Recitation 6 short-answer assignment Cystic fibrosis is an autosomal recessive d

Question

Recitation 6 short-answer assignment 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 (1 points each): p 0.027 q0.972 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. (1 point) Population: 4.8 million Frequency of heterozygous 2pq 0.052488 No. of carriers population size x frequeny 4.8 x10 6 x0.052488 251,942.2 0.25 million What is the expected number of healthy individuals in the population? Show your work.(1 point Reminder: only homozygous recessive individuals are affected by the disease Expected healthy p2+2pq 1 q42-0.999

Explanation / Answer

Let the alleles be C (dominant, represented by p) and c (recessive, represented by q)

Given, affected infant frequency (q2) = 1/1353 = 0.000739 ; q = 0.027

Now p + q =1, So,

p = 1-0.027 = 0.973

Frequency of carriers (heterozygotes) = 2 x p x q

Frequency of carriers (heterozygotes) = 2 x 0.973 x 0.027

Frequency of carriers (heterozygotes) =0.05254

Actual number of carriers = Frequency of carriers (heterozygotes) x Total population

= 0.05254 x 4.8 million

= 0.2522 million

Expected healthy individuals = Homozygous dominant individuals + Heterozygous individuals

= (p2 + 2pq) x total popultaion

Now, p2 = (0.973)2 = 0.9467

Expected healthy individuals = (0.94673+0.05254) x 4.8 million

= 0.99923 x 4.8 million

= 4.7965 million

Filling table and calculating p value

In 1910,Effected individuals = 65

Frequency of effected (q2) = 65/1,00,000 = 0.00065

q = (0.00065)1/2 = 0.0255

Normal individuals (p2 + 2pq) = 1,00,000 - 65 = 99935

p = 1-q = 1-0.0255 = 0.9745

2pq = 2 x 0.9745 x 0.0255 = 0.0497

Heterozygous individuals = 0.0497 x 1,00,000 = 4970

Homozygous dominant individuals = 99935 - 4970 = 94965

From chi square at df =2, value should be less than 5.99 (p = 0.05) which is not the case.

So, null hypothesis is rejected

Conclusion: Gene for cystic fibrosis is not under Hardy Weinberg equilibrium

Reason: Evolution is taking place and so, allele frequencies are varying

CC Cc cc Total Expected number of individuals based on present data set p2 x 1,00,000 = 0.94673 x 1,00,000 = 94673 2pq x 1,00,000 = 0.05254 x 1,00,000 = 5254 q2 x 1,00,000 = 0.00074x 1,00,000 = 74 1,00,000 Observed number of individuals in 1910 94965 4970 65 1,00,000 (O-E)2/E 0.9006 15.35 1.094 Chi square = 17.345

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