Fourteen substantial lessons that turn biological facts into explanations, experimental reasoning, evolutionary forces, Hardy-Weinberg calculations and energy flow.
Focused on the reasoning skills and selected misconceptions documented in the 2025 exam report. Not an exhaustive biology textbook or complete exam course.
Basic cells, DNA/RNA/protein vocabulary, ratios and graph reading.
Course outline
Read a biological graph as an argument
Connect axes and comparison to a bounded conclusion.
Controls answer a specific alternative
Explain what a control rules out in the actual experiment.
DNA, RNA and proteins are different levels
Track information flow without treating every molecule as interchangeable.
Pathways need causal links
Explain how an upstream change can propagate to an outcome.
Selection changes populations, not intentions
Distinguish variation, inheritance and differential reproduction.
Use evidence to distinguish models
Prefer a testable explanation over a story that fits everything.
Enzymes change rate, not equilibrium
Separate activation energy from reaction energy.
Transport follows gradients and membrane properties
Predict movement without confusing energy sources.
Inheritance probabilities are models
State assumptions before multiplying genotype chances.
Feedback maintains a range, not a perfect constant
Trace negative and positive feedback directions.
Chance matters most in small populations
Explain why genetic drift changes allele frequencies more in small populations than in large ones.
Which evolutionary force is acting?
Match a described change in a population to selection, drift, gene flow, mutation or nonrandom mating.
Hardy-Weinberg: from one frequency to the rest
Use p, q and the Hardy-Weinberg equation to calculate genotype frequencies and state what the model assumes.
Where the mass of a plant comes from
Use the photosynthesis equation to trace what goes in, what comes out, and where plant mass comes from.
Sources and curriculum note
Checked against the current official course description in October 2026. Toy examples are labelled and do not report real experiments. Lessons 11 to 14 use OpenStax Biology for AP Courses pages fetched October 2026.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Read a biological graph as an argument
Learning goal: Connect axes and comparison to a bounded conclusion.
A graph is a compressed argument. Before reading heights, identify the dependent variable and its units, the independent variable or categories, and whether values are raw, normalized or percentages. A bar labelled relative protein amount does not directly show absolute molecule count.
When a graph has grouped bars, hold one variable constant while comparing the other. For example, compare treated and control cells for protein A, then repeat for protein B. A difference for A but not B supports a selective effect under the tested conditions, not a universal statement about all proteins.
Error bars need a definition: standard deviation, standard error and confidence intervals describe different things. Overlap alone is not a universal significance test. Without a statistical procedure, describe the observed pattern and avoid claiming a tested difference. The 2025 biology report flags interpreting graphs with multiple independent variables.
Practice with an invented graph. A bar chart shows relative amount of protein A and protein B in control cells and in cells treated with a drug. Protein A drops to about half in treated cells, and protein B stays about the same. The y-axis says "relative amount", so the bars compare to a reference and not to a count of molecules. A fair sentence is: "In treated cells, the relative amount of protein A is lower than in control cells, while protein B is similar in both groups."
Notice what the sentence leaves out. It does not say the drug destroys protein A, because the graph does not show the mechanism. It does not call the gap significant unless a test is reported. Ask what the error bars represent and how many replicates were run. Good graph reading is slow and literal first, then interpretive.
Worked example
In an original teaching dataset, transport is 100% for A and B in control, 40% for A and 95% for B after treatment. State one supported claim.
Identify the outcome as normalized transport, not total protein abundance.
Protein A has lower measured transport under treatment than control.
Protein B changes much less in these illustrative values.
Do not claim statistical significance or a mechanism from these bars alone.
Practice problem and solution
Hypothetical experiment: three matched control readings are 70,80,90 units; treatment readings are 50,60,70. Enter treatment mean as a percentage of control mean. Show the averaging, calculate the percentage decrease, and limit your claim to transport.
Control mean=80, treatment mean=60. Remaining=60/80×100=75%; decrease=25%. Transport readings alone do not establish a change in protein production.
Mental model: Describe measured patterns before explaining mechanisms.
Common trap: Changing the outcome, denominator or comparison without noticing.
2. Controls answer a specific alternative
Learning goal: Explain what a control rules out in the actual experiment.
