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Test File
By M. Dennis Goode
to accompany
The Cell: A Molecular Approach, Fourth Edition
(Website Quizzes and Problems by Robert Mcgehee and Brian Storrie)
Chapter 6: Replication, Maintenance, and Rearrangements of
Genomic DNA
DNA Replication
Multiple Choice
1. DNA replication is
a. conservative.
b. semiconservative.
c. bidirectional.
d. liberal.
2. DNA polymerases can synthesize DNA
a. de novo, by catalyzing the polymerization of free dNTPs.
b. by adding dNTPs to complementary dNTPs on a single-stranded DNA.
c. by adding dNTPs to a hydroxyl group on the end of a growing polynucleotide chain
hydrogen-bonded to a strand of RNA.
d. by adding dNTPs to a hydroxyl group on the end of a growing polynucleotide chain
hydrogen-bonded to a strand of DNA.
3. Which eukaryotic DNA polymerase replicates the leading strand in a 5′ to 3′ direction?
a. α
b. ε
c. δ
d. All of the above
4. Which eukaryotic DNA polymerase replicates the lagging strand in a 3′ to 5′ to
direction?
a. α
b. γ
c. δ
d. ε
e. None of the above; the lagging strand is replicated in a 5′ to 3′ direction
5. In E. coli, the major enzyme responsible for DNA replication is DNA polymerase
a. Ι.
b. ΙΙ.
c. ΙΙΙ.
d. α.
6. Short segments of newly synthesized DNA on the lagging strand of DNA are called
a. Okazaki fragments.
b. replicons.
c. origins of replication.
d. lagging fragments.
7. Primase synthesizes short sequences of _______ complementary to the _______ strand.
a. DNA; leading
b. RNA; leading
c. DNA; lagging
d. RNA; lagging
8. In addition to synthesizing DNA, DNA polymerase I also has a second catalytic
activity—it can
a. synthesize short RNA sequences.
b. synthesize short polypeptide sequences.
c. remove RNA primers.
d. ligate short segments of DNA together.
9. In eukaryotic cells, RNA primers are removed by the combined action of 5′ to 3′
exonucleases and
a. RNase A.
b. RNase H.
c. DNA polymerase I.
d. DNA polymerase α
10. The short fragments of DNA produced during DNA replication are joined together by
a. RNA polymerase.
b. DNA polymerase.
c. DNA helicase.
d. DNA ligase.
11. Free rotation of one cut DNA strand around one uncut strand is the primary function
of
a. topoisomerase I.
b. topoisomerase II.
c. DNA helicase.
d. DNA polymerase.
12. The function of topoisomerase II is to
a. resolve DNA tangles.
b. allow DNA to swivel and unwind.
c. allow daughter chromatids to separate in anaphase.
d. All of the above
13. Proliferating cell nuclear antigen (PCNA) is a(n) _______ in eukaryotes.
a. DNA polymerase
b. sliding clamp protein
c. single-strand DNA-binding protein
d. origin-of-replication-binding protein
14. During DNA replication, the error frequency is 1 in _______ base pairs.
a. 104
b. 105
c. 106
d. 108
15. Origins of replication are the
a. sites where DNA transcription starts.
b. binding sites for the protein complex that initiates DNA synthesis.
c. loops with two replication forks seen in replicating DNA.
d. forks where DNA replication is occurring.
16. Eukaryotic DNA polymerase ε is active in nondividing as well as dividing cells. This
suggests that it
a. is the polymerase for leading-strand replication.
b. is the polymerase for lagging-strand replication.
c. fills in the gaps between Okazaki fragments.
d. functions primarily in repair of DNA damage.
17. The proofreading property of DNA polymerase is due to its _______ activity.
a. 3′ to 5′ exonuclease
b. 5′ to 3′ exonuclease
c. excision repair
d. mismatch repair
18. Autonomously replicating sequences are
a. yeast plasmids.
b. yeast telomeres.
c. bacterial plasmids.
d. yeast origins of replication.
19. Telomeres are the
a. midpoints of chromosomes.
b. microtubule attachment points on chromosomes.
c. end-sequences of chromosomes.
d. enzyme complexes that complete DNA replication.
20. Telomerase is
a. a reverse transcriptase.
b. the enzyme that adds a unique sequence onto the ends of chromosomes.
c. an enzyme first discovered in Tetrahymena.
d. All of the above
True/False
21. All known DNA polymerases synthesize DNA in the 5′ to 3′direction.
22. The leading and lagging strands at a replication fork are synthesized in opposite
directions, but both are synthesized in a continuous manner.
23. Some DNA polymerases have a nuclease activity that allows them to remove
mismatched nucleotides and repair a sequence.
