4. DNA replication Pages: Difficulty: 2 Ans: C Which one of the following statements about enzymes that interact with DNA is true?

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1 Chapter 25 DNA Metabolism Multiple Choice Questions 1. DNA replication Page: 977 Difficulty: 2 Ans: C The Meselson-Stahl experiment established that: A) DNA polymerase has a crucial role in DNA synthesis. B) DNA synthesis in E. coli proceeds by a conservative mechanism. C) DNA synthesis in E. coli proceeds by a semiconservative mechanism. D) DNA synthesis requires datp, dctp, dgtp, and dttp. E) newly synthesized DNA in E. coli has a different base composition than the preexisting DNA. 2. DNA replication Page: 978 Difficulty: 2 Ans: D When a DNA molecule is described as replicating bidirectionally, that means that it has two: A) chains. B) independently replicating segment. C) origins. D) replication forks. E) termination points. 3. DNA replication Page: 979 Difficulty: 2 Ans: D An Okazaki fragment is a: A) fragment of DNA resulting from endonuclease action. B) fragment of RNA that is a subunit of the 30S ribosome. C) piece of DNA that is synthesized in the 3' 5' direction. D) segment of DNA that is an intermediate in the synthesis of the lagging strand. E) segment of mrna synthesized by RNA polymerase. 4. DNA replication Pages: Difficulty: 2 Ans: C Which one of the following statements about enzymes that interact with DNA is true? A) E. coli DNA polymerase I is unusual in that it possesses only a 5' 3' exonucleolytic activity. B) Endonucleases degrade circular but not linear DNA molecules. C) Exonucleases degrade DNA at a free end. D) Many DNA polymerases have a proofreading 5' 3' exonuclease.

2 E) Primases synthesize a short stretch of DNA to prime further synthesis. 5. DNA replication Page: 982 Difficulty: 2 Ans: C E. coli DNA polymerase III: A) can initiate replication without a primer. B) is efficient at nick translation. C) is the principal DNA polymerase in chromosomal DNA replication. D) represents over 90% of the DNA polymerase activity in E. coli cells. E) requires a free 5'-hydroxyl group as a primer. 6. DNA replication Page: 982 Difficulty: 2 Ans: D The 5' 3' exonuclease activity of E. coli DNA polymerase I is involved in: Α) formation of a nick at the DNA replication origin. Β) formation of Okazaki fragments. Χ) proofreading of the replication process. ) removal of RNA primers by nick translation. Ε) sealing of nicks by ligase action. 7. DNA replication Pages: Difficulty: 2 Ans: C Prokaryotic DNA polymerase III: A) contains a 5' 3' proofreading activity to improve the fidelity of replication. B) does not require a primer molecule to initiate replication. C) has a subunit that acts as a circular clamp to improve the processivity of DNA synthesis. D) synthesizes DNA in the 3' 5' direction. E) synthesizes only the leading strand; DNA polymerase I synthesizes the lagging strand.

3 8. DNA replication Page: 988 Difficulty: 2 Ans: E At replication forks in E. coli: A) DNA helicases make endonucleolytic cuts in DNA. B) DNA primers are degraded by exonucleases. C) DNA topoisomerases make endonucleolytic cuts in DNA. D) RNA primers are removed by primase. E) RNA primers are synthesized by primase. 9. DNA recombination Page: 1012 Difficulty: 3 Ans: C The bacteriophage can lysogenize after infecting a bacterium, i.e. integrate into the host bacterial chromosome by site-specific recombination, and may reside there for many generations before an excision event regenerates the viral genome in an infective form. Which one of the following is not a component of these events? A) Excision requires two host proteins and two virally-encoded proteins. B) Integration requires a viral-specific protein, called integrase. C) RecA protein is required to catalyze the insertional recombination event. D) The excision event relies on different sequences than the integration event. E) The virus and the host DNAs share a 15 bp core region of perfect homology. Short Answer Questions 10. DNA replication Pages: Difficulty: 2 Describe briefly how equilibrium density gradient centrifugation was used to demonstrate that DNA replication in E. coli is semiconservative. Ans: Equilibrium density gradient centrifugation separates DNA molecules of slightly different buoyant density. For example, molecules containing 15 N-labeled ( heavy ) DNA are separable from identical molecules containing 14 N ( light ) DNA. Meselson and Stahl grew E. coli for many generations in a medium containing 15 N, producing cells in which all DNA was heavy. These cells were transferred to a medium containing 14 N, and the buoyant density of their DNA was determined (by equilibrium density gradient centrifugation) after 1, 2, 3, etc., generations. After one generation, all DNA was of a density intermediate between fully heavy and fully light, indicating that each double-stranded DNA molecule had one heavy (parental) and one light (newly synthesized) strand; replication was semiconservative. (See Fig. 25-2, p. 977.) 11. DNA replication Page: 979 Difficulty: 2 The DNA below is replicated from left to right. Label the templates for leading strand and lagging strand synthesis.

