BIO 171 Module 6 Exam Microbial Genetics & Biotechnology
Actual Exam 2026/2027 – Complete Test Bank with Detailed
Rationales | 100% Verified | Pass Guaranteed – A+ Graded
Section A: Microbial Genetics – DNA Replication & Mutation (Q1–Q7)
Q1: In semiconservative DNA replication, each resulting daughter DNA molecule
contains:
A. Two newly synthesized strands and no parental DNA
B. One parental (original) strand and one newly synthesized strand [CORRECT]
C. Two parental strands rejoined with random segments of new DNA
D. Only RNA segments copied from the template strand
Correct Answer: B
Rationale: Semiconservative replication means each daughter duplex conserves one
original template strand paired with one new complementary strand. Options A, C, and D
describe fully conservative, dispersive, or non-DNA models that were disproven by the
Meselson–Stahl experiment.
Q2: Which enzyme seals nicks in the DNA backbone by forming phosphodiester bonds
between adjacent nucleotide fragments?
A. Helicase
B. Primase
C. DNA ligase [CORRECT]
D. Topoisomerase
Correct Answer: C
Rationale: DNA ligase covalently joins DNA fragments (such as Okazaki fragments) by
catalyzing phosphodiester bond formation. Helicase unwinds the helix, primase
synthesizes RNA primers, and topoisomerase relieves supercoiling—none of these
ligate fragments.
Q3: During replication, a student observes short DNA fragments on one strand that must
later be joined together. These Okazaki fragments form on the lagging strand because:
A. DNA polymerase can only synthesize DNA in the 5′ to 3′ direction, so the lagging
strand must be built discontinuously away from the replication fork [CORRECT]
B. The lagging strand template is made of RNA and cannot be copied continuously
C. DNA ligase blocks continuous synthesis on one strand only
, D. Helicase moves in the opposite direction from DNA polymerase
Correct Answer: A
Rationale: Because all DNA polymerases synthesize exclusively 5′→3′, the strand whose
template runs toward the fork must be synthesized in short, discontinuous fragments.
RNA primers and ligase assist the process, and helicase directionality is not the cause.
Q4: A mutation changes the DNA coding sequence from TAC (mRNA AUG,
tyrosine/start region) to TAA (mRNA UAA, stop codon). This mutation is best classified
as:
A. Silent mutation
B. Missense mutation
C. Frameshift mutation
D. Nonsense mutation [CORRECT]
Correct Answer: D
Rationale: Conversion of a sense codon into a premature stop codon (UAA) is a
nonsense mutation, producing a truncated, usually nonfunctional protein. Silent
mutations leave the amino acid unchanged, missense substitutions replace one amino
acid with another, and frameshifts require insertions or deletions.
Q5: A single nucleotide is deleted near the beginning of a bacterial gene's coding region.
The most likely consequence is:
A. No change, since DNA repair will restore the exact original sequence
B. A frameshift mutation that alters every codon downstream, typically producing a
nonfunctional protein [CORRECT]
C. Only a single amino acid substitution at the deletion site
D. Immediate reversal of the mutation by DNA ligase
Correct Answer: B
Rationale: Deleting one nucleotide shifts the reading frame, changing the grouping of all
subsequent codons and usually introducing a premature stop. A single-base deletion
cannot cause merely one substitution, and ligase does not restore lost bases.
Q6: A bacterial culture is exposed to UV light, and a lab technician notes that adjacent
thymine bases on the same strand have bonded together, distorting the helix. The
primary repair pathway that excises and replaces the damaged segment is:
A. Nucleotide excision repair [CORRECT]
B. Mismatch repair
C. Base excision of a single depurinated nucleotide only
D. Proofreading by DNA polymerase III during replication
Correct Answer: A
Actual Exam 2026/2027 – Complete Test Bank with Detailed
Rationales | 100% Verified | Pass Guaranteed – A+ Graded
Section A: Microbial Genetics – DNA Replication & Mutation (Q1–Q7)
Q1: In semiconservative DNA replication, each resulting daughter DNA molecule
contains:
A. Two newly synthesized strands and no parental DNA
B. One parental (original) strand and one newly synthesized strand [CORRECT]
C. Two parental strands rejoined with random segments of new DNA
D. Only RNA segments copied from the template strand
Correct Answer: B
Rationale: Semiconservative replication means each daughter duplex conserves one
original template strand paired with one new complementary strand. Options A, C, and D
describe fully conservative, dispersive, or non-DNA models that were disproven by the
Meselson–Stahl experiment.
Q2: Which enzyme seals nicks in the DNA backbone by forming phosphodiester bonds
between adjacent nucleotide fragments?
A. Helicase
B. Primase
C. DNA ligase [CORRECT]
D. Topoisomerase
Correct Answer: C
Rationale: DNA ligase covalently joins DNA fragments (such as Okazaki fragments) by
catalyzing phosphodiester bond formation. Helicase unwinds the helix, primase
synthesizes RNA primers, and topoisomerase relieves supercoiling—none of these
ligate fragments.
Q3: During replication, a student observes short DNA fragments on one strand that must
later be joined together. These Okazaki fragments form on the lagging strand because:
A. DNA polymerase can only synthesize DNA in the 5′ to 3′ direction, so the lagging
strand must be built discontinuously away from the replication fork [CORRECT]
B. The lagging strand template is made of RNA and cannot be copied continuously
C. DNA ligase blocks continuous synthesis on one strand only
, D. Helicase moves in the opposite direction from DNA polymerase
Correct Answer: A
Rationale: Because all DNA polymerases synthesize exclusively 5′→3′, the strand whose
template runs toward the fork must be synthesized in short, discontinuous fragments.
RNA primers and ligase assist the process, and helicase directionality is not the cause.
Q4: A mutation changes the DNA coding sequence from TAC (mRNA AUG,
tyrosine/start region) to TAA (mRNA UAA, stop codon). This mutation is best classified
as:
A. Silent mutation
B. Missense mutation
C. Frameshift mutation
D. Nonsense mutation [CORRECT]
Correct Answer: D
Rationale: Conversion of a sense codon into a premature stop codon (UAA) is a
nonsense mutation, producing a truncated, usually nonfunctional protein. Silent
mutations leave the amino acid unchanged, missense substitutions replace one amino
acid with another, and frameshifts require insertions or deletions.
Q5: A single nucleotide is deleted near the beginning of a bacterial gene's coding region.
The most likely consequence is:
A. No change, since DNA repair will restore the exact original sequence
B. A frameshift mutation that alters every codon downstream, typically producing a
nonfunctional protein [CORRECT]
C. Only a single amino acid substitution at the deletion site
D. Immediate reversal of the mutation by DNA ligase
Correct Answer: B
Rationale: Deleting one nucleotide shifts the reading frame, changing the grouping of all
subsequent codons and usually introducing a premature stop. A single-base deletion
cannot cause merely one substitution, and ligase does not restore lost bases.
Q6: A bacterial culture is exposed to UV light, and a lab technician notes that adjacent
thymine bases on the same strand have bonded together, distorting the helix. The
primary repair pathway that excises and replaces the damaged segment is:
A. Nucleotide excision repair [CORRECT]
B. Mismatch repair
C. Base excision of a single depurinated nucleotide only
D. Proofreading by DNA polymerase III during replication
Correct Answer: A
