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Assortative mating

Mating by resemblance removes heterozygotes without changing allele frequency; mating by difference keeps them and moves the allele frequency.

Read first: Genotype frequencies and Hardy-Weinberg

Choosing partners by genotype

Random mating gives Hardy-Weinberg proportions in one generation, because partners are drawn without regard to genotype. Assortative mating breaks that rule. Individuals choose partners that resemble them (positive assortment) or differ from them (negative assortment), and the genotype frequencies change from one generation to the next.

The module follows one very large population at one autosomal locus with alleles A and a. The genotype frequencies are D for AA, H for Aa and R = 1 − D − H for aa, and the allele frequency is p = D + H/2. The defaults are D = 0.04 and H = 0.32, so R = 0.64 and p = 0.2. These are Hardy-Weinberg proportions for p = 0.2, which makes the starting point the neutral reference. Only the mating rule acts. There is no selection, mutation, migration or drift.

Three mating rules are offered.

Positive (dominance): phenotypes pair. A-carriers (AA and Aa) mate with A-carriers and aa mates with aa. Allowed crosses: AA×AA, AA×Aa, Aa×Aa, aa×aa.

Positive (no dom.): genotypes pair. Each genotype mates only with its own. Allowed crosses: AA×AA, Aa×Aa, aa×aa.

Negative: only unlike genotypes pair. Allowed crosses: AA×Aa, AA×aa, Aa×aa.

Each generation, a female picks among the allowed males in proportion to their frequency. Offspring follow Mendel for each cross, and the new D and H replace the old ones.

Positive assortment with dominance

Figure 1. D, H, R and p for 20 generations, with the A- and aa phenotype groups mating among themselves. H falls from 0.32 to 0.1778 in generation 1 and to 0.0188 in generation 20. D rises to 0.1906 and R to 0.7906. The p line is flat at 0.2000.

Heterozygotes thin out, p does not move

Under positive assortment every individual mates inside its own group. An AA or Aa individual never mates with aa, so no A copy moves into the aa group and none leaves. The allele frequency p = D + H/2 stays at 0.2000 in every generation. Only the way the copies are packed into genotypes changes: they sit more often in homozygotes.

For the dominance rule the next generation is D′ = (2D + H)²/(4(D + H)) and H′ = H(2D + H)/(2(D + H)). With D = 0.04 and H = 0.32 these give D′ = 0.16/1.44 = 0.1111 and H′ = 0.128/0.72 = 0.1778. The decline slows down as H shrinks. Aa×Aa and Aa×AA matings keep producing heterozygotes for as long as any remain, and an Aa individual cannot be told from an AA individual by phenotype. After 200 generations H is still 0.0020.

The genotype rule has no such hiding place. Each genotype mates with itself, and only Aa×Aa makes heterozygotes, a half of its offspring. So H′ = H/2 and D′ = D + H/4. The heterozygote frequency after t generations is 0.32/2^t: 0.16, 0.08, 0.04, 0.02 and 0.01 for generations 1 to 5, and 0.0003 at generation 10. This is the recursion for self-fertilisation, and it ends at D = p = 0.2 and R = q = 0.8.

Positive assortment by genotype

Figure 2. The same start with each genotype mating only with itself. H halves every generation: 0.16 at generation 1 and 0.01 at generation 5. By generation 20 the population is D = 0.2000, H = 0.0000, R = 0.8000. The p line stays at 0.2000.

Exercise

Compare the two positive rules

Open the module and run the defaults with Model Positive (dominance) and Gen 20. Read H at generation 20. Switch Model to Positive (no dom.) and run again.

What to expect. H is 0.0188 under dominance and 0.0000 under the genotype rule, with p at 0.2000 in both. The phenotype rule leaves 0.0188 of heterozygotes after 20 generations because Aa individuals still mate as A-carriers. Step 3 of the walkthrough runs the genotype rule.

Open the walkthrough step →

Mating by difference

Under negative assortment only unlike genotypes mate. The crosses AA×Aa, AA×aa and Aa×aa produce a lot of heterozygotes: AA×aa makes only Aa, and Aa×AA and Aa×aa make half. Aa×Aa is forbidden, so the half of its offspring that would be homozygous never appears.

Here the allele frequency moves. Every female must mate with an allowed male if one exists. A rare genotype, AA at 0.04, is therefore always paired, while under random mating it would meet an AA partner only 4% of the time. The matings of AA individuals are over-represented, and the matings of the common aa group are shared out among the few AA and Aa males. At generation 1, H is 0.5489 and p is 0.2905, up from 0.2.

After the first jump, the population settles slowly. At generation 20, D = 0.0001, H = 0.5001, R = 0.4998 and p = 0.2501. AA has all but disappeared, and what remains is Aa×aa matings that produce half Aa and half aa. In that population p = 0.25.

Negative assortment

Figure 3. Only unlike genotypes mate. H jumps to 0.5489 in generation 1 and then eases to 0.5001. The p line rises from 0.2000 to 0.2905 in generation 1 and ends at 0.2501. D falls to 0.0001 by generation 20.

Exercise

Start without heterozygotes

Set D (AA) 0.2 and H (Aa) 0, choose Model Negative and run. Then set D (AA) 1 and run again.

What to expect. With D = 0.2 and H = 0 only AA×aa matings are allowed, so after one generation H = 1 and D = R = 0. With D = 1 no allowed partner exists, so nothing changes. Step 5 of the walkthrough runs the negative rule from the default start.

Open the walkthrough step →

Summary

Positive assortment removes heterozygotes and leaves p at 0.2000, because every individual mates inside its own group.
With phenotype pairing H falls from 0.32 to 0.0188 in 20 generations. With genotype pairing H halves every generation and reaches 0.0003 by generation 10.
Negative assortment keeps H near 0.5 and raises p from 0.2 to 0.2905 in one generation, because rare genotypes always find a partner.
At generation 20 the negative rule gives D = 0.0001, H = 0.5001, R = 0.4998 and p = 0.2501.
All three rules are one-locus, infinite-population rules; real assortment acts only on the loci that affect the trait and the loci linked to them.