Chapter 4. DNA, RNA, and the Flow of Genetic Information

Question 1

Question 4.1

A t instead of an s? Differentiate between a nucleoside and a nucleotide.

Answer 1

1. A nucleoside is a base attached to a ribose or deoxyribose sugar. A nucleotide is a nucleoside with one or more phosphoryl groups attached to the ribose or deoxyribose.

Question 2

Question 4.2

A lovely pair. What is a Watson–Crick base pair?

Answer 2

2. Hydrogen-bond pairing between the base A and the base T as well as hydrogen-bond pairing between the base G and the base C in DNA.

Question 3

Question 4.3

Chargaff rules! Biochemist Erwin Chargaff was the first to note that, in DNA, [A] = [T] and [G] = [C], equalities now called Chargaff’s rule. Using this rule, determine the percentages of all the bases in DNA that is 20% thymine.

Answer 3

3. T is always equal to A, and so these two nucleotides constitute 40% of the bases. G is always equal to C, and so the remaining 60% must be 30% G and 30% C.

Question 4

Question 4.4

But not always. A single strand of RNA is 20% U. What can you predict about the percentages of the remaining bases?

Answer 4

4. Nothing, because the base-pair rules do not apply to single-stranded nucleic acids.

Question 5

Question 4.6

Compositional constraint. The composition (in mole-fraction units) of one of the strands of a double-helical DNA molecule is [A] = 0.30 and [G] = 0.24. (a) What can you say about [T] and [C] for the same strand? (b) What can you say about [A], [G], [T], and [C] of the complementary strand?

Answer 5

6. (a) [T] + [C] = 0.46. (b) [T] = 0.30, [C] = 0.24, and [A] + [G] = 0.46.

Question 6

Question 4.7

Size matters. Why are GC and AT the only base pairs permissible in the double helix?

Answer 6

7. Stable hydrogen bonding occurs only between GC and AT pairs. Moreover, two purines are too large to fit inside the double helix, and two pyrimidines are too small to form base pairs with each other.

Question 7

Question 4.8

Strong, but not strong enough. Why does heat denature, or melt, DNA in solution?

Answer 7

8. The thermal energy causes the chains to wiggle about, which disrupts the hydrogen bonds between base pairs and the stacking forces between bases and thereby causes the strands to separate.

Question 8

Question 4.10

Coming and going. What does it mean to say that the DNA strands in a double helix have opposite directionality or polarity?

Answer 8

10. One end of a nucleic acid polymer ends with a free 5′-hydroxyl group (or a phosphoryl group esterified to the hydroxyl group), and the other end has a free 3′-hydroxyl group. Thus, the ends are different. Two strands of DNA can form a double helix only if the strands are running in different directions—that is, have opposite polarity.

Question 9

Question 4.11

All for one. If the forces—hydrogen bonds and stacking forces—holding a helix together are weak, why is it difficult to disrupt a double helix?

Answer 9

11. Although the individual bonds are weak, the population of thousands to millions of such bonds provides much stability. There is strength in numbers.

Question 10

Question 4.12

Overcharged. DNA in the form of a double helix must be associated with cations, usually Mg2+. Why is this requirement the case?

Answer 10

12. There would be too much charge repulsion from the negative charges on the phosphoryl groups. These charges must be countered by the addition of cations.

Question 11

Question 4.13

Not quite from A to Z. Describe the three forms that a double helix can assume.

Answer 11

13. The three forms are the A-DNA, the B-DNA and the Z-DNA, with B-DNA being the most common. There are many differences (Table 4.2). Some key differences are: A-DNA and B-DNA are right-handed, whereas Z-DNA is left-handed. A-DNA forms in less-hydrated conditions than does B-DNA. The A form is shorter and wider than the B form.

Question 12

Question 4.16

Guide and starting point. Define template and primer as they relate to DNA synthesis.

Answer 12

16. A template is the sequence of DNA or RNA that directs the synthesis of a complementary sequence. A primer is the initial segment of a polymer that is to be extended during elongation.

Question 13

Question 4.17

An unseen pattern. What result would Meselson and Stahl have obtained if the replication of DNA were conservative (i.e., the parental double helix stayed together)? Give the expected distribution of DNA molecules after 1.0 and 2.0 generations for conservative replication.

Answer 13

17. In conservative replication, after 1.0 generation, 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 in conservative replication.

Question 14

Question 4.18

Which way? Explain, on the basis of nucleotide structure, why DNA synthesis proceeds in the 5′-to-3′ direction.

Answer 14

18. The nucleotides used for DNA synthesis have the triphosphate attached to the 5′-hydroxyl group with free 3′-hydroxyl groups. Such nucleotides can be utilized only for 5′-to-3′ DNA synthesis.

Question 15

Question 4.19

Tagging DNA.

Suppose that you want to radioactively label DNA but not RNA in dividing and growing bacterial cells. Which radioactive molecule would you add to the culture medium?

Suppose that you want to prepare DNA in which the backbone phosphorus atoms are uniformly labeled with 32P. Which precursors should be added to a solution containing DNA polymerase and primed template DNA? Specify the position of radioactive atoms in these precursors.

Answer 15

19. (a) Tritiated thymine or tritiated thymidine. (b) dATP, dGTP, dCTP, and TTP labeled with 32P in the innermost (α) phosphorus atom.

Question 16

Question 4.20

Finding a template. A solution contains DNA polymerase and the Mg2+ salts of dATP, dGTP, dCTP, and TTP. The following DNA molecules are added to aliquots of this solution. Which of them would lead to DNA synthesis?

A single-stranded closed circle containing 1000 nucleotide units.

A double-stranded closed circle containing 1000 nucleotide pairs.

A single-stranded closed circle of 1000 nucleotides base-paired to a linear strand of 500 nucleotides with a free 3′-OH terminus.

A double-stranded linear molecule of 1000 nucleotide pairs with a free 3′-OH group at each end.

Answer 16

20. Molecules in parts a and b would not lead to DNA synthesis, because they lack a 3′-OH group (a primer). The molecule in part d has a free 3′-OH group at one end of each strand but no template strand beyond. Only the molecule in part c would lead to DNA synthesis.

Question 17

Question 4.21

Retrograde. What is a retrovirus and how does information flow for a retrovirus differ from that for the infected cell?

Answer 17

21. A retrovirus is a virus that has RNA as its genetic material. However, for the information to be expressed, it must first be converted into DNA, a reaction catalyzed by the enzyme reverse transcriptase. Thus, at least initially, information flow is opposite that of a normal cell: RNA → DNA rather than DNA → RNA.

Question 18

Question 4.22

The right start. Suppose that you want to assay reverse transcriptase activity. If polyriboadenylate is the template in the assay, what should you use as the primer? Which radioactive nucleotide should you use to follow chain elongation?

Answer 18

22. A thymidylate oligonucleotide should be used as the primer. The poly(A) template specifies the incorporation of T; hence, radioactive thymidine triphosphate (labeled in the α phosphoryl group) should be used in the assay.

Question 19

Question 4.23

Essential degradation. Reverse transcriptase has ribonuclease activity as well as polymerase activity. What is the role of its ribonuclease activity?

Answer 19

23. The ribonuclease serves to degrade the RNA strand, a necessary step in forming duplex DNA from the RNA–DNA hybrid.

Question 20

Question 4.24

Virus hunting. You have purified a virus that infects turnip leaves. Treatment of a sample with phenol removes viral proteins. Application of the residual material to scraped leaves results in the formation of progeny virus particles. You infer that the infectious substance is a nucleic acid. Propose a simple and highly sensitive means of determining whether the infectious nucleic acid is DNA or RNA.

Answer 20

24. Treat one aliquot of the sample with ribonuclease and another with deoxyribonuclease. Test these nuclease-treated samples for infectivity.

Question 21

Question 4.25

Mutagenic consequences. Spontaneous deamination of cytosine bases in DNA takes place at low but measurable frequency. Cytosine is converted into uracil by loss of its amino group. After this conversion, which base pair occupies this position in each of the daughter strands resulting from one round of replication? Two rounds of replication?

Answer 21

25. Deamination changes the original G · C base pair into a G · U pair. After one round of replication, one daughter duplex will contain a G · C pair and the other duplex will contain an A · U pair. After two rounds of replication, there will be two G · C pairs, one A · U pair, and one A · T pair.

Question 22

Question 4.27

Key polymerases. Compare DNA polymerase and RNA polymerase from E. coli in regard to each of the following features: (a) activated precursors, (b) direction of chain elongation, (c) conservation of the template, and (d) need for a primer.

Answer 22

27.

(a) Deoxyribonucleoside triphosphates versus ribonucleoside triphosphates.
(b) 5′ → 3′ for both.
(c) Semiconserved for DNA polymerase I; conserved for RNA polymerase.
(d) DNA polymerase I needs a primer, whereas RNA polymerase does not.

