Question

[Briefly explain and answer 1-3]

1.Experimentally, using both kinetic and other approaches, how would you distinguish between an irreversible and a reversible inhibitor? Assume that the inhibitor is a low molecular weight compound.

2.In enzyme kinetics, a key point in the generation of meaningful data is the accurate measurement of the initial velocity. What are the problems associated with the determination of the initial rates? What approaches can be used to avoid these problems?

3.During a protein purification procedure, the following takes place:

a.The sample is prepared, homogenized, boiled, and ammonium sulfate is added the supernatant. The protein that is of interest precipitates along with 5 other proteins.

b.The ammonium sulfate pellet is resuspended in a 6M urea buffer.

c.Upon clarification of the resuspended pellet, the supernatant is loaded onto a CM column.

d.Several proteins come off during the column wash. Two proteins come off as we elute the proteins using a salt gradient.

What information can be gained from (a) relative to our protein of interest?

What information can be gained from (d) relative to our protein of interest?

Propose a method to separate the two proteins that eluted from (d).

Answer 1

1. In an enzyme assay, a reaction is conducted in presence of an inhibitor. With the addition of increasing substrate concentration, if the inhibition decreases, it is a reversible inhibitor else it is an irreversible inhibitor. This is because a reversible inhibitor can also be unbound from the enzyme.

2. It is crucial to get accurate initial velocity in enzyme kinetics. It is important to have a sensitive assay and an instrument (to read the output: spectrophotometer/colorimeter/fluorimeter) to detect the small amount of product formed. It is important to run the assay with as much enzyme and for enough time to allow less than 10% substrate conversion, else product inhibition will affect initial velocity measurements.

3. From step ‘a’, we know that the protein of interest is cytosolic as it was in the cell lysate fraction. It is probably of a smaller size as it is easily precipitated with ammonium sulfate. From step ‘d’, we know that the protein has a net positive charge as it binds the negatively charged CM matrix. The proteins in eluent can be subjected to gel filtration chromatography, to separate them on basis of size.