 So, you can see that the triple resonance experiments are basically driven by application to proteins, experiments driven by for proteins. In fact this has been an extremely useful feedback system that multidimensional NMR experiments developed but the proteins actually drove them very, very strongly. So the desire to obtain structural information on proteins has been the motivation for developing new and new pulse sequences and you see therefore all the triple resonance experiments which we will describe are primarily directed towards proteins and therefore we will be concentrating on that kind of a thing. Structural biology primarily has developed as a result of application of I mean NMR in structural biology has primarily been driven by application in proteins. Also of course nucleic acids and other systems but by and large methods development have occurred because of applications to proteins. So here I show this slide again to you for the indicating the strategy here. We have here a dipeptide NH C alpha CO NH C alpha CO and the transfer of magnetization happens through this polypeptide chain make either through the backbone alone or through the side chains as well accordingly we have different kinds of pulse sequences here and all these are based on transfers based on J correlations and therefore the J coupling constants are important here. These different J couplings are indicated here these are all one-bound couplings okay except these which is a two-bound coupling and even the two-bound coupling also does not vary too much as you can see and by and large it is about 4 to 9 hertz but mostly it is in a random coil situation they will be like 6 to 7 hertz and certain structural situations you may have 4 hertz and somewhere it is 9 hertz and things like that but mostly it is one-bound couplings and therefore there is no question of missing those peaks in your in three-dimensional experiments okay and how does this work I will illustrate this to you how the transfer happens in a particular in the protein in the pulse sequence and we will illustrate this is one taking one particular example which is called as the constant time HNCA I described to you the HNCA this is the say another version of the same one the magnetization transfer follows in this particular path. So it starts from the proton as you can see here and this is the amide proton this is here the amide proton which you label here the amide proton is transfer from here and then you transfer from this amide proton using the inept sequence. So you have the initial inept sequence here so the magnetization flows from here and comes to nitrogen nitrogen at this point it starts here and comes to nitrogen at this point up to and out of that part and then during the next period from here to here it flows in the nitrogen on the nitrogen path remains on nitrogen and then it is transferred to the alpha. So this is the pathway so the magnetization flows in this manner from the amide proton it goes to the nitrogen from the nitrogen it goes to this C alpha and now you see in the amide when it is in the amide portion you have this T1 by 2 increment the T1 increment is present here therefore and this total time from here to here it is T okay total time from here to here is T and that is why it is called as constant time but in the same one this T1 is incorporated okay from here to here it is T therefore it is called as constant time during that time the J evolution happens and during the T1 period is the chemical shift evolution which chemical shift that is nitrogen chemical shift because you have the T1 increment and the magnetization is on the nitrogen therefore during the T1 period we have the nitrogen chemical shift okay and now from this it is from the nitrogen it is transferred to the carbonyl to the C alpha and then here you see T2 by 2 T2 by 2 and T2 by 2 this is T2 so total is T2 this is again an increment this is the incrementable time period and therefore what is the chemical shift labeling here if there is an incrementable time period it means there is a chemical shift labeling that frequency is getting labeled with a particular evolution okay during that evolution that particular frequency is getting labeled so therefore during the T2 period I have this C alpha getting labeled in the T1 period N15 is getting labeled and during the T2 period C alpha is getting labeled now from here it goes flows here and then it goes to nitrogen again it goes to the nitrogen again and during the nitrogen there is no labeling here so this is the constant time again this is the same constant time as here T total period from here to here is T and during this time the whatever certain refocusing things like that will happen and it is on the nitrogen the magnetization is on the nitrogen and then it flows through the nitrogen until here and then it goes back to proton it goes back to proton here and then during this period there is the next effort there is a refocusing of the amide proton chemical shift which is anti-phase there and you produce at this point E in phase proton magnetization so that they can detect here and therefore what is in the T1 period T1 period I have the nitrogen chemical shift