Bio 180 Ch 15 (CSI) – Flashcards
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What does it mean to say that strands in a double helix are antiparallel? a. Their primary sequences consist of a sequence of complementary bases. b. They each have a sugar-phosphate backbone. c. They each have a 5′ S 3′ directionality. d. They have opposite directionality, or polarity.
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d
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Which of the following is not a property of DNA polymerase? a. It adds dNTPs only in the 5′ -> 3′ direction. b. It requires a primer to work. c. It is associated with a sliding clamp only on the leading strand. d. Its exonuclease activity is involved in proofreading.
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a
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The enzyme that removes twists in DNA ahead of the replication fork is _____.
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topoisomerase
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What is the function of primase? a. synthesis of the short section of double-stranded DNA required by DNA polymerase b. synthesis of a short RNA, complementary to single-stranded DNA c. closing the gap at the 3′ end of DNA after excision repair d. removing primers and synthesizing a short section of DNA to replace them
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b
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How are Okazaki fragments synthesized? a. using the leading strand template, and synthesizing 5′ S 3′ b. using the leading strand template, and synthesizing 3′ -> 5′ c. using the lagging strand template, and synthesizing 5′ -> 3′ d. using the lagging strand template, and synthesizing 3′ -> 5′
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c
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An enzyme that uses an internal RNA template to synthesize DNA is _____.
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telomerase
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Researchers design experiments so that only one thing is different between the treatments that are being compared. In the Hershey- Chase experiment, what was this single difference?
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Labeling DNA or labeling proteins.
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What is the relationship between defective DNA repair and cancer?
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DNA is constantly damaged, and many pathways have evolved to repair this onslaught of damage. If a DNA re- pair pathway is inactivated by mutation, damage is inef- ficiently repaired. Consequently mutation rates increase, and the increased number of mutations increases the probability that cancer-causing mutations will occur.
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Why is the synthesis of the lagging strand of DNA discontinuous? How is it possible for the synthesis of the leading strand to be continuous?
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On the lagging strand, DNA polymerase moves away from the replication fork. When helicase unwinds a new section of DNA, primase must build a new primer on the template for the lagging strand (closer to the fork) and another polymerase molecule must begin synthesis at this point. This makes the lagging-strand synthesis discontinuous. On the leading strand, DNA polymerase moves in the same direction as helicase, so synthesis can continue, without interrup- tion, from a single primer (at the origin of replication).
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Explain how telomerase prevents linear chromosomes from shortening during replication.
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Telomerase binds to the overhang at the end of a chromosome. Once bound, it begins catalyzing the addition of deoxyribonucleotides to the overhang in the 5' -> 3' direction, lengthening the overhang. This allows primase, DNA polymerase, and ligase to catalyze the addition of deoxyribonucleotides to the lagging strand in the 5' -> 3' direction, restoring the lagging strand to its original length.
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Predict what would occur in a bacterial mutant that lost the ability to chemically mark the template strand of DNA. a. The mutation rate would increase. b. The ability of DNA polymerase to discriminate between correct and incorrect base pairs would decrease. c. The energy differences between correct and incorrect base pairs would decrease. d. The energy differences between correct and incorrect base pairs would increase.
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a (Because the ability to dis- tinguish which strand contains the incorrect base would be lost).
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What aspect of DNA structure makes it possible for the enzymes of nucleotide excision repair to recognize many different types of DNA damage? a. the polarity of each DNA strand b. the antiparallel orientation of strands in the double helix c. the energy differences between correct and incorrect base pairs d. the regularity of DNA's overall structure
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d (The regularity of DNA's structure allows enzymes to recognize any type of damage that distorts this regular structure.)
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If you could engineer an activity into DNA polymerase to allow both strands to follow the replication fork, what would this additional activity be? a. the ability to begin DNA synthesis without a primer b. the ability to proofread in the 5′ S 3′ direction c. the ability to synthesize DNA in the 3′ S 5′ direction d. the ability to synthesize DNA without using a template
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c (If DNA polymerase could synthesize DNA 3¿ S 5¿ as well as the normal 5¿ S 3¿, then both newly synthesized DNA strands could be extended to follow the replication fork.)
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In the late 1950s, Herbert Taylor grew bean root-tip cells in a solution of radioactive thymidine and allowed them to undergo one round of DNA replication. He then transferred the cells to a solution without the radioactive deoxyribonucleotide, allowed them to replicate again, and examined their chromosomes for the presence of radioactivity. His results are shown in the following figure, where red indicates a radioactive chromatid. a. Draw diagrams explaining the pattern of radioactivity observed in the sister chromatids after the first and second rounds of replication. b. What would the results of Taylor's experiment be if eukaryotes used a conservative mode of DNA replication?
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(a) In FIGURE A15.4, the gray lines represent DNA strands containing radioactivity. (b) After one round of replication in radioactive solution, one double-stranded DNA would be radioactive in both strands and the other would not be radioactive in either strand. After another round of DNA synthesis, this time in nonradioactive solution, one of the four DNA molecules would be radioactive in both strands and the other three DNA molecules would contain no radioactivity in any strand.
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The graph that follows shows the survival of four different E. coli strains after exposure to increasing doses of ultraviolet light. The wild-type strain is normal, but the other strains have a mutation in either a gene called uvrA, a gene called recA, or both. a. Which strains are most sensitive to UV light? Which strains are least sensitive? b. What are the relative contributions of these genes to the repair of UV damage?
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(a) The double mu- tant of both uvrA and recA is most sensitive to UV light; the single mutants are in between; and the wild type is least sensitive. (b) The recA gene contributes more to UV repair through most of the UV dose levels. But at very high UV doses, the uvrA gene is somewhat more important than the recA gene.
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QUANTITATIVE Assuming that each replication fork moves at a rate of 500 base pairs per second, how long would it take to replicate the E. coli chromosome (with 4.6 million base pairs) from a single origin of replication?
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About 4600 seconds or 77 minutes. This answer comes from knowing that replication proceeds bidirectionally, so replication from each fork is predicted to replicate half the chromosome. This is 4.6 million base pairs/2 = 2.3 million base pairs. At 500 base pairs per second, this requires 2.3 million base pairs/500 base pairs per second = 4600 seconds. To obtain the time in minutes, divide 4600 seconds by 60 seconds per minute.
