Derive an equation for allele frequecy due to selection at a

Derive an equation for allele frequecy due to selection at a haploid locus with three alleles at frequencies p,q, and r and fitnesses w1,w2,w3. Remember that in this case, p+q+r=1. Simplify as much as possible and point out similarities to the two-allele haploid equation. (Hint: you should be able to recover the two allele equation by setting r=0)

Solution

Let us take that we we have a gene locus with three distinct molecular alleles, \"X\", \"Y\" and \"Z\".

Let us assume that the population is in Hardy Weinberg equilibrium), and by including the additional frequency variable (r), the genotypic frequencies can be calculated as:

(p + q + r)²⁼ p² + 2pq + q² + 2pr + 2qr + r² = 1.0

The following genotypes are possible at this locus:

Let us consider the genotypes of YY and ZZ and let us take that there 1000 individuals in total.

Therefore, out of the total 1000 individuals sequenced, 200 were of YY genotype (0.2, or 20%) and 50 were of ZZ genotype (0.05, or 5% of the population).

The square root of each value to will yield q and r:

square root of q² = 0.44

square root of r² = 0.22

Putting the values of r and q in the eqution p+q+r= 1, we can find the value of p

If p + q + r = 1.0, then 1.0 - 0.44 - 0.22 = p, or 0.34

q = 0.45, r = 0.22, p = 0.33

Applying the values of each genotype into the expanded form of (p+q+r=1), we can calculate the relative genotype frequencies of a toatl population of 1000 individuals.

p² + 2pq + q² + 2pr + 2qr + r² = 1.0

=> 0.33² + 2(0.33)(0.45) + 0.45² + 2(0.33)(0.22) + 2(0.45)(0.22) + 0.22²

Multiplying each frequency with 1000, we can find out the number of individuals having that particular genotype.

Thus, from 1000 individuals, the number of individuals having a particular genotype can be deducted as: