Question

In a population of skunks, there are striped and stripeless individuals. Stripes are dominant to the absence of stripes. In this population, p = 0.8 and the frequency of heterozygotes is 0.25. This shows that the population is in Hardy-Weinberg equilibrium

a) True
b) False

Answer 1

false, it is not in equilibrium.

Answer 2

False

Step-by-step explanation:

Given,

Striped population is dominant over stripeless individuals.

Let the allele for striped population be represented by "S" and stripeless population be represented by "s"

p denotes the frequency of dominant allele

As per hardy Weinberg, first equilibrium equation-

`p + q = 1\\0.8 + q = 1\\q = 1-0.8\\q= 0.2`

Now as per Hardy Weinberg, second equilibrium equation -

`p^2 + q^ 2+ 2pq = 1\\0.64 + 0.04 + 2pq = 1\\2pq = 1 - 0.64 -0.04\\2pq = 0.32`

But this does not match with the frequency of heterozygotes given in the question i.e 0.25.

Hence, the statement is false