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by: Jaden Stiedemann


Marketplace > Rice University > Biological Sciences > BIOS 481 > MOLECULAR BIOPHYSICS I
Jaden Stiedemann
Rice University
GPA 3.78


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This 12 page Class Notes was uploaded by Jaden Stiedemann on Monday October 19, 2015. The Class Notes belongs to BIOS 481 at Rice University taught by Staff in Fall. Since its upload, it has received 25 views. For similar materials see /class/225026/bios-481-rice-university in Biological Sciences at Rice University.

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Date Created: 10/19/15
VI Vibrational Spectroscopy Molecular vibrations can be detected either by Infrared spectroscopythe absorption of light in the range 2003000 cm391 330050000 nm or by Raman spectroscopythe inelastic scattering of visible light typically measured between 2801000 nm The two techniques are complementary because many vibrators are either only quotinfrared activequot or quotRaman activequot This is because Infrared absorption requires a change in electric dipole either in direction or magnitude while Raman scattering requires a change in electric polarizability these two requirements are usually mutually exclusive Examples of Utility in Biosciences Binding of small molecules ligands eg CO CN N3 to active centers Characterizing active center heterogeneity 3D conformation of macromolecules CD has better resolution but requires optically clear samples Oxidation state of Oxygen bound to Oxygen Carriers Hemoglobin 0239 Hemocyanin 02 Oxidation and spinstate of hemes Proton pathway in Bacteriorhodopsin Mechanism of Oxygen reduction by cytochrome oxidase General Background A classical model A simple diatomic molecule can be represented as two masses M1 and M2 connected by a spring Fig 1 At rest the equilibrium distance is x0 Upon elongation or compression the distance changes to x There is a restoring force f kxx0 k is the force constant the larger the value of k the stronger is the bond between M1 and M2 When released M1 and M2 oscillate harmonically with the amplitude of the oscillations decaying with time because of friction The A frequency of the oscillations v is given by Hook s Law k is a 5 expressed in millidyne per Angstrom mdyn A39l or X 105 gt W dynescm PAH a v 1 ku12 in Hz VI1 27 F1g VIl Or v39 1303 ku12 in cm391 is the reduced mass M M M 2 ifM M H H ml z l l 2 IfM1gtgt M2 then u M2 For the harmonic oscillation of two atoms connected by a bond the potential energy is V 05 k x x02 V Energy This is a parabola Fig VI2 k measures the curvature of this parabola a large k gives a steep curve and viceversa 25 Although the surface of the parabola is continuous quantum mechanics demands that only certain energies are allowedthe ener is uantized Flg v12 gy q E hv o 12 VI3 o is an integer O l 2 which speci es the vibrational quantum number v3 of the state These states are represented by the horizontal lines in Fig 3 3 V 2 For true harmonic motion the potential energy curve is a perfect 1 2 39 parabola and the energy levels would be equally spaced as shown In vo reality k is a function of XXO k gets smaller the bond gets weaker as XXo gets larger Obviously at large separation the molecule dissociates F1 g V13 Consequently the simple parabola is replaced by the Morse curve which recognizes the fact the atoms will dissociate at sufficiently large separation the energy levels get closer together Fig 4 Likewise k gets larger as XXo gets smaller obviously the two atoms cannot enter each other39s space Thus the potential curve is anharmom39c For infrared spectroscopy the selection rules are i The Vibration must be accompanied by Continuum a change in dipole momenteither in magnitude E stretch or in direction bend 15 A 03333quot ii A1 i 1 Zero 532 Vibrational energies are in the range 0 1003000 cm39l At room temperature thermal lnternuclear distancex energy CHI1 10K N CHI1 Thus Fig VI4 most molecules are in o0 Transition Relative Intensity co gt m gt fundamental frequency 1 co gt o2 gt f1rstovertone 01 co gt o3 gt second overtone 001 fundamental first harmonic f1rstovertone second harmonic etc 12 For a change of A1 1 AE hZTC ku The corresponding frequency is given by AB hv Hz hcv39 cm39l Because the masses of M1 and M2 are known a measurement of v allows the calculation of k Typical values of k rou k mdyn A39l Typical freguencies cm39l CC 5 CZC 10 15001700 CEC 15 20002700 CO 12 C H 5 The