AP Biology — Cheatsheet
Formulas, exam-day tips, and key terms on one page.
Formulas & relationships
Hydrogen bond
δ⁺H···O δ⁻
The dotted line is the hydrogen bond (an intermolecular attraction), not the covalent O–H bond within a molecule.
Dehydration synthesis / hydrolysis
monomer–OH + H–monomer ⇌ monomer–monomer + H₂O
Left-to-right removes water to bond (dehydration synthesis); right-to-left adds water to break (hydrolysis).
Peptide bond formation
–COOH + H₂N– → –CO–NH– + H₂O
A peptide bond joins the carboxyl carbon of one amino acid to the amino nitrogen of the next, releasing water.
Complementary base pairing (DNA)
A–T · G–C
Adenine pairs with thymine (2 hydrogen bonds); guanine pairs with cytosine (3 hydrogen bonds). In RNA, uracil replaces thymine, so A pairs with U.
Surface-area-to-volume ratio (cube)
SA:V = 6s² / s³ = 6 / s
For a cube of side length s, surface area is 6s² and volume is s³. As s increases, the ratio 6/s falls — bigger cells have relatively less membrane per unit of contents.
The secretory pathway
ribosome/rough ER → transport vesicle → Golgi → secretory vesicle → plasma membrane
The path a protein follows from synthesis to export. Each arrow is a vesicle budding off one compartment and fusing with the next.
Transport at a glance
passive: high → low (no ATP) · active: low → high (ATP)
The direction relative to the concentration gradient tells you whether energy is needed. "Down" the gradient is free; "up" the gradient costs ATP.
Water potential
Ψ = Ψp + Ψs
Ψ is water potential; Ψp is pressure potential and Ψs is solute potential. Water always moves from higher Ψ to lower Ψ. Pure water at atmospheric pressure has Ψ = 0; adding solute makes Ψs (and thus Ψ) negative.
Enzyme catalysis
enzyme + substrate ⇌ enzyme·substrate → enzyme + product
The enzyme binds substrate, forms an enzyme–substrate complex, releases product, and is regenerated — never used up.
Overall photosynthesis
6 CO₂ + 6 H₂O + light → C₆H₁₂O₆ + 6 O₂
Inputs: carbon dioxide, water, and light energy. Outputs: glucose and oxygen. This is the exact reverse of cellular respiration.
Overall aerobic respiration
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O + energy (ATP)
Inputs: glucose and oxygen. Outputs: carbon dioxide, water, and ATP. The reverse of photosynthesis.
ATP cycle
ATP + H₂O ⇌ ADP + Pᵢ + energy
Left-to-right (hydrolysis) releases energy for cellular work; right-to-left (regeneration) requires energy from catabolism.
Signal amplification cascade
1 ligand → few receptors → many kinases → millions of products
Each step multiplies the number of active molecules, so a small signal yields a large response.
The one-line test
response opposes change ⇒ negative · response amplifies change ⇒ positive
Ask only: does the loop push the variable back toward the set point, or further away?
Cell cycle at a glance
G1 → S → G2 → [ Prophase → Metaphase → Anaphase → Telophase ] → Cytokinesis
G1 · S · G2 make up interphase; PMAT is mitosis; cytokinesis splits the cytoplasm into two cells.
What drives a checkpoint transition
cyclin (rising) + Cdk (constant) → active cyclin-Cdk complex → advance
Cdk levels stay roughly constant; it is the cyclic rise and fall of cyclins that turns the complexes on and off.
Independent assortment combinations
2ⁿ possible gamete types
n is the haploid chromosome number. For humans (n = 23) this is 2²³ ≈ 8.4 million combinations — before crossing over adds even more.
Probability rules
P(A and B) = P(A) × P(B) · P(A or B) = P(A) + P(B)
Multiply for independent events happening together; add for mutually exclusive outcomes. Independent assortment is what licenses the multiplication rule across genes.
Incomplete dominance vs. codominance
Incomplete: Aᴿ + Aᵂ → blended intermediate · Codominant: Iᴬ + Iᴮ → both shown at once
The heterozygote is the giveaway: an in-between phenotype (pink) is incomplete dominance; two full phenotypes displayed together (type AB) is codominance.
