Use the following information to solve questions 16-19. A mixture of three pepti

Question

Use the following information to solve questions 16-19. A mixture of three peptides is provided to you to separate. The sequences of the three peptides are known I. KCWRR (pl 12.01) ll. DHEIE (pl = 4.13) IL. YWIFW (pl 5.52) of an SDS-PAGE gel is shown of the mixture of the three peptides before purification. the molecular weight of all three peptides is similar, only one wide band appears when the 16. A picture Because peptide mixtu mercaptoethanol was excluded appear in the absence of BM? re is ran on SDS-PAGE (+BM). However, a second band appears when from the loading solution (-BM). Why does this second band to Marker +BM -BM 2000 Da 1200 Da_ 660 Da 450 Da 300 Da 200 Da Peptide I can form a disulfide bond with itself in the absence of -mercaptoethanol. Peptide I can dimerize via its charged amino acid residues in the absence of - mercaptoethanol Peptide can dimerize via its charged amino acid residues in the absence of - mercaptoethanol Peptide III can dimerize via its hydrophobic residues in the absence of B-mercaptoethanol. None of the above; the lack of ß-mercaptoethanol removes the charge carrier necessary for the protein to migrate through the SDS-PAGE gel. a. b. c, d. e.

Explanation / Answer

Single letter abbreviation of amino acids has been written as three letter abbreviation for simplfication:

peptide I : -lys-cys-trp-arg-arg-

peptide II : -asp-his-glu-ile-glu-

petide III : -tyr-trp-ile-phe-trp-

The function of SDS-PAGE relies on denaturing the protein of interest, but denaturation is not the same as breaking the dilsulfide (S-S) bonds. A reducing SDS-PAGE gel will have beta mercaptoethanol (BM) added to the sample before boiling and will break any disulfide bonds in the protein. Hence, a reducing SDS gel will do both (denature and break disulfide bonds. On the contrary, non-reducing SDS-PAGE gel (no BM) will not break disulfide bonds.

Absence of -Mercaptoethanol has nothing to do with dimerization of peptide I, II via its charged amino acid residues or peptide III via its hydrophobic residues.

So, option (a) is correct.

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