contestada

An antibody X binds to an antigen Y. When the antigen concentration is 5.0 × 10 − 8 M , 5.0×10−8 M, θ θ is 0.55 . 0.55. What is the dissociation constant, K d , Kd, of the antibody for the antigen?

Respuesta :

Larrysammz

Answer:

The dissociation constant Kd of the antibody for the antigen is 4.09090909 ×10⁻⁸M

Explanation:

the formula for dissociation is  Ф = [L] / [L] + Kd

we want to find the dissociation constant Kd

in trying to do that we need to make Kd the subject of the formula

cross multiply

Ф( [L] + Kd) = [L]

open the bracket

Ф[L] + ФKd = [L]

subtracting Ф[L]  from both sides

ФKd = [L] -Ф[L]

ФKd = [L](1-Ф)

divide both sides by Ф

Kd = {[L](1-Ф)}/Ф

we want to find the dissociation constant Kd

[L] = 5.0 x 10⁻⁸M

Ф= 0.55

Kd = {5.0 x 10⁻⁸(1-0.55)}/0.55

Kd = {5.0 x 10⁻⁸(0.45)} / 0.55

Kd = (2.25x10⁻⁸)/0.55

kd= 4.09090909 ×10⁻⁸M

raphealnwobi

Ans:

Kd = 40.9 × 10⁻⁹ = 40.9nM

Explanation:

The equation to calculate the concentration of the antibody for the antigen is given by:

Θ = [tex]\frac{[L]}{[L]+ K_{d} }[/tex]

Cross multiplying, we get

Θ([L] + Kd) = [L]

Opening the bracket,

Θ[L] + ΘKd = [L]

Collecting like terms,

[L] - Θ [L] = ΘKd

[L](1 - Θ) = ΘKd

Making  Kd the subject of the formula,

Kd =  [L](1 - Θ) / Θ

Given that [L] = 0.55 × 5 × 10⁻⁸

Θ = 0.55

substituting values, we get Kd = 40.9 × 10⁻⁹ = 40.9nM

Kd = (5 × 10⁻⁸)(1 - 0.55) / 0.55

Kd = 2.25 × 10⁻⁸ / 0.55

Kd = (4.09 × 10⁻⁸) M

the dissociation constant, Kd is 40.9 nM

Otras preguntas