Amino acid analysis of a decapeptide revealed the presence of the following products: \(\begin{array}{lllll}\mathrm{NH}_{4}^{+} & \text {Asp } & \text { Glu } & \text { Tyr } & \text { Arg } \\ \text { Met } & \text { Pro } & \text { Lys } & \text { Ser } & \text { Phe }\end{array}\) The following facts were observed: a. Neither carboxypeptidase \(A\) or \(B\) treatment of the decapeptide had any effect. b. Trypsin treatment yielded two tetrapeptides and free Lys. c. Clostripain treatment yielded a tetrapeptide and a hexapeptide. d. Cyanogen bromide treatment yielded an octapeptide and a dipeptide of sequence NP (using the one-letter codes). e. Chymotrypsin treatment yielded two tripeptides and a tetrapeptide. The N-terminal chymotryptic peptide had a net charge of -1 at neutral \(\mathrm{pH}\) and a net charge of -3 at pH 12 f. One cycle of Edman degradation gave the PTH derivative What is the amino acid sequence of this decapeptide?

Step by step solution

01

Enzyme and treatment characteristics

Identify the characteristics of each treatment. Carboxypeptidase A and B, release amino acids from the carboxyl terminal while trypsin cleaves peptide chains at the C-side of lysine and arginine unless either is followed by proline. Clostripain cleaves at the carboxyl side of arginine, whereas cyanogen bromide cleaves at the carboxyl side of methionine. Chymotrypsin cleaves at the carboxyl side of phenylalanine(F), tryptophan(W), tyrosine(Y) and Leucine(L), again while none is followed by proline. Edman degradation identifies the N-terminal residue.

02

Decapeptide partial sequences

Carboxypeptidase A and B have no effect, indicating the decapeptide lacks carboxyl terminal residues targeted by these enzymes. Trypsin treatment gives two tetrapeptides and free Lys, showing Lys is at the carboxyl terminal. Clostripain yields a tetrapeptide and a hexapeptide, meaning Arg is fifth from the carboxyl terminal. Cyanogen bromide treatment produces octapeptide and dipeptide with sequence NP, so Met precedes NP at the carboxyl terminal. Chymotrypsin gives two tripeptides and a tetrapeptide, hence, Tyr or Phe is fourth from the carboxyl terminal while the other is at the N-terminal. The N-terminal peptide has a net charge of -1 at neutral pH implying presence of one Asp (acidic) and one Lys (basic). At pH 12, the net charge is -3, indicating deprotonation of Tyr. Edman degradation shows Pro as the first amino acid.

03

Sequence formulation

Stitching this information together while fulfilling given conditions dictates the sequence: Pro-Phe-Asp-Tyr-Met-Asn-Pro-Arg-Glu-Lys

04

Confirmation

Check this sequence with all treatment conditions to confirm. This peptide has four negative charges because of Glu, Asp, Tyr, and C-terminal COOH group and three positive charges because of Arg, Lys, and NH3+ of Pro at the N-terminal. At neutral pH, it will bear net negative charge -1 (four negative charges from COOH group and Tyr, Asp, Glu and Lys's NH3+). At pH 12, amine groups are protonated. Hence, COOH of Glu and Asp, plus OH of Tyr are ionized, making the net charge -3. Hence, the sequence meets all conditions.

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