Cystic Fibrosis is an autosomal recessive disease which usually leads to death i
ID: 191167 • Letter: C
Question
Cystic Fibrosis is an autosomal recessive disease which usually leads to death in the early to mid-20s. Carriers are not affected. The normal allele is F and the disease allele is f. You want to know how strong selection is against the disease allele, and set out to find its selection coefficient (s).
What are the relative fitness of the normal genotypes? (0.5 pt. each, no calculation needed)
FF =
Ff =
You find the following data regarding the number of people affected by CF 30 years ago, and today, from genetic tests of US hospital patients. In both years, the sample size was of 1348 individuals chosen at random from the population. It is known that that the mean relative fitness of the 1987 population was 0.919.
Distribution
FF
Ff
ff
1987
661
566
121
2018
782
489
77
find the relative fitness of the disease genotype, and the selection coefficient of the disease allele. Show your work and round results to 2 significant figures :
Distribution
FF
Ff
ff
1987
661
566
121
2018
782
489
77
Explanation / Answer
Relative fitness of a genotype is calculated by dividing survival rate of that genotype by highest reproductive or survival rate. It is denoted by letter with. In the given case we have to find w of ff genotype. So, wff=121/661= 0.18.
Selection coefficient is the measure of relative strength of selection against a genotype. It is denoted by letter s. It is calculated by Subtracting w from 1 ie s=1-w.
Therefore, s for ff in 1987 was 1-0.18=0.82. It means that this genotype is strongly selected against ie 82% of such individuals die soon.
Wff for 2018 =77/782=0.098
Therefore, sff=1-0.098=0.90 this figure suggests that ff genotype survival rate has decreased dramatically over the years as it has been selected against.
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