if 9 percentage (i.e q^2) of an african population is born with a severe sickle cell anemia, what percentage of population (2pq) will be more resistant to malaria because they are heterozygous for sickle cell gene?
0.3 percentage
0.7percentage
42 percentage
210 percentage
Answers
Explanation:
9 percentage (i.e q^2) of an african population is born with a severe sickle cell anemia, what percentage of population (2pq) will be more resistant to malaria because they are heterozygous for sickle cell gene?
0.3 percentage
0.7percentage
42 percentage
210 percentage
Answer:
Correct option - (c) 42 percentage.
42 percent of the population (2pq) will be more resistant to malaria because they are heterozygous for the sickle cell gene.
Explanation:
According to Hardy-Weinberg's principle, a randomly mating population is in equilibrium when there are no external forces acting upon it.
Let,
p = population without sickle cell anemia
q = Population with sickle cell anemia
pq = population heterozygous for sickle cell gene
According to Hardy and Weinberg,
Given,
% = 0.09
From equation (i),
From equation (iii)
(applying the value of
and
)
Thus, the percentage of heterozygous population = %