A simple example

To understand in more detail what happens when there is genetic drift, let's consider the simplest possible example: a haploid population consisting of 2 individuals.¹ Suppose that there are initially two alleles in this population $A_1$ and $A_2$. This implies that $p = q = 0.5$, but we'll ignore that numerical fact and simply imagine that the frequency of the $A_1$ allele is $p$.

We imagine the following scenario:

• Each individual in the population produces a very large number of offspring.

• Each offspring is an identical copy of its parent, i.e., $A_1$ begets $A_1$ and $A_2$ begets $A_2$. In other words, there's no mutation.

• The next generation is constructed by picking two offspring at random from the very large number of offspring produced by these two individuals.

Then it's not too hard to see that

$$\begin{eqnarray*} \mbox{Probability that both offspring are $A_1$} &=& p^2 \\ \... ...2pq \\ \mbox{Probability that both offspring are $A_2$} &=& q^2 \end{eqnarray*}$$

Of course $p = 1$ if both offspring sampled are $A_1$, $p = 1/2$ if one is $A_1$ and one is $A_2$, and $p = 0$ if both are $A_2$, so that set of equations is equivalent to this one:

$$P(p=1) = p^2$$ (1)

$$P(p=1/2) = 2pq$$ (2)

$$P(p=0) = q^2$$ (3)

In other words, we can no longer predict with certainty what allele frequencies in the next generation will be. We can only assign probabilities to each of the three possible outcomes. Of course, in a larger population the amount of uncertainty about the allele frequencies will be smaller,² but there will be some uncertainty associated with the predicted allele frequencies unless the population is infinite.

The probability of ending up in any of the three possible states obviously depends on the current allele frequency. In probability theory we express this dependence by writing equations (1)-(3) as conditional probabilities:

$$P(p_1=1\vert p_0) = p_0^2$$ (4)

$$P(p_1=1/2\vert p_0) = 2p_0q_0$$ (5)

$$P(p_1=0\vert p_0) = q_0^2$$ (6)

I've introduced the subscripts so that we can distinguish among various generations in the process. Why? Because if we can write equations (4)-(6), we can also write the following equations:³

$$\begin{eqnarray*} P(p_2=1\vert p_1) &=& p_1^2 \\ P(p_2=1/2\vert p_1) &=& 2p_1q_1 \\ P(p_2=0\vert p_1) &=& q_1^2 \end{eqnarray*}$$

Now if we stare at those a little while, we⁴ begin to see some interesting possibilities. Namely,

$$\begin{eqnarray*} P(p_2=1\vert p_0) &=& P(p_2=1\vert p_1=1)P(p_1=1\vert p_0) + P... ... &=& (1)(q_0^2) + (1/4)(2p_0q_0) \\ &=& q_0^2 + (1/2)p_0q_0 \end{eqnarray*}$$

It takes more algebra than I care to show,⁵ but these equations can be extended to an arbitrary number of generations.

$$\begin{eqnarray*} P(p_t=1\vert p_0) &=& p_0^2 + \left(1 - (1/2)^{t-1}\right)p_0q... ...P(p_t=0\vert p_0) &=& q_0^2 + \left(1 - (1/2)^{t-1}\right)p_0q_0 \end{eqnarray*}$$

Why do I bother to show you these equations?⁶ Because you can see pretty quickly that as $t$ gets big, i.e., the longer our population evolves, the smaller the probability that $p_t = 1/2$ becomes. In fact, it's not hard to verify two facts about genetic drift in this simple situation:

1. One of the two alleles originally present in the population is certain to be lost eventually.

2. The probability that $A_1$ is fixed is equal to its initial frequency, $p_0$, and the probability that $A_2$ is fixed is equal to its initial frequency, $q_0$.

Both of these properties are true in general for any finite population and any number of alleles.

1. Genetic drift will eventually lead to loss of all alleles in the population except one.⁷

2. The probability that any allele will eventually become fixed in the population is equal to its current frequency.