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03-871 Molecular Biophysics CD Spectroscopy September 25, 2009Optical Rotary Dispersion and Circular DichroismSuggested reading: Chapter 10 van Holde et al., pp. 465-496. : The Feynman Lectures on Physics. Vol 2, Chapter 32.Summary: • Circular dichroism (CD) arises because of the differential interaction of circularly polarized light with molecules which contain a chiral center or have coupled electric dipoles that are chiral. • CD is not observed unless an absorption band exists. • CD spectra of proteins and nucleic acids are much more sensitive to conformational changes than absorption spectra. • CD spectra of proteins can be used to obtain useful information on secondary structure.Introduction: Until now we have been studying the interaction of thetransition electric dipole of molecules with the electric field of light. Inaddition to this interaction, the magnetic field of light can also interactwith magnetic transition dipoles in molecules. In order for this interactionto be productive it is necessary that the molecule possess a center ofasymmetry (such as α-carbons) or to have coupled electronic dipoles.These centers will interact with polarized light. The consequences of thisinteraction are:λ >> λABS: • Birefringence: (nL - nR) • Optical Rotary Dispersion (ORD): φ = 180 l (nL - nR)/ λλ ≈ λABS: • Circular dichroism (CD) differential absorption of circularly polarized light: AL-AR • Ellipticity: θ = 2.303 (AL-AR) 180/ 4π or ∆ε = εL – εRLinear & Circularly Polarized Light: To understand how these effectsoccur when polarized light is passed through a sample we will decomposelinearly polarized light into right and left circular-polarized light. For awave propagating along the y axis (j), the electric field at the origin is: ˆE r = k cos ωt + iˆ sin ωt ˆ ˆ E l = k cos ωt − i sin ωtThe sum of these two fields gives linearly polarized light.For the case of Er the electric field vector will rotate in a right handdirection because the x-component of the electric field grows after t=0. Incontrast El will rotate in a left hand direction. Figure 1: Generation of circular polarized light.Figure 1: Generation of circular-polarized lightThe optical activity of a molecule is due to a differential interaction of themolecule with right or left circularly polarized light. To help understandthe origin of optical activity consider thefollowing two molecules, which are mirror images of each other:Figure 1: Some "lock-washer" molecules1
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009Origin of Optical Activity: A flat molecule has no optical activitybecause it possesses a plane of symmetry. In contrast, the helicalmolecules shown below have optical activity. When a molecule is placedin an electric field the electrons involved in the transition will oscillateaccording to the direction of the electric field. In the case of the helicalmolecule they can be driven in a helical path. This generates a currentloop in the helix and the current loop generates a magnetic dipole in thesame way an electromagnet generates a magnetic field. The inducedmagnetic dipole is parallel with the z-component of the electric field thatgenerated it. The induced magnetic dipole can interact with the magneticcomponent of the electric field. • No interaction is possible with the magnetic field associated from the z-component of the electric field. Why? • An interaction between the induced magnetic dipole and the electric field in the x-y plane is possible. Why?The phase of the electric field oscillation affects the coupling between theinduced magnetic dipole and the magnetic field. For a productiveinteraction between the induced magnetic dipole and the magnetic fieldfrom the x-y electric vector the two fields must oscillate with the samephase. Only one direction of the circularly polarized light (right or left)will possess the correct phase, hence the reason why one of thecomponents of the polarized light (Er or El) interacts more strongly withthe chromophore.2
