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AP Biology Practice Questions

16 original exam-style questions covering all of AP Biology. Answer each one to see the explanation — no account needed.

Question 1 of 16 · Enzyme Structure and Function

A student measures the activity of the enzyme amylase at various pH levels, keeping temperature and substrate concentration constant. The results are recorded in the table below.

Amylase Activity at Different pH Levels
pHReaction Rate (mg starch/min)Relative Activity (%)
20.34
42.128
67.5100
77.397
84.864
100.811
120.11
Which of the following best explains why amylase activity drops sharply below pH 4?
  1. A. The substrate starch is chemically degraded at low pH, leaving nothing for the enzyme to bind
  2. B. Acidic conditions cause the enzyme to form additional disulfide bonds that increase its activity
  3. C. Low pH increases the kinetic energy of the enzyme, causing it to move too quickly to bind substrate
  4. D. Excess hydrogen ions alter the ionization state of amino acid R-groups at the active site, changing its shape
Show answer and explanation

Correct answer: D

Extreme pH values protonate or deprotonate amino acid side chains at the active site, disrupting the precise three-dimensional shape needed for substrate binding and catalysis. Choice A is incorrect because starch is relatively stable at low pH under these experimental conditions.

Question 2 of 16 · Properties of Water

Which of the following properties of water is most directly responsible for water's role as a universal solvent for ionic and polar molecules?
  1. A. Water molecules form covalent bonds with solute ions
  2. B. Cohesion between water molecules creates surface tension that holds solutes in solution
  3. C. Water has a high specific heat capacity that prevents solutes from precipitating
  4. D. The polarity of water molecules allows them to surround and stabilize charged or polar solutes
Show answer and explanation

Correct answer: D

Water's partial negative charge on oxygen and partial positive charges on hydrogen atoms allow it to orient around dissolved ions and polar molecules, stabilizing them in solution via electrostatic interactions. High specific heat (C) and cohesion (D) are important water properties but do not explain solvent capacity.

Question 3 of 16 · Osmosis and Water Potential

A researcher investigates osmosis in plant cells by placing leaf discs in solutions of varying sucrose concentrations and measuring the change in mass after 30 minutes.

Mass Change of Leaf Discs in Sucrose Solutions
Sucrose Concentration (M)Initial Mass (g)Final Mass (g)% Change in Mass
0.02.002.24+12
0.22.002.12+6
0.42.002.01+0.5
0.62.001.88-6
0.82.001.74-13
Based on the data, what is the approximate solute concentration of the leaf cell cytoplasm?
  1. A. 0.0 M sucrose, because cells gain mass in pure water
  2. B. 0.2 M sucrose, because cells gain the most mass in that solution
  3. C. Approximately 0.4 M sucrose, because cells show almost no mass change in that solution
  4. D. 0.8 M sucrose, because cells lose mass in that solution
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Correct answer: C

When cell mass does not change, the cell's water potential equals that of the surrounding solution; the 0.4 M solution results in ~0.5% change (essentially no net water movement), indicating the cell's internal solute concentration is approximately 0.4 M. Choice A confuses mass gain (hypotonic solution) with isotonic conditions.

Question 4 of 16 · Membrane Transport

Cystic fibrosis is caused by a mutation in the CFTR gene encoding a chloride ion channel. In affected cells, the misfolded CFTR protein is retained in the endoplasmic reticulum and degraded by the proteasome before reaching the plasma membrane. Which of the following best explains how this leads to the symptoms of cystic fibrosis?
  1. A. Without functional CFTR at the cell surface, ClX−\ce{Cl-} cannot move across the apical membrane, leading to thick, dehydrated mucus secretions
  2. B. The misfolded protein triggers apoptosis of all epithelial cells, destroying the lining of the lungs and digestive tract
  3. C. CFTR degradation in the ER depletes cells of ATP, preventing active transport of nutrients across epithelial membranes
  4. D. Proteasomal degradation of misfolded CFTR releases toxic peptide fragments that damage the plasma membrane of neighboring cells
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Correct answer: A

CFTR functions as a ClX−\ce{Cl-} channel at the apical surface of epithelial cells; without functional CFTR, chloride secretion is impaired, reducing water movement into the mucus layer by osmosis and producing thick, sticky secretions that obstruct airways and ducts. Choice B is an extreme overstatement — CF patients have dysfunctional, not destroyed, epithelial cells.

