 So, let us continue the discussion using different kinds of experiments. So, we talked about the HNN experiment which is particularly useful for understanding or studying disorder proteins or unfolded proteins. Because the N15 dispersion was what very crucial there we make it made use of that. And in the we generated a kind of a triplet filter through the HSQC spectrum giving correlations between the amides and the N15s of residues i, i minus 1 and i plus 1. But often it is difficult to figure out which one is i minus 1 and which is i plus 1. So, a complementary experiment is what I am going to describe now and that is called as the HNCN experiment which is indicated here. So, the pathway of magnetization transfer is indicated here. Let us say you start from the amide proton of residue i. So, you go to the nitrogen 15 of residue i and then from here now we go to the carbonyl. You go to the carbonyl of residue i minus 1. So, this is the residue i minus 1. Now, this is i, the i is going NH C alpha CO and this is NH C alpha CO of residue i minus 1. So, from the residue and nitrogen of i you go to the carbonyl of i, i minus 1. And then from the carbonyl you will transfer to the C alpha of residue i minus 1. Now, from the C alpha we go back to the nitrogen here. You go back to the nitrogen of this residue i and also the nitrogen of this residue i minus 1. So, C alpha i minus 1 you go to the nitrogen i minus 1 and also to nitrogen i. So, from here onwards you go to the amide proton of i of residue i minus 1 and then you go to the amide proton of residue i. So, therefore, you get two correlations here. You do not have to the i plus 1. So, therefore, you will have two peaks coming from there and this will help us resolve as to which is the residue i and which is the residue i minus 1. Let us draw this picture here, draw this pattern of connectivities which are coming here. So, let us I showed you here the residue h n plus n i we go to the nitrogen i and then from here we go to the carbonyl of i, then from here we go C alpha of i. Now, your branch, so from the C alpha of i we go to the nitrogen of i and then the nitrogen of i minus 1. And then this one will give me h n of i, this will give me h n of i minus 1. This is my T 3, this is my T 2 and this is my T 1. So, I did not do anything in between. I did not label the carbonyl, chemical shift labeling was not done. I only label the nitrogen here. So, along the T 1 dimension that means along the F 1 dimension I will have nitrogen, F 2 dimension I will also have nitrogen and F 3 dimension I will have amide protons. Let me do the same for and I will write here also from let us say I start from h n i plus 1. So, what do I do? I go to nitrogen i plus 1, then I go to the carbonyl of 1 minute this was i minus 1 here. This is also h n of i, sorry I made a mistake here this is C alpha of i minus 1, yeah now this is correct. Then from nitrogen i minus 1 and this is n i, so this is fine. So, now C alpha CO of i and then I go from here to C alpha of i, then where do I go? From here I go to the nitrogen of i plus 1, I am writing in the same manner over there and this will be nitrogen of i, then I go here and this is h n of i plus 1 and this will be h n of i. Now, this is my T 2, this is my T 1 and this is my T 3. So, correspondingly this will be F 1, this will be F 2 and this will be F 3. Now, look where does this h n i appear, h n i comes here, if I draw this plane here let us say I draw here h n chemical shift of residue i and this side is the n 15 chemical shift in the F 1 dimension. So, this is the F 1 dimension and the F 2, F 3 dimension also have the nitrogen. Now, where we which of these two will give me peaks in the F 1 dimension, F 1 dimension n i is giving me h n of i, n i is giving me h n of i, I am getting this peak here. Now, n i plus 1 in the F 1 dimension is also giving me h n i. So, this is also giving me h n i therefore I will have another peak, this is i plus 1, this is i plus 1 and this is i. So, therefore and this is at a particular amide proton chemical shift, this is along i of 3, I am plotting the F 1, F 3 plane, F 1, F 3 plane these two starting points n i and i plus 1 they both resulted in h n of i. And therefore, if we were to draw of h n i minus 1 then I will also see like that. Suppose I want to take here h n i minus 1, what will I get? I will get two peaks again and this will be h n, this will be i minus 1 and this will be i. Now, what if I were to look at the F 2, F 3 plane? At a particular F 1 position, at a particular F 1 position what will be the peaks in the F 2 plane? Let us draw that here, this is F 3 and this is my F 2. And let us say I look at the h n of i, h n of i, this is i and where does it come from? h n of i comes from, I will have in the F 2, F 2 plane, F 2 plane which are the things which will I will see, I will see a peak which is corresponding to the n i, I will see a peak where it comes from to the n i because I am looking at the F 2. I should be looking here, I should be looking here, h n i is giving me n i but I also have h n i minus 1, so I will also have this is the peak and I also I have in the same experiment I also have the h n of i minus 1 originating from h n of i, nitrogen of i, I am getting h n i minus 1 and h n i both I am getting, therefore I also have h n i minus 1 here and I will have a peak which corresponds to its own peak. And this will be the n, this is again n 15, this is of i and this is of i minus 1. So you see in this experiment I will have both the directions established. So if I took the F 1, F 3 plane, this is F 1, F 1, F 3 plane I know the n 15 of the same