Hardy-Weinberg Equilibrium Calculator
From the recessive phenotype frequency, derive allele frequencies p and q, plus heterozygote (carrier) and homozygous dominant shares of the population.
Carriers — heterozygotes (%)
32
Detalhamento
- Recessive allele q
- 0.20
- Dominant allele p
- 0.80
- Homozygous dominant (%)
- 64
- Homozygous recessive (%)
- 4
- Carriers per affected individual
- 8
Última atualização:
FAQ
What are the Hardy-Weinberg equations?
p + q = 1 for allele frequencies and p² + 2pq + q² = 1 for genotype frequencies. Since the recessive phenotype (q²) is directly observable, you can take its square root to get q and derive everything else.
Why is the carrier frequency usually much higher than the disease frequency?
Because q is the square root of q². At 1 in 10,000 affected (q² = 0.0001), q = 0.01 and carriers 2pq ≈ 2% — 200 carriers for every affected person. Rare recessive traits hide in the heterozygotes.
When does Hardy-Weinberg equilibrium fail?
The model assumes no mutation, no selection, no migration, a large population, and random mating. Real populations drift from these ideals, so treat results as a baseline snapshot rather than an evolutionary prediction.
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