Recombination frequency
RF = (recombinant offspring ÷ total offspring) × 100%
A recombination frequency of 1% corresponds to roughly 1 map unit (centimorgan). The maximum is 50%, at which point genes assort as if unlinked.
Direction of synthesis
DNA polymerase builds 5′ → 3′ only
New nucleotides can only be added to a free 3′ end, so the new strand grows 5′→3′ while it is read off a 3′→5′ template.
Transcription base pairing
DNA A → RNA U · DNA T → RNA A · DNA G → RNA C · DNA C → RNA G
RNA has no thymine — wherever the DNA template has adenine, the new mRNA carries uracil (U), not thymine (T).
Codon → anticodon pairing
mRNA codon 5′–A U G–3′ pairs with tRNA anticodon 3′–U A C–5′
The anticodon is antiparallel and complementary to the codon (A–U, G–C), just like the two strands of DNA.
Mutation effects at a glance
silent = same aa · missense = different aa · nonsense = premature stop · frameshift = whole downstream frame altered
Substitutions cause silent/missense/nonsense; insertions and deletions cause frameshifts.
PCR amplification
copies after n cycles = starting copies × 2ⁿ
Each cycle doubles the DNA, so growth is exponential — 30 cycles turns one molecule into roughly a billion.
Relative fitness (w)
w = (offspring of a genotype) ÷ (offspring of the most successful genotype)
The fittest genotype is set to w = 1; every other genotype is scored relative to it. Fitness is always comparative, never an absolute number.
Molecular similarity and relatedness
more shared DNA/protein sequence → more recent common ancestor
Sequence differences accumulate over time, so the degree of molecular difference estimates how long ago two lineages diverged.
Allele frequencies
p + q = 1
For a gene with two alleles, p is the frequency of the dominant allele and q the frequency of the recessive allele; together they must account for 100% of the alleles.
Genotype frequencies
p² + 2pq + q² = 1
p² = homozygous dominant, 2pq = heterozygous, q² = homozygous recessive. This is just (p + q)² expanded, and the three genotype frequencies must sum to 1.
The engine of speciation
reduced gene flow + divergence (mutation, selection, drift) → reproductive isolation
Cut off gene flow long enough and independently accumulating genetic differences eventually make interbreeding impossible — that is a new species.
Relatedness on a tree
more recent shared node → more closely related
Closeness is judged by how recently two lineages share a common ancestor (branch point), not by physical resemblance or by how far apart the tips are drawn.
The 10% rule
energy at next level ≈ 0.10 × energy at current level
Roughly 90% of the energy at each trophic level is lost as heat (respiration), waste, and uneaten parts; only ~10% is stored as biomass the next level can eat.
Exponential growth
dN/dt = rN
dN/dt is the population’s growth rate (individuals per unit time); r is the per-capita rate of increase; N is the current population size. Bigger N → faster growth, with no upper limit.
Logistic growth
dN/dt = rN((K − N)/K)
The (K − N)/K factor slows growth as N approaches carrying capacity K. When N = K, the factor is 0 and growth stops; when N is small, the factor ≈ 1 and growth is nearly exponential.
Symbiosis by outcome
mutualism +/+ · commensalism +/0 · parasitism +/−
Read each pair as the effect on partner 1 / partner 2. Predation and herbivory are +/− interactions too, but they are brief feeding events, not long-term symbioses.
Heat transfer
q = m·c·ΔT
q = heat (J), m = mass (g), c = specific heat (4.18 J/g·°C for liquid water), ΔT = change in temperature (°C). Water’s large c means a large q is needed for a small ΔT.
Dehydration synthesis ⇌ hydrolysis
n monomers → (n − 1) bonds → (n − 1) H₂O released
Forming each bond in a linear chain removes one water (dehydration synthesis); breaking each bond consumes one water (hydrolysis). For n monomers linked into one chain there are (n − 1) bonds — the same rule for sugars, amino acids, and nucleotides.
Protein structure hierarchy
primary → secondary → tertiary → quaternary
Sequence → local backbone folds (H-bonds) → whole-chain 3-D shape (R-group interactions) → assembly of multiple chains. Denaturation unravels secondary–quaternary levels but leaves primary (peptide bonds) intact.
