH CH
D
CH CH
D
CH CH
3
2
3
2
3
2
COO: COO:
S-CoA
CH
CH
CH
3
transamination
3
ox. decarboxylation
3
O
OH
O
FAD
FADH2 H O D
CH CH
2
3
S-CoA
D H C
S-CoA
3
CH
CH
dehydrogenation
3
hydration
3
O
Acetyl-S-CoA
O
O H C
S-CoA
3
NAD; NADH
CoASH
O
D
D H C
S-CoA
3
CH
S-CoA
dehydrogenation
3
thiolysis
CH3
Propionyl-S-CoA
20. (a) PAN has no effect in the absence of nucleotides.
(b) ATP is required. Neither ADP nor AMP-PNP is effective in stimulating protein digestion.
(c) AMP-PNP is a nonhydrolyzable analogue of ATP. The finding that AMP-PNP does not stimulate the digestion suggests that ATP hydrolysis to ADP and Pi is required.
(d) The peptide digestion does not require PAN or ATP.
(e) See answer 1(b), above. Small peptides do not need to be unfolded or threaded through a superstructure in order to facilitate digestion.
CHAPTER 2
The Biosynthesis of Amino Acids
4
In this chapter, the biosynthetic origins of the amino acids are explained, beginning with the need for a source of nitrogen for the amino acids. This need is met by nitrogen fixation, which is the process of converting atmospheric nitrogen in the form of N
+
2 to NH4 . The enzyme that carries out this difficult task, nitrogenase, is discussed in detail, including the role of an unusual molybdenum-iron cofactor. The authors then explain how NH + 4 is incorporated into the amino acids glutamate and glutamine via the enzymes glutamate dehydrogenase and glutamine synthetase. These two amino acids are major nitrogen donors in a range of biosynthetic pathways, including those of the remaining amino acids whose synthesis is discussed next. While the pathways for the synthesis of the amino acids are diverse, they have in common the fact that their carbon skeletons come from intermediates in glycolysis, the pentose phosphate pathway or the citric acid cycle. This leads to a grouping of the amino acids into one of six biosynthetic families based on their starting material: oxaloacetate, pyruvate, ribose-5-phosphate, a-ketoglutarate, 3-phosphoglycerate, and phosphoenolpyruvate/erythrose 4-phosphate. The synthesis of the members of each family is examined in detail, including the role of three important cofactors involved in some of the syntheses: pyridoxal phosphate, tetrahydrofolate, and S-adenosylmethionine. The latter two are carriers of single carbon atoms in metabolism. The authors also explain that the lack of some biosynthetic pathways in humans has led to the dietary requirement for nine amino acids. The examination of amino acid synthesis concludes with a general discussion of how metabolic pathways are controlled via feedback inhibition, using examples from amino acid metabolism to illustrate the relevant principles.
425
426
CHAPTER 24
The chapter concludes with a look at the role of the amino acids as precursors of many important biomolecules. The synthesis of glutathione, a sulfhydryl buffer and detoxifying agent, and nitric oxide, a short-lived signal molecule, are examined. The final topic is the biosynthesis and degradation of the porphyrin heme. The multistep synthetic pathway beginning with glycine and succinyl CoA is examined as is the mechanism for degradation of excess heme. The physiological consequences of disorders in heme biosynthesis (collectively known as porphyrias) are discussed.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Recognize that nitrogen in the form of ammonia is the source of nitrogen for all amino acids.
Nitrogen Fixation: Microorganisms Use ATP and a Powerful Reductant
to Reduce Atmospheric Nitrogen to Ammonia (Text Section 24.1) 2. Define nitrogen fixation and name the groups of organisms that can carry out this conversion.
3. Describe the nitrogenase complex and explain the roles of its reductase and nitrogenase com ponents. Note the function of the FeMo-cofactor.
4. Explain the energy requirement for nitrogen fixation and write the equation giving the stoichiometry of the overall reaction.
5. Outline the key roles of glutamate and glutamine in the assimilation of NH +
4
into amino acids and describe the reactions of glutamate dehydrogenase, glutamine synthetase, and glutamate synthase. Recognize the functions of ATP and NADPH in these processes.
Amino Acids Are Made from Intermediates of the Citric Acid Cycle
and Other Major Pathways (Text Section 24.2) 6. Classify the amino acids into six biosynthetic families and identify their seven precursors.
Name the metabolic pathways from which these precursors originate.
7. Identify the essential amino acids for humans and explain why they are essential.
8. Describe the single-step biosyntheses of alanine, aspartate, and asparagine.
9. Outline the syntheses of glutamate, glutamine, proline, and arginine from a-ketoglutarate.
10. Outline the syntheses of serine, glycine, and cysteine from 3-phosphoglycerate.
11. Explain the roles of pyridoxal phosphate, tetrahydrofolate, and S-adenosylmethionine in amino acid biosyntheses.
12. Identify the structure of tetrahydrofolate and indicate the reactive part of the molecule.
Draw the structures of the single-carbon groups that can be carried on tetrahydrofolate and provide examples of reactions that generate and use them. Describe the sources of this cofactor in humans.
THE BIOSYNTHESIS OF AMINO ACIDS
427
13. Draw the structure of S-adenosylmethionine and describe its synthesis. Indicate the reac tive part of the molecule and describe the basis of its high methyl group–transfer potential.
14. Outline the activated methyl cycle and describe the roles of methylcobalamin and ATP in the cycle. Give examples of important derivatives from S-adenosylmethionine.
15. Describe the synthesis of cysteine from homocysteine and serine.
16. Outline the biosyntheses of phenylalanine, tyrosine, and tryptophan in E. coli. Describe the roles of phosphoenolpyruvate, erythrose 4-phosphate, and phosphoribosylpyrophosphate
in these reactions.
17. Describe the structure of tryptophan synthetase and the role of substrate channeling in its catalytic reaction.
Amino Acid Biosynthesis Is Regulated by Feedback Inhibition
(Text Section 24.3) 18. Define the committed step of a metabolic pathway and recognize that it is often the tar get of feedback regulation. Note the main features of control of branched pathways by feedback inhibition and activation, enzyme multiplicity, and cumulative feedback.
19. Describe the cumulative feedback control of glutamine synthetase from E. coli. Explain the mechanisms and functions of the reversible covalent modifications and describe the advantage of employing an enzymatic cascade in regulating this reaction.
Amino Acids Are Precursors of Many Biomolecules (Text Section 24.4) 20. Give examples of important biomolecules that are derived from amino acids.
21. Draw the structure of glutathione and describe its cycle of oxidation/reduction. Indicate the functions of glutathione and describe the involvement of selenium in the glutathione
peroxidase reaction.
22. Outline the synthesis of nitric oxide (NO) and explain its function.
23. Name the two molecular precursors of the porphyrins in mammals and outline the biosynthesis and degradation of heme.
24. Explain the molecular defects in congenital erythropoietic porphyria and acute intermittentporphyria.
SELF-TEST
Nitrogen Fixation: Microorganisms Use ATP and a Powerful Reductant
to Reduce Atmospheric Nitrogen to Ammonia 1. Define nitrogen fixation and explain why it is crucial to the maintenance of life on earth.
2. Place the following components, reactants, and products of the nitrogenase complex re action in their correct sequence during the electron transfers of nitrogen fixation: (a) oxidized ferredoxin (e) N2 (b) reductase component (f)
reduced ferredoxin (c) nitrogenase component (g) electron source (d) NH3 3. Write the net equation for nitrogen fixation and describe the sources of the electrons and ATP.
4. Match the structural components or features in the right column with the appropriate component of the nitrogenase reaction.
(a) reductase component (1) MoFe-cofactor (b) nitrogenase component (2) [4Fe-4S] cluster (3) ATP-ADP binding site (4) N2-binding site (5) a 2b2 tetramer (6) dimer of identical subunits 5. Match the enzyme with the reaction it catalyzes.
(a) glutamine synthetase (b) glutamate dehydrogenase (c) glutamate synthase
3
a-Ketoglutarate
Glutamine
1
2
Glutamate
6. Which of the reactions shown in question 5 require the following?
(a) NH +
4
(b) ATP (c) NADH (d) NADPH
7. All organisms can incorporate NH +
4
into glutamate and glutamine using glutamate de hydrogenase and glutamine synthetase. Why do prokaryotes have an additional enzyme, glutamate synthase, to perform this function?
Amino Acids Are Made from Intermediates of the Citric Acid Cycle
and Other Major Pathways 8. Which of the following amino acids are essential dietary components for an adult human?
(a) alanine (b) aspartate (c) histidine (d) tryptophan (e) leucine (f) phenylalanine (g) glutamine (h) asparagine (i) glutamate
(j)
threonine (k) methionine
THE BIOSYNTHESIS OF AMINO ACIDS
429
9. Which of the following amino acids are derived from pyruvate?
(a) phenylalanine (b) alanine (c) tyrosine (d) histidine (e) valine (f) leucine (g) cysteine (h) glycine
10. Which of the following amino acids are derived from a-ketoglutarate?
(a) glutamate (e) glutamine (b) proline (f)
arginine (c) cysteine (g) ornithine (d) aspartate (h) serine
11. Which of the following compounds provide the carbon skeletons of the six biosynthetic families of amino acids? Name the metabolic pathways from which each of the precursor compounds originates.
(a) pyruvate (b) oxaloacetate (c) a-ketoglutarate (d) succinate (e) 2-deoxyribose (f) 3-phosphoglycerate
(g) ribose 5-phosphate (h) glucose 6-phosphate (i)
phosphoenolpyruvate
(j) erythrose 4-phosphate (k) a-ketobutyrate
12. Three coenzymes are involved in carrying activated one-carbon units. Match the acti vated group in the right column with the appropriate coenzyme in the left column.
(a) tetrahydrofolate (1) −CH3 (b) S-adenosylmethionine (2) −CH −
2
(c) biotin (3) −CHO (4) −CHNH (5) −CH= (6) −CO −
2
13. Which of the following answers completes the sentence correctly? The major source of one-carbon units for the formation of the tetrahydrofolate derivative N5,N10methylenetetrahydrofolate is the conversion of (a) methionine to homocysteine.
(b) deoxyuridine 5′-phosphate to deoxythymidine 5′-phosphate.
(c) 3-phosphoglycerate to serine.
(d) serine to glycine.
14. Why does S-adenosylmethionine have a higher methyl group–transfer potential than N5 methyltetrahydrofolate?
430
CHAPTER 24
15. S-adenosylmethionine is involved directly in which of the following reactions?
(a) methyl transfer to phosphatidyl ethanolamine (b) synthesis of glycine from serine (c) DNA methylation (d) conversion of homocysteine into methionine (e) synthesis of ethylene in plants
16. How many high-energy bonds are expended during the synthesis of S-adenosylmethionine from ATP and methionine?
17. The conversion of homocysteine into methionine involves which of the following cofactors?
(a) N5-methyltetrahydrofolate (b) N5,N10-methylenetetrahydrofolate (c) methylcobalamin (d) pyridoxal phosphate
18. Which of the following are intermediates in the pathway for the biosynthesis of both phenylalanine and tryptophan?
(a) anthranilate (b) chorismate (c) shikimate (d) prephenate
19. The two binding sites for indole on tryptophan synthetase subunits a and b are about 25 Å apart. Explain how indole is transferred between these sites.
Amino Acid Biosynthesis Is Regulated by Feedback Inhibition 20. In the following biosynthetic pathway A
B
C
D
E
F G, which is likely to be the committed step? Which compound is likely to inhibit the committed step?
21. Since glutamine is an important source of nitrogen in biosynthetic reactions, the enzyme that synthesizes it is carefully regulated. Which of the following compounds act as inhibitors of glutamine synthetase in E. coli?
(a) tryptophan (b) histidine (c) carbamoyl phosphate (d) glucosamine 6-phosphate (e) AMP (f)
CTP (g) alanine (h) glycine
22. How does covalent modification contribute to the regulation of glutamine synthetase in E. coli?
THE BIOSYNTHESIS OF AMINO ACIDS
431
Amino Acids Are Precursors of Many Biomolecules 23. Glutathione is composed of which of the following amino acids?
(a) glutamine (b) glutamate (c) methionine (d) cysteine (d) glycine
24. Which of the following answers complete the sentence correctly? Glutathione
(a) cycles between oxidized and reduced forms in the cell.
(b) is involved in the detoxification of H2O2 and organic peroxides.
(c) donates amide groups from its g-glutamyl residue during biosynthetic reactions.
(d) contains an Se atom.
25. Which of the following statements about nitric oxide (NO) is correct?
(a) It is used as an inhalation anesthetic.
(b) It is a long-lived signal molecule.
(c) It is produced from asparagine.
(d) Its synthesis requires NADPH and O2.
(e) Its synthesis requires ATP and NH + 4 .
26, Which of the following are intermediates or precursors in the synthesis of heme?
(a) a-aminolevulinic acid (e) glycine (b) bilirubin (f)
succinyl CoA (c) porphobilinogen (g) Fe2+ (d) biliverdin ANSWERS TO SELF-TEST
1. Nitrogen fixation is the process by which nitrogen present in the atmosphere as N2 is enzymatically converted to NH3 by some bacteria and blue-green algae. This process is crucial to all other organisms because they can use only NH + 4 , and not N2, as the source of nitrogen for biosynthesis.
2. g, f, a, b, c, e, d
3. The net equation for nitrogen fixation is
N +
+
+
2
8 e− + 16 ATP + 16 H2O
2 NH3 16 ADP + 16 Pi
8 H+ + H2 The eight electrons needed to reduce N2 are supplied by oxidative processes in nonphotosynthetic nitrogen-fixing organisms and by light energy from the sun in photosynthetic nitrogen-fixing organisms. The ATP requirement is met by the usual oxidative or photosynthetic mechanisms of the cells.
432
CHAPTER 24
4. (a) 2, 3, 6 (b) 1, 2, 4, 5 5. (a) 2 (b) 1 (c) 3 6. (a) 1, 2 (b) 2 (c) None (d) 1, 3. Glutamate dehydrogenase uses NADPH when catalyz ing reductive aminations and NAD+ when carrying out oxidative deaminations.
7. Glutamate synthase catalyzes the reductive amination of a-ketoglutarate in a reaction with glutamine to form two glutamates. Glutamate can also be made from NH +
4
and a-ketoglutarate using glutamate dehydrogenase. However, this route requires high concentrations of NH +
+
4 because of the high KM of the enzyme for NH4 . Prokaryotes can use glutamine synthetase, which has a low K
+
+
M for NH4 , to form glutamine when NH4 concentrations are low. Thus, by using an additional enzyme, they can form glutamate from glutamine and a-ketoglutarate when NH + 4 is scarce.
8. c, d, e, f, j, k 9. b, e, f 10. a, b, e, f, g. Recall that ornithine is a precursor of arginine in the urea cycle and is derived from a-ketoglutarate.
11. The biosynthetic precursors are a, b, c, f, g, i, and j. The pathways they originate from are as follows:
Glycolysis: pyruvate (a) 3-phosphoglycerate (f) phosphoenolpyruvate (i)
Citric acid cycle: oxaloacetate (b) a-ketoglutarate (c) Pentose phosphate pathway: ribose 5-phosphate (g) erythrose 4-phosphate (j) 12. 1, 2, 3, 4, 5 (b) 1 (c) 6 13. d. Furthermore, since 3-phosphoglycerate can give rise to serine, you can see how carbohydrates can provide activated one-carbon units via a glucose 3-phosphoglycerate
serine
glycine pathway.
14. The positive charge on the sulfur atom of S-adenosylmethionine activates the methyl sul fonium bond and makes methyl group transfer from S-adenosylmethionine energetically more favorable than from N5-methyltetrahydrofolate.
15. a, c, e. The cofactor for reactions (b) and (d) is tetrahydrofolate.
16. Three high-energy bonds are expended. The adenosyl group of ATP is condensed with methionine to form a carbon-to-sulfur bond with the release of Pi and PPi, which is hydrolyzed to 2 Pi.
17. a, c. Homocysteine transmethylase uses a vitamin B12–derived cofactor.
18. b, c 19. Tryptophan synthetase contains a 25-Å-long channel between the active sites of adjacent a and b subunits. This channel allows the diffusion of the intermediate, indole, through the protein from one binding site to the other without diffusing away from the enzyme.
This alleviates the potential problem of the hydrophobic indole molecule diffusing across the plasma membrane and out of the cell were it allowed to leave the enzyme.
THE BIOSYNTHESIS OF AMINO ACIDS
433
20. A B. Control of the first step conserves the first compound, A, in the sequence and also saves metabolic energy by preventing subsequent reactions in the pathway.
Compound G would likely inhibit the committed step. The end product of a biosynthetic pathway often controls the committed step.
21. All the choices are correct. When all eight compounds are bound to the enzyme, it is al most completely inactive. The control of this enzyme is an excellent example of cumulative feedback inhibition.
22. Glutamine synthetase can be covalently modified by the attachment of an AMP to each of its 12 subunits. The more adenylylated the enzyme becomes, the more susceptible it is to feedback inhibition by the compounds listed in question 21. Thus, covalent modification modulates the sensitivity of the enzyme to its effectors. An added level of control exists in this system; adenylyltransferase, the enzyme that adenylylates glutamine synthetase, is itself covalently modified.
23. b, d, e 24. a, b. Answer (d) is incorrect because the enzyme glutathione peroxidase, rather than glu tathione, contains an Se analog of cysteine.
25. d. Note that (c) is incorrect because arginine rather than asparagine is the precursor of NO.
26. a, c, e, f, g
PROBLEMS
1. The glyA− mutation in Chinese hamster ovary cells in tissue culture makes these cells partially dependent on glycine. The mutation affects the mitochondrial form of serine transhydroxymethylase, which catalyzes the conversion of serine to glycine, with tetrahydrofolate serving as an acceptor of the hydroxymethyl group. Would you expect heme synthesis to be adversely affected in glyA− mutants? Why?
2. In early nutritional studies, cysteine was thought to be an essential amino acid. In 1937, Abraham White and E. F. Beach showed that cysteine could be removed from protein hydrolysates with cuprous mercaptide. Rats fed on such treated hydrolysates could grow, provided that sufficient methionine was supplied in the diet.
(a) What did this result reveal about cysteine metabolism?
(b) What would you expect the result to be if the rats were fed homocysteine along with the treated hydrolysates?
3. The essential amino acids are those that cannot be synthesized de novo in humans. Given an abundance of other amino acids in the diet, the a-keto acid analogs that correspond to the essential amino acids can substitute for these compounds in the diet.
(a) What do these observations tell you about the steps in the synthesis of essential amino acids that may be missing in humans?
(b) If 15N-labeled alanine is supplied in the diet, many other amino acids in the body will contain at least a small amount of the label within 48 hours. What enzymes are primarily responsible for this observation?
4. In muscle, glutamine synthetase is very active, catalyzing the formation of glutamine from glutamate and ammonia at the expense of a molecule of ATP. In the liver, the rate 434
CHAPTER 24 of formation of glutamine is very low, but a high level of glutaminase activity, which generates ammonia and glutamate, is observed. How would you explain the difference in the levels of enzyme activity in these two organs?
5. Consider three forms of bacterial glutamine synthetase: GS, the deadenylylated form; GS–(AMP)1, a form with one AMP unit per 12 subunits; and GS–(AMP)12, the fully adenylylated form.
(a) Which of these forms is most sensitive to feedback inhibition by several of the final products of glutamine metabolism, such as tryptophan or histidine? Why is it important that the activity of the most sensitive form not be completely inhibited by tryptophan?
(b) Which form has the lowest KM for ammonia?
(c) Why is it important that adenylyl transferase not carry out adenylylation and dead enylylation of glutamine synthetase at the same time?
(d) Glutamine synthetase in mammals is not subject to the same type of complex reg ulation that is seen in bacteria. Why?
6. Most of the proteins synthesized in mammals contain all 20 common amino acids. More protein is degraded than is synthesized when even one essential amino acid is missing from the diet.
(a) Under such conditions, how could an increase in the rate of protein degradation provide the missing amino acid?
(b) How does an increase in the rate of protein degradation contribute to increased levels of nitrogen excretion?
7. The diagram below outlines the biosynthesis of a compound that is required for the oxidation of fatty acids in the mitochondrion.
CH
CH
CH
+
3
3
3
+
+
+
NH
J
J
J
3
H CJNJCH
H CJNJCH
H CJNJCH
J
3
3
3
3
3
3
CH
J
J
J
2
CH
CH
CH
J
2
2
2
CH
J
J
a-Ketoglutanate Succinate+CO
J
2
CH
CH
2
HOJCJH
J 3 B
2
2
CH
D
J
DDD
J
J
2
CH
CH
CH
J
2
2
2
O
CH
J
J
J
2
2
CH COO: COO:
J
+
2
HJCJNH
J
+
D
3
HJCJNH
J
3
COO:
J COO:
A
C (a) Name compound D and briefly explain its role in fatty acid metabolism.
(b) Name compound A. Why is it considered essential in human diets?
(c) Three molecules of compound B are required for the formation of compound C. Its synthesis depends on the availability of an essential amino acid. Name that amino acid and then name compound B and compound C.
THE BIOSYNTHESIS OF AMINO ACIDS
435
8. A pathway for the synthesis of ornithine from glutamate is shown in Figure 24.1.
FIGURE 24.1 Biosynthesis of ornithine from glutamate.
COO:
O
H COO:
O
H COO:
J
O
K
J
J
K
J
J H ;NJCJH
K H CJCJNJCJH ;
H CJCJNJCJH
3
H CJCJSJCoA
3
NADPH NADP
3
J
3
J
J
CH
CH
CH
2
2
2
J
J
J
CH
CH
CH
2
2
2
J
J
J COO: COO:
CKO
Glutamate
N -Acetylglutamate H J
N -Acetylglutamate-g - semialdehyde
Glutamatea -Ketoglutarate COO:
O
H COO:
;
J
K
J
J H NJCJH
H CJCJNJCJH
3
3
J
J
:
CH
H CJCOO
3
H O
CH
2
2
2
J
J
CH
CH
2
2
J
J NH ; NH ;
3
3
Ornithine
N -Acetylornithine (a) Why can this pathway also be considered to be part of the de novo pathway for the synthesis of arginine?
(b) Inspect the pathway for proline biosynthesis given on page 674 of the text, and then explain why the N-acetylation of glutamate is needed for the synthesis of ornithine.
9. Elevated levels of ammonia in blood can result from deficiencies in one or another of the enzymes of the urea cycle. Measures taken to relieve hyperammonemia have included limiting intake of dietary proteins, administering a-keto analogs of several of the naturally occurring L-amino acids, or administering other compounds designed to exploit pathways of nitrogen metabolism and excretion.
(a) In trying to determine why a patient has hyperammonemia, which organ should you check first for normal function? Why?
(b) Why would limiting protein intake assist in relieving chronic hyperammonemia?
Why would eliminating dietary proteins altogether (without any other supplement) probably increase the level of hyperammonemia?
(c) Write a brief rationale for using a-keto acid analogs in treating hyperammonemia, mentioning a particular group of enzymes essential to your explanation. Would it be better to use a-keto analogs of essential or nonessential amino acids? Why?
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CHAPTER 24
10. Plants synthesize all 20 common amino acids de novo. Glyphosate, a weed killer sold under the trade name Roundup, is an analog of phosphoenolpyruvate that specifically inhibits 3-enolpyruvylshikimate 5-phosphate synthase, a key enzyme of the pathway for chorismate biosynthesis. This compound is a very effective plant herbicide, but has virtually no effect on mammals. Why?
11. In B. subtilis, the pathway from chorismate to tryptophan is feedback-inhibited by tryptophan, which suppresses anthranilate synthase activity. Mutant B. subtilis that lacks tryptophan synthetase can grow on minimal medium only when supplemented with exogenous tryptophan. Under these conditions, none of the intermediates in the tryptophan biosynthetic pathway from anthranilate to indole 3-glycerol phosphate are produced. However, when the bacteria have depleted the medium of tryptophan, the levels of those intermediates increase, even though there is no net production of tryptophan. Why?
12. Both genetic and biochemical methods have been used to establish the biosynthetic path ways for essential amino acids in bacteria and other microorganisms. One classic approach is isotope competition, which begins with the use of radioactive glucose as the sole source of carbon for growing bacteria. Under these conditions, all the intermediates in a particular pathway will be uniformly labeled, but if a nonradioactive intermediate in the pathway is added to the growing cells, it will reduce or dilute the radioactivity of that intermediate and others following it in a pathway.
Britten and his coworkers (R. B. Roberts, P. H. Abelson, D. B. Cowie, E. T. Bolton, and R. J. Britten, Studies in Biosyntheses in Escherichia coli. Carnegie Institution of Washington, Pub. No. 607, 1955), used isotope competition to examine the biosynthetic pathway for threonine, methionine, and other amino acids. They grew E. coli in a minimal medium containing labeled glucose and nonlabeled homoserine, which was known to relieve auxotrophy for several amino acids. Under these conditions, isoleucine, threonine, and methionine isolated from the cells had little or no radioactivity. In contrast, the radioactivity of aspartate and lysine was unchanged whether cells were grown with or without the addition of nonradioactive homoserine.
In a similar experiment, nonradioactive aspartate was used in the growth medium; as partate from protein hydrolysates was virtually nonradioactive, as were lysine, methionine, threonine, and isoleucine. When nonlabeled threonine was used along with radioactive glucose, only threonine and isoleucine from protein hydrolysates had reduced radioactivity. And when either nonradioactive isoleucine or methionine was used, they affected only their own levels of radioactivity in protein hydrolysates.
On the basis of these observations, write an outline of the biosynthetic pathway for the amino acids isoleucine, threonine, homoserine, methionine, and lysine.
ANSWERS TO PROBLEMS
1. Glycine is an obligatory precursor of heme; in the reaction catalyzed by d-aminolevuli nate synthase, glycine condenses with succinyl CoA to form d-aminolevulinate. The reduction in the concentration of glycine in the cell caused by the glyA− mutation will cause a decrease in the rate of heme synthesis.
2. (a) The studies led White and Beach to conclude that methionine, which is another sulfur-containing amino acid, is a biosynthetic precursor of cysteine. Later work elucidated the roles of methionine in the active methyl cycle and in the synthesis of cysteine and confirmed their conclusion.
THE BIOSYNTHESIS OF AMINO ACIDS
437
(b) In the active methyl cycle, S-adenosylhomocysteine is cleaved to yield adenosine and homocysteine. Although homocysteine can be converted to methionine, it can also condense with serine to form cystathionine, which is then cleaved to yield cysteine and a-ketobutyrate. Feeding rats homocysteine along with a treated hydrolysate will make supplementation with methionine unnecessary.
3. (a) The fact that a-keto acid analogs can substitute for essential amino acids means that the carbon skeletons of the essential amino acids are not synthesized in humans.
Many studies have shown that one or more of the enzymes needed for the synthesis of these structures are missing.
(b) The enzymes that are primarily responsible for the distribution of the 15N label among the other amino acids are the aminotransaminases, which catalyze the interconversions of amino acids and their corresponding a-keto acids. The redistribution of the label begins with the transamination of alanine, with a-ketoglutarate serving as the amino acceptor to yield pyruvate and glutamate. Glutamate then serves as an amino donor for other a-keto acids. In order for an essential amino acid to be labeled, you must postulate the transamination of that amino acid to yield the corresponding a-keto acid analog, followed by the donation of a labeled amino group from glutamate or another donor of amino groups.
4. Ammonia, which is generated as part of the process of amino acid catabolism in muscle, is toxic and must be removed from the cells. This could be done through the synthesis of urea, but that process occurs only in the liver. In muscle cells, therefore, glutamine synthetase catalyzes the formation of glutamine, which is an efficient and nontoxic carrier of ammonia. This accounts for the high activity of that enzyme in muscle. The glutamine is transported by the blood to the liver, where glutaminase and aspartate aminotransferase work together to generate aspartate and two molecules of ammonia from glutamine, hence, the high activity of glutaminase in the liver. Aspartate and ammonia are both used by the liver for the synthesis of urea, a nontoxic and disposable form of ammonia.
5. (a) The fully adenylylated form of glutamine synthetase is the most sensitive to mole cules like tryptophan and histidine. Because glutamine is utilized for the synthesis of a variety of compounds, complete inhibition of the enzyme by only one of those products, such as tryptophan, would inappropriately inhibit the synthesis of all the others. Thus, the enzyme is cumulatively inhibited by at least eight different nitrogen-containing compounds.
(b) The deadenylylated form, which is not subject to cumulative feedback inhibition, is generated in response to increases in the cellular concentrations of a-ketoglutarate (the precursor of glutamate) and ATP. These molecules signal the need for glutamine synthesis, even when other nitrogen-containing compounds are present.
Under these conditions, the deadenylylated form of the enzyme binds ammonia even when ammonia concentrations are relatively low; that is, it has a relatively low KM for ammonia.
(c) The simultaneous adenylylation and deadenylylation of glutamine synthetase would result in a loss of feedback control of the enzyme because the adenylylated form is subject to cumulative inhibition whereas the deadenylylated form is not. In addition, it would lead to the wasteful hydrolysis of ATP, since every round of adenylylation and deadenylylation generates AMP and inorganic pyrophosphate from ATP.
(d) Mammals acquire many nitrogen-containing compounds, such as tryptophan and histidine, in their diet rather than through de novo biosynthesis, so glutamine synthetase does not play so prominent a role in the nitrogen metabolism of mammals as it does in that of bacteria. Complex regulation of the enzyme is therefore not needed in mammals.
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CHAPTER 24
6. (a) Many experiments have shown that under normal conditions cells continuously synthesize and degrade proteins. Although both essential and nonessential amino acids are continuously recycled during these processes, reutilization is not completely efficient; thus, additional amino acids are needed. In mammals, there are no reservoirs of free amino acids; the only sources of essential amino acids are dietary proteins or the proteins of the body tissues. If an essential amino acid is not available from the diet, cells appear to accelerate the hydrolysis of their own proteins, in order to generate the missing essential amino acid. How the rate of cellular proteolysis is accelerated in response to a deficiency of an essential amino acid is not understood.
(b) An increased rate of protein degradation generates a higher concentration of free amino acids. During the oxidation of those amino acids not used for synthesis of other proteins, ammonia will be produced. An elevation in ammonia concentration in the body stimulates the formation of urea, causing the level of nitrogen excretion to increase.
7. (a) Compound D is carnitine, which, when esterified to the acyl group of a long-chain fatty acid, shuttles it from the cytosol to the matrix of the mitochondrion, where fatty acid oxidation takes place.
(b) Compound A is the essential amino acid lysine; it is termed essential because it can not be synthesized de novo in humans. Lysine and other essential amino acids must be obtained from the diet.
(c) Compound C is trimethyllysine, and the methyl groups that are attached to lysine are likely to be derived from compound B, S-adenosylmethionine, the major donor of methyl groups in biosynthetic reactions. The methyl group of S-adenosylmethionine is derived from methionine, an essential amino acid.
8. (a) Ornithine is a precursor of arginine, as part of the pathway for the synthesis of urea.
Thus the pathway for the synthesis of ornithine from glutamine, along with part of the urea cycle pathway, can together be considered as a de novo pathway for the synthesis of arginine. Arginine can in turn be used for the synthesis of urea, or it can serve instead as one of the amino acids used for polypeptide synthesis.
(b) In the pathway for proline biosynthesis, glutamic-g-semialdehyde cyclizes with the loss of water to form D1-pyrroline-5-carboxylate. However, in the pathway for the formation of ornithine shown in Figure 24.1, an N-acetylated derivative of the semialdehyde molecule is formed. The N-acetyl group blocks the condensation of the amino group with the aldehyde group, thereby preventing the formation of the pyrroline ring. This allows the pathway to proceed toward the synthesis of ornithine.
9. (a) You should assess liver function, because enzymes of the urea cycle are found pri marily in this organ. In addition, liver takes up amino acids such as alanine, glutamate, and glutamine, which are in effect nontoxic forms of ammonia generated by muscle and other tissues. Amino groups of these compounds, along with carbon dioxide from carbamoyl phosphate, are precursors of urea in the liver.
(b) During digestion, dietary proteins are hydrolyzed to their component amino acids.
Those amino acids that are not needed immediately for protein synthesis or for the biosynthesis of other nitrogen-containing compounds are degraded. One of the products of amino acid degradation is ammonia. Usually ammonia is metabolized through conversion to nitrogen carriers such as alanine, glutamate, and glutamine, and it is ultimately utilized for the synthesis of urea when the urea cycle is operat ing. Thus limiting protein intake in a patient with a deficiency in urea synthesis would be expected to reduce ammonia production in liver and other tissues.
Protein turnover and amino acid degradation constantly take place in the tissues, and essential amino acids (those that cannot be synthesized de novo in human tissues) must be generated either from dietary sources or from additional breakdown of body proteins. A complete restriction of dietary protein would accelerate body protein breakdown and would exacerbate the condition of hyperammonemia.
(c) The a-keto acid analogs can serve as acceptors for amino groups from glutamate and other amino acids, in reactions catalyzed by transaminases or aminotransferases.
Nonessential amino acids formed as the result of this process may be themselves eliminated, degraded (often generating more ammonia), or else used for biosynthesis of other nitrogen-containing compounds. Employing analogs of essential amino acids might be preferable, because tissues are more likely to require them for protein synthesis or other biosyntheses.
10. Chorismate is an intermediate in the biosynthesis of the aromatic amino acids trypto phan, phenylalanine, and tyrosine. Mammals do not synthesize these amino acids from chorismate. Instead, they obtain the essential aromatic amino acids tryptophan and phenylalanine from the diet, and they can synthesize tyrosine from phenylalanine.
Glyphosate is an effective herbicide because it prevents synthesis of aromatic amino acids in plants. But the compound has no effect on mammals because they have no active pathway for de novo aromatic amino acid synthesis.
11. Since anthranilate synthase is inhibited by tryptophan, exogenous tryptophan from the medium, inhibits the production of downstream intermediates in the biosynthetic pathway by halting the first step in the pathway. When exogenous tryptophan is depleted, intracellular levels of tryptophan decrease and anthranilate synthase activity is no longer inhibited. An increase in anthranilate production leads to an increase in the production of other intermediates in the pathway until equilibria are established among those compounds. The block at the step catalyzed by tryptophan synthetase prevents large accumulations of the intermediates, but under these conditions their concentrations are higher than in the presence of exogenous tryptophan.
12. The experiments show that aspartate is a key precursor for the five amino acids, because when it is included in the growth medium with labeled glucose, radioactivity of each of the amino acids is reduced. Homoserine affects the labeling of isoleucine, threonine, and methionine. Thus, homoserine is an intermediate in the pathway for those three amino acids but not lysine or aspartate. Dilution of threonine and isoleucine from cells grown in nonradioactive threonine suggests that those two amino acids are on the same pathway, separated from those of the other three. Methionine and isoleucine are affected only when they are used in growth experiments, indicating that they are not on other pathways (and, for isoleucine, that threonine must precede it in a biosynthetic pathway).
The overall pathway below shows that lysine and homoserine derived from aspartate.
Threonine and isoleucine are both derived from aspartate, and homoserine must be an intermediate in their synthesis, as well as in the synthesis of methionine.
Aspartate
Homoserine
Threonine
Isoleucine
Lysine
Methionine
440
CHAPTER 24 EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. Glucose + 2 ADP + 2 P +
i
2 NAD+ + 2 glutamate 2 alanine + 2 a-ketoglutarate + 2 ATP + 2 NADH + H+ Glucose is converted to pyruvate via glycolysis, and pyruvate is converted to alanine by transamination.
2. N
+
2
NH4
glutamate
serine
glycine
d-aminolevulinate
porphobilinogen
heme
3. (a) N5,N10-methylenetetrahydrofolate, (b) N5-methyltetrahydrofolate 4. g-Glutamyl phosphate is a likely reaction intermediate.
5. The administration of glycine leads to the formation of isovalerylglycine. This water-sol uble conjugate, in contrast with isovaleric acid, is excreted very rapidly by the kidneys.
See R. M. Cohn, M. Yudkoff, R. Rothman, and S. Segal. New Engl. J. Med. 299(1978):996.
6. They carry out nitrogen fixation. The absence of photosystem II provides an environ ment in which O2 is not produced. Recall that the nitrogenase is very rapidly inactivated by O2. See R. Y. Stanier, J. L. Ingraham, M. L. Wheelis, and P. R. Painter, The Microbial
World, 5th ed. (Prentice-Hall, 1986), pp. 356–359, for a discussion of heterocysts.
7. The cytosol is a reducing environment, whereas the extracellular milieu is an oxidizing environment.
8. The synthesis of d-aminolevulinate requires succinyl CoA, an intermediate in the citric acid cycle, which occurs in the mitochondrial matrix. Thus it makes sense to have the first step in porphyrin biosynthesis occur in the matrix.
9. Alanine and aspartate can be synthesized directly from glutamate by transamination of pyruvate and oxaloacetate, respectively. Glutamate is the common intermediate that serves as the amino group donor in these reactions. (a-Ketoglutarate, the side product, can condense with another molecule of ammonia to regenerate the glutamate for further transamination reactions.)
10. Each final product should inhibit the first unique step toward its synthesis, that is, the first step following the branch point of the pathway. Thus, Y should inhibit the conversion of C to D, and Z should inhibit the conversion of C to F. (To avoid wasteful accumulation of C or B if neither Y nor Z is needed, high levels of C should also inhibit the conversion of A to B at the beginning of the pathway.)
11. The effects would multiply, so that the net rate would equal (100 × 0.6 × 0.4 s−1), or 24−1 s.
12. The reaction will begin with the formation of a Schiff base between pyridoxal-5′-phos phate and the amino group of S-adenosylmethionine (external aldimine, as opposed to the internal aldimine in which PLP is connected to a lysine on the enzyme). In usual fashion for PLP enzymes, the a-hydrogen will be extracted from S-adenosylmethionine (step 2, below). Then the next series of p-electron transfers (step 3) will eliminate Smethylthioadenosine and form the three-membered ring. Finally, 1-aminocyclopropane1-carboxylate (ACC) will be released, and the original internal aldimine between PLP and lysine will be restored, so that the enzyme and cofactor are ready for another round of synthesis. The second product is S-methylthioadenosine.
THE BIOSYNTHESIS OF AMINO ACIDS
441
S -adenosyl-Met
CH
CH
3
3
J
J
+
H
adenosyl-S-CH -CH COO:
S
N
H
2
2
H C
adenosyl
+
2
+
H C G
H
N
G O:
2
COO:
+
H
PO
=O
3
O:
H
N
=O PO
+ H +
N
CH
3
3
O:
J
H
+
N
CH
=O PO
3
3
internal aldimine
J
H
N
CH
external aldimine
3
COO:
(Lys)
COO:
+ S -methyl-thioadenosine
NH
3
+
H
N
N
H
G
G
+ H
H
O: O: =O PO
=O PO
3
3
+
N
CH3
+
N
CH3
J
J
H
H
13. As in problem 15 in Chapter 23 of this manual, an external aldimine (protonated Schiff base) is formed between serine and pyridoxal-5′-phosphate, and the serine a-hydrogen is removed. The a-hydrogen then can be reattached from either of two faces to give the Schiff base of either D-serine or L-serine (see below). The equilibrium constant for the reaction will be one.
L-approach
H;
H
HOCH COO:
HOCH COO:
2
2
H
N
H
N
D-approach
+
H
+ H
O: O:
G
=O PO =O PO
3
3
+
N
CH
N
CH
3
3
J
J
H
H
14. Aspartate and glutamate would be synthesized from the citric acid cycle intermediates oxaloacetate and a-ketoglutarate. Increased synthesis of aspartate and glutamate therefore could begin to deplete citric acid cycle intermediates. The cell would need to respond by breaking down carbohydrates to replenish the supply by net synthesis of new cycle intermediates.
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CHAPTER 24
15. Because S-adenosylmethionine (SAM) is the methyl donor for the methylation of DNA (Figure 24.15), a deficiency in SAM could diminish the extent of the methylation of the mutated bacteria’s DNA. The lower level of methylation would render the DNA more susceptible to digestion by restriction enzymes.
16. (a) Asn, Gln, and Gly are affected. Asn is much more concentrated in dark-adapted plants, whereas Gln and Gly are present in somewhat elevated levels in lightadapted plants.
(b) The transcription of specific mRNA, translation or enzymatic activity of several en zymes, particularly nitrogen-utilizing enzymes such as asparagine synthetase and glutamine synthetase, could be regulated by light.
(c) From the graph, asparagine would appear to be a likely candidate. (The name of the plant would also suit this interpretation!)
CHAPTER 2
Nucleotide Biosynthesis
5
In this chapter, the authors complete their treatment of the biosyntheses of the major classes of macromolecular precursors by describing the synthesis of the purine and pyrimidine nucleotides. Besides being the precursors of RNA and DNA, these compounds serve a number of other important roles that are reviewed in the opening paragraph of the chapter. Nucleotide nomenclature is reviewed in the introduction to the chapter, as is an outline for the synthesis of nucleotides through de novo and salvage pathways.
The chapter begins with the synthesis of the pyrimidine nucleotides. The pyrim idine ring is synthesized de novo from bicarbonate, aspartate, and ammonia (usually from glutamine) prior to attachment to a ribose sugar. The authors go through the synthesis step-by-step, paying particular attention to the enzyme carbamoyl phosphate synthetase (CPS), which synthesizes carbamoyl phosphate from bicarbonate and ammonia and catalyzes the committed step in eukaryotic pyrimidine synthesis.
The next step in the synthesis is catalyzed by aspartate transcarbamoylase (ATCase), an enzyme that was discussed in Chapter 10. This reaction, the formation of carbamoylasparate from carbamoyl phosphate and aspartate, is the committed step in prokaryotic pyrimidine synthesis. A condensation and an oxidation reaction complete the formation of orotate, which is then coupled to a phosphoribose by reaction with 5-ribosyl-1-pyrophosphate (PRPP) to form the pyrimidine nucleotide orotidylate.
Decarboxylation of orotidylate gives uridine monophosphate (UMP), which can be phosphorylated by nucleoside mono- and diphosphate kinases to form UDP and UTP, respectively. Amination of UTP forms cytidine triphosphate (CTP) and completes the synthesis of the pyrimidine ribonucleotides.
Next the authors turn to synthesis of the purine nucleotides. Unlike synthesis of pyrimidines, synthesis of purine nucleotides builds upon the ribose ring. As in the
443
444
CHAPTER 25 pyrimidine ring system, the ribose sugar is donated by the activated form of ribose 5-phosphate, 5-phosphoribosyl-1-pyrophosphate (PRPP). The two purine nucleotides AMP and GMP have a common precursor, inosine 5′-monophosphate (IMP). The authors discuss the synthesis of this initial purine product and then formation of AMP and GMP. The reactions that allow cells to salvage free purines and the control of purine biosynthesis are presented. The regulation of nucleotide biosynthesis through feedback inhibition is discussed later in the chapter.
Two reactions that are required to form the precursors of DNA are described in detail: ribonucleotide reductase converts ribonucleotides to deoxyribonucleotides, and thymidylate synthase methylates dUMP to form dTMP. The authors present the mechanisms and cofactors of these enzymes and explain how some anticancer drugs and antibiotics function by inhibition of dTMP synthesis and thus the growth of cells. Nucleotides also serve important roles as constituents of NAD+, NADP+, FAD, and coenzyme A (CoA), so the syntheses of these cofactors are described briefly. The chapter concludes with an explanation of how the purines are catabolized and some of the pathological conditions that arise from defects in the catabolic pathway of the purines.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. List the major biochemical roles of the nucleotides.
2. Distinguish among the purine and pyrimidine nucleosides and nucleotides.
In de Novo Synthesis, the Pyrimidine Ring Is Assembled from Bicarbonate,
Aspartate, and Glutamine (Text Section 25.1) 3. Draw the structure of a pyrimidine ring and identify the precursors that provide each carbon and nitrogen atom of the ring. Note the numbering of the ring atoms.
4. Discuss the structure and mechanism of carbamoyl phosphate synthetase (CPS). Explain the role of carbamoyl phosphate in pyrimidine biosynthesis.
5. Write the aspartate transcarbamoylase reaction and outline the remaining reactions that form orotate. Outline the conversion of orotate to uridylate (UMP).
6. Contrast the enzymes of pyrimidine biosynthesis in E. coli with those of higher organisms and list the potential advantages of multifunctional enzymes.
7. Explain how nucleoside mono- and diphosphate kinases interconvert the nucleoside mono-, di-, and triphosphates.
8. Describe the reaction that converts UTP to CTP.
Purines Can Be Synthesized de Novo or Recycled by Salvage Pathways
(Text Section 25.2) 9. Outline the synthesis of purine nucleotides by the salvage reactions and explain why these reactions are energetically advantageous.
10. Draw the structure of a purine ring and identify the precursors that provide each carbon and nitrogen atom of the ring. Note the numbering of the ring atoms.
NUCLEOTIDE BIOSYNTHESIS
445
11. Describe the committed step in de novo purine biosynthesis and the enzyme that cat alyzes it.
12. Explain the mechanism for replacing a carbonyl oxygen with an amino group, and list the potential sources of the nitrogen atom in these reactions.
13. Outline the synthesis of inosine 5′-monophosphate (IMP), noting the sources of the atoms and the cofactors involved. List the steps that require ATP.
14. Describe the synthesis of adenylate (AMP) and guanylate (GMP) from IMP. List the cofac tors and intermediates of the reactions.
Deoxyribonucleotides Are Synthesized by the Reduction of Ribonucleotides
Through a Radical Mechanism (Text Section 25.3) 15. Describe the subunit structure of ribonucleotide reductase. Outline its reaction, mecha nism, and regulation. Explain the roles of NADH, thioredoxin, and thioredoxin reductase
in the reaction.
16. Describe the thymidylate synthase reaction. Account for the source of the methyl group and describe the change in the oxidation state of the transferred carbon atom that occurs during the reaction.
17. Explain the role of dihydrofolate reductase in the synthesis of deoxythymidylate.
18. Account for the ability of fluorouracil to act as a suicide inhibitor. Describe the inhibitory mechanism of methotrexate and aminopterin. Explain how these three compounds interfere with the growth of cancer cells. List the mechanisms by which a cell could become resistant to methotrexate. Explain the antibiotic activity of trimethoprim.
Key Steps in Nucleotide Biosynthesis Are Regulated by Feedback Inhibition
(Text Section 25.4) 19. Outline the regulation of the biosynthesis of the purine and pyrimidine nucleotides and name the committed steps in the pathways.
NAD+, FAD, and Coenzyme A Are Formed from ATP (Text Section 25.5) 20. Describe the synthesis of NAD+ and account for the sources of the nicotinamide portion of the molecule.
21. Outline the synthesis of FAD and explain the role of the PPi that is released during the synthesis of FAD, NAD+, and CoA.
Disruptions in Nucleotide Metabolism Can Cause Pathological Conditions
(Text Section 25.6) 22. Describe the reactions of the nucleotidases and nucleoside phosphorylases.
23. Outline the conversions of AMP and guanine to uric acid. Describe the role of xanthineoxidase in these processes.
24. Describe the major clinical findings in patients with gout and explain the rationale for the use of allopurinol to alleviate the symptoms of the disease.
25. Describe the antioxidant role of urate.
26. Name the biochemical lesion that leads to the Lesch-Nyhan syndrome and describe the symptoms of the disease.
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CHAPTER 25
SELF-TEST
Introduction 1. Describe the physiological roles of the nucleotides.
2. Which of the following answers completes the sentence correctly? Cytosine is a (a) purine base.
(c) purine nucleoside.
(b) pyrimidine base.
(d) pyrimidine nucleoside.
3. Which of the following are nucleotides?
(a) deoxyadenosine (c) deoxyguanylate (b) cytidine (d) uridylate In de Novo Synthesis, the Pyrimidine Ring Is Assembled from Bicarbonate,
Aspartate, and Glutamine 4. Which of the following statements about the carbamoyl phosphate synthetase of mam mals, which is used for pyrimidine biosynthesis, are true?
(a) It is located in the mitochondria.
(b) It is located in the cytosol.
(c) It uses NH +
4
as a nitrogen source.
(d) It uses glutamine as a nitrogen source.
(e) It requires N-acetylglutamate as a positive effector.
5. Which of the following statements about 5-phosphoribosyl-1-pyrophosphate (PRPP) are true?
(a) It is an activated form of ribose 5-phosphate.
(b) It is formed from ribose 1-phosphate and ATP.
(c) It has a pyrophosphate group attached to the C-1 atom of ribose in the a configuration.
(d) It is formed in a reaction in which PPi is released.
6. How is orotate, a free pyrimidine, converted into a nucleotide? Is this reaction consid ered to be a salvage reaction or a biosynthetic one?
7. Which of the following enzymes are involved in converting the nucleoside 5′ monophosphate (NMP) products of the purine or pyrimidine biosynthetic pathways into their 5′-triphosphate (NTP) derivatives?
(a) purine nucleotidase (b) nucleoside diphosphate kinase (c) nucleoside monophosphate kinases (d) nucleoside phosphorylase 8. How is the exocyclic amino group on the N-4 position of cytosine formed?
Purines Can Be Synthesized de Novo or Recycled by Salvage Pathways 9. Which of the following compounds directly provide atoms to form the purine ring?
(a) aspartate (e) CO2 (b) carbamoyl phosphate (f) N5, N10-methylenetetrahydrofolate (c) glutamine (g) N10-formyltetrahydrofolate (d) glycine (h) NH +
4
(c) IMP.
(b) GMP.
(d) xanthylate (XMP).
11. The conversion of IMP to AMP requires which of the following?
(a) ATP (d) glutamine (b) GTP (e) NAD+ (c) aspartate
12. The conversion of IMP to GMP requires which of the following?
(a) ATP (d) glutamine (b) GTP (e) NAD+ (c) aspartate
13. Describe the general mechanism used by cells to replace a carbonyl group with an amino group in nucleotide biosynthesis.
14. Which of the following reactants and products are involved in the salvage reactions of purine biosynthesis?
(a) IMP
AMP (c) adenine
AMP (b) IMP
GMP (d) guanine
GMP
15. During a purine salvage reaction, what is the source of the energy required to form the C–N glycosidic bond between the base and ribose?
16. Show which of the nucleotides in the right column regulate each of the conversions in the left column.
(a) ribose 5-phosphate
PRPP
(1) AMP (b) PRPP
_phosphoribosylamine
(2) GMP (c) phosphoribosylamine
IMP
(3) IMP (d) IMP
adenylosuccinate (e) IMP xanthylate (XMP) Deoxyribonucleotides Are Synthesized by the Reduction of Ribonucleotides
Through a Radical Mechanism 17. Which of the following statements about ribonucleotide reductase are true?
(a) It converts ribonucleoside diphosphates into 2′-deoxyribonucleoside diphosphates in humans.
(b) It catalyzes the homolytic cleavage of a bond.
(c) It accepts electrons directly from FADH2.
(d) It receives electrons directly from thioredoxin.
(e) It contains two kinds of allosteric regulatory sites—one for control of overall activity and another for control of substrate specificity.
18. Select from the following those compounds that are precursors of 2′-deoxythymidine 5-triphosphate (dTTP) in mammals and place them in their correct biosynthetic order.
(a) OMP (f)
dUDP (b) UMP (g) dUTP (c) UDP (h) dTMP (d) UTP (i)
dTDP (e) dUMP (j)
dTTP
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CHAPTER 25
19. Define suicide inhibitor and give an example from pyrimidine biosynthesis.
20. Methotrexate and trimethoprim are both inhibitors of dihyrofolate reductase. Why is trimethoprim the drug of choice in treating a human microbial infection?
Key Steps in Nucleotide Biosynthesis Are Regulated by Feedback Inhibition 21. What is the committed step in purine biosynthesis and which of the following com pounds are involved in the control of the purine biosynthetic pathway?
(a) IMP (b) AMP (c) OMP (d) GMP (e) PRPP NAD+, FAD, and Coenzyme A Are Formed from ATP 22. Which of the following can serve as a precursor of NAD+ in humans?
(a) riboflavin (d) tryptophan (b) pantothenate (e) niacin (c) tyrosine Disruptions in Nucleotide Metabolism Can Cause Pathological Conditions 23. Which of the following compounds would give rise to urate if they were catabolized completely in humans?
(a) ADP-glucose (e) CoA (b) GDP-mannose (f)
FAD (c) CDP-choline (g) UMP (d) UDP-galactose 24. What is the benefit of high serum levels of urate in humans given that too much urate leads to gout?
ANSWERS TO SELF-TEST
1. The nucleotides (1) are the activated precursors of DNA and RNA; (2) are the source of derivatives that are activated intermediates in many biosyntheses; (3) include ATP, the universal currency of energy in biological systems, and GTP, which powers many movements of macromolecules; (4) include the adenine nucleotides, which are components of the major coenzymes NAD+, FAD, and CoA; and (5) serve as metabolic regulators.
2. b 3. c, d. Nucleotides are nucleosides that contain one or more phosphate substituents on their ribose or deoxyribose moieties.
4. b, d. The mitochondrial carbamoyl phosphate synthetase used for urea synthesis is ac tivated by N-acetylglutamate and uses N +
4
as the nitrogen source.
NUCLEOTIDE BIOSYNTHESIS
449
5. a, c.
6. Orotate condenses with PRPP in a reaction catalyzed by orotate phosphoribosyl trans ferase to form the nucleotide orotidylate (OMP). Orotidylate decarboxylase converts OMP to the more abundant nucleotide UMP. The reaction occurs during de novo pyrimidine biosynthesis and is therefore not a salvage reaction.
7. b, c. Several different specific nucleoside monophosphate kinases phosphorylate dNMPs and NMPs, using ATP as the phosphoryl donor. A single enzyme, nucleoside diphosphate kinase, uses the phosphorylation potential of ATP to convert the dNDPs and NDPs to dNTPs and NTPs. The ubiquitous adenylate nucleotides are interconverted by adenylate kinase (myokinase).
8. CTP is formed by amination of UTP. The carbonyl oxygen at C-4 of UTP is replaced with an amino group via the formation of an enol phosphate ester intermediate. In E. coli,
NH +
4
serves as the source of the nitrogen atom that displaces the phosphate group, whereas the amide group of glutamine serves this purpose in mammals.
9. a, c, d, e, g 10. c 11. b, c 12. a, d, e 13. A carbonyl oxygen is converted into a phosphoryl ester or a substituted phosphoryl ester through a reaction with a high-energy phosphate (ATP or GTP) to form a mono- or diphosphate ester. The phosphate or pyrophosphate group can then be readily displaced by the nucleophilic attack of the nitrogen atom from NH3, the side-chain amide group of glutamine, or the a-amino group of aspartate. The resulting adduct is an amino group or can be converted into one.
14. c, d 15. The activated form of ribose 5-phosphate (R-5-P), PRPP, reacts with the purine base to form the nucleotide and release PPi. The displacement and subsequent hydrolysis of PPi drives the formation of the N-glycosyl bond. ATP ultimately provides the energy through its reaction with R-5-P to form PRPP.
16. (a) 1, 2, 3 (b) 1, 2, 3 (c) none (d) 1 (e) 2 17. a, b, d, e. NADPH provides electrons via thioredoxin.
18. a, b, c, f, g, e, h, i, j. In mammals, NDPs are converted to dNDPs by ribonucleotide re ductase. Thus, UDP is converted to dUDP, which is converted to dUTP by nucleoside diphosphate kinase. A specific pyrophosphatase hydrolyzes dUTP to dUMP, which is then converted to dTMP by thymidylate synthase. The dTMP is converted to dTTP. A priori, you might have expected the dUDP product of the ribonucleotide reductase reaction to be converted directly to dUMP. However, in fact, cells contain a dUTP pyrophosphatase to prevent dUTP from serving as a DNA precursor, and it is this enzyme that functions in the dTTP biosynthetic pathway.
19. A suicide inhibitor is a substrate that is converted by an enzyme into a substance that is capable of reacting with and inactivating the enzyme. In pyrimidine biosynthesis, thymidylate synthase converts fluorouracil into a derivative that becomes covalently attached to the enzyme and thereby inactivates it.
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CHAPTER 25
20. Since trimethoprim binds to the mammalian dihydrofolate reductase much less tightly than to the enzyme of susceptible microorganisms, it causes fewer deleterious effects to humans than does methotrexate.
21. The conversion of PRPP into phosphoribosylamine by glutamine phosphoryl amidotrans-ferase is the committed step in purine biosynthesis. Compounds a, b, and d are involved in the regulation.
22. d, e. Niacin becomes a dietary requirement if the supply of the essential amino acid tryp tophan is inadequate.
23. a, b, e, f. Each of these compounds contains a heterocyclic purine base.
24. Urate levels in humans are often close to the solubility limit, which leads to gout when salts of urate crystallize (resulting in damage to joints and kidneys). There is a significant benefit to high concentrations of urate, however, as urate is a highly effective scavenger of reactive oxygen species (ROS). ROS can cause damage in cells contributing to cancer and the effects of aging. Urate is about as effective as ascorbate (vitamin C) as an antioxidant.
PROBLEMS
1. Why might covalently linked (multifunctional) enzymes, such as those of the pyrimi dine biosynthetic pathway of mammals, be advantageous to an organism?
2. Mammalian lymphocytes that lack adenosine deaminase neither grow nor divide. The level of dATP in these cells is 100 times higher than that in normal lymphocytes, and the synthesis of DNA in the cells is impaired.
(a) How is adenosine converted to dATP? Assume that the first step is catalyzed by a specific nucleoside kinase.
(b) How does the elevation in dATP concentration in the abnormal lymphocytes affect the synthesis of DNA?
3. Clinicians who use F-dUMP and methotrexate together in cancer treatment find that the combined effects on cancer cells are not synergistic. Suggest how the administration of methotrexate could interfere with the action of F-dUMP.
4. Elevated levels of ammonia in the blood can be caused by a deficiency of mitochon drial carbamoyl phosphate synthetase or a deficiency of any of the urea cycle enzymes.
These two types of disorders can be distinguished by the presence of orotic acid or related metabolites in the urine.
(a) Why is it possible to determine the basis of hyperammonemia in this way?
(b) Why would a deficiency of cytoplasmic carbamoyl phosphate synthetase not cause hyperammonemia? What problems would such an enzyme deficiency cause? How would you treat a patient who has a deficiency in cytoplasmic carbamoyl phosphate synthetase?
NUCLEOTIDE BIOSYNTHESIS
451
5. The degradation of thymine yields b-aminoisobutyrate, as shown in the figure below, which can be converted to succinyl CoA and then degraded in the citric acid cycle. What cofactors are needed to convert b-aminoisobutyrate to succinyl CoA?
