Question

Cystic fibrosis is an autosomal recessive disorder caused by a loss-of-function mutation in a chlorine channel, which causes a thickening of the mucus, leading to respiratory failure. This disease affects approximately 1 in 1353 newborns in the Republic of Ireland every year.

Fill in the following values:

p =

q =

Current population of Ireland is approximately 4.8 million. Based on the allele frequencies that you calculated, what is the expected number of cystic fibrosis carriers (heterozygotes) in Ireland? Show your work.

What is the expected number of healthy individuals in the population? Show your work.

Reminder: only homozygous recessive individuals are affected by the disease.

You do some research and find that a census of Ireland in the 1910s found 65 individuals afflicted with cystic fibrosis out of 100,000 people. You want to investigate whether the gene involved in cystic fibrosis is under Hardy-Weinberg equilibrium. To do this, you use the frequencies observed in the present to determine the expected number of individuals in each class in the 1910 study.

Fill in the following table:

	CC	Cc	cc	Total
Expected number of individuals (i.e. number of individuals in the present dataset)
Observed number of individuals in 1910			65
(0-E)^2/E				?2 =

Use the ?² table provided below to infer the p-value of the test

Reminder: the p value is found by cross referencing the ?² value with the critical value. For example, if your ?² value is 5.34 and the degree of freedom is 1, you would say 0.05 > p-value > 0.01.

Can you conclude that the gene involved in Cystic fibrosis is in Hardy-Weinberg Equilibrium with 95% certainty? Explain why or why not in a sentence or two.

Please explain all.

Answer 1

1.The cystic fibrosis is an autosomal recessive disease.

So, q2 = 1/1353 = 0.00074

q = 0.00074 = 0.027 = 0.03

We know p+ q =1

So,p = 1-q = 1- 0.03 = 0.97

p = 0.97

q= 0.03

2.Carriers = 2pq

= 2*0.97*0.03

=.0582

Expected number of carriers = 0.0582 * 4.8

=0.28 million (ans)

Healthy individuals = Dominant phenotypes and carriers = p and 2pq

p=0.97 ans 2pq = 0.0582

Healthy individuals (p)= 0.97 * 4.8 =4.6 million

2pq= 0.28 million

Healthy individuals =4.6 + 0.28 = 4.88 million