AP Biology Practice Test — 20 Free Questions with Explanations (2026)

Free AP Biology practice test with 20 multiple-choice questions covering all 8 exam units. Each question includes the correct answer and a detailed explanation. Use this to find weak areas before your exam.

AP Biology Exam Format (2026)

The 2 long FRQ questions (8–10 points each) require you to analyze data, design experiments, and explain biological mechanisms. The 4 short FRQs (4 points each) test specific concepts.

AP Biology Practice Test: 20 Questions

Unit 1 — Chemistry of Life

Question 1. A researcher places red blood cells in a solution that causes them to swell and eventually burst. The solution is best described as:

A) Hypertonic — more solutes than inside the cell
B) Isotonic — equal solute concentration to inside the cell
C) Hypotonic — fewer solutes than inside the cell
D) Saturated — containing dissolved gases that expand inside the cell

Answer: C

In a hypotonic solution, solute concentration is lower outside than inside the cell. Water moves into the cell by osmosis (from low solute = high water concentration to high solute = low water concentration), causing the cell to swell (lyse in animal cells). Hypertonic solutions cause cells to shrink (crenation).

Question 2. Which property of water is primarily responsible for the ability of water to moderate temperature in large bodies such as oceans?

A) Water is a polar molecule that forms hydrogen bonds
B) Water has a high heat capacity due to hydrogen bonding
C) Water expands when it freezes, making ice less dense
D) Water is an effective solvent that dissolves many ionic compounds

Answer: B

Water has an unusually high specific heat capacity — it takes a large input of energy to raise its temperature. This is because hydrogen bonds must be broken before water molecules can move faster (heat up). This allows oceans and large lakes to absorb heat with minimal temperature change, moderating climate.

Unit 2 — Cell Structure and Function

Question 3. A liver cell contains a large number of smooth endoplasmic reticulum (SER) compared to a typical cell. The most likely function of this cell is:

A) Producing and secreting large amounts of protein antibodies
B) Detoxifying drugs and lipid metabolism
C) Performing photosynthesis
D) Rapid cell division and DNA replication

Answer: B

The smooth ER (lacking ribosomes) is involved in lipid synthesis, steroid hormone production, and detoxification of drugs and poisons. Liver cells have abundant SER specifically for these metabolic functions. Rough ER (with ribosomes) is used for protein synthesis and secretion — relevant to antibody-producing cells.

Question 4. The sodium-potassium pump moves 3 Na+ ions out and 2 K+ ions in for each ATP hydrolyzed. This is an example of:

A) Facilitated diffusion — proteins assist movement down a gradient
B) Active transport — movement against a concentration gradient using ATP
C) Cotransport — using one gradient to drive another
D) Exocytosis — membrane-enclosed transport of ions

Answer: B

Active transport uses energy (ATP) to move substances against their concentration gradient. Both Na+ and K+ are moved against their electrochemical gradients by the sodium-potassium pump. Facilitated diffusion does not require ATP. Cotransport (secondary active transport) uses a pre-existing gradient created by a pump like this one.

Unit 3 — Cellular Energetics

Question 5. During glycolysis, one molecule of glucose is converted into two molecules of pyruvate. What is the net ATP yield from this process?

A) 0 ATP (ATP used equals ATP produced)
B) 2 ATP net
C) 4 ATP net
D) 36–38 ATP net

Answer: B

Glycolysis produces 4 ATP (substrate-level phosphorylation) but uses 2 ATP in the energy investment phase, for a net yield of 2 ATP. It also produces 2 NADH. The 36–38 ATP yield includes the full aerobic respiration process (glycolysis + Krebs + electron transport chain).

Question 6. A plant is placed in the dark for 24 hours. Which of the following would you expect to observe?

A) Photosynthesis continues using stored ATP from the previous day
B) The plant undergoes only cellular respiration, consuming oxygen
C) The Calvin cycle continues but the light reactions stop
D) The plant stops producing glucose and CO2 levels inside the plant decrease

Answer: B

In darkness, the light reactions (which require photons) stop, halting photosynthesis. The plant continues cellular respiration, consuming glucose and oxygen and releasing CO2. The Calvin cycle requires ATP and NADPH from the light reactions, so it also stops without light. CO2 levels inside the plant rise (not decrease) as respiration continues.

