GATE 2020 BT Q.52: solution and answer

Official answer: 25Source: Official answer key, GATE 2020 BT (IIT Delhi)

1. Gel filtration chromatography is run under native (non-denaturing) conditions, so the 200 kDa peak is the mass of the intact, folded protein A–protein B assembly. This confirms a complex exists but is not itself needed to compute protein B's molecular weight.

2. SDS treatment (with heating) always denatures secondary, tertiary and quaternary (non-covalent) structure in an SDS-PAGE lane, whether or not a reducing agent is present. What differs between the two lanes is only whether the covalent bond linking A and B is broken.

3. Without beta-mercaptoethanol, no reducing agent is present, so the interchain disulfide bond that makes A and B a "covalent complex" survives SDS treatment. A and B therefore still run together as a single covalently linked species, and that band's apparent mass equals $\text{MW}(A)+\text{MW}(B)$.

4. A covalently linked A–B species cannot migrate as if it were lighter than either subunit alone, so of the two observed SDS-PAGE masses (50 kDa and 25 kDa), the larger one must belong to this non-reduced, still-linked species: $$\text{MW}(A)+\text{MW}(B)=50\text{kDa}$$

5. With beta-mercaptoethanol, the disulfide bond is reduced and A and B separate into free monomers. Since only a single band is seen in this lane, the two monomers must co-migrate at the same apparent mass, 25 kDa — consistent with the given $\text{MW}(A)=25\text{kDa}$.

6. Solving: $$\text{MW}(B)=50-\text{MW}(A)=50-25=25\text{kDa}$$

7. Cross-check: with $\text{MW}(A)=\text{MW}(B)=25\text{kDa}$, the reduced lane correctly shows a single 25 kDa band, matching the data. (A native complex made of four such covalent A–B pairs, $4\times 50\text{kDa}=200\text{kDa}$, is one stoichiometry consistent with the gel filtration peak, though this detail is not needed to answer for protein B.)