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please answer clearly so i can understand, thank you.
2. 10-points: The MN blood group in humans is under the control of a pair of co-dominant alleles LM and LN. I a group of 556
Note: the degrees of freedom is interestingly 1 since while there are genotypes.these are determined by only 2 allelic freq
P(X 5x) 0.100 0 .900 Xổ () Xổ to(1) 0.010 xổ sr) 0.000 0.020 0.115 0.297 0.554 0.872 1.239 1.646 2.088 0.050 Xổ (1) 0.004 0.1
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Answer #1

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.

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