2) A human blood group, related to the MN system, is also controlled by two alle

Question

2) A human blood group, related to the MN system, is also controlled by two alleles (S and s), and three distinct phenotypes can be identified by means of the appropriate reagents. In a sample of 1000 people, the number of different genotypes were as follows: 99 SS, 418 Ss, and 483 ss. Estimate the allele frequency of S (p) and s (q) and carry out a Chi square test of goodness of fit between the observed genotype frequencies and their Hardy-Weinberg expectations. Is there any reason to reject the hypothesis of Hardy-Weinberg proportions for this gene?

Explanation / Answer

The proportion of individuals with ss genotype = 483/ 1000 = 0.483 = q2

The frequency of s allele (q) = 0.694 (approximate;y 0.7)

Hence, the frequency of p = 1-q = 1-0.694 = 0.306 (approximately 0.3)

Expected frequency of SS = p2 = 0.3*0.3 = 0.09, expected number of individuals = 0.09* 1000 = 90

Expected requency of Ss = 2pq = 2* 0.3* 0.7 = 0.42, expected number of individuals = 0.42* 1000 = 420

Expected frequency of ss = q2 = 0.7* 0.7 = 0.49, expected number of individuals = 0.49* 1000 = 490

CHI - SQUARE (X2):

X2 = (O - E)2 / E

Where O = observed frequency, E = expected frequency

X2 = (99- 90)2 /90 + (418 -420)2 /420 + (483 -490)2 /490

Chisquare value = 0.9 + 0.0095 + 0.1 = 1.0095.

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