A control is not merely "something normal". It provides a comparison that addresses an alternative explanation. If a drug is delivered in a solvent, a solvent-only group tests whether the solvent rather than the drug caused the observed change.
A positive control shows that an assay can detect the expected effect under known effective conditions. A negative control checks the background response without the target treatment. Neither makes all confounding impossible. Keep other relevant variables as similar as feasible and include replication.
Write the explanation in context: "The solvent-only group allows the effect of solvent exposure to be separated from the effect of the drug." The 2025 report flags vague explanations such as "to establish a baseline" that never say what alternative is addressed.
A second protein unaffected by the targeted treatment can help test whether the effect is specific rather than a general change across all measured proteins; it does not by itself prove the mechanism.
Extend the solvent example with a second invented experiment. Researchers test whether a compound slows the growth of bacteria. One plate gets the compound dissolved in a small amount of ethanol. If only that plate is compared with an untreated plate, a smaller colony could be due to the compound or to the ethanol. An ethanol-only plate answers that specific alternative. An untreated plate shows normal growth, and a plate with a known antibiotic works as a positive control to show the method can detect growth inhibition.
Each control addresses a different doubt, and none replaces replicates. Running each condition several times shows whether differences are consistent. When you write, finish the sentence "This control rules out..." with the particular alternative in the scenario, because that phrase turns a generic statement into an explanation.
Worked example
A plant treatment is sprayed in water. Which comparison distinguishes chemical effect from spraying?
Use a water-only spray group with the same amount and schedule.
Keep plant age, light and soil conditions as comparable as possible.
Compare outcomes across replicated groups.
An untreated group alone cannot isolate spray exposure from chemical exposure.
Practice problem and solution
In an original model, control=100, solvent only=80, and Drug X plus solvent=50 units. What is Drug X’s decrease relative to the solvent-only control, as a percentage of that control? In your reasoning: Compute the solvent-only decrease from the untreated control, then distinguish the drug’s matched-control decrease from the combined decrease.
(80−50)/80 ×100=37.5%. Comparing with 100 confounds the treatment and solvent effects. Solvent decreases transport by (100−80)/100=20%. Drug plus solvent decreases it by 50% from untreated control, but only 37.5% relative to the solvent-only matched control.
Mental model: A control earns its value by addressing a named alternative explanation.
Common trap: Writing "baseline" without explaining what comparison means.
3. DNA, RNA and proteins are different levels
Learning goal: Track information flow without treating every molecule as interchangeable.
Transcription uses a DNA template to make RNA. Translation reads messenger RNA codons to assemble a polypeptide. Ribosomes synthesize proteins from amino acids; they do not manufacture amino acids. The 2025 Chief Reader report directly identifies this confusion.
Each codon consists of three nucleotides. A coding sequence’s nucleotide count can be converted to codon count only when its boundaries and reading frame are known. A stop codon does not encode an amino acid. Not every DNA nucleotide in a gene is part of the translated sequence: introns and untranslated regions matter.
A mutation’s effect depends on where it occurs and what changes. A substitution can be silent, missense or nonsense in a coding region. An insertion not divisible by three may shift a coding reading frame, but do not claim all insertions do so or every mutation changes phenotype.
Trace one invented example. A short stretch of DNA template reads 3'-TAC-5' and is transcribed to mRNA 5'-AUG-3'. The codon AUG specifies methionine, the usual start codon. A ribosome reads codons in order, and transfer RNAs bring the matching amino acids. If a substitution changed a later codon so that it became a stop codon, translation would end early and give a shorter polypeptide, which is a nonsense mutation. If the change produced a different codon for the same amino acid, the protein would be unchanged, which is a silent mutation.
To count amino acids from nucleotides, divide the coding length by three and subtract the stop codon, but only when you know where reading starts. Keep the molecules distinct: DNA stores the sequence, RNA carries it, and protein carries out many cell functions.
Worked example
One coding RNA includes 303 nucleotides from start codon through one stop codon. How many amino acids, assuming no other processing?
303/3 = 101 codons.
One is the stop codon, which contributes no amino acid.
The translated chain contains 100 amino acids.
This calculation assumes the specified segment contains no untranslated nucleotides.
