Quiz 6: CH 5

Which of the following repair pathways can accurately repair a double-strand break? a) base excision repair b) nucleotide excision repair c) direct chemical reversal d) homologous recombination e) nonhomologous end joining

d) homologous recombination feedback: Compared to the faithful homologous recombination pathway to repair double-strand breaks, nonhomologous end joining is a "quick and dirty" solution that often results in mutations at the site of repair.

Phase variation helps protect the bacterium Salmonella typhimurium against the immune system of its host by switching the orientation of a certain promoter. This process ... a) is carried out through a DNA transposition mechanism. b) is irreversible. c) can often result in the excision of the promoter from the chromosome altogether. d) is mediated by enzymes that form transient covalent bonds with the DNA.

d) is mediated by enzymes that form transient covalent bonds with the DNA. feedback: Phase variation in Salmonella is brought about by a conservative site-specific recombinase encoded by the bacterium. The recombinases that catalyze conservative site-specific recombination form transient, high-energy covalent bonds with the DNA and use this energy to complete the DNA rearrangements, an action reminiscent of topoisomerases.

DNA glycosylases constitute an enzyme family found in all three domains of life. They can ... a) add sugar moieties to DNA b) remove sugar moieties from DNA c) add a purine or pyrimidine base to DNA d) remove a purine or pyrimidine base from DNA e) remove a nucleotide from DNA

d) remove a purine or pyrimidine base from DNA feedback: These enzymes can recognize altered bases in DNA and catalyze their removal by hydrolyzing the glycosidic bond between the base and the deoxyribose sugar.

DNA-only transposons ... a) can be recognized by the presence of short inverted repeats at each end. b) often encode a transposase that mediates the transposition process. c) leave double-strand breaks in the donor chromosome. d) can move by a cut-and-paste mechanism. e) All of the above.

e) All of the above. feedback: DNA-only transposons often encode a transposase to move by a cut-and-paste mechanism that results in a broken donor chromosome. Short inverted repeats of DNA sequence are found at the ends of these transposons.

Upon heavy damage to the cell's DNA, the normal replicative DNA polymerases may stall when encountering damaged DNA, triggering the use of backup translesion polymerases. These backup polymerases ... a) lack 3′-to-5′ exonucleolytic proofreading activity. b) are replaced by the replicative polymerases after adding only a few nucleotides. c) can create mutations even on undamaged DNA. d) may recognize specific DNA damage and add the appropriate nucleotide to restore the original sequence. e) All of the above.

e) All of the above. feedback: The translesion polymerases come in different flavors and some of them are very efficient in repairing specific damage, but they generally have a low fidelity as they cannot proofread. The cell thus limits their usage to very few polymerization steps.

What group of mobile genetic elements is largely responsible for the resistance of the modern strains of pathogenic bacteria to common antibiotics? a) DNA-only transposons b) retroviral-like retrotransposons c) nonretroviral retrotransposons d) site-specific recombinases

a) DNA-only transposons feedback: DNA-only transposons that carry genes encoding antibiotic-inactivating enzymes are largely responsible for the spread of antibiotic resistance in bacterial strains.

Which of the following is NOT correct regarding long and short interspersed nuclear elements? a) Each of them encodes a reverse transcriptase b) They rely on the cellular transcription machinery to produce their RNA transcript c) They use one of the strands in the target DNA as a primer to synthesize their DNA. d) Together, they make up about 40% of our genome. e) They can move into new regions of genome that do not have any homology with their DNA.

a) Each of them encodes a reverse transcriptase feedback: Most short interspersed nuclear elements (SINEs) do not encode any proteins and spread by pirating enzymes encoded by other transposons, such as long interspersed nuclear elements (LINEs).

