AP Biology Practice Questions: 25 MCQs With Explanations
Most free AP Biology practice question sets give you the right answer letter and nothing else. That format tells you whether you were lucky; it does not tell you why your reasoning was wrong. Each question here includes a step-by-step explanation that addresses every distractor, so students who miss a question still walk away with a clearer understanding of the concept than they had before.
The 25 AP Biology sample questions below are distributed across the four big ideas proportional to their weight in the College Board AP Biology Course and Exam Description (CED). Units 7, 3, and 6 carry the most exam weight and get the most coverage here. No College Board released questions appear verbatim; every question is original and reviewed by AP-credentialed contributors for CED alignment and biological accuracy.
How to Use These AP Biology Practice Questions Effectively
A practice session that builds real exam skill looks different from one that just checks whether you already know the answer.
Format and Scoring Yourself
Cover the answer section of each question before reading it. Write your answer choice on paper or in a notes app. Only after committing to an answer should you click “Show Answer & Explanation.” This retrieval practice structure matters. Research by Roediger and Karpicke (2006) found that actively retrieving information produces significantly better long-term retention than reviewing material passively, even when the retrieval attempt is unsuccessful.
Score yourself out of 25. Each correct answer counts as 1 point. There is no guessing penalty on the AP Biology exam, so commit to every answer rather than leaving it blank. Target at least 17 correct (68%) before the exam to give yourself a realistic shot at a 4 or 5. For full exam-day logistics, see the AP Biology exam 2026 guide.
The AP Biology MCQ section gives you 90 minutes for 60 questions (90 seconds each). For this 25-question set, set a 38-minute timer. Timed practice trains the skill of committing to answers efficiently rather than cycling through options indefinitely.
Question Distribution by Big Idea
The questions follow the CED's four big ideas, which organize the entire AP Biology course. Treating this as a timed AP Biology practice test — rather than a casual read-through — will give you the most accurate read on where your reasoning gaps are. Understanding which big idea each unit belongs to helps you see the conceptual structure of the exam, not just a list of disconnected topics.
Big Idea 1: Evolution (7 Questions)
Evolution is the conceptual backbone of AP Biology. Questions tied to Big Idea 1 appear across multiple unit contexts: population genetics and Hardy-Weinberg in Unit 7, molecular evidence and phylogenetics also in Unit 7, and physiological adaptations connecting back to Unit 1. Expect roughly 1 in 8 exam questions to center on evolutionary reasoning.
Hardy-Weinberg and Population Genetics
Q1 · Accessible · Unit 7 · Evolution
In a population of 500 flowering plants, 320 have purple flowers (dominant phenotype) and 180 have white flowers (recessive phenotype). Assuming Hardy-Weinberg equilibrium, approximately what fraction of the population is heterozygous for flower color?
- A.0.36
- B.0.48
- C.0.16
- D.0.32
Show Answer & Explanation
Correct Answer: B
0.48
White-flowered plants are recessive homozygotes: q² = 180/500 = 0.36, so q = √0.36 = 0.6. Then p = 1 − 0.6 = 0.4. The heterozygote frequency is 2pq = 2(0.4)(0.6) = 0.48. Choice A (0.36) is q², the frequency of white-flowered plants, not heterozygotes. Choice C (0.16) is p², the frequency of purple homozygotes. Choice D (0.32) is a common arithmetic error of doubling p instead of calculating 2pq. Tests CED Unit 7, Science Practice 5 (quantitative analysis).
Q2 · Moderate · Unit 7 · Evolution
In a population of field mice, individuals with medium brown fur survive predation better than individuals with either very light or very dark fur. Which type of natural selection does this represent, and what is its effect on phenotypic variation over time?
- A.Directional selection; it shifts the distribution toward one extreme, reducing variation
- B.Disruptive selection; it favors both extremes and increases bimodal variation
- C.Stabilizing selection; it favors the intermediate phenotype and narrows phenotypic variation
- D.Artificial selection; it reduces variation by enforcing human-defined trait preferences
Show Answer & Explanation
Correct Answer: C
Stabilizing selection; it favors the intermediate phenotype and narrows phenotypic variation
Stabilizing selection acts against both phenotypic extremes and favors the intermediate form, narrowing the distribution and reducing variation over time. Choice A describes directional selection, where one extreme phenotype is favored. Choice B describes disruptive selection, which favors both extremes over the intermediate. Choice D is incorrect because artificial selection involves human-imposed pressure, not natural predation. CED Unit 7, Big Idea 1 (Evolution).
Natural Selection and Speciation
Questions on natural selection test whether you can identify which selective pressure applies to a given scenario, not just whether you can name the three types. Speciation questions typically hinge on the distinction between prezygotic and postzygotic barriers, and on whether geographic separation (allopatry) or co-occurrence (sympatry) is involved.
