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Conditional Probability

by mtc1973
Tags: conditional, probability
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mtc1973
#1
Jan29-13, 07:37 AM
P: 112
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
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mtc1973
#2
Jan29-13, 08:01 AM
P: 112
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!!
CompuChip
#3
Jan29-13, 08:07 AM
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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]

mtc1973
#4
Jan29-13, 08:33 AM
P: 112
Conditional Probability

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.
CompuChip
#5
Jan29-13, 10:34 AM
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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?)
mtc1973
#6
Jan29-13, 11:40 AM
P: 112
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?
CompuChip
#7
Jan29-13, 11:59 AM
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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).
mtc1973
#8
Jan29-13, 12:08 PM
P: 112
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.
mtc1973
#9
Jan29-13, 12:11 PM
P: 112
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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