Which of the following is NOT correct regarding homologous recombination and its regulation? a) Loss of heterozygosity can occur if a broken chromosome is repaired using a sister chromatid instead of its homologous chromosome. b) Repair of double-strand breaks by homologous recombination is favored during or soon after DNA replication. c) Homologous recombination can rescue broken or stalled replication forks in S phase. d) Excessive use of homologous recombination by human cells can lead to cancer. e) Low usage of homologous recombination by human cells can lead to cancer.

a) Loss of heterozygosity can occur if a broken chromosome is repaired using a sister chromatid instead of its homologous chromosome feedback: If a homologous chromosome is used (instead of a sister chromatid) to repair a double-strand break, one of the parental alleles of the affected gene can be replaced by the other allele, leading to loss of heterozygosity.

Which of the following spontaneous lesions in DNA occurs most frequently in a mammalian cell? a) depurination b) cytosine deamination c) guanine oxidation d) guanine alkylation e) depyrimidination

a) depurination feedback: Depurination (that is, the hydrolytic removal of purine bases from DNA) is by far the most common endogenous DNA lesion in our cells. Please refer to Table 5-3 for details.

This protein folds into a doughnut shape that can encircle DNA. It can load on the DNA only when the DNA is broken in both strands, so that the DNA can thread through the hole in the protein. Which of the following proteins do you think matches this description? a) PCNA, the sliding clamp for DNA polymerases at the replication forks b) Ku, the protein that recognizes DNA ends and can initiate nonhomologous end joining c) MCM, the helicase critical for the initiation and elongation of replication d) Topoisomerase II, which can create or relax superhelical tension in DNA e) RecA/Rad51, which carries out strand invasion in homologous recombination

b) Ku, the protein that recognizes DNA ends and can initiate nonhomologous end joining feedback: The Ku dimer forms a hole that accommodates a DNA double helix. In addition, its structure reflects the fact that it should only bind to free DNA ends (for example, from double-strand breaks) in a fashion similar to threading a needle.

In which phases of the eukaryotic cell cycle does homologous recombination often occur to repair DNA damage? a) G1 and S phase b) S and G2 phase c) G2 and M phase d) M and G1 phase e) G1 and G2 phase

b) S and G2 phase feedback: Homologous recombination is used to repair DNA damage during and shortly after the S phase of the cell cycle, when a sister chromatid is available for faithful repairs.

Which of the following is true regarding retroviral-like retrotransposons? a) They encode both a reverse transcriptase and an RNA polymerase. b) They have directly repeated long terminal repeats at their two ends when integrated into chromosomal DNA. c) Their genomic RNA can be translated to produce viral coat proteins. d) They leave double-strand breaks in the original donor DNA. e) The Alu element in our genome is an example of retroviral-like retrotransposons

b) They have directly repeated long terminal repeats at their two ends when integrated into chromosomal DNA. feedback: Retroviral-like retrotransposons resemble retroviruses in their replication, but lack the ability to produce a protein coat. They have directly repeated long terminal repeats at each end and are transcribed by cellular RNA polymerases. The transcript is then used as a template by a reverse transcriptase (encoded by the element) to make a double-stranded DNA copy that is then integrated into a new site on the chromosome using an integrase (also encoded by the element). This mechanism keeps the original copy unchanged. Alu elements are nonretroviral retrotransposons.

In meiosis, a crossover in one position is thought to inhibit crossing-over in the neighboring regions. This regulatory mechanism ... a) results in a very uneven distribution of crossover points along each chromosome. b) ensures that even small chromosomes undergo at least one crossover. c) controls how the Holliday junctions are resolved. d) All of the above.

b) ensures that even small chromosomes undergo at least one crossover. feedback: The poorly understood regulatory mechanism of crossover control ensures a roughly even distribution of crossover points along chromosomes. It also ensures that each chromosome undergoes at least one crossover every meiosis. For many organisms, roughly two crossovers per chromosome occur, one on each arm.

What are the products of deamination of cytosine and 5-methyl cytosine, respectively? a) thymine and uracil b) thymine in both cases c) uracil and thymine d) uracil in both cases e) xanthine and hypoxanthine

c) uracil and thymine feedback: The use of thymine instead of uracil by DNA has the advantage that deamination of cytosine (to uracil) can be readily detected as an abnormal base in DNA. However, deamination of 5-methyl cytosine (a modification that is found in vertebrate DNA) can produce thymine and lead to mutations at a higher rate.