PCB 3063 Exam 2 V2 | PCB 3063 Genetics
| Actual Q&A with Rationale (PCB3063
Exam 2) | University of Central Florida
1. Which enzyme is primarily responsible for synthesizing the leading strand during DNA
replication in E. coli?
A. DNA Polymerase I
B. Primase
C. DNA Polymerase III
D. Helicase
Answer: C
Rationale: DNA Polymerase III is the main replicative enzyme in prokaryotes that adds
nucleotides in the 5’ to 3’ direction. It exhibits high processivity and includes a 3’ to 5’
exonuclease activity for proofreading. DNA Polymerase I is mainly involved in primer
removal and gap filling.
2. In the Meselson-Stahl experiment, what observation led to the rejection of the
conservative model of DNA replication?
A. The presence of only light nitrogen (14N) after the first round.
B. The appearance of a single intermediate-density band after one generation.
C. The presence of only heavy nitrogen (15N) after the first round.
,D. The appearance of two separate bands (one heavy, one light) after one generation.
Answer: B
Rationale: After one generation of growth in 14N, the DNA showed a single band
representing a hybrid of 15N and 14N. If the conservative model were correct, two distinct
bands representing the original heavy and new light DNA would have appeared. This result
proved that replication is semiconservative.
3. What is the function of the Sigma factor in prokaryotic transcription?
A. Elongation of the RNA chain by adding ribonucleotides.
B. Recognition of the promoter sequence and initiation of transcription.
C. Splicing of introns from the primary transcript.
D. Termination of transcription at rho-dependent sites.
Answer: B
Rationale: The Sigma factor is a subunit of the RNA polymerase holoenzyme that targets
the enzyme to specific promoter regions like the -10 and -35 boxes. Once transcription has
been successfully initiated, the Sigma factor typically dissociates from the core enzyme.
This specificity is crucial for regulating which genes are expressed under different
environmental conditions.
4. Which molecule acts as the inducer for the lac operon in E. coli?
A. Allolactose
, B. cAMP
C. Glucose
D. Tryptophan
Answer: A
Rationale: Allolactose is an isomer of lactose that binds 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 of the operon.
5. During translation, which site on the ribosome does the incoming aminoacyl-tRNA first
occupy (except for the initiator tRNA)?
A. P site (Peptidyl site)
B. A site (Aminoacyl site)
C. E site (Exit site)
D. The 5’ cap site
Answer: B
Rationale: The A site is where the aminoacyl-tRNA, which carries the next amino acid to be
added, enters the ribosome. It pairs with the codon currently positioned in that site
through its anticodon. The initiator tRNA is the only tRNA that skips the A site and enters
directly into the P site.
| Actual Q&A with Rationale (PCB3063
Exam 2) | University of Central Florida
1. Which enzyme is primarily responsible for synthesizing the leading strand during DNA
replication in E. coli?
A. DNA Polymerase I
B. Primase
C. DNA Polymerase III
D. Helicase
Answer: C
Rationale: DNA Polymerase III is the main replicative enzyme in prokaryotes that adds
nucleotides in the 5’ to 3’ direction. It exhibits high processivity and includes a 3’ to 5’
exonuclease activity for proofreading. DNA Polymerase I is mainly involved in primer
removal and gap filling.
2. In the Meselson-Stahl experiment, what observation led to the rejection of the
conservative model of DNA replication?
A. The presence of only light nitrogen (14N) after the first round.
B. The appearance of a single intermediate-density band after one generation.
C. The presence of only heavy nitrogen (15N) after the first round.
,D. The appearance of two separate bands (one heavy, one light) after one generation.
Answer: B
Rationale: After one generation of growth in 14N, the DNA showed a single band
representing a hybrid of 15N and 14N. If the conservative model were correct, two distinct
bands representing the original heavy and new light DNA would have appeared. This result
proved that replication is semiconservative.
3. What is the function of the Sigma factor in prokaryotic transcription?
A. Elongation of the RNA chain by adding ribonucleotides.
B. Recognition of the promoter sequence and initiation of transcription.
C. Splicing of introns from the primary transcript.
D. Termination of transcription at rho-dependent sites.
Answer: B
Rationale: The Sigma factor is a subunit of the RNA polymerase holoenzyme that targets
the enzyme to specific promoter regions like the -10 and -35 boxes. Once transcription has
been successfully initiated, the Sigma factor typically dissociates from the core enzyme.
This specificity is crucial for regulating which genes are expressed under different
environmental conditions.
4. Which molecule acts as the inducer for the lac operon in E. coli?
A. Allolactose
, B. cAMP
C. Glucose
D. Tryptophan
Answer: A
Rationale: Allolactose is an isomer of lactose that binds 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 of the operon.
5. During translation, which site on the ribosome does the incoming aminoacyl-tRNA first
occupy (except for the initiator tRNA)?
A. P site (Peptidyl site)
B. A site (Aminoacyl site)
C. E site (Exit site)
D. The 5’ cap site
Answer: B
Rationale: The A site is where the aminoacyl-tRNA, which carries the next amino acid to be
added, enters the ribosome. It pairs with the codon currently positioned in that site
through its anticodon. The initiator tRNA is the only tRNA that skips the A site and enters
directly into the P site.
