BIOD 210
OBJECTIVE ASSESSMENT - EXAM
BIOD 210 Genetics Final Exam
Questions and Answers (2026/2027)
(Verified Answers) 2026/2027
75Exam
Certification 100% 2026/2027
QUESTIONS VERIFIED ANSWERS EDITION
TOPICS COVERED
DNA Structure, Replication, and Repair Mechanisms
Chromosomal Aberrations and Genetic Mutations
Transcription, Translation, and Gene Regulation Population Genetics and Hardy-Weinberg Equilibrium
Mendelian Inheritance Patterns and Pedigree Analysis
Molecular Techniques and Genetic Engineering
COVER PAGE - 1
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 1 of 42
,ar Genetics and DNA Replication | Q1-Q15 | BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers)
Q1 Question 1 of 75
A 28-year-old patient presents with extreme sensitivity to sunlight and multiple skin
cancers. Genetic testing reveals a defect in nucleotide excision repair. Which DNA
lesion is this patient most likely unable to repair effectively?
A. Pyrimidine dimers caused by UV light
B. Double-strand breaks induced by ionizing radiation
C. Deamination of cytosine to uracil
D. Oxidative damage to guanine bases
Correct Answer: A
Rationale:
Nucleotide excision repair specifically removes bulky helix-distorting lesions such as pyrimidine dimers
caused by UV radiation. Xeroderma pigmentosum results from defects in this pathway. Double-strand
breaks are repaired by homologous recombination or non-homologous end joining, not NER.
Q2 Question 2 of 75
During S phase in a eukaryotic cell, a replication fork encounters a region of DNA
wrapped tightly around histone octamers. Which protein complex is responsible for
repositioning these nucleosomes ahead of the advancing fork?
A. Helicase
B. Topoisomerase II
C. Chromatin remodeling complex
D. Single-strand binding protein
Correct Answer: C
Rationale:
Chromatin remodeling complexes use ATP to slide, eject, or restructure nucleosomes, allowing replication
machinery access to DNA. Helicase unwinds DNA but does not reposition nucleosomes. Topoisomerase II
relieves supercoiling tension but does not move histones.
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 2 of 42
, Q3 Question 3 of 75
A researcher observes that DNA polymerase III in E. coli synthesizes DNA at
approximately 1,000 nucleotides per second with very high fidelity. Which subunit of
Pol III provides the 3' to 5' exonuclease proofreading activity?
A. Alpha subunit
B. Beta subunit
C. Epsilon subunit
D. Gamma subunit
Correct Answer: C
Rationale:
The epsilon subunit of DNA polymerase III contains the 3' to 5' exonuclease activity responsible for
proofreading. The alpha subunit carries the polymerase activity, while the beta subunit forms the
processivity clamp. The gamma subunit is part of the clamp loader complex.
Q4 Question 4 of 75
A bacterial culture is treated with an antibiotic that inhibits DNA gyrase. Which
consequence would most directly result from this inhibition during DNA replication?
A. Accumulation of positive supercoils ahead of replication forks
B. Failure to initiate RNA primer synthesis
C. Premature termination of Okazaki fragments
D. Inability to ligate the final phosphodiester bond on the lagging strand
Correct Answer: A
Rationale:
DNA gyrase (a type II topoisomerase) introduces negative supercoils to relieve positive supercoiling ahead
of replication forks. Without gyrase activity, torsional stress accumulates and halts fork progression. RNA
primer synthesis depends on primase, not gyrase.
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 3 of 42
, Q5 Question 5 of 75
In an experiment measuring DNA replication in vitro, the addition of
dideoxynucleotides causes chain termination. What structural feature of
dideoxynucleotides prevents further elongation?
A. Lack of a 3' hydroxyl group on the sugar
B. Absence of a 5' phosphate group
C. Presence of a methyl group at the 2' carbon
D. Replacement of the nitrogenous base with a fluorescent analog
Correct Answer: A
Rationale:
Dideoxynucleotides lack the 3' OH group required for phosphodiester bond formation, causing chain
termination. They retain the 5' phosphate and normal bases. The 2' position is irrelevant to chain elongation
in DNA synthesis.
Q6 Question 6 of 75
A patient with Bloom syndrome shows elevated rates of sister chromatid exchange.
Which enzyme is defective in this disorder, leading to impaired resolution of DNA
replication intermediates?
A. RecQ helicase
B. DNA ligase IV
C. ATM kinase
D. FEN1 flap endonuclease
Correct Answer: A
Rationale:
Bloom syndrome results from mutations in the BLM gene encoding a RecQ helicase, which normally
resolves Holliday junctions and prevents aberrant recombination. DNA ligase IV deficiency causes ligase IV
syndrome. ATM mutations cause ataxia-telangiectasia.
