Question

According to the hardy-weinberg principle, if 75% of the alleles in the gene pool are a1 and 25% are a2, what is the proportion of individuals with genotype a1a2 in this population?

Answer 1

Frequency of allele a1 is 75%, p=0.75
Frequency of allele a2 is 25%, q=0.25
 If a population is in Hardy-Weinberg equilibrium then:
p2+2pq+q2=1
p2  is the frequency of genotype a1a1 (homozygous),
2pq is the frequency of genotype a1a2 (heterozygous),
q2 is the frequency of genotype a2a2
The proportion of individuals with genotype a1a2 is then 2*0.75*0.25=0.375 (37.5%).