Question

One hundred persons from a small town in Pennsylvania were tested for their MN blood types. The genotypic data are: MM, 41; MN, 38; and NN, 21. Perform a Chi-Square test to determine if the population they represent is in Hardy-Weinberg equilibrium. What is your Chi-square value?

Please show work.

Answer 1

Answer:

MM41 = 41*2 = 82 M alleles

MN 38 = 38 M & 38 N alleles

NN 21= 42 N alleles

Total alleles = 200

Total M alleles = 82+38 = 120

Frequency of M = 120 / 200 = 0.6

Total N alleles = 42+38 = 80

Frequency of N = 80 / 200 = 0.4

Total progeny = 100

Expected to MM = 0.6*0.6*100 = 36

Expected to MN = 2 x 0.6*0.4*100 = 48

Expected NN = 0.4*0.4*100 = 16

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
MM	41	36	5	25.0000	0.6944
MN	38	48	-10	100.0000	2.0833
NN	21	16	5	25.0000	1.5625
Total	100	100			4.3403

Chi-square value = 4.34

Degrees of freedom = number of categories – 1

Df = 3-1 = 2

p value = 5.99

The chi-square value of 4.34 is less than the critical value of 5.99. So we can accept the null hypothesis. Hence the population is in Hardy-Weinberg equilibrium