 So, we stopped at this slide last time, how to calculate the structure of the protein using this algorithm of distance geometry. You have a set of distance constraints derived from the nosy spectra which will tell you what a particular distance should be, within what range it should be. So this you have collected let us say about 1000 such kind of distance restraints, then you build a model in your computer and then algorithm called distance geometry which will optimize the various torsion angles along the polypeptide chain and you can get the final structure satisfying the distance constraints you might have. So that is the kind of an indication which I am just repeating what I said last time that you start with the different initial structures for a particular protein and you have a set of distance constraints and the as the computer gets starts optimizing it eventually you will reach this final structure here and the final structure is the same in every case regardless of where you started from. Therefore that adds confidence to the fact that your distance constraint set is complete and it is adequate to define a unique structure for a particular protein. Well that worked very well for relatively small size proteins like about 60, 70, 80 and determinized residues but the moment you go to even larger proteins then you start getting into difficulties because the nosy spectrum is extremely crowded. See this is a spectrum which has about 130 amino acid residues and then you see the number of peaks which are present here. This is enormously complex overlapping peaks and therefore to quantify each one of these peaks here it is going to be an extremely difficult task and there are many peaks which are close by here and then you cannot quantify these peak intensities to establish the distance constraints. Therefore you already started feeling the inadequacy of the two dimensional nosy spectrum. Now what do we do? This is all proton-proton spectrum. So therefore what was done was to make use of other nuclei which may be present in your protein in the sense that you have carbon-13 and the nitrogen-15. So here you have the nitrogen-15 proton correlation spectrum. So you have these amide proton region here and these are the N15 chemical shifts of the backbone amides and you see here you can count to the number of peaks. Basically we see correlation peaks here only you do not see anything else. So you have one peak per residue each amide group produces one peak. So there is a good dispersion of the N15 chemical shifts here and there is also good dispersion of the amide proton chemical shifts here. So therefore you can make use of the N15 chemical shift as well to display your spectrum in more than two dimensions then you can get more information about from the nosy spectrum. So that is the schematic indicated here. So you have the proton-proton correlation spectrum here as we saw in the earlier spectrum. Then you also have the N15 proton correlation spectrum here and then you pull these peaks apart along the N15 dimension making use of the N15 chemical shifts here. There are three N15 residues having the same amide proton chemical shift at this point. Therefore when you pull them apart then you will see they will get each one in the different plane. The three N15s they come out in different planes. So therefore the resolution has improved and you will be able to assign the individual peaks to individual N15 chemical shifts. Now to further improve the resolution we can go to C13 separation and make use of the C13 chemical shifts of the individual amide proton and these correlations here then you can spread them apart on every plane you have only one peak. So then this becomes easier to identify a larger number of peaks and quantify the intensities. So therefore this may be a proton-proton nosy spectrum for example here and you spread them apart in the different planes then you can quantify each one of those peaks then you can calculate the structure of the protein of a larger protein. So therefore this means that you have to have N15 labels in your protein, you have to have also C13 labels in your protein and this has to be done by recombinant molecular biological methods. So you express the protein inside bacterial cells, E.coli cells and which are grown in the presence of N15 labelled ammonium chloride and C13 labelled glucose. So they E.coli cells incorporate this nuclei in every protein they synthesize and then after that you isolate and purify your protein then you can study the individual proteins. So that is therefore you need three different site kind of nuclear and these are then called as triple resonance experiments because they make use of proton, nitrogen and carbon-13. So there are several sets of experiments which we have also seen this earlier when we are talking about the methods. So we will now go back in a little bit of those details and try and explain how these things work. Some of those things we will illustrate and the other ones can easily be followed thereafter. So you have this HNCA experiment, this is a triple resonance three dimensional experiment. So you have the amide proton here and the nitrogen here and the two C alphas here. The magnetization transfer happens from the amide proton to this nitrogen and from this nitrogen it goes to this C alpha and to this C alpha and from this C alpha it comes back to this nitrogen from here also it comes back to the nitrogen and it is transferred back to the amide proton and you detect it here. Now you label the nitrogens and you