Q3 · Moderate · Unit 7 · Evolution
A cladogram shows the following grouping: ((Lamprey, Shark), (Frog, (Lizard, Human))). Which statement is best supported by this cladogram?
- A.Sharks and frogs share a more recent common ancestor than sharks and lampreys do
- B.Lizards and humans are more closely related to each other than either is to frogs
- C.Lampreys are more primitive than sharks and therefore evolved later in geological time
- D.Frogs and humans are equally distantly related to sharks
Show Answer & Explanation
Correct Answer: B
Lizards and humans are more closely related to each other than either is to frogs
In a cladogram, closer branching indicates more recent common ancestry. Lizards and humans share a more recent node than the one connecting either to frogs, placing them in a nested clade together. Choice A is incorrect: sharks and lampreys share an outer node, not frogs and sharks. Choice C reflects a common misconception that cladograms rank organisms by advancement; they show relationships, not a progression from "primitive" to "advanced." Choice D is incorrect because humans share a more recent common ancestor with frogs than sharks do. CED Unit 7.
Q4 · Hard · Unit 7 · Evolution
Cytochrome c, a protein in the electron transport chain, shows only 1 amino acid difference between humans and chimpanzees, but 44 differences between humans and yeast. Which conclusion is best supported by this data?
- A.Cytochrome c is non-functional in yeast because of the high number of mutations
- B.Humans and chimpanzees diverged from a common ancestor more recently than humans and yeast did
- C.The rate of mutation in cytochrome c is higher in yeast than in humans
- D.Cellular respiration is less efficient in yeast than in humans, causing more mutations
Show Answer & Explanation
Correct Answer: B
Humans and chimpanzees diverged from a common ancestor more recently than humans and yeast did
Fewer amino acid differences in homologous proteins reflect more recent divergence from a common ancestor; more differences reflect ancient divergence. This is the logic of molecular phylogenetics. Choice A is incorrect: cytochrome c is functional in yeast and is essential to aerobic respiration. Choice C cannot be concluded from amino acid sequence comparisons alone, which measure cumulative change, not current mutation rates. Choice D conflates metabolic efficiency with mutation rate, two unrelated phenomena. CED Unit 7, Science Practice 6.
Q5 · Moderate · Unit 7 · Evolution
A small group of colonists settles an isolated island. Within 10 generations, the frequency of a recessive allele for blood type increases dramatically compared to the mainland population, even though the allele confers no selective advantage on the island. This scenario best illustrates:
- A.Natural selection acting on neutral alleles through environmental pressure
- B.Gene flow between the island and mainland populations
- C.The founder effect as a form of genetic drift
- D.Disruptive selection reducing genetic variation in the island population
Show Answer & Explanation
Correct Answer: C
The founder effect as a form of genetic drift
The founder effect occurs when a small group establishes a new population, carrying only a subset of the original gene pool. Random sampling causes some alleles to be overrepresented or lost entirely by chance, not by selection. Choice A is incorrect because the scenario specifies no selective advantage. Choice B describes gene flow, which would reduce differences between populations; isolation is described here. Choice D describes selection, which requires differential survival based on a trait. CED Unit 7.
Evidence for Evolution and Adaptation
Q6 · Accessible · Units 1 and 7 · Chemistry of Life and Evolution
Which of the following best represents a physiological adaptation that directly increases an organism's fitness in a cold, high-altitude environment where oxygen partial pressure is low?
- A.A larger body size relative to closely related lowland species (Bergmann's Rule)
- B.Behavioral avoidance of direct sunlight during midday hours
- C.A seasonal change in coat color to match snow cover
- D.Increased red blood cell production in response to lower atmospheric oxygen
Show Answer & Explanation
Correct Answer: D
Increased red blood cell production in response to lower atmospheric oxygen
At high altitude, lower partial pressure of oxygen drives increased erythropoietin production, stimulating the bone marrow to produce more red blood cells. This raises oxygen-carrying capacity directly, improving aerobic function and survival. Choice A represents a thermoregulatory trend across species, not a direct physiological response to altitude oxygen levels. Choice B describes behavioral thermoregulation. Choice C is seasonal camouflage, unrelated to altitude physiology. CED Units 1 and 7.
Q7 · Hard · Unit 7 · Evolution
Two populations of the same bird species are separated by a mountain range for thousands of years. Beak morphology diverges significantly. When the mountain range erodes and populations come into secondary contact, individuals from the two populations rarely mate successfully. This outcome most directly results from:
- A.Habitat isolation preventing any geographic overlap between the two populations
- B.Behavioral reproductive isolation that developed during allopatric speciation
- C.Ecological competition that excludes one population from the shared habitat
- D.Prezygotic mechanical isolation due to physical incompatibility of mating structures
Show Answer & Explanation
Correct Answer: B
Behavioral reproductive isolation that developed during allopatric speciation
Behavioral (ethological) isolation develops when populations diverge in mate-recognition signals during geographic separation. After secondary contact, the two groups rarely mate because courtship signals and preferences no longer align. Choice D would require physical incompatibility of mating structures, which is not described. Choice A describes the original isolation mechanism, not the result of secondary contact. Choice C is competitive exclusion, a different ecological concept with no mating-specific barrier. CED Unit 7.
