USABO Open Exam USA Biology Olympiad
Official Exam 2026/2027 Actual Exam Complete
Questions and Answers Detailed Rationales Pass
Guaranteed - A+ Graded
TABLE OF CONTENTS
Section 1 | Cell and Molecular Biology | Q1 – Q10
Section 2 | Genetics and Evolution | Q11 – Q20
Section 3 | Anatomy and Physiology | Q21 – Q30
Section 4 | Ecology and Systematics | Q31 – Q40
Section 5 | Plant Biology and Ethology | Q41 – Q50
Instructions: Choose the single best answer.
══════════════════════════════════════
SECTION 1: CELL AND MOLECULAR BIOLOGY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A graduate student is measuring oxygen consumption and ATP production in isolated
liver mitochondria. In one flask, she adds succinate, ADP, and inorganic phosphate. In a
second flask, she adds the same substrates plus cyanide. After 10 minutes, the first
flask has produced 15 μmol ATP while the second has produced less than 0.2 μmol.
She then measures the proton-motive force across the inner membrane. Which result is
most consistent with these findings?
A. Both flasks maintain equal proton-motive force because cyanide stimulates proton
leak
B. The first flask shows a steep proton gradient; the second shows a collapsed gradient
due to inhibited electron transport ✓ CORRECT
C. The second flask shows a higher proton gradient because cyanide blocks ATP
synthase and protons accumulate
,D. Both gradients are identical because the F1F0-ATPase can generate its own proton
gradient without electron transport
Correct Answer: B
Rationale: Cyanide inhibits cytochrome c oxidase, halting electron transport and
preventing proton pumping, so the proton-motive force collapses and ATP synthesis via
chemiosmosis stops. Choice C is tempting because inhibitors can cause proton
accumulation, but cyanide blocks the transfer of electrons to oxygen, which prevents
proton pumping rather than blocking ATP synthase directly. In practice, cyanide
poisoning rapidly depletes cellular ATP because oxidative phosphorylation ceases
entirely.
Question 2 of 50
A pathologist examines a tumor from a 14-year-old patient and finds cells with highly
abnormal chromosome numbers. Molecular testing reveals a mutation in a gene that
normally prevents premature progression to anaphase. Which gene is most likely
affected?
A. Separase
B. Cohesin
C. Mad2 ✓ CORRECT
D. APC/C
Correct Answer: C
Rationale: Mad2 is a critical spindle assembly checkpoint protein that inhibits the
anaphase-promoting complex until all kinetochores attach to spindle microtubules; loss
of Mad2 allows cells to enter anaphase prematurely, producing aneuploid daughter
cells. Choice A is incorrect because separase activation normally triggers anaphase
onset by cleaving cohesin, but its premature activity would be downstream of
,checkpoint failure rather than the primary cause. Aneuploidy is a hallmark of many
aggressive childhood cancers.
Question 3 of 50
A 29-year-old researcher is studying protein trafficking in cultured fibroblasts. She
introduces a temperature-sensitive mutant of vesicular stomatitis virus G protein
(tsO45) into cells. At 40°C, the protein is misfolded and retained; at 32°C, it folds
correctly and moves through the secretory pathway. She shifts cells from 40°C to 32°C
and adds a drug that blocks COPII vesicle formation. Where does the G protein
accumulate?
A. The endoplasmic reticulum ✓ CORRECT
B. The cis-Golgi network
C. The trans-Golgi network
D. The plasma membrane
Correct Answer: A
Rationale: COPII vesicles mediate anterograde transport from the ER to the cis-Golgi;
blocking COPII traps newly synthesized cargo proteins within the ER. Choice B is
tempting because COPI vesicles move retrograde from cis-Golgi to ER, but the drug
specifically blocks COPII, not COPI. Temperature-sensitive VSV-G trafficking is a classic
tool for dissecting vesicular transport in cell biology.
Question 4 of 50
A marine biologist is studying osmoregulation in gill cells of a euryhaline fish
transferred from seawater to freshwater. She measures ion concentrations and cell
volume over 24 hours. Initially, the cells swell slightly but then restore normal volume.
Which transport mechanism best explains the long-term volume recovery?
, A. Increased activity of the Na+/K+ pump alone, which pumps water directly out of the
cell
B. Opening of aquaporins that allow water to exit passively down its gradient
C. Increased endocytosis of the plasma membrane to remove excess surface area
D. Activation of K+/Cl- cotransporters and increased Na+/K+ pump activity, driving net
solute efflux ✓ CORRECT
Correct Answer: D
Rationale: Regulatory volume decrease in hypotonic conditions relies on solute efflux,
primarily through K+/Cl- cotransport and the Na+/K+-ATPase establishing gradients that
drive net osmolyte loss, followed osmotically by water. Choice B is incorrect because
opening aquaporins would increase water permeability but does not create the osmotic
gradient needed for sustained water efflux. Fish gill cells must constantly balance
osmotic challenges in changing salinities.
Question 5 of 50
A biochemist is characterizing a newly discovered enzyme from thermophilic bacteria.
At 25°C, the enzyme shows very low activity. As she increases temperature to 75°C,
activity rises dramatically until it suddenly drops at 85°C. She measures the Michaelis
constant at 70°C and finds Km = 2.5 mM. Which statement best describes what
happens to Km and Vmax as she raises the temperature from 25°C to 70°C?
