PROTEIN STRUCTURE DETERMIN
PROTEIN STRUCTURE DETERMIN BCBP 4870
Popular in Course
Popular in Biochemistry
This 216 page Class Notes was uploaded by Bartholome Goyette on Monday October 19, 2015. The Class Notes belongs to BCBP 4870 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/224815/bcbp-4870-rensselaer-polytechnic-institute in Biochemistry at Rensselaer Polytechnic Institute.
Reviews for PROTEIN STRUCTURE DETERMIN
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/19/15
Protein S trUCture De rmiquot ation 3 jray Crystallography 2x The method Crystals X rays Atoms wavelength z size of cell transmitted light lenses 2D image measures relative absorbance image composed of pixels sample is thin section wavelength z size of atom scattered light no lenses possible 0D images 3D construction measures relative e densHy composed of structure factors waves sample is thick crystal Experimental setup Xray source Xray detector beam stop Dimensions Xray detector Eel6 Gel Crystal thickness 010100 mm Beam width 020 mm W V 2x w 6 v W W W Y W ltgtltgtgt Unit cell 100A 000001 mm W 0 V W ll W 5 7 Xx 11W l V m I g Typical protein molecule 30A 0000003 mm 9 W Dimensions CC bond distance 152A ForCHs Wavelength of Cu Koc Xrays 15418A J Dimensions Angle of incidence8 090 Bragg plane separation distance resolution 0750A J l I I Carbon atom am nt an electron mo es in one xray cycle Xrays see 639 as if the v ere standin still Electromagnetic spectrum wavelength cm Wavelength of Xerays used in crystallography 1A 7 3A A 103910m most commonly 154 Cu Frequency cA 3x108ms 154X103910m 2x1018 5391 oscillating e39 scatter Xrays in all direction oscillation mWiV W emission fe s is scattered Xray diffractometer Reconstruction of e39 density The density at every point in the crystal is calculated by summing over all of the density waves Crystal growth Diffraction theory Symmetry Experimental methods Interpretation of data Software 2 f quot A uniU tr em cosa isina Eulers theorem n 2d sinH Bragg39s law g so S Reciprocol space A Xsym M 17 Symmetry Fwd 2 PXyz62mh ky z Fourier transform xyz pxyz 2Fhkle 2mh ky z Inverse Fourier transform hkl all terms defined physicalgeometric interpretation Supplementary reading Matrix algebra An Introduction to Matrices Sets and Groups for Science Students by G Stephenson 795 Wave physics Physics for Scientists and Engineers by Paul A Tipler Protein structure Introduction to Protein Structure by Carl Ivar Branden and John Tooze Introduction to Protein Architecture The Structural Biology of Proteins by Arthur M Lesk Materials graph paper straight edge protractor compass calculator Wtri g functions httpwwwbioinforpiedulbystrclcourseslbcbp4870lbcbp4870htm Protein Strucutre Determination lectire 6 Ewald sphere data co ec onscaHng Scattering factor of an atom An atom is a spherically symmetrical cloud of electron density Which is densest in the center By integrating over the electron cloud we get the Fourier transform of the atom 392 S39 f S f p 06 rdr atom If we de ne r to be a vector relative to the center of the atom then fS can be thought of as a single wave coming from the center of the atom Scattering from atoms and other centresymmetric objects has phase equal to either 0 or 180 The imaginary part cancels out because sin 27IS r sin27IS r 392 S39 e rccs2nS rzsm2 S r 392 S39 e r COS2JTS39r ZSIHZJTS I iZJ39ES39 r e eim 39r 2ccs2nS r net sine part is zero Temperature factor B The sharper the electron density distribution the broader the scattering factor The temperature factor B modi es the scattering factor by spreading out the electron density Correction factor for atomic Bsin2 9 scattering factor 6 A2 Fourier transform as a sum over all atoms sin2 6 FSgt 2fgeB Zeim rg atoms g One wave for each atom The amphtude of the scattering factor f g depends on how many electrons that atom has Each f g is positive and real ie not complex Using Miller indeces sin 2 6 Flthgt 2fltggt 3g7em xg atoms g What reflections can we see without moving the beam X ray source b661 Behavior of 8 versus 26 with the beam fixed crystal 280 p gt beam S0 S 28 30 M Elm Q 29 170 For a given direction of the incoming X rays the Ewald sphere is the set of possible scattering vectors S given so The radius of the sphere is 19 reciprocal units The Ewald Sphere is de ned as the sphere of radius 1 that is the locus of possible scattering vectors S When the beam s0 is xed X This s here is in scatterin quot cryStal pOSition p g quot39quot de nes the space reczprocal space coordinate axes If a scattered wave has S on the Ewald sphere it is Visible on the lmdetector Moving the beam moves the Ewald Sphere For a given direction of the incoming X rays the set of SO X possible scattering vectors S is the surface of a sphere of radius 1 passing through the crystallographic origin C Keeping the crystal xed we rotate the X ray source The Ewald Sphere moves in parallel With the X ray source The