Hardy–Weinberg Equilibrium Calculator
Enter the dominant allele frequency p to instantly compute the genotype frequencies p² (AA), 2pq (Aa), q² (aa) and the dominant/recessive phenotype proportions, with a reference table common in DSE Biology.
Input Data
Results
At a glance:Hardy–Weinberg equilibrium is the core law of population genetics: under ideal conditions (no mutation, migration, selection, random mating, and a large population), allele and genotype frequencies remain stable across generations. If the dominant allele frequency is p and recessive is q (p + q = 1), the three genotype frequencies necessarily satisfy p² + 2pq + q² = 1, corresponding to homozygous dominant (AA), heterozygous (Aa), and homozygous recessive (aa) respectively.
Formula
Allele-frequency relation: p + q = 1, so q = 1 − p.
Genotype frequencies: AA = p², Aa = 2pq, aa = q², summing to 1.
Phenotype: dominant phenotype = p² + 2pq, recessive phenotype = q².
$$p + q = 1$$$$p^2 + 2pq + q^2 = 1$$How to Use
- Enter the dominant allele frequency p (0–1); q is auto-computed as 1 − p.
- The tool shows genotype frequencies p², 2pq, q² and both phenotype proportions.
- Use it to check whether an observed population fits equilibrium or infer hidden carrier rates.
Common allele frequencies and genotype proportions
| p | q = 1−p | AA (p²) | Aa (2pq) | aa (q²) |
|---|---|---|---|---|
| 0.5 | 0.5 | 0.25 | 0.50 | 0.25 |
| 0.6 | 0.4 | 0.36 | 0.48 | 0.16 |
| 0.7 | 0.3 | 0.49 | 0.42 | 0.09 |
| 0.8 | 0.2 | 0.64 | 0.32 | 0.04 |
| 0.9 | 0.1 | 0.81 | 0.18 | 0.01 |
Sum of AA + Aa + aa = 1. As p rises, AA grows fastest (squared) while aa shrinks fastest; heterozygotes peak at p = 0.5.
Case Studies
Typical DSE example (p = 0.6)
Enter p = 0.6 → q = 0.4.
AA = p² = 0.36, Aa = 2pq = 0.48, aa = q² = 0.16 (sum 1).
Dominant phenotype = 0.36 + 0.48 = 0.84, recessive phenotype = 0.16. Heterozygotes are the largest group, larger than either homozygote.
Estimating carrier rate from a recessive disease
Suppose albinism (recessive) affects 1 in 10,000, so q² = 0.0001 → q = 0.01, p = 0.99.
Heterozygous carriers Aa = 2pq ≈ 2 × 0.99 × 0.01 = 0.0198 ≈ 2%.
Although the affected rate is only 0.01%, about 1 in 50 people are silent carriers — far higher, showing why carrier screening matters.
FAQ
What are the conditions for Hardy–Weinberg equilibrium?
Five ideal assumptions: no mutation, no migration (gene flow), no natural selection, random mating, and a sufficiently large population (no genetic drift). If all hold, allele and genotype frequencies stay constant across generations; deviation suggests one of these forces is acting.
How to get q from the recessive phenotype?
The recessive phenotype equals aa = q². So q = √(recessive phenotype frequency), then p = 1 − q, and heterozygous carrier rate = 2pq. This lets you infer allele frequencies from phenotype data alone, the classic application for recessive traits.
Why is the heterozygote often the largest group?
Because 2pq is a product and is maximized at p = 0.5 (2pq = 0.5). Unless p is very close to 0 or 1, heterozygotes usually outnumber either homozygote — important in carrier estimation, where most recessive-allele carriers show no phenotype.
What does deviation from equilibrium mean?
If observed genotype frequencies do not fit p², 2pq, q², the population is not in HWE, implying selection, non-random mating (e.g. inbreeding), migration, mutation, or drift. Comparing observed vs expected is a standard test of evolutionary forces.
Is the dominant phenotype p² + 2pq?
Yes. The dominant phenotype includes both homozygous dominant (AA = p²) and heterozygous (Aa = 2pq), since the dominant allele masks the recessive in heterozygotes. So dominant phenotype proportion = p² + 2pq = 1 − q², and recessive phenotype = q² only.
Related Tools
References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.