Practice problem and solution
Hypothetical coding regions contain 363 and 273 nucleotides, each including one three-nucleotide stop codon and no introns. Enter the difference in translated amino-acid counts. Show both counts and explain why stop codons do not contribute amino acids.
Translated lengths=(363−3)/3=120 and (273−3)/3=90 amino acids; difference=30. Stop codons terminate translation rather than encode amino acids.
Mental model: Follow the molecule and the level: template, message, codon, protein.
Common trap: Dividing all genomic nucleotides by three without checking what is translated.
4. Pathways need causal links
Learning goal: Explain how an upstream change can propagate to an outcome.
A pathway diagram describes a mechanism, not just a list of nouns. Identify what each arrow means: activation, inhibition, movement or production. An inhibitory line has the opposite effect from an activating arrow. Read the entire chain before predicting an outcome.
If A activates B and B activates C, inhibiting A can reduce C. If A inhibits B and B activates C, inhibiting A can increase C. This sign logic assumes the diagram captures the relevant dominant pathway. Parallel branches, feedback and saturation can change the response.
In an explanation, connect every step: treatment reduces receptor activity, so the downstream relay is less active, which reduces the measured response. A final outcome without intermediate reasoning leaves the mechanism unspecified. The 2025 report flags difficulty explaining downstream effects of pathway inhibition.
Work through a signal chain with invented names. A hormone binds a receptor on the cell surface, the receptor activates a relay protein, the relay protein activates an enzyme, and the enzyme produces a response such as opening a channel. If a drug blocks the receptor, the relay and enzyme are activated less, so the response falls. If instead the relay protein is normally inhibited by a repressor, and the drug blocks the repressor, then the relay becomes more active and the response rises.
Write the answer one link at a time: what changed first, what that does to the next component, and what the last link does to the measured outcome. Skipping the middle steps is the usual loss of credit. When a question asks about a pathway with feedback or parallel branches, say that the simple chain is a model and name the extra branch that could change the result.
Worked example
A inhibits B; B activates C. Predict the effect of removing A.
Removing A reduces inhibition of B.
B can become more active under the model.
Greater B activity activates C more strongly.
Therefore C is expected to increase, subject to the model assumptions.
Practice problem and solution
Hypothetical pathway: A inhibits B and B activates C. Inhibiting A while B remains functional predicts what change in C? Enter the direction. Then predict the result if B is also blocked and explain each causal link and the model’s limits.
Less A releases B; active B increases C. If B is blocked, this route no longer predicts an increase in C. These are conditional pathway predictions, not claims about all other pathways.
Mental model: Use signed arrows and explain intermediate steps.
Common trap: Assuming every upstream inhibition causes a downstream decrease.
5. Selection changes populations, not intentions
Learning goal: Distinguish variation, inheritance and differential reproduction.
Natural selection acts on existing variation when individuals with different traits have different reproductive success. An individual does not acquire the needed mutation because it needs to survive. Mutations arise without regard to future usefulness; the environment influences which heritable variants leave more offspring.
A population can evolve as allele frequencies change across generations. Acclimation or a behavioral change within one individual is not itself evidence of a population-level genetic change. A selected trait must affect reproductive outcomes in the relevant environment, not merely look "stronger".
Reproductive isolation can prevent gene flow and allow divergence. A geographic barrier may help create isolation, but a barrier alone does not immediately establish a new species. Explain how interbreeding or fertile offspring is prevented in the model being used.
Use a population example with invented numbers. A beetle population has a green form and a brown form that differ by an inherited trait. Birds eat the more visible form on a dark soil background, so more brown beetles survive and reproduce there. Over generations the allele for brown becomes more common. The individual beetles did not change color because they needed to. The change happened because brown beetles left more offspring on that background.
For speciation, suppose a river splits one population into two. If the groups later cannot produce fertile offspring when brought back together, they are reproductively isolated and may count as separate species. The river is the barrier, not the cause of the incompatibility itself. When asked about evolution, name the variation, the inheritance, the difference in reproductive success and the resulting change in frequency.
Worked example
Antibiotic exposure leaves more resistant bacteria reproducing. Explain without saying they "tried to adapt".
Resistance variants were already present or arose independently of need.
The antibiotic reduces reproduction of susceptible bacteria more strongly.
