 So, so far we discussed about the various aspects of the DNA structures. Now we are going to talk how NMR can be used in this. So, what is our first observation in NMR? We have to talk about the chemical shifts, proton chemical shifts. We talk about the proton chemical shifts and what are the protons which are possible. We already mentioned about the various protons in the DNA and in the RNA in various places and I will show you here I have just listed them here and the proton chemical shifts as we have already seen can go from 0 to 15 ppm with respect to particular reference typically we all we record in water typically we use a TSP as a reference then you have this ppm 0 to 15 ppm. Which protons appear where? Now what are the protons which we have? We have the sugar ring protons, sugar ring protons are here 1 prime, 2 prime, 2 double prime for the DNA, 3 prime, 4 prime, 5 prime, 5 double prime these are all part of the sugar ring okay 5 prime, 5 double prime lie outside in the backbone and then these are the basis H2, H8 and H6 which H2 and H8 are present in both G and A and H6 is present only in C okay H2, H8 are present in both G and A and H6 is present in C and these nomenclatures are important one has to remember this okay. Now with regard to the loops and these are the immunoprotons NH protons are the immunoprotons and these all on these in this area we also have the NH2s G, C, A we have the NH2s in all these 3 bases we have seen that okay when we are talked about the various base pairing schemes we saw that there are NH2s in each of one of these and they also appear in this a region and this is what is this ppm this is about 7.5 so 5, 7.5, 10, 12.5, 15. So around 7.5 area you have this 7.5 to 8, 8.5 you have these bases and then you also have the immunos these are the G, C, A and you come in this area from 10 ppm to 15 ppm you have the immunos of G, T and U okay G, T and U these immunos appear in this area and these are the ones which are present in the loops these are not hydrogen bonded these immunoprotons are not hydrogen bonded therefore these are mismatches okay this called as mismatches that are the loops these hydrogen bonded one this is the G, C pair and A, T pair G, C pair appears earlier okay so this is around 12, 12.5 and the A, T pair appears later of course this 15 ppm is little bit this is the wider range here typically you will find them around 14 things like that okay 14, 14.5 that is where you will find them and you have this G, C pairs appearing in this area and the A, T's appear in this area okay and these are all at pH 7 around pH 7 because this is important to mention the pH here because the pH plays an important role in determining what sort of a base pairing can occur okay because there is a protonation which a protons are exchanging and these are exchangeable protons remember these ones are exchangeable protons so if you want to observe these you will have to record spectra in H2O and not D2O okay these spectra these amino protons whereas these ones here these are the non exchangeable protons so these are attached to the carbons these which are attached to the carbons these can be observed in D2O have D2O solutions as well now if you come to the RNA RNA does not have RNA does not have H2 prime H2 double prime it has only one H2 prime H2 double prime does not exist in RNA therefore now you see because of that because there is an OH group at the 2 prime position this H2 prime gets shifted to this area okay it comes to this area H2 prime H3 prime H4 prime so this is quite a crowded area in the case of RNA the H1 prime is not affected so much this is also not affected so much and all the other things remain the same by and large therefore the proton chemical shift ranges differ between DNA and RNA in this area this becomes quite distinct in the case of DNA whereas RNA there is quite an overlap in this region alright now here is a typical spectrum of a nucleic acid the segment which is given here so you have a deoxy this is the DNA segment so it is a 14 or here GAATTCC GAATTC okay and this is the sugar ring all the nomenclatures are given once more here so you have the one prime and in the basis you have this H6 H8 H2 and H5 H6 and H5 are present in the cytosine H8 and H2 are present in adenines and on the gonin you only have H8 you do not have the H2 at the H2 position there is an amino group in the case of gonin now where do these ones appear in the spectrum this is the one-dimensional spectrum here and all these H8 H2 H6 they are present in this area and the H1 primes and the H5 they appear here and the H3 primes are here at around 4.5 ppm here and all these H4 prime H5 prime H5 double prime they all get crowded in this particular area the extremely crowded region you notice for all the 14 nucleotides they are all present in this similarly and all the 14 nucleotides 2 prime 2 double primes are present in this and the methyl here where are the methyl the methyl are coming from the thymines the thymines have at the H where in the case of cytosine at 5 position that is the H5 in the case of thymine there is a CH3 group here how many thymines are there here you have here thymines 1 2 3 and 4 there are 4 thymines and these 4 thymines are present in this area you can actually come because these are quite distinct you have 2 here and 2 more here okay so all the 4 thymines methyl group you can