Use the data below to predict the outcome of selection in th

Use the data below to predict the outcome of selection in this population

A1 = 0.6

A2= 0.4

w11= 0.5

w12= 1.0

w22=0.75

a) calculate the average fitness for the population. b) calculate the genotype frequency among surviving adults. c) calculate the frequency of the A1 allele after selection. d) calculate the frequency of the A2 allele after selection. e) calculate the change in the frequency of the A1 allele after selection. f) based on your calculations, briefly describe the outcome of selection on the frequency alleles and genotypes in the population.

Solution

a)average fitness for the population:

A1 =p= 0.6 ; A1A1 = p2 = 0.36 ; w11 = 0.5

A2=q=0.4 ; A2A2 = q2=0.16 ; w22 = 0.75

A1A2= 2pq = 0.48 ;w12 = 1

Average fitness = W = p2w11 + 2pq w12 + q2w22

=0.36* 0.5 + 0.48* 1+ 0.16*0.75

= 0.18+0.48+0.12

=0.78 ans

b) frequency of A1A1 genotype = p2w11 / W = 0.18 / 0.78= 0.23 ans

frequency of A1A2 genotype = 2pq w12 /W = 0.36/0.78 = 0.46 ans

frequency of A2A2 genotype = q2w22 / W = 0.16/0.78 = 0.205 ans

c)Frequency of allele A1 after selection

p\' =  p2w11 / W + pq w12 /W

= 0.23 + (0.24 * 1 / 0.78)

= 0.23 + 0.31

=0.54 ans

d)Frequency of allele A2 after selection

q\' =  q2w22 / W + pq w12 /W

=0.205 + (0.24 * 1 / 0.78)

= 0.205 + 0.31

=0.51 ans

e)Change in frequency of A1 after selection

  p = pq / W [ p *( wAa -wAA) + q* (wAa -waa )

= 0.6 *0.4 / 0.78 [ 0.6 ( 1- 0.5 ) + 0.4 ( 1 -0.75 )

= 0.31 [ 0.3 + 0.7]

= 0.31 ans

f) The selection is relatively weak.Recessive allele will be favoured and the favoured recessive genotype will be rare.There will be a slow change in allele frequency.