PCB 3063 Exam 3 V2 | PCB 3063 Genetics
| Actual Q&A with Rationale (PCB3063
Exam 3) | University of Central Florida
1. In DNA replication, which enzyme is responsible for relieving the torsional strain
(supercoiling) ahead of the replication fork?
A. DNA Helicase
B. Primase
C. DNA Ligase
D. DNA Gyrase (Topoisomerase)
Answer: D
Rationale: DNA Gyrase, a type of topoisomerase, acts to reduce the tension created by the
unwinding of the DNA double helix. It does this by creating transient breaks in the DNA
backbone to allow the strands to swivel. Without this activity, the DNA would become too
tightly coiled for the replication fork to continue moving forward.
2. Which subunit of the prokaryotic RNA polymerase holoenzyme is responsible for
recognizing the promoter sequence?
A. Sigma factor
B. Beta subunit
C. Alpha subunit
,D. Omega subunit
Answer: A
Rationale: The sigma factor is essential for the initiation of transcription in bacteria as it
provides promoter specificity to the RNA polymerase core enzyme. It recognizes the -10
and -35 consensus sequences to ensure the enzyme binds at the correct site. Once
transcription initiation is complete and elongation begins, the sigma factor typically
dissociates from the complex.
3. In the lac operon, what happens when both glucose and lactose are present in the
environment?
A. Low levels of transcription occur because glucose inhibits CAP binding.
B. High levels of transcription occur because lactose is present.
C. No transcription occurs because glucose acts as a repressor.
D. Transcription is maximal because allolactose is present.
Answer: A
Rationale: When glucose is present, cAMP levels are low, which prevents the Catabolite
Activator Protein (CAP) from binding to the promoter. Even if lactose is present and the
repressor is removed, the lack of CAP binding results in only basal (low) levels of
transcription. This mechanism ensures that the cell preferentially metabolizes glucose
before activating the lac operon for lactose metabolism.
, 4. Which of the following mutations would likely be the most deleterious to the protein
product?
A. Silent mutation in the middle of the gene
B. Missense mutation at the C-terminus
C. Frameshift mutation at the beginning of the coding sequence
D. Transition mutation in an intron
Answer: C
Rationale: A frameshift mutation early in the coding sequence alters every subsequent
codon in the mRNA transcript. This usually results in a completely different amino acid
sequence or the premature appearance of a stop codon. Such mutations almost always
result in a non-functional or severely truncated protein compared to single amino acid
changes or silent mutations.
5. During DNA replication, the leading strand is synthesized through:
A. Continuous synthesis in the 5’ to 3’ direction
B. Discontinuous synthesis of Okazaki fragments
C. Continuous synthesis in the 3’ to 5’ direction
D. Fragmented synthesis requiring multiple RNA primers
Answer: A
| Actual Q&A with Rationale (PCB3063
Exam 3) | University of Central Florida
1. In DNA replication, which enzyme is responsible for relieving the torsional strain
(supercoiling) ahead of the replication fork?
A. DNA Helicase
B. Primase
C. DNA Ligase
D. DNA Gyrase (Topoisomerase)
Answer: D
Rationale: DNA Gyrase, a type of topoisomerase, acts to reduce the tension created by the
unwinding of the DNA double helix. It does this by creating transient breaks in the DNA
backbone to allow the strands to swivel. Without this activity, the DNA would become too
tightly coiled for the replication fork to continue moving forward.
2. Which subunit of the prokaryotic RNA polymerase holoenzyme is responsible for
recognizing the promoter sequence?
A. Sigma factor
B. Beta subunit
C. Alpha subunit
,D. Omega subunit
Answer: A
Rationale: The sigma factor is essential for the initiation of transcription in bacteria as it
provides promoter specificity to the RNA polymerase core enzyme. It recognizes the -10
and -35 consensus sequences to ensure the enzyme binds at the correct site. Once
transcription initiation is complete and elongation begins, the sigma factor typically
dissociates from the complex.
3. In the lac operon, what happens when both glucose and lactose are present in the
environment?
A. Low levels of transcription occur because glucose inhibits CAP binding.
B. High levels of transcription occur because lactose is present.
C. No transcription occurs because glucose acts as a repressor.
D. Transcription is maximal because allolactose is present.
Answer: A
Rationale: When glucose is present, cAMP levels are low, which prevents the Catabolite
Activator Protein (CAP) from binding to the promoter. Even if lactose is present and the
repressor is removed, the lack of CAP binding results in only basal (low) levels of
transcription. This mechanism ensures that the cell preferentially metabolizes glucose
before activating the lac operon for lactose metabolism.
, 4. Which of the following mutations would likely be the most deleterious to the protein
product?
A. Silent mutation in the middle of the gene
B. Missense mutation at the C-terminus
C. Frameshift mutation at the beginning of the coding sequence
D. Transition mutation in an intron
Answer: C
Rationale: A frameshift mutation early in the coding sequence alters every subsequent
codon in the mRNA transcript. This usually results in a completely different amino acid
sequence or the premature appearance of a stop codon. Such mutations almost always
result in a non-functional or severely truncated protein compared to single amino acid
changes or silent mutations.
5. During DNA replication, the leading strand is synthesized through:
A. Continuous synthesis in the 5’ to 3’ direction
B. Discontinuous synthesis of Okazaki fragments
C. Continuous synthesis in the 3’ to 5’ direction
D. Fragmented synthesis requiring multiple RNA primers
Answer: A