Question 5 of 16 · ATP Synthesis Mechanisms

Which of the following correctly describes a key difference between substrate-level phosphorylation and oxidative phosphorylation in cellular respiration?
  1. A. Substrate-level phosphorylation requires oxygen, while oxidative phosphorylation can occur under anaerobic conditions
  2. B. Substrate-level phosphorylation directly transfers a phosphate group from a substrate to ADP, while oxidative phosphorylation uses the proton gradient to drive ATP synthesis
  3. C. Substrate-level phosphorylation occurs in the mitochondrial matrix only, while oxidative phosphorylation occurs in the cytoplasm
  4. D. Substrate-level phosphorylation produces more ATP per glucose than oxidative phosphorylation because it is a more direct process
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Correct answer: B

Substrate-level phosphorylation involves direct enzymatic transfer of a phosphate from a high-energy intermediate to ADP (e.g., in glycolysis and the Krebs cycle), while oxidative phosphorylation couples the proton motive force from the ETC to ATP synthase activity. Choice A inverts the requirement — it is oxidative phosphorylation (via the ETC) that requires OX2\ce{O2} as the terminal electron acceptor, not substrate-level phosphorylation.

Question 6 of 16 · Cellular Respiration

Students measure the rate of cellular respiration in germinating pea seeds using a respirometer that tracks O2O_2 consumption (mL) over time. Two sets of tubes are prepared: one with germinating seeds (living) and one with dry seeds (controls). Tubes are placed at two temperatures. The table below shows cumulative O2O_2 consumed after 20 minutes.

Table 2. O₂ Consumed (mL) by Germinating vs. Dry Pea Seeds
Condition10°C — Germinating10°C — Dry Seeds25°C — Germinating25°C — Dry Seeds
O₂ consumed (mL)0.80.02.40.0
Which of the following conclusions is best supported by the data in Table 2?
  1. A. Dry seeds consume more O2O_2 per unit mass than germinating seeds because their metabolic enzymes are more concentrated
  2. B. The dry seeds serve as a positive control, confirming that the respirometer accurately measures O2O_2 production
  3. C. Germinating seeds switch from aerobic respiration to fermentation when temperature drops to 10°C
  4. D. Germinating seeds carry out aerobic cellular respiration, and their metabolic rate increases with temperature
Show answer and explanation

Correct answer: D

Germinating seeds consume measurable O2O_2 at both temperatures (indicating aerobic respiration), and consumption at 25°C (2.4 mL) is three times higher than at 10°C (0.8 mL), demonstrating that metabolic rate increases with temperature. Choice C is incorrect because the data show that germinating seeds still consume O2O_2 at 10°C, which is inconsistent with a complete switch to fermentation — an anaerobic process that involves no O2O_2 consumption.

Question 7 of 16 · Cell Cycle

Students examined cells from a root tip meristem under a microscope and counted the number of cells in each phase of the cell cycle. Their results are summarized below.

Cell cycle phase distribution in onion root tip cells (n = 500 cells observed)
PhaseNumber of CellsPercentage of Cells
Interphase (G1, S, G2)44088%
Prophase326.4%
Metaphase112.2%
Anaphase81.6%
Telophase/Cytokinesis91.8%
A student claims that cells spend more time in mitosis than in interphase. Which of the following best evaluates this claim using the data?
  1. A. The claim is supported because mitotic figures are visible in 12% of cells.
  2. B. The claim cannot be evaluated because microscope snapshots do not reflect time spent in each phase.
  3. C. The claim is supported because prophase is the longest individual phase based on cell count.
  4. D. The claim is refuted because 88% of cells are in interphase, indicating cells spend far more time in interphase than in mitosis.
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Correct answer: D

The proportion of cells in each phase reflects the relative time spent there. Since 88% of cells are in interphase versus 12% in mitosis, cells spend roughly seven times longer in interphase. The claim is therefore refuted by the data.

Question 8 of 16 · Stages of Mitosis

In a diploid organism with 2n = 6, how many chromosomes are present in a cell at the START of anaphase of mitosis, and how many chromatids are being separated?
  1. A. 6 chromosomes, separating 12 chromatids (sister pairs)
  2. B. 12 chromosomes, each consisting of a single chromatid after separation
  3. C. 3 chromosomes moving to each pole, with 6 chromatids per pole
  4. D. 6 chromosomes, separating into 6 chromatids total
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Correct answer: A

At the start of anaphase, the cell still has 6 chromosomes (each composed of two sister chromatids), and cohesin cleavage separates all 12 chromatids simultaneously. After separation each individual chromatid is considered a chromosome, giving 12 chromosomes total moving to the poles — but at the START of anaphase, there are 6 chromosomes with 12 chromatids separating.

