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?

Answers

Answer 1

Answer:

0.079

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 = [tex]\sqrt{0.316[/tex]

r =  0.56

To find the frequency of the [tex]I^B[/tex] 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:

[tex]=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

where;

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

[tex]=\dfrac{-(1.12) \pm \sqrt{(1.12)^2-4(1)(-0.084)}}{2\times 1}[/tex]

[tex]=\dfrac{-(1.12) \pm \sqrt{1.2544-(-0.336)}}{2}[/tex]

[tex]=\dfrac{-(1.12) \pm \sqrt{1.5904}}{2}[/tex]

[tex]=\dfrac{-(1.12) \pm 1.261}{2}[/tex]

[tex]=\dfrac{-(1.12) + 1.261}{2} \ \ OR \ \ \dfrac{-(1.12) - 1.261}{2}[/tex]

[tex]=\dfrac{0.141}{2} \ \ OR \ \ \dfrac{-2.381}{2}[/tex]

[tex]\simeq 0.079 \ \ OR \ \ -1.1905[/tex]

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


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