BIO 250 MICROBIOLOGY MIDTERM EXAM
QUESTIONS AND ANSWERS A+ GRADED WITH
EXPERT SOLUTIONS - 130 Questions with Answers
Page 1
,Q1. A clinical isolate of Enterococcus faecium exhibits high-level vancomycin
resistance (MIC > 256 g/mL) but remains susceptible to daptomycin. Whole-genome
sequencing reveals a vanA operon on a transposon. Which molecular mechanism best
explains the initial acquisition of this resistance phenotype?
A. Conjugation-mediated transfer of the Tn1546 element from a co-infecting strain
B. Spontaneous point mutations in the chromosomal pbp5 gene
C. Transformation with naked DNA released by lysed vancomycin-resistant
staphylococci
D. Transduction by a bacteriophage carrying the vanA gene cluster
Correct Answer: A. Conjugation-mediated transfer of the Tn1546 element from a
co-infecting strain
Rationale: Vancomycin resistance in enterococci is typically acquired via conjugation, as
the vanA operon resides on transposon Tn1546, which is often part of a conjugative
plasmid. This allows efficient transfer between enterococcal strains. Point mutations in
pbp5 confer -lactam resistance, not vancomycin resistance. Transformation and
transduction are less likely mechanisms for this specific resistance spread in clinical
settings.
Why Wrong:
B - Mutations in pbp5 affect penicillin-binding proteins and are associated with
ampicillin resistance, not vancomycin resistance.
C - Transformation with naked DNA is possible but not the primary mechanism for
vanA dissemination; the operon is typically transferred via conjugation.
D - Transduction by bacteriophages can transfer genetic material, but vanA is
predominantly spread by conjugative plasmids in enterococci.
Reference: Murray, P.R., Rosenthal, K.S., & Pfaller, M.A. (2026). Medical Microbiology,
10th Ed., Ch. 21.
Q2. In a chemostat culture, a bacterial population is maintained at a dilution rate (D)
of 0.2 h¹. The substrate concentration in the reservoir is 10 mM, and the Monod
constant (K_s) is 0.5 mM. If the maximum specific growth rate (_max) is 0.8 h¹, what
is the steady-state substrate concentration in the culture vessel?
A. 0.17 mM
B. 0.50 mM
C. 0.83 mM
D. 1.67 mM
Correct Answer: A. 0.17 mM
Rationale: At steady state in a chemostat, D = ¼. Using the Monod equation, D = ¼_max *
S / (K_s + S). Solving for S: S = D * K_s / (_max - D) = (0.2 * 0.5) / (0.8 - 0.2) = 0..6
= 0.1667 mM, approximately 0.17 mM. The other options arise from misapplying the
formula or confusing K_s with substrate concentration.
Page 2
,Why Wrong:
B - 0.50 mM is the K_s value, not the steady-state substrate concentration.
C - 0.83 mM would result from incorrectly using S = K_s * (_max - D) / D.
D - 1.67 mM would result from using S = D * _max / K_s, which is not the correct
rearrangement.
Reference: Madigan, M.T., Bender, K.S., Buckley, D.H., Sattley, W.M., & Stahl, D.A.
(2024). Brock Biology of Microorganisms, 16th Ed., Ch. 5.
Q3. A patient develops a severe intestinal infection after antibiotic therapy. Stool
analysis reveals abundant Gram-positive rods that produce toxins. The organism is
strictly anaerobic and forms spores. Which virulence factor is most directly
responsible for the pseudomembranous colitis observed in this patient?
A. Toxin A (enterotoxin)
B. Toxin B (cytotoxin)
C. Binary toxin (CDT)
D. Flagellar motility
Correct Answer: B. Toxin B (cytotoxin)
Rationale: Clostridioides difficile toxin B (TcdB) is the primary cytotoxin that causes the
characteristic pseudomembranous colitis by disrupting the actin cytoskeleton of colonic
epithelial cells, leading to cell death and inflammation. Toxin A (TcdA) is also an
enterotoxin but is less potent in causing severe disease in humans. Binary toxin (CDT) is
an accessory toxin that may enhance virulence but is not essential. Flagellar motility is not
directly responsible for tissue damage.
Why Wrong:
A - Toxin A is an enterotoxin that contributes to disease but is not the primary driver
of pseudomembranous colitis; toxin B is more potent.
C - Binary toxin (CDT) is produced by some strains but is not the main cause of
pseudomembranous colitis; it may act as an adjuvant.
D - Flagellar motility aids in colonization but does not directly cause the cytopathic
effects seen in pseudomembranous colitis.
Reference: Carroll, K.C., Morse, S.A., Mietzner, T.A., & Miller, S. (2023). Jawetz, Melnick,
& Adelberg's Medical Microbiology, 28th Ed., Ch. 17.
Q4. A researcher is studying the regulation of the lac operon in Escherichia coli. They
observe that in a strain with a mutation in the crp gene (encoding CAP), the
expression of -galactosidase is very low even in the presence of lactose and absence of
glucose. Which of the following best explains this observation?
