Transcript LECT06 peptides

Peptides
1. Little Proteins
A. Dipeptides, tripeptides, …polypeptides
2. Basic Structure
rotation
N-terminal
+
H3N
H
H
C
H
O
N
C
C
O
rotation
Peptide bond
C
H
C-terminal
O
CH3
Glycylalanine
H
H
C
+
H3N
H
O
N
C
C
C
H
O
H
H
+
H3N
y
C
O
N
C
C
H
f
CH3
H
C
O
O
O
CH3
Primary Structure
Conformations of the Peptide Chain
O
H
C
C
C
R
Trans-conformation
N
H
R
Steric Interference
H
Cis-conformation
8 kJ more energy
(less stable) than trans
Torsion Angles
O
y
H
C
C
R
C
N
f
N
H
H
R
Clockwise
N
Clockwise
Most Favorable Conformation is
the one with least interactions
+180
-sheet
Allowable
Y
O
 helix
-180
-180
O
F
+180
Biochemical Properties of Peptides
Solving Peptide Mysteries
Preparing Peptides for Sequence Analysis
S-S
-COO-
H3NM
R
K
c
c
Trypsin
S-S
M
K
Performic
acid
CNBr
Homoserine lactone
3
R
SO3-
1
2
K
4
SO3-
A. Absorption of light at 280 nm. If the peptide in question absorbs uv light, this is an
indication of aromatic amino acids. Look for residues of phenylalanine, tyrosine, or
tryptophan. Failure to absorb uv light would indicate the absence of these residues.
B. Cleavage by Trypsin. Sensitivity to trypsin indicates the present of a basic amino
acid (lysine or arginine) at the cleavage site. Trypsin cleaves the bond and leaves the
basic amino acid as the C-terminus of the split product.
C. Cleavage by Chymotrypsin. The same as trypsin, only with an aromatic amino acid
residue is at the cleavage site.
D. Cleavage by Carboxypeptidase. The residue released by carboxypeptidase is the Cterminal residue of the peptide. This exopeptidase cleaves from the C-terminus one
amino acid residue at a time. The enzyme is used to identify residues at or near the
C-terminus of the peptide.
E. Cleavage by Aminopeptidase. The residue released by aminopeptidase is the Nterminal residue of the peptide. Like carboxypeptidase, only working from the other
end of the peptide.
F. Cleavage by CNBr. CNBr (cyanogen bromide) cleaves at methionine residues.
CNBr is one of the few chemical reagents that cleave at a specific location in the
peptide.
See p 42-43 in Strategies
A. Cleavage by Sulfhydryl Reagents. This indicates that the peptide contains a
disulfide bond. Reagents such as -mercaptoethanol, dithiothreitol, and glutathione
split the bond by reducing action. A disulfide bond is created by an interaction
between two cysteine –SH (sulfhydryl) groups. A disulfide bond may not be broken
by hydrolysis. Reagents such as performic acid can, however, oxidize it.
B. Residue split does not rotate light. This indicates that the glycine (the only amino
acid lacking a chiral center) was released in the procedure.
C. Residue split is an acidic amino acid. The residue is either aspartic acid or glutamic
acid.
D. Residue split is a basic amino acid. The residue is either lysine or arginine.
E. Residue split has a group with a pK = 6. The residue is histidine
F. Peptide releases NH3 when hydrolyzed by 6 N HCl. The amino acids asparagine or
glutamine are in the structure. Both contain an amide group that releases NH3 in
response to strong acid hydrolysis.
G. Peptide requires more than 2 equivalents of OH- to titrate. One or more residues
with charged R groups are present. If no charged R groups were present, only two
equivalent of base (one for the NH3 and one for –COOH) would be required.
H. Peptide does not react with N-terminal reagents. Peptide is either cyclic or has a
blocked N-terminus.
A dipeptide absorbs light at 280 nm. On hydrolysis two
amino acids are found. One does not react with ninhydrin.
The other contains a phenolic -OH group. The dipeptide
does not react with chymotrypsin.
CLUE 1: Absorb light means an aromatic amino acid is present
CLUE 2: Phenolic groups is found in tyrosine
CLUE 3: Only proline fails to react with ninhydrin
CHOICES: Prolyltyrosine or Tyrosylproline
CLUE 4: Chymotrypsin will only react when the C-terminal
of an aromatic amino acid is in peptide linkage.
ANSWER: Prolyltyrosine
HO
Chymotrypsin
CH2
H
C
H3N
N
C
O
N
H
Tyrosylproline
C
O
COO
H
N H COO
C
CH2
OH
Prolyltyrosine
A peptide fragment with a Mwt of about 300, upon
treatment with 6 N HCl gave 2 amino acids. One did
not rotate light. The N- and C- terminal residue had a
pKa of 6. What is the structure?
CLUE 1: 300 Mwt means 3 residues
CLUE 2 pKa 6 means Histidine
CLUE 3 Not able to rotate light means Glycine
ANSWER:
His-Gly-His
A tripeptide did not react with chymotrypsin. It did
react with trypsin, liberating a dipeptide. The
peptide required 4 equivalents of base to titrate all protons
It absorbed light a 280 nm. One residue was valine, and
one had a pKa around 12.
1 Chymotrypsin cleaves on -COO side of aromatic
groups
2 Trypsin cleaves on -COO side of lysine and arginine
3 4 equivalents means 2 charged amino acids
4 280 nm absorption means aromatic amino acid
5
pKa of 12 identifies arginine
ANSWER:
Arg-Val-Tyr