NOE

1. NOE • Transferring magnetization through scalar coupling is a “coherent” process. This means that all of the spins are doing the same thing at the same time. • Relaxation is an “incoherent” process, because it is caused by random fluxuations that are not coordinated. • The nuclear Overhauser effect (NOE) is in incoherent process in which two nuclear spins “cross-relax”. Recall that a single spin can relax by T1 (longitudinal or spin-latice) or T2 (transverse or spin-spin) mechanisms. Nuclear spins can also cross-relax through dipole-dipole interactions and other mechanisms. This cross relaxation causes changes in one spin through perturbations of the other spin. • The NOE is dependent on many factors. The major factors are molecular tumbling frequency and internuclear distance. The intensity of the NOE is proportional to r-6 where r is the distance between the 2 spins.

2. Qualitative Description 2 spins I and S Two nuclear spins within about 5 Å will interact with each other through space. This interaction is called cross-relaxation, and it gives rise to the nuclear Overhauser effect (NOE). Two spins have 4 energy levels, and the transitions along the edges correspond to transitions of one or the other spin alone. W2 and W0 are the cross-relaxation pathways, which depend on the tumbling of the molecule. bb n4(*) WI(2) WS(2) W2 (***)n2ab ban3(*) W0 WI(1) WS(1) aa n1(***)

3. dn1/dt = -WS(1)n1-WI(1)n1–W2n1 + WS(1)n2+WI(1)n3+W2n4 … etc for n2,3,4 using: Iz= n1-n3+n2-n4 Sz= n1-n2+n3-n4 2IzSz= n1-n3-n2+n4 One gets the ‘master equation’ or Solomon equation dIz/dt = -(WI(1)+WI(2)+W2+W0)Iz – (W2-W0)Sz –(WI(1)-WI(2))2IzSz dSz/dt = -(WS(1)+WS(2)+W2+W0)Sz – (W2-W0)Iz – (WS(1)-WS(2))2IzSz d2IzSz/dt = -(WI(1)+WI(2)+ WS(1)+WS(2))2IzSz - (WS(1)-WS(2))Sz - (WI(1)-WI(2))Iz (WI(1)+WI(2)+W2+W0) auto relaxation rate of Iz or rI (WS(1)+WS(2)+W2+W0) auto relaxation rate of Rz or rR (W2-W0) cross relaxation rate sIS Terms with 2IzSz can be neglected in many circumstances unless (WI/S(1)-WI/S(2)) (D-CSA ‘cross correlated relaxation’ etc …)

4. Spectral densities J(w) • W0gI2gS2rIS-6tc / [ 1 + (wI - wS)2tc2] • W2gI2gS2rIS-6tc / [ 1 + (wI + wS)2tc2] • WSgI2gS2rIS-6tc / [ 1 + wS2tc2] • WIgI2gS2rIS-6tc / [ 1 + wI2tc2] • Since the probability of a transition depends on the different • frequencies that the system has (the spectral density), the • W terms are proportional the J(w). • Also, since we need two magnetic dipoles to have dipolar • coupling, the NOE depends on the strength of the two • dipoles involved. The strength of a dipole is proportional to • rIS-3, and the Ws will depend on rIS-6: • for proteins only W0 is of importance W I,S,2 << • The relationship is to the inverse sixth power of rIS, which • means that the NOE decays very fast as we pull the two • nuclei away from each other. • For protons, this means that we can see things which are at • most 5 to 6 Å apart in the molecule (under ideal conditions…).

5. d(Iz – Iz0)/dt = - rI (Iz–Iz0) - sIS (Sz–Sz0) d(Sz – Sz0)/dt = - sIS (Iz–Iz0) - rS (Sz–Sz0) Note that in general there is no simple mono-exponential T1 behaviour !!

6. Steady State NOE Experiment For a ‘steady state’ with Sz saturation Sz=0 d(IzSS – Iz0)/dt = - rI (IzSS–Iz0) - sIS (0–Sz0) = 0 IzSS = sIS/rI Sz0 + Iz0 for the NOE enhancement h=(IzSS-Iz0)/ Iz0= sIS/rI Sz0/Iz0

7. NOE difference Ultrahigh quality NOE spectra: The upper spectrum shows the NOE enhancements observed when H 5 is irradiated. The NOE spectrum has been recorded using a new technique in which pulsed field gradients are used; the result is a spectrum of exceptional quality. In the example shown here, it is possible to detect the enhancement of H10 which comes from a three step transfer via H6 and H9. One-dimensional NOE experiments using pulsed field gradients, J. Magn. Reson., 1997, 125, 302.

8. Transient NOE experiment Solve the Solomon equation With the initial condition Iz(0)=Iz0 Sz(0)=-Sz0 For small mixing times tm the ‘linear approximation’ applies: d(Iz(t)– Iz0)/dt = -rI(Iz(t)–Iz0) - sIS(Sz–Sz0) ~ 2 sISSz0 Valid for tmrS and tmsIS << 1 (i.e. S is still inverted and very little transfer from S) h(tm) = (Iz(tm ) - Iz0)/ Iz0 = 2sIStm The NOE enhancement is proportional to sIS !

9. Longer mixing times a system of coupled differential equations can be solved by diagonalization or by numerical integration Multi-exponential solution: the exponentials are the Eigenvalues of the relaxation matrix

10. NOESY The selective S inversion is replaced with a t1evolution period Sz(0)=cosWSt1Sz0, Iz(0)=cosWIt1Iz0 (using the initial rate appx.) Sz(tm)=sIStmIz0 + rStmSz0 (a) +cosWIt1[sIStm]Iz0 (b) +cosWSt1[rStm-1]Sz0 (c)

11. Enhancement NOE goes through zero wtc NOE ~33kDa ~10 kDa Small peptides ~1 kDa NOE vs. ROE

12. ROESY 90s • wSL << wo, w * tc << 1 • The analysis of a 2D ROESY is pretty much the same than for a 2D NOESY, with the exception that all cross-peaks are the same sign (and opposite sign to peaks in the diagonal). Also, integration of volumes is not as accurate… 90 tm tm t1

13. 1H-1H (homonuclear) 2D 1H 1H 3D 1H 1H 1H 1H 1H 1H 1H 1H 1H 1H 1H 15N 13C 13C 15N 15N 15N 13C 13C 1H 1H 3D 4D Approaches to Identifying NOEs • 15N- or 13C-dispersed (heteronuclear)

14. 2D - 3D NOE 3D- NOESY-HSQC

15. 4D NH-NH NOE N1 – H1 H2 – N2 H1 H2 N1 N2