Genotype frequencies and Hardy-Weinberg

Two alleles, three genotypes

At one autosomal locus with two alleles, A and a, every diploid individual is one of three genotypes: AA, Aa or aa. Write p for the frequency of A among all gene copies and q = 1 − p for a. Counting is the whole of the first step: if a population has D individuals of AA, H of Aa and R of aa, then p = (2D + H)/2(D + H + R), because each AA carries two A copies and each Aa carries one.

The second step runs the other way. Suppose individuals mate at random, which means the gametes that meet are independent draws from the gene pool. An egg carries A with probability p and so does a sperm, so the zygote is AA with probability p². It is aa with probability q², and Aa with probability 2pq, the 2 because the A can come from either parent. These are the Hardy-Weinberg proportions, and they sum to 1 because (p + q)² = p² + 2pq + q².

The Genotype Frequencies module plots exactly these three expressions for every p from 0 to 1. There is nothing random in it and nothing to set: Run draws the curves, Reset clears them. Each vertical slice through the chart is one population at Hardy-Weinberg equilibrium.

Reading the curves

Two features of the chart carry most of the teaching. The first is the top of the hump: 2pq is largest at p = q = 0.5, where it equals 0.5 and the two homozygotes are 0.25 each. No population at Hardy-Weinberg equilibrium has more than half heterozygotes. A sample can exceed 0.5 by chance, which is what the Chi-Square HWE Test module is for, but the expectation never does.

The second feature is at the edges. Take p = 0.9, so the rare allele a has q = 0.1. Only q² = 0.01 of the population is aa, but 2pq = 0.18 is Aa. Of all the a copies, the fraction sitting in heterozygotes is 2pq/(2pq + 2q²) = p = 0.9. A rare allele lives almost entirely in heterozygotes, where a recessive phenotype does not show. This single fact explains why selection against a rare recessive is slow (see the Natural Selection tutorial) and why carrier frequencies of recessive diseases are far higher than disease frequencies.

A useful habit is to read any genotype count against this chart. If 1 in 10,000 people have a recessive condition, then q² = 0.0001, q = 0.01, and the carrier frequency 2pq is about 0.02: one person in fifty, two hundred times the frequency of the condition itself.

One generation is enough

The proportions do not require the parents to be in Hardy-Weinberg proportions. Whatever genotype frequencies a population starts with, one round of random mating produces p², 2pq and q² in the offspring, because the gamete frequencies are p and q regardless of how the copies were packaged into parents. From then on the proportions stay put as long as the assumptions hold: random mating, a very large population, and no selection, mutation or migration. The De Finetti Parabola module draws this one-generation jump as an arrow onto the equilibrium curve.

The one-generation rule needs the two sexes to carry the same allele frequency, which is automatic for an autosomal locus. An X-linked locus breaks it: males carry one copy and females two, so if the sexes start at different frequencies the population needs several generations to settle. Sons get their X from their mother, so the male frequency each generation equals the previous female frequency, while the female frequency is the average of the two.