Determination of the Stokes Radius by measuring the Rotational Diffusion Coefficient: Drot Drot = kT frot 8phRs3 = Isotropic rotation Dx ≠Dy ≠ Dz Dx = Dy = Dz Dx = Dy ≠ Dz kT Measure Drot Drot = 8phRs3 If you know Rs  h microviscosity If you know h  Rs molecular size, shape

Rotational Diffusion Rate can be determined by Fluorescence Spectroscopy Fluorescence polarization (anisotropy) can measure how rapidly a molecule is tumbling in solution sample excitation light t1 t2 I I fluorescent light

Fluorescence anisotropy For polarized excitation light, the probability of absorption of a photon depends on cos2 preferential excitation of molecules whose is parallel to the electric field vector of the excitation light Z q hu X Y After excitation non-random distribution Before (random)

Fluorescence anisotropy After formation of the excited state, the molecule can rotate during the time prior to photon emission Z orientation upon absorption q rot hu orientation at the time of emission X Y

Fluorescence polarization The probability of detecting the emitted photon depends on the orientation of the transition dipole moment of the excited state molecule AT THE TIME OF EMISSION and the orientation of the polarizer in the detection device q X IZ w z-component = x-component = y-component = Y IX probability of detecting the photon with polarization along the z-axis = ||2 cos2q x-axis = (sinq cos w)2 y-axis = (sinq sin w)2 Average over population IY proportional to < sin2q > < sin2w > Ix proportional to < sin2q > < cos2w > Iz proportional to < cos2q >

Fluorescence polarization q Define I = IZ X III = IZ w If we start with excitation light polarized along the z-axis, then Ix = Iy = I Y IX = I The total light intensity at the detector is Itot = III + 2I Define: polarization, P = III - I III + I anisotropy, A = III - I III + 2I A = 2 3 1 P - -1

Polarization Relate Anisotropy to < cos2q > A = III - I III + 2I = < cos2q > - < cos2w > < sin2q > < cos2q > + 2 < cos2w > < sin2q > Z ^ q Since we excite with light polarized along Z, we must have symmetry about this axis - so w will always be random => < cos2 w > = 1/2 X w Y A = 3 < cos2q > - 1 2 By measuring IX and IZ we can compute < cos2q > for the emitting molecules

m of excited state molecules What goes into < cos2q > ? m  Eex Z q I Photoselection Probability of absorption  cos2q The excited state population is not randomly oriented immediately after excitation m of excited state molecules   E Excitation beam After excitation non-random (III > I ) Before (random)

What goes into < cos2q > ? II Change in m for absorption + emission  1 2 hu1 m02 IC hu2 m01 abs em Fixed angle l photoselection Change m

What goes into < cos2q > ? III Rotational diffusion: Rotational diffusion coefficient Drot Rotate Change m photoselection rotation

photoselection change m rotation View each of these as a series of isotropic displacements of the average direction of the transition dipole moment Z m  q Step 1 Z Z l qrot Step 3 Step 2 photoselection change m rotation Goal: Find average displacement () in terms of known parameters

Measure this Find this |rot| Drot Stokes radius Use Soleillet’s equation - from geometry, describing a series of isotropic displacements of a vector 3 < cos2q > - 1 2 = 3 < cos2a1 > - 1 3 < cos2a2 > - 1 . ….. 3 < cos2aN > - 1 Z a1 a2 etc. Perrin’s equation: A = 3 < cos2q > - 1 2 = 3 < cos2q0 > - 1 3 < cos2l > - 1 . 3 < cos2qrot > - 1 Measure this Find this 3 1 |rot| Drot Stokes radius

Probability of absorption  cos2q0 Photoselection Photoselection: 3 < cos2q > - 1 3 < cos2q0 > - 1 3 < cos2l > - 1 3 < cos2qrot > - 1 . = . A = 2 2 2 2 3 2 Find this 1 Measure this Probability of absorption  cos2q0 < cos2q0 > = 3/5 This is derived in the 346 Class Notes So A = 0.4 P = 0.5 Maximum values of Anisotropy and Polarization 1 Z m  q Step 1

