A heptapeptide contained the following amino acids: Asp, Leu, Lys, Met, Phe, Tyr

Question

A heptapeptide contained the following amino acids: Asp, Leu, Lys, Met, Phe, Tyr. The following facts were observed during sequencing. Edman degradation resulted in a PTH-derivative, containing Phe side chain. Trypsin treatment had no apparent effect. Chymotrypsin treatment yielded several products, including a dipeptide and a tetrapeptide. The amino composition of tetrapeptide was Leu, Lys, and Met. Cyanogen bromide (CNBr) treatment yielded a dipeptide, a tetrapeptide. and free Lys. Determine the amino acid sequence of this peptide based on the above information. Elaborate your reasoning. Write the sequence in 1-letter abbreviations.

Explanation / Answer

1. The edmann degradtion gives the product with the N-terminal amino acid. So the N-terminal amino acid is Phe

2. Trypsin cleaves the peptide at the C-ter of Lys, Arg. So, the peptide does not have any Lys in the middle. The last amino acid is Lys

3. Chymotrypsin selectively cleaves C-ter of peptide bonds formed by aromatic residues (tyrosine, phenylalanine, and tryptophan). The dipeptide ends with Tyr.

4. CNBr cleaves at the C-ter of Met residues. The peptide has 2 Met residues as you get a dipeptide, tetrapeptide and free lys

The peptide sequence is Phe-Asp-Tyr-Met-Leu-Met-Lys

One leter abbreviation F-N-Y-M-L-M-K

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