Chapter 4: Exploring Proteins

Chapter 3 was extended by introducing the most important methods.

The principles of the structure and function presented in the preceding chapter were essential to discovering many of these.

These methods are used in modern biochemical research and underlie current developments.

The concept of the proteome is defined by the authors.
They outline principles for analysis and purification.
They explain why the knowledge revealed by these techniques is so important.
They begin with a discussion of antibodies as highly specific analytical reagents, followed by a discussion of the uses of peptides of defined sequence and how they are synthesised.
There is a discussion of the use of x-ray crystallography and nuclear magnetic resonance spectroscopy to determine the three-dimensional structures of proteins to close the chapter.

You should be able to complete the objectives once you have mastered this chapter.

Give examples of the important information that is contained in the sequence.

There are methods that use specific antibodies.

Compare the advantages and disadvantages of x-ray crystallography.

The sequence of the genome tells us what the organisms can make.

The following five are listed with their isoelec tric points.

Give the order of their migration from the top to the bottom.

electrophoresis on a paper support was once used for the separation and analysis of small peptides.

The separation was done on the basis of the charge on the peptide.
The direction of migration for the following peptides is predicted.
If the peptide remains stationary, use C for migration toward the negative pole, A for migration toward the anode, and O for migration toward the positive pole.

Evaluate Table 4.1 in the text, looking at a purification scheme.

After isolating and purifying a small enzyme from a culture ofbacteria, you are confused as to whether you grew a wild-type or a mutant strain that produced the same thing.

It cannot react with phenyl isothiocyanate.

One tries to generate overlap by using cleav ages at specific sites.

The sequence of the peptides was obtained.

Match the terms in the left column with the items in the right column.

You don't have enough of the pure protein to make antibodies because the antigenic determinants are predicted from the sequence.

Many genes that are pro ducing are not being expressed in any given cell.

Single-celled organisms tend to respond to their environment by producing enzymes to deal with the conditions in the area.

Different parts of the body need different types of cells.

Humans have different needs in different parts of the body.
The proteome is a description of the proteins present in a functioning cell.

The sum of all the charges is zero, regardless of the pH.

There is a net charge of + 1.5 at pH 2.0, since the pH coincides with the pK value.

The net charge is +0.5 and the terminal carboxyl is almost completely ionized.
The net charge is zero because of a +2 charge and a -2 charge.
The net charge is - 1.5 because of the deprotonated a-amino group and half-protonated lys side chain.

The movement of compounds would be slowed because of the increased viscosity of the solution, so more time would be required for the separation.

The speed at which a molecule moves is dependent on the elasticity of the solution.

Material is lost in each step of purification.

In a good purification, the totalProtein drops much faster, so that specific activity goes up dramatically.
Not all of the proteins can be isolated with affinity chromatography.
It's a good idea to leave out the ion exchange and the molecular exclusion steps when you can use affinity chromatography to work with a unique Substrate.

One-step purifications of 1,000-fold with nearly 100% recovery have been reported with this technique, according to a standard source.

The mass is determined by the SDS-polyacrylamide gel electrophoresis when the reducing agent in the medium causes disulfide bonds to be disrupted.

The sedimentation equilibrium method is used to determine the weight of a native protein.
Gel-filtration chromatography can be used to get approximate native weights.

A substitution of a valine for a valine could result in an extra 41 atomic mass units being contained in this approximate mass.

The intact fragment can be analyzed by the Edman degradation method, and it can be determined by labeling it with either fluorodinitrobenzene or dabsyl chloride.

The performic acid oxidizes the disulfide bonds.

Answer (d) may be correct in some cases where the difference between the two structures is small.

The polypep tide chain is derived from the DNA sequence.

Since the function of a protein depends on its mature structure, it is necessary to analyze it to determine if any changes have occurred after translation.

In the presence of many other substances, recognition and binding can occur because of the highly specific interaction of an antibody with its antigen.

You could combine the peptides with the antigenic macromolecule to make the synthetic peptides and then make the antibodies against it.

The antibodies prepared against the synthetic peptides should also react with theprotein if the same antigenic determinants are present in it.

x-ray crystallography requires crystals, but not for NMR.

Significant information can be derived from the three-dimensional shape if a highly concentrated solution is obtained.

x-rays, not electrons, are scattered in x-ray crystallography.

If the three-dimen sional structures of both are obtained using x-ray crystallography, the difference between the two structures should reveal how the substrate fits in its binding site and what kind of bonds are involved.

Many of the methods described in Chapter 4 are used to purify their native state.

Discuss the standards you would like to use in this technique.

The sequence analysis of peptides from 2 to 20 amino acids can be done with mass spectrometry.

The procedure is very sensitive and requires only a small amount of protein and cationic fragments.
The peptides are treated with triethylamine and acetic anhydride.
The modified peptide is prepared with a strong base.

The release of ammonium ion accompanies the formation of a Heterocyclic ring when a glutamine residue is spontaneously cyclized.

The structure of the ring is shown in the diagram.

There is apeptide composed of 12 amino acids that does not react with phenyl isothiocyanate.

The equivalent of aspartate and two peptides can be found in the staphylococcal protease.

A mouse polypeptide hormone is analyzed using a number of chemical and enzymatic methods.

The dipeptide contained in the intact hexapeptide is a result of treating it with trypsin.

The intact hexapeptide is treated with aidase A.
It's important to remember that the hexapeptide is isolated from a mouse, so write its sequence using both three-letter and one-letter abbreviations.

The basis of most pregnancy test kits is the production of a small acidic protein.

