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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

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Answer:

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

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