BIO 201 Cumulative Final Exam V1 and V2 |
Questions and Answers | 2026/2027 Updated
| 100% Correct - Straighterline.
Practice Question Bank & Answer Rationale
THIS PACK CONTAINS
147 exam-style multiple-choice questions. Each question includes the correct answer and a complete
rationale for focused revision.
COVERAGE
A comprehensive cumulative final examination covering the full BIO 201 curriculum, including biological
chemistry, cell structure and function, metabolism, cell division, molecular genetics, inheritance, evolution,
organismal physiology, and ecology. Questions emphasize application, data interpretation, and integration of
core biological principles at the sophomore level.
STUDY GUIDE
Recommended duration: 3 hours. Passing target: 70% (35/50).
HIGHP - Page 1 of 47
,1. A researcher treats cultured cells with a drug that inhibits the Na+/K+ ATPase. Which
immediate consequence is most likely?
[ ] A. Increased intracellular Na+ and decreased intracellular K+
[ ] B. Increased intracellular K+ and decreased intracellular Na+
[ ] C. No change in ion gradients but decreased ATP consumption
[ ] D. Rapid influx of Ca2+ due to direct pump inhibition
CORRECT: A. Increased intracellular Na+ and decreased intracellular K+
The Na+/K+ ATPase normally exports 3 Na+ and imports 2 K+ per ATP, maintaining low intracellular Na+ and
high intracellular K+. Inhibiting it causes Na+ to accumulate inside and K+ to leak out down their gradients.
Option B describes the normal pump action, C is false because gradients collapse, and D is not a direct
effect of pump inhibition.
2. In a dihybrid cross between two heterozygotes (AaBb × AaBb), what proportion of
offspring is expected to show both recessive phenotypes?
[ ] A. 1/16
[ ] B. 3/16
[ ] C. 9/16
[ ] D. 1/4
CORRECT: A. 1/16
For independently assorting genes, the probability of the recessive phenotype at each locus is 1/4.
Multiplying gives 1/4 × 1/4 = 1/16 for both recessive traits. 3/16 corresponds to one dominant and one
recessive phenotype, 9/16 to both dominant, and 1/4 would apply to a single locus.
3. Which observation best supports the endosymbiotic theory for the origin of
mitochondria?
[ ] A. Mitochondria contain circular DNA and divide independently of the cell cycle.
[ ] B. Mitochondria are surrounded by a single membrane.
[ ] C. Mitochondrial proteins are all encoded by nuclear genes.
[ ] D. Mitochondria lack ribosomes and rely on host translation.
CORRECT: A. Mitochondria contain circular DNA and divide independently of the cell cycle.
Circular DNA and independent division are bacterial features retained by mitochondria, supporting their
descent from an engulfed prokaryote. Mitochondria are double-membraned, not single, and they possess
their own ribosomes, so B and D are incorrect. Many mitochondrial proteins are nuclear-encoded, but that
does not refute endosymbiosis.
HIGHP - Page 2 of 47
,4. A mutation changes a codon from UAC to UAA in the middle of an mRNA. What is the
most likely effect on the protein?
[ ] A. Premature termination of translation
[ ] B. Substitution of one amino acid
[ ] C. Frameshift causing a completely different sequence
[ ] D. No effect because the genetic code is degenerate
CORRECT: A. Premature termination of translation
UAA is a stop codon, so the ribosome will terminate translation early, producing a truncated polypeptide. A
missense substitution would require a codon for a different amino acid, a frameshift requires an insertion or
deletion not a multiple of three, and degeneracy does not apply because UAA is not a sense codon.
5. In a population at Hardy-Weinberg equilibrium, the frequency of a recessive allele is 0.2.
What percentage of the population is expected to be heterozygous?
[ ] A. 32%
[ ] B. 16%
[ ] C. 64%
[ ] D. 4%
CORRECT: A. 32%
With q = 0.2, p = 0.8. Heterozygote frequency = 2pq = 2(0.8)(0.2) = 0.32, or 32%. 16% would be q²
(homozygous recessive), 64% is p² (homozygous dominant), and 4% is not a standard term in this
calculation.
6. Which statement best describes the role of the enzyme rubisco in photosynthesis?
[ ] A. It fixes CO2 onto ribulose-1,5-bisphosphate during the Calvin cycle.
[ ] B. It splits water to release oxygen in the light reactions.
[ ] C. It synthesizes ATP via chemiosmosis in the thylakoid membrane.
[ ] D. It transfers electrons from photosystem II to photosystem I.
CORRECT: A. It fixes CO2 onto ribulose-1,5-bisphosphate during the Calvin cycle.
Rubisco catalyzes the first step of carbon fixation in the Calvin cycle, attaching CO2 to RuBP. Water splitting
is performed by the oxygen-evolving complex of photosystem II, ATP synthesis by ATP synthase, and
electron transfer by the cytochrome complex and plastocyanin.
