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How do you calculate dissociation constant?

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How do you calculate dissociation constant?

Postby globy123 » Mon Sep 29, 2008 4:53 am

How do you calculate the dissociation constant if 6 mM ATP is present in solution with 10 mM protein, and 20% of the ATP is bound to the protein???
Last edited by globy123 on Tue Sep 30, 2008 2:00 am, edited 2 times in total.
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Postby alextemplet » Mon Sep 29, 2008 11:57 am

Calculate how much equals 10% of 100mM ATP and figure out how much protein it takes to bind to that amount. Then figure that number as a percentage of the total amount of protein and you'll have your answer.
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Postby globy123 » Mon Sep 29, 2008 2:25 pm

How do you find out how much protein it takes to bind to 10mM of ATP? Does 1 ATP bind to 1mM?
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Postby alextemplet » Mon Sep 29, 2008 3:53 pm

Look up the characteristics of the given protein to be sure, but in general I'd say that's a fair bet.
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Postby globy123 » Mon Sep 29, 2008 4:03 pm

Not very helpful
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Postby alextemplet » Mon Sep 29, 2008 4:14 pm

Just do the math. You'll be surprised how easy it is.
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Postby globy123 » Mon Sep 29, 2008 4:18 pm

I did the math as follows Kd=20(100)/[(10)(10)] and I got 0.02M which is not the right answer. What am I doing wrong?
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Re: How do you calculate dissociation constant?

Postby blcr11 » Mon Sep 29, 2008 5:33 pm

Kd = ( [P] [ATP])/([P-ATP])

I’m assuming a 1:1 complex of protein with ATP. 10% of your ATP represents 10 mM of bound ATP. But that means that your bound protein concentration is also 10 mM (the bound protein and bound ATP concentrations have to be equal assuming a 1:1 stoichiometry of binding) leaving you with 20-10 or 10 mM of free protein that is not complexed with any ATP, [P], at equilbrium. Since you used 10 mM of your total ATP to bind to protein, you now only have 100-10 mM or 90 mM of free ATP, [ATP], at equilibrium. The numbers you want to plug into the expression for Kd are the equilibrium numbers, not the total input concentrations:

[P] = 10 mM
[ATP] = 90 mM
[P-ATP] = 10 mM

Kd = (10)(90)/(10) = 90 mM or 0.09 M whichever you prefer.
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Postby blcr11 » Mon Sep 29, 2008 5:40 pm

It just so happens that [P] and [P-ATP] cancel in the expression for Kd the way this problem was set up. It won't always be the case. For example, if the total amount of protein had been 0.03 M, while the same 10% of ATP was bound, you would get [P] = 20 mM instead of 10 mM and you would have a different Kd even though [P-ATP] would still be 10 mM and [ATP] would still be 90 mM.
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Postby globy123 » Mon Sep 29, 2008 8:16 pm

I don't understand why the numbers you plug into the expression for Kd are the the equilibrium numbers and not the total input concentrations??
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Postby blcr11 » Tue Sep 30, 2008 1:22 am

Kd is one form of an equilibrium constant. You can write a reaction as either a dissociation or association reaction.

AB <--> A + B, Kd = [A][B]/[AB]

or

A + B <--> AB, Keq = 1/Kd = [AB]/[A][B]


Either way, the constants are determined by the equilibrium concentrations of the reactants and products. If you mix total and equilibrium concentrations together in the same expression the way you did, you aren't describing the situation correctly and the calculated equilibrium constant will be incorrect.
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Postby blcr11 » Tue Sep 30, 2008 2:40 am

Sorry, somehow I got off with the wrong set of total concentrations. The original problem had 6 mM ATP with 20 % of ATP bound in the presence of 10 mM protein. The reasoning is the same, though.

If 20% of the ATP is bound, then the concentration of bound ATP is 0.2x6 mM = 1.2 mM. This is also the concentration of bound Protein. If the bound ATP and Protein concentrations are 1.2 mM, then the free ATP concentration is 4.8 mM and the free Protein concentration is 8.8 mM.

Kd = ( 4.8 x 8.8 ) / 1.2 = 35.2 mM
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