Question

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.

Frequencies observed in the present: p=.97 , q=.03

Fill in the following table (2 points):

	CC	Cc	cc	Total
Expected number of individuals (i.e. numbers based on present dataset)				100,000
Observed number of individuals in 1910			65
				?2 =

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 ?²value 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)

Answer 1

Answer:

Based on the given information:

Present:

• Frequencies in present p = 0.97 and q = 0.03
• Expected number of CC (current) = p²*100000 = 0.97^2*100000 = 94090
• Expected number of cc (current) = q²*100000 = 0.03^2*100000 = 90
• Expected number of Cc (current) = 2*p*q*100000 = 2*0.97*0.03*100000 = 5820

1910:

• Frequencies in 1910 p = ? and q = ?
• Observed number of cc (1910) = 65
• Observed frequency cc (1910) = q² = 65/100000 = 0.00065
• Oberved frequency q = sqrt(0.00065) = 0.0255
• Observed frequency p = 1 - q = 1 - 0.025495 = 0.975
• Observed number of CC = p²*100000 = 0.975^2*100000 = 94966
• Observed number of Cc = 2*p*q*100000 = 2*0.975*0.0255*100000 = 4969

Chi Square Test:

Genotype	Observed No. in 1910 (O)	Expected Number current (E	(O-E)	(O-E)^2/E
CC	94966	94090	876	8.2
Cc	4969	5820	-851	124.4
cc	65	90	-25	6.9
Total	100000	100000	Chi square (sum)	139.5

Degree of freedom = Number of categories - 1 = 3 - 1 = 2

P-Value corresponding to Chi square value of 139.5 with 2 degree of freedom with 95% certainity (significance level 0.05) : P-Value is < 0.00001

Since P-value is less than the significance level (0.05), therefore Null Hypothesis is rejected. Based on the experimental evidence, there is statistically significant difference in the expected frequency (current) and observed frequency (1910) and the population deviates from Hardy Weinberg Equilibrium.