O
K
C
HN
CJCH3
OKC
CH
N
H
Thymine
O
K
C
H
J
HN
C J
CH3
OKC
CH
2
N
H
Dihydrothymine
H
H
J
O
K H NJCJ
NJ
CH JCJ
C
2
2
K
J
J
O:
O
CH3 N - Carbamoylisobutyrate
H
J
O
K HN ;+CO +;H NJ CH JCJ
C
4
2
3
2
J
J
O:
CH3
b-Aminoisobutyrate
6. You wish to prepare 14C-labeled purines by growing bacteria in a medium containing a suitably labeled precursor. The only precursors available are amino acids that are all uniformly labeled to the same specific activity per carbon atom. Which of the amino acids would you use to obtain purine rings that are labeled to the highest specific activity?
7. 6-Mercaptopurine (6-MP) can be converted to the corresponding nucleotide 6 thioinosine-5′-monophosphate (tIMP) through the purine salvage pathway. tIMP can then be converted to 6-thioguanine nucleotides (6-TNG) or methylated to form MetIMP. Methotrexate (MTX) increases incorporation of 6-TNGs through the salvage pathway (Bokkerink et al., Hematol. Blood Transf. 33(1990):110–117). Both 6-MP and MTX are clinically useful anticancer agents and have been used together for many years in the treatment of childhood leukemia in part because of the synergistic effect they have on each other.
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CHAPTER 25
SH
J
N
N
N
N
6-Mercaptopurine
(a) Briefly describe the salvage reactions required to convert 6-MP to the correspon ding nucleotide.
(b) What step in the de novo biosynthesis of purines is likely to be inhibited by tIMP?
(c) How could the presence of MTX increase the incorporation of 6-TNG into DNA and RNA?
8. Hydroxyurea, a potent chelator of ferric ions, has been shown to interfere with DNA synthesis, and it is used as an antitumor agent. What is a likely target enzyme for hydroxyurea?
O
K
HOHNJCJ NH3
Hydroxyurea
9. Methotrexate, a folate antagonist, interferes with nucleic acid biosynthesis. Would you expect it to inhibit purine or pyrimidine biosynthesis or both processes? Explain.
10. Nucleoside phosphorylases catalyze the interconversion of bases and nucleosides through the following reactions:
Ribose 1-phosphate + base ribonucleoside + Pi
or
Deoxyribose 1-phosphate + base deoxyribonucleoside + Pi
The equilibrium constant for each of these reactions is close to 1.
(a) The pathway for the incorporation of radioactive thymine into bacterial DNA in cludes a step catalyzed by nucleoside phosphorylase. It has often been observed that the incorporation of thymine into DNA is enhanced when deoxyadenosine or deoxyguanosine is added to the medium. Can you explain this observation? Why might deoxyguanosine be preferable to deoxyadenosine?
(b) In cells that cannot carry out de novo synthesis of IMP, inosine can be utilized to produce IMP but only through an indirect salvage route because of the absence of inosine kinase. Suggest an alternative pathway for the formation of IMP from inosine. Among the enzymes you will need are nucleoside phosphorylase and phosphoribomutase, which isomerizes ribose 1-phosphate to ribose 5-phosphate.
11. Many multivitamin preparations contain nicotinamide. Most mammalian cells contain cytosolic enzymes that convert nicotinamide directly to NAD+. What other substrates are required for the formation of NAD+ from nicotinamide? How could PRPP and ATP be used as sources of those substrates?
12. Hypoxanthine-guanine phosphoribosyl transferase (HGPRT), a salvage enzyme of nu cleotide metabolism, uses 5′-phosphoribosylpyrophosphate (PRPP) to convert hypox anthine to IMP and guanine to GMP. A deficiency of this enzyme can lead to an increased level of purine synthesis, excess formation of uric acid, and hyperuricemia, or gout.
(a) How might a deficiency in HGPRT stimulate purine synthesis?
(b) Under what conditions might one expect a deficiency of hypoxanthine-guanine phosphoribosyl transferase to affect the rate of pyrimidine nucleotide synthesis? How could you estimate the rate of pyrimidine nucleotide synthesis in humans?
13. In mammals, the committed step for pyrimidine synthesis is catalyzed by carbamoyl phosphate synthetase, while in bacteria, the committed step is the formation of N-carbamoylaspartate, catalyzed by aspartate transcarbamoylase.
(a) Account for these differences in mammals and bacteria.
(b) Bacterial carbamoyl phosphate synthetase is only partially inhibited by UMP. Why?
14. The synthesis of deoxythymidylate can proceed not only from dUMP but also from dCMP. The route from dCMP begins with the formation of dCDP from CDP, catalyzed by ribonucleotide reductase, followed by dephosphorylation to dCMP, and the deamination of dCMP to form dUMP, catalyzed by dCMP deaminase.
DCMP + H
+
2O + H+ dUMP + NH3 dCMP deaminase is an allosteric enzyme that is stimulated by dCTP and inhibited by dTTP. Account for these effects and relate them to the regulation of ribonucleotide reductase by deoxynucleoside triphosphates.
ANSWERS TO PROBLEMS
1. The clustering of two or more enzymes (active sites) in a single polypeptide chain en sures that their synthesis is coordinated and helps assure that they will assemble into a coherent complex. Also, the proximity of the active sites means that side reactions are minimized as substrates are channeled from one active site to another. Finally, a multifunctional complex with covalently linked active sites is likely to be more stable than a complex formed by noncovalent interactions.
2. (a) Adenosine is phosphorylated to AMP, with ATP serving as the phosphoryl donor, in a reaction carried out by a specific nucleoside kinase. The conversion of AMP to ADP through the action of a specific nucleoside monophosphate kinase is accomplished, with ATP again utilized as a phosphate donor. Ribonucleotide reductase catalyzes the reduction of ADP to dADP, which is then converted to dATP by nucleotide diphosphokinase.
(b) High concentrations of dATP displace ATP from the overall activity site on ribonu cleotide reductase, which lowers the rate of synthesis of all four deoxyribonucleoside diphosphates. This, in turn, leads to a depletion of deoxyribonucleoside triphosphates, which are the substrates for DNA synthesis.
3. Methotrexate blocks the regeneration of tetrahydrofolate from dihydrofolate, which is produced during the synthesis of thymidylate. The failure to regenerate tetrahydrofolate means that those biochemical reactions in the cell that depend on one-carbon metabolism cannot be carried out. One of the products of tetrahydrofolate is methylenete 454
CHAPTER 25 trahydrofolate, which is used as a substrate by thymidylate synthetase and is required for inhibition of the enzyme by F-dUMP. A deficiency of methylenetetrahydrofolate means that F-dUMP cannot irreversibly inactivate thymidylate synthetase. Conversely, F-dUMP prevents the formation of dihydrofolate, thereby abolishing the adverse effects caused by the depletion of tetrahydrofolate in the cell.
4. (a) The presence of orotic acid, a precursor of pyrimidines, in the urine suggests that carbamoyl phosphate synthesized in mitochondria is not utilized there. Instead, carbamoyl phosphate enters the cytosol, where it stimulates an increase in the rate of synthesis of precursors of pyrimidines, including orotic acid. An excess of carbamoyl phosphate arises in mitochondria whenever any of the urea cycle enzymes are deficient. Such a condition will lead to hyperammonemia, as well as to the accumulation of carbamoyl phosphate. Although a deficiency in mitochondrial carbamoyl phosphate synthetase leads to hyperammonemia, it cannot lead to an accumulation of mitochondrial carbamoyl phosphate and therefore does not stimulate pyrimidine synthesis in the cytosol.
(b) Cytoplasmic carbamoyl phosphate synthetase is involved primarily in the pathway for pyrimidine synthesis, not for the assimilation of ammonia. Recall that, in the cytosol, the substrate for the formation of carbamoyl phosphate is glutamine, not ammonia. A deficiency of carbamoyl phosphate synthesis in the cytosol would cause a depletion of pyrimidines. Such a deficiency is treated by administration of uracil or uridine, which are precursors of UMP and CMP.
5. The transamination of b-aminoisobutyrate to form methylmalonate semialdehyde re quires pyridoxal phosphate as a cofactor. This reaction is similar to the conversion of ornithine to glutamate g-semialdehyde. Then NAD+ serves as an electron acceptor for the oxidation of methylmalonate semialdehyde to methylmalonate. The conversion of methylmalonate to methylmalonyl CoA requires coenzyme A. The final reaction, in which methylmalonyl CoA is converted to succinyl CoA, is catalyzed by methylmalonyl CoA mutase, an enzyme that contains a derivative of vitamin B12 as its coenzyme.
6. Examination of the pathway for purine synthesis shows that only glycine is incorporated intact into the purine ring at the C-4 and C-5 positions. Therefore, glycine is a good choice as the radiolabeled precursor. Serine can also be considered because it is a precursor of glycine and the ultimate donor of C-1 groups to tetrahydrofolate, and activated tetrahydrofolate derivatives participate in two reactions in the formation of purines.
Whether serine is a better choice than glycine depends on the relative amounts of the two unlabeled amino acids in the cell.
7. (a) 6-Mercaptopurine is converted to the mononucleotide through the action of hypoxanthine-guanine phosphoribosyl transferase (HGPRT), which uses PRPP to add 5′-phosphoribose to the purine ring. The resulting compound is 6-thioinosine5′-monophosphate, an analog of IMP.
(b) IMP (and presumably its analog 6-MP) inhibits Gln-PRPP aminotransferase, which catalyzes the committed step of purine biosynthesis. The 6-MP metabolite Me-tIMP may be involved in the inhibition of Gln-PRPP aminotransferase as well (Stet et al., Biochem. Journal 304(1994):163–168).
(c) MTX inhibits folate-dependent enzymes in the de novo purine biosynthetic path way leading to an accumulation of PRPP. This increase in substrate availability for the salvage pathway leads to a greater conversion of 6-MP into 6-TNG and there fore a higher level of incorporation of 6-TNGs into DNA. Also, since 6-MP is a substrate for HGPRT—see part (a)—it can compete with endogenouse purine bases for HGPRT leading to a decrease in the synthesis of AMP and GMP.
8. Hydroxyurea inhibits ribonucleotide reductase. By sequestering ferric ions, hydroxyurea destabilizes the organic free radical in the R2 subunit of the enzyme. The inhibition of enzyme activity leads to a depletion of deoxyribonucleoside diphosphates, which are normally converted to deoxyribonucleoside triphosphates, the substrates for DNA synthesis.
9. Methotrexate and aminopterin, a similar compound, are analogs of dihydrofolate (DHF) and inhibitors of dihydrofolate reductase, an enzyme that converts DHF to tetrahydrofolate (THF). The thymidylate synthase reaction converts N5,N10-methylenetetrahydrofolate to DHF in the process of methylating dUMP to form dTMP. In the presence of one of the inhibitors, this reaction functions as a sink that reduces the THF level of the cell by converting THF to DHF. Since THF derivatives are substrates in two reactions of purine metabolism and one of pyrimidine metabolism, both pathways are affected by the inhibitor.
10. (a) Deoxyribonucleosides such as deoxyadenosine can be converted to the free base and deoxyribose 1-phosphate by nucleoside phosphorylase. Increased levels of deoxyribose 1-phosphate are then available for the formation of deoxythymidine from thymine in the reverse reaction catalyzed by nucleoside phosphorylase.
Deoxyguanosine might be preferable to deoxyadenosine because the conversion of elevated levels of deoxyadenosine to dAMP and then to dATP could lead to the inactivation of ribonucleotide reductase, which is sensitive to the concentration of dATP.
(b) Inosine is cleaved to produce hypoxanthine and ribose 1-phosphate through the action of nucleoside phosphorylase; note that inorganic phosphate is required for this reaction. Hypoxanthine-guanine phosphoribosyl transferase converts free hypoxanthine to IMP by condensation with PRPP. PRPP can be derived from ribose 1phosphate in two steps: (1) the conversion of ribose 1-phosphate to ribose 5-phosphate, which is catalyzed by phosphoribomutase; and (2) the formation of PRPP from ribose 5-phosphate and ATP, which is catalyzed by PRPP synthetase.
11. Both ribose phosphate and an AMP moiety must be added to nicotinamide in order to convert it to NAD+. The ribose phosphate is derived from PRPP when a phosphoribosyl transferase catalyzes the formation of nicotinamide ribonucleotide or nicotinamide mononucleotide. The final step utilizes ATP, which serves as an adenyl donor for the formation of the dinucleotide NAD+.
12. (a) When active, HGPRT consumes PRPP as it catalyzes the synthesis of GMP and IMP.
Decreased flux through this reaction raises the steady-state level of PRPP, thereby increasing the activity of PRPP amidotransferase, which catalyzes the initial step in purine synthesis. Increased activity may make the amidotransferase resistant to feedback inhibition by AMP and GMP, the end products of the purine biosynthetic pathway.
(b) As noted above, decreased activity of HGPRT increases the concentration of PRPP.
This increases the rate of pyrimidine synthesis at the orotate phosphoribosyl transferase reaction, if PRPP levels are normally subsaturating for that enzyme. Orotate incorporation into nucleotide pools or into nucleic acids would give a reasonable estimate of de novo pyrimidine nucleotide synthesis.
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13. (a) Bacteria use a single form of carbamoyl phosphate synthetase not only for the syn thesis of pyrimidines but also for the synthesis of arginine. Arginine biosynthesis begins with glutamate and includes formation of citrulline from ornithine and carbamoyl phosphate, a pathway that resembles urea formation in mammals. Two forms of carbamoyl phosphate synthetase, a cytoplasmic form for pyrimidine synthesis and a mitochondrial form for arginine and urea synthesis, are employed in mammals. Because the two pathways are compartmentalized, the formation of carbamoyl phosphate in the cytosol is regarded as the committed step for pyrimidine synthesis.
(b) UMP does not completely inhibit bacterial carbamoyl phosphate synthetase because that inhibition would interfere with arginine production.
14. An increase in dCTP levels signals that the cell has ample deoxynucleotides for DNA syn thesis and that there is a need for thymidylate synthesis. An increase in dTTP levels signals that the activity of thymidylate synthase can be decreased, and the inhibition of dCMP deaminase by dTTP reduces the input of dUMP into the pathway. While ribonucleotide reductase is subject to regulation by other deoxynucleotides, it is not subject to allosteric regulation by dCTP. Instead it appears that regulation of dCMP deaminase provides a second control point for the generation of deoxynucleotides in the cell.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. Glucose + 2 ATP + 2 NADP+ + H +
2O
PRPP + CO2 ADP + AMP + 2 NADPH + H+
This equation is the summation of the following conversions: glucose
glucose
6-phosphate
ribose 5-phosphate PRPP (see Chapter 16 in the text).
2. Glutamine + aspartate + CO +
+
2
2 ATP + NAD+ orotate + 2 ADP + 2 Pi glutamate + NADH + H+ 3. (a, c, d, e) PRPP, (b) carbamoyl phosphate
4. PRPP and formylglycinamide ribonucleotide accumulate because they are reactions 1 and 4 in the first stage of purine biosynthesis (Figure 25.7 in the text). If these glutaminerequiring amidotransferase reactions are inhibited, the precursor molecules will accumulate. However, since the synthesis of formylglycinamide requires that PRPP be converted to phosphoribosylamine, probably only PRPP will accumulate significantly.
5. dUMP + serine + NADPH + H+ dTMP + NADP+ + glycine
This is the summation of reactions shown in Figure 25.14 in the text. Note that the coenzyme, tetrahydrofolate, is regenerated; it acts catalytically and does not appear in the equation.
6. There is a deficiency of N10-formyltetrahydrofolate (see Figure 25.7 in the text).
Sulfanilamide inhibits the synthesis of folate by acting as an analog of p-aminobenzoate, one of the precursors of folate.
7. PRPP is the activated intermediate in the synthesis of (a) phosphoribosylamine in the de novo pathway of purine formation, (b) purine nucleotides from free bases by the salvage pathway, (c) orotidylate in the formation of pyrimidines, (d) nicotinate ribonucleotide, (e) phosphoribosyl-ATP in the pathway leading to histidine, and (f) phosphoribosylanthranilate in the pathway leading to tryptophan.
NUCLEOTIDE BIOSYNTHESIS
457
8. (a) Cell A cannot grow in a HAT medium, because it cannot synthesize dTMP either from thymidine or from dUMP. Cell B cannot grow in this medium, because it cannot synthesize purines by either the de novo pathway or the salvage pathway. Cell
C can grow in a HAT medium because it contains active thymidine kinase from cell B (enabling it to phosphorylate thymidine to dTMP) and hypoxanthine-guanine phosphoribosyl transferase from cell A (enabling it to synthesize purines from hypoxanthine by the salvage pathway).
(b) Transform cell A with a plasmid containing foreign genes of interest and a func tional thymidine kinase gene. The only cells that will grow in a HAT medium are those that have acquired a thymidylate kinase gene; nearly all these transformed cells will also contain the other genes on the plasmid.
9. These patients have a high level of urate because of the breakdown of nucleic acids.
Allopurinol prevents the formation of kidney stones and blocks other deleterious consequences of hyperuricemia by inhibiting the formation of urate.
10. Though often termed binding constants, the values given are really dissociation constants.
Therefore, to calculate the free energy of binding, one uses the reciprocals of the values given. Thus, for the wild type,
1
DG°′ = −1 3 . 6 log = −1 3 . 6 ×
10.15
×
−11
7
10
= −13 8 . kcal / mol Similar calculations for Asn 27 and Ser 27 will give the results in the answer in the text.
11. IMP is the product of the pathway for de novo purine biosynthesis and the precursor of AMP and GMP. With the de novo pathway for purines not working effectively, it would be helpful to stimulate the salvage pathway, perhaps with a diet that is rich in nucleotides that would then be a source of the preformed purine bases hypoxanthine, adenine, and guanine.
12. N1 in the purine ring of IMP, AMP, ATP, GMP, and GTP will be labeled, following the scheme below: COO:
O :OOC
K COO:
N
N
J
N*
H :OOC
J
+
D
H
H
NJ
H N
2
J H N N
:
+
OOC
N*H
ribose-P
2
3
ribose-P
15N*-aspartate
5-aminoimidazole 4-carboxylate
D
ribonucleotide
fumarate
O
O
K
K
N
N H*N 6
1
5
7
H N*
2
8
J
2
4
H
3
D
D
N
J
J
N
N
9
H N
2
ribose-P
ribose-P
*N1-IMP
458
CHAPTER 25
13. Allopurinol is an analog of hypoxanthine in which the N and C atoms at positions 7 and 8 are interchanged. Xanthine oxidase will hydroxylate C2 of allopurinol in similar manner to its normal reaction with hypoxanthine, but this C-2 hydroxylation of allopurinol gives the new inhibitor:
OH
OH
OH
H
H
J6
6J
6J
N
N
1
N
N
5
7
8
1
5
7
8
1
5
7
8
J
2
4
H
2
4
N
2
4
N
3
3
3
H N N
N
N
9 J H N
9 J
HO
N
9 J
H
H
H
Hypoxanthine
Allopurinol
New inhibitor
14. For the production of glycinamide ribonucleotide, an acyl phosphate (anhydride) inter mediate is formed by reaction of ATP with a carboxylic acid of an amino acid (glycine), whereas in the production of guanylate (GMP) a phosphoryl ester intermediate is formed by reaction of ATP with an enol alcohol on the purine ring of the nucleotide.
Glycinamide ribonucleotide. Step 1.
H
H
H
H
=
;
O:
;
OPO
H N
+
ATP D
H N
3
+
ADP
3
3
K
K
O
O
glycine
glycine acyl phosphate
Step 2.
; H N
3
CH
=
2
O POCH
NH
H
3
2
2
H
OK
=
=
;
OPO
+
H N
3
D O POCH
NH
3
2
+
P
3
i
K
O
HO
OH
5-phosphoribosyl-1-amine
HO
OH
glycinamide
ribonucleotide
Guanylate. Step 1.
O
O
K
H
K
H
J
N
J
N
N
N
JH
G
JH
+
ATP D ADP
K
O
N
N
K
HO
N
N
+
J
ribose-P
J
ribose-P
O
xanthylate tautomeric enol form
K H J N
N
JH
=
J O PO
N
N
3
J
ribose-P
NUCLEOTIDE BIOSYNTHESIS
459
Step 2.
O
O K
NH
K
2
H J N
H
N
+
J
D
H
glutamate (Glu) ;
O:
=
H N J
3
K
O PO
N
N
3
J
+
ribose-P
O
O
glutamine
K H J N
N
J
JH
H N
2
=
J O PO
N
N
3
J
J
ribose-P
H
D
HOPO =
3
O
K H J N
N
JH
J H N N
N
2
J
ribose-P
guanylate (GMP)
15. The reaction involves a dehydration and ring closure. The amino group that was intro duced from aspartate and that will become N-1 in inosinate (see problem 12, above) should be activated (at an enzyme active site) to make a nucleophilic attack on the nearby formyl carbonyl carbon to close the six-membered ring and give a tetrahedral intermediate. The tetrahedral intermediate can then lose water to render the six-membered ring aromatic and generate the inosinate (IMP):
O
O
K
K H N N
2
HN
N
JH
D
JH
H
J
H
N
J
N
N
N
K
J
O
J
:
O
J
ribose-P
J
ribose-P
H
H;
H
5-formyamidoinidazole
D
4-carboxamide
ribonucleotide
H O
2
O
K
N
HN
JH
J
H
N
N
inosinate (IMP)
J
ribose-P
16. (a) Cyclic-(5′-3′)-AMP and cyclic GMP influence many intracellular processes.
(b) ATP is the prototype energy-storage molecule with a high phosphoryl-group trans fer potential.
460
CHAPTER 25 (c) ATP is a phosphate donor for generating glucose-6-phosphate and fructose-1,6-bis phosphate.
(d) Flavin adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD+) participate in the production of acetyl-CoA (“active acetate”) from pyruvate by the pyruvate dehydrogenase complex.
(e) NADH and FADH2 are reduced molecules that have a high electron transfer po tential.
(f) Synthetic (2′,3′)-dideoxynucleoside triphosphates serve as chain terminators for DNA sequencing. (The four natural deoxynucleoside triphosphates—dGTP, dCTP, dATP, and dTTP—are the monomers that are precursors for chain elongation.)
(g) The thymine analogue 5-fluorouracil (converted to 5-fluorodeoxyuridylate in vivo) is a potent anticancer drug.
(h) ATP and CTP reciprocally regulate the activity of aspartate transcarbamoylase.
17. In vitamin B12 deficiency, methyltetrahydrofolate cannot donate its methyl group to ho mocysteine to regenerate methionine. Because the synthesis of methyltetrahydrofolate is irreversible (text, p. 675), the cell’s tetrahydrofolate ultimately will be converted into this form. No formyl or methylene tetrahydrofolate will be left for nucleotide synthesis.
Pernicious anemia illustrates the intimate connection between amino acid metabolism and nucleotide metabolism. The metabolism of fatty acids that have odd numbers of carbons also will be affected because methylmalonyl-CoA mutase requires vitamin B12 for the production of succinyl-CoA. A further connection is that methylmalonyl-CoA mutase also is involved in the degradation of valine and isoleucine.
18. The cytosolic level of ATP in liver falls and that of AMP rises above normal in all three conditions. The excess AMP is degraded to urate. See C. R. Scriver, A. L. Beaudet, W .S.
Sly, and D. Vale (eds.), The Metabolic Basis of Inherited Disease, 6th ed. (McGraw-Hill, 1989), pp. 984–988, for an illuminating discussion.
19. Succinate can be converted to oxaloacetate by the citric acid cycle. The oxaloacetate can then be transaminated to yield asparate, a key precursor of pyrimidines. The carbons of aspartate then will label positions 4, 5, and 6 in the pyrimidine rings:
:
O
O
O
K H O
NADH
H;
K
+
2
H;
H NAD;
NH
CH
J
N 4
CH
2
2
5
G
G
6
K
J
H
O
N
OK N
K
J COO:
J
H
H O:
N-carbamoylaspartate
20. (a) The ADP from muscle contraction is a ready source of additional ATP for additional contraction. Half of the ADP can be converted immediately to ATP at the expense of the other half (being converted to AMP).
(b) The reactants and products have the same number of high-energy phosphate bonds.
The interconversion of (2 ADP) with (ATP + AMP) therefore is essentially isoenergetic.
(c) In the reaction 2 ADP ATP + AMP, the removal of one of the products (AMP) will shift the equilibrium to the right and favor the production of additional ATP.
(d) By first removing and then replacing AMP, the cycle buys time (at the expense of GTP) until aerobic metabolism can “catch up” and reconvert available AMP as well as ADP back into ATP.
CHAPTER 2
The Biosynthesis
of Membrane Lipids and Steroids
6
This chapter describes the biosynthesis of membrane lipids, steroids, and other im portant lipid molecules, such as bile salts, vitamin D, and polymers of isoprene units. As background material for this chapter, you should review the earlier chap ters on cell membranes (Chapters 2 [Section 2.3.2] and 12) and fatty acid metabolism (Chapter 22), paying particular attention to the structure and properties of lipids and the central role of acetyl CoA in the metabolism of lipids. Chapter 26 begins with a discussion of the formation of triacylglycerols, phosphoglycerides, and sphingolipids from the simple precursors glycerol 3-phosphate, fatty acyl CoAs, polar alcohols, for example, choline, serine, and sugars. The text then describes the synthesis of cholesterol from acetyl CoA via the important intermediate isopentenyl pyrophosphate. The regulation of a key enzyme in the biosynthetic pathway as well as other modes of regulation of cholesterol metabolism is also outlined. Cholesterol is the precursor of bile salts as well as steroid hormones. The cholesterol and triacylglycerols synthesized in the liver and intestines are transported by lipoproteins to peripheral tissues. Cholesterol from dietary sources is moved from the intestine to the liver by lipoproteins. Therefore, the classification, the properties, and the mechanisms by which the lipoproteins deliver lipids to cells are discussed next. Finally, the text describes the synthesis of steroid hormones and vitamin D from cholesterol and introduces a variety of isoprenoid lipids that are derived from isopentenyl pyrophosphate.
461
462
CHAPTER 26
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Name the three major lipid-based components of biological membranes (Chapter 12).
2. Explain the biological significance of cholesterol.
Phosphatidate Is a Common Intermediate in the Synthesis
of Phospholipids and Triacylglycerols (Text Section 26.1) 3. Describe the roles of phosphatidate, glycerol 3-phosphate, lysophosphatidate, and diacylglycerol (DAG) in the synthesis of triacylglycerols and phospholipids.
4. List the primary biological functions of triacylglycerols and phospholipids.
5. Contrast the biosynthesis of phosphoglycerides in bacteria and mammals. Note the sig nificance of CDP–diacylglycerol and CDP–choline or CDP–ethanolamine, the activated precursors in these biosyntheses.
6. Describe the biosyntheses of phosphatidyl serine, phosphatidyl ethanolamine, phosphatidylcholine, and phosphatidyl inositol.
7. Restate the physiologic roles of phosphatidyl inositol and its degradation products, in-ositol 1,4,5-trisphosphate and diacylglycerol.
8. Compare the structures and biosynthetic pathways of the glyceryl ether phospholipids, including platelet-activating factor and plasmalogens, with those of the glyceryl ester
phospholipids.
9. Summarize the steps in the biosynthesis of sphingosine from palmitoyl CoA and serine.
10. Outline the synthesis of sphingomyelin, cerebrosides, and gangliosides from sphingosine.
Note the use of activated sugars and acidic sugars.
11. Provide examples of how sphingolipids confer diversity on lipid structure and function.
12. Discuss the general degradation pathway of gangliosides and the biochemical basis of Tay-Sachs disease and respiratory distress syndrome.
Cholesterol Is Synthesized from Acetyl Coenzyme A in Three Stages
(Text Section 26.2) 13. Describe the physiologic roles of cholesterol.
14. List the major stages in cholesterol biosynthesis and give the key intermediates.
15. Compare the synthetic paths leading from acetyl CoA to mevalonate and to the ketonebodies. Note the role of 3-hydroxy-3-methylglutaryl CoA reductase (HMG CoA reductase) as the major regulatory enzyme in cholesterol biosynthesis.
16. Describe the conversion of mevalonate into isopentenyl pyrophosphate.
17. Outline the condensation reactions leading from isopentenyl pyrophosphate to squalene.
Describe the mechanisms of these condensation reactions.
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
463
18. Discuss the cyclization of squalene and the formation of cholesterol from lanosterol. Note the role of O2 in the formation of cholesterol.
The Complex Regulation of Cholesterol Biosynthesis Takes Place
at Several Levels (Text Section 26.3) 19. List the sources of cholesterol and outline the mechanisms of regulation of cholesterol biosynthesis.
20. List the various classes of lipoproteins together with their lipid and protein components.
Describe their lipid transport functions.
21. Summarize the steps in the delivery of cholesterol to cells via the low-density-lipoprotein(LDL) receptor. Discuss the regulation of cellular functions by the LDL pathway.
22. Describe the proposed domain structure of the LDL receptor derived from the primary sequence of this protein. Define the term mosaic protein.
23. Discuss the biochemical defects of the LDL receptor that result in familial hypercholes-terolemia.
24. Summarize approaches used to reduce serum cholesterol.
Important Derivatives of Cholesterol Include Bile Salts
and Steroid Hormones (Text Section 26.4) 25. Describe the physiological roles and the general structures of the bile salts.
26. List the five major classes of steroid hormones, their physiological functions, and their sites of synthesis. Outline their biosynthetic relationships.
27. Give the numbering scheme for the carbon atoms of cholesterol and its derivatives, and distinguish between the a- and b-oriented groups and cis or trans ring fusions.
28. Describe the hydroxylation reactions involving cytochrome P450. Indicate the role of these monooxygenase reactions in steroid biosynthesis, the detoxification of xenobiotic compounds,
and the generation of carcinogens.
29. Describe the synthesis of pregnenolone from cholesterol, the conversion of pregnenolone into progesterone, and the subsequent reactions leading to cortisol and aldosterone.
30. Outline the synthesis of androgens and estrogens from progesterone.
31. Discuss the synthesis and the physiological role of vitamin D.
32. Give examples of biomolecules that contain isoprene units.
SELF-TEST
Introduction 1. Which of the following are components of biological membranes?
(a) free fatty acids (d) phospholipids (b) sphingolipids (e) cholesterol (c) triacylglycerols (f)
proteins
464
CHAPTER 26
Phosphatidate Is a Common Intermediate in the Synthesis
of Phospholipids and Triacylglycerols 2. Which of the following reactions are significant sources of glycerol 3-phosphate that is used in lipid synthesis?
(a) reduction of dihydroxyacetone phosphate (b) oxidation of glyceraldehyde 3-phosphate (c) phosphorylation of glycerol (d) dephosphorylation of 1,3-bisphosphoglycerate (e) reductive phosphorylation of pyruvate
3. Match the lipids in the left column with the major synthetic precursors or intermediates listed in the right column.
(a) triacylglycerol (1) phosphatidate (b) phosphatidyl ethanolamine (bacteria) (2) diacylglycerol (c) phosphatidyl ethanolamine (mammals) (3) acyl CoA (4) glycerol 3-phosphate (5) CDP-diacylglycerol (6) CDP-ethanolamine 4. Explain the role of the CDP derivatives in the synthesis of phosphoglycerides.
5. Calculate the number of “high-energy” phosphate bonds that are expended in the for mation of phosphatidyl choline from diacylglycerol and choline in mammals.
6. Which of the following is a common reaction used for the formation of phosphatidyl ethanolamine in bacteria?
(a) decarboxylation of phosphatidyl serine (b) reaction of CDP-ethanolamine with a diacylglycerol (c) demethylation of phosphatidyl choline (d) reaction of ethanolamine with CDP-diacylglycerol (e) reaction of CDP-ethanolamine with CDP-diacylglycerol 7. Which of the following is a lipid with a signal-transducing activity?
(a) phosphatidyl choline (b) phosphatidyl serine (c) plasminogen activator (d) phosphatidyl inositol 4,5-bisphosphate (e) phospholipase A2 8. For the lipid classes listed in the left column, select the characteristic structural compo nents or properties from the right column.
(a) glyceryl ester phospholipids (1) two long hydrocarbon chains (b) plasmalogens (2) acetyl group (c) platelet-activating factor (3) phosphate group (4) ether linkage (5) a,b-double bond (6) glycerol group (7) long fatty acyl chain (8) relatively high solubility in water 9. Which of the following phospholipases would you expect to cleave the R1-containing chain from the phospholipid shown in Figure 26.1?
(a) phospholipase A1 (b) phospholipase A2 (c) phospholipase C (d) phospholipase D (e) none of the above FIGURE 26.1 A phospholipid.
H
H
O H CJOJCKCJR
2
1
K
J R JCJOJCJH
O
2
+
J
K H CJOJPJOJCH JCH JN(CH )
2
2
2
3 3
J O: Synthesis of sphingolipids
10. Which of the following is NOT a precursor or intermediate in the synthesis of sphin gomyelin?
(a) palmitoyl CoA (b) lysophosphatidate (c) CDP-choline (d) acyl CoA (e) serine
11. Match the lipids in the left column with the appropriate activated precursors from the right column.
(a) sphingomyelin (1) acyl CoA (b) ganglioside (2) CDP-choline (c) phosphatidyl serine (3) CDP-diacylglycerol (4) CMP-N-acetylneuraminate (5) UDP-sugar 12. In which compartment of the cell does ganglioside GM2 accumulate in Tay-Sachs pa tients? What is the biochemical defect?
Cholesterol Is Synthesized from Acetyl Coenzyme A in Three Stages 13. From the following compounds, identify the intermediates in the synthesis of cholesterol and list them in their proper sequence: (a) geranyl pyrophosphate (b) squalene (c) isopentenyl pyrophosphate (d) mevalonate (e) cholyl CoA (f) farnesyl pyrophosphate (g) lanosterol (a) Both involve 3-hydroxy-3-methylglutaryl CoA (HMG CoA).
(b) Both require NADPH.
(c) Both require the HMG CoA cleavage enzyme.
(d) Both occur in the mitochondria.
(e) Both occur in liver cells.
15. Select the appropriate characteristics from the right column for the three stages in the synthesis of cholesterol in the left column.
(a) mevalonate to isopentenyl (1) releases PPi
pyrophosphate
(2) requires NADPH (b) isopentenyl pyrophosphate (3) requires O2 to squalene (4) releases CO2 (c) squalene to cholesterol (5) requires ATP 16. Yeast cells growing aerobically synthesize sterols and incorporate them into membranes.
However, under anaerobic conditions yeast cells do not survive unless they are provided with an exogenous source of sterols. Explain the metabolic basis for this nutritional requirement.
The Complex Regulation of Cholesterol Biosynthesis Takes Place
at Several Levels 17. The key step in cholesterol biosynthesis is the conversion of 3-hydroxy-3-methylglutaryl CoA to mevalonate. Which of the following are ways in which this reaction can be modulated?
(a) covalent modification HMG CoA reductase through phosphorylation (b) controlling the rate of translation of the mRNA encoding HMG CoA reductase (c) controlling the rate of transcription of the gene encoding HMG CoA reductase (d) proteolytic degradation of HMG CoA reductase (e) deletion and duplication of the gene encoding HMG CoA reductase 18. Match the appropriate components or properties in the right column with the lipopro teins in the left column.
(a) chylomicron (1) contains apoprotein B-100 (b) VLDL (2) contains apoprotein B-48 (c) LDL (3) contains apoprotein A (d) HDL (4) transports endogenous cholesterol esters (5) transports dietary triacylglycerols (6) transports endogenous triacylglycerols (7) is degraded by lipoprotein lipase (8) is taken up by cells via receptor mediated mechanisms (9) is a precursor of LDL (10) may remove cholesterol from cells 19. Which of the following events occur in the LDL pathway in fibroblasts? Place them in their proper sequential order.
(a) breakdown of LDL in lysosomes (b) endocytosis of LDL along with LDL receptors (c) degradation of LDL receptors in lysosomes (d) binding of LDL to LDL receptors (e) return of LDL receptors to the plasma membrane THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
467
20. Exons in the gene for the LDL receptor give rise to structurally diverse domains. What is the likely function of the cysteine-rich amino-terminal domain, which contains a cluster of negatively charged side chains?
(a) carbohydrate binding (d) growth-factor binding (b) membrane attachment (e) clathrin binding (c) Ca+ binding (f)
structure stabilization
21. Explain how LDL regulates the cholesterol content in fibroblasts.
22. Assume that LDL is produced normally in a patient but that the apoprotein B-100 do main that recognizes the receptor is functionally defective, which prevents the binding of LDL to its receptor. What outcome would this defect have on cholesterol metabolism in peripheral cells?
Important Derivatives of Cholesterol Include Bile Salts and Steroid Hormones 23. The physiological roles of bile salts include which of the following?
(a) They aid in the digestion of lipids.
(b) They aid in the digestion of proteins.
(c) They facilitate the absorption of sugars.
(d) They facilitate the absorption of lipids.
(e) They provide a means for excreting cholesterol.
24. Which of the following are common features in the structures of cholesterol and glyco cholate?
(a) Both have three hydroxyl groups.
(b) Both contain four fused rings.
(c) Both have a hydrocarbon side chain.
(d) Both contain a carboxylate group.
(e) Both contain double bonds.
(f) Both contain a sulfur atom.
25. Explain the structural characteristics of bile salts that make them effective biological detergents.
26. For the sterol structure in Figure 26.2, answer the following: FIGURE 26.2 A sterol.
H COH
2
J
CKO
HO
. J. .OH
J
J
J
K
O (a) Name this sterol.
(b) From what is it synthesized via three hydroxylation reactions?
(c) How many fewer carbon atoms does it have than cholesterol?
(d) Its concentration will be diminished if there is a deficiency of 21-hydroxylase— true or false? Explain why.
468
CHAPTER 26
27. Hydroxylation reactions involving cytochrome P450 have which of the following char acteristics?
(a) They require a proton gradient.
(b) They involve electron transport from NADPH to O2.
(c) They activate O2 by binding it to adrenodoxin.
(d) They transfer one oxygen atom from O2 to the substrate and form water from the other oxygen atom.
(e) They occur in adrenal mitochondria and liver microsomes.
28. Explain how foreign aromatic compounds are detoxified and excreted by mammals.
Describe a possible deleterious effect of this process.
29. Match the steroid hormones in the left column with the characteristics in the right col umn that distinguish them from one another.
(a) aldosterone (1) has 18 carbon atoms (b) estrogen (2) has 19 carbon atoms (c) testosterone (3) has 21 carbon atoms (4) contains an aromatic ring (5) contains an aldehyde group at C-18 30. Name the principle form of excreted cholesterol.
31. Which of the following statements about active vitamin D are INCORRECT?
(a) It has the same fused ring system as cholesterol.
(b) It requires hydroxylation reactions for its synthesis from cholecalciferol.
(c) It is important in the control of calcium and phosphorus metabolism.
(d) It can be synthesized from cholesterol in the presence of UV light.
(e) It can be derived from the diet.
32. Which of the following lipids does not contain isoprene units?
(a) coenzyme Q (b) carotene (c) vitamin K (d) arachidonate (e) phytol side chain of chlorophyll ANSWERS TO SELF-TEST 1. b, d, e, f. Answers (a) and (c) are incorrect because neither neutral fats nor free fatty acids appear in membranes.
2. a, c. Reduction of dihydroxyacetone phosphate is the primary route.
3. (a) 1, 2, 3, 4 (b) 1, 3, 4, 5 (c) 1, 2, 3, 4, 6. In mammals phosphatidyl ethanolamine can also be formed through an exchange reaction of ethanolamine with phosphatidyl serine.
4. CDP-diacylglycerol and the CDP-alcohols are activated intermediates that allow the for mation of phosphate ester bonds in phosphoglycerides, a process that is otherwise highly exergonic. ATP supplies the energy to form these compounds. Recall that UDP-sugars are used in a similar manner in the synthesis of carbohydrates (see text, pp. 589–590).
5. Summing the individual reactions:
Choline+ATPDphosphorylcholine+ADP
Phosphorylcholine+CTPDCDP–choline+PPi
CDP–choline+diacylglycerolDCMP+phosphatidyl choline
Choline+ATP+CTP+diacylglycerolDADP+PPi+CMP+phosphatidyl choline Two high-energy bonds (from ATP and CTP) are directly consumed in these reactions.
In addition, pyrophosphate is hydrolyzed by pyrophosphatase, driving the net reaction farther to the right. A total of three high-energy bonds would be consumed to regenerate ATP and CTP from ADP and CMP.
6. a
7. d
8. (a) 1, 3, 6, 7 (b) 1, 3, 4, 5, 6, 7 (c) 2, 3, 4, 6, 8 9. e. Phospholipases are specific for ester bonds; therefore, none will cleave the ether bond on the C-1 carbon of the plasmalogen. Phospholipases A2, C, and D cleave specific ester linkages: A2 cleaves the R2-containing chain to release the fatty acid, C cleaves the phophodiester bond to produce the R3-phosphate, and D cleaves the phosphodiester bond to produce the R3-alcohol. If the phospholipid had been a glycerol ester phospholipid, for example, phosphatidyl ethanolamine, phospholipase A1 would have cleaved the ester bond to yield the R1-contining fatty acid. Phospholipase C hydrolyzes phosphatidyl inositol 4, 5-bisphosphate to produce inositol 1, 4, 5-trisphosphate and diacylglycerol, which are intracellular second messengers (Section 15.2).
10. b
11. (a) 1, 2 (b) 1, 4, 5 (c) 1, 3 12. The degradative enzymes for gangliosides are located in lysosomes; therefore, ganglio side GM2 will accumulate in the lysosomes of Tay-Sachs patients. The enzyme that removes the terminal sugar, GalNAc, from the ganglioside is deficient in these people.
13. All the compounds given are intermediates in the biosynthesis of cholesterol except for (e) cholyl CoA, which is a catabolic derivative of cholesterol and a precursor of bile salts.
The proper sequence is d, c, a, f, b, g.
14. a, e 15. (a) 4, 5 (b) 1, 2 (c) 2, 3 16. A key intermediate in the biosynthesis of cholesterol and related sterols is squalene, an open-chain isoprenoid hydrocarbon. It is converted to squalene 2,3-epoxide, which in turn is converted to lanosterol. The conversion of squalene to the 2,3-epoxide is catalyzed by a monooxygenase, and molecular oxygen is a required component for this reaction. Under anaerobic conditions, yeast cells cannot synthesize sterols because they lack oxygen, a substrate for the monooxygenase reaction.
17. a, b, c, d 20. c, f 21. The main source of cholesterol for cells outside the liver and intestine is from circulating LDL. Cholesterol released during the degradation of LDL suppresses the formation of new LDL receptors, thereby decreasing the uptake of exogenous cholesterol by the cell.
22. A defect in apoprotein B-100 that prevents the binding of LDL to the cell-surface re ceptor would result in the stimulation of the synthesis of endogenous cholesterol and LDL receptors and a decrease in the synthesis of cholesterol esters via the ACAT reaction. Indeed, the cellular and physiological consequences of such a mutation may be similar to those seen in familial hypercholesterolemia.
23. a, d, e 24. b, e 25. Bile salts are effective detergents because they contain both polar and nonpolar re gions. They have several hydroxyl groups, all on one side of the ring system, and a polar side chain that allow interactions with water. The ring system itself is nonpolar and can interact with lipids or other nonpolar substances. Bile salts are planar, amphipathic molecules, in contrast with such detergents as sodium dodecyl sulfate (text, p. 84), which are linear.
26. (a) cortisol (b) progesterone (c) six (d) true. A deficiency of 21-hydroxylase will impair hydroxylation at C-21 of proges terone, which will prevent the normal synthesis of cortisol and mineralocorticoids from progesterone.
27. b, d, e 28. In mammals, foreign aromatic molecules are hydroxylated by the cytochrome P450 dependent monooxygenases that are present in the endoplasmic reticulum of the liver cells. The hydroxylated derivatives are more water-soluble and have functional groups for the attachment of very polar substances, such as glucuronate, that allow them to be excreted in urine. The action of the cytochrome P450 system sometimes converts potential carcinogenic compounds into highly carcinogenic derivatives.
29. (a) 3, 5 (b) 1, 4 (c) 2 30. The water-soluble bile salt glycocholate is a major cholesterol breakdown product.
31. a
32. d
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
471
PROBLEMS
1. Why is synthesis of cholesterol de novo dependent on the activity of ATP-citrate lyase?
2. An infant has an enlarged liver and spleen, cataracts, and anemia and exhibits general retardation of development. Mevalonate is found in the urine. Investigation reveals a deficiency of mevalonate kinase, which catalyzes the formation of 5-phosphomevalonate from mevalonate.
(a) Why is urinary excretion of mevalonate consistent with a deficiency of meval onate kinase?
(b) How would a deficiency of mevalonate kinase affect cholesterol synthesis in this infant?
(c) What level of activity, relative to normal, would you expect to find for HMG CoA reductase in cells isolated from the infant? Briefly explain your answer.
3. Normally, most of the bile acids that are secreted into the intestine undergo reabsorption and are returned to the liver. Cholestyramine is a positively charged resin that binds bile acids in the intestinal lumen and prevents their reabsorption.
(a) To examine the effects of cholestyramine on LDL metabolism, two fractions of LDL were prepared: one was covalently labeled on tyrosine residues with 125I; the other was similarly labeled with 131I and treated with cyclohexanedione, which interferes with LDL binding to the LDL receptor. When rabbits were given cholestyramine, hepatic uptake of 125I-labeled LDL was enhanced relative to normal, whereas the uptake of 131I-labeled LDL was unchanged relative to that in rabbits that had not been given cholestyramine. Briefly explain the relationship between the action of cholestyramine and LDL uptake in the liver.
(b) The administration of cholestyramine usually results in a 15 to 20% reduction in levels of circulating LDL, whereas the administration of a combination of cholestyramine and mevinolin (lovastatin) can often yield a 30 to 40% reduction. Why?
4. The presence of apoprotein E in lipoproteins enables them to be taken up by hepatic cells. Provide a brief explanation for each of the following observations made of a person with a deficiency in apoprotein E synthesis.
(a) elevated levels of plasma triacylglycerols and cholesterol, coupled with the presence of chylomicron remnants and IDL. These latter particles persist in the bloodstream much longer than in normal people.
(b) abnormally low levels of LDL in the blood (c) abnormally high levels of LDL receptors in liver cells (d) a marked reduction in levels of circulating chylomicron remnants and IDL when the diet is low in cholesterol and fat 5. Pregnant women often have increased rates of triacylglycerol breakdown and, as a result, have elevated levels of ketone bodies in their blood. Why do they also often exhibit an increase in plasma lipoprotein levels?
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6. Hopanoids are pentacyclic molecules that are found in bacteria and in some plants. As an example, a typical bacterial hopanoid is shown below. Organisms that make hopanoids use a pathway similar to that for cholesterol synthesis. The biosynthetic pathway for hopane includes the formation of squalene, followed by more steps to form the final product itself, a C30 compound. Hopane is similar to lanosterol (text, p. 726), but lacks the hydroxyl group.
H C
3
J
J
H C J
3
CH3
J
CH
3
J
H C
3
H C
3
H CJCH
3
J
CH2
J
CH2
J
HCJOH
J
HOJCH
J
HCJOH
J
CH2
J
CH OH
2
Bacteriohopanetetrol (a) How many molecules of mevalonic acid are required for the synthesis of hopane?
(b) Squalene can undergo concerted cyclization to form hopane in a reaction that is catalyzed by a unique type of squalene cyclase. The reaction is initiated by a proton and does not require oxygen. Compare this step with the formation of lanosterol from squalene. Why could it be argued that the synthesis of hopanoids preceded the synthesis of sterols in evolution?
7. Your colleague has discovered a compound that is a very powerful inhibitor of HMG CoA reductase, and she has evidence that the drug will completely block the synthesis of mevalonate in liver. Why is this compound unlikely to be useful as a drug?
8. The liver is the site of the synthesis of plasma phospholipids and lipoproteins. Rats main tained on a diet deficient in choline often develop fat deposits in liver tissue. How could choline deficiency be related to this aberration in lipid metabolism?
9. Among the sugar residues found in a blood group ganglioside is fucose. Experiments uti lizing isolated Golgi membranes and ribonucleoside triphosphates show that fucose can be incorporated into the ganglioside only when GTP is available. What is the role of GTP in fucose incorporation?
10. Suppose a cell is deficient in phosphatidate phosphatase, which catalyzes the formation of diacylglycerol from phosphatidate. What effects on lipid metabolism would you expect?
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
473
11. Desmolase is involved in the synthesis of pregnenol (text, p. 735).
(a) Would you expect virilization among patients who have desmolase deficiency?
(b) Why is enlargement of the adrenal glands common among such patients?
12. In the adult form of Gaucher’s disease, glucosylcerebrosides accumulate in liver, spleen, and bone-marrow cells. Although the common galactosylceramides and their derivatives are found in the tissues of affected patients, accumulations of galactosylcerebrosides or their metabolites are not found, nor do ceramides accumulate. What enzyme activity is probably deficient in patients with Gaucher’s disease?
13. Cells of the adrenal cortex have very high concentrations of LDL receptors. Why?
14. At low concentrations of phospholipid substrates in water, the reaction catalyzed by a phospholipase occurs at a rather low rate. The reaction rate accelerates when the concentrations of the phospholipid substrates increase to the point that micelles are formed.
How is this property of phospholipases related to their activity in the cell?
15. Glucagon has been shown to reduce the activity of HMG CoA reductase. Why is this ob servation consistent with the overall effect of glucagon on cellular metabolism?
16. People who have elevated levels of LDL in their serum can be treated in a number of ways. These include restriction of dietary intake of cholesterol, ingestion of positively charged resin polymers that inhibit intestinal reabsorption of bile salts, and administration of lovastatin, a competitive inhibitor of 3-hydroxy-3-methylglutaryl CoA reductase.
(a) Briefly explain how each of these treatments reduces serum LDL levels.
(b) None of the above treatments should be used for patients who are homozygous for a defect in LDL receptors? Why?
17. Currently there are two established methods for dealing with hypercholesterolemia in patients homozygous for LDL receptor deficiency.
(a) The first is plasma apheresis, whereby the plasma and blood cells of a patient are separated in a continuous flow device, and the plasma is passed over a column that removes lipoproteins containing apoprotein B-100. Which lipoproteins are removed by this procedure, and how could their removal lower plasma cholesterol levels?
(b) A more extreme method of dealing with extreme hypercholesterolemia in FH ho mozygotes is liver transplantation. The rationale for this procedure is based on the observation that over 70% of total body LDL receptors are in the liver. In the small group of patients who have undergone liver transplants, LDL cholesterol levels are substantially reduced. In one particular case, the rate of LDL turnover increased about threefold, and one patient became responsive to lovastatin after the transplant. Why did increased LDL turnover and lovastatin response indicate that the transplantation procedure was successful?
18. Provide a physiological rationale for each of the following responses to an increase in the rate of transport of unesterified cholesterol into a mammalian cell.
(a) stimulation of the synthesis of cholesteryl oleate esters.
(b) suppression of the activity of HMG CoA reductase.
(c) suppression of the synthesis of LDL receptors.
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19. Glycerol phosphate acyltransferase can convert 3,4-dihydroxybutyl-1-phosphonate, an analog of glycerol 3-phosphate, to diacylbutyl-1-phosphonate, which is an analog of phosphatidate. Would this analog be more likely to inhibit triglyceride synthesis or CDP–diacylglycerol synthesis? Briefly explain your answer. See the figure below.
H CJOH
2
J
HOJCJH
O
J
K CH JCH JPJO:
2
2
J O: 3, 4-Dihydroxybutyl-1-phosphate 20. During the uptake of LDL by a liver cell, LDL-receptor protein complexes are internal ized by endocytosis. The endosomes then fuse with lysosomes, where protein components of LDL are hydrolyzed to free amino acids, while cholesterol esters are hydrolyzed by a lysosomal acid lipase. The LDL receptor itself is not affected by lysosomal enzymes.
(a) Briefly describe what would happen to cholesterol metabolism in a cell deficient in lysosomal acid lipase.
(b) Why is it important that LDL receptors are not degraded by lysosomal enzymes?
21. Glycerol kinase catalyzes the conversion of free glycerol into glycerol 3-phosphate, using ATP as a phosphoryl donor. Although liver tissue has high levels of the enzyme, the activity of glycerol kinase in adipose tissue is low. How do these differences contribute to the balance between carbohydrate and triglyceride metabolism in mammals?
22. Niemann-Pick disease is an inherited disorder of sphingomyelin breakdown due to a deficiency of sphingomyelinase. This enzyme, found in all tissues and easily assayed in white blood cells collected from a patient, converts sphingomyelin to ceramide and phosphocholine. Phosphocholine is highly soluble in water, while sphingomyelin is more soluble in chloroform. Assuming that you can obtain sphingomyelin labeled with 14C in any desired carbon atoms, design an assay that would allow you to confirm a diagnosis of Niemann-Pick disease.
ANSWERS TO PROBLEMS
1. ATP-citrate lyase catalyzes the formation of acetyl CoA in the cytosol (text, p. 622). Acetyl CoA is used for the synthesis of HMG CoA, which gives rise to mevalonate for the synthesis of cholesterol in the cytosol.
2. (a) A deficiency of mevalonate kinase activity means that mevalonate cannot be uti lized as a precursor of 5-phosphomevalonate. If no other pathways can use mevalonate, its concentration in the liver will increase until it spills into the blood and then, in turn, into the urine. Furthermore, because the activity of HMG CoA reductase is increased in this infant—see answer (c)—the rate of mevalonate synthesis will be stimulated.
(b) You would expect the rate of cholesterol synthesis to be depressed because the path way is blocked at the step in which 5-phosphomevalonate is formed.
(c) You would expect to find a higher-than-normal level of HMG CoA reductase activ ity. A depressed rate of cholesterol synthesis lowers the amount of cholesterol in the cell. HMG CoA reductase activity increases because inhibition of its synthesis and its activity by cholesterol is reduced.
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
475
3. (a) The experiments show that rabbits given cholestyramine have higher rates of re moval of LDL from the blood and that hepatic uptake of LDL depends on the ability of the lipoprotein to bind to the LDL receptor. One explanation for this is that cholestyramine interferes with the return of bile acids to the liver, stimulating the synthesis of more bile acids from cholesterol. An increased demand for cholesterol stimulates the synthesis of LDL receptors, which take up more cholesterol-containing LDL particles from the blood.
(b) Although cholestyramine stimulates the hepatic uptake of cholesterol-contain ing LDL, it has no direct effect on cholesterol synthesis de novo in the liver.
Mevinolin inhibits HMG CoA reductase, thereby depressing the rate of cholesterol biosynthesis. The subsequent requirement for cholesterol leads to a further increase in the number of LDL receptors, which in turn can take up more LDL from the circulation.
4. (a) Chylomicrons, chylomicron remnants, and IDL normally contain apoprotein E. A deficiency of the apoprotein means that hepatic uptake of chylomicron remnants and IDL is impaired, so these particles persist in the circulation. Because both of these types of particles contain triacylglycerols and cholesterol, circulating levels of these compounds are also elevated.
(b) Both chylomicron remnants and IDL particles serve as precursors of VLDL in the liver. When the uptake of the VLDL precursors by hepatic tissue is impaired by an apoprotein E deficiency, the rate of synthesis and export of VLDL particles is reduced. Since VLDL are LDL precursors in circulation, LDL are reduced.
(c) As discussed in answer (b), VLDL synthesis in the liver is impaired. Additional LDL receptors are synthesized because their synthesis is no longer repressed by VLDLderived cholesterol.
(d) A diet low in cholesterol and fat will reduce the rate of formation of chylomicrons, which are precursors of chylomicron remnants.
5. Increased levels of ketone bodies, such as acetoacetate, imply that the levels of acetyl CoA and HMG CoA, both precursors of cholesterol, are also elevated. Cholesterol synthesis is stimulated by an increase in the availability of these substrates. The subsequent decreased demand for dietary cholesterol results in an elevation in cholesterol-containing lipoproteins.
6. (a) Mevalonic acid, a six-carbon compound, is a precursor of isopentenyl pyrophos phate (IPP), which contains five carbon atoms. IPP serves as the basic unit for the formation of squalene, a 30-carbon compound. Six molecules of IPP are needed for the synthesis of a molecule of squalene, which is in turn the precursor of hopane.
Thus, six molecules of mevalonic acid are required.
(b) Aerobic processes, such as the synthesis of sterols, probably evolved later than anaerobic processes and only after free oxygen became available. Thus, the synthesis of hopane from squalene, an anaerobic process, probably preceded the synthesis of sterols, such as lanosterol, from squalene.
7. The synthesis of mevalonate is required not only for the synthesis of cholesterol but also for the synthesis of a number of other important compounds derived from isopentenyl pyrophosphate, including ubiquinone (CoQ), an important component of the electron transport chain. Therefore, the complete blockage of mevalonate synthesis, even if adequate cholesterol is available in the diet, would be ill-advised.
8. Choline, which is ordinarily supplied by the diet and is synthesized only to a limited extent in mammals, is a constituent of phosphatidyl choline, an important component of membranes and lipoproteins. A deficiency in dietary choline, which could not be completely 476
CHAPTER 26 replaced by the methylation of phospatidyl ethanolamine, could interfere with the synthesis and export of lipoproteins like VLDL, which is a carrier of triacylglycerols to peripheral tissues. Failure to export fats such as triacylglycerols leads to their accumulation in the liver.
9. Nucleotide sugars, such as UDP-glucose, serve as donors during the incorporation of sugar residues into gangliosides. In this case, it appears that the donor of fucose residues is GDP-fucose, which is synthesized from fucose and GTP.
10. You would expect to see reduced rates of synthesis of triacylglycerols, which use diacyl glycerols as acceptors of activated acyl groups. In addition, phosphatidyl choline synthesis is dependent on the availability of diacylglycerols as acceptors of choline phosphate from CDP-choline.
11. (a) You would not expect desmolase deficiency to lead to virilization. An increase in androgen production, which causes virilization in both males and females, is due to elevated levels of 17 a-hydroxyprogesterone. The pathway from cholesterol to 17 a-hydroxyprogesterone includes a step catalyzed by desmolase, which cleaves the bond between C-20 and C-22 in 20a, 22b dihydroxycholesterol to form pregnenolone (see the text, p. 735). A deficiency of desmolase would therefore decrease the rate of androgen synthesis.
(b) Pregnenolone is a precursor of the glucocorticoids, which exert feedback control on the activity of the adrenal cortex. Desmolase deficiency leads to diminished production of glucocorticoids. Failure of the normal feedback mechanism leads to increased ACTH production and to enlargement of the adrenal glands.
12. Because glucosylcerebrosides accumulate but galactosylcerebrosides do not, you would suspect that the defect involves ganglioside breakdown rather than ganglioside synthesis. The defect involves the step that removes glucose from the cerebroside to yield free ceramide, or N-acyl sphingosine. The enzyme that carries out this step is a glycosyl hydrolase; it is also called b-glucosidase.
13. Cells of the adrenal cortex utilize cholesterol for the synthesis of a number of steroid hor mones, including cortisol. Although these cells can themselves synthesize cholesterol, it is often also necessary for additional cholesterol to be obtained from plasma lipoproteins.