24. DNA replication starts at sites called replication forks.
25. Telomere sequences form loops at the ends of eukaryotic chromosomes.
26. Telomerase extends the ends of linear chromosomes by making a copy that is
complementary to the other strand of DNA.
27. Telomerase is a reverse transcriptase.
28. Mitochondria have their own unique DNA polymerase.
29. Eukaryotic cells have many different DNA polymerases, one or more of which
function primarily in repair of damaged DNA.
Short Answer
30. How do RNA polymerases differ from DNA polymerases in terms of primer
requirement?
31. Why does the synthesis of Okazaki fragments require an RNA primer?
32. A mutation in this enzyme will block cells in metaphase of mitosis.
33. What are the functions of telomeres?
34. Why is telomerase important in cancer cells?
35. What are the functions of origins of replication?
36. How many (1) telomeres, (2) centromeres, and (3) origins of replication can be found
on a typical eukaryotic chromosome?
37. How many (1) telomeres (2), centromeres, and (3) origins of replication can be found
on a typical prokaryotic chromosome?
38. For each of the following, explain how a prokaryotic chromosome can replicate and
divide normally with the numbers of (1) telomeres, (2) centromeres, and (3) origins of
replication given in Question 37.
39. Why does DNA polymerase synthesize DNA only in the 5′ to 3′ direction?
DNA Repair
Multiple Choice
40. The most common cause of skin cancer is damage to DNA by
a. infrared light.
b. ultraviolet light.
c. β particle radiation.
d. chemical exposure.
41. Point mutations in DNA result from
a. incorporation of incorrect bases during replication.
b. changes as a result of chemical exposure.
c. changes as a result of radiation exposure.
d. All of the above
42. Pyrimidine dimers
a. block DNA replication and transcription.
b. can be repaired by photoreactivation.
c. can be repaired by nucleotide-excision repair.
d. All of the above
43. The human disease in which affected individuals are extremely sensitive to sunlight
and develop multiple skin cancers on exposed areas is called
a. melanoma.
b. zero pigment disease.
c. xeroderma pigmentosum.
d. Cockayne’s syndrome.
44. Cultured cells from xeroderma pigmentosum patients were unable to carry out
a. base-excision repair.
b. nucleotide-excision repair.
c. synthesis of melanin.
d. DNA synthesis.
45. The human genes that convey a susceptibility to hereditary nonpolyposis colorectal
cancer are genes coding proteins involved in the DNA repair mechanism called
a. mismatch repair.
b. nucleotide-excision repair.
c. photoreactivation.
d. All of the above
46. Patients with hereditary nonpolyposis colorectal cancer genes should have
a. frequent colonoscopies.
b. a colonectomy.
c. gene-replacement therapy.
d. an intraperitoneal injection of the normal enzyme.
47. Transcribed genes are repaired _______ when compared to non-transcribed genes.
a. at the same rate
b. more rapidly
c. less rapidly
d. not at all
48. E. coli DNA polymerase V
a. is induced in response to UV irradiation.
b. recognizes thymidine dimers and inserts AA on the opposite strand.
c. makes a high frequency of errors.
d. All of the above
49. The genes responsible for inherited breast cancer (BRCA1 and BRCA2) encode
proteins that are involved in
a. initiation of cell death by apoptosis.
b. regulation of cell proliferation
c. repair of double-strand DNA breaks by homologous recombination.
d. error-prone repair.
Recombination between Homologous DNA Sequences
Multiple Choice
50. Recombination of DNA strands is important because it can
a. rearrange DNA sequences to change gene expression during development.
b. repair damaged sequences.
c. increase genetic diversity in the next generation.
d. All of the above
51. A DNA recombination intermediate (before its resolution into two recombined
strands) is called a
a. RecABCD complex.
b. cross-over complex.
c. recombinase binding site.
d. Holliday junction.
52. The main eukaryotic protein involved in homologous recombination is
a. RecA.
b. Rad51.
c. RuvA.
d. RAD1.
53. The molecule involved in homologous DNA recombination that coats single-stranded
DNA and binds it to a second DNA strand is
a. RecA.
b. RecB.
c. RecC.
d. RecD.
54. Holliday junctions have been visualized by
a. fluorescence microscopy.
b. electron microscopy.
c. phase-contrast microscopy.
d. gel electrophoresis.
55. The specificity of homologous DNA recombination is due to
a. proteins that recognize specific DNA sequences.
b. RNA sequences in the recombinase complex that are complementary to the DNA
sequences.
c. the complementary base pairing of single strands of DNA to each other.
d. the recognition of homologous sequences in two double-stranded molecules.
