1559Tch26454-46910/20/0515:39Page462 462 chapter 26 AMINO ACIDS,PEPTIDES,PROTEINS,AND NUCLEI ACIDS:NITROGEN-CONTAINING POLYMERS IN NATURE determine the sequence of amino acids.Because ony nine are present.the entire chain can be sequenced in his way. 3 (a) -CHCH(CH)+Al-Cys NH H NH +Thr-Pro-Ly 44.Because the peptide is cyclic.the Edman process will not give a normal result.It will simply form thiourea derivatives at the two"extra"Om amino groups: Because there is no a-amino group available to react.mild acid treatment will cleave no bonds in the product,and the cyclic polypeptide structure will remain intact. 45.Complete hydrolysis indicates that a total of nine amino acid units are present.Examine the four fragments of incomplete hydrolysis.You know that the peptide begins with Arg (iv)so the fragmen Arg-PTO-PIO be overlapping all the Ag品5678 Gly-Phe-Ser Ser-Pro-Phe42. (1) Purify and cleave the disulfide bridge (Section 9-10). (2) On a portion of the sample, degrade the entire chain by amide hydrolysis (6 N HCl, 110°C, 24 h) to determine amino acid composition by using an amino acid analyzer. (3) On another portion of this material, apply repetitive Edman degradation to determine the sequence of amino acids. Because only nine are present, the entire chain can be sequenced in this way. 43. (a) (b) (c) (d) 44. Because the peptide is cyclic, the Edman process will not give a normal result. It will simply form thiourea derivatives at the two “extra” Orn amino groups: Because there is no
-amino group available to react, mild acid treatment will cleave no bonds in the product, and the cyclic polypeptide structure will remain intact. 45. Complete hydrolysis indicates that a total of nine amino acid units are present. Examine the four fragments of incomplete hydrolysis. You know that the peptide begins with Arg (given), so the fragment Arg-Pro-Pro-Gly must be first. There is only one Gly present, so the last Gly of this tetrapeptide must be the same one that is at the start of the tripeptide fragment Gly-Phe-Ser. You can use the same logic to start overlapping all the pieces to generate the whole solution. Thus, because only one Ser is present, the last Ser in the above tripeptide must be the same as the first one in Ser-Pro-Phe. So far, you have 123 4 5 67 8 Arg-Pro-Pro-Gly Gly-Phe-Ser Ser-Pro-Phe NHCNH(CH2)3 S CHCH2 S O N OH NH Gly-Gly-Phe-Leu CHCH2 S N O NH Thr-Pro-Lys N N H CHCH2OH S N O NH Asp CHCH(CH3)2 S N O NH Ala-Cys 462 • Chapter 26 AMINO ACIDS, PEPTIDES, PROTEINS, AND NUCLEIC ACIDS: NITROGEN-CONTAINING POLYMERS IN NATURE 1559T_ch26_454-468 10/20/05 15:39 Page 462