Partial self-fertilization

Many plants reproduce by a mixture of outcrossing and self-fertilization. To a population geneticist that means that they reproduce by a mixture of selfing and random mating. Now I'm going to pull a fast one and derive the equations that determine how allele frequencies change from one generation to the next without using a mating table. To do so, I'm going to imagine that our population consists of a mixture of two populations. In one part of the population all of the reproduction occurs through self-fertilization and in the other part all of the reproduction occurs through random mating. If you think about it for a while, you'll realize that this is equivalent to imagining that each plant reproduces some fraction of the time through self-fertilization and some fraction of the time through random mating. Let $\sigma$ be the fraction of progeny produced through self-fertilization, then

$$x_{11}' = p^2(1-\sigma) + (x_{11} + x_{12}/4)\sigma$$ (8)

$$x_{12}' = 2pq(1-\sigma) + (x_{12}/2)\sigma$$ (9)

$$x_{22}' = q^2(1-\sigma) + (x_{22} + x_{12}/4)\sigma$$ (10)

Notice that I use $p^2$, $2pq$, and $q^2$ for the genotype frequencies in the part of the population that's mating at random. Question: Why can I get away with that?

It takes a little more algebra than it did before, but it's not difficult to verify that the allele frequencies don't change between parents and offspring.

$$p' = p^2(1-\sigma) + (x_{11} + x_{12}/4)\sigma + pq(1-\sigma) + (x_{12}/4)\sigma$$ (11)

$$= p(p+q)(1-\sigma) + (x_{11} + x_{12}/2)\sigma$$ (12)

$$= p(1-\sigma) + p\sigma$$ (13)

$$= p$$ (14)

Because homozygous parents can always have heterozygous offspring (when they outcross), heterozygotes are never completely eliminated from the population as they are with complete self-fertilization. In fact, we can solve for the equilibrium frequency of heterozygotes, i.e., the frequency of heterozygotes reached when genotype frequencies stop changing.² By definition, an equilibrium for $x_{12}$ is a value such that if we put it in on the right side of equation 9 we get it back on the left side, or in equations

$$\hat x_{12} = 2pq(1-\sigma) + (x_{12}/2)\sigma$$ (15)

$$\hat x_{12}(1 - \sigma/2) = 2pq(1-\sigma)$$ (16)

$$\hat x_{12} = \frac{2pq(1-\sigma)}{(1-\sigma/2)}$$ (17)

It's worth noting several things about this set of equations:

1. I'm using $\hat x_{12}$ to refer to the equilibrium frequency of heterozygotes. I'll be using hats over variables to denote equilibrium properties throughout the course.³

2. I can solve for $\hat x_{12}$ in terms of $p$ because I know that $p$ doesn't change. If $p$ changed, the calculations wouldn't be nearly this simple.

3. The equilibrium is approached gradually (or asymptotically as mathematicians would say). A single generation of random mating will put genotypes in Hardy-Weinberg proportions (assuming all the other conditions are satisfied), but many generations may be required for genotypes to approach their equilibrium frequency with partial self-fertilization.