7. Which of the following best explains why the action potential travels in one direction
along an axon?
[ ] A. The refractory period prevents backward propagation.
[ ] B. Myelin sheaths only allow forward current.
[ ] C. Neurotransmitters are released only at the axon hillock.
HIGHP - Page 3 of 47
, [ ] D. Sodium channels are permanently inactivated behind the impulse.
CORRECT: A. The refractory period prevents backward propagation.
After an action potential, voltage-gated Na+ channels enter an inactivated state (absolute refractory period),
preventing the membrane just behind the impulse from re-firing and thus enforcing one-way travel. Myelin
increases speed but does not dictate direction, neurotransmitters are released at axon terminals, and Na+
channels recover from inactivation, not permanently.
8. A patient with a mutation in the CFTR gene has defective chloride transport. Which
cellular process is most directly impaired?
[ ] A. Regulation of water and ion balance across epithelial surfaces
[ ] B. Synthesis of ATP in mitochondria
[ ] C. Packaging of proteins in the Golgi apparatus
[ ] D. Replication of DNA during S phase
CORRECT: A. Regulation of water and ion balance across epithelial surfaces
CFTR is a chloride channel that regulates salt and water movement across epithelial cells; its dysfunction
leads to thick mucus in cystic fibrosis. ATP synthesis, Golgi packaging, and DNA replication are not directly
mediated by CFTR.
9. Which of the following is an example of a post-translational modification that can
regulate protein activity?
[ ] A. Phosphorylation of serine residues by a kinase
[ ] B. Alternative splicing of pre-mRNA
[ ] C. Binding of a transcription factor to an enhancer
[ ] D. Addition of a 5' cap to mRNA
CORRECT: A. Phosphorylation of serine residues by a kinase
Phosphorylation after translation covalently modifies a protein and can switch its activity on or off. Alternative
splicing, transcription factor binding, and 5' capping all occur before translation and are not post-translational
events.
10. A cell is treated with a drug that inhibits the enzyme flippase. Which membrane
property would be most directly disrupted?
[ ] A. Maintenance of asymmetric distribution of phospholipids between leaflets
[ ] B. Formation of lipid rafts enriched in cholesterol and sphingolipids
[ ] C. Fluidity of the membrane as governed by fatty acid saturation
[ ] D. Insertion of integral proteins into the lipid bilayer
HIGHP - Page 4 of 47
Questions and Answers | 2026/2027 Updated
| 100% Correct - Straighterline.
Practice Question Bank & Answer Rationale
THIS PACK CONTAINS
147 exam-style multiple-choice questions. Each question includes the correct answer and a complete
rationale for focused revision.
COVERAGE
A comprehensive cumulative final examination covering the full BIO 201 curriculum, including biological
chemistry, cell structure and function, metabolism, cell division, molecular genetics, inheritance, evolution,
organismal physiology, and ecology. Questions emphasize application, data interpretation, and integration of
core biological principles at the sophomore level.
STUDY GUIDE
Recommended duration: 3 hours. Passing target: 70% (35/50).
HIGHP - Page 1 of 47
,1. A researcher treats cultured cells with a drug that inhibits the Na+/K+ ATPase. Which
immediate consequence is most likely?
[ ] A. Increased intracellular Na+ and decreased intracellular K+
[ ] B. Increased intracellular K+ and decreased intracellular Na+
[ ] C. No change in ion gradients but decreased ATP consumption
[ ] D. Rapid influx of Ca2+ due to direct pump inhibition
CORRECT: A. Increased intracellular Na+ and decreased intracellular K+
The Na+/K+ ATPase normally exports 3 Na+ and imports 2 K+ per ATP, maintaining low intracellular Na+ and
high intracellular K+. Inhibiting it causes Na+ to accumulate inside and K+ to leak out down their gradients.
Option B describes the normal pump action, C is false because gradients collapse, and D is not a direct
effect of pump inhibition.
2. In a dihybrid cross between two heterozygotes (AaBb × AaBb), what proportion of
offspring is expected to show both recessive phenotypes?
[ ] A. 1/16
[ ] B. 3/16
[ ] C. 9/16
[ ] D. 1/4
CORRECT: A. 1/16
For independently assorting genes, the probability of the recessive phenotype at each locus is 1/4.
Multiplying gives 1/4 × 1/4 = 1/16 for both recessive traits. 3/16 corresponds to one dominant and one
recessive phenotype, 9/16 to both dominant, and 1/4 would apply to a single locus.
3. Which observation best supports the endosymbiotic theory for the origin of
mitochondria?
[ ] A. Mitochondria contain circular DNA and divide independently of the cell cycle.
[ ] B. Mitochondria are surrounded by a single membrane.
[ ] C. Mitochondrial proteins are all encoded by nuclear genes.
