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 = 0.60Â Â Â Â Â Â 2pq
= 2 X .78 X .22 = 0.34Â Â Â 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 |
                                                                                                              Â
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.