A laboratory group wants to make a monoclonal antibody that can be used to make a Western blotting sample.

There is a map of the electron density that is needed for the determination of the three-dimensional structure.

x-ray analyses of proteins can't see hydrogen atoms.
If you want to get an electron-density map, you should compare the structures of valine, threonine, and isoleucine.

The same chemical reactions can be carried out by some forms of enzymes that are catalytically active.

It is easier to get a structure with the help of the NMR.

It's not necessary for theProteins to be crystallised, just purified and dissolved.

This is a serious problem.

Although they are not enzymes, some proteins have some activity that can be utilized in an assays.
Maybe theProtein will be used for a reaction.
Gel electrophoresis can be used to detect a missing enzyme activity by looking for the right spot on the gel.
The GFP is produced by jellyfish.
The intensity of the fluorescent dye could be used as the basis for a test.

If the gene is known, there are ways to modify the sequence to make it more efficient.

There is no need for a test.
In his-tagging, six tandem histidines are added to the sequence of the gene.
It is possible to eluted the protein with imidazole after it is purified on a nickel column.

The renaturation of a purified protein is necessary to restore the activity of the enzyme, but it could be difficult or impossible because of the SDS disruption.

You should conduct small-scale pilot tests to find out if the activity of the enzyme would be lost.
Pilot experiments should be conducted when salting out with ammonium sulfate.
In many instances, concentrations of ammonium sulfate can be chosen to make it easy and quick to purify.

The easiest and most rapid method for providing an estimate of molecular weight is the use of SDS-polyacrylamide gel electrophoresis, which takes only a few hours and has a high degree of resolution.

Small samples can be detected on the gel.
If you want to use markers on the gel, you should use two or more that have higher or lower molecular weights.

When it comes to the logarithm of their mass, the covalently attached and embedded proteins do not migrate.

The reason for the anomalies is not clear, and may be related to the large ring of the carbohydrates in the polyacrylamide gel.

The gel system may not be fully compatible with the high proportion of hydrophobic amino acid residues.

acetic anhydride and triethylamine will yield acetylated groups.

The groups would be smilng with smilng with smilng with smilng with smilng with smilng with smilng with smilng with smilng with smilng with smilng Leucine and isoleucine have the same weights, so they can't be distinguished by mass spectrometry.

The Edman procedure starts with the reaction of phenyl isothiocyanate with the terminal a-amino group of the peptide.

The group is not available in the ring.

The Edman procedure needs to be used before the cyclized residue can be removed.

The sequence of the peptides produced by the two methods are circular.

The mouse's sequence is Met-Ile-Cys-Lys-Glu-Tyr.

The PTH-Met derivative places Met at the N-terminal end, with Ile next in the sequence.

Trypsin treatment cleaves the carboxyl side of Lys, so that it is next to Cys.
Tyrosine is located on the C-terminal end, as shown by the observation.

The majority of blood proteins don't show up in the urine.

It is produced very soon after the egg is fertilized, and then in increasing amounts as the pregnancy progresses.
Variations of this method are the basis of all home pregnancy test kits.

There is a better understanding of the use of ELISA in home pregnancy tests if you watch the Animated Technique: Elisa Method for Detecting HCG at www.whfreeman.com/biochem5.

The reacting proteins are denatured before they are used in a Western blotting.

Similar specificities can be achieved in the test with the use of an SDSdenatured antigen.

Even though individual atoms can be delineated at a resolution of 1.5 A, the structure of individual side chains that are similar in shape and size cannot be clearly established.

The primary structure of the chain needs to be available.

The path of the backbone can be traced.

Those that are similar in size and shape can be distinguished by using the primary structure as a guide as they are fitted by eye to the electron-density map.

The precise orientation of atoms involved in binding and acting on substrates is what determines the chemical reactions carried out by proteins like hemoglobin and lysozyme.

The structure of the enzyme must be the same as that found in solution in order for catalysis to occur.

High resolution can be obtained with x-ray crystallography.

The structures obtained are always approximate, even though it is true that NMR is an easier method.

This is not a group of related structures.
The results are "fuzzy" like this.
x-ray crystallography would be better for determining the active site dimensions than NMR.
It is possible to get approximate structures of small proteins with the use of NMR.

The reagent of choice is dab syl chloride.

Easy answers are provided by the known cleavage specificities of chymotrypsin.

The hydrazides can be separated by the use of anion exchange resin.

A plus charge has been added to the cysteine side chain.

It is similar to lysine in both size and charge, and therefore, its carboxyl peptide bonds are susceptible to hydrolysis by trypsin.

The solution of myoglobin is 17,800 kd.

100,000 x 1 x 5.62 x 10-5).
The incident light is transmitted by 14.4%.

When we say the ratio is 6.96, we mean 6.96/1.

The 1/6.96 is given by Inverting this ratio.

The myoglobin mol is assumed.

Myoglobins havemol.

spherical molecules have smaller coefficients than rod-shaped ones.

The rod-shaped tropomyosin has a lower sedimentation coefficient than the spherical hemoglobin, even though it has a higher molecular weight.

Imagine a metal pellet and a nail of equal weight and density in a syrup.

The pellet will sink in a straight line, while the nail will twist and turn.

The two-thirds root of the mass is proportional to the sedimentation coefficients.

The mobilities are usually proportional to the log of the weight.

If the patterns are essentially the same, the disulfide pair is the same in both of them.

If you can make a fluorescent-labeled derivative of a degradation product and use it to detect cells that have a receptors for this peptide, you might be able to use this as a screening tool.

The isoelectric points are divided by the pK values for ionizable groups from Table 3.1.

