Question

Suppose that genotypes A1A1, A1A2, and A2A2 have fitnesses 1, 0.5, and 1.5. The population is infinite and mates at random. What is the frequency of allele A1 at the two stable equilibria?

Answer 1

From the law of Hardy Weinberg equilibrium, p + q = 1. In this equation, p represents the dominant allele, q is the recessive allele, and pq is the heterozygotic allele.

The A allele can be represented by p, and the A allele frequency will be p² + pq or we can also write down it as A1A1 + A1A2.

The frequency of a particular allele can be calculated by the total number of particular allele divided by the total fitness of all the alleles.

A1A1 = 1 (fitness)

A1A2 = 0.5

A2A2 = 1.5

Calculation: (A1A1 ) + (A1A2)

                Total fitness of all the alleles

= 1 + 0.5/2

1+0.5+1.5

= 1.25/ 3

= 0.41, is the frequency of ‘A1’ allele in one stable population (first generation).

Frequency of A1 allele in two stable population or we can say the second generation:

As from the law, p+q=1, and q = 1-p, and the A1 frequency p² + pq, this equals to :

p2 + p (1-p)                (q = (1-p) )

= p² +p –p² = p.

From the calculation, it is cleared that p frequency will remain the same from one generation to another (second) generation until unless there is any alteration of the alleles due to mutations or any other evolutionary factors.

So the A1 frequency in the second generation (two stable) will remain the same that is 0.41.