HESI A2 Biology Exam Questions
(Latest 2026/2027 Update) Complete
Q&A with Verified Answers and
Detailed Rationales Cell Biology,
HESI A2 Biology Exam Questions (Latest 2026/2027 |
Comprehensive Examination
HESI A2 Biology Exam Questions (Latest 2026/2027 Update)
Complete Q&A with Verified Answers and Detailed
Rationales Cell Biology, Genetics, Photosynthesis A+ Graded
INTRODUCTION
This document is an important and very helpful preparation
resource for HESI A2 Biology Exam Questions (Latest
2026/2027 Update) Complete Q&A with Verified Answers and
Detailed Rationales Cell Biology, Genetics, Photosynthesis A+
Graded. Every item uses a focused exam stem, four response
options, a keyed response, and a clear rationale, making it
easier to understand the subject, spot distractors, and build real
exam confidence.
Page 1
, Question 1
In an experiment, a cell is treated with a drug that inhibits the Na+/K+ ATPase.
Which of the following would be the most immediate consequence for
secondary active transport of glucose in the intestinal epithelium?
A. Increased intracellular Na+ concentration, reducing the driving force
for glucose uptake.
B. Decreased intracellular K+ concentration, directly impairing glucose
transporter affinity.
C. Enhanced glucose absorption due to compensatory ATP production.
D. No effect, because glucose transport is independent of ion gradients.
Correct Answer: A - Increased intracellular Na+ concentration,
reducing the driving force for glucose uptake.
RATIONALE
Inhibition of Na+/K+ ATPase leads to intracellular Na+ accumulation,
which dissipates the Na+ gradient necessary for secondary active
transport of glucose via SGLT1. This reduces the driving force for
glucose uptake. Option B is incorrect because K+ concentration does
not directly affect glucose transporter affinity. Option C is incorrect as
ATP production would not be enhanced; inhibition would decrease it.
Option D is incorrect because glucose transport in the intestine is
Na+-dependent.
Question 2
A dihybrid cross is performed between two heterozygous individuals (AaBb).
If the genes are completely linked and no crossing over occurs, what
phenotypic ratio is expected in the offspring?
A. 9:3:3:1
B. 1:1:1:1
C. 3:1
D. 1:2:1
Correct Answer: C - 3:1
Page 2
, RATIONALE
With complete linkage and no crossing over, the parental gametes (AB
and ab) are produced in equal frequencies, resulting in offspring with
phenotypes AB and ab in a 3:1 ratio. Option A is the ratio for
independent assortment. Option B is the ratio for a testcross with
independent assortment. Option D is for a monohybrid cross with
incomplete dominance.
Question 3
In the light-dependent reactions of photosynthesis, the proton motive force
across the thylakoid membrane is primarily used for which purpose?
A. Direct synthesis of glucose via the Calvin cycle.
B. ATP synthesis by ATP synthase.
C. Reduction of NADP+ to NADPH.
D. Splitting of water to release oxygen.
Correct Answer: B - ATP synthesis by ATP synthase.
RATIONALE
The proton motive force generated by electron transport drives ATP
synthesis via ATP synthase (chemiosmosis). Option A is incorrect
because glucose synthesis occurs in the Calvin cycle, which uses ATP
and NADPH but is not directly driven by the proton motive force.
Option C is incorrect because NADP+ reduction occurs at the end of
the electron transport chain, not directly by the proton motive force.
Option D is incorrect because water splitting is carried out by the
oxygen-evolving complex of photosystem II.
Question 4
During glycolysis, the conversion of glucose to pyruvate yields a net gain of
ATP via which mechanism?
A. Oxidative phosphorylation
B. Substrate-level phosphorylation
C. Photophosphorylation
Page 3
, D. Chemiosmosis
Correct Answer: B - Substrate-level phosphorylation
RATIONALE
Glycolysis produces ATP through substrate-level phosphorylation,
where a phosphate group is directly transferred from a substrate to
ADP. Oxidative phosphorylation occurs in the mitochondria during the
electron transport chain. Photophosphorylation occurs in
photosynthesis. Chemiosmosis is the mechanism for ATP synthesis in
oxidative and photophosphorylation, not directly in glycolysis.
Question 5
A mutation in the gene encoding the CFTR chloride channel leads to cystic
fibrosis. Which of the following best describes the molecular consequence of
the most common mutation, F508?
A. Premature stop codon leading to truncated protein.
B. Deletion of a single amino acid causing misfolding and degradation.
C. Duplication of a codon leading to elongated protein.
D. Point mutation causing altered ion selectivity.
Correct Answer: B - Deletion of a single amino acid causing
misfolding and degradation.
RATIONALE
The F508 mutation is a deletion of three nucleotides, resulting in the
loss of phenylalanine at position 508. This causes protein misfolding
and premature degradation. Option A describes a nonsense mutation.
Option C describes a frameshift or insertion. Option D describes a
missense mutation affecting function rather than folding.