neW set of S vectors describe the phase vs direction of scatter for that position of the source gt4 K Moving the beam Crystal fixed By moving the X ray source relative to the crystal we can sample every possible S ignifg The visible part of reciprocal space The set of all vectors S red given all possible directions of the beam black arrows is called reciprocal space Remember in this vieW the crystal is xed center of image Where the X rays are pointed In real life we nd it easier to move the crystal not the source It doesn t matter which one you move the crystal or the source The results are the same Determining the space group 1 Find the high symmetry axes using precession photos 2 Are the axes all 90 apart orthorhombic tetragonal or cubic 90 90 and 120 trigonal or hexagonal 90 90 and 9 0 monoclinic none of the above triclinic Pl Precession camera geometry Seeing one plane of the reciprocal lattice at a time By using a screen all but one lattice plane is masked out If the angle of the beam With the a aXis is a then the correct ngle setting for the screen is o Ewald sphere This method NOT used for data collection Too slow for example bc plane at h0 path of beam relative to crystal Precession photography A precession photograph contains one complete reciprocal plane on one lm What crystal form is this between axes relative space of spots angle absences note systematic Precessi on photograph Determining the space group 3If trigonal of hexagonal take h01 recession photo 6 fold symmetry P6n W 6m symmetry P6112112 Q q 3 fold symmetry P3 P31 P32 3m symmetry P3n21 or P3n12 If orthorhombic or tetragonal Are cell dimens e a b c orthorhombic ab c tetragonal abc cubic Systematic absences 4To get screw aXis n look at systematic absences examples orthorhombic odd 1 re ections missing in 1100 line gt 21 trigonal only 13n re ections present in 001 line gt 31 or 32 hexagonal only 12n re ections present in 001 line gt63 tetragonal only 14n re ections present in 001 line gt 41 or 43 Enantiomeric space groups P41 and P43 can t be distinguished until the phases have been solved Data collection 2 Measuring the intensity amplitude squared of each re ection Output of data collection thousands of re ections each With 5 parameters h k 1 F sigma Diffractometers yesterday and today Counter moves in quot LL 26 V 21 39 Crystal moves in 3 X mys angles uK and 1p Single photon counter photo multiplier tube 1 widely used Raw images are scanned into digital images I quot Each image has three angle associated with it K34 and 00 A series of lms each With a different 00 angle are collected and digitized Today Image plates I 1 crystalline phosphorescent image plate material with color centers exposure to Xrays P fl39 II f II39II39 I39fl39n39l39f39f lfn39u39lI39I l I II39I IW39 39 I39II39HII39II39II39I39I39f I Image plates are ultrasensitive laser scans the plate reusable lms Data stored energy is photomultiplier measures collectlon IS done 1 re easeim Iiiii n W A r v r 7 same Way as for photographic lm erasure of residual image by exposure to white light Area detectors photons Window absorption gap I rst cathode amplL cauon gap anode second cathode Position sensitive X ray detectors give a 3D image of each spot Where ld or image plates give 2D images Moving the crystal is like moving the Ewald sphere By moving the X ray source relative to the crystal we can sample every possible S data coHec on frame 1 transform Ewald sphere Cl 93 B a 4 Ewald sphere data coHec on rotates rotates the reciprocal lattice As the crystal is Try this How far can you rotate the crystal before two spots DraW an Ewald sphere and a lattice fall on top of Put your pencil on a hkl each other on the position that is on the lm sphere Rotate sphere and pencil together until pencil hits 7 the next reciprocal lattice lane 410 p How far did you rotate Ewald sphere For this exercise to work in reality the lattice and sphere must be drawn to scale Ewald sphere radius 1 Xray diffractometer with area detector The detector or lm sits on a TWO theta arm that can swing out away from the beam to collect high resolution data Schematic diffractometer crystal beamStOP X ray 3011GC quotMW I 6 detector settlng goniostat The film is a window Ewald sphere onthe All lattice points that fall on the Ewald sphere are on meaning photons are being scattered that direction but most of those re ections are off camera they end up on the laboratory wall The detector or lm collects a Window of the Ewald sphere As the crystal moves the Window moves relative to the reciprocal lattice path of detector through reciprocal lattice One sweep through reciprocal space collects a donutshaped volume of data This is the volume so of reciprocal space i that has been seen by the detector 1 axis 2 machine axis center of donut Red circle shows the low resolution limit for this detector position A low angle setting of the detector would be necessary to collect