Now, we’ll introduce selection into the equation, which will violate condition #2 for Hardy-Weinberg equilibrium. If a population is not in equilibrium, the allele frequencies will change from generation to generation.

In the case of the peppered moth, we’ll assume that none of the light color moths survive. Therefore, in our next generation we’ll have the Aa genotype, and the AA genotype, but none of the aa genotype. In our example of a thousand offspring, that leaves us with 640 AA and 320 Aa. What happens in the next generation, after these remaining moths breed randomly?

There are now a total of 1920 alleles in the mating population. The proportion of the A alleles in the mating population
will be: (640 + 640 + 320) / 1920 = 0.83 and the proportion of a alleles
will be: 1 – 0.83 = 0.17. You can also calculate the proportion of a alleles by the following: 320/1920 = 0.17. Hence, *p* is now 0.83 and *q* is 0.17.

And so the mixture of genotypes in the next generation will be: (0.83A +
0.17a)^{2} =

0.689 (AA) + 0.282 (Aa)
+ 0.29 (aa)

Due to selection against the aa genotype, we’ve gone from allele a frequency of 0.2 to 0.167 in one generation.

Now, let’s assume the surviving genotypes AA and Aa reproduce, leading to 1000 individuals in the next (second) generation. Based on what we’ve just calculated, in the second generation, the proportion of:

AA is 0.689 X 1000 = 689

Aa is 0.282 X 1000 = 282

aa is 0.289 X 1000 = 29

Assuming none of the aa genotype moths survive to reproduction, the population will now consist of 689 AA and 282 Aa (total = 971) individuals. The proportion of A alleles in the mating population, *p’*, will be:

(689 + 689 + 282)/1942 = 0.855

The proportion of a alleles in the mating population, *q’*, will be:

1 – 0.855 = 0.145

In two generations, we’ve gone from allele a frequency of 0.2 to 0.167 and to 0.145, when selection against aa genotype is present.

In the illustration below, with selection for the A allele, its frequency will increase and approach 1.0, while the frequency of the a allele will approach 0 and eventually be lost.

The frequency of the a allele will continue to decline as long as the same selection is applied each subsequent generation. It is interesting to note that removing all individuals with the aa genotype from each parental generation does not remove the aa genotype completely from the population, because aa can arise from crossing the heterozygotes, Aa and Aa. Without removing the heterozygotes (which do not show the a allele phenotypically), it will take a long time for the a allele to disappear completely from the population.

**Let’s test your knowledge of this topic:**

A population of 100 diploid individuals originally experiencing Hardy-Weinberg equilibrium consists of 49 AA, 42 Aa and 9 aa individuals. Let’s assume an extreme scenario: A new disease targets and kills only those individuals who are homozygous recessive, aa, so all individuals with the aa genotype do not survive. What is the frequency of the allele A after this selection?

Frequency of A = ____________

__Click here for an explanation:__

Continuing on the same question, what will the frequency of allele A be if this generation of the population reproduces by random mating? You may assume 100 offspring were produced in the next generation.

Frequency of A = ______________

__Click here for an explanation:__