Emit from a different transition after internal conversion  2 Change in : Z Large l Z 02 X X 01 Y Y 0.4 from photoselection 3 < cos2l > - 1 A0 = 0.4 2 A0 is the anisotropy obtained in the absence of molecular rotation - e.g., using frozen solutions or solutions with high viscosity Depends on excitation + emission wavelengths Maximum value for l = 90 so limits on A0 and P0 are -0.2  A0  0.4 -0.33  P0  0.5

Rotational Diffusion: will always tend to bring A  0 or P  0 A = A0 3 < cos2qrot > - 1 2 qrot Anisotropy of frozen sample < cos2qrot > = 1 if there is no rotation: A = Ao < cos2qrot > =1/3 for random orientation: A = 0

Rotational Diffusion: will always tend to bring A  0 or P  0 3 Rotational Diffusion: will always tend to bring A  0 or P  0 A = A0 3 < cos2qrot > - 1 2 Ao value from photoselection plus change in  Break down rotation into a series of N isotropic steps of size dq 3 < cos2q > - 1 2 = 3 < cos2dq > - 1 . ….. Z dq etc. cos2dq = 1 - sin2dq  1 - dq2 For a small “step” Rotational Diffusion: < dq2 > = 4 Drot dt = 1 - 4 Drot dt 3 < cos2qrot > - 1 2 = (1 - 6 Drot dt)t/dt N

Rotational Diffusion But (1-x)  e-x for x << 1 3 < cos2qrot > - 1 2 = (1 - 6 Drot dt)t/dt N But (1-x)  e-x for x << 1 3 < cos2qrot > - 1 2 = e- 6 D t rot A = A0 e- 6 D t rot Time-resolved Anisotropy yields Drot A Calculate Drot Monitor anisotropy following a pulse of excitation light: time

Steady State Measurement of Fluorescence Anisotropy Fraction of photons emitted in (t, t+dt) = f(t)dt t t+t f(t)dt t A(t) Average anisotropy A = A0 (1 + 6 Drot t)-1 Perrin’s equation

Various forms of Perrin’s equation (1) (2) substitute : V  molecular volume (3) Perrin plot 1 P 3 - (4) Define: rotational diffusion time = T h (5)

Result for steady state fluorescence from a rotating molecule Fluorescence lifetime Measure this kT 8pRs3 Define: rotational diffusion time =

Protein Rotational Diffusion I. Extrinsic Probes + H2N Dansyl chloride t = 14 nsec II. Results r0 = (3hV) / (kT) = (3hV2M) / (NkT)  [(mol.wt) / 1.15] x 10-3 nsec (A0 / A) = 1 + (3t / r) Measure A0, A, t find r

Example: Fluorescence Anisotropy Measurement of Protein-Protein Interactions TATA-box Binding Protein (TBP) binds with high affinity to TBP-associated Factor subunit TAF130p 1. Label TBP with tetramethyl rhodamine at a reactive cysteine residue 2.To a solution of 100 nM TBP (labeled) titrate increasing amounts of TAF130p (0-255 nM) 3. Exitation at 540 nm; emission monitored at 575 nm Conclude: High affinity binding (Kd = 0.5 nM) Biological consequence: binding of TAF130p to TBP competes with DNA binding to TBP J. Biol. Chem., Vol. 276, Issue 52, 49100-49109, December 28, 2001

Fluorescence Anisotropy used to monitor Protein-DNA Interactions 1. 50 nM Rhodamine labeled DNA 2. Titrate TBP 3. Excitation at 580 nm/Emission at 630 nm SHOWS 1:1 COMPLEX 1. 1:1 DNA:TBP complex, with labeled DNA (50, 100, 250 nm TBP) 2. Titrate TAF130p SHOWS ELIMINATION OF TBP-DNA COMPLEX BY COMPETING TAF130p 1. 50 nm labeled DNA 2. Titrate with TAF130p SHOWS NO COMPLEX BETWEEN DNA AND TAF130p J. Biol. Chem., Vol. 276, Issue 52, 49100-49109, December 28, 2001