the lOW resolution data Multiple m sweeps are usually necessary Gray area volume of reciprocol space that has been seen by the Ewald sphere and thus the detector Intersecting donuts of data add up to the Whole Unique Set or more Most re ections have multiple copies Completeness What fraction of the unique set has been collected at least once Number of re ections depends on resolution cutoff raW images Data reduction ix 97 1ndex1ng backgr0und estimation integrati0n of spots hkl 200 210 201 220 221 222 merging of partials 99 65 78 19 88 HkowU39II O F 000000 NNI NUJNQ scaling merging of syms re ection data Structure factors Data reduction OindeXing nding the location of each reciprocol lattice point HKL Obackground estimation 2 like subtracting the baseline in 2D 0integration of spots 2 intensity is proportional to F2 Omerging of partials 2 One re ection may be split between two lms Scaling If there is signi cant decay then data is scaled in blocks of time Averaging of syms Symmetry related re ections are averaged IndeXIng the data A reciprocal lattice is initialized using the known cell dimensions sphere A systematic search rotation of the lattice is done until the predictions match the obsemtions Small re nements in the beam position might be required spot has an index hkl Calibrating the film or detector For photographic lm or any type of X ray counter a calibration curve has been pre calculated The pixels are counted multiplied by I from the calibration curve to get Ihkl for each spot absorbance Merging partials If lms were switched while a spot was on the Ewald sphere both copies partials are summed together to get Ihkl 001530 m1535 FirSt half 0f 100t hits the Ewald Other half of spot passes through sphere How can partials exist Re ections are points in reciprocal sp e ri t Wrong Re ections have size and shape I 11 th quot S directions a b c because the crystal latti is 9 erfect and in nite Re ections have additional s pe i e laboratory dimensions Xy on the lm r det tor because the beam is not in nitely small and 39 e or tal is not in nitel small 7A v The spot shape is partly the shape of the intersection of the beam and crystal Spot profiles in 3D1 Pro le of the average spot summed over all spots With similar W Xayszesw Intensity of each spot blue is the summed only within the Spot pro le limits grey This prevents counting spurious data like this Scaling within a dataset Re ections may have errors in amplitude Within a dataset because 0 Xray intensity varied 0 Filmdetector sensitivity varied 0 Crystal orientation cross section varied With W 0 Crystal decayed over time 0 Exposure time varied 0 Background radiation varied Scaling assumes 1 Symmetry related re ections have the same amplitude 2 Re ections that were collected together are scaled together ie applied the same scale factor Elwhk1Fhkl wRhklFR a sym op hkl Quality of the data set R W W Elwhlehkl hkl Should be lt 2 Example Structure Factor file datarlpkqsf auditrevisionid 10 auditcreationdate 2003 07 15 auditupdaterecord 39Initial release39 loop ref1nwave1engthid ref1ncrysta1id reflnindexh reflnindexk reflnindex1 reflnFmeasau reflnFmeassigmaau reflnstatus 1 1 39 0 26 70300 34700 0 1 1 39 0 27 158300 25740 0 1 1 39 1 1 156000 15800 0 1 1 39 1 25 54100 23690 0 1 1 39 1 26 201400 11450 0 1 1 39 2 25 151900 11970 0 1 1 39 3 22 202800 22730 0 1 1 38 0 26 75900 37400 0 Structure Factors are deposited in the PDB WWWrcsborg along With the atomic coordinates Lecture 3 Wave addition light Wave ength x Crest mpmude a V Tmugh Dvecmn of momn Electromagnetic spectrum Visible H Microwave wavelength cm Wavelength of Xerays used in crystallography 1A 7 3A A lO39lOm most commonly 154 Cu Frequency cA 3x108msl54x103910m 2x1018 5391 tagsEff angle Xray beam wmdow window Xray beam 39 39 anode cathode negative high voltage sealed tube copper anode Xray source l I K3 K6 l K 05 10 x A 15 15418A Ni lter absorbs everything up to here phase an ifilt i wavelength The instantaneous eeotrio field at time t oscillation rate in cyclessecond gtkalso an oscillating magnetic eld of the same frequency 90 degrees out of phase quot 3139s orbitgatesajlround llOiOth C z2X106ms So in one X ray cycle would travel 21161106111 2X10183 1 3910 12m 00113 not compared to the size of an atom 1A atom amount an electron moves in one xray cycle In other word Xrays see e39h were stand ktill l 39 emission That s scattering In practice this means all directions since X rays are not polarized TWO electrons oscill ting in the same place oscilate in phase and scatter twice as much V 2 electrons separated by M2 oscilale out of phase and cancel each other Rule 1 The sum of two waves with wavelength 9 always producesZ wave of wavelength 9 l Constructive interference amplitude increases A 50 on 60 or 437 radians phase 0L in the equation Acosootoc Where We are at t0 relative to a cosine wave The reference wave cosine has phase 2 0 A sine wave has