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Protein Sequencing
BL4010
09.19.06
Schematic amino acid R groups
A
C
D
E
F
G
H
I
K
L
M
N
P
Q
R
S
T
V
W
Y
Ala
Cys
Asp
Glu
Phe*
Gly
His*
Ile*
Lys*
Leu*
Met*
Asn
Pro
Gln
Arg*
Ser
Thr*
Val*
Trp*
Tyr
C
 N
 O
 S
Ionization of amino acids
pKa values are affected by
environment
Ionization R groups
D pKR = 3.7
E pKR = 4.3
H pKR = 6.0
C pKR = 8.2
Y pKR = 10.0
K pKR = 10.5
R pKR = 12.5
pKa values are affected by
environment
Isoelectric point
pI = ½(pK1 + pK2)
pI = ½(2.3 + 9.6)
pI = ½(11.9) = ~6
Isoelectric point
pK1 = 2.2
pK2 = 4.3
pK3 = 9.7
pI = ½(pK1 + pK2)
pI = ½(2.2 + 4.3)
pI = ½(6.5) = ~3
What is the pH of a glutamic acid solution
if the alpha carboxyl is 1/4 dissociated?
• pH = 2 + log10 [1]
¯¯¯¯¯¯¯
[3]
• pH = 2 + (-0.477)
• pH = 1.523
What is the pH of a lysine
solution if the side chain amino
group is 3/4 dissociated?
• pH = 10.5 + log10 [3]
¯¯
¯¯¯¯¯
[1]
• pH = 10.5 + (0.477)
• pH = 10.977 = 11.0
Amino acid detection
• All amino acids absorb in infrared region
• Only Phe, Tyr, and Trp absorb UV
• Absorbance at 280 nm is a good diagnostic
device for amino acids
• NMR spectra are characteristic of each
residue in a protein, and high resolution NMR
measurements can be used to elucidate
three-dimensional structures of proteins
Aromatics absorb UV
The ninhydrin reaction
Derivatization
• Detection vs. derivatization
• Several derivatization chemistries are in
common use:
– dansyl derivatives
– o-pthalaldehyde (OPA)
– phenylisothiocyanate (PITC)
– 9-fluorenylmethyl chloroformate
(Fmoc)
– 6-aminoquinolyl-Nhydroxysuccinimidyl carbamate
(AQC) .
Separation of amino acids
• Mikhail Tswett father of ‘chromatography’
• Chromatography
– Ion exchange chromatography (net ionic character)
– High-performance liquid chromatography (HPLC)
– Gas chromatography (derivatization for volatility)
• Electrophoresis(charge/size)
– Paper (ninhydrin)
– Capillary electrophoresis (fluorescence, UV, laser-induced
capillary vibration)
Chromatography
Detection: refractive index, circular dichroism, (MS/MS) vs.
derivitization for UV or fluorescence or MS
Chromatography
Elution order depends on affinity to the matrix!!
Sequence matters
• Because the N-terminus of a peptide chain is
distict from the C-terminus, a small peptide
composed of different aminoacids may have a
several constitutional isomers (e.g. Asp-Phe or
Phe-Asp).
• The methyl ester of the first dipeptide
(structure on the right) is the artificial sweetner
aspartame, which is nearly 200 times sweeter
than sucrose. Neither of the component amino
acids is sweet (Phe is actually bitter), and
derivatives of the other dipeptide (Phe-Asp)
are not sweet.
TRH thyrotropin releasing hormone
• Oxytocin (Hormone) 9 amino acids:
Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-GlyNH2
• Melittin (Bee Venom) 26 amino acids:
Gly-Ile-Gly-Ala-Val-Leu-Lys-Val-Leu-Thr-Thr-Gly-Leu-Pro~
~Ala-Leu-Ile-Ser-Trp-Ile-Lys-Arg-Lys-Arg-Gln-GlnNH2
Proteins (polypeptides)
Sequencing
• 100 amino acid protein has 20100 combinations
• 1953 Frederick Sanger sequenced the two
chains of insulin (21 aa)
• All of the molecules of a given protein have the
same sequence
• Proteins can be sequenced in two ways:
- direct amino acid sequencing
- indirect sequencing of the encoding gene (DNA)
Why does sequence matter?
• Sequences and composition often (not
always) reflect the function of the
protein (often proteins of similar function
will have similar sequences)
• Homologous proteins from different
organisms have homologous
sequences
Changes in protein sequence can be
used to infer evolutionary relationships
Conventional Sequencing
1. If more than one polypeptide chain,
separate.
