Mbio chapter 13 Exam Study Guide with
Complete Solutions New Update Rated
A+
1. Which of the following damaged templates encountered by a replication fork is
repaired by double-strand break repair?
A. The replication fork encounters a lesion and Trans lesion synthesis occurs.
B. The replication fork encounters a lesion and stalls.
C. The replication fork encounters a lesion and bypasses it, restarting synthesis on the
other side of the lesion.
D. The replication fork encounters a lesion at which repair has been initiated and the
fork collapses.
E. The replication fork encounters heteroduplex DNA. - Answers - ✔✔D
2. Which of the following damaged templates encountered by a replication fork is
repaired by gap repair?
A. A replication fork encounters a lesion and trans lesion synthesis occurs.
B. The replication fork encounters a lesion and stalls.
C. The replication fork encounters a lesion and bypasses it, restarting synthesis on the
other side of the lesion.
D. The replication fork encounters a lesion at which repair has been initiated and the
fork collapses.
E. The replication fork encounters heteroduplex DNA. - Answers - ✔✔C
3. Which of the following is not true of repairing double-strand breaks by recombination?
A. After invasion of the homologous chromosome, the 3' end of the invading strand acts
as a primer for DNA synthesis.
B. The broken ends are processed, with the 3'-ending strand selectively degraded to
create 5' overhangs.
C. The 3' single strand extensions invade the homologous chromosome, displacing the
homologous strand and base pairing with the complementary strand.
D. As a result of the two strand invasions, the two homologues are linked at two places
and are called a branched intermediate.
E. The enzymes required for strand invasion belong to a class of proteins called
recombinases. - Answers - ✔✔B
4. Which of the following is a benefit of recombination during meiosis in eukaryotes?
A. Branched intermediates aid in segregation by keeping the homologous
chromosomes together.
, B. Crossing over can result from recombination, decreasing the combinations of alleles
in each meiotic product.
C. Recombination introduces mutations that lead to increased variation in the
population.
D. Recombination does not occur during meiosis.
E. None of the choices given is correct. - Answers - ✔✔A
5. The branched intermediates generated by recombination can be resolved in a
number of ways. Which of the following is true of branched intermediates and their
resolution?
A. Resolution of the branched intermediate can occur by dissociation of the invading
strands and reannealing with their original complements. Gaps are filled in by DNA
polymerase and the DNA is sealed by ligase. This is known as synthesis-dependent
strand annealing (SDSA).
B. Resolution of the branched intermediate can proceed by double-strand break repair.
Replication is completed by ligating the strands while they are still linked. Cleaving and
rejoining the homologues can result in the original pair of chromosomes.
C. Resolution of the branched intermediate can proceed by double-strand break repair.
Replication is completed by ligating the strands while they are still linked. Cleaving and
rejoining the homologues can result in crossing over be - Answers - ✔✔E
6. Which of the following pathways does not require DNA ligase?
A. gap repair
B. fork regression and resolution of a stalled replication fork
C. synthesis-dependent strand annealing
D. double-strand break repair
E. All of the pathways given require DNA ligase. - Answers - ✔✔B
7. When the replication machinery encounters a single-strand break, the replication fork
collapses resulting in a double-strand break. Which of the following would be the fourth
step in repairing this damage?
A. The 3' single-strand overhang invades the intact portion of the chromosome, pairs
with its complementary strand, and displaces the other.
B. Once the strand invasion is complete, branched migration occurs.
C. The broken ends are processed, with the 5' ending strand selectively degraded to
create 3' overhangs.
D. Branch migration creates Holliday intermediates that are resolved by the enzyme
resolvase.
E. Resolution of the Holliday intermediate is followed by ligation and replication is
restarted. - Answers - ✔✔D
8. If a replication fork stalls at a lesion the usual first response is:
Complete Solutions New Update Rated
A+
1. Which of the following damaged templates encountered by a replication fork is
repaired by double-strand break repair?
A. The replication fork encounters a lesion and Trans lesion synthesis occurs.
B. The replication fork encounters a lesion and stalls.
C. The replication fork encounters a lesion and bypasses it, restarting synthesis on the
other side of the lesion.
D. The replication fork encounters a lesion at which repair has been initiated and the
fork collapses.
E. The replication fork encounters heteroduplex DNA. - Answers - ✔✔D
2. Which of the following damaged templates encountered by a replication fork is
repaired by gap repair?
A. A replication fork encounters a lesion and trans lesion synthesis occurs.
B. The replication fork encounters a lesion and stalls.
C. The replication fork encounters a lesion and bypasses it, restarting synthesis on the
other side of the lesion.
D. The replication fork encounters a lesion at which repair has been initiated and the
fork collapses.
E. The replication fork encounters heteroduplex DNA. - Answers - ✔✔C
3. Which of the following is not true of repairing double-strand breaks by recombination?
A. After invasion of the homologous chromosome, the 3' end of the invading strand acts
as a primer for DNA synthesis.
B. The broken ends are processed, with the 3'-ending strand selectively degraded to
create 5' overhangs.
C. The 3' single strand extensions invade the homologous chromosome, displacing the
homologous strand and base pairing with the complementary strand.
D. As a result of the two strand invasions, the two homologues are linked at two places
and are called a branched intermediate.
E. The enzymes required for strand invasion belong to a class of proteins called
recombinases. - Answers - ✔✔B
4. Which of the following is a benefit of recombination during meiosis in eukaryotes?
A. Branched intermediates aid in segregation by keeping the homologous
chromosomes together.
, B. Crossing over can result from recombination, decreasing the combinations of alleles
in each meiotic product.
C. Recombination introduces mutations that lead to increased variation in the
population.
D. Recombination does not occur during meiosis.
E. None of the choices given is correct. - Answers - ✔✔A
5. The branched intermediates generated by recombination can be resolved in a
number of ways. Which of the following is true of branched intermediates and their
resolution?
A. Resolution of the branched intermediate can occur by dissociation of the invading
strands and reannealing with their original complements. Gaps are filled in by DNA
polymerase and the DNA is sealed by ligase. This is known as synthesis-dependent
strand annealing (SDSA).
B. Resolution of the branched intermediate can proceed by double-strand break repair.
Replication is completed by ligating the strands while they are still linked. Cleaving and
rejoining the homologues can result in the original pair of chromosomes.
C. Resolution of the branched intermediate can proceed by double-strand break repair.
Replication is completed by ligating the strands while they are still linked. Cleaving and
rejoining the homologues can result in crossing over be - Answers - ✔✔E
6. Which of the following pathways does not require DNA ligase?
A. gap repair
B. fork regression and resolution of a stalled replication fork
C. synthesis-dependent strand annealing
D. double-strand break repair
E. All of the pathways given require DNA ligase. - Answers - ✔✔B
7. When the replication machinery encounters a single-strand break, the replication fork
collapses resulting in a double-strand break. Which of the following would be the fourth
step in repairing this damage?
A. The 3' single-strand overhang invades the intact portion of the chromosome, pairs
with its complementary strand, and displaces the other.
B. Once the strand invasion is complete, branched migration occurs.
C. The broken ends are processed, with the 5' ending strand selectively degraded to
create 3' overhangs.
D. Branch migration creates Holliday intermediates that are resolved by the enzyme
resolvase.
E. Resolution of the Holliday intermediate is followed by ligation and replication is
restarted. - Answers - ✔✔D
8. If a replication fork stalls at a lesion the usual first response is:
