Question

1) The Himalayan colour pattern in rabbits is controlled by a single locus with three alleles; F for full colour, H for Himlalayan and A for albino. The Fallele is dominant over the H and A alleles while the H allele is dominant over the A allele. If the number of individuals with the Himalayan colour pattern represent 35% of the population and the albinos represent 17% of the population, what is the frequency of the F allele? (rounded to 3 decimal places)

Your Answer: should be 0.279

Please guide me thought the steps to get the correct answer . I posted this question many times, but no one provided me the correct answer .Please do provide me correct answer this time.

Hint for solving the question : You need to assume random mating to do this, there isn't any other way. The frequency of the F allele is 1 - f(H) - f(A). f(A) = SQRT(f(AA)) and f (H) = SQRT(f(HH)+f(HA)+f(AA)) - SQRT(f(AA)).

Answer 1

Given: F>H>A

F+H+A=1 (According to H-W Equilibrium), i.e F²+H²+A²+2FH+2HA+2FA=1......................Eq 1

Himalayan colour population is 35% or 0.35 {(H²+2HA) or f(H.H)+f(H.A)} ...............According to equation 1 and himalayan only dominate over albino colour.

i.e f(H.H)+f(H.A)=0.35

then. A² or f(A.A)=17% or 0.17

so A=f(A)= SQRT (0.17)=0.412

from hints: f (H) = SQRT(f(HH)+f(HA)+f(AA)) - SQRT(f(AA)).

we can put the values in this equation, f(H)= SQRT[0.35+0.17]- SQRT[0.17]

= SQRT[0.52]-SQRT[0.17]

f(H)=0.721-0.412 = 0.309

from these calculations we knew that, f(H)=0.309, f(A)=0.412 and from these we can find f(F)=?

so, f(F) or frequency of F is

f(F)+f(H)+f(A)=1

f(F)+0.309+0.412=1

f(F)= 1- (0.309+0.412)= 1- 0.721

f(F)= 0.279

The frequency of F allele is 0.279.