Question

(Hardy weinberg equation)

You are examining a population that is under selection where the fitness of the heterozygotes is 0.34, homozygotes is 1. If there's 546 AA, 127 aa, and 982 Aa individuals in population, what will be the "p" value for the next generation?

Answer 1

Given:

AA = 546

aa = 127

Aa = 982

Total population size, N = (546+127+982) which is 1655.

Let the frequency of allele A be p and frequency of allele "a" be q.

We can calculate frequency of p using the formula,

p = (2AA + Aa) / (2N)

=[(2×546) +(982) ] / [ 2×1655]

= 0.63

Since the population is in Hardy-weinberg equilibrium, thus frequency of allele "a",

q=1-p

=1-0.63

=0.37

We need to calculate the value of p in next generation, given that, fitness of heterozygous individual, Aa is 0.34 and fitness of either homozygous individual, AA or aa is 1.

Thus, WAa=0.34, WAA = 1 and Waa =1.

We can calculate the frequency of p in next generation as follows:

Thus, value of p (frequency of allele A) in the next generation will be 0.6877.