4. (18 pts) After the polypeptide (45 residues) shown below was treated with mal

Question

4. (18 pts) After the polypeptide (45 residues) shown below was treated with maleic anhydride, trypsin was no longer able to cleave the polypeptide on the C-side of lysine. Gly-Ala-Asp-Ala-Leu-Pro-Gly-lle-Leu-Val-Arg-Asp-Val-Gly-Lys-Val-Glu-Val-Phe-Glu-Ala-Gly-Arg Ala-Glu-Phe-Lys-Glu-Pro-Arg-Leu-Val-Met-Lys-Val-Glu-Gly-Arg-Pro-Val-Gly-Ala-Gly-Leu-Trp (a) Why was trypsin no longer able to cleave the polypeptide on the C-side of lysine? Provide the chemical reaction to make your point if necessary (b) List all the fragments produced from the cleavage of the maleic anhydride modified polypeptide. (c) What is the net charge of each fragment at pH 5? Show your work. d) In what order would the fragments be eluted from an anion-exchange column using a buffer of pH 5? Explain.

Explanation / Answer

a)maliec anhydrate react with amino group and acetylate it as lysin have an extra amine group in its c terminal it would be modified to acetylated form and could not be reacted with Lysin cleavage.

b) 1. Gly-Ala-Asp-Ala-leu-Pro-Gly-Ile-Leu-Val-Arg-

2) Asp-Val-Gly-Lys-

3) Val-Glu-Val-Phe-Glu-Ala-Gly-Arg

4)Ala-Glu-Phe-Lys-Glu-Pro-Arg

5)Leu-Val-Met-Lys-

6)Val-Glu-Gly-Arg

7)Pro-val-Gly-Leu-Trp

Total seven fragments

C) NH3+ state as they reacts with free hidrogen ions

D)

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