Untitled
CHAPTER 5
DNA, RNA, and the Flow
of Genetic Information Chapters 2 and 3 introduced you to proteins. The authors now turn to a second class of macromolecules, the nucleic acids that serve as the storage forms of genetic information. First, they describe the structures of the nucleoside building blocks of DNA and the phosphodiester bond that links them together. Following this, the Watson-Crick DNA double helix is presented, an overview of how the strands of DNA separate for replication is given, and some of the various conformations and structures that nucleic acids can assume are described. The polymerases that form DNA chains are introduced next. The section describing the molecules that store genetic information ends by providing two examples of viruses in which the genetic material is not duplex DNA but rather single-strand RNA. The authors next describe the way in which RNA viruses replicate through double-strand nucleic-acid intermediates whose formation is directed by specific base pairing.
How the information stored in DNA or RNA directs the formation of the pro teins of a cell is discussed next. The authors start with descriptions of the basic structures and kinds of RNA and provide an explanation of the central roles of RNAs in the overall flow of genetic information. They then present the specific functions of messenger RNA, transfer RNA, and ribosomal RNA in protein synthesis, along with a description of the polymerase that synthesizes all cellular RNAs. The genetic code, which relates the nucleotide sequence of RNA to the amino acid sequence of proteins, is described. The collinear relationship between the sequences of nucleotides in the DNA and the amino acids of the encoded protein is compared in prokaryotes and in eukaryotes where some genes are interrupted by noncoding sequences (introns). The authors next describe the process by which these intron sequences are removed from the initial transcript to form functional messenger RNA and the biological consequences of such splicing.
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CHAPTER 5
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Nucleic Acids (Text Section 5.1) 1. Locate the structural components of DNA, namely, the nitrogenous bases, the sugar, and the phosphate group. Know the various conventions used to represent these components and the structure of DNA.
2. Differentiate purines, pyrimidines, ribonucleosides, deoxyribonucleosides, ribonucleotides, and deoxyribonucleotides.
3. Recognize the deoxyadenosine, deoxycytidine, deoxyguanosine, and deoxythymidine con stituents of DNA, and describe the phosphodiester bond that joins them together to form DNA.
4. Compare the phosphodiester backbones of RNA and DNA. Contrast the composition and structures of RNA and DNA. Distinguish thymine from uracil and 2′-deoxyribose
from ribose.
5. Relate the polarity of the DNA chain to the direction a DNA sequence abbreviation is written.
6. Compare the lengths of the DNA molecules in polyoma virus, the bacterium E. coli, and the average human chromosome.
Double-Helices (Text Section 5.2) 7. List the important features of the Watson-Crick DNA double helix. Relate the base pairing of adenine with thymine and of cytosine with guanine to the duplex structure of DNA and to the replication of the helix. Explain the molecular determinants of the specific base pairs in DNA.
8. Outline the Meselson-Stahl experiment and relate it to semiconservative replication. Define the melting temperature (Tm) for DNA and relate it to the separation of the strands of duplex DNA. Describe annealing.
9. Describe supercoiling and state its biological consequences.
10. Appreciate the variety of structures that single-strand nucleic acids can assume.
DNA Polymerases (Text Section 5.3) 11. List the substrates and the important enzymatic properties of DNA polymerases as they relate to replication. Distinguish between a primer and a template and describe their functions.
12. Define virus and appreciate that RNA is the genome of some viruses.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 13. Relate the catalytic activity of reverse transcriptase (an RNA-directed DNA polymerase) to the replication of retroviruses. Provide an overview of retroviral replication.
Gene Expression (Text Section 5.4) 14. State the role of DNA in protein synthesis. Outline the flow of genetic information during gene expression.
15. Define the terms transcription and translation and relate these processes to the flow of ge netic information.
16. Name the three major classes of RNA found in E. coli and explain their functions. Compare their sizes and their relative amounts in the cell.
17. List the substrates and important enzymatic properties of RNA polymerases (DNA-de-pendent RNA polymerases). Explain the roles of the DNA template, promoter, enhancer sequences, and terminator in transcription.
18. Describe the transcription of duplex DNA to form single-strand RNA. Relate the sequence of mRNA to that of the coding strand of the DNA template from which it is transcribed.
19. Describe the role of tRNA as the adaptor molecule acting between mRNA and amino acids during protein synthesis. Outline how specific amino acids are covalently attached to specific tRNA molecules. Explain the relationship of the codon and anticodon to the specific interaction between mRNA and tRNA.
Amino Acids and the Genetic Code (Text Section 5.5) 20. Explain what the genetic code is and list its major characteristics. Define the terms tripletcode (codon), nonoverlapping, degenerate, synonym, triplet, and reading frame as they apply to the genetic code. Recognize the initiation and termination codons.
21. Using the genetic code, predict the sequence of amino acids in a peptide encoded by a template DNA or mRNA sequence.
Introns and Exons (Text Section 5.6) 22. Discuss the universality of the genetic code. Describe the composition and function of spliceosomes.
23. Contrast the linear relationship between the sequence of DNA in a gene and the sequence of the amino acids in the protein it encodes in bacteria and in higher eukaryotes. Apply the terms intron and exons to these relationships.
24. Outline RNA processing in eukaryotes. Name the alterations made to the RNA after it is initially formed by RNA polymerase.
25. Recount a hypothesis relating exons and functional domains to the generation and evolu tion of protein diversity. Differentiate between nucleotide sequence rearrangements by genetic recombination at the DNA level and by RNA splicing.
CHAPTER 5
SELF-TEST
Nucleic AcidsFIGURE 5.1
NH
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H OKPJO: H
H
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A
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1. Which of the preceding structures (a) contains ribose?
in Figure 5.1 (b) contains deoxyribose?
(c) contains a purine?
(d) contains a pyrimidine?
(e) contains guanine?
(f)
contains a phosphate monoester?
(g) contains a phosphodiester?
(h) is a nucleoside?
(i) is a nucleotide?
(j) would be found in RNA?
(k) would be found in DNA?
Double Helices
2. Which of the following are characteristics of the Watson-Crick DNA double helix?
(a) The two polynucleotide chains are coiled about one another and about a common axis.
(b) Hydrogen bonds between A and C and between G and T help hold the two chains together.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (c) The helix makes one complete turn every 34Å because each base pair is rotated by 36º with respect to adjacent base pairs and is separated by 3.4Å from them along the helix axis.
(d) The purines and pyrimidines are on the inside of the helix and the phosphodiester linked backbones are on the outside.
(e) Base composition analyses of DNA duplexes isolated from many organisms show that the amounts of A and T are equal as are the amounts of G and C.
(f) The sequence in one strand of the helix varies independently of that in the other strand.
3. If a region of one strand of a Watson-Crick DNA double helix has the sequence ACG TAACC, what is the sequence of the complementary region of the other strand?
4. Explain why A T and G C are the only base pairs possible in normal double-strand DNA.
5. Match the appropriate characteristics in the right column with the structures of double strand or single-strand DNA.
(A) double-strand DNA (1) is a rigid rod (B) single-strand DNA (2) shows a greater hyperchromic effect upon heating (3) contains equal amounts of A and T bases (4) may contain different amounts of C and G bases (5) contains U rather than T bases (6) may contain stem-loop structures 6. Haploid human DNA has 3 p 106 kilobase pairs (a kilobase pair, abbreviated kb, is 1000 base pairs). What is the total length of human haploid DNA in centimeters?
7. Outline the basic process by which a Watson-Crick duplex replicates to give two iden tical daughter duplexes. Explain the reasons for the accuracy of the process.
8. The DNA in a bacterium is uniformly labeled with 15N, and the organism shifted to a growth medium containing 14N-labeled DNA precursors. After two generations of growth, the DNA is isolated and is subjected to density-gradient equilibrium sedimentation. What proportion of light-density DNA to intermediate-density DNA would you expect to find?
9. Purified duplex DNA molecules can be (a) linear.
(b) circular and supercoiled.
(c) linear and supercoiled.
(d) circular and relaxed, that is, not supercoiled.
10. You are given two solutions containing different purified DNAs. One is from the bac terium P. aeruqinosa and has a G + C composition of 68%, whereas the other is from a mammal and has a G + C composition of 42.5%.
(a) You measure the absorbance of ultraviolet light of each solution as a function of in creasing temperature. Which solution will yield the higher Tm value and why?
(b) After melting the two solutions, mixing them together, and allowing them to cool, what would you expect to happen?
(c) Would appreciable amounts of bacterial DNA be found associated in a helix with mammalian DNA? Explain.
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DNA Polymerases
11. DNA polymerase activity requires (a) a template.
(b) a primer with a free 5′-hydroxyl group.
(c) dATP, dCTP, dGTP, and dTTP.
(d) ATP.
(e) Mg2+.
12. Derive the polarity of the synthesis of a DNA strand by DNA polymerase from the mech anism for the formation of the phosphodiester bond.
13. You are provided with a long, single-strand DNA molecule having a base composition of C = 24.1%, G = 18.5%, T = 24.6%, and A = 32.8%; DNA polymerase; [a-32P]dATP (dATP with the innermost phosphate labeled), dCTP, dGTP, and dTTP; a short primer that is complementary to the single-strand DNA; and a buffer solution with Mg2+. What is the base composition of the radiolabeled product DNA after the completion of one round of synthesis?
14. For the virus in the left column, indicate the appropriate characteristics from the right column.
(a) Tobacco mosaic virus (1) linear genome (b) AIDS virus (2) genome contains U rather than T (3) single-strand nucleic acid genome (4) DNA intermediates are involved in replication (5) uses RNA-directed RNA polymerase to replicate (6) uses RNA-directed DNA polymerase to replicate
15. Propose how a single-strand DNA virus could replicate by incorporating semiconserva tive replication into the process.
16. From the following nucleic acids, select those that appear during the infection of a cell with a retrovirus, for example, the AIDS virus, and place them in the order in which genetic information flows during the process of forming a new progeny virus.
(a) double-strand DNA-RNA helix in the cell (b) single-strand RNA in the virus (c) single-strand RNA in the cell (d) double-strand DNA in the cell (e) double-strand RNA in the virus (f) double-strand RNA in the cell Gene Expression
17. Transcription is directly involved in which of the following possible steps in the flow of genetic information.
(a) DNA to RNA (d) RNA to protein (b) RNA to DNA (e) protein to RNA (c) DNA to DNA (a) DNA to RNA (b) RNA to DNA (c) DNA to DNA (d) RNA to protein (e) protein to RNA 19. Answer the following questions about RNA.
(a) What is the name of the bond joining the ribonucleoside components of RNA to one another?
(b) Is this bond between the 2′- or the 3′-hydroxyl group of one ribose and the 5′ hydroxyl of the next?
(c) Intramolecular base pairs form what kinds of structures in RNA molecules?
(d) What bases pair with one another in RNA?
(e) What are the three major classes of RNA in a cell and which is most abundant?
20. If you have samples of pure RNA and duplex DNA, how can you tell whether they have any complementary nucleotide sequences?
21. If all the RNA referred to in question 20 turns out to have sequences that were comple mentary to the DNA, will its percentage of G and C be identical to that of the DNA?
Explain.
22. If each of the three major classes of RNA found in a cell were hybridized to denatured DNA from the same cell and the presence of RNA-DNA hybrids were tested, which of the classes would be retained on the filter?
(a) mRNA (b) rRNA (c) tRNA RNA Polymerases and Transcription 23. Which of the following are required for the DNA-dependent RNA polymerase reaction to produce a unique RNA transcript?
(a) ATP (g) RNA (b) CTP (h) Mg2+ (c) GTP (i)
promoter sequence (d) dTTP (j)
operator sequence (e) UTP (k) terminator sequence (f)
DNA
24. What is the sequence of the mRNA that will be synthesized from a template strand of DNA having the following sequence:
…ACGTTACCTAGTTGC…?
25. Describe the mechanism of chain growth in RNA synthesis. What is the polarity of syn thesis and how is it related to the polarity of the template strand of DNA?
CHAPTER 5
The Genetic Code and Protein Synthesis 26. Which of the following are characteristics or functions of tRNA?
(a) It contains a codon.
(b) It contains an anticodon.
(c) It can become covalently attached to an amino acid.
(d) It interacts with mRNA to stimulate transcription.
(e) It can have any of a number of different sequences.
(f) It serves as an adaptor between the information in mRNA and an individual amino acid.
27. What is the minimum number of contiguous nucleotides in mRNA that can serve as a codon? Explain.
28. What is the sequence of the polypeptide that would be encoded by the DNA sequence given in Question 24? Assume that the reading frame starts with the 5′ nucleotide given.
The genetic code is given on page 134 of the text.
29. The following is a partial list of mRNA codons and the amino acids they encode: AGU = serine AGC = serine AAU = asparagine AAC = asparagine AUG = methionine AUA = isoleucine
Based on this list, which of the following statements are correct?
(a) The genetic code is degenerate.
(b) The alteration of a single nucleotide in the DNA directing the synthesis of these codons could lead to the substitution of a serine for an asparagine in a polypeptide.
(c) The alteration of a single nucleotide in the DNA directing the synthesis of these codons would necessarily lead to an amino acid substitution in the encoded polypeptide.
(d) A tRNA with the anticodon ACU would be bound by a ribosome in the presence of one of these codons.
30. Explain why mitochondria can use a genetic code that is different from the standard code used in the nucleus.
Introns and Splicing
31. Explain how genetic techniques and amino acid sequence analyses could be used to show the collinear relationship of a prokaryotic gene and the protein it encodes.
32. Answer the following questions about what was revealed when DNA encoding the gene for the b-chain of hemoglobin and the mRNA for the b-chain were compared.
(a) What was the major finding when the nucleotide sequence of the gene and the amino acid sequence of the b-chain were compared?
(b) What did hybridization between the partially denatured DNA and the mRNA for b-globin show?
(c) What must happen to the primary transcript from the b-globin gene before it can serve as an mRNA for protein synthesis?
33. How might the fact that some exons encode discrete functional domains in proteins be related to the evolution of new proteins?
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 34. Spliceosomes (a) recombine DNA sequences in a process called exon shuffling.
(b) are composed of RNA and proteins.
(c) recognize RNA sequences that signal for the removal of introns.
(d) can produce different mRNA molecules by splicing at alternative sites.
ANSWERS TO SELF-TEST
1. (a) B (b) A, C (c) A (d) B, C, D (e) A (f) C (g) A (h) B (i) C; strictly speaking, A is called a dinucleotide, not a nucleotide (j) B (k) A, C, D 2. a, c, d, and e. Answer (b) is not correct because A pairs with T and G pairs with C.
Answer (f) is not correct because the sequence of one strand determines the sequence of the other by base pairing.
3. GGTTACGT. The convention for indicating polarity is that the 5′-end of the sequence is written to the left. The two chains of the Watson-Crick double helix are antiparallel, so the correct complementary sequence is not TGCATTGG.
4. The space between the two deoxyribose-phosphodiester strands is precisely defined. This distance is not large enough for two purines to hydrogen-bond. Conversely, two pyrimidine bases would not be close enough to form stable hydrogen bonds. Furthermore, in the double-strand structure the hydrogen-bond donor and acceptor groups are not properly aligned to form stable G T or A C base pairs.
5. (a) 1, 2, 3 (b) 4, 6. Answer (5) does not apply because DNA does not contain uracil.
6. 102 cm. The math is as follows: (3 × 106 kb × 103 bases/kb × 3.4 Å/base × 10−8 cm/Å) = 102 cm 7. When replication occurs, the two strands of the Watson-Crick double helix must sepa rate so that each can serve as a template for the synthesis of its complement. Since the two strands are complementary to one another, each bears a definite sequence relationship to the other. When one strand acts as a template, it directs the synthesis of its complement. The product of the synthesis directed by each template strand is therefore a duplex molecule that is identical to the starting duplex. The process is accurate because of the specificity of base pairing and because the protein apparatus that catalyzes the replication can remove mismatched bases.
8. After two generations, you should expect to find equal amounts of light-density DNA, in which both strands of each duplex were synthesized from 14N precursors, and intermediate-density DNA, in which each duplex consists of a heavy 15N strand paired with a light 14N strand.
9. a, b, and d. Answer d is correct because, if at least one discontinuity exists in the phos phodiester backbone of either chain of a circular duplex molecule, the chains are free to rotate about one another to assume the relaxed circular form. Answer c is incorrect because supercoiling requires closed circular molecules. In a linear molecule, the ends of each strand are not constrained with respect to rotation about the helical axis; therefore, the molecule cannot be supercoiled.
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10. (a) The bacterial DNA solution has the higher Tm value because it has the higher G + C content and is therefore more stable to the thermal-induced separation of its strands because G C base pairs are more stable than A T base pairs.
(b) The complementary DNA strands from each species will anneal to form Watson Crick double helices as the solution cools.
(c) No; each strand will find its partner because the perfect match between the linear arrays of the bases of complementary strands is far more stable than the mostly imperfect matches in duplexes composed of one strand of bacterial and one strand of mammalian DNA would be.
11. a, c, and e. Answer (b) is not correct because, although the enzyme requires a primer, the nature of its 5′-end is irrelevant since dNMP residues are added to its 3′-end. A primer with a 3-OH is required. Answer (d) is not correct because the enzyme uses dNTP and not NTP molecules, where N means A, C, G, T, or U.
12. The 3′-hydroxyl of the terminal nucleotide of the primer makes a nucleophilic attack on the innermost phosphorus atom of the incoming dNTP that is appropriate for WatsonCrick base pairing to the template strand to form the phosphodiester bond. As a result, a dNMP residue is added onto the 3′-end of the primer with the concomitant release of PPi, and the chain grows in the 5′ 3′ direction.
13. After the completion of one round of synthesis, the template strand will have directed the polymerization of a complement in which C = 18.5%, G = 24.1%, T = 32.8%, and A = 24.6%. Since the primer is short with respect to the template, its contribution to the composition of the product strand can be neglected.
14. (a) 1, 2, 3, 5 (b) 1, 2, 3, 4, 5, 6 15. The single-strand DNA penetrates the cell, where it has converted by enzymes to a du plex replicative form through Watson-Crick base pairing. The replicative form is then reproduced by a mechanism similar to that used for the semiconservative replication of the duplex chromosome of the host cell. Finally, after this stage of replication, the mechanism shifts to one in which the replicative form serves as a template to produce copies of the single-strand DNA found in the mature virus.
16. Starting with the single-strand RNA in the virus and ending with the single-strand RNA in the progeny viruses, the order in which genetic information flows during the infection of a cell with a retrovirus is: b, c, a, d, c, b. Retroviruses use the enzyme reverse transcriptase to convert their single-strand genomes into a DNA-RNA replicative form that is subsequently converted into a duplex DNA replicative form prior to insertion into the host chromosome and ultimate reconversion into the single-strand viral RNA by a DNA-dependent RNA polymerase.
17. a and b. Answer (b) is correct because the reverse transcription of RNA sequences into DNA sequences occurs during the replication of retroviruses. The term transcription is usually used to describe the formation of RNA from a DNA duplex by RNA polymerase.
18. d
19. (a) The bond is called the phosphodiester bond.
(b) The bond joins the 3′-hydroxyl to the 5′-hydroxyl to form a 3′ 5′ phospho diester bond.
(c) Hairpin loops are formed when the RNA chain folds back upon itself and some of the bases become hydrogen bonded to form an antiparallel duplex stem with unpaired bases forming loop at one end.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (d) A pairs with U, and G pairs with C; G can also pair with U, but the association is weaker than that of the G C base pair.
(e) The three major classes of RNA found in a cell are mRNA, rRNA, and tRNA; the most abundant is rRNA.
