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 heterozygous carriers 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

F) 0.1

Sickle cell anemia is an autosomal recessive disease. Thus, individuals with homozygous recessive genes will have the disease, while heterozygous individuals will be carriers only and will not have the disease. The frequency of individuals having sickle cell anemia is given as 0.0025. Also, the population is in Hardy-Weinberg equilibrium. Thus, as per the principle of HW equilibrium,

p+q = 1 and p2 + q2 + 2pq = 1

where, p and q are allele frequencies of dominant and recessive allele respectively; p2 and q2 are homozygous doiminat and homozygous recessive frequencies respectively. 2pq is the heterozygous frequency.

In the question, q2 = 0.0025, hence q = 0.05. Therefore, p = 1-q = 1-0.05 = 0.95, Therefore, frequency of heterozygotes = 2pq = 2 x 0.05 x 0.95 = 0.095 which is approximately equal to 0.1 on rounding off the digits.