Genotype A2A2 is lethal and yet a population has an equilibr

Genotype A₂A₂ is lethal, and yet a population has an equilibrium frequency for A₂ of 0.4. If the fitness of A₁A₂ is 1.0, what is the fitness of the A₁A₁genotype, and how do you know?

Solution

A we know that frequency of (A1A1) = p2 = p × p = 0.6 × 0.6 = 0.36, or 36% of the population will be A1A1. Similiraly frequency of (A1A2) = 2 × p × q = 2 × 0.6 × 0.4 = 0.48, or 48% of the population will be A1A2 on the same hand frequency of (A2A2) = q2 = 0.4 × 0.4 = 0.16, or 16% of the population will be A2A2. Note -: that the genotype frequencies add to 1: p2 + 2pq + q2 = 1 E The numbers of individuals of each genotype that are expected in the population can be calculated by multiplying the genotype frequencies by the population size, N. Number of A1A1 individuals = p2 × N Number of A1A2 individuals = 2pq × N Equation 3 Number of A2A2 individuals = q2 × N If our population consists of 400 individuals, for example, 0.36 × 400 = 144 individuals are expected to be A1A1, 0.48 × 400 = 192 individuals are expected to be A1A2, and 0.16 × 400 = 64 individuals are expected to be A2A2.