A. CAP is required for RNA polymerase to bind the lac promoter efficiently
B. The mutation causes constitutive expression of the lac repressor
C. cAMP levels are insufficient to activate CAP
Page 3
, D. The lac operator is mutated, preventing repressor binding
Correct Answer: A. CAP is required for RNA polymerase to bind the lac promoter
efficiently
Rationale: The lac operon is under positive control by CAP (catabolite activator protein),
which binds to the CAP site upstream of the promoter when cAMP levels are high. This
binding facilitates RNA polymerase binding and transcription initiation. In a crp mutant,
CAP is nonfunctional, so even in the presence of lactose (which inactivates the repressor)
and absence of glucose (which ensures high cAMP), transcription is severely impaired
because RNA polymerase cannot efficiently bind the promoter. The other options do not
directly explain the low expression in a crp mutant.
Why Wrong:
B - A mutation in crp does not affect the lac repressor; the repressor is still functional
and would be inactivated by lactose.
C - cAMP levels are unaffected by the crp mutation; the problem is the lack of
functional CAP.
D - The lac operator mutation would affect repressor binding, but here the operator is
normal; the issue is CAP.
Reference: Snyder, L., Peters, J.E., Henkin, T.M., & Champness, W. (2020). Molecular
Genetics of Bacteria, 5th Ed., Ch. 10.
Q5. A researcher is investigating the microbial community of a deep-sea
hydrothermal vent. They discover a novel archaeon that thrives at 110°C and pH 4.5.
Which of the following adaptations is most critical for the stability of its cytoplasmic
membrane at this extreme temperature?
A. High content of ether-linked isoprenoid lipids
B. High proportion of unsaturated fatty acids
C. Thick peptidoglycan layer
D. Production of heat-shock proteins
Correct Answer: A. High content of ether-linked isoprenoid lipids
Rationale: Archaea, including hyperthermophiles, have membranes composed of
ether-linked isoprenoid lipids, which are more stable at high temperatures than the
ester-linked fatty acids found in bacteria and eukaryotes. The ether bonds and isoprenoid
chains resist hydrolysis and maintain membrane integrity under extreme heat. Unsaturated
fatty acids would increase fluidity and decrease stability at high temperatures.
Peptidoglycan is not present in archaea (they have pseudopeptidoglycan or other cell wall
components). Heat-shock proteins are involved in protein folding, not membrane stability.
Why Wrong:
B - Unsaturated fatty acids increase membrane fluidity, which would be detrimental at
high temperatures; hyperthermophiles have saturated or isoprenoid lipids.
C - Peptidoglycan is not a component of archaeal cell walls; they have other structural
polymers.
Page 4
QUESTIONS AND ANSWERS A+ GRADED WITH
EXPERT SOLUTIONS - 130 Questions with Answers
Page 1
,Q1. A clinical isolate of Enterococcus faecium exhibits high-level vancomycin
resistance (MIC > 256 g/mL) but remains susceptible to daptomycin. Whole-genome
sequencing reveals a vanA operon on a transposon. Which molecular mechanism best
explains the initial acquisition of this resistance phenotype?
A. Conjugation-mediated transfer of the Tn1546 element from a co-infecting strain
B. Spontaneous point mutations in the chromosomal pbp5 gene
C. Transformation with naked DNA released by lysed vancomycin-resistant
staphylococci
D. Transduction by a bacteriophage carrying the vanA gene cluster
Correct Answer: A. Conjugation-mediated transfer of the Tn1546 element from a
co-infecting strain
Rationale: Vancomycin resistance in enterococci is typically acquired via conjugation, as
the vanA operon resides on transposon Tn1546, which is often part of a conjugative
plasmid. This allows efficient transfer between enterococcal strains. Point mutations in
pbp5 confer -lactam resistance, not vancomycin resistance. Transformation and
transduction are less likely mechanisms for this specific resistance spread in clinical
settings.
Why Wrong:
B - Mutations in pbp5 affect penicillin-binding proteins and are associated with
ampicillin resistance, not vancomycin resistance.
C - Transformation with naked DNA is possible but not the primary mechanism for
vanA dissemination; the operon is typically transferred via conjugation.
D - Transduction by bacteriophages can transfer genetic material, but vanA is
predominantly spread by conjugative plasmids in enterococci.
Reference: Murray, P.R., Rosenthal, K.S., & Pfaller, M.A. (2026). Medical Microbiology,
10th Ed., Ch. 21.
Q2. In a chemostat culture, a bacterial population is maintained at a dilution rate (D)
of 0.2 h¹. The substrate concentration in the reservoir is 10 mM, and the Monod
constant (K_s) is 0.5 mM. If the maximum specific growth rate (_max) is 0.8 h¹, what
is the steady-state substrate concentration in the culture vessel?
A. 0.17 mM
B. 0.50 mM
C. 0.83 mM
D. 1.67 mM
Correct Answer: A. 0.17 mM
Rationale: At steady state in a chemostat, D = ¼. Using the Monod equation, D = ¼_max *
S / (K_s + S). Solving for S: S = D * K_s / (_max - D) = (0.2 * 0.5) / (0.8 - 0.2) = 0..6
= 0.1667 mM, approximately 0.17 mM. The other options arise from misapplying the
formula or confusing K_s with substrate concentration.