[ ] D. Mitochondria lack ribosomes and rely on host translation.
CORRECT: A. Mitochondria contain circular DNA and divide independently of the cell cycle.
Circular DNA and independent division are bacterial features retained by mitochondria, supporting their
descent from an engulfed prokaryote. Mitochondria are double-membraned, not single, and they possess
their own ribosomes, so B and D are incorrect. Many mitochondrial proteins are nuclear-encoded, but that
does not refute endosymbiosis.
HIGHP - Page 2 of 47
,4. A mutation changes a codon from UAC to UAA in the middle of an mRNA. What is the
most likely effect on the protein?
[ ] A. Premature termination of translation
[ ] B. Substitution of one amino acid
[ ] C. Frameshift causing a completely different sequence
[ ] D. No effect because the genetic code is degenerate
CORRECT: A. Premature termination of translation
UAA is a stop codon, so the ribosome will terminate translation early, producing a truncated polypeptide. A
missense substitution would require a codon for a different amino acid, a frameshift requires an insertion or
deletion not a multiple of three, and degeneracy does not apply because UAA is not a sense codon.
5. In a population at Hardy-Weinberg equilibrium, the frequency of a recessive allele is 0.2.
What percentage of the population is expected to be heterozygous?
[ ] A. 32%
[ ] B. 16%
[ ] C. 64%
[ ] D. 4%
CORRECT: A. 32%
With q = 0.2, p = 0.8. Heterozygote frequency = 2pq = 2(0.8)(0.2) = 0.32, or 32%. 16% would be q²
(homozygous recessive), 64% is p² (homozygous dominant), and 4% is not a standard term in this
calculation.
6. Which statement best describes the role of the enzyme rubisco in photosynthesis?
[ ] A. It fixes CO2 onto ribulose-1,5-bisphosphate during the Calvin cycle.
[ ] B. It splits water to release oxygen in the light reactions.
[ ] C. It synthesizes ATP via chemiosmosis in the thylakoid membrane.
[ ] D. It transfers electrons from photosystem II to photosystem I.
CORRECT: A. It fixes CO2 onto ribulose-1,5-bisphosphate during the Calvin cycle.
Rubisco catalyzes the first step of carbon fixation in the Calvin cycle, attaching CO2 to RuBP. Water splitting
is performed by the oxygen-evolving complex of photosystem II, ATP synthesis by ATP synthase, and
electron transfer by the cytochrome complex and plastocyanin.
7. Which of the following best explains why the action potential travels in one direction
along an axon?
[ ] A. The refractory period prevents backward propagation.
[ ] B. Myelin sheaths only allow forward current.
[ ] C. Neurotransmitters are released only at the axon hillock.
HIGHP - Page 3 of 47
, [ ] D. Sodium channels are permanently inactivated behind the impulse.
CORRECT: A. The refractory period prevents backward propagation.
After an action potential, voltage-gated Na+ channels enter an inactivated state (absolute refractory period),
preventing the membrane just behind the impulse from re-firing and thus enforcing one-way travel. Myelin
increases speed but does not dictate direction, neurotransmitters are released at axon terminals, and Na+
channels recover from inactivation, not permanently.
8. A patient with a mutation in the CFTR gene has defective chloride transport. Which
cellular process is most directly impaired?
[ ] A. Regulation of water and ion balance across epithelial surfaces
[ ] B. Synthesis of ATP in mitochondria
[ ] C. Packaging of proteins in the Golgi apparatus
[ ] D. Replication of DNA during S phase
CORRECT: A. Regulation of water and ion balance across epithelial surfaces
CFTR is a chloride channel that regulates salt and water movement across epithelial cells; its dysfunction
leads to thick mucus in cystic fibrosis. ATP synthesis, Golgi packaging, and DNA replication are not directly
mediated by CFTR.
9. Which of the following is an example of a post-translational modification that can
regulate protein activity?
[ ] A. Phosphorylation of serine residues by a kinase
[ ] B. Alternative splicing of pre-mRNA
[ ] C. Binding of a transcription factor to an enhancer
[ ] D. Addition of a 5' cap to mRNA
CORRECT: A. Phosphorylation of serine residues by a kinase
Phosphorylation after translation covalently modifies a protein and can switch its activity on or off. Alternative
splicing, transcription factor binding, and 5' capping all occur before translation and are not post-translational
events.
10. A cell is treated with a drug that inhibits the enzyme flippase. Which membrane
property would be most directly disrupted?
[ ] A. Maintenance of asymmetric distribution of phospholipids between leaflets
[ ] B. Formation of lipid rafts enriched in cholesterol and sphingolipids
[ ] C. Fluidity of the membrane as governed by fatty acid saturation
[ ] D. Insertion of integral proteins into the lipid bilayer
HIGHP - Page 4 of 47
