BIOS 242 Exam 3 Fundamentals of
Microbiology with Lab Guide ACTUAL
EXAM 2026/2027 | Verified Q&A | Pass
Guaranteed - A+ Graded
Section A: Microbial Genetics & Gene Regulation (Q1–15)
Q1: A mutation occurs in the lacI gene of E. coli, producing a repressor that cannot bind allolactose but
can still bind the operator. In the presence of lactose, what will happen?
A. Constitutive transcription
B. Normal regulation
C. Repressor degradation
D. No transcription of lac operon [CORRECT]
Rationale: Allolactose (the inducer) normally binds to the LacI repressor, causing a conformational
change that prevents the repressor from binding the operator. If the repressor cannot bind allolactose,
it remains permanently bound to the operator regardless of lactose presence, continuously blocking
RNA polymerase access and preventing transcription.
Q2: A clinical laboratory isolates a strain of Staphylococcus aureus that produces beta-lactamase
constitutively, even in the absence of penicillin. Which type of mutation most likely occurred in the
regulatory region of the bla gene?
A. Missense mutation in the structural gene
B. Nonsense mutation in the structural gene
C. Mutation in the operator sequence preventing repressor binding [CORRECT]
D. Frameshift mutation in the structural gene
Rationale: Constitutive expression of an inducible gene typically results from an operator mutation that
prevents the repressor protein from binding. When the repressor cannot attach to the operator, RNA
polymerase has uninterrupted access to the promoter, resulting in continuous transcription even
without inducer presence.
,Q3: During conjugation, an F⁺ E. coli cell transfers its F factor to an F⁻ cell. Which statement accurately
describes the outcome?
A. The recipient becomes F⁻ and the donor remains F⁺
B. Both cells become F⁻
C. The recipient becomes F⁺ and the donor remains F⁺ [CORRECT]
D. The recipient becomes Hfr and the donor becomes F⁻
Rationale: The F factor is a plasmid that replicates independently. During conjugation, the F factor is
copied via rolling circle replication while being transferred through the sex pilus. Because the original F
factor remains in the donor cell and a copy is transferred, both the donor and recipient end up
containing the F factor, making both F⁺.
Q4: A researcher performs generalized transduction using a virulent bacteriophage grown on Salmonella
typhimurium. The phage lysate is used to infect a recipient S. typhimurium that is his⁻ (histidine
auxotroph). After plating on minimal medium without histidine, several colonies grow. What does this
result indicate?
A. The phage carried a plasmid with the his gene
B. A host bacterial DNA fragment containing the his⁺ gene was packaged into a phage head and
transferred [CORRECT]
C. The recipient underwent spontaneous mutation at high frequency
D. Specialized transduction occurred
Rationale: Generalized transduction occurs when bacteriophage packaging machinery mistakenly
incorporates fragments of host bacterial DNA instead of viral DNA into phage heads. When these
transducing particles infect new bacteria, they inject bacterial DNA, which can recombine into the
recipient chromosome. The appearance of prototrophic colonies on minimal medium confirms that a
his⁺ gene was transferred via this mechanism.
Q5: In the trp operon of E. coli, tryptophan acts as a corepressor. A mutation in the trpR gene produces a
repressor that cannot bind tryptophan. What is the phenotypic effect?
A. The operon is always off, regardless of tryptophan levels
B. The operon is always on, regardless of tryptophan levels [CORRECT]
C. The operon functions normally
D. Tryptophan synthesis is permanently repressed
Rationale: The TrpR repressor is inactive on its own and requires tryptophan binding to achieve a
conformation that allows operator binding. If the repressor cannot bind tryptophan, it remains
, permanently inactive and cannot attach to the operator, so RNA polymerase continuously transcribes
the operon regardless of cellular tryptophan concentration.
Q6: A clinical isolate of Neisseria gonorrhoeae shows resistance to penicillin due to altered penicillin-
binding proteins (PBPs). Which mechanism of horizontal gene transfer most likely contributed to this
resistance?
A. Transformation [CORRECT]
B. Conjugation only
C. Transduction only
D. Vertical gene transfer only
Rationale: Neisseria species are naturally competent and readily take up naked DNA from the
environment through transformation. The penA gene encoding altered PBPs in penicillin-resistant N.
gonorrhoeae has been shown to have mosaic structures resulting from recombination with DNA
acquired from related commensal Neisseria species via transformation.
Q7: A frameshift mutation is introduced into the coding sequence of a bacterial gene. Which of the
following best describes the most likely consequence?
A. A single amino acid substitution
B. A premature stop codon or altered reading frame from the mutation point onward [CORRECT]
C. No change in the protein product
D. Enhanced protein function
Rationale: Frameshift mutations result from insertions or deletions of nucleotides not in multiples of
three, which shifts the translational reading frame. Every codon downstream of the mutation is altered,
typically producing a completely different amino acid sequence and frequently introducing a premature
stop codon due to the random generation of nonsense codons in the new reading frame.
Q8: A student observes that a mutant E. coli strain cannot grow on lactose but grows normally on
glucose. Genetic analysis reveals a mutation in the lacZ gene. Which enzyme activity is deficient?