A high concentration of LDL receptors enables cortical cells to take up LDL, which contains cholesterol, rapidly.
14. In the cell, a phospholipase would most often encounter substrates that are part of an aggregate, such as those phospholipids found in membranes. Thus, the enzyme should be expected to function at a higher rate with aggregates or assemblies of lipid molecules because their local concentrations would be higher than if they were individually free in solution.
15. The presence of glucagon is a signal that carbohydrate and triacylglycerol catabolism is needed to generate energy in the organism. Under such conditions, one would expect biosynthetic reactions to be suppressed because energy charge is low. Low energy charge means high AMP levels that would activate an AMP-dependent protein kinase leading to the phosphorylation of HMG CoA reductase.
16. (a) Restricting the level of dietary cholesterol lowers the input of exogenous cholesterol into lipoproteins, so that fewer LDL molecules are present in serum. Compounds that inhibit bile salt reabsorption from the gut stimulate additional synthesis of bile acids from cholesterol in the liver, decreasing concentrations of the sterol in liver cells and, by causing an increase in the number of receptors on the cell surface, stimulating LDL uptake from the circulation. Finally, mevinolin reduces the rate of cholesterol biosynthesis de novo, which can also stimulate uptake of LDL from the bloodstream. Any of these treatments could help in reducing circulating levels of cholesterol as LDL.
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477
(b) Homozygotes have virtually no functional LDL receptors. Such people are unable to internalize significant amounts of LDL, which means that circulating levels of that lipoprotein are elevated in the blood. In addition, an absence of LDL receptors means that endogenous cholesterol fails to enter the liver cell to suppress de novo synthesis. Dietary restriction could reduce exogenous LDL levels somewhat but would not prevent formation of LDL and other lipoproteins arising from cell turnover of cholesterol. Bile sequestrants of bile salts could cause some depletion of liver cell cholesterol, but again would not stimulate LDL uptake from the circulation. Mevinolin will suppress cholesterol synthesis de novo, but this would again not be compensated for by uptake from the circulation. Homozygotes with two nonfunctional receptor genes are therefore resistant to compounds that inhibit LDL synthesis and stimulate uptake. Thus, none of these measures has a great effect on reducing circulating LDL levels in these patients.
17. (a) Protein B-100 is found in VLDL, IDL, and LDL, and all three lipoproteins are re moved from the plasma. Each contains cholesterol, and, in addition, VLDL and IDL are regarded as precursors of LDL, so that total cholesterol concentration in the blood would be lowered by apheresis. This method can reduce LDL cholesterol levels by 70%. Treatment must be repeated about every two weeks, and the effects of long-term apheresis are problematic.
(b) Increased LDL turnover indicates that the transplanted liver is producing normal LDL receptors that can take up LDL from the circulation and can facilitate its conversion to other lipoproteins. A response to lovastatin also indicates that functional LDL receptors are available to accelerate uptake in response to diminished de novo cholesterol synthesis, which is inhibited by lovastatin. It should be noted that liver transplant operations are hazardous, especially in FH homozygotes who usually have advanced atherosclerosis. The fact that normal liver cells can contribute functional LDL receptors has stimulated interest in gene therapy designed to target a normal LDL gene to liver cells of FH homozygotes.
18. (a) Formation of cholesteryl esters provides the cell with a means of storing cholesterol until it is needed for membrane biosynthesis or other purposes.
(b) Suppression of the activity of HMG CoA reductase leads to a decrease in cholesterol synthesis de novo; this occurs when the cell has sufficient endogenous cholesterol so that it does not need to synthesize the steroid on its own.
(c) Suppression of LDL receptor synthesis leads to a gradual reduction in the number of LDL receptors in the cell, because the receptors undergo a relatively constant rate of degradation. Reduction in LDL receptor levels means that fewer LDL particles will be able to enter the cell, resulting in a decrease in the entry of endogenous cholesterol.
19. The conversion of phosphatidate to a triacylglycerol is initiated by hydrolysis of the phosphate group and the formation of diacylglycerol. The C-P bond in the phosphonate is much less likely to be cleaved by the phosphatase (also known as phosphatidate
phosphohydrolase), so it is likely that triglyceride synthesis could be impaired by the analog. On the other hand, formation of CDP-diacylglycerol involves the formation of an anhydride bond between the phosphates of phosphatidic acid and cytidylic acid (text, p. 717). Since the phosphonate is not cleaved in this reaction, formation of the phosphonyl analog of CDP-diacylglycerol seems possible, or at least the synthesis of normal substrates would not be impaired. Phosphonolipids are known in trace amounts in mammals, but are found more extensively in some invertebrates, including the protozoan Tetrahymena, where they may represent up to 25% of total phospholipid (see D. E. Vance and J. Vance [eds.], Biochemistry of Lipids, Lipoproteins and
Membranes. [Elsevier, 1991], p. 205).
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20. (a) Both the LDL receptor gene and the gene for HMG CoA reductase contain sterol regulatory elements that are responsive to free cholesterol. A reduction in free cholesterol release from the lysosome leads to an increase in LDL receptor production and to increased HMG CoA reductase activity. Both these consequences lead to an increase in cholesterol concentrations in the cell, through an increased rate of LDL entry and accelerated cholesterol synthesis. Lysosomal accumulation of cholesteryl esters and triglycerides can eventually destroy the cell. One form of this disorder, termed Wolman disease, is characterized by liver enlargement, digestive difficulties, and enlargement and deterioration of the adrenal glands. The disease is usually fatal within a year after birth.
(b) LDL receptors, after their release from lysosomes, return to the cell surface, where they take up other LDL particles and bring them back to lysosomes. A round trip takes about 10 minutes. Destruction of LDL receptors by lysosomal enzymes would make it necessary to synthesize the 115-kd glycoprotein at a much faster and energetically wasteful rate.
21. Free glycerol in mammals is produced in adipocytes when triglycerides are converted to free fatty acids and glycerol by hormone-sensitive lipases. Hormone-responsive lipolysis occurs when glucose levels are low and fatty acids are needed as fuels. Under these conditions, the low activity of glycerol kinase in adipocytes prevents unnecessary resynthesis of triacylglycerols from free fatty acids and glycerol 3-phosphate, via phosphatidic acid. Triacylglycerol synthesis is more likely to occur when glucose levels are high.
Adipocytes can then synthesize glycerol 3-phosphate from dihydroxyacetone phosphate produced from glucose during glycolysis. Thus, adipocytes are unable to synthesize triglycerides unless glucose or another suitable carbon source is available.
22. Extracts of white cells from the blood of a person suspected of having a sphingomyeli nase deficiency are incubated in a buffered solution with radioactive sphingomyelin labeled with 14C in the methyl groups of the phosphocholine moiety. Then the incubation mixture is extracted with chloroform. Any radioactive phosphocholine liberated by the enzyme will remain in the upper aqueous layer, while intact radioactive sphingomyelin will be extracted into the lower chloroform layer. Using white cells from patients with Neimann-Pick disease, incubation with cell extracts followed by chloroform extraction results in little or no radioactivity in the aqueous phase, confirming the deficiency of sphingomyelinase activity.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. Glycerol+4 ATP+3 fatty acids+4 H2ODtriacylglycerol+ADP+3 AMP+ 7 Pi+4 H;
One ATP is used in the formation of glycerol 3-phosphate and three ATPs are used to convert three fatty acids to acyl CoAs. The three PPi formed during fatty acid activation are converted to Pi, hence, the total of seven Pi in the equation above.
2. Glycerol+3 ATP+2 fatty acids+2 H2O+CTP+serineD phosphatidyl serine+CMP+ADP+2 AMP+6 Pi+3 H;
The equation above is a summation of equations shown in the text on pages 716–717.
3. (a) CDP-diacylglycerol, (b) CDP-ethanolamine, (c) acyl CoA, (d) CDP-choline, (e) UDP-glucose or UDP-galactose, (f) UDP-galactose, (g) geranyl pyrophosphate 4. (a and b) None, because the label is lost as CO2.
5. (a) No receptor is synthesized.
(b) Receptors are synthesized but do not reach the plasma membrane because they lack signals for intracellular transport or do not fold properly.
(c) Receptors reach the cell surface, but they fail to bind LDL normally because of a defect in the LDL-binding domain.
(d) Receptors reach the cell surface and bind LDL, but they fail to cluster in coated pits because of a defect in their carboxyl-terminal region.
6. Deamination of cytidine to uridine changes CAA (Gln) into UAA (stop).
7. Benign prostatic hypertrophy can be treated by inhibiting the 5 a-reductase. Finasteride, the 4-aza steroid analog of dihydrotestosterone, competitively inhibits the reductase but does not act on androgen receptors. Patients taking finasteride have a markedly lower plasma level of dihydrotestosterone and a nearly normal level of testosterone. The prostate becomes smaller, whereas testosterone-dependent processes such as fertility, libido, and muscle strength appear to be unaffected (see E. Stoner, Steroid Biochem. Molec.
Biol. 37(1990):375–378). Genetic deficiencies of 5a-reductase are discussed by J. E.
Griffin and J. D. Wilson in C. R. Scriver, A. L. Beaudet, W. S. Sly, and D. Valle (eds.), The
Metabolic Basis of Inherited Disease, 6th ed. (McGraw-Hill, 1989), pp. 1919–1944.
CONHC(CH ) 3 3
J
J
J
K
O
N
....
H
H
Finasteride
8. Patients who are most sensitive to debrisoquine have a deficiency of a liver P450 enzyme encoded by a member of the CYP2 subfamily. This characteristic is inherited as an autosomal recessive trait. The capacity to degrade other drugs may be impaired in people who hydroxylate debrisoquine at a slow rate because a single P450 enzyme usually handles a broad range of substrates. See W. B. Pratt and P. Taylor, Principles of Drug Action:
The Basis of Pharmacology, 3d ed. (Churchill Livingstone, 1990), pp. 496–500; and F. J.
Gonzalez and D. W. Nebert. Trends Genet. 6(1990):182.
9. Many hydrophobic odorants are deactivated by hydroxylation. O2 is activated by a cy tochrome P450 monooxygenase. NADPH serves as the reductant. One oxygen atom of O2 goes into the odorant substrate, whereas the other is reduced to water.
10. Propecia effectively lowers the plasma level of dihydrotestosterone (see also problem 7). But dihydrotestosterone is an important embryonic androgen that instigates the development and differeniation of the male phenotype. Pregnant women who had contact with Propecia therefore would risk developmental abnormalities for their unborn male children.
11. The various P450 isozymes probably serve two major categories of function in plants.
First, some of them will be important in the detoxification of foreign substances that may arise in the plant’s environment. Second, plants may use P450 enzymes for the synthesis of useful molecules such as toxins to fight pests, or pigments to attract organisms that aid in dispersing pollen or seeds.
12. Individual polymorphisms in some of the P450 isozyme genes likely would alter the rates of metabolic degradation (or conversely activation) or particular clinical drugs.
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Knowledge of the individual differences therefore would help in prescribing appropriately different clinical doses of particular medicines for different individual patients.
13. Phosphorylation of the serine will place a negative charge adjacent to the key histidine side chain, thereby stabilizing the positively charged form of the histidine and preventing (or markedly slowing) the donation of the proton to the thiolate. A key step of the reaction therefore will be inhibited.
14. In a classic monooxygenase fashion, one oxygen from O2 will hydroxylate a substrate methyl group, and the other oxygen from O2 will be reduced to water. The elimination of formyaldehyde will then lead to the products: methylamine and formyaldehyde (see J. Biol. Chem. 271[1996]:27321–27329).
H C
J CH
3
J
3
N + O + NADPH + H; D
2
J
H H C
J CH OH
3
J
2
N + H O + NADP;
2
J
H H C
J H
3
J
N
O
+
K
J
H
H
H
15. (a) Cholesterol feeding has no effect on the amount of mRNA for HMG-CoA reductase.
(b) The actin mRNA is a positive control to (1) verify that RNA can be effectively re covered from all samples, and (2) allow normalization of the results, if necessary, to correct for variations in the extent of overall RNA recovery from sample to sample.
(c) The amount of HMG-CoA reductase protein is greatly reduced when the animals are fed a cholesterol diet.
(d) Although the amount of specific HMG-CoA reductase mRNA is not affected by cholesterol, the level of HMG-CoA reductase protein decreases to near zero for the cholesterol-fed mice.
(e) Several mechanisms could explain the presence of the specific mRNA and yet the absence of the specific protein that the mRNA encodes: 1. HMG-CoA reductase could be subject to translational control, so that translation of the message and synthesis of HMG-CoA reductase by ribosomes are inhibited by cholesterol. 2.
Alternatively, the protein could be synthesized but then rapidly degraded in the cholesterol-fed mice.
CHAPTER 2DNA Structure, Replication,
and Repair
7
The text returns to the topic of the flow of genetic information and considers the detailed biochemical mechanisms underlying this complex process. Chapter 5 introduced you to DNA and RNA and outlined the storage, duplication, and expression of genetic information. In Chapter 6, the enzymes and techniques used to analyze, construct, and clone DNA were presented. You should review these chapters to prepare for studying Chapter 27. Pay particular attention to DNA structure, the supercoiling of DNA, and DNA polymerase I in Chapter 5 and to DNA ligase in Chapter 6.
Chapter 27 covers the biochemistry of the transmission of genetic information from parent to progeny by describing DNA and enzyme systems that replicate, recombine, and maintain it. The chapter opens with a presentation of the problems that a cell must overcome to duplicate its duplex DNA. The text expands upon earlier coverage of DNA with a more thorough description of the A, B, and Z forms it can assume and the underlying chemical determinants of these structural variations. The text also explains how the sequence-dependent variation of the structure of DNA provides a basis for its sequence-specific recognition by proteins. The previous general description of the biochemistry of DNA polymerases is expanded to explain the roles the template, primer, and metal ions play in their activities. The chemical basis for the fidelity of DNA chain extension by the polymerases is also presented. The helicases that unwind DNA are described next. The text then provides a detailed description of the topology of covalently closed circular DNAs and the topoisomerases that modulate their linking numbers. It then describes the replication fork; replication initiation; and RNA-primed, semidiscontinuous DNA elongation. The mechanisms and role of DNA ligases are also given. The structure and important roles of
481
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CHAPTER 27 DNA polymerase III in replication is described in detail. The special problems of replication arising from the amount of eukaryotic DNA in a cell and the structure of chromatin are introduced and the nature and functions of the telomeres and telomerase are described.
To provide the new sequences of nucleotides in DNA upon which evolution can act, not only mutation but also recombination between two different DNA molecules occurs. The breakage and joining of fragments of DNA with similar sequences also rearranges gene orders within the chromosome and is a mechanism for repairing damaged DNA and regulating gene expression. Review the material on plasmids and bacteriophages in Chapter 6 and recombination in Section 5.6.2 of Chapter 5 to better understand this section. A description of a key intermediate in recombination, the Holliday junction, and the recombinases that form and resolve it, are presented. A description of mutations—their nature, causes, and consequences— and their repair follows. The chapter concludes with examples of pathological deficiencies of DNA repair in humans, the relationship of repair impairments and mutation to carcinogenesis, and a test system for detecting potential carcinogens through their mutagenic action on bacteria.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Outline the problems facing a cell in creating a duplicate copy of a double-strand DNA molecule. Appreciate that enzymes and DNA-binding proteins play essential roles in solving the challenges of DNA replication.
DNA Can Assume a Variety of Structural Forms (Text Section 27.1) 2. Contrast the structural information provided by the x-ray diffraction analysis of DNA fibers and DNA crystals.
3. Indicate the role of deoxyribose puckering in determining the structural differences between A-DNA and B-DNA helices. Compare the structures of double-strand RNA
and RNA-DNA hybrids with that of the A-DNA helix.
4. Describe the major and minor grooves of the B-DNA helix. Distinguish between the four possible base pairs (A–T, T–A, C–G, and G–C) in terms of the unique arrays of hydrogen
bond acceptors and donors and methyl groups they present in the grooves of the DNA.
5. Explain why local DNA structure depends on base sequence and appreciate how proteins that bind DNA in a sequence-dependent manner exploit this sequence-dependent variation.
6. Describe Z-DNA in terms of the handedness of the helix and the shape of the path traced by its phosphodiester backbone.
DNA Polymerases Require a Primer and a Template (Text Section 27.2) 7. Outline the key features of the reactions catalyzed by DNA polymerases. Define template and primer as they relate to DNA polymerases.
8. Appreciate the common structures and the evolutionary relationships among DNA polymerases.
9. Explain the role of Mg 2; in the reaction catalyzed by DNA polymerases.
10. Account for the fidelity with which a DNA polymerase selects the correct incoming deoxyribonucleotide triphosphate (dNTP) substrate.
11. Relate the 3„D5„ nuclease activity of DNA polymerases to the fidelity of DNA replication.
12. Describe how helicases separate the strands of duplex DNA.
Double-Strand DNA Can Wrap Around Itself to Form Supercoiled Structures
(Text Section 27.3) 13. Define the linking number (Lk) of a circular DNA molecule and relate supercoiling to the electrophoretic and centrifugal mobility of the molecule.
14. Write the equation relating the linking number to the twisting number (Tw) and writhingnumber (Wr) of a DNA topoisomer. Explain how a negative superhelix density can facilitate the unwinding of the helix. Describe the partitioning of the free energy of a negatively supercoiled molecule into Tw and Wr.
15. Describe the three steps of the reaction catalyzed by topoisomerases; distinguish between type I and type II topoisomerases; and describe the substrates, products, and mechanisms of topoisomerase I and II.
DNA Replication of Both Strands Proceeds Rapidly from Specific Start Sites
(Text Section 27.4) 16. Draw a replication fork, and describe the reactions and the movements of the DNA strands that occur during replication.
17. Define continuous replication and discontinuous replication and relate these terms to the leading and lagging strands of replicating DNA. Describe an Okazaki fragment.
18. Describe the function and features of the nucleotide sequence of oriC and note that it is the unique site of bidirectional replication initiation in E. coli. List the proteins that interact with the DNA in this region of the chromosome, and give the reactions they catalyze and functions they serve.
19. Explain the roles of RNA in DNA replication. Describe the primosome, and describe how enzymes form and remove RNA primers from the genome.
20. List the distinctive features of the DNA polymerase III holoenzyme, and describe how an asymmetric dimer of the enzyme, along with other proteins, coordinates the synthesis of the leading and lagging strands of the daughter duplexes. Appreciate the structural complexity of the replication machinery.
21. Summarize the reactions and identify the proteins at the replication fork that carry out DNA replication.
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22. List the substrates and outline the reaction mechanisms of the DNA ligases.
23. Describe the special problems arising from DNA length and the cell cycle in eukary otes. Explain the roles of multiple replication origins and the telomeres in eukaryotic DNA replication.
24. Describe how telomerase makes DNA of defined sequence in the absence of an external template.
Double-Strand DNA Molecules with Similar Sequences Sometimes Recombine
(Text Section 27.5) 25. Describe the Holliday model for homologous recombination. Explain how resolution of the Holliday junction intermediate can form different recombinant DNA products.
26. Outline the reactions catalyzed by recombinases and note their mechanistic similarity to the topoisomerases.
Mutations Involve Changes in the Basic Sequence of DNA (Text Section 27.6) 27. Distinguish among substitution, insertion, and deletion mutations, and relate them to changes in DNA. Distinguish between transversion and transition substitution mutations and explain the origin of mutations that alter the reading frame in translation.
28. Give examples of some chemical mutagens and their mechanisms. Describe the structure of the pyrimidine dimer formed by ultraviolet light. Outline the mechanisms for the enzymatic repair of damaged DNA.
29. Explain why thymine rather than uracil is used in DNA.
30. Describe xeroderma pigmentosum and hereditary nonpolyposis colorectal cancer, their causes, pathological consequences, and relationships to DNA repair.
31. Relate Huntington disease to triplet expansions within DNA.
32. Relate mutagens and carcinogens. Outline the Ames Salmonella mutagen assay.
SELF-TEST
DNA Can Assume a Variety of Structural Forms 1. Which of the following statements about the double-helical structure of DNA are correct?
(a) It has adenine paired with thymine and cytosine paired with guanine.
(b) It can assume many different forms including A-DNA, B-DNA, and Z-DNA.
(c) In B-DNA, all the hydrogen-bonded base pairs lie in a plane perpendicular to the helix axis.
(d) Different puckering of their ribose residues occurs in A-DNA and B-DNA.
(e) It is rigid and static.
2. Figure 27.1 shows the structures of the adenine and thymine (A–T) and guanine and cy tosine (G–C) base pairs in B-DNA. Use the figure to answer the following questions:
(a) Which hydrogen-bond donors or acceptors of the A–T base pair are in the major DNA STRUCTURE, REPLICATION, AND REPAIR
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groove?
(b) Which hydrogen-bond donors or acceptors of the G–C base pair are in the minor groove?
(c) Using n to represent heterocyclic ring nitrogen atoms, o for oxy groups, and h for protons, give the patterns of hydrogen-bond acceptors and donors in the major groove for a G–C and a C–G base pair.
(d) Using the same symbols as in (c), give the patterns of hydrogen-bond acceptors and donors in the minor groove for an A–T and a T–A base pair.
(e) Explain the possible biological significance of the unique arrays of hydrogen-bond donors and acceptors in the grooves of B-DNA.
FIGURE 27.1 Watson-Crick base pairs.
H
N N J H . . . O
CH3
7
K
J
8
N 9 5 6
4 5
4
1 N . . . H J N 3
6
3 2
1
2
N
N
K
O
A–T
H
N O J H . . . N
7
K
J
8
N 9 5 6
4 5
4
1 N J H . . . N 3
6
3 2
1
2
N
N
K N J H . . . O
H
G–C
3. X-ray diffraction studies of DNA held at low relative humidity revealed the existence of
A-DNA. Since such arid conditions presumably never occur in the cell, what is the significance of the structure of this DNA?
4. Match the features or characteristics in the right column with the type of DNA helix in the left column.
(a) A-DNA (1) Phosphates in the backbone zigzag (b) B-DNA along the helix.
(c) Z-DNA (2) Formation is favored sequences of alternating purine and pyrimidines.
(3) has a relatively wide and deep major groove
(4) has a right-handed helix (5) has 10.4 base pairs per turn (6) has a structure similar to that of double-strand RNA (7) Strands in the helix have opposite polarities.
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DNA Polymerases Require a Primer and a Template 5. Which of the following statements about DNA polymerases are correct?
(a) They add deoxyribonucleotide units to the 3„-hydroxyl of a primer.
(b) They use the template strand to help choose which deoxyribonucleotide unit to add to the growing DNA chain.
(c) They contain a 3„D5„ nuclease that cleaves phosphodiester bonds of misin corporated deoxyribonucleotides.
(d) They check the size of an incoming deoxyribonucleotide triphosphate (dNTP) to help insure that the correct, complementary choice is made.
(e) They bind one complementary dNTP and add a second complementary dNTP to initiate a new DNA chain.
6. Mg2; serves which of the following functions in the reaction catalyzed by DNA polymerases?
(a) stabilizes the pentacoordinate transition state of the phosphodiester bond formation (b) precipitates inorganic phosphate (Pi) arising from the reaction (c) interacts with the 3„-hydroxyl of the incoming dNTP (d) forms a bridge between the 3„-hyroxyl of the primer and a phosphate in the dNTP (e) stabilizes the negative charge on the departing pyrophosphate, which is derived from the incoming dNTP
7. Why are helicases required during DNA replicaton? Is ATP required for their action?
Double-Strand DNA Can Wrap Around Itself to Form Supercoiled Structures 8. The topological features of circular DNA may affect which of the following?
(a) the electrophoretic mobility of the DNA (b) the sedimentation properties of the DNA (c) its affinities toward proteins that bind to the DNA (d) the susceptibility of the strands of the DNA to unwinding (e) the susceptibility of the DNA to the action of DNA ligase 9. Which of the following statements about DNA molecules that are topoisomers are correct?
(a) They are bound to topoisomerases.
(b) They differ from one another topologically only in that they have different link ing numbers.
(c) They may be separated from one another by electrophoresis.
(d) They have identical molecular weights.
(e) They are topological or spatial isomers.
10. Which of the following statements about topoisomerases are correct?
(a) They alter the linking numbers of topoisomers.
(b) They break and reseal phosphodiester bonds.
(c) They require NAD; as a cofactor to supply the energy to drive the conversion of a supercoiled molecule to its relaxed form.
(d) They form covalent intermediates with their DNA substrates.
(e) They can, in the case of a particular type of topoisomerase, use ATP to form nega tively supercoiled DNA from relaxed DNA in E. coli.
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DNA Replication of Both Strands Proceeds Rapidly from Specific Start Sites 11. Match the properties or functions in the right column with the DNA polymerase in the left column.
(a) DNA polymerase I (1) involved in replication (b) DNA polymerase III (2) requires a primer and a template (3) involved in DNA repair (4) makes most of the DNA phosphodi ester bonds during replication (5) removes the primer and fills in gaps during replication
12. Which of the following statements about DNA replication in E. coli are correct?
(a) It occurs at a replication fork.
(b) It starts at a unique locus on the chromosome.
(c) It proceeds with one replication fork per replicating molecule.
(d) It is bidirectional.
(e) It involves discontinuous synthesis on the leading strand.
(f) It uses RNA transiently as a template.
13. Which of the following statements about DNA polymerase III holoenzyme from E. coli are correct?
(a) It elongates a growing DNA chain hundreds of times faster than does DNA poly merase I.
(b) It associates with the parental template, adds a few nucleotides to the growing chain, and then dissociates before initiating another synthesis cycle.
(c) It maintains a high fidelity of replication, in part by acting in conjunction with a subunit containing a 3„D5„ exonuclease activity.
(d) When replicating DNA, it is a molecular assembly composed of at least 10 differ ent kinds of subunits.
14. Explain how the b2 subunit of DNA polymerase III holoenzyme contributes to the pro cessivity of the DNA synthesis machinery.
15. Which of the following statements about DNA ligase are correct?
(a) It forms a phosphodiester bond between a 5„-hydroxyl and a 3„-phosphate in duplex DNA.
(b) It requires a cofactor, either NAD; or ATP, depending on the source of the enzyme, to provide the energy to form the phosphodiester bond.
(c) It catalyzes its reaction by a mechanism that involves the formation of a covalently linked enzyme adenylate.
(d) It catalyzes its reaction by a mechanism that involves the activation of a DNA phos phate through the formation of a phosphoanhydride bond with AMP.
(e) It is involved in DNA replication, repair, and recombination.
16. Why is RNA synthesis essential to DNA synthesis in E. coli?
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17. Match the functions or features related to DNA replication in E. coli listed in the right column with the molecules or structures in the left column.
(a) replication fork (1) Synthesis direction is opposite that of (b) oriC replication fork movement.
(c) lagging strand (2) unwinds strands at the origin of repli (d) leading strand cation in association with dnaA and (e) Okazaki fragment dnaC proteins
(f)
dnaB helicase (3) is synthesized continuously (g) single-strand binding protein (ssb) (4) synthesizes most of the DNA (h) DNA gyrase (5) is synthesized discontinuously (i)
primase
(6) relieves positive supercoiling (j) DNA polymerase III holoenzyme (7) is the locus of DNA unwinding (k) ¯ subunit of DNA polymerase III (8) hydrolyzes ATP to reduce the linking (l) DNA polymerase I number of DNA (m) DNA ligase (9) binds dnaA, dnaB, and dnaC proteins (10) fills in gaps where RNA existed (11) is the point of initiation of synthesis (12) joins lagging strand pieces to each other (13) contains a 5„D3„exonuclease that removes RNA primers (14) is an RNA polymerase (15) performs “proofreading” on most of the DNA synthesized (16) stabilizes unwound DNA (17) uses NAD; to form phosphodiester bonds
18. Why are the antibiotics novobiocin, ciprofloxin, and nalidixic acid, which inhibit DNA gyrase, useful in treating bacterial infections in humans?
19. The duplication of the ends of linear, duplex DNA presents a problem to the replicative machinery of a human cell. What causes the problem and how does the cell overcome it?
Double-Strand DNA Molecules with Similar Sequences Sometimes Recombine 20. Which of the following statements about genetic recombination are correct?
(a) It generates new combinations of genes.
(b) It can move a segment of DNA from one chromosome to another, for example, from a virus to a host cell.
(c) It is mediated by the breakage of DNA and the rejoining of the resulting fragments.
(d) It generates genome sequence variability, upon which natural selection can act.
21. How many strands of DNA are present at the junction of a Holliday junction?
22. Recombinases (a) make and break phosphodiester bonds in a reaction requiring ATP.
(b) pair homologous DNA molecules prior to strand breakage to form a recombination synapse.
(c) form covalent complexes with their DNA substrates.
(d) have a reaction intermediate reminiscent of those of the DNA ligases.
(e) are related to type I topoisomerases by divergent evolution.
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Mutations Involve Changes in the Sequence of DNA 23. Which of the following nucleotide substitutions are transition mutations?
(a) G for A (c) C for T (b) A for C (d) T for G 24. Which of the substitutions in question 23 are transversion mutations?
25. How could the tautomerization of a keto group on a guanine residue in DNA to the enol form lead to a mutation?
26. Explain why most nucleotides that have been misincorporated during DNA synthesis in E. coli do not lead to mutant progeny.
27. Match the type of mutation or physiological consequence in the right column with the appropriate mutagen in the left column.
(a) 5-bromouracil (1) transversion (b) 2-aminopurine (2) transition (c) aflatoxin (3) insertion or deletion (d) acridines (4) translational frameshift (e) nitrous acid (5) block in replication (f)
ultraviolet light
28. What property of DNA allows the repair of some residues damaged through the action of mutagens?
29. Which of the following enzymes or processes can be involved in repairing DNA in E. coli damaged by UV-light-induced formation of a thymine dimer?
(a) DNA ligase seals the newly synthesized strand to undamaged DNA to form the in tact molecule.
(b) The UvrABC enzyme (excinuclease) hydrolyzes phosphodiester bonds on both sides of the thymine dimer.
(c) DNA polymerase I fills in the gap created by the removal of the oligonucleotide bearing the thymine dimer.
(d) The UvrABC enzyme recognizes a distortion in the DNA helix caused by the thymine dimer.
(e) A photoreactivating enzyme absorbs light and cleaves the thymine dimer to re-form two adjacent thymine residues.
30. Given that the base T requires more energy to synthesize than U, and A pairs equally well with U or T, why does DNA likely contain A–T base pairs instead of A–U base pairs?
31. Explain how mutations in genes encoding proteins likely to be involved in DNA repair, such as those defective in xeroderma pigmentosum and hereditary nonpolyposis colorectal cancer, may contribute to the onset of cancer.
32. Explain how some strains of Salmonella are used to detect carcinogens. How is an ex tract from human liver involved in this test?
ANSWERS TO SELF-TEST 1. a, b, d. Answer (c) is incorrect because B-DNA has local variations from the average struc ture, which was observed in DNA fibers. Structures of individual DNA crystals reveal that the hydrogen-bonded base pairs are often twisted and tilted out of the plane that is perpendicular to the helix axis.
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2. (a) In the major groove, N-7 of A is an acceptor, H-6 on the 6-exocyclic (i.e., not in the heterocycle ring) amino group of A is a donor, and O-4 on T is an acceptor.
(b) In the minor groove, N-3 of G is an acceptor, H-2 on the 2-exocyclic amino group of G is a donor, and O-2 on C is an acceptor.
(c) The patterns in the major groove are noh for G–C and hon for C–G.
(d) The patterns in the minor groove are no for A–T and on for T–A, with no donors being present.
(e) The patterns of hydrogen bond donors and acceptors in the major and minor grooves of B-DNA, recognized by complementary hydrogen bond donors and acceptors on the amino acid side chains of proteins, facilitate the sequence-specific interaction of proteins and DNA. Using such interactions proteins can recognize specific sequences in DNA without disrupting the DNA helix. Note that A–T base pairs have two symmetrically related acceptor atoms in the minor groove and thus have less “information” in this groove than do G–C base pairs. The methyl groups on T may also participate in the recognition of DNA through specific hydrophobic interactions with proteins. Finally, a given local DNA sequence gives rise to a particular local DNA shape or conformation. A protein or enzyme that binds sequencespecifically might recognize the shape rather than the sequence per se.
3. Double-strand RNA, RNA–DNA hybrids, and some short sequences of double-strand DNA embedded in B-DNA have structures like that of A-DNA. Thus, knowledge about the helical structure of A-DNA contributes to an understanding of other similar helices that are of physiological importance.
4. (a) 4 ,6, 7 (b) 3, 4, 5, 7 (c) 1, 2, 7. Relevant to answer (2) for (c), the B-DNA to Z-DNA transition would require the complete unwinding of the right-handed helix to form the left-handed one. Negative supercoiling promotes unwinding of B-DNA and thus facilitates the conversion of a segment of B-DNA into Z-DNA.
5. a, b, c, d 6. a, d, e.
7. Duplex B-DNA is a stable molecule at physiological temperature and helicases are re quired to unwind the two strands of the helix so that each can serve as a template for DNA polymerases. Because DNA is so stable, energy is required in the form of ATP hydrolysis to drive helicase action.
8. a, b, c, d. Regarding answer (e), supercoiling is ordinarily a property of covalently closed circular DNA—that is, of DNA in which there are no discontinuities in either strand of the helix. Hence, these molecules are not substrates for DNA ligase, because they lack ends.
9. b, c, d, e. Answer (a) is not correct, because the DNA topoisomer need not necessarily be bound by a topoisomerase.
10. a, b, d, e. Although all topoisomerases break and reseal phosphodiester bonds, an ex ternal energy source is not always required. Relieving the torsional stress in a negatively supercoiled DNA molecule by relaxing it with topoisomerase I is exergonic and requires no energy input, whereas introducing negative supercoils with DNA gyrase is endergonic and must be coupled to ATP hydrolysis. The particular catalytic mechanisms of given topoisomerases determine whether they are coupled to ATP hydrolysis.
11. (a) 1, 2, 3, 5 (b) 1, 2, 4 12. a, b, d. Although not explicitly stated in the text, answer (c) is incorrect because the repli cating E. coli chromosome has two replication forks that synthesize the DNA bidirec tionally from the unique oriC origin. Answer (e) is incorrect because only the lagging strand is synthesized discontinuously. Answer (f ) is incorrect because RNA serves as a primer and not as a template.
13. a, c, d. Answer (b) is incorrect because DNA polymerase III holoenzyme is a highly pro cessive enzyme that synthesizes extensively before dissociating from its template.
14. The b2 subunit forms a torus, with the duplex DNA in its aperture. The b2 ring acts as a sliding clamp that holds the replication machinery on the DNA.
15. b, c, d, e. Answer (a) is incorrect because the enzyme joins a 3„-hydroxyl to a 5„-phos phate. Answer (d) is correct; although not completely described in the text, DNA ligase activity is required to seal the discontinuities in DNA arising during DNA replication, repair, and recombination.
16. Because DNA polymerases are unable to initiate DNA chains de novo, and because they require a primer with a 3„-hydroxyl group, short RNA chains are used as primers to start DNA replication on the leading strand at the origin of replication and to initiate the Okazaki fragments of the lagging strand. RNA polymerases can start RNA chains by adding a nucleotide to an initiating NTP, but they do so with relatively low accuracy.
RNA-initiated DNA chains facilitate high-fidelity replication at the beginning sequences of new chains because they allow DNA polymerase I to replace the RNA with DNA, using the information in the complementary strand and both of its exonucleases in a nick translation reaction (text, p. 760).
17. (a) 7 (b) 7, 9, 11 (c) 1, 5 (d) 3 (e) 1, 3 (f) 2 (g) 16 (h) 6, 8 (i) 14 (j) 4 (k) 15 (l) 1, 10, 13 (m) 12, 17. Answer (5) is not a correct match with (e), because each Okazaki fragment is synthesized continuously.
18. These compounds interfere with the essential helix-destabilizing function of DNA gy rase in bacterial DNA replication. By inhibiting its action, they prevent the gyrase from relieving the positive supercoils that build up ahead of the moving replication fork.
Human cells lack an enzyme similar to gyrase, and thus they are relatively unharmed by these antibiotics.
19. Because DNA polymerases extend their growing polynucleotide chains only in the 5„D3„ direction, and because the two strands of the parental DNA duplex are antiparallel, removal of the RNA primer that is paired with the 3„ end of the parental template DNA would leave an overhanging 3„ DNA strand with no means of having its complement synthesized. Ordinary DNA polymerases are unable to initiate DNA chains de novo. Each round of replication would consequently shorten the DNA because a portion could not be copied. To circumvent this problem, human DNA chromosomes have a segment of repeating G-rich DNA (telomeres) at their ends. In addition, a special enzyme, telomerase, which is an RNA-dependent DNA polymerase (reverse transcriptase) that carries its own RNA template, can, in effect, extend the uncompleted end at each round of replication. The RNA template renews the repeating telomere sequence so that the DNA is not shortened (text, pp.765–766).
20. a, b, c, d
21. Four. The Holliday junction is formed from the four strands of two interacting duplex DNA molecules that, as a result of the initial reactions of recombination, become joined to form one molecule. The Holliday junction is resolved when recombination is completed and two separate duplexes are reformed. Although not mentioned in the text, the products can sometimes have regions of duplex where one strand of DNA is from one parent and the other strand from the other parent, that is, a heteroduplex is formed.
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CHAPTER 27 22. b, c, e. Recombinases have reaction mechanisms similar to those of the topoisomerases in that a covalent enzyme–DNA phosphodiester bond is formed. The bond linking the enzyme to the DNA has a high free energy of hydrolysis and is, thus, capable of resynthesizing the phosphodiester bond broken during its formation. Consequently, recombinases, unlike DNA ligases and DNA polymerases, do not require an external source of energy to form a phosphodiester bond.
23. a, c. Transition mutations are substitutions of a purine for a purine or a pyrimidine for a pyrimidine.
24. b, d. Transversion mutations substitute a purine for a pyrimidine or vice versa.
25. The rare enol tautomer of G could base-pair with a T in the template to allow its incor poration into a growing DNA strand during replication. If the proofreading process missed this erroneous incorporation, the resulting daughter DNA duplex would contain a G–T base pair. During the next round of replication, the T would direct the incorporation of an A into its complementary daughter strand. The final result would be the substitution of an A–T base pair for the original G–C base pair.
26. The proofreading 3„D5„ nuclease of the ¯ subunit of DNA polymerase III holoen zyme removes most of the misincorporated nucleotides that do not form a base pair with the template. The polymerase activity of the holoenzyme then has a second chance to incorporate the correct nucleotide. Additionally, DNA repair systems exist that can detect and repair a mismatched base pair resulting from a misincorporation during synthesis.
27. (a) 2 (b) 2 (c) 1 (d) 3, 4 (e) 2 (f) 5 28. Since DNA is double-strand, damage to one strand of the DNA can often be repaired by using the undamaged complementary strand as a template to direct new incorportion of correct deoxynucleotides in place of the removed incorrect ones.
29. a, b, c, d, e. Although not mentioned, the uvrD protein is a helicase that removes the 12-nucleotide-long oligonucleotide bearing the thymine dimer.
30. C spontaneously deaminates to form U in DNA. This change would lead to a mutation during the next round of replication of the DNA, since U would pair with A changing what was a C–G base pair into an U–A base pair. The repair machinery of a cell that used U normally in its DNA would be unable to distinguish the U in an A–U base pair arising from a C deamination from one formed during “normal” replication. The methyl group on T, which is almost universally used in DNA, distinguishes it from the uracils formed by deamination so that the uracils can be repaired.
31. The inability to effectively repair mutagenic lesions in DNA may lead to their accumu lation. As a consequence, genes regulating cellular proliferation may malfunction, causing cancer.
32. Special strains of Salmonella have been developed to detect substitution, insertion, and deletion mutations in their DNA as a result of exposure to exogenously supplied chemicals. Mutagens can alter the DNA in these strains and thus convert them from auxotrophs, which are unable to grow in the absence of histidine, to prototrophs. The revertants can grow on media lacking histidine and are detected with high sensitivity.
Since there is a correlation between mutagenicity and carcinogenicity, these strains are used as an inexpensive initial test of the carcinogenic potential of a compound. Because animals sometimes metabolize innocuous compounds and convert them to carcinogens, incubation of a suspect chemical with a human liver extract before using the bacterial test can sometimes mimic what would happen to the chemical in vivo. This adjunct to the test expands its capacity to detect potential human carcinogens.
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PROBLEMS
1. The major and minor grooves of B-DNA contain groups that are potential hydrogen-bond donors and acceptors and hence might be important in specific interactions between proteins and DNA. Suppose that a given region of B-DNA is G–C-rich. Will the number of potential hydrogen-bond-forming groups in the major groove and the minor groove differ from the number that would be present if the region were A–T-rich? Explain.
2. Which of the following nucleotide sequences when embedded in longer DNA might allow for the formation of Z-DNA? Explain your answer.
(a) CCTAGTTA (b) CATATATA (c) GCGCGCGT (d) TGAATTCA
3. Many proteins that bind to DNA consist of two a-helical recognition units that are re lated by a twofold axis of symmetry and are separated by a distance of ~34 Å.
(a) Are these proteins likely to bind to double-strand or to single-strand DNA? Explain.
(b) Would the proteins bind preferentially to A-DNA, B-DNA, or Z-DNA? Explain.
(c) Describe how these proteins likely interact with DNA.
4. Suppose that two polynucleotide chains are joined by DNA ligase in a reaction mixture to which ATP labeled with 32P in the a-phosphoryl (innermost) group has been added as an energy source. What products of the reaction would be expected to carry the radioactive label? Explain.
5. Suppose that negatively supercoiled DNA with Lk=23, Tw=25, and Wr=:2 is acted on by topoisomerase I. After one catalytic cycle, what would be the approximate values of Lk, Tw, and Wr?
6. Suppose that negatively supercoiled DNA with Lk=23, Tw=25, and Wr=:2 is acted on by DNA gyrase and ATP. After one catalytic cycle, what would be the approximate values of Lk, Tw, and Wr?
7. What property of DNA polymerase I led to the observation that polA1 mutants of E. coli are more sensitive to ultraviolet light than the wild-type cells? (Hint: The 5„D3„ exonuclease activity of DNA polymerase I can remove damaged nucleotides from DNA as well as destroying the RNA primers on Okazaki fragments.)
8. Suppose that a single-strand circular DNA with the base composition 30% A, 20% T, 15% C, and 35% G serves as the template for the synthesis of a complementary strand by DNA polymerase.
(a) Give the base composition of the complementary strand.
(b) Give the overall base composition of the resulting double-helical DNA.
9. An early and initially attractive mechanism proposed for genetic recombination was the copy-choice model. It suggested that recombination between two parental DNA duplexes occurs during DNA replication when DNA polymerase switches or jumps from one parental duplex to the other so as to produce a recombinant daughter DNA duplex that contains sequences derived from the templates of two different DNA duplexes. The copychoice model is now known to act infrequently. One experimental finding inconsistent with the copy-choice model is the observation that when E. coli bacteria are infected by T4 bacteriophage of two different genotypes—whose DNAs are distinctly marked, one by 32P and the other by bromouracil—recombinant DNAs containing both markers are 494
CHAPTER 27 found under conditions in which DNA synthesis is blocked. How are these findings inconsistent with the copy-choice model, but consistent with the breakage-reunion model for genetic recombination?
10. Relate genetic recombination to exon shuffling, that is, the rearrangement of exons to form new proteins.
11. Suppose that a plasmid with a single origin of replication on its circular chromosome and containing only genes A, B, C, and D begins to replicate rapidly at time t=0. At t=1, there are twice as many copies of genes B and C as there are copies of genes D and A. Is it possible to establish the order of the four genes on the plasmid? Explain.
12. Suppose that a bacterial mutant is found to replicate its DNA at a very low rate. Upon analysis, it is found to have normal levels of activity of DNA polymerases I and III, DNA gyrase, and DNA ligase. It also makes normal amounts of the wild-types of dnaA, dnaB, dnaC, and SSB proteins. The sequence of the oriC region of its chromosome is found to be wild type. What defect might account for the abnormally low rate of DNA replication in this mutant? Explain briefly.
13. Which of the following mutations in a polypeptide chain could have been induced by a single hit of the mutagen 5-bromouracil on the coding strand DNA? (See text, p. 769, and the genetic code on p. 134.)
(a) PheDGlu (c) PheDLeu (b) AspDAla (d) MetDLys 14. Hydroxylamine reacts readily with cytosine to yield a product that base pairs with ade nine. Which of the following mutations could result from the action of a single-hit reaction of hydroxylamine with DNA?
(a) GlnDAsn (c) HisDTyr (b) GluDLys (d) GlyDAsp 15. The drug fluorouracil is used as an anticancer agent. It irreversibly inactivates the en zyme thymidylate synthase. Explain how this treatment retards the growth of tumor tissue. Will the growth of normal cells be affected as well?
16. Mammalian cells of two differing genotypes can be fused together, often in the presence of Sendai virus, to form multinucleate cells (heterokaryons) containing nuclei of both genotypes. When fibroblasts from two patients suffering from xeroderma pigmentosum were fused, the resulting heterokaryons showed no deficiency in DNA repair. What conclusions can be drawn from this observation? Explain.
17. Physical studies on the interaction of the b2 subunit of DNA polymerase III holoen zyme show that the b2 subunit binds much more tightly to circular than to linear DNA molecules.
(a) Propose an explanation for this observation.
(b) What do you think would happen if the circular DNA were treated with a double strand hydrolyzing endonuclease?
18. Eukaryotic DNA can be highly methylated at the C-5 position of cytosine. The degree of methylation is inversely correlated with gene expression (text, pp. 878–879). Although the exact role of C-5 methylation in gene expression is not known, it is known that these C-5-methylated cytosines can cause mutations. How?
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19. Acyclovir is an antiviral agent used to reduce the pain and promote the healing of skin lesions resulting from adult chicken pox. Its structure is shown in Figure 27.2.
FIGURE 27.2 Structure of acyclovir.
O
K
N H N N
N H N
2
O O H (a) What nucleoside does this drug resemble?
(b) Why does the drug need to be administered in dephosphorylated form?
(c) Acyclovir has very few side effects because it inhibits DNA replication only in herpes infected cells. This is because all herpes viruses encode a thymidine kinase gene that is able to activate the drug. What is this activating reaction?
(d) Once acyclovir is activated, how does it inhibit DNA replication?
(e) Why are cells uninfected by virus relatively unaffected by acyclovir?
(f) The herpes virus can become insensitive to acyclovir therapy by mutations in either of two genes. What might they be?
ANSWERS TO PROBLEMS
1. Both G–C and A–T base pairs of B-DNA have one hydrogen bond donor and two hy drogen bond acceptors in the major groove. However, a G–C pair has one donor and two acceptors in the minor groove, whereas an A–T pair has only two acceptors. Thus, a G–C-rich region on DNA differs from an A–T-rich region with respect to the number of hydrogen-bond–forming groups in the minor groove (text, p. 748).
2. Both (b) and (c) are alternating sequences of purines and pyrimidines and are therefore sites of possible Z-DNA formation (text, p. 791). Sequences rich in GC dinucleotides (c) form Z-DNA more easily than those rich in (AT) dinucleotides (b). The latter can be driven into the Z conformation if they are flanked by GC-rich sequences or the DNA is highly negatively supercoiled.
3. (a) They bind to a region of double-strand DNA that also has a twofold axis of sym metry, such as a palindromic region.
(b) B-DNA. The distance between adjacent major grooves is 34 Å. Also these grooves are wide enough to accommodate the recognition helix of the binding protein.
(c) The helical recognition regions bind to two adjacent major grooves of B-DNA.
(See Section 9.3.3 of the text [pp. 248–251] for an example of an enzyme, EcoRV, that recognizes a palindromic DNA sequence. Note, however, EcoRV does not use a helices to contact the DNA. See p. 874 for examples of helix-turn-helix-containing proteins that bind adjacent major grooves of B-DNA.)
4. In the overall reaction, ATP is hydrolyzed to AMP and pyrophosphate. Only AMP would be labeled. The phosphate involved in the formation of the phosphodiester bond is furnished by the polynucleotide chain and does not arise from ATP.
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5. Lk=24, Tw=25, and Wr=:1. Topoisomerase I increases the linking number of DNA by 1 each catalytic cycle (text, p. 756). This increase comes about at the expense of unwinding the negative supercoil.
6. Lk=21, Tw=25, and Wr=:4. DNA gyrase catalyzes a reaction in which both DNA strands are broken, the linking number is decreased by 2, and the number of negative supercoils is correspondingly increased by 2. The answers given for this and the preceding question are approximate for the values of Tw and Wr because solution conditions affect the distribution between twists and supercoils.
7. The polA1 mutants are extraordinarily sensitive to ultraviolet irradiation because they are deficient in the 5„D3„ exonuclease activity of DNA polymerase I and are therefore impaired in DNA repair. They have only 1% of the activity of their wild-type counterpart and cannot efficiently remove thymine dimers formed by UV light. They can, however, replicate their DNA at normal rates because DNA polymerase III is the enzyme that is primarily responsible for DNA replication. An enzyme, RNaseH, which hydrolyzes RNA only when it is base paired to DNA, likely replaces the 5„D3„ exonuclease in processing the Okazaki fragments.
8. (a) 30% T, 20% A, 15% G, 35% C (b) 25% A, 25% T, 25% C, 25% G. The base composition of the double strand is the average of that of the two single strands.
9. First, if copy-choice were a correct model, no recombinant phage should be produced in the absence of new DNA synthesis. Second, according to that model, no recombinant DNA duplexes should contain both bromouracil and 32P. However, the Holliday model for homologous recombination (text, pp. 766–768) accounts for how different labels from different DNA molecules could occur in the same progeny molecule.
10. Exons often encode protein domains. Genetic recombination can lead to rearrangements in the order of exons in a gene. Upon expression, such rearranged genes could give rise to proteins with new domain orders and possibly new capabilities (text, pp. 137–138).
11. The order cannot be unambiguously established from the information given. Two pos sibilities are shown in Figure 27.3.
FIGURE 27.3 Two possible gene arrangements for problem 11.
o r i C o r i C
B
C
B
C
A
D
D
A
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12. A decrease in the activity of primase would account for the low rate of DNA replication.
Synthesis of DNA itself requires the prior synthesis of RNA primers. Also, decreased rates of dNTP synthesis could slow replication.
13. The mutagen 5-bromouracil changes A–T pairs to G–C pairs or G–C pairs to A–T pairs.
The mutation in (c) could be induced by 5-bromouracil. For example, the DNA sequence AAA, which codes for phenylalanine, could be changed to the sequence AAG, which codes for leucine. The other mutations could not arise from treatment with 5bromouracil. Remember that the genetic code presented in the text is expressed in terms of RNA. The sequence UUU on RNA corresponds to the sequence AAA on the informational strand of DNA. Leucine is encoded by the sequence CUU on RNA, which corresponds to the sequence AAG on the informational strand of DNA.
Remember also that, unless otherwise specified, nucleotide sequences are written in the 5„D3„ direction.
14. Hydroxylamine causes the unidirectional change of C–G pairs to T–A pairs. The muta tion in (a) cannot result from the action of hydroxylamine. Those in (b), (c), and (d) might. In (b), TTC (Glu) could change to TTT (Lys). In (c), ATG (His) could be converted to ATA (Tyr). In (d), ACC (Gly) could change to ATC (Asp), or GCC (Gly) could change to GTC (Asp).
15. Because thymidylate synthase (text, p. 706) is inactivated, the supply of dTTP is insuf ficient to support the synthesis of DNA at normal rates. If DNA synthesis is suppressed, so will be the rate of division of the tumor cells. This type of treatment takes advantage of the fact that tumor cells divide more rapidly than do normal cells. The dosage of the drug is adjusted so that it will primarily affect more rapidly dividing cells. However, the division of some normally rapidly dividing cells, for example those lining the intestinal tract and blood-forming cells, may be retarded as well.
16. The fibroblasts from the two patients show complementation (the defect in each is reme died by the other), so it is likely that the two patients suffer from different genetic variants of xeroderma pigmentosum. The action of several genes is likely responsible for the excision and subsequent repair of damaged DNA. One patient, for example, could have produced a normal nuclease that excises damaged DNA but could have been deficient in a ligase. The other patient could have produced normal ligase but could have been deficient in nuclease activity. There are at least nine different complementation groups among xeroderma patients.
17. (a) A possible explanation is that the b2 subunit falls off the end of a linear mole cule whereas it is trapped on the circular molecule because it forms a torus around the DNA.
(b) Treatment of circular DNA with an endonuclease cleaving both strands would con vert it to a linear DNA, and the b2 protein could dissociate from the free end, thereby decreasing its apparent affinity.
18. C-5 cytosine can spontaneously deaminate just as cytosine can. When C-5 cytosine deaminates, it forms thymidine, not uracil. Therefore uracil N-glycosylase, a DNA repair enzyme, will not recognize this product of deamination as an inappropriate base and will not remove it from the DNA, causing a transition mutation.
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19. (a) guanosine (b) The phosphorylated form could not cross cell membranes.
(c) The activation involves adding phosphates onto the distal :OH group at the ex pense of ATP hydrolysis by kinases to produce a compound resembling 5„-GTP.
(Although one of the enzymes responsible for the activation is called thymidine
kinase, it can activate other nucleosides and some of their analogues as well.)
(d) The triphosphate of acyclovir will serve as a substrate for DNA polymerase, and acyclovir nucleotide residues will be incorporated into growing polynucleotide chains in the place of guanosine resiues. Acyclovir will, however, cause premature termination of nascent polynucleotide chains because it lacks a free :OH group onto which further nucleotides can be linked by phosphodiester bonds.
(e) Thymidine kinase is encoded by a viral, not a host, gene. Therefore, uninfected cells will lack the susceptible enzyme.
(f) Mutations that impair the ability of either the viral thymidine kinase or DNA polymerase to use the analogue would render infected cells insensitive to the agent.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. DNA polymerase I uses deoxyribonucleoside triphosphates; pyrophosphate is the leav ing group. DNA ligase uses DNA-adenylate (AMP joined to the 5„-phosphate) as a reaction partner; AMP is the leaving group. Topoisomerase I uses a DNA-tyrosyl intermediate (5„-phosphate linked to the phenolic OH); the tyrosine residue of the enzyme is the leaving group.
2. FAD, CoA, and NADP; are plausible alternatives. Note that, like NAD; and ATP, all three of these molecules contain a pyrophosphate (PP) linkage.
3. Positive supercoiling resists the unwinding of DNA. The melting temperature of DNA increases in going from negatively supercoiled to relaxed to positively supercoiled DNA.
Positive supercoiling is probably an adaptation to high temperature, where the unwinding (melting, denaturing) of DNA is markedly increased.
4. (a) Long stretches of each occur because the transition is highly cooperative.
(b) B-Z junctions are energetically highly unfavorable.
(c) A-B transitions are less cooperative than B-Z transitions because the helix stays right handed at an A-B junction but not at a B-Z junction.
5. (a) 1000 nucleotides/s divided by 10.4 nucleotides/turn for B-DNA gives 96.2 revolu tions per second.
(b) 0.34 mm/s. (1000 nucleotides/s corresponds to 3400 Å/s because the axial distance between nucleotides in B-DNA is 3.4 Å.)
(c) The rate of rotation of DNA at a replication fork (96.2 rps) is nearly the same as that of a bacterial flagellum (100 rps) (for further detail, see Chapter 33 in the text).
Skeletal muscle sarcomere shortening is about 15-fold as rapid as polymerase movement at a replication fork (5 mm/s, compared with 0.34 mm/s) (for further detail, see Chapter 33 in the text). Smooth muscle sarcomere shortening (0.4 mm/s) occurs at about the same speed as replication fork movement.
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6. The unwinding of DNA to expose single-stranded regions at the replication fork causes overwinding (positive supercoils) ahead of the fork. The action of topoisomerase II overcomes this effect by introducing negative supercoils to compensate. Without topoisomerase II, the DNA would become too tightly wound ahead of the fork.
7. Telomerase is required to synthesize the ends of new linear chromosomes during cell di vision. Because cancer cells are dividing rapidly, it is likely that the telomerase gene must be activated for a cell to become cancerous.
8. The activity will be similar to the replacement of an RNA primer with DNA by DNA polymerase I. One makes use of the combined 5„D3„ exonuclease and 5„D3„ polymerase activities of DNA polymerase I. From the point of the internal nick (of only one strand) by the endonuclease, polymerase I will extend the free 3„-OH using radioactive dNTPs while at the same time digesting from the internal 5„-phosphate to make room for the newly synthesized DNA. The result is a “nick translation” event in which an unlabeled portion of one DNA strand is replaced with a radioactive stretch of DNA.
(Over the section of new synthesis, only one strand becomes labeled. The strand used as the “template” remains unlabeled.)
9. If replication were unidirectional, tracks with a low-grain density at one end and a high grain density at the other end would be seen. On the other hand, if replication were bidirectional, the middle of a track would have a low density, as shown in the margin.
For E. coli, the grain tracks are denser on both ends than in the middle, indicating that replication is bidirectional.
10. (a) Pro (CCC), Ser (UCC), Leu (CUC), and Phe (UUC). Alternatively, the last base of each of these codons could be U.
(b) These CDU mutations were produced by nitrous acid.
Unidirectional synthesis
Origin
Bidirectional synthesis
Origin
11. Potentially deleterious side reactions are avoided. The enzyme itself might be damaged by light if it could be activated by light in the absence of bound DNA harboring a pyrimidine dimer. The DNA-induced absorption band is reminiscent of the glucose-induced activation of the phosphotransferase activity of hexokinase.
12. DNA ligase relaxes supercoiled DNA by catalyzing the cleavage of a phosphodiester bond in a DNA strand. The attacking group is AMP, which becomes attached to the 5„-phosphoryl group at the site of scission. AMP is required because this reaction is the reverse of the final step in the joining of pieces of DNA (see Figure 27.29 on p. 762 in the text).
13. ATP hydrolysis is required to release DNA gyrase after it has acted on its DNA substrate.
Negative supercoiling requires only the binding of ATP, not its hydrolysis.
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14. (a) Different supercoiled forms (topological isomers) migrate through the gel at differ ent rates. The highly supercoiled DNA has a compact shape and migrates rapidly.
Relaxed DNA has a more extended shape (a larger radius of gyration) and moves more slowly through the gel matrix.
(b) The bands in lane B represent different supercoiled isomers of DNA. Neighboring bands differ from each other by ±1 superhelical turn.
(c) With longer exposure to topoisomerase 1 (a “relaxing” enzyme), the population of DNA molecules shifts toward a distribution that is near thermal equilibrium.
15. (a) The control plate indicates the extent of spontaneous reversion in the absence of an external mutagen.
(b) The known mutagen is a positive control to show that the procedures are correctly implemented and the test is working.
