Question
Asked Nov 24, 2019
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  1. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?
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Step 1

Sickle-cell anaemia is a genetic disease.

Homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, these individuals are susceptible to malaria.

Individuals homozygous for the sickle-cell trait (ss) have defective red blood cells and the oxygen binding ability of sickle-shaped RBCs are less. Although malarial parasite cannot complete its life cycle in these red blood cells, individuals often die because of the genetic disorder. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these "partially defective" red blood cells. Thus, heterozygotes (Ss) tend to be more resistant to malaria than either of the homozygous conditions.

Step 2

According to Hardy-Weinberg Principle: The Hardy–Weinberg principle, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

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p2 2pq 1andp + q 1 p frequency of the dominant allele in the population qfrequency of the recessive allele in the population p2 percentage of homozygous dominant individuals q2percentage of homozygous recessive individuals 2pq percentage of heterozygous individuals

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Step 3

Calculations according to the Hardy-Wei...

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In the given question the percentage of homozygous recessive individual is 9%, which means q2 is 9% q2 0.09, q 0.3 putting the value of q in Hardy-Weinberg equation: p+q 1 p+0.3 p 1 0.3 p 0.7 therefore, the percentage of homozygous dominant individual is 0.49%. We have to determine the value of 2pq (= percentage of heterozygous individuals, Ss) Value of 2pq 2 x 0.7 x 0.3 =0.42

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