Relationship between Kd and bonding affinity?

[τheory]

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?

 

[epenguin]

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]

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.

 

[epenguin]

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, 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.

 

[LudusRex]

In biochemistry, the dissociation constant (Kd) is a measure of the strength of the interaction between a protein and its ligand. It is defined as the concentration of ligand at which half of the binding sites on the protein are occupied. Therefore, a lower Kd value indicates a stronger binding affinity between the protein and its ligand.

In this context, the dissociation constants Ka and Kb represent the dissociation rates for two different dissociation reactions involving the same protein and two different ligands. The values of Ka and Kb are inversely proportional to the corresponding Kd values, as shown by the equations Ka = 1/Kd and Kb = 1/Kd. This means that a lower Ka or Kb value corresponds to a lower Kd value and therefore a higher binding affinity between the protein and its ligand.

In the given reactions, Ka = 4 x 10^-3 M and Kb = 2 x 10^-7 M. Comparing these values, we can see that Kb is significantly lower than Ka, indicating a higher binding affinity between the protein and ligand B. This means that the free protein will bind to ligand B more tightly than ligand A because the dissociation rate for the reaction involving ligand B is much lower than that of the reaction involving ligand A.

In summary, the lower dissociation constant (Kd) indicates a tighter binding or higher affinity between a protein and its ligand. In the given scenario, the free protein will bind to ligand B more tightly due to its lower dissociation constant (Kd) compared to ligand A.