Biology, asked by artil, 1 month ago

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

Answered by nikhilkelaskar464
8

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

Answered by abhijita6lm
0

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,

p+q = 1.....(i)\\(p+q)^{2} = 1.....(ii)\\p^{2} + 2pq + q^{2} = 1....(iii)

Given,

q^{2} = 9%% = 0.09

q = \sqrt{0.09} = 0.3

From equation (i),

p + q = 1\\p + 0.3 = 1\\p = 1-0.3\\p = 0.7\\\\p^{2} = 0.49

From equation (iii)

p^{2} +2pq+q^{2} = 1\\

0.49 +2pq + 0.09 = 1 (applying the value of p^{2 and q^{2})

0.58 + 2pq = 1\\2pq = 1-0.58\\2pq = 0.42

Thus, the percentage of heterozygous population = 0.42 X 100 = 42%

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