and T2 period I have the C13 C alpha chemical shift and during the T3 period I have the amide proton chemical shift because what I am going to detect here is coming from the nitrogen from the N15 it is coming to the proton and where does it come it comes to the proton which is attached to it only therefore it is on the amide proton and the amide proton is refocused here it is anti-phase it comes to in phase and then you detect this during the T3 period therefore what after 3D Fourier transformation what do you expect to get we expect to get a now this is let us say if I call this as F3 F1 and F2 what will be present in F1 F1 will be nitrogen 15 okay and therefore this will go to F1 F1 this will go to F2 and this will go to F3 okay so this is the way it goes of course we apply certain pulses on the carbonyl channel etc for the purpose of decoupling etc we may not want to go into those details here but this is the way the magnetization flows and therefore this is the kind of a schematic spectrum you will get okay now let us look at this spectrum how does it look this is what I indicated to you here and on this here you have the schematic spectrum of a particular sequence whatever randomly some 4 amino acids are taken here so how many amino acids are taken here there is one green there is one here red and then the sand and then of course there is another red here this is brown now each one of them has a strong peak here and then also a peak at this point each one of them has that so therefore it is indicating as I mentioned to you earlier you generate a self peak as well as the sequential peak because there this is called as the sequential peak of residue I-1 so if you let me go back there and show you what we had there HNCA so if you look at this so if you look at this we transferred we transferred from the amide proton to the nitrogen and nitrogen we transferred to both the C alpha this C alpha as well as this C alpha so this is the self residue I and residue I-1 therefore the both the things we will have to reflect in your spectrum and that is what is happening here so you see you have here the two peaks at each amide proton chemical shift two peaks one belongs to its own C alpha other one needs to the neighbouring C alpha which is I-1 residue same here therefore if you took a cross section at a particular N15 chemical shift and these ones are appearing at different N15 chemical shifts this is through the depth you can go through the N15 here depth so these ones are different depth these four ones are present at different depths okay so if I take a cross section at any particular N15 N any particular N15 this is the F1 axis I have two peaks this one of them is a residue I other one is a residue I-1 these ones are slightly different intensities by enlarge they may not have also but sometimes you do have slightly different intensities because this coupling constant is slightly larger than this coupling constant because this is dependent on the two bound coupling whereas this is dependent on the one bound coupling as we saw this coupling constant is of the order of 7 to 9 hertz whereas this 7 to 11 hertz this coupling constant is of the order of 4 to 9 hertz so depending upon what it is sometimes we cannot different intensities in fact it is helpful if it has different intensity it is helpful to figure out that which one is a self and which one is sequential okay once we have got this now this is this sequential which is I-1 residue now obviously at some other nitrogen chemical shift this will become the self peak okay when you move through the N15 plane at this particular C alpha this peak will become the self peak because it has to have a particular self peak also so therefore you search through the N15 plane here then you will find that okay at the particular chemical shift then of the N15 you find exactly at the same C alpha chemical shift you will you find a strong peak here this strong peak therefore is the self peak of residue I-1 and then correspondingly you have another peak which is the weaker peak which is of I-2 so again you search through the nitrogen 15 planes then you will find a peak which corresponds to the same C alpha chemical shift but at a different nitrogen amide proton chemical shift okay and also the different N15 chemical shift so therefore you will see a strong peak here and a weak peak here so you can continue like that so therefore you see you can walk through the polypeptide chain by going through the different N15 planes scan through the N15 planes identify where the C alphas are and you can start connecting the peaks in this manner okay once you have the C alphas identified like this obviously you can also identify the other protons through the toxic or you can also use the other ones where you can go to the C betas and so on so this is an experiment which is extremely useful okay now here the separation will depend upon how good your C alpha chemical shift is and there sometimes the C alpha chemical shifts may not be too too good and in that situation one has to use the some other strategy and here is an experiment which is called as HNN so here what you do is I will explain to you the first the strategy how