Learn to judge whether a specific binding interaction is selective enough to all

Question

Learn to judge whether a specific binding interaction is selective enough to allow for binding in a complex biomolecular environment.

An enzyme E is involved in a serious disease and the research community is searching for potential drugs that can bind to the active site of the enzyme E and inhibit its activity. The enzyme E is commonly found in the cytosol of liver cells at a concentration [E] = 10.0 nM (i.e. 1.00 x 10-8 mol/l).

A hopeful drug candidate L has been identified, which forms a complex with E with a dissociation constant KELdiss of 1.00 nM (i.e. 1.00 x 10-9 mol/l). However, the drug also forms complexes with another protein X with a dissociation constant of KXLdiss= 100 nM (i.e. 1.00 x 10-7 mol/l). The concentration of the protein X in the cytosol of liver cells is [X] = 10.0 µM (i.e. 1.00 x 10-5 mol/l).

Is the drug molecule L able to selectively bind the drug target E without significant formation of the competing complex XL upon increasing the concentration of L within the cytosol of liver cells? Is A or B correct?

a. The drug molecules will bind selectively to the drug target E to form the desired EL complex. Only at high total concentrations of L also the weaker XL complex forms, but essentially all E molecules will be bound in the more stable EL complexes then.

b. The drug molecules L will primarily bind to X to form the XL complex. Only for very high total concentrations of L, when a significant amount of XL is already formed, will the drug bind to E and form the EL complex. Selective binding to E is not possible under these conditions!

Explanation / Answer

In present case drug L is used to inhibit the harmful effect of enzyme E, but problem is that L can also bind with protein X due to which its efficiency to bind with E may get affected.

By their dissociation constants and concentrations we can predict behaviour of L toward E and X

Now dissociation constant of EL = 1nM

Dissociation constant of XL = 100nM

Smaller the dissociation constant, more will be the affinity between components to form complex.

This indicates EL is more stable than XL because tendency of XL to get dissociated is more than EL

If we see the concentrations [E] =10-8 M and [X]=10-5M

Conc. Of X is more than E which shows higher chances to form complex with L in the form of XL

But as L has higher affinity to bind with E, so it will form complex with E first . After that when all E gets converted in to EL and still L is left then weaker complex XL may form. So ( A) is correct.

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