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 4 of 42
OBJECTIVE ASSESSMENT - EXAM
BIOD 210 Genetics Final Exam
Questions and Answers (2026/2027)
(Verified Answers) 2026/2027
75Exam
Certification 100% 2026/2027
QUESTIONS VERIFIED ANSWERS EDITION
TOPICS COVERED
DNA Structure, Replication, and Repair Mechanisms
Chromosomal Aberrations and Genetic Mutations
Transcription, Translation, and Gene Regulation Population Genetics and Hardy-Weinberg Equilibrium
Mendelian Inheritance Patterns and Pedigree Analysis
Molecular Techniques and Genetic Engineering
COVER PAGE - 1
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 1 of 42
,ar Genetics and DNA Replication | Q1-Q15 | BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers)
Q1 Question 1 of 75
A 28-year-old patient presents with extreme sensitivity to sunlight and multiple skin
cancers. Genetic testing reveals a defect in nucleotide excision repair. Which DNA
lesion is this patient most likely unable to repair effectively?
A. Pyrimidine dimers caused by UV light
B. Double-strand breaks induced by ionizing radiation
C. Deamination of cytosine to uracil
D. Oxidative damage to guanine bases
Correct Answer: A
Rationale:
Nucleotide excision repair specifically removes bulky helix-distorting lesions such as pyrimidine dimers
caused by UV radiation. Xeroderma pigmentosum results from defects in this pathway. Double-strand
breaks are repaired by homologous recombination or non-homologous end joining, not NER.
Q2 Question 2 of 75
During S phase in a eukaryotic cell, a replication fork encounters a region of DNA
wrapped tightly around histone octamers. Which protein complex is responsible for
repositioning these nucleosomes ahead of the advancing fork?
A. Helicase
B. Topoisomerase II
C. Chromatin remodeling complex
D. Single-strand binding protein
Correct Answer: C
Rationale:
Chromatin remodeling complexes use ATP to slide, eject, or restructure nucleosomes, allowing replication
machinery access to DNA. Helicase unwinds DNA but does not reposition nucleosomes. Topoisomerase II
relieves supercoiling tension but does not move histones.
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 2 of 42
, Q3 Question 3 of 75
A researcher observes that DNA polymerase III in E. coli synthesizes DNA at
approximately 1,000 nucleotides per second with very high fidelity. Which subunit of
Pol III provides the 3' to 5' exonuclease proofreading activity?
A. Alpha subunit
B. Beta subunit
C. Epsilon subunit
D. Gamma subunit
Correct Answer: C
Rationale:
The epsilon subunit of DNA polymerase III contains the 3' to 5' exonuclease activity responsible for
proofreading. The alpha subunit carries the polymerase activity, while the beta subunit forms the
processivity clamp. The gamma subunit is part of the clamp loader complex.
Q4 Question 4 of 75
A bacterial culture is treated with an antibiotic that inhibits DNA gyrase. Which
consequence would most directly result from this inhibition during DNA replication?
A. Accumulation of positive supercoils ahead of replication forks
B. Failure to initiate RNA primer synthesis
C. Premature termination of Okazaki fragments
D. Inability to ligate the final phosphodiester bond on the lagging strand
Correct Answer: A
Rationale:
DNA gyrase (a type II topoisomerase) introduces negative supercoils to relieve positive supercoiling ahead
of replication forks. Without gyrase activity, torsional stress accumulates and halts fork progression. RNA
primer synthesis depends on primase, not gyrase.
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 3 of 42
, Q5 Question 5 of 75
In an experiment measuring DNA replication in vitro, the addition of
dideoxynucleotides causes chain termination. What structural feature of
dideoxynucleotides prevents further elongation?
A. Lack of a 3' hydroxyl group on the sugar
B. Absence of a 5' phosphate group
C. Presence of a methyl group at the 2' carbon
D. Replacement of the nitrogenous base with a fluorescent analog
Correct Answer: A
Rationale:
Dideoxynucleotides lack the 3' OH group required for phosphodiester bond formation, causing chain
termination. They retain the 5' phosphate and normal bases. The 2' position is irrelevant to chain elongation
in DNA synthesis.
Q6 Question 6 of 75
A patient with Bloom syndrome shows elevated rates of sister chromatid exchange.
Which enzyme is defective in this disorder, leading to impaired resolution of DNA
replication intermediates?
A. RecQ helicase
B. DNA ligase IV
C. ATM kinase
D. FEN1 flap endonuclease
Correct Answer: A
Rationale:
Bloom syndrome results from mutations in the BLM gene encoding a RecQ helicase, which normally
resolves Holliday junctions and prevents aberrant recombination. DNA ligase IV deficiency causes ligase IV
syndrome. ATM mutations cause ataxia-telangiectasia.
BIOD 210 Genetics Final Exam Questions and Answers (2026/2027) (Verified Answers) 2026/2027 - 2026/2027 | Passing Score: 80% | Page 4 of 42