also label the C alphas and then you detect the amide protons as the amide proton chemical use as the third dimension. Therefore you have three dimensions here that is why it is labeled as HNCA. So one dimension is proton, other dimension is nitrogen and the third dimension is C alpha. And here you see it correlates the two consecutive amino acid residues. If this is the residue of a particular amino acid, this is the NH of a particular amino acid. So this goes as NH C alpha CO and this is the previous residue. So we see correlation to the C alpha of the same residue and also to the C alpha of the previous residue at the N15 chemical shift of this particular residue. If we call this as I, this as I minus 1, so I will have the correlation of the amide proton and the nitrogen 15 of the residue I to the C alpha of I and I minus 1. So this is how the magnetization transfer goes, of course you will have both the peaks. Now this experiment actually provides a directionality here. So from this amide proton you go to this nitrogen and then from here you do not go to the C alpha directly but you go to the carbonyl and then from the carbonyl you go to the C alpha and then from C alpha you come back to the carbonyl and come back to the nitrogen then come back to the amide proton to record the signal. So therefore you label the amide protons and the nitrogen and the C alpha. Therefore it provides a directionality, you will get only correlation to the I minus 1 residue C alpha. Let us try and draw this picture in a little bit more explicit manner. We are going to talk about the HNCA, HNCA. So we start from the amide proton of residue I and we transfer the magnetization to the nitrogen of residue I. From this nitrogen of residue I, so here what we do is we put a time label here T1 which means we are generating an independent time domain. So magnetization is evolving on the nitrogen and this is T1 the time domain that time is incremented systematically to this provides an independent time variable. So now from here you transfer to C alpha of I and C alpha of I minus 1. Now you frequency label here also what that means is the magnetization comes on the C alpha you evolve as a function of time and this time is also incremented systematically. So this generates the second time variable. Now from this one you come back to the nitrogen of the same residue once more. You come back to the nitrogen of residue I but you do not do anything here. Of course one can do experiments in two different ways either you can use this as a particular time increment, incrementable time or you can also use this one but either way one can do this experiments. So once we have this magnetization back onto the nitrogen you transfer it back to the amide proton and then here of the residue I, so you get this is the time variable T3. So what chemical shifts are present during the time variable T1? So in the time variable T1 I have this nitrogen chemical shift of residue I and in T2 I have the C alpha chemical shifts of I and C alpha of I minus 1. I have both these chemical shifts and then in the T3 I have the chemical shift of amide proton chemical shift of I. So therefore what sort of a spectrum I will get here? Suppose I want to draw a three dimensional spectrum here, so this will go to F3 and this will be F1 in the frequency domain and this will be F2 in the frequency domain. Suppose I write here F3 this is the amide proton chemical shift and I write this as F1 and I write this as F2. So let us say I have the N15 of residue I here or let us say this is residue I. Then what I do? Suppose I take a cross section here, so I take this dotted line plane what does this contain and the corresponding amide proton, so I will draw that particular plane here. I am only drawing that particular plane the dotted line. What do I have along this axis? I have the F2 along this axis I have the F3. And this is the chemical shift of residue I, Hn of I, Hn this is Hn of I and what will I have in the F2 dimension? I will have two chemical shifts here it is C13 and here it is Hn, Hn of I and I will have two peaks there correct. So one of them is this is the C13, this is of I and this is of let us say I minus 1 it could have been either way also other way around also it does not matter but one of them is I other one is I minus 1. This is at a particular F1 is equal to Ni, N15 of residue I. So what I have here I have at the NH proton chemical shift amide proton chemical shift of residue I I have the C alpha of residue I and C alpha of I minus 1. Now suppose I go to the Ni chemical shift of residue I minus 1. So what I should see there if I go to the plane of Ni minus 1 suppose I go to the plane of Ni minus 1 I will draw here suppose I go suppose I am at the F1 is equal to Ni minus 1. Then add the N15 and the amide proton of I minus 1 amide proton of I minus 1 again I should see two peaks right and that will be I should see that here it should be it should be the same C alpha chemical shift it should be here this these two should match right it should be at the same place. Then I will have another peak which will correspond to I minus 2 let us say I draw that here this will be I minus 2 now this will be I I minus 2 and this will be I minus 1. Now this is the so-called in this plane this is the self peak and this is the sequential peak in this plane this is the self peak and this is the sequential peak right. So therefore but apriori you do not know which is the N15 of I minus 1. So therefore what one will have to do one will have to scan through this N15 planes