Sickle cell anemia is a homozygous recessive condition The f

Solution

As the frequency among African Anericans is given=0.0025

f(SS) = p²
f(Ss) = 2pq
f(ss) = q² = 0.0025 ==> q=0.05
so f(Ss) = 2(0.05)p

Hence using Hardy-Weinberg equation (p²+2pd+q²=1 )

we get ==> p² + 0.1p + 0.0025 = 1

==> p² + 0.1p - 0.9975 = 0

Solving the quadratic equation, roots are given by ==> (-b +/- SQRT(b²-4ac))/2a

(-0.1 +/- SQRT(0.1²-4(1)(-0.9975))/2(1) = (-0.1 +/- SQRT(4))/2

finally we get two values

= -2.1 / 2 and 1.9 / 2

Negative value can be droped as probability cant go in negative

Thus p= 1.9/2 or 0.95

f(Ss) = 2pq = 2(0.95)x(0.05) = 0.095

Ans) Thus frequency of Heterozygous= 0.095

Ans = 0.95