NOE Transferring magnetization through scalar coupling is a

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 T 1 (longitudinal or spin-latice) or T 2 (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.

Qualitative Description 2 spins I and S bb n 4(*) WI(2) (***) n 2 ab W 2 WS(2) ba W 0 WI(1) WS(1) aa n 1(***) 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 n 3(*) transitions along the edges correspond to transitions of one or the other spin alone. W 2 and W 0 are the cross-relaxation pathways, which depend on the tumbling of the molecule.

dn 1/dt = -WS(1)n 1 -WI(1)n 1–W 2 n 1 + WS(1)n 2+WI(1)n 3+W 2 n 4 … etc for n 2, 3, 4 using: Iz= n 1 -n 3+n 2 -n 4 Sz= n 1 -n 2+n 3 -n 4 2 Iz. Sz= n 1 -n 3 -n 2+n 4 One gets the ‘master equation’ or Solomon equation d. Iz/dt = -(WI(1)+WI(2)+W 2+W 0)Iz – (W 2 -W 0)Sz –(WI(1)-WI(2))2 Iz. Sz d. Sz/dt = -(WS(1)+WS(2)+W 2+W 0)Sz – (W 2 -W 0)Iz – (WS(1)-WS(2))2 Iz. Sz d 2 Iz. Sz/dt = -(WI(1)+WI(2)+ WS(1)+WS(2))2 Iz. Sz - (WS(1)-WS(2))Sz - (WI(1)-WI(2))Iz (WI(1)+WI(2)+W 2+W 0) auto relaxation rate of Iz or r. I (WS(1)+WS(2)+W 2+W 0) auto relaxation rate of Rz or r. R (W 2 -W 0) cross relaxation rate s. IS Terms with 2 Iz. Sz can be neglected in many circumstances unless (WI/S(1)-WI/S(2)) (D-CSA ‘cross correlated relaxation’ etc …)

Spectral densities J(w) W 0 g. I 2 g. S 2 r. IS-6 tc / [ 1 + (w. I - w. S)2 tc 2] W 2 g. I 2 g. S 2 r. IS-6 tc / [ 1 + (w. I + w. S)2 tc 2] WS g. I 2 g. S 2 r. IS-6 tc / [ 1 + w. S 2 tc 2] WI g. I 2 g. S 2 r. IS-6 tc / [ 1 + w. I 2 tc 2] • 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 r. IS-3, and the Ws will depend on r. IS-6: for proteins only W 0 is of importance W I, S, 2 << • The relationship is to the inverse sixth power of r. IS, 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…).

d(Iz – Iz 0)/dt = - r. I (Iz–Iz 0) - s. IS (Sz–Sz 0) d(Sz – Sz 0)/dt = - s. IS (Iz–Iz 0) - r. S (Sz–Sz 0) Note that in general there is no simple mono-exponential T 1 behaviour !!

Steady State NOE Experiment For a ‘steady state’ with Sz saturation Sz=0 d(Iz. SS – Iz 0)/dt = - r. I (Iz. SS–Iz 0) - s. IS (0–Sz 0) = 0 Iz. SS = s. IS/r. I Sz 0 + Iz 0 for the NOE enhancement h=(Iz. SS-Iz 0)/ Iz 0= s. IS/r. I Sz 0/Iz 0

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 H 10 which comes from a three step transfer via H 6 and H 9. . One-dimensional NOE experiments using pulsed field gradients, J. Magn. Reson. , 1997, 125, 302

Transient NOE experiment Solve the Solomon equation With the initial condition Iz(0)=Iz 0 Sz(0)=-Sz 0 For small mixing times tm the ‘linear approximation’ applies: d(Iz(t)– Iz 0)/dt = -r. I(Iz(t)–Iz 0) - s. IS(Sz–Sz 0) ~ 2 s. ISSz 0 Valid for tmr. S and tms. IS << 1 (i. e. S is still inverted and very little transfer from S) h(tm) = (Iz(tm ) - Iz 0)/ Iz 0 = 2 s. IS tm The NOE enhancement is proportional to s. IS !

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

NOESY The selective S inversion is replaced with a t 1 evolution period Sz(0)=cos. WSt 1 Sz 0, Iz(0)=cos. WIt 1 Iz 0 (using the initial rate appx. ) Sz(tm)=s. IStm. Iz 0 + r. Stm. Sz 0 (a) +cos. WIt 1[s. IStm]Iz 0 (b) +cos. WSt 1[r. Stm-1]Sz 0 (c)

Enhancement NOE vs. ROE NOE goes through zero wtc NOE Small peptides ~1 k. Da ~10 k. Da ~33 k. Da

ROESY 90 s 90 tm • • t 1 tm w. SL << wo, w * tc << 1 The analysis of a 2 D ROESY is pretty much the same than for a 2 D 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…

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

2 D - 3 D NOE 3 D- NOESY-HSQC

4 D NH-NH NOE N 1 – H 1 H 2 – N 2 H 1 N 1 H 2 N 2