If high-pressure liquid chromatography is used, either an ion-exchange or a molecular-exclusion approach should work.

An ion-exchange column should be used because all of the products are dipeptides.

There is a chance that theidase was removed during the purification step.

The activity will increase when the inhibitor is not in use.

The total activity is divided by the specific activity.

The initial specific activity is what determines the purification level.
The total activity is divided by the initial total activity to get the yield.
There are answers in the table.

There are two types of 15-kD subunits, one beginning with N-terminal Ala and the other with N-terminal Leu.

The small 15-kD subunits are linked by bonds of disulfide that are broken only by mercaptoethanol.
The 30-kD species has disulfide-linked subunits.
The 60-kD particle is formed by two 30-kD species and two 15-kD species.
The 60-kD particle is the native protein.

The final quaternary structure can be described as either (A-A)(B-B), or (A-B)2, where a hyphen indicates a disulfide bond.

One needs a method that separates the A-B and A-A dimers before the disulfide bonds are broken.

No method is certain to accomplish this, but methods based on nativegel electrophoresis or high-resolution ion-exchange chromatography will provide opportunities for a favorable outcome.
A good choice would be to use two-dimensional electrophoresis consisting of isoelectric focusing of 30-kD units, followed by mercaptoethanol/SDS-PAGE in the second direction.
Below is a diagram of the possible outcomes.

Light was used to make the peptides.

The a-amino group is protected by the Boc.
The release of the protecting group was caused by the illumination of selected regions of the solid support.
The ultimate products and their locations are defined by the pattern of masks used in these illuminations.

The molecule is called AVRYSR.

Trypsin and chymotrypsin cleave after R.

The full substance is: S-Y-G-K-L-S-I-F-T-M-S-W-S-L.

The C-terminal L is cleaved by the Carboxypeptidase.

S-Y-G-K-L-S-I-F-T-M* and S-W-S-L are cleaved after M.

The free a-amino group of alanine-amide reacts with phenylisothiocyanate to form a derivative.

The second step involves the use of anhydrous acid for a concerted cyclization of the phenylthiocarbamyl-peptide.
In step 3, the phenylthiazolinone is converted to the phenylthiohydantoin.

Chapters 2 and 3 introduced you to something.

The second class of macromolecules, the nucleic acids, are used to store genetic information.
The structures of the nucleoside building blocks of DNA are described first.
There is an overview of how the strands of DNA separate for replication and some of the various structures that nucleic acids can assume are described.
Next, the polymerases that form DNA chains are introduced.
The section that describes the molecule that stores genetic information ends with two examples of viruses in which the genetic material is not duplex DNA but rather single-strand RNA.
The way in which double-strand nucleic-acid intermediates are directed by specific base pairs is described by the authors.

Next, the formation of pro teins of a cell is discussed.

The authors give an explanation of the central roles ofRNAs in the overall flow of genetic information after describing the basic structures and kinds ofRNA.
The functions of messenger, transfer, and ribosomal RNA are presented along with a description of the polymerase that makes all cellular RNAs.
The genetic code relates the sequence of genes to the sequence of genes.
The relationship between the sequence of nucleotides in the DNA and the amino acids in the proteins is compared in pro and eukaryotes.
The process by which the intron sequence is removed from the initial transcript to form messenger RNA is described by the authors.

You should be able to complete the objectives once you have mastered this chapter.

Understand the various convention used to represent these components.

Predict the sequence of the amino acids using the genetic code.

The alterations to the RNA were made after it was formed.

The helix rotates by 36o with respect to adjacent base pairs and is separated from them by 3.4A.

Match the characteristics in the right column with the structures of double strand or single strand DNA.

A kilobase pair is 1000 base pairs.

The bacterium is labeled with 15N and the growth medium is 14N-labeled.

The DNA is isolated after two generations of growth.

Purified duplex DNA can be linear.

You are given two different solutions.

The activity requires a template.

The formation of the phosphodiester bond is dependent on the polarity of the synthesis of a DNA strand by DNA polymerase.

You are provided with a long, single-strand DNA molecule with a base composition of C, G, T, and A.

The appropriate characteristics for the virus in the left column can be found in the right column.

How a single-strand DNA virus could replicate by incorporating semiconserva tive replication into the process is the proposal.

Pick the nucleic acids that appear in a cell that has a retroviruses, and place them in the order in which genetic information flows during the process of forming a new progeny virus.

The flow of genetic information can be traced back to which of the following possible steps.

The reading frame should start with the 5' nucleotide.

The text has a genetic code on it.

When the b-chain of hemoglobin was compared, what was revealed was that the genes for the b-chain were different.

A process called exon shuffling involves the recombinement of DNA.

The answer is not correct because the sequence of one strand determines the sequence of the other.

The 5' end of the sequence is written to the left.

TGCATTGG is not the correct sequence because the two chains of the double helix are antiparallel.

The space between the strands is defined.

Two purines to hydrogen-bond is not enough for this distance.
Two pyrimidine bases would not be close enough to form stable hydrogen bonds.
The hydrogen-bond donor and acceptor groups are not aligned to form stable G T or A C base pairs.

The answer does not apply because the DNA does not contain uracil.

The formula is as follows: 3 x 106 kb x 103 bases/kb x 3.4 A/base x 10-8 cm/A.

The two strands of the double helix must be synchronized so that they can be used as a template for the synthesis of its complement.
The two strands are related to one another.
One strand directs the synthesis of its complement.
The process is accurate because of the specificity of base pairs and the fact that the replication can remove mismatches.

After two generations, you should 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 with a light 14N strand.

