Calculate the frequency of genotypes in a population in Hardy Weinberg equilibrium with the allele frequencies of A=0.6 and a=0.4. Thus calculate the number of individuals with each genotype in a population of 150

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