Question

Consider the shmoo (see image below). In a large population of shmoos, there is a gene with two alleles (B and b). The ‘B’ allele is the wild type allele and individuals exhibit a fairly normal shmoo-type phenotype. Individuals with two copies of the ‘b’ allele can, under some conditions, spontaneously combust.

                                                                          

After extensive (and very careful) research it was determined that the frequency of the “combustable” shmoos was 4 individual per 100 shmoos.

a. Given that there is strong directional selection against the ‘b’ allele, would you expect the ‘b’ allele to persist in this population? Why or why not? (Assume that the relative fitnesses are W_BB = 1.0, W_Bb = 1.0, and W_bb = 0.5)

b. It is also possible that the three genotypes (BB, Bb, bb) have unique relative fitnesses: (W_BB = 1.0, W_Bb = 0.75, and W_bb = 0.5). Would you expect the ‘b’ allele to persist in this population? Why or why not?

Answer 1

Genotype frequency of combustabe (bb) phenotype=q2=4/100=0.04

Allele frequency of b allele=q=√0.04= 0.2

Allele frequency of B allele=p= 1-0.2=0.8

p2=(0.8)^2=0.64

2pq=2x0.8x0.2=0.32

A) p2(WBB)+(2pqWBb)+q2(Wbb)=W*

W*=(0.64x1)+(2x0.8x0.2x1)+0.04x0.5=0.98

b allele frequency after selection=q'=[(pq WBb) +(q2Wbb)]/W*

q'=(0.2x0.8x1)+0.04x0.5/0.98=0.18

p'=(0.2x0.8x1)+0.64x1/0.98=0.82

Heterozygotes freq after selection=2p'q'=2x 0.18x 0.82=0.29

As shown b allele frequency exist in this population after selection in both heterozygotes and homozygous recessive.

B)

p2(WBB)+2pq(WBb)+q2(Wbb)=W*

W*= mean finess=(0.64x1)+(2x0.8x0.2x0.75)+(0.04x0.5)=0.9

b allele frequency after selection=q'=[(pqxWBb) +q²Wbb]/W*

q'=[(0.2x0.8x0.75)+(0.04x0.5)]/0.9=0.155

p'=(0.8x0.2x0.75)+(0.64x1)/0.9=0.844

Hetetozygote frequency =2p'q'=2x0.844x0.155=0.26

This b allele persist in population in both recessive homozygous and heterozygotes.