Question 23

Question 4.28

Different strands. Explain the difference between the coding strand and the template strand in DNA.

Answer 23

28. The template strand has a sequence complementary to that of the RNA transcript. The coding strand has the same sequence as that of the RNA transcript except for thymine (T) in place of uracil (U).

Question 24

Question 4.29

Family resemblance. Differentiate among mRNA, rRNA and tRNA.

Answer 24

29. Messenger RNA encodes the information that, on translation, yields a protein. Ribosomal RNA is the catalytic component of ribosomes, the molecular complexes that synthesize proteins. Transfer RNA is an adaptor molecule, capable of binding a specific amino acid and recognizing a corresponding codon. Transfer RNAs with attached amino acids are substrates for the ribosome.

Question 25

Question 4.30

A code you can live by. What are the key characteristics of the genetic code?

Answer 25

30. Three nucleotides encode an amino acid; the code is nonoverlapping; the code has no punctuation; the code exhibits directionality; the code is degenerate.

Question 26

Question 4.31

Encoded sequences.

Write the sequence of the mRNA molecule synthesized from a DNA template strand having the following sequence:

5′–ATCGTACCGTTA–3′

What amino acid sequence is encoded by the following base sequence of an mRNA molecule? Assume that the reading frame starts at the 5′ end.

5′–UUGCCUAGUGAUUGGAUG–3′

What is the sequence of the polypeptide formed on addition of poly(UUAC) to a cell-free protein-synthesizing system?

Answer 26

31. (a) 5′-UAACGGUACGAU-3′
    (b) Leu-Pro-Ser-Asp-Trp-Met
    (c) Poly(Leu-Leu-Thr-Tyr)

Question 27

Question 4.32

A tougher chain. RNA is readily hydrolyzed by alkali, whereas DNA is not. Why?

Answer 27

32. The 2′-OH group in RNA acts as an intramolecular nucleophile. In the alkaline hydrolysis of RNA, it forms a 2′-3′ cyclic intermediate.

Question 28

Question 4.33

A picture is worth a thousand words. Write a reaction sequence showing why RNA is more susceptible to nucleophilic attack than DNA.

Answer 28

Question 29

Question 4.34

Flowing information. What is meant by the phrase gene expression?

Answer 29

34. Gene expression is the process of expressing the information of a gene in its functional molecular form. For many genes, the functional information is a protein molecule. Thus, gene expression includes transcription and translation.

Question 30

Question 4.35

We can all agree on that. What is a consensus sequence?

Answer 30

35. A nucleotide sequence whose bases represent the most-common, but not necessarily the only, members of the sequence. A consensus sequence can be thought of as the average of many similar sequences.

Question 31

Question 4.36

A potent blocker. Cordycepin (3′-deoxyadenosine) is an adenosine analog. When converted into cordycepin 5′-triphosphate, it inhibits RNA synthesis. How does cordycepin 5′-triphosphate block the synthesis of RNA?

Answer 31

36. Cordycepin terminates RNA synthesis. An RNA chain containing cordycepin lacks a 3′-OH group.

Question 32

Question 4.37

Silent RNA. The code word GGG cannot be deciphered in the same way as can UUU, CCC, and AAA, because poly(G) does not act as a template. Poly(G) forms a triple-stranded helical structure. Why is it an ineffective template?

Answer 32

37. Only single-stranded RNA can serve as a template for protein synthesis.

Question 33

Question 4.38

Sometimes it is not so bad. What is meant by the degeneracy of the genetic code?

Answer 33

38. Degeneracy of the code refers to the fact that most amino acids are encoded by more than one codon.

Question 34

Question 4.39

In fact, it can be good. What is the biological benefit of a degenerate genetic code?

Answer 34

39. If only 20 of the 64 possible codons encoded amino acids, then a mutation that changed a codon would likely result in a nonsense codon, leading to termination of protein synthesis. With degeneracy, a nucleotide change might yield a synonym or a codon for an amino acid with similar chemical properties.

Question 35

Question 4.42

Two from one. Synthetic RNA molecules of defined sequence were instrumental in deciphering the genetic code. Their synthesis first required the synthesis of DNA molecules to serve as templates. Har Gobind Khorana synthesized, by organic-chemical methods, two complementary deoxyribonucleotides, each with nine residues: d(TAC)3and d(GTA)3. Partly overlapping duplexes that formed on mixing these oligonucleotides then served as templates for the synthesis by DNA polymerase of long, repeating double-helical DNA strands. The next step was to obtain long polyribonucleotide chains with a sequence complementary to only one of the two DNA strands. How did Khorana obtain only poly(UAC)? Only poly(GUA)?