Resistant bacteria contribute a larger share of descendants.
Resistance becomes more frequent in the population over generations.
Practice problem and solution
Hypothetical diploid population: before selection, 100 individuals include 4 AA,32 Aa,64 aa; the next generation of 100 includes 12 AA,46 Aa,42 aa. Enter the percentage-point change in A frequency. Show allele counts and explain why the data do not show individuals choosing mutations.
Before A=(2×4+32)/200=0.20; after A=(2×12+46)/200=0.35. Change=15 percentage points. Population frequencies changed; no individual intention is established.
Mental model: Selection sorts heritable variation through reproduction.
Common trap: Using the word "adapt" without identifying the mechanism.
6. Use evidence to distinguish models
Learning goal: Prefer a testable explanation over a story that fits everything.
A hypothesis is stronger when it predicts an observable difference between competing explanations. If two models predict exactly the same measured outcome, that outcome cannot distinguish them. Choose a measurement where their predictions diverge.
Biological conclusions are bounded by design and data. If inhibiting a protein reduces transport, it may support a role in transport, but it does not alone establish direct binding, sufficiency or exclusive responsibility. A rescue experiment can add evidence by restoring function after the perturbation.
Claims should name the result, the comparison and the link to a model. Distinguish "necessary" from "sufficient": a component is necessary if the outcome cannot occur without it under the tested conditions; sufficient if it can produce the outcome in the appropriate tested setting. Complex biology often requires careful qualification.
Compare two models with an invented question. Model one says a protein carries a substance across a membrane. Model two says the substance diffuses through on its own and the protein only senses it. A measurement of total uptake may look the same under both models, so it cannot tell them apart. A better test adds a molecule that blocks the protein: model one predicts uptake falls, model two predicts no change.
Now check how far the result goes. If uptake falls, the result supports a role for the protein but does not show it binds the substance directly, and it does not show the protein is the only route. A rescue experiment adds back a working copy of the protein and checks whether uptake returns, which strengthens the case. Write claims with the result, the comparison and a limited conclusion such as "supports" instead of "proves".
Worked example
Knocking out X removes an outcome, and restoring X rescues it. State a bounded conclusion.
The knockout reduces the outcome relative to the comparison group.
The rescue reduces concern that an unrelated knockout effect caused the loss.
The results support X as necessary for the outcome under these conditions.
They do not establish that X alone is sufficient in every context.
Practice problem and solution
Hypothetical signal readings: normal cells 100, X-knockout 0, restored X in knockout 95. Enter the requirement relationship supported for X. In reasoning compare loss and rescue, explain why sufficiency is not shown, and propose a matched test of sufficiency.
Loss and rescue support X being necessary under tested conditions, not sufficient alone. A sufficiency test would add X in a context lacking the activating input, with matched controls, and compare signal; other required components must be considered.
Mental model: Evidence favors models through comparisons and distinct predictions.
Common trap: Concluding a mechanism is proven from one compatible observation.
7. Enzymes change rate, not equilibrium
Learning goal: Separate activation energy from reaction energy.
Enzymes lower an activation-energy barrier and increase reaction rate under suitable conditions. They do not change the reaction’s free-energy difference or equilibrium constant. A catalyst can speed approach to equilibrium in both directions.
Substrate concentration, temperature, pH and enzyme amount can affect measured rates. Increasing substrate may increase rate until enzyme sites become limiting. A rate increase does not automatically mean the product is more energetically favored.
Inhibitor effects depend on mechanism. In a simplified competitive model, substrate competes for active-site binding; higher substrate can reduce the inhibitor’s relative effect. Real regulation can involve additional interactions, so keep model assumptions visible.
Link the energy diagram to the words. Picture a hill the reactants must climb before they can become products. The height of that hill is the activation energy, and an enzyme lowers it. The difference in height between reactants and products is the overall energy change, and the enzyme does not move it. Because the barrier is lower in both directions, the reaction reaches its equilibrium faster, but the amounts at equilibrium stay the same.
For rate graphs, a common pattern is that rate rises with substrate and then levels off when enzyme molecules are all busy. Adding more enzyme would raise the plateau. Extreme temperature or pH can change the enzyme's shape and lower its rate. When explaining an inhibitor, describe whether it competes for the active site, since that decides whether adding more substrate helps. These patterns are general models and real enzymes vary.
