Question

The MN blood group in humans is under the control of a pair of co-dominant alleles, M (we will call the frequency of M, p) and N (we will call the frequency of N, q). In a group of 556 individuals, the following numbers of individuals are found for each of the genotypes:
167 MM
280 MN
109 NN
a) What is the frequency of each allele? p= q=
b) What is the value of the Chi-square statistic test to find if in this particular case the genotypic frequencies conform to the Hardy-Weinberg distribution.
Chi square value =
c) What is the probability associated with you chi square statistic calculated above? Please complete the blanks below with the corresponding symbol, < OR > than the critical value, your conclusion with respect to the null hypothesis of Hardy Weinberg equilibrium, Retain or Reject.
P value __ than 0.05
Conclusion ___ the null hypothesis of HW equilibrium

Answer 1

Answer:

a) f(M) = p = 0.55

   f(N) = q = 0.45

b) X² = 12.12

c)  P₀.₀₅ = 5.991

d) P₀.₀₅ < X²

e) Reject the null hypothesis of HW equilibrium

Explanation:

Due to technical problems, you will find the complete explanation in the attached files

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