A human population in Africa living under a severe malaria selection has the fol

Question

A human population in Africa living under a severe malaria selection has the following relative fitnesses (w) of homozygous genotypes in beta-hemoglobin locus:

w11 = 0.75 for A1A1

w22 = 0.85 for A2A2 .

Assume that w12 = 1.00 for A1A2 .

Using heterozygote advantage concept, determine frequency of the mutant sickle-cell beta-hemoglobin allele (q) in this population assuming that genotype frequencies are at balance (?p = 0; balancing selection).

1) What is the average fitness of this population for locus (gene) A?

2)Explain why this balance is not the same as Hardy-Weinberg balance.

Explanation / Answer

A1A1. A1A2. A2A2

Relative fitness 0.75. 1. 0.85

Fitness in symbol. 1-s. 1-t

Selection coefficient s= 0.25. t= 0.15

s = selection coefficient of homozygous dominant genotype

t = selection coefficient of homozygous recessive genotype

Equilibrium frequency

p = t/s+t = 0.625

q = s/s+t = 0.375

When genotypic frequency are at balance, frequency of mutant sickle cell beta haemoglobin allele in this population is 0.375.

Mean fitness of this population for locus A

W = p2 w11 + 2pq w12 +q2 w22

= (0.625)2 x .75 + 2 (0.625) (0.375) x 1 + (0.375)2 x 0.85

= 0.8811

2. p = t/s+t = 0.15/0.25 + 0.15 = 0.75

     q= s/s+t = 0.25/0.25+0.15 = 1.15

     p+q = 1.9

     In Hardy-weinberg balance, the sum of p and q is always 1. Here, p + q is not equal to 1. So, this is not in Hardy-Weinberg's balance.

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