2. Cleave (reduce) disulfide bridges
3. Determine composition of each chain
4. Determine N- and C-terminal
residues
Conventional Sequencing
5. Cleave each chain into smaller
fragments and determine the
sequence of each chain
6. Repeat step 5, using a different
cleavage procedure to generate a
different set of fragments.
Conventional Sequencing
7. Reconstruct the sequence of the
protein from the sequences of
overlapping fragments
8. Determine the positions of the
disulfide crosslinks
Separation of chains
• Subunit interactions depend on weak
forces
• Separation is achieved with:
- extreme pH
- 8M urea
- 6M guanidine HCl
- high salt concentration (usually
ammonium sulfate)
Cleavage of Disulfide bonds
• Performic acid oxidation
• Sulfhydryl reducing agents
- mercaptoethanol
- dithiothreitol or dithioerythritol
• alkylate (iodoacetate) prevents
recombination
Fragmentation of the chains
• Enzymatic fragmentation
–
–
–
–
Trypsin (R or K)
Chymotrypsin (F or Y or W)
Clostripain (R>K) (can be incomplete for K)
Staphylococcal protease (D or E)
• Chemical fragmentation
– cyanogen bromide (Methomoserine lactone)
Trypsin digestion
Cyanogen bromide
N-terminal analysis (Edman)
– (PITC) phenylisothiocyanate
– PTH derivatives (phenylthiohydantions)
– Iterative application to short peptides yields
short sequences
C-terminal analysis
• Enzymatic analysis (carboxypeptidase)
– Carboxypeptidase A cleaves any residue
except Pro, Arg, and Lys
– Carboxypeptidase B (hog pancreas) only
works on Arg and Lys
– Carboxypeptidase C, Y any residue
Amino acid composition
• Hydrolysis (6M HCl, 2M TFA)
– Destroys W
– Partially destroys S, T
– Loss of amides ND, Q E (Asx, Glx)
– Incomplete cleavage hydrophobic residues
• Derivatization and chromatography
Mass spectrometry
Tandem MS-MS
Reconstructing the Sequence
• Use two or more fragmentation agents in
separate fragmentation experiments
• Sequence all the peptides produced (usually
by Edman degradation)
• Compare and align overlapping peptide
sequences to learn the sequence of the
original polypeptide chain
Compare cleavage by trypsin and
staphylococcal protease on a typical peptide:
• Trypsin cleavage:
A-E-F-S-G-I-T-P-K
L-V-G-K
• Staphylococcal protease:
F-S-G-I-T-P-K
L-V-G-K-A-E
Determine disulfide sites
Diagonal gel electrophoresis
-peptides off diagonal
are cross-linked
Example problem
1. Hydrolysis and amino acid analysis yields: K, M,
C, E, F
2. Treatment with phenylisothiocyanate yields: M
3. Treatment with carboxypeptidase Y yields: Q
4. Treatment with staphylcoccal protease yields two
products. Only the tripeptide contains sulfur.
5. Treatment with chymotrypsin yields two products.
One is a dipeptide.
Answer
1. Hydrolysis and amino acid analysis yields: K, M,
C, E, F
Assume pentapeptide (could be more) with one of
each amino acid. Remember: hydrolysis destroys
W and converts Q to E and N to D. So...E could
be Q.
2. Treatment with phenylisothiocyanate yields: M
PITC makes derivatives of N-terminus so structure is:
M-X-X-X-X
Answer (cont.)
3. Treatment with carboxypeptidase Y yields: Q
Carboxypeptidase cleaves C-terminus so
M-X-X-X-Q
(Also...that means E in hydrolysis was Q)
Answer (cont.)
4. Treatment with staphylcoccal protease yields two
products. Only the tripeptide contains sulfur.
Staph protease cleaves after D,E
Since a cleavage took place (peptide is at least a
pentapeptide...a D, or E must be present...hydrolysis
says E but we already found Q so that means there is
another residue...hexapeptide.
Since we know first residue is M (contains sulfur) that will
be the tripeptide. if a hexapeptide E must be in the
third position to get a tripeptide. Also, second residue
must by C because only one tripeptide contains sulfur.
M-X-E X-X-Q
Answer (cont.)
5. Treatment with chymotrypsin yields two products.
One is a dipeptide.
We think we have M-C-E-X-X-Q
residues left indclue K and F
cymotrypsin cleaves after F,W,Y
to get a dipeptide F must be after E and the other
residue is K
M-C-E-F-K-Q