phase 90 m n 1 I 433 sine part t0 4333 sin out t0 50 cosoot 1375 2 50 cos 13n cos out 50 sin 13n sin Dt 25 cos 0t 433 sin Dt cosoc 3 cos 0L cos 3 sin 06 sin 3 05th sinwi amplitude of amplitude of cosine part sine part i 60 Cosine parts and Sine parts can be summed independently like orthogonal coordinates 505 439 05Cosut 539 08639640866 Sin00t 05 cosoot 0866 sinut III39IIIIIII Illlullll reference sine wave reference cosme wave phase arctan 08F6605 60 amplitude B 2 050030 0505 10 For y 0 1 0CS to the calculator s answer Split waves into cosine and sine parts add them Bc0 sc0t3 2 cosoot 90 4 cosc0t 60 phase3 amplitudeB 1Sepafate each WaVe into sine and cosine parts A cos cot 10 A cosoc cosmt A sinioc sincut 2 Add cosine parts sine parts Alcosa1 Azcosa2cosoot Alsina1 Azsinaz sinoot coe cicient 0f sinwt term 3 Solve for new hase 3ICtan p 3 coe czent 0f cos cot term 4501V6 for new amplitUd63 Bzcoe icient 0f coswt termcosf 0rBcoe Cienl 0f sinoot termsinf H A2 cow2 4005 20 A2 smocz 4O O866 3 4 6 BCOS00t Alcosoc AZCOS OLZCOSDt Alsil lOL1 Azsincxz sinoot 3 arctan203462000 36 1 B20cos3 200808 247 quot amplitude of cosine part Reference wave sinoot amplitude of sine part cosine part RCfCI39Cl lCC WEIVC COS1t Bz25 ix 36 Proof write the exnansions and sum tth ea 1 05 a22 a33 a44 a55 em 1 ia oc22 ioc33 a44 was5 cosa 1 a22 a44I a66 isina ioc ioc33 ices5 ia77 cosine part i Argand diagram we may conveniently use Argand diagrams for waves cosine part real part sine part imaginary part Multiplying complex exponentials 2 phase shift 60 Alelou 6102 AlelOL1OLZ Try it Add these waves using a protractor and ruler Start at the origin Add head to tail 30 30 20 180 05 90 20 45 10 135 20 120 05 120 05 10 20 30 Use these bars to calibrate a makeshift ruler if necessary Schematic diffractometer crysta beamstop my beam x 12322 QGd Y goniostat yS length9 the number of oscillations completed When hitting the detector The Qhase is the noninteger part Exactly where the light turns the corner determines the phase VV Same for scattered path Difference in pathlength 2 rs r030 Relative phase on 275rs rSO a b alblcosff0 Difference in pathlength 2 rs ri s0 Relative phase on 27rs rs0t If e l scatters With amplitude A1 and 62 scatters With amplitude A2 then the sum of their scattered waves is A1 Azei and this Get the difference divide by the wavelength Multiply by 27c That s the phase difference Add the two waves using vectors In class Exercise Sum the scattered waves incoming xrays scattered xrays 2890 s0l00 820 1 0 26 4 Length of box 2 A gt Add 4 waves using the vector method Assume amplitudes are all identical Patterson Space and Heavy Atom Isomorphous Replacement MIR 2 Multiple heavy atom Isomorphous Replacement phasing SIR 2 Single heavy atom Isomorphous Replacement phasing Amplitude or The amplitude of a wave Fh kl scattered by just 2 atoms depends on the distance distance between them in the h kl direction Ignoring phase What happens to the amplitude as the two atoms slide perpendicular to the Bragg planes It stays the same Only the phase changes Largest amplitudes are when atoms are in phase If two atoms are in phase they have the largest amplitude If one of the atoms either one is at the origin then the phase is 0 Phase zero Fourier transform g as v Whatever the true phases of the F s if they are set to zero then the density goes to the points of intecsection Which are the interatom vectors The phase zero inverse Fourier is called a Patterson Map Patterson maps phallse s a s s i s q 4 Reverse transform 007 2 Fhe12 h39r F zZnh ra Wllh phases h h Reverse transform 1307 2 h F iZa r Without phases gt h K it uses the measured amplitude no phase This is the Patterson function i J Patterson space The Patterson map is the centrosymmetric projection real sp e F f 1 atterson space phaseOL Phase 2 0 Using observed amplitudes but setting all phases to 0 creates a centro symmetric image of the molecule Patterson map represents all inter atomic vectors To generate a centrosyrnrnetric projection in 2D draW all inter atomic vectors then move the tails to the origin The heads are Where peaks would be For example take glycine 5 Move each vector to the origin atoms not counting H s Patteron map for Gly in P1 m unit cell vector peak Can you reassemble glycine from this J translational For small molecules vector geometry problem can be solved if you know the geometry bond lengths angles of the molecule Patterson peaks generated by symmetry operations are on Harker sections 2 g Ln 39 Patterson space Real space Harker section 2205 I Harker sections tell us the location of atoms relative to the cell axes Ay YX Divide this vector The 2 position by tWO t0 get the IS found 0H XY coordinates other sectlons Non Harker sections tell us inter atomic vectors not related by symmetry If there is more than one atom in the asu you can get the vector between them by searching for peaks in non Harker sections of the