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Deduce the sequence of a heptapeptide that contains the amino acids Ala, Arg, Glu, Gly, Leu, Phe, and Ser, from the following experimental data. Edman degradation cleaves Leu from the heptapeptide, and carboxypeptidase forms Glu and a hexapeptide. Treatment of the heptapeptide with chymotrypsin forms a hexapeptide and a single amino acid. Treatment of the heptapeptide with trypsin forms a pentapeptide and a dipeptide. Partial hydrolysis forms Glu, Leu, Phe, and the tripeptidesGly–Ala–Ser and Ala–Ser–Arg

Short Answer

Expert verified

³¢±ð³Ü–G±ô²â–A±ô²¹â€“S±ð°ù–A°ù²µâ€Ñï³ó±ð–G±ô³Ü

The sequence of the given heptapeptide

Step by step solution

01

Edman degradation

The method of sequencing peptides without disturbing the peptide bonds is Edman degradation. During this method, N-terminal amino acid Leu is cleaved first.

The sequence becomes like this in the given heptapeptide:

³¢±ð³Ü–_³å³å–_³å³å–_³å³å–_³å³å–_³å³å–_³å³å

02

Chymotrypsin Cleavage

The treatment of heptapeptide with chymotrypsin forms a hexapeptide and a single amino acid since chymotrypsin cleaves at an amide bond with a carbonyl group from Phe, Tyr, or Trp.

The sequence becomes like this in the given heptapeptide:

³¢±ð³Ü–_³å³å–_³å³å–_³å³å–_³å³å–_³å³å–G±ô³Ü

03

Partial hydrolysis

Partial hydrolysis forms Glu, Leu, Phe, and the tripeptidesGly–Ala–Ser and Ala–Ser–Arg

Hence, the sequence becomes:

³¢±ð³Ü–G±ô²â–A±ô²¹â€“S±ð°ù–A°ù²µâ€“_³å³å–G±ô³Ü

or

³¢±ð³Ü–_³å³å–G±ô²â–A±ô²¹â€“S±ð°ù–A°ù²µâ€“G±ô³Ü

04

Cleavage by trypsin

Cleavage by trypsin is after Arg and yields a dipeptide; therefore, this must be the peptide:

³¢±ð³Ü–G±ô²â–A±ô²¹â€“S±ð°ù–A°ù²µâ€Ñï³ó±ð–G±ô³Ü

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