(c) The experimental compound by itself (plate C) gives results that are only margin ally above background (plate A). The experimental compound itself therefore should be classified as either non-mutagenic or only very slightly mutagenic.
However, a metabolic product derived from the experimental compound is mutagenic (plate D).
(d) One or more enzymes from the liver probably are responsible for the metabolic con version of the experimental compound into a mutagenic compound.
CHAPTER 2
RNA Synthesis and Splicing
8
The conversion of DNA nucleotide sequences into RNA sequences is an early step in the expression of genetic information (see Chapter 5). This chapter describes the DNA-dependent RNA polymerases that catalyze this reaction, and the ways in which the product RNA transcript must sometimes be cleaved and modified in various ways before it becomes functional.
The chapter begins with an overview of the three stages of RNA synthesis: initi ation, elongation, and termination. The subunit structure of RNA polymerase from E. coli is described and a distinction between the core and holoenzymes is made. The powerful technique of DNA footprinting is explained in the context of determining the binding sites of RNA polymerase. The nature of prokaryotic promoters is given, and the function of the s subunit of RNA polymerase in specific transcript initiation is explained. The important role that supercoiling plays in transcription initiation is emphasized. The authors next describe the structure of the ternary elongating complex consisting of the template DNA, RNA polymerase, and product RNA. Finally, two mechanisms of transcription termination, one of which requires the r protein, are detailed. The cleavage and modification reactions involved in the maturation of ribosomal and transfer RNAs are outlined. In addition, two antibiotics are described, rifampicin and actinomycin D, that inhibit transcription by different mechanisms.
The authors next turn to the more complex process of transcription in eukary otes. The impossibility of coupling transcription and translation in eukaryotes as it occurs in prokaryotes (because of eukaryotic subcellular separation in the nucleus and cytoplasm) is pointed out. The three eukaryotic RNA polymerases that carry out transcription are described and related to the kinds of RNA they synthesize. The role of the eukaryotic TATA box and the TATA-box-binding protein in basal transcription are explained, as are other eukaryotic promoters and enhancers and some of the proteins
501
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CHAPTER 28 that bind them. The reactions that modify the 5′ and 3′ ends of typical eukaryotic mRNA transcripts to cap and add a poly(A) tail to them are described as are ways in which the nucleotide sequence of certain mRNAs can be modified by base alterations and insertions, a process called RNA editing. The molecular machinery and reactions that remove introns from eukaryotic mRNA (known as splicing) are outlined along with the consequences of alternative splicing reactions. The chapter concludes with the discovery of catalytic RNA and the mechanism of RNAcatalyzed self-splicing in Tetrahymena.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Define transcription.
2. Name the three stages of RNA synthesis, and list the functions of RNA polymerase in these processes.
Transcription Is Catalyzed by RNA Polymerase (Text Section 28.1) 3. Describe the subunit structure of RNA polymerase from E. coli, and assign functions to the individual subunits.
4. Define promoter, and recount how RNA polymerase protects promoters from hydrolysis by DNAse. Explain how footprinting can be used to locate promoters.
5. Recognize the convention for numbering the nucleotides in the DNA template with re gard to the transcription start site. Distinguish between the template (or antisense) strand
and the coding (or sense) strand of the duplex DNA template.
6. Note the consensus sequences around the −35 and −10 positions of E. coli promoters.
Contrast the rates of transcript initiation on strong and weak promoters in E. coli.
7. Explain how the s factor enables RNA polymerase to recognize promoters. Distinguish be tween the s70 and s32 subunits, contrast the sequences of standard promoters and heatshock promoters, and provide examples of how s factors can determine which genes are expressed.
8. Describe how topoisomerase I was used to determine the number of promoter DNA basepairs that are unwound upon the binding of RNA polymerase. Relate negative supercoiling
to promoter efficiency. Distinguish between closed promoter and open promoter complexes.
9. Describe the model for the transcription bubble. State the number of base pairs in theRNA–DNA hybrid. Appreciate the rate of RNA chain elongation in terms of both the nucleotides added and the distance on the template traversed by RNA polymerase.
10. Contrast the fidelity of polymerization of RNA polymerase with that of DNA polymerase, and explain the difference.
11. Contrast r-dependent and r-independent transcription termination. Outline the mecha nisms of the r protein and explain the role of ATP hydrolysis in its function.
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12. Describe the mechanisms of inhibition of transcription by rifampicin and actinomycin D.
Eukaryotic Transcription and Translation Are Separated in Space and Time
(Text Section 28.2) 13. Note the spatial and temporal differences in transcription and translation between prokaryotes and eukaryotes. Consider the regulatory implications of these differences.
14. Contrast mRNA processing in prokaryotes and eukaryotes. Outline the major reactions that process the primary RNA transcripts of eukaryotes.
15. Describe the RNA polymerases of eukaryotes, locate them within the cell, and list the kinds of RNA they synthesize. Account for the toxic effects of a-amanitin, and describe how it can be used to differentiate the three eukaryotic RNA polymerases.
16. List the salient sequence elements of eukaryotic promoters. Contrast the nucleotide sequences and locations of the TATA box of eukaryotes and the −10 sequence of prokaryotes.
17. Explain the key role of TATA-box-binding protein in assembling active transcription complexes.
18. Outline the combinatorial activity of transcription factors and other DNA-binding regu latory proteins in directing RNA polymerase to specific genes.
The Transcription Products of All Three Eukaryotic Polymerases Are Processed
(Text Section 28.3) 19. Draw the structure of the 5′ end of a typical eukaryotic mRNA, and distinguish caps 0, 1, and 2. Outline the reactions required to cap the primary transcript.
20. Describe the events leading to the production of mRNA with a poly(A) tail.
21. Describe RNA editing and provide examples of its effects on gene expression.
22. Describe the splicing of eukaryotic mRNA, give the consensus sequences at the splice site junc-tions, and designate the other nucleotide sequence elements involved in the process. List the functions of this post-transcriptional modification, and explain the importance of alternative splicing in gene expression.
23. Describe the role of transesterification reactions in splicing and compare the number of phosphodiester bonds broken and formed during mRNA splicing.
24. Describe the spliceosome and detail the structural and catalytic involvement of smallnuclear ribonucleoprotein particles (snRNPs) in mRNA splicing.
The Discovery of Catalytic RNA Was Revealing with Regard to Both
Mechanism and Evolution (Text Section 28.4) 25. Describe the discovery of self-splicing RNA and outline the reactions that occur during the conversion of a ribosomal RNA (rRNA) precursor from Tetrahymena into the spliced, mature rRNA and other linear and circular products. Explain the role of guanosine or a guanylyl nucleotide in the self-splicing reaction.
26. Contrast the group I and group II self-splicing introns. Compare spliceosome-catalyzed splicing with self-splicing.
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SELF-TEST
Transcription Is Catalyzed by RNA Polymerase 1. Give the subunit composition of the RNA polymerase of E. coli for both the holoenzyme and the core enzyme.
2. Match the subunit of the RNA polymerase of E. coli in the left column with its putative function during catalysis from the right column.
(a) a (1) binds the DNA template (b) b (2) binds regulatory proteins and sequences (c) b′ (3) binds NTPs and catalyzes bond formation (d) s70 (4) recognizes the promoter and initiates synthesis 3. Which of the following statements about E. coli promoters are correct?
(a) They may exhibit different transcription efficiencies.
(b) For most genes they include variants of consensus sequences.
(c) They specify the start sites for transcription on the DNA template.
(d) They have identical and defining sequences.
(e) They are activated when C or G residues are substituted into their −10 regions by mutation.
(f) Those that have sequences that correspond closely to the consensus sequences and are separated by 17 base pairs are very efficient.
4. The sequence of a duplex DNA segment in a DNA molecule is 5′-ATCGCTTGTTCGGA-3′
3′-TAGCGAACAAGCCT-5′
When this segment serves as a template for E. coli RNA polymerase, it gives rise to a segment of RNA with the sequence 5′-UCCGAACAAGCGAU-3′
Which of the following statements about the DNA segment are correct?
(a) The top strand is the coding strand.
(b) The bottom strand is the sense strand.
(c) The top strand is the template strand.
(d) The bottom strand is the antisense strand.
5. Which of the following statements about the s subunit of RNA polymerase are correct?
(a) It enables the enzyme to transcribe asymmetrically.
(b) It confers on the core enzyme the ability to initiate transcription at promoters.
(c) It decreases the affinity of RNA polymerase for regions of DNA that lack promoter sequences.
(d) It facilitates the termination of transcription by recognizing hairpins in the transcript.
6. When growing E. coli are subjected to a rapid increase in temperature, a new and char acteristic set of genes is expressed. Explain how this alteration in gene expression occurs.
7. Match the regions of a r-independent transcription termination signal in a DNA tem plate in the left column with the structures or the functions performed by the encoded transcript segments in the right column.
(a) GC-rich palindromic region (1) oligo(U) stretch in RNA (b) AT-rich region (2) hairpin in RNA (3) promotes the dissociation of RNA–DNA hybrid helix (4) causes the enzyme to pause 8. Which of the following statements about the r protein of E. coli are correct?
(a) It is an ATPase that is activated by binding to single-stranded DNA.
(b) It recognizes specific sequences in single-stranded RNA.
(c) It recognizes sequences in the DNA template strand.
(d) It causes RNA polymerase to terminate transcription at template sites that are dif ferent from those that lead to r-independent termination.
(e) It acts as a RNA–DNA helicase.
9. Match the functions in the right column with the antibiotic inhibitors of E. coli tran scription in the left column.
(a) rifampicin (b) actinomycin D (1) interacts with the template (2) interacts with nascent mRNA polymerase (3) prevents initiation (4) prevents elongation (5) intercalates into mRNA hairpins 10. Explain how a mutation might give rise to an E. coli that is resistant to the antibiotic rifampicin.
Eukaryotic Transcription and Translation Are Separated in Space and Time 11. Which of the following statements about eukaryotic mRNAs are correct?
(a) They are derived from larger RNA precursors.
(b) They result from extensive processing of their primary transcripts before serving as translation components.
(c) They usually have poly(A) tails at their 5′ ends.
(d) They have a cap at their 3′ ends.
(e) They are often encoded by noncontiguous segments of template DNA.
12. Match the descriptions in the right column with the appropriate eukaryotic DNA dependent RNA polymerases in the left column.
(a) RNA polymerase I (1) is located in the nucleolus (b) RNA polymerase II (2) is located in the nucleoplasm (c) RNA polymerase III (3) makes mRNA precursors (4) makes tRNA precursors (5) makes 5S rRNA (6) makes 18S, 5.8S, and 28S rRNA precursors (7) is strongly inhibited by a-amanitin (8) synthesizes RNA in the 5′
3′
direction
(9) is composed of several subunits (10) has a subunit with repeated amino acid sequences subject to phosphorylation
13. List the major sequence features of promoters for eukaryotic mRNA genes.
14. Describe the role of TATA-box-binding protein (TBP, TFIID) in forming the basal tran scription apparatus. What properties does TBP confer on this apparatus?
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15. Which of the following statements about enhancers are correct?
(a) They function as promoters.
(b) They function when in either orientation in the DNA.
(c) They function when on either side of the activated promoter.
(d) They function even when located many base pairs away from the promoter.
(e) They function only in specific types of cells.
The Transcription Products of All Three Eukaryotic Polymerases Are Processed 16. Which of the following statements about the poly(A) tails that are found on most eu karyotic mRNAs are correct?
(a) They are added as preformed polyriboadenylate segments to the 3′ ends of mRNA precursors by an RNA ligase activity.
(b) They are encoded by stretches of polydeoxythymidylate in the template strand of the gene.
(c) They are added by RNA polymerase II in a template-independent reaction using ATP as the sole nucleotide substrate.
(d) They are added by poly(A) polymerase using dATP as the sole nucleotide substrate.
(e) They are cleaved from eukaryotic mRNAs by a sequence-specific endoribonuclease that recognizes the RNA sequence AAUAAA.
17. What functions are the caps and tails of mRNAs thought to perform?
18. Which of the following statements about apolipoprotein B (apo B) are correct?
(a) The apo B-48 form is formed by the proteolytic cleavage of the primary (apo B-100) translation product.
(b) Apo B-48 and apo B-100 are formed in different tissues.
(c) Apo B-48 arises from the expression of a form of the gene for apo B-100 that has been shortened by nonhomologous recombination.
(d) The transcript of the apo B-100 gene is spliced to remove a segment and form apo B-48.
(e) A specific nucleotide in the apo B-100 transcript is altered, thereby creating a stop codon in the mRNA.
19. Which of the following are important sequence elements in the splicing reactions that produce eukaryotic mRNAs?
(a) exon sequences located between 20 and 50 nucleotides from the 5′ splice site (b) exon sequences located between 20 and 50 nucleotides from the 3′ splice site (c) intron sequences located between 20 and 50 nucleotides from the 5′ splice site (d) intron sequences located between 20 and 50 nucleotides from the 3′ splice site (e) intron sequences at the 5′ splice site
(f) intron sequences at the 3′ splice site 20. Eukaryotic mRNA splicing involves which of the following?
(a) the formation of 2′ 5′ phosphodiester bonds (b) a sequence-specific endoribonuclease that hydrolyzes the phosphodiester bond at the junctions of the intron with the exon (c) the spliceosome (d) the coupling of phosphodiester bond formation to ATP hydrolysis (e) the formation of lariat intermediates RNA SYNTHESIS AND SPLICING
507
The Discovery of Catalytic RNA Was Revealing with Regard to Both
Mechanism and Evolution 21. Place the following events in the order in which they occur during the formation of ma ture rRNA in Tetrahymena.
(a) The 3′-hydroxyl of a guanine nucleoside attacks the phosphodiester bond at the 5′ splice site leaving the 5′ (upstream) exon with a free 3′-hydroxyl and attaching the guanine nucleoside to the 5′ end of the intron.
(b) The transcript from the rRNA gene folds into a specific structure.
(c) The rRNA transcript specifically binds a guanine nucleoside or nucleotide.
(d) Self-splicing occurs within the intron to form L19 RNA.
(e) The 3′-hydroxyl of the 5′ exon attacks the bonds at the 3′ splice junction to form the spliced rRNA and eliminate the intron.
22. Match the descriptions in the right column with the type of splicing in the left column.
(a) group I splicing (1) A 2′ 5′ phosphodiester bond is (b) group II splicing involved.
(c) nuclear mRNA splicing (2) Nuclease and ligase activities are required.
(3) Transesterification reactions break and form bonds.
(4) A guanine nucleoside or nucleotide is re quired.
(5) Spliceosomes are required.
(6) Lariat intermediates are involved.
(7) snRNPs are involved.
ANSWERS TO SELF-TEST
1. The holoenzyme has the subunit composition a2bb′s. The core enzyme lacks the s subunit.
2. (a) 2 (b) 3 (c) 1 (d) 4 3. a, b, c, f. The promoters of most E. coli genes include variants of defining, consensus se quences that are centered at about the −35 and −10 positions. The nearer the sequences of a promoter are to the consensus sequence and the nearer the separation between them is to the optimal 17-bp spacing, the more efficient the promoter. The −10 consensus sequence is TATAAT. The substitution of a C or G into the sequence would likely lower the efficiency of a promoter.
4. b, c. The sense strand (bottom) of the template DNA has the same sequence as the mRNA.
5. a, b, c. The s subunit recognizes promoter sites, decreases the affinity of the enzyme for regions of DNA lacking promoter sequences, and facilitates the specific, oriented initiation of transcription. Orienting the binding of the enzyme to the DNA results in only one of the two DNA strands functioning as a template for RNA transcription; that is, it gives rise to asymmetric transcription.
6. The temperature increase induces the synthesis of a new s factor, s32, which directs RNA polymerase to promoters that have −10 and −35 sequences which are different from those recognized by s70. Transcription from these promoters gives rise to characteristic heat-shock proteins.
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7. (a) 2, 4 (b) 1, 3 8. d, e. The r protein recognizes and binds stretches of RNA that are devoid of hairpins and are at least 72 nucleotides long. It acts to hydrolyze ATP and to unwind the RNA–DNA hybrid in the transcription bubble.
9. (a) 2, 3 (b) 1, 4. Actinomycin D intercalates only into duplex DNA.
10. Rifampicin must bind to the b subunit of RNA polymerase to inhibit the enzyme. A mu tation in the gene encoding this subunit that would interfere with the binding of the antibiotic but not with polymerization would produce a rifampicin-resistant cell.
11. a, b, e. Answers (c) and (d) are incorrect because the poly(A) tail is found at the 3′ end and a cap is found at the 5′ end of typical eukaryotic mRNA. Segments of the primary transcript are discarded by RNA splicing to bring RNA sequences encoded by noncontiguous regions of the template together; that is, introns are removed to form a product smaller than the primary transcript.
12. (a) 1, 6, 8, 9 (b) 2, 3, 7, 8, 9, 10 (c) 2, 4, 5, 8, 9. Both RNA polymerases I and III are involved in rRNA synthesis, with polymerase I synthesizing the 18S, 5.8S, and 28S precursor transcripts and polymerase III synthesizing the 5S rRNA.
13. A TATA box centered at about −25 from the start site of transcription and consisting of a variant of the consensus heptanucleotide sequence TATAAAA is essential for promoter activity. In addition, many genes have a CAAT box, a GC box, or both elements located between −40 and −110. These sequences can function in either orientation, that is, be in either DNA strand. Genes that are expressed constitutively, as distinct from those whose expression is regulated, tend to have GC boxes. Activating sequences farther upstream of these promoter elements are necessary for the functioning of most promoters.
14. TBP serves a critical role as an enucleating center for the assembly of the minimal molecu lar apparatus (the basal transcription assembly) required for transcription by RNA polymerase II. After TBP binds the TATA box, other TFII proteins and RNA polymerase II join the supramolecular complex to render it transcriptionally competent. In addition to enabling the assembly of the apparatus through its ability to recognize and bind the TATA box, TBP binds this asymmetric sequence in one orientation that “points” the RNA polymerase in the right direction, thereby defining the strand of DNA that will serve as the template.
15. b, c, d, e. Answer (a) is incorrect because enhancer sequences do not serve as promot ers per se. They will not, by themselves, enable RNA polymerase II to initiate a transcript, and will function only when the basal transcription apparatus exists.
16. None of the statements are correct. Poly(A) polymerase uses ATP, not dATP, to add a stretch of A residues to the 3′-hydroxyl formed by the cleavage of an mRNA precursor by a specific ribonuclease that recognizes the upstream sequence, AAUAAA, within particular mRNA sequence contexts.
17. The 5′ caps are thought to contribute to the stability and efficiency of translation of the mRNA. The poly(A) tails probably perform the same functions, albeit by different, unknown mechanisms.
18. b, e. The apo B system illustrates one kind of RNA editing. The pre-mRNA transcript has its sequence altered—in this case, a specific cytidine is deaminated to uracil to create the stop codon. Other editing mechanisms involve the addition and removal of U residues, an A-to-G change, and an adenosine-to-inosine change.
19. d, e, f 20. a, c, e. Answers (b) and (d) are incorrect because hydrolysis of phosphodiester bonds is not involved during mRNA splicing up to the point of intron removal and lariat formation; only transesterification reactions occur. Consequently, ATP is not required for the synthesis of phosphodiester bonds.
21. b, c, a, e, d
22. (a) 3, 4 (b) 1, 3, 6 (c) 1, 3, 5, 6, 7. The splicing of eukaryotic tRNAs requires protein nucleases to cut the phosphodiester bonds at the intron–exon junctions and protein RNA ligases to form the bonds joining the exons.
PROBLEMS
1. The technique of footprinting (p. 784 in text) involves, in part, the digestion of unpro tected DNA by DNAse I.
(a) Why is it necessary to use a nuclease that is not sequence-specific?
(b) What would be the consequences of an extensive digestion with DNAse I?
2. The rate constant for the binding of RNA polymerase holoenzyme to a promoter on a long DNA molecule is greater than that for the collision of two small molecules in solution. Since small molecules diffuse through solutions more rapidly than large ones, how can this be true?
3. One would expect an analog of 5′-ATP that lacks an oxygen at the 3′ position of its ribose (3′-deoxy-5′-ATP; see Figure 28.1) to interrupt RNA formation because it cannot form phosphodiester bonds at its 3′position. Could such a compound be used to ascertain the direction of chain growth in RNA synthesis? Explain. (Refer to figure on p. 787 of text.)
FIGURE 28.1 Structure of 3′-deoxy-5′-ATP.
NH
2
J
N
N
O
O
O
N
N
K
K
K :OJPJOJPJOJPJOJCH 2
J
J
J
O :O O: O:
J
J
H
H
H
3„
2„
J
J
H
OH
4. The ciliated protozoan Tetrahymena contains an enzyme that can synthesize 5′-pseudouri dine monophosphate from a mixture of PRPP and uracil. For a time it was thought that this enzyme was instrumental in the synthesis of transfer RNAs in Tetrahymena. Explain why this is not the case.
5. When mammalian genes are cloned, a strategy that is frequently followed involves the isolation of mRNA rather than DNA from a cell and the preparation of a complementary DNA (cDNA) by the enzyme reverse transcriptase. Suppose that mRNA isolated from a cell specialized for the production of protein X is used as a template for the production of cDNA. What major difference or differences would you expect to find between the structure of that cDNA and genomic DNA for protein X?
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6. Rifampicin specifically inhibits the initiation of transcription in prokaryotes and may therefore be used in humans as a therapeutic antibacterial agent. Would you expect actinomycin D to be useful in antibacterial therapy? Why or why not?
7. The mRNAs produced by mammalian viruses undergo modification at the 5′ and 3′ ends in a fashion similar to that of eukaryotic mRNA. Why do you think this is the case?
8. Sketch the most stable secondary structure that could be assumed by the oligonucleotide AAGGCCCUACGGGGCCG.
9. Suppose that human DNA is cleaved into fragments approximately the size of a given mature human messenger RNA, and that mRNA–DNA hybrids are then prepared. The corresponding procedure is then carried out for E. coli. When the mRNA–DNA hybrids from each species are examined with an electron microscope, which will show the greater degree of hybridization? Explain.
10. What are snRNPs, and how are they involved in the eukaryotic mRNA splicing reaction?
11. Figure 28.28 on page 800 of the text gives an example of how a base-change mutation within an intron in the b-globin gene can produce a 3′ splice site upstream from the normal 3′ splice site. The result is that five amino acids not normally present in the protein are inserted into the chain before synthesis is terminated by a stop codon.
(a) Corresponding mutants in exons are also known in which base change mutations introduce 5′ splice sites upstream from the normal 5′ sites (see Figure 28.2A; X marks the location of the new 5′ splice site). Sketch the resulting processed mRNA.
What changes in amino acid sequence would be expected to result?
FIGURE 28.2A Creation of a new 5′ splice site in exon 2 of β-globin.
5„
Exon 2
X
Exon 3 (b) In some forms of thalassemia, the creation of a new 5′ splice site within intron 2 activates a “cryptic” 3′ splice site upstream from the 5′ site (see Figure 28.2B; X marks the location of the new 5′ splice site and the cryptic 3′ side is labeled). Sketch the resulting processed mRNA. How would the resulting changes in amino acid sequence differ from those in (A)?
FIGURE 28.2B Activation of a cryptic 3′ splice site in intron 2 of β-globin.
3„
5„
Exon 2
X
Exon 3 (c) In antisense drug design, an oligonucleotide is designed to form a DNA–mRNA hybrid that will block transcription of targeted genes. Gorman et al. (PNAS
[1998]95:4929–4934) used antisense methods to block splicing of the cryptic and 5′ splice sites in (b) by altering nucleotides in snRNA sequences to allow them to form snRNA–DNA hybrids at the splice sites. What snRNA sequence would be needed to block the 3′ splice site shown in Figure 28.28 of the text? Use six nucleotides for your sequence. Would this snRNA also bind to the normal 3′ splice site? Why or why not?
RNA SYNTHESIS AND SPLICING
511
12. In an attempt to determine whether a given RNA was catalytically active in the cleavage of a synthetic oligonucleotide, the following experimental results were obtained. When the RNA and the oligonucleotide were incubated together, cleavage of the oligonucleotide occurred. When either the RNA or the oligonucleotide was incubated alone, there was no cleavage. When the RNA was incubated with higher concentrations of the oligonucleotide, saturation kinetics of the Michaelis-Menten type were observed. Do these results demonstrate that the RNA has catalytic activity? Explain.
ANSWERS TO PROBLEMS
1. (a) A nonspecific nuclease will cleave duplex DNA into short, random fragments. Thus potential cleavage sites would exist on either side of the protecting protein. If the nuclease was too specific, its cleavage sites would be too far apart to reveal the protein binding site on the DNA.
(b) In principle, one wants to time the digestion with DNAse so that there is, on the average, one cut per DNA molecule. Since there are many potential cleavage sites on each DNA molecule, the result will be a population of molecules of varying lengths (see Figure 28.3 on p. 784 of text). Extensive digestion with DNAse would produce a population of very short fragments.
2. RNA polymerase holoenzyme has lower affinity for nonspecific DNA sequences than for promoter sequences. The nonspecific affinity, however, allows the enzyme to bind to “random-sequence” DNA and then “slide” along the molecule in a unidimensional random walk until it encounters a promoter sequence, for which its binding affinity is higher. Diffusion in one dimension is much faster than diffusion in three dimensions, thereby explaining the observed rapid rate constant for the binding of RNA polymerase holoenzyme to promoter sequences. If one measured the encounter of the polymerase with the nonspecific regions of the DNA rather than with promoter sequences, the value of the rate constant would be much lower and would fit our expectations for a threedimensional, diffusion-limited reaction between macromolecules.
3. A 3′-deoxy analog of ATP could be used to establish the direction of chain growth. In
5′
3′ growth the analog would donate a nucleotide containing 3′-deoxyadenosine, and the polynucleotide chain would be terminated as a result. No additional nucleotides could be added because of the lack of a 3′-OH group on the terminal 3′-deoxyadenosine. In 5′ 3′ growth the nucleotide could not be added to the growing polynu cleotide chain because of the lack of the 3′-OH group.
4. Nascent polynucleotides formed by RNA polymerases contain only the four usual bases.
Subsequently, some of the bases are chemically modified. Were unusual nucleotides to be incorporated into a growing RNA chain, this would in turn require the presence of unusual bases on DNA. The pseudouridine found in transfer RNAs is formed by breaking the nitrogen–carbon bond linking uracil to ribose and forming a carbon–carbon bond instead. Only certain uracils are modified in this manner, owing to their position in the three-dimensional structure of the RNA and to the specificity of the enzymes that carry out the modification.
5. The cDNA prepared from mRNA would have a long poly(T) tail, unlike genomic DNA.
Remember that the poly(A) tail is added to the 3′ end of mammalian mRNA and that there is no counterpart on DNA. A second striking difference would be that the cDNA would contain no intervening sequences (introns) and would therefore be much shorter 512
CHAPTER 28 than the corresponding sections of genomic DNA. (Remember that most mammalian genes are mosaics of introns and exons.) A third difference would be found if any RNA editing were involved. An edited mRNA could generate a cDNA with nucleotides that did not correspond to those in genomic DNA.
6. In order to be useful as a therapeutic antibacterial agent, a compound must selectively inhibit processes in prokaryotes but leave the corresponding processes in eukaryotes (including those in mitochondria) largely unaffected. Because rifampicin selectively inhibits the initiation of transcription in prokaryotes but not in eukaryotes, it is useful as an antibacterial agent. Actinomycin D is an intercalating agent that binds to DNA duplexes and inhibits both DNA replication and transcription, although it has a greater inhibiting effect on transcription than on replication. It cannot discriminate between the duplex DNA of bacteria and that of humans, however, and will therefore bind to both. Because it disrupts eukaryotic as well as prokaryotic processes, it is not very useful as an antibacterial agent. It is sometimes used as an anticancer agent, however, because of its ability to slow the replication rate of human DNA.
7. Viruses use the host’s enzyme system to replicate their DNA and to synthesize their pro teins. Since eukaryotic translation systems must synthesize viral protein, the structure of viral mRNAs must mimic that of the host mRNA.
8. The structure is a stem-and-loop (“lollipop”) hairpin structure, as shown in Figure 28.3.
FIGURE 28.3 Stable secondary structure for the oligonucleotide in problem 8.
A
C
U
G
C
G
C
G
C
G
G
C
G
C
A
G
A
9. The mRNA–DNA hybrid of E. coli will show greater hybridization because it is produced continuously from a DNA template without processing. Because of the presence of intervening sequences in human DNA, there will be regions in the human RNA–DNA hybrids where no base pairing occurs.
10. The snRNPs are small ribonucleoprotein particles that occur in the nucleus. Each is com posed of a small RNA molecule and several characteristic proteins, some of which are common to different snRNPs. Distinct snRNPs recognize and bind to splice junctions and the branch site and are involved in assembling the spliceosome in an ATP-dependent manner. They are requisite components of the splicing apparatus, and the RNA components of some of them are probably catalytically active. The RNAs of some snRNPs form hydrogen bonds with sequences within introns and exons to help to juxtapose properly the reacting splice junctions.
11. (a) The spliced gene arising from the mutation would be shorter than the correctly spliced version (see below) and some amino acids would be omitted from the region of the protein encoded by nucleotides between the newly introduced and normal 5′ splicing sites. Also, the introduction of a new splicing site could lead to changes in the reading frame, and therefore to gross changes in amino acid sequence, polypeptide chain length, or both.
RNA SYNTHESIS AND SPLICING
513
Exon 2
Exon 3
(b) The resulting processed mRNA would be longer than the normal processed b-globin gene, introducing the possibility of a misfolded or malfunctioning protein or a truncated protein if a stop codon were introduced from the intron sequence. As in (a) a change in reading frame could also result.
Exon 2
Exon 3
(c) The sequence containing the mutated 3′ splice site is 5′-ATTAGT-3′. The com plementary snRNA would therefore be 3′-UAAUCA-5′. Since the normal site is 5′-CTTAGG-3′, which would be bound by 3′-GAAUCC-5′, the mutated snRNA should not bind to the normal 3′-splice site.
12. These results alone do not establish that the RNA has catalytic activity because a cata lyst must be regenerated. It is entirely possible that the results observed could be accounted for by a stoichiometric, as opposed to a catalytic, interaction between RNA and the oligonucleotide, in which the RNA may “commit suicide” as the oligonucleotide is cleaved. In such an interaction, a portion of the RNA would participate chemically in the cleavage of the oligonucleotide, but it would also be cleaved itself as a part of the reaction. Four reaction products would accumulate, two resulting from the cleavage of RNA and two from the cleavage of the oligonucleotide. To show that this particular RNA was catalytic, it would be necessary to demonstrate that it turns over and is regenerated in the course of the reaction.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. The sequence of the coding (+, sense) strand is 5′-ATGGGGAACAG CAAGAGTGGGGCCCTGTCCAAGGAG-3′ and the sequence of the template (−, antisense) strand is 3′-TACCCCTTGTCGTTCTCACCCCGGGAC AGGTTCCTC-5′
Note that the coding strand has the same sequence as mRNA (except U T), whereas the template strand is complementary to the coding strand.
2. RNA turns over in the cell, whereas DNA is the “nearly” permanent record of inherited in formation. The consequences of most errors in RNA synthesis will be short-lived, perhaps a protein synthesized with a mistake in its sequence for a short time during the life of the cell, but then the RNA in question may be degraded and the error will disappear. By contrast, an uncorrected error in DNA replication will be passed to the next generation.
3. RNA synthesis involves copying only short segments of chromosomes, whereas DNA replication must involve copying entire chromosomes at the time of cell division. The length of consecutive nucleotide sequence that must be copied is many times greater for DNA replication than for RNA transcription.
4. Heparin, a glycosaminoglycan, is highly anionic. Its negative charges, like the phospho diester bridges of DNA templates, bind to lysine and arginine residues of b′.
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CHAPTER 28
5. This mutant sigma would competitively inhibit the binding of holoenzyme and prevent the specific initiation of RNA chains at promoter sites.
6. The core enzyme without sigma binds more tightly to the DNA template than does the holoenzyme. The retention of sigma after chain initiation would make the mutant RNA polymerase less processive. Hence, RNA synthesis would be much slower than normal.
7. A 100-kd protein contains about 910 residues, which are encoded by 2730 nucleotides.
At a maximal transcription rate of 50 nucleotides per second, the protein would be synthesized in 54.6 seconds.
8. Initiation at strong promoters occurs every two seconds. In this interval, 100 nu cleotides are transcribed. Hence, centers of transcription bubbles are 34 nm (340 Å) apart (100 × 3.4 Å = 340 Å).
9. (a) The lowest band on the gel will be that of (i), whereas the highest will be that of (v). Band (ii) will be at the same position as (i) because the RNA is not complementary to the nontemplate strand, whereas band (iii) will be higher because a complex is formed between RNA and the template strand. Band (iv) will be higher than the others because strand 1 is complexed to 2, and strand 2 to 3. Band (v) is the highest because core polymerase associates with the three strands.
(b) None, because rifampicin acts before the formation of the open complex.
(c) RNA polymerase is processive. Once the template is bound, heparin cannot enter the DNA-binding site.
(d) The longest RNA product formed is a 36-mer rather than the full 72-mer. Because GTP is absent, synthesis stops when the first C downstream of the bubble is encountered in the template strand.
10. Segments of double helix that are shorter than tetranucleotides (having fewer than about 4 base pairs) are unstable at physiological temperatures due to their small extent of base stacking and small number of interstrand hydrogen bonds. Until the critical length of RNA is synthesized for a stable RNA/DNA double helix, the short initial di- and trinucleotides are susceptible to release.
11. (a) The lack of a 3′-OH group will cause cordycepin to be a chain terminator for RNA sysnthesis. Cordycepin will be incorporated at the 3′-end of a chain, and further elongation will be blocked.
(b) The substrate specificity of poly(A) polymerase is higher than that of RNA poly merase because the RNA polymerase uses four nucleotides (ATP, UTP, GTP, CTP), whereas poly(A) polymerase uses only ATP. The result suggests that poly(A) polymerase has a higher apparent affinity for 3′-deoxy-ATP than does RNA polymerase.
(c) Yes. It must receive a 5′-triphosphate in order to be a substrate for poly(A) poly merase or RNA polymerase.
12. Note that the DNA strands shown in Figure 28.28 in the text are, by convention, com plementary to the DNA strand coding for mRNA. If every T is changed to U, the polarity and base sequence of the DNA strands shown are identical with the mRNA synthesized from the complementary strand. Thus, TCT becomes UCU (Ser), and so forth.
13. A mutation that disrupted the normal AAUAA recognition sequence for the endonucle ase could account for this finding. In fact, a change from U to C in this sequence caused this defect in a thalassemic patient. Cleavage occurred at the AAUAAA 900 nucleotides downstream from this mutant AACAAA site. See S. H. Orkin, T.-C. Cheng, S. E.
Antonarakis, and H. H. Kazazian, Jr. EMBO J. 4(1985):453.
RNA SYNTHESIS AND SPLICING
515
14. One possibility is that the 3′ of the poly(U) donor strand cleaves the phosphodiester bond on the 5′ side of the insertion site. The newly formed 3′ terminus of the acceptor strand then cleaves the poly(U) strand on the 5′ side of the nucleotide that initiated the attack. In other words, a U could be added by two transesterification reactions. This postulated mechanism is akin to the one in RNA splicing (see Figure 28.29 in the text). See T. R. Cech. Cell 64(1991):667.
15. The possibilities of alternative splicing and of RNA editing allow the final protein prod ucts to be more complex and more highly varied than the genes that encode them.
Additionally, posttranslational modification of proteins (e.g., glycosylation, phosphorylation) will further enhance the complexity.
16. One could make use of the A–T base pairing potential of the poly-A sequence that is characteristic of eukaryotic mRNAs. One would construct an affinity chromatography column in which an oligo-dT nucleotide is covalently linked to a resin. Eukaroytic mRNAs would bind to this column, whereas other RNAs would not. After washing the other RNAs away from the column, one would elute the eukaroytic mRNAs by weakening the A–T hydrogen bonds, for example, by changing the temperature, or by washing the column with a solution containing an excess of soluble oligo-dA (to displace the mRNA).
17. (a) The different genes are expressed to differing extents. Only those genes for which mRNA is actively being transcribed will give positive hybridization signals.
(b) Gene expression patterns differ in the different tissues. Some of the mRNAs are tran scribed in some tissues but not others.
(c) The genes that are expressed in all three tissues could be essential for fundamental metabolic processes that are common to most if not all cells.
(d) Including the initiation inhibitor allows counting and comparison of the number of ongoing mRNA chains being synthesized among the different gene types and from tissue to tissue at the given moment in time when the cells are broken.
(Without such an inhibitor, the results could be skewed by differing initiation rates from gene to gene, or by the possibility of some artificial initiation events that could be induced when the cells are broken to isolate the nuclei.)
18. The long strand that goes entirely across the picture from left to right is the DNA. The strands of increasing length are molecules of mRNA that are beginning to be transcribed.
Transcription begins just ahead of the site on the DNA where shortest strands of mRNA are seen. Transcription ends just after the site on the DNA where the longest strands of mRNA are attached. (As RNA polymerase passes this site, the primary transcript mRNA is released.) On the page, the direction of RNA synthesis is from left to right. Many different enzymes are simultaneously making many different RNA molecules on a single gene.
Protein Turnover
and Amino Acid Catabolism
3
Organisms derive energy from both stored and exogenous fuels. The catabolism of carbohydrates (Chapter 16) and fats (Chapter 22) have been discussed in previous chapters. In Chapter 23, the authors explain the role of proteins in energy metabolism. Although proteins are not stored as fuels per se as carbohydrates and fats are, supplies of proteins in excess of those needed to provide biosynthetic precursors are degraded for energy or are converted into fats or carbohydrates. Most of the amino groups of surplus amino acids are converted into urea through the urea cycle, whereas their carbon skeletons are transformed into acetyl CoA, pyruvate, or one of the citric acid cycle intermediates. The chapter begins with a discussion of the process of ubiquination, by which proteins are targeted for degradation. The N-terminal amino acid of a protein strongly determines its half-life. When tagged by ubiquitin, a large protease complex called the proteosome carries out the degradation of the protein. The proteosome then cleaves off the ubiquitin intact, so it can be recycled.
Once a protein is cleaved into individual amino acids, the amino acids are ei ther incorporated into newly synthesized proteins or degraded to specific compounds for entry into an energy transduction pathway. If entry into an energy transduction pathway is its fate, the nitrogen(s) must first be removed. The a-amino groups of most amino acids are transferred to a-ketoglutarate to form glutamate by transamination (catalyzed by aminotransferases), and the a-amino group of glutamate is then converted to ammonia by an oxidative deamination. The authors describe the reaction mechanism of aminotransferases and the role the coenzyme pyridoxal phosphate (PLP) plays in this enzyme and others. The urea cycle is introduced next, which carries out the condensation of ammonia, the a-amino group of aspartate, and CO2 to
407
408
CHAPTER 23 form urea—a nontoxic excretory product of nitrogen in higher animals. The urea cycle is linked to the citric acid cycle (Chapter 17) due to its production of the citric acid cycle intermediate fumarate.
Because there are 20 amino acids, the catabolic pathways of their carbon skeletons are numerous and of varied types. The authors describe how the carbon atoms of each amino acid are funneled into one or more of seven primary products. Two of these, acetyl CoA and acetoacetyl CoA, can be converted to ketone bodies (Chapter 22), and the remaining five can be converted into glucose (Chapter 16) all of which can be oxidized in energy-generating pathways. The two groups of products lead to the glycogenic-ketogenic classification of the amino acids. The chapter ends emphasizing the importance of carrying out amino acid catabolism by examining the pathological consequences of defects or deficiencies in some of the enzymes involved in catabolism of amino acids and the synthesis of urea.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
INTRODUCTION
1. State the fate of exogenously supplied amino acids that are not used for protein synthesis.
Proteins Are Degraded to Amino Acids (Text Section 23.1) 2. Contrast the effect of different N-terminal amino acids on the half-lives of proteins in yeast.
Protein Turnover Is Tightly Regulated (Text Section 23.2) 3. Explain the functions of the three enzymes (E1, E2, and E3) that participate in the attachment of ubiquitin to proteins targeted for degradation.
4. List the advantages of poly-ubiquination of proteins targeted for degradation.
5. Describe the subunit structure and function of the 26S proteosome.
6. Explain how protein degradation plays a role in regulation of NF-kB.
The First Step in Amino Acid Degradation Is the Removal of Nitrogen
(Text Section 23.3) 7. Name the major organ of amino acid degradation in mammals.
8. Describe the reactions catalyzed by the aminotransferases (transaminases) and state the major function of these reactions.
9. Write the equations for the transamination reactions catalyzed by aspartate aminotrans-ferase and alanine aminotransferase.
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
409
10. Describe the reaction catalyzed by glutamate dehydrogenase, outline its regulation, and state its major function. Note that the participation of NAD+ in the reaction links nitrogen metabolism and energy generation.
11. Recognize the structures of pyridoxal phosphate (PLP) and pyridoxamine phosphate (PMP), indicate the reactive functional groups on each, and name the dietary precursor of PLP.
12. Describe the aminotransferase reaction mechanism and explain the involvement of a Schiff base and PLP. List the other kinds of reactions catalyzed by PLP and describe the common features of PLP catalysis. Define stereoelectronic control.
13. Explain how the a-amino groups of serine and threonine can be directly converted into NH +
4 .
Ammonium Ion Is Converted into Urea in Most Terrestrial Vertebrates
(Text Section 23.4) 14. Define the terms ureotelic, uricotelic, and ammonotelic.
15. Name the molecule that brings nitrogen and carbon into the urea cycle.
16. Name the enzymes and the other components of the urea cycle, note their intracellular locations, and indicate the molecular connection between this cycle and the citric acid cycle. Account for the ATP requirement of the cycle.
17. Distinguish the structural and functional properties of the two isozymes of carbamoylphosphate synthetase in mammals.
18. Explain how deficiencies in several different enzymes of the urea cycle give rise to hy-perammonemia.
Carbon Atoms of Degraded Amino Acids Emerge
as Major Metabolic Intermediates (Text Section 23.5) 19. State the strategy used by humans for catabolizing the carbon atom skeletons of amino acids and name the seven major metabolic products formed.
20. Describe the basis for the glycogenic-ketogenic designation of the amino acids and classify each amino acid accordingly. Appreciate the limitations of this classification.
21. List the amino acids that give rise to pyruvate, oxaloacetate, fumarate, succinyl CoA, acetylCoA, and acetoacetyl CoA respectively.
22. List the amino acids that give rise to a-ketoglutarate. Describe the role of tetrahydrofolate in one of the conversions.
Inborn Errors in Metabolism Can Disrupt Amino Acid Degradation
(Text Section 23.6) 23. Describe alcaptonuria. Relate Garrod’s hypothesis concerning this disorder.
24. Explain the biochemical bases of phenylketonuria and maple syrup urine disease. Describe some of the consequences of these diseases.
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SELF-TEST
Introduction 1. Which of the following answers complete the sentence correctly? Surplus dietary amino acids may be converted into (a) proteins.
(b) fats.
(c) ketone bodies.
(d) glucose.
(e) a variety of biomolecules for which they are precursors.
Proteins Are Degraded to Amino Acids 2. Order the following events in the correct sequence in the digestion of dietary proteins.
(a) proteolysis by peptidases (b) proteolysis by pepsin in stomach (c) release of free amino acids into bloodstream (d) proteolysis in lumen of intestine by aminopeptidases (e) acidic denaturation of proteins in stomach.
(f) transport of free amino acids and di- and tri-peptides into intestinal cells Protein Turnover Is Tightly Regulated 3. Which of the following are features of the 26S proteosome?
(a) consists of a 20S catalytic subunit and a 19S regulatory subunit (b) releases only free amino acids as products (c) digests ubiquitin along with the protein to which it is attached (d) contains six ATPases The First Step in Amino Acid Degradation Is the Removal of Nitrogen 4. Which of the following compounds serves as an acceptor for the amino groups of many amino acids during catabolism?
(a) glutamine (c) a-ketoglutarate (b) asparagine (d) oxalate
5. Which of the following answers completes the sentence correctly? The removal of a-amino groups from amino acids for conversion to urea in animals may occur by (a) transamination.
(c) oxidative deamination.
(b) reductive deamination.
(d) transamidation.
6. Which of the following amino acids have their a-amino groups removed by dehy dratases?
(a) histidine (d) glutamine (b) tryptophan (e) threonine (c) serine (b) aspartate and a-ketoglutarate.
(c) alanine and oxaloacetate.
(d) alanine and a-ketoglutarate.
8. Explain the role of pyridoxal phosphate in aminotransferase reactions. Be sure to de scribe the Schiff base and the ketimine that are involved in the mechanism.
Ammonium Ion Is Converted into Urea in Most Terrestrial Vertebrates 9. Considering all forms of life, which of the following are major excretory forms of the a-amino groups of amino acids?
(a) urea (c) ammonia (b) uracil (d) uric acid
10. How many moles of ATP are required to condense two moles of nitrogen and one mole of CO2 into one mole of urea via the urea cycle? How many high-energy bonds are used in this process? Do both atoms of nitrogen enter the cycle as NH + 4 ?
11. What would be the net effect of linking the following two transamination reactions, and why would such coupling be useful when excess proteins were being catabolized for energy generation?
Alanine + a-ketoglutarate pyruvate + glutamate
Oxaloacetate + glutamate aspartate + a-ketoglutarate
12. Describe the role of ornithine in the urea cycle.
13. Explain how hyperammonemia, the increased concentration of NH +
4
in the serum, can arise from defects in more than one enzyme of the pathway that forms urea.
Carbon Atoms of Degradaed Amino Acids Emerge as Major Metabolic
Intermediates 14. Match the catabolic products in the right column with the amino acids in the left col umn from which they can be derived.
(a) alanine (1) succinyl
CoA
(b) aspartate (2) acetoacetate (c) glutamine (3) a-ketoglutarate (d) phenylalanine (4) oxaloacetate (e) leucine (5) pyruvate (f)
valine
(6) acetyl CoA (7) fumarate 15. Classify the following amino acids as glycogenic (G), ketogenic (K), or both (GK).
(a) leucine (e) histidine (b) alanine (f)
isoleucine (c) tyrosine (g) aspartate (d) serine (h) phenylalanine (a) histidine (b) phenylalanine (c) tyrosine (d) isoleucine (e) glutamine Inborn Errors in Metabolism Can Disrupt Amino Acid Degradation 18. Which of the following statements is true of the metabolic disease phenylketonuria?
(a) The disease is caused by an inability to synthesize phenylalanine.
(b) The disease can be caused by a deficiency in phenylalanine hydroxylase.
(c) The disease can be caused by a deficiency in tetrahydrobiopterin.
(d) The disease is treated with a high phenylalanine diet.
(e) The disease leads to a build-up of phenylalanine in the body.
ANSWERS TO SELF-TEST
1. All are correct. The amino acids that are needed for protein synthesis and as precursors for other biomolecules are used directly for those purposes. The carbon skeletons of any in excess can be converted into acetyl CoA or glucose, depending on the particular amino acid, and thus into products that are derivable from these two basic molecules.
2. The correct order is e, b, d, f, a, and c.
3. The correct answers are (a) and (d). (b) is incorrect because the proteosome also releases small peptides and (c) is incorrect because ubiquitin is spared degradation so it can be recycled.
4. c. The transamination of several different amino acids with a-ketoglutarate forms gluta mate and a-keto acids that can subsequently be catabolized.
5. a, c 6. c, e. Serine and threonine are deaminated by dehydratases that take advantage of the b-hydroxyl of these amino acids to carry out a dehydration followed by a rehydration to release NH + 4 .
7. d. A five-carbon amino acid will yield a five-carbon e-keto acid as a result of a transam ination; in this case, glutamate yields e-ketoglutarate. The other partner in the reaction, pyruvate, will yield alanine as a product.
8. Pyridoxal phosphate (PLP) acts as an amino carrier in transamination reactions. It is co valently bound to the a-amino group of a lysine residue in the enzyme by a Schiff base or aldimine bond, that is, by a carbon-nitrogen double bond between the a-amino group of the lysine and a carbon of PLP (see p. 641 of text). The enzyme catalyzes a displacement of the e-amino group of the lysine of the enzyme and forms an analogous aldimine bond with the a-amino group of an amino acid. Through a deprotonation and a reprotonation, the aldimine is isomerized to a ketimine in which the carbon-nitrogen bond is between the a-amino nitrogen and the a-carbon of the amino acid. The addition of H2O to the ketimine releases the carbon skeleton of the amino acid as an a-keto acid and leaves the coenzyme in the enzyme-bound pyridoxamine form, which now contains the a-amino group of the amino acid. After dissociation of the a-keto acid, a different a-keto acid binds to the enzyme to form a new ketimine, and the overall process reverses, resulting in the transfer of the enzyme-bound amino group to the keto acid to form a new amino acid and regenerate the pyridoxal phosphate.
9. a, c, d. Uracil is a pyrimidine component of RNA and is not a nitrogen excretory prod uct. Uric acid is a purine derivative excreted by uricotelic organisms.
10. Three moles of ATP are directly involved in the synthesis of urea. Two are converted to ADP and Pi by carbamoyl phosphate synthetase, and one is converted to AMP and PPi by argininosuccinate synthetase. Two ATP would be required to convert AMP back into ATP, so a total of four high-energy bonds are used. Only one molecule of nitrogen enters as NH + 4 ; the other enters in the a-amino group of aspartate, which can be formed by a transamination between oxaloacetate and glutamate.
11. The sum of the two reactions would be alanine + oxaloacetate aspartate + pyru vate, and the net effect would be that the a-amino group of alanine would appear as the a-amino group of aspartate, one of the two direct donors of nitrogen into the urea cycle—the other being carbamoyl phosphate. The a-amino groups of many other amino acids can be similarly collected on aspartate and fed into the cycle.
12. Ornithine serves as the carrier on which the urea molecule is constructed. The a-amino group of ornithine has a carbamoyl group added to it by ornithine transcarbamoylase to form citrulline. Subsequent steps of the cycle add aspartate to bring in the second nitrogen atom and also regenerate ornithine when arginase cleaves arginine to form urea.
13. A defect in carbamoyl phosphate synthesis would cause increased NH + 4 concentrations, as would a defect in any of the four reactions that condense carbamoyl phosphate with ornithine and regenerate the ornithine. Essentially, blocking a biochemical pathway at any of its steps may lead to increased concentrations of any of the precursors or members of the pathway.
14. (a) 5 (b) 4, 7; aspartate can be converted to fumarate via the urea cycle. (c) 3 (d) 2, 7 (e) 2, 6 (f) 1 15. (a) K (b) G (c) GK (d) G (e) G (f) GK (g) G (h) GK
16. Both give rise to methylmalonyl CoA, which can, in turn, be converted into succinyl CoA and ultimately into glucose.
17. b, c. Monoxygenases and dioxygenases are involved in the conversion of phenylalanine to tyrosine and the subsequent opening of the aromatic ring during tyrosine catabolism.
18. Answers (b), (c), and (e) are correct. Phenylketonuria is a metabolic disorder arising from an absence or deficiency in the enzyme phenylalanine hydroxylase or (more rarely) its cofactor tetrahydrobiopterin. It results in the build-up of phenylalanine in the body and is treated with a diet low in phenylalanine.
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PROBLEMS
1. Three enzymes are needed for the attachment of ubiquitin to proteins targeted for degra dation: E1, E2, and E3. For each enzyme state its name and give a brief description of its role in ubiquitin attachment.
2. Birds require arginine in their diet. Would you expect to find the production of urea in these animals? Explain.
3. Which would you expect to have a greater effect on the rate of urea biosynthesis, a de fect in fumarase activity or a defect in alanine aminotransferase?
4. Pyridoxal phosphate or related metabolites are required growth factors for Lactobacillus species (Morishita et al., J. Bacteriol. 148[1981]:64–71). For example, when amino acids such as alanine or glutamate are used as the sole source of nutrition, these bacilli do not grow nor do they generate metabolic energy unless pyridoxal phosphate or its metabolites are supplied. Explain these observations.
5. A male infant six weeks of age exhibited symptoms of pronounced hyperammone mia, which included vomiting, fever, irritability, and screaming episodes interspersed with periods of lethargy. The infant had low levels of blood urea, elevated serum transaminase, and generalized hyperaminoacidemia and aminoaciduria. The levels of citrulline, argininosuccinic acid, and arginine were relatively low in both blood and urine. Enzymatic assays of liver tissue established that the level of mitochondrial carbamoyl phosphate synthetase (CPS) was approximately 20% of normal; the enzyme was active only in the presence of relatively high concentrations of N-acetylglutamate.
All other urea cycle enzyme levels were relatively normal. The infant was treated with a supplement containing arginine, pyridoxine, and a-keto analogs of essential amino acids (those that cannot be synthesized in humans). As part of the therapy, dietary protein was also restricted.
(a) Why were blood levels of ammonia high in this infant?
(b) Explain why infants with CPS deficiency normally exhibit relatively low blood lev els of citrulline, argininosuccinic acid, arginine, and urea.
(c) Why is the administration of supplemental arginine recommended for this infant?
(d) Why were a-keto analogs of essential amino acids and pyridoxine administered?
(e) Why was dietary protein restricted?
(f) Why would you expect to see elevated concentrations of glutamate, glutamine, and alanine in the blood and urine of this infant?
6. Why is glutamate dehydrogenase a logical point for the control of ammonia production in cells?
7. During the process of glomerular filtration in the kidney, amino acids, as well as other metabolites, enter the lumen of the kidney tubule. Normally, a large portion of these amino acids are reabsorbed into the blood through the action of membrane-bound carrier systems that are specific for different classes of amino acids. Cystinuria is a disorder whose symptoms include urinary excretion with unusually high concentrations of cystine as well as excess amounts of ornithine, lysine, and arginine. Cystine is a dibasic amino acid, composed of two cysteine molecules joined by a disulfide linkage.
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
415
Patients with this disorder often have urinary tract stones, which are caused by the limited solubility of cystine. A related disorder found in other people is characterized by the appearance of ornithine, lysine, and arginine in the urine, although the levels of urinary cystine are normal.
(a) What is the most likely source of cystine in cells?
(b) What common structural feature of the four amino acids—cystine, ornithine, ly sine, and arginine—is recognized by the carrier in the kidney tubule membrane?
(c) How many carrier systems may exist for these molecules?
(d) Other amino acidurias are due to a deficiency in one or more of the enzymes in the catabolic pathway for an amino acid. This leads to higher concentrations of the amino acid in the blood and a corresponding increase in the concentrations in the glomerular filtrate. In this case, the capacity of the reabsorption system is surpassed, causing some amino acid to be lost in the urine. How could you distinguish between a defect in amino acid metabolism and a defect in a renal transport system?
8. Why is the catabolism of isoleucine said to be both glucogenic and ketogenic?
9. Brain cells take up tryptophan, which is then converted to 5-hydroxytryptophan by tryp tophan hydroxylase, an enzyme whose activity is similar to that of phenylalanine hydroxylase. Aromatic amino acid decarboxylase then catalyzes the formation of the potent neurotransmitter 5-hydroxytryptamine, also called serotonin. In the blood, tryptophan is bound to serum albumin, with an affinity such that about 10% of the tryptophan is freely diffusable. The rate of tryptophan uptake by brain cells depends on the concentration of free tryptophan. In these cells, tryptophan concentration is normally well below that of the KM for tryptophan hydroxylase. Aspirin and other drugs displace tryptophan from albumin, thereby increasing the concentration of free tryptophan.
(a) What cofactor is required for the activity of tryptophan hydroxylase?
(b) Dietary deficiencies in pyridoxin and related metabolites can induce a number of symptoms, including those that appear to be related to derangements in serotonin metabolism. What enzyme could be affected by a deficiency of vitamin B6?
(c) What effect does aspirin have on tryptophan metabolism in brain cells?
10. In many microorganisms, glutamate dehydrogenase (GDH) participates in the catabo lism of glutamate by generating ammonia and a-ketoglutarate, which undergoes oxidation in the citric acid cycle. However, when E. coli is grown with glutamate as the sole source of carbon, the synthesis of GDH protein is strongly repressed. Under these conditions, aspartase, an enzyme that catalyzes the removal of ammonia from aspartate to form fumarate, is required in order for the cell to grow in glutamate. Propose a cyclic pathway for the catabolism of glutamate that includes aspartate.
11. When E. coli is grown in glucose and ammonia, GDH synthesis is accelerated and the en zyme is active. Under these conditions, what role does GDH play in bacterial metabolism?
12. After an overnight fast, muscle tissue proteolysis generates free amino acids, many of which pass into the blood. Among the amino acids that are found in the blood are alanine, glutamate, and glutamine, all of which are rapidly taken up by the liver. What happens to these amino acids when they enter hepatic cells?
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CHAPTER 23
13. Propionyl CoA and methylmalonyl CoA both inhibit N-acetylglutamate synthase activ ity in slices of liver tissue. What clinical symptom would you expect to see in patients suffering from methylmalonic aciduria as a result of this inhibition?
14. Dialysis of purified glutamate aminotransferase can be used to remove pyridoxal phos phate from the enzyme, but the dissociation of the cofactor from the enzyme is very slow.
Why would the addition of glutamate to the enzyme solution increase the rate of dissociation of the cofactor from the enzyme?
15. Early work by Esmond Snell on the enzymes that employ pyridoxal phosphate in cluded experiments in which free pyridoxal was heated with amino acids like glutamate. Snell found that the (-amino group of glutamate was transferred to pyridoxal, generating pyridoxamine. Why was this an important clue to the function of pyridoxal as a cofactor?
16. The enzyme serine dehydratase employs pyridoxal phosphate (PLP) in the direct deam ination of serine. In this reaction, the a-carbon of serine undergoes a two-electron oxidation through a-elimination. Show how PLP participates in the process by writing a mechanism for serine dehydration and deamination.
17. A small number of infants who have phenylketonuria have normal levels of phenylala nine hydroxylase activity, but on normal diets they continue to accumulate phenylalanine as well as other metabolites, including phenylpyruvate, phenyllactate, and phenylacetate. They also have high levels of quinonoid dihydrobiopterin.
(a) What is the probable enzyme deficiency in these infants? Rationalize the deficiency with the observed clinical symptoms.
(b) Write brief pathways for the formation of the phenylalanine metabolites found in these infants.
J
CH
OH
2
J
J
CKO
JCH J CJ COO: JCH J COO:
2
2
J
J
COO:
H
Phenylpyruvate
Phenyllactate
Phenylacetate
ANSWERS TO PROBLEMS
1. E1, ubiquitin-activating enzyme: adenylates ubiquitin on its terminal carboxylate with release of PPi. Transfers ubiquitin to a sulfhydryl on the enzyme with release of AMP, forming a thioester bond. E2, ubiquitin-conjugating enzyme: receives ubiquitin from E1.
