Question

In a population where skin color gene can be determined by 4 alleles B(black)=0.412561 o(olive) = 0.312450 y(yellow) = 0.112689 w(white) =0.123540. The gene for skin type can be determined by 3 alleles N(normal)=0.425789 d(dry)= 0.345689 x(oily)=0.214130. The gene for skin elasticity can be determined by 2 alleles E(elastic)= 0.456820 and e(non-elastic)= 500123. Assume Hardy-Weinberg equlibrium, what is the frequency of getting yellow, normal, elastic skin? (p² + 2pq + q² + 2pr + 2qr + r² = 1). ?

Answer 1

According to Hardy weinberg equilibrium, genetic variation remain constant from one generation to another in absence of any disturbing factor. Here forces like mutation, natural selection, gene flow ,non random matings are considered as disturbing factors. These disturbing factors may bring changes in gene frequencies and thus responsible for genetic variation in population.

This genetic variation at equilibrium is calculated by Hardy weinberg equation. Which is expressed as:

p2+2pq+q2=1

To understand this let us consider two allele namely A and a. Now p is the frequency of allele 'A' and q is the frequency of allele 'a'. Further p2 express the frequency f homozygous genotype 'AA' and q2 for homozygous genotype 'aa'. '2pq' is the hetrozygous genotype 'Aa'.

Now some of allele frequencies for all allele at locus is considered to be 1.

so, p+q=1

similarly for three allele

p+q+r=1

p2+q2+r2+2pq+2pr+2qr=1

now for 4 allele p=0.412561,q=0.412561,r=0.312450 and s=0.123540

then p+q+r+s=1

p2+2pq+2pr+2ps+q2+2qr+2qs+r2+2rs+s2

0.17+0.257+0.092+0.135+0.097+0.07+0.102+0.013+0.037+0.02

r=0.112

so frequency of getting yellow is 0.112

.(p=q)²=p²+q²+2pq