Worked example
An enzyme doubles reaction rate. Did it necessarily change equilibrium?
Catalysis changes the path and activation barrier.
The reaction’s free-energy difference is unchanged.
No equilibrium shift follows from the rate increase alone.
Practice problem and solution
Hypothetical reversible reaction has the same equilibrium ratio with and without an enzyme. Starting away from equilibrium, the enzyme speeds both directions. Enter whether a changed final equilibrium composition is predicted. In reasoning compare early and equilibrium measurements and design a two-time-point comparison that separates speed from final composition.
No. An early measurement can differ because catalysis speeds approach to equilibrium; an equilibrium measurement should converge to the same composition under the same conditions. Use matched mixtures, one early time and a verified equilibrium time, without changing temperature or starting amounts.
Mental model: Enzyme = kinetic facilitator, not free-energy rewrite.
Common trap: Calling faster reactions more energetically favorable.
8. Transport follows gradients and membrane properties
Learning goal: Predict movement without confusing energy sources.
Diffusion is net movement down a concentration gradient arising from molecular motion. At equilibrium molecules still move, but there is no net flux under the defined conditions. Facilitated diffusion uses membrane proteins without directly requiring metabolic energy to move a substance down its electrochemical gradient.
Active transport moves a substance against its relevant gradient using an energy source, directly or through a coupled gradient. A membrane protein’s presence alone does not make transport active.
Osmosis concerns water movement across a selectively permeable membrane. Predicting volume change requires the effective solute gradient and permeability. Simple classroom predictions assume nonpenetrating solutes and otherwise comparable conditions.
Predict with an invented cell. A cell sits in a solution that has a lower concentration of solutes than the cell's interior, and the membrane does not let those solutes through. Water moves toward the higher solute concentration, which is into the cell, so the cell tends to swell. An animal cell may burst if the swelling continues. A plant cell with a wall becomes firm because the wall resists expansion. In a solution with a higher solute concentration, the opposite happens and water leaves.
For ions and larger molecules, ask three questions: which direction is down the gradient, does the membrane allow the substance through directly, and is a protein or an energy source involved? A protein channel moving ions down a gradient is facilitated diffusion. A pump moving them against it with an energy source is active transport. Use the gradient and the energy to decide, not the presence of a protein alone.
Worked example
A transporter moves solute from low to high concentration using ATP. Classify it.
The solute moves against its concentration gradient.
ATP supplies energy.
This is active transport under the given conditions.
Practice problem and solution
Hypothetical uncharged solute equilibrates across a freely permeable membrane: concentration 8 in an outside volume of 1, and concentration 2 in an inside volume of 1. Enter final concentration on each side. Calculate conserved amount and predict the result if the outside volume were 2 instead, assuming fixed volumes and no reactions or active transport.
Equal-volume total=8+2=10 over volume 2, giving 5 each. With outside volume 2, total=16+2=18 over volume 3, giving 6 each. Equilibrium does not stop molecular motion.
Mental model: Track gradient, permeability and energy separately.
Common trap: Calling all protein-mediated transport active.
9. Inheritance probabilities are models
Learning goal: State assumptions before multiplying genotype chances.
A Mendelian cross models gamete production and combination under defined assumptions. A heterozygote Aa produces A and a gametes in equal proportions in a simple segregation model. Crossing two heterozygotes yields genotype probabilities 1/4 AA, 1/2 Aa and 1/4 aa.
A phenotype ratio requires assumptions about dominance and penetrance. Independent assortment of two loci is a model for unlinked or effectively independently inherited genes; tightly linked loci can violate the simple product rule.
Probabilities predict long-run proportions, not exact counts in every small family. Four offspring need not include one of each predicted category. Distinguish expected counts from guaranteed outcomes.
For two independently inherited heterozygous loci, P(aa and bb)=P(aa)P(bb)=1/4×1/4=1/16. This product is a model result. If the loci are linked, obtain gamete probabilities from the linkage information instead of assuming all four gametes have equal frequency.
