6. (5 pts) Determining the sequence of proteins is a multistep process that requ

Question

6. (5 pts) Determining the sequence of proteins is a multistep process that requires the usage of a variety of tools, such as chemicals and specific enzymes, to determine the amino acid sequence. a) In three separate experiments, identify the fragments produced by treatment of peptide "A" with (1) chymotrypsin, (2) trypsin and (3) cyanogen bromide (CNBr). Peptide A: AQKGHMTDLRFPHSCHWTGYH (1) Chymotrypsin (2) Trypsin: (3) CNBr: b) To a solution of peptide "A" you add peptide "B." Upon running SDS-PAGE analysis, you observe the results below. Explain the results. Peptide B: QWPCRRCGAHSDR A B no DTT A B DTT

Explanation / Answer

(a) Sequence A: AQKGHMTDLRFPHSCHWTGYH

We need to know some of the amino acides given in the sequence i.e Lysine - K, Arginine - R, Proline - P, Tyrosine - Y, Tryptophan - W, Phenylalanine - F, Methionine - M

(1) Fragments produced by chymotrypsin: Chymotrypsin cleaves bond at carboxyl side of aromatic amino acid provided the next amino acid should not be Proline. In the given AQKGHMTDLRFPHSCHWTGYH sequence, bond next to F,W and Y will be cleaved but as P is next to F, this bond won't be cleaved. As Y is towards the end and chymotrypsin is endopeptidase, even this bond won't be cleaved. Only bond next to W will be cleaved. so 2 fragments obtained - AQKGHMTDLRFPHSCH and TGYH.

(2) Fragments produced by trypsin: Trypsin cleaves bond at carboxyl side of Lysine and Arginine provided the next amino acid should not be Proline. In the given AQKGHMTDLRFPHSCHWTGYH sequence, bond next to K and R will be cleaved so 3 fragments are obtained - AQK, GHMTDLR and FPHSCHTGYH.

(3) Fragments produced by Cyanogen Bromide: It breaks C terminal bond on Methionine. It is selective non enzymatic process, so 2 fragments obtained - AQKGHM and TDLRFPHSCHTGYH.

(b) On running SDS PAGE, Column A has only A protein with linear sequence so we get only 1 band, similarly in column B, as only B protein is loaded, we only get 1 band.

Column C has both A protein, B protein and no DTT (Cleland's Reagent). So Because of no DTT. Disulphide bond is formed between Cystein residues of Protein A and Protein B. therefore only 1 band is seen which is of higher molecular weight.

Column D has both A protein, B protein and DTT so DTT cleaves the disulphide bond between A and B and therefore two band of parental size are obtained

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