Answer 35

42. Incubation with RNA polymerase and only UTP, ATP, and CTP led to the synthesis of only poly(UAC). Only poly(GUA) was formed when GTP was used in place of CTP.

Question 36

Question 4.44

A new translation. A transfer RNA with a UGU anticodon is enzymatically conjugated to 14C-labeled cysteine. The cysteine unit is then chemically modified to alanine (with the use of Raney nickel, which removes the sulfur atom of cysteine). The altered aminoacyl-tRNA is added to a protein-synthesizing system containing normal components except for this tRNA. The mRNA added to this mixture contains the following sequence:

5′–UUUUGCCAUGUUUGUGCU–3′

What is the sequence of the corresponding radiolabeled peptide?

Answer 36

44. Phe-Cys-His-Val-Ala-Ala

Question 37

Question 4.45

A tricky exchange. Define exon shuffling and explain why its occurrence might be an evolutionary advantage.

Answer 37

45. Exon shuffling is a molecular process that can lead to the generation of new proteins by the rearrangement of exons within genes. Because many exons encode functional protein domains, exon shuffling is a rapid and efficient means of generating new genes.

Question 38

Question 4.46

From one, many. Explain why alternative splicing increases the coding capacity of the genome.

Answer 38

46. Alternate splicing allows one gene to code for several different but related proteins.

Question 39

Question 4.47

The unity of life. What is the significance of the fact that human mRNA can be accurately translated in E. coli?

Answer 39

47. It shows that the genetic code and the biochemical means of interpreting the code are common to even very distantly related life forms. It also testifies to the unity of life: that all life arose from a common ancestor.

Question 40

Question 4.48

Back to the bench. A protein chemist told a molecular geneticist that he had found a new mutant hemoglobin in which aspartate replaced lysine. The molecular geneticist expressed surprise and sent his friend scurrying back to the laboratory. (a) Why did the molecular geneticist doubt the reported amino acid substitution? (b) Which amino acid substitutions would have been more palatable to the molecular geneticist?

Answer 40

48. (a) A codon for lysine cannot be changed to one for aspartate by the mutation of a single nucleotide. (b) Arg, Asn, Gln, Glu, Ile, Met, or Thr.

Question 41

Question 4.49

Eons apart. The amino acid sequences of a yeast protein and a human protein having the same function are found to be 60% identical. However, the corresponding DNA sequences are only 45% identical. Account for this differing degree of identity.

Answer 41

49. The genetic code is degenerate. Of the 20 amino acids, 18 are specified by more than one codon. Hence, many nucleotide changes (especially in the third base of a codon) do not alter the nature of the encoded amino acid. Mutations leading to an altered amino acid are usually more deleterious than those that do not and hence are subject to more stringent selection.

Question 42

Question 4.50

3 is greater than 2. The adjoining illustration graphs the relation between the percentage of GC base pairs in DNA and the melting temperature. Suggest a plausible explanation for these results.

Answer 42

50. GC base pairs have three hydrogen bonds compared with two for AT base pairs. Thus, the higher content of GC means more hydrogen bonds and greater helix stability.

Question 43

Question 4.51

Blast from the past. The illustration below is a graph called a C0t curve (pronounced “cot”). The y-axis shows the percentage of DNA that is double stranded. The x-axis is the product of the concentration of DNA and the time required for the double-stranded molecules to form. Explain why the mixture of poly(A) and poly(U) and the three DNAs shown vary in the C0t value required to completely anneal. MS2 and T4 are bacterial viruses (bacteriophages) with genome sizes of 3569 and 168,903 bp, respectively. The E. coli genome is 4.6 × 106 bp.

Answer 43

51. C0t value essentially corresponds to the complexity of the DNA sequence—in other words, how long it will take for a sequence of DNA to find its complementary strand to form a double helix. The more complex the DNA, the slower it reassociates to make the double-stranded form.

Question 44

Question 4.52

Salt to taste. The graph below shows the effect of salt concentration on melting temperature of bacterial DNA. How does salt concentration affect the melting temperature of DNA? Account for this effect.

Answer 44

52. Increasing salt increases the melting temperature. Because the DNA backbone is negatively charged, there is a tendency for charge repulsion to destabilize the helix and cause it to melt. The addition of salt neutralizes the charge repulsion, thereby stabilizing the double helix. The results show that within the parameters of the experiment, more salt results in more stabilization, which is reflected as a higher melting temperature.