Answer:
167 MM = 2 * 167 = 334 “M” alleles
109 NN = 218 “N” alleles
MN 280 = 280 “M” and 280“N” alleles
Total alleles = 1112
Frequency of “M” allele = 334+280 / 1112 =0.55
Frequency of “N” allele = 280+218/1112 = 0.45
Three expected genotypes are produced as follows.
Total population = 556
MM individuals = 0.55*0.55*556 = 168
NN individuals = 0.45*0.45*556 = 113
MN individuals = 2*0.55*0.45*556 = 275
Phenotype |
Observed(O) |
Expected (E) |
O-E |
(O-E)2 |
(O-E)2/E |
MM |
167 |
168 |
-1 |
1.00 |
0.0060 |
MN |
280 |
275 |
5 |
25.00 |
0.0909 |
NN |
109 |
113 |
-4 |
16.00 |
0.1416 |
Total |
556 |
556 |
0.2385 |
Ch-square value = 0.2385
Degrees of Freedom = 1
Critical value = 3.84
The chi-square value of 0.2385 is less than the critical value of 3.84. Hence, the null hypothesis is accepted and the data is in Hardy-Weinberg equilibrium.
please answer clearly so i can understand, thank you. 2. 10-points: The MN blood group in...
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