2. The persistence of the sickle-cell anemia allele (Ho) at the Hemoglobin locus

Question

2. The persistence of the sickle-cell anemia allele (Ho) at the Hemoglobin locus is explained by the fact that being heterozygous (Hb /Hb) provides protection against malaria. However, this leads to sickle- cell anemia in individuals with the Hb Hb genotype. In Equatorial Guinea, the frequency of the Hb allele has been estimated to be 0.1 Recent efforts in some countries have reduced the population of the malaria mosquitos. Based on this success, a researcher proposes that the different genotypes have the following relative fitness values: Relative fitness valuesA a) What is the expected equilibrium frequency for the Hb allele after mosquito control reduces the mosquito population? (6 points) hiwita the empctend cuibrzum freaueney (ealulated in art a at h population mean e) Assuming nothing else changes, what is the expected frequency of Hb one generation after equilibrium is achieved? (4 points)

Explanation / Answer

Given,

Frequency of HbS = p = 0.1
Frequency of HbA = q = 1 - p = 0.9
Fitness of HbA/HbA = W(AA) = 0.95
Fitness of HbA/HbS = W(AS) = 1.0
Fitness of HbS/HbS = W(SS) = 0.0

a)
at equilibrium let the expected frequency of the HbS allele be p'
p' = p2*W(SS) + 0.5*2*p*q*W(AS)
p' = 0.1*0.1*0 + 0.5*2*0.1*0.9*1
p' = 0 + 0.0855 = 0.09

b)
p' = 0.09
q' = 1 - p' = 0.91

mean fitness = (p' )2*W(SS) + 2*(p' )*(q' )*W(AS) + (q' )2*W(AA)
= 0 + 2*0.09*0.91*1 + 0.91*0.91*0.95
= 0.1638 + 0.7866
= 0.9504

c) Let the expected frequency of HbS allele one generation after equilibrium be p''
p'' = (p' )2 * W(SS) + 0.5*2*(p' )*(q' )*W(AS)
p'' = 0 + 0.5*2*0.09*0.91*1
p'' = 0.0819

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