Question

A proportion of the individuals in a given population exhibit a deleterious trait that is controlled by a single gene. We know that this gene has only two allelic variants segregating and one of them is deleterious in the homozygous condition. Heterozygous individuals and individuals homozygous for the other variant are indistinguishable from each other and do not exhibit the trait. The population consists of 345 normal individuals and 155 affected individuals. The selection coefficient for the recessive homozygote is 0.1.

a. Assuming Hardy-Weinberg equilibrium, what are the allelic frequencies in this population?

b. What kind of selection pressure is being exerted on this gene and how will the allelic frequencies change over time (you can also use a figure to illustrate)?

c. We assumed that this population is in Hardy-Weinberg equilibrium. Do you believe this is an appropriate assumption for this scenario? Justify

Answer 1

Ans: As given in the question that the deleterious trait is controlled by the allele of a single gene and this allele is recessive which control this trait and that is why homozygous recessive individuals are the ones who are affected and homozygous dominant and heterozygous are not affected and are indistinguishable from each other phenotypically.

Now for homozygous recessive the selection coefficeint is 0.1 and selection coefficient is equal to 1 - relative fitness, so for homozygous recessive individuals their relative fitness is equal to 0.9.

Now relative fitness for homozygous recessive is the ratio of individuals with homozygous recessive genotype and homozygous dominant genotype individuals and it is given that the affected individuals are 155 in number and lets suppose that homozyous dominant individuals are equal to X so putting the values in the equation of relative fitness we get 0.9 = (155/X) and X is approximately equal to 172.

So individuals with homozygous dominant genotype is equal to 172, individuals with heterozygous genotype is equal to 345-172 which is 173 and individuals with homozygous recessive genotype is 155.

1) Now we are assuming that the population is in Hardy Weinberg equilibrium which means that this population follows p2+q2+2pq=1 where p2 is the frequency of homozygous dominant q2 is the frequency of homozygous recessive and 2pq is the frequency of heterozygotes and p is the frequency of Dominant allele and q is the frequency of recessive allele.

So p2 = 172/500 = 0.344 and p is equal to 0.586 and similarly q2 is equal to 155/500 = 0.31 and q is equal to 0.557.

Hence the frequency of dominant allele is 0.586 and frequency of recessive allele is 0.557.

2) Since the trait is deleterious the selection pressure which exerted on this gene is natural selection and natural selection is a negative selection in which selective removal of deleterious allele takes place. This can result in stabilizing selection through the purging of deleterious genetic polymorphisms.

Now as for the allelic frequencies, the carrier of the deleterious trait have fewer offspring each generation, reducing the frequency of this trait in the gene pool. So the frequency of recessive allele will decrease over time and frequency of dominant allele will increase over time.

3) If a population is in Hardy Weinberg equilibrium than in this population there should be no mutation, random mating, no gene flow, infinite population size, and no selection. But in the question it is given that one of the allele is deleterious allele hence it will be selected out over time which is not a key feature of Hardy Weinberg equilibrium. Hence assuming that this population is in Hardy Weinberg equilibrium is not appropriate for this scenario.