BIOL 3200 Exam 3 Actual Exam V2 | BIOL 3200 General Microbiology
(BIOL 3200 Exam 3) | Auburn University
1. Which enzyme is responsible for relieving torsional stress (supercoiling) ahead of the
replication fork in bacteria?
A. DNA Gyrase (Topoisomerase II)
B. DNA Helicase
C. DNA Ligase
D. Primase
Answer: A
Rationale: DNA Gyrase is essential for reducing the tension created by the unwinding of
the DNA double helix. It introduces negative supercoils to counteract the positive
supercoiling caused by helicase. Without this enzyme, the replication fork would stall due
to excessive physical strain on the DNA molecule.
2. In the context of bacterial transcription, what is the specific function of the sigma factor?
A. To catalyze the formation of phosphodiester bonds
B. To recognize and bind to the promoter region
C. To unwind the DNA at the replication origin
D. To terminate transcription at rho-independent sites
Answer: B
Rationale: The sigma factor is a subunit of the RNA polymerase holoenzyme that provides
specificity for transcription initiation. It recognizes the -10 and -35 consensus sequences
within the promoter to ensure the enzyme starts at the correct site. Once transcription
begins, the sigma factor usually dissociates from the core enzyme.
3. During translation, which site on the ribosome does the incoming aminoacyl-tRNA typically
enter first (except for the initiator tRNA)?
A. P (Peptidyl) site
B. A (Aminoacyl) site
C. E (Exit) site
D. S (Shine-Dalgarno) site
Answer: B
,Rationale: The A site serves as the entry point for all charged tRNAs during the elongation
phase of translation. The ribosome checks for the correct codon-anticodon match before
allowing the tRNA to stay. After peptide bond formation, the tRNA shifts from the A site to
the P site.
4. A mutation that changes a codon from UAC (Tyrosine) to UAA (Stop) is classified as a:
A. Silent mutation
B. Missense mutation
C. Frameshift mutation
D. Nonsense mutation
Answer: D
Rationale: A nonsense mutation results in the premature termination of protein synthesis
by creating a stop codon where an amino acid was previously encoded. This usually leads
to a non-functional, truncated protein product. It is distinct from missense mutations,
which only change one amino acid to another.
5. What is the primary role of cAMP in the regulation of the lac operon?
A. It acts as a direct inducer by binding to the repressor
B. It binds to CAP to signal that glucose levels are low
C. It increases the rate of DNA replication
D. It inhibits the transport of lactose into the cell
Answer: B
Rationale: cAMP levels are inversely proportional to glucose levels in the bacterial cell.
When glucose is absent, cAMP binds to the Catabolite Activator Protein (CAP), allowing it to
bind the lac promoter. This interaction recruits RNA polymerase and significantly boosts
transcription of the operon.
6. Which process involves the uptake of free, ‘naked’ DNA from the environment by a
competent bacterial cell?
A. Conjugation
B. Transduction
C. Transposition
D. Transformation
Answer: D
Rationale: Transformation is a mechanism of horizontal gene transfer where bacteria take
up DNA fragments released by dead cells. The cell must be in a state of ‘competence’ to
, successfully transport this DNA across its membrane. Once inside, the DNA can be
integrated into the host genome via homologous recombination.
7. In the trp operon, what occurs when tryptophan levels are very high?
A. The repressor dissociates from the operator
B. Transcription is initiated by cAMP binding
C. The ribosome stalls at the leader sequence
D. Tryptophan binds the repressor, which then binds the operator
Answer: D
Rationale: The trp operon is a repressible system where tryptophan acts as a corepressor.
When tryptophan is abundant, it binds to the inactive aporepressor to form a functional
holorepressor. This complex binds to the operator, physically blocking RNA polymerase
from transcribing the structural genes.
8. Generalized transduction is characterized by which of the following events?
A. The transfer of only specific genes near the prophage insertion site
B. The random packaging of host DNA into a phage head during the lytic cycle
C. Direct cell-to-cell contact via a sex pilus
D. The use of F-prime plasmids to transfer metabolic traits
Answer: B
Rationale: Generalized transduction occurs when a bacteriophage accidentally packages
bacterial chromosomal DNA instead of viral DNA. Because the packaging is random, any
part of the bacterial genome can be transferred to a recipient cell. This happens during the
assembly phase of a purely lytic or lysogenic phage cycle.
9. An Hfr (High Frequency of Recombination) cell is a bacterial cell that has:
A. A separate, circular F plasmid
B. Lost the ability to form a pilus
C. The F plasmid integrated into its chromosome
D. Multiple copies of R-plasmids
Answer: C
Rationale: An Hfr cell results from the integration of the F factor into the bacterial
chromosome via site-specific recombination. During conjugation, the Hfr cell attempts to
transfer its entire chromosome to the recipient. This allows for the transfer of large
amounts of genetic information, though the bridge usually breaks before the entire
chromosome is moved.
