Question

Calculate allele frequencies and use the Chi squared and G statistic to determine whether the population is in HW equilibrium. Blood Type Frequency O 44.00 A 42.00 B 10.00 AB 4.00

Answer 1

According to Hardy-Weinberg equilibrium, the allelic frequency in a give set of population with allele frequency p2, q2 and 2pq for homozygous individual of either types and heterozygotes is given by:

p2 + q2 + 2pq = 1 and p+q = 1.

In case of three phenotypes, we would have p+q+r = 1

Thus, here we have:

Allelic frequency, p = square root of (Number of individuals / total individuals)

Thus, for individuals with blood type A: 42 / 42 + 44 + 4 + 10 or 42/100 or 0.42

Thus, allelic frequency for type A = sq. root of 0.42 = 0.648

Similarly for type B = sq. root of ( 10/100) or 0.316

Similarly for type AB = sq. root of (4/100) or 0.2

Similarly for type O = sq. root of (44/100) or 0.663

On adding all allelic frequencies, i.e. 0.316 + 0.2 + 0.663 + 0.648 = 1.827

Since this number is more than unity, it can be clearly stated that the population does not follow Hardy-Weinberg equilibrium.

Reason: The human blood group type shows complete dominance between A/B and O whereas the alleles A and B show co-dominance. This fact violates the assumption of Hardy-Weinberg equilibrium and hence human blood system does not follow this equilibrium.