In: Biology
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:
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