Suppose the eggtoadult survival of A1A1 is 80 percent as gre

Suppose the egg-to-adult survival of A1A1 is 80 percent as great as that of A1A2, and the survival of A2A2 is 95% as great.

What is the frequency (p) of A1 at equilibrium?

What are the genotype frequencies among zygotes at equilibrium?

Now suppose the population has reached this equilibrium, but that the environment changes so that the relative survival rates of A1A1, A1A2, and A2A2 become 1.0, 0.95, and 0.90 respectively.

What will the frequency of A1 be after one generation in the new environment?

Solution

A₁A₁ – 0.8 – W₁₁-survival rates or fitness

A₁A₂ – 1 – W₁₂ - survival rates or fitness

A₂A₂ – 0.95 – W₂₂ - survival rates or fitness

According to the Hardy-Weinberg equilibrium,

p² W₁₁ +2pq W₁₂ + q² W₂₂ = W (average fitness)

0.8p² +2pq + 0.95q² = average fitness

p² +2pq + q² = 1 is the equation for genotypic frequency

Selection coefficient of A₁A₁ genotype is 1-0.8 = 0.2

Selection coefficient of A₁A₂ genotype is 1-1=0

Selection coefficient of A₂A₂ genotype is 1-0.95=0.05

There is no selection against A₁A₂ genotype

The selection against A₁A₁ genotype is stronger than that of the A₂A₂ genotype

It can be predicted that A₁ allele will reduce in frequency over time in case there is a steady relationship between genotype and fitness (if selection coefficient is constant).

1. From the data that is provided in the question, A₁ frequency can be said to be lower than that of A₂. But, it cannot be determined from the available data.

2. It is same with genotypic frequencies.

3.If the survival rate for A1A1 is 1, A₂A₂ is 0.9 and that of A₁A₂ is 0.95,

Selection coefficient, s for A₁A₁ will be 1-1=0, A₂A₂ = 0.1, and for A₁A₂ = 0.05

It means that no selection happens against A₁A₁ genotype

Selection against A₂A₂ is stronger than A₁A₂ .

So, it can be predicted that gradually A₂ allele frequency will lower.

4.The frequency of A₁ after one generation in new environment will be the same.