I did the first portion, the disease affects approx. 1 in 1353 : p = (1-q) = .97

Question

I did the first portion, the disease affects approx. 1 in 1353 : p = (1-q) = .97 q2= 1/1353 = .00074 = .027

Current pop. is approx. 4.8 million. The expected # of CF carriers (hetero)

Carriers (2pq) = 2(.027)(.97) = .052 x 480,000,000 = 25 million

Expected Healthy individuals

Healthy (p) = (.97)2 =.94 x 480,000,000 = 4.5 million

Hopefully I did the first part right, if not please correct me! Thank You!

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.

Use the ?2 table provided below to infer the p-value of the test (0.5 points)

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. (0.5 point)

Total 100,000 Cc of individuals (i.e numbers based on present dataset) Observed number of individuals in 1910 65 (o - E)2

Explanation / Answer

Let us assume q2 is frequency of effected (homozygous recessive), p2 is frequency of uneffected (homozygous dominant) and 2pq is frequency of carriers (heterozygous). The heterozygotes are also uneffected.

In first portion

q2 = 1/1343 = 0.00074

Now q = (0.00074)1/2 = 0.027

So, p = 1-0.027 = 0.973

p2 = (0.973)2 = 0.9467

2pq = 2 x 0.027 x 0.973 = 0.05254

Effected individuals (homozygous cc) = q2 x total population = 0.00074 x 4.8 million = 0.003552 million

Uneffected individuals (homozygous dominant CC) = p2 x total population = 0.9467 x 4.8 million = 4.544 million

Uneffected individuals (Heterozygous Cc) =2pq x total population = 0.05254 x 4.8 million = 0.252 million

Now, we can calculate the expected individuals in 1910,

Expected (CC) p2= 0.9467 x 100,000 = 94670 individuals

Expected (Cc) 2pq= 0.05254 x 100,000 = 5254 individuals

Expected (cc) q2= 0.00074 x 100,000 = 74 individuals

Chi square = 1.09

Degrees of freedom = 1

Now, by following the chi square table with degrees of freedom =1

Critical value at 0.05 is 3.84 and at 0.01 is 6.64 but our value is 1.09 which is less than critical value so, there exists Hardy weinberg equilibrium (95% certainity) for cystic fibrosis in given population.

Such a population will show almost same gene frequencies in upcoming generations and no evolution is taking place in the population. Only changes will be observed if the mutations occur in the concerned gene.

CC Cc cc Total Expected (E) 94,670 5254 74 100,000 Observed in 1910 (O) 65 100,000 O-E -9 (O-E)2/E 1.09

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