Cystic Fibrosis is an autosomal recessive disease which usually leads to death i
ID: 191381 • 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) 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
1) Relative fitness is defined as the survival or reproductive rate of a genotype relative to the maximum survival or reproductive rate of other genotypes in the population. Cystic fibrosis is an autosomal recessive disease. So, the relative fitness of the genotypes FF and Ff are both 1.
2) In the year 1987, the total number of people are 1348.
Among them, the number of persons with normal allele are (661 + (566/2)) = 945
So, the frequency of normal allele in 1987 = 945/1348 = 0.7
The frequency of diseased allele in 1987 = (121 + (566/2)) = 405/1348 = 0.3
In the year 2018, the total number of people are 1348.
Among them, the persons with normal allele are (782 + (489/2)) = 1027
So, the frequency of normal allele = 1027/1348 = 0.762
The frequency of diseased allele in 2018 = (77 + (489/2)) = 0.238
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.