Determine the amino acid sequence of a polypeptide from the following data: Complete hydrolysis of the peptide yields Arg, 2 Gly, Ile, 3 Leu, 2 Lys, 2 Met, 2 Phe, Pro, Ser, 2 Tyr, and Val. Treatment with Edman's reagent releases PTH-Gly.
Carboxypeptidase A releases Phe.
Treatment with cyanogen bromide yields the following three peptides:
1. Gly-Leu-Tyr-Phe-Lys-Ser-Met
2. Gly-Leu-Tyr-Lys-Val-Ile-Arg-Met
3. Leu-Pro-Phe
Treatment with trypsin yields the following four peptides:
1. Gly-Leu-Tyr-Phe-Lys
2. Ser-Met-Gly-Leu-Tyr-Lys
3. Val-Ile-Arg 4. Met-Leu-Pro-Phe

Solution :

1. It is given that when Edman's reagent is used, it releases PTH-Glycin. The degradation of Edmans is used for the amino acids to sequencing of the peptide/protein and is also removes N terminal amino acid and thus gives us the PTH amino acid product.

Therefore, the first amino acid is Glycine.

2. The Carboxypeptidase A releases the C terminal amino acid, and therefore the last amino acid is Phe.

3. The Cyanogen Bromide helps to chop after the Med residue, therefore the sequence is 1 - 2 - 3 or 2 - 1 - 3.

4. The trysin chops after the basic amino acid - Arg and the Lys unless followed by Proline and so the sequence of the peptide is 1 - 2 - 3 4.

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