Protein Sequencing BL4010 09.19.06

Schematic amino acid R groups A Ala C Cys D Asp E Glu F Phe* G Gly H His* I Ile* K Lys* L Leu* M Met* N Asn P Pro Q Gln R Arg* S Ser T Thr* V Val* W Trp* Y 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)

Isoelectric point 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-N-hydroxysuccinimidyl 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 Hydrolysis and amino acid analysis yields: K, M, C, E, F Treatment with phenylisothiocyanate yields: M Treatment with carboxypeptidase Y yields: Q Treatment with staphylcoccal protease yields two products. Only the tripeptide contains sulfur. Treatment with chymotrypsin yields two products. One is a dipeptide.

Answer 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. Treatment with phenylisothiocyanate yields: M PITC makes derivatives of N-terminus so structure is: M-X-X-X-X

Answer (cont.) 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.) 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.) 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