Cystic Fibrosis is an autosomal recessive disease which usually leads to death i

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) 1. What are the relative fitness of the normal genotypes? (0.5 pt. each, no calculation needed) 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 1987 2018 Ff 566 489 661 121 782 2. From these data, calculate the following quantities. Show you work, and round results to 2 significant figures (0.5 pt. each). Hint: you don't need the Hardy-Weinberg formula here p (frequency of normal allele in 1987 q (frequency of disease allele in 1987)- p, (frequency of normal allele in 2018)- q' (frequency of disease allele in 2018) - Do your p and q add up to 1? Do p' and q' add up to 1? If the answer is no, check your calculations! Use the two equations below and the data above to 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 (0.5 pt. each) 3.

Explanation / Answer

A1) relative fitness would be 100% for both the cases since the disease is autosomal recessive.

A2) p(1987) = ((661+283)/1348)x100 = 70.02% or 0.7002

q(1987) = 29.97% or 0.2997

p'= 76.14% or 0,7614

q'= 23.85% or 0.2385

this is all i know

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