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AP Biology

College Board · 8 units · exam-weighted · high school

AP Biology: every unit with its share of the exam, as College Board publishes it, with practice questions per unit and a countdown to the exam.

Set up AP Biology → One set per unit, 8 of them pre-written and answer-checked. Free, no account needed.

Units and exam weighting

Unit 1: Chemistry of Life 8–11% of the exam

Water and its properties, biological macromolecules (carbohydrates, lipids, proteins, nucleic acids), monomers and polymers, and how structure determines function.

3 of 13 questions in this unit, with the worked answer

Which property of water is most directly responsible for its ability to moderate temperature changes in aquatic environments?

Answer. High specific heat, caused by the many hydrogen bonds that must absorb or release energy before water's temperature changes

Hydrogen bonds absorb heat energy when they break and release it when they form, so a large amount of energy is needed to change water's temperature, buffering aquatic habitats against rapid swings.

Explain at the molecular level why ice floats on liquid water.

Answer. Hydrogen bonds in ice lock water molecules into a fixed, spread-out crystalline lattice, making ice less dense than the more randomly packed molecules in liquid water.

In liquid water, hydrogen bonds constantly break and reform, allowing molecules to pack closely. In ice, each molecule forms four stable hydrogen bonds in a rigid hexagonal lattice that takes up more space, lowering density.

A biologist observes water molecules clinging to the walls of a plant's xylem vessel as water is pulled upward. This clinging behavior is best explained by which property?

Answer. Adhesion

Adhesion is the attraction of water molecules to a different polar substance (here, the polar cellulose walls of the xylem), which along with cohesion drives capillary action.

All 13 questions for Unit 1, with the working →

What people get wrong

  1. Reversing dehydration synthesis and hydrolysis: students say water is released during digestion. Fix: hydrolysis (breaking polymers, e.g., digestion) adds water; dehydration synthesis (building polymers) removes water.
  2. Treating cellulose and starch as chemically identical because both are glucose polymers. Fix: identify the glycosidic bond type — alpha-1,4 (starch, digestible, helical) versus beta-1,4 (cellulose, indigestible by most animals, rigid fibers).
  3. Assuming every protein has quaternary structure. Fix: quaternary structure only exists if the functional protein is made of more than one polypeptide chain (e.g., hemoglobin); a single-chain protein like myoglobin stops at tertiary structure.
  4. Attributing a bulk property of water (like high specific heat) to one single hydrogen bond. Fix: explain these properties as emerging from the collective, constantly reforming network of many hydrogen bonds, not one bond's strength.

Unit 2: Cells 10–13% of the exam

Cell structure and organelles, membrane structure and transport, osmosis and water potential, surface-area-to-volume, compartmentalisation, and the origins of cells.

3 of 13 questions in this unit, with the worked answer

Explain why water moves from a hypotonic solution to a hypertonic solution across a selectively permeable membrane, in terms of water potential.

Answer. Because the hypertonic side has more dissolved solute, its solute potential (and thus overall water potential) is lower; water always moves from higher to lower water potential, so it flows from the hypotonic (higher Ψ) side to the hypertonic (lower Ψ) side.

Adding solute lowers Ψs, which lowers total Ψ if pressure potential is unchanged. Water follows the water potential gradient, not the solute gradient directly, ending up moving toward the more concentrated solution.

A red blood cell is placed in a 0.5 M NaCl solution, which is hypertonic relative to the cell's cytoplasm. Predict and explain what happens to the cell.

Answer. The cell will lose water by osmosis and shrink (crenate), because water moves from the cell's higher water potential to the solution's lower water potential.

Since the surrounding solution has more solute, its water potential is lower than that of the cytoplasm. Water exits the cell down this gradient, causing the cell to shrivel.

A plant cell is placed in distilled water. Describe what happens to the cell's pressure potential and volume as it reaches equilibrium.

Answer. Water enters the cell by osmosis, increasing cell volume and building up pressure potential (Ψp) against the rigid cell wall until Ψp rises enough to balance the negative solute potential, bringing total Ψ to equilibrium with the surrounding water (Ψ=0).

