A genetics researcher finds that a form of human albinism in a certain population is controlled by the recessive allele, a, for the absence of skin pigment and the completely do…

Question

asked Jul 11, 2019 19.8k views

2 Answers

July · Answer 1 · 2019-07-16T10:30:59+0000

Answer:

Option A

Percent of the population that has skin pigment but also carries the recessive allele is
$21\\$ percent

Step-by-step explanation:

As per hardy-weinberg equation,

$p^2+q^2+2pq = 1\\$ ------------- Eq(A)

where

$p^2= \\$ frequency of the homozygous genotype (dominant)

$q^2= \\$ frequency of the homozygous genotype (recessive)

$pq= \\$ frequency of the heterozygous genotype (recessive)

Also, sum of the allele frequencies at the locus is equal to one.

Thus,
$p + q = 1\\$ ------------- Eq(B)

Here frequency of recessive allele i.e
$q = 30\\$ %

Substituting this in equation B, we get

$p + 0.3 = 1\\p = 1-0.3\\p= 0.7\\$

frequency of dominant allele i.e
$q = 70\\$ %

Substituting the value of frequency of both dominant and recessive allele in equation A, we get

$(0.3^2) + (0.7^2) + 2pq = 1\\pq = (1-0.3^2-0.7^2)/(2) \\pq = (0.42)/(2) \\pq = 0.21\\$

Thus, percent of the population that has skin pigment but also carries the recessive allele is
$21\\$ percent