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?

Table:

Genotype:	A1A1	A1A2	A2A2
Juveniles:	795	267	138
Adults:	596	242	83

Answer 1

Part A.

Absolute fitness (W) is the change in abundance of a genotype over one generation attributable to selection. Its symbol is ‘W’. This is also known as survival rate.

W = N_(t+1) / N_(t) = number of juveniles divided by number of adults

So, absolute fitness of A1A1 = (795/596) = 1.33

Absolute fitness of A1A2 = (267/242) = 1.1

Absolute fitness of A2A2 = (138/83) = 1.66

Part B.

Relative fitness (w) is survival rate of the genotype divided by maximum survival rate.

Relative fitness of A1A1 = (1.33/1.66) = 0.80

Relative fitness of A1A2 = (1.1/1.66) = 0.66

Relative fitness of A2A2 = (1.66/1.66) = 1

Part C.

Selection coefficient (s) is given by the formula s = (1-w)

Note that if the relative fitness of the fittest genotype (A2A2) is 1, then we define the selection coefficient to be the difference between that and the other homozygote.

“s” = (1 - 0.80) = 0.20

Part D.

Heterozygous effect (h) is defined as the fitness of the heterozygote.

“h*s” = (1 - fitness of heterozygote) = (1 - 0.66) = 0.34

“h” = (0.34 / 0.20) = 1.7

Part E.

Note that,

• (h < 0) in case of over dominance;
• (0 < h < 1) in case of directional selection;
• (h >1) in case of under-dominance.

So, this is a case of under-dominance.

**Note: Please post rest of the parts separately.