Unit 4 — Cell Communication and Cell Cycle

Question 7. A signal molecule binds to a G-protein coupled receptor (GPCR) on the cell surface. The signal molecule itself does not enter the cell. This process is an example of:

A) Direct communication via gap junctions
B) Signal transduction — an extracellular signal triggers intracellular changes
C) Intracrine signaling — the cell responds to its own secreted signal
D) Autocrine signaling through nuclear receptors

Answer: B

Signal transduction converts an extracellular signal (the ligand binding the receptor) into an intracellular response through a relay of molecular events (G-protein activation → second messengers → cellular response). The signal molecule (first messenger) does not need to enter the cell — its binding triggers internal changes.

Question 8. During mitosis, sister chromatids separate during which phase?

A) Prophase
B) Metaphase
C) Anaphase
D) Telophase

Answer: C

In anaphase of mitosis, the cohesin proteins holding sister chromatids together are cleaved, and the chromatids are pulled to opposite poles by spindle fibers (kinetochore microtubules). In meiosis I, homologous chromosomes separate; in meiosis II, sister chromatids separate (also in anaphase II).

Unit 5 — Heredity

Question 9. In guinea pigs, black fur (B) is dominant to white fur (b). A cross between two heterozygous parents (Bb × Bb) produces 80 offspring. How many would you expect to have white fur?

A) 10
B) 20
C) 40
D) 60

Answer: B

From a Bb × Bb cross: 1/4 BB, 2/4 Bb, 1/4 bb. White fur = bb = 1/4 of offspring. Expected white offspring = 80 × 1/4 = 20.

Question 10. A woman who is a carrier for hemophilia (X^H X^h) has children with a man who does not have hemophilia (X^H Y). What is the probability that their son will have hemophilia?

A) 0%
B) 25%
C) 50%
D) 100%

Answer: C

Sons receive the Y chromosome from dad and either X^H or X^h from mom (carrier). 50% chance son gets X^h (hemophilia) and 50% chance son gets X^H (unaffected). So 50% of sons will have hemophilia. Daughters cannot be hemophiliac from this cross (they get X^H from dad), but 50% will be carriers.

Unit 6 — Gene Expression and Regulation

Question 11. A mutation in the promoter region of a gene prevents RNA polymerase from binding. The most likely result is:

A) The gene is transcribed more rapidly than normal
B) A non-functional mRNA is produced
C) Transcription of the gene does not occur
D) Translation is blocked but transcription continues normally

Answer: C

The promoter is the DNA sequence where RNA polymerase binds to initiate transcription. If RNA polymerase cannot bind, transcription does not begin and no mRNA is produced. This is distinct from mutations in the coding sequence (which affect the mRNA/protein) or regulatory sequences that affect transcription rate.

Question 12. In bacteria, the lac operon is turned on when:

A) Glucose is present and lactose is absent
B) Lactose is present and glucose is absent (or low)
C) The repressor protein binds to the operator
D) cAMP levels are low

Answer: B

The lac operon in E. coli is regulated by two signals: lactose availability and glucose availability. Lactose (allolactose) inactivates the repressor (allowing transcription), and low glucose increases cAMP which activates CAP (catabolite activator protein), further stimulating transcription. The operon is fully induced only when lactose is present AND glucose is absent/low.

Unit 7 — Natural Selection and Evolution

Question 13. A population of beetles lives in a forest. Green beetles survive better on green leaves (camouflage) and are more likely to reproduce. Over many generations, the proportion of green beetles increases. This is an example of:

A) Genetic drift
B) Artificial selection
C) Natural selection
D) Gene flow

Answer: C

Natural selection requires: heritable variation (color differences are inherited), differential reproductive success based on that variation (green beetles survive better), and environmental pressure (predation). The increase in green beetles over generations due to survival advantage is textbook natural selection.

Question 14. Which of the following provides evidence for common ancestry between humans and chimpanzees?

A) Humans and chimpanzees live in overlapping geographic ranges
B) Both humans and chimpanzees have opposable thumbs for similar environmental reasons
C) Comparing DNA sequences shows about 98–99% similarity between human and chimpanzee genomes
D) Both species have similar diets

Answer: C

Molecular evidence — especially DNA sequence comparisons — is among the strongest evidence for evolutionary relationships. A 98–99% genomic similarity between humans and chimpanzees indicates very recent common ancestry. Geographic overlap and similar anatomy could result from convergent evolution rather than common descent.

