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.