20. You could sequence the RNA and DNA and compare the sequences of each to see if the two are complementary; this method provides definitive evidence of identity. An easier but less precise way would be to use hybridization. You would mix the samples, heat the mixture to melt the double-strand DNA and RNA hairpins, slowly cool the solution, and then examine it to see if it contains double-strand DNA-RNA hybrids. Such hybrids would indicate that the RNA and DNA sequences are complementary.
21. Not necessarily; RNA synthesis is asymmetric, and only one strand of any region of the DNA serves as a template. This can lead to RNA with a G + C composition different from that of the duplex DNA.
22. a, b, c. All cellular RNA is encoded by the DNA of the cell.
23. a, b, c, e, f, h, i, and k. Answer (f) is correct because DNA is needed to serve as the tem plate. Answers (k) and (i) are correct because the promoter and terminator sequences are needed to specify the precise start and stop points, respectively, for the transcription.
24. The mRNA sequence will be …GCAACUAGGUAACGU…, written in the 5′ 3′ di rection.
25. The 3′-hydroxyl terminus of the growing RNA chain makes a nucleophilic attack on the a-phosphate (the innermost phosphate) of the ribonucleoside triphosphate that has been selected by base pairing to the template strand of the duplex DNA. RNA polymerase catalyzes the reaction. A ribonucleoside monophosphate residue is added to the chain as a result, and the chain has grown in the 5′ 3′ direction; that is, the chain has grown at its 3′ end. As with all Watson-Crick base pairing, the strands are antiparallel; that is, the RNA chain is assembled in the 3′ 5′ direction with respect to the polarity of the template strand of the DNA.
26. b, c, e, and f. Answer (d) is incorrect because the interaction of tRNA with mRNA takes place during translation, not transcription.
27. Three contiguous nucleotides is the minimum that can serve as a codon. There are four kinds of nucleotides in mRNA. A codon consisting of only two nucleotides (either of which could be any of the four possible nucleotides) allows only 16 possible combinations (4 × 4 = 16). This would not be sufficient to specify all 20 of the amino acids. A codon consisting of three nucleotides, however, allows 64 possible combinations (4 × 4 × 4 = 64), more than enough to specify the 20 amino acids.
28. The sequence of the polypeptide would be Ala-Thr-Arg. The reading frame is set by the nucleotide at the 5′ end of the mRNA transcript; the fourth codon of the mRNA transcript is UAA, which is a translation termination codon.
29. a, b, and d. Answer (a) is correct because both AGU and AGC specify serine; since more than one codon can specify the same amino acids, the genetic code is said to be degenerate. Answer (b) is correct because the alteration of a single nucleotide in the DNA could change a codon on the mRNA transcript from AGU, which specifies serine, to AAU, which specifies asparagine. Answer (d) is correct because the anticodon ACU would base-pair with the codon AGU. Answer (c) is not correct because the alteration of a single nucleotide in the DNA could result in another codon that specifies the same amino acid; for example, a codon changed from AGU to AGC would continue to specify serine.
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30. Mitochondria can use a genetic code that differs from the standard code because mito chondrial DNA encodes a distinct set of tRNAs that are matched to the genetic code used in their mRNAs.
31. The mutations in a given gene of E. coli could be mapped by recombination analysis. The proteins encoded by the wild-type and the mutant genes could then be sequenced, and the location and nature of the amino acid substitution for each mutation identified. The result would be that the order of the mutations on the genetic map is the same as the order of the corresponding changes in the amino acid sequence of the polypeptide produced by the gene; these experiments established that genes and their polypeptide products are collinear in prokaryotes.
32. (a) The number of nucleotides in the gene was significantly greater than three times the number of amino acids in the protein. There were two stretches of extra nucleotides between the exon sequences that encode the amino acids in the b-chain.
(b) The mRNA hydridized to the DNA under conditions where DNA-RNA hybrids are more stable than DNA-DNA hybrids, but there were sections of duplex DNA between the hybrid regions. This indicated that there are intron sequences in the DNA that have no corresponding sequences in the mRNA. (See Figure 5.33 in the text.)
(c) The intervening sequences (introns) in the nascent or primary transcript, which are complementary to the template strand of the DNA of the gene but do not encode amino acids in the protein, must be removed by splicing to generate the mRNA that functions in translation.
33. The shuffling of exons that encode discrete functional domains, such as catalytic sites, binding sites, or structural elements, preserves the functional units but allows them to interact in new ways, thereby generating new kinds of proteins.
34. b, c, and d. (a) is incorrect because exon shuffling takes place at the DNA level through breakage and rejoining of DNA not RNA.
PROBLEMS
1. The genome of the mammalian virus SV40 is a circular DNA double helix containing 5243 base pairs. When a solution containing intact DNA molecules is heated, one observes an increase in the absorbance of ultraviolet light at 260 nm. When the solution is then cooled slowly, a decrease in absorbance is observed. If one or more breaks are made in the sugar-phosphate backbones of the SV40 double-strand circles, heating causes a similar hyperchromic effect. However, when the solution of nicked molecules is cooled, the reduction in absorbance is much slower than that observed in the solution containing intact molecules. Why do the two types of molecules behave differently when they are cooled after heating?
2. A number of factors influence the behavior of a linear, double-strand DNA molecule in a 0.25M sodium chloride solution. Considering this, explain each of the following observations.
(a) The Tm increases in proportion to length of the molecule.
(b) As the concentration of sodium chloride decreases, the Tm decreases.
(c) Renaturation of single strands to form double strands occurs more rapidly when the DNA concentration is increased.
(d) The Tm value is reduced when urea is added to the solution.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 3. (a) Many proteins that interact with double-strand DNA bind to specific sequences in the molecule. Why is it unlikely that these enzymes operate by sensing differences in the diameter of the helix?
(b) What other features of the double-strand helix might be recognized by the protein?
4. You have a double-strand linear DNA molecule, the appropriate primers, all the enzymes required for DNA replication, four 32P-labeled deoxyribonucleoside triphosphates, Mg2+ ion, and the means to detect newly synthesized radioactive DNA. Why is this system not sufficient to distinguish between conservative and semiconservative replication of the DNA molecule?
5. Certain deoxyribonucleases cleave any sequence of single-strand DNA to yield nucleo side monophosphates; these enzymes do not hydrolyze base-paired DNA sequences.
What products would you expect when you incubate a solution containing a singlestrand specific deoxyribonuclease and the following oligodeoxyribonucleotide?
5′-ApGpTpCpGpTpApTpCpCpTpCpTpApCpGpApCpTp-3′
6. Formaldehyde reacts with amino groups to form hydroxymethyl derivatives. Would you expect formaldehyde to react with bases in DNA? Suppose you have a solution that contains separated complementary strands of DNA. How would the addition of formaldehyde to the solution affect reassociation of the strands?
7. When double-strand DNA is placed in a solution containing tritiated water, hydrogens associated with the bases readily exchange with protons in the solution. The greater the percentage of AT base pairs in the DNA, the greater the rate of exchange. Why?
8. While many experiments were suggesting that DNA in chromosomes is very long and continuous, it was established that DNA polymerase adds deoxyribonucleotides to the 3′-hydroxyl terminus of a primer chain and that a DNA template is essential. Why did those investigators interested in DNA replication focus a great deal of attention on determining whether chromosomal DNA contained breaks in the sugar-phosphate backbone?
9. The value of the Tm for DNA in degrees Celsius can be calculated using the formula,
T =
m
69.3 + 0.41(G + C), where G + C is the mole percentage of guanine plus cytosine.
(a) A sample of DNA from E. coli contains 50 mole percent G + C. At what tempera ture would you expect this DNA molecule to melt?
(b) The melting curves for most naturally occurring DNA molecules reveal that their Tm values are normally greater than 65ºC. Why is this important for most organisms?
10. During early studies of the denaturation of double-strand DNA, it was not known whether the two strands unwind and completely separate from each other. Suppose that you have double-strand DNA in which one strand is labeled with 14N and the other is labeled with 15N. If density-gradient equilibrium sedimentation can be used to distinguish between both double- and single-strand molecules of different densities, how can you determine whether DNA strands separate completely after denaturation?
11. Under strongly acidic conditions, several atoms of DNA bases are protonated; these include the N-1 and N-7 of adenine, the N-7 of guanine, the N-3 of cytosine, and the O-4 of thymine. Predict the effects of such protonations occurring at low pH on the stability of double-strand DNA.
12. Sol Spiegelman found that some types of single-strand RNA can associate with single strand DNA to form double-strand molecules. What is the most important condition that must be satisfied in order to allow the formation of these hybrid molecules?
CHAPTER 5
13. Many cells can synthesize deoxyuridine 5′-triphosphate (dUTP). Can dUTP be used as a substrate for DNA polymerase? If so, with which base will uracil pair in newly replicated DNA?
14. The DNA of bacteriophage l is a linear double-strand molecule that has complementary single-strand ends. These molecules can form closed circular molecules when two “cohesive” ends on the same molecule join, and they can form linear dimers, trimers, or longer molecules when sites on different molecules are joined.
(a) What conditions should be chosen to insure that l phage DNA molecules form closed-circular monomers?
(b) Under certain conditions, l phage DNA molecules are infective. When a very low concentration of l phage DNA is incubated with DNA polymerase I and the four deoxyribonucleoside triphosphates, the infectious activity of l phage DNA is destroyed. Brief treatment of l phage DNA with bacterial exonuclease III, an enzyme that removes 5′-mononucleotides from the 3′-ends of double-strand DNA molecules also destroys infectivity, but subsequent treatment of the DNA with DNA polymerase I and nucleotide substrates can restore infectivity. Describe more completely the structure of l phage DNA, and provide an interpretation of the action of the two enzymes on the molecule.
15. The isolation of viral DNA from animal cells that have been infected with adenovirus yields linear double-strand molecules that, when denatured and allowed to reassociate under conditions favoring intramolecular annealing, form single-strand circles.
Although circular molecules can be detected using the electron microscope, resolution is not sufficient to visualize the ends of the molecule. Other analyses of the singlestrand molecule show that each end has a sequence that allows the structure shown in Figure 5.2 to form.
FIGURE 5.2 A single-strand circle formed by intra-molecular annealing of adenovirus DNA.
Double-stranded
region (a) Suppose that the base sequence at one end of the single-strand molecule is 5′-AC TACGTA…. What is the corresponding sequence at the other end? Show how these sequences would allow full-length, double-strand linear molecules to be formed.
(b) An alternate suggestion for the formation of the single-strand molecules was also proposed; it is shown in Figure 5.3. Why is this proposed pairing scheme unlikely?
FIGURE 5.3 Another proposal for formation of single-stranded molecules of adenovirus DNA.
a
b
c a„
b„ c„
3„
5„
5„
3„ a„
b„ c„ a
b
c
Denature
Anneal
a
b
c
c„
b„ a„ 3„
5„
+ 5„
3„ a„
b„ c„ c
b
a
16. Thermoacidophilic bacteria can grow in volcanic sulfur springs at pH 2 and at tempera tures as high as 85ºC. DNA polymerase purified from the thermophile Snifolobus acidocaldarius has an optimal activity at 70ºC and is stable at 80ºC. When incubated with a circular DNA template at 100ºC, the isolated polymerase can extend a 20-nucleotide primer by more than 100 nucleotides. These experiments require that enzyme-to-primer concentration be at least 1:1. The Tm value for the double-strand DNA used in the experiment is about 60ºC. Unlike DNA polymerase I from E. coli, DNA polymerase from S. acidocaldarius has no demonstrable exonuclease activity to correct mistakes in DNA by removing mismatched nucleotides (such an enzyme activity is often referred to as proofreading).
(a) Why should the ratio of enzyme to primer be 1 in order for primer extension to take place at 100ºC?
(b) Would you expect to find an auxiliary proofreading enzyme in S. acidocaldarius? Why?
(c) Would you expect DNA from the genome of S. acidocaldarius to have a G + C con tent higher or lower than that from a bacterium that grows at a more normal temperature? Why?
17. Terminal deoxynucleotidyl transferase (TdT), an enzyme found in bone marrow and thy mus tissue, can extend a DNA primer by 5′ 3′ polymerization using deoxyri bonucleoside triphosphates as substrates. The primer must be at least three nucleotides CHAPTER 5 in length and must have a free 3′-OH end. The enzyme does not require a template nor does it copy one.
(a) Compare TdT with DNA polymerase I.
(b) Would TdT be useful for synthesizing DNA molecules that carry genetic informa tion? Why?
18. The 2′,3′-dideoxynucleosides can be used as reagents to inhibit DNA replication. These analogs must be converted to dideoxynucleoside triphosphates in order to have a measurable effect on DNA synthesis. When incorporated into a growing DNA chain, a single dideoxyribonucleoside residue can effectively block subsequent chain extension.
(a) Why must a 2′,3′-dideoxyribonucleoside be converted to a dideoxyribonucleoside triphosphate to be incorporated into DNA?
(b) What feature of a 2′,3′-dideoxynucleoside is most likely to account for inhibition of DNA chain extension?
FIGURE 5.4
HOCH
Base
2
O
J
H
H
J
H
H
J
J
H
H
2„, 3„-Dideoxyribonucleoside 19. In each chain-elongation reaction catalyzed by DNA polymerase, a phosphodiester bond is formed and pyrophosphate is concomitantly released. Hydrolysis of pyrophosphate to two molecules of inorganic phosphate occurs rapidly because most cells have a potent pyrophosphorylase. By removing one of the products of the chain-elongation reaction, pyrophosphate cleavage in the cell is partially responsible for the forward progress of polymerization. However, isolated DNA polymerases can efficiently carry outchain extension in the absence of pyrophosphate cleavage, so long as the double-strand helix is allowed to form during elongation. What forces resulting from DNA helix formation might contribute to driving polymerization forward?
20. In his studies of DNA in the late 1940s, Erwin Chargaff established that DNA from all organisms has equal numbers of adenine and thymine bases and equal numbers of guanine and cytosine bases. Considering that thymine and uracil are equivalent in their abilities to form hydrogen bonds with adenine, state whether you would expect similar constraints on base composition to be found in the following: (a) single-strand RNA from tobacco mosaic virus.
(b) the DNA-RNA hybrid molecule synthesized by reverse transcriptase.
(c) RNA from a virus in the reovirus family, which have large genomes composed of double-strand RNA molecules.
21. Certain DNA endonucleases degrade double-strand DNA to yield mononucleotides and dinucleotides, but these enzymes do not degrade those duplex sequences to which other proteins are tightly bound.
(a) How can you use such a DNA endonuclease and RNA polymerase to locate a promoter site?
(b) Why should this process be performed in the absence of ribonucleoside triphosphates?
22. The amino acid at position 102 in the primary sequence of a bacterial enzyme is valine, and the corresponding codon in the mRNA sequence for the enzyme is GUU. Suppose a mutation that alters the codon to GCU has no effect on the activity of the enzyme, but another mutation that changes the codon to GAU completely inactivates the enzyme.
Briefly explain these observations.
23. It is essential for spliceosomes to remove introns precisely, that is, between the terminal nucleotide of an intron and the first nucleotide of an exon. To see why, suppose that the sequence at the normal junction in a pre-spliced mRNA between an intron and an exon is …UUAG GCUAACGG… Suppose further that a spliceosome occasionally miscleaves the
Intron
Exon pre-mRNA transcript between the C and U residues in the exon sequence to yield the following two splicing intermediates:
…UUAGGC
UAACGG…
What would be the consequence of this cleavage?
24. Although nearly all the proteins synthesized by a bacterial cell after it has been infected with T2 bacteriophage are determined by the viral genome, some bacterial proteins are also required for successful infection. What bacterial enzyme is needed to initiate viral infection when T2 DNA first enters the cell?
25. (a) The genome of bacteriophage FX174 is a single strand of DNA containing 5386 nucleotides. If only one AUG in the genome were used as an initiation signal, how many amino acids could be encoded by the genome? If the average molecular weight of an amino acid is 112, what is the maximum molecular weight of protein encoded by the genome?
(b) Studies have shown that the FX174 genome can encode a larger number of proteins than expected. One reason for this increased encoding capacity is that some of the genes overlap each other. For example, the coding sequence for gene B is located entirely within the sequence that codes for gene A. However, the amino acid sequences of the two proteins specified by these genes are entirely different. How is this possible?
26. In contrast to DNA polymerase, RNA polymerase has no nuclease capability to excise mismatched nucleotides. Suggest why the two enzymes are different in this respect.
27. The genome of bacteriophage G4 is a small, single-strand circle of DNA. Replication of the circle is initiated when an RNA polymerase, a product of the dnaG gene of E. coli, synthesizes a small segment of RNA that binds to a unique sequence on the G4 chromosome.
Initiation of G4 DNA synthesis does not occur in bacterial dnaG gene mutants, which have an inactive RNA polymerase. Suggest a function for the small segment of RNA.
28. (a) In E. coli, a tRNA that carries tyrosine is composed of 85 nucleotides. However, transcription of the gene that codes for tyrosine tRNA yields an RNA molecule consisting of 350 nucleotides. At least three ribonuclease enzymes cooperate in removing a 41-base segment on the 5′ side of the tRNA sequence and a 224-base segment that extends from the 3′ terminus of the tRNA sequence. The tRNA sequence in the primary transcript is continuous, and no nucleotides are removed from that part of the transcript during processing. How does this type of RNA processing differ from splicing?
(b) Another primary transcript that is synthesized in bacteria contains 6500 nucleotides, including sequences for the 23S, 16S, and 5S RNA molecules found in ribosomes.
This primary transcript has sequences on either side of the set of rRNA sequences, as well as “spacer” sequences between each of them. Suggest a reason for the synthesis of a transcript containing all three rRNA sequences.
CHAPTER 5
29. The codons UAA, UAG, and UGA are signals for chain termination in protein synthesis because none of these codons are read by tRNA molecules. These codons are normally found at the ends of coding sequences for proteins. However, single-base mutations in certain codons can also cause premature termination of the protein chain.
(a) Which codons can be converted to the chain-termination codon UAA by a single base change?
(b) Suppose a mutation creates a UAA codon that is three codons away from the 3′ end of the normal mRNA coding sequence. Why might you assume that the prematurely terminated protein might still be functional?
(c) Revertants of chain-termination mutants include those in which a single-base sub stitution changes a termination codon to one that can again be read by a tRNA molecule. For example, a UAG codon can mutate to UCG. What amino acid would then be found at the corresponding position in the protein?
(d) Other revertants retain the original termination codon at the premature termination site, but an amino acid is inserted at the corresponding site in the protein so that the protein has the same length as the nonmutant protein would have. These revertants are due to another mutation in which the anticodon of a tRNA molecule is altered so that the tRNA molecule can read a termination codon. These tRNA molecules are called suppressor tRNAs because they suppress the effect of a chain-termination mutation. Suppose you have a chain-termination mutation that is due to the presence of a UAG codon in the normal coding sequence. If the effect of the UAG codon is suppressed by a tRNA mutation, which amino acids could be found at the site corresponding to the premature termination signal? Assume that a single base change occurs in each case.
30. Polynucleotide phosphorylase, which polymerizes ribonucleoside diphosphates (NDP) to form RNA and Pi, was used in the laboratory to synthesize polyribonucleotides that were useful in determining the genetic code. Why is it unlikely that this enzyme synthesizes RNA in the cell? Suggest how the cell uses this enzyme.
31. (a) When an experiment was done to form hybrids between mRNA produced after bac teriophage T2 infection and the denatured T2 genomic DNA (mRNA was in molar excess over DNA strands), significantly less than 100% of the DNA could form a DNA-RNA hybrid. What did this suggest about whether transcription takes place on one or both of the two DNA strands at any location on the chromosome?