Hardy-Weinberg equilibrium requires five conditions: a large population, random mating, no mutation, no gene flow, and no natural selection. Any AP question describing a real population violates at least one. Your job is to identify which condition is violated and what that predicts about allele frequency change.
Big Idea 2: Energetics (7 Questions)
Energetics covers two major pathways that interlock: photosynthesis (Units 3) and cellular respiration (Unit 3), plus the cell cycle and signaling systems that regulate energy expenditure (Unit 4). The AP exam tests these at the mechanism level, not just the equation level. Knowing that photosynthesis produces oxygen is not enough; you need to know where in the pathway, why, and what experiment would measure it.
Photosynthesis and Cellular Respiration
Q8 · Accessible · Unit 3 · Cellular Energetics
Which of the following correctly describes what happens to water molecules during the light-dependent reactions of photosynthesis?
- A.Water is reduced at Photosystem I, providing electrons to generate NADPH
- B.Water is oxidized at Photosystem II, releasing electrons, protons, and oxygen gas
- C.Water provides the carbon skeleton used as a substrate in the Calvin cycle
- D.Water is synthesized from H⁺ ions during ATP production by ATP synthase
Show Answer & Explanation
Correct Answer: B
Water is oxidized at Photosystem II, releasing electrons, protons, and oxygen gas
Photolysis at Photosystem II oxidizes water: 2H₂O → 4H⁺ + 4e⁻ + O₂. The electrons replace those lost from P680, H⁺ ions contribute to the proton gradient driving ATP synthesis, and O₂ is released as a byproduct. Choice A is incorrect on two counts: water provides electrons to PSII, not PSI, and the process is oxidation, not reduction. Choice C is incorrect; the Calvin cycle uses CO₂ as its carbon source, not water. Choice D reverses the chemistry entirely. CED Unit 3.
Q9 · Moderate · Unit 3 · Cellular Energetics
A student measures the products of glucose oxidation under anaerobic conditions in a yeast culture. After 2 hours, which set of products would most likely be detected in the growth medium?
- A.CO₂ and H₂O only
- B.Ethanol, CO₂, and regenerated NAD⁺
- C.Lactic acid and ATP only
- D.Pyruvate and NADH only, with no further conversion
Show Answer & Explanation
Correct Answer: B
Ethanol, CO₂, and regenerated NAD⁺
Under anaerobic conditions, yeast perform alcoholic fermentation: pyruvate is converted to acetaldehyde (releasing CO₂), then acetaldehyde is reduced to ethanol, regenerating NAD⁺ from NADH. NAD⁺ regeneration allows glycolysis to continue. Choice A describes aerobic respiration products. Choice C describes lactic acid fermentation, which occurs in animal muscle cells and certain bacteria, not yeast. Choice D is incomplete; pyruvate and NADH are intermediate products that are consumed during fermentation. CED Unit 3.
Q10 · Hard · Unit 3 · Cellular Energetics
A researcher adds a chemical that makes the inner mitochondrial membrane freely permeable to H⁺ ions, bypassing ATP synthase. Which correctly predicts the effect on ATP production and oxygen consumption?
- A.Both ATP production and oxygen consumption decrease
- B.ATP production decreases; oxygen consumption increases
- C.ATP production increases because protons flow more freely toward ATP synthase
- D.Both ATP production and oxygen consumption remain unchanged
Show Answer & Explanation
Correct Answer: B
ATP production decreases; oxygen consumption increases
This describes an uncoupler. By making the membrane permeable to H⁺, the proton gradient is constantly dissipated without driving ATP synthase, so ATP production falls. But the electron transport chain accelerates to try to restore the lost gradient, consuming more O₂ and releasing energy as heat instead. This is how thermogenin (brown adipose tissue) and certain chemical uncouplers like 2,4-dinitrophenol operate. Choice A is wrong because O₂ consumption increases, not decreases. Choice C is wrong; free permeability bypasses ATP synthase rather than directing protons through it. CED Unit 3.
Q11 · Moderate · Unit 3 · Cellular Energetics
An enzyme has a Michaelis constant (Km) of 2.5 mM for its substrate. At what substrate concentration would the reaction rate equal approximately 50% of maximum velocity (Vmax)?