A. Both Km and Vmax decrease because the enzyme becomes more specific at lower
temperatures
B. Vmax increases and Km generally increases as kinetic energy and collision frequency
rise ✓ CORRECT
C. Vmax decreases and Km decreases because the enzyme-substrate complex
becomes too stable
D. Vmax remains constant and Km decreases because thermal energy improves binding
affinity
Correct Answer: B
Official Exam 2026/2027 Actual Exam Complete
Questions and Answers Detailed Rationales Pass
Guaranteed - A+ Graded
TABLE OF CONTENTS
Section 1 | Cell and Molecular Biology | Q1 – Q10
Section 2 | Genetics and Evolution | Q11 – Q20
Section 3 | Anatomy and Physiology | Q21 – Q30
Section 4 | Ecology and Systematics | Q31 – Q40
Section 5 | Plant Biology and Ethology | Q41 – Q50
Instructions: Choose the single best answer.
══════════════════════════════════════
SECTION 1: CELL AND MOLECULAR BIOLOGY Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A graduate student is measuring oxygen consumption and ATP production in isolated
liver mitochondria. In one flask, she adds succinate, ADP, and inorganic phosphate. In a
second flask, she adds the same substrates plus cyanide. After 10 minutes, the first
flask has produced 15 μmol ATP while the second has produced less than 0.2 μmol.
She then measures the proton-motive force across the inner membrane. Which result is
most consistent with these findings?
A. Both flasks maintain equal proton-motive force because cyanide stimulates proton
leak
B. The first flask shows a steep proton gradient; the second shows a collapsed gradient
due to inhibited electron transport ✓ CORRECT
C. The second flask shows a higher proton gradient because cyanide blocks ATP
synthase and protons accumulate
,D. Both gradients are identical because the F1F0-ATPase can generate its own proton
gradient without electron transport
Correct Answer: B
Rationale: Cyanide inhibits cytochrome c oxidase, halting electron transport and
preventing proton pumping, so the proton-motive force collapses and ATP synthesis via
chemiosmosis stops. Choice C is tempting because inhibitors can cause proton
accumulation, but cyanide blocks the transfer of electrons to oxygen, which prevents
proton pumping rather than blocking ATP synthase directly. In practice, cyanide
poisoning rapidly depletes cellular ATP because oxidative phosphorylation ceases
entirely.
Question 2 of 50
A pathologist examines a tumor from a 14-year-old patient and finds cells with highly
abnormal chromosome numbers. Molecular testing reveals a mutation in a gene that
normally prevents premature progression to anaphase. Which gene is most likely
affected?
A. Separase
B. Cohesin
C. Mad2 ✓ CORRECT
D. APC/C
Correct Answer: C
Rationale: Mad2 is a critical spindle assembly checkpoint protein that inhibits the
anaphase-promoting complex until all kinetochores attach to spindle microtubules; loss
of Mad2 allows cells to enter anaphase prematurely, producing aneuploid daughter
cells. Choice A is incorrect because separase activation normally triggers anaphase
onset by cleaving cohesin, but its premature activity would be downstream of
,checkpoint failure rather than the primary cause. Aneuploidy is a hallmark of many
aggressive childhood cancers.
Question 3 of 50
A 29-year-old researcher is studying protein trafficking in cultured fibroblasts. She
introduces a temperature-sensitive mutant of vesicular stomatitis virus G protein
(tsO45) into cells. At 40°C, the protein is misfolded and retained; at 32°C, it folds
correctly and moves through the secretory pathway. She shifts cells from 40°C to 32°C
and adds a drug that blocks COPII vesicle formation. Where does the G protein
accumulate?
A. The endoplasmic reticulum ✓ CORRECT
B. The cis-Golgi network
C. The trans-Golgi network
D. The plasma membrane
Correct Answer: A
Rationale: COPII vesicles mediate anterograde transport from the ER to the cis-Golgi;
blocking COPII traps newly synthesized cargo proteins within the ER. Choice B is
tempting because COPI vesicles move retrograde from cis-Golgi to ER, but the drug
specifically blocks COPII, not COPI. Temperature-sensitive VSV-G trafficking is a classic
tool for dissecting vesicular transport in cell biology.
Question 4 of 50
A marine biologist is studying osmoregulation in gill cells of a euryhaline fish
transferred from seawater to freshwater. She measures ion concentrations and cell
volume over 24 hours. Initially, the cells swell slightly but then restore normal volume.
Which transport mechanism best explains the long-term volume recovery?
, A. Increased activity of the Na+/K+ pump alone, which pumps water directly out of the
cell
B. Opening of aquaporins that allow water to exit passively down its gradient
C. Increased endocytosis of the plasma membrane to remove excess surface area
D. Activation of K+/Cl- cotransporters and increased Na+/K+ pump activity, driving net
solute efflux ✓ CORRECT
Correct Answer: D
Rationale: Regulatory volume decrease in hypotonic conditions relies on solute efflux,
primarily through K+/Cl- cotransport and the Na+/K+-ATPase establishing gradients that
drive net osmolyte loss, followed osmotically by water. Choice B is incorrect because
opening aquaporins would increase water permeability but does not create the osmotic
gradient needed for sustained water efflux. Fish gill cells must constantly balance
osmotic challenges in changing salinities.
Question 5 of 50
A biochemist is characterizing a newly discovered enzyme from thermophilic bacteria.
At 25°C, the enzyme shows very low activity. As she increases temperature to 75°C,
activity rises dramatically until it suddenly drops at 85°C. She measures the Michaelis
constant at 70°C and finds Km = 2.5 mM. Which statement best describes what
happens to Km and Vmax as she raises the temperature from 25°C to 70°C?
A. Both Km and Vmax decrease because the enzyme becomes more specific at lower
temperatures
B. Vmax increases and Km generally increases as kinetic energy and collision frequency
rise ✓ CORRECT
C. Vmax decreases and Km decreases because the enzyme-substrate complex
becomes too stable
D. Vmax remains constant and Km decreases because thermal energy improves binding
affinity
Correct Answer: B
