SELECTION AGAINST THE RECESSIVE HOMOZYGOTE

The standard protocol used to build models of population genetics are...

• define the initial conditions of the selection action and population
• allow selection to act on the population
• calculate the allelic frequency after selection has occurred (qn+1)
• calculate the change in the frequency of the allele (change in q from one generation to the next)
• when the change in q has become zero, we say the equilibrium frequency has been reached.

Let's consider a simple population in which

• we consider a single, autosomal locus
• individuals are diploid and sexually reproducing
• selection results in direct change in genotypes at this locus

After selection has occurred, the individuals of the population will mate at random, and the new generation will have different Hardy Weinberg proportions of genotypes than the original population.

Here's our initial population. Both the dominant homozygote and the heterozygote have the same phenotype, and so the same fitness (W = 1.0). The recessive homozygote, however, is being selected against (W = 1 - s).

Note that the initial population is in Hardy Weinberg equilibrium. Even when selection acts on the organism (let's say it's a fruit fly), Hardy-Weinberg proportions will be re-established every time there's a round of random mating--but each time this happens after selection has occurred, the new Hardy Weinberg proportions will be different from those of the initial population.

After selection has occurred, the ratio of each genotype is determined by multiplying its frequency by its fitness. (This makes sense, since fitness is a measure of relative survival).

This means that for every one of the AA and Aa individuals that has survived, only (1-s) of the aa genotype individuals has survived.

For example, if the selection coefficient of the aa genotype was 0.4, then its fitness would be 1 - 0.4, or 0.6.

This means that for every ten AA or Aa individuals that survived to reproduce, only 6 aa individuals did so.

The total of the three genotypes after selection is

1 - sq2

That is...
p2 + 2pq + q2(1-2)
= p2 + 2pq + q2 - sq2
= 1 = sq2
(and you thought your old 7th grade algebra would never come in handy...)

Overall, the selection and its results can be calculated like so:

Initial genotype frequencies:

• AA = p2
• Aa = 2pq
• aa = q2
• total = 1.0 (100% of genotypes)

Fitness of each genotype:

• AA = 1.0
• Aa = 1.0
• aa = 1 - s

Ratio in population after natural selection (one generation):

• AA = p2
• Aa = 2pq
• aa = q2(1-s)
• total = 1 - sq2 (= Mean Fitness of the Population, or Wmean)

Genotypic frequencies after natural selection:

• AA = p2/Wmean
• Aa = 2pq/Wmean
• aa = q2(1-s)/Wmean
• total = 1.0

The MEAN FITNESS OF THE POPULATION is the sum of the fitnesses of the genotypes multiplied by the frequency at which they occur. Essentially, this is just weighting each genotype's fitness by the frequency with which it occurs in the population.

(By way of explanation...Since any set of numbers can be converted to proportions of unity by dividing them by their sum, you can return the genotypeic frequencies to their actual populational ratios by dividing them by the mean fitness of the population.)

One can calculate the new frequency of the recessive allele after selection (qn + 1) as the proportion of aa homozygotes plus HALF the proportion of the heterozygotes:

qn + 1 = [q2(1-s)/1 - sq2] + [pq/1 - sq2]

which distills down to...

q(1 - sq)/1 - sq2 (Call this Equation #1)

If we say that the recessive allele is lethal (simplest case!), then s = 1.0, and the above simplifies to:

qn + 1 = [q2(1 - s)/1 - q2] + [pq/1 - q2] (We'll call this Equation #2)

If you factor out 1 - q2, it becomes (1 - q)(1 + q)

and this, in turn, will change Equation 2 into:

q(1-q)/(1-q)(1-q)

Which, cancelling out properly, will give you...

q/(1 + q)

To calculate the change in allele frequency from one generation to the next, use the above and subtract the allele frequency:

[q/(1 + q)] - q

Which you can't solve until you provide a common denominator, so multiply the pesky thing by (1 + q)/(1 + q)

This will give you...

change in allele frequency = [q - q(1 + q)]/1 + q

Which distills down into...

-q2/(1 + q) (We'll call this Equation #3)

Equation #3 is what you use to calculate the change in allellic frequencies caused by natural selection between two generations! (Are you tired now?)

Note two things from Equation #3...
• The frequency of the recessive allele is declining: the fraction is negative.
• The change in the allele frequency (q) is proportional to the genotype frequency (q2. In hosrt, the more homozygous recessives there are in the population, the more rapidly the recessive allele will be selected out of the population. (Intuitively clear, but mathematically shown.)

EQUILIBRIUM CONDITIONS

When the change in q is zero, there will be no further change in allelic frequencies from one generation to the next. This means that under equilibrium conditions, q2 (the numerator of the equation) = 0.

In plain language: if there is selection against the a allele, the allele frequencies will reach equilibrium only once a is completely removed from the population. Only mutation will return a to the pool, resulting in further selection.

It can take many generations for a deleterious allele to be removed from the population, which means there may be time for mutations to return a certain number of those alleles to the population. The SELECTION-MUTATION EQUILIBRIUM is a measure of how quickly selection is removing the allele from the population relative to how quickly mutation is returning it. We could derive this equation, too. But for now, we'll give it a rest.

There. Aren't you glad we didn't have time to cover this?