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Hardy-Weinberg equilibrium

Allele and genotype frequencies (p² + 2pq + q² = 1) from the recessive phenotype, the allele or observed counts.

Parameters

What you know
1 affected in 2,500: enter 0.0004.

Abacus

the law over the whole run of p
Hardy-Weinberg · p = 0.98 · q = 0.02AA (p²) 0.96 · aa (q²) 4 × 10⁻⁴000.250.50.751frequency of allele A (p)genotype frequency2pq maximum = 0.5 at p = 0.5Aa (2pq) 0.0392partitionq²2pqp²
p = 0.98

Playback only moves the entry of the chosen mode, p directly, or q² = (1 − p)² when the known datum is the recessive phenotype: p travels 0 → 1 in ten seconds, and every frame is fully recomputed.

AA (p²)Aa (2pq)aa (q²)

Calculation
q = √(q²) = √(4 × 10⁻⁴) = 0.02
Partition
p² + 2pq + q² = 0.9604 + 0.0392 + 4 × 10⁻⁴ = 1
Carriers
2pq / (p² + 2pq) = 0.0392 / 0.9996 = 0.039 216
Note

2pq peaks at p = 0.5.

Result

continuous computation
Heterozygotes
2pq0.03927.84 % of the peak of the parabola
Allele frequency p
0.98
Allele frequency q
0.02
Dominant homozygotes (p²)
0.9604
Heterozygotes (2pq)
0.0392
Recessive homozygotes (q²)
4 × 10⁻⁴
Carriers among the unaffected
3.9216 %

Calculation, assumptions and sources

OverviewAllele and genotype frequencies of a randomly mating population at equilibrium.

Allele and genotype frequencies of a randomly mating population at equilibrium. When the allele is recessive, only aa homozygotes are visible: q = √(q²) gives all the other frequencies, healthy carriers included. Possible starting data: recessive phenotype, allele frequency, or counts of the three genotypes, then compared with predicted frequencies.

Method & formulasp + q = 1
p + q = 1p² + 2pq + q² = 1q = √(recessive phenotype frequency)p = (2·NAA + NAa) / 2N (allele counting)

At equilibrium, the genotypes AA, Aa and aa follow the binomial expansion of (p + q)². Direct allele counting (two per individual) gives p without assuming equilibrium.

Allele frequency
the share of an allele in the gene pool: p for A, q for a, and p + q = 1 for a two-allele gene.
Genotype frequency
the share of AA, Aa or aa individuals. At equilibrium: p², 2pq, q².
Healthy carrier
the heterozygote Aa of a recessive allele, which transmits it without expressing it. Half its gametes carry a.
Panmixia
random mating, with no genotype-linked choice: one of the five conditions of equilibrium.
Validity domainThe theorem describes an autosomal two-allele gene in an ideal population: no selection, mutation or migration, random mating, large numbers.

The theorem describes an autosomal two-allele gene in an ideal population: no selection, mutation or migration, random mating, large numbers. A departure from equilibrium points to an unmet condition. Sex-linked genes, multi-allele systems and inbreeding call for extensions of the model.

Frequency of healthy carriersFor a recessive disease affecting one birth in 2,500, q = 1/50 and 2pq ≈ 1/25, a hundred times the frequency of the affected.

For a recessive disease affecting one birth in 2,500, q = 1/50 and 2pq ≈ 1/25, a hundred times the frequency of the affected. Rare recessive alleles sit almost entirely in heterozygotes, with no effect on the phenotype.

Reading the resultA population of 500 AA and 500 aa has p = 0.5 and no heterozygotes, where equilibrium predicts 50%: the model conditions are not met.

A population of 500 AA and 500 aa has p = 0.5 and no heterozygotes, where equilibrium predicts 50%: the model conditions are not met. The chi-squared test (neighbouring tool) says whether such a gap is significant.