- A.1.25 mM
- B.5.0 mM
- C.2.5 mM
- D.Any concentration above 0 mM, since Vmax/2 is always reached early
Show Answer & Explanation
Correct Answer: C
2.5 mM
By definition, Km is the substrate concentration at which v = Vmax/2. This is the foundational relationship from Michaelis-Menten kinetics. Choice A (1.25 mM) is half the Km, giving a rate below Vmax/2. Choice B (5.0 mM) is twice the Km, giving a rate above Vmax/2 but still below Vmax. Choice D is incorrect; the rate depends on [S] according to the hyperbolic Michaelis-Menten equation, not a generic threshold. CED Unit 3, Science Practice 1.
Experimental Reasoning and Cell Cycle
The AP Biology exam tests experimental design heavily in both the MCQ and FRQ sections. For MCQs, the most common pattern is a controlled experiment setup where you must identify the correct control condition, predict an outcome, or evaluate whether a conclusion follows from the data. Questions 12-14 below practice that reasoning pattern.
Q12 · Accessible · Unit 3 · Cellular Energetics
Which statement correctly explains why fermentation is necessary even though it yields far less ATP per glucose molecule than aerobic respiration?
- A.Fermentation produces more pyruvate per glucose than aerobic respiration does
- B.Fermentation regenerates NAD⁺, allowing glycolysis to continue producing ATP when oxygen is absent
- C.Fermentation directly generates ATP through substrate-level phosphorylation in the mitochondria
- D.Fermentation converts NADH into NADPH, which cells require for anabolic biosynthesis
Show Answer & Explanation
Correct Answer: B
Fermentation regenerates NAD⁺, allowing glycolysis to continue producing ATP when oxygen is absent
Glycolysis converts NAD⁺ to NADH while producing a net gain of 2 ATP. Without oxygen, NADH cannot be reoxidized by the electron transport chain. Fermentation regenerates NAD⁺ by passing electrons from NADH to an organic acceptor (pyruvate or acetaldehyde), allowing glycolysis to continue. Without this, the cell would produce no ATP at all. Choice A is incorrect: both aerobic and anaerobic pathways start with the same glycolysis step, producing the same pyruvate. Choice C is wrong; fermentation does not produce ATP directly, and it occurs in the cytoplasm. Choice D confuses NADH with a biosynthetic coenzyme. CED Unit 3.
Q13 · Hard · Unit 3 · Cellular Energetics
A student compares photosynthetic rates in aquatic plants under red versus green light at equal intensities. Oxygen sensors track O₂ production in sealed chambers. Which chamber serves as the most appropriate negative control?
- A.A plant chamber placed in complete darkness
- B.A plant chamber exposed to white light at the same intensity
- C.A sealed chamber containing only water and no plant, exposed to red light
- D.A plant chamber exposed to ultraviolet light to increase mutation rate
Show Answer & Explanation
Correct Answer: C
A sealed chamber containing only water and no plant, exposed to red light
A negative control confirms that the measured signal (O₂ change) is caused by the treatment variable (photosynthesis in the plant) rather than the medium itself. A water-only chamber under red light tests whether the light-water system alone changes O₂ readings, isolating plant activity as the source. Choice A is a procedural control for respiration, not a negative control for the measurement system. Choice B is a positive control, confirming the system works with white light. Choice D introduces an unrelated confounding variable. CED Unit 3, Science Practice 3.
Q14 · Moderate · Unit 4 · Cell Communication and Cell Cycle
A mutation inactivates the tumor suppressor gene RB (retinoblastoma protein), which normally prevents cell cycle progression by binding to transcription factor E2F. What is the most likely direct consequence of this mutation in affected cells?
- A.Cells arrest in G₂ and cannot complete mitosis regardless of growth signals
- B.Cells progress through the G₁/S checkpoint without the normal growth factor requirement
- C.Cells undergo immediate apoptosis because RB normally prevents programmed cell death
- D.DNA replication occurs twice per cell cycle, producing tetraploid daughter cells
Show Answer & Explanation
Correct Answer: B
Cells progress through the G₁/S checkpoint without the normal growth factor requirement
Unphosphorylated RB binds E2F and prevents transcription of S-phase genes. Growth factors signal CDK-cyclin complexes to phosphorylate RB, releasing E2F and permitting S-phase entry. When RB is inactivated by mutation, E2F remains constitutively free, driving cells through the G₁/S checkpoint even without growth factor signals. This is a key mechanism in many cancers. Choice A describes a G₂/M checkpoint failure. Choice C is incorrect; RB promotes cell cycle arrest, not apoptosis. Choice D would require a failure in replication licensing, not checkpoint control. CED Unit 4.
Big Idea 3: Information Storage and Transmission (6 Questions)
Big Idea 3 covers how genetic information is stored in DNA, expressed through transcription and translation, regulated by cellular machinery, and transmitted through heredity. Units 5 and 6 together account for 20-27% of the exam. Gene regulation questions consistently appear on released exams as multi-part problems because they require understanding both the molecular mechanism and the logic of when a gene should be expressed.