identify in this area of course you do not know sequence specifically which one is which but thymines are easier to identify because they appear at the methyl appear at a very distinct position in the chemical shift range in the proton. So this is the typical one dimensional spectrum okay what about the immunoprodons now the immunoprodons as I mentioned to you they appear between 12 to 15 ppm okay so you have this immunoprodons for these protons these ones had to be observed in water the experiment will have to be performed in water H2O otherwise in D2O these ones exchange out and you will not be able to see those signals and you see this particular sequence here this is it goes GATC okay TTCCCCGGAA this is the sequence okay so this sequence has the kind of a 2 molecules form a stem of a duplex and then it forms a loop here this sees and these sees will not be observable to you because they exchange out with water these 4 sees will not be observable to you because they exchange with water and this forms a symmetrical duplex here and with the Watson Crick base pairs the Watson Crick base pairs are indicated here the thymine adenine and cytosine guanine which are the protons which are observable and those are the immunoprodons the immunoprodons appear in this range okay as I mentioned the immunoprodons appear around 10 ppm so 9.5 to 10 ppm if though and those ones exchange more rapidly with water therefore often you do not see these immunoprodons you will basically see the immunoprodons and immunoprodons will produce one immunoprodon one signal per base pair because there is only one immunoprodons whether you take this one whether you take this base pair or this base pair there is only one immunoprodon in this case that is this one here that is at N3 position and in this case there is a guanine which is at the N1 position okay so now we can see here how many signals are there so you can actually count this is at different temperatures okay this is at 278 degree Kelvin and you slowly small temperature you can go up in the temperature why do you do this temperature dependence okay this is the duplex here right this is the duplex the duplex with the stability of the duplex is the highest in the interior and as you go further towards the end it becomes fragile so it is less stable so therefore when we and initially at the lowest temperature you will see all the signals you will see how many CGs you see here you see 3 Gs these all 3 these belong to the Gs okay there is a symmetrical structure here so you see this G this G and one of these Gs so you see 3 Gs okay because this become at the base of this loop this is not so stable okay so you will see that one exchanges out so you see these 3 Gs which are present this belong to the interior of the duplex okay and these are the T's which are present that is this one you have 1, 2 and 3 okay 3 T's which are present here you have all of them present at this point okay they are pretty well seen now as you start increasing the temperature you see that the duplex which is present here starts melting it starts melting slowly this immunopronomous melting meaning what the base pairs start separating out once the base pairs separate out then they become exposed to the solvent to the exposed to the water then they will start disappearing so at this temperature this you see the everything is exposed all of them are exposed they are exchanging with water that the completely the lines are broadened out if you go still higher there will be nothing seen everything will be the no signal will be present in this therefore to find out whether your structure has a duplex or the base pairing has happened or not you have to do the experiments at low temperature and monitor the immunopronoms. And here we show the thermal stability how you can mention determine the thermal stability of a DNA hairpin you have this hairpin here G A A T T C and then you have the X n there are many other bases here which are in the loop and you will not find signals from there but the ones which are paired here you will find the signals for this this is the one particular molecule which is going around and forming a hairpin structure you will see it peaks for these ones you will see peak for this G and you will see peaks for the T's and those ones are the ones which are present here and you can see as the temperature is increased this is 5 degree centigrade 25 degree centigrade and you have this 35 degree centigrade and so this the positions are indicated at 1, 2, 3, 4, 5, 6 so these are the 1, 2, 3, 4, 5, 6 these are the ones which are present here so you have the G's which are present here at position this is at position 1 and so this see the sequence is going like this this is the single sequence and this G and this G are not the same they are different this comes through the loop here and this G and this is the position 6 G that is this one there and then you have the other 2, 3, 4, 5 these are the 80 base pairs these are 2, 3, 4, 5 these are the 80 base pairs so therefore these belong to the thymine thymine and 3 immunos so those ones are distinctly present here and these ones are the aminos so these belong to the aminos and various places and these ones you do not and you do not see them very clearly even at 5 degree centigrade you see they are all quite broad as increase the temperature these ones will disappear