 Welcome to today's class. So we were discussing polarization transfer and today we will focus on how we can measure distance using NOE. So we will connect the concept of distance and NOE into this class. So what we had done previously, we explained the origin of NOE, what is a physical process behind NOE and as we said that it is a nuclear relaxation which cause nuclear overhouser effect. So how it is done? So we are saying that there are if two couple spins, they are coupled by dipolar coupling or through a space. What we are doing? We are applying a saturation pulse on one of the signal. So this saturation pulse actually equalizes the population across this transition and then by irradiation we equalize this population and then effect of that, of that equalization of population by irradiation of a particular signal, we look at what is happening on other spin. So if two spins here we are equalizing population and we are looking effect of this on another coupled spin. So NOE essentially is manifestation of attempt of the system to come back to the equilibrium and that is we had defined at nJs is i minus i0 divided by i. So this is after perturbation, this is the equilibrium magnetization and that ratio with equilibrium magnetization that is called NOE enhancement factor. So we discussed the NOE effect and just to remind you what we had discussed earlier. So we had two spin, say spin A and spin X, these are say weakly coupled spins A and X. So there are four energy level, here both spins are say up, up, alpha, alpha, here one spin is up, another spin is down, here one spin is down, another spin is up and here both spins are down. So now there are four transitions possible, these are called omega 1 single quantum transition that means only one spin is flipping, so this spin is flipping that is why alpha is becoming better, this spin is flipping alpha again becoming better. So these two belongs to omega 1, omega 1 of A spin and these two belongs to X spin, so these are single quantum transition. Then we had looked here both spins are flipping, X and A both spin are flipping, this is called double quantum transition and here both spins are flipping simultaneously, if you look at alpha goes to beta, beta goes to alpha, so net magnetization change is 0 that is why this is 0 quantum transition. So resonance line intensity changes caused by this dipolar cross relaxation, so this is happening of various transition and then there is a dipolar cross relaxation because A and X spins are connected by dipolar coupling. So from the neighboring spin with a perturbation energy level population that is what now it changes and we have seen previously because of this relaxation the population redistribution happens. So omega 1, 2 only these two spins are flipping and omega 2 alpha goes to beta, beta for a one spin and another spin is this, so that is omega 2 and here both spins are going. So actually that is what is the transition probability for omega 1, omega 0 and omega 2 that is what we had discussed in the previous class. So now to give you more clearer picture that what is actually happening here, so I draw a population diagram. So let us say we have two spins again A and X spin, here are alpha state and beta state and here again in X spin alpha state and beta state. So if you saturate one of the population say A, there are two mechanism that will relax and equalize the population and both of these contributes to the NOE, one we had discussed is omega 0 and another we have discussed that is omega 2. So we also discussed that positive NOE that comes because of this omega 2 mechanism, this we have shown quantitatively in the previous class and negative NOE which happens for bigger molecule that is coming because of omega 0 mechanism, right. So mechanism which cause unperturbed spin say X to deviate from Boltzmann equilibrium distribution a decrease in alpha beta population difference happens and that actually cause what is positive NOE or negative NOE. To make it even more simpler let me try to consider these spins as a magnet. So omega 1 spin and omega 2 spin is what like if you say we have H1 frequency. So say we are doing proton resonance at 500 MHz and transition is happening at 500 MHz for a proton. So that means one transition here say alpha is this alpha is flipping to beta here this one. One spin is flipping so that means that 500 MHz induces a transition. Now for both spin flip you see this 500 MHz and this 500 MHz so total transition say will be equal to field 100 MHz to flip and for 0 quantum you just need couple of hertz difference because both spins are at 500 MHz but their chemical shift is little bit different and that difference in terms of hertz so that will be around 102,000 hertz. So omega 1 requires 500 MHz say omega 2 will require 1000 MHz however omega 0 will require approximately 102,000 hertz. So this is 0 quantum transition here what is so bar magnet kind of a simplistic cartoon representation we have shown. So flipping is happening just from spin here to spin here both are up down happening. In omega 1 transition flipping is happening from one state to another state and here double quantum flipping is happening from this state to this state. So that is what we require for these flippings to happen but this is a static case whatever I showed from here to here 0 quantum this transition is single quantum this transition is double quantum. But this is happening for like here it is cartoon is showing for static case but suppose spins are reorienting in solution. So they may generate some frequency and we can