DNA Replication and Gene Expression
Q15 · Accessible · Unit 6 · Gene Expression and Regulation
During DNA replication, which enzyme synthesizes new DNA strands in the 5’ to 3’ direction, reading the template strand in the 3’ to 5’ direction?
- A.DNA ligase
- B.Helicase
- C.Primase
- D.DNA polymerase
Show Answer & Explanation
Correct Answer: D
DNA polymerase
DNA polymerase reads the template 3’ to 5’ and synthesizes the new strand in the 5’ to 3’ direction by adding deoxyribonucleotides. Helicase (Choice B) unwinds the double helix but does not synthesize DNA. Primase (Choice C) synthesizes short RNA primers needed to initiate synthesis but cannot produce DNA itself. Ligase (Choice A) joins Okazaki fragments on the lagging strand after primer removal and does not elongate the primary strand. CED Unit 6.
Q16 · Moderate · Unit 6 · Gene Expression and Regulation
The initiator tRNA recognizes the start codon 5’-AUG-3’. After the first peptide bond forms and the ribosome translocates, which codon occupies the A site if the mRNA reading frame is 5’-AUG-GCC-UAA-3’?
- A.UAA
- B.GCC
- C.AUG
- D.GGA
Show Answer & Explanation
Correct Answer: B
GCC
During initiation, AUG sits in the P site. After the first peptide bond forms and the ribosome translocates one codon in the 5’ to 3’ direction, the next codon (GCC) moves into the A site. UAA (Choice A) is the stop codon that will enter the A site after the next translocation. AUG (Choice C) was the start codon in the P site. GGA (Choice D) does not appear in this reading frame. CED Unit 6.
Q17 · Hard · Unit 6 · Gene Expression and Regulation
In the lac operon of E. coli, a repressor protein normally blocks transcription by binding the operator. Lactose derivatives inactivate the repressor. Transcription also requires cAMP-CAP binding upstream of the promoter. Which condition produces the HIGHEST level of lac operon transcription?
- A.High glucose, high lactose
- B.Low glucose, high lactose
- C.Low glucose, low lactose
- D.High glucose, low lactose
Show Answer & Explanation
Correct Answer: B
Low glucose, high lactose
Maximum transcription requires two conditions: the repressor must be inactivated (by allolactose from lactose) and cAMP-CAP must bind (requiring low glucose, which raises cAMP). With low glucose and high lactose, both conditions are met. Choice A fails the positive-regulation requirement; high glucose keeps cAMP low, so cAMP-CAP does not bind even though the repressor is released. Choice C has no lactose, so the repressor blocks transcription. Choice D satisfies neither condition. CED Unit 6.
Students often memorize that the lac operon needs lactose to be turned on, and stop there. The College Board expects you to know both regulatory systems: the repressor system (negative regulation, requires allolactose to inactivate the repressor) and the cAMP-CAP system (positive regulation, requires low glucose to activate). Miss the second layer and you get Questions 17-type problems wrong every time.
Genetics, Heredity, and Epigenetics
Unit 5 (Heredity) tests Mendelian genetics, non-Mendelian patterns (incomplete dominance, codominance, sex-linkage), and quantitative problems like dihybrid crosses. The College Board has consistently included at least one chi-square or probability question on released exams. Unit 6 extends into epigenetics, where gene expression changes without DNA sequence changes. Epigenetics questions have increased in frequency since the 2020 CED redesign. For a quick-reference overview of all genetics patterns organized by exam weight, see the AP Biology visual reference guide.
Q18 · Moderate · Unit 5 · Heredity
Two pea plants heterozygous for both seed color (Yy, yellow dominant) and seed shape (Rr, round dominant) are crossed. What proportion of offspring would be expected to have green, round seeds?
- A.9/16
- B.3/16
- C.1/16
- D.6/16
Show Answer & Explanation
Correct Answer: B
3/16
A YyRr × YyRr dihybrid cross yields the expected phenotypic ratio 9:3:3:1. Green round seeds require genotype yy (frequency 1/4) and R_ (frequency 3/4): 1/4 × 3/4 = 3/16. Choice A (9/16) is the proportion of yellow round seeds, the most common class. Choice C (1/16) is the proportion of green wrinkled seeds. Choice D (6/16) does not correspond to any standard Mendelian ratio and reflects a calculation error. CED Unit 5.
Q19 · Hard · Unit 5 · Heredity
In snapdragons, red-flowered plants (R¹R¹) crossed with white-flowered plants (R²R²) produce all pink-flowered offspring (R¹R²), demonstrating incomplete dominance. If two pink-flowered plants are crossed, what phenotypic ratio is expected among the offspring?
