The MN blood group in humans is under the control of a pair of co-dominant allel

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

167

MM

280

MN

109

NN

Explanation / Answer

(a). Hardy and Weinberg also described all the possible genotypes for a gene with two alleles. The binomial expansion representing this is, p2 + 2pq + q2 = 1.0

Where,

p2 = proportion of homozygous dominant individuals

q2 = proportion of homozygous recess

2pq = proportion of heterozygotes.

p2 = 167/ 556 = 0.30; p = 0.547

q2 = 109/ 556 = 0.196; q = 0.442

2 pq = 280/ 556 = 0.503

(b) C2 = (Observed freq- Expected freq)2 / Expected freq

Expected MM = 0.30* 556 = 166.8

Expected MN = 0.196* 556 = 108.97

Expected NN = 0.503* 556 = 279.6

C2 (Chi-square value) = (167-166.8)/ 166.8+ (280-279.6)/ 279.6+ (109-108.97)/ 108.97 = 0.00119 + 0.00143 + 0.00027 = 0.00289.

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