residue and also of the next residue the i plus 1. But if I took the F 2, F 3 plane then in the particular plane I will have the peaks at the individual HSQC positions that is i and its own peak and i minus 1 also its own peak. This is the F 2, F 3 plane. So therefore this produces an indication of which direction one is walking in the polyprop along the polyprop type chain. So this is an extremely useful technique then you do not you can straight away jump to which n 15 plane you should go, there is no need of scanning because immediately that information is already available to you. So that is that saves a lot of trouble with regard to the degeneracies in various other chemical shifts. This is particularly useful for IDPs or intrinsically unfolded proteins. Now this is the same thing I indicated here. You can see in the F 1, F 3 plane I have i and i plus 1 at the H n chemical shift of residue i and in the orthogonal plane at the F 2, F 3, F 2, F 3 plane I will have the self peaks here i and i minus 1 and these are at a respective chemical shifts amide proton chemical shifts as well. That is what I showed you in the explicitly by the calculation. Now there is one other important feature. If you noticed there is in the previous slides I had used different colors. Use different colors for this the self peak and the sequential peak. Similarly also I had used different colors in the H n n experiment also. For the self peaks and the sequential peaks I had used different colors. So there is a particular significance to that. The significance is the following. These indicate positive and negative peaks. If I take the black peaks as the positive peaks the red peaks are the negative peaks. So typically these are the, this is the in the H n n spectrum. In the H n n spectrum we showed already that there is going to be 3 peaks, there is going to be 3 peaks and as the squares are the self peaks for example I took this particular sequence g g prime y x z, g g prime y this is the particular sequence which I am looking at. This is the protein sequence is going from g g prime there are 2 g's here and y here and x here and z there. This is the sequence what you are looking at. So in this case if I make, if I take the residue x what are its neighbors? The neighbors are y and z. What are y and z? They can be any residue other than the glycine, it can be any residue. So then what I see for the residue x the self peaks it is a positive peak and the 2 sequential peaks which are there are negative peaks. Now the situation changes if there is a glycine somewhere. The glycine actually produces a whole different lot of patterns. For example if I look here if I like this one now y for the pattern if there is a y g z suppose there is a y g z. The sequence as a y g z and then I will have here for the g I will have this is the central residue. The central residue I will have the self peaks which is negative and the self peak which is y this will also be negative and g z which is on the i plus 1 side that will be positive. Therefore you see here the self peak is negative the sequential peak on the i minus 1 side is negative but the 1 on the i plus 1 side is positive. Now we look at g x z, g x z if there is a triplet sequence like this then the x is the central residue for the triplet. The central residue x has a positive diagonal means we say the self peak and the positive i minus 1 peak this is g is i minus 1 that is a positive peak and the sequential i plus 1 peak is negative. If I have this sort of a thing g prime g z suppose this is the kind of a sequence I will have then the central residue is g i minus 1 residue is also g and the i plus 1 residue is z. In this case the g in the center has a negative peak which is the square and it is 1 which is on the i minus 1 side is positive and the i plus 1 side is also positive. Of course if there is a proline then you do not have a peak because proline does not have the amide proton you will not see that and here you will have the x the central residue has a positive peak and the 1 which is on the i plus 1 side that is negative. If I have a P g z then the central residue is g that is again negative and the z is positive. So one thing we have generalized here is that the self peak for all the glycines no matter where it is it will always be negative. The self peak of g will be negative and the sequential peaks will depend what the residue on the i minus 1 is on the i plus 1 is. So you will get therefore different kinds of peak factors here combinations of positive and negative peaks in this kind of a sequence. So this is a typically indicated here how these ones are appearing. Now in the case of h and c n this will also have similar kind of a features. Now as I said I will have the i and the i plus 1 in the case of h and c n suppose I have this y x z as the same as here in this case y x z gives me the self peak as positive and the sequential peak is negative which is the similar to this except that I do not have the i minus 1. But now you see here this fellow the y g z the g is in the middle g is in the middle I will see only i to i plus 1 right. So i to i plus 1 these both the peaks are positive in this sequence the g is in the middle the both the self peak and the i plus 1 peak they are both positive. Now if it is g x z if g is on the i minus 1 side x is in this middle and z is on the i plus 1 side both these peaks are negative x is also negative and this is g that is also z i plus 1 I do not see to the i minus 1 you see to the i plus 1 therefore the i plus 1 is here and this both will be negative this is very interesting here you see here these two