In a population of cats, the phenotypic frequency of black cats is 91%, and the phenotypic frequency of white cats is 9%. assuming that black (b) is dominant and white (b) is recessive, identify the appropriate frequency for each of the following for a population in hardy-weinberg equilibrium. match each frequency with the correct value.1. allele frequency of b.
2. allele frequency of b.
3. genotype frequency of bb.
4. genotype frequency of bb

1. Allele frequency of b = 0.09 (or 9%)

2. Allele frequency of B = 0.91 (0.91%)

3. Genotype frequency of BB = 0.8281 (or 82.81%)

4. Genotype frequency of Bb = 0.1638 (or 16.38%)

Explanation:

Given that:

p = the frequency of the dominant allele (represented here by B)  = 0.91

q = the frequency of the recessive allele (represented here by b)  = 0.09

For a population in genetic equilibrium:

p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)

(p + q)^2 = 1

Therefore:

p^2 + 2pq + q^2 = 1

in which:  

p^2 = frequency of BB (homozygous dominant)

2pq = frequency of Bb (heterozygous)

q^2 = frequency of bb (homozygous recessive)

p^2 = 0.91^2 = 0.8281

2pq = 2(0.91)(0.9) = 0.1638

answer: - b cells

explanation: -

so, the b cells mediate their action by identifying the antigens present on the surface of the pathogens and then producing antibodies in return to mediate its effect.