Question

If 72 out of 200 individuals express the recessive phenotype, what percentage will be homozygous dominant? Please Help!

Answer 1

If the recessive phenotype has the genotype nn, then the frequency of nn in the population is 72/200 = 0.37. If the frequency of nn = 0.37, then the frequency of n is sqrt(0.37)=0.6083. According to the Hardy-Weinberg equation #1, the frequency of the dominant allele, N, is 1 - 0.6083 = 0.3917. Therefore, the frequency of the dominant homozygous, NN = 0.3917↑2 = 0.15342889 or 15.342889%

Answer 2

Answer:

16%

Explanation:

If the population is in Hardy Weinberg equilibrium:

p + q = 1

p² + 2pq + q² = 1

where, p= frequency of dominant allele

q= frequency of recessive allele

p² = frequency of homozygous dominant population

q² = frequency of homozygous recessive population

2pq= frequency of heterozygous dominant population

Given, q²= 72/200 = 0.36

q= √0.36 = 0.6

Hence, p = 1-q

= 0.4

p² = 0.16 = 16%

Hence, 16% of population will be homozygous dominant.