Question

3 pts] At a locus with three alleles in a population of voles, you accumulate the following genotypic data:

genotype	observed # individuals
A1A1	19
A2A2	11
A3A3	14
A1A2	56
A1A3	79
A2A3	51
total	230

Is this population in Hardy Weinberg equilibrium at this locus? Determine using a chi-square test.

Answer 1

Answer:

A1A1-19 = 19*2=38 A1 alleles

A2A2-11 = 22 A2 alleles

A3A3-14 = 28 A3 alleles

A1A2-56 = 56A1 & 56 A2 alleles

A1A3-79= 79A1 & 79 A3

A2A3-51 = 51 A2 & 51 A3 alleles

Total alleles = 460

Total A1 alleles = 38+56+79 = 173

Frequency of A1 allele = 173/460 = 0.376

Total A2 alleles = 22+56+51 = 129

Frequency of A2 allele = 129/460 = 0.28

Total A3 alleles = 28+79+51 = 158

Frequency of A3 allele = 158/460 = 0.344

Expected number of Frequency of each genotype:

A1A1 =0.376 * 0.376 * 230 = 32.52

A2A2 = 0.28 * 0.28 * 230 = 18.03

A3A3 = 0.344 * 0.344 * 230 = 27.22

A1A2 = 2* 0.376 * 0.28 *230 = 48.43

A1A3 = 2 * 0.376 * 0.344 *230 = 50.49

A2A3 = 2 * 0.28 * 0.344 *230 = 44.31

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
A1A1	19	32.52	-13.52	182.79	5.621
A2A2	11	18.03	-7.03	49.42	2.741
A3A3	14	27.22	-13.22	174.77	6.421
A1A2	56	48.43	7.57	57.30	1.183
A1A3	79	59.49	19.51	380.64	6.398
A2A3	51	44.31	6.69	44.76	1.010
Total	230	230.00	0		23.374

Chi-square value = 23.374

Degrees of freedom = number of phenotypes – 1

Df = 6-1=5

P value = 11.07

The X^2 value of 23.374 is greater than the critical (p) value of 11.07. We can reject null hypothesis and the population is not in H-W equilibrium