Ubiquitin is also bound to a sulfhydryl of E2 via a thioester bond. E3, ubiquitin-protein ligase: transfers ubiquitin from E2 to the e-amino group of a target protein.
2. In the urea cycle, arginine is cleaved to yield urea and ornithine. The fact that birds require arginine in their diet indicates that they are unable to synthesize it for utilization in protein synthesis. As a result, they are also unable to synthesize urea to dispose of ammonia; instead, they synthesize uric acid. Birds do have carbamoyl phosphate synthetase activity; however, it is located in the cytosol, and it catalyzes the formation of carbamoyl phosphate, which is then utilized for pyrimidine synthesis.
3. Fumarase activity has an effect on the urea cycle because it is needed, along with malate dehydrogenase, for the regeneration of oxaloacetate, which in turn undergoes transamination to form aspartate. The amino group of aspartate contains one of the two nitrogen atoms that are used to synthesize urea. Alanine aminotransferase is one of a number of aminotransferases that can transfer amino groups from amino acids to a-ketoglutarate to generate glutamate. Subsequent deamination of glutamate provides ammonia for the urea cycle. If all the other aminotransferases in the cell are active, then alanine aminotransferase would not be particularly essential. Thus, a defect in fumarase activity would have the greater effect on the rate of urea biosynthesis.
4. In order to utilize amino acids as sources of oxidative energy or to generate glucose through gluconeogenesis, the bacilli must carry out transamination reactions to dispose of ammonia (or to use it for the biosynthesis of other nitrogen-containing compounds) as well as to generate a-keto acids that can be used in the citric acid cycle or other pathways. Pyridoxal phosphate is a required cofactor for the aminotransferase enzymes, and in bacteria in which it cannot be synthesized, it must be derived from the growth medium. Otherwise, amino acids cannot be metabolized. In this case, where an amino acid is the only source of carbon and of nitrogen, the bacilli will not be able to introduce the amino acid into any catabolic pathway. Pyridoxal phosphate also functions as a cofactor for a large number of other enzymes, including decarboxylases, racemases, aldolases, and deaminases. Therefore, deficiencies in pyridoxal phosphate would also adversely affect a large number of other pathways.
5. (a) Carbamoyl phosphate synthetase utilizes ammonia and bicarbonate to synthesize carbamoyl phosphate, which enters the urea cycle. A deficiency in the activity of this enzyme leads to an accumulation of ammonia in blood and in urine.
(b) Because carbamoyl phosphate condenses with ornithine to form citrulline, it is in effect a precursor of all the components of the urea cycle. When its synthesis is depressed, the rate of synthesis of other urea cycle components will be decreased.
(c) Arginine is needed as a component of most proteins; it must therefore be supplied to avoid arginine deficiency, which would result in a decrease in the rate of protein synthesis. In the urea cycle, arginine serves as a precursor of ornithine, which in turn serves as an acceptor of the relatively small number of carbamoyl groups that enter the urea cycle. Thus, supplemental arginine could accelerate the rate of urea synthesis in an attempt to drive the reaction catalyzed by the partially active CPS enzyme toward the net formation of carbamoyl phosphate. Finally, recent findings suggest that arginine is a feed-forward activator for the synthesis of N-acetyl glutamate, which activates CPS. Recall that the deficient enzyme appeared to be heavily dependent on levels of N-acetyl glutamate. Supplying arginine indirectly activates CPS, thereby increasing the rate of ammonia utilization.
(d) The administration of a-keto analogs of essential amino acids serves two functions.
First, these substrates undergo transamination with glutamate as the amino donor.
The corresponding increase in the synthesis of glutamate, through the action of glutamate dehydrogenase, utilizes more ammonia, removing at least some of it from the blood. Second, the transamination of the a-keto acids generates essential amino acids, which are used for protein synthesis. This is especially important in a situation in 418
CHAPTER 23 which dietary protein intake is restricted. Since the cofactor pyridoxyl phosphate is required for aminotransferases to function, adding its vitamin precursor (pyridoxine) to the diet will ensure sufficient quantities to process the added a-keto analogs into essential amino acids.
(e) When dietary protein is hydrolyzed to its component amino acids, subsequent cata bolic steps include transamination and deamination, which produce free ammonia.
The less protein consumed, the lower the production of ammonia from the breakdown of amino acids. A reduction in the blood ammonia level is the primary goal of the therapy for this infant.
(f) Increased levels of ammonia in the cells will drive the reaction catalyzed by glutamate dehydrogenase toward the net formation of glutamate and will also stimulate the synthesis of glutamine, which is in effect a carrier of two molecules of ammonia. With the high concentrations of glutamate, the equilibria for most transamination reactions would be shifted toward the net formation of other amino acids, such as alanine, a transaminated form of pyruvate. Concentrations of these and other amino acids would therefore be increased in both the blood and the urine.
6. The deamination of amino acids occurs through the action of transaminases as well as through the action of glutamate dehydrogenase (GDH). GDH is the only enzyme that catalyzes the oxidative deamination of an L amino acid. It deaminates glutamate, whose precursor, a-ketoglutarate, is the ultimate acceptor of amino groups from almost all the amino acids. In addition, while the reactions catalyzed by the transaminases are freely reversible, the GDH reaction is far from equilibrium; it is therefore a logical activity to control because large changes in its velocity can be achieved with small changes in the concentrations of allosteric effectors like ATP or NADH. Thus, GDH is the enzyme of choice for the control of ammonia synthesis.
7. (a) Disulfide linkages exist in many proteins, and when they are hydrolyzed by pro teases to yield free amino acids, cystine is often one of the products.
(b) Like cystine, the other three amino acids—arginine, ornithine, and lysine—have two basic groups. The carrier systems probably recognize and bind these groups for transport.
(c) From the disorders described, it is likely that there are two transport systems. One carries all four dibasic acids; when it is defective, reabsorption of all four species fails, allowing all four to spill over into the urine. Another system transports ornithine, arginine, and lysine but not cystine; its failure to function accounts for the appearance of these three amino acids but not cystine in the urine.
(d) A defect in the catabolic pathway for a particular amino acid causes elevation in the concentration of that single amino acid in blood and in urine as well, unless other amino acids carried by the same kidney transport system are lost to the urine. A defect in renal reabsorption means that all those amino acids that share the affected carrier system will be lost to the urine. Their concentration in blood will be lower than normal, while their concentration in urine will be higher.
8. The catabolic pathway for isoleucine leads to the formation of acetyl CoA and propionyl CoA. Acetyl CoA can be utilized for the net synthesis of fatty acids or ketone bodies, but it cannot be used for the net synthesis of glucose; thus, it is said to be ketogenic. In contrast to acetyl CoA, propionyl CoA is converted to succinyl CoA, which can be utilized through part of the citric acid cycle and the gluconeogenic pathway to give the net formation of glucose. The distinction between the two types of substrates is somewhat arbitrary, however. For example, succinyl CoA can also be converted via pyruvate and acetyl CoA to citrate, which, when transported to the cytosol, serves as the source of carbons for the synthesis of fatty acids. Thus, glucogenic substrates can, under certain conditions, be ketogenic; however, ketogenic substrates cannot be glucogenic, unless a cell has a functional glyoxylate pathway.
9. (a) Tetrahydrobiopterin is utilized as a reductant by many hydroxylase enzymes, in cluding phenylalanine hydroxylase and tryptophan hydroxylase.
(b) Pyridoxal phosphate, which is derived from dietary pyridoxine (vitamin B6), is a cofactor for a number of enzymatic reactions that occur at the a carbon of an amino acid including decarboxylations (see text, p. 642). In this case, the cofactor participates in the decarboxylation of 5-hydroxytryptophan by the enzyme aromatic amino acid decarboxylase to form serotonin. A deficiency of vitamin B6 would lead to a reduction in the rate of synthesis of serotonin.
(c) The higher the concentration of free tryptophan, the greater the rate of uptake of the amino acid by the brain cells. Because the normal concentration of tryptophan in these cells is below that of the K for tryptophan hydroxylase, an influx of more
M tryptophan into the cells provides more substrate for the enzyme. You would therefore expect an increase in the level of 5-hydroxytryptophan production.
10. A possible catabolic pathway for glutamate that includes aspartate is as follows:
Glutamate+OxaloacetateDAspartate+a-Ketoglutarate
NADH
Oxidation in the +H;
citric acid cycle NAD;
NH4 H O
2
MalateEFumarate
Glutamate undergoes transamination with oxaloacetate to generate a-ketoglutarate and aspartate. Oxidation of a-ketoglutarate is carried out in the citric acid cycle, whereas aspartate is cleaved to yield ammonia and fumarate. Fumarate is converted to malate, which is then oxidized to oxaloacetate in the citric acid cycle so that it can be regenerated to serve as an acceptor of the amino group from glutamate. It should be noted that glutamate will also be used as a source of amino groups by the aminotransferases and other enzymes involved in biosynthesis.
11. In cells grown in glucose and ammonia, GDH catalyzes the assimilation of ammonia by incorporating it into a-ketoglutarate to yield glutamate, which serves as a source of amino groups for other biosynthetic reactions.
12. In the liver, alanine, glutamate, and glutamine are utilized as sources of carbon for glu coneogenesis. Glutamine is deaminated to yield glutamate and ammonia. Glutamate undergoes oxidative deamination to form ammonia and a-ketoglutarate, a substrate for gluconeogenesis. Pyruvate is generated from the transamination of alanine, and carboxylation of the a-keto acid yields oxaloacetate, another source of carbon for gluconeogenesis. The ammonia generated by the conversion of the amino acids to their corresponding a-keto acids is used for the synthesis of urea. Glucose synthesized by liver can be returned through the blood to the muscle, where it serves as a source of energy.
Thus, muscle uses the amino acids as a means of contributing to the generation of glucose in liver as well as a means of transporting ammonia to the liver, where the synthesis of urea can be carried out.
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13. The activity of mitochondrial carbamoyl phosphate synthetase (CPS) depends on the availability of N-acetylglutamate, which is generated from glutamate and acetyl CoA. A reduction in the availability of the activating molecule will lead to a decrease in the activity of CPS, which utilizes ammonia and bicarbonate for the synthesis of carbamoyl phosphate. This, in turn, leads to an increase in the level of ammonia in blood and urine.
Over two-thirds of patients with methylmalonic aciduria are hyperammonemic.
14. In the native enzyme, the dissociation of the cofactor from the enzyme during dialysis is very slow because PLP is covalently bound to a lysine residue. The addition of glutamate, a substrate for the transamination reaction, leads to the formation of a Schiff base between glutamate and PLP, which means that the cofactor is no longer covalently attached to the enzyme molecule. Although PLP is bound to the enzyme by noncovalent forces, these forces are not as strong as a covalent bond, so during dialysis, the rate of dissociation of the cofactor from the enzyme is increased.
15. Snell’s observations of the formation of pyridoxamine by heating with a-amino acids suggested that pyridoxal is involved in transamination reactions. As an enzyme cofactor it could transfer an a-amino group from an amino acid to the a-keto group of an a-keto acid. It is now established that the action of pyridoxal phosphate in aminotransferase enzymes includes formation of pyridoxamine phosphate during the catalytic cycle (see text, p. 641) 16. In the accompanying figure, PLP in Schiff base linkage with a lysine residue in the enzyme forms a new Schiff base link with serine. A hydrogen atom is removed from the a-carbon, and then the hydroxyl group is removed by elimination from the b-carbanion, generating aminoacrylate attached to pyridoxal phosphate. Hydrolysis of the Schiff base to give aminoacrylate and PLP is followed by tautomerization to the imino form. This compound hydrolyzes spontaneously to form pyruvate and ammonia. PLP is once again linked covalently to the enzyme.
ENZ
HJO
J
J (CH ) ;
HJO
H CH JC:JCOO: :
H CKCJCOO: 2 4
H
2
OH
2
J
J
J
J
J
N
+
CH JCJCOO:
NJ
N
2
J
H
K
J
K
H
b-Elimination
K
HJC
NH
HJC
HJC
3
J
OH
J O:
J O:
J
Serine
J
J
PLP
Carbanion
H O
2
H O CH JCJCOO:
2
H CJCJCOO: H CKCJCOO:
3
3
2
K
Pyruvate
G
K
J
O
+NH
NH
2
2
+
+
Aminoacrylate
Ammonia
NH
Imino Form
PLP
3
17. (a) The enzyme that is deficient in these infants is dihydropteridine reductase, which converts quinonoid dihydrobiopterin to tetrahydrobiopterin, using NADH as a substrate. In the phenylalanine hydroxylase reaction, tetrahydrobiopterin, the reductant in the conversion of phenylalanine to tyrosine, is oxidized to quinonoid dihydrobiopterin. The reductase enzyme regenerates tetrahydrobiopterin so that it can be used for further use in tyrosine formation. Cells that are deficient in the reductase cannot carry out efficient conversion of phenylalanine to tyrosine because they cannot regenerate tetrahydrobiopterin.
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
421
(b) Phenylpyruvate can be generated from phenylalanine by transamination, and re duction of phenylpyruvate by NADH or NADPH generates phenyllactate, in a reaction similar to that catalyzed by lactate dehydrogenase. Phenylacetate can be generated by oxidative decarboxylation of phenylpyruvate, in a reaction reminiscent of the conversion of pyruvate to acetyl CoA, catalyzed by pyruvate dehydrogenase.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. (a) The energy from ATP hydrolysis could be used for maintaining the specificity of peptide-bond hydrolysis, unfolding the target protein, threading the target protein through the barrel, or other related functions.
(b) Small peptides would not need to be unfolded or threaded, and would not need to have any of their bonds protected (e.g., remember that the ubiquitin tags are not hydrolyzed).
2. (a) pyruvate, (b) oxaloacetate, (c) a-ketoglutarate, (d) a-ketoisocaproate, (e) phenylpyru vate, and (f) hydroxyphenylpyruvate
3. (a) Aspartate + a-ketoglutarate + GTP + ATP + 2 H2O + NADH + H+ 1⁄2 glucose + glutamate + CO +
2
ADP + GDP + NAC+ + 2Pi
The glucogenic route for aspartate involves transamination to oxaloacetate, con version of the latter to phosphoenolpyruvate, which is then converted to glucose (see the text, Chapter 20). PLP and NADH participate as coenzymes in the conversion of aspartate to glucose.
(b) Aspartate + CO + + +
2
NH4 3 ATP + NAD+ + 4 H2O oxaloacetate + urea + 2 ADP + 4 P +
i
AMP + NADH + H+
This equation represents the summation of the stoichiometry of urea synthesis, the hydrolysis of PPi, and the conversion of fumarate to oxaloacetate in the citric acid cycle.
4. Different sites could specialize in the hydrolysis of different categories of peptide bonds (e.g., those that join hydrophobic/hydrophilic, hydrophobic/hydrophobic, or hydrophilic/hydrophilic amino acids, etc.) and thereby optimize the overall kinetics of degrading diverse protein sequences.
5. By analogy with the 20S proteasome (Figure 23.7), the overall architecture could be conserved. Perhaps the six different AAA ATPase subunits from the 19S regulatory complex associate into a hexamer with pseudo six-fold symmetry. If these hexamers could be separated from the 19S complexes, they might be visulaized by electron microscopy or crystallography; suitable crosslinking experiments might then reveal which pairs of AAA ATPase subunits border each other.
6. thiamine pyrophosphate
7. It acts as an electron sink. See C. Walsh (1979), Enzymatic Reaction Mechanisms, p. 178 (W. H. Freeman).
8. CO + + +
2
NH4 3 ATP + NAD+ + 3H2O + glutamate urea + 2 ADP + 2 P +
+
i
AMP + PPi NADH + H+ + a-ketoglutarate
The answer in the text is for the conversion of fumarate to oxaloacetate rather than to aspartate. The equation above gives the correct stoichiometry for the synthesis of urea from NH +
4
and glutamate. It includes a non-energy-requiring transamination reaction.
Hence, the number of ~P spent remains at four. Note that aspartate does not appear in this equation, since it is resynthesized. Thus aspartate can be considered a nitrogencarrying cofactor in the synthesis of urea.
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CHAPTER 23
9. The compound should inhibit ornithine transcarbamoylase because it appears to be a nonhydrolyzable analogue of an intermediate that should be formed between ornithine and carbamoyl phosphate. (The CH2 group in compound A will prevent the release of the phosphate.)
10. High concentrations of ammonia could increase the ratio of glutamate/a-ketoglutarate (glutamate dehydrogenase reaction) and increase the level of glutamate in the brain.
Ammonia also could increase the ratio of asparatate/oxaloacetate; the resulting lower level of oxaloacetate would decrease the availability of all citric acid cycle intermediates.
11. The mass spectrometric analysis strongly suggests that three enzymes—pyruvate dehy drogenase, a-ketoglutarate dehydrogenase, and the branched-chain a-keto dehydrogenase—are deficient. Most likely, the common E3 component of these enzymes is missing or defective. This proposal could be tested by purifying these three enzymes and assaying their capacity to catalyze the regeneration of lipoamide.
12. Benzoate, phenylacetate, and arginine would be given in a protein-restricted diet.
Nitrogen would emerge in hippurate, phenylacetylglutamine, and citrulline. See S. W.
Brusilow and A. L. Horwich in C. R. Scriver, A. L. Beaudet, W. S. Sly, and D. Valle (eds.)
(1989), The Metabolic Basis of Inherited Disease, 6th ed., pp. 629–663 (McGraw-Hill), for a detailed discussion of the therapeutic rationale.
This therapy is designed to facilitate nitrogen excretion in the absence of urea synthesis.
Note that much of the nitrogen would be excreted as glycine (hippurate is benzoylglycine), glutamine, and citrulline.
13. Aspartame, a dipeptide ester (aspartylphenylalanine methyl ester), is hydrolyzed to L-as partate and L-phenylalanine. High levels of phenylalanine are harmful in phenylketonurics.
14. N-acetylglutamate is synthesized from acetyl CoA and glutamate. Once again, acetyl CoA serves as an activated acetyl donor. This reaction is catalyzed by N-acetylglutamate synthase.
15. First serine forms a protonated Schiff base (external aldimine) with pyridoxal-5′-phos phate. Removal of the serine a-hydrogen leads to a quinonoid intermediate, which then can eliminate the b-OH to generate the Schiff base of aminoacrylate. The reaction is com H
HOCH COO:
2
H
H
HOCH COO:
OH
+
NH
2
3
COO:
(Lys)
H
N
H (Lys)
Ser
H
N
+
N
H
H
G
G O:
+ H
PO
O:
=O
3
=O PO
O:
3
=O PO
3
+
N
CH
3
N
CH
J
+
3
H
N
CH
J
3
J
H
H
Lys H CK COO: H C K COO:
2
2
(Lys)
+
+
+
H
N
NH
3
N
H
H
H
aminoacrylate
G
G O: O:
=O PO =O PO
3
3
+
N
CH3
+
N
CH3
J
J
H
H
PROTEIN TURNOVER AND AMINO ACID CATABOLISM
423
pleted by the transfer of pyridoxal phosphate back to a Schiff base with a lysine of the enzyme (internal aldimine), with the concomitant release of aminoacrylate. (The aminoacrylate will react with water to give pyruvate and ammonia.)
(Note: Another family of serine dehydratases does not use pyridoxal phosphate, but rather an iron-sulfur cluster as the cofactor. These enzymes may use a mechanism similar to the dehydration of citrate catalyzed by aconitase. See Trends Biochem. Sci.
18[1993]:297–300.)
16. As in problem 15, an external aldimine (protonated Schiff base) is formed between serine and pyridoxal-5′-phosphate, and the serine a-hydrogen is removed. The a-hydrogen then can be reattached from either of two faces to give the Schiff base of either D-serine or L-serine (see below). The equilibrium constant for the reaction will be one.
L-approach
H;
H
HOCH COO:
HOCH COO:
2
2
H
N
H
N
D-approach
+
H
+ H
O: O:
G
=O PO =O PO
3
3
+
N
CH
N
CH
3
3
J
J
H
H
17. Protein-protein interactions often relate directly to biological function. Two or more as pects of these interactions may be relevant for degradation: (1) If an interaction domain becomes chemically damaged so that interaction with a partner protein is no longer feasible, then it would be appropriate to turn over (degrade) the protein. (2) If a partner protein is in short supply within the cell, then the interacting protein may not be needed and should be degraded. (Without the partner protein, the interaction domain could be exposed as a possible signal for degradation.)
18. (a) The initial surge originates from a need to supply glucose to the brain. When car bohydrates are depleted, mammals cannot resupply glucose from fatty acids.
Glycerol provides a small source of carbohydrate, but mammals also must break down amino acids to meet the short-term demands of the brain for glucose. The ammonia byproduct from amino acid degradation accounts for the initial surge of nitrogen excretion.
(b) Over time, the liver begins to metabolize acetyl-CoA (from fats) to ketone bodies, and the brain adapts to using ketone bodies. During this period, fats can provide many of the energy needs of the brain, and little nitrogen is excreted.
(c) When lipid stores have been depleted, the organism once again must metabolize amino acids to provide glucose for the brain, and nitrogen excretion again increases.
19. Isoleucine can give its amino group to a-ketoglutarate in a transamination reaction and then be oxidatively decarboxylated and dehydrogenated to form the corresponding (a,b)-unsaturated acyl-CoA derivative. Further reactions (see the figure on p. 424) then are identical to fatty acid oxidation until the carbon skeleton is split into acetylS-CoA and propionyl-S-CoA. The three subsequent steps for the conversion of the (odd-chain) propionyl-S-CoA to succinyl-S-CoA have been discussed for the oxidation of odd-chain fatty acids (see Chapter 22).
424
CHAPTER 23
NADH, CO
2
+
NH3
O
NAD;,
O a-KG Glu
CoASH
CH CH
D
CH CH
D
CH CH
3
2
3
2
3
2
COO: COO:
S-CoA
CH
CH
CH
3
transamination
3
ox. decarboxylation
3
O
OH
O
FAD
FADH2 H O D
CH CH
2
3
S-CoA
D H C
S-CoA
3
CH
CH
dehydrogenation
3
hydration
3
O
Acetyl-S-CoA
O
O H C
S-CoA
3
NAD; NADH
CoASH
O
D
D H C
S-CoA
3
CH
S-CoA
dehydrogenation
3
thiolysis
CH3
Propionyl-S-CoA
20. (a) PAN has no effect in the absence of nucleotides.
(b) ATP is required. Neither ADP nor AMP-PNP is effective in stimulating protein digestion.
(c) AMP-PNP is a nonhydrolyzable analogue of ATP. The finding that AMP-PNP does not stimulate the digestion suggests that ATP hydrolysis to ADP and Pi is required.
(d) The peptide digestion does not require PAN or ATP.
(e) See answer 1(b), above. Small peptides do not need to be unfolded or threaded through a superstructure in order to facilitate digestion.
CHAPTER 2
The Biosynthesis of Amino Acids
4
In this chapter, the biosynthetic origins of the amino acids are explained, beginning with the need for a source of nitrogen for the amino acids. This need is met by nitrogen fixation, which is the process of converting atmospheric nitrogen in the form of N
+
2 to NH4 . The enzyme that carries out this difficult task, nitrogenase, is discussed in detail, including the role of an unusual molybdenum-iron cofactor. The authors then explain how NH + 4 is incorporated into the amino acids glutamate and glutamine via the enzymes glutamate dehydrogenase and glutamine synthetase. These two amino acids are major nitrogen donors in a range of biosynthetic pathways, including those of the remaining amino acids whose synthesis is discussed next. While the pathways for the synthesis of the amino acids are diverse, they have in common the fact that their carbon skeletons come from intermediates in glycolysis, the pentose phosphate pathway or the citric acid cycle. This leads to a grouping of the amino acids into one of six biosynthetic families based on their starting material: oxaloacetate, pyruvate, ribose-5-phosphate, a-ketoglutarate, 3-phosphoglycerate, and phosphoenolpyruvate/erythrose 4-phosphate. The synthesis of the members of each family is examined in detail, including the role of three important cofactors involved in some of the syntheses: pyridoxal phosphate, tetrahydrofolate, and S-adenosylmethionine. The latter two are carriers of single carbon atoms in metabolism. The authors also explain that the lack of some biosynthetic pathways in humans has led to the dietary requirement for nine amino acids. The examination of amino acid synthesis concludes with a general discussion of how metabolic pathways are controlled via feedback inhibition, using examples from amino acid metabolism to illustrate the relevant principles.
425
426
CHAPTER 24
The chapter concludes with a look at the role of the amino acids as precursors of many important biomolecules. The synthesis of glutathione, a sulfhydryl buffer and detoxifying agent, and nitric oxide, a short-lived signal molecule, are examined. The final topic is the biosynthesis and degradation of the porphyrin heme. The multistep synthetic pathway beginning with glycine and succinyl CoA is examined as is the mechanism for degradation of excess heme. The physiological consequences of disorders in heme biosynthesis (collectively known as porphyrias) are discussed.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Recognize that nitrogen in the form of ammonia is the source of nitrogen for all amino acids.
Nitrogen Fixation: Microorganisms Use ATP and a Powerful Reductant
to Reduce Atmospheric Nitrogen to Ammonia (Text Section 24.1) 2. Define nitrogen fixation and name the groups of organisms that can carry out this conversion.
3. Describe the nitrogenase complex and explain the roles of its reductase and nitrogenase com ponents. Note the function of the FeMo-cofactor.
4. Explain the energy requirement for nitrogen fixation and write the equation giving the stoichiometry of the overall reaction.
5. Outline the key roles of glutamate and glutamine in the assimilation of NH +
4
into amino acids and describe the reactions of glutamate dehydrogenase, glutamine synthetase, and glutamate synthase. Recognize the functions of ATP and NADPH in these processes.
Amino Acids Are Made from Intermediates of the Citric Acid Cycle
and Other Major Pathways (Text Section 24.2) 6. Classify the amino acids into six biosynthetic families and identify their seven precursors.
Name the metabolic pathways from which these precursors originate.
7. Identify the essential amino acids for humans and explain why they are essential.
8. Describe the single-step biosyntheses of alanine, aspartate, and asparagine.
9. Outline the syntheses of glutamate, glutamine, proline, and arginine from a-ketoglutarate.
10. Outline the syntheses of serine, glycine, and cysteine from 3-phosphoglycerate.
11. Explain the roles of pyridoxal phosphate, tetrahydrofolate, and S-adenosylmethionine in amino acid biosyntheses.
12. Identify the structure of tetrahydrofolate and indicate the reactive part of the molecule.
Draw the structures of the single-carbon groups that can be carried on tetrahydrofolate and provide examples of reactions that generate and use them. Describe the sources of this cofactor in humans.
THE BIOSYNTHESIS OF AMINO ACIDS
427
13. Draw the structure of S-adenosylmethionine and describe its synthesis. Indicate the reac tive part of the molecule and describe the basis of its high methyl group–transfer potential.
14. Outline the activated methyl cycle and describe the roles of methylcobalamin and ATP in the cycle. Give examples of important derivatives from S-adenosylmethionine.
15. Describe the synthesis of cysteine from homocysteine and serine.
16. Outline the biosyntheses of phenylalanine, tyrosine, and tryptophan in E. coli. Describe the roles of phosphoenolpyruvate, erythrose 4-phosphate, and phosphoribosylpyrophosphate
in these reactions.
17. Describe the structure of tryptophan synthetase and the role of substrate channeling in its catalytic reaction.
Amino Acid Biosynthesis Is Regulated by Feedback Inhibition
(Text Section 24.3) 18. Define the committed step of a metabolic pathway and recognize that it is often the tar get of feedback regulation. Note the main features of control of branched pathways by feedback inhibition and activation, enzyme multiplicity, and cumulative feedback.
19. Describe the cumulative feedback control of glutamine synthetase from E. coli. Explain the mechanisms and functions of the reversible covalent modifications and describe the advantage of employing an enzymatic cascade in regulating this reaction.
Amino Acids Are Precursors of Many Biomolecules (Text Section 24.4) 20. Give examples of important biomolecules that are derived from amino acids.
21. Draw the structure of glutathione and describe its cycle of oxidation/reduction. Indicate the functions of glutathione and describe the involvement of selenium in the glutathione
peroxidase reaction.
22. Outline the synthesis of nitric oxide (NO) and explain its function.
23. Name the two molecular precursors of the porphyrins in mammals and outline the biosynthesis and degradation of heme.
24. Explain the molecular defects in congenital erythropoietic porphyria and acute intermittentporphyria.
SELF-TEST
Nitrogen Fixation: Microorganisms Use ATP and a Powerful Reductant
to Reduce Atmospheric Nitrogen to Ammonia 1. Define nitrogen fixation and explain why it is crucial to the maintenance of life on earth.
2. Place the following components, reactants, and products of the nitrogenase complex re action in their correct sequence during the electron transfers of nitrogen fixation: (a) oxidized ferredoxin (e) N2 (b) reductase component (f)
reduced ferredoxin (c) nitrogenase component (g) electron source (d) NH3 3. Write the net equation for nitrogen fixation and describe the sources of the electrons and ATP.
4. Match the structural components or features in the right column with the appropriate component of the nitrogenase reaction.
(a) reductase component (1) MoFe-cofactor (b) nitrogenase component (2) [4Fe-4S] cluster (3) ATP-ADP binding site (4) N2-binding site (5) a 2b2 tetramer (6) dimer of identical subunits 5. Match the enzyme with the reaction it catalyzes.
(a) glutamine synthetase (b) glutamate dehydrogenase (c) glutamate synthase
3
a-Ketoglutarate
Glutamine
1
2
Glutamate
6. Which of the reactions shown in question 5 require the following?
(a) NH +
4
(b) ATP (c) NADH (d) NADPH
7. All organisms can incorporate NH +
4
into glutamate and glutamine using glutamate de hydrogenase and glutamine synthetase. Why do prokaryotes have an additional enzyme, glutamate synthase, to perform this function?
Amino Acids Are Made from Intermediates of the Citric Acid Cycle
and Other Major Pathways 8. Which of the following amino acids are essential dietary components for an adult human?
(a) alanine (b) aspartate (c) histidine (d) tryptophan (e) leucine (f) phenylalanine (g) glutamine (h) asparagine (i) glutamate
(j)
threonine (k) methionine
THE BIOSYNTHESIS OF AMINO ACIDS
429
9. Which of the following amino acids are derived from pyruvate?
(a) phenylalanine (b) alanine (c) tyrosine (d) histidine (e) valine (f) leucine (g) cysteine (h) glycine
10. Which of the following amino acids are derived from a-ketoglutarate?
(a) glutamate (e) glutamine (b) proline (f)
arginine (c) cysteine (g) ornithine (d) aspartate (h) serine
11. Which of the following compounds provide the carbon skeletons of the six biosynthetic families of amino acids? Name the metabolic pathways from which each of the precursor compounds originates.
(a) pyruvate (b) oxaloacetate (c) a-ketoglutarate (d) succinate (e) 2-deoxyribose (f) 3-phosphoglycerate
(g) ribose 5-phosphate (h) glucose 6-phosphate (i)
phosphoenolpyruvate
(j) erythrose 4-phosphate (k) a-ketobutyrate
12. Three coenzymes are involved in carrying activated one-carbon units. Match the acti vated group in the right column with the appropriate coenzyme in the left column.
(a) tetrahydrofolate (1) −CH3 (b) S-adenosylmethionine (2) −CH −
2
(c) biotin (3) −CHO (4) −CHNH (5) −CH= (6) −CO −
2
13. Which of the following answers completes the sentence correctly? The major source of one-carbon units for the formation of the tetrahydrofolate derivative N5,N10methylenetetrahydrofolate is the conversion of (a) methionine to homocysteine.
(b) deoxyuridine 5′-phosphate to deoxythymidine 5′-phosphate.
(c) 3-phosphoglycerate to serine.
(d) serine to glycine.
14. Why does S-adenosylmethionine have a higher methyl group–transfer potential than N5 methyltetrahydrofolate?
430
CHAPTER 24
15. S-adenosylmethionine is involved directly in which of the following reactions?
(a) methyl transfer to phosphatidyl ethanolamine (b) synthesis of glycine from serine (c) DNA methylation (d) conversion of homocysteine into methionine (e) synthesis of ethylene in plants
16. How many high-energy bonds are expended during the synthesis of S-adenosylmethionine from ATP and methionine?
17. The conversion of homocysteine into methionine involves which of the following cofactors?
(a) N5-methyltetrahydrofolate (b) N5,N10-methylenetetrahydrofolate (c) methylcobalamin (d) pyridoxal phosphate
18. Which of the following are intermediates in the pathway for the biosynthesis of both phenylalanine and tryptophan?
(a) anthranilate (b) chorismate (c) shikimate (d) prephenate
19. The two binding sites for indole on tryptophan synthetase subunits a and b are about 25 Å apart. Explain how indole is transferred between these sites.
Amino Acid Biosynthesis Is Regulated by Feedback Inhibition 20. In the following biosynthetic pathway A
B
C
D
E
F G, which is likely to be the committed step? Which compound is likely to inhibit the committed step?
21. Since glutamine is an important source of nitrogen in biosynthetic reactions, the enzyme that synthesizes it is carefully regulated. Which of the following compounds act as inhibitors of glutamine synthetase in E. coli?
(a) tryptophan (b) histidine (c) carbamoyl phosphate (d) glucosamine 6-phosphate (e) AMP (f)
CTP (g) alanine (h) glycine
22. How does covalent modification contribute to the regulation of glutamine synthetase in E. coli?
THE BIOSYNTHESIS OF AMINO ACIDS
431
Amino Acids Are Precursors of Many Biomolecules 23. Glutathione is composed of which of the following amino acids?
(a) glutamine (b) glutamate (c) methionine (d) cysteine (d) glycine
24. Which of the following answers complete the sentence correctly? Glutathione
(a) cycles between oxidized and reduced forms in the cell.
(b) is involved in the detoxification of H2O2 and organic peroxides.
(c) donates amide groups from its g-glutamyl residue during biosynthetic reactions.
(d) contains an Se atom.
25. Which of the following statements about nitric oxide (NO) is correct?
(a) It is used as an inhalation anesthetic.
(b) It is a long-lived signal molecule.
(c) It is produced from asparagine.
(d) Its synthesis requires NADPH and O2.
(e) Its synthesis requires ATP and NH + 4 .
26, Which of the following are intermediates or precursors in the synthesis of heme?
(a) a-aminolevulinic acid (e) glycine (b) bilirubin (f)
succinyl CoA (c) porphobilinogen (g) Fe2+ (d) biliverdin ANSWERS TO SELF-TEST
1. Nitrogen fixation is the process by which nitrogen present in the atmosphere as N2 is enzymatically converted to NH3 by some bacteria and blue-green algae. This process is crucial to all other organisms because they can use only NH + 4 , and not N2, as the source of nitrogen for biosynthesis.
2. g, f, a, b, c, e, d
3. The net equation for nitrogen fixation is
N +
+
+
2
8 e− + 16 ATP + 16 H2O
2 NH3 16 ADP + 16 Pi
8 H+ + H2 The eight electrons needed to reduce N2 are supplied by oxidative processes in nonphotosynthetic nitrogen-fixing organisms and by light energy from the sun in photosynthetic nitrogen-fixing organisms. The ATP requirement is met by the usual oxidative or photosynthetic mechanisms of the cells.
432
CHAPTER 24
4. (a) 2, 3, 6 (b) 1, 2, 4, 5 5. (a) 2 (b) 1 (c) 3 6. (a) 1, 2 (b) 2 (c) None (d) 1, 3. Glutamate dehydrogenase uses NADPH when catalyz ing reductive aminations and NAD+ when carrying out oxidative deaminations.
7. Glutamate synthase catalyzes the reductive amination of a-ketoglutarate in a reaction with glutamine to form two glutamates. Glutamate can also be made from NH +
4
and a-ketoglutarate using glutamate dehydrogenase. However, this route requires high concentrations of NH +
+
4 because of the high KM of the enzyme for NH4 . Prokaryotes can use glutamine synthetase, which has a low K
+
+
M for NH4 , to form glutamine when NH4 concentrations are low. Thus, by using an additional enzyme, they can form glutamate from glutamine and a-ketoglutarate when NH + 4 is scarce.
8. c, d, e, f, j, k 9. b, e, f 10. a, b, e, f, g. Recall that ornithine is a precursor of arginine in the urea cycle and is derived from a-ketoglutarate.
11. The biosynthetic precursors are a, b, c, f, g, i, and j. The pathways they originate from are as follows:
Glycolysis: pyruvate (a) 3-phosphoglycerate (f) phosphoenolpyruvate (i)
Citric acid cycle: oxaloacetate (b) a-ketoglutarate (c) Pentose phosphate pathway: ribose 5-phosphate (g) erythrose 4-phosphate (j) 12. 1, 2, 3, 4, 5 (b) 1 (c) 6 13. d. Furthermore, since 3-phosphoglycerate can give rise to serine, you can see how carbohydrates can provide activated one-carbon units via a glucose 3-phosphoglycerate
serine
glycine pathway.
14. The positive charge on the sulfur atom of S-adenosylmethionine activates the methyl sul fonium bond and makes methyl group transfer from S-adenosylmethionine energetically more favorable than from N5-methyltetrahydrofolate.
15. a, c, e. The cofactor for reactions (b) and (d) is tetrahydrofolate.
16. Three high-energy bonds are expended. The adenosyl group of ATP is condensed with methionine to form a carbon-to-sulfur bond with the release of Pi and PPi, which is hydrolyzed to 2 Pi.
17. a, c. Homocysteine transmethylase uses a vitamin B12–derived cofactor.
18. b, c 19. Tryptophan synthetase contains a 25-Å-long channel between the active sites of adjacent a and b subunits. This channel allows the diffusion of the intermediate, indole, through the protein from one binding site to the other without diffusing away from the enzyme.
This alleviates the potential problem of the hydrophobic indole molecule diffusing across the plasma membrane and out of the cell were it allowed to leave the enzyme.
THE BIOSYNTHESIS OF AMINO ACIDS
433
20. A B. Control of the first step conserves the first compound, A, in the sequence and also saves metabolic energy by preventing subsequent reactions in the pathway.
Compound G would likely inhibit the committed step. The end product of a biosynthetic pathway often controls the committed step.
21. All the choices are correct. When all eight compounds are bound to the enzyme, it is al most completely inactive. The control of this enzyme is an excellent example of cumulative feedback inhibition.
22. Glutamine synthetase can be covalently modified by the attachment of an AMP to each of its 12 subunits. The more adenylylated the enzyme becomes, the more susceptible it is to feedback inhibition by the compounds listed in question 21. Thus, covalent modification modulates the sensitivity of the enzyme to its effectors. An added level of control exists in this system; adenylyltransferase, the enzyme that adenylylates glutamine synthetase, is itself covalently modified.
23. b, d, e 24. a, b. Answer (d) is incorrect because the enzyme glutathione peroxidase, rather than glu tathione, contains an Se analog of cysteine.
25. d. Note that (c) is incorrect because arginine rather than asparagine is the precursor of NO.
26. a, c, e, f, g
PROBLEMS
1. The glyA− mutation in Chinese hamster ovary cells in tissue culture makes these cells partially dependent on glycine. The mutation affects the mitochondrial form of serine transhydroxymethylase, which catalyzes the conversion of serine to glycine, with tetrahydrofolate serving as an acceptor of the hydroxymethyl group. Would you expect heme synthesis to be adversely affected in glyA− mutants? Why?
2. In early nutritional studies, cysteine was thought to be an essential amino acid. In 1937, Abraham White and E. F. Beach showed that cysteine could be removed from protein hydrolysates with cuprous mercaptide. Rats fed on such treated hydrolysates could grow, provided that sufficient methionine was supplied in the diet.
(a) What did this result reveal about cysteine metabolism?
(b) What would you expect the result to be if the rats were fed homocysteine along with the treated hydrolysates?
3. The essential amino acids are those that cannot be synthesized de novo in humans. Given an abundance of other amino acids in the diet, the a-keto acid analogs that correspond to the essential amino acids can substitute for these compounds in the diet.
(a) What do these observations tell you about the steps in the synthesis of essential amino acids that may be missing in humans?
(b) If 15N-labeled alanine is supplied in the diet, many other amino acids in the body will contain at least a small amount of the label within 48 hours. What enzymes are primarily responsible for this observation?
4. In muscle, glutamine synthetase is very active, catalyzing the formation of glutamine from glutamate and ammonia at the expense of a molecule of ATP. In the liver, the rate 434
CHAPTER 24 of formation of glutamine is very low, but a high level of glutaminase activity, which generates ammonia and glutamate, is observed. How would you explain the difference in the levels of enzyme activity in these two organs?
5. Consider three forms of bacterial glutamine synthetase: GS, the deadenylylated form; GS–(AMP)1, a form with one AMP unit per 12 subunits; and GS–(AMP)12, the fully adenylylated form.
(a) Which of these forms is most sensitive to feedback inhibition by several of the final products of glutamine metabolism, such as tryptophan or histidine? Why is it important that the activity of the most sensitive form not be completely inhibited by tryptophan?
(b) Which form has the lowest KM for ammonia?
(c) Why is it important that adenylyl transferase not carry out adenylylation and dead enylylation of glutamine synthetase at the same time?
(d) Glutamine synthetase in mammals is not subject to the same type of complex reg ulation that is seen in bacteria. Why?
6. Most of the proteins synthesized in mammals contain all 20 common amino acids. More protein is degraded than is synthesized when even one essential amino acid is missing from the diet.
(a) Under such conditions, how could an increase in the rate of protein degradation provide the missing amino acid?
(b) How does an increase in the rate of protein degradation contribute to increased levels of nitrogen excretion?
7. The diagram below outlines the biosynthesis of a compound that is required for the oxidation of fatty acids in the mitochondrion.
CH
CH
CH
+
3
3
3
+
+
+
NH
J
J
J
3
H CJNJCH
H CJNJCH
H CJNJCH
J
3
3
3
3
3
3
CH
J
J
J
2
CH
CH
CH
J
2
2
2
CH
J
J
a-Ketoglutanate Succinate+CO
J
2
CH
CH
2
HOJCJH
J 3 B
2
2
CH
D
J
DDD
J
J
2
CH
CH
CH
J
2
2
2
O
CH
J
J
J
2
2
CH COO: COO:
J
+
2
HJCJNH
J
+
D
3
HJCJNH
J
3
COO:
J COO:
A
C (a) Name compound D and briefly explain its role in fatty acid metabolism.
(b) Name compound A. Why is it considered essential in human diets?
(c) Three molecules of compound B are required for the formation of compound C. Its synthesis depends on the availability of an essential amino acid. Name that amino acid and then name compound B and compound C.
THE BIOSYNTHESIS OF AMINO ACIDS
435
8. A pathway for the synthesis of ornithine from glutamate is shown in Figure 24.1.
FIGURE 24.1 Biosynthesis of ornithine from glutamate.
COO:
O
H COO:
O
H COO:
J
O
K
J
J
K
J
J H ;NJCJH
K H CJCJNJCJH ;
H CJCJNJCJH
3
H CJCJSJCoA
3
NADPH NADP
3
J
3
J
J
CH
CH
CH
2
2
2
J
J
J
CH
CH
CH
2
2
2
J
J
J COO: COO:
CKO
Glutamate
N -Acetylglutamate H J
N -Acetylglutamate-g - semialdehyde
Glutamatea -Ketoglutarate COO:
O
H COO:
;
J
K
J
J H NJCJH
H CJCJNJCJH
3
3
J
J
:
CH
H CJCOO
3
H O
CH
2
2
2
J
J
CH
CH
2
2
J
J NH ; NH ;
3
3
Ornithine
N -Acetylornithine (a) Why can this pathway also be considered to be part of the de novo pathway for the synthesis of arginine?
(b) Inspect the pathway for proline biosynthesis given on page 674 of the text, and then explain why the N-acetylation of glutamate is needed for the synthesis of ornithine.
9. Elevated levels of ammonia in blood can result from deficiencies in one or another of the enzymes of the urea cycle. Measures taken to relieve hyperammonemia have included limiting intake of dietary proteins, administering a-keto analogs of several of the naturally occurring L-amino acids, or administering other compounds designed to exploit pathways of nitrogen metabolism and excretion.
(a) In trying to determine why a patient has hyperammonemia, which organ should you check first for normal function? Why?
(b) Why would limiting protein intake assist in relieving chronic hyperammonemia?
Why would eliminating dietary proteins altogether (without any other supplement) probably increase the level of hyperammonemia?
(c) Write a brief rationale for using a-keto acid analogs in treating hyperammonemia, mentioning a particular group of enzymes essential to your explanation. Would it be better to use a-keto analogs of essential or nonessential amino acids? Why?
436
CHAPTER 24
10. Plants synthesize all 20 common amino acids de novo. Glyphosate, a weed killer sold under the trade name Roundup, is an analog of phosphoenolpyruvate that specifically inhibits 3-enolpyruvylshikimate 5-phosphate synthase, a key enzyme of the pathway for chorismate biosynthesis. This compound is a very effective plant herbicide, but has virtually no effect on mammals. Why?
11. In B. subtilis, the pathway from chorismate to tryptophan is feedback-inhibited by tryptophan, which suppresses anthranilate synthase activity. Mutant B. subtilis that lacks tryptophan synthetase can grow on minimal medium only when supplemented with exogenous tryptophan. Under these conditions, none of the intermediates in the tryptophan biosynthetic pathway from anthranilate to indole 3-glycerol phosphate are produced. However, when the bacteria have depleted the medium of tryptophan, the levels of those intermediates increase, even though there is no net production of tryptophan. Why?
12. Both genetic and biochemical methods have been used to establish the biosynthetic path ways for essential amino acids in bacteria and other microorganisms. One classic approach is isotope competition, which begins with the use of radioactive glucose as the sole source of carbon for growing bacteria. Under these conditions, all the intermediates in a particular pathway will be uniformly labeled, but if a nonradioactive intermediate in the pathway is added to the growing cells, it will reduce or dilute the radioactivity of that intermediate and others following it in a pathway.
Britten and his coworkers (R. B. Roberts, P. H. Abelson, D. B. Cowie, E. T. Bolton, and R. J. Britten, Studies in Biosyntheses in Escherichia coli. Carnegie Institution of Washington, Pub. No. 607, 1955), used isotope competition to examine the biosynthetic pathway for threonine, methionine, and other amino acids. They grew E. coli in a minimal medium containing labeled glucose and nonlabeled homoserine, which was known to relieve auxotrophy for several amino acids. Under these conditions, isoleucine, threonine, and methionine isolated from the cells had little or no radioactivity. In contrast, the radioactivity of aspartate and lysine was unchanged whether cells were grown with or without the addition of nonradioactive homoserine.
In a similar experiment, nonradioactive aspartate was used in the growth medium; as partate from protein hydrolysates was virtually nonradioactive, as were lysine, methionine, threonine, and isoleucine. When nonlabeled threonine was used along with radioactive glucose, only threonine and isoleucine from protein hydrolysates had reduced radioactivity. And when either nonradioactive isoleucine or methionine was used, they affected only their own levels of radioactivity in protein hydrolysates.
On the basis of these observations, write an outline of the biosynthetic pathway for the amino acids isoleucine, threonine, homoserine, methionine, and lysine.
ANSWERS TO PROBLEMS
1. Glycine is an obligatory precursor of heme; in the reaction catalyzed by d-aminolevuli nate synthase, glycine condenses with succinyl CoA to form d-aminolevulinate. The reduction in the concentration of glycine in the cell caused by the glyA− mutation will cause a decrease in the rate of heme synthesis.
2. (a) The studies led White and Beach to conclude that methionine, which is another sulfur-containing amino acid, is a biosynthetic precursor of cysteine. Later work elucidated the roles of methionine in the active methyl cycle and in the synthesis of cysteine and confirmed their conclusion.
THE BIOSYNTHESIS OF AMINO ACIDS
437
(b) In the active methyl cycle, S-adenosylhomocysteine is cleaved to yield adenosine and homocysteine. Although homocysteine can be converted to methionine, it can also condense with serine to form cystathionine, which is then cleaved to yield cysteine and a-ketobutyrate. Feeding rats homocysteine along with a treated hydrolysate will make supplementation with methionine unnecessary.
3. (a) The fact that a-keto acid analogs can substitute for essential amino acids means that the carbon skeletons of the essential amino acids are not synthesized in humans.
Many studies have shown that one or more of the enzymes needed for the synthesis of these structures are missing.
(b) The enzymes that are primarily responsible for the distribution of the 15N label among the other amino acids are the aminotransaminases, which catalyze the interconversions of amino acids and their corresponding a-keto acids. The redistribution of the label begins with the transamination of alanine, with a-ketoglutarate serving as the amino acceptor to yield pyruvate and glutamate. Glutamate then serves as an amino donor for other a-keto acids. In order for an essential amino acid to be labeled, you must postulate the transamination of that amino acid to yield the corresponding a-keto acid analog, followed by the donation of a labeled amino group from glutamate or another donor of amino groups.
4. Ammonia, which is generated as part of the process of amino acid catabolism in muscle, is toxic and must be removed from the cells. This could be done through the synthesis of urea, but that process occurs only in the liver. In muscle cells, therefore, glutamine synthetase catalyzes the formation of glutamine, which is an efficient and nontoxic carrier of ammonia. This accounts for the high activity of that enzyme in muscle. The glutamine is transported by the blood to the liver, where glutaminase and aspartate aminotransferase work together to generate aspartate and two molecules of ammonia from glutamine, hence, the high activity of glutaminase in the liver. Aspartate and ammonia are both used by the liver for the synthesis of urea, a nontoxic and disposable form of ammonia.
5. (a) The fully adenylylated form of glutamine synthetase is the most sensitive to mole cules like tryptophan and histidine. Because glutamine is utilized for the synthesis of a variety of compounds, complete inhibition of the enzyme by only one of those products, such as tryptophan, would inappropriately inhibit the synthesis of all the others. Thus, the enzyme is cumulatively inhibited by at least eight different nitrogen-containing compounds.
(b) The deadenylylated form, which is not subject to cumulative feedback inhibition, is generated in response to increases in the cellular concentrations of a-ketoglutarate (the precursor of glutamate) and ATP. These molecules signal the need for glutamine synthesis, even when other nitrogen-containing compounds are present.
Under these conditions, the deadenylylated form of the enzyme binds ammonia even when ammonia concentrations are relatively low; that is, it has a relatively low KM for ammonia.
(c) The simultaneous adenylylation and deadenylylation of glutamine synthetase would result in a loss of feedback control of the enzyme because the adenylylated form is subject to cumulative inhibition whereas the deadenylylated form is not. In addition, it would lead to the wasteful hydrolysis of ATP, since every round of adenylylation and deadenylylation generates AMP and inorganic pyrophosphate from ATP.
(d) Mammals acquire many nitrogen-containing compounds, such as tryptophan and histidine, in their diet rather than through de novo biosynthesis, so glutamine synthetase does not play so prominent a role in the nitrogen metabolism of mammals as it does in that of bacteria. Complex regulation of the enzyme is therefore not needed in mammals.
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CHAPTER 24
6. (a) Many experiments have shown that under normal conditions cells continuously synthesize and degrade proteins. Although both essential and nonessential amino acids are continuously recycled during these processes, reutilization is not completely efficient; thus, additional amino acids are needed. In mammals, there are no reservoirs of free amino acids; the only sources of essential amino acids are dietary proteins or the proteins of the body tissues. If an essential amino acid is not available from the diet, cells appear to accelerate the hydrolysis of their own proteins, in order to generate the missing essential amino acid. How the rate of cellular proteolysis is accelerated in response to a deficiency of an essential amino acid is not understood.
(b) An increased rate of protein degradation generates a higher concentration of free amino acids. During the oxidation of those amino acids not used for synthesis of other proteins, ammonia will be produced. An elevation in ammonia concentration in the body stimulates the formation of urea, causing the level of nitrogen excretion to increase.
7. (a) Compound D is carnitine, which, when esterified to the acyl group of a long-chain fatty acid, shuttles it from the cytosol to the matrix of the mitochondrion, where fatty acid oxidation takes place.
(b) Compound A is the essential amino acid lysine; it is termed essential because it can not be synthesized de novo in humans. Lysine and other essential amino acids must be obtained from the diet.
(c) Compound C is trimethyllysine, and the methyl groups that are attached to lysine are likely to be derived from compound B, S-adenosylmethionine, the major donor of methyl groups in biosynthetic reactions. The methyl group of S-adenosylmethionine is derived from methionine, an essential amino acid.
8. (a) Ornithine is a precursor of arginine, as part of the pathway for the synthesis of urea.
Thus the pathway for the synthesis of ornithine from glutamine, along with part of the urea cycle pathway, can together be considered as a de novo pathway for the synthesis of arginine. Arginine can in turn be used for the synthesis of urea, or it can serve instead as one of the amino acids used for polypeptide synthesis.
(b) In the pathway for proline biosynthesis, glutamic-g-semialdehyde cyclizes with the loss of water to form D1-pyrroline-5-carboxylate. However, in the pathway for the formation of ornithine shown in Figure 24.1, an N-acetylated derivative of the semialdehyde molecule is formed. The N-acetyl group blocks the condensation of the amino group with the aldehyde group, thereby preventing the formation of the pyrroline ring. This allows the pathway to proceed toward the synthesis of ornithine.
9. (a) You should assess liver function, because enzymes of the urea cycle are found pri marily in this organ. In addition, liver takes up amino acids such as alanine, glutamate, and glutamine, which are in effect nontoxic forms of ammonia generated by muscle and other tissues. Amino groups of these compounds, along with carbon dioxide from carbamoyl phosphate, are precursors of urea in the liver.
(b) During digestion, dietary proteins are hydrolyzed to their component amino acids.
Those amino acids that are not needed immediately for protein synthesis or for the biosynthesis of other nitrogen-containing compounds are degraded. One of the products of amino acid degradation is ammonia. Usually ammonia is metabolized through conversion to nitrogen carriers such as alanine, glutamate, and glutamine, and it is ultimately utilized for the synthesis of urea when the urea cycle is operat ing. Thus limiting protein intake in a patient with a deficiency in urea synthesis would be expected to reduce ammonia production in liver and other tissues.
Protein turnover and amino acid degradation constantly take place in the tissues, and essential amino acids (those that cannot be synthesized de novo in human tissues) must be generated either from dietary sources or from additional breakdown of body proteins. A complete restriction of dietary protein would accelerate body protein breakdown and would exacerbate the condition of hyperammonemia.
(c) The a-keto acid analogs can serve as acceptors for amino groups from glutamate and other amino acids, in reactions catalyzed by transaminases or aminotransferases.
Nonessential amino acids formed as the result of this process may be themselves eliminated, degraded (often generating more ammonia), or else used for biosynthesis of other nitrogen-containing compounds. Employing analogs of essential amino acids might be preferable, because tissues are more likely to require them for protein synthesis or other biosyntheses.
10. Chorismate is an intermediate in the biosynthesis of the aromatic amino acids trypto phan, phenylalanine, and tyrosine. Mammals do not synthesize these amino acids from chorismate. Instead, they obtain the essential aromatic amino acids tryptophan and phenylalanine from the diet, and they can synthesize tyrosine from phenylalanine.
Glyphosate is an effective herbicide because it prevents synthesis of aromatic amino acids in plants. But the compound has no effect on mammals because they have no active pathway for de novo aromatic amino acid synthesis.
11. Since anthranilate synthase is inhibited by tryptophan, exogenous tryptophan from the medium, inhibits the production of downstream intermediates in the biosynthetic pathway by halting the first step in the pathway. When exogenous tryptophan is depleted, intracellular levels of tryptophan decrease and anthranilate synthase activity is no longer inhibited. An increase in anthranilate production leads to an increase in the production of other intermediates in the pathway until equilibria are established among those compounds. The block at the step catalyzed by tryptophan synthetase prevents large accumulations of the intermediates, but under these conditions their concentrations are higher than in the presence of exogenous tryptophan.
12. The experiments show that aspartate is a key precursor for the five amino acids, because when it is included in the growth medium with labeled glucose, radioactivity of each of the amino acids is reduced. Homoserine affects the labeling of isoleucine, threonine, and methionine. Thus, homoserine is an intermediate in the pathway for those three amino acids but not lysine or aspartate. Dilution of threonine and isoleucine from cells grown in nonradioactive threonine suggests that those two amino acids are on the same pathway, separated from those of the other three. Methionine and isoleucine are affected only when they are used in growth experiments, indicating that they are not on other pathways (and, for isoleucine, that threonine must precede it in a biosynthetic pathway).
The overall pathway below shows that lysine and homoserine derived from aspartate.
Threonine and isoleucine are both derived from aspartate, and homoserine must be an intermediate in their synthesis, as well as in the synthesis of methionine.
Aspartate
Homoserine
Threonine
Isoleucine
Lysine
Methionine
440
CHAPTER 24 EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. Glucose + 2 ADP + 2 P +
i
2 NAD+ + 2 glutamate 2 alanine + 2 a-ketoglutarate + 2 ATP + 2 NADH + H+ Glucose is converted to pyruvate via glycolysis, and pyruvate is converted to alanine by transamination.
2. N
+
2
NH4
glutamate
serine
glycine
d-aminolevulinate
porphobilinogen
heme
3. (a) N5,N10-methylenetetrahydrofolate, (b) N5-methyltetrahydrofolate 4. g-Glutamyl phosphate is a likely reaction intermediate.
5. The administration of glycine leads to the formation of isovalerylglycine. This water-sol uble conjugate, in contrast with isovaleric acid, is excreted very rapidly by the kidneys.
See R. M. Cohn, M. Yudkoff, R. Rothman, and S. Segal. New Engl. J. Med. 299(1978):996.
6. They carry out nitrogen fixation. The absence of photosystem II provides an environ ment in which O2 is not produced. Recall that the nitrogenase is very rapidly inactivated by O2. See R. Y. Stanier, J. L. Ingraham, M. L. Wheelis, and P. R. Painter, The Microbial
World, 5th ed. (Prentice-Hall, 1986), pp. 356–359, for a discussion of heterocysts.
7. The cytosol is a reducing environment, whereas the extracellular milieu is an oxidizing environment.
8. The synthesis of d-aminolevulinate requires succinyl CoA, an intermediate in the citric acid cycle, which occurs in the mitochondrial matrix. Thus it makes sense to have the first step in porphyrin biosynthesis occur in the matrix.
9. Alanine and aspartate can be synthesized directly from glutamate by transamination of pyruvate and oxaloacetate, respectively. Glutamate is the common intermediate that serves as the amino group donor in these reactions. (a-Ketoglutarate, the side product, can condense with another molecule of ammonia to regenerate the glutamate for further transamination reactions.)
10. Each final product should inhibit the first unique step toward its synthesis, that is, the first step following the branch point of the pathway. Thus, Y should inhibit the conversion of C to D, and Z should inhibit the conversion of C to F. (To avoid wasteful accumulation of C or B if neither Y nor Z is needed, high levels of C should also inhibit the conversion of A to B at the beginning of the pathway.)