along the F1 dimension and find out where this peak appears exactly at the same place as this along the C alpha chemical shift where this is present and you should be present at the same place. The amide proton chemical shift can be different of course because amide proton chemical shift of I minus 1 and I can be different. Therefore when you search through the N15 planes look for the same C alpha chemical shift which are present in the two planes then wherever this is present that will be the amide proton chemical shift of I minus 1 and that is what is indicated here. So you go from the amide proton to the nitrogen here then you go to the C alpha of I and C alpha of I minus 1 then you go to the N9 that is how this kind of a pattern will appear. Once you have this when you scan through the N15 planes you will go to the amide proton of I minus 1 then it will continue. So you can keep drawing this kind of planes so you will have so if I draw those planes here one after the other. So this one is F1 is equal to Ni this is F1 is equal to Ni minus 1 this F1 is equal to Ni minus 2 and this is equal to Ni minus 3 and so on. So I will have here let us say I have two peaks here one peak here I am writing this as self, self is a field one I am writing although the sequential I am writing with open and then and this will be the Hn of I. Now I come here I will get this is a chemical shift of Hn I minus 1 then I will also get here a sequential peak this will be I minus 2. So then from here I will go let us say I get a peak here and this will be the Hn of I minus 2 and there can be another sequence there will be sequential peak in the same plane and that will be of I minus 3. So then from here I go I will find let us say I go here and this is Hn I minus 3 and I may again find a sequential peak here. So therefore then from here you go now you notice here the empty circle connects to the field circle from one plane to the other plane. So this is how we walk along the polypeptide chain walk along the polypeptide chain sequentially connecting one residue to the next residue only you will have to scan through the N15 planes. So this is the strategy that is used. What happens in the HnCOCA experiment there we did not go to the residue I we only went to the I minus 1. Therefore in that case what will happen is in HnCOCA experiment then you will see only the sequential peak. So in the HnCOCA experiment if I do this here I draw here this is for the HnCA and if I do the HnCOCA experiment in the same plane if I do I am only going to see this one I will only see this. In fact this is the strategy which is used to distinguish between which is the self peak and which is the sequential peak and this is of Hn I and this is of I minus 1 C alpha and these are all C alphas. Suppose there is a confusion in this HnCO which one is the self and which is the sequential peak C alpha peak you make use of this HnCOCA experiment and you can get the identification of which is the sequential peak. So that is how you distinguish between you can walk along the polypeptide chain. Now suppose I want to look at another experiment these two therefore form a pair of experiments. So let us go back and see what other experiments we looked at these two experiments here HnCA and HnCOCA these are the pairs and the HnCO I may not draw want to draw these experiments in the same manner as I do earlier. So therefore here you have from the amide proton you go to the nitrogen you go to the carbonyl and you increment the carbonyl increment the nitrogen increment the amide therefore in the three dimensional spectrum here you will have amide proton chemical shoot on one nitrogen on the other and the carbonyl of the previous residue on the third dimension. So therefore this is a different experiment which will give you carbonyl assignment these ones will give you amide nitrogen and C alpha assignments but once you know the amide and the nitrogen assignments you can also get the carbonyl assignment. So in the same manner you can go to these experiments this is HnCA and this goes little bit more drawn out goes much longer in the sequence okay we can see here also the way it goes so you have from the amide proton to this here the nitrogen from this to the C alpha also to the C alpha then it can also go to the C beta here but you now label the beta okay you label the C beta you are not labeling the C alpha that means when it is on the C beta you systematically increment the chemical shift however often the C alpha and the C betas may not be separable and you may have some sometimes both the peaks present. So you have this amide proton to this C alpha to the C beta you label this then come back here come back here then you from here you come back here come back here and here therefore you will get both these kinds of connections here. So there are a whole lot of such experiments where you can record different kinds of chemical shifts along the different axis. So this has allowed an enormous variety in the experimental sequences and you can calculate the structure of larger proteins okay. A 3 dimensional typical spectrum it is shown how the 3 dimensional spectrum will look like here and you have this amide protons on one axis N50 on one axis and C13 on the other axis. Now this is actually a stereo picture if you have stereo glasses you can actually see this as one stereo picture where all the peaks are present here indicated now the projections are also shown here depending upon which plane which projection you take you will get those correlations if you