BIOD 210
OBJECTIVE ASSESSMENT - EXAM
BIOD 210 Genetics Final Exam
Questions and Answers
(2026/2027) (Verified Answers)
2026/2027
Genetics Final Exam
75 100% 2026/2027
QUESTIONS VERIFIED ANSWERS EDITION
TOPICS COVERED
DNA Structure, Replication & Repair Population Genetics & Hardy-Weinberg
Mechanisms Equilibrium
Mendelian Inheritance & Pedigree Analysis Gene Expression, Mutation & Epigenetics
Chromosome Structure, Mapping & Genetic Disorders, CRISPR & Biotech
Recombination Applications
COVER PAGE - 1
, SECTION 1 | Molecular Genetics & DNA Replication | Q1-Q15 | BIOD 210 Genetics Final Exam Questions and Answer
Q1 Question 1 of 75
Q1. A 24-year-old laboratory technician accidentally spills a small amount of ethidium
bromide on a gel containing purified human DNA. She recalls that this intercalating
agent inserts itself between adjacent base pairs. Which cellular consequence would
most likely result from prolonged exposure of living cells to this compound?
A. Increased rate of silent mutations in coding regions
B. Inhibition of DNA polymerase III proofreading activity
C. Frameshift mutations due to DNA unwinding and replication errors
D. Activation of telomerase reverse transcriptase
Correct Answer: C
Rationale:
Ethidium bromide intercalates between base pairs, distorting the DNA helix and causing unwinding during
replication. This leads to insertions or deletions (frameshift mutations) as DNA polymerase slips on the
distorted template. Choice A is incorrect because silent mutations are point mutations, not caused by
intercalation. Choice B is wrong because proofreading is a separate exonuclease function. Choice D is
unrelated to intercalating agents.
Q2 Question 2 of 75
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 2 of 45
, Q2. During a replication fork assay, a researcher observes that the leading strand is
synthesized continuously while the lagging strand forms short Okazaki fragments. In E.
coli, which enzyme is directly responsible for removing the RNA primers from these
fragments before ligation?
A. DNA polymerase I
B. DNA polymerase III
C. DNA ligase
D. Primase
Correct Answer: A
Rationale:
DNA polymerase I possesses 5' to 3' exonuclease activity that removes RNA primers from Okazaki fragments
and simultaneously fills the gap with DNA nucleotides. DNA polymerase III (B) is the main replicative
polymerase but lacks this 5' exonuclease activity. DNA ligase (C) seals nicks after primer removal but does not
remove primers. Primase (D) synthesizes primers, not removes them.
Q3 Question 3 of 75
Q3. A biochemistry student is studying the mechanism of action of the antibiotic
ciprofloxacin. She learns that this drug inhibits bacterial DNA gyrase, a type II
topoisomerase. What is the primary function of DNA gyrase that is blocked by this
antibiotic, leading to bacterial cell death?
A. Unwinding of the DNA double helix ahead of the replication fork
B. Introduction of negative supercoils to relieve torsional stress during replication
C. Separation of sister chromatids during bacterial cell division
D. Repair of thymine dimers caused by UV radiation
Correct Answer: B
Rationale:
DNA gyrase introduces negative supercoils into bacterial DNA to relieve the torsional strain generated ahead of
the replication fork. Ciprofloxacin stabilizes the DNA-gyrase complex, preventing this relaxation and causing
replication fork stalling. Choice A describes helicase function. Choice C describes bacterial chromosome
segregation, not gyrase. Choice D describes photolyase or nucleotide excision repair.
Q4 Question 4 of 75
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 3 of 45
, Q4. In a clinical genetics lab, a technician is performing a PCR-based assay to detect a
point mutation in the CFTR gene. She notices that the Taq polymerase being used lacks
3' to 5' exonuclease activity. What is the most significant consequence of this
enzymatic deficiency during PCR amplification?
A. The polymerase cannot initiate synthesis without a primer
B. The error rate of DNA synthesis increases due to lack of proofreading
C. The amplification yield decreases because the enzyme dissociates rapidly
D. The reaction requires higher magnesium concentrations to proceed
Correct Answer: B
Rationale:
Taq polymerase lacks 3' to 5' exonuclease (proofreading) activity, meaning it cannot remove mismatched
nucleotides after incorporation. This results in a higher error rate (~1 in 9,000 bases) compared to high-fidelity
polymerases. Choice A is incorrect because all DNA polymerases require primers. Choice C describes
processivity issues, not proofreading. Choice D is unrelated to exonuclease activity.
Q5 Question 5 of 75
Q5. A researcher is investigating the mechanism of xeroderma pigmentosum (XP), a
disorder characterized by extreme sensitivity to UV light. She discovers that cells from
an XP patient cannot repair thymine dimers efficiently. Which DNA repair pathway is
defective in these patients?
A. Base excision repair
B. Nucleotide excision repair
C. Mismatch repair
D. Homologous recombination repair
Correct Answer: B
Rationale:
Nucleotide excision repair (NER) removes bulky DNA lesions such as thymine dimers caused by UV radiation.
XP results from mutations in NER genes (XPA through XPG). Base excision repair (A) fixes small base
modifications. Mismatch repair (C) corrects replication errors. Homologous recombination (D) repairs
double-strand breaks, not pyrimidine dimers.
