Sickle-cell anaemia is an example of how fitness depends on environmental condit

Question

Sickle-cell anaemia is an example of how fitness depends on environmental conditions. In places where malaria, caused by Plasmodium falciparum is common, Hb^A Hb^S heterozygotes have a selective advantage compared to individuals who are homozygotes for the wild type allele. However, Hb^S Hb^S homozygotes suffer from sickle-cell anaemia and usually die before producing offspring. The relative fitness (w) of the three genotypes in Nigeria are: a. What is the mathematical relationship between relative fitness and selection coefficient? b. What is the selection coefficient (s) for each genotype? c. Calculate the frequency of the three genotypes after one generation of selection, if the genotype frequencies before selection (i.e. at birth) are 0.4 for Hb^A Hb^A homozygotes, 0.46 for heterozygotes and 0.14 for Hb^S Hb^S homozygotes. d. Is it likely that over the next 100 years the Hb^S allele will be lost in the population?

Explanation / Answer

a. Selection coefficient is an estimate of the relative strength of a given genotype to get selected by acting against the most fit form of the genotype. Since the most fit form of the genotype has a fitness value of 1, hence

Selection coefficient (s) = 1- w

A Selection coefficient value of 0 indicates that the genotype is not being selected against, i.e they are most favourably selected.

b. Selection Coefficient for each genotype:

1. HbAHbA, s= 1- 0.78 = 0.22

2. HbAHbS, s= 1- 1 =0

3. HbSHbS, s= 1- 0.07 = 0.93

c. We have, for the given forms of genotypes (simplified as AA, AS, and SS considering only the variation notations)

Observed genotype frequency for:

AA= 0.4, no of individuals with this genotype = 40

AS= 0.46, no of individuals= 46

SS= 0.14, no of individuals = 14

(considering a total of 100 individuals, and a gene pool of 1).

Now,

Considering all as diploid genotypes,

Allele count for A= 40+40+46= 126

Allele count for S= 46+14+14=74

Since total no. of alleles is 200,

Allele frequency for A= 126/200= 0.63 (p)

Allele frequency for B= 74/200=0.37 (q)

Using Hardy-Weinberg Equilibrium principle, expected frequency of genotypes after one generation using Punnet square:

For AA, p2= 0.632=0.3969~ 0.4

For AS, 2pq= 2*0.63*0.37= 0.4662~ 0.46

For SS, q2= 0.372= 0.1369~ 0.14

d. The Hardy-Weinberg principle of equilibrium is based on a Null model, including the assumptions of random mating, and no mutation, migration or selection. According to this principle, the allele frequency is conserved after every generation.

However, in real biological conditions, several evolutionary forces come into play, such as mutation, genetic drift, non-random mating, and thus all the assumptions of the null model cannot hold true.

Therefore, it can be inferred that in 100 years, the allele HbShas a chance to be lost from the population.

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