effect of reduced mass Compare CH with CD 121 122 121 122 1213 2414 0923 171 u CH 093 uCD 171 AE M210 1W1 2 09317112 073 Because k is essentially unchanged by the substitution of H by D the frequency will be lowered by 30 Isotopic substitutions are an important tool in con rming assignments by verifying that the frequency shift follows that predicted from the reduced mass formula Number Of Vibrations A diatomic N 2 has only one vibrationthe symmetric stretch An important example is the stretch of CO at 2143 cm39 1 For N gt 2 the No Of Vibrations 3N6 or 3N5 if the molecule linear The 65 eliminate net translations and rotationsiwe are only interested in the motion of atoms relative to each other These are called the fundamental or normal modesthey are independent selfrepeating motions in which all of the atoms move at the same frequency but with amplitudes which depend on the relative masses Example C02 OCO Fig 5 left N 3 and it is linear so number of vibrations 335 4 They are1 The symmetric stretch v1 1340 cm1 Oxygens move together The asymmetric stretch v3 2349 cm1 Carbon moves towards oxygen A pair ofbends v2 667 cm1 There are two at rightangles In C02 the ir spectrum detects the 2349 and 667 cm391 modes BUT the mode at 1340 cm391 cannot be observed Why Note the charge distributionthe dipole moment is unchanged by stretching lefthand side cancels righthand side Fig 11 pVI9 IT IS OBSERVED IN RAMAN see below 1The subscripts are assigned as follows First the totally symmetric modes are numbered starting with the one at highest energy 3 1 etc Ifthe molecule is linear the bending modes are considered next 3 2 etc Then the asymmetric stretches are considered 3 3 v1 4 adw q 9 y R I Ryp Sio39 M5de a 5 R H y 9 o o gtO39quot Q 77 0 p lt5 7 451 Ina campl x malecule mm m m wbnmmsbm mastu zmmms pudm this manual mndzs mm m lacumumymm large spluemzmsa wu adjacemamms andue m ng candypmn nedby hz snnmmdmgs Grou Fre uen nan am OH I n Cy 9 NH 5H Emltco cooH mm 000 me u The problem with water Fig 6 00W I I l I l l I l I l I l l I l X H20 I I I II 02 I quotquot 03 I If l I I l 04 39 I I I v i 05 I I 06 I F I I r39 quot 08 I xx I I I I Nv Q 10 II quot 0quot 8 o 1 g 115 1 V 1 t I 1 A 1 l l 1 1I 1 1 1 I1 x e 1 I l I l 39 l l I l T l l l H 1 O r 1 w quot D lt II 39 01 39 A II I 3 I I 9 I 1 o L 02 1 H III I I II I U I g I i l I I a O I I 5 03 I I E O I I V kl 04 II II 3 H 1 39 quot 13 I Mn 8 I W I O 5 c F O 9 I 2 I 5 O r 06quot 1 39 l 4 U i I I I I I I 39 08 O O 39 39 I I I I I I I 0 NH NH bondII j 15 1quot 1 1 1 1 1 1 I II 1 1 1 l 1 quotr 3500 II I I I I 3000 2500 2000 1800 1600 1400 1200 1000 m F1g 6 InstrumentTraditional Fig 7 versus Modern Fig 8 Reference cell Detector Ugh source Monochromator Recorder Sample cell Fig 7 A digression The FTIR Machine and how it works We start with a light source which emits a single wavelength 7 1 of 2000 cm391 note that wavelength is commonly quoted as either length e g nm or reciprocal length e g cm39l even though the latter is the wavenumberThe light is split at the beam splitter One half goes to the xed mirror and the other half goes to the moving mirror The two beams re ect are recombined at the beam splitter and go to the detector VI6 Beam splitter Fixed Mirror Moving Mirror i R Source i I Computer 7 IR Detector Sample I Fig 8 When the moving mirror is at its middle null position both light beams traverse paths that are the same length and so when they recombine they are in phase and maximum brightness is observed However when the mirror has moved by 025 it the length of the light path traversed by the second beam is changed by 05 it and there will be a 180 phase difference between the two beams and so the recombined beam will show total destructive interferenceit will be black The detector will see variations in the brightness of the light source which occur at 4000 Hz if the velocity of the mirror is l cmsec see Table below Thus variations in the intensity of the light beam which occur at 4000 Hz label this light as being 2000 cm39l Likewise if the incident beam is of wavelength 1000 cm391 the variations in intensity will occur at 2000 Hz Wavenumber 9 nm 025 9 cm Time sec Fre uenc Hz 2000 cm391 5000 nm 125 x 104 125 x 104 4000 1000 cm391 10000 nm 25 x10394 25 x10394 2000 Time to traverse 25 for a mirror with velocity of l cmsec If both wavelengths are present simultaneously the brightness of the detected light will vary in time in a way which is the result of combining the 2000 Hz and 4000 