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Selection

Selection is defined as any act that restricts the random mating of individuals in a population, sometimes just called non-random mating. Consider a locus with two alleles with frequencies, p and q. The population is in Hardy-Weinberg equilibrium. Now suppose that a fraction, u, of the A₁A₁ genotypes, a fraction, v, of the A₁A₂ genotypes, and a fraction, s, of the A₂A₂ genotypes have been culled from this population. The resulting parent population can be summarized as follows:

Genotype	Initial	Fitness	After
	Frequency		Culling
A1A1	p2	1-u	p2(1-u)
A1A2	2pq	1-v	2pq(1-v)
A2A2	q2	1-s	q2(1-s)
	1		1-p2u-2pqv-q2s

Assuming that culling has been applied equally to both sexes, then the new frequency of the A₁ allele in the parent generation is

$$p_{c} = \frac{p^{2}(1-u)+pq(1-v)}{1-p^{2}u-2pqv-q^{2}s}.$$

The remaining animals, to be used for mating, are not in Hardy-Weinberg equilibrium. The remaining animals are mated randomly to produce the progeny generation. The progeny generation will be in Hardy-Weinberg equilibrium with p²_c of the A₁A₁ genotypes, 2p_cq_c of the A₁A₂ genotypes, and q²_c of the A₂A₂genotypes.

Culling could have been applied differentially to males and females, so that the allele frequencies in the remaining males and females could be different. Their progeny would not be in Hardy-Weinberg equilibrium.

The genotypic variance in the initial population is given by

$$\sigma^{2}_{G} = 2pq[a+d(q-p)]^{2}+[2pqd]^{2}.$$

The genotypic variance after culling will be different, and the genotypic variance in the progeny generation after random mating of the selected parents will be

$$\sigma^{2}_{G} = 2p_{c}q_{c}[a+d(q_{c}-p_{c})]^{2} +[2p_{c}q_{c}d]^{2}.$$

If q_c has changed towards .5, then genotypic variance should be increased. If a=d, (complete dominance), then genotypic variance will be maximized when $q_{c} = \sqrt{.5}$. As q_c changes towards 0 or 1, then the genotypic variance decreases towards 0, which it reaches when the alleles become fixed. If d > a, (overdominance), then fixation does not occur, except by chance, and the genotypic variance stays greater than 0.

1. Assortative Mating

Consider an initial population in Hardy-Weinberg equilibrium, as before. Below is a mating diagram in which the frequency of the A₁allele can be different in the male and female population. The frequencies in the table are the expected frequencies of each type of mating - in a random mating population.

Female		Male Genotypes
Genotypes	Frequencies	Frequencies
		A1A1	A1A2	A2A2
		pm2	2 pmqm	qm2
A1A1	pf2	pf2pm2	pf22pmqm	pf2qm2
A1A2	2pfqf	2pfqfpm2	4pfqfpmqm	2pfqfqm2
A2A2	qf2	qf2pm2	qf22pmqm	qf2qm2

Non-random mating means that the frequencies in the above table are altered from their expected values. An example is assortative mating, which would give a mating diagram as follows:

Female		Male Genotypes
Genotypes	Frequencies	Frequencies
		A1A1	A1A2	A2A2
		pm2	2 pmqm	qm2
A1A1	pf2	pf2pm2	0	0
A1A2	2pfqf	0	4pfqfpmqm	0
A2A2	qf2	0	0	qf2qm2

Note that the resulting frequencies do not sum to one. Assuming that p_m=p_f to simplify the discussion, then the sum of the non-zero elements is

$$\begin{eqnarray*}S & = & p^{4} + 4p^{2}q^{2} + q^{4} \\ & = & (p^{2}+q^{2})^{2... ...p+q)(p-q)]^{2} + 2p^{2}q^{2} \\ & = & (p-q)^{2} + 2p^{2}q^{2} \end{eqnarray*}$$

The progeny genotypic frequencies would be as follows:

Parent		Progeny Genotypic
Genotypes		Frequencies
Males	Females	A1A1	A1A2	A2A2
A1A1	A1A1	p4/S	0	0
A1A2	A1A2	p2q2/S	2p2q2/S	p2q2/S
A2A2	A2A2	0	0	q4/S

The frequency of the A₁ allele in the progeny is

p_c = [p⁴ + 2p²q²]/S.

The progeny genotypes are not in Hardy-Weinberg equilibrium. However, if the progeny generation is mated randomly, then the next generation will be in Hardy-Weinberg equilibrium again.

