BIOLOGY 251 LECTURE EXAM 3 | FULL QUESTIONS,
CORRECT ANSWERS, AND RATIONALES | 2026/27
UPDATE | 100% CORRECT - WSU.
116 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
BIOLOGY 251 LECTURE EXAM 3 | FULL QUESTIONS, CORRECT ANSWERS, AND RATIONALES | 2026/27
UPDATE | 100% CORRECT - WSU.. It contains 116 carefully selected questions that reflect the most current
exam content and testing strategies. Each question is accompanied by a correct answer and a detailed rationale
that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 116 Questions
Foundations - Application - Biology 251 Lecture 3 FULL Correct AND Rationales 2026/27 Update 100
Correct WSU Biology 251 CELL AND Molecular Biology Undergraduate YEAR 3 / Upper Division
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Nervous System Structure 1-20 Likely, Cells, Directly, Binds, Mutation
AND Function
Sensory Systems AND 21-40 Likely, Factor, Protein, Cells, Effect
Perception
Motor Systems AND Muscle 41-60 Protein, Mutation, Likely, Researcher, Mutant
Contraction
Endocrine System Hormones 61-80 Selection, Researcher, Transcription, Likely, Experimental
AND Glands
Cardiovascular System Heart 81-100 Researcher, Directly, Identify, Synaptic, Sequence
AND Blood Vessels
Respiratory System GAS 101-116 STOP Codon, Likely, Explains, Identifies, Factor
Exchange AND Regulation
TOTAL 116 All questions include answers and detailed rationales
,Section A - Nervous System Structure AND Function
Q1.
A novel kinase inhibitor is found to block EGF-induced cell proliferation. In cells treated
with the inhibitor, EGF still triggers receptor dimerization and trans-autophosphorylation,
but downstream ERK phosphorylation is abolished. Which molecular event is most likely
directly inhibited?
A. GRB2 SH2 domain binding to the B. EGFR tyrosine kinase activity
receptor
C. Ras GTPase activity D. MEK phosphorylation by Raf
Correct: A - GRB2 SH2 domain binding to the receptor
Rationale:Since receptor autophosphorylation is intact, the inhibitor likely blocks the
interaction between the phosphotyrosine residues and the SH2 domain of GRB2, preventing
Ras activation. EGFR kinase activity (B) would impair autophosphorylation. Ras GTPase
activity (C) would not be directly affected by a GRB2 inhibitor. MEK phosphorylation (D)
occurs downstream of Ras and would be blocked only if Raf or MEK were inhibited.
Q2.
In an in vitro motility assay, kinesin-1 is observed to move microtubules at a rate of 0.8
m/s. When the ATP concentration is reduced to 1 M, the velocity decreases to 0.2 m/s, but
the microtubules remain attached. Which property of kinesin's mechanochemical cycle
does this best illustrate?
A. Processivity is ATP-dependent B. The duty ratio is low
C. The power stroke is independent of ATP D. The ATPase rate limits the stepping cycle
hydrolysis
Correct: D - The ATPase rate limits the stepping cycle
Rationale:Velocity of motor proteins is proportional to the rate of ATP hydrolysis; at low ATP,
the waiting time for ATP binding becomes rate-limiting, slowing stepping. Processivity (A) is
not directly indicated by velocity change; attachment persists because kinesin remains bound.
Duty ratio (B) is a measure of fraction of time attached, not velocity. Power stroke (C) is
coupled to ATP hydrolysis and release of products.
Q3.
During apoptosis, cytochrome c released from mitochondria binds Apaf-1, triggering
apoptosome formation. Which of the following best explains why this event alone does
not guarantee cell death in all cell types?
Page 3
, Section A - Nervous System Structure AND Function
A. Caspase-9 must be cleaved by B. XIAP can bind and inhibit caspase-9,
caspase-3 to become active blocking the cascade
C. Apaf-1 requires ATP hydrolysis to D. Cytochrome c is rapidly degraded by the
oligomerize ubiquitin-proteasome pathway
Correct: B - XIAP can bind and inhibit caspase-9, blocking the cascade
Rationale:XIAP (X-linked inhibitor of apoptosis) directly inhibits caspase-9 and caspase-3,
providing a brake on the intrinsic pathway. Caspase-9 is activated by conformational change
in the apoptosome, not by cleavage (A). ATP is required for apoptosome function but is
generally available (C). Cytochrome c is not degraded by proteasome in the cytosol (D).
Q4.
A tumor suppressor gene (TSG) shows loss of heterozygosity (LOH) in a sporadic cancer.
The remaining allele carries a missense mutation that abolishes its function. Which of the
following additional events is most likely to have occurred in the same cell?
A. A gain-of-function mutation in an B. Silencing of the TSG promoter by
oncogene hypermethylation
C. Amplification of a proto-oncogene D. A chromosomal translocation creating a
fusion protein
Correct: A - A gain-of-function mutation in an oncogene
Rationale:Cancer progression typically requires both activation of oncogenes and inactivation
of tumor suppressors. The question describes a two-hit inactivation of a TSG; the presence of
an oncogenic mutation is a common cooperative event. Silencing of the TSG promoter (B)
would be an alternative second hit, but the remaining allele is already mutated. Amplification
(C) and translocation (D) are possible but not specifically linked to LOH.
Q5.
In a yeast strain with a temperature-sensitive mutation in the gene encoding separase,
what is the predicted phenotype at the restrictive temperature?
A. Cells will arrest in metaphase with sister B. Cells will undergo premature sister
chromatids still paired chromatid separation
C. Cells will fail to replicate DNA D. Cells will exit mitosis but fail cytokinesis
Correct: A - Cells will arrest in metaphase with sister chromatids still paired
Page 4
CORRECT ANSWERS, AND RATIONALES | 2026/27
UPDATE | 100% CORRECT - WSU.
