Question

In a randomly mating laboratory population of Droso- phila, 4 percent of the flies have black bodies (encoded by the autosomal recessive b), and 96 percent have brown bodies (the wild type, encoded by B). If this pop- ulation is assumed to be in Hardy–Weinberg equilibri- um, what are the allele frequencies of B and b and the genotypic frequencies of B/B and B/b?

Answer 1

The frequencies are:

B=80%

b=20%

BB=64%

bb=4%

B/b=32%

Step-by-step explanation:

The Hardy-Weinberg equation to find the frequency of alleles is: p^{2}+2pq+q^{2} = 1 and we know as well that p+q=1.

p representes the B allele

q represents the b allele

p^{2} represents BB

q^{2} represents bb and

pq represents Bb.

From the problem we know that 4% of the population is bb, so we know that q^{2}=4%=0.04. By getting the square root of q^{2} we obtain the value of q, q=0.2=20%.

We know p+q=1. That means p=1-q=1-0.2=0.8. The frequency of B is 80%. The value of p^{2}=(0.8)(0.8)=0.64, the frequency of BB is 64%.

The Value of Aa=2pq=2(0.8)(0.2)=0.32, the frequency of Bb is 32%.

If we substitute the values in the original formula we have: p^{2}+2pq+q^{2} = 0.64+0.32+0.04=1.