You met the Hardy-Weinberg principle in Hardy-Weinberg Equilibrium: in a population that is not evolving, allele and genotype frequencies stay the same from one generation to the next. Here we use it as a tool: from one known number, find all the others, and compare real data with the prediction, as in chi-square.
Key idea
For a gene with two alleles, with frequencies $p$ (dominant) and $q$ (recessive): $$p + q = 1 \qquad p^2 + 2pq + q^2 = 1$$ $p^2$, $2pq$ and $q^2$ are the frequencies of AA, Aa and aa. Start from what you know: usually $q^2$, the share showing the recessive phenotype. Then $q = \sqrt{q^2}$, $p = 1 - q$, and carriers $= 2pq$.
Why it works
Why squares? If parents pair up at random, each new individual gets one allele from a gamete of each parent, drawn independently from the gene pool. The chance of A then A is $p \times p$; of a then a is $q \times q$; and a heterozygote can arise two ways (A then a, or a then A), so $2pq$. The equilibrium holds only under five conditions: a large population, parents pairing at random, no mutation, no migration (gene flow) and no natural selection. When data break the prediction, at least one of these is failing.
Worked examples
Worked example: From the recessive phenotype
In a population of moths, 16% are white, the recessive phenotype. Find p, q and the share of heterozygous carriers.
- $q^2 = 0.16$, so $q = \sqrt{0.16} = 0.4$.
- $p = 1 - 0.4 = 0.6$.
- $2pq = 2(0.6)(0.4) = 0.48$.
- Check: $p^2 = 0.36$, and $0.36 + 0.48 + 0.16 = 1$.
Answer: p = 0.6, q = 0.4, and 48% are carriers.
Worked example: From genotype counts
A sample of 200 plants has 90 AA, 80 Aa and 30 aa. Find p and q, and the genotype counts expected at equilibrium.
- Alleles: 400 in all. A alleles: $2(90) + 80 = 260$, so $p = 260/400 = 0.65$; $q = 0.35$.
- Expected AA: $0.65^2 \times 200 = 84.5$; Aa: $2(0.65)(0.35) \times 200 = 91$; aa: $0.35^2 \times 200 = 24.5$.
- Observed 90, 80, 30 against expected 84.5, 91, 24.5: fewer heterozygotes than expected.
Answer: p = 0.65, q = 0.35. A chi-square test would tell whether the shortage of heterozygotes is more than chance. (When p is estimated from the same data, statisticians use 1 degree of freedom for three genotypes; follow what a problem tells you.)
Common mistake
Two classic slips. Computing $2pq$ as $2(p + q)$, which is always 2. And treating the dominant phenotype share as $p^2$: the dominant phenotype is AA and Aa, so it equals $p^2 + 2pq = 1 - q^2$. That is why you start from the recessive phenotype: only there do genotype and phenotype match.