Question

Explain this question please!

Name (4 ofer to the Table of Amino Acids at the beginning of the exam to help solve this problem. Polypeptide I is a 12-mer and has the following amino acid composition: Ala, Arg, 2 His, Leu, 2 Lys, 2 Phe, Ser, Thr, Trp Edman degradation of I shows that its N-terminal amino acid is His. Chymotrypsin cleavage of I yields peptide fragments A-D.Trypsin cleavage of I gives peptide fragments E-G. Shown below is the amino acid composition of each fragment A-G (in alphabetical order). The Chymotrypsin Cleavage: A: Ala, Arg, Phe, Ser B: Leu, Lys, Th C: His, Lys, Trp D: His, Phe E: Ala, Arg, His Phe, Ser F: Leu, Lys, Thr, Trp G: His, Lys, Phe Trypsin Cleavage: Determine the amino acid sequence of polypeptide I and write the sequence in the answer boxes below. Use the three letter code for cach amino acid. Only what is written in the boxes will be graded. C-terminus N-terminus His

Answer 1

Chymotrypsin is an enzyme which cleaves on the C-terminal of the aromatic amino acids (Phe,Tyr,Trp) of other proteins. The serine residue present on the enzyme cleaves at hydrophobic pocket of the aromatic amino acids.

Chymotrypsin cleavage: the N terminal is given and the C terminal will be aromatic acid (because of cleavage)

A: Ala-Arg-Phe-Ser: N terminal- Ser, C terminal- Phe, so the fragment will be: Ser-Ala/Arg-Arg/Ala-Phe

B: Leu-Lys-Thr: N terminal- Thr, C terminal- ...., so the fragment will be: Thr-Leu/Lys-Lys/Leu

C: His-Lys-Trp: N terminal- His, C terminal- Trp, so the fragment will be: His-Lys-Trp

D: His-Phe: N terminal- His, C terminal- Phe, so the fragment will be: His-Phe

Trypsin cleaves the peptide bond between the carboxyl group of arginine or the carboxyl group of lysine and the amino group of the adjacent amino acid.

Trypsin cleavage: the N terminal is given and the C terminal will be arginine or lysine (because of cleavage)

E: Ala-Arg-His-Phe-Ser: N terminal- His, C terminal- Arg, so the fragment will be: His-Ala/Ser/Phe-Ser/Phe/Ala-Phe/Ala/Ser-Arg

F: Leu-Lys-Thr-Trp: N terminal- Trp, C terminal- Lys, so the fragment will be: Trp-Leu/Thr-Thr/Leu-Lys

G: His-Lys-Phe: N terminal- Phe, C terminal- Lys, so the fragment will be: Phe-His-Lys

Since 1 Histidine is at N terminal (given), so the fragment D's His will be N terminal His of Peptide. It clears the second position in fragment E, because the His in fragment G is not the terminal.

So, the fragment E: His-Phe-Ser/Ala-Ala/Ser-Arg

Now, considering fragments C and G, the combined fragment we get: Phe-His-Lys-Trp

Now, considering fragments A, D and E, the combined fragment we get: His-Phe-Ser-Ala-Arg-Phe

Now, considering fragments A,C, D, E and G, the combined fragment we get: His-Phe-Ser-Ala-Arg-Phe-His-Lys-Trp

Combining all the fragments, we get the whole peptide of 12 amino acid as:

His-Phe-Ser-Ala-Arg-Phe-His-Lys-Trp-Thr-Leu-Lys

So, the answer is: His-Phe-Ser-Ala-Arg-Phe-His-Lys-Trp-Thr-Leu-Lys