DNA Rearrangements
Multiple Choice
56. DNA rearrangements occur commonly during
a. mitosis of somatic cells.
b. meiosis of germ cells.
c. development of immune-system cells.
d. meiosis of germ cells and development of immune-system cells.
57. The discovery that genes can move from one chromosome location to another was
made by
a. Barbara McClintock, in studies on corn genetics.
b. Howard Temin, in studies on retrotransposon viruses.
c. Gerald Rubin, in studies on transposable elements in Drosophila.
d. Susumu Tonegawa, in studies on rearrangement of immunoglobulin genes in mice.
58. Retrovirus and retrotransposon DNA sequences insert into the DNA of the host with
the aid of sequences at their ends. These virus end sequences are called
a. telomeres.
b. inverted repeats.
c. short terminal repeats.
d. long terminal repeats.
59. The major class of highly repetitive transposable elements in the human genome
consists of elements called
a. SINEs.
b. TINEs.
c. LINEs.
d. KINEs.
60. Gene amplification is
a. seen in ribosomal genes of developing amphibian oocytes.
b. seen in the genes for muscle proteins of developing muscle cells.
c. seen in the genes for immunoglobulins during B lymphocyte development.
d. never seen in eukaryotic cells, since the amount of DNA per cell is always constant
except during S phase of the cell cycle.
61. Recombining the different heavy chain genes in B lymphocytes results in how many
different heavy chains (not counting mutations)?
a. 240
b. 2400
c. 24,000
d. 2,400,000
62. A major difference between immunoglobulin heavy chains and light chains is that
heavy chains contain one of
a. several hundred V-regions.
b. 12 D-regions.
c. 4 J-regions.
d. 2 C-regions.
63. The mechanism that produces most of immunoglobulin diversity is
a. alteration during protein folding in the endoplasmic reticulum.
b. alternative splicing of transcripts.
c. heavy-chain switching.
d. site-specific recombination within heavy- and light-chain genes.
64. Somatic hypermutation occurs in what portion of the immunoglobulin locus?
a. V
b. D
c. J
d. C
True/False
65. The rate of DNA replication in mammalian cells is tenfold faster than in prokaryotic
cells.
66. Both topoisomerase I and II allow DNA to relieve torsion.
67. Telomerase carries its own template DNA.
68. Proofreading is done by DNA polymerases.
69. Yeast origin-of-recognition complexes are autonomously replicating sequences.
70. DNA polymerases cannot replicate DNA across from a site of DNA damage.
71. Somatic hypermutation is focused in the diversity region of the immunoglobulin gene.
72. Most transposons in eukaryotic cells move by reverse transcription of RNA
intermediates and insertion into the genome.
Fill in the Blank
73. The genes that encode the immunoglobulin light chain consist of regions that are
_______, _______, and _______.
74. IgM antibodies contain _______ constant regions in their heavy chains.
75. T-cell receptors resemble _______ in many ways.
76. T-cell receptors bind to _______ during immune responses.
77. The BRCA genes responsible for inherited breast cancer are involved in the repair of
_______ DNA breaks by homologous recombination.
Short Answer
78. Explain Tonegawa’s key conclusion about the development of immunoglobulin genes.
79. What evidence led Hozumi and Tonegawa to the conclusion above?
80. Explain the mechanism of somatic hypermutation in immunoglobulin genes.
81. How are pseudogenes, which are similar to normal genes but lack introns, thought to
be produced?
82. Explain how gene amplification can contribute to tumor development.
Website Multiple Choice Quiz
1. In E. coli, the major enzyme responsible for DNA replication is DNA polymerase
a. I.
b. II.
c. III.
d. α.
2. Which of the following statements is true of all known DNA polymerases?
a. They synthesize DNA in the 5′ to 3′ direction, and they require a preformed primer
hydrogen-bonded to the template.
b. They synthesize DNA in the 5′ to 3′ direction, and they possess primase activity.
c. They require a preformed primer, and they possess helicase activity.
d. They synthesize DNA in the 3′ to 5′ direction, and they possess exonuclease activity.
3. The twisting of the parental DNA strands around each other ahead of a replication fork
is relieved by enzymes called
a. DNA helicases.
b. topoisomerases.
c. DNA ligases.
d. DNA polymerases.
4. How many replication origins are there in the human genome?
a. 1
b. 30,000
c. 350
d. 350,000
5. Which of the following statements is false about pyrimidine dimers?
a. They are lesions in DNA caused by UV radiation.
b. They are formed between adjacent pyrimidines on a DNA strand.
c. Their formation blocks DNA replication and transcription.
d. They can be repaired by photoreactivation in human cells.
