Question

Consider an experiment where your goal is to isolate Bruton’s tyrosine kinase (BTK) enzyme from a whole‑cell lysate. You have an affinity chromatography column with a tyrosine kinase inhibitor molecule covalently attached to the beads. The tyrosine kinase inhibitor binds and inhibits BTK. As a result of the experiment, you are able to elute BTK from the column, but in a mixture of other tyrosine kinases.

Why are tyrosine kinases other than BTK present in the eluate?
(A) The kinase inhibitor has low binding affinity.
(B) BTK is rapidly degraded during cell lysis.
(C) The kinase inhibitor has low specificity.
(D) BTK is inactive in the cell.

Answer 1

(C) The kinase inhibitor has low specificity

Step-by-step explanation:

In the affinity chromatography usually there is a ligand immobilized in the column matrix that can binds the protein or macromolecule of interest.

In this case, it was chosen to use an inhibitor of the Bruton's tyrosine kinase (BTK) enzyme, that it is supposed to bind the protein strongly enough to separate it from the mixture.

However, BTK is eluted along with other tyrosine kinases. This suggest that this inhibitor could be able to bind other kinases.

Note that if the BTK was degraded in the cell, then the enzyme wouldn't be in the whole-cell lysate.

Also, if the inhibitor was of low binding affinity, then it wouldn't be suitable for an affinity chromatography, as the protein would unbind from the matrix during the washes. So, BTK wouldn't be found in the eluted fraction, or would be in low concentration.

Finally, if the kinase was inactive in the cell, it would already have an inhibitor molecule bound in the corresponding site. Then it wouldn't attach to the matrix and wouldn't be found in the eluted fraction.

In summary, the most likely reason for not being able to isolate BTK alone, is that the inhibitor can bind multiple kinases, so a mixture of enzyme is obtained after the chromatography. In other words, the kinase inhibitor has low specificity.