Relationship between Kd and bonding affinity?


by τheory
Tags: affinity, bonding, relationship
τheory
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#1
Oct19-13, 06:32 PM
P: 43
From a biochemical context, considering the following two dissociation reactions and their respective dissociation constants for a protein-ligand complex:

P*A + B ⇔ P + A + B, this contains the a dissociation constant called Ka = 4 x 10^-3 M

P*B + A ⇔ P + A + B, this contains the a dissociation constant called Kb = 2 x 10^-7 M

Compare the values of Ka and Kb; Does the free protein bind ligand A or ligand B more tightly? I know that a lower dissociation value means tighter bonding or "higher affinity," so the free protein should bind to ligand B more tightly, but can someone explain to me why this is the case?
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Oct20-13, 01:23 AM
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Quote Quote by τheory View Post
From a biochemical context, considering the following two dissociation reactions and their respective dissociation constants for a protein-ligand complex:

P*A + B ⇔ P + A + B, this contains the a dissociation constant called Ka = 4 x 10^-3 M

P*B + A ⇔ P + A + B, this contains the a dissociation constant called Kb = 2 x 10^-7 M

Compare the values of Ka and Kb; Does the free protein bind ligand A or ligand B more tightly? I know that a lower dissociation value means tighter bonding or "higher affinity," so the free protein should bind to ligand B more tightly, but can someone explain to me why this is the case?
You seen to have answered your own question, so I don't know really what it is. But to explain to yourself "why this is the case" just write down the equation for the equilibrium, then ask yourself what is the ligand concentration when the protein is half-saturated, i.e. bound protein = free protein.

Your first equation contains a B on both sides, and your second an A which if they were not there would make no difference and might confuse you less.
τheory
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Oct20-13, 03:02 AM
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So the question, of why one ligand binds more strongly than the other, pertains to the idea that in a system, ligand B will have a higher concentration of bound vs unbound proteins compared to ligand A at the half saturation level?

Edit: To your previous statement, I did answer my own question implicitly due to the way I phrased the original post, that was my mistake. What I meant to say was that I uncovered a principle (lower Kd = tighter bonding) from my textbook that I didn't fundamentally understand. As a result, I wanted someone to try to explain this concept.

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Oct20-13, 05:37 AM
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Relationship between Kd and bonding affinity?


Quote Quote by τheory View Post
So the question, of why one ligand binds more strongly than the other, pertains to the idea that in a system, ligand B will have a higher concentration of bound vs unbound proteins compared to ligand A at the half saturation level?
More like at half saturation with B, [B] will be at lower concentration than [A] will be at half saturation with A.

It does not sound like you have written the equilibrium equation as I suggested; if you do you should see what's what. To use the half-saturation point is a convenience, but a pretty essential one you will meet constantly.

It will also help you see how this works if you calculate the ratio of liganded to unliganded protein at 10-5M free ligand in cases A and B which have very different affinities.


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