Disruptive selection

If we plot $\bar w$ as a function of $p$ when $w_{11} > w_{12}$ and $w_{22} > w_{12}$, we wee a very different pattern (Figure 2). Since the Fundamental Theorem tells us that $\bar w' \ge \bar w$, we know that if the population starts with an allele on one side of the bowl $A_1$, will be lost. If it starts on the other side of the bowl, $A_2$ will be lost.¹¹

Figure 2: With disruptive selection ( $w_{11} > w_{12} < w_{22}$) viability selection may lead either to an increasing frequency of the $A$ allele or to a decreasing frequency. Ultimately, the population will be monomorphic for one of the homozygous genotypes. Which homozygous genotype comes to predominate, however, depends on the initial allele frequencies in the population.

Let's explore this example a little further. To do so, I'm going to set $w_{11} = 1 + s_1$, $w_{12} = 1$, and $w_{22} = 1+ s_2$.¹² When fitnesses are written this way $s_1$ and $s_2$ are referred to as selection coefficients. Notice also with these definitions that the fitnesses of the homozygotes is greater than 1.¹³ Using these definitions and plugging them into (1),

$$p' = \frac{p^2(1+s_1) + pq}{p^2(1+s_1) + 2pq + q^2(1+s_2)}$$

$$= \frac{p(1 + s_1p)}{1 + p^2s_1 + q^2s_2} \quad .$$ (3)

We can use equation (3) to find the equilibria of this system, i.e., the values of $p$ such that $p' = p$.

$$\begin{eqnarray*} p &=& \frac{p(1 + s_1p)}{1 + p^2s_1 + q^2s_2} \\ p(1 + p^2s_1... ... \\ p(-pqs_1 + q^2s_2) &=& 0 \\ pq(-ps_q + qs_2) &=& 0 \quad . \end{eqnarray*}$$

So $p' = p$ if $p=0$, $q=0$, or $ps_1 = qs_2$. We can simplify that last one a little further, too.

$$\begin{eqnarray*} ps_1 &=& qs_2 \\ ps_1 &=& (1-p)s_2 \\ p(s_1 + s_2) &=& s_2 \\ p &=& \frac{s_2}{s_1+s_2} \quad . \end{eqnarray*}$$

Fisher's Fundamental Theorem tells us which of these equilibria matter. I've already mentioned that depending on which side of the bowl you start, you'll either lose the $A_1$ allele or the $A_2$ allele. But suppose you happen to start exactly at the bottom of the bowl. That corresponds to the equilibrium with $p = s_2/(s_1+s_2)$. What happens then?

Well, if you start exactly there, you'll stay there forever (in an infinite population). But if you start ever so slightly off the equilibrium, you'll move farther and farther away. It's what mathematicians call an unstable equilibrium. Any departure from that equilibrium gets larger and larger. For evolutionary purposes, we don't have to worry about a population getting to an unstable equilibrium. It never will. Unstable equilibria are ones that populations evolve away from.

When a population has only one allele present it is said to be fixed for that allele. Since having only one allele is also an equilibrium (in the absence of mutation), we can also call it a monomorphic equilibrium. When a population has more than one allele present, it is said to be polymoprhic. If two or more alleles are present at an equilibrium, we can call it a polymorphic equilibrium. Thus, another way to describe the results of disruptive selection is to say that the monomorphic equilibria are stable, but the polymorphic equilibrium is not.¹⁴