(b) Later the principle of transcription on only one strand of DNA was established firmly by studies with DNA from the virus SP8, which infects the bacterium Bacillus
subtilis. Because the two complementary strands of SP8 DNA have very different base compositions, they can be easily separated by density gradient centrifugation.
How could you use these separated strands to show that the transcription of SP8 DNA occurred on one strand only?
32. You are studying the effects of amino acid replacements on the stabilbity of a particu lar a helix that is buried in the protein myoglobin. You carry out a series of replacements of a particular leucine residue located in the helix, using site-specific mutagenesis, a technique described in detail in Section 6.4 of the text. The replacements are as follows: (i)
leucine
arginine (ii) leucine
valine (iii) leucine
proline (iv) leucine
glycine (v) leucine
alanine (a) For each replacement, predict whether the change would stabilize, destabilize, or have no effect on the structure of the a helix. Briefly explain each of your predictions.
(b) For each replacement, write the most likely mRNA codon required to code for the particular amino acid. Which of the replacements can be carried out by single-base changes? Which of the replacements can be produced only by altering two bases in the mRNA codon?
33. Cordycepin (3′-deoxyadenosine) is a compound that can block the synthesis of RNA, because a cordycepin residue in an RNA chain lacks the 3′-OH end needed for chain extension by RNA polymerase. The structure of cordycepin is shown below.
NH
2
J
C
N
N
C
CH
HC
C
N
N
HOCH
2
O
J
H
H
J
H
H
J
J
H O H (a) Cordycepin does not inhibit the growth of bacteria, but it does inhibit growth and division of mammalian cells. Consider the reactions that are required for cordycepin to be converted into a substrate for RNA polymerase and then propose a reason for its ineffectiveness in bacteria.
(b) Would you expect cordycepin to block DNA synthesis as well? Why?
34. Raney nickel can convert cysteinyl-tRNACys to alanyl-tRNACys. When this altered aminoa cyl-tRNA is used in a protein-synthesizing system in vitro, alanyl residues are placed in the position normally occupied by cysteinyl residues in the protein. What does this experiment tell you about the ability of the protein-synthesizing machinery to recognize an inappropriate aminoacyl-tRNA like alanyl-tRNACys.
35. The products of the cleavage of RNA in dilute alkali include 2′- and 3′-monophosphates.
What does this observation reveal about the mechanism of cleavage of RNA? How might sensitivity to base-catalyzed cleavage have affected the choice of DNA or RNA as the primary carrier of genetic information?
36. Many steps in the flow of genetic information are subject to regulation. Stringent con trol of the production of macromolecules limits expenditure of energy by the cell, permitting the synthesis of particular proteins only as they are required. Consider the steps in storage and transmission of genetic information, and describe which one, when regulated, makes it possible to achieve the greatest economy in energy expenditure by a mature cell.
37. In 1971, David Baltimore was investigating whether polymerase activities were con tained in the Rauscher murine leukemia virus. This virus has an RNA genome and causes leukemia in mice. He disrupted purified virus particles and incubated the resulting mixture with Mg2+ and either the four dNTPs or the four NTPs in a buffered solution. One of the dNTPs or one of the NTPs was radiolabeled. After allowing time for a reaction to occur, the mixtures were treated with strong acid to precipitate nucleic acids while leaving unreacted nucleoside triphosphates in solution. By measuring the precipitated radioactivity, this assay allowed him to detect the formation of the product of a putative polymerase. He found the following: (1) NTPs were not incorporated into CHAPTER 5 product; (2) dNTPs were incorporated into product; (3) the isolated, radiolabeled product was destroyed by DNase (an enzyme that hydrolyzes DNA) but not by RNase (an enzyme that hydrolyzes RNA); (4) the isolated product was not destroyed by NaOH; (5) pretreatment of the disrupted virus extract with DNase did not prevent the formation of product whereas pretreatment with RNase did.
Do these experiments suggest the presence of a polymerase? Why? What kind of poly merase is likely present? What is its template and what is the product formed? What did these experiments indicate, for the first time, about the flow of genetic information?
38. Radioisotopes have been critical for identifying specific molecules involved in biochemi cal processes. John Hershey and Martha Chase carried out an experiment in 1952 with bacteriophage T2 that had been radiolabeled by being grown in either 32PO2−
2−
4 - or 35SO4 -containing medium. Bacteriophage T2 has a DNA genome. After infecting the bacterial cells in separate cultures with the two different labeled virus preparations for a time short enough to ensure that newly made viruses did not develop to the point of lysing the cells, they put the culture of infected cells in a blender to strip off any part of the virus that did not enter the cell. They next collected the infected stripped cells by centrifugation and compared the amounts of radioisotope in the cells to that remaining in the supernatant.
What do you think they observed and why? Why was this experiment important?
ANSWERS TO PROBLEMS
1. When the intact double-strand circular DNA molecule is heated in solution, its base pairs are disrupted and an increase in the absorbance of light at 260nm is observed.
However, the two resulting single-strand circles are so tangled about one another that they remain closely associated. When the molecules are cooled, the interlocked strands move relative to each other until their base sequences are properly aligned and a double-strand molecule is reformed. This molecule absorbs less ultraviolet light at 260nm (hypochromism) than does the pair of denatured single strands. Breaks in one or both strands of a double-strand DNA molecule allow the two strands to separate completely from one another during denaturation. In order to form a doublestrand molecule, the separate strands collide randomly until at least a small number of correct base pairs is formed (nucleation); then the remaining bases pairs form to generate a completely double-strand molecule. Reassociation of a pair of separate strands in solution is slower than that of a pair of interlocked circles because the local concentration of strands is lower, so a corresponding difference in the reduction of absorbance will be observed.
2. (a) The longer the DNA molecule, the larger the number of base pairs it contains. As a result, more thermal energy is required to disrupt entirely the helical structure of the longer DNA molecule. Experiments show that such a relationship is true for molecules up to ~4000 base pairs in length.
(b) Sodium ions neutralize the negative charges of the phosphate groups in both strands. As the concentration of NaCl decreases, repulsion between the negatively charged phosphate groups increases, making it easier to separate the two strands.
The tendency for the strands to separate more easily means that dissociation occurs at a lower temperature, which is reflected in a lower Tm value of the molecule.
(c) The reassociation of single strands begins when a short sequence of bases in one strand forms hydrogen bonds with a complementary sequence in another. Once a short stretch of base pairs is formed, reassociation to form the longer double-strand molecule occurs rapidly. The higher the concentration of DNA, the greater the number of complementary sequences in the solution, and thus the quicker the complementary sequences will find and pair with each other.
(d) Urea, which contains hydrogen bond donors (−NH2) and hydrogen bond acceptors ( > C = O), disrupts the hydrogen bonds between bases. Because hydrogen bonds are partly responsible for the stability of the double helix, the disruption of these bonds makes the structure more sensitive to denaturation by thermal energy and thereby reduces the Tm value. In addition to hydrogen bonding, the tendency of bases to stack also contributes significantly to the stability of the helix. Base stacking minimizes the contact of the relatively insoluble bases with water, and it also allows the sugar-phosphate chain to be located on the outside of the helix, where it can be highly solvated. Urea may also cause destabilization of the helix by allowing bases to associate more readily with water by disrupting its structure.
3. (a) The four base pairs found in the DNA double helix are almost identical in size and shape, so the diameter of the double helix is essentially uniform all along its length.
It is therefore unlikely that a protein can identify a specific sequence by sensing differences in the diameter of the helix.
(b) Proteins that interact with specific sequences might do so by forming hydrogen bonds with the bases; in some cases, it might be necessary for the double strand to undergo local unwinding or melting in order for the bases to form hydrogen bonds with a protein. However, hydrogen-bond donors and acceptors are also found in the grooves of the intact helix. A protein could also bind to a specific location on DNA by forming hydrogen bonds with a particular group of atoms in one of the grooves of the helix. Hydrophobic interactions between amino acid side chains and the methyl group of thymine or the edges of the bases can also contribute to the specificity of the interaction.
4. Although the system described could yield 32P-labeled daughter DNA molecules, chem ical methods cannot distinguish DNA in which both strands are radioactively labeled from DNA in which one strand is labeled and one strand is unlabeled. In their experiments, Meselson and Stahl used a physical technique, density gradient equilibrium sedimentation, to separate the labeled molecules according to their content of 14N and 15N, which differ in their specific densities.
5. In solution, the oligodeoxyribonucleotide forms an interchain double-strand molecule with flush ends and a small single-strand loop containing the sequence 5′pTpCpCpTpCp-3′. The deoxyribonuclease hydrolyzes the phophodiester bonds in this single-strand region to form nucleoside monophosphates, leaving a small double-strand linear molecule remnant containing seven base pairs.
6. Formaldehyde could react with the exocyclic amino groups on the C-6 carbon of adenine, the C-2 of guanine, and the C-4 of cytosine to form hydroxymethyl derivatives. Because these derivatives cannot form hydrogen bonds with complementary bases, formaldehyde-treated single strands would reassociate to a lesser extent than would untreated single strands. The actual sites of the reaction of formaldehyde with DNA are not precisely known; these sites may also include the ring nitrogen atoms in pyrimidines.
7. This experiment suggests that the hydrogen bonds of base-paired regions of double strand DNA may undergo reversible dissociation to form single-strand regions, often known as bubbles. The transient disruption of these hydrogen bonds allows the exchange of protons with the tritiated water. A T pairs open more easily than G C pairs. Thus, the greater the percentage of AT pairs, the greater the rate of proton exchange.
CHAPTER 5
8. A continuous, linear double-strand DNA molecule has only two 3′-OH groups available for the initiation of DNA synthesis by DNA polymerase; because each is located at opposite ends of the molecule, no template sequence is available. In order to construct a relatively simple mechanism for chromosomal replication, one could postulate that the enzyme initiates DNA replication at a number of breaks along the chromosome, with each of the breaks offering the 3′-OH group required for the initiation of the new DNA strand. The template required for replication would then be located on the strand opposite the break, thus ensuring that DNA synthesis could continue. It is now well established that DNA in chromosomes is very long and continuous. The fact that there are initially no breaks in the molecule makes the mechanism of replication complex. It involves a number of enzyme activities, as well as the use of RNA to prime the synthesis of DNA. For details, see page 760 of the text.
9. (a) The expected melting temperature for E. coli DNA containing 50% GC base pairs is
T =
m
69.3 + 0.41(G + C) = 69.3 + 0.41(50) = 69.3 + 20.5 = 89.8ºC (b) Most organisms live at temperatures that are considerably lower than 65ºC. Because both the transmission and expression of genetic information depends on the integrity of the double-strand DNA molecule, it is important that the molecule not be disrupted by thermal energy.
10. First, you must determine the temperature at which the hydrogen bonds are disrupted and single strands are formed. You can do this by heating the double-strand DNA to various temperatures and measuring the extent of hyperchromicity. Once the DNA has been melted, centrifuge the sample using the density-gradient equilibrium sedimentation technique to attempt to separate the 14N-labeled DNA strands from the 15N-labeled DNA strands, which will be the denser of the two. If you are successful, this would suggest that the strands separate completely during thermal denaturation.
11. The protonation of the N-1 and N-7 of adenine, the N-7 of guanine, the N-3 of cyto sine, and the O-4 of thymine makes normal hydrogen bonding at these locations impossible because the atoms can no longer serve as hydrogen-bond acceptors. Therefore, at low pH, where proton concentrations are high and protonation of these atoms occurs, double-strand DNA is less stable than at neutral pH values. At high pH values ( > 11) DNA is also denatured by deprotonation of other ring atoms.
12. The association of a molecule of RNA with a molecule of DNA to form a hybrid mole cule depends primarily on the two molecules having complementary sequences of bases.
The formation of hydrogen bonds between complementary bases will allow the formation of a double helix composed of RNA and DNA.
13. The deoxyribonucleoside triphosphate dUTP can be used as a substrate for DNA poly merase during DNA replication because the structure and hydrogen-bonding properties of uracil are very similar to those of thymine. When incorporated into a double-strand DNA polymer, uracil pairs with adenine, as does thymine. For a discussion of the reasons uracil is not normally incorporated into DNA, see page 771 of the text.
14. (a) To insure that l phage DNA molecules form closed circular monomers, the con centration of l phage DNA should be relatively low so that the intrachain formation of hydrogen bonds is favored. At higher concentrations, the probability of interchain joining to form multimers is enhanced.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (b) The most reasonable model for the structure of the l phage DNA molecule is a dou ble-strand molecule having single-strand protrusions at the 5′-ends, as illustrated in the margin. The 3′-ends have hydroxyl groups, which allow them to serve as primers for DNA synthesis catalyzed by DNA polymerase I. This enzyme fills in the single-strand regions of the molecule, producing a molecule with flush ends. Such a molecule no longer has cohesive ends that can form the required circular molecule needed for infectivity.
3„
5„
5„
3„
Molecules treated with exonuclease III have a longer single-strand sequence at both ends; they may not be infective because the newly exposed bases may not be fully complementary to each other, which would mean that the ends could no longer be joined. When the exonuclease-treated DNA is treated with DNA polymerase I, the single-strand regions are sufficiently filled in to reform a molecule that has protruding single strands that are approximately the same length as those in the native molecule. Hence, the molecule once again becomes infective. Further treatment of the molecule with DNA polymerase I will once again produce a molecule that has flush ends and is no longer infective.
15. (a) The sequence at the other end of the single-strand molecule must be composed of complementary bases. It must therefore be
…TACGTAGT-3′
The structure of the full-length, double-strand, linear molecule would be 5′-ACTACGTA———TACGTAGT-3′
3′-TGATGCAT———ATGCATCA-5′
Each single strand has a pair of inverted repeats.
(b) The formation of double-strand helical segments depends upon hydrogen bond for mation between bases in nucleotide chains that are antiparallel, as follows:
5′…PuPyPuPyPu…3′
3′…PyPuPyPuPy…5′ where Py = pyrimidine and Pu = purine. When the suggested structure is labeled using this scheme, as in Figure 5.5, it can be seen that it would require the formation of base pairs between parallel chains, and such pairing cannot readily take place.
FIGURE 5.5 Base pairing between parallel nucleotide chains, required to form the circular structures shown in Figure 5.3, is unlikely in DNA.
(3„) 5„
3„ (5„) 16. (a) Each enzyme molecule probably tightly clamps the primer as well as the extended, newly synthesized chain to the template, thereby protecting the helix from denaturation at high temperature. One primer would be required for each template strand. The association between primer and enzyme may protect the enzyme from thermal denaturation as well. There are many examples in biochemistry where substrate binding stabilizes protein structure.
(b) During polymerization of a new DNA chain, the chance that base incorporation er rors can occur will increase at high temperatures. Even though a proofreading or error-correcting activity is not found in the S. acidocaldarius polymerase polypeptide, you would expect an auxiliary enzyme to be present in the cells in order for the bacterial genome to be accurately replicated so that the genetic integrity of the organism is maintained.
(c) The higher the G + C content of double-strand DNA, the higher the melting point at which the helix is denatured. Therefore, you might expect DNA from a thermophile to have a higher G + C composition. Surprisingly, DNA base ratios are sometimes not very different from those in bacteria living at lower temperatures, so that DNA isolated from these thermophilic bacteria melts at temperatures lower than those encountered in the hot springs where they grow. There must be proteins or other molecules in the bacterial cells that protect the genomic DNA of these thermophiles from thermal denaturation.
17. (a) Like DNA polymerase I, TdT can extend a DNA primer by using deoxynucleoside triphosphates as substrates. However, TdT does not use a template and cannot copy from one, so that the base composition of the newly synthesized single strand of DNA will depend solely on the relative concentrations of the deoxynucleoside triphosphate substrates. For chains synthesized by DNA polymerase I, the base composition will be complementary to that of the template strand. Although DNA polymerase has an exonuclease activity that removes mismatched bases from newly synthesized strands, TdT has no such activity and does not need one.
(b) DNA molecules that carry genetic information must be synthesized as faithful copies of template strands; TdT cannot copy a template and would not be useful in genomic DNA synthesis. TdT is sometimes used to introduce sequence variation into DNA during antibody formation.
18. (a) DNA polymerase requires deoxyribonucleoside triphosphates as substrates for DNA chain extension. Nucleosides like the 2′,3′-dideoxy analogs must be converted to nucleoside triphosphates in order to serve as substrates for DNA polymerase.
Studies on inhibition of DNA synthesis in living cells involve incubating those cells with the nucleoside forms of the analogs instead of their nucleoside triphosphate forms, because negatively charged phosphate anions cannot pass across the plasma membrane, while relatively neutral nucleosides can. Once inside the cell, nucleoside analogs are phosphorylated by cellular enzymes that normally function to “salvage” nucleosides generated by turnover of nucleotides from RNA and DNA.
(b) Dideoxynucleosides lack a free 3′-hydroxyl group, which would normally serve as an acceptor for incorporation of the next nucleotide into the growing polynucleotide chain. The lack of a 3′-OH group also interferes with excision by the error-correcting exonuclease activity of some DNA polymerases, so that chain extension is blocked.
19. Noncovalent forces also contribute to driving the reaction forward. Hydrogen bonds form between opposing A and T bases and between G and C molecules in the antiparallel chains. There are also significant hydrophobic interactions between adjacent bases on the same strand, and these stacking interactions may in fact contribute significantly to helix formation and stability. In vivo, it is likely that these noncovalent forces, along with the cleavage of the product, account for the forward progress of chain elongation.
20. (a) The ratios observed by Chargaff can be attributed to the requirement that in a dou ble-strand polynucleotide, only certain bases can form hydrogen bonds with one another. Although a single-strand polynucleotide might form some hydrogen bonds between bases as it folds, the overall base ratios will not conform to Chargaff’s established rules. This is also true for single-strand DNA in a virus like FX174.
(b) Because hydrogen bonding between A and T (or U, in the case of RNA) and be tween G and C can occur in a duplex molecule formed by one strand of DNA and a complementary strand of RNA, you would expect to see base ratios like those observed by Chargaff.
(c) Double-strand RNA molecules form hydrogen bonds between bases in a manner similar to those in DNA helices. Therefore, bearing in mind that U, not T, would normally be found in RNA, the number of uracil residues would equal the number of adenines, and the number of guanine bases would be expected to be the same as those of cytosine.
21. (a) To locate a promoter site, you would first incubate the double-strand DNA with RNA polymerase; the RNA polymerase will bind tightly to the promoter site. Next, you would add the DNA endonuclease, which will degrade the DNA that is not protected by the bound RNA polymerase. Electrophoresis can then be used to determine the size of the protected fragments of DNA, and the base sequence can be determined using methods discussed in Chapter 6 of the text.
(b) Ribonucleoside triphosphates are substrates for RNA polymerase transcription. If present when this process is performed, they would allow the polymerase molecule to move from the promoter site to the site on the template where transcription begins, as well as beyond, as transcription progresses. As a result, the promoter site would no longer be protected from endonuclease degradation.
22. The GCU codon in the first mutation corresponds to a substitution of alanine for valine at position 102. Although the side chain of alanine is smaller than that of valine, both are aliphatic amino acids, so the alteration in the structure of the enzyme does not necessarily affect the enzyme activity. The GAU codon of the second mutation specifies the substitution of aspartate for valine at position 102. This substitution could have a detrimental effect because the side-chain carboxyl group of aspartate has a negative charge at neutral pH. The charged group could disrupt the native conformation of the enzyme, thereby inactivating it.
23. If the spliceosome cleaves the initial transcript within the normal exon sequence as shown, the exon coding sequence in the spliced mRNA will be altered because of the loss of two bases. Instead of beginning with the codon GCU in the normal exon, the reading frame will begin with the codon UAA in the altered exon. This codon is, in fact, a termination signal for protein synthesis, which means that the translation of the polypeptide specified by the spliced messenger RNA would be terminated prematurely.