Page 2
,Why Wrong:
B - 0.50 mM is the K_s value, not the steady-state substrate concentration.
C - 0.83 mM would result from incorrectly using S = K_s * (_max - D) / D.
D - 1.67 mM would result from using S = D * _max / K_s, which is not the correct
rearrangement.
Reference: Madigan, M.T., Bender, K.S., Buckley, D.H., Sattley, W.M., & Stahl, D.A.
(2024). Brock Biology of Microorganisms, 16th Ed., Ch. 5.
Q3. A patient develops a severe intestinal infection after antibiotic therapy. Stool
analysis reveals abundant Gram-positive rods that produce toxins. The organism is
strictly anaerobic and forms spores. Which virulence factor is most directly
responsible for the pseudomembranous colitis observed in this patient?
A. Toxin A (enterotoxin)
B. Toxin B (cytotoxin)
C. Binary toxin (CDT)
D. Flagellar motility
Correct Answer: B. Toxin B (cytotoxin)
Rationale: Clostridioides difficile toxin B (TcdB) is the primary cytotoxin that causes the
characteristic pseudomembranous colitis by disrupting the actin cytoskeleton of colonic
epithelial cells, leading to cell death and inflammation. Toxin A (TcdA) is also an
enterotoxin but is less potent in causing severe disease in humans. Binary toxin (CDT) is
an accessory toxin that may enhance virulence but is not essential. Flagellar motility is not
directly responsible for tissue damage.
Why Wrong:
A - Toxin A is an enterotoxin that contributes to disease but is not the primary driver
of pseudomembranous colitis; toxin B is more potent.
C - Binary toxin (CDT) is produced by some strains but is not the main cause of
pseudomembranous colitis; it may act as an adjuvant.
D - Flagellar motility aids in colonization but does not directly cause the cytopathic
effects seen in pseudomembranous colitis.
Reference: Carroll, K.C., Morse, S.A., Mietzner, T.A., & Miller, S. (2023). Jawetz, Melnick,
& Adelberg's Medical Microbiology, 28th Ed., Ch. 17.
Q4. A researcher is studying the regulation of the lac operon in Escherichia coli. They
observe that in a strain with a mutation in the crp gene (encoding CAP), the
expression of -galactosidase is very low even in the presence of lactose and absence of
glucose. Which of the following best explains this observation?
A. CAP is required for RNA polymerase to bind the lac promoter efficiently
B. The mutation causes constitutive expression of the lac repressor
C. cAMP levels are insufficient to activate CAP
Page 3
, D. The lac operator is mutated, preventing repressor binding
Correct Answer: A. CAP is required for RNA polymerase to bind the lac promoter
efficiently
Rationale: The lac operon is under positive control by CAP (catabolite activator protein),
which binds to the CAP site upstream of the promoter when cAMP levels are high. This
binding facilitates RNA polymerase binding and transcription initiation. In a crp mutant,
CAP is nonfunctional, so even in the presence of lactose (which inactivates the repressor)
and absence of glucose (which ensures high cAMP), transcription is severely impaired
because RNA polymerase cannot efficiently bind the promoter. The other options do not
directly explain the low expression in a crp mutant.
Why Wrong:
B - A mutation in crp does not affect the lac repressor; the repressor is still functional
and would be inactivated by lactose.
C - cAMP levels are unaffected by the crp mutation; the problem is the lack of
functional CAP.
D - The lac operator mutation would affect repressor binding, but here the operator is
normal; the issue is CAP.
Reference: Snyder, L., Peters, J.E., Henkin, T.M., & Champness, W. (2020). Molecular
Genetics of Bacteria, 5th Ed., Ch. 10.
Q5. A researcher is investigating the microbial community of a deep-sea
hydrothermal vent. They discover a novel archaeon that thrives at 110°C and pH 4.5.
Which of the following adaptations is most critical for the stability of its cytoplasmic
membrane at this extreme temperature?
A. High content of ether-linked isoprenoid lipids
B. High proportion of unsaturated fatty acids
C. Thick peptidoglycan layer
D. Production of heat-shock proteins
Correct Answer: A. High content of ether-linked isoprenoid lipids
Rationale: Archaea, including hyperthermophiles, have membranes composed of
ether-linked isoprenoid lipids, which are more stable at high temperatures than the
ester-linked fatty acids found in bacteria and eukaryotes. The ether bonds and isoprenoid
chains resist hydrolysis and maintain membrane integrity under extreme heat. Unsaturated
fatty acids would increase fluidity and decrease stability at high temperatures.
Peptidoglycan is not present in archaea (they have pseudopeptidoglycan or other cell wall
components). Heat-shock proteins are involved in protein folding, not membrane stability.
Why Wrong:
B - Unsaturated fatty acids increase membrane fluidity, which would be detrimental at
high temperatures; hyperthermophiles have saturated or isoprenoid lipids.
C - Peptidoglycan is not a component of archaeal cell walls; they have other structural
polymers.
Page 4