Fluorescence anisotropy used to measure microviscosity The rotational diffusion coefficient can be related to Molecular volume: V V  molecular volume Measure this If you know V You can determine 

Probe dissolved in membrane bilayer Fluorescent Probes dissolved are used to measure Membrane microviscosity Diphenyl hexatriene (DPH) Perylene Hydrophobic probes: Probe dissolved in membrane bilayer partition into membrane bilayer do not bind to protein rotation reports local viscosity Measure for specific probe Molecular constant: determine for probe in known viscosity (h) Then: A => h

Typically: for biological membranes h  1 Poise 100-times the viscosity of water Note: (1 / h) = fluidity of membrane

Note: (1 / h) = fluidity of membrane Typically: for biological membranes, h  1 Poise (~100X water viscosity) Note: (1 / h) = fluidity of membrane Two points: (1) Do not know where probe is located in the bilayer - maybe regions of differing fluidity. Where is probe? (2) Probes are not spherical. Hence, the specific model of how it moves will influence final interpretation

in Arabidopsis thaliana Example: The effect of deletion of the gene encoding -6-oleate desaturase in Arabidopsis thaliana Changes the fatty acid composition of the mitochondrial membrane: mostly oleic acid is present in the mutant (fad2) J. Biol. Chem., Vol. 276, Issue 8, 5788-5794, February 23, 2001

Membrane Fluidity monitored by the fluorescence Anisotropy of anthroyloxy fatty acid derivatives: the fluorophore is located at different depths in the membrane bilayer mutant wild-type mitochondrial membranes CONCLUDE: Decrease in unsaturation of the fatty acid components in the membrane results in increased anisoptropy, or increase in membrane viscosity Biological Consequence: decreased respiration and altered bioenergetics of the mitochondria Extracted lipids 2 18 depth in the membrane bilayer/position in fatty acid chain)

Fluorescence Polarization Example: Enolase dissociation + dimer 2 x 45,000 mol. wt. Can monitor dissociation: (1) Polarization of tryptophan emission (t = 4.5 nsec) (Due to changes in local motions) (2) Polarization of bound co-valent probe E.g. Dansyl (t = 14 nsec) => Probe can alter equilibrium constant!!!

Let C0 = total protein concentration as dimer + D 2M Let C0 = total protein concentration as dimer [M]2 K = = 4C0 [a2 / (1 - a)] [D] [M] [D] fM = fD = 2C0 C0 Where a = degree of dissociation [M] [M] a = = [M] + 2[D] 2C0 AD ID + AM IM Anisotropy: A = ID + IM Fluorescence intensity from Monomer from Dimer From A => a => K

monomer Enolase: D 2M upon dilution Dilute to 1.55 M (140 g/mL) Decrease in polarization from tryptophans : changes due to local structural changes that increase local motion of tryptophans PNAS (1982)79, 5268 -

Enolase: Pressure - Induced dissociation Biochemistry (1981)20, 2587 - Dansyl conjugate: [C0] = 5.6 mM + Favored at high pressure Lower molar volume due to more efficient packing of water at protein - interface region

Fluorescein (donor) and Rhodamine (acceptor)

Longer DNA double strand results in lower FRET (Energy Transfer) model DNA double helix as a cylinder

Fluorescence Anisotropy goes down when double strand helix “melts” to form single strand DNA Rhodomine-labeled DNA double strand single strand

DNA Melting Curves obtained from Fluorescence intensity measurements (longer DNA has steeper melting transition and higher Tm)

Salt induced conformational transition of the four-way DNA junction monitored by the Perrin Plot of rhodamine anisotropy Less compact: slower rotation larger molecular volume smaller slope in Perrin Plot More compact: faster rotation smaller molecular volume larger slope in Perrin Plot

