Natural Selection
(version 31 July 2002)
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The fitness of an individual is the number of offspring it leaves. There are several components to the total fitness
In population genetics, we look for differences in the fitness of different genotypes at a particular locus (or set of loci).
In quantitative genetics, we look for differences in the fitness of different characters ( phenotypes).
In predicting the response to selection, relative fitnesses are all that we need. Here, we can set any particular fitness to one and consider the fitness of other genotypes/phenotypes relative to our standard.
Selection occurs when different classes of individuals have (on average) different fitnesses.
There is a response to selection if there are transmittable genetic differences between these groups.
Types of selection
Consider a locus with two alleles, A and a, and let p = freq(A) . What is the change in p after a single generation of selection?
We can also express this more compactly using marginal fitnesses . Let
Consider a locus with two alleles, with genotype fitnesses
If p = freq(A) = 1/4, what is the change in the frequency of allele A following one generation of selection?
This is an example of directional selection where one homozygote genotype has the highest fitness. Here, the population becomes fixed for that allele --- its frequency goes to one.
The fitnesses here are WAA = 1, WAa = 1, Waa = 1-s
Here allele A is ultimately fixed (if s > 0) or lost (if s < 0), but selection is very slow, especially as the allele become rarer. For example, suppose we have a recessive lethal, so that s = 1, so that Waa = 0, and we start at freq(A) = 0.1
| Change in A | number of generations required | change in freq(aa) |
| 10% to 50% | 3 generations | 81% to 25% |
| 50% to 90% | 10 generations | 25% to 1% |
| 90% to 99% | 93 generations | 1% to 0.01% |
| 99% to 99.9% | 902 generations | 0.01% to 0.0001% |
Why does selection slow down so much as freq(a) becomes small?
Key: Only the recessive homozygote is selected against, with freq(aa) = freq(a)*freq(a) << 1
Here, the heterozygote has the highest fitness:
The net result (instead of fixation) is a selective equilibrium, where the allele frequencies converge to an equilibrium frequency of
At the equilibrium point, the marginal fitnesses of both alleles are equal (homework problem).

Note that here, at equilibrium, the mean fitness is at its maximum value.
This is a stable equilibrium point. If we deviate slightly from it, the allele frequencies return to the equilibrium frequency.
Suppose the heterozygote has the lowest fitness.
The result is an unstable equilibrium , wherein if the allele frequency starts below the equilibrium point, it is lost, while the allele is fixed if its frequency starts above the equilibrium point.

Here at equilibrium the mean fitness is at its minimum value.
| Fitnesses | Outcome |
| WAa > WAA, Waa | Stable polymorphic equilibrium with A and a |
| WAa < WAA = Waa | Unstable equilibrium, which allele fixed depends on the initial frequencies |
| WAA > WAa , Waa | Allele A fixed |
| WAA = WAa > Waa | Allele A fixed |
By using marginal fitnesses, the above results easily extend to multiple alleles at one locus. Let pi be the frequency of allele i.

As above, the mean population fitness Wbar is simply the average of the marginal fitnesses, so that with k alleles,
Hence, those alleles whose marginal fitness exceed the mean fitness will increase, while those alleles whose marginal fitness is less than the mean fitness will decrease
Note that the marginal fitness changes as the frequencies of allele change (i.e. they are not constant), as they depend on the distribution of genotypes in the population.
At an equilibrium point, the mean fitness of all segregating alleles is equal.
When selection acts on only a single locus, then generally mean population fitness increases each generation.
This is often not the case when selection acts on multiple loci. Why? Linkage disequilibrium
The analysis of even two-locus selection models is extremely complicated, and we will not cover this further.
Suppose the genotype AaBb has fitness 1.1, while all other genotypes have fitness 1.
If we cross AABB and aabb parents, the F1 are all AaBb, and hence the mean population fitness is 1.1. However, each generation the mean population fitness declines each generation to an equilibrium value of 1.025 (if the two loci are unlinked).