Question 9 of 16 · Dihybrid Crosses and Testcross Design

A genetics student performed a series of crosses to investigate the inheritance of two traits in pea plants: seed color (yellow, Y dominant over green, y) and seed texture (round, R dominant over wrinkled, r). The student crossed two true-breeding strains to obtain an F1 dihybrid, then performed a testcross with a homozygous recessive plant. The progeny counts and chi-square analysis are shown below.

Testcross Progeny: F1 (YyRr) × yyrr
PhenotypeObserved (O)Expected (E)(O−E)2/E(O - E)^2 / E
Yellow, Round2182001.62
Yellow, Wrinkled1862000.98
Green, Round1922000.32
Green, Wrinkled2042000.08
**Total**800800χ2=3.00\chi^2 = 3.00

Critical value for chi-square at p=0.05p = 0.05 with 3 degrees of freedom: χcrit2=7.815\chi^2_{crit} = 7.815

Which of the following best explains why the student used a testcross rather than an F1 × F1 cross to analyze whether these two genes assort independently?
  1. A. A testcross directly reveals the gamete types produced by the F1 parent, because the homozygous recessive parent contributes only recessive alleles to all offspring
  2. B. A testcross eliminates the need for statistical analysis because all four offspring phenotype classes will appear in exactly equal numbers
  3. C. A testcross produces a 9:3:3:1 phenotype ratio that is more diagnostic of independent assortment than other cross types
  4. D. A testcross ensures the recessive parent contributes dominant alleles, so offspring phenotype depends entirely on the recessive parent's genotype
Show answer and explanation

Correct answer: A

The yyrr testcross parent contributes only recessive alleles, so each offspring phenotype directly reflects which gamete the F1 parent (YyRr) produced; a 1:1:1:1 phenotype ratio is expected if the two loci assort independently. Choice C reverses the cross types — the 9:3:3:1 ratio comes from an F1 × F1 cross, not a testcross.

Question 10 of 16 · Patterns of Inheritance

In a human genetic study, a trait appears in every generation, affects both males and females equally, and an affected individual always has at least one affected parent. Which of the following inheritance patterns is most consistent with these observations?
  1. A. Autosomal recessive, because the trait can appear in both sexes
  2. B. X-linked recessive, because the trait is passed from affected mothers to all sons
  3. C. Autosomal dominant, because every affected individual has an affected parent and both sexes are equally affected
  4. D. X-linked dominant, because the trait appears in every generation without skipping
Show answer and explanation

Correct answer: C

Autosomal dominant traits appear in every generation, affect both sexes equally, and every affected individual has an affected parent — all consistent with the observations. X-linked recessive (B) would predominantly affect males and often skip generations in females, which contradicts the data.

Question 11 of 16 · Epigenetic Regulation

Researchers used chromatin immunoprecipitation (ChIP) to study histone modification patterns near the promoters of three genes in cancer cells compared to normal cells. H3K27me3 is a repressive mark (gene silencing); H3K4me3 is an activating mark (gene expression).

Histone modification enrichment at gene promoters (fold-enrichment relative to control)
GeneH3K27me3 (Normal)H3K27me3 (Cancer)H3K4me3 (Normal)H3K4me3 (Cancer)
CDKN2A (tumor suppressor)1.08.36.21.1
MYC (proto-oncogene)0.90.81.59.4
BRCA1 (DNA repair)1.16.95.80.9
Based on the ChIP data, which of the following conclusions about gene expression in cancer cells is best supported?
  1. A. Tumor suppressor genes are upregulated and proto-oncogenes are silenced in cancer cells.
  2. B. DNA sequence mutations in the promoters of these genes explain the altered expression patterns.
  3. C. Tumor suppressors (CDKN2A, BRCA1) are epigenetically silenced while MYC is overactivated in cancer cells.
  4. D. Histone modifications have no effect on gene expression; only transcription factors control cancer cell proliferation.
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Correct answer: C

CDKN2A and BRCA1 gain repressive H3K27me3 marks and lose activating H3K4me3 marks in cancer cells, indicating epigenetic silencing. MYC gains H3K4me3 (activating), indicating overexpression. This pattern is epigenetic — ChIP detects histone marks, not DNA sequence changes, ruling out option C.

