2. In a population of haploid, asexual yeast living in your left ear, a benefici

Question

2. In a population of haploid, asexual yeast living in your left ear, a beneficial mutation (W) arises that allows mutant individuals to consume your ear wax a higher rate that those lacking the mutation. Because these yeasts are asexual and haploid, we can assume that the two alleles, W and w, represent the genotypes of the population. If the mutant genotype (W) is able to reproduce asexually at twice the rate of the non-mutant genotype (w), what is the selection coefficient of each genotype? (W = 4 and w = 2)

a) If the population at time t=0 is composed of 98 w individuals and 2 W individuals, calculate the following for the population at time t=1: change in genotype numbers, change in genotype frequencies and q (recall: q always represents the inferior genotype).

b) Using of the entire population as a measure of the population's fitness, carry out the calculations you performed above for a total of 10 generations (t=10). How has changed between t=1 and t=10? What does this mean for the population in regards to their environment?

c) Ignoring the number of yeast cells that have died, what is the estimated size of the yeast population in your ear at time t=10?

Explanation / Answer

Two alleles, W and w, represent the genotypes of the population. Selection coefficient of each genotype, (W = 4 and w = 2).

The selection coefficient (s) is a measure of the relative fitness of a phenotype i population genetics. It is comparison of one phenotype to another favored phenotype. In this context the W genotype is carrying fitness advantage. Here w = 2W, the double reproduction rate.

Generally s = 1 - (WXX / WXx) ,

Here for W genotype,

s = 1- (WW / WW ) = 1 - (1/ 1) = 1. 100 % survival fitness,

In case of w phenotype

s = 1- (Ww / WW ) = 1 - (2/ 4) = 0.5, 50% survival fitness.

a) Parental generation, w = 98, W = 2, total 100 individuals.

Genotype number w 98(t=0). While W reproduces asxually in double rate of the w. Here we can assume that population is mentioned as in p value of 9.8 for w allele, and q value for W allele is 0.2.

When they are freely mix with each other with allele in the same proportion. Probability of w is meeting with an w is 9.8 * 9.8 = 96.04,

The probability of w meeting with a W is 9.8 * 0.2 = 1.96, as W generates double the rate of w, so here the number will be 1.96*2= 3.92,

Also the probability of W meeting W is 0.2 *0.2 = 0.04, for the double generation, 0.04*2 = 0.08.

Therefore in the following generation, we would expect to have the following proportion of genotypes:

96.04 ww, 3.92 wW,  0.08 WW.

If there are 10000 offspring’s, there will be 9604 ww individuals, 392 wW individuals and 8 WW individuals.

For 100 offspring’s, there will be 96 ww individuals, 3 wW individuals and 1 WW individual after time t1.

As per the Hardy Weinberg equation,

The frequency of w alleles will be p2 + pq = 96 + 2= 98,

The frequency of W alleles will be q2 + pq = 0 + 2 = 2.

Altogether p + q makes 1, or 100, as per the Hardy Weinberg Equation.

Finally here inferior genotype is 0-1.

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.