In a population of 500 snarks 300 individuals are homozygous

In a population of 500 snarks, 300 individuals are homozygous dominant at the purple-spotted locus (SS), 150 are heterozygous (Ss), and 50 have no spots (ss). Determine the allelic frequencies.

a.) p=0.15, q=0.50

b.) p=0.75, q=0.25

c.) p=0.68, q=0.32

d.) p=0.5, q=0.5

e.) p=0.82, q=0.18

Solution

If gene frequency of dominant allele \'S\' is represented by P and gene frequency of recessive allele \'s\' is represenred by q.

Then gene frequency of allele S = p = f(S) = (2 × 300) + 150/ 2 × 500 = 600 + 150/ 1000 = 750 / 1000 = 0.75.

Gene frequency of s = q = f (s) = (2 × 50 ) + 150 / 2 × 500

= 100 + 150 / 1000 = 250 / 1000 = 0.25

Answer = b.) p=0.75, q=0.25