Question

In a population of rabbits, there are 496 black rabbits and 27 white rabbits. Fur color is determined by a pair of alleles where (B) is the dominant allele which produces black fur and (b) the recessive allele which produces white fur. The frequency of the dominant allele for black fur (B) is 0.8. What is the frequency of the black fur phenotype in the population (the population is in Hardy-Weinberg Equilibrium)? Round your answer to the nearest hundredth. (Hint: remember the numeric format of frequency).

Answer 1

Answer:

$0.96$

Explanation:

It is given that

B is the dominant allele which represents the black color

and b is the recessive allele which represents the white fur.

B being dominant will result into black color fur for genotype "Bb"

Given -

Frequency of black fur allele (p) is $0.8$

As per Hardy Weinberg's  first law of equilibrium

$p + q = 1$

Substituting the value of p in above equation, we get -

$q = 1-p\\q = 1-0.8\\q= 0.2$

q represents the frequency for white fur allele

Frequency of white fur phenotype is

$q^2\\= 0.2^2\\= 0.04$

Frequency of homozygous black fur phenotype (BB) is

$p^2\\= 0.8^2\\= 0.64$

As per Hardy Weinberg's second law of equilibrium -

$p^2 + q^2 + 2pq = 1\\0.64 + 0.04 + 2pq = 1\\2pq = 1 - 0.68\\2pq = 0.32$

Combined frequency of homozygous and heterozygous black fur phenotype is

$0.64 + 0.32\\= 0.96$