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Solutions for Amino Acids, Peptides, and Proteins
C
HC
O
CH2SH
HN
S S
COOH
NH
CHHSCH2
C O
SS
R
RC
O
H
(CH2)4COOH
SS
CH COOH
CH2
H2N
N
NH
HN
HC CH2S
CO
RC OH
O
COOH
SH SH
R
SH SH
C
CH
O
SCH2
NH
SH SH
(CH2)4COOH
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(a) There are two possible sources of ammonia in the hydrolysate. The C-terminus could have been 
present as the amide instead of the carboxyl, or the glutamic acid could have been present as its amide, 
glutamine.
(b) The C-terminus is present as the amide. The N-terminus is present as the lactam (cyclic amide) 
combining the amino group with the carboxyl group of the glutamic acid side chain.
(c) The fact that hydrolysis does not release ammonia implies that the C-terminus is not an amide. Yet, 
carboxypeptidase treatment gives no reaction, showing that the C-terminus is not a free carboxyl group. 
Also, treatment with phenyl isothiocyanate gives no reaction, suggesting no free amine at the N-terminus. 
The most plausible explanation is that the N-terminus has reacted with the C-terminus to produce a cyclic 
amide, a lactam. (These large rings, called macrocycles, are often found in nature as hormones or 
antibiotics.)
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(a) Lipoic acid is a mild oxidizing agent. In the process of oxidizing another reactant, lipoic acid is reduced.
oxidized form reduced form
(b)
(c)
H2O + + +
basic not basic: pyrrole-type N
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(a) histidine:
See the solution to Problem 6.
pyridine-type N
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