are negative y g z g x z so these are negative and if you have the g prime g z so the middle one is g middle one g g that is always negative this is negative here and the i plus 1 here this is now positive i plus 1 this is z that is positive. So like that you get different kinds of sequences here and that is what is illustrated with this particular experiment here these are experimental spectrum this is the sequential walk through the experimental spectrum. So what is the sequence here v 35 i 36 g 37 this is q 38 d 39 s 40 s 41 e 42 i 43 h 44 f 45 k 46 v 47 k. So therefore there is a g here there is a g here which makes these are called as check points whenever you see a g the glycine it provides a peak pattern change with regard to the combinations of positive and negative peaks therefore these become check points along their polypeptide chain they will allow you to identify the peaks in an ambiguous manner sequence. So let us look at that here let us say we start from this so we are what we have here this v 35 i 36 g 37 now this fellow at this v 35 I see two peaks and this shows a self peak and the sequential peak the sequential peak is negative. So what is the sequence is this is very similar to this y x z this like the y x z so I have here a positive self peak and a negative sequential peak here this is like the x and z here because the next residue is also i v and i now we come here to this so this is now i and i plus 1 is a glycine i plus 1 is a glycine where how does this correspond to which is that situation is here. So this is i 36 so and this is v what is this pattern here so this pattern is where we should take that is this one here g is in the middle g is not in the middle so we have to take actually this one only y x z y x z this is now the next one is glycine so we should take y x g so we should see where is y x g p x g g x z this pattern is not shown in this y g z is there y x g when it is there now that is this is like a y x g so this fellow has a positive peak for the self and a negative peak for the i plus 1 so i plus 1 is g and therefore that is here in this situation this i plus 1 peak will be negative if it is that pattern is not shown here but anyway so that we can see here this pattern will be negative the self peak is positive and the sequential peak is negative now you go to the next one the next one is g next one is g here and what will be its peak pattern so this will be this will be similar to y g z the central residue is g and the other two residues are non-gs and therefore what I should get I should get this sort of a pattern y g z and you see you have both the peaks positive the self peak is positive sequential peak is also positive and the immediate next ones will be negative negative negative that is the same as here g x z now this becomes a g x z therefore this will be negative negative therefore you see you can walk along the polypeptide chain positive negative you check here go to this pattern of positive positive and negative negative here then after that it will all follow positive negative positive negative and so on so forth okay so you can walk along this way and reach up to the end point so t 50 here and once again we see for the t 50 we see set of a positive because the next residue is similar to what is the pattern which is present here so this is how you walk along the polypeptide chain by making use of this pattern of peaks therefore these various glycines which are present these provide extremely useful check points in your sequential assignment procedure and there can be no errors here because if you made an error then you will not hit the proper peak pattern okay now you can do to the step forward and suppose you took projections you took projections of this factor these are three dimensional spectra they are three dimensional spectra and if you took projections this is the way they will look what these projections are I have this projection the f 2 f 3 plane f 2 f 3 plane that means I go down the f 1 plane I go down the f 1 plane I take the projection in the f 2 f 3 plane I plot if I do that because of the various cancellations that will happen you will only see peaks coming from the G's and the residues next to them okay for example here you see G 47 K 48 G 10 E 11 okay and then you also have this R R 54 and these are also coming G 75 and G 76 G 75 G 76 okay so therefore you will see only this I and I plus 1 P projections which are coming out here and they also have different signs so therefore by looking at this alone you can figure out where your glycines are this is f 2 f 3 plane projection on the other hand if I took f 1 f 3 plane projection I have two kinds of things here one is the normal HSQC normal HSQC spectrum then I will have the sequential peaks the sequential peaks to the I plus 1 for a from every residue I see to the I plus 1 and they have different sign patterns as in the case here therefore all of them will be present in one plane they are all present in one plane therefore this will allow you to walk along this entire sequence of the polypeptide chain in one spectrum and this spectrum can be recorded very rapidly okay this is because this will be two dimensional spectrum what you need to do is if I want to record only the f 1 f 3 plane I do not increment the t 2 time period therefore the t 2 time period is kept constant you do not make an increment of the t 2 so I will record directly the f 1 f 3 plane only if I want to record f 2 f 3 plane directly then I do not do the t 1 increment I will get only the f 2 f 3 plane now this is an illustration of the sequential connectivity is using this spectrum