look at this projection what is present here on one axis you have the amide protons other one is a nitrogen. So you will be like in proton nitrogen HSQC spectrum okay. So this axis is sorry this axis is carbon this axis is proton. So this spectrum here is the proton C13 correlation spectrum suppose I want to take a projection of here this projection suppose I take this projection this plane this plane here what does it contain along this axis you have the proton and this axis is the nitrogen and therefore this peak spectrum will be the proton nitrogen 15 HSQC spectrum. So this different projections you can take of a such a 3 dimensional spectrum you will get those correlations. If you take the third projection here along this down there which is there so what is present along this you have the C13 chemical shift and here is a nitrogen 15 chemical shift. So therefore this projection shows you the C13 N15 correlations. So in one experiment you have all these possibilities of obtaining correlations of different types in a 2 dimensional way when you take the projections you will get this sort of data in your spectrum okay. As a result of all of this of course you are able to handle larger and larger protein systems and you are also able to do a larger number of protein systems in a small amount of time because these ones go a few experiments are there and there is no question of ambiguities over the overlaps of peaks because the dispersion is very good you are using the chemical shift dispersion of both nitrogen 15 and Cs carbon 13 and therefore a larger number of protein structures have been determined and these have been determined have been deposited in the PDB data bank and this is a data which is of 2019 you can see how many protein structures have been deposited here. This initially was taking several years to determine one protein structures but with the development of these methods this number of data deposition has increased quite rapidly and you can see how the progress has happened over the years and at this point you see the total number is something like about 12,000 protein structure at up to 2019 so this will continue to grow and now with the number of experimental sequence developing you get better and better structures. Now the emphasis is also getting towards larger protein systems and also complexes and the interactions because those ones will be separate issues and we can talk about that separately. Now there are other kinds of protein systems which are which pose challenges and that is the so called intrinsically unfolded proteins or it is also called as IDPs intrinsically disordered proteins and these proteins have they pose a challenge because this is exemplified here you can see this is a folded protein this is HIV protein and this protein if you de-assure it becomes an unfolded protein then you see the HSCU spectrum looks like this. So the pigs dispersion which was present here so much dispersion which was present here has been lost. What is lost? The amide proton dispersion is lost. Amide proton dispersion is lost the N15 dispersion is still very good. Now when this happens actually the C alpha dispersion also will get lost when the amide proton dispersion is lost the often it turns out that the C alpha dispersion is also becomes very poor in that in that case so the HNCA based experiments will have difficulties because there will be lot of degeneracy in the C alpha chemical shifts why that has happened? Let me go back and see show you here so what we showed here see you suppose we have this sort of a chemical shift we are we are searching through the nitrogen 15 planes suppose I have this sort of a peak present here and more than one nitrogen planes identified a particular nitrogen chemical shift but suppose there are some 3 or 4 where the same peak appears same 3 alpha chemical shift peak appears then you do not know which one to peak as Ni minus 1. So that is where if the C alpha chemical shift dispersion disappears you will have difficulty in identifying which one to choose and therefore one will have a difficulty there. So what one should do we have to devise some other methods where we make use of the nitrogen 15 chemical shift more than the C alpha chemical shift because the nitrogen chemical shift dispersion is very good even in this situation as well. So several people have put effort in this direction and we also did some work in this direction and we published a series of papers and we will look at these series of experiments which we called as HNN, HNC and series of experiments. So how does these experiments work? Basically it is very similar to the HNCA but it goes a step forward to label the nitrogen 15 more than the C alpha okay. So let us look here. So let us say I start from this particular amide proton and I will have I transfer the magnetization to this nitrogen as in the case of HNCA and I label this as T1 from this nitrogen I go to the C alpha here and also to the C alpha in the HNCA experiment this C alpha and the C alpha were labeled as T2 the time was incremented here but I will not do that I will not do that what I will do instead of going labeling here C alpha I transfer the magnetization to the nitrogen here this nitrogen. Similarly from this C alpha to this nitrogen of course part of it goes from this C alpha to this nitrogen as well from this C alpha also it goes to this nitrogen as well. So when I transfer from the C alpha to the nitrogen what transfers will happen the transfer will happen one from here to here it will also happen from here