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    Two experiments were performed on a peptide sample ‘X’. In experiment 1, treatment of ‘X’ with dithiothreitol (DTT), followed by blocking of free sulp
    Question

    Two experiments were performed on a peptide sample ‘X’. In experiment 1, treatment of ‘X’ with dithiothreitol (DTT), followed by blocking of free sulphydryl groups, yielded two polypeptides whose amino acid sequences are as shown below:

    I. Ala-Phe-CysA³-Met-Tyr-CysA⁶-Leu-Trp-CysA⁹-Asn
    II. Val-CysB²-Trp-Val-Ile-Phe-Gly-CysB⁸-Lys

    In experiment 2, ‘X’ was treated with chymotrypsin, a protease that cleaves the carboxy-terminal of aromatic residues. The amino acid composition of five peptides obtained from this experiment are shown below:

    I. ([Ala], [Phe])
    II. ([Asn], 2[Cys], [Met], [Tyr])
    III. ([Cys], [Gly], [Lys])
    IV. (2[Cys], [Leu], 2[Trp], [Val])
    V. ([Ile], [Phe], [Val])

    Based on the above results, which cysteine/s are linked by disulfide bond in peptide ‘X’?

    A.

    A6-B2 and B8-A3

    B.

    A3-A9 and B2-A6

    C.

    A3-B2 only

    D.

    A3-B8 only

    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.


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