24. Because only DNA and no protein from T2 enters the cell, the synthesis of viral-directed proteins cannot begin until T2 messenger RNA has been made. Transcription of the T2 DNA must therefore be carried out by bacterial RNA polymerase using ribonucleoside triphosphates synthesized by the bacterial cell.
CHAPTER 5
25. (a) Three consecutive bases are required to encode an amino acid, so up to 1794 amino acids could be specified by the FX174 genome. The molecular weight of this much protein would be approximately 201,000 (b) Overlapping genes can yield proteins with different primary amino acid sequences only if each of the protein coding sequences is read in a different reading frame.
The critical feature for the establishment of the proper reading frame is the location of the AUG initiation signal. As an example, consider the following mRNA sequence, which contains two AUG codons that are in overlapping but different reading frames:
Reading frame 1
A U G C C U A G A U G C A G U U C G
Reading frame 2
When an initiator tRNA binds to the first AUG codon, reading frame 1 is established; similarly, reading frame 2 is established when an initiator tRNA binds to the second AUG codon. The polypeptides specified by the two different mRNA sequences will necessarily have different amino acid sequences.
26. DNA polymerase is responsible for the duplication of the DNA of chromosomes, which is the repository of the genetic information that is passed on to progeny cells. Any error that occurs in the copying of a DNA template will be transmitted not only to the duplicated chromosome but also to all the messenger RNA molecules transcribed from the miscopied DNA template. Therefore, it is crucial that DNA polymerase be able to correct errors that occur due to the incorporation of mismatched nucleotides. RNA polymerase makes many copies of mRNA, but these molecules have relatively brief lives in the cell, and very few are passed on to progeny cells. Occasional errors in transcription can result in the production of defective proteins, but it appears that the cell can tolerate such errors provided that not too many occur.
27. The small RNA molecule serves as the primer for the initiation of DNA synthesis by DNA polymerase. To synthesize a DNA chain DNA polymerase requires a primer nucleotide with a 3′-hydroxyl terminus along with a template. Studies show that either an oligodeoxyribonucleotide or an oligoribonucleotide can serve as a primer. DNA polymerase cannot synthesize a primer sequence on a closed DNA circle, but the RNA polymerase produced by the dnaG gene can synthesize a short primer sequence, which is then extended as DNA by DNA polymerase. When replication has extended around the circle and the 5′ terminus of the RNA primer is reached, the primer is hydrolyzed and the small gap is filled in with a DNA sequence. Bacterial dnaG mutants cannot support G4 infection because they cannot synthesize the RNA primer.
28. (a) The processing of the primary tRNA transcript removes RNA on either side of the uninterrupted tRNA sequence of 85 nucleotides, whereas protein splicing operations remove RNA sequences that are located within regions of pre-mRNA that code for a continuous polypeptide; these sequences must be removed to ensure that the protein specified by the RNA will have the correct amino acid sequence. In some organisms, tRNA are produced by RNA splicing, which removes sequences from within the pre-tRNA transcript.
(b) The most obvious reason for transcribing all three rRNA sequences simultaneously is that it ensures that an equal number of ribosomal RNA molecules will be avail able for the assembly of ribosomes. In addition, only one promoter, rather than three, is required for rRNA synthesis.
29. (a) The protein-encoding codons that could be mutated to UAA by a single base change are CAA, GAA, AAA, UCA, UUA, UAU, and UAC.
(b) Chain termination near the 3′ end of the normal coding sequence could allow the prematurely terminated protein to be functional because most of the polypeptide sequence would be intact. Removing a few amino acids from the C-terminal end of many, but not all, proteins does not appreciably affect their normal function.
(c) When a UAG codon reverts to UCG, a serine residue will be incorporated at the site in the protein that corresponds to the chain-termination site; it will be linked by a peptide bond to the next amino acid in the polypeptide.
(d) To determine which amino acids would be carried by suppressor tRNAs to a UAG codon site, you should identify all those tRNA molecules having an anticodon that, by a single base change, can read a UAG codon. Each such tRNA molecule would suppress the premature termination of the chain by inserting an amino acid at the corresponding site in the protein. The amino acids that could be found at the site (along with their codons) are Glu (GAG), Gln (CAG), Leu (UUG), Lys (AAG), Ser (UCG), Trp (UGG), and Tyr (UAC and UAU). When a cell contains a suppressor tRNA, proteins whose mRNA sequence normally ends with a single stop codon may not be efficiently terminated. Although the extension of such proteins could be lethal, most cells tolerate suppression. One explanation is that other proteins involved in chain termination may recognize a stop codon even though a tRNA that reads the codon is present. More work is needed to develop a full understanding of the reasons for toleration of suppression.
30. Equilibrium for the reaction catalyzed by polynucleotide phosphorylase lies toward the direction of RNA degradation rather than synthesis. High concentrations of ribonucleoside diphosphates are required to achieve the net synthesis of RNA; and it is likely that their concentrations in the cell are not sufficient to drive net polynucleotide synthesis.
Also, polynucleotide phosphorylase does not use a template, so the polyribonucleotides it synthesizes contain random sequences, which makes them of no value for protein synthesis. The cell uses polynucleotide phosphorylase as a degradative enzyme in conjunction with other nucleases that regulate the lifetimes of RNA molecules, including mRNA.
In bacteria mRNA lifetimes are relatively short.
31. (a) If transcription occurred simultaneously on both DNA strands, the excess, comple mentary mRNA molecules synthesized from each template strand could form double-strand structures with all the DNA strands. The fact that less than all the DNA could form hybrids indicated that some of the DNA strands lacked sequences complementary to the mRNA, that is, only one of the two strands at a given location along the DNA was being transcribed into DNA. In rare cases in some organisms over limited regions, RNA is synthesized from both strands of the template DNA.
(b) To establish whether one or both strands of SP8 DNA are used for transcription, you can carry out hybridization experiments with the separate strands using radioactive RNA synthesized during the infection of Bacillus with SP8. The results show that such RNA hybridizes to only one of the two strands, which means that only one of the two strands of the DNA of the SP8 virus is transcribed. In most other organisms, different regions of each strand are used for transcription; SP8 virus is exceptional in that one strand is used exclusively for all mRNA synthesis.
CHAPTER 5
32. (a) (i) Leucine arginine destabilizes the helix because you substitute a hy drophilic residue for a hydrophobic residue (most likely in a hydrophobic region of the helix).
(ii) Leucine valine has no effect (a conservative replacement of one hy drophobic residue for another).
(iii) Leucine proline destabilizes the helix because proline does not allow ro tation about its peptide bond and no hydrogen atom in this amino acid is available for hydrogen bonding.
(iv) Leucine glycine destabilizes because glycine is a very flexible residue and can act as a swivel, disrupting the helix (the frequency of glycines in a helix is as rare as the frequency of prolines).
(v) Leucine alanine has no effect (a conservative replacement of one hy drophobic residue for another).
(b) (i) CUU
CGU
single-base change required (ii) CUU
GUU
single-base change required (iii) CUU
CCU
single-base change required (iv) CUU
GGU
two-base alteration required (v) CUU
GCU
two-base alteration required
33. (a) Cordycepin is a nucleoside, and it must be converted to the triphosphate form be fore it can be incorporated (as cordycepin monophosphate) by RNA polymerase into a growing polynucleotide chain. The conversion of cordycepin to the triphosphate form is carried out by a number of kinase enzymes (see Chapter 25, p. 697 on nucleotide metabolism in the text) that utilize ATP as a phosphate donor. Bacteria probably cannot phosphorylate cordycepin efficiently, which makes them less susceptible to inhibition of RNA and DNA synthesis.
(b) You would not expect cordycepin to inhibit DNA polymerase because although a cordycepin residue that had been added to DNA would also lack a 3′-OH and act as a chain terminator, the presence of the 2′-OH on the ribose of the triphosphate form of cordycepin would be discriminated against by the DNA polymerase. You will learn later that DNA synthesis requires an RNA primer, and cordycepin might inhibit DNA synthesis by inhibiting RNA primer formation.
34. The ribosomal complex that carries out protein synthesis is unable to recognize alanyl tRNACys as an inappropriate or erroneous form of tRNA. An amino acid that is attached to a transfer RNA molecule will be transferred into a growing polypeptide chain solely on the basis of recognition between the anticodon in tRNA and the codon in the messenger RNA molecule. Once an aminoacyl-tRNA has been formed, accurate translation does not depend on recognition of the attached amino acid. This important point was established by Dintzis and von Ehrenstein, who carried out the incisive experiments using Raney nickel to reduce the cysteinyl residue on cysteinyl-tRNACys to an alanyl residue, then analyzing the resulting protein using an in vitro hemoglobin-synthesizing system.
35. The base-catalyzed generation of 2′- and 3′-monophosphates argues that a cyclic 2′,3′ phosphodiester is formed during cleavage of RNA. It is likely that a hydroxyl anion abstracts the hydrogen atom of the 2′-OH of RNA, leaving a 2′-O− that attacks the phosphorus atom, cleaving the 5′ phosphodiester bond and generating a cyclic 2′,3′phosphodiester. This cyclic derivative is unstable and decomposes by hydrolysis to form either a 2′- or a 3′-phosphate ester. DNA has no 2′-OH groups and is therefore not susceptible to alkali degradation. Because it is more stable than RNA, it may have been selected as the primary carrier of genetic information to future generations.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 36. The step that would afford the maximum economy is probably transcription, through the control of the activity of RNA polymerase. Transcription by RNA polymerase to form RNA is the first step in the expression of genetic information. It follows that controlling messenger RNA production, by stimulating or inhibiting the activity of RNA polymerase, allows the cell to make particular types of mRNA and to synthesize the encoded proteins only when required. A cell that could not regulate RNA polymerase activity would produce unneeded mRNA molecules, and the energy required to produce those polynucleotides would be wasted, even if translation were stringently regulated.
37. Yes, the incorporation of nucleoside triphosphates into an acid-insoluble form is in dicative of the presence of a polymerase. The polymerase is likely a DNA polymerase because dNTPs, and not NTPs, were used to form product. Further evidence for a DNA polymerase was that the radiolabeled product was destroyed by a nuclease, DNase, specific for hydrolyzing DNA, and not by one specific for RNA hydrolysis.
Additionally, NaOH, which destroys RNA but not DNA, did not destroy the radiolabeled product. Pretreatment of the extract with the two hydrolytic enzymes demonstrated that the enzyme depends on an RNA and not a DNA template for its activity.
Thus, this enzyme is an RNA-dependent DNA polymerase. No such enzyme had been observed previously in a cell, and this demonstration, along with similar findings by Howard Temin, of its existence in an RNA tumor virus caused a revision of Francis Crick’s central dogma of molecular biology, which stated that information flowed from DNA to RNA to proteins. The demonstration of this RNA-dependent DNA polymerase suggested that in some cases information could flow from RNA to DNA. (This question was derived from D. Baltimore. Viral RNA-dependent DNA polymerase. Nature
226:[1971]1209–1213.)
38. Hershey and Chase observed that most of the 32P was associated with the cells and most of the 35S was in the supernatant. Since nucleic acids are rich in phosphorus and DNA does not contain sulfur, they concluded that DNA had entered the cell. The sulfur of sulfate is incorporate into the amino acids cysteine and methionine so that 35S is a good marker for proteins. The experiment indicated that protein did not enter the cells. Recalling that bacteriophage T2 displays heredity, that is, passes genetic traits to its progeny, they concluded that DNA, not protein, is likely the genetic information because it entered the cells and was replicated. At the time these experiments were performed, they helped solidify the view that the genetic material was DNA, not protein. (This question was derived from A. D. Hershey & M. Chase. Independent functions of viral protein and nucleic acid in growth of bacteriophage. J. Gen. Physiol.
36:[1952]39–56.)
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. By convention, when polynucleotide sequences are written, left to right means 5′ 3′.
Since complementary strands are antiparallel, if one wishes to write the complementary sequence without specifically labeling the ends, the order of the bases must be reversed.
(a) TTGATC (b) GTTCGA (c) ACGCGT (d) ATGGTA (b) Due to base pairing (A:T, G:C) in the complementary strand, [T] = 0.30, [C] = 0.24, and [A] + [G] = 0.46.
3. To answer this question one must know that 2mm = 2 × 10−6, that one Å = 10−10 m, and that the distance between the base pairs is 3.4 Å. The length of a DNA segment (in this case 2 × 10−6 m) divided by the distance between the base pairs (3.4 × 1−10 m) gives the answer, 5.88 × 103 base pairs.
4. After 1.0 generation, one-half of the molecules would be 15N-15N, the other half 14N-14N. After 2.0 generations, one-quarter of the molecules would be 15N-15N, the other three-quarters 14N-14N. Hybrid 14N-15N molecules would not be observed.
5. (a) Thymine is the molecule of choice because it occurs in DNA, is not a component of RNA, and is not readily converted to cytosine or uracil. If they enter the cell, labeled deoxythymidine or dTTP would also be useful molecules. Its large negative charge prevents dTTP from entering most cells.
(b) During DNA synthesis, the b- and g-phosphorus atoms of the nucleoside triphosphates are lost as pyrophosphate. Since the a-phosphorous atom is incorporated into DNA, one should use dATP, dGTP, dTTP, and dCTP labeled with 32P in the a position.
6. Only (c) would lead to DNA synthesis because (a) and (b) have no primer or open end to build on and (d) has no template extending beyond a free 3′-OH. Note: Singlestranded linear DNA can be used as a template for DNA synthesis because it can prime synthesis through hairpin formation at its 3′ end.
7. A short polythymidylate chain would serve as a primer because T base pairs with A.
Radioactive dTTP labeled in any position except the b- and g-phosphates would be useful for following chain elongation.
8. After the synthesis of the complementary (–) DNA on the RNA template, the RNA must be disposed of by hydrolysis prior to the completion of the synthesis of the DNA duplex.
9. One should treat the infectious nucleic acid with either highly purified ribonuclease or deoxyribonuclease and then determine its infectivity. RNAse will destroy the infectivity if it is RNA; DNAse will destroy it if it is DNA.
10. Ultimately, this mutation results in half the daughter DNA duplexes being normal and half having a TA pair that had been CG. The first two rounds of replication at the mutant site will be as follows:
CG
CG
CG
CG
UG
TA
UA
UA
11. (a) From the 4 mononucleotides one can formulate 16 different dinucleotides. If you don’t believe it, try it! From these dinucleotides you can make 64 different trinucleotides. Note that 64 is 43. There will be 44 (256) tetranucleotides. Proceeding in this manner we get to 48 (65,536) different octonucleotides (8-mers).
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (b) A bit specifies two bases (say A and C), and a second bit specifies the other two (G and T). Hence, two bits are needed to specify a single nucleotide (or base pair) in DNA. An 8-mer stores 16 bits (216 = 65,535), the E. coli genome (4 × 106 bp) stores 8 × 106 bits, and the human genome (2.9 × 109 bases) stores 5.8 × 109 bits of genetic information.
(c) A high-density diskette stores about 1.5 megabytes, which is equal to 1.2 × 107 bits.
A large number of 8-mer sequences could be stored on such a diskette. The DNA sequence of E. coli, once known, could be written on a single diskette. Nearly 500 diskettes would be needed to record the human DNA sequence.
12. (a) Deoxyribonucleoside triphosphates versus ribonucleoside triphosphates.
(b) 5′ 3′ for both.
(c) DNA serves as the template for both polymerases. During DNA replication by poly merase I each parent strand acts as a template for the formation of a new complimentary strand. Since each daughter molecule receives one strand from the parent DNA molecule, the template is said to be semiconserved. However, after guiding the synthesis of RNA by RNA polymerase, the DNA double helix remains intact.
Hence the template is said to be conserved.
(d) DNA polymerase I requires a primer, whereas RNA polymerase does not.
13. (a) Because mRNA is synthesized antiparallel to the DNA template and A pairs with U and T pairs with A, the correct sequence is 5′-UAACGGUACGAU-3′.
(b) Since the 5′ end of an mRNA molecule codes for the amino terminus, appropriate use of the genetic code (see text, p. 109) leads to Leu-Pro-Ser-Asp-Trp-Met.
(c) Since one has a repeating tetramer (UUAC) and a 3-base code, repetition will be observed at a 12-base interval (3 × UUAC). Comparison of this 12-base sequence with the genetic code leads to the conclusion that a polymer with a repeating tetrapeptide (Leu-Leu-Thr-Tyr) unit will be formed.
14. The instability of RNA in alkali is due to its 2′-OH group. In the presence of OH− the 2′-OH group of RNA is converted to an alkoxide ion (RO−) by removal of a proton. Intramolecular attack by the 2′-alkoxide on the phosphodiester in RNA gives a 2′,3′-cyclic nucleotide, cleaving the phosphodiester bond in the process. Further attack by OH− on the 2′,3′-cyclic nucleotide produces a mixture of 2′ and 3′-nucleotides.
Note that the mechanism for ribonuclease action is quite similar (see Figure 9.18, p. 216). Since DNA lacks a 2′-OH group, it is quite stable in alkali.
15. Apparently cordycepin is converted to its 5′-triphosphate and incorporated into the growing RNA chain. This chain containing cordycepin now lacks a 3′-OH group; hence, RNA synthesis is terminated.
16. Only single-stranded mRNAs can serve as templates for protein synthesis. Since poly(G) forms a triple-stranded helix, it cannot serve as a template for protein synthesis.
17. Note that each complimentary strand is missing one of the four bases; d(TAC) lacks G and d(GTA) lacks C. Thus, incubation with RNA polymerase and only UTP, ATP, and CTP led to the synthesis of only poly(UAC), the RNA complement of d(GTA). When GTP was used in place of CTP, the complement of d(TAC), poly(GUA), was formed.
18. In a nonoverlapping code, each individual nucleotide mutation would change at most one amino acid in the protein sequence. (Because of codon degeneracy, some individual mutations will not change the amino acid sequence.) In an overlapping code, some mutations (but not all) will change the identity of two consecutive amino acids in the CHAPTER 5 protein sequence. Therefore several experiments will be needed. If one makes a series of individual nucleotide mutations, determines the resulting protein sequences, and never finds that two consecutive amino acids are changed, then one could reasonably conclude that the code is nonoverlapping.
One can also note that the putative overlapping code described in the problem would have a four-nucleotide stretch (e.g., ABCD) encoding a dipeptide. The maximum possible number of different dipeptides that could be encoded by four different nucleotides in this scenario would be only 44 = 256. However, the 20 amino acids can be used to make 202 = 400 different dipeptides, all of which are represented in known protein sequences. Therefore, this numerical analysis of naturally occurring dipeptide sequences also would argue against a completely overlapping triplet code. (See also Crick, Barnett, Brenner, & Watts-Tobin. Nature 192:[1961]1227–1232.)
19. Since three different polypeptides are synthesized, the synthesis must start from three different reading frames. One of these will be in phase with the AAA in the sequence shown in the problem and will therefore have a terminal lysine, since UGA is a stop signal. The reading frame in phase with AAU will result in a polypeptide having an AsnGlu sequence in it, and the reading frame in phase with AUG will have a Met-Arg sequence in it.
DNA, RNA, and the Flow
of Genetic Information Chapters 2 and 3 introduced you to proteins. The authors now turn to a second class of macromolecules, the nucleic acids that serve as the storage forms of genetic information. First, they describe the structures of the nucleoside building blocks of DNA and the phosphodiester bond that links them together. Following this, the Watson-Crick DNA double helix is presented, an overview of how the strands of DNA separate for replication is given, and some of the various conformations and structures that nucleic acids can assume are described. The polymerases that form DNA chains are introduced next. The section describing the molecules that store genetic information ends by providing two examples of viruses in which the genetic material is not duplex DNA but rather single-strand RNA. The authors next describe the way in which RNA viruses replicate through double-strand nucleic-acid intermediates whose formation is directed by specific base pairing.
