DAT Biology Exam 2026 160+ High-Yield
Practice Questions & Verified Answers
1. A researcher treats isolated mitochondria with oligomycin and then measures
oxygen consumption and ATP synthesis. Which of the following best explains the
observed decrease in oxygen consumption?
A. Oligomycin directly inhibits Complex IV, preventing electron transfer to oxygen.
B. Oligomycin blocks ATP synthase proton flow, causing proton gradient buildup that
thermodynamically back-pressures the electron transport chain.
C. Oligomycin uncouples the proton gradient by transporting protons across the inner
membrane.
D. Oligomycin inhibits the malate-aspartate shuttle, reducing NADH delivery to the
matrix.
Answer: B. Oligomycin blocks ATP synthase proton flow, causing proton gradient
buildup that thermodynamically back-pressures the electron transport chain.
Rationale: Oligomycin inhibits the Fo subunit of ATP synthase, blocking proton
translocation and causing the proton motive force to rise; this back-pressure slows electron
transport and thus oxygen consumption. It does not inhibit Complex IV, uncouple, or affect
shuttle activity.
2. In a eukaryotic mRNA, a silent mutation occurs within an exonic splicing enhancer
(ESE) sequence. Which of the following is the most likely consequence?
A. No effect on protein sequence because silent mutations do not alter amino acids.
B. Exon skipping or alternative splicing leading to a truncated or aberrant protein.
C. Frameshift mutation downstream of the ESE.
D. Premature termination of transcription by RNA polymerase II.
Answer: B. Exon skipping or alternative splicing leading to a truncated or aberrant
protein.
Rationale: Silent mutations can disrupt ESEs, altering splicing and causing exon skipping
or aberrant isoforms, which may produce nonfunctional proteins. They do not cause
frameshifts or affect transcription termination.
Page 1 of 66
,3. A population of birds on an island has a mean beak depth of 10 mm. After a
drought, only birds with beak depths >12 mm survive. The next generation's mean
beak depth is 12.5 mm. What is the selection differential (S) and response to selection
(R)?
A. S = 2 mm; R = 0.5 mm
B. S = 0.5 mm; R = 2 mm
C. S = 2.5 mm; R = 2 mm
D. S = 10 mm; R = 12.5 mm
Answer: A. S = 2 mm; R = 0.5 mm
Rationale: Selection differential S is the difference between the mean of selected parents
(12 mm) and the overall population mean (10 mm), so S = 2 mm. Response R is the
difference between offspring mean (12.5 mm) and original population mean (10 mm), so R
= 0.5 mm.
4. Which of the following best explains why the oxygen dissociation curve of fetal
hemoglobin (HbF) is left-shifted relative to maternal hemoglobin (HbA)?
A. HbF has a higher affinity for 2,3-BPG than HbA.
B. HbF binds 2,3-BPG less effectively due to its gamma chains, increasing oxygen
affinity.
C. HbF has a lower affinity for oxygen, facilitating oxygen transfer to maternal blood.
D. HbF exhibits a Bohr effect that is more pronounced than that of HbA.
Answer: B. HbF binds 2,3-BPG less effectively due to its gamma chains, increasing
oxygen affinity.
Rationale: HbF has gamma chains that interact poorly with 2,3-BPG, reducing its
stabilizing effect on the T state and increasing oxygen affinity, which facilitates oxygen
transfer from mother to fetus. The other options misstate the mechanism or direction of the
shift.
Page 2 of 66
,5. During a laboratory experiment, a cell is treated with a drug that inhibits the
Na+/K+ ATPase. Which of the following immediate consequences is most likely?
A. Increased intracellular Na+ and decreased intracellular K+
B. Decreased intracellular Na+ and increased intracellular K+
C. Hyperpolarization of the cell membrane
D. Increased secondary active transport of glucose
Answer: A. Increased intracellular Na+ and decreased intracellular K+
Rationale: Inhibition of Na+/K+ ATPase prevents extrusion of Na+ and uptake of K+,
leading to intracellular Na+ accumulation and K+ depletion, and depolarization.