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009Circular Birefringence & Optical Rotary Dispersion (ORD)Optically active material will show a different refractive index for rightand left circular polarized light. The index of refraction of a material is aconsequence of the generation of an electric field by the dipoles within thematerial which oscillate due to the applied light. • At frequencies far removed from the absorption band the oscillations of the applied field induce oscillations of charges in the material which are in phase with the applied field. • As the frequency of the applied light approaches the absorption maximum there is an increasingly larger phase shift in the induced oscillations. • At higher frequencies, the induced oscillations change sign, reversing the effects seen at low frequencies.The oscillating electrons generate an electric field that contributes to thefield of the transmitted light. The induced field generates a difference inrefractive index for the two forms of light. The difference in refractiveindex is called circular birefringence: nl - nrOptical Rotary Dispersion (ORD) arises because of a difference in theindex of refraction for left and right circularly polarized light. Due to thedifference in refractive index the wavelength of the light for each directionof circularly polarized light is different while the light is in the media.Consequently, a phase shift will develop between the two circularcomponents. This phase shift will cause a rotation of the linearly polarizedlight when it leaves the media. 180l (nl − nr ) φ= [degrees] λThe optical rotation as a function of wavelength is referred to the opticalrotary dispersion. The ORD curve changes sign at the absorptionmaximum. This occurs because the phase of the oscillations of theelectrons becomes 180 degrees out of phase from the incident light. Thisimplies that the polarized light (right or left) which was retarded by thematerial at longer wavelength now becomes advanced with respect to theother polarized direction at shorter wavelengths.Circular Dichroism occurs as the wavelength of the incident lightapproaches that of the absorption band. In this case the oscillation ofcharges in the material is damped as energy is removed from the field bythe absorption process. If the absorption is different for right and lefthanded circular-polarized light then the linearly polarized light willbecome elliptically polarized. The ellipticity (θ) of the light is defined bythe arc tangent of the ratio of the major axis to the minor axis of thetransmitted light. Usually, the actual absorption of each component oflight is measured and the difference in absorption is called the circulardichroism (CD):CD = Al – ArWhich is related to the ellipticity by: 2.303( Al − Ar )180θ= = 32.98 × ( Al − Ar ) 4π3
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009 ER − ELtan θ = ER + EL I R/ 2 − I L/ 2 1 1 Small angle approximation.θ = 1/ 2 1/ 2 IR + ILI = I 0 e − A ln10 Beer’s law A=-log(I/Io) − AR − AL ∆A ln x = ln10 log10 x ln 10 ln 10 ln 10 e 2 −e 2 e 2 −1θ= − AR − AL = ∆A + AL ln 10 ln10 ln10 ln 10 2 +1 Multiply by e e 2 +e 2 e 2 ∆A ln 10 ∆Ae2 ≈1+ ln 10 2 Series expansion of ex ≈1 + x ∆A Apply series expansion ln 10θ≈ 2 ∆A ln 10 + 2 2 ∆A Assume ∆A <<1 ln 10θ≈ 2 2 Convert to degrees. ∆A 360 θ≈ ln 10 × 4 2π The molar ellipticity of the sample is given by:[θ ] = 100θ ClWhere C is the concentration in moles/liter and l is the path-length in cm,and θ is the measured ellipticity. The historical units of [θ] are deg cm2dmole-1 instead of deg M-1 cm-1. Although these units seem odd, they canbe easily derived: L mol 1000cm 3 moldeg M −1cm −1 = deg × cm −1 = deg × = deg100cm 2 dmol −1 mol 10dmol mol 10dmolThe molar circular dichroism can also be related to the difference in theextinction coefficient for right and left circularly polarized light:θ = 32.98 × (ε l − ε r ) deg M −1cm[θ ] = 3,298∆ε = 3,298 (ε l − ε r ) deg cm2 dmol-1It is becoming more common to express CD spectra in terms of the muchmore straight-forward difference in molar extinction coefficient:∆ε = ε l − ε r4
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009Applications of Circular Dichroism:Determining Secondary Structure by CD: Various secondarystructures have characteristic CD spectra (Figure 3). helix beta Figure 3: CD spectra of polypeptides (left) and CD spectra extracted from proteins. The scale on the left has been multiplied by 10-3, i.e. 80 = 80,000.Note that the CD spectrum (or optical activity) of α-helical residues issignificantly greater than either random coil or β-sheet. This is aconsequence of the interaction between transition dipoles oriented by thehelical structure of peptide. This interaction is weaker in β-sheet andrandom coil. A similar interaction will also occur with nucleic acids (seebelow).The