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Conditional Probability

  1. Jan 29, 2013 #1
    Okay so I have a complex setup that I hope I can convey.

    I have 9 sites to which X can bind. 6 out of the 9 sites are active and 3 out of the 9 sites are inactive. I need 3 of the active sites to be bound to get the response I am looking for - which we will call EMAX.

    So when I add a single X - the chance of it binding to an active site is 6/9 the chance of it binding to an inactive site is 3/9.

    My probability knowledge is shaky - bear with me.

    Assume that the binding is irreversible. So how many X do I need to add to be sure I have activated 3 active sites. Or more precisely, how many X do I need to add to get a >95% chance that 3 active sites are bound.

    Then I want to go more complicated. Say I add 3 Y - which inactivates the sites. The chances are that 2 active sites will be inactivated and 1 inactive site will still be inactive with Y bound.

    Now under these new conditions - how much X do I need to add to be sure 3 remaining active sites are occupied?

    So I know it will be a probability - so I guess lets say that how much X do I need to add to have a greater than 95% chance that 3 active sites are now bound with X to get EMAX
  2. jcsd
  3. Jan 29, 2013 #2
    PS - this ain't homework! On an intrinsic level its clear to me that in the new condition more X has to be added to ensure that 3 active sites are bound (I hope my intrinsic thoughts are correct!) - but I want to be able to put a number on it.

    FYI - this is a real word problem I am trying to figure. The problem comes from trying to explain why the potency of a drug is affected initially (but not the efficacy) when you add a small concentration of antagonist to a population of receptors that have spare receptors included. Then as you increase antagonist concentration you finally see a fall in EMAX - as there are not enough active sites left for agonist to work at.

    My probability maths does not extend beyond coin flipping and dice rolling - hence I hope any explanations are at a level I can appreciate!!
  4. Jan 29, 2013 #3


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    For probability 1, it's fairly easy to see that you need at least 6 X. After all, the worst case scenario is that you bind all the inactive sites first, and then the next 3 X that you place are active.

    For the rest, this is precisely the model that I (and probably a lot of other people too) use when thinking about hypergeometric probabilities. Suppose that you place n X on arbitrary sites. You can do this in [itex]\binom{9}{n}[/itex] ways (the number of ways you can choose n sites to bind to from the 9 available. The probability that k are bound to any of the 3 inactive sites and n - k to active sites is then
    [tex]\frac{\binom{3}{k} \binom{6}{n - k}}{\binom{9}{n}}[/tex]
  5. Jan 29, 2013 #4
    OKay - there are stupid questions even though we tell everyone there ain't - so just so I am totally clear and making no assumptions - define k.
  6. Jan 29, 2013 #5


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    k is the number (out of the n sites that you are binding to an X) of X that are bound to an inactive site.

    For example, the probability that if you place 4 X's and exactly 1 is on an inactive site and 3 are on an active site can be calculated by plugging in n = 4, k = 1.

    Of course, you are not interested in a single value of k but in all possible values (question back: what are the allowed values?)
  7. Jan 29, 2013 #6
    well in situation 1 k = 3 i guess, but then in situation 2 when we add the antagonist - it is more difficult - since antagonist can either bind to an inactive receptor and do nothing (for the first antagonist particle there is a 3/9 chance of that), but then there is a possibility that the antagonist binds to an active receptor and inactivates it - and thus the pool of inactive receptors (k) will increase. Then we also have to think about the fact that I have 3 sequential antagonists binding and so that then changes the k pool too?

    Am i making sense?
  8. Jan 29, 2013 #7


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    I didn't really get that far yet, I thought we'd solve the simpler problem first.
    If the binding of X's and Y's is independent you can find the probability that not 3 but 3 + d sites are deactivated and for every value of d you can find the probability that the X's will activate 3 sites (where the case d = 0 is the simpler one that I was looking at).
  9. Jan 29, 2013 #8
    I should add though that your worst case scenario take on the problem will let me explain it beautifully. As you say in instance 1 you need 6 agonist - 3 to use up the inactive sites and the next 3 are guaranteed to fill active sites.
    Extending that into the second scenario - when i first add 3 antagonist particles the likeliest result is 2 active receptors are bound to 1 inactive receptor (probability is 3 inactive being bound 1/84, 3 active being bound is 20/84 - not sure how to discriminate 2 active 1 inactive or 2 inactive 1 active - think it is 45/84 that we have 2 active 1 inactive and 18/84 that 2 inactive and 1 active) - thus the total pool of inactive receptors is now 5. Therefore 5+3 agonist is required worst case to get the same response.
    That doesn't actually calculate the probabilities as I initially thought I would have to do - but it explains far more simply why we need more agonist in the presence of antagonist to get the same response as when no antagonist is present.
  10. Jan 29, 2013 #9
    And yes - the binding of X and Y is independent and each event does not change the binding of the other. So a single receptor can have an X and a Y bound. But would not be active.
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