The endosymbiotic sequence
free-living aerobic prokaryote → engulfed by host cell → retained, undigested endosymbiont → modern organelle (double membrane · circular DNA · 70S ribosomes · binary fission)
Mitochondria trace to an aerobic bacterium; chloroplasts to a later-engulfed cyanobacterium. The nucleus and ER, by contrast, arose autogenously by infolding of the host membrane — not by engulfment.
Water potential
Ψ = Ψp + Ψs
Ψ is water potential, Ψp is pressure potential, Ψs is solute potential (all in MPa). Water always moves toward lower (more negative) Ψ. Pure water at atmospheric pressure has Ψ = 0.
Solute potential
Ψs = −iCRT
i = ionization constant (1 for sucrose/glucose, 2 for NaCl); C = molarity; R = 0.00831 L·MPa·mol⁻¹·K⁻¹; T = temperature in kelvin (°C + 273). Ψs is always 0 or negative — solutes lower water potential.
Calvin cycle per net G3P
3 CO₂ + 9 ATP + 6 NADPH → 1 G3P + 9 ADP + 8 Pᵢ + 6 NADP⁺
Three turns per G3P; double everything (6 CO₂, 18 ATP, 12 NADPH) for one glucose. ATP > NADPH because RuBP regeneration costs extra ATP.
Carriers delivered to the ETC (per glucose)
10 NADH + 2 FADH₂ → ETC → ~26–28 ATP
Approx. 2.5 ATP per NADH and 1.5 ATP per FADH₂: (10 × 2.5) + (2 × 1.5) = 25 + 3 = 28, trimmed by shuttle costs to ~26–28.
Amplification factor of a catalytic cascade
total amplification ≈ (product molecules per enzyme)^(number of catalytic steps)
Amplification compounds *multiplicatively* — each catalytic step multiplies the previous total, so factors grow as a power of the step count, not a sum.
The G1 → S decision, in one line
growth signal → cyclin D-Cdk4/6 → Rb phosphorylated → E2F released → S-phase genes ON
Rb *restrains* the cycle when unphosphorylated; phosphorylating it *removes* the brake. Losing Rb removes the brake permanently.
Chi-square goodness-of-fit statistic
χ² = Σ (O − E)² / E
O = observed count, E = expected count. Sum the term (O − E)² / E over every phenotype category. Use raw counts, never percentages.
Degrees of freedom
df = (number of categories) − 1
A 9:3:3:1 cross has 4 categories → df = 3. A monohybrid 3:1 cross has 2 categories → df = 1. Degrees of freedom depend on the categories, NOT on the sample size.
Recombination frequency
RF = (recombinant offspring ÷ total offspring) × 100%
Count both recombinant classes. 1% RF ≈ 1 map unit (centimorgan). Maximum RF is 50%, at which point linked genes assort as if independent.
Additivity of map distances
For gene order A — B — C: distance(A,C) ≈ distance(A,B) + distance(B,C)
The largest pairwise distance spans the two outer genes. Double crossovers make the measured outer distance slightly less than the exact sum, so short intervals give the most accurate maps.
lac operon: two switches, one output
strong transcription ⇔ (lactose present → repressor OFF) AND (glucose absent → cAMP-CAP ON)
Negative control = the repressor senses lactose; positive control = CAP senses glucose (via cAMP). Fail either switch and expression is low.
PCR amplification
copies after n cycles = starting copies × 2ⁿ
Each cycle doubles the target, so growth is exponential. Examples: 2³ = 8, 2⁴ = 16, 2⁸ = 256, 2¹⁰ = 1024, 2²⁰ ≈ 1 million.
Allele frequencies
p + q = 1
p is the frequency of the dominant allele, q the frequency of the recessive allele. By gene-counting, p = (2·AA + Aa) ÷ (2N).
Genotype frequencies
p² + 2pq + q² = 1
p² = homozygous dominant, 2pq = heterozygous, q² = homozygous recessive. Only a population at equilibrium is expected to match these three values.
Maximum parsimony
preferred tree = the tree requiring the fewest character-state changes
Fewer changes means less assumed homoplasy (convergence or reversal). Among competing trees, the simplest explanation is favored.