Secondary active transport that depends on the Na+ gradient would be impaired, not
increased.
6. A novel antibiotic inhibits bacterial DNA gyrase. Which of the following best
describes the immediate effect on DNA replication?
A. Replication fork stalls due to positive supercoiling ahead of the fork.
B. RNA primers cannot be synthesized, preventing replication initiation.
C. Okazaki fragments are not ligated, resulting in fragmented DNA.
D. The sliding clamp cannot load onto DNA, reducing processivity.
Answer: A. Replication fork stalls due to positive supercoiling ahead of the fork.
Rationale: DNA gyrase relieves positive supercoils ahead of the replication fork; inhibition
causes fork stalling. The other options describe defects in primase, ligase, or clamp loader,
respectively.
Page 3 of 66
, 7. In a species of flowering plant, flower color is controlled by two genes. A cross
between two true-breeding parents with white flowers yields all purple F1. When F1
are selfed, the F2 shows a 9:7 ratio of purple:white. Which genetic phenomenon
explains this?
A. Recessive epistasis
B. Duplicate recessive epistasis (complementary gene action)
C. Dominant epistasis
D. Incomplete dominance
Answer: B. Duplicate recessive epistasis (complementary gene action)
Rationale: A 9:7 ratio in F2 indicates complementary gene action, where both dominant
alleles at two loci are required for pigment production; homozygous recessive at either
locus yields white. The other options produce different ratios.
8. Which of the following best explains the role of the enzyme telomerase in stem cells
and cancer cells?
A. It removes RNA primers and fills gaps with DNA.
B. It extends the 3' end of the template strand using an RNA template, maintaining
telomere length.
C. It adds repetitive sequences to the 5' end of the lagging strand.
D. It repairs double-strand breaks by non-homologous end joining.
Answer: B. It extends the 3' end of the template strand using an RNA template,
maintaining telomere length.
Rationale: Telomerase is a reverse transcriptase that carries its own RNA template and
extends the 3' end of the DNA strand, allowing completion of lagging strand synthesis and
preventing telomere shortening. The other options describe other enzymes or processes.
Page 4 of 66
Practice Questions & Verified Answers
1. A researcher treats isolated mitochondria with oligomycin and then measures
oxygen consumption and ATP synthesis. Which of the following best explains the
observed decrease in oxygen consumption?
A. Oligomycin directly inhibits Complex IV, preventing electron transfer to oxygen.
B. Oligomycin blocks ATP synthase proton flow, causing proton gradient buildup that
thermodynamically back-pressures the electron transport chain.
C. Oligomycin uncouples the proton gradient by transporting protons across the inner
membrane.
D. Oligomycin inhibits the malate-aspartate shuttle, reducing NADH delivery to the
matrix.
Answer: B. Oligomycin blocks ATP synthase proton flow, causing proton gradient
buildup that thermodynamically back-pressures the electron transport chain.
Rationale: Oligomycin inhibits the Fo subunit of ATP synthase, blocking proton
translocation and causing the proton motive force to rise; this back-pressure slows electron
transport and thus oxygen consumption. It does not inhibit Complex IV, uncouple, or affect
shuttle activity.
2. In a eukaryotic mRNA, a silent mutation occurs within an exonic splicing enhancer
(ESE) sequence. Which of the following is the most likely consequence?
A. No effect on protein sequence because silent mutations do not alter amino acids.
B. Exon skipping or alternative splicing leading to a truncated or aberrant protein.
C. Frameshift mutation downstream of the ESE.
D. Premature termination of transcription by RNA polymerase II.
Answer: B. Exon skipping or alternative splicing leading to a truncated or aberrant
protein.
Rationale: Silent mutations can disrupt ESEs, altering splicing and causing exon skipping
or aberrant isoforms, which may produce nonfunctional proteins. They do not cause
frameshifts or affect transcription termination.
Page 1 of 66
,3. A population of birds on an island has a mean beak depth of 10 mm. After a
drought, only birds with beak depths >12 mm survive. The next generation's mean
beak depth is 12.5 mm. What is the selection differential (S) and response to selection
(R)?
