Question

There are two alleles, B and b, at a gene locus in a population at Hardy-Weinberg equilibrium. The relative frequency of the dominant (B) allele is 0.7. What is the expected number of heterozygous individuals if the total population size is 500 individuals?

Answer 1

Answer:

210

Explanation:

Frequency of dominant allele is represented by "p"

and frequency of recessive allele is represented by "q"

Given

$p = 0.7$

Thus,

$q = 1-p\\q = 1 - 0.7 \\q = 0.3$

As per Hardy Weinberg's second equation of equilibrium -

$p^2 + q^ 2 + 2pq = 1$

Substituting the given values in above equation, we get -

$0.7^ 2+ 0.3^2 + 2pq = 1\\2pq = 1- 0.49-0.09\\2pq = 0.42$

So the total number of heterozygous individual is equal to

$0.42 * 500\\= 210$