Question

A: 56, B: 11, AB: 8, O: 25 . Draw Punnett Square:

Assume the population is in Hardy-Weinberg equilibrium

Calculate the allele frequencies:   p =      

q =       

r =

Algebraic equation    Numerical freq.

Expected freq of   A:

B:

AB:

O:

Answer 1

Punnett Square :

Alleles	A(p)	B (q)	O (r)
A (p)	AA (p2)	AB (pq)	AO (pr)
B (q)	AB ( pq)	BB (q2)	BO (qr)
O (r)	AO (pr)	BO (qr)	OO (r2)

Number of individuals

A-type = 56

B-type = 11

AB type = 8

O type = 25

Total = 100

Frequency of blood types :

A type = F[A] = 56 / 100 = 0.56

B-type = F[B] = 11/100 = 0.11

AB type = 8/100 = 0.08

O type = F[O] = 25/100 = 0.25

Let frequency of A, B and O allele be p, q and r respectively.

Frequency of alleles :

Algaebraic Equation : Applying Bernstein's equation

p = 1 - (F[B] + F[O])1/2
q = 1 - (F[A] + F[O])1/2
r = (F[O])1/2

p = 1 - (0.11 + 0.25)1/2 = 1 - 0.6 = 0.4

q = 1 - (0.56 + 0.25)1/2 = 1 - 0.9 = 0.1

r = (0.25)1/2 = 0.5

Expected frequency :

A type :

Since invidivual with two A alleles and individuals with one A allele and one O allele can have A blood type, the expected frequency will be p2 + 2pr = (0.4)2 + 2 x 0.4 x 0.5 = 0.16 + 0.4 = 0.56

B type :

Since invidivual with two B alleles and individuals with one B allele and one O allele can have B blood type, the expected frequency will be q2 + 2qr = (0.1)2 + 2 x 0.1 x 0.5 = 0.01 + 0.1 = 0.11

O-type :

Since invidivual with two O alleles can have O blood type, the expected frequency will be r2 = (0.5)2 = 0.25

AB type

Since individuals with one A allele and one B allele can have AB blood type, the expected frequency is = 2pq = 2 x 0.4 x 0.1 = 0.08