Gene frequencies of populations Not in Hardy-Weinberg equilibrium

Question

Hardy-Weinberg equilibrium assumes p^2+2pq+q^2 of a single gene, generation after generation ie. gene frequencies DO NOT change over time. If this was true in nature populations would never evolve ( never change).

If I put a selection pressure of 0.2 against the AA genotype (which means I only allow 0.8 of the AA genotype to reproduce to leave the next generation)

Answer 1

given (p^2)AA (2pq)Aa (q^2)aa ( a population in H-W equilibrium)

Applying a 0.2 selection pressure (or alternatively allowing 0.8 to reproduce) we would have this

(0.8)(p^2)AA + 2pq + q^2

If initial population is p=0.5 and q=0.5 then

0.8(.25)AA + .50(Aa) + .25(aa)

therefore 0.2(AA) + 0.5(Aa) + .25(aa) ( total population is 0.95 so we must divide each value by 0.95 to have the whole population equal to 1.0)

new population after selection of 0.2 is applied is

0.21(AA) + 0.53(Aa) + 0.26(aa)

New gene frequencies A= 0.21 + 0.265 (this number is 1/2 the heterozygote frequency) = 0.475 = 48%

a= 0.265 + 0.25 = .51.5 = 52%

Gene frequencies changed therefore evolution is occurring!