Answer d is correct because if there is at least one discontinuity 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 it requires closed circular molecules.
The molecule cannot be supercoiled because the ends of each strand are not constrained with respect to rotation.

The nature of the 5'-end is irrelevant since dNMP is added to its 3'-end.

A primer with a 3-OH is needed.
The answer is not correct because the enzyme uses dNTP and not NTP molecule, where N means A, C, G, T, or U.

The terminal nucleotide of the primer makes a nucleophilic attack on the innermost atom of the incoming dNTP that is appropriate for the template strand to form the phosphodiester bond.

When PPi is released, a dNMP residue is added onto the 3'-end of the primer, and the chain grows in the 5' 3' direction.

The template strand will have directed the polymerization of a complement in which C, G, T, and A are all equal.

The primer's contribution to the product strand can be neglected since it is short with respect to the template.

The du plex replicative form of the single-strand DNA is created by converting it to a du plex through the use of Watson-Crick base pairs.

A mechanism similar to that used for the semiconservative replication of the duplex chromosome of the host cell is used to reproduce the replicative form.
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.

The order in which genetic information flows during the infection of a cell with a retroviruses is: b, c, a, d, c, b.

Answer (b) is correct because the reverse transcription ofRNA into DNA occurs during the replication of retroviruses.

The term "transcription" is used to describe the process of making RNA from a DNA duplex.

G can also pair with U, but the association is weaker than that of the G C base pair.

This method provides definitive evidence of identity if you compare the genomes of the two to see if they are related.

It would be easier to use hybridization.
You would mix the samples, heat the mixture to melt the double-strand DNA and then slowly cool the solution to see if it contains double-strand DNA-RNA hybrid.
The hybrid would show that the genes are related.

There is only one strand of the DNA that can be used as a template.

The duplex DNA has a G + C composition.

The cell's genes are contained in theRNA of the cell.

Answer (f) is correct because it is necessary for the plate to function.

The promoter and terminator sequence are required to specify the precise start and stop points for the transcription.

The sequence will be written in the 3' di rection.

The a-phosphate of the ribonucleoside triphosphate is selected by base pair to the template strand of the duplex DNA.

The reaction is made by the use of the RNA polymerase.
A ribonucleoside monophosphate is added to the chain and it has grown in the 3' direction.
The strands are antiparallel, that is, they are assembled in the 3' 5' direction with respect to the polarity of the template strand of the DNA.

Answer (d) is incorrect because the interaction of the two genes takes place during translation.

The minimum that can serve as a codon is three contiguous nucleotides.

There are four different types of nucleotides.
Only 16 possible combinations can be made with a codon consisting of only two nucleotides.
It's not enough to specify all 20 of the amino acids.
A codon with three nucleotides allows 64 combinations, more than enough to specify the 20 amino acids.

There is a sequence of the polypeptide.

The fourth codon of the transcript is UAA, which is a translation codon, and the reading frame is set at the 5' end of the transcript.

The answer is correct because both AGU and AGC specify serine.

Answer (b) is correct because the codon on the transcript from AGU could be changed by the change of a single nucleotide in the DNA.
The anticodon ACU would base-pair with the codon AGU.

Mitochondria can use a genetic code that is different from the standard code because they have a distinct set of tRNAs that are matched to the genetic code used in their mRNAs.

The location and nature of the substitution of the amino acid for each of the identified genes could be deduced from the sequence of the wild-type and the mutant genes.

The result would be that the order of the genes on the map is the same as the order of the polypeptides produced by them.

There were two stretches of extra nucleotides between exons.

There are intron sequences in the DNA that don't have the same sequence in the mRNA.

The shuffling of exons allows them to interact in new ways while preserving the functional units.

exon shuffling takes place at the DNA level through breakage and rejoining of DNA notRNA.

There are 5243 base pairs in the genomes of mammals.

One can observe an increase in the absorbance of ultraviolet light when a solution containing intact DNA is heated.
A decrease in absorbance is observed when the solution is cooled slowly.
heating causes a similar effect if breaks are made in the sugar-phosphate backbones.
The reduction in absorbance is slower when the solution is cooled than it is when it is intact.

There are a number of factors that affect the behavior of a linear, double-strand DNA molecule.

Explain each of the observations considering this.

You have a double-strand linear DNA molecule, the appropriate primers, all the enzymes required for DNA replication, four 32P-labeled deoxyribonucleoside triphosphates, and the means to detect newly synthesized radioactive DNA.

The deoxyribonucleases do not hydrolyze base-paired DNA.

Formaldehyde forms hydroxymethyl derivatives.

If you have a solution with separated strands of DNA, what should it look like?

When double-strand DNA is placed in a solution containing tritiated water, hydrogens associated with the bases readily exchange with protons.

While many experiments were suggesting that the chromosomes are very long and continuous, it was found that the primer chain has to have a DNA template in order for it to function.

The mole percentage is G + C.

Early studies of the denaturation of double-strand DNA did not know if the strands separated from each other.

If you have two strands of double- strand DNA, one strand is labeled with 14N and the other is labeled with 15N.

The N-1 and N-7 of adenine, the N-7 of guanine, the N-3 of cytosine, and the O-4 of thymine are all protonsated under strongly acidic conditions.

Predict the effects of low pH on double-stranded DNA.

Some types of single- strandRNA can be used to form double- strand molecules.

There are many cells that can make deoxyuridine 5'-triphosphate.

The DNA of bacteriophage l is a linear double-strand molecule.

When two "cohesive" ends on the same molecule join, they can form a closed circular molecule.