Fluorescence Anisotropy used to study the Dynamics of a Transcription Initiation Complex Fos-Jun: transcription regulatory complex can bind to target palindromic DNA (AP-1 site) in either orientation NFAT1 protein: regulatory element that binds to Fos-Jun heterodimer only in one orientation on the DNA 180o rotation Question: Does Fos-Jun dissociate from the DNA and then re-bind when NFAT1 interacts with it? PNAS (2001) 98, 4893-4898

Fluorescence Anisotropy used to study the Dynamics of a Transcription Initiation Complex Label the DNA on both strands with 1) Cy3 and 2) Fluorescein Jun is labeled with Texas Red, which is a FRET acceptor from both Cy3 and Fluorescein Adding NFAT1 to the Complex causes the Fos-Jun protein to reverse its orientation on the DNA PNAS (2001) 98, 4893-4898

Binding of Fos-Jun to DNA site is primarily oriented with Texas Red label near Cy3 End of the DNA Emission spectra: excite fluorescein at 460 nm (solid lines) and Cy3 at 530 nm (dashed lines) FRET NO FRET PNAS (2001) 98, 4893-4898

Binding of NFAT1 to the Fos-Jun-DNA complex causes the 180o re-orientation of the protein The Texas Red label on Jun is now near the Fluorescein End of the DNA Emission spectra: excite fluorescein (solid lines) at 460 nm and Cy3 at 530 nm (dashed lines) PNAS (2001) 98, 4893-4898

Time-course of Fluorescence Changes shows rate of re-orientation of Fos-Jun complex after addition of NFAT1 PNAS (2001) 98, 4893-4898

Time-course of Fluorescence Anisotropy shows that binding of NFAT1 precedes reorientation of Fos-Jun complex fast binding to form complex with higher anisotropy due to larger size slow reorientaion fast binding slow re-orientation of the protein while remaining on the DNA monitored by FRET PNAS (2001) 98, 4893-4898

Probing DNA structure and dynamics Fluorescein (donor) and Rhodamine (acceptor) http://lfd.uiuc.edu/staff/gohlke/poster/#DNA junction

Can model as a series of discontinuous jumps : <dq2> = 4Drot dt Rotational Diffusion Can model as a series of discontinuous jumps : <dq2> = 4Drot dt units of sec-1 start end The “orientation vector” for a molecule is provided by the transition moment dipole moi  Optical experiments - Absorption + Fluorescence provide information about molecular orientation moi q  E Probability of Absorption (or emission)  cos2q

Rotational Diffusion Isotropic - for a sphere Dx = Dy = Dz 2 starting points 1 Discontinuous jumps : <dq2> = 4Drot dt units of sec-1 f(q,t) 2 f(q, t) 2 Ficks 2nd law : = Drot t q2 f(q, t) = fraction of molecules oriented with m in angular interval (q, q + dq) q q + dq The “orientation vector” for a molecule is provided by the transition moment dipole moi  Optical experiments - Absorption + Fluorescence provide information about molecular orientation moi q  E Probability of Absorption (or emission)  cos2q

The basis for measuring Drot is photoselection Outer segment Disk membrane Rhodopsin The basis for measuring Drot is photoselection We can see how this works by reviewing transient absorption changes in the molecule rhodopsin that were used to determine the motion of this membrane protein Inner segment Nuclear region Synaptic region Axial view 60 mm Transverse view 11-cis retinal 5 mm

Rhodopsin Retina provides oriented sample Z X Y Experiments ^ ^ ^ (1)Direct light (500 nm) along Y, polarized in X or Z directions Z AX AZ > 4 X ^ ^ ^ (1)Direct light (500 nm) along Z, polarized in X or Y directions AX AY = 1 “random” Top view