Molecular clock
sequence divergence ≈ rate × time → time = divergence ÷ rate
Calibrate the rate using a node whose age is known from the fossil record, then apply it to date other divergences.
Exponential growth
dN/dt = rN
Whole-population growth rate = per-capita rate r × current size N. The per-capita rate itself, (1/N)(dN/dt), is simply the constant r — it never changes no matter how large N gets.
Logistic growth
dN/dt = rN((K − N)/K)
The (K − N)/K term is the density-dependent brake. The per-capita rate here is (1/N)(dN/dt) = r((K − N)/K), which shrinks toward 0 as N → K. This is the equation form of "competition intensifies with crowding."
Maximum growth rate
peak dN/dt = rK/4, at N = K/2
Substituting N = K/2 into the logistic equation: r(K/2)((K − K/2)/K) = r(K/2)(1/2) = rK/4. This is the maximum sustainable yield point exploited in fisheries and wildlife management.
Interaction outcomes
competition −/− · predation +/− · mutualism +/+ · commensalism +/0
Read each pair as the effect on species 1 / species 2. Competitive exclusion applies specifically to the −/− case when two species share one limiting resource with fully overlapping niches.
Simpson’s Diversity Index
D = 1 − Σ(n/N)²
n = number of individuals of one species; N = total individuals of all species; Σ sums the squared proportions across every species. D ranges from 0 (one species dominates, no diversity) to nearly 1 (many equally abundant species). Higher D = greater diversity.
On the exam
- On the AP exam, almost every water question traces back to one root cause: **polarity → hydrogen bonding**. When asked to explain a property, name that chain explicitly rather than just stating the property — the mechanism earns the point.
- The AP throughline for proteins is "**sequence → shape → function**." When a free-response asks why a mutation or a temperature change matters, walk that chain explicitly: the altered structure changes the shape, and the changed shape changes what the protein can do.
- Two facts win most nucleic-acid questions: complementary pairing (**A–T/U, G–C**) lets you fill in any partner strand, and **antiparallel** orientation means you must flip direction when you write it out. State both explicitly on free-response answers.
- When an AP question links cell size to function, argue from **SA:V**: small cells and highly folded membranes (microvilli, root hairs, alveoli) maximize surface area per volume for faster exchange. Naming the ratio and its direction earns the point.
- Know the three plant-only structures cold: **cell wall, chloroplast, and central vacuole**. A frequent exam prompt gives an unlabeled cell and asks you to identify it as plant or animal — the wall and chloroplasts are the giveaways.
- For any transport question, run a two-step check: (1) Which way relative to the gradient? Down = passive/no ATP, up = active/ATP. (2) Does it need a protein? Small nonpolar = simple diffusion; ion or polar = protein-mediated. Those two answers name the mechanism.
- On free response, state the rule explicitly — "**water moves from higher Ψ to lower Ψ**" — then plug in the numbers, remembering that a **less negative** value is the higher potential. Adding solute makes Ψs more negative and lowers Ψ, pulling water in.
- The AP throughline for enzymes is "**shape → function.**" Any factor that changes the enzyme’s shape — temperature, pH, or a noncompetitive inhibitor — changes the active site and thus the rate. When you explain a rate change, name the effect on shape and the active site explicitly.
- Master the hand-off: the **light reactions** (thylakoid) turn light + water into **ATP, NADPH, and O₂**; the **Calvin cycle** (stroma) spends that ATP and NADPH to fix **CO₂** into **sugar**, returning ADP and NADP⁺. Tracing which product feeds the next stage earns free-response points.
- Keep the tally straight: **glycolysis** (cytoplasm) → 2 ATP + 2 NADH; **pyruvate oxidation + Krebs cycle** (matrix) → 2 ATP + more NADH/FADH₂ + CO₂; **oxidative phosphorylation** (inner membrane) → the bulk of ATP, with O₂ as final electron acceptor forming H₂O. Anaerobically, only glycolysis’s 2 ATP remain.
- Zoom out for the big picture: **photosynthesis and respiration form one coupled cycle.** Photosynthesis captures energy and stores it in glucose (exergonic light harvest driving endergonic sugar synthesis); respiration releases that energy to make ATP. Energy flows one way (sunlight → heat), but matter (C, H, O) cycles.