How the information stored in DNA or RNA directs the formation of the pro teins of a cell is discussed next. The authors start with descriptions of the basic structures and kinds of RNA and provide an explanation of the central roles of RNAs in the overall flow of genetic information. They then present the specific functions of messenger RNA, transfer RNA, and ribosomal RNA in protein synthesis, along with a description of the polymerase that synthesizes all cellular RNAs. The genetic code, which relates the nucleotide sequence of RNA to the amino acid sequence of proteins, is described. The collinear relationship between the sequences of nucleotides in the DNA and the amino acids of the encoded protein is compared in prokaryotes and in eukaryotes where some genes are interrupted by noncoding sequences (introns). The authors next describe the process by which these intron sequences are removed from the initial transcript to form functional messenger RNA and the biological consequences of such splicing.
51
CHAPTER 5
LEARNING OBJECTIVES
When you have mastered this chapter, you should be able to complete the following objectives.
Nucleic Acids (Text Section 5.1) 1. Locate the structural components of DNA, namely, the nitrogenous bases, the sugar, and the phosphate group. Know the various conventions used to represent these components and the structure of DNA.
2. Differentiate purines, pyrimidines, ribonucleosides, deoxyribonucleosides, ribonucleotides, and deoxyribonucleotides.
3. Recognize the deoxyadenosine, deoxycytidine, deoxyguanosine, and deoxythymidine con stituents of DNA, and describe the phosphodiester bond that joins them together to form DNA.
4. Compare the phosphodiester backbones of RNA and DNA. Contrast the composition and structures of RNA and DNA. Distinguish thymine from uracil and 2′-deoxyribose
from ribose.
5. Relate the polarity of the DNA chain to the direction a DNA sequence abbreviation is written.
6. Compare the lengths of the DNA molecules in polyoma virus, the bacterium E. coli, and the average human chromosome.
Double-Helices (Text Section 5.2) 7. List the important features of the Watson-Crick DNA double helix. Relate the base pairing of adenine with thymine and of cytosine with guanine to the duplex structure of DNA and to the replication of the helix. Explain the molecular determinants of the specific base pairs in DNA.
8. Outline the Meselson-Stahl experiment and relate it to semiconservative replication. Define the melting temperature (Tm) for DNA and relate it to the separation of the strands of duplex DNA. Describe annealing.
9. Describe supercoiling and state its biological consequences.
10. Appreciate the variety of structures that single-strand nucleic acids can assume.
DNA Polymerases (Text Section 5.3) 11. List the substrates and the important enzymatic properties of DNA polymerases as they relate to replication. Distinguish between a primer and a template and describe their functions.
12. Define virus and appreciate that RNA is the genome of some viruses.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 13. Relate the catalytic activity of reverse transcriptase (an RNA-directed DNA polymerase) to the replication of retroviruses. Provide an overview of retroviral replication.
Gene Expression (Text Section 5.4) 14. State the role of DNA in protein synthesis. Outline the flow of genetic information during gene expression.
15. Define the terms transcription and translation and relate these processes to the flow of ge netic information.
16. Name the three major classes of RNA found in E. coli and explain their functions. Compare their sizes and their relative amounts in the cell.
17. List the substrates and important enzymatic properties of RNA polymerases (DNA-de-pendent RNA polymerases). Explain the roles of the DNA template, promoter, enhancer sequences, and terminator in transcription.
18. Describe the transcription of duplex DNA to form single-strand RNA. Relate the sequence of mRNA to that of the coding strand of the DNA template from which it is transcribed.
19. Describe the role of tRNA as the adaptor molecule acting between mRNA and amino acids during protein synthesis. Outline how specific amino acids are covalently attached to specific tRNA molecules. Explain the relationship of the codon and anticodon to the specific interaction between mRNA and tRNA.
Amino Acids and the Genetic Code (Text Section 5.5) 20. Explain what the genetic code is and list its major characteristics. Define the terms tripletcode (codon), nonoverlapping, degenerate, synonym, triplet, and reading frame as they apply to the genetic code. Recognize the initiation and termination codons.
21. Using the genetic code, predict the sequence of amino acids in a peptide encoded by a template DNA or mRNA sequence.
Introns and Exons (Text Section 5.6) 22. Discuss the universality of the genetic code. Describe the composition and function of spliceosomes.
23. Contrast the linear relationship between the sequence of DNA in a gene and the sequence of the amino acids in the protein it encodes in bacteria and in higher eukaryotes. Apply the terms intron and exons to these relationships.
24. Outline RNA processing in eukaryotes. Name the alterations made to the RNA after it is initially formed by RNA polymerase.
25. Recount a hypothesis relating exons and functional domains to the generation and evolu tion of protein diversity. Differentiate between nucleotide sequence rearrangements by genetic recombination at the DNA level and by RNA splicing.
CHAPTER 5
SELF-TEST
Nucleic AcidsFIGURE 5.1
NH
O
O
2
J
K
K
N
N
J
N
HJN
HJN
H
J
J
H
H
J
HJ
H NJ
OK
H
N
2
N
N
N
N
HOCH
OCH
HOCH
2
2
2
O
O
O
J
H
H
J
J
H
H
J
J
H
H
J
H
H OKPJO: H
H
H
H
J
J
J
J
J
J
O
H
HO
H
HO
OH
A
B
NH
O
2
J
K
J
N
H
J
HJN
CH
3
OK
J H
OK
J H
O
N
N
K
:
J
OJPJOCH
2
H
J
O O:
J
H
H
J
H
H
J
J
HO
H
C
D
1. Which of the preceding structures (a) contains ribose?
in Figure 5.1 (b) contains deoxyribose?
(c) contains a purine?
(d) contains a pyrimidine?
(e) contains guanine?
(f)
contains a phosphate monoester?
(g) contains a phosphodiester?
(h) is a nucleoside?
(i) is a nucleotide?
(j) would be found in RNA?
(k) would be found in DNA?
Double Helices
2. Which of the following are characteristics of the Watson-Crick DNA double helix?
(a) The two polynucleotide chains are coiled about one another and about a common axis.
(b) Hydrogen bonds between A and C and between G and T help hold the two chains together.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (c) The helix makes one complete turn every 34Å because each base pair is rotated by 36º with respect to adjacent base pairs and is separated by 3.4Å from them along the helix axis.
(d) The purines and pyrimidines are on the inside of the helix and the phosphodiester linked backbones are on the outside.
(e) Base composition analyses of DNA duplexes isolated from many organisms show that the amounts of A and T are equal as are the amounts of G and C.
(f) The sequence in one strand of the helix varies independently of that in the other strand.
3. If a region of one strand of a Watson-Crick DNA double helix has the sequence ACG TAACC, what is the sequence of the complementary region of the other strand?
4. Explain why A T and G C are the only base pairs possible in normal double-strand DNA.
5. Match the appropriate characteristics in the right column with the structures of double strand or single-strand DNA.
(A) double-strand DNA (1) is a rigid rod (B) single-strand DNA (2) shows a greater hyperchromic effect upon heating (3) contains equal amounts of A and T bases (4) may contain different amounts of C and G bases (5) contains U rather than T bases (6) may contain stem-loop structures 6. Haploid human DNA has 3 p 106 kilobase pairs (a kilobase pair, abbreviated kb, is 1000 base pairs). What is the total length of human haploid DNA in centimeters?
7. Outline the basic process by which a Watson-Crick duplex replicates to give two iden tical daughter duplexes. Explain the reasons for the accuracy of the process.
8. The DNA in a bacterium is uniformly labeled with 15N, and the organism shifted to a growth medium containing 14N-labeled DNA precursors. After two generations of growth, the DNA is isolated and is subjected to density-gradient equilibrium sedimentation. What proportion of light-density DNA to intermediate-density DNA would you expect to find?
9. Purified duplex DNA molecules can be (a) linear.
(b) circular and supercoiled.
(c) linear and supercoiled.
(d) circular and relaxed, that is, not supercoiled.
10. You are given two solutions containing different purified DNAs. One is from the bac terium P. aeruqinosa and has a G + C composition of 68%, whereas the other is from a mammal and has a G + C composition of 42.5%.
(a) You measure the absorbance of ultraviolet light of each solution as a function of in creasing temperature. Which solution will yield the higher Tm value and why?
(b) After melting the two solutions, mixing them together, and allowing them to cool, what would you expect to happen?
(c) Would appreciable amounts of bacterial DNA be found associated in a helix with mammalian DNA? Explain.
CHAPTER 5
DNA Polymerases
11. DNA polymerase activity requires (a) a template.
(b) a primer with a free 5′-hydroxyl group.
(c) dATP, dCTP, dGTP, and dTTP.
(d) ATP.
(e) Mg2+.
12. Derive the polarity of the synthesis of a DNA strand by DNA polymerase from the mech anism for the formation of the phosphodiester bond.
13. You are provided with a long, single-strand DNA molecule having a base composition of C = 24.1%, G = 18.5%, T = 24.6%, and A = 32.8%; DNA polymerase; [a-32P]dATP (dATP with the innermost phosphate labeled), dCTP, dGTP, and dTTP; a short primer that is complementary to the single-strand DNA; and a buffer solution with Mg2+. What is the base composition of the radiolabeled product DNA after the completion of one round of synthesis?
14. For the virus in the left column, indicate the appropriate characteristics from the right column.
(a) Tobacco mosaic virus (1) linear genome (b) AIDS virus (2) genome contains U rather than T (3) single-strand nucleic acid genome (4) DNA intermediates are involved in replication (5) uses RNA-directed RNA polymerase to replicate (6) uses RNA-directed DNA polymerase to replicate
15. Propose how a single-strand DNA virus could replicate by incorporating semiconserva tive replication into the process.
16. From the following nucleic acids, select those that appear during the infection of a cell with a retrovirus, for example, the AIDS virus, and place them in the order in which genetic information flows during the process of forming a new progeny virus.
(a) double-strand DNA-RNA helix in the cell (b) single-strand RNA in the virus (c) single-strand RNA in the cell (d) double-strand DNA in the cell (e) double-strand RNA in the virus (f) double-strand RNA in the cell Gene Expression
17. Transcription is directly involved in which of the following possible steps in the flow of genetic information.
(a) DNA to RNA (d) RNA to protein (b) RNA to DNA (e) protein to RNA (c) DNA to DNA (a) DNA to RNA (b) RNA to DNA (c) DNA to DNA (d) RNA to protein (e) protein to RNA 19. Answer the following questions about RNA.
(a) What is the name of the bond joining the ribonucleoside components of RNA to one another?
(b) Is this bond between the 2′- or the 3′-hydroxyl group of one ribose and the 5′ hydroxyl of the next?
(c) Intramolecular base pairs form what kinds of structures in RNA molecules?
(d) What bases pair with one another in RNA?
(e) What are the three major classes of RNA in a cell and which is most abundant?
20. If you have samples of pure RNA and duplex DNA, how can you tell whether they have any complementary nucleotide sequences?
21. If all the RNA referred to in question 20 turns out to have sequences that were comple mentary to the DNA, will its percentage of G and C be identical to that of the DNA?
Explain.
22. If each of the three major classes of RNA found in a cell were hybridized to denatured DNA from the same cell and the presence of RNA-DNA hybrids were tested, which of the classes would be retained on the filter?
(a) mRNA (b) rRNA (c) tRNA RNA Polymerases and Transcription 23. Which of the following are required for the DNA-dependent RNA polymerase reaction to produce a unique RNA transcript?
(a) ATP (g) RNA (b) CTP (h) Mg2+ (c) GTP (i)
promoter sequence (d) dTTP (j)
operator sequence (e) UTP (k) terminator sequence (f)
DNA
24. What is the sequence of the mRNA that will be synthesized from a template strand of DNA having the following sequence:
…ACGTTACCTAGTTGC…?
25. Describe the mechanism of chain growth in RNA synthesis. What is the polarity of syn thesis and how is it related to the polarity of the template strand of DNA?
CHAPTER 5
The Genetic Code and Protein Synthesis 26. Which of the following are characteristics or functions of tRNA?
(a) It contains a codon.
(b) It contains an anticodon.
(c) It can become covalently attached to an amino acid.
(d) It interacts with mRNA to stimulate transcription.
(e) It can have any of a number of different sequences.
(f) It serves as an adaptor between the information in mRNA and an individual amino acid.
27. What is the minimum number of contiguous nucleotides in mRNA that can serve as a codon? Explain.
28. What is the sequence of the polypeptide that would be encoded by the DNA sequence given in Question 24? Assume that the reading frame starts with the 5′ nucleotide given.
The genetic code is given on page 134 of the text.
29. The following is a partial list of mRNA codons and the amino acids they encode: AGU = serine AGC = serine AAU = asparagine AAC = asparagine AUG = methionine AUA = isoleucine
Based on this list, which of the following statements are correct?
(a) The genetic code is degenerate.
(b) The alteration of a single nucleotide in the DNA directing the synthesis of these codons could lead to the substitution of a serine for an asparagine in a polypeptide.
(c) The alteration of a single nucleotide in the DNA directing the synthesis of these codons would necessarily lead to an amino acid substitution in the encoded polypeptide.
(d) A tRNA with the anticodon ACU would be bound by a ribosome in the presence of one of these codons.
30. Explain why mitochondria can use a genetic code that is different from the standard code used in the nucleus.
Introns and Splicing
31. Explain how genetic techniques and amino acid sequence analyses could be used to show the collinear relationship of a prokaryotic gene and the protein it encodes.
32. Answer the following questions about what was revealed when DNA encoding the gene for the b-chain of hemoglobin and the mRNA for the b-chain were compared.
(a) What was the major finding when the nucleotide sequence of the gene and the amino acid sequence of the b-chain were compared?
(b) What did hybridization between the partially denatured DNA and the mRNA for b-globin show?
(c) What must happen to the primary transcript from the b-globin gene before it can serve as an mRNA for protein synthesis?
33. How might the fact that some exons encode discrete functional domains in proteins be related to the evolution of new proteins?
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 34. Spliceosomes (a) recombine DNA sequences in a process called exon shuffling.
(b) are composed of RNA and proteins.
(c) recognize RNA sequences that signal for the removal of introns.
(d) can produce different mRNA molecules by splicing at alternative sites.
ANSWERS TO SELF-TEST
1. (a) B (b) A, C (c) A (d) B, C, D (e) A (f) C (g) A (h) B (i) C; strictly speaking, A is called a dinucleotide, not a nucleotide (j) B (k) A, C, D 2. a, c, d, and e. Answer (b) is not correct because A pairs with T and G pairs with C.
Answer (f) is not correct because the sequence of one strand determines the sequence of the other by base pairing.
3. GGTTACGT. The convention for indicating polarity is that the 5′-end of the sequence is written to the left. The two chains of the Watson-Crick double helix are antiparallel, so the correct complementary sequence is not TGCATTGG.
4. The space between the two deoxyribose-phosphodiester strands is precisely defined. This distance is not large enough for two purines to hydrogen-bond. Conversely, two pyrimidine bases would not be close enough to form stable hydrogen bonds. Furthermore, in the double-strand structure the hydrogen-bond donor and acceptor groups are not properly aligned to form stable G T or A C base pairs.
5. (a) 1, 2, 3 (b) 4, 6. Answer (5) does not apply because DNA does not contain uracil.
6. 102 cm. The math is as follows: (3 × 106 kb × 103 bases/kb × 3.4 Å/base × 10−8 cm/Å) = 102 cm 7. When replication occurs, the two strands of the Watson-Crick double helix must sepa rate so that each can serve as a template for the synthesis of its complement. Since the two strands are complementary to one another, each bears a definite sequence relationship to the other. When one strand acts as a template, it directs the synthesis of its complement. The product of the synthesis directed by each template strand is therefore a duplex molecule that is identical to the starting duplex. The process is accurate because of the specificity of base pairing and because the protein apparatus that catalyzes the replication can remove mismatched bases.
8. After two generations, you should expect to find equal amounts of light-density DNA, in which both strands of each duplex were synthesized from 14N precursors, and intermediate-density DNA, in which each duplex consists of a heavy 15N strand paired with a light 14N strand.
9. a, b, and d. Answer d is correct because, if at least one discontinuity exists in the phos phodiester backbone of either chain of a circular duplex molecule, the chains are free to rotate about one another to assume the relaxed circular form. Answer c is incorrect because supercoiling requires closed circular molecules. In a linear molecule, the ends of each strand are not constrained with respect to rotation about the helical axis; therefore, the molecule cannot be supercoiled.
CHAPTER 5
10. (a) The bacterial DNA solution has the higher Tm value because it has the higher G + C content and is therefore more stable to the thermal-induced separation of its strands because G C base pairs are more stable than A T base pairs.
(b) The complementary DNA strands from each species will anneal to form Watson Crick double helices as the solution cools.
(c) No; each strand will find its partner because the perfect match between the linear arrays of the bases of complementary strands is far more stable than the mostly imperfect matches in duplexes composed of one strand of bacterial and one strand of mammalian DNA would be.
11. a, c, and e. Answer (b) is not correct because, although the enzyme requires a primer, the nature of its 5′-end is irrelevant since dNMP residues are added to its 3′-end. A primer with a 3-OH is required. Answer (d) is not correct because the enzyme uses dNTP and not NTP molecules, where N means A, C, G, T, or U.
12. The 3′-hydroxyl of the terminal nucleotide of the primer makes a nucleophilic attack on the innermost phosphorus atom of the incoming dNTP that is appropriate for WatsonCrick base pairing to the template strand to form the phosphodiester bond. As a result, a dNMP residue is added onto the 3′-end of the primer with the concomitant release of PPi, and the chain grows in the 5′ 3′ direction.
13. After the completion of one round of synthesis, the template strand will have directed the polymerization of a complement in which C = 18.5%, G = 24.1%, T = 32.8%, and A = 24.6%. Since the primer is short with respect to the template, its contribution to the composition of the product strand can be neglected.
14. (a) 1, 2, 3, 5 (b) 1, 2, 3, 4, 5, 6 15. The single-strand DNA penetrates the cell, where it has converted by enzymes to a du plex replicative form through Watson-Crick base pairing. The replicative form is then reproduced by a mechanism similar to that used for the semiconservative replication of the duplex chromosome of the host cell. Finally, after this stage of replication, the mechanism shifts to one in which the replicative form serves as a template to produce copies of the single-strand DNA found in the mature virus.
16. Starting with the single-strand RNA in the virus and ending with the single-strand RNA in the progeny viruses, the order in which genetic information flows during the infection of a cell with a retrovirus is: b, c, a, d, c, b. Retroviruses use the enzyme reverse transcriptase to convert their single-strand genomes into a DNA-RNA replicative form that is subsequently converted into a duplex DNA replicative form prior to insertion into the host chromosome and ultimate reconversion into the single-strand viral RNA by a DNA-dependent RNA polymerase.
17. a and b. Answer (b) is correct because the reverse transcription of RNA sequences into DNA sequences occurs during the replication of retroviruses. The term transcription is usually used to describe the formation of RNA from a DNA duplex by RNA polymerase.
18. d
19. (a) The bond is called the phosphodiester bond.
(b) The bond joins the 3′-hydroxyl to the 5′-hydroxyl to form a 3′ 5′ phospho diester bond.
(c) Hairpin loops are formed when the RNA chain folds back upon itself and some of the bases become hydrogen bonded to form an antiparallel duplex stem with unpaired bases forming loop at one end.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (d) A pairs with U, and G pairs with C; G can also pair with U, but the association is weaker than that of the G C base pair.
