A sample of protein is divided into two, and one sample istreated with cyanogen

Question

A sample of protein is divided into two, and one sample istreated with cyanogen bromide. The other sample is treated with trypsin. Each set of products is sequenced by Edmandegradation. What is the sequence of the entire polypeptide?

CNBr fragments:1. Asp 2. Lys-Phe-Met 3. Tyr-Arg-Gly-Met

Trypsin fragments:4. Lys 5. Gly-Met-Asp 6. Phe-Met-Tyr-Arg

CNBr cleaves after MethionineTrypsin cuts after Arg or Lys

I have no idea how this is supposed to be done and a similar problem will be on my upcoming exam, help?

Explanation / Answer

Cyanogen bromide cleaves on the C-side of Met residues.

CNBr fragments:

1. Asp

2.Lys-Phe-Met

3. Tyr-Arg-Gly-Met

N-Tyr-Arg-Gly-Met-Lys-Phe-Met- Asp-C ……………………… Combination 1

N-Lys-Phe-Met-Tyr-Arg-Gly-Met-Asp-C   ……………………… Combination 2

Trypsin cleaves the Carboxy side (C-side) of Arg or Lys.

Trypsin fragments:

4. Lys

5. Gly-Met-Asp

6. Phe-Met-Tyr-Arg

This indicates that Lys is either not on C-terminal region and Arg is neither C-terminal nor N-terminal region. So, the two possible combinations are:

N-Phe-Met-Tyr-Arg-Lys- Gly-Met-Asp-C ………………………… combination3

N-Lys- Phe-Met-Tyr-Arg-Gly-Met-Asp-C   …………………………..combination4

Combination 2 and combination 4 give the same peptide sequence. So the peptide is:

N-Lys-Phe-Met-Tyr-Arg-Gly-Met-Asp-C

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