Hz oscillations A Fourier transform of the variations in brightness will yield the values of two wavelengths and the amplitudes of each of them This is the principle of the FTIR machine The only additional feature is that the light source is whiteall wavelengths within a spectral range are presentand thus the detected beam has no obvious frequency components quotto the eye the detected brightness Fig 9 upper varies in a complex apparently undecipherable manner but the Fourier transform calculation will still extract the frequency components and their amplitudes Fig 9 lower VI7 AL cm 001 000 001 002 l l l Interferogram b 39E39 2 g 3 VP Fourier Transform 8 g Spectrum 4000 l 2000 Wavenumbor cm1 Fig 9 This intensity variation as a function of mirror position is the raw data It is rst recorded for a blank B e g solvent and then for the sample S The Fourier transform of these data yields the infrared transmission of the blank FTb amp sample FTS and log10FTbFTS gives the infrared absorbance as a function of wavenumber The function of the mirror has been to move the frequency information in the irradiating light from video frequencies to audio frequencies where it can be handled by standard data acquisition systems VI8 Vibration N H Amide I CO stretch Amide II CN stretch N H bend Modes of the Peptide Bond Ca lt H gt 6 Frequency H bonding ca 3400 3300 1690 1650 01 1640 random 1630 5 lt 1520 1540 01 1525 5 Amide I is used to study conformation Amide II hydrogendeuterium exchange COOH COO39 Conformation xheliX Antiparallel 5 Parallel l Unordered 5 Turns The Carbonyl Group CO stretch 1710 1570 IR Characteristics of Peptide Bond H20 D20 1653 1650 1632 s 1632 163036 16281638 1690 w 1630 s 1632 1656 1643 166080 VI9 Raman Spectroscopy An alternative to IR Basis The inelastic scattering of incident radiation Model Fig VIlO Incident radiation drives the electron cloud into oscillations Two possible fates i The electron cloud quotbroadcastsquot at the same frequency elastic Rayleighintensity relative to incident radiation 10393 and ii During the broadcast event the molecule undergoes a vibration Raman scattering relative intensity 10396 This can either i decrease the energy of emission Stokes or ii increase the energy of emission antiStokes 0m Rayleigh Raman hv V39 IR INFRARED RAMAN hvo I Rayleigh Abs R Raman A JIL l jl V Vvv V0 V0 Vvv39 Fig 10 RequirementThere must be a change in electric polarizability ie a deformation in electron distribution in response to an applied eld This requires that the relative positions of nuclei change and that there must be regions of high electron density between the atoms atoms of similar electonegativities will usually show large changes in polarizability during bond stretching eg CC CC SS NN CS etc Conversely asymmetric charge distribution eg CO give weak Raman and strong IR Instrumentation Qualitatively similar to that used in recording uorescence emission spectra except that a laser is normally used as the light source Fluorescence competes with the Stokes Raman and is normally so much more intense that it is dif cult if not impossible to record Stokes VIlO Raman in a uorescent molecule In such cases the very much weaker antiStokes scattering must be studied The Raman of C02 Fig 11 The symmetric stretch symmetric stretching 70 displacements extended equilibrium corresponding polarizability ellipsoid 171 39 i t l 2 e 2 i 5 Fig 11 O m C O compressed The upper row shows the position of the nuclei during the symmetric stretch and the lower row shows the associated deformation of the electron cloud By contrast the asymmetric stretch stretches one bond and compresses the other so that the net change in polarizability is zero With the exception of totally symmetric and totally asymmetric stretches prediction of Raman actiVity usually requires resort to theory Re s 0 n a 11 cc Ram an An enormous enhancement of the intensity of Raman scattering found when the frequency of the exciting light matches a G gt S transition In practice it is resonance Raman that is utilized in biological work Example Raman Frequencies for Oxygen derivatives Molecule Bond Order v cml 02 2 1560 0239 1 12 1100 02 1 850 02 2 12 1865 Hemerythrin 1602 844 1802 798 Reaction FeII Fe11 02 gt FeIHFeIH02 VIll Z Possible Modes of Binding 0 o O Fe Fe Fe Fe O O 0 Fe F I o 11 111 e o o Fe Fe I Fe quotquot quotFe 0 IV V 160180 825 and 819 cm1 So must be asymmetric either 111 or IV VI12


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