calculate how much frequency is generated by this random rotation of a spins. So if we calculate then we can find an expression for something called spectral density at any frequency omega and spectral density is given by j omega at any frequency omega 0 which will be given by tau c and the frequency of that particular spins. So tau c is the tumbling time of that particular system or spins. So that is given by 2 tau c divided by 1 plus omega square tau c square. So this j omega is called spectral density and we can calculate how much certain frequency is generated by this random rotation of spins because of all these relaxation that is coming. Now so that has been calculated and plotted spectral density that is j omega versus log of omega here. So this is the frequency. One can see if a molecule is large, large molecule has a slow motion and that is how the spectral density will be plotted like this so it decreases fast. Now an intermediate motion which is for intermediate molecule you have a decrease which is relatively slow and for a fast motion for a small molecule you have a quite broad range of j omega. So j omega value is less but it varies for a long omega t. So one can see for any particular omega t for all these molecules for intermediate molecular weight you have value here and for a small molecule you have j omega value here but for a large molecule you have very short j omega. So we have a maximum at omega equal to 0. So that is a spectral density and it is obvious intuitive that big molecule will relax fast and small molecule will relax slow. So tau c here whatever we have defined is a molecular rotation correlation time. So correlation time in short it is a time that that molecule takes to rotate in the solution. A long correlation time is for sluggish rotation and it fast drops. So if you go here if you look at here what is happening for a large molecule j omega spectral density is rapidly actually decreasing that is what we mean that it is a sluggish rotation and it drops fast. So j value very small at high frequency contribution for short tau c that means fast rotation causing much wider frequency distribution or more frequency contribution if you look at here. So for a small molecule you have a broad range of frequency at a particular j omega. So that is the spectral density with correlating with your tau c or omega. So give some value for this rotation. So larger molecule generally as we say slow reorientation time therefore longer tau c. So say molecular weight of 100 Dalton that is 1 kilo Dalton will corresponds to typically like 0.5 nanosecond. If molecular size increases like it becomes say 10,000 Dalton that is 10 kg tau c will correspond to 5 nanoseconds. So you see as molecular size increases the correlation time increases. So for a typical Spiegel size protein the correlation time is 5 nanosecond however for a small organic molecule correlation time is 0.5 nanosecond and that is what we see that spectral density accordingly changes and for any particular frequency. Like tau c correlation time is not fixed for a molecule it can also vary depending upon temperature. So like if temperature is higher like if you increase the temperature molecule tumbles fast. So even a bigger molecule can give you sharp line or actually different spectral density if you record a spectrum at higher temperature and the other thing is viscosity. If you keep increasing the viscosity of solvent now small molecules can also experience slower motion. So all those spectral density and the frequency depends upon temperature and solvent viscosity because more viscous longer tau c if you have longer tau c like we have for larger molecule spectral density decreases fast. Now we correlated your spectral density with the frequency as well as with the correlation time. Next move and let us see we have seen this earlier as well that NOE. Now NOE enhancement coming back to now NOE, NOE we have seen that depends upon molecular motion which is tau c and spectral density J are dependent of the reorientational correlation time it depends upon omega. So we had seen earlier that correlation times and NOE enhancement is like this. So for large molecule we have a negative enhancement for small molecule we have a positive enhancement and correlation time tau c in nanosecond typical protein like large molecules comes here and typical organic molecule comes here for a small molecule that we had seen earlier as well. So just to refresh you again how we are doing these experiments in case of a steady state we were taking two coupled spin one spin was perturbed with a CW like weak irradiating pulse then we are applying a 90 degree pulse and we are acquiring and we are seeing what is the signal change and comparing that with a non irradiated or reference signal. So we have seen that this long low power CW irradiation which is shown here actually induces deviation from Boltzmann population distribution and then relaxation happens that relaxation because of omega 0 and omega 2 this also we have seen and this happens because of T1 relaxation. So we have seen that T1 relaxation is the main cause of these two relaxation omega 0 and omega 2 and it causes population distribution in the neighboring spin. So that population distribution because of irradiation is compared with reference spectrum and then we get positive and negative energy. This we have understood earlier as well. So now refreshing you. So observable signal that is determined is given by the omega 0 frequency and a spectral density at particular omega 0 because this is the double quantum