the magnetization transfer happens in this okay in this experiment the magnetization goes through starts with the N15 amide proton starts to the amide proton and is transferred to the nitrogen of the same residue and it is we can see labeled with a T1 there is a T1 increment there now from this T1 you transfer to the C alphas of the two residues I and I minus 1 just as we did in the case of HNCA okay from Ni you transfer to the C alpha of I and also to the C alpha of I minus 1 okay now in the earlier case you label this C alpha but here we do not label the C alpha we do not make it as a T2 part what we do is we transfer this further to the next nitrogen there from this C alpha during a particular time period here it called 2 times tau CN you transfer again to the nitrogen of I minus 1 partly to the nitrogen of I minus 1 and partly to the nitrogen of I itself so this is Ni minus 1 and this is Ni plus 1 here and this will be I plus 1 so here again this will be I minus 1 because the transfer happens from the Ni to the next residue Ni of C alpha I from the C alpha I it can go to the I plus 1 okay and C alpha I minus 1 its I plus 1 will be I okay so therefore it can go to I here and part of it remains as I minus 1 here okay so this is the way the transfer happens and after this you have got the proton you have the magnetization on the nitrogens of 3 residues I I minus 1 and I plus 1 and this is then transferred to the corresponding amide protons this will be I plus 1 this will be I and this will be I minus 1 so these ones are detected during the T3 period okay and now if you do a Fourier transformation what you get here you get here a three-dimensional spectrum F3 axis has amide protons F1 has nitrogen and F2 also has nitrogen F2 also has nitrogen okay now if you were to take a cross section if you take a cross section through the through this 3d spectrum how does it look like suppose I take a F1 cross section at a particular F2 chemical shift at a particular F2 chemical shift I take this plane I take this plane and show it here so this is the F1 F3 plane at a particular F2 chemical shift okay at a particular F2 chemical shift that is I have here a particular F2 is equal to ni and this is the ni of the F2 then I will see three peaks these three peaks I I minus 1 and I plus 1 all the three we see here and correspondingly if I were to take a F1 cross section that means I take cross section through this here through this plane at this particular chemical shift then I will get here the F2 F3 cross section the F2 F3 cross section has the same three peaks but at the respective amide proton chemical shifts see these are three different chemical shifts of the individual three residues these three peaks so what do you see of these this is like a triplet filter through the HSQC what do I mean by that so suppose I have a HSQC spectrum this is N15 and this is HN you have lots of peaks here okay your many peaks for the 2d spectrum but now out of these three consecutive residues whichever are the three consecutive residues and you will figure filter them out in this particular plane you will filter that in a particular plane okay so therefore it is called as at the particular F2 F1 chemical shift the particular N15 chemical shift you get three peaks which are neighbors okay the self peaks I minus 1 peak and the I plus 1 peak all the three peaks you are seeing in this therefore this is called as a triplet filter through the HSQC spectrum and this is extremely useful because you immediately know which plane you have to go to identify the next residue you do not have to scan through the N15 planes as in the case of HNCA to figure out where the C alpha becomes a self peaks here you do not worry about the C alphas and the N15 dispersion is always very good compared to the C alpha chemical shifts in most proteins and the particularly in the case of disorder proteins intrinsically disorder proteins or loop segments in proteins you have more chemical shift dispersion for the C alphas but the N15 dispersion is always good therefore you can see this kind of peak separation in every case okay and another important thing you have to notice and that is I have used different colors here see I have used one particular color for the self peaks the residue of the same residue and I have one particular color for the two others I have two different colors because they have negative sign if this is positive sign this is the negative sign okay and vice versa whichever one you want to choose that way so they have opposite signs and that also shows up here and this is extremely useful in a generalized sense but another important feature of this is this sign pattern what we are having depends upon what is the sequence in the triplet what kind of a sequence do you have in the triplet is there a glycine in the triplet sequence or not now if there is a glycine in the triplet sequence this color combinations will change and that is an extremely useful factor because you you will be able to figure out where you are along the polypeptide chain as you are walking along the polypeptide chain if we hit the glycine you hit a different kind of a peak pattern therefore that will appear as a