- A.All pink
- B.3 red : 1 white
- C.1 red : 2 pink : 1 white
- D.1 red : 1 white
Show Answer & Explanation
Correct Answer: C
1 red : 2 pink : 1 white
With incomplete dominance, the heterozygote shows an intermediate phenotype. A cross of R¹R² × R¹R² yields genotypic ratio 1 R¹R¹ : 2 R¹R² : 1 R²R², which maps directly to 1 red : 2 pink : 1 white. Choice A would result only if pink were completely dominant. Choice B (3:1) describes standard Mendelian dominance/recessiveness, not incomplete dominance. Choice D describes a test cross with a recessive homozygote. CED Unit 5.
Q20 · Moderate · Unit 6 · Gene Expression and Regulation
Identical twins with the same DNA sequence develop different rates of breast cancer as they age. The twin who develops cancer earlier shows heavier methylation of histone tails at the promoter regions of tumor suppressor genes. Which conclusion is best supported?
- A.The affected twin has acquired new mutations in tumor suppressor gene coding sequences
- B.Epigenetic histone modifications that condense chromatin can silence gene expression without altering DNA sequence
- C.Environmental exposure caused new alleles to be inserted into the genome by horizontal gene transfer
- D.mRNA stability of tumor suppressor genes decreased because of altered codon usage in the affected twin
Show Answer & Explanation
Correct Answer: B
Epigenetic histone modifications that condense chromatin can silence gene expression without altering DNA sequence
Histone methylation promotes chromatin condensation, reducing transcription factor access and silencing nearby genes. Silenced tumor suppressor genes increase cancer risk without any change to the underlying DNA sequence. This is epigenetic regulation. Choice A contradicts the premise that both twins have identical DNA. Choice C incorrectly applies horizontal gene transfer, a mechanism that does not operate this way in humans. Choice D confuses histone modification with mRNA stability, which are distinct post-transcriptional mechanisms. CED Unit 6.
Big Idea 4: Systems and Their Interactions (5 Questions)
Big Idea 4 covers how cells, organs, populations, and ecosystems function as interconnected systems. This spans membrane transport (Unit 2), signal transduction cascades (Unit 4), and community ecology (Unit 8). A student who studied Units 7, 3, and 6 deeply but skipped Unit 8 ecology will lose 10-15% of their possible score to just five of the eight units.
Cell Membranes, Signaling, and Osmosis
Q21 · Accessible · Unit 2 · Cell Structure and Function
Which of the following molecules would most readily cross a phospholipid bilayer by simple diffusion, without a transport protein?
- A.Glucose (large, polar, hydroxyl-rich)
- B.Na⁺ ions (charged, hydrophilic)
- C.CO₂ (small, nonpolar gas)
- D.Large glycoproteins (charged macromolecules embedded in the membrane)
Show Answer & Explanation
Correct Answer: C
CO₂ (small, nonpolar gas)
The hydrophobic interior of the bilayer permits only small, nonpolar molecules to pass by simple diffusion. CO₂ is small and nonpolar, crossing membranes freely during gas exchange. Glucose (Choice A) requires GLUT transporter proteins because it is large and polar. Na⁺ ions (Choice B) require ion channels or pumps because they carry a charge. Large glycoproteins (Choice D) cannot cross the bilayer by diffusion; they are integral or peripheral membrane proteins by definition. CED Unit 2.
Q22 · Moderate · Unit 4 · Cell Communication and Cell Cycle
A G-protein-coupled receptor (GPCR) is activated by a first messenger. The active G-protein stimulates adenylyl cyclase, which produces cAMP. cAMP activates protein kinase A (PKA). A drug specifically inhibits phosphodiesterase, the enzyme that degrades cAMP. What would be the expected cellular effect?
- A.Decreased activation of protein kinase A due to cAMP depletion
- B.Prolonged elevation of cAMP levels and an extended cellular response to the original signal
- C.Reduced cAMP production by adenylyl cyclase via negative feedback
- D.Immediate termination of the cellular response because the signaling cascade is interrupted
Show Answer & Explanation
Correct Answer: B
Prolonged elevation of cAMP levels and an extended cellular response to the original signal
Phosphodiesterase degrades cAMP to 5’-AMP, terminating the signal. Inhibiting phosphodiesterase prevents cAMP degradation, causing levels to remain elevated and PKA to stay active longer. This prolongs the downstream cellular response. This is the mechanism by which caffeine, theophylline, and sildenafil operate. Choice A is the opposite of the actual effect. Choice C is incorrect; phosphodiesterase does not regulate adenylyl cyclase directly. Choice D is also the opposite of what occurs. CED Unit 4.
Q23 · Hard · Unit 2 · Cell Structure and Function
A plant cell has an osmotic potential of −0.8 MPa and a pressure potential of +0.4 MPa. It is placed in a solution with a water potential of −0.6 MPa. In which direction will net osmosis occur, and what is the initial water potential of the plant cell?