frequency and this is zero quantum frequency. So this is cross correlation rate depends upon our double quantum frequency and zero quantum frequency. So if this term is higher than GA0 we get a positive energy and that is typical in a small molecular case. So their double quantum frequency is higher than zero quantum frequency and cross relaxation rate actually depends upon that and it is also inversely proportional to distance between them. For a small molecule this is higher therefore we have a positive energy. So we can have that for a small molecule where there is a relatively high spectral density at 2 omega 0 that is double quantum and we have a short tau c. For a long tau c which is typically seen in a macro molecule omega 0 dominates this whole thing can be negative and if that happens this is negative then cross correlation rate which depends upon these four this gives you negative enhancement. So that is what we have seen. So typically there is a crossing point between positive and negative energy that is given by omega 0 tau c which is typically 1. So if you have seen in previous slide here so this is a crossing point which is typically around 1, omega 0 tau c is typically around 1. At some cases like where omega 0 and tau c is typically 1. So omega 0 relaxation rate and omega 2 relaxation rate will cancel each other and there will be no energy. So let me summarize again omega 2 which is dominant mechanism of relaxation for small molecules gives positive energy. Omega 0 which is dominant relaxation mechanism for macro molecules gives negative energy. So maybe some cases comes where both of them cancel and that will cancel when omega 0 and tau c becomes almost 1. So there will be no effect. So effect from omega 0 and omega 2 will cancel each other and there will be no net annoying enhancement for some cases. So we had done detailed calculation to find it out that enhancement between A and X spin will be given by the gyromagnetic ratio of A and X spin. So like we have seen if two couple spin proton and carbon this is the distance between them Rij if they are coupled proton and carbon so if this is 13c and this is proton the gyromagnetic ratio of this is 4 so we have enhancement of 2. If they are proton and nitrogen then NOE enhancement will be 5 and if they are proton-proton then NOE enhancement that we get is 0.5. So we have seen STD STD NOE does not depend upon distance but if these two spins are less than 5 to 6 angstrom you get an enhancement looking at these formula. However as you said NOE is very very important parameter to measure the distance quantitatively between two coupled spins. So for doing that experiment we had seen earlier that there was a another experiment transient NOE but just look at two spins how much enhancement we get if you do STD STD NOE. So if you have two spin system like A and X so suppose we are saturating spin A we will get a 50% enhancement in spin X and we have seen what are these two spins say proton and proton so their gamma is equal so we get half 1 divided by 1 that is a 50%. So this is generally for small molecules because omega 2 is dominating now suppose we are saturating X and we are looking at A again here enhancement will be 50% so both of these is for small molecules. Now suppose we are saturating here say A and for a large molecule where omega 0 is dominating you can get negative NOE enhancement of almost 100% because the mechanism is very different and similarly X spin is saturated then you can get enhancement of 100% because for this omega 0 dominating. Now let us take linear combination of three spins AMX now here we are seeing the distance between M to X and A to M is same if we are saturating A now what is happening now we saturated A by applying a RF pulse and M and X are like their population is changed and they are going undergoing relaxation now. So now they are competing both of these are competing mechanism for the relaxation so they have to compete for two neighboring protons and therefore that NOE enhancement will be distributed so for such cases one can do calculation so that M will get 25% enhancement and X will get negative enhancement of 11.5%. Similarly if we saturate M now M is sitting in the center A, X and A are located so both of this will get approximately 50% enhancement however if we saturate X now again for X is located at one side and we are saturated here so now enhancement will be similar like this 25% for M and minus 11.5% for X so in such case this is three spin system but now there is a competition of relaxation or redistribution of population between M and X therefore this is the saturation 25% and 11% for both of this. But let us see what happens if the distances are different between Mx so say here Mx distance to time more than M distance so now if we saturate A so M will get around 50% and X will get around minus 18% negative enhancement and if we saturate M then both will get 50-50% and if we saturate again X very less enhancement will happen for M and A here only 0.8% and here minus 0.4% and this is the case when omega 2 is much much higher than omega 0 for a small molecule case. So in this case AM are closer and relax with each other very efficiently therefore you see enhancement is quite a bit. If you saturate A almost 50% enhancement, if you saturate M here 50% enhancement. So now one thing is clear because here we change the distance enhancement are going to be different. So NOE buildup can be measured depending upon distance and that is done in another way which is called transient NOE. So NOE requires a significant