Hardy weinberg and Chi-square For Human blood, there are two alleles(called S an

Question

Hardy weinberg and Chi-square

For Human blood, there are two alleles(called S and s) and three distinct phenotypes that can be identified by means of the appropriate regents. The following data was taken from people in Britain. Among the 1000 people sampled, the following genotype freqencies were observed SS=99, Ss= 418 and ss=483. Calculate the freqency of S and s in this population and carry out a X^2 test. Is there any reason to reject the hypothesis of Hardy-weinberg proportions in this population?

Explanation / Answer

The proportion of SS individuals in the total population, p2 = 99/ 1000 = 0.099

The proportion of sS individuals in the total population, 2pq = 418/ 1000 = 0.418

The proportion of ss individuals in the total population, q2 = 483/ 1000 = 0.483

The frequency of S (or p) allele = p2+ 1/2(2pq) = 0.099 + 0.209 = 0.308

The frequency of s (or q) allele = q2+ 1/2(2pq) = 0.483 + 0.209 = 0.692

Calculation of expected frequencies:

Expected frequency of MM individuals = p2 = 0.308*0.308 = 0.094 or 94/1000

Expected frequency of MN individuals = 2pq = 2*0.308* 0.692 = 0.426 or 426/1000

Expected frequency of NN individuals = q2 = 0.692*0.692 = 0.478 or 478/1000

Formula for chi-square test = Sum of (observed value - expected value)2 / Expected value.

Chi-square value = (99- 94)2 / 94 + (418- 426)2/ 426 + (483- 478)2/ 478 = 0 (approx)+ = 0.265+ 0.15 + 0.05 = 0.465. The population obeys H-W equilibrium,

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