, Question 1
A Gram-negative rod from a wound infection grows on MacConkey agar as
lactose-fermenting pink colonies and tests positive for indole, but negative for
citrate utilization and urease. Which organism is most consistent with these
results?
A. Klebsiella pneumoniae
B. Escherichia coli
C. Proteus mirabilis
D. Enterobacter aerogenes
Correct Answer: B - Escherichia coli
RATIONALE
E. coli ferments lactose (pink on MacConkey), is indole-positive,
citrate-negative, and urease-negative, matching the profile. Klebsiella
is indole-negative, citrate-positive, and urease-positive; Proteus is
urease-positive and lactose-negative; Enterobacter is indole-negative
and citrate-positive.
Question 2
An E. coli strain is engineered with a nonsense mutation in the gene encoding
the sigma factor RpoS. Which phenotype is most likely to emerge under
stationary-phase stress?
A. Constitutive expression of heat-shock proteins due to loss of negative
regulation
B. Impaired transition to stationary phase and reduced stress resistance
C. Increased transcription of rRNA operons and rapid growth
D. Failure to initiate DNA replication because RpoS is an origin-binding
protein
Correct Answer: B - Impaired transition to stationary phase and
reduced stress resistance
Page 2
, RATIONALE
RpoS (^S) is the master regulator of the general stress response and
stationary-phase gene expression; its loss impairs survival under
starvation, oxidative, and osmotic stress. It is not a replication
initiator, nor does it directly repress heat-shock genes or drive rRNA
synthesis.
Question 3
A patient with prolonged neutropenia develops a pulmonary infection. Biopsy
reveals broad, sparsely septate hyphae invading blood vessels. Which organism
is the most likely cause?
A. Aspergillus fumigatus
B. Candida albicans
C. Rhizopus species
D. Pneumocystis jirovecii
Correct Answer: C - Rhizopus species
RATIONALE
Rhizopus (order Mucorales) produces broad, ribbon-like, sparsely
septate hyphae with wide-angle branching and is angioinvasive,
causing rhinocerebral/pulmonary mucormycosis in neutropenic or
diabetic patients. Aspergillus has narrow, septate, acute-angle
branching hyphae; Candida is yeast/pseudohyphae; Pneumocystis does
not form hyphae.
Question 4
In a batch culture, a facultative anaerobe is grown aerobically on glucose. At
the point of diauxic shift, which regulatory event best explains the lag before
lactose utilization?
A. Inducer exclusion and catabolite repression lower cAMP-CRP,
blocking lac operon transcription
B. Lactose permease is inhibited by ATP, preventing uptake
Page 3
, C. The lac repressor is degraded by Lon protease, allowing transcription
D. Ribosomal mutations reduce translation of -galactosidase
Correct Answer: A - Inducer exclusion and catabolite repression
lower cAMP-CRP, blocking lac operon transcription
RATIONALE
During diauxie, glucose lowers cAMP and prevents CRP activation
and also causes inducer exclusion (via EIIA^Glc), so the lac operon is
not transcribed until glucose is depleted. This produces the classic
two-phase growth curve. The other options misstate the molecular
mechanism.
Question 5
A previously healthy adult presents with fever, hypotension, and a diffuse
erythematous rash during menstruation. Blood cultures grow Gram-positive
cocci in clusters that are catalase-positive and coagulase-positive. Which
virulence factor most directly explains the hypotension?
A. Protein A binding Fc of IgG
B. TSST-1 acting as a superantigen
C. Alpha-toxin forming pores in neutrophils
D. Coagulase converting fibrinogen to fibrin
Correct Answer: B - TSST-1 acting as a superantigen
RATIONALE
The presentation is toxic shock syndrome (TSS) due to
Staphylococcus aureus. TSST-1 is a superantigen that cross-links
MHC II and TCR V, causing massive cytokine release and
hypotension. Protein A, alpha-toxin, and coagulase contribute to
immune evasion or local pathology but do not directly cause the
systemic hypotension of TSS.