Q6 Question 6 of 75
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 4 of 45
OBJECTIVE ASSESSMENT - EXAM
BIOD 210 Genetics Final Exam
Questions and Answers
(2026/2027) (Verified Answers)
2026/2027
Genetics Final Exam
75 100% 2026/2027
QUESTIONS VERIFIED ANSWERS EDITION
TOPICS COVERED
DNA Structure, Replication & Repair Population Genetics & Hardy-Weinberg
Mechanisms Equilibrium
Mendelian Inheritance & Pedigree Analysis Gene Expression, Mutation & Epigenetics
Chromosome Structure, Mapping & Genetic Disorders, CRISPR & Biotech
Recombination Applications
COVER PAGE - 1
, SECTION 1 | Molecular Genetics & DNA Replication | Q1-Q15 | BIOD 210 Genetics Final Exam Questions and Answer
Q1 Question 1 of 75
Q1. A 24-year-old laboratory technician accidentally spills a small amount of ethidium
bromide on a gel containing purified human DNA. She recalls that this intercalating
agent inserts itself between adjacent base pairs. Which cellular consequence would
most likely result from prolonged exposure of living cells to this compound?
A. Increased rate of silent mutations in coding regions
B. Inhibition of DNA polymerase III proofreading activity
C. Frameshift mutations due to DNA unwinding and replication errors
D. Activation of telomerase reverse transcriptase
Correct Answer: C
Rationale:
Ethidium bromide intercalates between base pairs, distorting the DNA helix and causing unwinding during
replication. This leads to insertions or deletions (frameshift mutations) as DNA polymerase slips on the
distorted template. Choice A is incorrect because silent mutations are point mutations, not caused by
intercalation. Choice B is wrong because proofreading is a separate exonuclease function. Choice D is
unrelated to intercalating agents.
Q2 Question 2 of 75
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 2 of 45
, Q2. During a replication fork assay, a researcher observes that the leading strand is
synthesized continuously while the lagging strand forms short Okazaki fragments. In E.
coli, which enzyme is directly responsible for removing the RNA primers from these
fragments before ligation?
A. DNA polymerase I
B. DNA polymerase III
C. DNA ligase
D. Primase
Correct Answer: A
Rationale:
DNA polymerase I possesses 5' to 3' exonuclease activity that removes RNA primers from Okazaki fragments
and simultaneously fills the gap with DNA nucleotides. DNA polymerase III (B) is the main replicative
polymerase but lacks this 5' exonuclease activity. DNA ligase (C) seals nicks after primer removal but does not
remove primers. Primase (D) synthesizes primers, not removes them.
Q3 Question 3 of 75
Q3. A biochemistry student is studying the mechanism of action of the antibiotic
ciprofloxacin. She learns that this drug inhibits bacterial DNA gyrase, a type II
topoisomerase. What is the primary function of DNA gyrase that is blocked by this
antibiotic, leading to bacterial cell death?
A. Unwinding of the DNA double helix ahead of the replication fork
B. Introduction of negative supercoils to relieve torsional stress during replication
C. Separation of sister chromatids during bacterial cell division
D. Repair of thymine dimers caused by UV radiation
Correct Answer: B
Rationale:
DNA gyrase introduces negative supercoils into bacterial DNA to relieve the torsional strain generated ahead of
the replication fork. Ciprofloxacin stabilizes the DNA-gyrase complex, preventing this relaxation and causing
replication fork stalling. Choice A describes helicase function. Choice C describes bacterial chromosome
segregation, not gyrase. Choice D describes photolyase or nucleotide excision repair.
Q4 Question 4 of 75
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 3 of 45
, Q4. In a clinical genetics lab, a technician is performing a PCR-based assay to detect a
point mutation in the CFTR gene. She notices that the Taq polymerase being used lacks
3' to 5' exonuclease activity. What is the most significant consequence of this
enzymatic deficiency during PCR amplification?
A. The polymerase cannot initiate synthesis without a primer
B. The error rate of DNA synthesis increases due to lack of proofreading
C. The amplification yield decreases because the enzyme dissociates rapidly
D. The reaction requires higher magnesium concentrations to proceed
Correct Answer: B
Rationale:
Taq polymerase lacks 3' to 5' exonuclease (proofreading) activity, meaning it cannot remove mismatched
nucleotides after incorporation. This results in a higher error rate (~1 in 9,000 bases) compared to high-fidelity
polymerases. Choice A is incorrect because all DNA polymerases require primers. Choice C describes
processivity issues, not proofreading. Choice D is unrelated to exonuclease activity.
Q5 Question 5 of 75
Q5. A researcher is investigating the mechanism of xeroderma pigmentosum (XP), a
disorder characterized by extreme sensitivity to UV light. She discovers that cells from
an XP patient cannot repair thymine dimers efficiently. Which DNA repair pathway is
defective in these patients?
A. Base excision repair
B. Nucleotide excision repair
C. Mismatch repair
D. Homologous recombination repair
Correct Answer: B
Rationale:
Nucleotide excision repair (NER) removes bulky DNA lesions such as thymine dimers caused by UV radiation.
XP results from mutations in NER genes (XPA through XPG). Base excision repair (A) fixes small base
modifications. Mismatch repair (C) corrects replication errors. Homologous recombination (D) repairs
double-strand breaks, not pyrimidine dimers.
Q6 Question 6 of 75
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 4 of 45
