Question

In a population of mosquitos living in my backyard, I genotyped a representative sample of 100 at the buzz gene which has two alleles, B and b. Each copy of B makes the mosquito louder. Of my sample, 20 individuals are genotype BB (loud buzz), 40 are Bb (weak buzz) and 40 are bb (silent).

A) What is the frequency of the b allele?

B) What is the observed frequency of the Bb genotype?

C) What is the Hardy-Weinberg Equilibrium expected count of the Bb genotype?

Answer 1

Number of Individuals of BB genotype = 20
Number of Individuals of Bb genotype = 40
Number of Individuals of bb genotype = 40
Total Number of Individuals = 20 + 40 + 40 = 100

A)

Frequency of the b allele (q)
= No. of Individuals of bb Genotype + (1/2) * No. of individuals of Bb genotype / Total No. of Individuals
= (40 + 0.5 * 40)/100 = 60/100 = 0.6

B)

Observed Frequency of the Bb genotype = No. of individuals of Bb genotype/Total No. of Individuals
    = 40/100 = 0.4

C)

Hardy-Weinberg Expected Count of Bb genotype
= Hardy-Weinberg Expected Bb genotype Frequency * Total No. of individuals
= 2pq * 100 = 2 * (1 - q) * q * 100 = 2 * (1 - 0.6) * 0.6 * 100 = 2 * 0.4 * 0.6 * 100 = 48

The Expected count of Bb individuals is 48.