BIOLOGY 1000 ACTUAL FINAL EXAM COMPLETE
QUESTIONS WITH EXPERT SOLUTIONS 2026 LATEST
UPDATED GET A+ - 109 Questions with Answers
Page 1
,Q1. In a classic Meselson-Stahl experiment, E. coli were grown in heavy nitrogen
(15N) for many generations, then shifted to light nitrogen (14N) for exactly one
generation. DNA was then centrifuged. Assuming semiconservative replication, what
would be the expected distribution of DNA bands?
A. A single band at the heavy position
B. A single band at an intermediate position
C. Two bands: one heavy and one light
D. Two bands: one intermediate and one light
Correct Answer: B. A single band at an intermediate position
Rationale: After one generation in 14N, each DNA molecule has one old heavy strand and
one new light strand, producing a single intermediate band. Conservative replication
would yield two bands (heavy and light), while dispersive would also yield a single
intermediate band but would differ after a second generation.
Why Wrong:
A - This would be the result before any replication in light medium.
C - This would indicate conservative replication, where original heavy duplex remains
intact.
D - Two bands (intermediate and light) would appear after two generations, not one.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 5
Q2. A mutation in the gene encoding the lac repressor (lacI) produces a repressor that
cannot bind allolactose. In a lacI+ (wild-type) background, what will be the
expression level of the lac operon in the presence of high lactose and low glucose?
A. High expression because lactose is present
B. Low expression because the repressor remains bound to the operator
C. High expression because glucose is low, activating CAP
D. Low expression because lactose cannot induce the mutant repressor
Correct Answer: D. Low expression because lactose cannot induce the mutant
repressor
Rationale: The mutant repressor cannot bind allolactose, so it remains bound to the
operator, blocking transcription even in the presence of lactose. CAP activation is
irrelevant because RNA polymerase cannot access the promoter.
Why Wrong:
A - Lactose is present but cannot inactivate the mutant repressor.
B - While the repressor is bound, the reason is specifically the inability to bind
allolactose.
C - CAP activation occurs, but the operator block prevents transcription.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 7
Page 2
,Q3. In a population of 10,000 individuals, the frequency of a recessive lethal allele is
0.02. Assuming Hardy-Weinberg equilibrium and complete selection against
homozygotes, what is the expected reduction in allele frequency after one generation
of selection?
A. 0.0004
B. 0.00196
C. 0.000392
D. 0.02
Correct Answer: C. 0.000392
Rationale: Initial allele frequency q=0.02. After selection, q' = q/(1+q) = 0.02/1.02 "H
0.019608. Reduction = 0.02 - 0.019608 = 0.000392. This small change reflects the low
frequency of the allele.
Why Wrong:
A - This is q^2, the frequency of homozygotes, not the change in allele frequency.
B - This is the new allele frequency, not the reduction.
D - This is the original allele frequency, not the change.
Reference: Futuyma, Evolution, 4th ed., Ch. 6
Q4. Which of the following best explains the observation that the mutation rate per
genome per generation is roughly constant across diverse species, from bacteria to
mammals?
A. Natural selection optimizes mutation rates to balance deleterious and beneficial
mutations.
B. DNA replication fidelity is inherently limited by the error rate of DNA polymerase.
C. The number of cell divisions per generation is similar across species.
D. The mutation rate per base pair per cell division is higher in larger genomes.
Correct Answer: A. Natural selection optimizes mutation rates to balance deleterious
and beneficial mutations.
Rationale: The constant per-genome mutation rate suggests that selection tunes mutation
rates to an optimal level, where the cost of deleterious mutations balances the benefit of
occasional beneficial ones (the drift-barrier hypothesis). DNA polymerase error rates are
not fixed; they are modified by repair systems.
Why Wrong:
B - DNA polymerase error rates vary and are subject to selection, not fixed.
C - Cell division counts vary widely; this is not the explanation.
D - Per-base mutation rates are generally lower in larger genomes, not higher.
Reference: Lynch, The Origins of Genome Architecture, Ch. 4
Page 3
, Q5. A researcher measures the rate of oxygen consumption in isolated mitochondria
under state 3 (ADP present) and state 4 (no ADP). They then add an uncoupler.
Which of the following outcomes is expected?