(e) The three major classes of RNA found in a cell are mRNA, rRNA, and tRNA; the most abundant is rRNA.
20. You could sequence the RNA and DNA and compare the sequences of each to see if the two are complementary; this method provides definitive evidence of identity. An easier but less precise way would be to use hybridization. You would mix the samples, heat the mixture to melt the double-strand DNA and RNA hairpins, slowly cool the solution, and then examine it to see if it contains double-strand DNA-RNA hybrids. Such hybrids would indicate that the RNA and DNA sequences are complementary.
21. Not necessarily; RNA synthesis is asymmetric, and only one strand of any region of the DNA serves as a template. This can lead to RNA with a G + C composition different from that of the duplex DNA.
22. a, b, c. All cellular RNA is encoded by the DNA of the cell.
23. a, b, c, e, f, h, i, and k. Answer (f) is correct because DNA is needed to serve as the tem plate. Answers (k) and (i) are correct because the promoter and terminator sequences are needed to specify the precise start and stop points, respectively, for the transcription.
24. The mRNA sequence will be …GCAACUAGGUAACGU…, written in the 5′ 3′ di rection.
25. The 3′-hydroxyl terminus of the growing RNA chain makes a nucleophilic attack on the a-phosphate (the innermost phosphate) of the ribonucleoside triphosphate that has been selected by base pairing to the template strand of the duplex DNA. RNA polymerase catalyzes the reaction. A ribonucleoside monophosphate residue is added to the chain as a result, and the chain has grown in the 5′ 3′ direction; that is, the chain has grown at its 3′ end. As with all Watson-Crick base pairing, the strands are antiparallel; that is, the RNA chain is assembled in the 3′ 5′ direction with respect to the polarity of the template strand of the DNA.
26. b, c, e, and f. Answer (d) is incorrect because the interaction of tRNA with mRNA takes place during translation, not transcription.
27. Three contiguous nucleotides is the minimum that can serve as a codon. There are four kinds of nucleotides in mRNA. A codon consisting of only two nucleotides (either of which could be any of the four possible nucleotides) allows only 16 possible combinations (4 × 4 = 16). This would not be sufficient to specify all 20 of the amino acids. A codon consisting of three nucleotides, however, allows 64 possible combinations (4 × 4 × 4 = 64), more than enough to specify the 20 amino acids.
28. The sequence of the polypeptide would be Ala-Thr-Arg. The reading frame is set by the nucleotide at the 5′ end of the mRNA transcript; the fourth codon of the mRNA transcript is UAA, which is a translation termination codon.
29. a, b, and d. Answer (a) is correct because both AGU and AGC specify serine; since more than one codon can specify the same amino acids, the genetic code is said to be degenerate. Answer (b) is correct because the alteration of a single nucleotide in the DNA could change a codon on the mRNA transcript from AGU, which specifies serine, to AAU, which specifies asparagine. Answer (d) is correct because the anticodon ACU would base-pair with the codon AGU. Answer (c) is not correct because the alteration of a single nucleotide in the DNA could result in another codon that specifies the same amino acid; for example, a codon changed from AGU to AGC would continue to specify serine.
CHAPTER 5
30. Mitochondria can use a genetic code that differs from the standard code because mito chondrial DNA encodes a distinct set of tRNAs that are matched to the genetic code used in their mRNAs.
31. The mutations in a given gene of E. coli could be mapped by recombination analysis. The proteins encoded by the wild-type and the mutant genes could then be sequenced, and the location and nature of the amino acid substitution for each mutation identified. The result would be that the order of the mutations on the genetic map is the same as the order of the corresponding changes in the amino acid sequence of the polypeptide produced by the gene; these experiments established that genes and their polypeptide products are collinear in prokaryotes.
32. (a) The number of nucleotides in the gene was significantly greater than three times the number of amino acids in the protein. There were two stretches of extra nucleotides between the exon sequences that encode the amino acids in the b-chain.
(b) The mRNA hydridized to the DNA under conditions where DNA-RNA hybrids are more stable than DNA-DNA hybrids, but there were sections of duplex DNA between the hybrid regions. This indicated that there are intron sequences in the DNA that have no corresponding sequences in the mRNA. (See Figure 5.33 in the text.)
(c) The intervening sequences (introns) in the nascent or primary transcript, which are complementary to the template strand of the DNA of the gene but do not encode amino acids in the protein, must be removed by splicing to generate the mRNA that functions in translation.
33. The shuffling of exons that encode discrete functional domains, such as catalytic sites, binding sites, or structural elements, preserves the functional units but allows them to interact in new ways, thereby generating new kinds of proteins.
34. b, c, and d. (a) is incorrect because exon shuffling takes place at the DNA level through breakage and rejoining of DNA not RNA.
PROBLEMS
1. The genome of the mammalian virus SV40 is a circular DNA double helix containing 5243 base pairs. When a solution containing intact DNA molecules is heated, one observes an increase in the absorbance of ultraviolet light at 260 nm. When the solution is then cooled slowly, a decrease in absorbance is observed. If one or more breaks are made in the sugar-phosphate backbones of the SV40 double-strand circles, heating causes a similar hyperchromic effect. However, when the solution of nicked molecules is cooled, the reduction in absorbance is much slower than that observed in the solution containing intact molecules. Why do the two types of molecules behave differently when they are cooled after heating?
2. A number of factors influence the behavior of a linear, double-strand DNA molecule in a 0.25M sodium chloride solution. Considering this, explain each of the following observations.
(a) The Tm increases in proportion to length of the molecule.
(b) As the concentration of sodium chloride decreases, the Tm decreases.
(c) Renaturation of single strands to form double strands occurs more rapidly when the DNA concentration is increased.
(d) The Tm value is reduced when urea is added to the solution.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 3. (a) Many proteins that interact with double-strand DNA bind to specific sequences in the molecule. Why is it unlikely that these enzymes operate by sensing differences in the diameter of the helix?
(b) What other features of the double-strand helix might be recognized by the protein?
4. You have a double-strand linear DNA molecule, the appropriate primers, all the enzymes required for DNA replication, four 32P-labeled deoxyribonucleoside triphosphates, Mg2+ ion, and the means to detect newly synthesized radioactive DNA. Why is this system not sufficient to distinguish between conservative and semiconservative replication of the DNA molecule?
5. Certain deoxyribonucleases cleave any sequence of single-strand DNA to yield nucleo side monophosphates; these enzymes do not hydrolyze base-paired DNA sequences.
What products would you expect when you incubate a solution containing a singlestrand specific deoxyribonuclease and the following oligodeoxyribonucleotide?
5′-ApGpTpCpGpTpApTpCpCpTpCpTpApCpGpApCpTp-3′
6. Formaldehyde reacts with amino groups to form hydroxymethyl derivatives. Would you expect formaldehyde to react with bases in DNA? Suppose you have a solution that contains separated complementary strands of DNA. How would the addition of formaldehyde to the solution affect reassociation of the strands?
7. When double-strand DNA is placed in a solution containing tritiated water, hydrogens associated with the bases readily exchange with protons in the solution. The greater the percentage of AT base pairs in the DNA, the greater the rate of exchange. Why?
8. While many experiments were suggesting that DNA in chromosomes is very long and continuous, it was established that DNA polymerase adds deoxyribonucleotides to the 3′-hydroxyl terminus of a primer chain and that a DNA template is essential. Why did those investigators interested in DNA replication focus a great deal of attention on determining whether chromosomal DNA contained breaks in the sugar-phosphate backbone?
9. The value of the Tm for DNA in degrees Celsius can be calculated using the formula,
T =
m
69.3 + 0.41(G + C), where G + C is the mole percentage of guanine plus cytosine.
(a) A sample of DNA from E. coli contains 50 mole percent G + C. At what tempera ture would you expect this DNA molecule to melt?
(b) The melting curves for most naturally occurring DNA molecules reveal that their Tm values are normally greater than 65ºC. Why is this important for most organisms?
10. During early studies of the denaturation of double-strand DNA, it was not known whether the two strands unwind and completely separate from each other. Suppose that you have double-strand DNA in which one strand is labeled with 14N and the other is labeled with 15N. If density-gradient equilibrium sedimentation can be used to distinguish between both double- and single-strand molecules of different densities, how can you determine whether DNA strands separate completely after denaturation?
11. Under strongly acidic conditions, several atoms of DNA bases are protonated; these include the N-1 and N-7 of adenine, the N-7 of guanine, the N-3 of cytosine, and the O-4 of thymine. Predict the effects of such protonations occurring at low pH on the stability of double-strand DNA.
12. Sol Spiegelman found that some types of single-strand RNA can associate with single strand DNA to form double-strand molecules. What is the most important condition that must be satisfied in order to allow the formation of these hybrid molecules?
CHAPTER 5
13. Many cells can synthesize deoxyuridine 5′-triphosphate (dUTP). Can dUTP be used as a substrate for DNA polymerase? If so, with which base will uracil pair in newly replicated DNA?
14. The DNA of bacteriophage l is a linear double-strand molecule that has complementary single-strand ends. These molecules can form closed circular molecules when two “cohesive” ends on the same molecule join, and they can form linear dimers, trimers, or longer molecules when sites on different molecules are joined.
(a) What conditions should be chosen to insure that l phage DNA molecules form closed-circular monomers?
(b) Under certain conditions, l phage DNA molecules are infective. When a very low concentration of l phage DNA is incubated with DNA polymerase I and the four deoxyribonucleoside triphosphates, the infectious activity of l phage DNA is destroyed. Brief treatment of l phage DNA with bacterial exonuclease III, an enzyme that removes 5′-mononucleotides from the 3′-ends of double-strand DNA molecules also destroys infectivity, but subsequent treatment of the DNA with DNA polymerase I and nucleotide substrates can restore infectivity. Describe more completely the structure of l phage DNA, and provide an interpretation of the action of the two enzymes on the molecule.
15. The isolation of viral DNA from animal cells that have been infected with adenovirus yields linear double-strand molecules that, when denatured and allowed to reassociate under conditions favoring intramolecular annealing, form single-strand circles.
Although circular molecules can be detected using the electron microscope, resolution is not sufficient to visualize the ends of the molecule. Other analyses of the singlestrand molecule show that each end has a sequence that allows the structure shown in Figure 5.2 to form.
FIGURE 5.2 A single-strand circle formed by intra-molecular annealing of adenovirus DNA.
Double-stranded
region (a) Suppose that the base sequence at one end of the single-strand molecule is 5′-AC TACGTA…. What is the corresponding sequence at the other end? Show how these sequences would allow full-length, double-strand linear molecules to be formed.
(b) An alternate suggestion for the formation of the single-strand molecules was also proposed; it is shown in Figure 5.3. Why is this proposed pairing scheme unlikely?
FIGURE 5.3 Another proposal for formation of single-stranded molecules of adenovirus DNA.
a
b
c a„
b„ c„
3„
5„
5„
3„ a„
b„ c„ a
b
c
Denature
Anneal
a
b
c
c„
b„ a„ 3„
5„
+ 5„
3„ a„
b„ c„ c
b
a
16. Thermoacidophilic bacteria can grow in volcanic sulfur springs at pH 2 and at tempera tures as high as 85ºC. DNA polymerase purified from the thermophile Snifolobus acidocaldarius has an optimal activity at 70ºC and is stable at 80ºC. When incubated with a circular DNA template at 100ºC, the isolated polymerase can extend a 20-nucleotide primer by more than 100 nucleotides. These experiments require that enzyme-to-primer concentration be at least 1:1. The Tm value for the double-strand DNA used in the experiment is about 60ºC. Unlike DNA polymerase I from E. coli, DNA polymerase from S. acidocaldarius has no demonstrable exonuclease activity to correct mistakes in DNA by removing mismatched nucleotides (such an enzyme activity is often referred to as proofreading).
(a) Why should the ratio of enzyme to primer be 1 in order for primer extension to take place at 100ºC?
(b) Would you expect to find an auxiliary proofreading enzyme in S. acidocaldarius? Why?
(c) Would you expect DNA from the genome of S. acidocaldarius to have a G + C con tent higher or lower than that from a bacterium that grows at a more normal temperature? Why?
17. Terminal deoxynucleotidyl transferase (TdT), an enzyme found in bone marrow and thy mus tissue, can extend a DNA primer by 5′ 3′ polymerization using deoxyri bonucleoside triphosphates as substrates. The primer must be at least three nucleotides CHAPTER 5 in length and must have a free 3′-OH end. The enzyme does not require a template nor does it copy one.
(a) Compare TdT with DNA polymerase I.
(b) Would TdT be useful for synthesizing DNA molecules that carry genetic informa tion? Why?
18. The 2′,3′-dideoxynucleosides can be used as reagents to inhibit DNA replication. These analogs must be converted to dideoxynucleoside triphosphates in order to have a measurable effect on DNA synthesis. When incorporated into a growing DNA chain, a single dideoxyribonucleoside residue can effectively block subsequent chain extension.
(a) Why must a 2′,3′-dideoxyribonucleoside be converted to a dideoxyribonucleoside triphosphate to be incorporated into DNA?
(b) What feature of a 2′,3′-dideoxynucleoside is most likely to account for inhibition of DNA chain extension?
FIGURE 5.4
HOCH
Base
2
O
J
H
H
J
H
H
J
J
H
H
2„, 3„-Dideoxyribonucleoside 19. In each chain-elongation reaction catalyzed by DNA polymerase, a phosphodiester bond is formed and pyrophosphate is concomitantly released. Hydrolysis of pyrophosphate to two molecules of inorganic phosphate occurs rapidly because most cells have a potent pyrophosphorylase. By removing one of the products of the chain-elongation reaction, pyrophosphate cleavage in the cell is partially responsible for the forward progress of polymerization. However, isolated DNA polymerases can efficiently carry outchain extension in the absence of pyrophosphate cleavage, so long as the double-strand helix is allowed to form during elongation. What forces resulting from DNA helix formation might contribute to driving polymerization forward?
20. In his studies of DNA in the late 1940s, Erwin Chargaff established that DNA from all organisms has equal numbers of adenine and thymine bases and equal numbers of guanine and cytosine bases. Considering that thymine and uracil are equivalent in their abilities to form hydrogen bonds with adenine, state whether you would expect similar constraints on base composition to be found in the following: (a) single-strand RNA from tobacco mosaic virus.
(b) the DNA-RNA hybrid molecule synthesized by reverse transcriptase.
(c) RNA from a virus in the reovirus family, which have large genomes composed of double-strand RNA molecules.
21. Certain DNA endonucleases degrade double-strand DNA to yield mononucleotides and dinucleotides, but these enzymes do not degrade those duplex sequences to which other proteins are tightly bound.
(a) How can you use such a DNA endonuclease and RNA polymerase to locate a promoter site?
(b) Why should this process be performed in the absence of ribonucleoside triphosphates?
22. The amino acid at position 102 in the primary sequence of a bacterial enzyme is valine, and the corresponding codon in the mRNA sequence for the enzyme is GUU. Suppose a mutation that alters the codon to GCU has no effect on the activity of the enzyme, but another mutation that changes the codon to GAU completely inactivates the enzyme.
Briefly explain these observations.
23. It is essential for spliceosomes to remove introns precisely, that is, between the terminal nucleotide of an intron and the first nucleotide of an exon. To see why, suppose that the sequence at the normal junction in a pre-spliced mRNA between an intron and an exon is …UUAG GCUAACGG… Suppose further that a spliceosome occasionally miscleaves the
Intron
Exon pre-mRNA transcript between the C and U residues in the exon sequence to yield the following two splicing intermediates:
…UUAGGC
UAACGG…
What would be the consequence of this cleavage?
24. Although nearly all the proteins synthesized by a bacterial cell after it has been infected with T2 bacteriophage are determined by the viral genome, some bacterial proteins are also required for successful infection. What bacterial enzyme is needed to initiate viral infection when T2 DNA first enters the cell?
25. (a) The genome of bacteriophage FX174 is a single strand of DNA containing 5386 nucleotides. If only one AUG in the genome were used as an initiation signal, how many amino acids could be encoded by the genome? If the average molecular weight of an amino acid is 112, what is the maximum molecular weight of protein encoded by the genome?
(b) Studies have shown that the FX174 genome can encode a larger number of proteins than expected. One reason for this increased encoding capacity is that some of the genes overlap each other. For example, the coding sequence for gene B is located entirely within the sequence that codes for gene A. However, the amino acid sequences of the two proteins specified by these genes are entirely different. How is this possible?
26. In contrast to DNA polymerase, RNA polymerase has no nuclease capability to excise mismatched nucleotides. Suggest why the two enzymes are different in this respect.
27. The genome of bacteriophage G4 is a small, single-strand circle of DNA. Replication of the circle is initiated when an RNA polymerase, a product of the dnaG gene of E. coli, synthesizes a small segment of RNA that binds to a unique sequence on the G4 chromosome.
Initiation of G4 DNA synthesis does not occur in bacterial dnaG gene mutants, which have an inactive RNA polymerase. Suggest a function for the small segment of RNA.
28. (a) In E. coli, a tRNA that carries tyrosine is composed of 85 nucleotides. However, transcription of the gene that codes for tyrosine tRNA yields an RNA molecule consisting of 350 nucleotides. At least three ribonuclease enzymes cooperate in removing a 41-base segment on the 5′ side of the tRNA sequence and a 224-base segment that extends from the 3′ terminus of the tRNA sequence. The tRNA sequence in the primary transcript is continuous, and no nucleotides are removed from that part of the transcript during processing. How does this type of RNA processing differ from splicing?
(b) Another primary transcript that is synthesized in bacteria contains 6500 nucleotides, including sequences for the 23S, 16S, and 5S RNA molecules found in ribosomes.
This primary transcript has sequences on either side of the set of rRNA sequences, as well as “spacer” sequences between each of them. Suggest a reason for the synthesis of a transcript containing all three rRNA sequences.
CHAPTER 5
29. The codons UAA, UAG, and UGA are signals for chain termination in protein synthesis because none of these codons are read by tRNA molecules. These codons are normally found at the ends of coding sequences for proteins. However, single-base mutations in certain codons can also cause premature termination of the protein chain.
(a) Which codons can be converted to the chain-termination codon UAA by a single base change?
(b) Suppose a mutation creates a UAA codon that is three codons away from the 3′ end of the normal mRNA coding sequence. Why might you assume that the prematurely terminated protein might still be functional?
(c) Revertants of chain-termination mutants include those in which a single-base sub stitution changes a termination codon to one that can again be read by a tRNA molecule. For example, a UAG codon can mutate to UCG. What amino acid would then be found at the corresponding position in the protein?
(d) Other revertants retain the original termination codon at the premature termination site, but an amino acid is inserted at the corresponding site in the protein so that the protein has the same length as the nonmutant protein would have. These revertants are due to another mutation in which the anticodon of a tRNA molecule is altered so that the tRNA molecule can read a termination codon. These tRNA molecules are called suppressor tRNAs because they suppress the effect of a chain-termination mutation. Suppose you have a chain-termination mutation that is due to the presence of a UAG codon in the normal coding sequence. If the effect of the UAG codon is suppressed by a tRNA mutation, which amino acids could be found at the site corresponding to the premature termination signal? Assume that a single base change occurs in each case.
30. Polynucleotide phosphorylase, which polymerizes ribonucleoside diphosphates (NDP) to form RNA and Pi, was used in the laboratory to synthesize polyribonucleotides that were useful in determining the genetic code. Why is it unlikely that this enzyme synthesizes RNA in the cell? Suggest how the cell uses this enzyme.