Unlike an animal cell, a plant cell doesn't burst because the cell wall resists expansion, generating turgor pressure that opposes further net water entry.

All 13 questions for Unit 2, with the working →

What people get wrong

  1. Assuming water moves toward the hypotonic (low-solute) side because 'water follows water.' Instead, remember water moves toward the side with lower water potential, which is the side with MORE solute (hypertonic) when pressure potential is zero.
  2. Treating pressure potential (Ψp) as always positive or zero. In real plant systems, especially xylem, Ψp can be negative due to tension; don't drop it from the equation just because it 'should' be positive.
  3. Assuming facilitated diffusion requires ATP because it uses a protein channel or carrier. It is still passive transport — it moves solutes down their concentration gradient and needs no energy input, unlike active transport.
  4. Thinking bigger cells are more efficient because they have more total surface area. Compare the ratio, not the raw numbers: SA:V shrinks as cells grow, which is why cells stay small or fold their membranes (microvilli, cristae) instead of simply enlarging.

Unit 3: Cellular Energetics 12–16% of the exam

Enzyme structure, catalysis and regulation; energy and coupled reactions; photosynthesis; cellular respiration and fermentation.

3 of 14 questions in this unit, with the worked answer

A researcher adds a molecule to an enzymatic reaction. Data show the reaction's Vmax decreases while Km stays the same. What type of regulation does this molecule most likely represent?

Answer. A noncompetitive (allosteric) inhibitor.

Noncompetitive inhibitors bind a site distinct from the active site, changing the enzyme's shape and reducing its maximum catalytic capacity (Vmax) regardless of substrate concentration. Because it doesn't compete for the active site, adding more substrate cannot restore full activity, so Km (substrate affinity) is unaffected.

Explain, in terms of activation energy and free energy change (ΔG), why enzymes speed up reactions without changing whether a reaction is spontaneous.

Answer. Enzymes lower the activation energy needed to reach the transition state, increasing the rate at which the reaction proceeds, but they do not alter the ΔG of the reaction.

Rate and spontaneity are separate properties: spontaneity is set by the intrinsic free energy difference between reactants and products, which enzymes cannot change. Enzymes only reduce the energy barrier (Ea) that must be overcome, so a reaction that was already spontaneous just happens faster, and a non-spontaneous one still won't proceed on its own.

A cell uses ATP hydrolysis to phosphorylate a membrane transport protein, allowing it to pump ions against their concentration gradient. What term describes this linking of an exergonic process to an endergonic one?

Answer. Energy coupling (a coupled reaction).

ATP hydrolysis (exergonic, releases free energy) is coupled to the endergonic process of pumping ions uphill against their gradient. The phosphate group transferred to the transport protein raises its energy state, allowing the otherwise unfavorable transport to proceed.

All 14 questions for Unit 3, with the working →

What people get wrong

  1. Thinking enzymes change whether a reaction is spontaneous. Fix: enzymes only lower activation energy and speed up the approach to equilibrium; ΔG and equilibrium position are unchanged.
  2. Mixing up competitive and noncompetitive inhibitor graphs. Fix: memorize that competitive inhibition raises Km only (Vmax unchanged, overcome by more substrate); noncompetitive/allosteric inhibition lowers Vmax only (Km unchanged, cannot be overcome by more substrate).
  3. Swapping the locations of the light reactions and the Calvin cycle. Fix: light reactions happen in the thylakoid membrane (photosystems, ETC, chemiosmosis); the Calvin cycle happens in the stroma (carbon fixation, reduction, RuBP regeneration).
  4. Assuming fermentation replaces cellular respiration entirely when oxygen is absent. Fix: fermentation only regenerates NAD+ so glycolysis (which always occurs in the cytoplasm) can continue; it does not run the Krebs cycle or ETC.

Unit 4: Cell Communication and Cell Cycle 10–15% of the exam

Cell signalling pathways, signal transduction, feedback, the cell cycle and its regulation, mitosis and checkpoints.

3 of 14 questions in this unit, with the worked answer

A signaling cell secretes a local chemical that diffuses through extracellular fluid and affects only nearby cells. What type of signaling is this?

