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?

The frequencies are:

B=80%

b=20%

BB=64%

bb=4%

B/b=32%

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.

both male and female gametes are produced by meiosis of the germ cells. therefore, both male and female gametes are haploid. one male and one female gamete unite during the fertilization to form the diploid zygote. in seed bearing plants, pollen grain is the male gametophyte, which produces sperm cells.