Unit 8 — Ecology

Question 15. In a food web, energy is lost between trophic levels. If a producer captures 10,000 kcal of solar energy, how much energy is available to a tertiary consumer (assuming 10% energy transfer efficiency at each level)?

A) 1,000 kcal
B) 100 kcal
C) 10 kcal
D) 1 kcal

Answer: C

At 10% efficiency: Producer (10,000 kcal) → Primary consumer (1,000 kcal) → Secondary consumer (100 kcal) → Tertiary consumer (10 kcal). Energy is lost as heat and cellular respiration at each level, which is why food chains rarely exceed 4–5 trophic levels.

Question 16. Which of the following is an example of a density-dependent limiting factor on a rabbit population?

A) A drought that reduces water availability for all rabbits equally
B) A wildfire that destroys habitat regardless of population size
C) Competition among rabbits for food that intensifies as population grows
D) A cold winter that kills a fixed number of rabbits

Answer: C

Density-dependent factors become more intense as population density increases. Competition for food, disease transmission, and predation pressure increase as populations grow. Density-independent factors (drought, wildfire, extreme temperature) affect populations regardless of their density — they exert the same pressure whether there are 10 or 10,000 rabbits.

Bonus Questions — Mixed Topics

Question 17. A scientist discovers a new unicellular organism that has a nucleus, mitochondria, and chloroplasts. This organism should be classified as:

A) Prokaryote
B) Bacterium
C) Archaeon
D) Eukaryote

Answer: D

The presence of a nucleus and membrane-bound organelles (mitochondria, chloroplasts) defines eukaryotes. Bacteria and archaea are prokaryotes — they lack a nucleus and membrane-bound organelles. This organism is likely a photosynthetic protist (such as algae).

Question 18. The diagram below shows an enzyme reaction rate vs. substrate concentration. The curve levels off at Vmax. Adding more enzyme molecules to the reaction would:

A) Have no effect because Vmax is already reached
B) Shift Vmax upward, allowing a higher maximum reaction rate
C) Decrease Km, making the enzyme more efficient
D) Change the shape of the active site

Answer: B

Vmax is determined by the total amount of enzyme present. Adding more enzyme increases the number of active sites available, increasing Vmax (more substrate can be converted per unit time at saturation). Km reflects enzyme-substrate affinity and is not changed by adding more enzyme.

Question 19. Which of the following best explains why the inner membrane of mitochondria is highly folded into cristae?

A) Folding protects the electron transport chain from damage
B) Increased surface area allows more ATP synthase and ETC proteins, increasing ATP output
C) Cristae store the pyruvate used in the Krebs cycle
D) Folding allows the mitochondria to fit inside the cell

Answer: B

The cristae dramatically increase the surface area of the inner mitochondrial membrane, allowing more electron transport chain complexes and ATP synthase to be embedded. More ATP synthase means more proton gradient can be harnessed to produce ATP. This is the same principle as intestinal villi — surface area maximizes functional capacity.

Question 20. A student finds that a new chemical inhibits acetylcholinesterase (the enzyme that breaks down acetylcholine in synapses). What effect would this have on a synapse?

A) Action potentials would be blocked from reaching the synapse
B) Acetylcholine would accumulate in the synaptic cleft, causing prolonged stimulation
C) The postsynaptic neuron would stop producing receptors for acetylcholine
D) Less acetylcholine would be released from the presynaptic terminal

Answer: B

Acetylcholinesterase normally breaks down acetylcholine in the synaptic cleft, terminating the signal. If this enzyme is inhibited, acetylcholine accumulates and continues binding to receptors on the postsynaptic cell, causing sustained stimulation. This is the mechanism of nerve agents (organophosphates) and some pesticides.

Score Estimation

Use our AP Biology Score Calculator to estimate your full exam score based on actual MC and FRQ raw scores.

AP Biology Score Cutoffs (2026)

Most-Tested AP Biology Topics

Units 3, 6, and 7 together account for 37–52% of the exam — master these first.

AP Biology FRQ Tips

• Long FRQ (8–10 pts): Usually involves data analysis, experimental design, or explaining a mechanism. Show all reasoning.
• Short FRQ (4 pts): Direct concept application — often "describe and explain" or "predict and justify."
• Data interpretation: Practice reading graphs, tables, and experimental setups and drawing biological conclusions.
• Experimental design: Know how to identify independent/dependent variables, controls, and potential sources of error.
• Never leave a part blank — partial credit is available.

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