When denatured and allowed to reassociate under certain conditions, single-strand circles can be formed.

Resolution is not enough to see the ends of the molecule.

A circle formed by adenoviruses.

There are volcanic sulfur springs with a pH 2 and tempera tures as high as 85oC.

Incubation with a circular DNA template at 100oC can extend a 20-nucleotide primer by more than 100 nucleotides.
The concentration of the enzyme-to-primer must be at least 3:1.

TdT can be used to extend a DNA primer by 5' 3' using deoxyri Bonucleoside triphosphates.

The primer must have a free 3'-OH end.
Theidase does not need a template or copy one.

The 2',3'-dideoxynucleoside can be used as a reagent.

To have a measurable effect on DNA synthesis, these analogs must be converted to dideoxynucleoside triphosphates.
A single dideoxyribonucleoside can effectively block subsequent chain extension when incorporated into a growing DNA chain.

In each chain-elongation reaction, a phosphodiester bond is formed and pyrophosphate is released.

Most cells have a potent pyrophosphorylase, which makes it easy tolysis of pyrophosphate.
One of the products of the chain-elongation reaction is partially responsible for the forward progress of the polymerization.
If the double-strand helix is allowed to form, isolated DNA polymerases can efficiently carry out chain extension.

In his studies of DNA in the late 1940s, Chargaff found that all organisms have the same number of bases.

Considering that the ability to form hydrogen bonds with adenine is the same as the ability to form single-strandedRNA from tobacco mosaic virus, do you expect similar constraints on base composition to be found in the following?

Some DNA endonucleases degrade double-strand DNA to yield mononucleotides and dinucleotides, but they don't degrade the duplex sequence to which other proteins are tightly bound.

The codon in the mRNA sequence for the enzyme is GUU, which is the same as the valine in the primary sequence.

If the codon to GCU is changed, the activity of the enzyme is unaffected, but if the codon to GAU is changed, the activity of the enzyme is completely inactivated.

Explain the observations.

To remove introns precisely, it's important to remove them between the first and second nucleotides of an exon.

The sequence at the normal junction in a pre-spliced mRNA between an intron and an exon is...

The viral genome is the most important factor in determining the amount of the proteins synthesized by a cell after it has been exposed to T2 bacteriophage.

Some of the genes overlap, that's one reason for the increased capacity.

The coding sequence for genes B and A is located in the same sequence.
The two proteins specified by these genes are completely different.

There is no nuclease capability in RNA polymerase.

Tell us why the two enzymes are different.

The genome of G4 is a small circle of DNA.

There is a small segment ofRNA.

An RNA molecule consisting of 350 nucleotides can be found in the transcript of the gene that codes for tyrosine tRNA.

A 41-base segment on the 5' side of the tRNA sequence and a 224-base segment on the 3' side of the tRNA sequence are removed by at least three ribonuclease enzymes.
The primary transcript's tRNA sequence is continuous, and no nucleotides are removed from that part of the transcript during processing.

This primary transcript has a sequence on one side of the rRNA sequence and another on the other side.

There is a transcript with all three rRNA sequences.

The codons UAA, UAG, and UGA are not read by the tRNA molecule.

These codons are found at the end of the coding sequence.
Premature termination of the protein chain can be caused by single-base mutations in certain codons.

A codon can change to UCG.

There is a change in the anticodon of the tRNA molecule that allows it to read a terminated codon.

Suppressive tRNAs suppress the effect of a chain-termination change.
The normal coding sequence has a UAG codon in it.
Each case will have a single base change.

In the laboratory, poly ribonucleotides were used to determine the genetic code.

Tell us how the cell uses this enzyme.

density gradient centrifugation can be used to separate the two strands of the same DNA.

The stabilbity of a particu lar a helix that is buried in the myoglobin is being studied.

Section 6.4 of the text describes site-specific mutagenesis, a technique used to replace leucine in the helix.

Explain your predictions.

There is a compound called cordycepin that can block the synthesis of RNA because it doesn't have the 3'-OH end needed for chain extension.

The structure of cordycepin is shown.

H O H (a) Cordycepin does not affect the growth ofbacteria, but it does affect the growth and division of mammals.

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 inbacteria.

Raney nickel can be used to convert cysteinyl-tRNACys to alanyl-tRNACys.

This experiment tells you about the ability of the machinery to recognize alanyl-tRNACys.

2'- and 3'-monophosphates are included in the products of the cleavage of RNA.

The flow of genetic information is subject to regulation.

The production of macromolecules limits the amount of energy the cell can use.
It is possible to achieve the greatest economy in energy expenditure by a mature cell by regulating the steps in storage and transmission of genetic information.

David Baltimore was looking into the activity of the polymerase in the leukemia virus.

The virus causes leukemia in mice.
He created a mixture with either the four dNTPs or the four NTPs in a buffered solution after disrupting the virus particles.
Radiolabeled was one of the dNTPs.
After allowing time for a reaction to occur, the mixture was treated with strong acid to make nucleic acids and leave unreacted triphosphates in solution.
He was able to detect the formation of the product of a polymerase by measuring the precipitated radioactivity.
CHAPTER 5 product was not incorporated into it, dNTPs were incorporated into it, and the isolated radiolabeled product was destroyed by RNase.

Radioisotopes can be used to identify specific molecule involved in biochemi cal processes.

The medium is 4 or 35SO4.

The Bacteriophage T2 has a genome.
After infecting the cells with the two different virus preparations, they put the culture of the cells in a blender to remove any leftover virus.
They collected the stripped cells and compared the amount of radioisotope in them to the amount left in the supernatant.

