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Competitive antagonists vs any agonist

  1. Jul 23, 2015 #1
    Is it true that it should always be possible to design a competitive antagonist with higher affinity for the receptor than the affinity of any agonist?
     
  2. jcsd
  3. Jul 23, 2015 #2
    I don't know if I would say 'always' and 'any agonist', but it is possible to find orthosteric antagonists with higher affinity than a given agonist, why not?
    E.g. if your antagonist forms better/additional binding interactions and/or displaces 'unhappy' water, it can form a more stable complex with the receptor, i.e. one with higher affinity. Not sure that it's 'easy' to find such a molecule, but I don't think anyone can prove that it's definitely not possible.
    Note that the natural agonists of some receptors are not very potent, and antagonists (and agonists) with much higher affinity were found by doing drug discovery.

    At least, this is what I know - if someone has better information on this, please do correct me.
     
  4. Jul 23, 2015 #3
    Sorry, forgot to add that its in the context of reversible binding. By any agonist I mean any type. By higher affinity I mean lower Kd, so I believe some equation would be involved with the answer.

    Is there some sort of mathematical way to prove this?
     
  5. Jul 24, 2015 #4
    I was indeed thinking of reversible binding, because with irreversible binding the concept of affinity has to be modified a bit.
    And indeed, lower dissociation constant means greater affinity.

    I'm not sure I understand what equation you're looking for. The classic pharmacology of receptor binding usually obeys a 'logistic-like' pattern.
    You can read more on this here:
    http://www.sciencedirect.com/science/article/pii/S1056871914002470
    (article kindly suggested to me by member Stephen Tashi)
    and in this thread that I actually started myself, dealing with the error in IC50 determinations:
    https://www.physicsforums.com/threads/error-on-biological-assay-results.792633/

    Concerning your original question, by appropriate experiments you can measure the actual dissociation constant of the agonist and of the antagonist, and from what I know, in principle there is nothing stopping Kdantagonist from being smaller than Kdagonist for at least one [agonist, antagonist] pair.

    A mathematical proof... I don't think so. I guess to do that you should be able to describe your receptor-ligand complex, in a biological environment, fully and exactly by a system of equations. I may be wrong, but I don't think we're there yet. Several computer programmes are available, which take the crystal structure of a receptor and the structures of several ligands (e.g. designed by you or combinatorially), and give you estimates of their affinity for the receptor. Some of them give very good predictions, and it may be that you can say with a good confidence margin that antagonist X is likely to have a better affinity than agonist A. But I don't suppose that'd count as a proof; you would have to make the molecules and test them :smile:
     
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