Under the stated complete-dominance model, Aa×Aa predicts genotype probabilities 1/4 AA, 1/2 Aa, 1/4 aa, giving a 3:1 dominant-to-recessive phenotype ratio over many offspring.
Add a worked check of the model. A dominant allele A and a recessive allele a segregate in a cross of two heterozygotes. Out of 400 offspring, the model predicts about 300 with the dominant phenotype and about 100 with the recessive one. Real results might be 292 and 108. That is a normal difference caused by chance and is not evidence against the model by itself. A statistical test such as chi-square is the tool for judging whether a difference is larger than chance would give, but only when you follow a taught procedure and state the null model.
Always state the assumptions before calculating: one gene, complete dominance, equal gamete production and equal survival of offspring. If one assumption fails, such as lower survival of aa offspring, the observed ratio can shift. Naming the assumption is part of a strong answer, because it shows you know what the probability applies to.
Worked example
For Aa×Aa with complete dominance, find P(recessive phenotype).
Each parent contributes a with probability 1/2.
Both must contribute a for aa.
Multiply: 1/2×1/2=1/4.
The result is a probability, not a guaranteed one-in-four family pattern.
Practice problem and solution
Hypothetical cross Aa×Aa: two offspring are independent. Enter the probability exactly one is heterozygous. Derive the one-offspring probability, count both orders and state the model assumptions.
Each offspring has P(Aa)=1/2 and P(not Aa)=1/2. Exactly one Aa has probability 2×(1/2)×(1/2)=1/2, assuming the Mendelian segregation and independence model.
Mental model: Inheritance predictions require a model and do not dictate every small sample.
Common trap: Multiplying across linked loci without checking independence.
10. Feedback maintains a range, not a perfect constant
Learning goal: Trace negative and positive feedback directions.
Negative feedback opposes a deviation from a regulated range. If a variable rises, the response tends to reduce it, and vice versa. The process may involve delay and fluctuation; it does not guarantee an exact constant every instant.
Positive feedback amplifies a change, often until an external endpoint or limit stops the loop. "Positive" is not synonymous with beneficial, and "negative" is not harmful. These terms describe the loop direction.
A useful explanation identifies the regulated variable, sensed change, response and effect on the original variable. A diagram with only arrows between organ names may omit the feedback relationship.
Take a thermoregulation example. The regulated variable is body temperature. If it rises above the set range, sensors detect the rise and send a signal to an effector response, such as sweating or widening of surface blood vessels. These responses release heat, and the temperature falls back toward the range. If it falls too low, shivering and narrowing of surface vessels help warm the body. Because the response reverses the deviation, this is negative feedback.
A positive-feedback example is the increase of contractions during childbirth, in which a signal strengthens the contractions that in turn increase the signal until the baby is born. The loop ends when an outside event stops it. When writing, list the variable, the sensor, the response and the direction of change on the original variable. A response that merely names an organ without describing the loop will leave out the central idea.
Worked example
A rise in blood glucose triggers a response that lowers blood glucose. Classify the feedback.
The original variable rises.
The response opposes that rise.
This is negative feedback in the stated simplified model.
Practice problem and solution
Hypothetical feedback model: initial deviation +4 triggers response −3; the remaining deviation then triggers response −0.75. Enter the deviation immediately after the first response. In reasoning calculate the second result and identify the feedback direction in both steps.
After first response=4−3=1; after second=1−0.75=0.25. Both responses oppose positive deviation, so both represent negative feedback.
Mental model: Feedback signs describe direction, not goodness.
Common trap: Naming organs without explaining the loop.
11. Chance matters most in small populations
Learning goal: Explain why genetic drift changes allele frequencies more in small populations than in large ones.
Natural selection is not the only thing that changes a population. OpenStax Biology for AP Courses explains that genetic drift stems from the chance occurrence that some individuals have more offspring than others and so pass on more of their genes. Drift is random with respect to which allele is helpful. It does not move a population toward a better fit.
The size of the population matters. OpenStax states that small populations are more susceptible to genetic drift and that large populations are buffered against the effects of chance. Its example: if one individual in a population of 10 dies young before leaving any offspring, 1/10 of the gene pool is suddenly lost. In a population of 100, the same loss is only 1 percent of the gene pool, which is much less important to the population's genetic structure. Fractions like these are a quick way to show the effect in an answer.
