In a population of rabbits, there are 496 black rabbits and 27 white rabbits. Fur color is determined by a pair of alleles where "B" is the dominant allele which produces black fur, and "b" the recessive allele which produces white fur. The frequency of the dominant allele for B is 0.8. What is the frequency of the black fur phenotype in the population (the population is in Hardy-Weinberg Equilibrium)? 

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Since the population is in Hardy-Weinberg equilibrium, the dominant allele displays complete dominance. This means that rabbits with black fur have a genotype of either BB or Bb. Since the numbers of both black and white rabbits are given, we can easily calculate the frequency of the black fur phenotype.

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Since the population is in Hardy-Weinberg equilibrium, the dominant allele displays complete dominance. This means that rabbits with black fur have a genotype of either BB or Bb. Since the numbers of both black and white rabbits are given, we can easily calculate the frequency of the black fur phenotype.

Black fur phenotype frequency = number of black rabbits/(number of black rabbits + number of white rabbits)

                                            = 496/(496 + 27)

                                            = 496/523

                                            = 0.95

The Hardy-Weinberg equation assumes that a population is infinitely large, which is not the case in this question. Therefore, the observed frequency of 0.95 may differ from the theoretical frequency. To calculate the theoretical frequency of the black fur phenotype, we simply use the frequency given for B.

Since there are only two alleles in this population, the sum of both allele frequencies equals one. To write this out mathematically: B + b = 1. Since we already know B, we must solve for b.

b = 1 – B

b = 1 – 0.8

b = 0.2

To calculate the theoretical frequency, we need to use the Hardy-Weinberg equation: B^2 + 2 Bb + b^2 = 1. As noted earlier, black rabbits will have a genotype of either BB or Bb. Therefore, we are solving the equation for B^2 + 2 Bb.

B^2 + 2 Bb + b^2 = 1

                – b^2 =  – b^2

       B^2 + 2 Bb = 1 – b^2

Since b = 0.2, the theoretical frequency of black fur will equal 1 – b^2.

Black phenotype = 1 – b^2

                        = 1 – (0.2)2

                        = 1 – 0.04

                        = 0.96

The theoretical frequency of black fur is close, but not quite equal to the observed frequency of 0.95.

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