An increase in the absorbance of light is observed when the intact double-strand circular DNA molecule is heated in solution.

The circles are so intertwined that they remain closely associated.

When the molecule is cooled, the interlocked strands move relative to each other until their bases are aligned.
The molecule absorbs less light than the pair of strands.
During denaturation, breaks in one or both strands of a double-strand DNA molecule allow the two strands to separate from one another.
In order to form a doublestrand molecule, the separate strands collide randomly until at least a small number of correct base pairs is formed.
A difference in the reduction of absorbance will be observed if a pair of interlocked circles are compared to a pair of separate strands in solution.

The longer the DNA molecule, the more thermal energy is required to disrupt it.

Experiments show that there is a relationship between the length of a molecule and the number of base pairs.

It is easier to separate the two strands when the concentration of NaCl decreases.

Once a short stretch of base pairs is formed, reassociation to form the longer double-strand molecule occurs quickly.

The higher the concentration of DNA, the quicker the solution will find and pair with each other.

The tendency of bases to stack contributes to the stability of the helix.

Base stacking allows the sugar-phosphate chain to be located on the outside of the helix, where it can besolvated, and it also reduces the contact of the relatively insoluble bases with water.
The helix may be destabilized by the presence of urea, which may allow bases to associate more with water.

The diameter of the helix is not likely to be a factor in determining a specific sequence.

The grooves of the intact helix contain hydrogen-bond donors and acceptors.

A hydrogen bond can be formed between a group of atoms in one of the grooves of the helix.
The edges of the bases can contribute to the specificity of the interaction between the side chains.

Although the system described could yield 32P-labeled daughter DNA molecule, CHEMICAL methods can't distinguish between strands of the same molecule in which one strand is labeled and the other is unlabeled.

In their experiments, Meselson and Stahl used a physical technique to separate the labeled molecules according to their content of 14N and 15N, which differ in their specific densities.

In solution, the oligodeoxyribonucleotide forms an interchain double-strand molecule with flush ends and a small single-strand loop.

A small double-strand linear molecule remnant containing seven base pairs is left after the deoxyribonuclease hydrolyzes the phophodiester bonds.

Formaldehyde could react with the exocyclic groups on the carbon of adenine, the C-2 of guanine, and the C-4 of cytosine to form hydroxymethyl derivatives.

Because these derivatives can't form hydrogen bonds with bases, single strands would reassociate to a lesser extent.
The ring nitrogen atoms in pyrimidines may be present at the actual sites of the reaction of formaldehyde with DNA.

The experiment suggests that the hydrogen bonds of base-paired regions of double strand DNA may become bubbles.

The tritiated water allows the exchange of protons.

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 create a relatively simple mechanism for chromosomal replication, one could postulate that each of the breaks offered the 3'-OH group required for the initiation of the new DNA strand.
The strand opposite the break would be the location of the template required for replication.
It has been established that the DNA in the chromosomes is long and continuous.
The fact that there are no breaks in the molecule makes the mechanism of replication complexample.
See page 760 for details.

Most organisms live at temperatures that are considerably lower than 65oC.

The integrity of the double-strand DNA molecule is important for the transmission and expression of genetic information.

Determine the temperature at which the hydrogen bonds are disrupted and single strands are formed.

You can measure the extent of hyperchromicity by heating the double-strand DNA to different temperatures.
If you want to separate the 14N-labeled strands from the 15N-labeled strands, you have to use the density-gradient equilibrium sedimentation technique.
It would suggest that the strands separate during thermal denaturation if you are successful.

Normal hydrogen bonding is impossible because the atoms can no longer serve as hydrogen-bond acceptors.

Double-strand DNA is less stable at neutral pH values than it is at low pH ones.
Deprotonation of other ring atoms denatured DNA at high pH values.

A hybrid mole cule can be formed by the association of a molecule ofRNA with a molecule of DNA.

The formation of hydrogen bonds between bases will allow the creation of a double helix.

The structure and hydrogen-bonding properties of uracil are very similar to those of thymine and can be used as a base for DNA poly merase.

uracil and adenine pair with each other when incorporated into a double-strand DNA polymer.
The reasons uracil is not normally incorporated into DNA can be discussed on page 771 of the text.

Interchain joining to form multimers is more likely at higher concentrations.

The most reasonable model for the structure of the l phage DNA molecule is a dou ble-strand molecule with single-strand protrusions at the 5'-ends, as illustrated in the margin.

The 3'-ends have groups that allow them to be used as a primer for DNA synthesis.
The molecule with flush ends is created by the filling of the single-strand regions of the molecule.
The required circular molecule can no longer be formed with cohesive ends.

Molecules treated with exonuclease III may not be infective because they have a longer single-strand sequence at both ends, which means that the ends could no longer be joined.

When the exonuclease-treated DNA is treated with DNA polymerase I, the single-strand regions are filled in to reform a molecule that has protruding single strands that are approximately the same length as those in the native molecule.
The molecule becomes infective again.
I will once again produce a molecule with flush ends that is no longer infective after further treatment of the molecule.

There are a pair of inverted repeats on each strand.

Each strand would need a primer.

The association between primer and the enzyme may protect it from thermal denaturation.
There are many examples where the binding of the Substrate is stable.

You might expect a higher G + C composition from a thermophile.

It's not unusual for the base ratios of the DNA to be the same as those found in the hot springs where they grow.
The thermophiles must be protected from thermal denaturation with the help of the bacterial cells.

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 on the relative concentrations of the deoxynucleoside triphosphate substrates.