Patterson like the glycine example Then combining knowledge from Harker sections giving absolute positions and non Harker sections giving relative positions we can get the atomic coordinates Simple case 2 atoms P21 y The xy position is X found relative to the 2 fold axis for each atom The relative Z position is found for one atom relative to the other Protein Patterson maps are a mess Even if atomic resolution data is available no individual atom atom peaks can be identi ed because of overlap The number of atom atom peaks goes up as the square of the number of atoms In class exercise make a Patterson map Using the Escher Web Sketch Set the space group to P4mm Place one atom at 04 025 DraW the Patterson vectors Place a second atom different color at 02 01 DraW the Patterson vectors Multiple isomorphous replacement Tuming proteins into small molecules by soaking in heavy atoms The Fourier transform ie diffraction pattern of a heavy atom deriVitiVe is the Vector sum of the transforms of the protein and the heavy atoms NOTE protein and pmbeimheav tom crystals must be isonwrplwux Subtracting amplitudes is like subtracting density almost FH the Fourier transform of the heavy atoms is the Vector difference of FPH an FH w le 7 lel Subtracting Fourier transforms We can t do FPH FP 2 PH i directly because we don t know the phases Assume for the moment that IFHI ltlt IFPI IFPHI Then the phases of FP and FPH are approximately the same So IFPHFP z IFPHI IFPI FF FH I FPH In that case the Patterson map based on IFPHl IFPI Will be approximately equal to Patterson map based on the true IFHI A difference Patterson is dominated by the heavy atoms self peaks cross peaks In class exercise solving a simple Patterson Patterson peak 05 01 0333 Space group is P31 3 Where are the 3 heavy atoms 1 Draw a trigonal unit cell 2 Draw an equilateral triangle around the cell origin such that XY from these two vectors are the sides 3 Estimate the coordinates of the triangle vectices In class exercise solving a Patterson Patterson peaks at 04 02 0333 0l 05 0333 0503333 Space group is P31 Where are the 3 heavy atoms 1 Draw a trigonal unit cell 2 Draw an equilateral triangle around the cell origin such that XY from these two vectors are the sides 3 Estimate the coordinates of the triangle vectices Calculating FH from the heavy atom coordinates FHUY E gVMZ39r gall heavy atoms Once we know the heavy atom coordinates we can calculate amplitude and phase for all re ections FH We can represent a structure factor of unknown phase as a Circle 1 Radius of the Circle is the amplitude The true F lies somewhere on the Circle Harker diagram method for discovering phase from amplitudes We know only amplitude We know amplitude and phase One heavy atom SIR There are two ways to make the vector sums add up TWO heavy atom derivatives MIR unambigous phases IFP FHII FPH1 IFP FHZI IFPHZI Or IFPIZIFPH1FH1 IFPH2FH2 In class exercise Solve the phase problem for one F using two FH s IFPI 290 Draw three c1rcles W1th the three dlameters scale doesn tmatter IFlel 260 Offset the PHl circle from the P circle by FH1 IFPHZI 320 Offset the PH2 c1rcle from the P Cerle by FH2 Find the intersection of the circles 1HZ 110 aH19 Additional topics Crystal packing solvent content Mathews number reciprocal space symmetry amplitude error and phase error Crystal packing Protein crystal packing interactions are saltibridges and H7 bonds mostly These are much weaker than the hydrophobic interactions that hold proteins together This means thatlpr0tein crystals arefmgile and 2 proteins in crystals are probably not signi cantly distorted from their native conformations Remmder The asymmetnc unlt the smallest region of the unit cell that is suf cient to generate the whole unit cell by symmetry For space group P1 the whole cell is the asymmetric unit For other space groups it is the fraction of the unit cell corresponding to lZ where Z the number of equivalent positions For 13212121 2 4 W1 P212121 7Xry127z12 7x12y1272 x12ryz12 So the asymmetric unit will be 14 of the unit cell How many molecules are in the asu 1First nd the total volume of the unit cell V0611 2Divide by the number of equivalent positions Z this is the volume of the asymmetric unit VasuVcell Z 3Calculate the approximate volume of your molecule using the molecular weight and the density of protein about 135 g ml vmol MWl 35gml Nav 1024A3m1 4Assume the number of molecules in the asu is n Calculate percent solvent in the crystal Psol100 Vasu n vmol v asu Quicker and easier Matthews number Matthews number is VM Vcell MW Z Molecular weight is roughly the number of residues in the sequence X 110 Z is the number of equivalent positions Matthews number for proteins varies from 166 to 40 for crystals with 30 to 75 solvent If VM is too high there must be more molecules in the asu than you thought In class exercise how many molecules are in the asymmetric unit Your molecule has 