HESI A2 Biology Practice Questions
Chamberlain University (Latest
2026/2027 Update) Q&A with Verified
Answers and Detailed Rationales Cell
HESI A2 Biology Practice Questions Chamberlain University |
Comprehensive Examination
HESI A2 Biology Practice Questions Chamberlain University
(Latest 2026/2027 Update) Q&A with Verified Answers and
Detailed Rationales Cell Biology, Genetics, Metabolism A+
INTRODUCTION
This document is an important and very helpful preparation
resource for HESI A2 Biology Practice Questions Chamberlain
University (Latest 2026/2027 Update) Q&A with Verified
Answers and Detailed Rationales Cell Biology, Genetics,
Metabolism 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
A researcher treats cultured human cells with a drug that inhibits the enzyme
flippase. Which of the following outcomes is most likely to be observed
immediately after treatment?
A. Loss of phospholipid asymmetry between the inner and outer leaflets
of the plasma membrane.
B. Complete detachment of the glycocalyx from the extracellular
membrane surface.
C. Rapid disintegration of the nuclear envelope due to loss of lamin
phosphorylation.
D. Inhibition of ATP synthesis because flippase directly transports
protons across the inner mitochondrial membrane.
Correct Answer: A - Loss of phospholipid asymmetry between the
inner and outer leaflets of the plasma membrane.
RATIONALE
Flippase maintains phospholipid asymmetry by translocating specific
phospholipids from the outer to the inner leaflet; inhibition disrupts
this asymmetry. The glycocalyx is composed of carbohydrates and is
not directly affected by flippase. Lamin phosphorylation is regulated
by kinases/phosphatases, not flippase, and flippase does not transport
protons in mitochondria.
Question 2
In a pedigree analysis, a rare autosomal recessive disorder appears in
individuals across three generations. The proband's parents are phenotypically
normal but have an affected child. Which of the following statements best
explains the recurrence risk for future siblings?
A. The recurrence risk is 25% because both parents must be carriers, and
each child has a 1/4 chance of inheriting both mutant alleles.
B. The recurrence risk is 50% because the disorder exhibits incomplete
Page 2
, penetrance, making one parent a manifesting heterozygote.
C. The recurrence risk is 0% because the affected child must have a de
novo mutation, and parents are not carriers.
D. The recurrence risk is 100% because the disorder is autosomal
recessive and both parents are affected.
Correct Answer: A - The recurrence risk is 25% because both
parents must be carriers, and each child has a 1/4 chance of
inheriting both mutant alleles.
RATIONALE
For an autosomal recessive disorder with unaffected parents, both are
obligate carriers (heterozygous). Each pregnancy carries a 25% chance
of an affected child. Incomplete penetrance does not change the
fundamental Mendelian risk, and de novo mutations are unlikely given
the rare disorder and normal parents.
Question 3
During intense exercise, skeletal muscle relies heavily on anaerobic glycolysis.
Which of the following best explains the metabolic advantage of converting
pyruvate to lactate in this context?
A. It regenerates NAD+ from NADH, allowing glycolysis to continue and
produce ATP rapidly.
B. It produces additional ATP through oxidative phosphorylation in the
mitochondria.
C. It shuttles pyruvate into the citric acid cycle for complete oxidation.
D. It increases the proton gradient across the inner mitochondrial
membrane to drive ATP synthase.
Correct Answer: A - It regenerates NAD+ from NADH, allowing
glycolysis to continue and produce ATP rapidly.
Page 3
, RATIONALE
Lactate fermentation oxidizes NADH to NAD+, which is essential for
glyceraldehyde-3-phosphate dehydrogenase in glycolysis, sustaining
ATP production when oxygen is limited. It does not produce ATP
directly, nor does it enhance mitochondrial oxidation or the proton
gradient.
Question 4
A cell is treated with a drug that prevents the assembly of the mitotic spindle.
Which phase of the cell cycle will be most directly arrested?
A. Metaphase
B. G1 phase
C. S phase
D. Telophase
Correct Answer: A - Metaphase
RATIONALE
The mitotic spindle is required for chromosome alignment at the
metaphase plate and subsequent segregation. Without spindle
assembly, cells arrest in metaphase due to the spindle assembly
checkpoint. G1 and S phases occur before mitosis, and telophase
requires prior anaphase.
Question 5
Which of the following best describes the role of the enzyme telomerase in
eukaryotic cells?
A. It extends the 3' end of the DNA strand using an RNA template,
preventing shortening of chromosome ends.
B. It removes RNA primers and replaces them with DNA during lagging
strand synthesis.
C. It proofreads newly synthesized DNA and excises mismatched base
pairs.
