Question

3) Imagine a mutation occurs which introduces an intron into the coding sequence of a gene. Relative to the intron-free wild type allele, the selection coefficient against this mutant allele with an intron will be approximately s = -2.5 x 10-8.   Now consider what happens if this mutation occurs in the yeast Saccharomyces cerevisiae which has an estimated effective population size of Ne = 1 x 107 versus the mutation occurring in a different species of yeast Kluyveromyces lactis. The estimated effective population size for K. lactis is Ne = 3 x 107, which is substantially larger than S. cerevisiae. Using this information, answer the following questions.

(a) What is the probability that a newly introduced, single copy of the mutant allele with the intron will go to fixation due to drift alone in these two species. (In other words, assume that s = 0 and then calculate the probability that the mutant allele will eventually reach 100% frequency in S. cerevisiae and then repeat this calculation for K. lactis.)

(b) Now consider the fact that s is actually not equal to 0, but slightly less than 0 as stated above. Using the equation presented in class calculate θ for both species. i.e. what is the fixation probability of this new mutant allele given drift and natural selection against it relative to probability of fixation probability under drift alone.

Please show all work

Answer 1

Answer
In a breeding population of variable size. Let us suppose the mutation give rise in the gene copy and produce different allele of a gene say A. Then probability for this mutation is
P= 1 – e-2s/1- e-4Ns
P = probability of the mutant gene in a population
s = Fixation of the mutant allele in the population
N = Number of individual in the population.

According to the first situation S=0 he probability of the mutant gene population in . Saccharomyces cerevisiae
a)   If we put s= 0 in a above value
Then,
P = 1- e0/1-e0
P= 1 I.e. 100% population covers with this mutant allele.
Same as calculate for S.lactis.