IndietroTreatment of the nonapeptide from the previous problem (1-1) with CNBr produces a tetrapeptide containing the N-terminal amino acid and a pentapeptide. After one round of Edman degradation on the pentapeptide, a product is produced that contained a nonpolar, aliphatic R group, meaning that the pentapeptide has:
Deduce the entire sequence of the original nonapeptide using the following hint and the information from the previous four practice problems (1-1 through 1-4). Hint: The sequence of the nonapeptide using the one-letter amino acid codes reveals a relevant academic subject.
The peptide Leu—Cys—Arg—Ser—Gln is subject to Edman degradation. Products of the 1st cycle include:
Suppose you isolated a nonapeptide (9 amino acid residues) from a patient's blood. Reaction of the nonapeptide with FDNB followed by acid hydrolysis produces a DNP-product with a sulfhydryl R-group, indicating that:
Hydrazinolysis of the same pentapeptide from the problems above produced modified amino acids & a free α-amino acid with an aromatic-alcohol group. Combining the info from these problems, the pentapeptide is most likely:
The second and third rounds of Edman degradation on the same pentapeptide from the problem above (1-2) produced products with aliphatic alcohol groups, meaning that the pentapeptide had: