Question

Consider the table below giving the numbers of juveniles and adults for one generation.

Calculate fitnesses (viabilities), relative fitnesses, the selection coefficient, and the heterozygous

effect. What type of selection is this? What will the allele frequencies be for the adults in the

next two generations assuming random mating? What will the frequency of the A1 allele be at

equilibrium?

Genotype: A1A1 A1A2 A2A2

Juveniles: 795 267 138

Adults: 596 242 83

Answer 1

Genotype: A1A1 A1A2 A2A2

Juveniles: 795 267 138

Adults: 596 242 83

Viability:
A1A1 = (596/795)X100 = 75%
A1A2 = (242/267)X100 = 90%
A1A1 = (83/138)X100 = 60%
The heterozygotes have maximum fitness, So we will compare it with others.

Relative fitness(w) :
A1A1 = 75/90 = 0.84
A1A2 = 90/90 = 1
A2A2 = 60/90 = 0.66

Selection coefficeint (s):
S = 1-w
A1A1 = 1 - 0.84 = 0.16 (16% were eliminated)
A1A2 = 1-1 = 0   (no member was eliminated)
A2A2 = 1-0.66 = 0.33   (33% were eliminated)

This is the case of heterozygous advantage. The heterozygous has more survival chances than the homozygotes.

Allele frequency in next generation:
The alleles are getting selected and with each generation, the frequencies will change. First we need the allele frequency from the given genotypic numbers.
Total number of allels = total of all adults * 2 = 1842

Let p = allele frequency of A1A1 = 2*596 + 242 = 1434/1842= 0.78
Let q = allele frequency of A2A2 = 2*83+242 = 408 / 1842 = 0.22

genotype frequency of A1A1 = p^2 = 0.60
2pq = A1A2 = 0.06
q^2= A2A2 = 0.05

Mean fitness = p^2W11 + 2pqW12 + q^2W22
(W_) = 0.60*0.84 + 0.06*1 + 0.05*0.66
      = 0.5 + 0.06+ 0.033
      = 0.593

Allele frequency in next generation =
p' = [p^2W11+ 2pq/2] / W_
   = [0.5+0.03]/0.593
   = 0.89
q = [q^2W22+ 2pq/2] / W_
= [0.03 + 0.06] / 0.593
= 0.15.

The frequency of A1 at equilibriu m = 0.9