A. Lactose permease
B. Beta-galactosidase [CORRECT]
C. Transacetylase
D. Allolactose isomerase
Microbiology with Lab Guide ACTUAL
EXAM 2026/2027 | Verified Q&A | Pass
Guaranteed - A+ Graded
Section A: Microbial Genetics & Gene Regulation (Q1–15)
Q1: A mutation occurs in the lacI gene of E. coli, producing a repressor that cannot bind allolactose but
can still bind the operator. In the presence of lactose, what will happen?
A. Constitutive transcription
B. Normal regulation
C. Repressor degradation
D. No transcription of lac operon [CORRECT]
Rationale: Allolactose (the inducer) normally binds to the LacI repressor, causing a conformational
change that prevents the repressor from binding the operator. If the repressor cannot bind allolactose,
it remains permanently bound to the operator regardless of lactose presence, continuously blocking
RNA polymerase access and preventing transcription.
Q2: A clinical laboratory isolates a strain of Staphylococcus aureus that produces beta-lactamase
constitutively, even in the absence of penicillin. Which type of mutation most likely occurred in the
regulatory region of the bla gene?
A. Missense mutation in the structural gene
B. Nonsense mutation in the structural gene
C. Mutation in the operator sequence preventing repressor binding [CORRECT]
D. Frameshift mutation in the structural gene
Rationale: Constitutive expression of an inducible gene typically results from an operator mutation that
prevents the repressor protein from binding. When the repressor cannot attach to the operator, RNA
polymerase has uninterrupted access to the promoter, resulting in continuous transcription even
without inducer presence.
,Q3: During conjugation, an F⁺ E. coli cell transfers its F factor to an F⁻ cell. Which statement accurately
describes the outcome?
A. The recipient becomes F⁻ and the donor remains F⁺
B. Both cells become F⁻
C. The recipient becomes F⁺ and the donor remains F⁺ [CORRECT]
D. The recipient becomes Hfr and the donor becomes F⁻
Rationale: The F factor is a plasmid that replicates independently. During conjugation, the F factor is
copied via rolling circle replication while being transferred through the sex pilus. Because the original F
factor remains in the donor cell and a copy is transferred, both the donor and recipient end up
containing the F factor, making both F⁺.
Q4: A researcher performs generalized transduction using a virulent bacteriophage grown on Salmonella
typhimurium. The phage lysate is used to infect a recipient S. typhimurium that is his⁻ (histidine
auxotroph). After plating on minimal medium without histidine, several colonies grow. What does this
result indicate?
A. The phage carried a plasmid with the his gene
B. A host bacterial DNA fragment containing the his⁺ gene was packaged into a phage head and
transferred [CORRECT]
C. The recipient underwent spontaneous mutation at high frequency
D. Specialized transduction occurred
Rationale: Generalized transduction occurs when bacteriophage packaging machinery mistakenly
incorporates fragments of host bacterial DNA instead of viral DNA into phage heads. When these
transducing particles infect new bacteria, they inject bacterial DNA, which can recombine into the
recipient chromosome. The appearance of prototrophic colonies on minimal medium confirms that a
his⁺ gene was transferred via this mechanism.
Q5: In the trp operon of E. coli, tryptophan acts as a corepressor. A mutation in the trpR gene produces a
repressor that cannot bind tryptophan. What is the phenotypic effect?
A. The operon is always off, regardless of tryptophan levels
B. The operon is always on, regardless of tryptophan levels [CORRECT]
C. The operon functions normally
D. Tryptophan synthesis is permanently repressed
Rationale: The TrpR repressor is inactive on its own and requires tryptophan binding to achieve a
conformation that allows operator binding. If the repressor cannot bind tryptophan, it remains
, permanently inactive and cannot attach to the operator, so RNA polymerase continuously transcribes
the operon regardless of cellular tryptophan concentration.
Q6: A clinical isolate of Neisseria gonorrhoeae shows resistance to penicillin due to altered penicillin-
binding proteins (PBPs). Which mechanism of horizontal gene transfer most likely contributed to this
resistance?
A. Transformation [CORRECT]
B. Conjugation only
C. Transduction only
D. Vertical gene transfer only
Rationale: Neisseria species are naturally competent and readily take up naked DNA from the
environment through transformation. The penA gene encoding altered PBPs in penicillin-resistant N.
gonorrhoeae has been shown to have mosaic structures resulting from recombination with DNA
acquired from related commensal Neisseria species via transformation.
Q7: A frameshift mutation is introduced into the coding sequence of a bacterial gene. Which of the
following best describes the most likely consequence?
A. A single amino acid substitution
B. A premature stop codon or altered reading frame from the mutation point onward [CORRECT]
C. No change in the protein product
D. Enhanced protein function
Rationale: Frameshift mutations result from insertions or deletions of nucleotides not in multiples of
three, which shifts the translational reading frame. Every codon downstream of the mutation is altered,
typically producing a completely different amino acid sequence and frequently introducing a premature
stop codon due to the random generation of nonsense codons in the new reading frame.
Q8: A student observes that a mutant E. coli strain cannot grow on lactose but grows normally on
glucose. Genetic analysis reveals a mutation in the lacZ gene. Which enzyme activity is deficient?
A. Lactose permease
B. Beta-galactosidase [CORRECT]
C. Transacetylase
D. Allolactose isomerase