this is just an expansion of that which is from the same here so you can see you can make a walk through the polypeptide chain here from l 8 to t 9 okay and you see those thing here is the line is drawn at the center of the column so you go from here to here this positive negative you go here positive to negative you go here now you see positive positive this is because of the glycine you got the glycine here you got a positive positive peak pattern then you go immediately after that you have the negative negative peak pattern for that is what it should be right so you got the negative negative and then you go to the next once again you have the positive positive this is the sequence goes in the normal manner there is no glycine. So wherever there is a glycine the one, the same number and the one which is after that these two have very characteristic peak patterns. You have the negative negative here and positive positive there. So this will allow you to identify where you are along the polypeptide chain in a simple manner. So therefore you can obtain such large number of assignments very rapidly for unfolded proteins and that is we call it as the simultaneously the speed barrier you can get the assignments in a few hours. So now I think we are going into a slightly different topic that once we have done this analysis of the individual protein chains then you have to study what is the kind of a we are talking about the intrinsically disordered proteins. Intrinsically disordered proteins have various kinds of dynamics present all other proteins also have the dynamics present there and protein dynamics becomes an important part to study that. So there are protein dynamics will have various time scales of operations and these are the typical time regions for molecular motions and that is indicated here. So what we have done here is classified the various time scales. The various time scales are indicated here. So we may not go into the entire details as I only indicate here what sort of a time scales appear in the proteins and the disordered proteins will have all of these kinds of time scales that is the reason I am showing here. So we have been talking about the intrinsically disordered proteins various kinds of motions are possible there and the conformational fluctuations can happen but this general slide which will show you what kind of time scales are available for what kind of processes. So typically the time scale can run from this 10 to the power 3 seconds to 10 to the minus 12 seconds. So these are 10 to the power 3 is extremely slow and this is extremely fast. So what sort of time scales are there these 10 to the minus 12 of our bond vibrations 10 to the minus 9 or overall tumbling this is the nanosecond time scale and then you have in the microsecond to millisecond time scale you will have slow loop reorientations this will happen in your unfolded proteins then you can have local unfolding this is the local fluctuations there this also this will also happen in the 10 to the power millisecond time scale motions and global unfolding entire protein to unfold this will happen in seconds to hours time scales and typically of course chemical kinetic reactions also happen in this area and SS flipping and side chain rotation etc these will happen in this time scale. Now how do we measure these ones this goes into different kinds of experiments and those are various tools available and those are typically indicated here which it is time scales can be studied by what NMR technique you have the hydrogen exchange possible then you have the relaxation time measurements there and these are heteronuclear NOEs and relaxation time measurements and the coupling constant that the chemical shifts these are the various parameters which one uses to obtain this kind of information. A simple example is indicated here so if you have the chemical exchange happening on a slow exchange between conformational states if you have such kind of an you have one population here another population there and these ones are two different signals for the same proton or the same which is undergoing chemical exchange you will have two different intensity the populations here and here may not be the same so it is that depends on the energy difference between the two conformational states so you have different chemical shifts for the two you have this and if the exchange between these two is slow compared to the chemical shift difference between these two this we have studied earlier also and then you will have this kind of a peak pattern which is present. But if there is exchange rate increases it goes faster then you start seeing changes in this peak patterns and that is what is is indicated here so protein dynamics so depending upon the what sort of exchange rate is present you will have the peak patterns appearing like this this is slow exchange this is also slow exchange but the lines are broadened increase the exchange rate by increase in the temperature or different things then they start slowly start merging here then the coalescence temperature from the coalescence temperature you can calculate the exchange rates here and becomes fast exchange everything will merge and you will find a single line here in the middle and that will be at the weighted average chemical shift of the two populations there. So this is what will happen in the case of intrinsically disordered proteins or even in the folded proteins there can be certain domains where there are slow exchanges there is a domains where can be fast exchanges and all of these can be studied by using NMR. I think we can we can stop here.