The base composition of the chains will be similar to the template strand.
TdT does not need an exonuclease activity that removes bases from newly synthesized strands.

TdT can be used to introduce sequence variation.

The 2',3'-dideoxy analogs must be converted to nucleoside triphosphates in order to be used as a base for DNA polymerase.

In studies on the inhibition of DNA synthesis in living cells, cells with the neutral forms of the analogs are used instead of the negatively charged triphosphate forms.

Once inside the cell, nucleoside analogs are phosphorylated by cellular enzymes that normally function to "salvage" nucleosides.

The error-correcting exonuclease activity of some DNA polymerases is blocked by the lack of a 3'-OH group.

The reaction is driven by noncovalent forces.

There are hydrogen bonds between A and T bases in the antiparallel chains.
Stacking interactions between adjacent bases on the same strand may contribute to helix formation and stability.
The noncovalent forces are likely to account for the forward progress of the chain.

The base ratios will not conform to Chargaff's established rules because a single-strand polynucleotide might form hydrogen bonds between bases.

This is also true for single-stranded DNA.

The number of uracil and adenines would be the same as those of cytosine, and the number of guanine bases would be the same as well.

Adding the DNA endonuclease will degrade the DNA that is not protected.

The size of the protected fragments of DNA and the base sequence can be determined using methods discussed in Chapter 6 of the text.

When this process is performed, they will allow the polymerase molecule to move from the promoter site to the template where transcription begins, as well as beyond.

The promoter site was no longer protected from endonuclease degradation.

Alterations in the structure of the enzyme do not affect the activity of the enzyme.

The substitution of aspartate for valine is specified in the GAU codon.
The side-chain carboxyl group of aspartate has a negative charge at neutral pH.
The charged group could disrupt the native structure of the enzyme.

The exon coding sequence will be altered because of the loss of two bases if the spliceosome cleaves the initial transcript.

Instead of beginning with the codon GCU in the normal exon, the reading frame will begin with the codon UAA.
This codon is a signal for the end of the translation of the polypeptide specified by the messenger RNA.

The synthesis of viral-directed proteins can't begin until T2 messenger RNA is made.

The ribonucleoside triphosphates synthesized by the bacterial cell must be used for the Transcription of the T2 DNA.

If each of the coding sequence is read in a different frame, the weight of the molecule can be 201,000.

The location of the AUG initiation signal is critical to the establishment of the proper reading frame.

Both reading frames 1 and 2 are established when the first AUG codon is binding to the second AUG codon.

The two different genes specified by the two different genes will have different sequence of the same genes.

The genetic information that is passed on to progeny cells comes from the duplication of the DNA of the chromosomes.

Errors that occur in the copying of a DNA template will be transmitted to the duplicated chromosome and all the messenger RNA molecule that was transcribed from it.
Errors that occur due to the inclusion of mismatched nucleotides can be corrected by DNA polymerase.
Many copies of mRNA are made, but they have relatively brief lives in the cell, and very few are passed on to offspring cells.
It appears that the cell can tolerate errors in transcription if not many occur.

A primer nucleotide and a template are required for the synthesis of a DNA chain.

A primer can be served by either an oligodeoxyribonucleotide or an oligoribonucleotide.
When replication has extended around the circle and the 5' end of the primer is reached, the primer is hydrolyzed and the small gap is filled in with a DNA sequence.

In some organisms, the pre-tRNA transcript is removed in order to make tRNA.

Only one promoter is required for rRNA synthesis.

The removal of a few amino acids from the C-terminal end of many, but not all, does not affect their normal function.

The premature end of the chain would be suppressed by each tRNA molecule's placement of an amino acid at the corresponding site.

Along with their codons, the amino acids that could be found at the site are GAG, Trp, Leu, AAG, Ser, and Tyr.
When a cell contains a suppressor tRNA, it may not be possible to end it with a single stop codon.
Most cells are tolerant of suppression, even if the extension could be lethal.
One explanation is that the stop codon may be seen by other proteins that are involved in chain termination.
A full understanding of the reasons for toleration of suppression is needed.

The equilibrium for the reaction is in the direction of the degradation of the RNA.

It is likely that the concentrations of ribonucleoside diphosphates in the cell are not enough to drive net polynucleotide synthesis.

The polyribonucleotides it synthesises contain random sequences, which makes them useless for protein synthesis.

The nucleases that regulate the lifetimes of the genes are used by the cell.

The lifetimes of mRNAs inbacteria are short.

Only one of the two strands at a given location along the DNA was being transcribed into DNA if less than all the DNA could form a hybrid.

In rare cases, theRNA is made from both strands of the template DNA.

The results show that only one of the two strands of the SP8 virus is transcribed.

In most other organisms, different regions of each strand are used for different things, but in the case of the SP8 virus, only one strand is used.

The conversion of cordycepin to the triphosphate form is carried out by a number of kinases.

It's likely thatbacteria can't phosphorylate cordycepin efficiently, which makes them less susceptible to inhibition of RNA and DNA synthesis.

You will learn later that cordycepin might affect the synthesis of DNA.

The ribosomal complex can't recognize alanyl tRNACys as an inappropriate form of tRNA.

The recognition of the anticodon in the messenger RNA molecule and the codon in the transferRNA molecule will lead to the creation of a growing polypeptide chain.
Accurate translation does not depend on the recognition of the attached amino acid.

The base-catalyzed generation of 2'- and 3'-monophosphates claims that a 2',3' phosphodiester is formed during the cleavage of RNA.