11. The effects would multiply, so that the net rate would equal (100 × 0.6 × 0.4 s−1), or 24−1 s.
12. The reaction will begin with the formation of a Schiff base between pyridoxal-5′-phos phate and the amino group of S-adenosylmethionine (external aldimine, as opposed to the internal aldimine in which PLP is connected to a lysine on the enzyme). In usual fashion for PLP enzymes, the a-hydrogen will be extracted from S-adenosylmethionine (step 2, below). Then the next series of p-electron transfers (step 3) will eliminate Smethylthioadenosine and form the three-membered ring. Finally, 1-aminocyclopropane1-carboxylate (ACC) will be released, and the original internal aldimine between PLP and lysine will be restored, so that the enzyme and cofactor are ready for another round of synthesis. The second product is S-methylthioadenosine.
THE BIOSYNTHESIS OF AMINO ACIDS
441
S -adenosyl-Met
CH
CH
3
3
J
J
+
H
adenosyl-S-CH -CH COO:
S
N
H
2
2
H C
adenosyl
+
2
+
H C G
H
N
G O:
2
COO:
+
H
PO
=O
3
O:
H
N
=O PO
+ H +
N
CH
3
3
O:
J
H
+
N
CH
=O PO
3
3
internal aldimine
J
H
N
CH
external aldimine
3
COO:
(Lys)
COO:
+ S -methyl-thioadenosine
NH
3
+
H
N
N
H
G
G
+ H
H
O: O: =O PO
=O PO
3
3
+
N
CH3
+
N
CH3
J
J
H
H
13. As in problem 15 in Chapter 23 of this manual, an external aldimine (protonated Schiff base) is formed between serine and pyridoxal-5′-phosphate, and the serine a-hydrogen is removed. The a-hydrogen then can be reattached from either of two faces to give the Schiff base of either D-serine or L-serine (see below). The equilibrium constant for the reaction will be one.
L-approach
H;
H
HOCH COO:
HOCH COO:
2
2
H
N
H
N
D-approach
+
H
+ H
O: O:
G
=O PO =O PO
3
3
+
N
CH
N
CH
3
3
J
J
H
H
14. Aspartate and glutamate would be synthesized from the citric acid cycle intermediates oxaloacetate and a-ketoglutarate. Increased synthesis of aspartate and glutamate therefore could begin to deplete citric acid cycle intermediates. The cell would need to respond by breaking down carbohydrates to replenish the supply by net synthesis of new cycle intermediates.
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CHAPTER 24
15. Because S-adenosylmethionine (SAM) is the methyl donor for the methylation of DNA (Figure 24.15), a deficiency in SAM could diminish the extent of the methylation of the mutated bacteria’s DNA. The lower level of methylation would render the DNA more susceptible to digestion by restriction enzymes.
16. (a) Asn, Gln, and Gly are affected. Asn is much more concentrated in dark-adapted plants, whereas Gln and Gly are present in somewhat elevated levels in lightadapted plants.
(b) The transcription of specific mRNA, translation or enzymatic activity of several en zymes, particularly nitrogen-utilizing enzymes such as asparagine synthetase and glutamine synthetase, could be regulated by light.
(c) From the graph, asparagine would appear to be a likely candidate. (The name of the plant would also suit this interpretation!)
CHAPTER 2
Nucleotide Biosynthesis
5
In this chapter, the authors complete their treatment of the biosyntheses of the major classes of macromolecular precursors by describing the synthesis of the purine and pyrimidine nucleotides. Besides being the precursors of RNA and DNA, these compounds serve a number of other important roles that are reviewed in the opening paragraph of the chapter. Nucleotide nomenclature is reviewed in the introduction to the chapter, as is an outline for the synthesis of nucleotides through de novo and salvage pathways.
The chapter begins with the synthesis of the pyrimidine nucleotides. The pyrim idine ring is synthesized de novo from bicarbonate, aspartate, and ammonia (usually from glutamine) prior to attachment to a ribose sugar. The authors go through the synthesis step-by-step, paying particular attention to the enzyme carbamoyl phosphate synthetase (CPS), which synthesizes carbamoyl phosphate from bicarbonate and ammonia and catalyzes the committed step in eukaryotic pyrimidine synthesis.
The next step in the synthesis is catalyzed by aspartate transcarbamoylase (ATCase), an enzyme that was discussed in Chapter 10. This reaction, the formation of carbamoylasparate from carbamoyl phosphate and aspartate, is the committed step in prokaryotic pyrimidine synthesis. A condensation and an oxidation reaction complete the formation of orotate, which is then coupled to a phosphoribose by reaction with 5-ribosyl-1-pyrophosphate (PRPP) to form the pyrimidine nucleotide orotidylate.
Decarboxylation of orotidylate gives uridine monophosphate (UMP), which can be phosphorylated by nucleoside mono- and diphosphate kinases to form UDP and UTP, respectively. Amination of UTP forms cytidine triphosphate (CTP) and completes the synthesis of the pyrimidine ribonucleotides.
Next the authors turn to synthesis of the purine nucleotides. Unlike synthesis of pyrimidines, synthesis of purine nucleotides builds upon the ribose ring. As in the
443
444
CHAPTER 25 pyrimidine ring system, the ribose sugar is donated by the activated form of ribose 5-phosphate, 5-phosphoribosyl-1-pyrophosphate (PRPP). The two purine nucleotides AMP and GMP have a common precursor, inosine 5′-monophosphate (IMP). The authors discuss the synthesis of this initial purine product and then formation of AMP and GMP. The reactions that allow cells to salvage free purines and the control of purine biosynthesis are presented. The regulation of nucleotide biosynthesis through feedback inhibition is discussed later in the chapter.
Two reactions that are required to form the precursors of DNA are described in detail: ribonucleotide reductase converts ribonucleotides to deoxyribonucleotides, and thymidylate synthase methylates dUMP to form dTMP. The authors present the mechanisms and cofactors of these enzymes and explain how some anticancer drugs and antibiotics function by inhibition of dTMP synthesis and thus the growth of cells. Nucleotides also serve important roles as constituents of NAD+, NADP+, FAD, and coenzyme A (CoA), so the syntheses of these cofactors are described briefly. The chapter concludes with an explanation of how the purines are catabolized and some of the pathological conditions that arise from defects in the catabolic pathway of the purines.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. List the major biochemical roles of the nucleotides.
2. Distinguish among the purine and pyrimidine nucleosides and nucleotides.
In de Novo Synthesis, the Pyrimidine Ring Is Assembled from Bicarbonate,
Aspartate, and Glutamine (Text Section 25.1) 3. Draw the structure of a pyrimidine ring and identify the precursors that provide each carbon and nitrogen atom of the ring. Note the numbering of the ring atoms.
4. Discuss the structure and mechanism of carbamoyl phosphate synthetase (CPS). Explain the role of carbamoyl phosphate in pyrimidine biosynthesis.
5. Write the aspartate transcarbamoylase reaction and outline the remaining reactions that form orotate. Outline the conversion of orotate to uridylate (UMP).
6. Contrast the enzymes of pyrimidine biosynthesis in E. coli with those of higher organisms and list the potential advantages of multifunctional enzymes.
7. Explain how nucleoside mono- and diphosphate kinases interconvert the nucleoside mono-, di-, and triphosphates.
8. Describe the reaction that converts UTP to CTP.
Purines Can Be Synthesized de Novo or Recycled by Salvage Pathways
(Text Section 25.2) 9. Outline the synthesis of purine nucleotides by the salvage reactions and explain why these reactions are energetically advantageous.
10. Draw the structure of a purine ring and identify the precursors that provide each carbon and nitrogen atom of the ring. Note the numbering of the ring atoms.
NUCLEOTIDE BIOSYNTHESIS
445
11. Describe the committed step in de novo purine biosynthesis and the enzyme that cat alyzes it.
12. Explain the mechanism for replacing a carbonyl oxygen with an amino group, and list the potential sources of the nitrogen atom in these reactions.
13. Outline the synthesis of inosine 5′-monophosphate (IMP), noting the sources of the atoms and the cofactors involved. List the steps that require ATP.
14. Describe the synthesis of adenylate (AMP) and guanylate (GMP) from IMP. List the cofac tors and intermediates of the reactions.
Deoxyribonucleotides Are Synthesized by the Reduction of Ribonucleotides
Through a Radical Mechanism (Text Section 25.3) 15. Describe the subunit structure of ribonucleotide reductase. Outline its reaction, mecha nism, and regulation. Explain the roles of NADH, thioredoxin, and thioredoxin reductase
in the reaction.
16. Describe the thymidylate synthase reaction. Account for the source of the methyl group and describe the change in the oxidation state of the transferred carbon atom that occurs during the reaction.
17. Explain the role of dihydrofolate reductase in the synthesis of deoxythymidylate.
18. Account for the ability of fluorouracil to act as a suicide inhibitor. Describe the inhibitory mechanism of methotrexate and aminopterin. Explain how these three compounds interfere with the growth of cancer cells. List the mechanisms by which a cell could become resistant to methotrexate. Explain the antibiotic activity of trimethoprim.
Key Steps in Nucleotide Biosynthesis Are Regulated by Feedback Inhibition
(Text Section 25.4) 19. Outline the regulation of the biosynthesis of the purine and pyrimidine nucleotides and name the committed steps in the pathways.
NAD+, FAD, and Coenzyme A Are Formed from ATP (Text Section 25.5) 20. Describe the synthesis of NAD+ and account for the sources of the nicotinamide portion of the molecule.
21. Outline the synthesis of FAD and explain the role of the PPi that is released during the synthesis of FAD, NAD+, and CoA.
Disruptions in Nucleotide Metabolism Can Cause Pathological Conditions
(Text Section 25.6) 22. Describe the reactions of the nucleotidases and nucleoside phosphorylases.
23. Outline the conversions of AMP and guanine to uric acid. Describe the role of xanthineoxidase in these processes.
24. Describe the major clinical findings in patients with gout and explain the rationale for the use of allopurinol to alleviate the symptoms of the disease.
25. Describe the antioxidant role of urate.
26. Name the biochemical lesion that leads to the Lesch-Nyhan syndrome and describe the symptoms of the disease.
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CHAPTER 25
SELF-TEST
Introduction 1. Describe the physiological roles of the nucleotides.
2. Which of the following answers completes the sentence correctly? Cytosine is a (a) purine base.
(c) purine nucleoside.
(b) pyrimidine base.
(d) pyrimidine nucleoside.
3. Which of the following are nucleotides?
(a) deoxyadenosine (c) deoxyguanylate (b) cytidine (d) uridylate In de Novo Synthesis, the Pyrimidine Ring Is Assembled from Bicarbonate,
Aspartate, and Glutamine 4. Which of the following statements about the carbamoyl phosphate synthetase of mam mals, which is used for pyrimidine biosynthesis, are true?
(a) It is located in the mitochondria.
(b) It is located in the cytosol.
(c) It uses NH +
4
as a nitrogen source.
(d) It uses glutamine as a nitrogen source.
(e) It requires N-acetylglutamate as a positive effector.
5. Which of the following statements about 5-phosphoribosyl-1-pyrophosphate (PRPP) are true?
(a) It is an activated form of ribose 5-phosphate.
(b) It is formed from ribose 1-phosphate and ATP.
(c) It has a pyrophosphate group attached to the C-1 atom of ribose in the a configuration.
(d) It is formed in a reaction in which PPi is released.
6. How is orotate, a free pyrimidine, converted into a nucleotide? Is this reaction consid ered to be a salvage reaction or a biosynthetic one?
7. Which of the following enzymes are involved in converting the nucleoside 5′ monophosphate (NMP) products of the purine or pyrimidine biosynthetic pathways into their 5′-triphosphate (NTP) derivatives?
(a) purine nucleotidase (b) nucleoside diphosphate kinase (c) nucleoside monophosphate kinases (d) nucleoside phosphorylase 8. How is the exocyclic amino group on the N-4 position of cytosine formed?
Purines Can Be Synthesized de Novo or Recycled by Salvage Pathways 9. Which of the following compounds directly provide atoms to form the purine ring?
(a) aspartate (e) CO2 (b) carbamoyl phosphate (f) N5, N10-methylenetetrahydrofolate (c) glutamine (g) N10-formyltetrahydrofolate (d) glycine (h) NH +
4
(c) IMP.
(b) GMP.
(d) xanthylate (XMP).
11. The conversion of IMP to AMP requires which of the following?
(a) ATP (d) glutamine (b) GTP (e) NAD+ (c) aspartate
12. The conversion of IMP to GMP requires which of the following?
(a) ATP (d) glutamine (b) GTP (e) NAD+ (c) aspartate
13. Describe the general mechanism used by cells to replace a carbonyl group with an amino group in nucleotide biosynthesis.
14. Which of the following reactants and products are involved in the salvage reactions of purine biosynthesis?
(a) IMP
AMP (c) adenine
AMP (b) IMP
GMP (d) guanine
GMP
15. During a purine salvage reaction, what is the source of the energy required to form the C–N glycosidic bond between the base and ribose?
16. Show which of the nucleotides in the right column regulate each of the conversions in the left column.
(a) ribose 5-phosphate
PRPP
(1) AMP (b) PRPP
_phosphoribosylamine
(2) GMP (c) phosphoribosylamine
IMP
(3) IMP (d) IMP
adenylosuccinate (e) IMP xanthylate (XMP) Deoxyribonucleotides Are Synthesized by the Reduction of Ribonucleotides
Through a Radical Mechanism 17. Which of the following statements about ribonucleotide reductase are true?
(a) It converts ribonucleoside diphosphates into 2′-deoxyribonucleoside diphosphates in humans.
(b) It catalyzes the homolytic cleavage of a bond.
(c) It accepts electrons directly from FADH2.
(d) It receives electrons directly from thioredoxin.
(e) It contains two kinds of allosteric regulatory sites—one for control of overall activity and another for control of substrate specificity.
18. Select from the following those compounds that are precursors of 2′-deoxythymidine 5-triphosphate (dTTP) in mammals and place them in their correct biosynthetic order.
(a) OMP (f)
dUDP (b) UMP (g) dUTP (c) UDP (h) dTMP (d) UTP (i)
dTDP (e) dUMP (j)
dTTP
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CHAPTER 25
19. Define suicide inhibitor and give an example from pyrimidine biosynthesis.
20. Methotrexate and trimethoprim are both inhibitors of dihyrofolate reductase. Why is trimethoprim the drug of choice in treating a human microbial infection?
Key Steps in Nucleotide Biosynthesis Are Regulated by Feedback Inhibition 21. What is the committed step in purine biosynthesis and which of the following com pounds are involved in the control of the purine biosynthetic pathway?
(a) IMP (b) AMP (c) OMP (d) GMP (e) PRPP NAD+, FAD, and Coenzyme A Are Formed from ATP 22. Which of the following can serve as a precursor of NAD+ in humans?
(a) riboflavin (d) tryptophan (b) pantothenate (e) niacin (c) tyrosine Disruptions in Nucleotide Metabolism Can Cause Pathological Conditions 23. Which of the following compounds would give rise to urate if they were catabolized completely in humans?
(a) ADP-glucose (e) CoA (b) GDP-mannose (f)
FAD (c) CDP-choline (g) UMP (d) UDP-galactose 24. What is the benefit of high serum levels of urate in humans given that too much urate leads to gout?
ANSWERS TO SELF-TEST
1. The nucleotides (1) are the activated precursors of DNA and RNA; (2) are the source of derivatives that are activated intermediates in many biosyntheses; (3) include ATP, the universal currency of energy in biological systems, and GTP, which powers many movements of macromolecules; (4) include the adenine nucleotides, which are components of the major coenzymes NAD+, FAD, and CoA; and (5) serve as metabolic regulators.
2. b 3. c, d. Nucleotides are nucleosides that contain one or more phosphate substituents on their ribose or deoxyribose moieties.
4. b, d. The mitochondrial carbamoyl phosphate synthetase used for urea synthesis is ac tivated by N-acetylglutamate and uses N +
4
as the nitrogen source.
NUCLEOTIDE BIOSYNTHESIS
449
5. a, c.
6. Orotate condenses with PRPP in a reaction catalyzed by orotate phosphoribosyl trans ferase to form the nucleotide orotidylate (OMP). Orotidylate decarboxylase converts OMP to the more abundant nucleotide UMP. The reaction occurs during de novo pyrimidine biosynthesis and is therefore not a salvage reaction.
7. b, c. Several different specific nucleoside monophosphate kinases phosphorylate dNMPs and NMPs, using ATP as the phosphoryl donor. A single enzyme, nucleoside diphosphate kinase, uses the phosphorylation potential of ATP to convert the dNDPs and NDPs to dNTPs and NTPs. The ubiquitous adenylate nucleotides are interconverted by adenylate kinase (myokinase).
8. CTP is formed by amination of UTP. The carbonyl oxygen at C-4 of UTP is replaced with an amino group via the formation of an enol phosphate ester intermediate. In E. coli,
NH +
4
serves as the source of the nitrogen atom that displaces the phosphate group, whereas the amide group of glutamine serves this purpose in mammals.
9. a, c, d, e, g 10. c 11. b, c 12. a, d, e 13. A carbonyl oxygen is converted into a phosphoryl ester or a substituted phosphoryl ester through a reaction with a high-energy phosphate (ATP or GTP) to form a mono- or diphosphate ester. The phosphate or pyrophosphate group can then be readily displaced by the nucleophilic attack of the nitrogen atom from NH3, the side-chain amide group of glutamine, or the a-amino group of aspartate. The resulting adduct is an amino group or can be converted into one.
14. c, d 15. The activated form of ribose 5-phosphate (R-5-P), PRPP, reacts with the purine base to form the nucleotide and release PPi. The displacement and subsequent hydrolysis of PPi drives the formation of the N-glycosyl bond. ATP ultimately provides the energy through its reaction with R-5-P to form PRPP.
16. (a) 1, 2, 3 (b) 1, 2, 3 (c) none (d) 1 (e) 2 17. a, b, d, e. NADPH provides electrons via thioredoxin.
18. a, b, c, f, g, e, h, i, j. In mammals, NDPs are converted to dNDPs by ribonucleotide re ductase. Thus, UDP is converted to dUDP, which is converted to dUTP by nucleoside diphosphate kinase. A specific pyrophosphatase hydrolyzes dUTP to dUMP, which is then converted to dTMP by thymidylate synthase. The dTMP is converted to dTTP. A priori, you might have expected the dUDP product of the ribonucleotide reductase reaction to be converted directly to dUMP. However, in fact, cells contain a dUTP pyrophosphatase to prevent dUTP from serving as a DNA precursor, and it is this enzyme that functions in the dTTP biosynthetic pathway.
19. A suicide inhibitor is a substrate that is converted by an enzyme into a substance that is capable of reacting with and inactivating the enzyme. In pyrimidine biosynthesis, thymidylate synthase converts fluorouracil into a derivative that becomes covalently attached to the enzyme and thereby inactivates it.
450
CHAPTER 25
20. Since trimethoprim binds to the mammalian dihydrofolate reductase much less tightly than to the enzyme of susceptible microorganisms, it causes fewer deleterious effects to humans than does methotrexate.
21. The conversion of PRPP into phosphoribosylamine by glutamine phosphoryl amidotrans-ferase is the committed step in purine biosynthesis. Compounds a, b, and d are involved in the regulation.
22. d, e. Niacin becomes a dietary requirement if the supply of the essential amino acid tryp tophan is inadequate.
23. a, b, e, f. Each of these compounds contains a heterocyclic purine base.
24. Urate levels in humans are often close to the solubility limit, which leads to gout when salts of urate crystallize (resulting in damage to joints and kidneys). There is a significant benefit to high concentrations of urate, however, as urate is a highly effective scavenger of reactive oxygen species (ROS). ROS can cause damage in cells contributing to cancer and the effects of aging. Urate is about as effective as ascorbate (vitamin C) as an antioxidant.
PROBLEMS
1. Why might covalently linked (multifunctional) enzymes, such as those of the pyrimi dine biosynthetic pathway of mammals, be advantageous to an organism?
2. Mammalian lymphocytes that lack adenosine deaminase neither grow nor divide. The level of dATP in these cells is 100 times higher than that in normal lymphocytes, and the synthesis of DNA in the cells is impaired.
(a) How is adenosine converted to dATP? Assume that the first step is catalyzed by a specific nucleoside kinase.
(b) How does the elevation in dATP concentration in the abnormal lymphocytes affect the synthesis of DNA?
3. Clinicians who use F-dUMP and methotrexate together in cancer treatment find that the combined effects on cancer cells are not synergistic. Suggest how the administration of methotrexate could interfere with the action of F-dUMP.
4. Elevated levels of ammonia in the blood can be caused by a deficiency of mitochon drial carbamoyl phosphate synthetase or a deficiency of any of the urea cycle enzymes.
These two types of disorders can be distinguished by the presence of orotic acid or related metabolites in the urine.
(a) Why is it possible to determine the basis of hyperammonemia in this way?
(b) Why would a deficiency of cytoplasmic carbamoyl phosphate synthetase not cause hyperammonemia? What problems would such an enzyme deficiency cause? How would you treat a patient who has a deficiency in cytoplasmic carbamoyl phosphate synthetase?
NUCLEOTIDE BIOSYNTHESIS
451
5. The degradation of thymine yields b-aminoisobutyrate, as shown in the figure below, which can be converted to succinyl CoA and then degraded in the citric acid cycle. What cofactors are needed to convert b-aminoisobutyrate to succinyl CoA?
O
K
C
HN
CJCH3
OKC
CH
N
H
Thymine
O
K
C
H
J
HN
C J
CH3
OKC
CH
2
N
H
Dihydrothymine
H
H
J
O
K H NJCJ
NJ
CH JCJ
C
2
2
K
J
J
O:
O
CH3 N - Carbamoylisobutyrate
H
J
O
K HN ;+CO +;H NJ CH JCJ
C
4
2
3
2
J
J
O:
CH3
b-Aminoisobutyrate
6. You wish to prepare 14C-labeled purines by growing bacteria in a medium containing a suitably labeled precursor. The only precursors available are amino acids that are all uniformly labeled to the same specific activity per carbon atom. Which of the amino acids would you use to obtain purine rings that are labeled to the highest specific activity?
7. 6-Mercaptopurine (6-MP) can be converted to the corresponding nucleotide 6 thioinosine-5′-monophosphate (tIMP) through the purine salvage pathway. tIMP can then be converted to 6-thioguanine nucleotides (6-TNG) or methylated to form MetIMP. Methotrexate (MTX) increases incorporation of 6-TNGs through the salvage pathway (Bokkerink et al., Hematol. Blood Transf. 33(1990):110–117). Both 6-MP and MTX are clinically useful anticancer agents and have been used together for many years in the treatment of childhood leukemia in part because of the synergistic effect they have on each other.
452
CHAPTER 25
SH
J
N
N
N
N
6-Mercaptopurine
(a) Briefly describe the salvage reactions required to convert 6-MP to the correspon ding nucleotide.
(b) What step in the de novo biosynthesis of purines is likely to be inhibited by tIMP?
(c) How could the presence of MTX increase the incorporation of 6-TNG into DNA and RNA?
8. Hydroxyurea, a potent chelator of ferric ions, has been shown to interfere with DNA synthesis, and it is used as an antitumor agent. What is a likely target enzyme for hydroxyurea?
O
K
HOHNJCJ NH3
Hydroxyurea
9. Methotrexate, a folate antagonist, interferes with nucleic acid biosynthesis. Would you expect it to inhibit purine or pyrimidine biosynthesis or both processes? Explain.
10. Nucleoside phosphorylases catalyze the interconversion of bases and nucleosides through the following reactions:
Ribose 1-phosphate + base ribonucleoside + Pi
or
Deoxyribose 1-phosphate + base deoxyribonucleoside + Pi
The equilibrium constant for each of these reactions is close to 1.
(a) The pathway for the incorporation of radioactive thymine into bacterial DNA in cludes a step catalyzed by nucleoside phosphorylase. It has often been observed that the incorporation of thymine into DNA is enhanced when deoxyadenosine or deoxyguanosine is added to the medium. Can you explain this observation? Why might deoxyguanosine be preferable to deoxyadenosine?
(b) In cells that cannot carry out de novo synthesis of IMP, inosine can be utilized to produce IMP but only through an indirect salvage route because of the absence of inosine kinase. Suggest an alternative pathway for the formation of IMP from inosine. Among the enzymes you will need are nucleoside phosphorylase and phosphoribomutase, which isomerizes ribose 1-phosphate to ribose 5-phosphate.
11. Many multivitamin preparations contain nicotinamide. Most mammalian cells contain cytosolic enzymes that convert nicotinamide directly to NAD+. What other substrates are required for the formation of NAD+ from nicotinamide? How could PRPP and ATP be used as sources of those substrates?
12. Hypoxanthine-guanine phosphoribosyl transferase (HGPRT), a salvage enzyme of nu cleotide metabolism, uses 5′-phosphoribosylpyrophosphate (PRPP) to convert hypox anthine to IMP and guanine to GMP. A deficiency of this enzyme can lead to an increased level of purine synthesis, excess formation of uric acid, and hyperuricemia, or gout.
(a) How might a deficiency in HGPRT stimulate purine synthesis?
(b) Under what conditions might one expect a deficiency of hypoxanthine-guanine phosphoribosyl transferase to affect the rate of pyrimidine nucleotide synthesis? How could you estimate the rate of pyrimidine nucleotide synthesis in humans?
13. In mammals, the committed step for pyrimidine synthesis is catalyzed by carbamoyl phosphate synthetase, while in bacteria, the committed step is the formation of N-carbamoylaspartate, catalyzed by aspartate transcarbamoylase.
(a) Account for these differences in mammals and bacteria.
(b) Bacterial carbamoyl phosphate synthetase is only partially inhibited by UMP. Why?
14. The synthesis of deoxythymidylate can proceed not only from dUMP but also from dCMP. The route from dCMP begins with the formation of dCDP from CDP, catalyzed by ribonucleotide reductase, followed by dephosphorylation to dCMP, and the deamination of dCMP to form dUMP, catalyzed by dCMP deaminase.
DCMP + H
+
2O + H+ dUMP + NH3 dCMP deaminase is an allosteric enzyme that is stimulated by dCTP and inhibited by dTTP. Account for these effects and relate them to the regulation of ribonucleotide reductase by deoxynucleoside triphosphates.
ANSWERS TO PROBLEMS
1. The clustering of two or more enzymes (active sites) in a single polypeptide chain en sures that their synthesis is coordinated and helps assure that they will assemble into a coherent complex. Also, the proximity of the active sites means that side reactions are minimized as substrates are channeled from one active site to another. Finally, a multifunctional complex with covalently linked active sites is likely to be more stable than a complex formed by noncovalent interactions.
2. (a) Adenosine is phosphorylated to AMP, with ATP serving as the phosphoryl donor, in a reaction carried out by a specific nucleoside kinase. The conversion of AMP to ADP through the action of a specific nucleoside monophosphate kinase is accomplished, with ATP again utilized as a phosphate donor. Ribonucleotide reductase catalyzes the reduction of ADP to dADP, which is then converted to dATP by nucleotide diphosphokinase.
(b) High concentrations of dATP displace ATP from the overall activity site on ribonu cleotide reductase, which lowers the rate of synthesis of all four deoxyribonucleoside diphosphates. This, in turn, leads to a depletion of deoxyribonucleoside triphosphates, which are the substrates for DNA synthesis.
3. Methotrexate blocks the regeneration of tetrahydrofolate from dihydrofolate, which is produced during the synthesis of thymidylate. The failure to regenerate tetrahydrofolate means that those biochemical reactions in the cell that depend on one-carbon metabolism cannot be carried out. One of the products of tetrahydrofolate is methylenete 454
CHAPTER 25 trahydrofolate, which is used as a substrate by thymidylate synthetase and is required for inhibition of the enzyme by F-dUMP. A deficiency of methylenetetrahydrofolate means that F-dUMP cannot irreversibly inactivate thymidylate synthetase. Conversely, F-dUMP prevents the formation of dihydrofolate, thereby abolishing the adverse effects caused by the depletion of tetrahydrofolate in the cell.
4. (a) The presence of orotic acid, a precursor of pyrimidines, in the urine suggests that carbamoyl phosphate synthesized in mitochondria is not utilized there. Instead, carbamoyl phosphate enters the cytosol, where it stimulates an increase in the rate of synthesis of precursors of pyrimidines, including orotic acid. An excess of carbamoyl phosphate arises in mitochondria whenever any of the urea cycle enzymes are deficient. Such a condition will lead to hyperammonemia, as well as to the accumulation of carbamoyl phosphate. Although a deficiency in mitochondrial carbamoyl phosphate synthetase leads to hyperammonemia, it cannot lead to an accumulation of mitochondrial carbamoyl phosphate and therefore does not stimulate pyrimidine synthesis in the cytosol.
(b) Cytoplasmic carbamoyl phosphate synthetase is involved primarily in the pathway for pyrimidine synthesis, not for the assimilation of ammonia. Recall that, in the cytosol, the substrate for the formation of carbamoyl phosphate is glutamine, not ammonia. A deficiency of carbamoyl phosphate synthesis in the cytosol would cause a depletion of pyrimidines. Such a deficiency is treated by administration of uracil or uridine, which are precursors of UMP and CMP.
5. The transamination of b-aminoisobutyrate to form methylmalonate semialdehyde re quires pyridoxal phosphate as a cofactor. This reaction is similar to the conversion of ornithine to glutamate g-semialdehyde. Then NAD+ serves as an electron acceptor for the oxidation of methylmalonate semialdehyde to methylmalonate. The conversion of methylmalonate to methylmalonyl CoA requires coenzyme A. The final reaction, in which methylmalonyl CoA is converted to succinyl CoA, is catalyzed by methylmalonyl CoA mutase, an enzyme that contains a derivative of vitamin B12 as its coenzyme.
6. Examination of the pathway for purine synthesis shows that only glycine is incorporated intact into the purine ring at the C-4 and C-5 positions. Therefore, glycine is a good choice as the radiolabeled precursor. Serine can also be considered because it is a precursor of glycine and the ultimate donor of C-1 groups to tetrahydrofolate, and activated tetrahydrofolate derivatives participate in two reactions in the formation of purines.
Whether serine is a better choice than glycine depends on the relative amounts of the two unlabeled amino acids in the cell.
7. (a) 6-Mercaptopurine is converted to the mononucleotide through the action of hypoxanthine-guanine phosphoribosyl transferase (HGPRT), which uses PRPP to add 5′-phosphoribose to the purine ring. The resulting compound is 6-thioinosine5′-monophosphate, an analog of IMP.
(b) IMP (and presumably its analog 6-MP) inhibits Gln-PRPP aminotransferase, which catalyzes the committed step of purine biosynthesis. The 6-MP metabolite Me-tIMP may be involved in the inhibition of Gln-PRPP aminotransferase as well (Stet et al., Biochem. Journal 304(1994):163–168).
(c) MTX inhibits folate-dependent enzymes in the de novo purine biosynthetic path way leading to an accumulation of PRPP. This increase in substrate availability for the salvage pathway leads to a greater conversion of 6-MP into 6-TNG and there fore a higher level of incorporation of 6-TNGs into DNA. Also, since 6-MP is a substrate for HGPRT—see part (a)—it can compete with endogenouse purine bases for HGPRT leading to a decrease in the synthesis of AMP and GMP.
8. Hydroxyurea inhibits ribonucleotide reductase. By sequestering ferric ions, hydroxyurea destabilizes the organic free radical in the R2 subunit of the enzyme. The inhibition of enzyme activity leads to a depletion of deoxyribonucleoside diphosphates, which are normally converted to deoxyribonucleoside triphosphates, the substrates for DNA synthesis.
9. Methotrexate and aminopterin, a similar compound, are analogs of dihydrofolate (DHF) and inhibitors of dihydrofolate reductase, an enzyme that converts DHF to tetrahydrofolate (THF). The thymidylate synthase reaction converts N5,N10-methylenetetrahydrofolate to DHF in the process of methylating dUMP to form dTMP. In the presence of one of the inhibitors, this reaction functions as a sink that reduces the THF level of the cell by converting THF to DHF. Since THF derivatives are substrates in two reactions of purine metabolism and one of pyrimidine metabolism, both pathways are affected by the inhibitor.
10. (a) Deoxyribonucleosides such as deoxyadenosine can be converted to the free base and deoxyribose 1-phosphate by nucleoside phosphorylase. Increased levels of deoxyribose 1-phosphate are then available for the formation of deoxythymidine from thymine in the reverse reaction catalyzed by nucleoside phosphorylase.
Deoxyguanosine might be preferable to deoxyadenosine because the conversion of elevated levels of deoxyadenosine to dAMP and then to dATP could lead to the inactivation of ribonucleotide reductase, which is sensitive to the concentration of dATP.
(b) Inosine is cleaved to produce hypoxanthine and ribose 1-phosphate through the action of nucleoside phosphorylase; note that inorganic phosphate is required for this reaction. Hypoxanthine-guanine phosphoribosyl transferase converts free hypoxanthine to IMP by condensation with PRPP. PRPP can be derived from ribose 1phosphate in two steps: (1) the conversion of ribose 1-phosphate to ribose 5-phosphate, which is catalyzed by phosphoribomutase; and (2) the formation of PRPP from ribose 5-phosphate and ATP, which is catalyzed by PRPP synthetase.
11. Both ribose phosphate and an AMP moiety must be added to nicotinamide in order to convert it to NAD+. The ribose phosphate is derived from PRPP when a phosphoribosyl transferase catalyzes the formation of nicotinamide ribonucleotide or nicotinamide mononucleotide. The final step utilizes ATP, which serves as an adenyl donor for the formation of the dinucleotide NAD+.
12. (a) When active, HGPRT consumes PRPP as it catalyzes the synthesis of GMP and IMP.
Decreased flux through this reaction raises the steady-state level of PRPP, thereby increasing the activity of PRPP amidotransferase, which catalyzes the initial step in purine synthesis. Increased activity may make the amidotransferase resistant to feedback inhibition by AMP and GMP, the end products of the purine biosynthetic pathway.
(b) As noted above, decreased activity of HGPRT increases the concentration of PRPP.
This increases the rate of pyrimidine synthesis at the orotate phosphoribosyl transferase reaction, if PRPP levels are normally subsaturating for that enzyme. Orotate incorporation into nucleotide pools or into nucleic acids would give a reasonable estimate of de novo pyrimidine nucleotide synthesis.
456
CHAPTER 25
13. (a) Bacteria use a single form of carbamoyl phosphate synthetase not only for the syn thesis of pyrimidines but also for the synthesis of arginine. Arginine biosynthesis begins with glutamate and includes formation of citrulline from ornithine and carbamoyl phosphate, a pathway that resembles urea formation in mammals. Two forms of carbamoyl phosphate synthetase, a cytoplasmic form for pyrimidine synthesis and a mitochondrial form for arginine and urea synthesis, are employed in mammals. Because the two pathways are compartmentalized, the formation of carbamoyl phosphate in the cytosol is regarded as the committed step for pyrimidine synthesis.
(b) UMP does not completely inhibit bacterial carbamoyl phosphate synthetase because that inhibition would interfere with arginine production.
14. An increase in dCTP levels signals that the cell has ample deoxynucleotides for DNA syn thesis and that there is a need for thymidylate synthesis. An increase in dTTP levels signals that the activity of thymidylate synthase can be decreased, and the inhibition of dCMP deaminase by dTTP reduces the input of dUMP into the pathway. While ribonucleotide reductase is subject to regulation by other deoxynucleotides, it is not subject to allosteric regulation by dCTP. Instead it appears that regulation of dCMP deaminase provides a second control point for the generation of deoxynucleotides in the cell.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. Glucose + 2 ATP + 2 NADP+ + H +
2O
PRPP + CO2 ADP + AMP + 2 NADPH + H+
This equation is the summation of the following conversions: glucose
glucose
6-phosphate
ribose 5-phosphate PRPP (see Chapter 16 in the text).
2. Glutamine + aspartate + CO +
+
2
2 ATP + NAD+ orotate + 2 ADP + 2 Pi glutamate + NADH + H+ 3. (a, c, d, e) PRPP, (b) carbamoyl phosphate
4. PRPP and formylglycinamide ribonucleotide accumulate because they are reactions 1 and 4 in the first stage of purine biosynthesis (Figure 25.7 in the text). If these glutaminerequiring amidotransferase reactions are inhibited, the precursor molecules will accumulate. However, since the synthesis of formylglycinamide requires that PRPP be converted to phosphoribosylamine, probably only PRPP will accumulate significantly.
5. dUMP + serine + NADPH + H+ dTMP + NADP+ + glycine
This is the summation of reactions shown in Figure 25.14 in the text. Note that the coenzyme, tetrahydrofolate, is regenerated; it acts catalytically and does not appear in the equation.
6. There is a deficiency of N10-formyltetrahydrofolate (see Figure 25.7 in the text).
Sulfanilamide inhibits the synthesis of folate by acting as an analog of p-aminobenzoate, one of the precursors of folate.
7. PRPP is the activated intermediate in the synthesis of (a) phosphoribosylamine in the de novo pathway of purine formation, (b) purine nucleotides from free bases by the salvage pathway, (c) orotidylate in the formation of pyrimidines, (d) nicotinate ribonucleotide, (e) phosphoribosyl-ATP in the pathway leading to histidine, and (f) phosphoribosylanthranilate in the pathway leading to tryptophan.
NUCLEOTIDE BIOSYNTHESIS
457
8. (a) Cell A cannot grow in a HAT medium, because it cannot synthesize dTMP either from thymidine or from dUMP. Cell B cannot grow in this medium, because it cannot synthesize purines by either the de novo pathway or the salvage pathway. Cell
C can grow in a HAT medium because it contains active thymidine kinase from cell B (enabling it to phosphorylate thymidine to dTMP) and hypoxanthine-guanine phosphoribosyl transferase from cell A (enabling it to synthesize purines from hypoxanthine by the salvage pathway).
(b) Transform cell A with a plasmid containing foreign genes of interest and a func tional thymidine kinase gene. The only cells that will grow in a HAT medium are those that have acquired a thymidylate kinase gene; nearly all these transformed cells will also contain the other genes on the plasmid.
9. These patients have a high level of urate because of the breakdown of nucleic acids.
Allopurinol prevents the formation of kidney stones and blocks other deleterious consequences of hyperuricemia by inhibiting the formation of urate.
10. Though often termed binding constants, the values given are really dissociation constants.
Therefore, to calculate the free energy of binding, one uses the reciprocals of the values given. Thus, for the wild type,
1
DG°′ = −1 3 . 6 log = −1 3 . 6 ×
10.15
×
−11
7
10
= −13 8 . kcal / mol Similar calculations for Asn 27 and Ser 27 will give the results in the answer in the text.
11. IMP is the product of the pathway for de novo purine biosynthesis and the precursor of AMP and GMP. With the de novo pathway for purines not working effectively, it would be helpful to stimulate the salvage pathway, perhaps with a diet that is rich in nucleotides that would then be a source of the preformed purine bases hypoxanthine, adenine, and guanine.
12. N1 in the purine ring of IMP, AMP, ATP, GMP, and GTP will be labeled, following the scheme below: COO:
O :OOC
K COO:
N
N
J
N*
H :OOC
J
+
D
H
H
NJ
H N
2
J H N N
:
+
OOC
N*H
ribose-P
2
3
ribose-P
15N*-aspartate
5-aminoimidazole 4-carboxylate
D
ribonucleotide
fumarate
O
O
K
K
N
N H*N 6
1
5
7
H N*
2
8
J
2
4
H
3
D
D
N
J
J
N
N
9
H N
2
ribose-P
ribose-P
*N1-IMP
458
CHAPTER 25
13. Allopurinol is an analog of hypoxanthine in which the N and C atoms at positions 7 and 8 are interchanged. Xanthine oxidase will hydroxylate C2 of allopurinol in similar manner to its normal reaction with hypoxanthine, but this C-2 hydroxylation of allopurinol gives the new inhibitor:
OH
OH
OH
H
H
J6
6J
6J
N
N
1
N
N
5
7
8
1
5
7
8
1
5
7
8
J
2
4
H
2
4
N
2
4
N
3
3
3
H N N
N
N
9 J H N
9 J
HO
N
9 J
H
H
H
Hypoxanthine
Allopurinol
New inhibitor
14. For the production of glycinamide ribonucleotide, an acyl phosphate (anhydride) inter mediate is formed by reaction of ATP with a carboxylic acid of an amino acid (glycine), whereas in the production of guanylate (GMP) a phosphoryl ester intermediate is formed by reaction of ATP with an enol alcohol on the purine ring of the nucleotide.
Glycinamide ribonucleotide. Step 1.
H
H
H
H
=
;
O:
;
OPO
H N
+
ATP D
H N
3
+
ADP
3
3
K
K
O
O
glycine
glycine acyl phosphate
Step 2.
; H N
3
CH
=
2
O POCH
NH
H
3
2
2
H
OK
=
=
;
OPO
+
H N
3
D O POCH
NH
3
2
+
P
3
i
K
O
HO
OH
5-phosphoribosyl-1-amine
HO
OH
glycinamide
ribonucleotide
Guanylate. Step 1.
O
O
K
H
K
H
J
N
J
N
N
N
JH
G
JH
+
ATP D ADP
K
O
N
N
K
HO
N
N
+
J
ribose-P
J
ribose-P
O
xanthylate tautomeric enol form
K H J N
N
JH
=
J O PO
N
N
3
J
ribose-P
NUCLEOTIDE BIOSYNTHESIS
459
Step 2.
O
O K
NH
K
2
H J N
H
N
+
J
D
H
glutamate (Glu) ;
O:
=
H N J
3
K
O PO
N
N
3
J
+
ribose-P
O
O
glutamine
K H J N
N
J
JH
H N
2
=
J O PO
N
N
3
J
J
ribose-P
H
D
HOPO =
3
O
K H J N
N
JH
J H N N
N
2
J
ribose-P
guanylate (GMP)
15. The reaction involves a dehydration and ring closure. The amino group that was intro duced from aspartate and that will become N-1 in inosinate (see problem 12, above) should be activated (at an enzyme active site) to make a nucleophilic attack on the nearby formyl carbonyl carbon to close the six-membered ring and give a tetrahedral intermediate. The tetrahedral intermediate can then lose water to render the six-membered ring aromatic and generate the inosinate (IMP):
O
O
K
K H N N
2
HN
N
JH
D
JH
H
J
H
N
J
N
N
N
K
J
O
J
:
O
J
ribose-P
J
ribose-P
H
H;
H
5-formyamidoinidazole
D
4-carboxamide
ribonucleotide
H O
2
O
K
N
HN
JH
J
H
N
N
inosinate (IMP)
J
ribose-P
16. (a) Cyclic-(5′-3′)-AMP and cyclic GMP influence many intracellular processes.
(b) ATP is the prototype energy-storage molecule with a high phosphoryl-group trans fer potential.
460
CHAPTER 25 (c) ATP is a phosphate donor for generating glucose-6-phosphate and fructose-1,6-bis phosphate.
(d) Flavin adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD+) participate in the production of acetyl-CoA (“active acetate”) from pyruvate by the pyruvate dehydrogenase complex.
(e) NADH and FADH2 are reduced molecules that have a high electron transfer po tential.
(f) Synthetic (2′,3′)-dideoxynucleoside triphosphates serve as chain terminators for DNA sequencing. (The four natural deoxynucleoside triphosphates—dGTP, dCTP, dATP, and dTTP—are the monomers that are precursors for chain elongation.)
(g) The thymine analogue 5-fluorouracil (converted to 5-fluorodeoxyuridylate in vivo) is a potent anticancer drug.
(h) ATP and CTP reciprocally regulate the activity of aspartate transcarbamoylase.
17. In vitamin B12 deficiency, methyltetrahydrofolate cannot donate its methyl group to ho mocysteine to regenerate methionine. Because the synthesis of methyltetrahydrofolate is irreversible (text, p. 675), the cell’s tetrahydrofolate ultimately will be converted into this form. No formyl or methylene tetrahydrofolate will be left for nucleotide synthesis.
Pernicious anemia illustrates the intimate connection between amino acid metabolism and nucleotide metabolism. The metabolism of fatty acids that have odd numbers of carbons also will be affected because methylmalonyl-CoA mutase requires vitamin B12 for the production of succinyl-CoA. A further connection is that methylmalonyl-CoA mutase also is involved in the degradation of valine and isoleucine.
18. The cytosolic level of ATP in liver falls and that of AMP rises above normal in all three conditions. The excess AMP is degraded to urate. See C. R. Scriver, A. L. Beaudet, W .S.
Sly, and D. Vale (eds.), The Metabolic Basis of Inherited Disease, 6th ed. (McGraw-Hill, 1989), pp. 984–988, for an illuminating discussion.
19. Succinate can be converted to oxaloacetate by the citric acid cycle. The oxaloacetate can then be transaminated to yield asparate, a key precursor of pyrimidines. The carbons of aspartate then will label positions 4, 5, and 6 in the pyrimidine rings:
:
O
O
O
K H O
NADH
H;
K
+
2
H;
H NAD;
NH
CH
J
N 4
CH
2
2
5
G
G
6
K
J
H
O
N
OK N
K
J COO:
J
H
H O:
N-carbamoylaspartate
20. (a) The ADP from muscle contraction is a ready source of additional ATP for additional contraction. Half of the ADP can be converted immediately to ATP at the expense of the other half (being converted to AMP).
(b) The reactants and products have the same number of high-energy phosphate bonds.
The interconversion of (2 ADP) with (ATP + AMP) therefore is essentially isoenergetic.
(c) In the reaction 2 ADP ATP + AMP, the removal of one of the products (AMP) will shift the equilibrium to the right and favor the production of additional ATP.
(d) By first removing and then replacing AMP, the cycle buys time (at the expense of GTP) until aerobic metabolism can “catch up” and reconvert available AMP as well as ADP back into ATP.
CHAPTER 2
The Biosynthesis
of Membrane Lipids and Steroids
6
This chapter describes the biosynthesis of membrane lipids, steroids, and other im portant lipid molecules, such as bile salts, vitamin D, and polymers of isoprene units. As background material for this chapter, you should review the earlier chap ters on cell membranes (Chapters 2 [Section 2.3.2] and 12) and fatty acid metabolism (Chapter 22), paying particular attention to the structure and properties of lipids and the central role of acetyl CoA in the metabolism of lipids. Chapter 26 begins with a discussion of the formation of triacylglycerols, phosphoglycerides, and sphingolipids from the simple precursors glycerol 3-phosphate, fatty acyl CoAs, polar alcohols, for example, choline, serine, and sugars. The text then describes the synthesis of cholesterol from acetyl CoA via the important intermediate isopentenyl pyrophosphate. The regulation of a key enzyme in the biosynthetic pathway as well as other modes of regulation of cholesterol metabolism is also outlined. Cholesterol is the precursor of bile salts as well as steroid hormones. The cholesterol and triacylglycerols synthesized in the liver and intestines are transported by lipoproteins to peripheral tissues. Cholesterol from dietary sources is moved from the intestine to the liver by lipoproteins. Therefore, the classification, the properties, and the mechanisms by which the lipoproteins deliver lipids to cells are discussed next. Finally, the text describes the synthesis of steroid hormones and vitamin D from cholesterol and introduces a variety of isoprenoid lipids that are derived from isopentenyl pyrophosphate.
461
462
CHAPTER 26
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Name the three major lipid-based components of biological membranes (Chapter 12).
2. Explain the biological significance of cholesterol.
Phosphatidate Is a Common Intermediate in the Synthesis
of Phospholipids and Triacylglycerols (Text Section 26.1) 3. Describe the roles of phosphatidate, glycerol 3-phosphate, lysophosphatidate, and diacylglycerol (DAG) in the synthesis of triacylglycerols and phospholipids.
4. List the primary biological functions of triacylglycerols and phospholipids.
5. Contrast the biosynthesis of phosphoglycerides in bacteria and mammals. Note the sig nificance of CDP–diacylglycerol and CDP–choline or CDP–ethanolamine, the activated precursors in these biosyntheses.
6. Describe the biosyntheses of phosphatidyl serine, phosphatidyl ethanolamine, phosphatidylcholine, and phosphatidyl inositol.
7. Restate the physiologic roles of phosphatidyl inositol and its degradation products, in-ositol 1,4,5-trisphosphate and diacylglycerol.
8. Compare the structures and biosynthetic pathways of the glyceryl ether phospholipids, including platelet-activating factor and plasmalogens, with those of the glyceryl ester
phospholipids.
9. Summarize the steps in the biosynthesis of sphingosine from palmitoyl CoA and serine.
10. Outline the synthesis of sphingomyelin, cerebrosides, and gangliosides from sphingosine.
Note the use of activated sugars and acidic sugars.
11. Provide examples of how sphingolipids confer diversity on lipid structure and function.
12. Discuss the general degradation pathway of gangliosides and the biochemical basis of Tay-Sachs disease and respiratory distress syndrome.
Cholesterol Is Synthesized from Acetyl Coenzyme A in Three Stages
(Text Section 26.2) 13. Describe the physiologic roles of cholesterol.
14. List the major stages in cholesterol biosynthesis and give the key intermediates.
15. Compare the synthetic paths leading from acetyl CoA to mevalonate and to the ketonebodies. Note the role of 3-hydroxy-3-methylglutaryl CoA reductase (HMG CoA reductase) as the major regulatory enzyme in cholesterol biosynthesis.
16. Describe the conversion of mevalonate into isopentenyl pyrophosphate.
17. Outline the condensation reactions leading from isopentenyl pyrophosphate to squalene.
Describe the mechanisms of these condensation reactions.
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
463
18. Discuss the cyclization of squalene and the formation of cholesterol from lanosterol. Note the role of O2 in the formation of cholesterol.
The Complex Regulation of Cholesterol Biosynthesis Takes Place
at Several Levels (Text Section 26.3) 19. List the sources of cholesterol and outline the mechanisms of regulation of cholesterol biosynthesis.
20. List the various classes of lipoproteins together with their lipid and protein components.
Describe their lipid transport functions.
21. Summarize the steps in the delivery of cholesterol to cells via the low-density-lipoprotein(LDL) receptor. Discuss the regulation of cellular functions by the LDL pathway.
22. Describe the proposed domain structure of the LDL receptor derived from the primary sequence of this protein. Define the term mosaic protein.
23. Discuss the biochemical defects of the LDL receptor that result in familial hypercholes-terolemia.
24. Summarize approaches used to reduce serum cholesterol.
Important Derivatives of Cholesterol Include Bile Salts
and Steroid Hormones (Text Section 26.4) 25. Describe the physiological roles and the general structures of the bile salts.
26. List the five major classes of steroid hormones, their physiological functions, and their sites of synthesis. Outline their biosynthetic relationships.
27. Give the numbering scheme for the carbon atoms of cholesterol and its derivatives, and distinguish between the a- and b-oriented groups and cis or trans ring fusions.
28. Describe the hydroxylation reactions involving cytochrome P450. Indicate the role of these monooxygenase reactions in steroid biosynthesis, the detoxification of xenobiotic compounds,
and the generation of carcinogens.
29. Describe the synthesis of pregnenolone from cholesterol, the conversion of pregnenolone into progesterone, and the subsequent reactions leading to cortisol and aldosterone.
30. Outline the synthesis of androgens and estrogens from progesterone.
31. Discuss the synthesis and the physiological role of vitamin D.
32. Give examples of biomolecules that contain isoprene units.
SELF-TEST
Introduction 1. Which of the following are components of biological membranes?
(a) free fatty acids (d) phospholipids (b) sphingolipids (e) cholesterol (c) triacylglycerols (f)
proteins
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CHAPTER 26
Phosphatidate Is a Common Intermediate in the Synthesis
of Phospholipids and Triacylglycerols 2. Which of the following reactions are significant sources of glycerol 3-phosphate that is used in lipid synthesis?
(a) reduction of dihydroxyacetone phosphate (b) oxidation of glyceraldehyde 3-phosphate (c) phosphorylation of glycerol (d) dephosphorylation of 1,3-bisphosphoglycerate (e) reductive phosphorylation of pyruvate
3. Match the lipids in the left column with the major synthetic precursors or intermediates listed in the right column.
(a) triacylglycerol (1) phosphatidate (b) phosphatidyl ethanolamine (bacteria) (2) diacylglycerol (c) phosphatidyl ethanolamine (mammals) (3) acyl CoA (4) glycerol 3-phosphate (5) CDP-diacylglycerol (6) CDP-ethanolamine 4. Explain the role of the CDP derivatives in the synthesis of phosphoglycerides.
5. Calculate the number of “high-energy” phosphate bonds that are expended in the for mation of phosphatidyl choline from diacylglycerol and choline in mammals.
6. Which of the following is a common reaction used for the formation of phosphatidyl ethanolamine in bacteria?
(a) decarboxylation of phosphatidyl serine (b) reaction of CDP-ethanolamine with a diacylglycerol (c) demethylation of phosphatidyl choline (d) reaction of ethanolamine with CDP-diacylglycerol (e) reaction of CDP-ethanolamine with CDP-diacylglycerol 7. Which of the following is a lipid with a signal-transducing activity?
(a) phosphatidyl choline (b) phosphatidyl serine (c) plasminogen activator (d) phosphatidyl inositol 4,5-bisphosphate (e) phospholipase A2 8. For the lipid classes listed in the left column, select the characteristic structural compo nents or properties from the right column.
(a) glyceryl ester phospholipids (1) two long hydrocarbon chains (b) plasmalogens (2) acetyl group (c) platelet-activating factor (3) phosphate group (4) ether linkage (5) a,b-double bond (6) glycerol group (7) long fatty acyl chain (8) relatively high solubility in water 9. Which of the following phospholipases would you expect to cleave the R1-containing chain from the phospholipid shown in Figure 26.1?
(a) phospholipase A1 (b) phospholipase A2 (c) phospholipase C (d) phospholipase D (e) none of the above FIGURE 26.1 A phospholipid.
H
H
O H CJOJCKCJR
2
1
K
J R JCJOJCJH
O
2
+
J
K H CJOJPJOJCH JCH JN(CH )
2
2
2
3 3
J O: Synthesis of sphingolipids
10. Which of the following is NOT a precursor or intermediate in the synthesis of sphin gomyelin?
(a) palmitoyl CoA (b) lysophosphatidate (c) CDP-choline (d) acyl CoA (e) serine
11. Match the lipids in the left column with the appropriate activated precursors from the right column.
(a) sphingomyelin (1) acyl CoA (b) ganglioside (2) CDP-choline (c) phosphatidyl serine (3) CDP-diacylglycerol (4) CMP-N-acetylneuraminate (5) UDP-sugar 12. In which compartment of the cell does ganglioside GM2 accumulate in Tay-Sachs pa tients? What is the biochemical defect?
Cholesterol Is Synthesized from Acetyl Coenzyme A in Three Stages 13. From the following compounds, identify the intermediates in the synthesis of cholesterol and list them in their proper sequence: (a) geranyl pyrophosphate (b) squalene (c) isopentenyl pyrophosphate (d) mevalonate (e) cholyl CoA (f) farnesyl pyrophosphate (g) lanosterol (a) Both involve 3-hydroxy-3-methylglutaryl CoA (HMG CoA).
(b) Both require NADPH.
(c) Both require the HMG CoA cleavage enzyme.
(d) Both occur in the mitochondria.
(e) Both occur in liver cells.
15. Select the appropriate characteristics from the right column for the three stages in the synthesis of cholesterol in the left column.
(a) mevalonate to isopentenyl (1) releases PPi
pyrophosphate
(2) requires NADPH (b) isopentenyl pyrophosphate (3) requires O2 to squalene (4) releases CO2 (c) squalene to cholesterol (5) requires ATP 16. Yeast cells growing aerobically synthesize sterols and incorporate them into membranes.
However, under anaerobic conditions yeast cells do not survive unless they are provided with an exogenous source of sterols. Explain the metabolic basis for this nutritional requirement.
The Complex Regulation of Cholesterol Biosynthesis Takes Place
at Several Levels 17. The key step in cholesterol biosynthesis is the conversion of 3-hydroxy-3-methylglutaryl CoA to mevalonate. Which of the following are ways in which this reaction can be modulated?
(a) covalent modification HMG CoA reductase through phosphorylation (b) controlling the rate of translation of the mRNA encoding HMG CoA reductase (c) controlling the rate of transcription of the gene encoding HMG CoA reductase (d) proteolytic degradation of HMG CoA reductase (e) deletion and duplication of the gene encoding HMG CoA reductase 18. Match the appropriate components or properties in the right column with the lipopro teins in the left column.
(a) chylomicron (1) contains apoprotein B-100 (b) VLDL (2) contains apoprotein B-48 (c) LDL (3) contains apoprotein A (d) HDL (4) transports endogenous cholesterol esters (5) transports dietary triacylglycerols (6) transports endogenous triacylglycerols (7) is degraded by lipoprotein lipase (8) is taken up by cells via receptor mediated mechanisms (9) is a precursor of LDL (10) may remove cholesterol from cells 19. Which of the following events occur in the LDL pathway in fibroblasts? Place them in their proper sequential order.
(a) breakdown of LDL in lysosomes (b) endocytosis of LDL along with LDL receptors (c) degradation of LDL receptors in lysosomes (d) binding of LDL to LDL receptors (e) return of LDL receptors to the plasma membrane THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
467
20. Exons in the gene for the LDL receptor give rise to structurally diverse domains. What is the likely function of the cysteine-rich amino-terminal domain, which contains a cluster of negatively charged side chains?
(a) carbohydrate binding (d) growth-factor binding (b) membrane attachment (e) clathrin binding (c) Ca+ binding (f)
structure stabilization
21. Explain how LDL regulates the cholesterol content in fibroblasts.
22. Assume that LDL is produced normally in a patient but that the apoprotein B-100 do main that recognizes the receptor is functionally defective, which prevents the binding of LDL to its receptor. What outcome would this defect have on cholesterol metabolism in peripheral cells?
Important Derivatives of Cholesterol Include Bile Salts and Steroid Hormones 23. The physiological roles of bile salts include which of the following?
(a) They aid in the digestion of lipids.
(b) They aid in the digestion of proteins.
(c) They facilitate the absorption of sugars.
(d) They facilitate the absorption of lipids.
(e) They provide a means for excreting cholesterol.
24. Which of the following are common features in the structures of cholesterol and glyco cholate?
(a) Both have three hydroxyl groups.
(b) Both contain four fused rings.
(c) Both have a hydrocarbon side chain.
(d) Both contain a carboxylate group.
(e) Both contain double bonds.
(f) Both contain a sulfur atom.
25. Explain the structural characteristics of bile salts that make them effective biological detergents.
26. For the sterol structure in Figure 26.2, answer the following: FIGURE 26.2 A sterol.
H COH
2
J
CKO
HO
. J. .OH
J
J
J
K
O (a) Name this sterol.
(b) From what is it synthesized via three hydroxylation reactions?
(c) How many fewer carbon atoms does it have than cholesterol?