Question 12 of 16 · Epigenetic Regulation

A scientist discovers that inhibiting histone acetyltransferase (HAT) enzymes dramatically reduces transcription of many genes. Which of the following best explains this finding?
  1. A. Histone acetylation compacts chromatin, preventing transcription factors from accessing promoters.
  2. B. Histone acetylation neutralizes the positive charge of histone tails, relaxing chromatin and allowing transcription factors to bind.
  3. C. HAT enzymes directly transcribe mRNA from the DNA template, so inhibiting them stops transcription.
  4. D. Histone acetylation degrades mRNA after transcription, so inhibiting HAT stabilizes mRNA and increases protein levels.
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Correct answer: B

Acetylation of lysine residues on histone tails by HAT removes their positive charge, weakening histone-DNA electrostatic interactions and decondensing chromatin into euchromatin. This open chromatin configuration is accessible to transcription factors. Inhibiting HAT keeps chromatin compacted (hypoacetylated), reducing transcription. Histone deacetylation (not acetylation) condenses chromatin.

Question 13 of 16 · Hardy-Weinberg Equilibrium

A population of 10,000 beetles has two alleles for body color: BB (black, dominant) and bb (brown, recessive). The frequency of the recessive allele bb is 0.30. Assuming Hardy-Weinberg equilibrium, which of the following is the expected frequency of black beetles that are heterozygous carriers?
  1. A. 0.09
  2. B. 0.42
  3. C. 0.49
  4. D. 0.70
Show answer and explanation

Correct answer: B

With q=0.30q = 0.30, then p=0.70p = 0.70. Heterozygous (BbBb) frequency =2pq=2(0.70)(0.30)=0.42= 2pq = 2(0.70)(0.30) = 0.42. A common error is calculating p2=0.49p^2 = 0.49 (homozygous dominant) and mistaking it for the heterozygote frequency, or reporting q2=0.09q^2 = 0.09 (homozygous recessive).

Question 14 of 16 · Reproductive Isolating Mechanisms

Reproductive isolation between two sympatric species is maintained by the fact that one species flowers in March while the other flowers in June. This is an example of which type of reproductive isolating mechanism?
  1. A. Mechanical isolation, because the flower structures are physically incompatible.
  2. B. Behavioral isolation, because the plants learned to flower at different times.
  3. C. Temporal isolation, because the two species are reproductively active at different times.
  4. D. Habitat isolation, because the species occupy different soil microhabitats within the same forest.
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Correct answer: C

Temporal isolation is a prezygotic reproductive barrier in which two species are sexually active or mature at different times (seasons, times of day), preventing gamete encounter; different flowering seasons in plants is the classic example. Mechanical isolation specifically refers to incompatible reproductive morphology, not timing.

Question 15 of 16 · Top-Down Control and Trophic Cascades

Which of the following best explains why adding a top predator to an ecosystem often increases plant biomass?
  1. A. Top predators directly consume herbivores, reducing herbivory pressure and allowing plant biomass to increase.
  2. B. Top predators compete with plants for abiotic resources, stimulating plant growth through competition.
  3. C. Top predators release nutrients through their excretions, acting as a fertilizer for plant growth.
  4. D. Top predators replace plants in the food web, reducing competition for sunlight among producers.
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Correct answer: A

This is the top-down control mechanism underlying trophic cascades: top predators suppress herbivore populations, reducing the amount of plant material consumed; the resulting release from herbivory allows plants to grow to higher biomass. This three-trophic-level cascade (predator → herbivore → plant) is well-documented in terrestrial, freshwater, and marine systems.

Question 16 of 16 · Community Ecology

Two species of barnacles, Chthamalus and Semibalanus, live in the rocky intertidal zone. Chthamalus occupies the upper intertidal zone, while Semibalanus occupies the lower zone. When Semibalanus is experimentally removed, Chthamalus expands into the lower zone; when Chthamalus is removed, Semibalanus does not expand upward. Which of the following BEST explains these observations?
  1. A. Chthamalus has a broader fundamental niche than Semibalanus, allowing it to survive in both zones physiologically.
  2. B. Chthamalus is excluded from the lower zone by interspecific competition with Semibalanus, which is the superior competitor in that zone.
  3. C. Semibalanus cannot survive desiccation in the upper intertidal zone, and competition with Chthamalus further prevents its upward movement.
  4. D. Both competitive exclusion and physical tolerance limit each species to its realized niche in the intertidal zone.
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Correct answer: D

Chthamalus is physiologically capable of living lower (broader fundamental niche) but is excluded by Semibalanus competition — its realized niche is restricted. Semibalanus cannot survive upper-zone desiccation regardless of competition, so its realized and fundamental niches overlap in the lower zone. Both mechanisms operate simultaneously.

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