interaction as we have seen between the magnetic dipole and between two spins so dipolar is interaction that we have seen also varies upon distance so if these are two dipoles it depends upon how much distance between them. So one can measure NOE between two protons up to 526 angstrom that we have seen earlier and this is the enhancement that depends upon this distance. So one can vary in this case of transient NOE that we have explained earlier so it depends upon tau M the mixing time. So to remind you how this experiment was done there are two couple spins say I and S or A and X selectively we are inverting one spin by applying a 180 degree pulse then we are varying this tau M which is mixing time we apply a 90 degree pulse and that brings to XY plane and we detect it here. In the other experiment we are not doing this just applying a 90 degree pulse this is reference experiment and we are detecting here. So because of the selective perturbation of coupled spin and varying this TM time we can see difference in the NOE enhancement because this NOE enhancement directly depends upon our mixing time and also directly depends upon cross relaxation rate. So if they are both are coupled the two parameters that actually dictates one is distance between these two spins and another is mixing time. So if we vary say mixing time we can get a build up curve of NOE enhancement. So if we here TM and we have a NOE so we can get a build up and from this build up one can measure how much the distance between these two spins. So that is what is done and that is what I will explain it here. So cross relaxation rate as we know tau this term depends upon the distance between these two spins and it is inversely proportional to the R2 power 6. That means this term is inversely proportional to R2 power 6. Now in transient NOE experiment we can use this to estimate internuclear distance. So here is our enhancement factor and this is tau M. For short TM it is very linear so that is what in this linear regime we can vary and measure the distance and that is used for structural determinants. However, if you keep increasing this mixing time then it becomes nonlinear here if you look at it keeps increasing and then it starts decreasing that is a nonlinear behavior. Now nonlinearity and decay of NOE at higher mixing time is a consequence of leakage of the magnetization because of the lattice contribution and that is because of various cross relaxation rate comes into picture. So to simplify you if we keep increasing the mixing time the NOE enhancement is not going linearly starts decreasing. What is the reason for decrease that is various relaxation rate sticks over and this mechanism is called spin diffusion means spin is diffusing and losing out its magnetization because we have kept it too long so that is called spin diffusion. So in this case measurement is not very precise so for all the distance measurement we must stick to this linear regime of the tau M. And precisely these experiments are used for measuring the 2D experiment which is called nuclear overhouser effect and this nuclear overhouser effect becomes the holy grail of structural biology because if you do same experiment this is 2D so little bit of I just introduce it here for doing 2D you added this module which was not there and in the 1D but essentially it works on same formula you put up one of these spins and then you mix it at various time and then you apply a y pulse to bring it to xy plane and you detect it. If you have this mixing time spin through dipolar coupling it starts mixing it and the cross peak intensity that comes in this spectrum so there is overlay of 2 nosier spectrum you can find that there are various peaks that appears or this is the representation of distance between 2 coupled spins. So by measuring the intensity of each of these 2D plots one can get an accurate estimate of distance between these 2 spins and this is used in 2 dimensional spectroscopy for getting the distance in protein structure and used for structural biology. However as we saw that for small molecules this nosy is not very efficient way the rosy is the more efficient way so rosy is little variation from the nosy what you do here in case of rosy you apply a mixing pulse isotropic mixing pulse so this is 2D experiment so when we go to 2D we will explain little more detail so here is just for introducing one time domain for 2D and then you apply a weak RF pulse for mixing these spins similar like NOG concept but here it is done in rotating frame and you mix this spin and you get the magnetization transfer through dipolar coupling and that is what the cross peaks gives the distance information. So if you look at the same curve nosy is applied for these large molecules and rosy is applied for small molecules in this correlation time. So here interesting thing happens that cross peaks comes of opposite sign and then the diagonal peak and this is again used for distance information for small molecules and this distance information comes because these spins are coupled by dipolar coupling in small molecules. So thumb up rule rosy is applied for small molecule distance instrumentation nosy is applied for large molecule distance instrumentation so with this I would like to stop here and in polarization transfer next class we will discuss about the selective population inversion and how we do it and what we can achieve by doing this in case of heteronuclear system. So we will continue with that if you have any question please write to us or ask us we will be happy to answer you. Thank you very much.