Unit 7 · Topic 7.5

7.5 Hardy–Weinberg Equilibrium

The Hardy–Weinberg equations, p + q = 1 and p² + 2pq + q² = 1, describe a population that isn't evolving. Such a population would need five conditions that are never fully met, so the model works as a null hypothesis: if real genotype frequencies differ from the prediction, something is changing the population.

Key terms

Hardy–Weinberg equilibrium
allele frequency
genotype frequency
null hypothesis
random mating

The model and its five conditions

Hardy–Weinberg equilibrium describes a population in which allele and genotype frequencies stay the same from generation to generation. That only happens if nothing is changing them, which requires five conditions:

A large population, so genetic drift is negligible.
No migration, so there's no gene flow in or out.
No new mutations.
Random mating, so individuals don't choose mates by genotype.
No natural selection, so all genotypes survive and reproduce equally well.

The two equations

For a gene with two alleles, p is the frequency of one allele (often the dominant one) and q is the frequency of the other. Every copy of the gene is one or the other, so p + q = 1.

If mating is random, the chance that an offspring gets two copies of the first allele is p × p = p², two copies of the second is q², and one of each is 2pq (2 because it can come from either parent). So the genotype frequencies are p² (homozygous dominant) + 2pq (heterozygous) + q² (homozygous recessive) = 1. Both equations are on the AP formula sheet.

Where to start a problem

If you know the genotype counts, calculate allele frequencies directly: p = (2 × number of homozygous dominant + number of heterozygotes) ÷ (2 × total individuals). This works whether or not the population is in equilibrium.

If you know only phenotypes and one allele is completely dominant, start from the recessive phenotype. Recessive individuals are the only group whose genotype you know for sure (q²). Take the square root to get q, then p = 1 − q. You can't start from the dominant phenotype, because it lumps together p² and 2pq. This step assumes the population is in Hardy–Weinberg equilibrium.

Using the model as a null hypothesis

Real populations never meet all five conditions perfectly, but the model is still useful. It tells you what genotype frequencies to expect if the population is not evolving at this gene. Compare observed genotype counts with the expected counts (often with a chi-square test). If they're close, you have no evidence of evolution at this gene. If they're very different, at least one condition is being violated, and the pattern gives clues: too few heterozygotes can suggest nonrandom mating such as inbreeding, or selection against heterozygotes; a steady shift in allele frequency over generations suggests selection, drift or gene flow.

Hardy–Weinberg doesn't mean allele frequencies can't change. It describes what happens when the forces that change them are absent, and so it helps identify which force is at work.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Carriers of a recessive condition
A recessive genetic condition affects 1 in 2,500 newborns in a population assumed to be in Hardy–Weinberg equilibrium. What fraction of the population are carriers (heterozygous)?
Show the solutionHide the solution
1. Step 1: Affected people are homozygous recessive, so q² = 1/2,500 = 0.0004.
2. Step 2: q = √0.0004 = 0.02.
3. Step 3: p = 1 − q = 0.98.
4. Step 4: Carriers: 2pq = 2 × 0.98 × 0.02 = 0.0392.
5. Step 5: That's about 3.9%, or roughly 1 in 25 people, far more than the 1 in 2,500 who are affected. Most copies of a rare recessive allele are hidden in carriers.
Answer: 2pq ≈ 0.039, so about 3.9% of the population (about 1 in 25) are carriers.
Example 2Calculator allowed
Is the population in equilibrium?
In a population of 500 plants, 210 are AA, 180 are Aa and 110 are aa. Calculate the allele frequencies, the expected genotype counts under Hardy–Weinberg, and decide whether the population is in equilibrium at this gene.
Show the solutionHide the solution
1. Step 1: Count alleles: there are 2 × 500 = 1,000 copies. A copies = 2(210) + 180 = 600, so p = 600 ÷ 1,000 = 0.6 and q = 0.4.
2. Step 2: Expected genotype frequencies: p² = 0.36, 2pq = 0.48, q² = 0.16.
3. Step 3: Expected counts out of 500: AA = 180, Aa = 240, aa = 80.
4. Step 4: Chi-square: (210 − 180)²/180 = 5.0; (180 − 240)²/240 = 15.0; (110 − 80)²/80 = 11.25. Sum: χ² = 31.25.
5. Step 5: Using AP's rule (categories − 1), df = 2 and the critical value at p = 0.05 is 5.99. (Statisticians often use 1 df for this test because p was estimated from the data, giving 3.84; either way the conclusion is the same.)
6. Step 6: 31.25 is far above the critical value, so reject the null hypothesis of equilibrium. There are too few heterozygotes and too many homozygotes, which fits nonrandom mating such as inbreeding or self-pollination.
Answer: p = 0.6, q = 0.4; expected 180 AA, 240 Aa, 80 aa. χ² = 31.25, well above the critical value, so the population is not in Hardy–Weinberg equilibrium; the heterozygote shortage suggests nonrandom mating.
Example 3Calculator allowed
The dominant-phenotype trap
In a population in Hardy–Weinberg equilibrium, 84% of individuals show the dominant phenotype. Find p, q and the frequency of heterozygotes.
Show the solutionHide the solution
1. Step 1: Trap: setting p = 0.84 or p² = 0.84. The 84% includes both AA (p²) and Aa (2pq), so neither is correct.
2. Step 2: Start with the recessive phenotype instead: q² = 1 − 0.84 = 0.16.
3. Step 3: q = √0.16 = 0.4, and p = 1 − 0.4 = 0.6.
4. Step 4: Heterozygotes: 2pq = 2 × 0.6 × 0.4 = 0.48. Check: p² = 0.36, and 0.36 + 0.48 = 0.84, which matches the dominant phenotype.
Answer: p = 0.6, q = 0.4, heterozygotes = 0.48 (48%).

Common mistakes

Using q² (a genotype frequency) where q (an allele frequency) is needed, or the reverse. Label each number as an allele or genotype frequency.
Starting from the dominant phenotype. Only the recessive phenotype gives you a genotype frequency directly (q²).
Writing 'large population' and 'no selection' but forgetting the other three conditions: no migration, no mutation, random mating.
Thinking Hardy–Weinberg says allele frequencies never change. It's a model of what happens when no evolutionary forces act, used as a null hypothesis.

On the exam

Hardy–Weinberg calculations show up in both sections. Show the equation, the numbers you substitute and the result, and round sensibly.
When asked whether a population is evolving, compare observed and expected genotype frequencies and name which condition is likely violated, with a reason from the data.

Connected topics

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