checkpoint and that is demonstrated here in this particular slide okay what we are of course if there is a proline if there is a proline it will not produce a peak therefore you will have only two peaks not three peaks because there is no amide proton on the proline but whereas in a generalized sense you will have the self peak which is positive and the two sequential peaks will have negative sign I represent this as zx z prime x is the central peak which is the self residue and z and z prime are the i minus 1 and i plus 1 residue okay these ones have negative sign as indicated in the previous one if this is positive the square indicates that is the self peak or the of the same residue and the circle indicates it is a sequential peaks now similarly if I have here a proline with a glycine with a glycine there now the glycine has a very distinct chemical shift and with regard to the n15 and those ones will appear at the top in the n15 chemical shift okay therefore this will produce a distinct pattern to immediately identify that this is the glycine here now if there is a glycine in the middle of a sequence if the glycine is here at the end then of course the the x residue will show a peak here and the glycine sequential peak will appear here if the pgx that means the glycine in the middle then you will have the square which is on the top but this will have a negative sign see unlike this x which is a positive sign the glycine self peak will have a negative sign and you see everywhere where there is a glycine in the middle it has a negative sign and these are extremely useful factor because you will immediately figure out that where you are along the polypactite chain now these are the others where there is no proline and each one of them has three peaks each one of the peaks if there is a glycine in the middle then that self peak is negative and then you have a positive and a negative combination here and if the glycine is the i minus 1 position then the self peak is positive and you will have a negative and positive combination negative and a positive combinations here therefore you see this peak patterns which you are seeing is extremely important in the walking along the polypactite chain you immediately identify okay are you doing a right thing or not you go from one residue to another residue and if you hit a glycine in the middle then you must get this sort of peak patterns positive and negative combinations and that will be extremely useful for rapid assignments of your polypactite chain now there is another complementary experiment of this which is called as HNCN okay now this experiment the magnetization transfer pathway goes like this so here you start from the again the amide proton and you go to the nitrogen 15 here from the nitrogen 15 you go to the carbonyl you do not go to the C alpha you go to the carbonyl of residue i minus 1 from this carbonyl of i minus 1 you go to the C alpha of i minus 1 during this period and then from the C alpha of i minus 1 where do you go you go to the nitrogen of i minus 1 and nitrogen of i and here you have frequency label you call this as T2 and the T1 is here the T1 therefore along the N15 axis you have the T1 and here you also have the T2 okay so therefore then you have here and the from the amide from the nitrogen you go to the amide proton and you have here the system therefore what you have here and also in the same case in the previous one also so T1 T1 is N15 T2 is also N15 and T3 is amide this was all the also the case in the in the previous one maybe I should also write that here yeah so here also T1 is N15 T2 is also N15 and T3 is amide so therefore you will have three dimensional spectra in this manner okay so now if I take three dimensional spectrum of this what is the what is its content okay so let us look here you take the what are the correlations we will get see I am not going to the three residues there I only have two okay so if I take a cross section here through the at a particular F2 at a particular F2 position that means I have the F1 F3 cross section F1 F3 cross section as two peaks one particular to this own residue and then to the I minus 1 okay start from the hn of I hn of I I am seeing to I minus 1 and I correct so if you here hn of I I residue to I and I minus 1 I and I plus 1 okay sorry this will give you I and I plus 1 because you have to write down similarly for the hi of I plus 1 also hn I where it will come you have to combine these you have to write such a transfer pathways for hn I hn I minus 1 and I hn I plus 1 when you do that you will see that the particular hn I you will see two peaks one corresponding to the I of the N15 of I and N15 of I plus 1 you will see two peaks in the F1 F3 plane whereas if you look at the F2 F3 plane F2 F3 plane that is at a particular F1 which is the I that is this F1 I what are you seeing you are seeing 2 N nitrogen which are I and I minus 1 okay therefore and they will appear at the respective hn chemical shifts and that is what is shown here at the respective hn chemical shifts I minus 1 and I you are seeing two different things now this is like a doublet filter I see here two peaks which correspond to the I and I minus 1 residues