Genetics Chapter 12

40. Why did the Meselson and Stahl experiment require two rounds of replication?

The first round of replication, which produced a single band of intermediate density, was enough to rule out the conservative model of replication. However, a second round was necessary to show that now two bands of equal intensity were observed, consistent with semiconservative replication and inconsistent with dispersive replication.

7. In the diagram below, which letter indicates the 5′ end of the leading strand?

c. C

17. What is the function of DNA gyrase?

c. Reduces the torsional strain that builds up ahead of the replication fork as a result of unwinding

35. Which of the following is TRUE of DNA polymerases of eukaryotic cells?

c. Some DNA polymerases have the ability to function in DNA repair mechanisms.

30. Telomerase activity is most likely to be found in which cells in humans?

d. Germ line

38. Which of the following is a necessary step in the Holliday model of recombination?

e. A single-strand break occurs in the DNA molecule

28. If ribonucleotides were depleted from a cell during S phase, how would DNA synthesis be affected? (Ignore energetic considerations.)

e. Replication would cease because ribonucleotides are required to initiate DNA synthesis.

45. Which different DNA polymerases are found in eukaryotic cells? What are their functions?

(1) DNA polymerase : initiation of nuclear DNA synthesis and DNA repair (2) DNA polymerase : completes replication on lagging strands (5′ 3′ polymerase activity) (3) DNA polymerase : replication and repair of mitochondrial DNA (4) DNA polymerase : leading strand synthesis (5) Other eukaryotic DNA polymerases (and allow DNA replication to proceed past damaged regions of DNA (DNA lesions), or to play various roles in DNA repair processes. Section 12.3

54. Discuss the main differences in the initiation of recombination proposed by the Holliday model and the double-strand-break model.

(1) Holliday model: Recombination is initiated with single-strand breaks at identical positions in homologous chromosomes. The free ends then invade the other (homologous) DNA molecules, and the broken ends join and displace the original complementary strands. (2) Double-strand-break model: Recombination is initiated with double-strand breaks in one of the two aligned chromosomes. Nucleotides on either side of the breaks are degraded, and the partially degraded strand invades the other, intact strand and displaces the homologous strand as it is elongated.

43. Are Okazaki fragments formed on the leading strand during DNA replication? Explain your answer.

: No, they are not. Okazaki fragments form because DNA polymerase adds nucleotides in only one direction (5′ to 3′), yet it must copy two DNA templates—the leading strand and the lagging strand—that are antiparallel. Once an RNA primer is synthesized 5′ to 3′ on the leading strand, DNA synthesis may begin in the 5′ to 3′ direction along the exposed template. The replication fork moves in the same physical direction that the DNA polymerase on the leading strand does, so exposed template is continuously supplied to the polymerase, and it may continue uninterrupted without a need to form Okazaki fragments. However, for the lagging strand, when the RNA primers are synthesized 5′ to 3′ and the polymerase begins synthesizing DNA, the polymerase moves in a direction that is physically opposite to the overall movement of the replication fork along the template. Therefore, additional RNA primers must be added at interrupted intervals as replication fork movement exposes new template DNA behind the polymerase.

42. Why do eukaryotic chromosomes have multiple origins of replication, whereas prokaryotic chromosomes typically have only one origin?

: The large genome size of a typical eukaryotic chromosome, compared to that of a prokaryotic chromosome, necessitates multiple origins of replication so that the genetic material can be copied in a reasonable amount of time. Also, replication is much slower in eukaryotes (the average is about 10-100 nucleotides/second) than in prokaryotes (the average is about 1000 nucleotides/second). For example, the human haploid genome complement is estimated to be about 3.3 × 109 nucleotide base pairs. Divide this number by 23 (the haploid number of human chromosomes) to get an average number of 3.3 × 109/23 = 1.4 × 108, or about 140 million nucleotides (nt) for a single human chromosome. Even at 100 nucleotides/second, it would take (140,000,000 nt) × (100 nt/sec)−1 = (1,400,000 sec) × (1 min/60 sec) × (60 min/hr)−1 × (1 day/24 hr) = 16.2 days to replicate one eukaryotic chromosome if it contained only one origin.

