BIOL 302 1st Edition Lecture 8 Outline of Last Lecture I. Protein Misfolding II. Protein Sequence DeterminationOutline of Current Lecture I. Protein sequence determinationII. High resolution protein structure determinationCurrent LectureIII. Edman’s Method for SequencingA. Edman degradation procedure - Determining one residue at a time from the N-terminus 1. Treat peptide with phenylisothiocyanate (PITC) at pH 9.0 which reacts with the N-terminus to form a phenythiocarbonyl (PTC)-peptide.2. Treat the PTC-peptide with anh. trifluoroacetic acid (TFA) to selectively cleave the N-terminal peptide bond and form a thiazolinone derivative.3. Rearrange to a phenylthiohydantoin (PTH)-amino acid with aq. HCl then chromatograph.IV. Sequencing proteinA. Edman degradationThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.V. CNBr Cleavage at MetCNBr preferentially cleaves at the C-terminal side of methionineVI. Protein cleavage by enzymesA. Protein sequencing is most manageable with small polypeptides.B. Therefore, in order to sequence a large protein, it must be cleaved into smaller pieces.C. Cleavage is conducted most routinely using enzymes called proteases.D. The cleaved fragments must be separated and purified before sequencing.VII.Enzymatic cleavage by TrypsinTrypsin preferentially cleaves at the C-terminal side of lysine or arginine.VIII. Chemical and Enzymatic CleavageIX. An Example, Peptide overlapX. Ordering the peptide fragmentsXIIXI. iClicker questionYou are trying to determine the sequence of a protein that you know is pure. Give the most likely explanation for the following experimental observations. 1. The Edman reagent (PITC, phenylisothiocyanate) identifies Ala and Leu as amino-terminal residues, in roughly equal amounts.2. Your protein has an apparent Mr of 80,000 Da, as determined by SDS-polyacrylamide gel electrophoresis. After treatment of the protein with thereductant DTT, the same technique reveals two proteins with apparent Mr 35,000 Da and 45,000 Da.3. Gel-filtration chromatography experiments indicate that the native protein has an apparent Mr of 160,000 Da.Answer: a) The protein forms a homo-tetramer A4. Half of the protein got degraded from the N-terminus during purification, resulting in two masses. Half of the subunits A engage in a disulfide bond.b) The protein has two subunits, A and B, which form a hetero-tetramer A2B2, in which A and B are connected by disulfide bonds. c) The protein has two subunits, A and B, which form a hetero-tetramer A2B2, in which either A and A or B and B are connected by disulfide bonds. XII. MALDI Mass Spectrometer for intact mass determination and modern sequencingA. Matrix Assisted Laser Desorption IonisationB. Peptides are mixed with matrix and then applied to wells on a target plateC. Peptide ions are generated by a LASER firing at the target plateD. The time of firing of the LASER and the arrival time of the ions at the detector are known, the relative masses can then be calculatedE. We have this at BUXIII. One type of protein mass spectrometry: MALDI-TOFXIV. MALDI-TOF spectrum: intact, undigested proteinsXV. MS/MS AnalysisA. Carried out on nanospray/electrospray mass spectrometers Rather than spotted on atarget plate, the sample is introduced through an inlet from a capillaryB. Peptides are further fragmented inside the mass spectrometer and the resultant “daughter” ions observedC. We have this type of instrument at BUXVI. Protein Identification Using Peptide Mass FingerprintingA. Produce a theoretical digest of all the proteins in a database with a specific enzymeB. Compare these theoretical masses with experimentally observed masses for the digested peptidesC. Assign a score to matching peptides/proteinsD. Students do this in the Bioinformatics course (offered fall semesters)High resolution protein structure determinationXVII. NMRA. Absorption of radiofrequency radiation that lead to excitation of nuclear spin-states in target moleculesB. Need high magnetic fields so nuclear energy levels are sufficiently differentiated for adetectable NMR responseXVIII. Basis of NMR spectroscopyXIX. NOE effects identify pairs of protons that are <5 Angstroms apartA. Off diagonal peaks produced by nearby protons, providing structural informationXX. NOESY spectrum for simple 55 amino acid protein XXI. Protein NMRA. Can be performed in aqueous solutions under in vivo conditions, with protein concentrations of 1 mM B. The premier technique for studying protein dynamicsC. NMR provides inter-atomic distances (not absolute positions of atoms)XXII. Limitations with NMRA. Overcrowded spectraB. Limited to small proteins (less than 150-200 residues)C. Provides local distance constraints onlyXXIII. Overview of X-ray CrystallographyA. Protein crystalB. Diffraction patternC. Electron densityD. Atomic ModelXXIV. Interference of two wavesXXV. Bragg’s law for diffractionA. nl = 2d sin q1. d- Spacing between two atoms2. q- Angle of incidence of X-ray3. l- Wavelength of X-rayB. Bragg’s law implies that the diffraction pattern is determined by the spacing betweenthe atoms and therefore we can calculate the structure from the diffraction pattern.XXVI. The Raw DataA. Every atom in a unit cell contributes to every reflection in the diffraction pattern. Thespacing of the spots is due to the size and symmetry of your crystal lattice.B. Two Pieces of Data1. The position of a reflection point on the reciprocal lattice, given by coordinates h,k,l. 2. The intensity of the reflection.XXVII. Frozen crystal in cryoloopXXVIII. An example of a home source (Rigaku)A. X-rays are generated by bombarding electrons on an metallic anode. B. Emitted X-ray has a characteristic wavelength depending upon which metal is present.C. e.g. Wavelength of X-rays from Cu-anode = 1.54178 Å.XXIX. 1st step: space group determination to get crystal symmetryA. By applying symmetry operations of a particular space group to the ‘asymmetric unit’ we can generate the entire crystal lattice. Therefore, we only need to determinethe structure of the asymmetric unit and the space group in order to describe the entire crystal lattice.XXX. From diffraction to electron density mapA. To calculate the electron density from the diffraction pattern, you have to use a Fourier Transform.XXXI. Phaseproblem in X-rayCrystallographyA. Fourier transform to calculate an electron density map:B.