Question

The EC50 of drug A is 4*10^-5, when an antagonist is added at the concentration of 1*10_^-6M the EC50 value becomes 1*10^-3. What is the pKB? (topic: pharmacology)

Answer 1

This question is a straightforward application of Schild's Equation. Let me write down the equation and explain.

The schild's equation is given by:-

. Let us see what each term represents. The term ,EC_50a represents the concentration of the agonist (such that half the maximal resposne is produced) IN the PRESENCE of the antagonist. While the term EC_50c is the concentration of the agonist that produces half the maximal response, IN the ABSENCE of the antagonist.

The term K_b represents the dissociation constant of the antagonist while [B] represents the concentration of the antagonist used.

Now that the terms in the Schild's equation have been clearly defined, all that is left if putting in the values into the equation and solving the equation.

Lets put in the initial values into the equation:

. It is clearly seen that all we have done so far is simply subsitute the given values into the equation.

Let us now solve the equation and try to obtain the value of K_b.

We get

ie,

.

We have thus obtained the value of K_b.

_{Using this value of Kb, we can now find the value of pKb as}

_{pKb = - log[kB]}

_{which gives us}

_{pKb = -log (4.16*10^-8) = 7.3809.}

I hope that has helped. In case the answer is not clear, do let me know in the comment section. I will be more than glad to help you out.