Page 4
A Gram-negative rod from a wound infection grows on MacConkey agar as
lactose-fermenting pink colonies and tests positive for indole, but negative for
citrate utilization and urease. Which organism is most consistent with these
results?
A. Klebsiella pneumoniae
B. Escherichia coli
C. Proteus mirabilis
D. Enterobacter aerogenes
Correct Answer: B - Escherichia coli
RATIONALE
E. coli ferments lactose (pink on MacConkey), is indole-positive,
citrate-negative, and urease-negative, matching the profile. Klebsiella
is indole-negative, citrate-positive, and urease-positive; Proteus is
urease-positive and lactose-negative; Enterobacter is indole-negative
and citrate-positive.
Question 2
An E. coli strain is engineered with a nonsense mutation in the gene encoding
the sigma factor RpoS. Which phenotype is most likely to emerge under
stationary-phase stress?
A. Constitutive expression of heat-shock proteins due to loss of negative
regulation
B. Impaired transition to stationary phase and reduced stress resistance
C. Increased transcription of rRNA operons and rapid growth
D. Failure to initiate DNA replication because RpoS is an origin-binding
protein
Correct Answer: B - Impaired transition to stationary phase and
reduced stress resistance
Page 2
, RATIONALE
RpoS (^S) is the master regulator of the general stress response and
stationary-phase gene expression; its loss impairs survival under
starvation, oxidative, and osmotic stress. It is not a replication
initiator, nor does it directly repress heat-shock genes or drive rRNA
synthesis.
Question 3
A patient with prolonged neutropenia develops a pulmonary infection. Biopsy
reveals broad, sparsely septate hyphae invading blood vessels. Which organism
is the most likely cause?
A. Aspergillus fumigatus
B. Candida albicans
C. Rhizopus species
D. Pneumocystis jirovecii
Correct Answer: C - Rhizopus species
RATIONALE
Rhizopus (order Mucorales) produces broad, ribbon-like, sparsely
septate hyphae with wide-angle branching and is angioinvasive,
causing rhinocerebral/pulmonary mucormycosis in neutropenic or
diabetic patients. Aspergillus has narrow, septate, acute-angle
branching hyphae; Candida is yeast/pseudohyphae; Pneumocystis does
not form hyphae.
Question 4
In a batch culture, a facultative anaerobe is grown aerobically on glucose. At
the point of diauxic shift, which regulatory event best explains the lag before
lactose utilization?
A. Inducer exclusion and catabolite repression lower cAMP-CRP,
blocking lac operon transcription
B. Lactose permease is inhibited by ATP, preventing uptake
Page 3
, C. The lac repressor is degraded by Lon protease, allowing transcription
D. Ribosomal mutations reduce translation of -galactosidase
Correct Answer: A - Inducer exclusion and catabolite repression
lower cAMP-CRP, blocking lac operon transcription
RATIONALE
During diauxie, glucose lowers cAMP and prevents CRP activation
and also causes inducer exclusion (via EIIA^Glc), so the lac operon is
not transcribed until glucose is depleted. This produces the classic
two-phase growth curve. The other options misstate the molecular
mechanism.
Question 5
A previously healthy adult presents with fever, hypotension, and a diffuse
erythematous rash during menstruation. Blood cultures grow Gram-positive
cocci in clusters that are catalase-positive and coagulase-positive. Which
virulence factor most directly explains the hypotension?
A. Protein A binding Fc of IgG
B. TSST-1 acting as a superantigen
C. Alpha-toxin forming pores in neutrophils
D. Coagulase converting fibrinogen to fibrin
Correct Answer: B - TSST-1 acting as a superantigen
RATIONALE
The presentation is toxic shock syndrome (TSS) due to
Staphylococcus aureus. TSST-1 is a superantigen that cross-links
MHC II and TCR V, causing massive cytokine release and
hypotension. Protein A, alpha-toxin, and coagulase contribute to
immune evasion or local pathology but do not directly cause the
systemic hypotension of TSS.
Page 4