along the polypeptide chain okay now here again you will see interesting peak patterns peak patterns once again you will see different kinds of positive and negative combinations and again when there is a problem of course you will see only two peaks here but you will see two peaks in every case here okay but the patterns of positive and negative combinations are quite distinct compared to the previous one therefore this will allow you to identify which are the sequential peaks which are the self peaks and which direction you are going the important thing is this will provide you a direction in the earlier case it was not easy to figure out which is I minus 1 and which is I plus 1 but now you see you have only I minus 1 coming in this plane and in this way you are counting only I plus 1 okay so therefore it provides a directionality for your walking through the polypeptide chain and that is an important observation and this will have a significant value in your sequential resonance assignment okay what is it this positive positive here very unique this positive positive for a p g g sequence and for a x g z sequence also we see positive positive here if you have glycine in the middle and you have x and g again you have positive positive but if glycine is in the n I minus 1 positive you have negative negative this is very interesting so if you have a z g g then you have positive positive on the top here so if you have this situation negative negative so these are extremely important combinations of peak science and which will you can cross check therefore these appear as check points in your walk through the polypeptide chain so this shows experimental illustration of this spectra application here a particular protein is probably f k b p or ubiquity and I do not remember but you can see here so sequential correlations how they are appearing positive negative combinations see this is 5 5 so this is shown on the top the sequence is given here 5 6 7 8 9 10 11 12 13 14 for the same three sequence here through the h n n spectrum how you are walking and how you are doing it here for the h n c n you can do the sequential walk through both the spectra whether you want to do it through h n c n or h n n you can do that so you see you go from 5 to here see the appear sign then you go from here to here here to here here to here your same positive negative combinations but now as soon as you come here to the g 10 see as soon as you come to the g 10 you have to have two positive peaks there see it is t g k sequence t g k g is in the middle and that must give you positive in a positive positive combinations and this is what we saw here so see this is this this portion x z so this g is in the middle and x z can be any other residue and as soon as you reach there you must get a positive positive peak and in fact you see at this point you see a positive positive peak and the immediate next one should be negative negative because that one is g k g k t that is g x z so therefore you will have here negative negative combinations and then after that again the g is gone so you have positive negative positive negative positive negative and go on things like that the same feature will show up here in this particular case as well so you have here in the channel as well you go through the sequential walk from 5 to 6 6 to 7 and 7 to 8 and so on so forth you can walk through the polypeptide chain along that and notice once more here so as soon as you come here you see two negative peaks and this is the g 10 just as you had here two positive peaks and you will get two negative peaks and the positive peak here and the immediate next one will be two positive peaks and one negative peak and this is the guarantee that you have not made any mistake in your sequential walk all through the polypeptide chain so this is therefore extremely useful for obtaining rapid assignments in proteins regardless of the size especially in the intrinsically disorder proteins where there are flexible domains and is very difficult to obtain assignments without this sort of a dispersion so n 15 dispersion is always good so you can get resonance assignments in a rapid manner and in an unambiguous manner. This therefore experiments have been very successful in analyzing unfolding pathways or folding pathways of proteins or aggregation pathways of proteins this strategies have been used and this was possibly we will discuss later when we actually discuss more of the applications of these methods and to draw more information about the biology of that one the structural biology of Lawson here our focus is on discussing the methods and how the methods are useful to derive the information and how they have been developed to circumvent some problems of peak dispersion or difficulties with regard to the assignments or with regard to the structures in the proteins where there are unfolded regions in the folded regions together you will have difficulties. So the methods development has been driven by such kind of problems and the applications of these we will also discuss more as we go into particular kinds of questions with regard to the folding unfolding aggregation etc. So I think we can stop here.