Two situations produce strong drift. In the founder effect, a portion of a population leaves to start a new population, or a population is divided by a barrier. Those individuals are unlikely to be representative of the whole, so the new population's genetic structure differs from the original. A bottleneck occurs after an event, such as a natural disaster, that kills individuals at random and leaves a small group of survivors.
When an exam question describes a change in allele frequency, ask whether the change favors a trait that helps survival or whether it looks random. Random loss with no link to fitness, especially in a small group, points to drift. A change that consistently favors a trait points to selection. Name the evidence you used, such as the population size or the random nature of the event, instead of only naming the term.
Worked example
A population of 10 and a population of 100 each lose one individual that had no offspring. Compare the fraction of each gene pool lost and explain what this says about drift.
In the population of 10, one individual holds 1/10 of the gene pool, which is 10 percent.
In the population of 100, one individual holds 1/100 of the gene pool, which is 1 percent.
The same chance event changes the small population ten times more.
Drift therefore has a larger effect on small populations.
Practice problem and solution
In a hypothetical population of 20, one individual with no offspring dies. Enter the percentage of the gene pool lost, and say whether drift would affect this population more or less than a population of 200.
1/20 = 5 percent. Drift would affect the population of 20 more, because a single chance event is a larger share of a small gene pool.
Mental model: Drift is chance change in allele frequency, and it is stronger in small populations.
Common trap: Calling every change in allele frequency natural selection.
12. Which evolutionary force is acting?
Learning goal: Match a described change in a population to selection, drift, gene flow, mutation or nonrandom mating.
OpenStax lists the forces that can change allele frequencies: natural selection, genetic drift, gene flow, mutation, nonrandom mating and environmental variation. Exam questions rarely name the force. They describe a situation, and you must choose the force and justify the choice with evidence from the description.
Gene flow is the flow of alleles in and out of a population through the migration of individuals or gametes. OpenStax gives pollen carried far by wind or birds as an example. Look for movement between populations. Mutations are changes to an organism's DNA, and OpenStax calls the appearance of new mutations the most common way to introduce novel genotypic and phenotypic variability. Look for a new allele that was not in the population before.
Nonrandom mating occurs when individuals do not pair by chance. OpenStax gives the example of peahens preferring peacocks with bigger, brighter tails, and describes assortative mating, a preference for partners who are phenotypically similar. Geographic distance can also make mating nonrandom, since some individuals have much easier access to each other. Look for a pattern in who mates with whom.
Natural selection requires a trait that gives some individuals an advantage, so that they leave more offspring. Genetic drift is a chance change, strongest in small populations. To decide between them, ask whether the change is linked to a trait that affects survival or reproduction. If the question gives a reason related to the trait, choose selection. If it describes a random event or a small group, choose drift.
When you justify the choice, quote the evidence word for word, for example "migrants arrived", and link it to the definition of the force. Do not choose a force only because the story feels familiar.
Worked example
Pollen from a nearby field is carried by wind into a plant population, and a new flower-color allele appears at low frequency. Which force is described, and what is the evidence?
Pollen carried from another population moves alleles into this population.
OpenStax defines gene flow as the flow of alleles in and out of a population through migration of individuals or gametes.
Pollen is a gamete, so the pollen movement is the evidence.
The force is gene flow. The new allele arrived by migration of gametes, not by a change in DNA inside this population.
Practice problem and solution
Hypothetical: a hurricane kills 90 percent of a lizard population at random, and the survivors differ in allele frequencies from the original. Name the force in one or two words, and state the evidence.
Genetic drift (a bottleneck). The event killed individuals at random, not by trait, leaving a small group that is not representative of the original population.
Mental model: Choose the force from the clue: movement, new allele, mating pattern, advantage, or chance in a small group.
Common trap: Picking natural selection when the description gives no link between a trait and survival.
13. Hardy-Weinberg: from one frequency to the rest
Learning goal: Use p, q and the Hardy-Weinberg equation to calculate genotype frequencies and state what the model assumes.
The Hardy-Weinberg model describes a population that is not evolving at a gene with two alleles. Let p be the frequency of the dominant allele and q the frequency of the recessive allele. Because there are only two alleles, p + q = 1. If mating is random, the genotype frequencies in the next generation are p² for homozygous dominant, 2pq for heterozygous and q² for homozygous recessive. These add up to 1.