4 (5')ACTTCGGATCGTTAAGGCCGCTTTCTGT(3') (3')TGAAGCCTAGCAATTCCGGCGAAAGACA(5') Ans: The polarity of the strands indicates that the top strand is the template for lagging strand synthesis, and the bottom strand is the template for leading strand synthesis. (See Fig. 25-4, p. 979.) 12. DNA replication Page: 979 Difficulty: 2 All known DNA polymerases catalyze synthesis only in the 5' 3' direction. Nevertheless, during semiconservative DNA replication in the cell, they are able to catalyze the synthesis of both daughter chains, which would appear to require synthesis in the 3' 5' direction. Explain the process that occurs in the cell that allows for synthesis of both daughter chains by DNA polymerase. Ans: During DNA replication, one strand is synthesized continuously and the other is synthesized by a discontinuous mechanism. The daughter chain, which appears to be growing in the 3' 5' direction (the lagging strand ), is actually being synthesized by continual initiation of new chains and their elongation in the 5' 3' direction. 13. DNA replication Pages: 979, 987 Difficulty: 2 What is an Okazaki fragment? What enzyme(s) is (are) required for its formation in E. coli? Ans: An Okazaki fragment is an intermediate in DNA replication in E. coli. It is a short fragment of newly synthesized DNA, attached to the 3' end of a short RNA primer. Such fragments are produced by the combined action of primase (part of the primosome) and DNA polymerase III during replication of the lagging strand. (See Fig , p. 987.) 14. DNA replication Pages: Difficulty: 2 A suitable substrate for DNA polymerase is shown below. Label the primer and template, and indicate which end of each strand must be 3' or 5'. To observe DNA synthesis on this substrate in vitro, what additional reaction components must be added? Ans: The top strand (the primer) has its 5' end to the left; the bottom (template) strand has the opposite polarity. For DNA synthesis with this substrate in vitro, one would have to add DNA polymerase, the four deoxynucleoside triphosphates, Mg 2+, and a suitable buffer.

5 15. DNA replication Pages: 981, 984 Difficulty: 2 All known DNA polymerases can only elongate a preexisting DNA chain (i.e., require a primer), but cannot initiate a new DNA chain. Nevertheless, during semiconservative DNA replication in the cell, entirely new daughter DNA chains are synthesized. Explain the process that occurs in the cell that allows for the synthesis of daughter chains by DNA polymerase. Ans: In the cell, initiation of DNA chains occurs via the synthesis of an RNA primer by an RNA polymerase type of enzyme (primase). This primer is elongated by DNA polymerase to produce the daughter DNA chain. The RNA is removed by 5' exonucleolytic hydrolysis before replication is completed. 16. DNA replication Pages: Difficulty: 2 DNA replication in E. coli begins at a site in the DNA called the (a). At the replication fork the (b) strand is synthesized continuously while the (c) strand is synthesized discontinuously. On the strand synthesized discontinuously, the short pieces are called (d) fragments. An RNA primer for each of the fragments is synthesized by an enzyme called (e), and this RNA primer is removed after the fragment is synthesized by the enzyme (f), using its (g) activity. The nicks left behind in this process are sealed by the enzyme (h). Ans: (a) origin; (b) leading; (c) lagging; (d) Okazaki; (e) primase; (f) DNA pol I; (g) 5' 3' exonuclease; (h) DNA ligase 17. DNA repair Page: Difficulty: 2 The high fidelity of DNA replication is due primarily to immediate error correction by the 3' > 5' exonuclease (proofreading) activity of the DNA polymerase. Some incorrectly paired bases escape this proofreading, and further errors can arise from challenges to the chemical integrity of the DNA. List the four classes of repair mechanisms that the cell can use to help correct such errors. Ans: The four classes are listed in Table 25-5 (p. 994), and consist of (1) mismatch repair, (2) base-excision repair, (3) nucleotide-excision repair, and (4) direct repair. 18. DNA Recombination Pages: Difficulty: 2 Outline the four key features of the current model for homologous recombination during meiosis in a eukaryotic cell.

6 Ans: (1) Homologous chromosomes are aligned. (2) A double-strand break is enlarged by an exonuclease, leaving a single-strand extension with a free 3' hydroxyl end. (3) The exposed 3' ends invade the homologous intact duplex DNA, followed by branch migration to create a Holliday junction. (4) Cleavage of the two crossover products creates the two recombinant products. (See Fig , p ) 19. DNA recombination Pages: Difficulty: 2 What distinguishes the simple from the complex class of bacterial transposon? Ans: The simple class called insertion sequences contains only the information needed for transposition and the genes for proteins (transposases) that carry out the process. Those in the class of complex transposons carry additional genes, such as those for antibiotic resistance, a property they confer upon any host bacterium that harbors them. 20. DNA recombination Pages: Difficulty: 2 Briefly describe the role of recombination in the generation of antibody (immunoglobin) diversity. Ans: The genes for immunoglobin polypeptide chains are divided into segments, with multiple versions of each segment (which code for slightly different amino acid sequences). Recombination results in the joining of individual versions of each segment to generate a complete gene. Antibody diversity results from the very large number of different combinations that are possible. (See Fig , p )