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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

Respuesta :

meerkat18
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:

Frequency of black allele is [tex]0.4\\[/tex]

Frequency of brown allele is [tex]0.6\\[/tex]

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 [tex]q^2\\[/tex]

Frequency of AA is represented by [tex]p^2\\[/tex]

Frequency of AA is represented by [tex]pq\\[/tex]

And the mathematical equation is as follows -

[tex]p^ 2 + q^ 2 + 2pq = 1\\[/tex]

Substituting the given values in above equation we get

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

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

[tex]p + q = 1\\p + 0.6 = 1\\p = 0.4\\[/tex]

Frequency of black allele is [tex]0.4\\[/tex]

Frequency of brown allele is [tex]0.6\\[/tex]