- For AP free-response, name the stages in order — **reception → transduction → response** — and tie the cascade explicitly to **amplification**. Saying "the signal is passed along" earns little; explaining that each step activates many molecules at the next step earns the point.
- On the AP exam, always state the loop’s *direction* explicitly: does the response **counteract** the change (negative) or **amplify** it (positive)? Anchor the claim to a set point for negative feedback, or to a runaway endpoint (birth, clotting) for positive feedback.
- The AP framing for cancer is "**accelerators and brakes**." Tie any loss of control back to a specific failure: an **oncogene** stuck on (gain of function) or a **tumor suppressor** switched off (loss of function), and name the checkpoint (often **G1**, guarded by **p53**) that fails as a result.
- On free-response questions about "how sexual reproduction increases variation," name all three mechanisms explicitly: **crossing over** (prophase I), **independent assortment** (metaphase I), and **random fertilization**. Listing the mechanism *and* its stage is what earns the point.
- For "what fraction of offspring…" questions, do **not** draw a 16-box dihybrid grid. Solve each gene as a small monohybrid cross (¾ or ¼) and multiply. It is faster and far less error-prone under time pressure.
- For sex-linked pedigree questions, remember the two shortcuts: a son inherits his single X-linked allele from his **mother**, and X-linked recessive traits are **more common in males** because their one X has no partner to mask a recessive allele.
- The AP throughline for this unit is **phenotype = genotype + environment**. When a question shows one genotype producing different phenotypes (fur color, flower color, height), point to an environmental factor acting on gene expression — not to a mutation or a genotype change.
- A reliable free-response earner: match each enzyme to its job in one clean chain — **helicase unwinds → primase primes → polymerase extends 5′→3′ → ligase seals**. Naming the enzyme *and* its specific action is what scores the point.
- Alternative splicing is a favorite "how can one gene make many proteins?" answer. By keeping different combinations of exons, a single pre-mRNA yields multiple distinct mRNAs — a major reason humans have far more proteins than genes.
- Practice the full pipeline until it is automatic: **DNA template → (transcribe, A–U) → mRNA codons → (read code 5′→3′) → amino acids**. On the exam, show the mRNA and the codon splits explicitly — partial credit often lives in those middle steps.
- Nail the epigenetics distinction: methylation and histone modification change *whether* a gene is expressed, **not the DNA sequence**. If a question describes a heritable change in expression with an unchanged sequence, the answer is epigenetic regulation.
- Keep the tools straight by their jobs: **PCR** amplifies (makes more copies), **gel electrophoresis** separates and visualizes by size, and **CRISPR-Cas9** edits a specific sequence. Questions often hinge on picking the right tool for the stated goal.
- On free-response questions, do not just name the type of selection — justify it by pointing to which part of the trait distribution is favored (one extreme = directional, the middle = stabilizing, both extremes = disruptive) and tie the advantage to reproductive success in that specific environment.
- The AP exam treats evidence for evolution as convergent lines pointing to one conclusion: fossils, homologies, molecular sequences, and biogeography independently agree on the same tree of life. In free response, cite more than one line of evidence and state what each independently demonstrates.
- On the AP exam, always begin Hardy–Weinberg problems from the recessive phenotype: set it equal to q², square-root to get q, then p = 1 − q. Show every step. Graders award points for the correct setup (q² = recessive frequency) even if arithmetic slips later.
- For any isolation scenario on the exam, ask two questions in order: (1) Did a hybrid zygote form? No → prezygotic; Yes → postzygotic. (2) Then name the specific mechanism. Being able to justify the pre-/post- split earns more than just labeling it.
- For cladogram questions, translate "most closely related" into "shares the most recent common ancestor," and read shared derived characters from the branch points. When asked to justify relatedness on free response, cite the specific shared derived character and the node it defines.
- For any energy-flow free-response, lead with the one-way principle: **energy enters as sunlight, flows up one direction, and is lost as heat at every step — it is never recycled.** Then apply the 10% rule quantitatively. Naming that ~90% loss (respiration, heat, waste) is what earns the explanation point.
- On the exam, tie the pieces together: **exponential = J-curve, no limits (dN/dt = rN); logistic = S-curve leveling at K (dN/dt = rN((K − N)/K)).** The (K − N)/K term *is* the density-dependent brake in equation form — link the graph, the equation, and the limiting factors in one answer.
