The De Finetti diagram

Three frequencies, one point

At one locus with alleles A and a, a population has three genotype frequencies. Write D = P(AA), H = P(Aa) and R = P(aa). They sum to 1, so two of them fix the third: R = 1 − D − H. The module takes D and H and works out the rest.

The allele frequency follows by counting gene copies. Each AA carries two A copies and each Aa carries one, so p = D + H/2, and q = 1 − p. The module plots one point per population. Its horizontal position is p. Its height is H.

The points live inside a triangle. The corners are the populations that are all aa (p = 0, H = 0), all AA (p = 1, H = 0) and all Aa (p = 0.5, H = 1). A population cannot sit outside it, because H can never exceed 2 min(p, q). You can read the genotypes back from a point: D = p − H/2 and R = q − H/2.

This page is not a simulation. There is no population size, no time and no randomness. The same D and H always give the same point.

The parabola

Let gametes meet at random. An egg carries A with probability p and so does a sperm. The zygote is AA with probability p², aa with probability q², and Aa with probability 2pq. These are the Hardy-Weinberg proportions. One generation of random mating produces them, whatever the starting genotype frequencies were, and p does not change.

In the diagram those populations satisfy H = 2p(1 − p). The curve is a parabola with its top at (0.5, 0.5). Every population at Hardy-Weinberg equilibrium is on it. A point above it has more heterozygotes than random mating gives. A point below it has fewer. The signed gap is ΔH = H − 2pq, and the hover panel shows it.

The arrow on each point shows one generation of random mating. Because p is unchanged, the arrow is vertical. It ends on the parabola at (p, 2pq). For the second population in the figure, H rises from 0.2 to 0.48.

How far a point can be from it

The largest excess is the top corner, D = 0 and H = 1. Then p = 0.5, 2pq = 0.5 and ΔH = +0.5. Every individual is a heterozygote. The largest deficit is a population of homozygotes only, such as D = 0.5, H = 0 and R = 0.5. That is also p = 0.5, and ΔH = −0.5.

The corners where p = 0 or p = 1 are on the parabola. With one allele fixed there is nothing to be heterozygous for, and 2pq is 0.

The diagram shows frequencies, not counts. It says nothing about whether a gap is larger than sampling error would give. The Chi-Square HWE Test module answers that.

A lasting departure needs a cause, because random mating removes one in a single generation. Inbreeding and assortative mating give a deficit of heterozygotes. Selection acting on genotypes can move a population either way. Pooling subpopulations that have different p also gives a deficit, which is the Wahlund effect.