3. Based on the following data, write the sequence for the following decapeptide

Question

3. Based on the following data, write the sequence for the following decapeptide (10 aa's). a. Complete hydrolysis gave: Asp Glu Tyr Arg Met Pro Lys(2) Ser Phe b. trypsin digestion yielded 2 tetrapeptides, Lys, Lys c. Edman degradation released the PTH derivative of Ser d. chymotrypsin digestion yielded 2 tripeptides and a tetrapeptide. The composition of the tetrapeptide was Phe, Lys, Lys, Arg. e. CNBr treatment yielded an octapeptide and a dipeptide (Asn-Pro) f. clostripain (enzyme that selectively cuts after Arg) yielded a tetrapeptide and hexapeptide

Explanation / Answer

Trypsin cleaves at the C-terminal of Lys or Arg residues.

One round of Edman degradation reveals the identity of the N-terminal amino acid.

Chymotrypsin cleaves at peptide bonds formed by aromatic amino acids such as Phe, Trp, and Tyr.

Cyanogen bromide cleaves after Met residues.

Clostripain cleaves after Arg

Edman degradation produces PTH-Ser

N-terminal amino acid = Ser

CNBr treatment yielded an octapeptide and a dipeptide.

So, the 8th amino acid is Met.

The last two amino acids are Asp-Pro

Clostripain yielded a tetrapeptide and a hexapeptide.

So, the fourth amino acid = Arg

Trypsin digestion produces two tetrapeptides, Lys and Lys. So,

Ser-J1-J2-J3, J4-J5-J6-J7, Lys and Lys

J3 = Arg (because two free lys are already present)

So,

Ser-J1-J2-Arg-Lys, J4-J5-J6-J7 and Lys

Eighth amino acid = J6 = Met

So,

Ser-J1-J2-Arg-Lys, J4-J5-Met-J7 and Lys

Chymotrypsin digestion produces two tripeptides and a tetrapeptide

The tetrapeptide sequence = Phe-Lys-Lys-Arg

So,

Ser-J1-Tyr-Arg-Lys-Lys-Phe-Met-Asp-Pro

The remaining amino acid = Glu

So,

The sequence of the peptide = Ser-Glu-Tyr-Arg-Lys-Lys-Phe-Met-Asp-Pro

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