Answer:
42%
Explanation:
When allele frequency and genotype frequency stay constant in a randomly mated population, the population is said to obey Hardy Weinberg equilibrium.
By using Hardy Weinberg Equilibrium for the given information; we have the following equations:
p + q = 1; &
[tex]\mathbf{p^2+2pq+q^2 =1 }[/tex]
Here; allele a = p
allele b = q
the heterozygous allele ab = 2pq
Since; a = 70%, then the percentage of allele a = 0.70
Now;
p + q = 1
0.7 + q = 1
q = 1 - 0.7
q = 0.3
So, allele b = 0.3
Finally, to determine the percentage of the heterozygous population:
2pq = (2 × 0.7 × 0.3)
= 0.42
= 42%
