Question

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

Answer 1

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),
q² - for bb genotype (recessive homozygote with brown hair).


It is given:
p² = 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