Page 4
(Latest 2026/2027 Update) Complete
Q&A with Verified Answers and
Detailed Rationales Cell Biology,
HESI A2 Biology Exam Questions (Latest 2026/2027 |
Comprehensive Examination
HESI A2 Biology Exam Questions (Latest 2026/2027 Update)
Complete Q&A with Verified Answers and Detailed
Rationales Cell Biology, Genetics, Photosynthesis A+ Graded
INTRODUCTION
This document is an important and very helpful preparation
resource for HESI A2 Biology Exam Questions (Latest
2026/2027 Update) Complete Q&A with Verified Answers and
Detailed Rationales Cell Biology, Genetics, Photosynthesis A+
Graded. Every item uses a focused exam stem, four response
options, a keyed response, and a clear rationale, making it
easier to understand the subject, spot distractors, and build real
exam confidence.
Page 1
, Question 1
In an experiment, a cell is treated with a drug that inhibits the Na+/K+ ATPase.
Which of the following would be the most immediate consequence for
secondary active transport of glucose in the intestinal epithelium?
A. Increased intracellular Na+ concentration, reducing the driving force
for glucose uptake.
B. Decreased intracellular K+ concentration, directly impairing glucose
transporter affinity.
C. Enhanced glucose absorption due to compensatory ATP production.
D. No effect, because glucose transport is independent of ion gradients.
Correct Answer: A - Increased intracellular Na+ concentration,
reducing the driving force for glucose uptake.
RATIONALE
Inhibition of Na+/K+ ATPase leads to intracellular Na+ accumulation,
which dissipates the Na+ gradient necessary for secondary active
transport of glucose via SGLT1. This reduces the driving force for
glucose uptake. Option B is incorrect because K+ concentration does
not directly affect glucose transporter affinity. Option C is incorrect as
ATP production would not be enhanced; inhibition would decrease it.
Option D is incorrect because glucose transport in the intestine is
Na+-dependent.
Question 2
A dihybrid cross is performed between two heterozygous individuals (AaBb).
If the genes are completely linked and no crossing over occurs, what
phenotypic ratio is expected in the offspring?
A. 9:3:3:1
B. 1:1:1:1
C. 3:1
D. 1:2:1
Correct Answer: C - 3:1
Page 2
, RATIONALE
With complete linkage and no crossing over, the parental gametes (AB
and ab) are produced in equal frequencies, resulting in offspring with
phenotypes AB and ab in a 3:1 ratio. Option A is the ratio for
independent assortment. Option B is the ratio for a testcross with
independent assortment. Option D is for a monohybrid cross with
incomplete dominance.
Question 3
In the light-dependent reactions of photosynthesis, the proton motive force
across the thylakoid membrane is primarily used for which purpose?
A. Direct synthesis of glucose via the Calvin cycle.
B. ATP synthesis by ATP synthase.
C. Reduction of NADP+ to NADPH.
D. Splitting of water to release oxygen.
Correct Answer: B - ATP synthesis by ATP synthase.
RATIONALE
The proton motive force generated by electron transport drives ATP
synthesis via ATP synthase (chemiosmosis). Option A is incorrect
because glucose synthesis occurs in the Calvin cycle, which uses ATP
and NADPH but is not directly driven by the proton motive force.
Option C is incorrect because NADP+ reduction occurs at the end of
the electron transport chain, not directly by the proton motive force.
Option D is incorrect because water splitting is carried out by the
oxygen-evolving complex of photosystem II.
Question 4
During glycolysis, the conversion of glucose to pyruvate yields a net gain of
ATP via which mechanism?
A. Oxidative phosphorylation
B. Substrate-level phosphorylation
C. Photophosphorylation
Page 3
, D. Chemiosmosis
Correct Answer: B - Substrate-level phosphorylation
RATIONALE
Glycolysis produces ATP through substrate-level phosphorylation,
where a phosphate group is directly transferred from a substrate to
ADP. Oxidative phosphorylation occurs in the mitochondria during the
electron transport chain. Photophosphorylation occurs in
photosynthesis. Chemiosmosis is the mechanism for ATP synthesis in
oxidative and photophosphorylation, not directly in glycolysis.
Question 5
A mutation in the gene encoding the CFTR chloride channel leads to cystic
fibrosis. Which of the following best describes the molecular consequence of
the most common mutation, F508?
A. Premature stop codon leading to truncated protein.
B. Deletion of a single amino acid causing misfolding and degradation.
C. Duplication of a codon leading to elongated protein.
D. Point mutation causing altered ion selectivity.
Correct Answer: B - Deletion of a single amino acid causing
misfolding and degradation.
RATIONALE
The F508 mutation is a deletion of three nucleotides, resulting in the
loss of phenylalanine at position 508. This causes protein misfolding
and premature degradation. Option A describes a nonsense mutation.
Option C describes a frameshift or insertion. Option D describes a
missense mutation affecting function rather than folding.
Page 4