62 residues Your crystal cell dimensions are 40X50X60A orthorhombic P212121 24 Use Matthews number to get n VM vcell nMW Z Get molecular volume vmol MWl 35gml Nav 1024A3m1 z 1 23MW Percent solvent Psol100 Vasu n vmol v asu answer Molecular weight is roughly 62110 6820grnol Vmol 123 6820 8390133 Vasu 4050604 30000 3 if n21 VM 2 1200004 6820 439 too high if n22 VM 2 1200004 26820 22 just right so n2 Percent solvent 2 30000 2839030000 44 Oh no We can t measure phases Xray detectors lm photomultiplier tubes CCDs etc can measure only the intensity of the Xrays which is the amplitude squared but we need the full wave equations Aeio for each re ection to do the reverse Fourier transform And because it is called the phase problem the process of getting the phases is called a solution That s why we say we solved the crystal structure instead of measured or determined it Phase is more important than amplitude a Duck and duck FT b Cat and cat FT colorzphase angle darknesszamplitude 0 Duck intensities l Backlmnsfarm of c 1 phases and ca Reciprocal space symmetry Rotating the crystal rotates the reciprocal lattice by the same amount Applying a mirror or inversion does the same to the reciprocal lattice So the symmetry in the reciprocal lattice is the same as the symmetry in the crystal With the addition of Friedel symmetry but without translational symmetry The addition of Friedel symmetry causes mirror planes to appear in reciprocal space When there is 2 fold rotational symmetry in the crystal h k1 hll gt hkl Fgt hll mirr0r 2fold Frledel hk 1 Reciprocal space symmetry operators are the transpose of the real space symmetry operators The forward transform is 127152 F h g summed over all atoms 2 fge in the asu and all symmetry operators Z So the reverse transform is summed over the 00 Z 2FEe i2nliZF Z h unique set times Z Transposing the matrix is the same as reversing the order of multiplication a b c x y zLd e fxaydzg xbyezh xc3fzi g h 139 Flip the matrix and change the order and you get the same thing written as a column instead of a row a d gx xaydzg Lb e hJLyJbeyeth c f i z xc3fzi The Unique set is the asymmetric unit of reciprocal space Unique K data Initial phases Phases are not measured exactly because amplitudes are not measured exactly Error bars on FF and FPH create a distribution of possible phase values oc Width of circle is 10 deviation derived from data collection statistics Protein Crystallography Crystals X rays From protein to protein structure Crystals and Xrays Protein crystal growth ws indie Cp D different diffusion g C experients us 3 ABDFG Vapor g diffusion V 8 A E Bulk g F G 6 E C Micmdialysis n B i m Lliqud L Ssolid mmetzstzble state precipitant concentmtion CR supmam39w blue line saturation of protein Crysmlgmwvh occurs between red line supersaturation limit these two limits Above the supersatumu39on limit proteins form only disordered precipitate vapor diffusion setup Hanging Drop with rtein 1 quotr w 6 5quot 3 y quot g A 39zs eiuquot 2 am A 1 I Reservoirwith Precipitant Linbro plate Volatiles ie water evaporate from one surface and condence on the other Drop has higher water concentration than reservoir so drop slowly evaporates Crystallization theory Nucleation takes higher concentration than crystal growth I quot I I 1 I x I a ll K I x slow After nucleation the large size of a face makes the weak bond more likely Q R RRRRRRRRRRRRRRRR RRRRRRRRRRRRRRRR RRRRRRRRRRRRRRRR RRRRRRRRRRRRRRRR RRRRRRRRRRRRRRRR not so slow fast Crystallization theory Bonds A3 are stronger than PQ Crystal packing u Protein crystal packing interactions are saltibridges and H7 bonds mostly These are much weaker than the hydrophobic interactions that hold proteins together This means that 1pr0tein aystals arefmgile and 2 proteins in crystals are probably not signi cantly distorted from their native conformations Crystallization robot U j Highithroughput crystallography labs I J39 use pipeting robots to explore 39 thousnds of conditions Each condition is a formulation of the crystal drop and the reservoir solution Conditions can have different protein concentration pH precipitant precipitant concentration detergents organic coisolvents rnetal ions ligands concentration gradient protein crystals lmm cellulase subtilisin The color you see is birefringence the wavelength dependent rotation of polarized light Crystal mounting If not freezing LOW melting hard wax is used to glue the rod or capillary here Small wrenches t here here here and here Must freeze immediately or lm wax will dry out eucentric goniometer head Crystal must be kept at proper humidity and temperature Very fragile made NO HIUS machine center is the intersection of the beam and the two goniostat rotation axes Must be set by manufacturer ng the crystal in the b it s off center Fix it 5256 To place crystal at machine center rotate it and K and watch the crystal If it moves from side to side it is off