Question

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

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) =

Answer 1

Answer:

For the year 1987:

p² = frequency of homozygous dominant genotype (FF)= 661/ 1348 = 0.49 ......(1)

q² = frequency of homozygous recessive genotype (ff)= 121/ 1348 = 0.09 ........(2)

From eqtn (1), p² = 0.49; therefore p = 0.7.

From eqtn (2), q² = 0.09; therefore q = 0.3.

Hence,

p (frequency of normal allele in 1987) = 0.7

q (frequency of disease allele in 1987) = 0.3

For the year 2018:

p'² = frequency of homozygous dominant genotype (FF)= 782/ 1348 = 0.58 ......(1)

q'² = frequency of homozygous recessive genotype (ff)= 77/ 1348 = 0.057 ........(2)

From eqtn (1), p'² =0.58; therefore p' = 0.76.

From eqtn (2), q'² =  0.057; therefore q' = 0.24.

Hence,

p' (frequency of normal allele in 2018)  = 0.76.

q' (frequency of disease allele in 2018) = 0.24.