A= 0.6 a= 0.4 p+q=1 where p=A and q=a p2+2pq+q2=1where AA=P2 Thus AA frequency in the population = 0.62= 0.36Aa=2pq Aa frequency in the population = 2(0.4 X 0.6) = 0.48aa=q2 aa frequency in the population= 0.42=0.16
Since 1 is the frequency of alleles in the whole gene pool the total number of members in the population multiplies by the genotypes relative frequency will give a number of individuals with that genotype.
AA= 150 X 0.36 = 54 individuals Aa= 150 X 0.48 = 72 individuals aa= 150 X 0.16 = 24 individuals