It is possible that a hydroxyl anion leaves a 2'-O- that is 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- A 2'- or a 3'-phosphate ester can be formed from this unstable derivative.
The DNA is not susceptible to alkali degradation because it has no 2'-OH groups.
It may have been chosen as the primary carrier of genetic information because it is more stable.

The control of the activity of RNA polymerase is the step that would afford the maximum economy.

The first step in the expression of genetic information is transcription by the RNA polymerase.
It follows that controlling messenger RNA production, by stimulating or prohibiting the activity of RNA polymerase, allows the cell to make specific types of messenger RNA and to make only the ones that are required.
Even if translation were stringently regulated, the energy required to produce the polynucleotides would be wasted if the cell could not regulate the activity of the polymerase.

The inclusion of nucleoside triphosphates into an acid-insoluble form is dicative of the presence of a polymerase.

The product was formed using dNTPs and not NTPs.
There was evidence that the radiolabeled product was destroyed by a nuclease specific to hydrolyzing DNA and not by one specific for RNA hydrolysis.

NaOH did not destroy the radiolabeled product.

The extract was pre-treatment with the two hydrolytic enzymes and it was shown that the enzyme depends on an RNA and not a DNA template for its activity.

This is anRNA- dependent DNA polymerase.

A revision of Francis Crick's central dogma ofmolecular biology was caused by the discovery of a new enzyme in a tumor virus and by the fact that information flowed from the genes to the proteins.
In some cases information could flow fromRNA to DNA if the demonstration is correct.

Hershey and Chase observed that most of the 32P was associated with the cells and most of the 35S was in the supernatant.

They concluded that DNA had entered the cell because the nucleic acids are rich in phosphorus and sulfur.
35S is a good marker for proteins because the sulfur of sulfate is incorporated into the cysteine and methionine.
The experiment showed that the cells didn't have anyprotein in them.
The experiments helped solidify the view that the genetic material was DNA.

Left to right means 5' 3'.

The order of the bases must be reversed if one wishes to write a sequence without labeling the ends.

The distance between the base pairs is 3.4 A and the answer to this question is 2mm.

The answer is 5.88 x 103 base pairs if you divide the length of a DNA segment by the distance between the base pairs.

One-half of the molecule would be 15N-15N and the other half 14N-14N after 1.0 generation.

One-quarter of the molecule would be 15N-15N, the other three-quarters 14N-14N.
The 14N-15N molecule would not be observed.

It would be useful if they entered the cell.

The negative charge prevents dTTP from entering most cells.

One should use dATP, dGTP, dTTP, and dCTP labeled with 32P since the a-phosphorous atom is incorporated into DNA.

Only (c) synthesis would lead to DNA because (a) and (b) have no primer or open end to build on and (d) has no template beyond a free 3'-OH.

Single stranded linear DNA can be used as a template for DNA synthesis because it can prime synthesis through the 3' end.

T base pairs with A, so a short polythymidylate chain would serve as a primer.

The b- and g-phosphates are radioactive dTTP labeled in any position.

The infectious nucleic acid should be treated with either highly purified ribonuclease or deoxyribonuclease.

If it is DNA, it will be destroyed by the RNAse.

Half of the daughter DNA duplexes are normal and the other half have a CG pair.

64 different trinucleotides can be made from these dinucleotides.

It's important to note that 64 is 43.
There will be 44 of them.
Proceeding in this way will get us to 48 different octonucleotides.

The flow of genetic information has two bases, A and C, and a second base, G and T. Two bits are needed to specify a base pair.

A lot of 8-mer sequence could be stored on a diskette.

A lot of diskettes would be needed to record the sequence.

Each parent strand acts as a template for the creation of a new strand.

The template is semiconserved since each daughter molecule gets one strand from the parent.
The double helix of the DNA remains intact after the synthesis of RNA.

The template is said to be stable.

The conclusion is that a repeating tetrapeptide (Leu-Leu-Thr-Tyr) unit will be formed after a comparison of this 12-base sequence with the genetic code.

The 2'-OH group is responsible for the instability of RNA.

In the presence of OH- the 2'- OH group ofRNA is converted to an alkoxide ion.
The 2'-alkoxide attack leaves a 2',3'-cyclic nucleotide in the phosphodiester bond.
A mixture of 2' and 3'-nucleotides is produced by OH-.

Since it lacks a 2'-OH group, it is stable in alkali.

Apparently cordycepin is converted to 5'-triphosphate and incorporated into the growing RNA chain.

This chain contains cordycepin, but it lacks a 3'-OH group.

Only single-stranded mRNAs can be used as templates.

Since poly(G) is a triple-stranded helix, it cannot be used as a template.

The complimentary strands are missing one of the four bases; d(TAC), d(GTA), and d(GTA) are all missing C. The complement of d(TAC), poly(GUA) was formed when GTP was used.

In a nonoverlapping code, each individual nucleotide change would be the same as the last one.

In an overlap code, there will be some changes to the identity of two consecutive amino acids.
There will be several experiments needed.
One could reasonably conclude that the code is non-overlapping if one makes a series of individual nucleotide changes, determines the resulting sequence, and never finds that two consecutive amino acids are changed.

One can note that the code described in the problem has a four-nucleotide stretch.

The maximum number of different dipeptides that could be used in this scenario is only 44.
The 20 amino acids can be used to make 400 different dipeptides, all of which are represented in the same sequence.
This numerical analysis of naturally occurring dipeptide sequences would argue against a triplet code.

The synthesis must start from different reading frames.

One of these will have a terminal lysine since UGA is a stop signal.
The reading frame in phase with AUG will have a Met-Arg sequence in it, and the reading frame in phase with AAU will have an AsnGlu sequence in it.