In a population of 200 an allele Z has a frequency of 80 wha

In a population of 200, an allele Z has a frequency of 80%. what is the frequency of allele z? using the Hardy-Weinberg equation, estimate the numbers of homozygous dominant, heterozygous, and homozygous recessive genotypes.

Solution

Answer :

Answer:

Given p=80%,

q will be =20%

Since p+q=1

p=0.8, q=0.2

Hardy weinberg equation

p^2 +2pq+q^2=1

p^2 = number of homozygous dominants

q^2=number of homozygous recessive

2pq=number of heterozygotes

p^2=0.8 x 0.8 = 0.64

q^2 = 0.2 x 0.2 =0.04

2pq= 2 x 0.8 x 0.2 = 0.32

Among 200 individuals

64% are p^2, that means 200 x 0.64 =128 are homozygous dominant

(2) 200 x 0.04 = 8 are homozygous recessive

(3) 200 x 0.32 = 64 are heterozygotes