Educerie · AP · Biology
AP Biology 7.5 Hardy–Weinberg Equilibrium
In one breath
- p + q = 1 for alleles; p² + 2pq + q² = 1 for genotypes, because random gametes combine like the areas of a square.
- A population stays in Hardy–Weinberg equilibrium only if it is large, closed to migration, free of new mutations, mating at random and not under selection; real populations never are, so the equilibrium is a null hypothesis of no evolution.
- From a recessive phenotype, q² → q → p → 2pq; never take the root of the dominant phenotype.
- From a full genotype count, count allele copies and compare observed with expected numbers.
- Too few heterozygotes points to non-random mating.
- A change in allele frequency between generations is evolution, even if each generation fits the equation.
What you must be able to do
- Describe the five conditions under which allele and genotype frequencies stay constant, and use Hardy–Weinberg as a null hypothesis"Identify which condition of Hardy–Weinberg equilibrium is violated" (1 point); "State the null hypothesis" (1 point)
- Calculate allele frequencies from genotype frequencies (and back), and use the results to judge whether a population is evolving"Calculate the expected number of heterozygotes" (1–2 points); "Using the data, justify the claim that the population is evolving" (2 points)
Once you have read the note, tick off each one you can do in Check your understanding.
In the exam. Section I multiple choice, nearly every year: calculations from a count of a recessive phenotype or a table of genotypes, questions on which condition a scenario breaks, and graphs of allele frequency over generations. The calculator is allowed and the equations are on the formula sheet, so the marks are in choosing the right term. In Section II, Hardy–Weinberg can be the calculation in Interpreting and evaluating experimental results (Q1), the null hypothesis in Scientific investigation (Q3), or the whole of an Analyze data (Q6) question where you calculate, compare with expected values and justify a claim about evolution.
1The two equations, and the difference between them
The formula sheet gives:
where p is the frequency of allele 1 and q the frequency of allele 2. The CED warns that students mix up the two kinds of frequency, so keep them apart:
| Symbol | What it is a frequency of | In a population with dominant allele A and recessive allele a |
|---|---|---|
| p | an allele | allele A |
| q | an allele | allele a |
| p² | a genotype | AA individuals |
| 2pq | a genotype | Aa individuals (carriers) |
| q² | a genotype | aa individuals |
| p² + 2pq | a phenotype | everyone showing the dominant trait |
Figure 1 shows where the second equation comes from. Draw a square of side 1. Split the top edge into the fraction of sperm or pollen carrying A (p = 0.7) and carrying a (q = 0.3), and split the side edge the same way for eggs. Each small rectangle's area is the chance of that pair meeting: 0.49 for AA, 0.21 twice for Aa, and 0.09 for aa. Together they fill the square, so they add to 1.
Allele frequencies (p, q) and genotype frequencies (p², 2pq, q²) are different numbers: to get q from a count of recessive individuals, take the square root of q², never of anything else.
2The five conditions, and what breaking each does
Allele frequencies stay constant only if none of the forces of topic 7.4 is acting.
| Condition | What happens if it is broken |
|---|---|
| 1. A large population | Genetic drift changes allele frequencies by chance. |
| 2. No migration | Gene flow adds or removes alleles. |
| 3. No new mutations | New alleles appear and others become less frequent. |
| 4. Random mating | Genotype frequencies shift, for example fewer heterozygotes if relatives mate or if individuals choose mates like themselves. |
| 5. No natural selection | Alleles that improve survival or reproduction increase. |
Two exam habits help. First, match the scenario to one condition: "females mate only with males that have the longest tails" breaks random mating (and is also a form of selection on males, but the condition named is random mating). "A storm reduces the population to 15 individuals" breaks large size. "Seeds blow in from a neighbouring valley" breaks no migration. Second, remember that breaking condition 4 changes genotype frequencies but not, on its own, allele frequencies; the other four change allele frequencies.
3Worked example: from a recessive phenotype
This is the commonest exam route, because only the recessive homozygote can be identified by eye.
An invented population of 1,000 garden peas has purple flowers (dominant allele A) and white flowers (recessive, aa). There are 160 white-flowered plants. Assume the population is in Hardy–Weinberg equilibrium.
The trap: purple plants number 840, a frequency of 0.84. That is p² + 2pq, not p². Taking √0.84 = 0.92 and calling it p is the most common Hardy–Weinberg error. Always start from the recessive phenotype.
The same method for a human condition. An invented recessive disorder affects 1 in 2,500 newborns in a population.
A disorder that affects one baby in 2,500 is carried, unseen, by roughly one person in 26. Recessive alleles mostly sit hidden in heterozygotes, which is why selection against them is slow.
4Worked example: is this population in equilibrium?
Here all three genotypes can be identified (for example, by a gel that separates the two protein forms). An invented sample of 400 field voles, gene with alleles F and S:
| Genotype | FF | FS | SS |
|---|---|---|---|
| Observed number | 210 | 140 | 50 |
Step 1. Allele frequencies, by counting copies.
Step 2. Expected genotype numbers if the population were in equilibrium.
Step 3. Compare, then make a claim with evidence and reasoning.
| Genotype | FF | FS | SS |
|---|---|---|---|
| Observed | 210 | 140 | 50 |
| Expected under Hardy–Weinberg | 196 | 168 | 36 |
Claim: the population is not in Hardy–Weinberg equilibrium. Evidence: there are 28 fewer heterozygotes than expected (140 against 168) and more of both homozygotes (210 against 196, and 50 against 36). Reasoning: a shortage of heterozygotes with an excess of both homozygotes is what non-random mating produces, for instance if voles tend to mate with close relatives or with partners of their own genotype, so condition 4 may be broken.
A formal test would use chi-square, also on the formula sheet, to decide whether the gap is bigger than sampling chance would explain. Multiple-choice questions usually ask only whether observed numbers match expected ones.
5Worked example: allele frequencies over time
Hardy–Weinberg predicts a flat line: plot the frequency of an allele against generation, and a population in equilibrium gives a horizontal line. A line that rises or falls means evolution.
An invented population of beetles was sampled in two years.
| Year | AA | Aa | aa | Total |
|---|---|---|---|---|
| 2020 | 90 | 420 | 490 | 1,000 |
| 2025 | 250 | 500 | 250 | 1,000 |
The frequency of allele a fell from 0.70 to 0.50 in five years, so the null hypothesis of no change is rejected: the population is evolving at this gene. Notice that each year's genotypes happen to fit p², 2pq and q² for that year's allele frequencies. A population can fit the equation at every census and still be evolving between them. The test for evolution is a change in allele frequency over time.
Practice questions
9 questions on this note, from easy to hard. Each one you get right lights a star of this topic's constellation.