Km And Binding Constant (Mcat 2015)?

[medicalchanel]

Hi everyone!

 

I understand that Km is the substrate concentration at which the velocity of the reaction is at half its maximum (vmax/2). However, I'm just confused in biochemistry, when a passage states that a substrate has a Ka (for example) of a certain value for a specific material.

 

Am I right in assuming that a low value K would mean a decreased binding affinity (and a high K would be increased binding affinity?)

 

Thank you!

[Username100]

Can you post the question about the binding constant? 

[Paggles01]

Depends on the context. That's why it's important to read questions and passages very carefully. In receptor/ligand biochemistry, we usually talk about affinity in terms of Ka - the association constant (aka binding constant). Loosely speaking, the higher a constant, the more likely a system is to be found in the state that the constant primarily describes. If ligand A and enzyme B combine into AB: A + B --> AB, then at equilibrium Ka = k(on)/k(off) = [AB]/[A], where k(on) and k(off) are the rate constants for the forward and reverse reactions, respectively. Stated more simply, a higher Ka means that equilibrium favors a higher ratio of receptor-ligand complexes to free substrates.

 

Now, here's where it's important to consider context. Dissociation constant can be calculated as 1/Ka for the same reaction, but primarily refers to unbound substrates. Thus, a relatively higher Kd describes a greater ratio of unbound substrate to complexed receptor-ligand at equilibrium.

TL;DR: Yes, a relatively lower Ka would imply lower binding affinity.  

[doc123]

Be careful though! High Km means low affinity of enzyme to substrate! This makes sense as it implies that more substrate is needed to reach 1/2Vmax. Whereas, low Km implies high affinity 

[doc123]

I have only come across Ka in the context of acids and bases. I do remember learning about Ka in biochemsitry as being opposite of Kd