116 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
BIOLOGY 251 LECTURE EXAM 3 | FULL QUESTIONS, CORRECT ANSWERS, AND RATIONALES | 2026/27
UPDATE | 100% CORRECT - WSU.. It contains 116 carefully selected questions that reflect the most current
exam content and testing strategies. Each question is accompanied by a correct answer and a detailed rationale
that explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 116 Questions
Foundations - Application - Biology 251 Lecture 3 FULL Correct AND Rationales 2026/27 Update 100
Correct WSU Biology 251 CELL AND Molecular Biology Undergraduate YEAR 3 / Upper Division
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Nervous System Structure 1-20 Likely, Cells, Directly, Binds, Mutation
AND Function
Sensory Systems AND 21-40 Likely, Factor, Protein, Cells, Effect
Perception
Motor Systems AND Muscle 41-60 Protein, Mutation, Likely, Researcher, Mutant
Contraction
Endocrine System Hormones 61-80 Selection, Researcher, Transcription, Likely, Experimental
AND Glands
Cardiovascular System Heart 81-100 Researcher, Directly, Identify, Synaptic, Sequence
AND Blood Vessels
Respiratory System GAS 101-116 STOP Codon, Likely, Explains, Identifies, Factor
Exchange AND Regulation
TOTAL 116 All questions include answers and detailed rationales
,Section A - Nervous System Structure AND Function
Q1.
A novel kinase inhibitor is found to block EGF-induced cell proliferation. In cells treated
with the inhibitor, EGF still triggers receptor dimerization and trans-autophosphorylation,
but downstream ERK phosphorylation is abolished. Which molecular event is most likely
directly inhibited?
A. GRB2 SH2 domain binding to the B. EGFR tyrosine kinase activity
receptor
C. Ras GTPase activity D. MEK phosphorylation by Raf
Correct: A - GRB2 SH2 domain binding to the receptor
Rationale:Since receptor autophosphorylation is intact, the inhibitor likely blocks the
interaction between the phosphotyrosine residues and the SH2 domain of GRB2, preventing
Ras activation. EGFR kinase activity (B) would impair autophosphorylation. Ras GTPase
activity (C) would not be directly affected by a GRB2 inhibitor. MEK phosphorylation (D)
occurs downstream of Ras and would be blocked only if Raf or MEK were inhibited.
Q2.
In an in vitro motility assay, kinesin-1 is observed to move microtubules at a rate of 0.8
m/s. When the ATP concentration is reduced to 1 M, the velocity decreases to 0.2 m/s, but
the microtubules remain attached. Which property of kinesin's mechanochemical cycle
does this best illustrate?
A. Processivity is ATP-dependent B. The duty ratio is low
C. The power stroke is independent of ATP D. The ATPase rate limits the stepping cycle
hydrolysis
Correct: D - The ATPase rate limits the stepping cycle
Rationale:Velocity of motor proteins is proportional to the rate of ATP hydrolysis; at low ATP,
the waiting time for ATP binding becomes rate-limiting, slowing stepping. Processivity (A) is
not directly indicated by velocity change; attachment persists because kinesin remains bound.
Duty ratio (B) is a measure of fraction of time attached, not velocity. Power stroke (C) is
coupled to ATP hydrolysis and release of products.
Q3.
During apoptosis, cytochrome c released from mitochondria binds Apaf-1, triggering
apoptosome formation. Which of the following best explains why this event alone does
not guarantee cell death in all cell types?
Page 3
, Section A - Nervous System Structure AND Function
A. Caspase-9 must be cleaved by B. XIAP can bind and inhibit caspase-9,
caspase-3 to become active blocking the cascade
C. Apaf-1 requires ATP hydrolysis to D. Cytochrome c is rapidly degraded by the
oligomerize ubiquitin-proteasome pathway
Correct: B - XIAP can bind and inhibit caspase-9, blocking the cascade
Rationale:XIAP (X-linked inhibitor of apoptosis) directly inhibits caspase-9 and caspase-3,
providing a brake on the intrinsic pathway. Caspase-9 is activated by conformational change
in the apoptosome, not by cleavage (A). ATP is required for apoptosome function but is
generally available (C). Cytochrome c is not degraded by proteasome in the cytosol (D).
Q4.
A tumor suppressor gene (TSG) shows loss of heterozygosity (LOH) in a sporadic cancer.
The remaining allele carries a missense mutation that abolishes its function. Which of the
following additional events is most likely to have occurred in the same cell?
A. A gain-of-function mutation in an B. Silencing of the TSG promoter by
oncogene hypermethylation
C. Amplification of a proto-oncogene D. A chromosomal translocation creating a
fusion protein
Correct: A - A gain-of-function mutation in an oncogene
Rationale:Cancer progression typically requires both activation of oncogenes and inactivation
of tumor suppressors. The question describes a two-hit inactivation of a TSG; the presence of
an oncogenic mutation is a common cooperative event. Silencing of the TSG promoter (B)
would be an alternative second hit, but the remaining allele is already mutated. Amplification
(C) and translocation (D) are possible but not specifically linked to LOH.
Q5.
In a yeast strain with a temperature-sensitive mutation in the gene encoding separase,
what is the predicted phenotype at the restrictive temperature?
A. Cells will arrest in metaphase with sister B. Cells will undergo premature sister
chromatids still paired chromatid separation
C. Cells will fail to replicate DNA D. Cells will exit mitosis but fail cytokinesis
Correct: A - Cells will arrest in metaphase with sister chromatids still paired
Page 4