(BIOL 3200 Exam 3) | Auburn University
1. Which enzyme is responsible for relieving torsional stress (supercoiling) ahead of the
replication fork in bacteria?
A. DNA Gyrase (Topoisomerase II)
B. DNA Helicase
C. DNA Ligase
D. Primase
Answer: A
Rationale: DNA Gyrase is essential for reducing the tension created by the unwinding of
the DNA double helix. It introduces negative supercoils to counteract the positive
supercoiling caused by helicase. Without this enzyme, the replication fork would stall due
to excessive physical strain on the DNA molecule.
2. In the context of bacterial transcription, what is the specific function of the sigma factor?
A. To catalyze the formation of phosphodiester bonds
B. To recognize and bind to the promoter region
C. To unwind the DNA at the replication origin
D. To terminate transcription at rho-independent sites
Answer: B
Rationale: The sigma factor is a subunit of the RNA polymerase holoenzyme that provides
specificity for transcription initiation. It recognizes the -10 and -35 consensus sequences
within the promoter to ensure the enzyme starts at the correct site. Once transcription
begins, the sigma factor usually dissociates from the core enzyme.
3. During translation, which site on the ribosome does the incoming aminoacyl-tRNA typically
enter first (except for the initiator tRNA)?
A. P (Peptidyl) site
B. A (Aminoacyl) site
C. E (Exit) site
D. S (Shine-Dalgarno) site
Answer: B
,Rationale: The A site serves as the entry point for all charged tRNAs during the elongation
phase of translation. The ribosome checks for the correct codon-anticodon match before
allowing the tRNA to stay. After peptide bond formation, the tRNA shifts from the A site to
the P site.
4. A mutation that changes a codon from UAC (Tyrosine) to UAA (Stop) is classified as a:
A. Silent mutation
B. Missense mutation
C. Frameshift mutation
D. Nonsense mutation
Answer: D
Rationale: A nonsense mutation results in the premature termination of protein synthesis
by creating a stop codon where an amino acid was previously encoded. This usually leads
to a non-functional, truncated protein product. It is distinct from missense mutations,
which only change one amino acid to another.
5. What is the primary role of cAMP in the regulation of the lac operon?
A. It acts as a direct inducer by binding to the repressor
B. It binds to CAP to signal that glucose levels are low
C. It increases the rate of DNA replication
D. It inhibits the transport of lactose into the cell
Answer: B
Rationale: cAMP levels are inversely proportional to glucose levels in the bacterial cell.
When glucose is absent, cAMP binds to the Catabolite Activator Protein (CAP), allowing it to
bind the lac promoter. This interaction recruits RNA polymerase and significantly boosts
transcription of the operon.
6. Which process involves the uptake of free, ‘naked’ DNA from the environment by a
competent bacterial cell?
A. Conjugation
B. Transduction
C. Transposition
D. Transformation
Answer: D
Rationale: Transformation is a mechanism of horizontal gene transfer where bacteria take
up DNA fragments released by dead cells. The cell must be in a state of ‘competence’ to
, successfully transport this DNA across its membrane. Once inside, the DNA can be
integrated into the host genome via homologous recombination.
7. In the trp operon, what occurs when tryptophan levels are very high?
A. The repressor dissociates from the operator
B. Transcription is initiated by cAMP binding
C. The ribosome stalls at the leader sequence
D. Tryptophan binds the repressor, which then binds the operator
Answer: D
Rationale: The trp operon is a repressible system where tryptophan acts as a corepressor.
When tryptophan is abundant, it binds to the inactive aporepressor to form a functional
holorepressor. This complex binds to the operator, physically blocking RNA polymerase
from transcribing the structural genes.
8. Generalized transduction is characterized by which of the following events?
A. The transfer of only specific genes near the prophage insertion site
B. The random packaging of host DNA into a phage head during the lytic cycle
C. Direct cell-to-cell contact via a sex pilus
D. The use of F-prime plasmids to transfer metabolic traits
Answer: B
Rationale: Generalized transduction occurs when a bacteriophage accidentally packages
bacterial chromosomal DNA instead of viral DNA. Because the packaging is random, any
part of the bacterial genome can be transferred to a recipient cell. This happens during the
assembly phase of a purely lytic or lysogenic phage cycle.
9. An Hfr (High Frequency of Recombination) cell is a bacterial cell that has:
A. A separate, circular F plasmid
B. Lost the ability to form a pilus
C. The F plasmid integrated into its chromosome
D. Multiple copies of R-plasmids
Answer: C
Rationale: An Hfr cell results from the integration of the F factor into the bacterial
chromosome via site-specific recombination. During conjugation, the Hfr cell attempts to
transfer its entire chromosome to the recipient. This allows for the transfer of large
amounts of genetic information, though the bridge usually breaks before the entire
chromosome is moved.