Most problems start from the one frequency you can observe directly: the recessive phenotype. Its frequency equals q², since only homozygous recessive individuals show it. Take the square root to find q, then use p = 1 − q and then calculate 2pq and p². The most common mistake is to treat the frequency of the recessive phenotype as q instead of q². Another is forgetting to double pq for the heterozygotes.
For example, if 4 percent of a population shows a recessive trait, q² = 0.04, so q = 0.2 and p = 0.8. Carriers are 2pq = 2(0.8)(0.2) = 0.32, so 32 percent. Homozygous dominant individuals are p² = 0.64. Check: 0.64 + 0.32 + 0.04 = 1.
The model works only when conditions hold: a large population, random mating, no mutation, no gene flow and no natural selection. These match the forces OpenStax lists as changing allele frequencies. OpenStax also notes that small populations are more affected by drift. If observed genotype frequencies differ from the calculated ones, at least one condition is not met. A good answer names the likely force and gives evidence from the question. The calculation tells you what to expect, not why the data deviate.
Worked example
In an invented population, 4 percent of individuals show a recessive phenotype. Find the frequency of carriers.
The recessive phenotype is q² = 0.04.
q = √0.04 = 0.2, so p = 0.8.
Carriers are heterozygous: 2pq = 2(0.8)(0.2).
2pq = 0.32, so 32 percent of the population are carriers.
Practice problem and solution
Invented population in Hardy-Weinberg equilibrium: q² = 0.09. Enter the frequency of heterozygotes (2pq) as a decimal.
q = 0.3, p = 0.7, so 2pq = 2(0.7)(0.3) = 0.42.
Mental model: Recessive phenotype gives q², then q, then p, then 2pq.
Common trap: Using the recessive phenotype frequency as q.
14. Where the mass of a plant comes from
Learning goal: Use the photosynthesis equation to trace what goes in, what comes out, and where plant mass comes from.
OpenStax gives the overall equation for photosynthesis: six molecules of carbon dioxide and six of water, using sunlight as an energy source, produce one sugar molecule and six molecules of oxygen. The sugar is glucose, C6H12O6. It is made from two three-carbon molecules during the process. OpenStax notes that the equation is deceptively simple, because the process takes many steps with intermediate reactants and products.
Use the equation to trace matter. The six carbon atoms in glucose come from the carbon dioxide. Water is the only input containing hydrogen, so the hydrogen in glucose must come from water, and the total number of atoms of each element is the same on both sides. The equation is balanced, so no atoms are created or destroyed. That means the mass a plant gains as sugar and other molecules built from sugar comes mainly from the carbon dioxide and water it takes in, not from the soil. Soil supplies water and other materials, but the sugar itself is built from the gas.
Light is an energy input, not a material input. The energy of sunlight is converted into the chemical energy stored in the bonds of sugar. OpenStax describes photosynthesis as the process that lets organisms access free energy from the sun and transform it into the chemical energy of sugars.
In plants, photosynthesis generally occurs in leaves, in a middle layer of cells called the mesophyll. Carbon dioxide and oxygen move through small regulated openings called stomata, which also help regulate water balance. In an answer about a change, such as closing stomata, trace the effect: less carbon dioxide enters, so less sugar can be made.
Worked example
A plant makes one molecule of glucose. Using the overall equation, how many CO2 molecules were used, and where do the carbon atoms of the glucose come from?
The overall equation uses 6 CO2 and 6 H2O to make 1 glucose and 6 O2.
Glucose has 6 carbon atoms.
Six CO2 molecules contain exactly 6 carbon atoms.
So the carbon atoms of glucose come from carbon dioxide taken in from the air.
Practice problem and solution
Using the overall equation, how many molecules of oxygen are released when a plant makes 3 molecules of glucose? Enter the number.
Each glucose comes with 6 O2. For 3 glucose, 3 times 6 = 18 oxygen molecules.
Mental model: 6 CO2 + 6 H2O + light energy gives glucose + 6 O2; plant sugar carbon comes from carbon dioxide.
Common trap: Saying plants get their sugar mass from the soil.