Answer:
2a).
Observed genotypic frequencies:
A1A1 = 13/32=0.406
A1A2 = 7/32=0.219
A2A2 = 12/32= 0.375
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
A1A1 |
0.406 |
0.436 |
-0.03 |
0.001 |
0.002 |
A1A2 |
0.219 |
0.449 |
-0.23 |
0.053 |
0.118 |
A2A2 |
0.375 |
0.115 |
0.26 |
0.068 |
0.588 |
1 |
1 |
0.708 |
X^2 value = 0.708
b).
Degrees of freedom = number of genotype – 1
Df = 3-1=2
c).
Ther critical X^2 value is 5.99 for df of 2.
d).
The observed values are consistent with the expected values. Because, the X^value of 0.708 is less than the critical value of 5.99. So we can accept the null hypothesis.
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