A. Oxygen consumption in state 4 increases to the state 3 rate, and ATP synthesis stops.
B. Oxygen consumption in state 3 decreases, and ATP synthesis continues.
C. Oxygen consumption in both states increases, and ATP synthesis is unaffected.
D. Oxygen consumption in state 3 remains high, but ATP synthesis is abolished.
Correct Answer: A. Oxygen consumption in state 4 increases to the state 3 rate, and
ATP synthesis stops.
Rationale: Uncouplers dissipate the proton gradient, so oxygen consumption is no longer
coupled to ATP synthesis. State 4 (no ADP) has a low rate due to the absence of ATP
synthase activity; uncoupler relieves this backpressure, increasing oxygen consumption to
the maximal rate. ATP synthesis stops because the proton gradient is gone.
Why Wrong:
B - Uncouplers do not inhibit oxygen consumption; they may increase it.
C - ATP synthesis is abolished, not unaffected.
D - State 3 oxygen consumption may increase slightly, but ATP synthesis stops, not
continues.
Reference: Lehninger, Principles of Biochemistry, 7th ed., Ch. 19
Q6. A plant species has a C3 photosynthetic pathway. If atmospheric CO2
concentration doubles (from 400 ppm to 800 ppm) with no change in temperature or
light, what is the most likely immediate effect on photorespiration and net CO2
fixation?
A. Photorespiration increases; net CO2 fixation decreases.
B. Photorespiration decreases; net CO2 fixation increases.
C. Photorespiration is unchanged; net CO2 fixation is unchanged.
D. Photorespiration decreases; net CO2 fixation is unchanged.
Correct Answer: B. Photorespiration decreases; net CO2 fixation increases.
Rationale: Higher CO2 favors the carboxylase activity of Rubisco over the oxygenase
activity, reducing photorespiration and increasing net CO2 fixation. This is the basis for
the CO2 fertilization effect in C3 plants.
Why Wrong:
A - Higher CO2 suppresses oxygenase activity, not increases it.
C - Higher CO2 changes the substrate saturation of Rubisco.
D - Net CO2 fixation increases because of reduced photorespiration and increased
carboxylation.
Reference: Taiz & Zeiger, Plant Physiology, 6th ed., Ch. 9
Page 4
QUESTIONS WITH EXPERT SOLUTIONS 2026 LATEST
UPDATED GET A+ - 109 Questions with Answers
Page 1
,Q1. In a classic Meselson-Stahl experiment, E. coli were grown in heavy nitrogen
(15N) for many generations, then shifted to light nitrogen (14N) for exactly one
generation. DNA was then centrifuged. Assuming semiconservative replication, what
would be the expected distribution of DNA bands?
A. A single band at the heavy position
B. A single band at an intermediate position
C. Two bands: one heavy and one light
D. Two bands: one intermediate and one light
Correct Answer: B. A single band at an intermediate position
Rationale: After one generation in 14N, each DNA molecule has one old heavy strand and
one new light strand, producing a single intermediate band. Conservative replication
would yield two bands (heavy and light), while dispersive would also yield a single
intermediate band but would differ after a second generation.
Why Wrong:
A - This would be the result before any replication in light medium.
C - This would indicate conservative replication, where original heavy duplex remains
intact.
D - Two bands (intermediate and light) would appear after two generations, not one.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 5
Q2. A mutation in the gene encoding the lac repressor (lacI) produces a repressor that
cannot bind allolactose. In a lacI+ (wild-type) background, what will be the
expression level of the lac operon in the presence of high lactose and low glucose?
A. High expression because lactose is present
B. Low expression because the repressor remains bound to the operator
C. High expression because glucose is low, activating CAP
D. Low expression because lactose cannot induce the mutant repressor
Correct Answer: D. Low expression because lactose cannot induce the mutant
repressor
Rationale: The mutant repressor cannot bind allolactose, so it remains bound to the
operator, blocking transcription even in the presence of lactose. CAP activation is
irrelevant because RNA polymerase cannot access the promoter.
Why Wrong:
A - Lactose is present but cannot inactivate the mutant repressor.
B - While the repressor is bound, the reason is specifically the inability to bind
allolactose.
C - CAP activation occurs, but the operator block prevents transcription.
Reference: Alberts et al., Molecular Biology of the Cell, 7th ed., Ch. 7
Page 2
,Q3. In a population of 10,000 individuals, the frequency of a recessive lethal allele is
0.02. Assuming Hardy-Weinberg equilibrium and complete selection against
homozygotes, what is the expected reduction in allele frequency after one generation
of selection?