(d) Its concentration will be diminished if there is a deficiency of 21-hydroxylase— true or false? Explain why.
468
CHAPTER 26
27. Hydroxylation reactions involving cytochrome P450 have which of the following char acteristics?
(a) They require a proton gradient.
(b) They involve electron transport from NADPH to O2.
(c) They activate O2 by binding it to adrenodoxin.
(d) They transfer one oxygen atom from O2 to the substrate and form water from the other oxygen atom.
(e) They occur in adrenal mitochondria and liver microsomes.
28. Explain how foreign aromatic compounds are detoxified and excreted by mammals.
Describe a possible deleterious effect of this process.
29. Match the steroid hormones in the left column with the characteristics in the right col umn that distinguish them from one another.
(a) aldosterone (1) has 18 carbon atoms (b) estrogen (2) has 19 carbon atoms (c) testosterone (3) has 21 carbon atoms (4) contains an aromatic ring (5) contains an aldehyde group at C-18 30. Name the principle form of excreted cholesterol.
31. Which of the following statements about active vitamin D are INCORRECT?
(a) It has the same fused ring system as cholesterol.
(b) It requires hydroxylation reactions for its synthesis from cholecalciferol.
(c) It is important in the control of calcium and phosphorus metabolism.
(d) It can be synthesized from cholesterol in the presence of UV light.
(e) It can be derived from the diet.
32. Which of the following lipids does not contain isoprene units?
(a) coenzyme Q (b) carotene (c) vitamin K (d) arachidonate (e) phytol side chain of chlorophyll ANSWERS TO SELF-TEST 1. b, d, e, f. Answers (a) and (c) are incorrect because neither neutral fats nor free fatty acids appear in membranes.
2. a, c. Reduction of dihydroxyacetone phosphate is the primary route.
3. (a) 1, 2, 3, 4 (b) 1, 3, 4, 5 (c) 1, 2, 3, 4, 6. In mammals phosphatidyl ethanolamine can also be formed through an exchange reaction of ethanolamine with phosphatidyl serine.
4. CDP-diacylglycerol and the CDP-alcohols are activated intermediates that allow the for mation of phosphate ester bonds in phosphoglycerides, a process that is otherwise highly exergonic. ATP supplies the energy to form these compounds. Recall that UDP-sugars are used in a similar manner in the synthesis of carbohydrates (see text, pp. 589–590).
5. Summing the individual reactions:
Choline+ATPDphosphorylcholine+ADP
Phosphorylcholine+CTPDCDP–choline+PPi
CDP–choline+diacylglycerolDCMP+phosphatidyl choline
Choline+ATP+CTP+diacylglycerolDADP+PPi+CMP+phosphatidyl choline Two high-energy bonds (from ATP and CTP) are directly consumed in these reactions.
In addition, pyrophosphate is hydrolyzed by pyrophosphatase, driving the net reaction farther to the right. A total of three high-energy bonds would be consumed to regenerate ATP and CTP from ADP and CMP.
6. a
7. d
8. (a) 1, 3, 6, 7 (b) 1, 3, 4, 5, 6, 7 (c) 2, 3, 4, 6, 8 9. e. Phospholipases are specific for ester bonds; therefore, none will cleave the ether bond on the C-1 carbon of the plasmalogen. Phospholipases A2, C, and D cleave specific ester linkages: A2 cleaves the R2-containing chain to release the fatty acid, C cleaves the phophodiester bond to produce the R3-phosphate, and D cleaves the phosphodiester bond to produce the R3-alcohol. If the phospholipid had been a glycerol ester phospholipid, for example, phosphatidyl ethanolamine, phospholipase A1 would have cleaved the ester bond to yield the R1-contining fatty acid. Phospholipase C hydrolyzes phosphatidyl inositol 4, 5-bisphosphate to produce inositol 1, 4, 5-trisphosphate and diacylglycerol, which are intracellular second messengers (Section 15.2).
10. b
11. (a) 1, 2 (b) 1, 4, 5 (c) 1, 3 12. The degradative enzymes for gangliosides are located in lysosomes; therefore, ganglio side GM2 will accumulate in the lysosomes of Tay-Sachs patients. The enzyme that removes the terminal sugar, GalNAc, from the ganglioside is deficient in these people.
13. All the compounds given are intermediates in the biosynthesis of cholesterol except for (e) cholyl CoA, which is a catabolic derivative of cholesterol and a precursor of bile salts.
The proper sequence is d, c, a, f, b, g.
14. a, e 15. (a) 4, 5 (b) 1, 2 (c) 2, 3 16. A key intermediate in the biosynthesis of cholesterol and related sterols is squalene, an open-chain isoprenoid hydrocarbon. It is converted to squalene 2,3-epoxide, which in turn is converted to lanosterol. The conversion of squalene to the 2,3-epoxide is catalyzed by a monooxygenase, and molecular oxygen is a required component for this reaction. Under anaerobic conditions, yeast cells cannot synthesize sterols because they lack oxygen, a substrate for the monooxygenase reaction.
17. a, b, c, d 20. c, f 21. The main source of cholesterol for cells outside the liver and intestine is from circulating LDL. Cholesterol released during the degradation of LDL suppresses the formation of new LDL receptors, thereby decreasing the uptake of exogenous cholesterol by the cell.
22. A defect in apoprotein B-100 that prevents the binding of LDL to the cell-surface re ceptor would result in the stimulation of the synthesis of endogenous cholesterol and LDL receptors and a decrease in the synthesis of cholesterol esters via the ACAT reaction. Indeed, the cellular and physiological consequences of such a mutation may be similar to those seen in familial hypercholesterolemia.
23. a, d, e 24. b, e 25. Bile salts are effective detergents because they contain both polar and nonpolar re gions. They have several hydroxyl groups, all on one side of the ring system, and a polar side chain that allow interactions with water. The ring system itself is nonpolar and can interact with lipids or other nonpolar substances. Bile salts are planar, amphipathic molecules, in contrast with such detergents as sodium dodecyl sulfate (text, p. 84), which are linear.
26. (a) cortisol (b) progesterone (c) six (d) true. A deficiency of 21-hydroxylase will impair hydroxylation at C-21 of proges terone, which will prevent the normal synthesis of cortisol and mineralocorticoids from progesterone.
27. b, d, e 28. In mammals, foreign aromatic molecules are hydroxylated by the cytochrome P450 dependent monooxygenases that are present in the endoplasmic reticulum of the liver cells. The hydroxylated derivatives are more water-soluble and have functional groups for the attachment of very polar substances, such as glucuronate, that allow them to be excreted in urine. The action of the cytochrome P450 system sometimes converts potential carcinogenic compounds into highly carcinogenic derivatives.
29. (a) 3, 5 (b) 1, 4 (c) 2 30. The water-soluble bile salt glycocholate is a major cholesterol breakdown product.
31. a
32. d
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
471
PROBLEMS
1. Why is synthesis of cholesterol de novo dependent on the activity of ATP-citrate lyase?
2. An infant has an enlarged liver and spleen, cataracts, and anemia and exhibits general retardation of development. Mevalonate is found in the urine. Investigation reveals a deficiency of mevalonate kinase, which catalyzes the formation of 5-phosphomevalonate from mevalonate.
(a) Why is urinary excretion of mevalonate consistent with a deficiency of meval onate kinase?
(b) How would a deficiency of mevalonate kinase affect cholesterol synthesis in this infant?
(c) What level of activity, relative to normal, would you expect to find for HMG CoA reductase in cells isolated from the infant? Briefly explain your answer.
3. Normally, most of the bile acids that are secreted into the intestine undergo reabsorption and are returned to the liver. Cholestyramine is a positively charged resin that binds bile acids in the intestinal lumen and prevents their reabsorption.
(a) To examine the effects of cholestyramine on LDL metabolism, two fractions of LDL were prepared: one was covalently labeled on tyrosine residues with 125I; the other was similarly labeled with 131I and treated with cyclohexanedione, which interferes with LDL binding to the LDL receptor. When rabbits were given cholestyramine, hepatic uptake of 125I-labeled LDL was enhanced relative to normal, whereas the uptake of 131I-labeled LDL was unchanged relative to that in rabbits that had not been given cholestyramine. Briefly explain the relationship between the action of cholestyramine and LDL uptake in the liver.
(b) The administration of cholestyramine usually results in a 15 to 20% reduction in levels of circulating LDL, whereas the administration of a combination of cholestyramine and mevinolin (lovastatin) can often yield a 30 to 40% reduction. Why?
4. The presence of apoprotein E in lipoproteins enables them to be taken up by hepatic cells. Provide a brief explanation for each of the following observations made of a person with a deficiency in apoprotein E synthesis.
(a) elevated levels of plasma triacylglycerols and cholesterol, coupled with the presence of chylomicron remnants and IDL. These latter particles persist in the bloodstream much longer than in normal people.
(b) abnormally low levels of LDL in the blood (c) abnormally high levels of LDL receptors in liver cells (d) a marked reduction in levels of circulating chylomicron remnants and IDL when the diet is low in cholesterol and fat 5. Pregnant women often have increased rates of triacylglycerol breakdown and, as a result, have elevated levels of ketone bodies in their blood. Why do they also often exhibit an increase in plasma lipoprotein levels?
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CHAPTER 26
6. Hopanoids are pentacyclic molecules that are found in bacteria and in some plants. As an example, a typical bacterial hopanoid is shown below. Organisms that make hopanoids use a pathway similar to that for cholesterol synthesis. The biosynthetic pathway for hopane includes the formation of squalene, followed by more steps to form the final product itself, a C30 compound. Hopane is similar to lanosterol (text, p. 726), but lacks the hydroxyl group.
H C
3
J
J
H C J
3
CH3
J
CH
3
J
H C
3
H C
3
H CJCH
3
J
CH2
J
CH2
J
HCJOH
J
HOJCH
J
HCJOH
J
CH2
J
CH OH
2
Bacteriohopanetetrol (a) How many molecules of mevalonic acid are required for the synthesis of hopane?
(b) Squalene can undergo concerted cyclization to form hopane in a reaction that is catalyzed by a unique type of squalene cyclase. The reaction is initiated by a proton and does not require oxygen. Compare this step with the formation of lanosterol from squalene. Why could it be argued that the synthesis of hopanoids preceded the synthesis of sterols in evolution?
7. Your colleague has discovered a compound that is a very powerful inhibitor of HMG CoA reductase, and she has evidence that the drug will completely block the synthesis of mevalonate in liver. Why is this compound unlikely to be useful as a drug?
8. The liver is the site of the synthesis of plasma phospholipids and lipoproteins. Rats main tained on a diet deficient in choline often develop fat deposits in liver tissue. How could choline deficiency be related to this aberration in lipid metabolism?
9. Among the sugar residues found in a blood group ganglioside is fucose. Experiments uti lizing isolated Golgi membranes and ribonucleoside triphosphates show that fucose can be incorporated into the ganglioside only when GTP is available. What is the role of GTP in fucose incorporation?
10. Suppose a cell is deficient in phosphatidate phosphatase, which catalyzes the formation of diacylglycerol from phosphatidate. What effects on lipid metabolism would you expect?
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
473
11. Desmolase is involved in the synthesis of pregnenol (text, p. 735).
(a) Would you expect virilization among patients who have desmolase deficiency?
(b) Why is enlargement of the adrenal glands common among such patients?
12. In the adult form of Gaucher’s disease, glucosylcerebrosides accumulate in liver, spleen, and bone-marrow cells. Although the common galactosylceramides and their derivatives are found in the tissues of affected patients, accumulations of galactosylcerebrosides or their metabolites are not found, nor do ceramides accumulate. What enzyme activity is probably deficient in patients with Gaucher’s disease?
13. Cells of the adrenal cortex have very high concentrations of LDL receptors. Why?
14. At low concentrations of phospholipid substrates in water, the reaction catalyzed by a phospholipase occurs at a rather low rate. The reaction rate accelerates when the concentrations of the phospholipid substrates increase to the point that micelles are formed.
How is this property of phospholipases related to their activity in the cell?
15. Glucagon has been shown to reduce the activity of HMG CoA reductase. Why is this ob servation consistent with the overall effect of glucagon on cellular metabolism?
16. People who have elevated levels of LDL in their serum can be treated in a number of ways. These include restriction of dietary intake of cholesterol, ingestion of positively charged resin polymers that inhibit intestinal reabsorption of bile salts, and administration of lovastatin, a competitive inhibitor of 3-hydroxy-3-methylglutaryl CoA reductase.
(a) Briefly explain how each of these treatments reduces serum LDL levels.
(b) None of the above treatments should be used for patients who are homozygous for a defect in LDL receptors? Why?
17. Currently there are two established methods for dealing with hypercholesterolemia in patients homozygous for LDL receptor deficiency.
(a) The first is plasma apheresis, whereby the plasma and blood cells of a patient are separated in a continuous flow device, and the plasma is passed over a column that removes lipoproteins containing apoprotein B-100. Which lipoproteins are removed by this procedure, and how could their removal lower plasma cholesterol levels?
(b) A more extreme method of dealing with extreme hypercholesterolemia in FH ho mozygotes is liver transplantation. The rationale for this procedure is based on the observation that over 70% of total body LDL receptors are in the liver. In the small group of patients who have undergone liver transplants, LDL cholesterol levels are substantially reduced. In one particular case, the rate of LDL turnover increased about threefold, and one patient became responsive to lovastatin after the transplant. Why did increased LDL turnover and lovastatin response indicate that the transplantation procedure was successful?
18. Provide a physiological rationale for each of the following responses to an increase in the rate of transport of unesterified cholesterol into a mammalian cell.
(a) stimulation of the synthesis of cholesteryl oleate esters.
(b) suppression of the activity of HMG CoA reductase.
(c) suppression of the synthesis of LDL receptors.
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CHAPTER 26
19. Glycerol phosphate acyltransferase can convert 3,4-dihydroxybutyl-1-phosphonate, an analog of glycerol 3-phosphate, to diacylbutyl-1-phosphonate, which is an analog of phosphatidate. Would this analog be more likely to inhibit triglyceride synthesis or CDP–diacylglycerol synthesis? Briefly explain your answer. See the figure below.
H CJOH
2
J
HOJCJH
O
J
K CH JCH JPJO:
2
2
J O: 3, 4-Dihydroxybutyl-1-phosphate 20. During the uptake of LDL by a liver cell, LDL-receptor protein complexes are internal ized by endocytosis. The endosomes then fuse with lysosomes, where protein components of LDL are hydrolyzed to free amino acids, while cholesterol esters are hydrolyzed by a lysosomal acid lipase. The LDL receptor itself is not affected by lysosomal enzymes.
(a) Briefly describe what would happen to cholesterol metabolism in a cell deficient in lysosomal acid lipase.
(b) Why is it important that LDL receptors are not degraded by lysosomal enzymes?
21. Glycerol kinase catalyzes the conversion of free glycerol into glycerol 3-phosphate, using ATP as a phosphoryl donor. Although liver tissue has high levels of the enzyme, the activity of glycerol kinase in adipose tissue is low. How do these differences contribute to the balance between carbohydrate and triglyceride metabolism in mammals?
22. Niemann-Pick disease is an inherited disorder of sphingomyelin breakdown due to a deficiency of sphingomyelinase. This enzyme, found in all tissues and easily assayed in white blood cells collected from a patient, converts sphingomyelin to ceramide and phosphocholine. Phosphocholine is highly soluble in water, while sphingomyelin is more soluble in chloroform. Assuming that you can obtain sphingomyelin labeled with 14C in any desired carbon atoms, design an assay that would allow you to confirm a diagnosis of Niemann-Pick disease.
ANSWERS TO PROBLEMS
1. ATP-citrate lyase catalyzes the formation of acetyl CoA in the cytosol (text, p. 622). Acetyl CoA is used for the synthesis of HMG CoA, which gives rise to mevalonate for the synthesis of cholesterol in the cytosol.
2. (a) A deficiency of mevalonate kinase activity means that mevalonate cannot be uti lized as a precursor of 5-phosphomevalonate. If no other pathways can use mevalonate, its concentration in the liver will increase until it spills into the blood and then, in turn, into the urine. Furthermore, because the activity of HMG CoA reductase is increased in this infant—see answer (c)—the rate of mevalonate synthesis will be stimulated.
(b) You would expect the rate of cholesterol synthesis to be depressed because the path way is blocked at the step in which 5-phosphomevalonate is formed.
(c) You would expect to find a higher-than-normal level of HMG CoA reductase activ ity. A depressed rate of cholesterol synthesis lowers the amount of cholesterol in the cell. HMG CoA reductase activity increases because inhibition of its synthesis and its activity by cholesterol is reduced.
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
475
3. (a) The experiments show that rabbits given cholestyramine have higher rates of re moval of LDL from the blood and that hepatic uptake of LDL depends on the ability of the lipoprotein to bind to the LDL receptor. One explanation for this is that cholestyramine interferes with the return of bile acids to the liver, stimulating the synthesis of more bile acids from cholesterol. An increased demand for cholesterol stimulates the synthesis of LDL receptors, which take up more cholesterol-containing LDL particles from the blood.
(b) Although cholestyramine stimulates the hepatic uptake of cholesterol-contain ing LDL, it has no direct effect on cholesterol synthesis de novo in the liver.
Mevinolin inhibits HMG CoA reductase, thereby depressing the rate of cholesterol biosynthesis. The subsequent requirement for cholesterol leads to a further increase in the number of LDL receptors, which in turn can take up more LDL from the circulation.
4. (a) Chylomicrons, chylomicron remnants, and IDL normally contain apoprotein E. A deficiency of the apoprotein means that hepatic uptake of chylomicron remnants and IDL is impaired, so these particles persist in the circulation. Because both of these types of particles contain triacylglycerols and cholesterol, circulating levels of these compounds are also elevated.
(b) Both chylomicron remnants and IDL particles serve as precursors of VLDL in the liver. When the uptake of the VLDL precursors by hepatic tissue is impaired by an apoprotein E deficiency, the rate of synthesis and export of VLDL particles is reduced. Since VLDL are LDL precursors in circulation, LDL are reduced.
(c) As discussed in answer (b), VLDL synthesis in the liver is impaired. Additional LDL receptors are synthesized because their synthesis is no longer repressed by VLDLderived cholesterol.
(d) A diet low in cholesterol and fat will reduce the rate of formation of chylomicrons, which are precursors of chylomicron remnants.
5. Increased levels of ketone bodies, such as acetoacetate, imply that the levels of acetyl CoA and HMG CoA, both precursors of cholesterol, are also elevated. Cholesterol synthesis is stimulated by an increase in the availability of these substrates. The subsequent decreased demand for dietary cholesterol results in an elevation in cholesterol-containing lipoproteins.
6. (a) Mevalonic acid, a six-carbon compound, is a precursor of isopentenyl pyrophos phate (IPP), which contains five carbon atoms. IPP serves as the basic unit for the formation of squalene, a 30-carbon compound. Six molecules of IPP are needed for the synthesis of a molecule of squalene, which is in turn the precursor of hopane.
Thus, six molecules of mevalonic acid are required.
(b) Aerobic processes, such as the synthesis of sterols, probably evolved later than anaerobic processes and only after free oxygen became available. Thus, the synthesis of hopane from squalene, an anaerobic process, probably preceded the synthesis of sterols, such as lanosterol, from squalene.
7. The synthesis of mevalonate is required not only for the synthesis of cholesterol but also for the synthesis of a number of other important compounds derived from isopentenyl pyrophosphate, including ubiquinone (CoQ), an important component of the electron transport chain. Therefore, the complete blockage of mevalonate synthesis, even if adequate cholesterol is available in the diet, would be ill-advised.
8. Choline, which is ordinarily supplied by the diet and is synthesized only to a limited extent in mammals, is a constituent of phosphatidyl choline, an important component of membranes and lipoproteins. A deficiency in dietary choline, which could not be completely 476
CHAPTER 26 replaced by the methylation of phospatidyl ethanolamine, could interfere with the synthesis and export of lipoproteins like VLDL, which is a carrier of triacylglycerols to peripheral tissues. Failure to export fats such as triacylglycerols leads to their accumulation in the liver.
9. Nucleotide sugars, such as UDP-glucose, serve as donors during the incorporation of sugar residues into gangliosides. In this case, it appears that the donor of fucose residues is GDP-fucose, which is synthesized from fucose and GTP.
10. You would expect to see reduced rates of synthesis of triacylglycerols, which use diacyl glycerols as acceptors of activated acyl groups. In addition, phosphatidyl choline synthesis is dependent on the availability of diacylglycerols as acceptors of choline phosphate from CDP-choline.
11. (a) You would not expect desmolase deficiency to lead to virilization. An increase in androgen production, which causes virilization in both males and females, is due to elevated levels of 17 a-hydroxyprogesterone. The pathway from cholesterol to 17 a-hydroxyprogesterone includes a step catalyzed by desmolase, which cleaves the bond between C-20 and C-22 in 20a, 22b dihydroxycholesterol to form pregnenolone (see the text, p. 735). A deficiency of desmolase would therefore decrease the rate of androgen synthesis.
(b) Pregnenolone is a precursor of the glucocorticoids, which exert feedback control on the activity of the adrenal cortex. Desmolase deficiency leads to diminished production of glucocorticoids. Failure of the normal feedback mechanism leads to increased ACTH production and to enlargement of the adrenal glands.
12. Because glucosylcerebrosides accumulate but galactosylcerebrosides do not, you would suspect that the defect involves ganglioside breakdown rather than ganglioside synthesis. The defect involves the step that removes glucose from the cerebroside to yield free ceramide, or N-acyl sphingosine. The enzyme that carries out this step is a glycosyl hydrolase; it is also called b-glucosidase.
13. Cells of the adrenal cortex utilize cholesterol for the synthesis of a number of steroid hor mones, including cortisol. Although these cells can themselves synthesize cholesterol, it is often also necessary for additional cholesterol to be obtained from plasma lipoproteins.
A high concentration of LDL receptors enables cortical cells to take up LDL, which contains cholesterol, rapidly.
14. In the cell, a phospholipase would most often encounter substrates that are part of an aggregate, such as those phospholipids found in membranes. Thus, the enzyme should be expected to function at a higher rate with aggregates or assemblies of lipid molecules because their local concentrations would be higher than if they were individually free in solution.
15. The presence of glucagon is a signal that carbohydrate and triacylglycerol catabolism is needed to generate energy in the organism. Under such conditions, one would expect biosynthetic reactions to be suppressed because energy charge is low. Low energy charge means high AMP levels that would activate an AMP-dependent protein kinase leading to the phosphorylation of HMG CoA reductase.
16. (a) Restricting the level of dietary cholesterol lowers the input of exogenous cholesterol into lipoproteins, so that fewer LDL molecules are present in serum. Compounds that inhibit bile salt reabsorption from the gut stimulate additional synthesis of bile acids from cholesterol in the liver, decreasing concentrations of the sterol in liver cells and, by causing an increase in the number of receptors on the cell surface, stimulating LDL uptake from the circulation. Finally, mevinolin reduces the rate of cholesterol biosynthesis de novo, which can also stimulate uptake of LDL from the bloodstream. Any of these treatments could help in reducing circulating levels of cholesterol as LDL.
THE BIOSYNTHESIS OF MEMBRANE LIPIDS AND STEROIDS
477
(b) Homozygotes have virtually no functional LDL receptors. Such people are unable to internalize significant amounts of LDL, which means that circulating levels of that lipoprotein are elevated in the blood. In addition, an absence of LDL receptors means that endogenous cholesterol fails to enter the liver cell to suppress de novo synthesis. Dietary restriction could reduce exogenous LDL levels somewhat but would not prevent formation of LDL and other lipoproteins arising from cell turnover of cholesterol. Bile sequestrants of bile salts could cause some depletion of liver cell cholesterol, but again would not stimulate LDL uptake from the circulation. Mevinolin will suppress cholesterol synthesis de novo, but this would again not be compensated for by uptake from the circulation. Homozygotes with two nonfunctional receptor genes are therefore resistant to compounds that inhibit LDL synthesis and stimulate uptake. Thus, none of these measures has a great effect on reducing circulating LDL levels in these patients.
17. (a) Protein B-100 is found in VLDL, IDL, and LDL, and all three lipoproteins are re moved from the plasma. Each contains cholesterol, and, in addition, VLDL and IDL are regarded as precursors of LDL, so that total cholesterol concentration in the blood would be lowered by apheresis. This method can reduce LDL cholesterol levels by 70%. Treatment must be repeated about every two weeks, and the effects of long-term apheresis are problematic.
(b) Increased LDL turnover indicates that the transplanted liver is producing normal LDL receptors that can take up LDL from the circulation and can facilitate its conversion to other lipoproteins. A response to lovastatin also indicates that functional LDL receptors are available to accelerate uptake in response to diminished de novo cholesterol synthesis, which is inhibited by lovastatin. It should be noted that liver transplant operations are hazardous, especially in FH homozygotes who usually have advanced atherosclerosis. The fact that normal liver cells can contribute functional LDL receptors has stimulated interest in gene therapy designed to target a normal LDL gene to liver cells of FH homozygotes.
18. (a) Formation of cholesteryl esters provides the cell with a means of storing cholesterol until it is needed for membrane biosynthesis or other purposes.
(b) Suppression of the activity of HMG CoA reductase leads to a decrease in cholesterol synthesis de novo; this occurs when the cell has sufficient endogenous cholesterol so that it does not need to synthesize the steroid on its own.
(c) Suppression of LDL receptor synthesis leads to a gradual reduction in the number of LDL receptors in the cell, because the receptors undergo a relatively constant rate of degradation. Reduction in LDL receptor levels means that fewer LDL particles will be able to enter the cell, resulting in a decrease in the entry of endogenous cholesterol.
19. The conversion of phosphatidate to a triacylglycerol is initiated by hydrolysis of the phosphate group and the formation of diacylglycerol. The C-P bond in the phosphonate is much less likely to be cleaved by the phosphatase (also known as phosphatidate
phosphohydrolase), so it is likely that triglyceride synthesis could be impaired by the analog. On the other hand, formation of CDP-diacylglycerol involves the formation of an anhydride bond between the phosphates of phosphatidic acid and cytidylic acid (text, p. 717). Since the phosphonate is not cleaved in this reaction, formation of the phosphonyl analog of CDP-diacylglycerol seems possible, or at least the synthesis of normal substrates would not be impaired. Phosphonolipids are known in trace amounts in mammals, but are found more extensively in some invertebrates, including the protozoan Tetrahymena, where they may represent up to 25% of total phospholipid (see D. E. Vance and J. Vance [eds.], Biochemistry of Lipids, Lipoproteins and
Membranes. [Elsevier, 1991], p. 205).
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20. (a) Both the LDL receptor gene and the gene for HMG CoA reductase contain sterol regulatory elements that are responsive to free cholesterol. A reduction in free cholesterol release from the lysosome leads to an increase in LDL receptor production and to increased HMG CoA reductase activity. Both these consequences lead to an increase in cholesterol concentrations in the cell, through an increased rate of LDL entry and accelerated cholesterol synthesis. Lysosomal accumulation of cholesteryl esters and triglycerides can eventually destroy the cell. One form of this disorder, termed Wolman disease, is characterized by liver enlargement, digestive difficulties, and enlargement and deterioration of the adrenal glands. The disease is usually fatal within a year after birth.
(b) LDL receptors, after their release from lysosomes, return to the cell surface, where they take up other LDL particles and bring them back to lysosomes. A round trip takes about 10 minutes. Destruction of LDL receptors by lysosomal enzymes would make it necessary to synthesize the 115-kd glycoprotein at a much faster and energetically wasteful rate.
21. Free glycerol in mammals is produced in adipocytes when triglycerides are converted to free fatty acids and glycerol by hormone-sensitive lipases. Hormone-responsive lipolysis occurs when glucose levels are low and fatty acids are needed as fuels. Under these conditions, the low activity of glycerol kinase in adipocytes prevents unnecessary resynthesis of triacylglycerols from free fatty acids and glycerol 3-phosphate, via phosphatidic acid. Triacylglycerol synthesis is more likely to occur when glucose levels are high.
Adipocytes can then synthesize glycerol 3-phosphate from dihydroxyacetone phosphate produced from glucose during glycolysis. Thus, adipocytes are unable to synthesize triglycerides unless glucose or another suitable carbon source is available.
22. Extracts of white cells from the blood of a person suspected of having a sphingomyeli nase deficiency are incubated in a buffered solution with radioactive sphingomyelin labeled with 14C in the methyl groups of the phosphocholine moiety. Then the incubation mixture is extracted with chloroform. Any radioactive phosphocholine liberated by the enzyme will remain in the upper aqueous layer, while intact radioactive sphingomyelin will be extracted into the lower chloroform layer. Using white cells from patients with Neimann-Pick disease, incubation with cell extracts followed by chloroform extraction results in little or no radioactivity in the aqueous phase, confirming the deficiency of sphingomyelinase activity.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. Glycerol+4 ATP+3 fatty acids+4 H2ODtriacylglycerol+ADP+3 AMP+ 7 Pi+4 H;
One ATP is used in the formation of glycerol 3-phosphate and three ATPs are used to convert three fatty acids to acyl CoAs. The three PPi formed during fatty acid activation are converted to Pi, hence, the total of seven Pi in the equation above.
2. Glycerol+3 ATP+2 fatty acids+2 H2O+CTP+serineD phosphatidyl serine+CMP+ADP+2 AMP+6 Pi+3 H;
The equation above is a summation of equations shown in the text on pages 716–717.
3. (a) CDP-diacylglycerol, (b) CDP-ethanolamine, (c) acyl CoA, (d) CDP-choline, (e) UDP-glucose or UDP-galactose, (f) UDP-galactose, (g) geranyl pyrophosphate 4. (a and b) None, because the label is lost as CO2.
5. (a) No receptor is synthesized.
(b) Receptors are synthesized but do not reach the plasma membrane because they lack signals for intracellular transport or do not fold properly.
(c) Receptors reach the cell surface, but they fail to bind LDL normally because of a defect in the LDL-binding domain.
(d) Receptors reach the cell surface and bind LDL, but they fail to cluster in coated pits because of a defect in their carboxyl-terminal region.
6. Deamination of cytidine to uridine changes CAA (Gln) into UAA (stop).
7. Benign prostatic hypertrophy can be treated by inhibiting the 5 a-reductase. Finasteride, the 4-aza steroid analog of dihydrotestosterone, competitively inhibits the reductase but does not act on androgen receptors. Patients taking finasteride have a markedly lower plasma level of dihydrotestosterone and a nearly normal level of testosterone. The prostate becomes smaller, whereas testosterone-dependent processes such as fertility, libido, and muscle strength appear to be unaffected (see E. Stoner, Steroid Biochem. Molec.
Biol. 37(1990):375–378). Genetic deficiencies of 5a-reductase are discussed by J. E.
Griffin and J. D. Wilson in C. R. Scriver, A. L. Beaudet, W. S. Sly, and D. Valle (eds.), The
Metabolic Basis of Inherited Disease, 6th ed. (McGraw-Hill, 1989), pp. 1919–1944.
CONHC(CH ) 3 3
J
J
J
K
O
N
....
H
H
Finasteride
8. Patients who are most sensitive to debrisoquine have a deficiency of a liver P450 enzyme encoded by a member of the CYP2 subfamily. This characteristic is inherited as an autosomal recessive trait. The capacity to degrade other drugs may be impaired in people who hydroxylate debrisoquine at a slow rate because a single P450 enzyme usually handles a broad range of substrates. See W. B. Pratt and P. Taylor, Principles of Drug Action:
The Basis of Pharmacology, 3d ed. (Churchill Livingstone, 1990), pp. 496–500; and F. J.
Gonzalez and D. W. Nebert. Trends Genet. 6(1990):182.
9. Many hydrophobic odorants are deactivated by hydroxylation. O2 is activated by a cy tochrome P450 monooxygenase. NADPH serves as the reductant. One oxygen atom of O2 goes into the odorant substrate, whereas the other is reduced to water.
10. Propecia effectively lowers the plasma level of dihydrotestosterone (see also problem 7). But dihydrotestosterone is an important embryonic androgen that instigates the development and differeniation of the male phenotype. Pregnant women who had contact with Propecia therefore would risk developmental abnormalities for their unborn male children.
11. The various P450 isozymes probably serve two major categories of function in plants.
First, some of them will be important in the detoxification of foreign substances that may arise in the plant’s environment. Second, plants may use P450 enzymes for the synthesis of useful molecules such as toxins to fight pests, or pigments to attract organisms that aid in dispersing pollen or seeds.
12. Individual polymorphisms in some of the P450 isozyme genes likely would alter the rates of metabolic degradation (or conversely activation) or particular clinical drugs.
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Knowledge of the individual differences therefore would help in prescribing appropriately different clinical doses of particular medicines for different individual patients.
13. Phosphorylation of the serine will place a negative charge adjacent to the key histidine side chain, thereby stabilizing the positively charged form of the histidine and preventing (or markedly slowing) the donation of the proton to the thiolate. A key step of the reaction therefore will be inhibited.
14. In a classic monooxygenase fashion, one oxygen from O2 will hydroxylate a substrate methyl group, and the other oxygen from O2 will be reduced to water. The elimination of formyaldehyde will then lead to the products: methylamine and formyaldehyde (see J. Biol. Chem. 271[1996]:27321–27329).
H C
J CH
3
J
3
N + O + NADPH + H; D
2
J
H H C
J CH OH
3
J
2
N + H O + NADP;
2
J
H H C
J H
3
J
N
O
+
K
J
H
H
H
15. (a) Cholesterol feeding has no effect on the amount of mRNA for HMG-CoA reductase.
(b) The actin mRNA is a positive control to (1) verify that RNA can be effectively re covered from all samples, and (2) allow normalization of the results, if necessary, to correct for variations in the extent of overall RNA recovery from sample to sample.
(c) The amount of HMG-CoA reductase protein is greatly reduced when the animals are fed a cholesterol diet.
(d) Although the amount of specific HMG-CoA reductase mRNA is not affected by cholesterol, the level of HMG-CoA reductase protein decreases to near zero for the cholesterol-fed mice.
(e) Several mechanisms could explain the presence of the specific mRNA and yet the absence of the specific protein that the mRNA encodes: 1. HMG-CoA reductase could be subject to translational control, so that translation of the message and synthesis of HMG-CoA reductase by ribosomes are inhibited by cholesterol. 2.
Alternatively, the protein could be synthesized but then rapidly degraded in the cholesterol-fed mice.
CHAPTER 2DNA Structure, Replication,
and Repair
7
The text returns to the topic of the flow of genetic information and considers the detailed biochemical mechanisms underlying this complex process. Chapter 5 introduced you to DNA and RNA and outlined the storage, duplication, and expression of genetic information. In Chapter 6, the enzymes and techniques used to analyze, construct, and clone DNA were presented. You should review these chapters to prepare for studying Chapter 27. Pay particular attention to DNA structure, the supercoiling of DNA, and DNA polymerase I in Chapter 5 and to DNA ligase in Chapter 6.
Chapter 27 covers the biochemistry of the transmission of genetic information from parent to progeny by describing DNA and enzyme systems that replicate, recombine, and maintain it. The chapter opens with a presentation of the problems that a cell must overcome to duplicate its duplex DNA. The text expands upon earlier coverage of DNA with a more thorough description of the A, B, and Z forms it can assume and the underlying chemical determinants of these structural variations. The text also explains how the sequence-dependent variation of the structure of DNA provides a basis for its sequence-specific recognition by proteins. The previous general description of the biochemistry of DNA polymerases is expanded to explain the roles the template, primer, and metal ions play in their activities. The chemical basis for the fidelity of DNA chain extension by the polymerases is also presented. The helicases that unwind DNA are described next. The text then provides a detailed description of the topology of covalently closed circular DNAs and the topoisomerases that modulate their linking numbers. It then describes the replication fork; replication initiation; and RNA-primed, semidiscontinuous DNA elongation. The mechanisms and role of DNA ligases are also given. The structure and important roles of
481
482
CHAPTER 27 DNA polymerase III in replication is described in detail. The special problems of replication arising from the amount of eukaryotic DNA in a cell and the structure of chromatin are introduced and the nature and functions of the telomeres and telomerase are described.
To provide the new sequences of nucleotides in DNA upon which evolution can act, not only mutation but also recombination between two different DNA molecules occurs. The breakage and joining of fragments of DNA with similar sequences also rearranges gene orders within the chromosome and is a mechanism for repairing damaged DNA and regulating gene expression. Review the material on plasmids and bacteriophages in Chapter 6 and recombination in Section 5.6.2 of Chapter 5 to better understand this section. A description of a key intermediate in recombination, the Holliday junction, and the recombinases that form and resolve it, are presented. A description of mutations—their nature, causes, and consequences— and their repair follows. The chapter concludes with examples of pathological deficiencies of DNA repair in humans, the relationship of repair impairments and mutation to carcinogenesis, and a test system for detecting potential carcinogens through their mutagenic action on bacteria.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Outline the problems facing a cell in creating a duplicate copy of a double-strand DNA molecule. Appreciate that enzymes and DNA-binding proteins play essential roles in solving the challenges of DNA replication.
DNA Can Assume a Variety of Structural Forms (Text Section 27.1) 2. Contrast the structural information provided by the x-ray diffraction analysis of DNA fibers and DNA crystals.
3. Indicate the role of deoxyribose puckering in determining the structural differences between A-DNA and B-DNA helices. Compare the structures of double-strand RNA
and RNA-DNA hybrids with that of the A-DNA helix.
4. Describe the major and minor grooves of the B-DNA helix. Distinguish between the four possible base pairs (A–T, T–A, C–G, and G–C) in terms of the unique arrays of hydrogen
bond acceptors and donors and methyl groups they present in the grooves of the DNA.
5. Explain why local DNA structure depends on base sequence and appreciate how proteins that bind DNA in a sequence-dependent manner exploit this sequence-dependent variation.
6. Describe Z-DNA in terms of the handedness of the helix and the shape of the path traced by its phosphodiester backbone.
DNA Polymerases Require a Primer and a Template (Text Section 27.2) 7. Outline the key features of the reactions catalyzed by DNA polymerases. Define template and primer as they relate to DNA polymerases.
8. Appreciate the common structures and the evolutionary relationships among DNA polymerases.
9. Explain the role of Mg 2; in the reaction catalyzed by DNA polymerases.
10. Account for the fidelity with which a DNA polymerase selects the correct incoming deoxyribonucleotide triphosphate (dNTP) substrate.
11. Relate the 3„D5„ nuclease activity of DNA polymerases to the fidelity of DNA replication.
12. Describe how helicases separate the strands of duplex DNA.
Double-Strand DNA Can Wrap Around Itself to Form Supercoiled Structures
(Text Section 27.3) 13. Define the linking number (Lk) of a circular DNA molecule and relate supercoiling to the electrophoretic and centrifugal mobility of the molecule.
14. Write the equation relating the linking number to the twisting number (Tw) and writhingnumber (Wr) of a DNA topoisomer. Explain how a negative superhelix density can facilitate the unwinding of the helix. Describe the partitioning of the free energy of a negatively supercoiled molecule into Tw and Wr.
15. Describe the three steps of the reaction catalyzed by topoisomerases; distinguish between type I and type II topoisomerases; and describe the substrates, products, and mechanisms of topoisomerase I and II.
DNA Replication of Both Strands Proceeds Rapidly from Specific Start Sites
(Text Section 27.4) 16. Draw a replication fork, and describe the reactions and the movements of the DNA strands that occur during replication.
17. Define continuous replication and discontinuous replication and relate these terms to the leading and lagging strands of replicating DNA. Describe an Okazaki fragment.
18. Describe the function and features of the nucleotide sequence of oriC and note that it is the unique site of bidirectional replication initiation in E. coli. List the proteins that interact with the DNA in this region of the chromosome, and give the reactions they catalyze and functions they serve.
19. Explain the roles of RNA in DNA replication. Describe the primosome, and describe how enzymes form and remove RNA primers from the genome.
20. List the distinctive features of the DNA polymerase III holoenzyme, and describe how an asymmetric dimer of the enzyme, along with other proteins, coordinates the synthesis of the leading and lagging strands of the daughter duplexes. Appreciate the structural complexity of the replication machinery.
21. Summarize the reactions and identify the proteins at the replication fork that carry out DNA replication.
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22. List the substrates and outline the reaction mechanisms of the DNA ligases.
23. Describe the special problems arising from DNA length and the cell cycle in eukary otes. Explain the roles of multiple replication origins and the telomeres in eukaryotic DNA replication.
24. Describe how telomerase makes DNA of defined sequence in the absence of an external template.
Double-Strand DNA Molecules with Similar Sequences Sometimes Recombine
(Text Section 27.5) 25. Describe the Holliday model for homologous recombination. Explain how resolution of the Holliday junction intermediate can form different recombinant DNA products.
26. Outline the reactions catalyzed by recombinases and note their mechanistic similarity to the topoisomerases.
Mutations Involve Changes in the Basic Sequence of DNA (Text Section 27.6) 27. Distinguish among substitution, insertion, and deletion mutations, and relate them to changes in DNA. Distinguish between transversion and transition substitution mutations and explain the origin of mutations that alter the reading frame in translation.
28. Give examples of some chemical mutagens and their mechanisms. Describe the structure of the pyrimidine dimer formed by ultraviolet light. Outline the mechanisms for the enzymatic repair of damaged DNA.
29. Explain why thymine rather than uracil is used in DNA.
30. Describe xeroderma pigmentosum and hereditary nonpolyposis colorectal cancer, their causes, pathological consequences, and relationships to DNA repair.
31. Relate Huntington disease to triplet expansions within DNA.
32. Relate mutagens and carcinogens. Outline the Ames Salmonella mutagen assay.
SELF-TEST
DNA Can Assume a Variety of Structural Forms 1. Which of the following statements about the double-helical structure of DNA are correct?
(a) It has adenine paired with thymine and cytosine paired with guanine.
(b) It can assume many different forms including A-DNA, B-DNA, and Z-DNA.
(c) In B-DNA, all the hydrogen-bonded base pairs lie in a plane perpendicular to the helix axis.
(d) Different puckering of their ribose residues occurs in A-DNA and B-DNA.
(e) It is rigid and static.
2. Figure 27.1 shows the structures of the adenine and thymine (A–T) and guanine and cy tosine (G–C) base pairs in B-DNA. Use the figure to answer the following questions:
(a) Which hydrogen-bond donors or acceptors of the A–T base pair are in the major DNA STRUCTURE, REPLICATION, AND REPAIR
485
groove?
(b) Which hydrogen-bond donors or acceptors of the G–C base pair are in the minor groove?
(c) Using n to represent heterocyclic ring nitrogen atoms, o for oxy groups, and h for protons, give the patterns of hydrogen-bond acceptors and donors in the major groove for a G–C and a C–G base pair.
(d) Using the same symbols as in (c), give the patterns of hydrogen-bond acceptors and donors in the minor groove for an A–T and a T–A base pair.
(e) Explain the possible biological significance of the unique arrays of hydrogen-bond donors and acceptors in the grooves of B-DNA.
FIGURE 27.1 Watson-Crick base pairs.
H
N N J H . . . O
CH3
7
K
J
8
N 9 5 6
4 5
4
1 N . . . H J N 3
6
3 2
1
2
N
N
K
O
A–T
H
N O J H . . . N
7
K
J
8
N 9 5 6
4 5
4
1 N J H . . . N 3
6
3 2
1
2
N
N
K N J H . . . O
H
G–C
3. X-ray diffraction studies of DNA held at low relative humidity revealed the existence of
A-DNA. Since such arid conditions presumably never occur in the cell, what is the significance of the structure of this DNA?
4. Match the features or characteristics in the right column with the type of DNA helix in the left column.
(a) A-DNA (1) Phosphates in the backbone zigzag (b) B-DNA along the helix.
(c) Z-DNA (2) Formation is favored sequences of alternating purine and pyrimidines.
(3) has a relatively wide and deep major groove
(4) has a right-handed helix (5) has 10.4 base pairs per turn (6) has a structure similar to that of double-strand RNA (7) Strands in the helix have opposite polarities.
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DNA Polymerases Require a Primer and a Template 5. Which of the following statements about DNA polymerases are correct?
(a) They add deoxyribonucleotide units to the 3„-hydroxyl of a primer.
(b) They use the template strand to help choose which deoxyribonucleotide unit to add to the growing DNA chain.
(c) They contain a 3„D5„ nuclease that cleaves phosphodiester bonds of misin corporated deoxyribonucleotides.
(d) They check the size of an incoming deoxyribonucleotide triphosphate (dNTP) to help insure that the correct, complementary choice is made.
(e) They bind one complementary dNTP and add a second complementary dNTP to initiate a new DNA chain.
6. Mg2; serves which of the following functions in the reaction catalyzed by DNA polymerases?
(a) stabilizes the pentacoordinate transition state of the phosphodiester bond formation (b) precipitates inorganic phosphate (Pi) arising from the reaction (c) interacts with the 3„-hydroxyl of the incoming dNTP (d) forms a bridge between the 3„-hyroxyl of the primer and a phosphate in the dNTP (e) stabilizes the negative charge on the departing pyrophosphate, which is derived from the incoming dNTP
7. Why are helicases required during DNA replicaton? Is ATP required for their action?
Double-Strand DNA Can Wrap Around Itself to Form Supercoiled Structures 8. The topological features of circular DNA may affect which of the following?
(a) the electrophoretic mobility of the DNA (b) the sedimentation properties of the DNA (c) its affinities toward proteins that bind to the DNA (d) the susceptibility of the strands of the DNA to unwinding (e) the susceptibility of the DNA to the action of DNA ligase 9. Which of the following statements about DNA molecules that are topoisomers are correct?
(a) They are bound to topoisomerases.
(b) They differ from one another topologically only in that they have different link ing numbers.
(c) They may be separated from one another by electrophoresis.
(d) They have identical molecular weights.
(e) They are topological or spatial isomers.
10. Which of the following statements about topoisomerases are correct?
(a) They alter the linking numbers of topoisomers.
(b) They break and reseal phosphodiester bonds.
(c) They require NAD; as a cofactor to supply the energy to drive the conversion of a supercoiled molecule to its relaxed form.
(d) They form covalent intermediates with their DNA substrates.
(e) They can, in the case of a particular type of topoisomerase, use ATP to form nega tively supercoiled DNA from relaxed DNA in E. coli.
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DNA Replication of Both Strands Proceeds Rapidly from Specific Start Sites 11. Match the properties or functions in the right column with the DNA polymerase in the left column.
(a) DNA polymerase I (1) involved in replication (b) DNA polymerase III (2) requires a primer and a template (3) involved in DNA repair (4) makes most of the DNA phosphodi ester bonds during replication (5) removes the primer and fills in gaps during replication
12. Which of the following statements about DNA replication in E. coli are correct?
(a) It occurs at a replication fork.
(b) It starts at a unique locus on the chromosome.
(c) It proceeds with one replication fork per replicating molecule.
(d) It is bidirectional.
(e) It involves discontinuous synthesis on the leading strand.
(f) It uses RNA transiently as a template.
13. Which of the following statements about DNA polymerase III holoenzyme from E. coli are correct?
(a) It elongates a growing DNA chain hundreds of times faster than does DNA poly merase I.
(b) It associates with the parental template, adds a few nucleotides to the growing chain, and then dissociates before initiating another synthesis cycle.
(c) It maintains a high fidelity of replication, in part by acting in conjunction with a subunit containing a 3„D5„ exonuclease activity.
(d) When replicating DNA, it is a molecular assembly composed of at least 10 differ ent kinds of subunits.
14. Explain how the b2 subunit of DNA polymerase III holoenzyme contributes to the pro cessivity of the DNA synthesis machinery.
15. Which of the following statements about DNA ligase are correct?
(a) It forms a phosphodiester bond between a 5„-hydroxyl and a 3„-phosphate in duplex DNA.
(b) It requires a cofactor, either NAD; or ATP, depending on the source of the enzyme, to provide the energy to form the phosphodiester bond.
(c) It catalyzes its reaction by a mechanism that involves the formation of a covalently linked enzyme adenylate.
(d) It catalyzes its reaction by a mechanism that involves the activation of a DNA phos phate through the formation of a phosphoanhydride bond with AMP.
(e) It is involved in DNA replication, repair, and recombination.
16. Why is RNA synthesis essential to DNA synthesis in E. coli?
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17. Match the functions or features related to DNA replication in E. coli listed in the right column with the molecules or structures in the left column.
(a) replication fork (1) Synthesis direction is opposite that of (b) oriC replication fork movement.
(c) lagging strand (2) unwinds strands at the origin of repli (d) leading strand cation in association with dnaA and (e) Okazaki fragment dnaC proteins
(f)
dnaB helicase (3) is synthesized continuously (g) single-strand binding protein (ssb) (4) synthesizes most of the DNA (h) DNA gyrase (5) is synthesized discontinuously (i)
primase
(6) relieves positive supercoiling (j) DNA polymerase III holoenzyme (7) is the locus of DNA unwinding (k) ¯ subunit of DNA polymerase III (8) hydrolyzes ATP to reduce the linking (l) DNA polymerase I number of DNA (m) DNA ligase (9) binds dnaA, dnaB, and dnaC proteins (10) fills in gaps where RNA existed (11) is the point of initiation of synthesis (12) joins lagging strand pieces to each other (13) contains a 5„D3„exonuclease that removes RNA primers (14) is an RNA polymerase (15) performs “proofreading” on most of the DNA synthesized (16) stabilizes unwound DNA (17) uses NAD; to form phosphodiester bonds
18. Why are the antibiotics novobiocin, ciprofloxin, and nalidixic acid, which inhibit DNA gyrase, useful in treating bacterial infections in humans?
19. The duplication of the ends of linear, duplex DNA presents a problem to the replicative machinery of a human cell. What causes the problem and how does the cell overcome it?
Double-Strand DNA Molecules with Similar Sequences Sometimes Recombine 20. Which of the following statements about genetic recombination are correct?
(a) It generates new combinations of genes.
(b) It can move a segment of DNA from one chromosome to another, for example, from a virus to a host cell.
(c) It is mediated by the breakage of DNA and the rejoining of the resulting fragments.
(d) It generates genome sequence variability, upon which natural selection can act.
21. How many strands of DNA are present at the junction of a Holliday junction?
22. Recombinases (a) make and break phosphodiester bonds in a reaction requiring ATP.
(b) pair homologous DNA molecules prior to strand breakage to form a recombination synapse.
(c) form covalent complexes with their DNA substrates.
(d) have a reaction intermediate reminiscent of those of the DNA ligases.
(e) are related to type I topoisomerases by divergent evolution.
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Mutations Involve Changes in the Sequence of DNA 23. Which of the following nucleotide substitutions are transition mutations?
(a) G for A (c) C for T (b) A for C (d) T for G 24. Which of the substitutions in question 23 are transversion mutations?
25. How could the tautomerization of a keto group on a guanine residue in DNA to the enol form lead to a mutation?
26. Explain why most nucleotides that have been misincorporated during DNA synthesis in E. coli do not lead to mutant progeny.
27. Match the type of mutation or physiological consequence in the right column with the appropriate mutagen in the left column.
(a) 5-bromouracil (1) transversion (b) 2-aminopurine (2) transition (c) aflatoxin (3) insertion or deletion (d) acridines (4) translational frameshift (e) nitrous acid (5) block in replication (f)
ultraviolet light
28. What property of DNA allows the repair of some residues damaged through the action of mutagens?
29. Which of the following enzymes or processes can be involved in repairing DNA in E. coli damaged by UV-light-induced formation of a thymine dimer?
(a) DNA ligase seals the newly synthesized strand to undamaged DNA to form the in tact molecule.
(b) The UvrABC enzyme (excinuclease) hydrolyzes phosphodiester bonds on both sides of the thymine dimer.
(c) DNA polymerase I fills in the gap created by the removal of the oligonucleotide bearing the thymine dimer.
(d) The UvrABC enzyme recognizes a distortion in the DNA helix caused by the thymine dimer.
(e) A photoreactivating enzyme absorbs light and cleaves the thymine dimer to re-form two adjacent thymine residues.
30. Given that the base T requires more energy to synthesize than U, and A pairs equally well with U or T, why does DNA likely contain A–T base pairs instead of A–U base pairs?
31. Explain how mutations in genes encoding proteins likely to be involved in DNA repair, such as those defective in xeroderma pigmentosum and hereditary nonpolyposis colorectal cancer, may contribute to the onset of cancer.
32. Explain how some strains of Salmonella are used to detect carcinogens. How is an ex tract from human liver involved in this test?
ANSWERS TO SELF-TEST 1. a, b, d. Answer (c) is incorrect because B-DNA has local variations from the average struc ture, which was observed in DNA fibers. Structures of individual DNA crystals reveal that the hydrogen-bonded base pairs are often twisted and tilted out of the plane that is perpendicular to the helix axis.
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2. (a) In the major groove, N-7 of A is an acceptor, H-6 on the 6-exocyclic (i.e., not in the heterocycle ring) amino group of A is a donor, and O-4 on T is an acceptor.
(b) In the minor groove, N-3 of G is an acceptor, H-2 on the 2-exocyclic amino group of G is a donor, and O-2 on C is an acceptor.
(c) The patterns in the major groove are noh for G–C and hon for C–G.
(d) The patterns in the minor groove are no for A–T and on for T–A, with no donors being present.
(e) The patterns of hydrogen bond donors and acceptors in the major and minor grooves of B-DNA, recognized by complementary hydrogen bond donors and acceptors on the amino acid side chains of proteins, facilitate the sequence-specific interaction of proteins and DNA. Using such interactions proteins can recognize specific sequences in DNA without disrupting the DNA helix. Note that A–T base pairs have two symmetrically related acceptor atoms in the minor groove and thus have less “information” in this groove than do G–C base pairs. The methyl groups on T may also participate in the recognition of DNA through specific hydrophobic interactions with proteins. Finally, a given local DNA sequence gives rise to a particular local DNA shape or conformation. A protein or enzyme that binds sequencespecifically might recognize the shape rather than the sequence per se.
3. Double-strand RNA, RNA–DNA hybrids, and some short sequences of double-strand DNA embedded in B-DNA have structures like that of A-DNA. Thus, knowledge about the helical structure of A-DNA contributes to an understanding of other similar helices that are of physiological importance.
4. (a) 4 ,6, 7 (b) 3, 4, 5, 7 (c) 1, 2, 7. Relevant to answer (2) for (c), the B-DNA to Z-DNA transition would require the complete unwinding of the right-handed helix to form the left-handed one. Negative supercoiling promotes unwinding of B-DNA and thus facilitates the conversion of a segment of B-DNA into Z-DNA.
5. a, b, c, d 6. a, d, e.
7. Duplex B-DNA is a stable molecule at physiological temperature and helicases are re quired to unwind the two strands of the helix so that each can serve as a template for DNA polymerases. Because DNA is so stable, energy is required in the form of ATP hydrolysis to drive helicase action.
8. a, b, c, d. Regarding answer (e), supercoiling is ordinarily a property of covalently closed circular DNA—that is, of DNA in which there are no discontinuities in either strand of the helix. Hence, these molecules are not substrates for DNA ligase, because they lack ends.
9. b, c, d, e. Answer (a) is not correct, because the DNA topoisomer need not necessarily be bound by a topoisomerase.
10. a, b, d, e. Although all topoisomerases break and reseal phosphodiester bonds, an ex ternal energy source is not always required. Relieving the torsional stress in a negatively supercoiled DNA molecule by relaxing it with topoisomerase I is exergonic and requires no energy input, whereas introducing negative supercoils with DNA gyrase is endergonic and must be coupled to ATP hydrolysis. The particular catalytic mechanisms of given topoisomerases determine whether they are coupled to ATP hydrolysis.
11. (a) 1, 2, 3, 5 (b) 1, 2, 4 12. a, b, d. Although not explicitly stated in the text, answer (c) is incorrect because the repli cating E. coli chromosome has two replication forks that synthesize the DNA bidirec tionally from the unique oriC origin. Answer (e) is incorrect because only the lagging strand is synthesized discontinuously. Answer (f ) is incorrect because RNA serves as a primer and not as a template.
13. a, c, d. Answer (b) is incorrect because DNA polymerase III holoenzyme is a highly pro cessive enzyme that synthesizes extensively before dissociating from its template.
14. The b2 subunit forms a torus, with the duplex DNA in its aperture. The b2 ring acts as a sliding clamp that holds the replication machinery on the DNA.
15. b, c, d, e. Answer (a) is incorrect because the enzyme joins a 3„-hydroxyl to a 5„-phos phate. Answer (d) is correct; although not completely described in the text, DNA ligase activity is required to seal the discontinuities in DNA arising during DNA replication, repair, and recombination.
16. Because DNA polymerases are unable to initiate DNA chains de novo, and because they require a primer with a 3„-hydroxyl group, short RNA chains are used as primers to start DNA replication on the leading strand at the origin of replication and to initiate the Okazaki fragments of the lagging strand. RNA polymerases can start RNA chains by adding a nucleotide to an initiating NTP, but they do so with relatively low accuracy.
RNA-initiated DNA chains facilitate high-fidelity replication at the beginning sequences of new chains because they allow DNA polymerase I to replace the RNA with DNA, using the information in the complementary strand and both of its exonucleases in a nick translation reaction (text, p. 760).
17. (a) 7 (b) 7, 9, 11 (c) 1, 5 (d) 3 (e) 1, 3 (f) 2 (g) 16 (h) 6, 8 (i) 14 (j) 4 (k) 15 (l) 1, 10, 13 (m) 12, 17. Answer (5) is not a correct match with (e), because each Okazaki fragment is synthesized continuously.
18. These compounds interfere with the essential helix-destabilizing function of DNA gy rase in bacterial DNA replication. By inhibiting its action, they prevent the gyrase from relieving the positive supercoils that build up ahead of the moving replication fork.
Human cells lack an enzyme similar to gyrase, and thus they are relatively unharmed by these antibiotics.
19. Because DNA polymerases extend their growing polynucleotide chains only in the 5„D3„ direction, and because the two strands of the parental DNA duplex are antiparallel, removal of the RNA primer that is paired with the 3„ end of the parental template DNA would leave an overhanging 3„ DNA strand with no means of having its complement synthesized. Ordinary DNA polymerases are unable to initiate DNA chains de novo. Each round of replication would consequently shorten the DNA because a portion could not be copied. To circumvent this problem, human DNA chromosomes have a segment of repeating G-rich DNA (telomeres) at their ends. In addition, a special enzyme, telomerase, which is an RNA-dependent DNA polymerase (reverse transcriptase) that carries its own RNA template, can, in effect, extend the uncompleted end at each round of replication. The RNA template renews the repeating telomere sequence so that the DNA is not shortened (text, pp.765–766).
20. a, b, c, d
21. Four. The Holliday junction is formed from the four strands of two interacting duplex DNA molecules that, as a result of the initial reactions of recombination, become joined to form one molecule. The Holliday junction is resolved when recombination is completed and two separate duplexes are reformed. Although not mentioned in the text, the products can sometimes have regions of duplex where one strand of DNA is from one parent and the other strand from the other parent, that is, a heteroduplex is formed.