Answer. Paracrine signaling

Paracrine signaling involves local regulators that act on nearby target cells, unlike endocrine signaling which uses hormones traveling through the bloodstream to distant cells. Synaptic signaling requires a synapse and neurotransmitter release, and autocrine signaling means the cell signals itself.

Describe the sequence of events after a ligand binds a G protein-coupled receptor (GPCR), up to activation of an intracellular enzyme.

Answer. Ligand binding changes the GPCR's shape, allowing it to bind and activate a G protein inside the membrane; the activated G protein exchanges GDP for GTP, dissociates, and binds to and activates (or inhibits) a target enzyme, such as adenylyl cyclase.

GPCRs work through an intermediary G protein rather than acting as enzymes themselves. The conformational change caused by ligand binding is transmitted to the G protein, which then diffuses along the membrane to reach and regulate its target enzyme.

Adenylyl cyclase, activated by a G protein, converts ATP into a second messenger. Name this second messenger and the kinase it typically activates.

Answer. Cyclic AMP (cAMP), which activates protein kinase A (PKA)

cAMP is a common second messenger produced from ATP by adenylyl cyclase. Once formed, cAMP binds to and activates PKA, which then phosphorylates various target proteins to carry out the cellular response.

All 14 questions for Unit 4, with the working →

What people get wrong

  1. Confusing the G1 checkpoint with the G2/M checkpoint on exam questions. Fix: attach a specific question to each checkpoint (G1 asks 'is there enough growth signal/nutrients?', G2/M asks 'is DNA replication complete and undamaged?', M asks 'are all chromosomes attached to the spindle?') rather than memorizing them as an unordered list.
  2. Treating mitosis and the full cell cycle as the same thing, and including S phase (DNA replication) inside 'mitosis.' Fix: remember mitosis (M phase) is only the division of the nucleus and cell, occurring after interphase (G1, S, G2) is already complete.
  3. Assuming a phosphorylation cascade only transmits the signal rather than amplifying it. Fix: state explicitly that one active kinase phosphorylates many copies of the next kinase, so the signal grows in magnitude at each step, not just gets passed along unchanged.
  4. Mixing up proto-oncogenes and tumor suppressor genes when explaining a cancer scenario. Fix: ask whether the mutation makes a gene more active (proto-oncogene to oncogene, gain of function, one allele suffices) or less active (tumor suppressor, loss of function, usually needs both alleles hit).

Unit 5: Heredity 8–11% of the exam

Meiosis and genetic diversity, Mendelian genetics, non-Mendelian inheritance, chromosomal inheritance, probability and pedigrees.

3 of 14 questions in this unit, with the worked answer

During meiosis I, which event directly increases genetic diversity by exchanging genetic material between homologous chromosomes?

Answer. Crossing over between non-sister chromatids of homologous chromosomes during prophase I.

In prophase I, homologous chromosomes pair up (synapsis) to form a tetrad, and non-sister chromatids exchange segments of DNA at chiasmata. This physically shuffles alleles between maternal and paternal chromosomes, creating new allele combinations. Independent assortment and random fertilization also add diversity but do not involve physical exchange of DNA.

In a cross between two dihybrid pea plants, YyRr x YyRr, where yellow (Y) is dominant to green (y) and round (R) is dominant to wrinkled (r), what is the probability that an offspring will be green and wrinkled (yyrr)?

Answer. 1/16

Because the two genes assort independently, calculate each gene's probability separately using the product rule. For Yy x Yy, $P(yy)=1/4$. For Rr x Rr, $P(rr)=1/4$. Multiply the independent probabilities: $1/4 \times 1/4 = 1/16$.

A pea plant with round seeds has an unknown genotype (RR or Rr). Describe a testcross that would determine its genotype, and state the expected results for each possibility.

Answer. Cross the round-seeded plant with a homozygous recessive wrinkled plant (rr). If all offspring have round seeds, the unknown parent is RR; if about half the offspring are round and half are wrinkled, the unknown parent is Rr.

