BIOD 210 Genetics Module 4 Exam – (2026) Actual Questions & Answers –
Portage Learning.
SECTION 1: DNA Structure and Replication (Questions 1-25)
Question 1
In a biochemical assay, you measure the DNA content of a rapidly dividing cell
population. After inhibiting DNA ligase, you observe an accumulation of short
DNA fragments. Which of the following best explains this observation?
A) Primase is unable to synthesize RNA primers on the lagging strand.
B) Okazaki fragments cannot be joined due to lack of phosphodiester bond
formation between adjacent fragments.
C) Helicase activity is stalled, preventing fork progression.
D) Topoisomerase fails to relieve supercoiling, causing replication fork collapse.
Answer: B
Rationale: DNA ligase catalyzes the formation of phosphodiester bonds between
Okazaki fragments on the lagging strand. Without ligase, these fragments
accumulate. Primase (A) still functions; helicase (C) and topoisomerase (D) are not
directly affected by ligase inhibition .
,Question 2
A researcher introduces a mutation into the gene encoding the sliding clamp
(PCNA) that reduces its affinity for DNA polymerase δ. Which of the following
replication defects is most likely to be observed?
A) Decreased processivity of leading strand synthesis.
B) Increased frequency of RNA primer misincorporation.
C) Failure to unwind the DNA duplex at the origin.
D) Accumulation of supercoils ahead of the replication fork.
Answer: A
Rationale: PCNA acts as a sliding clamp that tethers DNA polymerase δ to DNA,
enhancing processivity. Reduced affinity leads to frequent dissociation of the
polymerase, especially on the leading strand. Primase (B) is unaffected; helicase
(C) and topoisomerase (D) are separate functions .
Question 3
In an in vitro replication system using a circular double-stranded DNA template,
you add a specific inhibitor that blocks the 3'→5' exonuclease activity of DNA
polymerase III. Which of the following is the most immediate consequence?
,A) Increased mutation rate due to loss of proofreading.
B) Stalling of replication fork due to inability to remove RNA primers.
C) Accumulation of nicked DNA due to lack of ligation.
D) Premature termination of leading strand synthesis.
Answer: A
Rationale: The 3'→5' exonuclease activity of DNA polymerase III is responsible for
proofreading. Without it, misincorporated nucleotides are not removed, leading
to a higher mutation rate. Primer removal (B) is done by RNase H or FEN1; ligation
(C) by ligase; leading strand synthesis (D) continues but with errors .
Question 4
You are studying a mutant strain of E. coli that lacks the DnaA initiator protein.
When you attempt to replicate its chromosome in vitro, which of the following is
most likely to occur?
A) Replication initiates normally but elongation is defective.
B) The origin of replication remains bound by histones and is inaccessible.
C) No replication bubbles form because origin unwinding cannot begin.
D) Okazaki fragments are produced but remain unligated.
, Answer: C
Rationale: DnaA binds to the origin (oriC) and promotes local unwinding, allowing
helicase loading. Without DnaA, the origin remains double-stranded, and
replication cannot initiate. Elongation (A) and Okazaki fragment ligation (D) are
downstream processes; histones (B) are not present in E. coli .
Question 5
A scientist adds ethidium bromide, an intercalating agent, to a replicating DNA
sample. Which of the following effects on DNA structure is most directly
responsible for inhibiting replication?
A) It covalently crosslinks complementary strands, preventing unwinding.
B) It introduces positive supercoils by increasing the linking number.
C) It inserts between base pairs, altering the twist and reducing helical stability.
D) It cleaves the phosphodiester backbone, causing strand breaks.
Answer: C
Rationale: Ethidium bromide intercalates between base pairs, which changes the
helical twist and unwinds the DNA, making it difficult for replication machinery to
Portage Learning.
SECTION 1: DNA Structure and Replication (Questions 1-25)
Question 1
In a biochemical assay, you measure the DNA content of a rapidly dividing cell
population. After inhibiting DNA ligase, you observe an accumulation of short
DNA fragments. Which of the following best explains this observation?
A) Primase is unable to synthesize RNA primers on the lagging strand.
B) Okazaki fragments cannot be joined due to lack of phosphodiester bond
formation between adjacent fragments.
C) Helicase activity is stalled, preventing fork progression.
D) Topoisomerase fails to relieve supercoiling, causing replication fork collapse.
Answer: B
Rationale: DNA ligase catalyzes the formation of phosphodiester bonds between
Okazaki fragments on the lagging strand. Without ligase, these fragments
accumulate. Primase (A) still functions; helicase (C) and topoisomerase (D) are not
directly affected by ligase inhibition .
,Question 2
A researcher introduces a mutation into the gene encoding the sliding clamp
(PCNA) that reduces its affinity for DNA polymerase δ. Which of the following
replication defects is most likely to be observed?
A) Decreased processivity of leading strand synthesis.
B) Increased frequency of RNA primer misincorporation.
C) Failure to unwind the DNA duplex at the origin.
D) Accumulation of supercoils ahead of the replication fork.
Answer: A
Rationale: PCNA acts as a sliding clamp that tethers DNA polymerase δ to DNA,
enhancing processivity. Reduced affinity leads to frequent dissociation of the
polymerase, especially on the leading strand. Primase (B) is unaffected; helicase
(C) and topoisomerase (D) are separate functions .
Question 3
In an in vitro replication system using a circular double-stranded DNA template,
you add a specific inhibitor that blocks the 3'→5' exonuclease activity of DNA
polymerase III. Which of the following is the most immediate consequence?
,A) Increased mutation rate due to loss of proofreading.
B) Stalling of replication fork due to inability to remove RNA primers.
C) Accumulation of nicked DNA due to lack of ligation.
D) Premature termination of leading strand synthesis.
Answer: A
Rationale: The 3'→5' exonuclease activity of DNA polymerase III is responsible for
proofreading. Without it, misincorporated nucleotides are not removed, leading
to a higher mutation rate. Primer removal (B) is done by RNase H or FEN1; ligation
(C) by ligase; leading strand synthesis (D) continues but with errors .
Question 4
You are studying a mutant strain of E. coli that lacks the DnaA initiator protein.
When you attempt to replicate its chromosome in vitro, which of the following is
most likely to occur?
A) Replication initiates normally but elongation is defective.
B) The origin of replication remains bound by histones and is inaccessible.
C) No replication bubbles form because origin unwinding cannot begin.
D) Okazaki fragments are produced but remain unligated.
, Answer: C
Rationale: DnaA binds to the origin (oriC) and promotes local unwinding, allowing
helicase loading. Without DnaA, the origin remains double-stranded, and
replication cannot initiate. Elongation (A) and Okazaki fragment ligation (D) are
downstream processes; histones (B) are not present in E. coli .
Question 5
A scientist adds ethidium bromide, an intercalating agent, to a replicating DNA
sample. Which of the following effects on DNA structure is most directly
responsible for inhibiting replication?
A) It covalently crosslinks complementary strands, preventing unwinding.
B) It introduces positive supercoils by increasing the linking number.
C) It inserts between base pairs, altering the twist and reducing helical stability.
D) It cleaves the phosphodiester backbone, causing strand breaks.
Answer: C
Rationale: Ethidium bromide intercalates between base pairs, which changes the
helical twist and unwinds the DNA, making it difficult for replication machinery to