to here then from here it will transfer will here and also partly to this. Now I am back on nitrogen now at this nitrogen all these nitrogen three these three nitrogen are present. Now I label this so this becomes my time incremented T2 when I do this time increment T2 along the T2 dimension I will again have the nitrogen but this time the nitrogen of residue I minus 1 residue I and also residue I plus 1 these three nitrogen will get labeled during the T2 time period. From then I transfer from the nitrogen to the amide proton so it goes to this amide from here goes to this amide from here it goes to this amide from here. So and now I detect as a function of time T3 period I have the amide proton chemical shifts so I have the amide proton chemical shifts of all the three residues of I plus 1 I and I minus 1 all the three will get labeled. So the result will be this is what will happen so I get a three dimensional spectrum where on the F3 dimension I will have NH chemical shifts F1 I have also N15 chemical shift F2 also I have N15 chemical shift. So what will be the details of this how does the detail look like in this that is shown here if I took this cross section here at a particular F2 chemical shift say this is the particular F2 chemical shift I take this plane I take this plane that is what do I have there then that is my F1 F3 plane F1 F3 plane this is my F1 axis this is my F3 axis and what I should get because starting from the particular amide proton of I I had transferred to the nitrogen of I minus 1 and also to the I plus 1. So therefore at the particular F2 chemical shift of residue I this is residue I F2 chemical shift of residue I so we can go back and see there the F2 chemical shift F2 chemical shift of residue I because this is F2 here F2 here and F2 here and I detect I started from T3 here so I will have correlation to this nitrogen this nitrogen and this nitrogen all the three will be present at the particular F2 chemical shift of residue I similarly go to the residue I minus 1 I will have different ones. So let us go back there and see so if I look at the F2 chemical shift of residue I I will have three peaks NH of residue I this is the N15 of residue I N15 of I minus 1 and N15 of I plus 1 as well okay. But if I now take the orthogonal cross section orthogonal cross section is this if I take this cross section here this cross section that is at the F1 chemical shift of the residue I this is the F1 chemical shift of residue I so what will I have I will have this at particular the chemical shift. So this axis is F2 and this axis is F3 so F3 is here F2 is here and this is at the N particular N F1 chemical shift what will I have I will have three peaks the residue self peak I2 its own N15 then I will have the I minus 1 NH to its own N15 I plus 1 NH to its own N15. So this is how I will get these three peaks here. So these three peaks will tell me that these are the three peaks fixed from the entire HSQC spectrum therefore we call this as the triplet filter through the HSQC spectrum. Let me try and draw this schematic here a little bit more for you to understand that. So let us say I have here Hn of I I go to N of I then I go this side let us say I call it as T1 here then I go from here to C alpha of I and C alpha of I minus 1. So from this now I go N of I then N of I plus 1 from this I go N of I minus 1 and N of I then now then I go from here NH Hn of I and this will go to Hn of I plus 1 this will go to Hn of I minus 1 and this will go to Hn of I. Now this is my T3 this is my T2 therefore you can see here Hn of I appears at these two places I chemical shift appears at these two places. Where does it get contribution from? It gets contribution from this Ni that is coming from this pathway. So you can write similarly coming from Hn of I minus 1 Hn of I plus 1 where does that thing comes from and this also showing you Ni plus 1 and you are also showing Ni minus 1. Hn of I is showing me to Ni and is also coming to Ni but it is this pathway is also generating this Ni plus 1 and Ni minus 1. Suppose I started from Hn I plus 1 here then I will also get Hn I plus 1 and Hn I I will show you that here suppose I want to start from Hn Hn I plus 1. So what will I get? N of I plus 1 then I go to C alpha of I plus 1 and C alpha of I then I go from here to Ni I plus 1 then Ni I plus 2 then from here what do I get Ni and Ni plus 1. So you see now this one will give me Hn I plus 1 this will give me Hn I plus 2 this will give me Hn I this will give me Hn I plus 1. Now you see here also I am getting Hn of I starting from this is my T1 here this is my T2 and this is my T3. So in this also I am getting Hn of I that is why I said here from this peaks here you can see Hn of I I see I 2 I and I plus 1. Similarly if I started from HI minus 1 then also I will get Hn of I that is why I get these 3 peaks there. So this is how this thing works these 3 working and you will get in this particular plane you have all these 3 peaks in the same nitrogen amide proton chemical shift immediately you know which is the nitrogen 15 plane of residue I minus 1 and which is the nitrogen 15 plane of I plus 1. You do not have to scan through the nitrogen 15 planes as you have to do in the HNCA and if you look at the oxagonal cross section you also get the NH chemical shifts of those 3 residues. So this is how this experiment works and this simplifies your analysis much better there is no searching through process and there is no degeneracy problem immediately you can identify the correlations. So I think we can stop here.