Hardy-Weinberg Equilibrium

Suppose we want to study the allele frequencies of the gene for coloration in a population of moths. The allele for the dark color pattern, B, is dominant to the allele for the light color pattern, b. In a certain population, the frequency of B is found to be 0.9, and that of b is 0.1 (we will see, below, how to determine these frequencies by studying the moths themselves). This means that 90 percent of all the alleles are B, and 10 percent are b.

The Hardy-Weinberg equilibrium states that, given the above conditions, allele frequencies will not change from one generation to the next. To show this is true, we need some algebra.

Random mating means each allele has an equal chance of being paired with each other allele. During random mating, the likelihood that a B allele from a mother will unite with a B allele from a father is given by

B × B = 0.9 × 0.9 = 0.81.

The genotype of this offspring will be BB.

Similarly, the likelihoods of other combinations:

B × b = 0.9 × 0.1 = 0.09 for genotype Bb;

b × B = 0.1 × 0.9 = 0.09 for genotype Bb; and

b × b = 0.1 × 0.1 = 0.01 for genotype bb.

Note that the two Bb genotypes are the same. Therefore the frequency of BB is 0.81, the frequency of Bb is 0.18, and the frequency of bb is 0.01. These add to 1, just as we would expect, since they represent all the members of the next generation.

Are the allele frequencies still 0.9 and 0.1? For simplicity, imagine we're looking at one hundred individuals, so that eighty-one are BB, eighteen are Bb, and one is bb. Since each individual has two alleles, there are 200 alleles in all.

The number of B alleles is given by (81 × 2) + (18 × 1) = 180.

The number of b alleles is given by (1 × 2) + (18 × 1) = 20.

By comparing 180 to 20, you can see the frequency of B is still 0.9 and that of b is still 0.1.