Question

Consider the following population:

Blood Type A B AB O
Number of individuals 533 84 67 316

What is the frequency of the IB allele in this population?

Answer 1

0.079

Step-by-step explanation:

Using Hardy Weinberg equilibrium for blood group;

p² + 2pq + q² + 2pr + r² + 2qr = 1

where;

p = allele frequency for A

q = allele frequency for B

r = allele frequency for O

N(p² + 2pr) = the number of allele for type A people

N(q² + 2qr) = the number of allele for type B people

N(2pq) = the number of allele for AB people

N(r²) = the number of allele for O people

N(r²) = 316

r² = 316/1000

r² = 0.316

r =
`\sqrt{0.316`

r = 0.56

To find the frequency of the
`I^B` allele;

N(q² + 2qr) = 84

q² + 2qr = 84/1000

q² + 2qr = 0.084

where;

r = 0.56

q² + 2q(0.56) = 0.084

q² + 1.12q = 0.084

q² + 1.12q - 0.084 = 0

Using the quadratic formula:

`=(-b \pm √(b^2-4ac))/(2a)`

where;

a = 1, b = 1.12, c = -0.084

`=(-(1.12) \pm √((1.12)^2-4(1)(-0.084)))/(2* 1)`

`=(-(1.12) \pm √(1.2544-(-0.336)))/(2)`

`=(-(1.12) \pm √(1.5904))/(2)`

`=(-(1.12) \pm 1.261)/(2)`

`=(-(1.12) + 1.261)/(2) \ \ OR \ \ (-(1.12) - 1.261)/(2)`

`=(0.141)/(2) \ \ OR \ \ (-2.381)/(2)`

`\simeq 0.079 \ \ OR \ \ -1.1905`

Since our value can be negative; then the frequency of the
`I^B` allele in this population is 0.079.