A. S = 2 mm; R = 0.5 mm
B. S = 0.5 mm; R = 2 mm
C. S = 2.5 mm; R = 2 mm
D. S = 10 mm; R = 12.5 mm
Answer: A. S = 2 mm; R = 0.5 mm
Rationale: Selection differential S is the difference between the mean of selected parents
(12 mm) and the overall population mean (10 mm), so S = 2 mm. Response R is the
difference between offspring mean (12.5 mm) and original population mean (10 mm), so R
= 0.5 mm.
4. Which of the following best explains why the oxygen dissociation curve of fetal
hemoglobin (HbF) is left-shifted relative to maternal hemoglobin (HbA)?
A. HbF has a higher affinity for 2,3-BPG than HbA.
B. HbF binds 2,3-BPG less effectively due to its gamma chains, increasing oxygen
affinity.
C. HbF has a lower affinity for oxygen, facilitating oxygen transfer to maternal blood.
D. HbF exhibits a Bohr effect that is more pronounced than that of HbA.
Answer: B. HbF binds 2,3-BPG less effectively due to its gamma chains, increasing
oxygen affinity.
Rationale: HbF has gamma chains that interact poorly with 2,3-BPG, reducing its
stabilizing effect on the T state and increasing oxygen affinity, which facilitates oxygen
transfer from mother to fetus. The other options misstate the mechanism or direction of the
shift.
Page 2 of 66
,5. During a laboratory experiment, a cell is treated with a drug that inhibits the
Na+/K+ ATPase. Which of the following immediate consequences is most likely?
A. Increased intracellular Na+ and decreased intracellular K+
B. Decreased intracellular Na+ and increased intracellular K+
C. Hyperpolarization of the cell membrane
D. Increased secondary active transport of glucose
Answer: A. Increased intracellular Na+ and decreased intracellular K+
Rationale: Inhibition of Na+/K+ ATPase prevents extrusion of Na+ and uptake of K+,
leading to intracellular Na+ accumulation and K+ depletion, and depolarization.
Secondary active transport that depends on the Na+ gradient would be impaired, not
increased.
6. A novel antibiotic inhibits bacterial DNA gyrase. Which of the following best
describes the immediate effect on DNA replication?
A. Replication fork stalls due to positive supercoiling ahead of the fork.
B. RNA primers cannot be synthesized, preventing replication initiation.
C. Okazaki fragments are not ligated, resulting in fragmented DNA.
D. The sliding clamp cannot load onto DNA, reducing processivity.
Answer: A. Replication fork stalls due to positive supercoiling ahead of the fork.
Rationale: DNA gyrase relieves positive supercoils ahead of the replication fork; inhibition
causes fork stalling. The other options describe defects in primase, ligase, or clamp loader,
respectively.
Page 3 of 66
, 7. In a species of flowering plant, flower color is controlled by two genes. A cross
between two true-breeding parents with white flowers yields all purple F1. When F1
are selfed, the F2 shows a 9:7 ratio of purple:white. Which genetic phenomenon
explains this?
A. Recessive epistasis
B. Duplicate recessive epistasis (complementary gene action)
C. Dominant epistasis
D. Incomplete dominance
Answer: B. Duplicate recessive epistasis (complementary gene action)
Rationale: A 9:7 ratio in F2 indicates complementary gene action, where both dominant
alleles at two loci are required for pigment production; homozygous recessive at either
locus yields white. The other options produce different ratios.
8. Which of the following best explains the role of the enzyme telomerase in stem cells
and cancer cells?
A. It removes RNA primers and fills gaps with DNA.
B. It extends the 3' end of the template strand using an RNA template, maintaining
telomere length.
C. It adds repetitive sequences to the 5' end of the lagging strand.
D. It repairs double-strand breaks by non-homologous end joining.
Answer: B. It extends the 3' end of the template strand using an RNA template,
maintaining telomere length.
Rationale: Telomerase is a reverse transcriptase that carries its own RNA template and
extends the 3' end of the DNA strand, allowing completion of lagging strand synthesis and
preventing telomere shortening. The other options describe other enzymes or processes.
Page 4 of 66
