Drugs can be provided as a competive or non-competitive form of agonists. But the effect of non-competitive antagonists is much more than the other, so what is the cause of that? And are there some exceptions?


1 Answer 1


Personally, I think the statement "the effect of a competitive antagonist is less than the effect of a non-competitive antagonist" is too general to have much meaning. There's a large number of systems, each with a potentially large number of competitive and non-competitive antagonists, each with a potentially large range of activities. Saying a priori that a non-competitive antagonist is more effective than a competitive one is presumptuous.

However, there is a bit of sense in there, for a particular defined sense of circumstances.

By definition, a competitive antagonist is only active against the non-liganded state of the receptor. The native ligand (agonist) and the competitive antagonist compete for the receptor. In contrast, a non-competitive antagonist can generally work against both the liganded (with agonist bound) and ligand-free versions of the receptor. (How active it is against each depends on the particulars about the antagonist and receptor.)

So in a situation where the antagonist isn't at saturating concentrations (say at a concentration near it's IC50), saturating amounts of native ligand (agonist) will completely reverse the effects of a competitive antagonist, as the agonist will compete off the antagonist. However, the non-competitive antagonist will still have some remaining inhibitory function, even at saturating concentrations of native ligand, as it can bind to and inhibit the ligand-bound (agonist-bound) form of the receptor.

However, there's an apples-to-oranges comparison embedded in there. That comparison assumes that the competitive and non-competitive are each near their respective IC50s. This value can change greatly between drugs, so you can have a competitive inhibitor with a really low IC50, and a non-competitive inhibitor with a really high one, such that if you used identical molar concentrations of each antagonist, the competitive antagonist would be successful in almost completely shutting off receptor activity, whereas the non-competitive antagonist would show barely noticeable activity.

  • $\begingroup$ You said 'a non-competitive antagonist can generally work against both the liganded and ligand-free versions of the receptor' Do you mean by liganded version, the agonist binding site? Because in this movie at min. 3:43 is said the opposite: youtube.com/watch?v=7mJut-vVZRs $\endgroup$
    – Marijn
    Mar 25, 2016 at 18:56
  • $\begingroup$ @Marijn By "liganded" I meant "agonist-bound form". As your video indicates from 3:50-4:30, the non-competitive antagonist can be bound to (and be inhibitory towards) the receptor both in the presence and absence of agonist binding. $\endgroup$
    – R.M.
    Mar 25, 2016 at 19:06

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