Supose Caspase-1 is allosterically inhibited. Since the inhibitor is not binding in the active site but instead in the allosteric binding site, can I conclude it is a non-competitive inhibitor?

  • $\begingroup$ Could you give me examples, please? The link you have provided is too technical, and these concepts are still somewhat confusing. $\endgroup$ Commented Apr 23, 2017 at 16:33
  • $\begingroup$ The concepts are clearer now. I'll definitely read your links. Thank you very much. $\endgroup$ Commented Apr 23, 2017 at 17:08
  • $\begingroup$ Why are you asking this question? Why do you want to use another name for an allosteric inhibitor? In any introductory text book you will see the kinetics of what are called non-competitive inhibitors described as being quite different from allosteric inhibitors. $\endgroup$
    – David
    Commented Apr 23, 2017 at 22:13
  • $\begingroup$ @tomd — perhaps you could formulate your comments into an answer. At the moment we seem to have an accepted answer that is at variance with them. $\endgroup$
    – David
    Commented Apr 23, 2017 at 22:14
  • $\begingroup$ Actually, this was about distinguishing concepts which were confusing, and applying them. $\endgroup$ Commented Apr 23, 2017 at 22:20

3 Answers 3


No, you cannot assume that an allosteric inhibitor is non-competitive. "Allosteric" refers to the location of binding, whereas terms like competitive, non-competitive, uncompetitive, or mixed inhibition refer to how (or if) the binding of the inhibitor affects binding with the substrate.

The Wikipedia page on competitive inhibition is a reasonable source for this question.

A competitive inhibitor is one that prevents binding of the normal substrate(s) at the active site. It is possible to achieve this either by literally getting in the way by binding in the active site or by binding elsewhere, causing a conformational change that prevents binding the active site. Both cases are considered competitive inhibition because of how they influence substrate binding.

  • $\begingroup$ I am puzzled by your answer. The question is about non-competitive inhibition, but you quote a link to competitive to an article on competitive inhibition. Are you concluding that if an inhibitor is not a competitive inhibitor then it must be a non-competitive inhibitor? I am no enzyme kineticist, but I thought the usage of the term in enzyme kinetics was much more restricted. I cannot see the utility of calling an allosteric inhibitor a non-competitive inhibitor. I would just say it is not a competitive inhibitor. Surely it is better to use definitions related to the kinetics of inhibitors. $\endgroup$
    – David
    Commented Apr 23, 2017 at 22:09
  • $\begingroup$ @David I think the title of the OP's question might be a little unclear, I read it to be "If an inhibitor is allosteric, does that mean it is non-competitive?" I answered that no, that is not always true, and used the definition for competitive inhibition as an example proving that you can have allosteric inhibition that is competitive, rather than non-competitive. I could have also gone into mixed or uncompetitive inhibition but I thought one example was sufficient. $\endgroup$
    – Bryan Krause
    Commented Apr 24, 2017 at 0:13
  • 1
    $\begingroup$ The question was formulated this way because I didn't know an allosteric inhibitor could be competitive or even mixed. $\endgroup$ Commented Apr 24, 2017 at 2:18

It may be technically correct to say that an allosteric inhibitor can be (or is) a “competitive inhibitor”, but, in terms of communication and education, I would advise a more measured approach. Most students encounter the terms “competitive inhibitor” and “non-competitive inhibitor” in the context of reactions exhibiting Michaelis–Menten kinetics. Their mental image of such an inhibitor will therefore be of the type:

Michaelis-Menten competitive inhibition

In discussing allosteric inhibitors I would therefore focus on the different mental model, below, in which substrate (or positive effector) and negative effector (use of this different term being deliberate) have opposing effects on the equilibrium between tense and relaxed states:

Allosteric enzyme: equilibrium between tense and relaxed states

I would then go on to say something to the effect that because the relative concentrations of substrate and negative effector determine the rate of the reaction “the negative effector indirectly competes with the substrate”.


First of all, I take your point about the previous post being too technical. On this occasion, there will be no equations and no diagrams.

When dealing with reversible enzyme inhibition, we are mainly concerned with the meaning of four terms: competitive, uncompetitive, non-competitive and mixed. As I said above, the literature is very confusing to the point of making relatively simple concepts appear very complex, and the confusion often arises from the meaning of the term 'non-competitive'.

First things first: reversible inhibitors are best classified by their effects on two kinetic constants rather than by the mechanism that gives rise to a particular inhibition pattern. The two kinetic constants are kcat and kcat/ Km. (So why not kcat and Km? This is discussed in a previous post, but the short answer is that it is simpler: Km is often a very complex kinetic constant). We can distinguish two 'limiting' cases. (i) A competitive inhibitor effects kcat/ Km but not kcat and (ii) an uncompetitive inhibitor effects kcat but not kcat / Km. (iii) We can also now distinguish a third case where both kinetic constants are affected. Let's call this one non-competitive inhibition.

The confusion in the literature arises because some authorities call non-competitive inhibition as defined above mixed inhibition, and (to really screw things up), they consider non-competitive inhibition a special case of mixed inhibition where both kinetic constants are changed to the same extent! For the purposes of this answer, I am not going there as I think what you mean by non-competitive is as defined in (iii) above. In any event, this is how Cleland, the leading authority in this field, defines non-competitive inhibition. In other words, 'mixed' inhibition is banished from this answer from this point on.

We could be a little bit more adventurous here and say that a competitive inhibitor only effects binding of the substrate to the enzyme and that an uncompetitive inhibitor effects only the catalytic step. Or, we could say that a competitive inhibitor effects only the apparent second order rate constant for the combination of enzyme with substrate (which the enzymologists call kcat/ Km) and that an uncompetitive inhibitor effects only the apparent first order rate constant (which the enzymologists term kcat). But that is becoming too abstract. But perhaps you can see where I am going? There are only two limiting cases, one affecting substrate binding and one affecting catalysis, and a combination of both. It really is as simple as that.

Having got the nomenclature mess out of the way, I will now attempt to answer your question: if an inhibitor does not bind to the active site can we conclude it is a non-competitive inhibitor?

IMO, the answer is an emphatic no. An allosteric inhibitor (one which binds to a site other than the active site) may be competitive, uncompetitive or non-competitive.

As Bryan Krause put it , a competitive inhibitor prevents binding of the normal substrate(s) at the active site, but there is no requirement that the inhibitor bind to the active site. Binding of the inhibitor at the allosteric site may cause a conformational change that prevents substrate binding, for example.

The same argument applies to uncompetitive inhibition. A simple mechanism that gives rise to uncompetitive inibition is where the inhibitor cannot bind to the 'free' enzyme but binds to the enzyme-substrate complex. And of course binding to the allosteric site might not be possible unless substrate is bound.

  • $\begingroup$ The confusion had arisen from the unawareness of those constants. I didn't know an allosteric inhibitor could be competitive or even mixed. Thanks a lot. $\endgroup$ Commented Apr 24, 2017 at 2:17

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