To illustrate the above, let p=.4 and let the genotypic values be a=2 and d=1. The genotypic mean and variance in the original random mating population would be

$$\mu = .16(2) + .48(1) + .36(-2) \ = \ 0.08,$$

and the genetic variance would be

$$\sigma^{2}_{G} = .16(4)+.48(1) + .36(4) - \mu^{2} \ = \ 2.5536.$$

The mating diagram for assortative mating would be

Female		Male Genotypes
Genotypes	Frequencies	Frequencies
		A1A1	A1A2	A2A2
		.16	.48	.36
A1A1	.16	.0256	0	0
A1A2	.48	0	.2304	0
A2A2	.36	0	0	.1296

Note that S=.3856. The progeny figures would be

Parent			Progeny Genotypic
Genotypes			Frequencies
Males	Females	Freq.	A1A1	A1A2	A2A2
A1A1	A1A1	.0664	.0664	0	0
A1A2	A1A2	.5976	.1494	.2988	.1494
A2A2	A2A2	.3360	0	0	.3360
		1.000	.2158	.2988	.4854

The new genetic mean and variance in the progeny generation is then

$$\mu = .2158(2) + .2988(1) + .4854(-2) \ = \ -.2406,$$

and

$$\sigma^{2}_{G} = .2158(4) + .2988(1) + .4854(4) - \mu^{2} \ = \ 3.0461.$$

Thus, the mean has decreased and the variance has increased. The new frequency of the A₁ allele in the progeny generation has decreased to

$$p_{c} = .2158 + .1494 \ = \ .3652.$$

2. Disassortative Mating

Disassortative mating is the mating of unlike genotypes. The mating diagram would look like

Female		Male Genotypes
Genotypes	Frequencies	Frequencies
		A1A1	A1A2	A2A2
		pm2	2 pmqm	qm2
A1A1	pf2	0	0	pf2qm2
A1A2	2pfqf	0	4pfqfpmqm	0
A2A2	qf2	qf2pm2	0	0

Note that the mating of A₁A₂ genotypes appears in both assortative and disassortative mating. Another possibility in both cases would be to remove these matings as well. The sum of the frequencies of the matings that were made is

S = 6p²q²,

assuming the male and female allele frequencies were equal. The progeny genotypic frequencies would be

Parent			Progeny Genotypic
Genotypes			Frequencies
Males	Females	Freq.	A1A1	A1A2	A2A2
A1A1	A2A2	p2q2/S	0	1/6	0
A1A2	A1A2	4p2q2/S	1/6	1/3	1/6
A2A2	A1A1	p2q2/S	0	1/6	0

The new frequency of A₁ is 0.5, regardless of the starting frequency, and the frequency will remain at 0.5 in future generations whether doing disassortative matings of the same type or random matings. The frequency of homozygous genotypes will diminish in future generations of disassortative matings.

3. Reality

Both types of selection most likely occur simultaneously in the real world, plus other types of selection. Some animals are culled, perhaps for reasons other than their genotype or phenotype, but the reasons may be associated with the genotypes resulting in a change in the parental allele frequencies. After culling, matings may not be random, which will alter the progeny allele frequencies. If these processes occur each generation, then it is unlikely that Hardy-Weinberg equilibrium is ever achieved. Also, genotypic variation could be altered by many different factors. If selection is on phenotypes (because genotypes are masked), then the effects of selection may be reduced if the residual variation is large relative to genotypic variation. However, the effects of selection will still influence the genotypic variance.

Suppose the bottom 25% are culled, and that this affects males and females equally. Let p=0.4 as before, then the new frequencies of parental genotypes would be

Genotype	Initial	After	Re-scaled
	Frequency	Culling
A1A1	.16	.16	.21333
A1A2	.48	.48	.64000
A2A2	.36	.11	.14667
	1	.75	1.00000

The new progeny distribution after random mating of the selected parents would be

Parent			Progeny Genotypic
Genotypes			Frequencies
Males	Females	Freq.	A1A1	A1A2	A2A2
A1A1	A1A1	.045511	.045511	0	0
A1A1	A1A2	.136533	.068267	.068267	0
A1A1	A2A2	.031289	0	.031289	0
A1A2	A1A1	.136533	.068267	.068267	0
A1A2	A1A2	.409600	.102400	.204800	.102400
A1A2	A2A2	.093867	0	.046933	.046933
A2A2	A1A1	.031289	0	.031289	0
A2A2	A1A2	.093867	0	.046933	.046933
A2A2	A2A2	.021511	0	0	.021511
		1.000000	.284444	.497778	.217778

If another 25% were culled from the bottom, then the new frequencies of the parent genotypes would be

Genotype	Initial	After	Re-scaled
	Frequency	Culling
A1A1	.284444	.284444	.379259
A1A2	.497778	.465556	.620741
A2A2	.217778	.000000	.000000
	1	.75	1.00000

All of the A₂A₂ genotypes would be culled and part of the heterozygotes. Eventually fixation to the A₁ allele would occur with continued selection.

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