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CHAPTER 27 22. b, c, e. Recombinases have reaction mechanisms similar to those of the topoisomerases in that a covalent enzyme–DNA phosphodiester bond is formed. The bond linking the enzyme to the DNA has a high free energy of hydrolysis and is, thus, capable of resynthesizing the phosphodiester bond broken during its formation. Consequently, recombinases, unlike DNA ligases and DNA polymerases, do not require an external source of energy to form a phosphodiester bond.
23. a, c. Transition mutations are substitutions of a purine for a purine or a pyrimidine for a pyrimidine.
24. b, d. Transversion mutations substitute a purine for a pyrimidine or vice versa.
25. The rare enol tautomer of G could base-pair with a T in the template to allow its incor poration into a growing DNA strand during replication. If the proofreading process missed this erroneous incorporation, the resulting daughter DNA duplex would contain a G–T base pair. During the next round of replication, the T would direct the incorporation of an A into its complementary daughter strand. The final result would be the substitution of an A–T base pair for the original G–C base pair.
26. The proofreading 3„D5„ nuclease of the ¯ subunit of DNA polymerase III holoen zyme removes most of the misincorporated nucleotides that do not form a base pair with the template. The polymerase activity of the holoenzyme then has a second chance to incorporate the correct nucleotide. Additionally, DNA repair systems exist that can detect and repair a mismatched base pair resulting from a misincorporation during synthesis.
27. (a) 2 (b) 2 (c) 1 (d) 3, 4 (e) 2 (f) 5 28. Since DNA is double-strand, damage to one strand of the DNA can often be repaired by using the undamaged complementary strand as a template to direct new incorportion of correct deoxynucleotides in place of the removed incorrect ones.
29. a, b, c, d, e. Although not mentioned, the uvrD protein is a helicase that removes the 12-nucleotide-long oligonucleotide bearing the thymine dimer.
30. C spontaneously deaminates to form U in DNA. This change would lead to a mutation during the next round of replication of the DNA, since U would pair with A changing what was a C–G base pair into an U–A base pair. The repair machinery of a cell that used U normally in its DNA would be unable to distinguish the U in an A–U base pair arising from a C deamination from one formed during “normal” replication. The methyl group on T, which is almost universally used in DNA, distinguishes it from the uracils formed by deamination so that the uracils can be repaired.
31. The inability to effectively repair mutagenic lesions in DNA may lead to their accumu lation. As a consequence, genes regulating cellular proliferation may malfunction, causing cancer.
32. Special strains of Salmonella have been developed to detect substitution, insertion, and deletion mutations in their DNA as a result of exposure to exogenously supplied chemicals. Mutagens can alter the DNA in these strains and thus convert them from auxotrophs, which are unable to grow in the absence of histidine, to prototrophs. The revertants can grow on media lacking histidine and are detected with high sensitivity.
Since there is a correlation between mutagenicity and carcinogenicity, these strains are used as an inexpensive initial test of the carcinogenic potential of a compound. Because animals sometimes metabolize innocuous compounds and convert them to carcinogens, incubation of a suspect chemical with a human liver extract before using the bacterial test can sometimes mimic what would happen to the chemical in vivo. This adjunct to the test expands its capacity to detect potential human carcinogens.
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PROBLEMS
1. The major and minor grooves of B-DNA contain groups that are potential hydrogen-bond donors and acceptors and hence might be important in specific interactions between proteins and DNA. Suppose that a given region of B-DNA is G–C-rich. Will the number of potential hydrogen-bond-forming groups in the major groove and the minor groove differ from the number that would be present if the region were A–T-rich? Explain.
2. Which of the following nucleotide sequences when embedded in longer DNA might allow for the formation of Z-DNA? Explain your answer.
(a) CCTAGTTA (b) CATATATA (c) GCGCGCGT (d) TGAATTCA
3. Many proteins that bind to DNA consist of two a-helical recognition units that are re lated by a twofold axis of symmetry and are separated by a distance of ~34 Å.
(a) Are these proteins likely to bind to double-strand or to single-strand DNA? Explain.
(b) Would the proteins bind preferentially to A-DNA, B-DNA, or Z-DNA? Explain.
(c) Describe how these proteins likely interact with DNA.
4. Suppose that two polynucleotide chains are joined by DNA ligase in a reaction mixture to which ATP labeled with 32P in the a-phosphoryl (innermost) group has been added as an energy source. What products of the reaction would be expected to carry the radioactive label? Explain.
5. Suppose that negatively supercoiled DNA with Lk=23, Tw=25, and Wr=:2 is acted on by topoisomerase I. After one catalytic cycle, what would be the approximate values of Lk, Tw, and Wr?
6. Suppose that negatively supercoiled DNA with Lk=23, Tw=25, and Wr=:2 is acted on by DNA gyrase and ATP. After one catalytic cycle, what would be the approximate values of Lk, Tw, and Wr?
7. What property of DNA polymerase I led to the observation that polA1 mutants of E. coli are more sensitive to ultraviolet light than the wild-type cells? (Hint: The 5„D3„ exonuclease activity of DNA polymerase I can remove damaged nucleotides from DNA as well as destroying the RNA primers on Okazaki fragments.)
8. Suppose that a single-strand circular DNA with the base composition 30% A, 20% T, 15% C, and 35% G serves as the template for the synthesis of a complementary strand by DNA polymerase.
(a) Give the base composition of the complementary strand.
(b) Give the overall base composition of the resulting double-helical DNA.
9. An early and initially attractive mechanism proposed for genetic recombination was the copy-choice model. It suggested that recombination between two parental DNA duplexes occurs during DNA replication when DNA polymerase switches or jumps from one parental duplex to the other so as to produce a recombinant daughter DNA duplex that contains sequences derived from the templates of two different DNA duplexes. The copychoice model is now known to act infrequently. One experimental finding inconsistent with the copy-choice model is the observation that when E. coli bacteria are infected by T4 bacteriophage of two different genotypes—whose DNAs are distinctly marked, one by 32P and the other by bromouracil—recombinant DNAs containing both markers are 494
CHAPTER 27 found under conditions in which DNA synthesis is blocked. How are these findings inconsistent with the copy-choice model, but consistent with the breakage-reunion model for genetic recombination?
10. Relate genetic recombination to exon shuffling, that is, the rearrangement of exons to form new proteins.
11. Suppose that a plasmid with a single origin of replication on its circular chromosome and containing only genes A, B, C, and D begins to replicate rapidly at time t=0. At t=1, there are twice as many copies of genes B and C as there are copies of genes D and A. Is it possible to establish the order of the four genes on the plasmid? Explain.
12. Suppose that a bacterial mutant is found to replicate its DNA at a very low rate. Upon analysis, it is found to have normal levels of activity of DNA polymerases I and III, DNA gyrase, and DNA ligase. It also makes normal amounts of the wild-types of dnaA, dnaB, dnaC, and SSB proteins. The sequence of the oriC region of its chromosome is found to be wild type. What defect might account for the abnormally low rate of DNA replication in this mutant? Explain briefly.
13. Which of the following mutations in a polypeptide chain could have been induced by a single hit of the mutagen 5-bromouracil on the coding strand DNA? (See text, p. 769, and the genetic code on p. 134.)
(a) PheDGlu (c) PheDLeu (b) AspDAla (d) MetDLys 14. Hydroxylamine reacts readily with cytosine to yield a product that base pairs with ade nine. Which of the following mutations could result from the action of a single-hit reaction of hydroxylamine with DNA?
(a) GlnDAsn (c) HisDTyr (b) GluDLys (d) GlyDAsp 15. The drug fluorouracil is used as an anticancer agent. It irreversibly inactivates the en zyme thymidylate synthase. Explain how this treatment retards the growth of tumor tissue. Will the growth of normal cells be affected as well?
16. Mammalian cells of two differing genotypes can be fused together, often in the presence of Sendai virus, to form multinucleate cells (heterokaryons) containing nuclei of both genotypes. When fibroblasts from two patients suffering from xeroderma pigmentosum were fused, the resulting heterokaryons showed no deficiency in DNA repair. What conclusions can be drawn from this observation? Explain.
17. Physical studies on the interaction of the b2 subunit of DNA polymerase III holoen zyme show that the b2 subunit binds much more tightly to circular than to linear DNA molecules.
(a) Propose an explanation for this observation.
(b) What do you think would happen if the circular DNA were treated with a double strand hydrolyzing endonuclease?
18. Eukaryotic DNA can be highly methylated at the C-5 position of cytosine. The degree of methylation is inversely correlated with gene expression (text, pp. 878–879). Although the exact role of C-5 methylation in gene expression is not known, it is known that these C-5-methylated cytosines can cause mutations. How?
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19. Acyclovir is an antiviral agent used to reduce the pain and promote the healing of skin lesions resulting from adult chicken pox. Its structure is shown in Figure 27.2.
FIGURE 27.2 Structure of acyclovir.
O
K
N H N N
N H N
2
O O H (a) What nucleoside does this drug resemble?
(b) Why does the drug need to be administered in dephosphorylated form?
(c) Acyclovir has very few side effects because it inhibits DNA replication only in herpes infected cells. This is because all herpes viruses encode a thymidine kinase gene that is able to activate the drug. What is this activating reaction?
(d) Once acyclovir is activated, how does it inhibit DNA replication?
(e) Why are cells uninfected by virus relatively unaffected by acyclovir?
(f) The herpes virus can become insensitive to acyclovir therapy by mutations in either of two genes. What might they be?
ANSWERS TO PROBLEMS
1. Both G–C and A–T base pairs of B-DNA have one hydrogen bond donor and two hy drogen bond acceptors in the major groove. However, a G–C pair has one donor and two acceptors in the minor groove, whereas an A–T pair has only two acceptors. Thus, a G–C-rich region on DNA differs from an A–T-rich region with respect to the number of hydrogen-bond–forming groups in the minor groove (text, p. 748).
2. Both (b) and (c) are alternating sequences of purines and pyrimidines and are therefore sites of possible Z-DNA formation (text, p. 791). Sequences rich in GC dinucleotides (c) form Z-DNA more easily than those rich in (AT) dinucleotides (b). The latter can be driven into the Z conformation if they are flanked by GC-rich sequences or the DNA is highly negatively supercoiled.
3. (a) They bind to a region of double-strand DNA that also has a twofold axis of sym metry, such as a palindromic region.
(b) B-DNA. The distance between adjacent major grooves is 34 Å. Also these grooves are wide enough to accommodate the recognition helix of the binding protein.
(c) The helical recognition regions bind to two adjacent major grooves of B-DNA.
(See Section 9.3.3 of the text [pp. 248–251] for an example of an enzyme, EcoRV, that recognizes a palindromic DNA sequence. Note, however, EcoRV does not use a helices to contact the DNA. See p. 874 for examples of helix-turn-helix-containing proteins that bind adjacent major grooves of B-DNA.)
4. In the overall reaction, ATP is hydrolyzed to AMP and pyrophosphate. Only AMP would be labeled. The phosphate involved in the formation of the phosphodiester bond is furnished by the polynucleotide chain and does not arise from ATP.
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5. Lk=24, Tw=25, and Wr=:1. Topoisomerase I increases the linking number of DNA by 1 each catalytic cycle (text, p. 756). This increase comes about at the expense of unwinding the negative supercoil.
6. Lk=21, Tw=25, and Wr=:4. DNA gyrase catalyzes a reaction in which both DNA strands are broken, the linking number is decreased by 2, and the number of negative supercoils is correspondingly increased by 2. The answers given for this and the preceding question are approximate for the values of Tw and Wr because solution conditions affect the distribution between twists and supercoils.
7. The polA1 mutants are extraordinarily sensitive to ultraviolet irradiation because they are deficient in the 5„D3„ exonuclease activity of DNA polymerase I and are therefore impaired in DNA repair. They have only 1% of the activity of their wild-type counterpart and cannot efficiently remove thymine dimers formed by UV light. They can, however, replicate their DNA at normal rates because DNA polymerase III is the enzyme that is primarily responsible for DNA replication. An enzyme, RNaseH, which hydrolyzes RNA only when it is base paired to DNA, likely replaces the 5„D3„ exonuclease in processing the Okazaki fragments.
8. (a) 30% T, 20% A, 15% G, 35% C (b) 25% A, 25% T, 25% C, 25% G. The base composition of the double strand is the average of that of the two single strands.
9. First, if copy-choice were a correct model, no recombinant phage should be produced in the absence of new DNA synthesis. Second, according to that model, no recombinant DNA duplexes should contain both bromouracil and 32P. However, the Holliday model for homologous recombination (text, pp. 766–768) accounts for how different labels from different DNA molecules could occur in the same progeny molecule.
10. Exons often encode protein domains. Genetic recombination can lead to rearrangements in the order of exons in a gene. Upon expression, such rearranged genes could give rise to proteins with new domain orders and possibly new capabilities (text, pp. 137–138).
11. The order cannot be unambiguously established from the information given. Two pos sibilities are shown in Figure 27.3.
FIGURE 27.3 Two possible gene arrangements for problem 11.
o r i C o r i C
B
C
B
C
A
D
D
A
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12. A decrease in the activity of primase would account for the low rate of DNA replication.
Synthesis of DNA itself requires the prior synthesis of RNA primers. Also, decreased rates of dNTP synthesis could slow replication.
13. The mutagen 5-bromouracil changes A–T pairs to G–C pairs or G–C pairs to A–T pairs.
The mutation in (c) could be induced by 5-bromouracil. For example, the DNA sequence AAA, which codes for phenylalanine, could be changed to the sequence AAG, which codes for leucine. The other mutations could not arise from treatment with 5bromouracil. Remember that the genetic code presented in the text is expressed in terms of RNA. The sequence UUU on RNA corresponds to the sequence AAA on the informational strand of DNA. Leucine is encoded by the sequence CUU on RNA, which corresponds to the sequence AAG on the informational strand of DNA.
Remember also that, unless otherwise specified, nucleotide sequences are written in the 5„D3„ direction.
14. Hydroxylamine causes the unidirectional change of C–G pairs to T–A pairs. The muta tion in (a) cannot result from the action of hydroxylamine. Those in (b), (c), and (d) might. In (b), TTC (Glu) could change to TTT (Lys). In (c), ATG (His) could be converted to ATA (Tyr). In (d), ACC (Gly) could change to ATC (Asp), or GCC (Gly) could change to GTC (Asp).
15. Because thymidylate synthase (text, p. 706) is inactivated, the supply of dTTP is insuf ficient to support the synthesis of DNA at normal rates. If DNA synthesis is suppressed, so will be the rate of division of the tumor cells. This type of treatment takes advantage of the fact that tumor cells divide more rapidly than do normal cells. The dosage of the drug is adjusted so that it will primarily affect more rapidly dividing cells. However, the division of some normally rapidly dividing cells, for example those lining the intestinal tract and blood-forming cells, may be retarded as well.
16. The fibroblasts from the two patients show complementation (the defect in each is reme died by the other), so it is likely that the two patients suffer from different genetic variants of xeroderma pigmentosum. The action of several genes is likely responsible for the excision and subsequent repair of damaged DNA. One patient, for example, could have produced a normal nuclease that excises damaged DNA but could have been deficient in a ligase. The other patient could have produced normal ligase but could have been deficient in nuclease activity. There are at least nine different complementation groups among xeroderma patients.
17. (a) A possible explanation is that the b2 subunit falls off the end of a linear mole cule whereas it is trapped on the circular molecule because it forms a torus around the DNA.
(b) Treatment of circular DNA with an endonuclease cleaving both strands would con vert it to a linear DNA, and the b2 protein could dissociate from the free end, thereby decreasing its apparent affinity.
18. C-5 cytosine can spontaneously deaminate just as cytosine can. When C-5 cytosine deaminates, it forms thymidine, not uracil. Therefore uracil N-glycosylase, a DNA repair enzyme, will not recognize this product of deamination as an inappropriate base and will not remove it from the DNA, causing a transition mutation.
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19. (a) guanosine (b) The phosphorylated form could not cross cell membranes.
(c) The activation involves adding phosphates onto the distal :OH group at the ex pense of ATP hydrolysis by kinases to produce a compound resembling 5„-GTP.
(Although one of the enzymes responsible for the activation is called thymidine
kinase, it can activate other nucleosides and some of their analogues as well.)
(d) The triphosphate of acyclovir will serve as a substrate for DNA polymerase, and acyclovir nucleotide residues will be incorporated into growing polynucleotide chains in the place of guanosine resiues. Acyclovir will, however, cause premature termination of nascent polynucleotide chains because it lacks a free :OH group onto which further nucleotides can be linked by phosphodiester bonds.
(e) Thymidine kinase is encoded by a viral, not a host, gene. Therefore, uninfected cells will lack the susceptible enzyme.
(f) Mutations that impair the ability of either the viral thymidine kinase or DNA polymerase to use the analogue would render infected cells insensitive to the agent.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. DNA polymerase I uses deoxyribonucleoside triphosphates; pyrophosphate is the leav ing group. DNA ligase uses DNA-adenylate (AMP joined to the 5„-phosphate) as a reaction partner; AMP is the leaving group. Topoisomerase I uses a DNA-tyrosyl intermediate (5„-phosphate linked to the phenolic OH); the tyrosine residue of the enzyme is the leaving group.
2. FAD, CoA, and NADP; are plausible alternatives. Note that, like NAD; and ATP, all three of these molecules contain a pyrophosphate (PP) linkage.
3. Positive supercoiling resists the unwinding of DNA. The melting temperature of DNA increases in going from negatively supercoiled to relaxed to positively supercoiled DNA.
Positive supercoiling is probably an adaptation to high temperature, where the unwinding (melting, denaturing) of DNA is markedly increased.
4. (a) Long stretches of each occur because the transition is highly cooperative.
(b) B-Z junctions are energetically highly unfavorable.
(c) A-B transitions are less cooperative than B-Z transitions because the helix stays right handed at an A-B junction but not at a B-Z junction.
5. (a) 1000 nucleotides/s divided by 10.4 nucleotides/turn for B-DNA gives 96.2 revolu tions per second.
(b) 0.34 mm/s. (1000 nucleotides/s corresponds to 3400 Å/s because the axial distance between nucleotides in B-DNA is 3.4 Å.)
(c) The rate of rotation of DNA at a replication fork (96.2 rps) is nearly the same as that of a bacterial flagellum (100 rps) (for further detail, see Chapter 33 in the text).
Skeletal muscle sarcomere shortening is about 15-fold as rapid as polymerase movement at a replication fork (5 mm/s, compared with 0.34 mm/s) (for further detail, see Chapter 33 in the text). Smooth muscle sarcomere shortening (0.4 mm/s) occurs at about the same speed as replication fork movement.
DNA STRUCTURE, REPLICATION, AND REPAIR
499
6. The unwinding of DNA to expose single-stranded regions at the replication fork causes overwinding (positive supercoils) ahead of the fork. The action of topoisomerase II overcomes this effect by introducing negative supercoils to compensate. Without topoisomerase II, the DNA would become too tightly wound ahead of the fork.
7. Telomerase is required to synthesize the ends of new linear chromosomes during cell di vision. Because cancer cells are dividing rapidly, it is likely that the telomerase gene must be activated for a cell to become cancerous.
8. The activity will be similar to the replacement of an RNA primer with DNA by DNA polymerase I. One makes use of the combined 5„D3„ exonuclease and 5„D3„ polymerase activities of DNA polymerase I. From the point of the internal nick (of only one strand) by the endonuclease, polymerase I will extend the free 3„-OH using radioactive dNTPs while at the same time digesting from the internal 5„-phosphate to make room for the newly synthesized DNA. The result is a “nick translation” event in which an unlabeled portion of one DNA strand is replaced with a radioactive stretch of DNA.
(Over the section of new synthesis, only one strand becomes labeled. The strand used as the “template” remains unlabeled.)
9. If replication were unidirectional, tracks with a low-grain density at one end and a high grain density at the other end would be seen. On the other hand, if replication were bidirectional, the middle of a track would have a low density, as shown in the margin.
For E. coli, the grain tracks are denser on both ends than in the middle, indicating that replication is bidirectional.
10. (a) Pro (CCC), Ser (UCC), Leu (CUC), and Phe (UUC). Alternatively, the last base of each of these codons could be U.
(b) These CDU mutations were produced by nitrous acid.
Unidirectional synthesis
Origin
Bidirectional synthesis
Origin
11. Potentially deleterious side reactions are avoided. The enzyme itself might be damaged by light if it could be activated by light in the absence of bound DNA harboring a pyrimidine dimer. The DNA-induced absorption band is reminiscent of the glucose-induced activation of the phosphotransferase activity of hexokinase.
12. DNA ligase relaxes supercoiled DNA by catalyzing the cleavage of a phosphodiester bond in a DNA strand. The attacking group is AMP, which becomes attached to the 5„-phosphoryl group at the site of scission. AMP is required because this reaction is the reverse of the final step in the joining of pieces of DNA (see Figure 27.29 on p. 762 in the text).
13. ATP hydrolysis is required to release DNA gyrase after it has acted on its DNA substrate.
Negative supercoiling requires only the binding of ATP, not its hydrolysis.
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14. (a) Different supercoiled forms (topological isomers) migrate through the gel at differ ent rates. The highly supercoiled DNA has a compact shape and migrates rapidly.
Relaxed DNA has a more extended shape (a larger radius of gyration) and moves more slowly through the gel matrix.
(b) The bands in lane B represent different supercoiled isomers of DNA. Neighboring bands differ from each other by ±1 superhelical turn.
(c) With longer exposure to topoisomerase 1 (a “relaxing” enzyme), the population of DNA molecules shifts toward a distribution that is near thermal equilibrium.
15. (a) The control plate indicates the extent of spontaneous reversion in the absence of an external mutagen.
(b) The known mutagen is a positive control to show that the procedures are correctly implemented and the test is working.
(c) The experimental compound by itself (plate C) gives results that are only margin ally above background (plate A). The experimental compound itself therefore should be classified as either non-mutagenic or only very slightly mutagenic.
However, a metabolic product derived from the experimental compound is mutagenic (plate D).
(d) One or more enzymes from the liver probably are responsible for the metabolic con version of the experimental compound into a mutagenic compound.
CHAPTER 2
RNA Synthesis and Splicing
8
The conversion of DNA nucleotide sequences into RNA sequences is an early step in the expression of genetic information (see Chapter 5). This chapter describes the DNA-dependent RNA polymerases that catalyze this reaction, and the ways in which the product RNA transcript must sometimes be cleaved and modified in various ways before it becomes functional.
The chapter begins with an overview of the three stages of RNA synthesis: initi ation, elongation, and termination. The subunit structure of RNA polymerase from E. coli is described and a distinction between the core and holoenzymes is made. The powerful technique of DNA footprinting is explained in the context of determining the binding sites of RNA polymerase. The nature of prokaryotic promoters is given, and the function of the s subunit of RNA polymerase in specific transcript initiation is explained. The important role that supercoiling plays in transcription initiation is emphasized. The authors next describe the structure of the ternary elongating complex consisting of the template DNA, RNA polymerase, and product RNA. Finally, two mechanisms of transcription termination, one of which requires the r protein, are detailed. The cleavage and modification reactions involved in the maturation of ribosomal and transfer RNAs are outlined. In addition, two antibiotics are described, rifampicin and actinomycin D, that inhibit transcription by different mechanisms.
The authors next turn to the more complex process of transcription in eukary otes. The impossibility of coupling transcription and translation in eukaryotes as it occurs in prokaryotes (because of eukaryotic subcellular separation in the nucleus and cytoplasm) is pointed out. The three eukaryotic RNA polymerases that carry out transcription are described and related to the kinds of RNA they synthesize. The role of the eukaryotic TATA box and the TATA-box-binding protein in basal transcription are explained, as are other eukaryotic promoters and enhancers and some of the proteins
501
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CHAPTER 28 that bind them. The reactions that modify the 5′ and 3′ ends of typical eukaryotic mRNA transcripts to cap and add a poly(A) tail to them are described as are ways in which the nucleotide sequence of certain mRNAs can be modified by base alterations and insertions, a process called RNA editing. The molecular machinery and reactions that remove introns from eukaryotic mRNA (known as splicing) are outlined along with the consequences of alternative splicing reactions. The chapter concludes with the discovery of catalytic RNA and the mechanism of RNAcatalyzed self-splicing in Tetrahymena.
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Introduction 1. Define transcription.
2. Name the three stages of RNA synthesis, and list the functions of RNA polymerase in these processes.
Transcription Is Catalyzed by RNA Polymerase (Text Section 28.1) 3. Describe the subunit structure of RNA polymerase from E. coli, and assign functions to the individual subunits.
4. Define promoter, and recount how RNA polymerase protects promoters from hydrolysis by DNAse. Explain how footprinting can be used to locate promoters.
5. Recognize the convention for numbering the nucleotides in the DNA template with re gard to the transcription start site. Distinguish between the template (or antisense) strand
and the coding (or sense) strand of the duplex DNA template.
6. Note the consensus sequences around the −35 and −10 positions of E. coli promoters.
Contrast the rates of transcript initiation on strong and weak promoters in E. coli.
7. Explain how the s factor enables RNA polymerase to recognize promoters. Distinguish be tween the s70 and s32 subunits, contrast the sequences of standard promoters and heatshock promoters, and provide examples of how s factors can determine which genes are expressed.
8. Describe how topoisomerase I was used to determine the number of promoter DNA basepairs that are unwound upon the binding of RNA polymerase. Relate negative supercoiling
to promoter efficiency. Distinguish between closed promoter and open promoter complexes.
9. Describe the model for the transcription bubble. State the number of base pairs in theRNA–DNA hybrid. Appreciate the rate of RNA chain elongation in terms of both the nucleotides added and the distance on the template traversed by RNA polymerase.
10. Contrast the fidelity of polymerization of RNA polymerase with that of DNA polymerase, and explain the difference.
11. Contrast r-dependent and r-independent transcription termination. Outline the mecha nisms of the r protein and explain the role of ATP hydrolysis in its function.
RNA SYNTHESIS AND SPLICING
503
12. Describe the mechanisms of inhibition of transcription by rifampicin and actinomycin D.
Eukaryotic Transcription and Translation Are Separated in Space and Time
(Text Section 28.2) 13. Note the spatial and temporal differences in transcription and translation between prokaryotes and eukaryotes. Consider the regulatory implications of these differences.
14. Contrast mRNA processing in prokaryotes and eukaryotes. Outline the major reactions that process the primary RNA transcripts of eukaryotes.
15. Describe the RNA polymerases of eukaryotes, locate them within the cell, and list the kinds of RNA they synthesize. Account for the toxic effects of a-amanitin, and describe how it can be used to differentiate the three eukaryotic RNA polymerases.
16. List the salient sequence elements of eukaryotic promoters. Contrast the nucleotide sequences and locations of the TATA box of eukaryotes and the −10 sequence of prokaryotes.
17. Explain the key role of TATA-box-binding protein in assembling active transcription complexes.
18. Outline the combinatorial activity of transcription factors and other DNA-binding regu latory proteins in directing RNA polymerase to specific genes.
The Transcription Products of All Three Eukaryotic Polymerases Are Processed
(Text Section 28.3) 19. Draw the structure of the 5′ end of a typical eukaryotic mRNA, and distinguish caps 0, 1, and 2. Outline the reactions required to cap the primary transcript.
20. Describe the events leading to the production of mRNA with a poly(A) tail.
21. Describe RNA editing and provide examples of its effects on gene expression.
22. Describe the splicing of eukaryotic mRNA, give the consensus sequences at the splice site junc-tions, and designate the other nucleotide sequence elements involved in the process. List the functions of this post-transcriptional modification, and explain the importance of alternative splicing in gene expression.
23. Describe the role of transesterification reactions in splicing and compare the number of phosphodiester bonds broken and formed during mRNA splicing.
24. Describe the spliceosome and detail the structural and catalytic involvement of smallnuclear ribonucleoprotein particles (snRNPs) in mRNA splicing.
The Discovery of Catalytic RNA Was Revealing with Regard to Both
Mechanism and Evolution (Text Section 28.4) 25. Describe the discovery of self-splicing RNA and outline the reactions that occur during the conversion of a ribosomal RNA (rRNA) precursor from Tetrahymena into the spliced, mature rRNA and other linear and circular products. Explain the role of guanosine or a guanylyl nucleotide in the self-splicing reaction.
26. Contrast the group I and group II self-splicing introns. Compare spliceosome-catalyzed splicing with self-splicing.
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SELF-TEST
Transcription Is Catalyzed by RNA Polymerase 1. Give the subunit composition of the RNA polymerase of E. coli for both the holoenzyme and the core enzyme.
2. Match the subunit of the RNA polymerase of E. coli in the left column with its putative function during catalysis from the right column.
(a) a (1) binds the DNA template (b) b (2) binds regulatory proteins and sequences (c) b′ (3) binds NTPs and catalyzes bond formation (d) s70 (4) recognizes the promoter and initiates synthesis 3. Which of the following statements about E. coli promoters are correct?
(a) They may exhibit different transcription efficiencies.
(b) For most genes they include variants of consensus sequences.
(c) They specify the start sites for transcription on the DNA template.
(d) They have identical and defining sequences.
(e) They are activated when C or G residues are substituted into their −10 regions by mutation.
(f) Those that have sequences that correspond closely to the consensus sequences and are separated by 17 base pairs are very efficient.
4. The sequence of a duplex DNA segment in a DNA molecule is 5′-ATCGCTTGTTCGGA-3′
3′-TAGCGAACAAGCCT-5′
When this segment serves as a template for E. coli RNA polymerase, it gives rise to a segment of RNA with the sequence 5′-UCCGAACAAGCGAU-3′
Which of the following statements about the DNA segment are correct?
(a) The top strand is the coding strand.
(b) The bottom strand is the sense strand.
(c) The top strand is the template strand.
(d) The bottom strand is the antisense strand.
5. Which of the following statements about the s subunit of RNA polymerase are correct?
(a) It enables the enzyme to transcribe asymmetrically.
(b) It confers on the core enzyme the ability to initiate transcription at promoters.
(c) It decreases the affinity of RNA polymerase for regions of DNA that lack promoter sequences.
(d) It facilitates the termination of transcription by recognizing hairpins in the transcript.
6. When growing E. coli are subjected to a rapid increase in temperature, a new and char acteristic set of genes is expressed. Explain how this alteration in gene expression occurs.
7. Match the regions of a r-independent transcription termination signal in a DNA tem plate in the left column with the structures or the functions performed by the encoded transcript segments in the right column.
(a) GC-rich palindromic region (1) oligo(U) stretch in RNA (b) AT-rich region (2) hairpin in RNA (3) promotes the dissociation of RNA–DNA hybrid helix (4) causes the enzyme to pause 8. Which of the following statements about the r protein of E. coli are correct?
(a) It is an ATPase that is activated by binding to single-stranded DNA.
(b) It recognizes specific sequences in single-stranded RNA.
(c) It recognizes sequences in the DNA template strand.
(d) It causes RNA polymerase to terminate transcription at template sites that are dif ferent from those that lead to r-independent termination.
(e) It acts as a RNA–DNA helicase.
9. Match the functions in the right column with the antibiotic inhibitors of E. coli tran scription in the left column.
(a) rifampicin (b) actinomycin D (1) interacts with the template (2) interacts with nascent mRNA polymerase (3) prevents initiation (4) prevents elongation (5) intercalates into mRNA hairpins 10. Explain how a mutation might give rise to an E. coli that is resistant to the antibiotic rifampicin.
Eukaryotic Transcription and Translation Are Separated in Space and Time 11. Which of the following statements about eukaryotic mRNAs are correct?
(a) They are derived from larger RNA precursors.
(b) They result from extensive processing of their primary transcripts before serving as translation components.
(c) They usually have poly(A) tails at their 5′ ends.
(d) They have a cap at their 3′ ends.
(e) They are often encoded by noncontiguous segments of template DNA.
12. Match the descriptions in the right column with the appropriate eukaryotic DNA dependent RNA polymerases in the left column.
(a) RNA polymerase I (1) is located in the nucleolus (b) RNA polymerase II (2) is located in the nucleoplasm (c) RNA polymerase III (3) makes mRNA precursors (4) makes tRNA precursors (5) makes 5S rRNA (6) makes 18S, 5.8S, and 28S rRNA precursors (7) is strongly inhibited by a-amanitin (8) synthesizes RNA in the 5′
3′
direction
(9) is composed of several subunits (10) has a subunit with repeated amino acid sequences subject to phosphorylation
13. List the major sequence features of promoters for eukaryotic mRNA genes.
14. Describe the role of TATA-box-binding protein (TBP, TFIID) in forming the basal tran scription apparatus. What properties does TBP confer on this apparatus?
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15. Which of the following statements about enhancers are correct?
(a) They function as promoters.
(b) They function when in either orientation in the DNA.
(c) They function when on either side of the activated promoter.
(d) They function even when located many base pairs away from the promoter.
(e) They function only in specific types of cells.
The Transcription Products of All Three Eukaryotic Polymerases Are Processed 16. Which of the following statements about the poly(A) tails that are found on most eu karyotic mRNAs are correct?
(a) They are added as preformed polyriboadenylate segments to the 3′ ends of mRNA precursors by an RNA ligase activity.
(b) They are encoded by stretches of polydeoxythymidylate in the template strand of the gene.
(c) They are added by RNA polymerase II in a template-independent reaction using ATP as the sole nucleotide substrate.
(d) They are added by poly(A) polymerase using dATP as the sole nucleotide substrate.
(e) They are cleaved from eukaryotic mRNAs by a sequence-specific endoribonuclease that recognizes the RNA sequence AAUAAA.
17. What functions are the caps and tails of mRNAs thought to perform?
18. Which of the following statements about apolipoprotein B (apo B) are correct?
(a) The apo B-48 form is formed by the proteolytic cleavage of the primary (apo B-100) translation product.
(b) Apo B-48 and apo B-100 are formed in different tissues.
(c) Apo B-48 arises from the expression of a form of the gene for apo B-100 that has been shortened by nonhomologous recombination.
(d) The transcript of the apo B-100 gene is spliced to remove a segment and form apo B-48.
(e) A specific nucleotide in the apo B-100 transcript is altered, thereby creating a stop codon in the mRNA.
19. Which of the following are important sequence elements in the splicing reactions that produce eukaryotic mRNAs?
(a) exon sequences located between 20 and 50 nucleotides from the 5′ splice site (b) exon sequences located between 20 and 50 nucleotides from the 3′ splice site (c) intron sequences located between 20 and 50 nucleotides from the 5′ splice site (d) intron sequences located between 20 and 50 nucleotides from the 3′ splice site (e) intron sequences at the 5′ splice site
(f) intron sequences at the 3′ splice site 20. Eukaryotic mRNA splicing involves which of the following?
(a) the formation of 2′ 5′ phosphodiester bonds (b) a sequence-specific endoribonuclease that hydrolyzes the phosphodiester bond at the junctions of the intron with the exon (c) the spliceosome (d) the coupling of phosphodiester bond formation to ATP hydrolysis (e) the formation of lariat intermediates RNA SYNTHESIS AND SPLICING
507
The Discovery of Catalytic RNA Was Revealing with Regard to Both
Mechanism and Evolution 21. Place the following events in the order in which they occur during the formation of ma ture rRNA in Tetrahymena.
(a) The 3′-hydroxyl of a guanine nucleoside attacks the phosphodiester bond at the 5′ splice site leaving the 5′ (upstream) exon with a free 3′-hydroxyl and attaching the guanine nucleoside to the 5′ end of the intron.
(b) The transcript from the rRNA gene folds into a specific structure.
(c) The rRNA transcript specifically binds a guanine nucleoside or nucleotide.
(d) Self-splicing occurs within the intron to form L19 RNA.
(e) The 3′-hydroxyl of the 5′ exon attacks the bonds at the 3′ splice junction to form the spliced rRNA and eliminate the intron.
22. Match the descriptions in the right column with the type of splicing in the left column.
(a) group I splicing (1) A 2′ 5′ phosphodiester bond is (b) group II splicing involved.
(c) nuclear mRNA splicing (2) Nuclease and ligase activities are required.
(3) Transesterification reactions break and form bonds.
(4) A guanine nucleoside or nucleotide is re quired.
(5) Spliceosomes are required.
(6) Lariat intermediates are involved.
(7) snRNPs are involved.
ANSWERS TO SELF-TEST
1. The holoenzyme has the subunit composition a2bb′s. The core enzyme lacks the s subunit.
2. (a) 2 (b) 3 (c) 1 (d) 4 3. a, b, c, f. The promoters of most E. coli genes include variants of defining, consensus se quences that are centered at about the −35 and −10 positions. The nearer the sequences of a promoter are to the consensus sequence and the nearer the separation between them is to the optimal 17-bp spacing, the more efficient the promoter. The −10 consensus sequence is TATAAT. The substitution of a C or G into the sequence would likely lower the efficiency of a promoter.
4. b, c. The sense strand (bottom) of the template DNA has the same sequence as the mRNA.
5. a, b, c. The s subunit recognizes promoter sites, decreases the affinity of the enzyme for regions of DNA lacking promoter sequences, and facilitates the specific, oriented initiation of transcription. Orienting the binding of the enzyme to the DNA results in only one of the two DNA strands functioning as a template for RNA transcription; that is, it gives rise to asymmetric transcription.
6. The temperature increase induces the synthesis of a new s factor, s32, which directs RNA polymerase to promoters that have −10 and −35 sequences which are different from those recognized by s70. Transcription from these promoters gives rise to characteristic heat-shock proteins.
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7. (a) 2, 4 (b) 1, 3 8. d, e. The r protein recognizes and binds stretches of RNA that are devoid of hairpins and are at least 72 nucleotides long. It acts to hydrolyze ATP and to unwind the RNA–DNA hybrid in the transcription bubble.
9. (a) 2, 3 (b) 1, 4. Actinomycin D intercalates only into duplex DNA.
10. Rifampicin must bind to the b subunit of RNA polymerase to inhibit the enzyme. A mu tation in the gene encoding this subunit that would interfere with the binding of the antibiotic but not with polymerization would produce a rifampicin-resistant cell.
11. a, b, e. Answers (c) and (d) are incorrect because the poly(A) tail is found at the 3′ end and a cap is found at the 5′ end of typical eukaryotic mRNA. Segments of the primary transcript are discarded by RNA splicing to bring RNA sequences encoded by noncontiguous regions of the template together; that is, introns are removed to form a product smaller than the primary transcript.
12. (a) 1, 6, 8, 9 (b) 2, 3, 7, 8, 9, 10 (c) 2, 4, 5, 8, 9. Both RNA polymerases I and III are involved in rRNA synthesis, with polymerase I synthesizing the 18S, 5.8S, and 28S precursor transcripts and polymerase III synthesizing the 5S rRNA.
13. A TATA box centered at about −25 from the start site of transcription and consisting of a variant of the consensus heptanucleotide sequence TATAAAA is essential for promoter activity. In addition, many genes have a CAAT box, a GC box, or both elements located between −40 and −110. These sequences can function in either orientation, that is, be in either DNA strand. Genes that are expressed constitutively, as distinct from those whose expression is regulated, tend to have GC boxes. Activating sequences farther upstream of these promoter elements are necessary for the functioning of most promoters.
14. TBP serves a critical role as an enucleating center for the assembly of the minimal molecu lar apparatus (the basal transcription assembly) required for transcription by RNA polymerase II. After TBP binds the TATA box, other TFII proteins and RNA polymerase II join the supramolecular complex to render it transcriptionally competent. In addition to enabling the assembly of the apparatus through its ability to recognize and bind the TATA box, TBP binds this asymmetric sequence in one orientation that “points” the RNA polymerase in the right direction, thereby defining the strand of DNA that will serve as the template.
15. b, c, d, e. Answer (a) is incorrect because enhancer sequences do not serve as promot ers per se. They will not, by themselves, enable RNA polymerase II to initiate a transcript, and will function only when the basal transcription apparatus exists.
16. None of the statements are correct. Poly(A) polymerase uses ATP, not dATP, to add a stretch of A residues to the 3′-hydroxyl formed by the cleavage of an mRNA precursor by a specific ribonuclease that recognizes the upstream sequence, AAUAAA, within particular mRNA sequence contexts.
17. The 5′ caps are thought to contribute to the stability and efficiency of translation of the mRNA. The poly(A) tails probably perform the same functions, albeit by different, unknown mechanisms.
18. b, e. The apo B system illustrates one kind of RNA editing. The pre-mRNA transcript has its sequence altered—in this case, a specific cytidine is deaminated to uracil to create the stop codon. Other editing mechanisms involve the addition and removal of U residues, an A-to-G change, and an adenosine-to-inosine change.
19. d, e, f 20. a, c, e. Answers (b) and (d) are incorrect because hydrolysis of phosphodiester bonds is not involved during mRNA splicing up to the point of intron removal and lariat formation; only transesterification reactions occur. Consequently, ATP is not required for the synthesis of phosphodiester bonds.
21. b, c, a, e, d
22. (a) 3, 4 (b) 1, 3, 6 (c) 1, 3, 5, 6, 7. The splicing of eukaryotic tRNAs requires protein nucleases to cut the phosphodiester bonds at the intron–exon junctions and protein RNA ligases to form the bonds joining the exons.
PROBLEMS
1. The technique of footprinting (p. 784 in text) involves, in part, the digestion of unpro tected DNA by DNAse I.
(a) Why is it necessary to use a nuclease that is not sequence-specific?
(b) What would be the consequences of an extensive digestion with DNAse I?
2. The rate constant for the binding of RNA polymerase holoenzyme to a promoter on a long DNA molecule is greater than that for the collision of two small molecules in solution. Since small molecules diffuse through solutions more rapidly than large ones, how can this be true?
3. One would expect an analog of 5′-ATP that lacks an oxygen at the 3′ position of its ribose (3′-deoxy-5′-ATP; see Figure 28.1) to interrupt RNA formation because it cannot form phosphodiester bonds at its 3′position. Could such a compound be used to ascertain the direction of chain growth in RNA synthesis? Explain. (Refer to figure on p. 787 of text.)
FIGURE 28.1 Structure of 3′-deoxy-5′-ATP.
NH
2
J
N
N
O
O
O
N
N
K
K
K :OJPJOJPJOJPJOJCH 2
J
J
J
O :O O: O:
J
J
H
H
H
3„
2„
J
J
H
OH
4. The ciliated protozoan Tetrahymena contains an enzyme that can synthesize 5′-pseudouri dine monophosphate from a mixture of PRPP and uracil. For a time it was thought that this enzyme was instrumental in the synthesis of transfer RNAs in Tetrahymena. Explain why this is not the case.
5. When mammalian genes are cloned, a strategy that is frequently followed involves the isolation of mRNA rather than DNA from a cell and the preparation of a complementary DNA (cDNA) by the enzyme reverse transcriptase. Suppose that mRNA isolated from a cell specialized for the production of protein X is used as a template for the production of cDNA. What major difference or differences would you expect to find between the structure of that cDNA and genomic DNA for protein X?
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6. Rifampicin specifically inhibits the initiation of transcription in prokaryotes and may therefore be used in humans as a therapeutic antibacterial agent. Would you expect actinomycin D to be useful in antibacterial therapy? Why or why not?
7. The mRNAs produced by mammalian viruses undergo modification at the 5′ and 3′ ends in a fashion similar to that of eukaryotic mRNA. Why do you think this is the case?
8. Sketch the most stable secondary structure that could be assumed by the oligonucleotide AAGGCCCUACGGGGCCG.
9. Suppose that human DNA is cleaved into fragments approximately the size of a given mature human messenger RNA, and that mRNA–DNA hybrids are then prepared. The corresponding procedure is then carried out for E. coli. When the mRNA–DNA hybrids from each species are examined with an electron microscope, which will show the greater degree of hybridization? Explain.
10. What are snRNPs, and how are they involved in the eukaryotic mRNA splicing reaction?
11. Figure 28.28 on page 800 of the text gives an example of how a base-change mutation within an intron in the b-globin gene can produce a 3′ splice site upstream from the normal 3′ splice site. The result is that five amino acids not normally present in the protein are inserted into the chain before synthesis is terminated by a stop codon.
(a) Corresponding mutants in exons are also known in which base change mutations introduce 5′ splice sites upstream from the normal 5′ sites (see Figure 28.2A; X marks the location of the new 5′ splice site). Sketch the resulting processed mRNA.
What changes in amino acid sequence would be expected to result?
FIGURE 28.2A Creation of a new 5′ splice site in exon 2 of β-globin.
5„
Exon 2
X
Exon 3 (b) In some forms of thalassemia, the creation of a new 5′ splice site within intron 2 activates a “cryptic” 3′ splice site upstream from the 5′ site (see Figure 28.2B; X marks the location of the new 5′ splice site and the cryptic 3′ side is labeled). Sketch the resulting processed mRNA. How would the resulting changes in amino acid sequence differ from those in (A)?
FIGURE 28.2B Activation of a cryptic 3′ splice site in intron 2 of β-globin.
3„
5„
Exon 2
X
Exon 3 (c) In antisense drug design, an oligonucleotide is designed to form a DNA–mRNA hybrid that will block transcription of targeted genes. Gorman et al. (PNAS
[1998]95:4929–4934) used antisense methods to block splicing of the cryptic and 5′ splice sites in (b) by altering nucleotides in snRNA sequences to allow them to form snRNA–DNA hybrids at the splice sites. What snRNA sequence would be needed to block the 3′ splice site shown in Figure 28.28 of the text? Use six nucleotides for your sequence. Would this snRNA also bind to the normal 3′ splice site? Why or why not?
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511
12. In an attempt to determine whether a given RNA was catalytically active in the cleavage of a synthetic oligonucleotide, the following experimental results were obtained. When the RNA and the oligonucleotide were incubated together, cleavage of the oligonucleotide occurred. When either the RNA or the oligonucleotide was incubated alone, there was no cleavage. When the RNA was incubated with higher concentrations of the oligonucleotide, saturation kinetics of the Michaelis-Menten type were observed. Do these results demonstrate that the RNA has catalytic activity? Explain.
ANSWERS TO PROBLEMS
1. (a) A nonspecific nuclease will cleave duplex DNA into short, random fragments. Thus potential cleavage sites would exist on either side of the protecting protein. If the nuclease was too specific, its cleavage sites would be too far apart to reveal the protein binding site on the DNA.
(b) In principle, one wants to time the digestion with DNAse so that there is, on the average, one cut per DNA molecule. Since there are many potential cleavage sites on each DNA molecule, the result will be a population of molecules of varying lengths (see Figure 28.3 on p. 784 of text). Extensive digestion with DNAse would produce a population of very short fragments.
2. RNA polymerase holoenzyme has lower affinity for nonspecific DNA sequences than for promoter sequences. The nonspecific affinity, however, allows the enzyme to bind to “random-sequence” DNA and then “slide” along the molecule in a unidimensional random walk until it encounters a promoter sequence, for which its binding affinity is higher. Diffusion in one dimension is much faster than diffusion in three dimensions, thereby explaining the observed rapid rate constant for the binding of RNA polymerase holoenzyme to promoter sequences. If one measured the encounter of the polymerase with the nonspecific regions of the DNA rather than with promoter sequences, the value of the rate constant would be much lower and would fit our expectations for a threedimensional, diffusion-limited reaction between macromolecules.
3. A 3′-deoxy analog of ATP could be used to establish the direction of chain growth. In
5′
3′ growth the analog would donate a nucleotide containing 3′-deoxyadenosine, and the polynucleotide chain would be terminated as a result. No additional nucleotides could be added because of the lack of a 3′-OH group on the terminal 3′-deoxyadenosine. In 5′ 3′ growth the nucleotide could not be added to the growing polynu cleotide chain because of the lack of the 3′-OH group.
4. Nascent polynucleotides formed by RNA polymerases contain only the four usual bases.
Subsequently, some of the bases are chemically modified. Were unusual nucleotides to be incorporated into a growing RNA chain, this would in turn require the presence of unusual bases on DNA. The pseudouridine found in transfer RNAs is formed by breaking the nitrogen–carbon bond linking uracil to ribose and forming a carbon–carbon bond instead. Only certain uracils are modified in this manner, owing to their position in the three-dimensional structure of the RNA and to the specificity of the enzymes that carry out the modification.
5. The cDNA prepared from mRNA would have a long poly(T) tail, unlike genomic DNA.
Remember that the poly(A) tail is added to the 3′ end of mammalian mRNA and that there is no counterpart on DNA. A second striking difference would be that the cDNA would contain no intervening sequences (introns) and would therefore be much shorter 512
CHAPTER 28 than the corresponding sections of genomic DNA. (Remember that most mammalian genes are mosaics of introns and exons.) A third difference would be found if any RNA editing were involved. An edited mRNA could generate a cDNA with nucleotides that did not correspond to those in genomic DNA.
6. In order to be useful as a therapeutic antibacterial agent, a compound must selectively inhibit processes in prokaryotes but leave the corresponding processes in eukaryotes (including those in mitochondria) largely unaffected. Because rifampicin selectively inhibits the initiation of transcription in prokaryotes but not in eukaryotes, it is useful as an antibacterial agent. Actinomycin D is an intercalating agent that binds to DNA duplexes and inhibits both DNA replication and transcription, although it has a greater inhibiting effect on transcription than on replication. It cannot discriminate between the duplex DNA of bacteria and that of humans, however, and will therefore bind to both. Because it disrupts eukaryotic as well as prokaryotic processes, it is not very useful as an antibacterial agent. It is sometimes used as an anticancer agent, however, because of its ability to slow the replication rate of human DNA.
7. Viruses use the host’s enzyme system to replicate their DNA and to synthesize their pro teins. Since eukaryotic translation systems must synthesize viral protein, the structure of viral mRNAs must mimic that of the host mRNA.
8. The structure is a stem-and-loop (“lollipop”) hairpin structure, as shown in Figure 28.3.
FIGURE 28.3 Stable secondary structure for the oligonucleotide in problem 8.
A
C
U
G
C
G
C
G
C
G
G
C
G
C
A
G
A
9. The mRNA–DNA hybrid of E. coli will show greater hybridization because it is produced continuously from a DNA template without processing. Because of the presence of intervening sequences in human DNA, there will be regions in the human RNA–DNA hybrids where no base pairing occurs.
10. The snRNPs are small ribonucleoprotein particles that occur in the nucleus. Each is com posed of a small RNA molecule and several characteristic proteins, some of which are common to different snRNPs. Distinct snRNPs recognize and bind to splice junctions and the branch site and are involved in assembling the spliceosome in an ATP-dependent manner. They are requisite components of the splicing apparatus, and the RNA components of some of them are probably catalytically active. The RNAs of some snRNPs form hydrogen bonds with sequences within introns and exons to help to juxtapose properly the reacting splice junctions.
11. (a) The spliced gene arising from the mutation would be shorter than the correctly spliced version (see below) and some amino acids would be omitted from the region of the protein encoded by nucleotides between the newly introduced and normal 5′ splicing sites. Also, the introduction of a new splicing site could lead to changes in the reading frame, and therefore to gross changes in amino acid sequence, polypeptide chain length, or both.
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Exon 2
Exon 3
(b) The resulting processed mRNA would be longer than the normal processed b-globin gene, introducing the possibility of a misfolded or malfunctioning protein or a truncated protein if a stop codon were introduced from the intron sequence. As in (a) a change in reading frame could also result.
Exon 2
Exon 3
(c) The sequence containing the mutated 3′ splice site is 5′-ATTAGT-3′. The com plementary snRNA would therefore be 3′-UAAUCA-5′. Since the normal site is 5′-CTTAGG-3′, which would be bound by 3′-GAAUCC-5′, the mutated snRNA should not bind to the normal 3′-splice site.
12. These results alone do not establish that the RNA has catalytic activity because a cata lyst must be regenerated. It is entirely possible that the results observed could be accounted for by a stoichiometric, as opposed to a catalytic, interaction between RNA and the oligonucleotide, in which the RNA may “commit suicide” as the oligonucleotide is cleaved. In such an interaction, a portion of the RNA would participate chemically in the cleavage of the oligonucleotide, but it would also be cleaved itself as a part of the reaction. Four reaction products would accumulate, two resulting from the cleavage of RNA and two from the cleavage of the oligonucleotide. To show that this particular RNA was catalytic, it would be necessary to demonstrate that it turns over and is regenerated in the course of the reaction.
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. The sequence of the coding (+, sense) strand is 5′-ATGGGGAACAG CAAGAGTGGGGCCCTGTCCAAGGAG-3′ and the sequence of the template (−, antisense) strand is 3′-TACCCCTTGTCGTTCTCACCCCGGGAC AGGTTCCTC-5′
Note that the coding strand has the same sequence as mRNA (except U T), whereas the template strand is complementary to the coding strand.
2. RNA turns over in the cell, whereas DNA is the “nearly” permanent record of inherited in formation. The consequences of most errors in RNA synthesis will be short-lived, perhaps a protein synthesized with a mistake in its sequence for a short time during the life of the cell, but then the RNA in question may be degraded and the error will disappear. By contrast, an uncorrected error in DNA replication will be passed to the next generation.
3. RNA synthesis involves copying only short segments of chromosomes, whereas DNA replication must involve copying entire chromosomes at the time of cell division. The length of consecutive nucleotide sequence that must be copied is many times greater for DNA replication than for RNA transcription.
4. Heparin, a glycosaminoglycan, is highly anionic. Its negative charges, like the phospho diester bridges of DNA templates, bind to lysine and arginine residues of b′.
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5. This mutant sigma would competitively inhibit the binding of holoenzyme and prevent the specific initiation of RNA chains at promoter sites.
6. The core enzyme without sigma binds more tightly to the DNA template than does the holoenzyme. The retention of sigma after chain initiation would make the mutant RNA polymerase less processive. Hence, RNA synthesis would be much slower than normal.
7. A 100-kd protein contains about 910 residues, which are encoded by 2730 nucleotides.
At a maximal transcription rate of 50 nucleotides per second, the protein would be synthesized in 54.6 seconds.
8. Initiation at strong promoters occurs every two seconds. In this interval, 100 nu cleotides are transcribed. Hence, centers of transcription bubbles are 34 nm (340 Å) apart (100 × 3.4 Å = 340 Å).
9. (a) The lowest band on the gel will be that of (i), whereas the highest will be that of (v). Band (ii) will be at the same position as (i) because the RNA is not complementary to the nontemplate strand, whereas band (iii) will be higher because a complex is formed between RNA and the template strand. Band (iv) will be higher than the others because strand 1 is complexed to 2, and strand 2 to 3. Band (v) is the highest because core polymerase associates with the three strands.
(b) None, because rifampicin acts before the formation of the open complex.
(c) RNA polymerase is processive. Once the template is bound, heparin cannot enter the DNA-binding site.
(d) The longest RNA product formed is a 36-mer rather than the full 72-mer. Because GTP is absent, synthesis stops when the first C downstream of the bubble is encountered in the template strand.
10. Segments of double helix that are shorter than tetranucleotides (having fewer than about 4 base pairs) are unstable at physiological temperatures due to their small extent of base stacking and small number of interstrand hydrogen bonds. Until the critical length of RNA is synthesized for a stable RNA/DNA double helix, the short initial di- and trinucleotides are susceptible to release.
11. (a) The lack of a 3′-OH group will cause cordycepin to be a chain terminator for RNA sysnthesis. Cordycepin will be incorporated at the 3′-end of a chain, and further elongation will be blocked.
(b) The substrate specificity of poly(A) polymerase is higher than that of RNA poly merase because the RNA polymerase uses four nucleotides (ATP, UTP, GTP, CTP), whereas poly(A) polymerase uses only ATP. The result suggests that poly(A) polymerase has a higher apparent affinity for 3′-deoxy-ATP than does RNA polymerase.
(c) Yes. It must receive a 5′-triphosphate in order to be a substrate for poly(A) poly merase or RNA polymerase.
12. Note that the DNA strands shown in Figure 28.28 in the text are, by convention, com plementary to the DNA strand coding for mRNA. If every T is changed to U, the polarity and base sequence of the DNA strands shown are identical with the mRNA synthesized from the complementary strand. Thus, TCT becomes UCU (Ser), and so forth.
13. A mutation that disrupted the normal AAUAA recognition sequence for the endonucle ase could account for this finding. In fact, a change from U to C in this sequence caused this defect in a thalassemic patient. Cleavage occurred at the AAUAAA 900 nucleotides downstream from this mutant AACAAA site. See S. H. Orkin, T.-C. Cheng, S. E.
Antonarakis, and H. H. Kazazian, Jr. EMBO J. 4(1985):453.
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14. One possibility is that the 3′ of the poly(U) donor strand cleaves the phosphodiester bond on the 5′ side of the insertion site. The newly formed 3′ terminus of the acceptor strand then cleaves the poly(U) strand on the 5′ side of the nucleotide that initiated the attack. In other words, a U could be added by two transesterification reactions. This postulated mechanism is akin to the one in RNA splicing (see Figure 28.29 in the text). See T. R. Cech. Cell 64(1991):667.
15. The possibilities of alternative splicing and of RNA editing allow the final protein prod ucts to be more complex and more highly varied than the genes that encode them.
Additionally, posttranslational modification of proteins (e.g., glycosylation, phosphorylation) will further enhance the complexity.
16. One could make use of the A–T base pairing potential of the poly-A sequence that is characteristic of eukaryotic mRNAs. One would construct an affinity chromatography column in which an oligo-dT nucleotide is covalently linked to a resin. Eukaroytic mRNAs would bind to this column, whereas other RNAs would not. After washing the other RNAs away from the column, one would elute the eukaroytic mRNAs by weakening the A–T hydrogen bonds, for example, by changing the temperature, or by washing the column with a solution containing an excess of soluble oligo-dA (to displace the mRNA).
17. (a) The different genes are expressed to differing extents. Only those genes for which mRNA is actively being transcribed will give positive hybridization signals.
(b) Gene expression patterns differ in the different tissues. Some of the mRNAs are tran scribed in some tissues but not others.
(c) The genes that are expressed in all three tissues could be essential for fundamental metabolic processes that are common to most if not all cells.
(d) Including the initiation inhibitor allows counting and comparison of the number of ongoing mRNA chains being synthesized among the different gene types and from tissue to tissue at the given moment in time when the cells are broken.
(Without such an inhibitor, the results could be skewed by differing initiation rates from gene to gene, or by the possibility of some artificial initiation events that could be induced when the cells are broken to isolate the nuclei.)
18. The long strand that goes entirely across the picture from left to right is the DNA. The strands of increasing length are molecules of mRNA that are beginning to be transcribed.
Transcription begins just ahead of the site on the DNA where shortest strands of mRNA are seen. Transcription ends just after the site on the DNA where the longest strands of mRNA are attached. (As RNA polymerase passes this site, the primary transcript mRNA is released.) On the page, the direction of RNA synthesis is from left to right. Many different enzymes are simultaneously making many different RNA molecules on a single gene.