Autosomal locus
Genotype counts give allele frequencies, and one generation of random mating moves any sample onto the Hardy-Weinberg parabola.
Read first: Genotype frequencies and Hardy-Weinberg
From counts to frequencies
This module is a calculation, not a simulation. You enter the numbers of AA, Aa and aa individuals in a sample and it computes frequencies. Nothing is random, so the same counts always give the same result.
With n = N(AA) + N(Aa) + N(aa) individuals, the genotype frequencies are D = N(AA)/n, H = N(Aa)/n and R = N(aa)/n. Each AA carries two A copies and each Aa carries one, so the frequency of A is p = D + H/2, and q = 1 − p. The defaults are 50 AA, 40 Aa and 10 aa. Then n = 100, D = 0.5, H = 0.4, R = 0.1 and p = 0.5 + 0.2 = 0.7. Only proportions count: 500, 400 and 100 give the same answers.
The default sample
Compared with Hardy-Weinberg
Under Hardy-Weinberg proportions the genotypes are p², 2pq and q². At p = 0.7 these are 0.49, 0.42 and 0.09. The sample has 0.50, 0.40 and 0.10, which is two heterozygotes fewer per hundred than expected.
The module summarizes the gap with F = 1 − H/(2pq). Here F = 1 − 0.40/0.42 = 0.0476. F = 0 means Hardy-Weinberg proportions. F = 1 means no heterozygotes. F < 0 means an excess of heterozygotes. The opposite case is a sample with too many heterozygotes. For 20 AA, 80 Aa and 0 aa, p = 0.6 and 2pq = 0.48, but H = 0.8. Then F = 1 − 0.8/0.48 = −0.667, and the sample sits above the parabola. The extreme is 0 AA, 50 Aa and 0 aa, with p = 0.5, H = 1 and F = −1. Reading the diagram: a point below the parabola has fewer heterozygotes than expected, and a point above has more. At p = 0.7 the highest possible H is 2q = 0.6, reached when every a copy sits in a heterozygote (40 AA, 60 Aa, 0 aa), and that sample has 0.6 against 0.42 expected. With F, the expected genotype frequencies are P(AA) = p² + Fpq, P(Aa) = 2pq(1 − F) and P(aa) = q² + Fpq.
No heterozygotes
Exercise
Test the extremes
Enter N(AA) 60, N(Aa) 0 and N(aa) 40, and run. Then enter 20, 80 and 0 and run again. Read F from the results panel.
What to expect. F is 1 for 60/0/40 and about −0.667 for 20/80/0. p is 0.6 in both, and in both the offspring land on the parabola at 2pq = 0.48.
Open the walkthrough step →Why one generation is enough
Random mating pools the gametes of the parents. Whatever the genotypes, the fraction of A gametes in the pool is p. AA parents give only A, Aa parents give A half the time, and aa parents give none. Two gametes join independently, so the offspring are AA with probability p², Aa with 2pq and aa with q².
The offspring depend only on p. The samples 60/0/40 and 20/80/0 differ by 0.8 in H, yet both give 0.36, 0.48 and 0.16, because p = 0.6 in both. The allele frequency does not change: p is 0.6 before and after. For the default sample, 100 offspring would be about 49 AA, 42 Aa and 9 aa, against 50, 40 and 10 in the parents. The gametes of the parents carry p = 0.7 either way.
One generation is enough only under the stated conditions: one autosomal locus, random union of gametes, an infinite population, no selection, mutation or migration, and the same p in both sexes. At an X-linked locus the sexes can differ in p and convergence takes several generations. The X-Linked module shows that.
Exercise
Check that only proportions matter
Enter 500, 400 and 100 and run. Then enter 25, 50 and 25 and run.
What to expect. The first sample reproduces p = 0.7 and F = 0.0476 from the defaults. The second gives p = 0.5 and F = 0, and sits on the parabola.
Open the walkthrough step →What to remember
- The allele frequency is p = D + H/2. The default sample of 50, 40 and 10 gives p = 0.7.
- F = 1 − H/2pq: 0.0476 for the defaults, 1 for 60/0/40 and −0.667 for 20/80/0.
- One generation of random mating gives p², 2pq and q², which is 0.49, 0.42 and 0.09 at p = 0.7, with p unchanged.
- The offspring depend only on p, so samples with different H reach the same genotype frequencies in one generation.