Question

g A gene has two alleles a and b. In the gene pool, the a allele accounts for 70 percent of the alleles at this locus. If the population is in Hardy-Weinberg equilibrium, what percentage of the population is heterozygotes with respect to this locus

Answer 1

42%

Step-by-step 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; &

`\mathbf{p^2+2pq+q^2 =1 }`

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%