Question

If the frequency of people with sickle-cell anemia in a village in Africa is 0.0025, and the population is at Hardy-Weinberg equilibrium, which is closest to the frequency of the sickle-cell allele?

	A) 0
	B) 0.0025
	C) 0.005
	D) 0.025
	E) 0.05
	F) 0.1
	G) 0.2
	H) 0.3
	I) 0.4
	J) 0.45
	K) 0.5
	L) 0.6
	M) 0.7
	N) 0.8
	O) 0.9
	P) 0.95
	Q) 0.975
	R) 0.9975
	S) 1

Answer 1

Sickle cell Anemia is an Autosomal recessive disorder.

An individual having sickle cell Anemia, the genotype must be aa.

Here,

Frequency of people with sickle cell Anemia, F(aa) = 0.0025

Since F(aa) = q​​​​​​2

=> q = √ 0.0025 = 0.05

We know that in Hardy Weinberg Equilibrium,

p + q = 1

p = 1 - q

p = 1 - 0.05 = 0.95

So now we calculate the frequency of each genotype :-

F (AA) = p​​​​​​2  = 0.95 2 = 0.9025

F (Aa) = 2pq = 2 × 0.95 × 0.05 = 0.095

F (aa) = q​​​​​​2 = 0.05 2 = 0.0025

The sickel cell allele is "a" ,

The frequency of a ,

F(a) = F (aa) + 1/2 F (Aa)

= 0.0025 + 1/2 ( 0.095)

= 0.0025 + 0.0475

= 0.05

Therefore the closest frequency of sickle Cell allele is 0.05.

So, the correct answer is (e) 0.05