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

False

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-

Now as per Hardy Weinberg, second equilibrium equation -

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

Hence, the statement is false

i need an image of the diagram

substitution is the answer in this case