center If it is off center we adjust the screws on the goniometer lnead Xray diffractometer with area detector The detector or lm sits on a TWO theta arm that can swing out away from the beam to collect high resolution data Synchrotrons provide tunable monochromatc Xrays A Crystals must be flash frozen Water must be frozen to lt 70 C very fast to prevent the formation of hexagonal ice crystals Water glass forms How Crystals mounted on loops are ash frozen by dipping in liquid propane or freon at 70 or by instant eXposure to N2 gas at 70 C hexagonal ice Precessi on photograph 0 symmetry 0 spacing of spots 0 systematic absences 0 angle between axes Data collection Measure the intensity amplitude squared of each re ection Output of data collection thousands of re ections each With 5 parameters h k 1 F sigma From protein to data 1 determine space group and cell dimensions l collect 1000 s of images i index and integrate spot intensities scale intensity data V Xray scattering amp the Fourier tranform light Wave ength x Crest mpmude a V Tmugh Dvecmn of momn The general equation for wave Photons are m oscillating E gt Amplitude electric a W t elds Phase Et A cosoot 0t wavelength The instantaneous electric eld at time t oscillation rate in cyclessecond gtkalso an oscillating magnetic eld of the same frequency 90 degrees out of phase An electric field accelerates charged particles e39 oscillates in an electric field 06 oscillation is the same frequency as the X rays Oe oscillation is much faster that orbiting motion of 6 around nucleus no signi cant Doppler effect The amplitude of the e oscillation is large because the mass of an equot is small Atomic nuclei don t oscillate much oscillating 939 create photons in all directions J to the oscillation of incoming oscillation xiv emission That s scattering X ray sources may be partially polarized Wave addition tii TR A Sum the electric elds at each p01nt in time Wk A The sum of two always produces wave of wavelength 9 Tii Th m W Rule 1 Eves with wavelength A Cons tructive interference amplitude increases Adding two waves by parts cosine sine Add amplitudes of cosine and sine parts then recombine them Cosine parts and Sine parts can be summed independently like orthogonal coordinates 10 m 60 The sum of angles rule cosoc 3 cos 0L cos 3 sin 0c sin 3 Applying the sum of angles rule to the wave equation decomposes it into sine and cosine parts Et A cosoot 0t Using the sum of angles rule A cosoot ct A cos0t cosoot A Sant sinoot amplitude of amplitude of cosine part sine part sine part cosine part Reference wave sinoot RCfCI Cl lCC WEIVC cosoot A waves can be represented as a complex exponential Euler s Theorem ela PI OOfI write the expansions and sum them ea 1oc a22 a33 a44 a55 em 1 ia 0522 ia33 144 was5 cosa 1 a22 a44I a66 isinoc ioc ioc33 ia55 ioc77 A wave as a complex exponen al Et A cosut 0L 1 i ZACOSOLCO t Asinaw Aei0 i Adding waves using vector addition WW4 mm W t 90 Bz25 A120 a1 90 A W V A2240 12 60 Phase depends on the distance traveled VA A A A A V V V V Phase 2 D 9 nearest integerD VV Same for scattered path definition Structure factor F is a single scattered wave photon F having an amplitude phase and direction relative to the crystallographic reference frame F is the sum of all scatter from electrons in thesarfle F m In class exercise sum the scattered waves Measure the distance traveled from the source to the detector for each electron Calculate phase Sum the waves as vectors S O u g r s0 gt f C e Fel0t1 elOt2 Path difference to get phase H H r i fill in the dot products 6 Phase difference length difference divided by the wavelength multiplied by 27c Fourier Transform an integral of waves in nitesimal volumes unit cell F S fprei2 5 rdr I pelectron density Inverse Fourier Transform There exists an inverse Fourier Transform which when substituted into the FT produces an identity forward transform FS fp r W39rdr reverse transform r fF s e im39rcKS note the minus sign Bragg s Law and Diffraction Although sometimes drawn as a vector X rays are plane waves Reflection plane same angle 8 With beam and scattered Xray 9 OOOOOOOOOOOOOOOOOOOOO O QOOO All points on the re ection plane scatter in ase That s Why a single direction of scatter is called a re ection Sir Lawrence Bragg Winner of 1914 Nobel Prize Bragg s law nk2d sine path length for this plane is shorter by A oocoocooco 4 Re ection planes separated by d scatter in phase If 8 is larger d is smaller dzkZsine Scattering by Bragg planes planes extend throughout the crystal All these electrons scatter in phase A is proportional to the total number of e39 on all of these planes Integrating planes separated by d i Planes shifteddldaree Ax All these electrons scatter in phase and the phase is shifted by 2316 0r 60 Integrating planes separated by d amplitude asa EEEEEEEEEEEEE functionofphase EEEEEEEEEEEEE i The total F is the wave 1 R inte ratin com lex sum over