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Answer :
Answer:
The total number of alleles in the population is 1000 x 2 = 2000.
Let's calculate the frequency of the "F" allele:
~The 480 heterozygous ferrets each carry one "F" allele, so there are 480 x 1 = 480 "F" alleles
~The 160 homozygous ferrets each carry two "F" alleles, so there are 160 x 2 = 320 "F" alleles
~The total number of "F" alleles in the population is 480 + 320 = 800
The frequency of the "F" allele (p) is therefore:
p = (number of "F" alleles / total number of alleles) = 800 / 2000 = 0.4
Let's calculate the frequency of the "f" allele:
~ The 360 ferrets with black feet are homozygous for the "f" allele, so there are 360 x 2 = 720 "f" alleles
~The 480 heterozygous ferrets each carry one "f" allele, so there are 480 x 1 = 480 "f" alleles
~The total number of "f" alleles in the population is 720 + 480 = 1200
The frequency of the "f" allele (q) is therefore:
q = (number of "f" alleles / total number of alleles) = 1200 / 2000 = 0.6
Therefore, the answer is (A) F = 0.4 and f = 0.6
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