Question

In a sample of 200 individuals, the frequency of the BB genotype is 0.4, the Bb genotype is 0.4, and the bb genotype is 0.2. Perform a Chi-square test to test whether this population is in Hardy-Weinberg equilibrium. For full credit, be sure to include the null hypothesis, Chi-square number, degrees of freedom, p-value, and conclusion.

Answer 1

The proportion of BB is 0.4, so the frequency of BB in 200 population is= 200*0.4 = 80

The proportion of Bb is 0.4, so the frequency of BB in 200 population is= 200*0.4 = 80

The proportion of BB is 0.2, so the frequency of BB in 200 population is= 200*0.2 = 40

Allele frequency estimation:

Genotype	Freequency	Allele B	Allele b	Total
BB	80	160	0	160
Bb	80	80	80	160
bb	40	0	80	80
Total	200	240	160	400

Allele B	= 240 /400 = 0.60
Allele b	= 160/ 400 = 0.40

Expected frequencies estimation:

BB	0.360	72
Bb	0.480	96
bb	0.160	32

Null hypothesis: The observed values are not deviating from the expected values.

Test static:

Category	BB	Bb	bb
Observed values	80	80	40	200
Exprected Values	72	96	32	200
Deviation	8	-16	8
D^2	64	256	64	384
D^2/E	0.89	2.67	2.00	5.555556
X^2	5.56
Degrees of freedom	1

Inference: The calculated chisquare value i.e. 5.56 is greater than the table value i.e.3.84 at 3 DF and 0.05 probability, hence the null hypothesis is rejected. Which means the population is not in HW equilibrium.