The rate of molecular evolution

We're now going to calculate the rate of molecular evolution, i.e., the rate of allelic substitution, under the hypothesis that mutations are selectively neutral. To get that rate we need two things: the rate at which new mutations occur and the probability with which new mutations are fixed. In a word equation

$$\begin{eqnarray*} \mbox{rate of substitution} &=& \mbox{rate of mutation}\times\mbox{probability of fixation} \\ \lambda &=& \mu_0p_0 \quad . \end{eqnarray*}$$

Now I've actually fudged a bit on my words here. When I wrote ``rate of mutation,'' what I really meant was ``the number of new mutations that occur in any one generation.'' That's pretty easy to calculate. In a diploid population of size $N$, there are $2N$ gametes. The probability that any one of them mutates is just the mutation rate, $\mu$, so

$$\mu_0 = \mu \quad .$$ (1)

To calculate the probability of fixation, we have to say something about the dynamics of alleles in populations. Let's suppose that we're dealing with a single population, to keep things simple. Now, you have to remember a little of what you learned about the properties of genetic drift. If the current frequency of an allele is $p_0$, what's the probability that is eventually fixed? $p_0$. When a new mutation occurs there's only one copy of it,¹ so the frequency of a newly arisen mutation is

$$p_0 = \frac{1}{2N} \quad .$$ (2)

Putting (1) and (2) together we find

$$\begin{eqnarray*} \lambda &=& \mu_0p_0 \\ &=& (2N\mu)\left(\frac{1}{2N}\right) \\ &=& \mu \quad . \end{eqnarray*}$$

In other words, if mutations are selectively neutral, the substitution rate is equal to the mutation rate. Since mutation rates are (mostly) governed by physical factors that remain relatively constant, mutation rates should remain constant, implying that substitution rates should remain constant. The prediction of a molecular clock follows directly from the hypothesis that mutations are selectively neutral.