determination of secondary structure by CD is based on the fact thatthe CD spectra of a protein is well approximated by the CD of eachpeptide linkage: Nθ λ = ∑θ iλ i =1Where θiλ is the CD of a single peptide bond.There are three methods are currently used to determine the secondarystructure of a protein from its CD spectrum.Method 1: The simplest method is to use the CD spectra of polyaminoacids. The CD spectrum of a protein can be taken to be the linear sum ofthe CD spectra from these various secondary structures: θ λ = fαθ λ + f βθ λβ + f cθ λc αWhere fα are the fraction of the number of residues in the protein that arein an α-helical conformation. fβ is the fraction in the β-sheet, etc. Theexperimental data are fit to the reference spectra to determine the amountsof each secondary structure in the protein. A significant problem with thisapproach is that the model compounds are usually infinite in length, thusthey do not mimic secondary structures in proteins, which are of finite5
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009length. In addition, the following problems also occur with any method ofusing CD to obtain secondary structure: • random coils are seldom random • Phe, Tyr, His, and Trp can contribute to peptide CD spectra • Left handed helical structures can occur • Disulfide bonds are very active • Prosthetic groups are also very activeMethod 2: A more reliable approach is to use proteins of known structure(and CD spectra) to define the basis set. For each of the known proteinsthe following is assumed: 3θ λ = ∑ χ jθ λj j =1For each of the known proteins, the fractions of residues in varioussecondary structures are known ( e.g. χ α ). From these data it is possibleto obtain basis spectra for the three types of secondary structure. Thesebasis spectra should now reflect the influence of the protein structure ofthe CD spectra of residues in various secondary structures. As with thefirst method, the CD spectra of the unknown protein is the weighted sumof the reference CD spectra: θ λ = fαθ λ + f βθ λβ + f cθ λc αThe hope here is that the new reference spectra will be more accurate thanthose based on homo-polymers.A flaw in the above approach is the assumption that there are only threeclasses of secondary structure and that each class has a unique CDspectrum.Method 3 [Hennessey & Johnson, Biochemistry, 20, 1085.]: A moreunbiased method of approaching this problem is to extract, usingmathematical methods, a generalized basis set from the CD spectra ofproteins with known structures without regard to the actual secondarystructure of the protein. That is, we now assume the CD spectra of aprotein is: Nθ λ = ∑ aiθ λi iWhere N are the total number of basis spectra, ai is the weight of the ithspectra to the total CD and θλi is the CD of the ith basis spectra at λ.Determining the basis spectra and the coefficients, ai is accomplished by ageneral technique called singular value decomposition.SVD: Singular Value Decomposition: This is a general mathematicaltechnique that can be used to extract out the principle components of acomplex mixture of different spectra. Principle components are analogousto the three basis vectors that can be used to define any vector in a three-dimensional space.6
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009SVD can be used in many different situations. As an example, say CDspectra (or IR spectra, or UV-Vis spectra...) were acquired at 3 differentligand concentrations and at five different wavelengths and we wanted toextract out the part of the spectra that was affected by ligand binding.The raw experimental data can be written in matrix form (each columnrepresents a different ligand concentration): aλ 1 aλ 1 aλ 1 a aλ 2 aλ 2 λ2 A = aλ 3 aλ 3 aλ 3 aλ 4 aλ 4 aλ 4 aλ 5 aλ 5 aλ 5 This matrix can be decomposed into a product of three matrices:A = USV TU: The columns of this matrix are the basis spectra, ui. A weighted linearsum of these spectra can be used to generate the original raw data. u1 1 λ 2 uλ 1 uλ1 3 1 2 3 uλ 2 uλ 2 uλ 2 U = uλ 3 1 2 uλ 3 uλ 3 3 1 2 3 uλ 4 uλ 4 uλ 4 u1 2 uλ 5 uλ 5 3 λ5 S: This is a diagonal matrix: s1 0 0 S = 0 s2 0 0 0 s3 Each entry in this matrix is associated with a basis spectrum in the Umatrix (e.g. s2 is associated with u2. The size of each si determines theimportance, or size of contribution, of this basis to the experimental data.This can be seen by taking the product of U and S: uλ 1 u λ 1 1 2 uλ 1 3 s1uλ1 1 2 s 2 uλ 1 s3uλ1 3 1 3 1 3 2 uλ 2 uλ 2 uλ 2 s1 0 0 s1uλ 2 2 s 2 uλ 2 s3uλ 2 U × S = uλ 3 u λ 3 0 = s1uλ 3 1 2 uλ 3 × 0 3 s2 1 2 s 2 uλ 3 s3uλ 3 3 1 3 3 2 uλ 4 uλ 4 uλ 4 0 0 s3 s1uλ 4 1 2 s 2 uλ 4 s3uλ 4 u1 u 2 3 uλ 5 s u1 2 s 2 uλ 5 s3uλ 5 3 λ5 λ5 1 λ5 VT Is a square matrix, each entry gives the contribution of each basisvector to the signal at a particular wavelength: s1uλ1 1 s2uλ12 s3uλ1 3 1 3 s1uλ 2 2 s 2 uλ 2 s3uλ 2 V11 V12 V13 U × S × V = s1uλ 3 s3uλ 3 × V21 V22 V23 T 1 2 3 s 2 uλ 3 1 3 s1uλ 4 2 s 2 uλ 4 s3uλ 4 V31 V32 V33 s u1 2 s2uλ 5 3 s3uλ 5 1 λ57
