Correct option is B
Explanation-
Experiment 1 (with DTT - reducing agent):
Revealed two polypeptides after breaking disulfide bonds:
Polypeptide I:
Ala–Phe–CysA³–Met–Tyr–CysA⁶–Leu–Trp–CysA⁹–Asn
Polypeptide II:
Val–CysB²–Trp–Val–Ile–Phe–Gly–CysB⁸–Lys
So, we have Cys residues at A³, A⁶, A⁹ and B², B⁸
These cysteines are potential candidates for disulfide bond formation before DTT treatment.
Experiment 2 (Chymotrypsin digestion):
Chymotrypsin cleaves after aromatic residues (Phe, Tyr, Trp).
Peptide fragments obtained:
1. [Ala, Phe]
2. [Asn], 2[Cys], [Met], [Tyr]
3. [Cys], [Gly], [Lys]
4. 2[Cys], [Leu], 2[Trp], [Val]
5. [Ile], [Phe], [Val]
Step-by-step interpretation:
Map the fragments from chymotrypsin digestion onto the original sequences to find linked cysteines.
Fragment II: [Asn], 2[Cys], [Met], [Tyr]
Contains Asn and two Cys residues — likely from A³–Met–Tyr–CysA⁶–Leu–Trp–CysA⁹–Asn
Suggests a fragment involving CysA⁶ and CysA⁹, which are not cleaved apart, hence likely linked by a disulfide bond.
Fragment IV: 2[Cys], [Leu], 2[Trp], [Val]
Corresponds well with CysA³–Leu–Trp–CysA⁹ from polypeptide I and CysB²–Trp–Val from polypeptide II
Again indicates inter-chain disulfide bonding, possibly CysA³–CysB²
So from the above:
1. CysA⁶–CysA⁹ are linked
2. CysA³–CysB² are also likely linked
Correct Option: b - A3–A9 and B2–A6
This matches the disulfide bond pattern deduced.


