What the Hardy-Weinberg principle says
In 1908, mathematician G. H. Hardy and physician Wilhelm Weinberg independently showed that, in a large population with random mating, the frequencies of alleles and genotypes stay constant from one generation to the next. If a gene has two alleles with frequencies p and q, the genotypes appear in fixed proportions: p², 2pq and q². This baseline lets geneticists estimate how many people carry a recessive allele and detect when something, such as selection or migration, is changing the population.
The formulas
- p + q = 1
- p² + 2pq + q² = 1: AA, Aa and aa
- From the recessive trait frequency: q = √(share of aa)
- From genotype counts: q = (Aa + 2 × aa) ÷ (2 × total)
- Carriers: 2pq, or one person in 1 ÷ 2pq
- Chi-square test: Σ (observed − expected)² ÷ expected, with 1 degree of freedom
Worked example: estimating carriers
A recessive condition affects 1 person in 2,500.
- q² = 1 ÷ 2,500 = 0.0004, so q = 0.02.
- p = 1 − 0.02 = 0.98.
- Carriers: 2 × 0.98 × 0.02 = 3.92%, about 1 person in 26.
So for every affected person there are about 98 healthy carriers. This is why recessive conditions persist even when they are rare.
Rare traits, common carriers
| Trait frequency | q | Carriers (2pq) |
|---|---|---|
| 1 in 100 | 0.100 | 18.0%, 1 in 6 |
| 1 in 1,000 | 0.032 | 6.1%, 1 in 16 |
| 1 in 2,500 | 0.020 | 3.9%, 1 in 26 |
| 1 in 10,000 | 0.010 | 2.0%, 1 in 51 |
| 1 in 20,000 | 0.007 | 1.4%, 1 in 71 |
Worked example: testing the equilibrium
In a sample of 1,000 individuals you count 360 AA, 480 Aa and 160 aa.
- q = (480 + 2 × 160) ÷ 2,000 = 0.40, p = 0.60.
- Expected: 360 AA, 480 Aa, 160 aa.
- The observed counts match exactly, so chi-square is 0 and the population is consistent with equilibrium.
A p-value below 0.05 would suggest that one of the assumptions does not hold.
The assumptions
- A large population, so random drift is small
- Random mating with respect to the gene
- No new mutations, no migration in or out
- No natural selection on the genotypes
Real populations break these rules to some degree, but for many genes the predictions are close. Departures are informative: an excess of homozygotes can point to inbreeding or to population structure.
Limits
The calculation assumes the trait is fully recessive and that everyone with the aa genotype shows it. Carrier screening in medicine relies on direct genetic testing rather than these estimates, which also vary between populations.