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009For example, the measured spectra at λ1 for the 2nd ligand concentration is: • A C2 = s1uλ1V12 + s2uλ1V22 + s3uλ1V33 λ1 1 2 31 Figure 4: Basis sets used to fit CD data (left). Fit to CD spectrum of papain (right).SVD of Protein CD spectra:Determining Secondary Structure by SVD:We would like to obtain the secondary structure of a protein from its CDspectrum. In matrix form, this relationship is: F= XAwhere F is a vector that is the fraction of each secondary structure, A is theCD spectrum of the unknown protein, and X relates the two. Since weknow from SVD that there are only five significant components to the CDspectrum of a protein we will assume that we can represent a protein interms of five secondary structures:8
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009 θ λ 1 f H X 11 . . . . . f θλ 2 ⊥β . . . . . . θ λ3 f||β = . . . . . . × θλ4 fT . . . . . . θ λ 5 fO . . . . . . θ λ 6 If we can determine X, then the fraction of each secondary structure for anunknown protein can be found.If we begin with a large set of proteins with known secondary structure (F)and CD spectra (A) we can obtain X, using SVD. For a set of 3 proteinsand CD data collected at six wavelengths, the matrices would look like: f1 θ 1 θ 2 θ 3 fH 2 fH X 3 . . . . . λ 1 λ1 λ1 3 H 1 3 11 1 2 2 f⊥ β f⊥ β f ⊥ β . . . . . . θ λ 2 θ λ 2 θ λ 2 1 2 3 θ 1 θ 2 θ 3 f ||β f f = . . . . . . × λ 31 λ3 2 λ3 3 || β ||β θ θλ 4 θ λ 4 1 3 . . . . . . λ 4 θ 1 θλ 5 θ λ 5 2 fT fT fT 2 3 fO 1 2 3 . . . . . . λ 5 fO fO 1 2 3 θ λ 6 θ λ 6 θ λ 6 For this set of known proteins:F = XA = X (USV T ) , where A has been subject to SVD, i.e.A=USVTX can be solved by multiplying F by V, S-1, and U:F × V × S −1 × U T = X (USV T ) × V × S −1 × U T= X (US ) × S −1 × U T = X (U ) × U T = XNote:U×UT = 1 since all U vectors are orthonormal.V×VT = 1 from the theory of SVD. 1 / s 0 S × S-1 =1 if we define S-1 = 1 0 1 / s2 Secondary Structure of Membrane Proteins:CD can also be used to determine the secondary structure ofmembrane proteins (or other aggregated systems). However, theanalysis is far from simple. Two optical effects occur insuspensions of particles which distort CD spectra. First, theoverall absorption of the system decreases because proteins inthe particle can be "shaded" by others in the same particle.Second, the scattering of right and left circularly polarized lightis not equal. The former problem affects the overall amplitude ofthe spectrum, while the latter affects the shape of the CD Figure 5: Effect of particle size on CD Spectrumspectrum. These effects can be corrected to produce rather9
03-871 Molecular Biophysics CD Spectroscopy September 25, 2009reasonable CD spectra of membrane bound proteins. However, it is clearthat IR is the preferred technique for obtaining information on thesecondary structure membrane proteins.Other Application of CD Spectroscopy to Proteins.CD can also be used in an empirical manner to determine proteinunfolding, ligand binding, etc. CD is much more sensitive to changes insecondary structure than absorption spectroscopy and is thus an excellentmethod of following protein denaturation.Application of CD to Nucleic acids:The major application of CD to the study of nucleic acids is to determinethe degree of base stacking. The CD of a dimer is very dependent on theinteraction of the monomers. For example: poly C has the followingspectral properties: Solvent Ellipticity A260 Water 35,000 1.0 Ethylene glycol 7,000 1.3In this case both the CD and the hyper-chromicity show that polyC is ahelix in water and that this helix is due to base stacking.DNA-Protein Interactions:Since proteins have weak CD bands at 250nm, CD is well suited forfollowing protein induced changes in nucleic acid structure. These changescan be convenient for monitoring binding of proteins to nucleic acids.10