A testcross uses a homozygous recessive individual because it can only contribute a recessive allele, so any dominant phenotype appearing in offspring must come from the unknown parent's dominant allele. An RR parent produces only R gametes, giving all Rr (round) offspring, while an Rr parent produces R and r gametes equally, giving a 1:1 round-to-wrinkled ratio.

All 14 questions for Unit 5, with the working →

What people get wrong

  1. Thinking crossing over happens between sister chromatids — it actually happens between non-sister chromatids of homologous chromosomes during prophase I; sister chromatids are genetically identical and swapping between them does nothing.
  2. Applying the standard 9:3:3:1 dihybrid ratio to an epistasis problem without adjusting it — instead, identify which genotype classes are masked by the epistatic gene and combine those categories (e.g., 9:3:4).
  3. Using the product rule ('and') when a question asks for 'either/or' outcomes, or vice versa — decide first whether the question wants both events to occur together (multiply) or one of several possible mutually exclusive outcomes (add).
  4. Assuming nondisjunction can only occur in meiosis I — it can also occur in meiosis II, and the two cases produce gametes with different chromosome compositions (non-identical homologs vs. identical sister chromatids).

Unit 6: Gene Expression and Regulation 12–16% of the exam

DNA and RNA structure, replication, transcription and translation, gene regulation, mutations, biotechnology.

3 of 15 questions in this unit, with the worked answer

In DNA replication, which enzyme synthesizes short RNA primers to give DNA polymerase a starting point?

Answer. Primase

DNA polymerase cannot begin synthesis on a bare template; primase lays down a short RNA primer that provides a free 3' end for DNA polymerase to extend.

A double-stranded DNA molecule is 30% adenine. What percentage of its bases is guanine?

Answer. 20%

By Chargaff's rule, A equals T, so T is also 30%, accounting for 60% of bases. The remaining 40% is split equally between G and C, giving 20% guanine.

Which statement correctly describes RNA polymerase's movement along the template strand and the mRNA it produces?

Answer. The template strand is read 3' to 5', and mRNA is synthesized 5' to 3'

RNA polymerase reads the template strand in the 3' to 5' direction and builds the new mRNA strand antiparallel to it, in the 5' to 3' direction, matching how DNA polymerase works during replication.

All 15 questions for Unit 6, with the working →

What people get wrong

  1. Copying the coding (non-template) strand as if it were read by RNA polymerase. Always identify the template strand first (read 3' to 5'), then build the mRNA 5' to 3' as its antiparallel complement.
  2. Assuming any nucleotide change automatically damages the protein. Always translate before and after the change and compare amino acids — the redundancy of the genetic code makes many substitutions silent.
  3. Treating all insertions and deletions like point mutations. Only insertions or deletions that are NOT a multiple of three shift the reading frame; check the number of bases added or removed before calling something a frameshift.
  4. Reversing the logic of the lac operon. In the absence of lactose the operon is OFF by default because the repressor is bound to the operator; lactose (via allolactose) inactivates the repressor to turn transcription ON.

Unit 7: Natural Selection 13–20% of the exam

Natural and artificial selection, population genetics and Hardy-Weinberg, evidence for evolution, phylogeny, speciation, extinction, origin of life.

3 of 14 questions in this unit, with the worked answer

In a population of 500 wildflowers, 45 plants show the recessive white-flower phenotype (aa). Assuming Hardy-Weinberg equilibrium, what is the frequency of the dominant allele A?

Answer. p = 0.7

q² = 45/500 = 0.09, so q = √0.09 = 0.3. Then p = 1 − q = 1 − 0.3 = 0.7. Only the homozygous recessive frequency can be square-rooted directly to find q.

Human birth weight data show that mortality is lowest for babies at intermediate weights and higher for both very low and very high birth weights. What type of natural selection does this illustrate?

Answer. Stabilizing selection

Stabilizing selection favors intermediate phenotypes and reduces variation at both extremes, unlike directional selection (favors one extreme) or disruptive selection (favors both extremes over the middle).

