The steps involved here
Step 1: Calculation number of alleles
Genotype |
Freequency |
Allele A |
Allele a |
Total |
AA |
150 |
300 |
0 |
300 |
Aa |
120 |
120 |
120 |
240 |
aa |
90 |
0 |
180 |
180 |
Total |
360 |
420 |
300 |
720 |
Step 2: Calculation of allele frequencies
Frequency = No of a alleles/Total no of alleles
Allele A |
= 420/720 |
= 0.58 |
Allele a |
= 300/720 |
= 0.42 |
Step 3: Expected genotype frequencies estimation
AA |
= 0.58*0.58= 0.340 |
*360= 123 |
Aa |
= 2*0.58* 0.42= 0.486 |
*360= 175 |
aa |
0.42*0.42= 0.174 |
*360= 63 |
Step 4: Chi-square test
Null hypothesis: The observed values are not deviating from the expected values.
Category |
AA |
Aa |
aa |
|
Observed values |
150 |
120 |
90 |
|
Exprected Values |
123 |
175 |
63 |
|
Deviation |
28 |
-55 |
28 |
|
D^2 |
756.25 |
3025 |
756.25 |
|
D^2/E |
6.17 |
17.29 |
12.10 |
35.56 |
X^2 |
35.56 |
|||
Degrees of freedom |
2 |
Inference: The calculated chisquar value i.e. 35.56 is greater than the table value i.e.5.99 at 2 DF and 0.05 probability, hence the null hypothesis is rejected. Which means the population in not in HW equilibrium.
use hardy wineberg equation to find equilbrium. AA 150. Aa 120. aa 90
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Evolution
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