Question

The trait for 'black feet in ferrets' is a recessive trait encoded for by "f". White feet is encoded for by a dominant allele encoded for by the letter "F".

An ecological study in ferret country conducted by Mrs. Wright found that out of 1000 ferrets:


360 had black feet

480 had white feet and were heterozygous

160 had white feet and were homozygous.


Using this information find the allele frequencies for F and f.


Remember "F" = p and "f" = q




a

F = 0. 4 and f =0. 6

b

F = 0. 7 and f = 1

c

F = 1 and f =0. 7

d

F = 0. 7 and f =0. 3

Answer 1

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