Page 4
Chamberlain University (Latest
2026/2027 Update) Q&A with Verified
Answers and Detailed Rationales Cell
HESI A2 Biology Practice Questions Chamberlain University |
Comprehensive Examination
HESI A2 Biology Practice Questions Chamberlain University
(Latest 2026/2027 Update) Q&A with Verified Answers and
Detailed Rationales Cell Biology, Genetics, Metabolism A+
INTRODUCTION
This document is an important and very helpful preparation
resource for HESI A2 Biology Practice Questions Chamberlain
University (Latest 2026/2027 Update) Q&A with Verified
Answers and Detailed Rationales Cell Biology, Genetics,
Metabolism 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
A researcher treats cultured human cells with a drug that inhibits the enzyme
flippase. Which of the following outcomes is most likely to be observed
immediately after treatment?
A. Loss of phospholipid asymmetry between the inner and outer leaflets
of the plasma membrane.
B. Complete detachment of the glycocalyx from the extracellular
membrane surface.
C. Rapid disintegration of the nuclear envelope due to loss of lamin
phosphorylation.
D. Inhibition of ATP synthesis because flippase directly transports
protons across the inner mitochondrial membrane.
Correct Answer: A - Loss of phospholipid asymmetry between the
inner and outer leaflets of the plasma membrane.
RATIONALE
Flippase maintains phospholipid asymmetry by translocating specific
phospholipids from the outer to the inner leaflet; inhibition disrupts
this asymmetry. The glycocalyx is composed of carbohydrates and is
not directly affected by flippase. Lamin phosphorylation is regulated
by kinases/phosphatases, not flippase, and flippase does not transport
protons in mitochondria.
Question 2
In a pedigree analysis, a rare autosomal recessive disorder appears in
individuals across three generations. The proband's parents are phenotypically
normal but have an affected child. Which of the following statements best
explains the recurrence risk for future siblings?
A. The recurrence risk is 25% because both parents must be carriers, and
each child has a 1/4 chance of inheriting both mutant alleles.
B. The recurrence risk is 50% because the disorder exhibits incomplete
Page 2
, penetrance, making one parent a manifesting heterozygote.
C. The recurrence risk is 0% because the affected child must have a de
novo mutation, and parents are not carriers.
D. The recurrence risk is 100% because the disorder is autosomal
recessive and both parents are affected.
Correct Answer: A - The recurrence risk is 25% because both
parents must be carriers, and each child has a 1/4 chance of
inheriting both mutant alleles.
RATIONALE
For an autosomal recessive disorder with unaffected parents, both are
obligate carriers (heterozygous). Each pregnancy carries a 25% chance
of an affected child. Incomplete penetrance does not change the
fundamental Mendelian risk, and de novo mutations are unlikely given
the rare disorder and normal parents.
Question 3
During intense exercise, skeletal muscle relies heavily on anaerobic glycolysis.
Which of the following best explains the metabolic advantage of converting
pyruvate to lactate in this context?
A. It regenerates NAD+ from NADH, allowing glycolysis to continue and
produce ATP rapidly.
B. It produces additional ATP through oxidative phosphorylation in the
mitochondria.
C. It shuttles pyruvate into the citric acid cycle for complete oxidation.
D. It increases the proton gradient across the inner mitochondrial
membrane to drive ATP synthase.
Correct Answer: A - It regenerates NAD+ from NADH, allowing
glycolysis to continue and produce ATP rapidly.
Page 3
, RATIONALE
Lactate fermentation oxidizes NADH to NAD+, which is essential for
glyceraldehyde-3-phosphate dehydrogenase in glycolysis, sustaining
ATP production when oxygen is limited. It does not produce ATP
directly, nor does it enhance mitochondrial oxidation or the proton
gradient.
Question 4
A cell is treated with a drug that prevents the assembly of the mitotic spindle.
Which phase of the cell cycle will be most directly arrested?
A. Metaphase
B. G1 phase
C. S phase
D. Telophase
Correct Answer: A - Metaphase
RATIONALE
The mitotic spindle is required for chromosome alignment at the
metaphase plate and subsequent segregation. Without spindle
assembly, cells arrest in metaphase due to the spindle assembly
checkpoint. G1 and S phases occur before mitosis, and telophase
requires prior anaphase.
Question 5
Which of the following best describes the role of the enzyme telomerase in
eukaryotic cells?
A. It extends the 3' end of the DNA strand using an RNA template,
preventing shortening of chromosome ends.
B. It removes RNA primers and replaces them with DNA during lagging
strand synthesis.
C. It proofreads newly synthesized DNA and excises mismatched base
pairs.
Page 4
