A certain form of congenital glaucoma is caused by an autosomal recessive allele

Question

A certain form of congenital glaucoma is caused by an autosomal recessive allele. Assume that the mutation rate is 10-5 and that people with this condition produce, on average, only about 80% of the offspring produced by people who do not have glaucoma. At equilibrium between mutation and selection, what will the frequency of the gene for congenital glaucoma be? (two significant digits) Number What will the frequency of the disease be in a randomly mating population that is at equilibrium? (two significant digits) Number

Explanation / Answer

Firstly we look at some definitions:

In genetics, fitness (W) is the ability to survive or reproduce.

Selection coefficient (s) is the measure of any changes in the fitness.

Fitness of people with glaucoma (W) = 0.8 ( given in the question that 80% offsprings produced)

Selection coefficient (s) = 1-W = 1-0.8 = 0.2

According to Hardy-Weinberg's law,

p2 + q2 +2 pq= 1

where q2 is the genotype frequency of recessive allele.

Mutation rate (m) is given as 10-5

q2 = m/s = 10-5 /0.2 = 0.00005

Thus the frequency of gene for congenital glaucoma,

q = 0.0071

Now looking at the second part of the question

The frequency of the disease will be given by the frequency of homozygotes( identical alleles) .

This is given by q2, which is equal to 0.00005

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