Hardy-Weinberg Calculator

It assumes a large, randomly mating population with no mutation, migration, or natural selection. Under these idealized conditions, allele and genotype frequencies stay constant from one generation to the next.

With dominant allele frequency p and recessive frequency q, the genotype frequencies come from expanding p plus q squared. This gives p squared for homozygous dominant, two p q for heterozygous, and q squared for homozygous recessive.

The recessive phenotype corresponds to the homozygous recessive genotype, whose frequency is q squared. Take the square root to get q, then p = 1 minus q. For example, if 9 percent of the population shows the recessive trait, q squared is 0.09, so q is 0.3 and p is 0.7.

For a recessive disorder, the incidence equals the homozygous recessive frequency q squared. Take its square root to get q, set p = 1 minus q, and the carrier frequency is 2pq. Cystic fibrosis, for instance, affects about 1 in 2,500 people of European descent, so q squared = 0.0004, giving q = 0.02 and p = 0.98. The carrier frequency is 2 x 0.98 x 0.02 = 0.039, or roughly 1 in 26 - far more common than the disease itself because most carriers are healthy heterozygotes.