Question

Jeffrey Mitton and his colleagues found three genotypes (R2R2, R2R3, and R3R3) at a locus encoding the enzyme peroxidase in ponderosa pine trees growing at Glacier Lake, Colorado. The observed numbers of these genotypes were: Genotypes: Number observed: R2R2 135 R2R3 44 R3R3 11 a. Determine the allele frequencies for the R2 and R3 alleles. b. Use the Chi square test to determine if this population is in Hardy-Weinberg equilibrium for the peroxidase trait. Generate a Chi square value. c. Interpret the Chi square value from part (b). If the population is not in Hardy-Weinberg equilibrium, suggest a reason why.

Answer 1

R2R2 P2 = 135

R2R3 2Pq = 44

R3R3 q2 = 11

Total = 190

R3R3 or q2 = 11 /190 = 0.05

Allelic frequency R3 or q = 0.22

P + q = 1

P = 1-0.22 = 0.78

P2 = 0.78 X 0.78 = 0.60       2pq = 2 X .78 X .22 = 0.34    q2 = .22 X .22 = 0.04

Chi square test

	Observed	Expected	O-E	(O-E)2	(O-E)2/ E
R2R2	135	115	25	625	5.43
R2R3	44	64	-20	400	6.25
R3R3	11	9	2	4	0.44

                                                                                                                12.12

Degree of freedom 1 at 95% chi square table valu = 3.814

12.12>3.814

we conclude that the genotype frequencies in this population are significantly different than what would be expected so this is not Hardy-Weinberg equilibrium.