Question

A protein was purified to homogeneity. Determination of the molecular weight by molecular exclusion
chromatography yields 60 kDa. Chromatography in the presence of 6 M urea yields a 30 kDa species. When
the chromatography is repeated in the presence of 6 M urea and 10 mM B-mercaptoethanol, a single molecular
species of 15 kDa results. The protein is a
a. homodimer of 30 kDa
b. homodimer of 15 kDa
c. homotetramer of 30 kDa
d. homotetramer of 15 kDa

solve this...​

Answer 1

Answer:

As urea is a denaturing agent, it will denature the protein-protein interaction and also the oligomeric state by disrupting all weak interactions forces but not the di-sulfide (covalent) bonds. But in the presence of beta-mercaptoethanol, the disulphide bond will get cleaved and we will get the purified monomer in chromatography. That's what happens in presence of 6 M urea + 10 mM beta-mercaptoethanol.

So, the protein is 15 KDa subunit, forms a dimer with another 15 KDa subunit through disulfide bond and this dimer interacts with the same dimer complex through weak interaction forces (for e.g. Hydrophobic or electrostatic) and makes the dimer of the dimer. Thus native condition yields 60 Kda, denaturing condition gives 30 Kda, and beta-mercaptoethanol gives yield at 15 kDa.

So, overall 4 subunits this protein have

Explanation: