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

Given:

p + q = 1 ; p² + 2pq + q² = 1

let B substitute p ; let b substitute q.

B + b = 1 ; B² + 2Bb + b² = 1

B² = 16% or 0.16

B = √0.16 = 0.40

B + b = 1
0.40 + b = 1
b = 1 - 0.40
b = 0.60

answer is:

C) p = 0.4 ; q = 0.6

Answer Link

podgorica podgorica · Answer 2 · 2019-05-20T10:12:49-04:00

Answer:

Frequency of black allele is \(0.4\\\)

Frequency of brown allele is \(0.6\\\)

Explanation:

As per Hardy-Weinberg equation,

Frequency of the homozygous genotype is represented by AA and aa

Frequency of the heterozygous genotype is represented by Aa

In the Hardy-Weinberg mathematical equation

Frequency of AA is represented by \(q^2\\\)

Frequency of AA is represented by \(p^2\\\)

Frequency of AA is represented by \(pq\\\)

And the mathematical equation is as follows -

\(p^ 2 + q^ 2 + 2pq = 1\\\)

Substituting the given values in above equation we get

\(0.16 + q^2 + 2 (0.24) = 1\\q^2 = 1 - 0.16 - 2 (0.24)\\q = \sqrt{0.36} \\q = 0.6\\\)

Also, sum of the allele frequencies for all the alleles at the locus is equal to 1 which means

\(p + q = 1\\p + 0.6 = 1\\p = 0.4\\\)

Frequency of black allele is \(0.4\\\)

Frequency of brown allele is \(0.6\\\)