Question

A population starts out 10% BB and 90% Bb. The population size is large, mating is random, there are no new mutations, and no migrations. B is dominant.

Let t = the allele frequency of B and d = the allele frequency of b. Calculate t and d.

What would you predict would be the frequency of individuals with the recessive phenotype many generations later with no selective advantage?

What would you predict would to be the equilibrium frequency of t and d many generations later if the recessive allele had a selective advantage?

Answer 1

Given,

The population size is large and mating is random. With no mutations and migrations are ideal conditions for a hardy Weinberg equilibrium model to work. In these conditions the genotypic and gene frequencies remain constant.

Population is said to start with 10% BB and 90% Bb (given in question).

Below is the hardy Weinberg equation for genotypic frequencies.

t2(BB)+2tb(Bb)+d2(bb)= 1

Frequency of allele B is t and that of b is d.

(P² is frequency of genotpe BB and so as given in brackets.)

d²=0 as there no individuals with this genotypes at start.

t²=10%=0.1

t=√0.1=0.3162

And 2tb=90%=0.9

Solving this for b

We get b=1.4321.

With no selective advantage and no mutations the individual genotypes frequencies remain same.

Individual phenotypic frequencies depend on the genotpyes . The recessive phenotype being expressed only in a homozygous allelic state. The frequency is too low or is not expressed.

Recessive allele having selective advantage increase the frequency d and slowly with generations it could shift the frequencies toward the higher d.