6. How does nucleotide-excision repair differ from base-excision repair?
a. Base-excision repair recognizes and removes single damaged bases, whereas
nucleotide-excision repair is more general, recognizing many different kinds of lesions
that distort the DNA molecule.
b. Nucleotide-excision repair reverses the chemical reaction that caused the lesion,
whereas base-excision repair removes the damaged bases and replaces them with normal
ones.
c. Only the base is removed in base-excision repair, whereas the entire nucleotide is
removed in nucleotide-excision repair.
d. Base-excision repair requires no protein components and can occur by simple
absorption of UV light, whereas nucleotide-excision repair requires several enzymes.
7. During mismatch repair in E. coli, the parental strand is recognized by
a. single-stranded breaks.
b. glycosylated adenines.
c. methylated adenines.
d. methylation of the O6 position of guanine residues.
8. A DNA recombination intermediate before its resolution into two recombined strands
is called a(n)
a. Holliday junction.
b. RecBCD complex.
c. cross-over complex.
d. attachment site.
9. Which of the following statements is false about integrase, the enzyme that catalyzes
the integration of bacteriophage DNA into the E. coli genome?
a. It is necessary for the process of lysogeny.
b. It is necessary for excision of the phage DNA from the bacterial genome.
c. It is encoded by the bacterial genome.
d. It possesses both nuclease and ligase activities.
10. Which of the following does not contribute to the large variety of antigen-binding
specificities found among the immunoglobulins?
a. Recombination between different versions of V, D, and J segments
b. Somatic hypermutation
c. Imprecise joining of immunoglobulin segments
d. Retrotransposition
11. Retrovirus and retrotransposon DNA sequences insert into the DNA of the host with
the aid of sequences at their ends called
a. telomeres.
b. long terminal repeats (LTRs).
c. inverted repeats.
d. J (joining) segments.
12. Which of the following statements about short interspersed elements (SINEs) is false?
a. The major family of SINEs contains the Alu sequence.
b. They arose by reverse transcription of small RNAs.
c. They are transposable elements.
d. They encode reverse transcriptase.
13. Which of the following statements concerning elongation of DNA at the replication
fork is false?
a. The leading strand is synthesized continuously in the direction of replication fork
movement.
b. The lagging strand is synthesized in Okazaki fragments backward from the overall
direction of replication.
c. The Okazaki fragments are joined by the action of DNA ligase.
d. Both strands are synthesized continuously at the replication fork.
14. DNA polymerase requires a primer and cannot initiate synthesis de novo. What serves
as a primer for DNA replication?
a. Short fragments of DNA complementary to the template strand
b. A protein with a free OH group
c. Short fragments of RNA complementary to the template strand
d. The DNA forms a loop resulting in the formation of double-stranded hairpins at the
end of the DNA molecule and these hairpins serve as primers.
15. Estimates of mutation rates for a variety of genes indicate that the frequency of errors
during replication is much lower than would be predicted on the basis of complementary
base pairing. Which of the following mechanisms accounts for the higher degree of
fidelity?
a. Conformational changes in DNA polymerase
b. 3′ to 5′ exonuclease activity of DNA polymerase
c. Requirement of a primer for DNA synthesis by DNA polymerase
d. All of the above
16. Which of the following statements about gene amplification is false?
a. Gene amplification is responsible for amplification of ribosomal RNA genes in
amphibian oocytes.
b. Amplified DNA sequences can be found as free extrachromosomal molecules.
c. Gene amplification occurs as an abnormal event in cancer cells.
d. Amplified DNA sequences are a common occurrence in virally infected cells.
17. What is the function of viral integrase in the life cycle of a retrovirus?
a. Integrase synthesizes a DNA molecule from the template of the RNA viral genome.
b. Integrase describes duplicate sites on viral RNA at which primers bind to initiate DNA
synthesis.
c. Integrase integrates linear viral DNA into the host chromosome for subsequent
transcription.
d. Integrase introduces strand breaks in front of the unwound helix to release tension.
18. Which of the following statements regarding somatic hypermutation is false?
a. The enzyme activation-induced deaminase (AID) is a key player in somatic
hypermutation.
b. Somatic hypermutation is thought to be the result of a high frequency of errors during
DNA repair.
c. Somatic hypermutation is thought to control the proliferation of B lymphocytes by
rendering their genome irreplicable.
d. Somatic hypermutation substantially increases affinity for antigen.
19. Which of the following proteins plays a role in homologous recombination in E. coli?
a. RecA
b. RuvA
c. RuvC
d. All of the above.
20. Mutation in the MutS gene is most commonly associated with
a. colorectal cancer.
b. melanoma.
c. squamous cell carcinoma of the skin.
d. small-cell carcinoma.