Question

A binding protein binds to a ligand L with a Kd of 400 nM. What is the concentration of ligand when Y is (a) 0.25, (b) 0.6, (c) 0.95?
Answers: (a) 133.33nm (b) 600nm (c) 7,600nm

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Answer 1

Answer:

(a) 133.33nm

(b) 600nm

(c) 7,600nm

Explanation:

The concentration of Y can be determined by using the formula:

$Y = \dfrac{[L]}{k_d+[L]}$

where;

[L] = concentration of the binding ligand.

kd = 400 nm

Thus:

When Y = 0.25; we get :

$0.25 = \dfrac{[L]}{400+[L]}$

0.25 (400 + [L]) = [L]

100 + 0.25[L] = [L]

100 = [L] - 0.25 [L]

100 = 0.75 [L]

[L] = 100/0.75

[L] = 133.33 nm

At, Y = 0.6

$0.6 = \dfrac{[L]}{400+[L]}$

0.6 (400 + [L]) = [L]

240 + 0.6[L] = [L]

240 = [L] - 0.6 [L]

240 = 0.4 [L]

[L] = 240/0.4

[L] = 600 nm

At, Y = 0.95

$0.95 = \dfrac{[L]}{400+[L]}$

0.95 (400 + [L]) = [L]

380 + 0.95[L] = [L]

380 = [L] - 0.95 [L]

380 = 0.05 [L]

[L] = 380/0.05

[L] = 7600 nm