50. Explain the need for and mechanism of licensing of DNA replication.

In order for the large genomes of eukaryotes to be replicated in a timely manner, replication occurs from thousands of origins scattered along the chromosomes. However, the use of multiple origins creates a problem of timing of replication: the entire genome must be replicated once, and only once, each cell cycle. To accomplish this, initiation of replication begins with "licensing" of replication origins early in the cell cycle. Licensing is accomplished by binding of a complex called MCM to the origin early in the cell cycle. After replication has begun at an origin during S phase, a protein called Geminin prevents MCM from binding to DNA and reinitiating replication at that origin. Thus, replication can occur from each origin only once each cell cycle. Geminin degrades after mitosis, allowing MCM to bind again and relicense each origin for the next round of replication.

44. What does "proofreading" refer to with regard to DNA replication?

Proofreading is the ability of DNA polymerases to back up, remove a newly added nucleotide, and replace it. Proofreading enables the cell to correct mistakes if the wrong base is added during 5′ to 3′ DNA polymerization. If a noncomplementary base is incorporated, then base pairing with the template is disrupted, leading to altered conformation of the DNA polymerase on the DNA. These conformational perturbations stall the polymerase, which allows its 3′ to 5′ enzymatic activity to remove the mispaired base. Removal restores a normal conformation so that the DNA polymerase resumes 5′ to 3′ activity, incorporates the correct nucleotide in the vacant site, and continues replicating the template

53. You have discovered a special dye that reveals the position of recombination sites on meiotic chromosomes. You use this dye to count the number of recombination sites and then compare this to the number of genetic exchanges that you can detect by looking at the segregation of markers across the genome. You find many more recombination sites as compared to genetic exchanges. Explain this result.

Recombination, whether by the Holliday model or by the double-strand-break model, involves Holliday junctions as intermediates. These junctions can be resolved into either crossovers (exchanges) or noncrossovers. The excess of recombination sites must be due to noncrossovers.

10. You are studying a new virus with a DNA genome of 12 Kb. It can synthesize DNA at a rate of 400 nucleotides per second. If the virus uses rolling-circle replication, how long will it take to replicate its genome?

c. 30 seconds

6. What type of synthesis occurs on the leading strand?

c. Continuous

25. DNA primase requires a ____ template and _____ nucleotides to initiate primer synthesis.

c. DNA; RNA

9. You learn that a Mars lander has retrieved a bacterial sample from the polar ice caps. You obtain a sample of these bacteria and perform the same kind of experiment that Meselson and Stahl did to determine how the Mars bacteria replicate their DNA. Based on the following equilibrium density gradient centrifugation results, what type of replication would you propose for these new bacteria?

c. Dispersive

15. Which of the following enzymes do NOT aid in unwinding of DNA for replication?

c. Primase

21. DNA synthesis during replication is initiated from

c. RNA primers.

20. Which of the following is a protein that facilitates the termination of replication in E. coli?

c. Tus

22. DNA polymerases require all of the following for DNA replication, EXCEPT

d. 3′ to 5′ polymerase activity.

16. DNA polymerase I and DNA polymerase III both have ______but only DNA polymerase I has _______.

d. 3′→5′ exonuclease activity; 5′→ 3′ exonuclease activity

27. Which activity is NOT associated with DNA polymerases?

d. Ability to synthesize a DNA from scratch without a primer

34. Which of the following is TRUE regarding nucleosome formation during DNA replication?

d. The addition of newly synthesized histones is a part of nucleosome assembly.

26. Which one of the following statements is not true for all E. coli DNA polymerases?

d. They possess 5′ 3′ exonuclease activity.

5. Meselson and Stahl showed that DNA is replicated by a __________ system.

b. semiconservative

24. The proofreading function of DNA polymerases involves

b. 3′ → 5′ exonuclease activity.

