Exam 6: DNA Replication and Repair

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In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes.During which process does swapping occur?

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Sometimes, chemical damage to DNA can occur just before DNA replication begins, not giving the repair system enough time to correct the error before the DNA is duplicated.This gives rise to mutation.If the cytosine in the sequence TCAT is deaminated and not repaired, which of the following is the point mutation you would observe after this segment has undergone two rounds of DNA replication?

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Which of the following statements about sequence proofreading during DNA replication is FALSE?

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Which of the following statements correctly explains what it means for DNA replication to be bidirectional?

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How were Meselson and Stahl able to separate "light" DNA from "heavy" DNA.Explain how this experimental approach was used to rule out two of the three models for DNA replication.

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Which scientists first proposed a general strategy for DNA replication.How did they imagine it would work and what was their reasoning?

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Most cells in the body of an adult human lack the telomerase enzyme because its gene is turned off and is therefore not expressed.An important step in the conversion of a normal cell into a cancer cell, which circumvents normal growth control, is the resumption of telomerase expression.Explain why telomerase might be necessary for the ability of cancer cells to divide over and over again.

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Human beings with the inherited disease xeroderma pigmentosum have serious problems with lesions on their skin and often develop skin cancer with repeated exposure to sunlight.What type of DNA damage is not being recognized in the cells of these individuals?

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You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation.These extracts provide the proteins required for DNA replication.Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site.The termination site is on the opposite side of the plasmid from the origin. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation.These extracts provide the proteins required for DNA replication.Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site.The termination site is on the opposite side of the plasmid from the origin.   Figure 6-11 What part of the DNA replication process would be most directly affected if a strain of bacteria lacking helicase were used to make the cell extracts? Figure 6-11 What part of the DNA replication process would be most directly affected if a strain of bacteria lacking helicase were used to make the cell extracts?

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You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation.These extracts provide the proteins required for DNA replication.Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site.The termination site is on the opposite side of the plasmid from the origin. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation.These extracts provide the proteins required for DNA replication.Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site.The termination site is on the opposite side of the plasmid from the origin.   Figure 6-11 What part of the DNA replication process would be most directly affected if a strain of bacteria lacking the exonuclease activity of DNA polymerase were used to make the cell extracts? Figure 6-11 What part of the DNA replication process would be most directly affected if a strain of bacteria lacking the exonuclease activity of DNA polymerase were used to make the cell extracts?

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A bacterial chromosome introduced into a yeast cell will not be duplicated along with yeast DNA chromosomes during cell division.Explain this observation.

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The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process.Which choice below describes the known sequence of events in this process?

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You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation.These extracts provide the proteins required for DNA replication.Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site.The termination site is on the opposite side of the plasmid from the origin. You prepare bacterial cell extracts by lysing the cells and removing insoluble debris via centrifugation.These extracts provide the proteins required for DNA replication.Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site.The termination site is on the opposite side of the plasmid from the origin.   Figure 6-11 What part of the DNA replication process would be most directly affected if a strain of bacteria lacking single-strand binding protein were used to make the cell extracts? Figure 6-11 What part of the DNA replication process would be most directly affected if a strain of bacteria lacking single-strand binding protein were used to make the cell extracts?

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Even though DNA polymerase has a proofreading function, it still introduces errors in the newly synthesized strand at a rate of 1 per 107 nucleotides.To what degree does the mismatch repair system decrease the error rate arising from DNA replication?

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Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25? Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?      (a)    (b)    (c)    (d)    Figure 6-25 (a) Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?      (a)    (b)    (c)    (d)    Figure 6-25 (b) Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?      (a)    (b)    (c)    (d)    Figure 6-25 (c) Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?      (a)    (b)    (c)    (d)    Figure 6-25 (d) Which of the choices below represents the correct way to repair the mismatch shown in Figure 6-25?      (a)    (b)    (c)    (d)    Figure 6-25 Figure 6-25

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Select the option that best completes the following statement: Nonhomologous end joining is a process by which a double-stranded DNA end is joined

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Researchers have isolated a mutant strain of E.coli that carries a temperature-sensitive variant of the enzyme DNA ligase.At the permissive temperature, the mutant cells grow just as well as the wild-type cells.At the nonpermissive temperature, all of the cells in the culture tube die within 2 hours.DNA from mutant cells grown at the nonpermissive temperature for 30 minutes is compared with the DNA isolated from cells grown at the permissive temperature.The results are shown in Figure 6-59, where DNA molecules have been separated by size by means of electrophoresis (P, permissive; NP, nonpermissive).Explain the appearance of a distinct band with a size of 200 base pairs (bp) in the sample collected at the nonpermissive temperature. Researchers have isolated a mutant strain of E.coli that carries a temperature-sensitive variant of the enzyme DNA ligase.At the permissive temperature, the mutant cells grow just as well as the wild-type cells.At the nonpermissive temperature, all of the cells in the culture tube die within 2 hours.DNA from mutant cells grown at the nonpermissive temperature for 30 minutes is compared with the DNA isolated from cells grown at the permissive temperature.The results are shown in Figure 6-59, where DNA molecules have been separated by size by means of electrophoresis (P, permissive; NP, nonpermissive).Explain the appearance of a distinct band with a size of 200 base pairs (bp) in the sample collected at the nonpermissive temperature.    Figure 6-59 Figure 6-59

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You have made a collection of mutant fruit flies that are defective in various aspects of DNA repair.You test each mutant for its hypersensitivity to three DNA-damaging agents: sunlight, nitrous acid (which causes deamination of cytosine), and formic acid (which causes depurination).The results are summarized in Table 6-61, where a "yes" indicates that the mutant is more sensitive than a normal fly, and blanks indicate normal sensitivity. sunlight nitrous acid formic acid Dracula Faust Mole Mr Self-destruct Marguerite yes yes yes yes yes yes yes Table 6-61 A.Which mutant is most likely to be defective in the DNA repair polymerase? Why? B.What aspect of repair is most likely to be affected in the other mutants?

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Indicate whether the following statements are TRUE or FALSE.If a statement is false, explain why it is false.
The sliding clamp is loaded once on each DNA strand, where it remains associated until replication is complete.
True
Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication.
False
Telomerase is a DNA polymerase that carries its own RNA molecule to use as a template at the end of the lagging strand.
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The sliding clamp is loaded once on each DNA strand, where it remains associated until replication is complete.
True
Primase requires a proofreading function that ensures there are no errors in the RNA primers used for DNA replication.
False
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The synthesis of DNA in living systems occurs in the 5′-to-3′ direction.However, scientists synthesize short DNA sequences needed for their experiments on an instrument dedicated to this task. A.The chemical synthesis of DNA by this instrument proceeds in the 3′-to-5′ direction.Draw a diagram to show how this is possible and explain the process. B.Although 3′-to-5′ synthesis of DNA is chemically possible, it does not occur in living systems.Why not?

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