- The whole primary-vs-secondary distinction hinges on one word: **soil**. No soil to start (bare rock, lava) → **primary**, slow, begins with pioneer lichens/mosses. Soil already present (after fire or farming) → **secondary**, faster. State the soil condition explicitly and the classification follows.
- The unifying thread of this unit: **diversity buys stability.** High species and genetic diversity lets ecosystems absorb disturbance and lets populations adapt to change; losing it — through invasion, habitat destruction, or pollution — makes systems fragile. On free-response, connect a specific disruption to reduced biodiversity to reduced resilience.
- Free-response graders reward the **mechanism**, not the label. "Water has a high specific heat" earns less than "heat energy is absorbed to break hydrogen bonds before molecular motion (temperature) increases, so water resists temperature change." Always connect the property back to hydrogen bonding.
- Build a mental four-column table before the exam: **macromolecule | monomer | bond | function**. Carbohydrate | monosaccharide | glycosidic | energy/structure. Protein | amino acid | peptide | enzymes/structure. Nucleic acid | nucleotide | phosphodiester | information. Lipid | (glycerol + fatty acids, not a polymer) | ester | storage/membranes. Most Unit 1 identification questions collapse the moment you can recall this grid.
- Memorize the **four evidences** as a set — **double membrane, circular DNA, 70S (bacteria-like) ribosomes, binary fission** — and be ready to say what each demonstrates. Free-response prompts often show organelle data and ask you to argue *for* endosymbiosis; naming the evidence and tying it to a bacterial ancestor earns the points.
- For a water-potential free-response, show the pipeline explicitly: convert to kelvin, compute **Ψs = −iCRT**, set **Ψp = 0** for an open solution (or solve for it at equilibrium), add to get **Ψ = Ψp + Ψs**, then state the rule — **water moves from higher Ψ to lower Ψ**. Carry the MPa units and the negative signs through every line; graders award the reasoning, not just the final number.
- Free-response scoring hinges on the hand-off and the numbers: the **light reactions** (thylakoid membrane) turn light + H₂O into **ATP, NADPH, and O₂**; the **Calvin cycle** (stroma) spends **18 ATP + 12 NADPH** per glucose across **6 turns / 6 CO₂**. If asked why the ATP:NADPH demand is unequal, cite **RuBP regeneration** consuming the extra ATP.
- Lock in the ledger: **glycolysis** → 2 ATP + 2 NADH (cytosol); **pyruvate oxidation** → 2 NADH; **Krebs** → 2 ATP + 6 NADH + 2 FADH₂ (matrix); **oxidative phosphorylation** → ~26–28 ATP from 10 NADH + 2 FADH₂ via the proton-motive force, with **O₂ as final electron acceptor → H₂O**. Total **~30–32**; strip away O₂ and you are left with glycolysis’s **2**.
- For AP free-response, treat **termination as a required half of the answer**: name a specific off-switch (G protein hydrolyzes GTP → GDP, phosphodiesterase degrades cAMP, phosphatases remove phosphates) and note that a *sustained* response needs *continued* ligand. Losing an off-switch causes constant, unregulated signaling — a common cancer mechanism.
- Cancer answers should name **which control failed and how**: an **oncogene** stuck on (gain of function, one copy) or a **tumor suppressor** lost (loss of function, two hits — e.g. **p53** at G1 or **Rb**), plus loss of **density-dependent inhibition**. Stress that cancer needs **several mutations accumulating together**, not one.
- The single most common χ² mistake on the AP exam is miscounting degrees of freedom. **df = number of phenotype categories − 1**, never the sample size. Four categories (9:3:3:1) → df = 3; two categories (3:1 or 1:1) → df = 1. Pick the wrong row of the table and a correct χ² still yields the wrong conclusion.
- Expect a question where the directly measured distance between the two outer genes is **smaller** than the sum of the inner intervals. The cause is **double crossovers**: a second crossover flips the middle marker back but leaves the outer alleles in their parental arrangement, so those offspring are never scored as outer-gene recombinants — long spans underestimate true distance.
