Sickle-cell anemia is caused by a mutation in the ß-globin gene. Heterozygotes f

Question

Sickle-cell anemia is caused by a mutation in the ß-globin gene. Heterozygotes for the Hbs disease allele have some protection from malaria relative to wild-type, but Hbs homozygotes rarely live to produce offspring. 50 individuals homozygous for the normal (non-disease) allele produced a total of 47 children. 50 heterozygous individuals produced a total of 55 children. 50 Hbs homozygous individuals produced a total of 0 children. What is the average fitness per individual of the heterozygotes? What is the relative fitness of the homozygous normal individuals?

Explanation / Answer

Let's start by laying out the given data 50 Parents Produced - 47 HH (non-disease) 50 Parents Produced - 55 HHbs (One disease Hbs allele) 50 Parents Produced - 0 HbsHbs (two disease alleles) This makes sense as we expect 0 individuals to survive early ontogeny (development) with such a deadly combination of alleles. Absolute fitness of a genotype is defined as the ratio between the number of individuals with that genotype after selection to those before selection. Absolute Fitness: Non-disease = (47/50) = 0.94 One Disease Hbs allele = 55/50 = 1.1 Two Disease Alleles = 0/50 = 0 From this absolute fitness information, we expect to see the heterozygote persist, whereas the homozygote non-disease will decrease in frequency(as it's fitness is less than 1) Relative fitness is quantified as the average number of surviving progeny of a particular genotype compared with average number of surviving progeny of competing genotypes after a single generation Relative Fitness: Non-disease = 0.94/1.1 = 0.85 One Disease Hbs allele = 1.1/1.1 = 1 Two Disease Alleles = 0 /1.1 = 0

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