A population of insects has two color morphs, dark and light, each camouflaged against a different substrate; intermediate-colored individuals are more visible to predators and rarely survive. What type of selection is acting, and what happens to the population's phenotypic distribution?

Answer. Disruptive selection; the distribution becomes bimodal, with both extremes favored over the intermediate phenotype.

Disruptive (diversifying) selection favors both extreme phenotypes at the expense of intermediates, which can eventually contribute to speciation if the morphs stop interbreeding.

All 14 questions for Unit 7, with the working →

What people get wrong

  1. Taking the square root of an observed phenotype frequency that is NOT homozygous recessive — only q² (the fully recessive phenotype) can be square-rooted directly; the dominant phenotype frequency is p² + 2pq and cannot be square-rooted to get p.
  2. Assuming a population is automatically in Hardy-Weinberg equilibrium without checking whether the listed conditions (random mating, no selection, no drift, no gene flow, no mutation, large population) are actually met in the scenario.
  3. Treating genetic drift and natural selection as the same process — drift is random and has the largest effect in small populations (founder effect, bottleneck), while selection is non-random and driven by fitness differences; a question describing a small isolated population after a storm or colonization event is testing drift, not selection.
  4. Reading a phylogenetic tree left-to-right as if the order of the tip labels indicates evolutionary relatedness — relatedness is determined only by the position of branch points (nodes), so two tips can look far apart on the page yet share a very recent common ancestor.

Unit 8: Ecology 10–15% of the exam

Responses to the environment, energy flow, population ecology, community ecology, biodiversity, disruptions to ecosystems.

3 of 15 questions in this unit, with the worked answer

A population has a per capita growth rate r = 0.04/year and current size N = 800. Using the exponential growth model, what is dN/dt?

Answer. 32 individuals per year

Using $\frac{dN}{dt}=rN$, substitute r = 0.04 and N = 800: 0.04 × 800 = 32. This is the instantaneous rate of population increase assuming unlimited resources.

Explain why logistic growth produces an S-shaped curve while exponential growth produces a J-shaped curve.

Answer. Logistic growth includes the term (K−N)/K, which approaches zero as N nears K, slowing growth until it levels off at carrying capacity; exponential growth has no limiting term, so growth accelerates continuously.

In $\frac{dN}{dt}=rN\frac{(K-N)}{K}$, as N approaches K, (K−N)/K shrinks toward 0, driving dN/dt toward 0 and flattening the curve (S-shape). Without this term, as in $\frac{dN}{dt}=rN$, growth rate keeps increasing with N, producing an unchecked J-shape.

A forest's producers have a gross primary productivity of 30,000 kcal/m²/yr and use 10,000 kcal/m²/yr for respiration. What is the net primary productivity?

Answer. 20,000 kcal/m²/yr

NPP = GPP − R = 30,000 − 10,000 = 20,000 kcal/m²/yr. NPP represents the energy actually available to be consumed by primary consumers.

All 15 questions for Unit 8, with the working →

What people get wrong

  1. Confusing GPP with NPP: students plug GPP straight into the 10% rule and forget to subtract the producers' own respiration first. Always compute NPP = GPP − R before doing any trophic transfer calculation.
  2. Applying the 10% rule by subtraction instead of multiplication (e.g., subtracting 10% instead of keeping only 10%). Remember that only about 10% is *retained*, so you multiply by 0.10 at each step, not subtract 10%.
  3. Mislabeling symbiotic relationships by only considering one species' benefit. Always assign a +, −, or 0 to *each* species before naming the interaction — this is how you correctly distinguish commensalism (+/0) from mutualism (+/+) or parasitism (+/−).
  4. Treating exponential growth as the normal, expected pattern in nature. In reality most populations are limited by resources and follow logistic growth; exponential growth only appears briefly, e.g., after colonizing a new habitat or during a population crash's absence of predators.

How it works

Choosing this curriculum creates a course in Chalkset with one study set per unit and the exam date set to the start of the next testing window. Readiness is weighted the way the exam is, so a weak unit that carries 18% of the score pulls harder than one that carries 6%. Pre-written units come with worked explanations and an independently checked answer key; the rest are written by the agent when you open them.

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