A. 0.0004
B. 0.00196
C. 0.000392
D. 0.02
Correct Answer: C. 0.000392
Rationale: Initial allele frequency q=0.02. After selection, q' = q/(1+q) = 0.02/1.02 "H
0.019608. Reduction = 0.02 - 0.019608 = 0.000392. This small change reflects the low
frequency of the allele.
Why Wrong:
A - This is q^2, the frequency of homozygotes, not the change in allele frequency.
B - This is the new allele frequency, not the reduction.
D - This is the original allele frequency, not the change.
Reference: Futuyma, Evolution, 4th ed., Ch. 6
Q4. Which of the following best explains the observation that the mutation rate per
genome per generation is roughly constant across diverse species, from bacteria to
mammals?
A. Natural selection optimizes mutation rates to balance deleterious and beneficial
mutations.
B. DNA replication fidelity is inherently limited by the error rate of DNA polymerase.
C. The number of cell divisions per generation is similar across species.
D. The mutation rate per base pair per cell division is higher in larger genomes.
Correct Answer: A. Natural selection optimizes mutation rates to balance deleterious
and beneficial mutations.
Rationale: The constant per-genome mutation rate suggests that selection tunes mutation
rates to an optimal level, where the cost of deleterious mutations balances the benefit of
occasional beneficial ones (the drift-barrier hypothesis). DNA polymerase error rates are
not fixed; they are modified by repair systems.
Why Wrong:
B - DNA polymerase error rates vary and are subject to selection, not fixed.
C - Cell division counts vary widely; this is not the explanation.
D - Per-base mutation rates are generally lower in larger genomes, not higher.
Reference: Lynch, The Origins of Genome Architecture, Ch. 4
Page 3
, Q5. A researcher measures the rate of oxygen consumption in isolated mitochondria
under state 3 (ADP present) and state 4 (no ADP). They then add an uncoupler.
Which of the following outcomes is expected?
A. Oxygen consumption in state 4 increases to the state 3 rate, and ATP synthesis stops.
B. Oxygen consumption in state 3 decreases, and ATP synthesis continues.
C. Oxygen consumption in both states increases, and ATP synthesis is unaffected.
D. Oxygen consumption in state 3 remains high, but ATP synthesis is abolished.
Correct Answer: A. Oxygen consumption in state 4 increases to the state 3 rate, and
ATP synthesis stops.
Rationale: Uncouplers dissipate the proton gradient, so oxygen consumption is no longer
coupled to ATP synthesis. State 4 (no ADP) has a low rate due to the absence of ATP
synthase activity; uncoupler relieves this backpressure, increasing oxygen consumption to
the maximal rate. ATP synthesis stops because the proton gradient is gone.
Why Wrong:
B - Uncouplers do not inhibit oxygen consumption; they may increase it.
C - ATP synthesis is abolished, not unaffected.
D - State 3 oxygen consumption may increase slightly, but ATP synthesis stops, not
continues.
Reference: Lehninger, Principles of Biochemistry, 7th ed., Ch. 19
Q6. A plant species has a C3 photosynthetic pathway. If atmospheric CO2
concentration doubles (from 400 ppm to 800 ppm) with no change in temperature or
light, what is the most likely immediate effect on photorespiration and net CO2
fixation?
A. Photorespiration increases; net CO2 fixation decreases.
B. Photorespiration decreases; net CO2 fixation increases.
C. Photorespiration is unchanged; net CO2 fixation is unchanged.
D. Photorespiration decreases; net CO2 fixation is unchanged.
Correct Answer: B. Photorespiration decreases; net CO2 fixation increases.
Rationale: Higher CO2 favors the carboxylase activity of Rubisco over the oxygenase
activity, reducing photorespiration and increasing net CO2 fixation. This is the basis for
the CO2 fertilization effect in C3 plants.
Why Wrong:
A - Higher CO2 suppresses oxygenase activity, not increases it.
C - Higher CO2 changes the substrate saturation of Rubisco.
D - Net CO2 fixation increases because of reduced photorespiration and increased
carboxylation.
Reference: Taiz & Zeiger, Plant Physiology, 6th ed., Ch. 9
Page 4
