Question

Action of papain on IgG molecule results in

• A. Two identical fragments
• B. Three fragments and two of which are identical
• C. Three identical fragments
• D. Three different fragments

Hint:

Antibody Enzymatic Cleavage:
Papain $\rightarrow 2\text{ Fab (identical)} + 1\text{ Fc}$ (3 fragments total).
Pepsin $\rightarrow 1\text{ F(ab')}_2 + \text{degraded Fc}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Proteolytic digestion of immunoglobulins by papain vs pepsin cleaves specific peptide bonds relative to the hinge disulfide bonds.
Key Formula or Approach:
\[ \text{IgG} \xrightarrow{\text{Papain (cleaves above hinge)}} 2\text{ Fab (identical)} + 1\text{ Fc} \quad \text{[3 fragments total]} \]

Step 2: Detailed Explanation:
Proteolytic cleavage of IgG:
1. Papain Digestion (Rodney Porter): Cleaves the heavy chains on the N-terminal side (above) the interchain hinge disulfide bonds, producing three separate fragments:
- Two identical Fab (Fragment antigen-binding) fragments (each monovalent, $\approx 50\text{ kDa}$).
- One Fc (Fragment crystallizable) fragment ($pprox 50\text{ kDa}$).
2. Pepsin Digestion: Cleaves below the hinge, yielding one bivalent $\text{F(ab')}_2$ and degraded Fc subfragments.

Step 3: Final Answer:
Therefore, papain cleavage produces Three fragments and two of which are identical, corresponding to option (B).