47. List and describe the key events of DNA replication in their relative order for E. coli.

(1) Initiator proteins recognize and bind DNA at specific A/T-rich sites, called origins of replication, and denature (or "melt") the double strands. (2) Single-strand DNA-binding proteins associate with the denatured DNA to stabilize the single strands and form an initial replication bubble. (3) DNA helicase binds to the double-strand DNA ahead of the replication bubble and begins breaking the inter-base hydrogen bonds. (4) DNA primase synthesizes RNA primers using the single-strand DNA templates. (5) DNA gyrase (a topoisomerase) associates with double-strand DNA in front of DNA helicase and breaks/rejoins DNA strands to relieve supercoils generated in the double helix as the replication fork progresses. (6) DNA polymerase III extends the RNA primers. Continuous DNA polymerization occurs for the leading strand; discontinuous DNA polymerization occurs for the lagging strand. Okazaki fragments (short RNA/DNA hybrid sequences) are formed on the lagging strand. (7) DNA polymerase I "repairs" the RNA primers and replaces RNA with DNA. (8) For the lagging strand only, DNA ligase seals the "nicks" (breaks in the phosphodiester backbone) remaining between adjacent nucleotides to produce a new, continuous DNA daughter strand. Section 12.3

41. Summarize the similarities and differences in rolling-circle replication, theta replication, and linear eukaryotic replication.

(1) Rolling-circle replication • Initiated by cleavage in a single DNA strand • Uncleaved strand used as template • New nucleotides added to 3′ end of cleaved strand • Linear single-strand DNA produced; subsequently circularized (2) Theta replication • Two replication forks • Initiated by DNA denaturation at single origin • Single, expanding replication bubble • Bidirectional replication • Circular DNA molecule produced (3) Linear eukaryotic replication • Two replication forks • Initiated by DNA denaturation at multiple origins • Multiple, expanding replication bubbles • Bidirectional replication • Linear DNA molecule produced

51. (a) Explain the significance of telomerase in cells of the germ line. (b) Explain the significance of telomerase in cancer cells.

(a) Telomerase allows for the ends of chromosomes to be replicated. In the absence of telomerase, the ends of chromosomes are clipped after each round of replication, causing the telomeres to be eventually lost and the ends of the chromosomes to become unstable. In order for the chromosomes to be maintained for use in the next generation, telomerase must be expressed in cells of the germ line. (b) Most somatic cells normally do not express telomerase. Because of this, somatic cells normally can divide only a limited number of times before the telomeres are completely removed and the chromosomes become unstable. However, in most cancers, mutations occur that allow telomerase to be expressed. Because the telomeres are restored after replication, these cells are able to divide an unlimited number of times, allowing cancer cells to reproduce themselves without limit and allowing tumors to grow to large size and cancer to spread within the body.

52. You discover a drug that specifically inhibits DNA synthesis. You apply the drug to yeast cells after S phase but before prophase of meiosis. You find a drastic reduction in recombination in the meiotic products. How can you explain this?

According to the double-strand-break model of recombination, after strand invasion DNA synthesis occurs at the 3′ ends of the DNA molecule that experienced the break. An inhibitor of DNA synthesis would block this and interrupt recombination.

29. If a deletion occurs in a gene that encodes DNA polymerase I and no functional DNA polymerase I is produced. What will be the most likely consequence of this mutation?

a. The DNA strands would contain pieces of RNA.

48. In the diagram below, the arrow indicates the direction of movement of a replication fork. a. Which strand, top or bottom, is the leading strand? Explain. b. On which strand, top or bottom, would you expect to find Okazaki fragments? Explain.

a. The bottom strand, which has 5′ to 3′ synthesis in the same direction as movement of the replication fork. b. The top strand, which experiences discontinuous synthesis because the direction of synthesis is opposite that of the movement of the replication fork.

2. Okazaki fragments are found in all of the following EXCEPT

a. leading strand.

55. What does the term "noncrossover recombinant" mean with respect to homologous recombination?

The terms "crossover" and "noncrossover" refer to the arrangement of DNA near to, but not participating in, a homologous recombination event. In the process of recombination, two double-strand DNA molecules join at a point of four-strand DNA called the Holliday junction. When the Holliday junction is cleaved and DNA is rejoined into two separate, double-strand DNA molecules, the composition of the resulting DNA molecules in adjacent regions—for example, a flanking pair of genes with one proximal and one distal—depends on which of two possible orientations the DNA strands are cleaved in. In one cleavage orientation, the segments of DNA that flank the region involved in recombination will be exchanged and emerge in a new, recombinant composition, like that discussed for single crossovers in meiosis in Chapter 5. Such a recombinant is therefore said to be of the crossover type. However, if the Holliday junction is resolved by cleavage in the other orientation, then all DNA segments that flank each side of the recombination region will emerge with the same composition of flanking DNA that was present before the recombination event. Recombinant DNA strands are still formed because a short segment of DNA has been exchanged, but the exchange is limited to the site of the initial DNA break(s) and the range of Holliday-junction migration. Such a DNA strand is therefore called a noncrossover recombinant.

11. You are studying a new virus with a DNA genome of 12 Kb. It can synthesize DNA at a rate of 400 nucleotides per second. If the virus uses theta replication, how long will it take to replicate its genome?

b. 15 seconds

36. What would be a likely result of expressing telomerase in somatic cells?

b. Cancer

39. The Holliday model describes which process?

b. Homologous recombination

31. For which of the following is the "end-replication problem" relevant?

b. Linear chromosomes

23. What type of bonds does DNA ligase create between adjacent nucleotides?

b. Phosphodiester

3. Which of the following does NOT utilize bidirectional replication?

b. Rolling circle model

33. _______ are tandemly repeated DNA sequences located at the ends of eukaryotic chromosomes.

b. Telomeres

12. The nuclear genome of a single human cell (i.e., the entire diploid complement) contains about 6.6 billion (6.6 × 109) base pairs of DNA. If synthesis at each replication fork occurs at an average rate of 50 nucleotides per second, all the DNA is replicated in 5 minutes. Assume that replication is initiated simultaneously at all origins. How many origins of replication exist in a human diploid genome?

a. 220,000

1. All DNA polymerases synthesize new DNA by adding nucleotides to the _____ of the growing DNA chain.

a. 3′ OH

13. The nuclear genome of a single human cell (i.e., the entire diploid complement) contains about 6.6 billion (6.6 × 109) base pairs of DNA. If synthesis at each replication fork occurs at an average rate of 50 nucleotides per second, all the DNA is replicated in 5 minutes. Assume that replication is initiated simultaneously at all origins. Assuming that the origins are approximately equally distributed across the chromosomes, what is the average number of origins per human chromosome?

a. 4783

37. Which of the following best describes heteroduplex DNA?

a. A single-stranded DNA molecule of one chromosome pairs with a single-stranded DNA molecule of another chromosome.

19. What is the function of DNA ligase?

a. Connects Okazaki fragments by sealing nicks in the sugar- phosphate backbone

49. Explain the effect on DNA replication of mutations that destroy each of the following activities in DNA polymerase. Also, for each kind of mutation, how might you detect the effect in an in vitro replication reaction? a. 5′ to 3′ polymerase b. 5′ to 3′ exonuclease c. 3′ to 5′ exonuclease

a. If the 5′ to 3′ polymerase activity is missing, no dNTPs will be incorporated into DNA. This could be detected using methods, such as density gradient centrifugation or electrophoresis, which would allow one to see that no new DNA polymers are being synthesized.b. 5′ to 3′ exonuclease is needed for DNA polymerase to remove RNA primers from the 5′ ends of Okazaki fragments. If the activity is missing, the primers will not be removed from newly synthesized DNA. This could be detected by including radiolabeled ribonucleotides in the reaction mix and seeing that the radioactivity remains with DNA polymers at the end of the reaction. c. 3′ to 5′ exonuclease is needed for proofreading. It allows incorrectly incorporated nucleotides to be removed before proceeding to the next nucleotide. Lack of this activity would decrease the accuracy of replication and allow many more replication errors to occur. This could be detected by sequencing the products of replication and determining the number of errors. Section 12.3

4. Which of the following typically only has one origin of replication?

a. Prokaryotes

8. Suppose that some cells are grown in culture in the presence of radioactive nucleotides for many generations so that both strands of every DNA molecule include radioactive nucleotides. The cells are then harvested and placed in new medium with nucleotides that are not radioactive so that newly synthesized DNA will not be radioactive. What proportion of DNA molecules will contain radioactivity after two rounds of replication?

e. 1/2

18. What is the function of single-strand-binding proteins?

e. Prevents the formation of secondary structures within single-stranded DNA

14. Suppose Meselson and Stahl had obtained the following results in their experiment. These results would be consistent with which model of replication?

e. Semiconservative and dispersive replication

32. Telomerase uses _____________ to synthesize new DNA.

e. an RNA template