- Map each control to its logic: **lac = inducible** (off by default, inducer turns it on), **trp = repressible** (on by default, corepressor turns it off). For eukaryotes, remember that epigenetic marks change **expression, not sequence**, and alternative splicing is the go-to answer for "one gene, many proteins."
- Match the tool to the goal: **PCR** amplifies (2ⁿ copies), **restriction enzymes + ligase** build recombinant DNA, **gel electrophoresis** separates and sizes fragments (small = far), and **CRISPR-Cas9** edits one specific sequence guided by its RNA. Free-response prompts usually hinge on naming the right tool *and* its mechanism.
- On the exam, state your setup explicitly before calculating: write "recessive phenotype = q²" (or "affected males = q" for X-linked), then show p = 1 − q, then p², 2pq, q². When asked whether a population is at equilibrium, compute allele frequencies by gene-counting and compare the observed genotypes to the predicted p²/2pq/q² — a mismatch is the evidence that the population is evolving.
- For tree questions, work in this order: use the outgroup to decide which states are derived, group taxa by shared *derived* characters (synapomorphies), and pick the tree with the fewest changes (parsimony). Treat a morphology-versus-molecule conflict as a homoplasy problem — name convergent evolution explicitly and explain why the character misleads.
- A quantitative FRQ often hands you r, K, and N and asks for dN/dt. Write the logistic equation, substitute, and **show the (K − N)/K step explicitly** — that is where points live. Then interpret: is N below, at, or above K/2? Below K/2, growth is still accelerating; above it, growth is decelerating toward the S-curve plateau at K.
- Nail the two "big impact" categories: a **keystone species** shapes the community through *interactions* despite *low* abundance (remove it → diversity collapses); a **foundation species** shapes it by *building habitat* and is *highly abundant*. And on any diversity calculation, show the full path — proportions → squared → summed → subtracted from 1 — because partial credit tracks each step.
How to get a 5
- Answer the verb precisely: "Describe" = details, "Explain" = cause-and-effect, "Justify" = evidence-based support, "Calculate" = show work.
- Always connect structure to function and cite a mechanism (e.g., "the proton gradient drives ATP synthase") rather than restating the phenomenon.
- Master the formula sheet: chi-square, Hardy-Weinberg, standard error, and water potential (Ψ = Ψp + Ψs) are near-annual easy points.
- For experimental design, identify the independent & dependent variables, a control, and change only one variable; sketch a testable prediction.
Key terms
Hardy–Weinberg equations — p + q = 1 and p² + 2pq + q² = 1. p² = homozygous dominant, 2pq = heterozygous, q² = homozygous recessive.
Five conditions for Hardy–Weinberg equilibrium — No mutation, no natural selection, no gene flow, random mating, and large population (no genetic drift).
Net products of glycolysis (per glucose) — 2 pyruvate, 2 net ATP, 2 NADH. Occurs in the cytoplasm; does not require oxygen.
Where does the ETC occur, and the final electron acceptor? — Inner mitochondrial membrane (cristae); O₂ is the final acceptor, forming H₂O. Produces most ATP (~26–28).
Light reactions vs Calvin cycle — Light reactions: thylakoid membrane, make ATP + NADPH + O₂. Calvin cycle: stroma, uses ATP/NADPH + CO₂ to make G3P/sugar.
Fluid mosaic membrane — Phospholipid bilayer with embedded proteins; hydrophilic heads out, hydrophobic tails in. Semi-permeable; regulates transport.
Osmosis and tonicity — Water moves from high to low water potential. Hypertonic → cell shrinks; hypotonic → swells/bursts; isotonic → no net movement.
Feedback (negative) inhibition — End product binds an early enzyme's allosteric site, shutting the pathway off — maintains homeostasis.
Central dogma — DNA → (transcription) → mRNA → (translation) → protein. Transcription in nucleus; translation at ribosomes.
Signal transduction steps — Reception (signal binds receptor) → Transduction (relay cascade, often phosphorylation) → Response (cellular change).
Mitosis vs Meiosis outcome — Mitosis: 1 division → 2 diploid identical cells. Meiosis: 2 divisions → 4 haploid varied gametes.
Sources of genetic variation in meiosis — Crossing over (prophase I), independent assortment (metaphase I), and random fertilization.