Question

What is the amino acid sequence of bradykinin? (5 pts) 2. Given four D-aldopentoses A, B, C and D. Bis epimeric with A at Cz. If B is subjected to the Kiliani-Fischer synthesis, two aldohexoses are produced - aldohexose E and D-glucose. If E is subjected to Ruff degradation, a D- aldopentose is obtained which is epimeric with Dat C. When C is reacted with HNO, the resulting aldaric acid is optically inactive. Draw the Fischer structures of A, B, C, D. (10 pts)

Answer 1

Solution: 27 A. B, C, D => D-Aldo pentores killiane- В -диск + aldoherose -E Fischer оно сно к CHO он н он Но H Н H ФН ОН Ио H H тон но- ч и о Н M + и ви сно E Сн, ой ся, оң со оно Ruff D-Aldo pentore son Degradation 4 on epimer и к до n ои во - ион Н и Гот И Сң, и i choy D HNO3 с сно СФОН ОН H »'c' will be Я НЯ и ТЕ ои го и Н он и: ои СИ сҢ, он optically Inactive Thus: A= epimer at C₂ of B = И СНО ОЙ Ч И СНИ сно но сно В. Он то Hence A = ф И M M H СНО өИ он ѕИ Н.РИ 에 н би ОН И ГИ и и- ch of chon СИИ solution 1) Bradykinin sequence RPP GFSPFR Ang-Pro-Pro-Gly- Phe-ser-Pro- phe- ke - Ат? C-Terminus N-Terminus