A sample of an unknown peptide was divided into two aliquots. One aliquot was tr

Question

A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin and the other was treated with cyanogen bromide. Given the following sequences (N-terminal to C- terminal) of the resulting fragments, deduce the sequence of the original peptide. Ala-Met-Gln Glu-Ile-His-Lys Cys-Glu Glu-Glu-Ala-Met-Val-Ala-Lys Ala -Tyr-Asn-Arg Cyanog Ala -Tyr Asn-Arg-Ala-Met Val-Ala-Lys-Cys-Glu Gln-Glu-lle-His-Lys-Glu Glu-Ala-Met lu-Glu-Ala-Met-Ala-Tyr-Asn-Arg-Ala-Met lu-Ala-Tyr-Asn-Arg-Ala-Met- Gln- Glu-Ile-His o Cys- Glu-Glu-Glu-Ala-Met-Val-Ala-Lys-Ala-Met-Gln-Glu-Ile-His Val-Ala-Lys- Cys Glu Lys- Glu-Glu-Ala-Met Arg

Explanation / Answer

Peptide chain longer than 50-70 amino acid are digested either by endopeptidase such as trypsin or by chemical reagent such as cyanogen bromibe.Endopeptidase means that cleavage occur within the polypeptide chain rather than at the terminal of amino acid located at the end of polypeptide chains.

Trypsin is formed in small intestine.It cleaves peptide chains at the C-terminal side of the amino acid lysin and arginine except when either is followed by proline.

Cyanogen bromide (CN-Br) hydrolyses or cleave the peptide bonds at the C-terminal of methionine.

Option 6 is the correct peptide that will give us all these given sequences on the treatment of trypsin and cyanogen bromide

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.