all Bragg ovegphasgegivelso F planes l 39 Bragg planes are always perpendicular to S I V S S 9 e Since s0 and s are the same length and have the same angle to the reflection plane S ss0gt is normal to the plane The length of S is 1d So sinex s S e e sinex The length of S is 23ine times the lengths ofs and so which is 10 So S 23in89 1d Crystal 3D lattice V V A vector expression that de nes a crystal lattice gt pr pr ta ub vc Where aband c are the unit cell axes and tu and V are integers pelectron density O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O 0 Crystal plane numbers Miller indeces Starting from the Origin and moving to the next plane if it intersects the a axis at lh the b axis at lk and the c axis at 11 then the planes are called h k 1 Each set of crystal planes de nes a re ection in scattering space also numbered using h k 1 NOTE h k and 1 must be integers proof later Examples of 3D crystal planes 2 3 3 crystal planes 4 6 6 crystal planes planes intersect the cell axes at fractional coordinates Ih00 0k0001l Proof The only Bragg planes that diffract X rays are those that match crystal planes In other words if we have seen a re ection on the lm that re ection corresponds to a set of crystal planes Since all crystal planes pass through the unit cell origins and since the phase of the Origin can be set to zero all observable Bragg planes of phase zero pass through the Origins Proof All Bragg planes of phase zero pass through the Origins 1 Bragg planes are either aligned with the Unit Cell Origins or they are not aligned If t phase at or1gin0 distance to the Bragg plane Proof All Bragg planes of phase zero pass through the Origins 2 All planes that pass through the Origins have the same number of electrons The angle and intercept with the Unit Cell determine with atoms are on the plane 3 All planes that pass through the Origins contribute the same amplitude because amplitude is proportional to number of electrons and statement 2 Proof All Bragg planes of phase zero pass through the Origins 4 Total amplitude is the sum of the amplitudes of the planes if the planes have the same phase l Amplitude contributed by origin planes is 10K times the amplitude of one such plane if there are 10K unit cells L L L L L L L L g 5 Total amplitude is approximately zero if the planes have different phases Phase shifts by a constant for each unit cell Vectors sum in a circle Summed over 10K unit cells vector K length is small f K x f V Proof All Bragg planes of phase zero pass through the Origins 6 Any point in the Unit Cell can be the Origin 7 All equivalent positions by lattice symmetry have the same phase Because of statement 6 statements 1 5 apply to any point in the Unit cell 8 If the Bragg planes do not pass through all Origins the diffraction amplitude is zero Because the total diffraction amplitude is the wave sum over all points in the Unit Cell Conclusion Bragg planes that pass through all of the Origins diffract Xrays Bragg planes that do not pass through all of the origins do not diffract Xrays Definition of the Reciprocal Lattice Let s de ne the reciprocal lattice as the subset of points in S for Which the Laue conditions hold S 2 ha kb lc then b b 1 See Drenth Ch 4 p 86 Table 41 for how to o c 1 calculate the reciprocal lattice vectors abc C reciprocal lattice axes Complete Laue conditions a39a1 b39a0 c39a0 ab0 bb1 cb0 a39c0 b39c0 c39c Real cell relationship to reciprocal cell 2D for simplicity Ifaltbthe agtltgtbgtllt aJb aJc bJa bJc b a CJa CJb Adding unit cells For unit cell scatter to add constructively scattering vector the dot product of the displacement and Sha kb 1c scatter must be an integer multiple of the wavelength unit cell displacement vector rtaubvc S r hata kbub lcvc 5 000 a Reciprocal lattice A periodic delta function in three dimensions 5amp1 U results in points of diffraction in S space Ewald Sphere Visible reciprocal space with the beam fixed crystal beam W 29 0 28 30 M Elm g 29 170 In 3D a spherical shell of space is Visible Radius of the sphere is 19 Ewald sphere intersecting the reciprocal lattice For a given orientation of the crystal With respect to the X rays only those value of S that are on the Ewald sphere are Visible to the detector 1quot I Iquot Iquot 39I I 40 1quot 0quot Iquot39 I I 3903939 Iquot 39Iquot39I 39I I 390 I quotquotI39 quot0 39I I I 0 I I I 39I I 0 quotI 39c I I I39quot o 39Iquot39I39 Ewald sphere J reciprocal lattice centered on crystal In class exercise index these spots is r39 a the beam answer the Beam Phasing maps To get p invert the FT forward transform electron density gt re ections 2prei2nrhxkylz r reverse transform re ections gt electron density pr 2Fhkle i2nhxkylz hkl Oh no We can t measure phases Xray detectors lm photomultiplier tubes CCDS etc can measure only the intensity of the Xrays Fhkl Fhkleiahkl We can only measure this part
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'