31. (a) When an experiment was done to form hybrids between mRNA produced after bac teriophage T2 infection and the denatured T2 genomic DNA (mRNA was in molar excess over DNA strands), significantly less than 100% of the DNA could form a DNA-RNA hybrid. What did this suggest about whether transcription takes place on one or both of the two DNA strands at any location on the chromosome?
(b) Later the principle of transcription on only one strand of DNA was established firmly by studies with DNA from the virus SP8, which infects the bacterium Bacillus
subtilis. Because the two complementary strands of SP8 DNA have very different base compositions, they can be easily separated by density gradient centrifugation.
How could you use these separated strands to show that the transcription of SP8 DNA occurred on one strand only?
32. You are studying the effects of amino acid replacements on the stabilbity of a particu lar a helix that is buried in the protein myoglobin. You carry out a series of replacements of a particular leucine residue located in the helix, using site-specific mutagenesis, a technique described in detail in Section 6.4 of the text. The replacements are as follows: (i)
leucine
arginine (ii) leucine
valine (iii) leucine
proline (iv) leucine
glycine (v) leucine
alanine (a) For each replacement, predict whether the change would stabilize, destabilize, or have no effect on the structure of the a helix. Briefly explain each of your predictions.
(b) For each replacement, write the most likely mRNA codon required to code for the particular amino acid. Which of the replacements can be carried out by single-base changes? Which of the replacements can be produced only by altering two bases in the mRNA codon?
33. Cordycepin (3′-deoxyadenosine) is a compound that can block the synthesis of RNA, because a cordycepin residue in an RNA chain lacks the 3′-OH end needed for chain extension by RNA polymerase. The structure of cordycepin is shown below.
NH
2
J
C
N
N
C
CH
HC
C
N
N
HOCH
2
O
J
H
H
J
H
H
J
J
H O H (a) Cordycepin does not inhibit the growth of bacteria, but it does inhibit growth and division of mammalian cells. Consider the reactions that are required for cordycepin to be converted into a substrate for RNA polymerase and then propose a reason for its ineffectiveness in bacteria.
(b) Would you expect cordycepin to block DNA synthesis as well? Why?
34. Raney nickel can convert cysteinyl-tRNACys to alanyl-tRNACys. When this altered aminoa cyl-tRNA is used in a protein-synthesizing system in vitro, alanyl residues are placed in the position normally occupied by cysteinyl residues in the protein. What does this experiment tell you about the ability of the protein-synthesizing machinery to recognize an inappropriate aminoacyl-tRNA like alanyl-tRNACys.
35. The products of the cleavage of RNA in dilute alkali include 2′- and 3′-monophosphates.
What does this observation reveal about the mechanism of cleavage of RNA? How might sensitivity to base-catalyzed cleavage have affected the choice of DNA or RNA as the primary carrier of genetic information?
36. Many steps in the flow of genetic information are subject to regulation. Stringent con trol of the production of macromolecules limits expenditure of energy by the cell, permitting the synthesis of particular proteins only as they are required. Consider the steps in storage and transmission of genetic information, and describe which one, when regulated, makes it possible to achieve the greatest economy in energy expenditure by a mature cell.
37. In 1971, David Baltimore was investigating whether polymerase activities were con tained in the Rauscher murine leukemia virus. This virus has an RNA genome and causes leukemia in mice. He disrupted purified virus particles and incubated the resulting mixture with Mg2+ and either the four dNTPs or the four NTPs in a buffered solution. One of the dNTPs or one of the NTPs was radiolabeled. After allowing time for a reaction to occur, the mixtures were treated with strong acid to precipitate nucleic acids while leaving unreacted nucleoside triphosphates in solution. By measuring the precipitated radioactivity, this assay allowed him to detect the formation of the product of a putative polymerase. He found the following: (1) NTPs were not incorporated into CHAPTER 5 product; (2) dNTPs were incorporated into product; (3) the isolated, radiolabeled product was destroyed by DNase (an enzyme that hydrolyzes DNA) but not by RNase (an enzyme that hydrolyzes RNA); (4) the isolated product was not destroyed by NaOH; (5) pretreatment of the disrupted virus extract with DNase did not prevent the formation of product whereas pretreatment with RNase did.
Do these experiments suggest the presence of a polymerase? Why? What kind of poly merase is likely present? What is its template and what is the product formed? What did these experiments indicate, for the first time, about the flow of genetic information?
38. Radioisotopes have been critical for identifying specific molecules involved in biochemi cal processes. John Hershey and Martha Chase carried out an experiment in 1952 with bacteriophage T2 that had been radiolabeled by being grown in either 32PO2−
2−
4 - or 35SO4 -containing medium. Bacteriophage T2 has a DNA genome. After infecting the bacterial cells in separate cultures with the two different labeled virus preparations for a time short enough to ensure that newly made viruses did not develop to the point of lysing the cells, they put the culture of infected cells in a blender to strip off any part of the virus that did not enter the cell. They next collected the infected stripped cells by centrifugation and compared the amounts of radioisotope in the cells to that remaining in the supernatant.
What do you think they observed and why? Why was this experiment important?
ANSWERS TO PROBLEMS
1. When the intact double-strand circular DNA molecule is heated in solution, its base pairs are disrupted and an increase in the absorbance of light at 260nm is observed.
However, the two resulting single-strand circles are so tangled about one another that they remain closely associated. When the molecules are cooled, the interlocked strands move relative to each other until their base sequences are properly aligned and a double-strand molecule is reformed. This molecule absorbs less ultraviolet light at 260nm (hypochromism) than does the pair of denatured single strands. Breaks in one or both strands of a double-strand DNA molecule allow the two strands to separate completely from one another during denaturation. In order to form a doublestrand molecule, the separate strands collide randomly until at least a small number of correct base pairs is formed (nucleation); then the remaining bases pairs form to generate a completely double-strand molecule. Reassociation of a pair of separate strands in solution is slower than that of a pair of interlocked circles because the local concentration of strands is lower, so a corresponding difference in the reduction of absorbance will be observed.
2. (a) The longer the DNA molecule, the larger the number of base pairs it contains. As a result, more thermal energy is required to disrupt entirely the helical structure of the longer DNA molecule. Experiments show that such a relationship is true for molecules up to ~4000 base pairs in length.
(b) Sodium ions neutralize the negative charges of the phosphate groups in both strands. As the concentration of NaCl decreases, repulsion between the negatively charged phosphate groups increases, making it easier to separate the two strands.
The tendency for the strands to separate more easily means that dissociation occurs at a lower temperature, which is reflected in a lower Tm value of the molecule.
(c) The reassociation of single strands begins when a short sequence of bases in one strand forms hydrogen bonds with a complementary sequence in another. Once a short stretch of base pairs is formed, reassociation to form the longer double-strand molecule occurs rapidly. The higher the concentration of DNA, the greater the number of complementary sequences in the solution, and thus the quicker the complementary sequences will find and pair with each other.
(d) Urea, which contains hydrogen bond donors (−NH2) and hydrogen bond acceptors ( > C = O), disrupts the hydrogen bonds between bases. Because hydrogen bonds are partly responsible for the stability of the double helix, the disruption of these bonds makes the structure more sensitive to denaturation by thermal energy and thereby reduces the Tm value. In addition to hydrogen bonding, the tendency of bases to stack also contributes significantly to the stability of the helix. Base stacking minimizes the contact of the relatively insoluble bases with water, and it also allows the sugar-phosphate chain to be located on the outside of the helix, where it can be highly solvated. Urea may also cause destabilization of the helix by allowing bases to associate more readily with water by disrupting its structure.
3. (a) The four base pairs found in the DNA double helix are almost identical in size and shape, so the diameter of the double helix is essentially uniform all along its length.
It is therefore unlikely that a protein can identify a specific sequence by sensing differences in the diameter of the helix.
(b) Proteins that interact with specific sequences might do so by forming hydrogen bonds with the bases; in some cases, it might be necessary for the double strand to undergo local unwinding or melting in order for the bases to form hydrogen bonds with a protein. However, hydrogen-bond donors and acceptors are also found in the grooves of the intact helix. A protein could also bind to a specific location on DNA by forming hydrogen bonds with a particular group of atoms in one of the grooves of the helix. Hydrophobic interactions between amino acid side chains and the methyl group of thymine or the edges of the bases can also contribute to the specificity of the interaction.
4. Although the system described could yield 32P-labeled daughter DNA molecules, chem ical methods cannot distinguish DNA in which both strands are radioactively labeled from DNA in which one strand is labeled and one strand is unlabeled. In their experiments, Meselson and Stahl used a physical technique, density gradient equilibrium sedimentation, to separate the labeled molecules according to their content of 14N and 15N, which differ in their specific densities.
5. In solution, the oligodeoxyribonucleotide forms an interchain double-strand molecule with flush ends and a small single-strand loop containing the sequence 5′pTpCpCpTpCp-3′. The deoxyribonuclease hydrolyzes the phophodiester bonds in this single-strand region to form nucleoside monophosphates, leaving a small double-strand linear molecule remnant containing seven base pairs.
6. Formaldehyde could react with the exocyclic amino groups on the C-6 carbon of adenine, the C-2 of guanine, and the C-4 of cytosine to form hydroxymethyl derivatives. Because these derivatives cannot form hydrogen bonds with complementary bases, formaldehyde-treated single strands would reassociate to a lesser extent than would untreated single strands. The actual sites of the reaction of formaldehyde with DNA are not precisely known; these sites may also include the ring nitrogen atoms in pyrimidines.
7. This experiment suggests that the hydrogen bonds of base-paired regions of double strand DNA may undergo reversible dissociation to form single-strand regions, often known as bubbles. The transient disruption of these hydrogen bonds allows the exchange of protons with the tritiated water. A T pairs open more easily than G C pairs. Thus, the greater the percentage of AT pairs, the greater the rate of proton exchange.
CHAPTER 5
8. A continuous, linear double-strand DNA molecule has only two 3′-OH groups available for the initiation of DNA synthesis by DNA polymerase; because each is located at opposite ends of the molecule, no template sequence is available. In order to construct a relatively simple mechanism for chromosomal replication, one could postulate that the enzyme initiates DNA replication at a number of breaks along the chromosome, with each of the breaks offering the 3′-OH group required for the initiation of the new DNA strand. The template required for replication would then be located on the strand opposite the break, thus ensuring that DNA synthesis could continue. It is now well established that DNA in chromosomes is very long and continuous. The fact that there are initially no breaks in the molecule makes the mechanism of replication complex. It involves a number of enzyme activities, as well as the use of RNA to prime the synthesis of DNA. For details, see page 760 of the text.
9. (a) The expected melting temperature for E. coli DNA containing 50% GC base pairs is
T =
m
69.3 + 0.41(G + C) = 69.3 + 0.41(50) = 69.3 + 20.5 = 89.8ºC (b) Most organisms live at temperatures that are considerably lower than 65ºC. Because both the transmission and expression of genetic information depends on the integrity of the double-strand DNA molecule, it is important that the molecule not be disrupted by thermal energy.
10. First, you must determine the temperature at which the hydrogen bonds are disrupted and single strands are formed. You can do this by heating the double-strand DNA to various temperatures and measuring the extent of hyperchromicity. Once the DNA has been melted, centrifuge the sample using the density-gradient equilibrium sedimentation technique to attempt to separate the 14N-labeled DNA strands from the 15N-labeled DNA strands, which will be the denser of the two. If you are successful, this would suggest that the strands separate completely during thermal denaturation.
11. The protonation of the N-1 and N-7 of adenine, the N-7 of guanine, the N-3 of cyto sine, and the O-4 of thymine makes normal hydrogen bonding at these locations impossible because the atoms can no longer serve as hydrogen-bond acceptors. Therefore, at low pH, where proton concentrations are high and protonation of these atoms occurs, double-strand DNA is less stable than at neutral pH values. At high pH values ( > 11) DNA is also denatured by deprotonation of other ring atoms.
12. The association of a molecule of RNA with a molecule of DNA to form a hybrid mole cule depends primarily on the two molecules having complementary sequences of bases.
The formation of hydrogen bonds between complementary bases will allow the formation of a double helix composed of RNA and DNA.
13. The deoxyribonucleoside triphosphate dUTP can be used as a substrate for DNA poly merase during DNA replication because the structure and hydrogen-bonding properties of uracil are very similar to those of thymine. When incorporated into a double-strand DNA polymer, uracil pairs with adenine, as does thymine. For a discussion of the reasons uracil is not normally incorporated into DNA, see page 771 of the text.
14. (a) To insure that l phage DNA molecules form closed circular monomers, the con centration of l phage DNA should be relatively low so that the intrachain formation of hydrogen bonds is favored. At higher concentrations, the probability of interchain joining to form multimers is enhanced.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (b) The most reasonable model for the structure of the l phage DNA molecule is a dou ble-strand molecule having single-strand protrusions at the 5′-ends, as illustrated in the margin. The 3′-ends have hydroxyl groups, which allow them to serve as primers for DNA synthesis catalyzed by DNA polymerase I. This enzyme fills in the single-strand regions of the molecule, producing a molecule with flush ends. Such a molecule no longer has cohesive ends that can form the required circular molecule needed for infectivity.
3„
5„
5„
3„
Molecules treated with exonuclease III have a longer single-strand sequence at both ends; they may not be infective because the newly exposed bases may not be fully complementary to each other, which would mean that the ends could no longer be joined. When the exonuclease-treated DNA is treated with DNA polymerase I, the single-strand regions are sufficiently filled in to reform a molecule that has protruding single strands that are approximately the same length as those in the native molecule. Hence, the molecule once again becomes infective. Further treatment of the molecule with DNA polymerase I will once again produce a molecule that has flush ends and is no longer infective.
15. (a) The sequence at the other end of the single-strand molecule must be composed of complementary bases. It must therefore be
…TACGTAGT-3′
The structure of the full-length, double-strand, linear molecule would be 5′-ACTACGTA———TACGTAGT-3′
3′-TGATGCAT———ATGCATCA-5′
Each single strand has a pair of inverted repeats.
(b) The formation of double-strand helical segments depends upon hydrogen bond for mation between bases in nucleotide chains that are antiparallel, as follows:
5′…PuPyPuPyPu…3′
3′…PyPuPyPuPy…5′ where Py = pyrimidine and Pu = purine. When the suggested structure is labeled using this scheme, as in Figure 5.5, it can be seen that it would require the formation of base pairs between parallel chains, and such pairing cannot readily take place.
FIGURE 5.5 Base pairing between parallel nucleotide chains, required to form the circular structures shown in Figure 5.3, is unlikely in DNA.
(3„) 5„
3„ (5„) 16. (a) Each enzyme molecule probably tightly clamps the primer as well as the extended, newly synthesized chain to the template, thereby protecting the helix from denaturation at high temperature. One primer would be required for each template strand. The association between primer and enzyme may protect the enzyme from thermal denaturation as well. There are many examples in biochemistry where substrate binding stabilizes protein structure.
(b) During polymerization of a new DNA chain, the chance that base incorporation er rors can occur will increase at high temperatures. Even though a proofreading or error-correcting activity is not found in the S. acidocaldarius polymerase polypeptide, you would expect an auxiliary enzyme to be present in the cells in order for the bacterial genome to be accurately replicated so that the genetic integrity of the organism is maintained.
(c) The higher the G + C content of double-strand DNA, the higher the melting point at which the helix is denatured. Therefore, you might expect DNA from a thermophile to have a higher G + C composition. Surprisingly, DNA base ratios are sometimes not very different from those in bacteria living at lower temperatures, so that DNA isolated from these thermophilic bacteria melts at temperatures lower than those encountered in the hot springs where they grow. There must be proteins or other molecules in the bacterial cells that protect the genomic DNA of these thermophiles from thermal denaturation.
17. (a) Like DNA polymerase I, TdT can extend a DNA primer by using deoxynucleoside triphosphates as substrates. However, TdT does not use a template and cannot copy from one, so that the base composition of the newly synthesized single strand of DNA will depend solely on the relative concentrations of the deoxynucleoside triphosphate substrates. For chains synthesized by DNA polymerase I, the base composition will be complementary to that of the template strand. Although DNA polymerase has an exonuclease activity that removes mismatched bases from newly synthesized strands, TdT has no such activity and does not need one.
(b) DNA molecules that carry genetic information must be synthesized as faithful copies of template strands; TdT cannot copy a template and would not be useful in genomic DNA synthesis. TdT is sometimes used to introduce sequence variation into DNA during antibody formation.
18. (a) DNA polymerase requires deoxyribonucleoside triphosphates as substrates for DNA chain extension. Nucleosides like the 2′,3′-dideoxy analogs must be converted to nucleoside triphosphates in order to serve as substrates for DNA polymerase.
Studies on inhibition of DNA synthesis in living cells involve incubating those cells with the nucleoside forms of the analogs instead of their nucleoside triphosphate forms, because negatively charged phosphate anions cannot pass across the plasma membrane, while relatively neutral nucleosides can. Once inside the cell, nucleoside analogs are phosphorylated by cellular enzymes that normally function to “salvage” nucleosides generated by turnover of nucleotides from RNA and DNA.
(b) Dideoxynucleosides lack a free 3′-hydroxyl group, which would normally serve as an acceptor for incorporation of the next nucleotide into the growing polynucleotide chain. The lack of a 3′-OH group also interferes with excision by the error-correcting exonuclease activity of some DNA polymerases, so that chain extension is blocked.
19. Noncovalent forces also contribute to driving the reaction forward. Hydrogen bonds form between opposing A and T bases and between G and C molecules in the antiparallel chains. There are also significant hydrophobic interactions between adjacent bases on the same strand, and these stacking interactions may in fact contribute significantly to helix formation and stability. In vivo, it is likely that these noncovalent forces, along with the cleavage of the product, account for the forward progress of chain elongation.
20. (a) The ratios observed by Chargaff can be attributed to the requirement that in a dou ble-strand polynucleotide, only certain bases can form hydrogen bonds with one another. Although a single-strand polynucleotide might form some hydrogen bonds between bases as it folds, the overall base ratios will not conform to Chargaff’s established rules. This is also true for single-strand DNA in a virus like FX174.
(b) Because hydrogen bonding between A and T (or U, in the case of RNA) and be tween G and C can occur in a duplex molecule formed by one strand of DNA and a complementary strand of RNA, you would expect to see base ratios like those observed by Chargaff.
(c) Double-strand RNA molecules form hydrogen bonds between bases in a manner similar to those in DNA helices. Therefore, bearing in mind that U, not T, would normally be found in RNA, the number of uracil residues would equal the number of adenines, and the number of guanine bases would be expected to be the same as those of cytosine.
21. (a) To locate a promoter site, you would first incubate the double-strand DNA with RNA polymerase; the RNA polymerase will bind tightly to the promoter site. Next, you would add the DNA endonuclease, which will degrade the DNA that is not protected by the bound RNA polymerase. Electrophoresis can then be used to determine the size of the protected fragments of DNA, and the base sequence can be determined using methods discussed in Chapter 6 of the text.
(b) Ribonucleoside triphosphates are substrates for RNA polymerase transcription. If present when this process is performed, they would allow the polymerase molecule to move from the promoter site to the site on the template where transcription begins, as well as beyond, as transcription progresses. As a result, the promoter site would no longer be protected from endonuclease degradation.
22. The GCU codon in the first mutation corresponds to a substitution of alanine for valine at position 102. Although the side chain of alanine is smaller than that of valine, both are aliphatic amino acids, so the alteration in the structure of the enzyme does not necessarily affect the enzyme activity. The GAU codon of the second mutation specifies the substitution of aspartate for valine at position 102. This substitution could have a detrimental effect because the side-chain carboxyl group of aspartate has a negative charge at neutral pH. The charged group could disrupt the native conformation of the enzyme, thereby inactivating it.
