BIO 171 Module 6 Review Exam Microbial Genetics & Biotechnology Actual Exam
2026/2027 – Complete 35 Questions with Detailed Rationales | 100% Verified | Pass
Guaranteed – A+ Graded
SECTION A: Microbial Genetics – DNA Replication & Mutation (Q1–Q7)
Q1: During DNA replication, which enzyme is responsible for unwinding and separating
the two parental strands of the double helix?
A. DNA ligase
B. DNA polymerase III
C. Helicase [CORRECT]
D. RNA primase
Correct Answer: C
Rationale: Helicase breaks the hydrogen bonds between complementary bases to
unwind the parental double helix at the replication fork. DNA ligase seals nicks between
fragments, polymerase III synthesizes new DNA, and primase lays down RNA primers.
Q2: While reviewing a diagram of the E. coli replication fork in lab, a student notes short,
discontinuous DNA segments being synthesized on one strand in the direction away
from the fork. These segments, which are later joined into a continuous strand, are
known as:
A. RNA primers
B. Okazaki fragments [CORRECT]
C. Replication bubbles
D. Leading strand segments
Correct Answer: B
Rationale: Okazaki fragments are short pieces of DNA synthesized discontinuously on
the lagging strand and later sealed by DNA ligase. RNA primers are removed before
ligation, and replication bubbles refer to regions of unwinding, not lagging-strand
products.
Q3: A point mutation that changes a codon specifying one amino acid into a codon
specifying a different amino acid is best classified as a:
A. Silent mutation
B. Nonsense mutation
C. Missense mutation [CORRECT]
, D. Back mutation
Correct Answer: C
Rationale: Missense mutations alter a codon so a different amino acid is incorporated
into the protein. Silent mutations leave the amino acid unchanged, nonsense mutations
create a premature stop codon, and back mutations revert to the original sequence.
Q4: A culture of bacteria exposed to a chemical mutagen acquires a single-base
insertion in the middle of a gene encoding a metabolic enzyme. What is the most likely
consequence for the protein produced?
A. The protein will be unaffected because the genetic code is redundant
B. A frameshift will alter every codon downstream, likely producing a nonfunctional
protein [CORRECT]
C. Only the single amino acid at the insertion site will change
D. The insertion will always produce an immediate stop codon
Correct Answer: B
Rationale: Insertions shift the reading frame, changing all downstream codons and
usually generating a nonfunctional or truncated protein. Codon redundancy only
protects against some base substitutions, not insertions or deletions.
Q5: A petri plate of bacteria is briefly placed under a UV germicidal lamp; colonies later
show an elevated mutation rate. Which DNA lesion is UV radiation most
characteristically responsible for?
A. Double-strand breaks
B. Insertion of base analogs
C. Spontaneous deamination
D. Thymine (pyrimidine) dimers [CORRECT]
Correct Answer: D
Rationale: UV light covalently links adjacent thymine bases into thymine dimers, which
distort the DNA helix and cause replication errors. Double-strand breaks are typical of
ionizing radiation, and base analogs and deamination are chemical mechanisms.
Q6: E. coli cells are grown in medium containing abundant glucose but no lactose.
Which statement best describes the state of the lac operon in these cells, and why?
A. The operon is OFF because the repressor protein remains bound to the operator in
the absence of lactose [CORRECT]
B. The operon is ON because glucose directly activates CAP
C. The operon is constitutively expressed because RNA polymerase ignores operators
D. The operon is ON because glucose is converted into allolactose
Correct Answer: A
2026/2027 – Complete 35 Questions with Detailed Rationales | 100% Verified | Pass
Guaranteed – A+ Graded
SECTION A: Microbial Genetics – DNA Replication & Mutation (Q1–Q7)
Q1: During DNA replication, which enzyme is responsible for unwinding and separating
the two parental strands of the double helix?
A. DNA ligase
B. DNA polymerase III
C. Helicase [CORRECT]
D. RNA primase
Correct Answer: C
Rationale: Helicase breaks the hydrogen bonds between complementary bases to
unwind the parental double helix at the replication fork. DNA ligase seals nicks between
fragments, polymerase III synthesizes new DNA, and primase lays down RNA primers.
Q2: While reviewing a diagram of the E. coli replication fork in lab, a student notes short,
discontinuous DNA segments being synthesized on one strand in the direction away
from the fork. These segments, which are later joined into a continuous strand, are
known as:
A. RNA primers
B. Okazaki fragments [CORRECT]
C. Replication bubbles
D. Leading strand segments
Correct Answer: B
Rationale: Okazaki fragments are short pieces of DNA synthesized discontinuously on
the lagging strand and later sealed by DNA ligase. RNA primers are removed before
ligation, and replication bubbles refer to regions of unwinding, not lagging-strand
products.
Q3: A point mutation that changes a codon specifying one amino acid into a codon
specifying a different amino acid is best classified as a:
A. Silent mutation
B. Nonsense mutation
C. Missense mutation [CORRECT]
, D. Back mutation
Correct Answer: C
Rationale: Missense mutations alter a codon so a different amino acid is incorporated
into the protein. Silent mutations leave the amino acid unchanged, nonsense mutations
create a premature stop codon, and back mutations revert to the original sequence.
Q4: A culture of bacteria exposed to a chemical mutagen acquires a single-base
insertion in the middle of a gene encoding a metabolic enzyme. What is the most likely
consequence for the protein produced?
A. The protein will be unaffected because the genetic code is redundant
B. A frameshift will alter every codon downstream, likely producing a nonfunctional
protein [CORRECT]
C. Only the single amino acid at the insertion site will change
D. The insertion will always produce an immediate stop codon
Correct Answer: B
Rationale: Insertions shift the reading frame, changing all downstream codons and
usually generating a nonfunctional or truncated protein. Codon redundancy only
protects against some base substitutions, not insertions or deletions.
Q5: A petri plate of bacteria is briefly placed under a UV germicidal lamp; colonies later
show an elevated mutation rate. Which DNA lesion is UV radiation most
characteristically responsible for?
A. Double-strand breaks
B. Insertion of base analogs
C. Spontaneous deamination
D. Thymine (pyrimidine) dimers [CORRECT]
Correct Answer: D
Rationale: UV light covalently links adjacent thymine bases into thymine dimers, which
distort the DNA helix and cause replication errors. Double-strand breaks are typical of
ionizing radiation, and base analogs and deamination are chemical mechanisms.
Q6: E. coli cells are grown in medium containing abundant glucose but no lactose.
Which statement best describes the state of the lac operon in these cells, and why?
A. The operon is OFF because the repressor protein remains bound to the operator in
the absence of lactose [CORRECT]
B. The operon is ON because glucose directly activates CAP
C. The operon is constitutively expressed because RNA polymerase ignores operators
D. The operon is ON because glucose is converted into allolactose
Correct Answer: A
