The Hardy-Weinberg equations link allele frequencies to genotype frequencies in a population in equilibrium. For a gene with two alleles, where p is the frequency of the dominant allele and q the frequency of the recessive allele:p + q = 1
p2 + 2pq + q2 = 1 In the second equation:• p2 = frequency of the homozygous dominant genotype
• 2pq = frequency of the heterozygous genotype
• q2 = frequency of the homozygous recessive genotypeIt comes from random fertilisation: the chance of two recessive gametes meeting is q × q, and a heterozygote can form in two ways (p × q + q × p).The key use is estimating allele frequencies when heterozygotes cannot be identified. Always start from the homozygous recessive phenotype, as it is the only genotype you can recognise.Worked example: 1 in 10 000 people in a population has a recessive condition.q2 = 1 / 10 000 = 0.0001
q = √0.0001 = 0.01
p = 1 − 0.01 = 0.99
Carriers: 2pq = 2 × 0.99 × 0.01 = 0.0198, about 2%So about 1 person in 50 is a carrier, even though only 1 in 10 000 is affected. This shows how many recessive alleles are hidden in heterozygotes.Common error: do not take q as the proportion of affected people; that is q2.
• 2pq = frequency of the heterozygous genotype
• q2 = frequency of the homozygous recessive genotypeIt comes from random fertilisation: the chance of two recessive gametes meeting is q × q, and a heterozygote can form in two ways (p × q + q × p).The key use is estimating allele frequencies when heterozygotes cannot be identified. Always start from the homozygous recessive phenotype, as it is the only genotype you can recognise.Worked example: 1 in 10 000 people in a population has a recessive condition.q2 = 1 / 10 000 = 0.0001
q = √0.0001 = 0.01
p = 1 − 0.01 = 0.99
Carriers: 2pq = 2 × 0.99 × 0.01 = 0.0198, about 2%So about 1 person in 50 is a carrier, even though only 1 in 10 000 is affected. This shows how many recessive alleles are hidden in heterozygotes.Common error: do not take q as the proportion of affected people; that is q2.