Question

Consider a population of sheep to be in Hardy-Weinberg’s equilibrium. If the allele for white wool (w) has an allele frequency of 0.81 while the allele for brown wool (W) has an allele frequency of 0.19, then the percentage of heterozygous individuals in the population is

• A. 4%
• B. 15%
• C. 31%
• D. 66%

Answer 1

The correct option is C 31%

According to Hardy-Weinberg principle, allele frequency in a population is expressed by the equation

$p^2+2\mathrm{pq}+q^2=1$

$p^2$ represent homozygous dominant,

$q^2$ represent homozygous recessive and

2pq represent heterozygous dominant individuals

Given, p = 0.81 and q = 0.19

Therefore, frequency of heterozygous dominant is:

$2\times 0.81\times 0.19=0.3078i.e.,30.78\%$

So, the percentage of heterozygous individuals is 31%.