23. If the spliceosome cleaves the initial transcript within the normal exon sequence as shown, the exon coding sequence in the spliced mRNA will be altered because of the loss of two bases. Instead of beginning with the codon GCU in the normal exon, the reading frame will begin with the codon UAA in the altered exon. This codon is, in fact, a termination signal for protein synthesis, which means that the translation of the polypeptide specified by the spliced messenger RNA would be terminated prematurely.
24. Because only DNA and no protein from T2 enters the cell, the synthesis of viral-directed proteins cannot begin until T2 messenger RNA has been made. Transcription of the T2 DNA must therefore be carried out by bacterial RNA polymerase using ribonucleoside triphosphates synthesized by the bacterial cell.
CHAPTER 5
25. (a) Three consecutive bases are required to encode an amino acid, so up to 1794 amino acids could be specified by the FX174 genome. The molecular weight of this much protein would be approximately 201,000 (b) Overlapping genes can yield proteins with different primary amino acid sequences only if each of the protein coding sequences is read in a different reading frame.
The critical feature for the establishment of the proper reading frame is the location of the AUG initiation signal. As an example, consider the following mRNA sequence, which contains two AUG codons that are in overlapping but different reading frames:
Reading frame 1
A U G C C U A G A U G C A G U U C G
Reading frame 2
When an initiator tRNA binds to the first AUG codon, reading frame 1 is established; similarly, reading frame 2 is established when an initiator tRNA binds to the second AUG codon. The polypeptides specified by the two different mRNA sequences will necessarily have different amino acid sequences.
26. DNA polymerase is responsible for the duplication of the DNA of chromosomes, which is the repository of the genetic information that is passed on to progeny cells. Any error that occurs in the copying of a DNA template will be transmitted not only to the duplicated chromosome but also to all the messenger RNA molecules transcribed from the miscopied DNA template. Therefore, it is crucial that DNA polymerase be able to correct errors that occur due to the incorporation of mismatched nucleotides. RNA polymerase makes many copies of mRNA, but these molecules have relatively brief lives in the cell, and very few are passed on to progeny cells. Occasional errors in transcription can result in the production of defective proteins, but it appears that the cell can tolerate such errors provided that not too many occur.
27. The small RNA molecule serves as the primer for the initiation of DNA synthesis by DNA polymerase. To synthesize a DNA chain DNA polymerase requires a primer nucleotide with a 3′-hydroxyl terminus along with a template. Studies show that either an oligodeoxyribonucleotide or an oligoribonucleotide can serve as a primer. DNA polymerase cannot synthesize a primer sequence on a closed DNA circle, but the RNA polymerase produced by the dnaG gene can synthesize a short primer sequence, which is then extended as DNA by DNA polymerase. When replication has extended around the circle and the 5′ terminus of the RNA primer is reached, the primer is hydrolyzed and the small gap is filled in with a DNA sequence. Bacterial dnaG mutants cannot support G4 infection because they cannot synthesize the RNA primer.
28. (a) The processing of the primary tRNA transcript removes RNA on either side of the uninterrupted tRNA sequence of 85 nucleotides, whereas protein splicing operations remove RNA sequences that are located within regions of pre-mRNA that code for a continuous polypeptide; these sequences must be removed to ensure that the protein specified by the RNA will have the correct amino acid sequence. In some organisms, tRNA are produced by RNA splicing, which removes sequences from within the pre-tRNA transcript.
(b) The most obvious reason for transcribing all three rRNA sequences simultaneously is that it ensures that an equal number of ribosomal RNA molecules will be avail able for the assembly of ribosomes. In addition, only one promoter, rather than three, is required for rRNA synthesis.
29. (a) The protein-encoding codons that could be mutated to UAA by a single base change are CAA, GAA, AAA, UCA, UUA, UAU, and UAC.
(b) Chain termination near the 3′ end of the normal coding sequence could allow the prematurely terminated protein to be functional because most of the polypeptide sequence would be intact. Removing a few amino acids from the C-terminal end of many, but not all, proteins does not appreciably affect their normal function.
(c) When a UAG codon reverts to UCG, a serine residue will be incorporated at the site in the protein that corresponds to the chain-termination site; it will be linked by a peptide bond to the next amino acid in the polypeptide.
(d) To determine which amino acids would be carried by suppressor tRNAs to a UAG codon site, you should identify all those tRNA molecules having an anticodon that, by a single base change, can read a UAG codon. Each such tRNA molecule would suppress the premature termination of the chain by inserting an amino acid at the corresponding site in the protein. The amino acids that could be found at the site (along with their codons) are Glu (GAG), Gln (CAG), Leu (UUG), Lys (AAG), Ser (UCG), Trp (UGG), and Tyr (UAC and UAU). When a cell contains a suppressor tRNA, proteins whose mRNA sequence normally ends with a single stop codon may not be efficiently terminated. Although the extension of such proteins could be lethal, most cells tolerate suppression. One explanation is that other proteins involved in chain termination may recognize a stop codon even though a tRNA that reads the codon is present. More work is needed to develop a full understanding of the reasons for toleration of suppression.
30. Equilibrium for the reaction catalyzed by polynucleotide phosphorylase lies toward the direction of RNA degradation rather than synthesis. High concentrations of ribonucleoside diphosphates are required to achieve the net synthesis of RNA; and it is likely that their concentrations in the cell are not sufficient to drive net polynucleotide synthesis.
Also, polynucleotide phosphorylase does not use a template, so the polyribonucleotides it synthesizes contain random sequences, which makes them of no value for protein synthesis. The cell uses polynucleotide phosphorylase as a degradative enzyme in conjunction with other nucleases that regulate the lifetimes of RNA molecules, including mRNA.
In bacteria mRNA lifetimes are relatively short.
31. (a) If transcription occurred simultaneously on both DNA strands, the excess, comple mentary mRNA molecules synthesized from each template strand could form double-strand structures with all the DNA strands. The fact that less than all the DNA could form hybrids indicated that some of the DNA strands lacked sequences complementary to the mRNA, that is, only one of the two strands at a given location along the DNA was being transcribed into DNA. In rare cases in some organisms over limited regions, RNA is synthesized from both strands of the template DNA.
(b) To establish whether one or both strands of SP8 DNA are used for transcription, you can carry out hybridization experiments with the separate strands using radioactive RNA synthesized during the infection of Bacillus with SP8. The results show that such RNA hybridizes to only one of the two strands, which means that only one of the two strands of the DNA of the SP8 virus is transcribed. In most other organisms, different regions of each strand are used for transcription; SP8 virus is exceptional in that one strand is used exclusively for all mRNA synthesis.
CHAPTER 5
32. (a) (i) Leucine arginine destabilizes the helix because you substitute a hy drophilic residue for a hydrophobic residue (most likely in a hydrophobic region of the helix).
(ii) Leucine valine has no effect (a conservative replacement of one hy drophobic residue for another).
(iii) Leucine proline destabilizes the helix because proline does not allow ro tation about its peptide bond and no hydrogen atom in this amino acid is available for hydrogen bonding.
(iv) Leucine glycine destabilizes because glycine is a very flexible residue and can act as a swivel, disrupting the helix (the frequency of glycines in a helix is as rare as the frequency of prolines).
(v) Leucine alanine has no effect (a conservative replacement of one hy drophobic residue for another).
(b) (i) CUU
CGU
single-base change required (ii) CUU
GUU
single-base change required (iii) CUU
CCU
single-base change required (iv) CUU
GGU
two-base alteration required (v) CUU
GCU
two-base alteration required
33. (a) Cordycepin is a nucleoside, and it must be converted to the triphosphate form be fore it can be incorporated (as cordycepin monophosphate) by RNA polymerase into a growing polynucleotide chain. The conversion of cordycepin to the triphosphate form is carried out by a number of kinase enzymes (see Chapter 25, p. 697 on nucleotide metabolism in the text) that utilize ATP as a phosphate donor. Bacteria probably cannot phosphorylate cordycepin efficiently, which makes them less susceptible to inhibition of RNA and DNA synthesis.
(b) You would not expect cordycepin to inhibit DNA polymerase because although a cordycepin residue that had been added to DNA would also lack a 3′-OH and act as a chain terminator, the presence of the 2′-OH on the ribose of the triphosphate form of cordycepin would be discriminated against by the DNA polymerase. You will learn later that DNA synthesis requires an RNA primer, and cordycepin might inhibit DNA synthesis by inhibiting RNA primer formation.
34. The ribosomal complex that carries out protein synthesis is unable to recognize alanyl tRNACys as an inappropriate or erroneous form of tRNA. An amino acid that is attached to a transfer RNA molecule will be transferred into a growing polypeptide chain solely on the basis of recognition between the anticodon in tRNA and the codon in the messenger RNA molecule. Once an aminoacyl-tRNA has been formed, accurate translation does not depend on recognition of the attached amino acid. This important point was established by Dintzis and von Ehrenstein, who carried out the incisive experiments using Raney nickel to reduce the cysteinyl residue on cysteinyl-tRNACys to an alanyl residue, then analyzing the resulting protein using an in vitro hemoglobin-synthesizing system.
35. The base-catalyzed generation of 2′- and 3′-monophosphates argues that a cyclic 2′,3′ phosphodiester is formed during cleavage of RNA. It is likely that a hydroxyl anion abstracts the hydrogen atom of the 2′-OH of RNA, leaving a 2′-O− that attacks the phosphorus atom, cleaving the 5′ phosphodiester bond and generating a cyclic 2′,3′phosphodiester. This cyclic derivative is unstable and decomposes by hydrolysis to form either a 2′- or a 3′-phosphate ester. DNA has no 2′-OH groups and is therefore not susceptible to alkali degradation. Because it is more stable than RNA, it may have been selected as the primary carrier of genetic information to future generations.
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION 36. The step that would afford the maximum economy is probably transcription, through the control of the activity of RNA polymerase. Transcription by RNA polymerase to form RNA is the first step in the expression of genetic information. It follows that controlling messenger RNA production, by stimulating or inhibiting the activity of RNA polymerase, allows the cell to make particular types of mRNA and to synthesize the encoded proteins only when required. A cell that could not regulate RNA polymerase activity would produce unneeded mRNA molecules, and the energy required to produce those polynucleotides would be wasted, even if translation were stringently regulated.
37. Yes, the incorporation of nucleoside triphosphates into an acid-insoluble form is in dicative of the presence of a polymerase. The polymerase is likely a DNA polymerase because dNTPs, and not NTPs, were used to form product. Further evidence for a DNA polymerase was that the radiolabeled product was destroyed by a nuclease, DNase, specific for hydrolyzing DNA, and not by one specific for RNA hydrolysis.
Additionally, NaOH, which destroys RNA but not DNA, did not destroy the radiolabeled product. Pretreatment of the extract with the two hydrolytic enzymes demonstrated that the enzyme depends on an RNA and not a DNA template for its activity.
Thus, this enzyme is an RNA-dependent DNA polymerase. No such enzyme had been observed previously in a cell, and this demonstration, along with similar findings by Howard Temin, of its existence in an RNA tumor virus caused a revision of Francis Crick’s central dogma of molecular biology, which stated that information flowed from DNA to RNA to proteins. The demonstration of this RNA-dependent DNA polymerase suggested that in some cases information could flow from RNA to DNA. (This question was derived from D. Baltimore. Viral RNA-dependent DNA polymerase. Nature
226:[1971]1209–1213.)
38. Hershey and Chase observed that most of the 32P was associated with the cells and most of the 35S was in the supernatant. Since nucleic acids are rich in phosphorus and DNA does not contain sulfur, they concluded that DNA had entered the cell. The sulfur of sulfate is incorporate into the amino acids cysteine and methionine so that 35S is a good marker for proteins. The experiment indicated that protein did not enter the cells. Recalling that bacteriophage T2 displays heredity, that is, passes genetic traits to its progeny, they concluded that DNA, not protein, is likely the genetic information because it entered the cells and was replicated. At the time these experiments were performed, they helped solidify the view that the genetic material was DNA, not protein. (This question was derived from A. D. Hershey & M. Chase. Independent functions of viral protein and nucleic acid in growth of bacteriophage. J. Gen. Physiol.
36:[1952]39–56.)
EXPANDED SOLUTIONS TO TEXT PROBLEMS 1. By convention, when polynucleotide sequences are written, left to right means 5′ 3′.
Since complementary strands are antiparallel, if one wishes to write the complementary sequence without specifically labeling the ends, the order of the bases must be reversed.
(a) TTGATC (b) GTTCGA (c) ACGCGT (d) ATGGTA (b) Due to base pairing (A:T, G:C) in the complementary strand, [T] = 0.30, [C] = 0.24, and [A] + [G] = 0.46.
3. To answer this question one must know that 2mm = 2 × 10−6, that one Å = 10−10 m, and that the distance between the base pairs is 3.4 Å. The length of a DNA segment (in this case 2 × 10−6 m) divided by the distance between the base pairs (3.4 × 1−10 m) gives the answer, 5.88 × 103 base pairs.
4. After 1.0 generation, one-half of the molecules would be 15N-15N, the other half 14N-14N. After 2.0 generations, one-quarter of the molecules would be 15N-15N, the other three-quarters 14N-14N. Hybrid 14N-15N molecules would not be observed.
5. (a) Thymine is the molecule of choice because it occurs in DNA, is not a component of RNA, and is not readily converted to cytosine or uracil. If they enter the cell, labeled deoxythymidine or dTTP would also be useful molecules. Its large negative charge prevents dTTP from entering most cells.
(b) During DNA synthesis, the b- and g-phosphorus atoms of the nucleoside triphosphates are lost as pyrophosphate. Since the a-phosphorous atom is incorporated into DNA, one should use dATP, dGTP, dTTP, and dCTP labeled with 32P in the a position.
6. Only (c) would lead to DNA synthesis because (a) and (b) have no primer or open end to build on and (d) has no template extending beyond a free 3′-OH. Note: Singlestranded linear DNA can be used as a template for DNA synthesis because it can prime synthesis through hairpin formation at its 3′ end.
7. A short polythymidylate chain would serve as a primer because T base pairs with A.
Radioactive dTTP labeled in any position except the b- and g-phosphates would be useful for following chain elongation.
8. After the synthesis of the complementary (–) DNA on the RNA template, the RNA must be disposed of by hydrolysis prior to the completion of the synthesis of the DNA duplex.
9. One should treat the infectious nucleic acid with either highly purified ribonuclease or deoxyribonuclease and then determine its infectivity. RNAse will destroy the infectivity if it is RNA; DNAse will destroy it if it is DNA.
10. Ultimately, this mutation results in half the daughter DNA duplexes being normal and half having a TA pair that had been CG. The first two rounds of replication at the mutant site will be as follows:
CG
CG
CG
CG
UG
TA
UA
UA
11. (a) From the 4 mononucleotides one can formulate 16 different dinucleotides. If you don’t believe it, try it! From these dinucleotides you can make 64 different trinucleotides. Note that 64 is 43. There will be 44 (256) tetranucleotides. Proceeding in this manner we get to 48 (65,536) different octonucleotides (8-mers).
DNA, RNA, AND THE FLOW OF GENETIC INFORMATION (b) A bit specifies two bases (say A and C), and a second bit specifies the other two (G and T). Hence, two bits are needed to specify a single nucleotide (or base pair) in DNA. An 8-mer stores 16 bits (216 = 65,535), the E. coli genome (4 × 106 bp) stores 8 × 106 bits, and the human genome (2.9 × 109 bases) stores 5.8 × 109 bits of genetic information.
(c) A high-density diskette stores about 1.5 megabytes, which is equal to 1.2 × 107 bits.
A large number of 8-mer sequences could be stored on such a diskette. The DNA sequence of E. coli, once known, could be written on a single diskette. Nearly 500 diskettes would be needed to record the human DNA sequence.
12. (a) Deoxyribonucleoside triphosphates versus ribonucleoside triphosphates.
(b) 5′ 3′ for both.
(c) DNA serves as the template for both polymerases. During DNA replication by poly merase I each parent strand acts as a template for the formation of a new complimentary strand. Since each daughter molecule receives one strand from the parent DNA molecule, the template is said to be semiconserved. However, after guiding the synthesis of RNA by RNA polymerase, the DNA double helix remains intact.
Hence the template is said to be conserved.
(d) DNA polymerase I requires a primer, whereas RNA polymerase does not.
13. (a) Because mRNA is synthesized antiparallel to the DNA template and A pairs with U and T pairs with A, the correct sequence is 5′-UAACGGUACGAU-3′.
(b) Since the 5′ end of an mRNA molecule codes for the amino terminus, appropriate use of the genetic code (see text, p. 109) leads to Leu-Pro-Ser-Asp-Trp-Met.
(c) Since one has a repeating tetramer (UUAC) and a 3-base code, repetition will be observed at a 12-base interval (3 × UUAC). Comparison of this 12-base sequence with the genetic code leads to the conclusion that a polymer with a repeating tetrapeptide (Leu-Leu-Thr-Tyr) unit will be formed.
14. The instability of RNA in alkali is due to its 2′-OH group. In the presence of OH− the 2′-OH group of RNA is converted to an alkoxide ion (RO−) by removal of a proton. Intramolecular attack by the 2′-alkoxide on the phosphodiester in RNA gives a 2′,3′-cyclic nucleotide, cleaving the phosphodiester bond in the process. Further attack by OH− on the 2′,3′-cyclic nucleotide produces a mixture of 2′ and 3′-nucleotides.
Note that the mechanism for ribonuclease action is quite similar (see Figure 9.18, p. 216). Since DNA lacks a 2′-OH group, it is quite stable in alkali.
15. Apparently cordycepin is converted to its 5′-triphosphate and incorporated into the growing RNA chain. This chain containing cordycepin now lacks a 3′-OH group; hence, RNA synthesis is terminated.
16. Only single-stranded mRNAs can serve as templates for protein synthesis. Since poly(G) forms a triple-stranded helix, it cannot serve as a template for protein synthesis.
17. Note that each complimentary strand is missing one of the four bases; d(TAC) lacks G and d(GTA) lacks C. Thus, incubation with RNA polymerase and only UTP, ATP, and CTP led to the synthesis of only poly(UAC), the RNA complement of d(GTA). When GTP was used in place of CTP, the complement of d(TAC), poly(GUA), was formed.
18. In a nonoverlapping code, each individual nucleotide mutation would change at most one amino acid in the protein sequence. (Because of codon degeneracy, some individual mutations will not change the amino acid sequence.) In an overlapping code, some mutations (but not all) will change the identity of two consecutive amino acids in the CHAPTER 5 protein sequence. Therefore several experiments will be needed. If one makes a series of individual nucleotide mutations, determines the resulting protein sequences, and never finds that two consecutive amino acids are changed, then one could reasonably conclude that the code is nonoverlapping.
One can also note that the putative overlapping code described in the problem would have a four-nucleotide stretch (e.g., ABCD) encoding a dipeptide. The maximum possible number of different dipeptides that could be encoded by four different nucleotides in this scenario would be only 44 = 256. However, the 20 amino acids can be used to make 202 = 400 different dipeptides, all of which are represented in known protein sequences. Therefore, this numerical analysis of naturally occurring dipeptide sequences also would argue against a completely overlapping triplet code. (See also Crick, Barnett, Brenner, & Watts-Tobin. Nature 192:[1961]1227–1232.)
19. Since three different polypeptides are synthesized, the synthesis must start from three different reading frames. One of these will be in phase with the AAA in the sequence shown in the problem and will therefore have a terminal lysine, since UGA is a stop signal. The reading frame in phase with AAU will result in a polypeptide having an AsnGlu sequence in it, and the reading frame in phase with AUG will have a Met-Arg sequence in it.