Evolution__3rd_Edition

122 PART 2 / Evolutionary Genetics
We construct a model of the
gene frequency of a recurrent
disadvantageous mutation
The disadvantageous mutation has
a low equilibrium frequency ...
equilibrium, natural selection removes more A genes than mutation creates and the
frequency decreases; vice versa if the frequency is lower than the equilibrium. At the
equilibrium, the rate of loss of A genes by selection equals their rate of gain by mutation.
We can use that statement to calculate the equilibrial gene frequency p*. What is the
rate per generation of creation of A genes by mutation? Each new A gene originates by
mutation from an a gene and the chance that any one a gene mutates to an A gene is the
mutation rate m. A proportion (1 − p) of the genes in the population are a genes.
Therefore:
Total rate of creation of A genes by mutation = m(1 − p)
And what is the rate at which A genes are eliminated? Each A gene has a (1 − s) chance of
surviving, or an s chance of dying. A proportion p of the genes in the population are A.
Therefore:
Total rate of loss of A genes by selection = ps
At the equilibrium gene frequency (p*):
Rate of gain of A gene = rate of loss of A gene
m(1 − p*) = p*s (5.8)
Which can be multiplied out:
m − mp* = p*s
p* = m/(s + m)
Of the two terms in the denominator, the mutation rate (maybe 10 −6 , Section 2.6, p. 32)
will usually be much less than the selection coefficient (perhaps 10 −1 or 10 −2 ). With
these values s + m ≈ s and the expression is therefore usually given in the approximate
form:
p* = m/s (5.9)
The simple result is that the equilibrium gene frequency of the mutation is equal to
the ratio of its mutation rate to its selective disadvantage. The result is intuitive: the
equilibrium is the balance between the rates of creation and elimination of the gene. To
obtain the result, we used an argument about an equilibrium. We noticed that at the
equilibrium the rate of loss of the gene equals the rate of gain and used that to work out
the exact result. This is a powerful method for deriving equilibria, and we shall use an
analogous argument in the next section.
The expression p = m/s allows a rough estimate of the mutation rate of a harmful
mutation just from a measurement of the mutant gene’s frequency. If the mutation is
rare, it will be present mainly in heterozygotes, which at birth will have frequency 2pq.
If p is small, q ≈ 1 and 2pq ≈ 2p. N is defined as the frequency of mutant bearers, which
equals the frequency of heterozygotes: i.e., N = 2p. As p = m/s, m = sp; if we substitute
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