Three amino acids — Tryptophan (A), Methionine (B), and Proline (C) — can form dipeptides. A dipeptide Y contains exactly one sulfur atom and two nitrogen atoms. Let P = number of oxygen atoms in Y, and Q = number of chiral centers in Y. Find (P + Q).

Question

Accepted Answer

5. Sulfur -> Methionine (B) must be present. Methionine has 1 N. For Y to have 2 N total, the second amino acid must contribute 1 N. Tryptophan has 2 N (indole + alpha-amino), Proline has 1 N. If Y = Methionine + Proline (B+C): total N = 1+1 = 2. Good. Structure of dipeptide Met-Pro (or Pro-Met): In Met: alpha-C is chiral. In Pro: alpha-C is chiral (proline's N is in ring). Peptide bond formation: -COOH + H2N- -> -CO-NH- + H2O. Met oxygen atoms: 2 (COOH gives 1 CO and 1 OH... in peptide: one COOH at C-terminus = 2O, one peptide bond CO = 1O). Let me count directly: Met-Pro dipeptide: H2N-CH(R_met)-CO-N(R_pro_ring)-CH(R_pro)-COOH where R_pro closes back. Oxygens: peptide C=O (1), C-terminal C

Three amino acids — Tryptophan (A), Methionine (B), and Proline (C) — can form dipeptides. A dipeptide Y contains exactly one sulfur atom and two nitrogen atoms. Let P = number of oxygen atoms in Y, and Q = number of chiral centers in Y. Find (P + Q).

Solution

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