A population of mice has black or brown hair. The gene for black hair color (B) is dominant over brown (b). If 16% of the mice are homozygous black and 24% are heterozygous black, what would be the frequency of each allele?

(p + q = 1, p2 + 2pq + q2 = 1)

is it (A. p=0.65, q=0.35 (B. p=0.25, q=0.75 (C. p=0.4, q=0.6

The answer is C. p=0.4, q=0.6.

If:

p - the frequency of dominant allele B,

q - the frequency of recessive allele b,

the frequencies of the genotypes are:

p² - for BB genotype (dominant homozygote with black hair),

2pq - for Bb genotype (heterozygote with black hair),

q ² - for bb genotype (recessive homozygote with brown hair).

It is given:

p ² = 16% = 0.16

2pq = 24% = 0.24

p = ?

q = ?

p = √p² = √0.16 = 0.4

Using the formula

p + q = 1

⇒ q = 1 - p = 1 - 0.4 = 0.6.

Therefore, the frequencies of the alleles are:

p = 0.4

q = 0.6