How effective is an antagonist?

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Presentation transcript:

How effective is an antagonist? Often find values of pA2 or pKb quoted The higher the value the more effective the antagonist pKb = -log Kb (Kb is the equilibrium constant) i.e an antagonist with a pKb value of 9 would be approximately 1000 times more effective than one with a value of 6 - note log scale i.e. need a much lower concentration to get an effective block Of course antagonists are SELECTIVE

Calculation of Kb This can be done in 2 ways Using the Schild equation Using a Schild plot Produce log concentration curves for the agonist alone and in the presence of a known concentration of antagonist From these data calculate a DOSE RATIO see graph

Calculation of Dose Ratio CARB + ATR 10-7 M 150 “EC50” 100 Response (mm) 50 -2 -1 1 2 3 4 Log conc carbachol (µM)

Schild Equation First need to calculate DOSE RATIO (DR) DR = EC50 (with anagonist) EC50 (agonist only) Schild equation: DR = Xb + 1 Kb Xb = antagonist conc So Kb = Xb DR-1

Schild Plot This time need to use a series of antagonist concentrations (at least 3, the more the better) Again have to calculate DR for each antagonist concentration Also calculate (DR-1) and log (DR-1) The Schild plot is a plot of log [antagonist] vs log (DR-1) see graph

For competitive antagonism Schild plot II For competitive antagonism slope = 1 2.5 2.0 Kb Log (DR-1) 1.5 1.0 0.5 0.0 -10.0 -9.5 -9.0 -8.5 -8.0 -7.5 -7.0 Log conc atropine (M)

pKb and pA2 pKb is a convenient way of expressing Kb pKb = -log Kb Note similar to pH When slope of Schild plot is 1 and ONLY when slope of Schild plot is 1 Then pKb = pA2

Recommended Reading pKb and pA2 Pharmacology, 5th Ed: Pages 16 & 17 Basic and Clinical Pharmacology Online Book: Section I, Chapter 2.