Hello fellow scientists,
I have some additional questions about the dissociation constant for a protein binding ligand.
My main questions is if the protein concentration is much greater than the KD and the ligand concentration is still in excess of the K_D how much protein is bound?
For a Protein binding a ligand, we have the relationship:
P + L ⇌ PL . This has forward and reverse rates of binding, and ...
KD = {[P] * [L]} / [PL],
The fraction of protein bound (FPL) will be:
FPL = [PL] / ([P] + [PL]) = {[L] / ([L] + KD)}
If the ligand concentration is some multiple integer n of the KD we get this cool relationship:
[L] = n * KD ; n = 1, 2, 3, ... etc.
FPL= (n * KD) / (n*KD + KD)
= n / (n + 1)
What they teach you in school is that if the ligand concentration is at the KD, n = 1 then half the protein is bound: 1 / (1 + 1) = 1 / 2. This relationship also tells you that if [L] = 9 * KD then 90% of the protein is bound.
One can plot this relationship as I have.
So I have some questions assuming a reasonable good ligand interaction with KD = 10 µM,
(A) If [L] = 9 * KD, and fixed protein concentration = KD = 10 µM, how much protein is bound?
I think that 90% of the protein is bound.
B) Say you are a structural biologist and you need more ligand and protein than the KD at 10 µM,
If [L] = 500 µM (that is 50KD) and initial protein concentration is 100 µM (that is 10KD), then how much protein is bound?
Given the relationship I derived, {50 / (50 + 1)} = 98% of the protein should be bound.
But the protein concentration is beyond the KD so can that even be correct?
Here is the problem. In the math above, [L] is not the total ligand concentration. It is the free ligand concentration. If [P] is near or above the Kd, and [L] is not >> [P], then the free and total ligand concentrations will be significantly different, and the mathematical relationships above will give incorrect results.
In the so-called tight-binding domain, which is the situation just described, you should calculate [PL] using the tight-binding (Morrison) equation.
FPL = {(Kd + PT + LT) - square root[(Kd + PT + LT)2 - 4PTLT]}/(2PT)
where PT andLT are the total concentrations of P and L.
Here is the problem. In the math above, [L] is not the total ligand concentration. It is the free ligand concentration. If [P] is near or above the Kd, and [L] is not >> [P], then the free and total ligand concentrations will be significantly different, and the mathematical relationships above will give incorrect results.
In the so-called tight-binding domain, which is the situation just described, you should calculate [PL] using the tight-binding (Morrison) equation.
FPL = {(Kd + PT + LT) - square root[(Kd + PT + LT)2 - 4PTLT]}/(2PT)
where PT andLT are the total concentrations of P and L.
Adam B Shapiro , thank you very much. I find the Morrison Equation very interesting. I have been trying to search for the derivation of that equation, and I see forms of that equation in kinetics which substitute Michealis Constant KM for KD. How is using KD appropriate? Could you guide me an explanation please or direct me to the simplest derivation? Thank you!
Adam B Shapiro, with a little big of googling I found the derivation of the equation you used on some educational slides from the University of Texas.
http://hackert.cm.utexas.edu/courses/bch370/fall2014/Ligand%20Binding/Lig_binding_f14.pdf
Seems that the derivation defines the fractional saturation of protein the same way I do. The total ligand concentration is defined which is what is known by the investigator and this is substituted into the equation for the free substrate concentration. Grouping [ES] or [PL] terms gives a quadratic with a,b,c terms.
Those terms are inputted into the quadratic equation, and the Morrison Equation arises. So there is no assumptions involved in this Morrison Equation, right?
Given my conditions, [P]Total = 100 µM, [L]total =500 µM, KD = 10 µM then this equation predicts that 97.6% of my protein should be bound. Excluding the black magic of crystallography,. there is a 2.4% chance that my ligand won't be bound in a co-crystallization.
97.6% of the protein would be bound to ligand. Most of the ligand would be free, since it is in excess over protein.
In a crystal of the protein, the local protein concentration is much higher than the average concentration in the well, so the equilibrium would be shifted. It's complicated, because it's a two-phase system, with crystalline protein competing with dissolved protein for the pool of ligand.
The smaller the dissociation constant, the more tightly bound the ligand is, or the higher the affinity between ligand and protein. For example, a ligand with a nanomolar (nM) dissociation constant binds more tightly to a particular protein than a ligand with a micromolar (μM) dissociation constant. https://en.m.wikipedia.org/wiki/Dissociation_constant
The dissociation constant is commonly used to describe the affinity between a ligand (such as a drug) and a protein; i.e., how tightly a ligand binds to a particular protein. Ligand-protein affinities are influenced by non-covalent intermolecular interactions between the two molecules such as hydrogen bonding, electrostatic interactions, hydrophobic and van der Waals forces. Affinities can also be affected by high concentrations of other macromolecules, which causes macromolecular crowding
https://en.m.wikipedia.org/wiki/Dissociation_constant
The smaller the dissociation constant, the more tightly bound the ligand is, or the higher the affinity between ligand and protein. For example, a ligand with a nanomolar (nM) dissociation constant binds more tightly to a particular protein than a ligand with a micromolar (μM) dissociation constant.
https://en.m.wikipedia.org/wiki/Dissociation_constant
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