- A.Water moves into the cell; cell water potential = −0.4 MPa
- B.Water moves out of the cell; cell water potential = −0.4 MPa
- C.No net movement occurs; cell water potential = −0.6 MPa
- D.Water moves into the cell; cell water potential = −0.8 MPa
Show Answer & Explanation
Correct Answer: B
Water moves out of the cell; cell water potential = −0.4 MPa
Water potential = osmotic potential + pressure potential = −0.8 + 0.4 = −0.4 MPa. The surrounding solution has ψ = −0.6 MPa. Water moves from higher water potential (less negative) to lower water potential (more negative). Because the cell (−0.4 MPa) has higher water potential than the solution (−0.6 MPa), water moves out of the cell. Choice A correctly calculates the cell water potential but predicts the wrong direction. Choice C mistakes equilibrium conditions for the initial state. Choice D uses osmotic potential alone, omitting pressure potential. CED Unit 2.
Ecology and Community Dynamics
Unit 8 (Ecology) draws on every previous unit. Population dynamics apply Hardy-Weinberg logic. Community succession involves energetics principles. Biogeochemical cycles connect cellular respiration to ecosystem carbon flux. Students who treat ecology as an isolated memorization unit tend to miss the Big Idea 4 questions that cross unit lines.
Q24 · Moderate · Unit 8 · Ecology
A sudden disease eliminates 80% of a rabbit population in a grassland ecosystem. Based on Lotka-Volterra predator-prey dynamics, which sequence of events is most likely to follow over the subsequent several years?
- A.Fox population rises due to less competition; rabbit population continues declining to extinction
- B.Fox population declines due to reduced prey availability; rabbit population recovers; fox population later recovers
- C.Rabbit population immediately rebounds; fox population is unaffected due to dietary switching
- D.Both populations decline simultaneously to extinction due to the destabilizing feedback loop
Show Answer & Explanation
Correct Answer: B
Fox population declines due to reduced prey availability; rabbit population recovers; fox population later recovers
Lotka-Volterra dynamics predict oscillating cycles: prey decline reduces predator food supply, causing predator decline through starvation. Reduced predation pressure then allows prey to recover, which subsequently supports predator recovery. Choice A would apply only if predators had no food dependency on prey. Choice C ignores the resource-consumer relationship. Choice D confuses negative feedback, which stabilizes systems, with a runaway collapse. The standard Lotka-Volterra model predicts oscillations, not extinction, within normal parameter ranges. CED Unit 8.
Q25 · Hard · Unit 8 · Ecology
After a volcanic eruption covers an island with lava, lichens are the first organisms to colonize the bare rock. Which correctly explains why lichens are pioneer species in this scenario, and what mechanism drives subsequent succession?
- A.Lichens reproduce rapidly, and succession proceeds by random immigration of species from nearby islands
- B.Lichens tolerate extreme conditions and produce acids that weather rock into soil; each successional community facilitates conditions for the next
- C.Lichens require nutrient-rich soil to colonize first because they import nutrients from ocean spray
- D.Primary succession occurs because animal communities arrive before plant communities and prepare the habitat
Show Answer & Explanation
Correct Answer: B
Lichens tolerate extreme conditions and produce acids that weather rock into soil; each successional community facilitates conditions for the next
Lichens are mutualistic associations of fungi and photosynthetic partners (algae or cyanobacteria) that tolerate desiccation, extreme temperatures, and a mineral-only substrate. Their carbonic and oxalic acid secretions begin weathering rock into rudimentary soil, enabling later colonizers. Facilitation, where pioneer species improve conditions for subsequent species, is the primary driver of primary succession. Choice A is partially correct on tolerance but incorrect that succession is random. Choice C is ecologically backward; lichens do not require nutrient-rich soil. Choice D is incorrect; plants and lichens, not animals, are primary colonizers. CED Unit 8.
What Your Practice Score Means
Score your answers, then use this table to interpret where you stand relative to the AP Biology score scale. Keep in mind: this set covers only MCQ-style reasoning. The actual AP Biology exam weights multiple-choice and free-response equally at 50% each. A student who scores 24/25 here but has never practiced FRQs will still cap around a 3.
| Raw Score (out of 25) | Approximate Performance Level | What to Do Next |
|---|---|---|
| 23-25 | Strong readiness for a 4 or 5 | Focus FRQ practice on Units 3, 6, 7. Time yourself on a full released exam. |
| 19-22 | Solid foundation, minor gaps remain | Review explanations for every missed question. Drill the specific units where you lost points. |
| 14-18 | Moderate preparation, clear unit gaps | Revisit CED unit outlines for units below 60% accuracy. Add one week of targeted review before full practice exams. |
| 0-13 | Early-stage preparation | Return to primary content review using the AP Biology CED before doing additional practice questions. |
Interpret your raw score as a directional signal, not a final verdict. FRQ performance shifts outcomes significantly.
For the full picture of what your AP Biology practice performance predicts, use our AP Biology resources hub to find additional study materials, including links to every released College Board FRQ from the past five years. To understand how the 5-rate shifted from 7.4% in 2021 to 18.9% in 2025 and what score earns college credit at most schools, the AP Biology score distribution analysis covers the complete five-year picture.
