Hardy-Weinberg Equilibrium Calculator
Enter the share of people with the recessive trait, or genotype counts, to get allele and carrier frequencies.
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Hardy-Weinberg Equilibrium Calculator by TheCalcTools
How the Hardy-Weinberg Equilibrium Calculator works
In a large, randomly mating population with no selection, mutation or migration, allele frequencies p and q stay constant and genotype frequencies are p², 2pq and q².
For a recessive condition, the share of affected people is q², so q = √(affected share). The carrier frequency 2pq is usually far larger than the affected frequency.
With genotype counts, allele frequencies come straight from the counts, and the chi-square value (1 degree of freedom) tests whether the population fits equilibrium: above 3.84 is significant at the 5% level.
The formula
Worked example
Using the default values shown in the calculator:
| I know | Frequency of the recessive phenotype (q²) |
|---|---|
| Recessive phenotype frequency | 0.0004 |
| Homozygous dominant (AA) count | 360 |
| Heterozygous (Aa) count | 480 |
| Homozygous recessive (aa) count | 160 |
- q = √0.0004 = 0.02, p = 0.98.
- Carriers: 2pq = 0.0392, about 1 in 26 people.
Result:
| Dominant allele frequency p | 0.9800 |
|---|---|
| Recessive allele frequency q | 0.0200 |
| AA frequency p² | 0.9604 |
| Carriers 2pq | 0.0392 |
| aa frequency q² | 0.0004 |
| About 1 in … people is a carrier | 26 |
| Chi-square vs equilibrium (counts only) | — |
Frequently asked questions
How common are cystic fibrosis carriers?
In populations where about 1 in 2,500 babies is affected, q ≈ 0.02 and about 1 in 25 people is a carrier.
Sources and further reading
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This calculator gives estimates for information only. Check important results independently before you rely on them. See our disclaimer.