Question

A mutation in one of the hemoglobin genes causes sickle cell anemia. The sickle cell allele, S, severely reduces fitness in people who are homozygotes, SS. In contrast, people with at least one normal hemoglobin allele, A, do not suffer the effects of sickle cell anemia, even if the individual is a heterozygote, AS. Interestingly, in areas with high rates of malaria, a single Sallele confers some resistance to malarial infection. Suppose there is a population with the observed and expected number of individuals shown in the table. Calculate the chi-square, X, value for this population. Express the X2 value to one decimal place. Observed 448 Genotype AA AS SS Expected 478 270 38 330 8 Xvalue: Compare the Xvalue to those in the X² table. What conclusions can you draw from these data? Selection is reducing the number of individuals with the AS genotype. This population is not in Hardy-Weinberg equilibrium. The Xvalue is much larger than that found at a p-value of 0.05. The observed and expected values are not significantly different from each other.

Answer 1

Anse) Genotyße AA AS SS Observed © 448 330 Expected E) 478 270 38 18-786] (6 - E2/E 1.893 13.333 23.684 x? 38.900 Jabbrox. 12-786] x² zalue = 138.91 @ The x² aralue is much larger than that found at p value 0.05. [At prout for a degree of freedom, dabulas Value is 5.941 which is much lower than our Calculated (x² = 38.9) ralue.]