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.

prophase: we can observe the chromosomes to coil up and condense. they become visible. nuclear membrane and envelope disintigrate.s

metaphase: chromosomes align at the center, the metaphase plate. spindle fibers attach to centromeres of the sister chromatids. they meet at kinetochores.

anaphase: sister chromatids are pulled towards each opposing centromere, set on different poles.

telophase: chromosomes are now clustered up only on either sides of the cell. nuclear membranes reform, and a new cell wall forms along the cell plate. this will lead to cytokinesis, the process that divides the cell membrane and forms the cell wall.

the answer is 1. photosynthesis. glad to