Please help me with this!! I am lost! Please show how to do this. Thank you!! An

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Please help me with this!! I am lost! Please show how to do this. Thank you!!

An HEXAPEPTIDE was sequenced using differing enzyme and chemical reagents. The results of these experiments, along with an explanation of how to interpret the results are provided below. Using this information, provide the 1° structure for this HEXAPEPTIDE. (10 points) 2. be sure to provide WORK/ANSWER for Points B-F! Observations from Sequencina Experiments A. The amino acid content in the peptide was: Y, R, M, K, G, D B. Reactions with the peptide and 1-fluoro-2,4-dinitrobenzene yielded DNP-G. C. Treatment of the peptide with chymotrypsin yielded two tripeptides. D. Treatment of the peptide with trypsin yielded three dipeptides. E. Treatment of the peptide with carboxypeptidase yielded methionine. F. N-terminal sequencing of one the chymotrypsin-produced tripeptides (from part C) yielded DNP-K Explanation and Summary of Protein Sequencing Techniques 1. 1-fluoro-2,4-dinitrobenzene (DNFB) is used to identify the N-terminal amino acid of a peptide, and provides the following product: DNP-AA (where AA = the N-terminal amino 2· 3. 4. acid of the peptide). Trypsin is an enzyme that cleaves a peptide at the C-terminal side of R and K. Chymotrypsin is an enzyme that cleaves a peptide at the C-terminal side of F, Y, and W. Carboxypeptidase is an enzyme that cleaves off the C-terminal amino acid of any peptide.

Explanation / Answer

A. The hexapeptide sequence should be HOOC-MDKYRG-NH2

B. Treatment with DNFB giave DNP-G, so the N-terminal end of the peptide should be Glycine.

C. Treatment with Chymotrypsin gave 2 tryipeptides and since it cleaves the c terminal of Y, the 4th peptide should be Y, as that position corresponds to the C-terminal end.

F. N terminal of the one of the tripeptides from C gave DNP-K, indicating that the 3rd position should be K. as the N-terminal of the other tripeptide is Glycine (from B)

D. Treatment with Trypsin yileded 3 dipeptides and Trypsin cleaves C terminals of K and R, confirming the positions of K and R as 3 and 5 respectively. (position 3 already conformed for K, from F)

E. Treatment with carboxypeptidase would cleave the c-terminal of the peptide, which yielded Methionine, leaving Methionine at the c-terminal

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