The AP Biology Difficulty Decoded post breaks down the 2025 pass rate (70.4%), 5-rate (18.9%), and what those numbers mean for how you should spend your remaining study time.
Estimate Your AP Biology Score
Once you have worked through the questions above and checked your answers, the AP Score Predictor translates your practice performance into an estimated AP Biology score range. Enter your MCQ accuracy and any available FRQ practice data to see where you would likely land on the 1-5 scale.
AP Score Predictor
Enter your AP Biology practice results to see your projected score range and what you need to improve to reach the next band.
Key Takeaways
- All 25 questions are original and grounded in the AP Biology CED. No College Board released questions appear verbatim. Every question was reviewed for biological accuracy and CED alignment.
- Big Ideas 1 and 2 (Evolution and Energetics) get the most questions because their corresponding units carry the highest exam weight. Unit 7 alone accounts for 13-20% of your score.
- The explanation for each wrong answer matters as much as the explanation for the right one. Students who read only the correct answer explanation miss half the learning value of practice.
- Retrieval practice produces significantly better retention than passive re-reading. Commit to each answer before expanding the explanation, even when you are unsure. The struggle is where the learning happens.
- The AP Biology exam is 50% free-response. MCQ practice is necessary but not sufficient. Pair this set with FRQ practice using College Board's released free-response questions and scoring guidelines.
- Gene regulation (Unit 6) and Hardy-Weinberg (Unit 7) appear on every released AP Biology exam in some form. Both reward students who understand the underlying mechanism, not just the vocabulary.
- Ecology (Unit 8) connects to all four big ideas. Students who study it as isolated content miss the cross-unit questions that test whether you see the biological system as a whole.
Frequently Asked Questions
Are these AP Biology practice questions from the College Board?
No. These are original questions written by Classeva's team and reviewed by AP-credentialed contributors for biological accuracy and CED alignment. They are inspired by the patterns and concepts in released College Board material but are not copied from any official source. College Board releases free-response questions from prior years at AP Central, which are the gold standard for FRQ practice.
How many AP Biology practice questions should I do before the exam?
Most students who earn a 4 or 5 work through at least 150-200 multiple-choice questions during their final 8-week study block, in addition to completing 5-7 full free-response sets using College Board's released FRQs and official scoring guidelines. Quality of review matters more than raw question count. Always read the explanation for every question you miss.
How does the difficulty of these questions compare to the real AP Biology exam?
The distribution mirrors the actual exam: roughly one-third accessible (similar to low-to-mid difficulty AP questions), one-third moderate, and one-third hard. The hardest questions here require multi-step reasoning across two or more concepts, matching the demand you'll see on the real MCQ section, particularly on questions tied to Units 3, 6, and 7.
What score on these AP Biology practice questions predicts a 5 on the exam?
Scoring above 20 out of 25 (80%+) on this set, while genuinely attempting each question before checking the answer, is a reasonable proxy for strong exam readiness. However, the real AP Biology exam also includes 6 free-response questions worth 50% of your score. Practice MCQ results alone cannot predict your final AP score without also assessing FRQ performance.
Should I time myself when doing AP Biology practice questions?
Yes, but not for short sets. The full AP Biology MCQ section gives you 90 minutes for 60 questions (90 seconds per question). When practicing a 25-question set, target 37-38 minutes. Timed practice trains you to commit to answers efficiently rather than second-guessing, which is a specific skill the exam rewards.
Which AP Biology units appear most often on the exam?
According to the AP Biology Course and Exam Description, Unit 7 (Natural Selection and Evolution) carries 13-20% of the exam weight, the highest of any single unit. Units 3 (Cellular Energetics) and 6 (Gene Expression and Regulation) each carry 12-16%. Together these three units account for 37-52% of your total exam score.
Is AP Biology multiple choice hard?
The MCQ section is challenging because questions rarely test isolated facts. Most questions require applying a concept to an experimental scenario, interpreting data from a graph or table, or reasoning across two connected biological systems. Students who practice with full explanations rather than just checking answer letters consistently perform better on these applied reasoning questions.
Where can I find more free AP Biology practice?
College Board publishes free AP Biology released free-response questions with scoring guidelines at AP Central. AP Classroom (accessible through your school) includes progress check MCQs for each unit. For concept review alongside practice, the AP Biology Course and Exam Description is the authoritative source for every topic that can appear on the exam.
AP Biology exam structure and unit weightings sourced from the AP Biology Course and Exam Description published by the College Board. Score distribution data from College Board AP Score Distributions (2025). For official FRQ practice, see the AP Biology Exam page at AP Central. Related reading: AP Biology exam 2026 guide, score distribution five-year breakdown, and the AP Biology visual reference guide.