Formation of a transient intermediate upon light absorption by rhodopsin 1) Photochemistry: intermediate 200 msec 11-trans retinal (opsin) 11-cis retinal (on rhodopsin) 580 nm 500 nm Rotate? falls off protein 580 nm Before bleaching “intermediate” Abs 400 500 600 l (nm)

? Does rhodopsin rotate in plane of the membrane? side view 1. Direct polarized light down to the membrane surface 2. After a delay, measure the absorption parallel and perpendicular to the original direction of the polarized excitation pulse top view arrows represent the transition dipole moments of the rhodopsin molecules

Pulse-Probe Experiment A looking down on retinal surface 500 nm 580 nm 500 nm (before) (after) A polarization of excitation light Selective photobleaching of the species absorbing at 500 nm leaving the remaining molecules preferentially oriented perpendicular to the axis of the direction of polarization of the excitation light Generation of intermediate absorbing at 580 nm oriented along the axis of the direction of polarization of the excitation light AII A > 1 A II < 1 A

 Photoselection   moi  q E Probe beam Selectively bleach molectules whose m are parallel to E   A II Excitation beam E A A A II < 1, II > 1 A A moi Before  q  E Absorption probability  cos2q A (t) Measure dichroic ratio = II = R(t) A (t) 1) Does it change? Yes rotation 2) How fast does it change? Drot Rs or 

Rate of decay depends on Drot Rhodopsin Dichroic ratio = at 580 nm R(t) 2 + e- 4Dt t = 0, R = 3 t = , R = 1 R(t) = ; Rate of decay depends on Drot 2 - e - 4Dt 2 Fresh sample R(t) 1 2 Crosslinked sample - cannot rotate 1 40 80 msec kT 8pRs3Drot h = Result: 1 4Drot = trot = 20 msec h = 2 Poise For Rs = 20 Angstroms 200X value for H2O

Photoselection is also the basis of fluorescence polarization The direction of the electric field vector of the emitted radiation is determined by the orientation of transition dipole moment of the molecule at the time of the emission E sample excitation light I I measure the parallel and perpendicular components of the emitted radiation

Photoselection and Fluorescence Polarization The excited state population is not randomly oriented immediately after excitation. Before (random) Excitation beam After excitation non-random (III > I ) m of excited state molecules  E Z E  in YZ plane X I III Y Monitor emitted light Emission will have preferentially along Z (III) (III > I ) ^ Polarized fluorescence Fluorescence Excitation light

Fluorescence polarization X Z Y q w III = IZ IX = I hu Probability of absorption (or emission) a cos2q For Z-polarized excitation : IX = IY = I ; Itot = III + 2I Define: polarization, P = III - I III + I anisotropy, A = III + 2I ; better to use since it is normalized to Itot A = 2 3 1 P - -1

Measured anisotropy from multiple fluorescing species is a weighted sum of the anisotropies of the component species A = III - I III + 2I Itotal = If multiple species contribute to fluorescence: I1 A1 + I2 A2 + I3 A3 A = I1 + I2 + I3 A = f1 A1 + f2 A2 + f3 A3 A simple weighted average fraction of fluorescence from species 1

Polarization m  Eex Relate Anisotropy to < cos2q > A = III - I < cos2q > - < cos2w > < sin2q > < cos2q > + 2 < cos2w > < sin2q > Z ^ Since we excite with light polarized along Z, we must have symmetry about this axis - so w will always be random => < cos2 w > = 1/2 q X A = 3 < cos2q > - 1 2 w So, Y By measuring IX and IZ we can compute < cos2q > for the emitting molecules 2) What goes into < cos2q > ? I Photoselection m  Eex Z q Probability of absorption  cos2q

Polarization   II Change in m for absorption + emission 1 2 hu1 m02 1 2 hu1 m02 IC hu2 m01 abs em Fixed angle l III Rotational diffusion: hu1 hu2 Rotate Rotational diffusion coefficient Drot View each of these as a series of isotropic displacements Z m  q l qro Step 1 Step 2 Step 3 Goal: Find average displacement in terms of known parameters