PCB 3063 Exam 3 V3 | PCB 3063 Genetics
| Actual Q&A with Rationale (PCB3063
Exam 3) | University of Central Florida
1. Which enzyme is responsible for unwinding the DNA double helix at the replication fork?
A. Helicase
B. DNA Polymerase III
C. Topoisomerase
D. Primase
Answer: A
Rationale: Helicase breaks the hydrogen bonds between the parental DNA strands to
create a template for replication. This process requires energy derived from ATP hydrolysis
to move along the DNA molecule. The unzipping of the strands creates the replication fork
structure necessary for other enzymes to bind.
2. In prokaryotic transcription, which component of the RNA polymerase holoenzyme is
responsible for promoter recognition?
A. Sigma factor
B. Alpha subunit
C. Beta subunit
D. Rho protein
,Answer: A
Rationale: The sigma factor is an essential protein that guides RNA polymerase to the
specific promoter consensus sequences, such as the -10 and -35 boxes. Once transcription
is initiated and a short stretch of RNA is synthesized, the sigma factor usually dissociates
from the core enzyme. Without this factor, RNA polymerase would initiate transcription at
random sites along the DNA.
3. What is the function of the 5’ cap added to eukaryotic mRNA?
A. It facilitates ribosome binding and protects the mRNA from degradation.
B. It identifies the location of the intron.
C. It marks the end of transcription.
D. It acts as a template for the poly-A tail.
Answer: A
Rationale: The 5’ cap consists of a 7-methylguanosine residue attached via a unique 5’-to-
5’ phosphate linkage. This structure is recognized by initiation factors that help the small
ribosomal subunit find the start codon. Additionally, it prevents 5’ to 3’ exonucleases from
prematurely breaking down the transcript.
4. In the lac operon, which molecule acts as the inducer?
A. Glucose
B. cAMP
, C. Galactose
D. Allolactose
Answer: D
Rationale: Allolactose is an isomer of lactose that binds directly to the lac repressor
protein. This binding causes a conformational change in the repressor, preventing it from
binding to the operator site. Consequently, RNA polymerase can access the promoter and
transcribe the structural genes.
5. Which DNA repair mechanism specifically fixes thymine dimers caused by UV light?
A. Nucleotide excision repair
B. Mismatch repair
C. Base excision repair
D. Homologous recombination
Answer: A
Rationale: Nucleotide excision repair (NER) involves the removal of a short segment of
DNA containing the bulky lesion. UV-induced pyrimidine dimers distort the DNA backbone,
which is recognized by the NER complex. Once the damaged segment is excised, DNA
polymerase fills the gap using the undamaged strand as a template.
6. During DNA replication, which enzyme removes RNA primers and fills the gaps with DNA?
A. DNA Ligase
| Actual Q&A with Rationale (PCB3063
Exam 3) | University of Central Florida
1. Which enzyme is responsible for unwinding the DNA double helix at the replication fork?
A. Helicase
B. DNA Polymerase III
C. Topoisomerase
D. Primase
Answer: A
Rationale: Helicase breaks the hydrogen bonds between the parental DNA strands to
create a template for replication. This process requires energy derived from ATP hydrolysis
to move along the DNA molecule. The unzipping of the strands creates the replication fork
structure necessary for other enzymes to bind.
2. In prokaryotic transcription, which component of the RNA polymerase holoenzyme is
responsible for promoter recognition?
A. Sigma factor
B. Alpha subunit
C. Beta subunit
D. Rho protein
,Answer: A
Rationale: The sigma factor is an essential protein that guides RNA polymerase to the
specific promoter consensus sequences, such as the -10 and -35 boxes. Once transcription
is initiated and a short stretch of RNA is synthesized, the sigma factor usually dissociates
from the core enzyme. Without this factor, RNA polymerase would initiate transcription at
random sites along the DNA.
3. What is the function of the 5’ cap added to eukaryotic mRNA?
A. It facilitates ribosome binding and protects the mRNA from degradation.
B. It identifies the location of the intron.
C. It marks the end of transcription.
D. It acts as a template for the poly-A tail.
Answer: A
Rationale: The 5’ cap consists of a 7-methylguanosine residue attached via a unique 5’-to-
5’ phosphate linkage. This structure is recognized by initiation factors that help the small
ribosomal subunit find the start codon. Additionally, it prevents 5’ to 3’ exonucleases from
prematurely breaking down the transcript.
4. In the lac operon, which molecule acts as the inducer?
A. Glucose
B. cAMP
, C. Galactose
D. Allolactose
Answer: D
Rationale: Allolactose is an isomer of lactose that binds directly to the lac repressor
protein. This binding causes a conformational change in the repressor, preventing it from
binding to the operator site. Consequently, RNA polymerase can access the promoter and
transcribe the structural genes.
5. Which DNA repair mechanism specifically fixes thymine dimers caused by UV light?
A. Nucleotide excision repair
B. Mismatch repair
C. Base excision repair
D. Homologous recombination
Answer: A
Rationale: Nucleotide excision repair (NER) involves the removal of a short segment of
DNA containing the bulky lesion. UV-induced pyrimidine dimers distort the DNA backbone,
which is recognized by the NER complex. Once the damaged segment is excised, DNA
polymerase fills the gap using the undamaged strand as a template.
6. During DNA replication, which enzyme removes RNA primers and fills the gaps with DNA?
A. DNA Ligase