and you have only the immunos which are present in this 80 G C pairs and these are possibly the T's X probably is a T here so you have this T's which are appearing in this which are not clearly observed see the DNA hairpin and this is for the duplex melting of the duplex DNA indicated another example of the duplex DNA here you go from this particular sequence you have the C G C G A A T T C G C G this is the self complementary sequence you write the other way around here so it forms a symmetrical duplex with the 12 base pairs so the 12 base pairs you have so you have how many you can see 6 the 4 G's C G C G and C G C G there are 4 G's and how many T's there are 2 T's A A T T there are 2 T's because we are going to see signals in the immunoproton spectral region you will see either from the G or from the T okay so therefore you have 2 T's at 0 degrees you can see both of them and you can see all the 4 G's these are the 4 G's which are present here at 0 degree centigrade and as you start increase in the temperature see they start melting which goes first you see it is this one goes first right this one is going first what does that mean this will tell you that this is the terminal G okay so this terminal G is actually going first so slowly this is how you identify the as the DNA melts as looking at how the immunoprotons are disappearing you will identify them as belonging to which T and which G so at 60 degrees almost all of them are vanished and you can see this continues C G 1 C G 2 C G 3 C G 4 these are the base pairs these are the base pairs which are indicated there there is a symmetry there there is a symmetry so from here to here and from here to here this is the duplicate because it repeats itself in this manner right so various base pair what is indicated here is the base pair which base pair is appearing the first base pair will disappear this is the first base pair okay this is extreme end so that is the one which will first disappear then followed by two one then followed by three followed by four okay so the four is quite in the interior and that will be the last one to go so this stays here you see all the way up to 50 degrees centigrade and this is how melting state these were the standard techniques used earlier to find out what is the melting temperature of the DNA so here then you plot here the melting temperatures of the DNA so you have various kinds of modulation modifications done if you make a modifications in your DNA does it increase the stability of the DNA or decrease the stability of the DNA this you will figure out by looking at the melting temperature what is the meaning of melting temperature if you plot this normalized chemical shifts of the immunoprotons against the temperature here you will get a sigmoidal curve in this manner and the midpoint of this will give you the melting temperature so at that point of course you will have very poor immunoproton signals and by looking at the melting temperature you can decide whether the DNA is stabilized by a particular modification this is unmodified here and this is modified here at the particular place these are indicated here at which place the modification is done and and then of course you have another one and the thiop means is actually modified by a sulfur group instead of the oxygen you have a sulfur group here you see this is p2s2s so normally you have an oxygen but these are all modified here so all the sulfur groups are present here and okay here is a normal one and these are the modified modifications at this point you have a sulfur group here and then you have two sulfur groups at this point so both to the oxygens are replaced by the sulfur and how does that change the temperature so you see when you have oxygen replaced by the sulfur group the temperature has come down quite a lot here see it is almost 10 degrees so it is very unstable the normal DNA is of course the most stable one because it has the highest temperature the melting temperature is almost about 65 degrees okay so as you have one sulfur one oxygen you replace by the sulfur you have this temperature comes down to 35 degrees or 40 degrees two sulfur equally one goes even further okay so this is this was all that could be done earlier with single one-dimensional NMR spectra so melting temperatures see the stabilities of the modifications stabilities of the DNA and what sort of a structure the particular molecule is having one also study the interactions with various other small molecules depending upon the from the one-dimensional spectra but then there is the revolution with this two-dimensional spectroscopy which allowed us to identify the individual protons in the DNA segments non-exchangeable protons we could monitor because in the normal case these non-exchangeable protons were not very difficult to monitor because of the extensive overlap of the signals okay but in this two-dimensional spectroscopy this came in the 1970s it became possible to study the non-exchangeable protons as well okay so this the theories of these ones we have already discussed we are not going to go into that so here I am showing you one particular two-dimensional spectrum this is a nosy spectrum the so-called nosy spectrum which of a particular DNA segment that is indicated here so a CGCG AG TTGT CGCG so this is a for teamer you have this two-dimensional nosy spectrum so you have the diagonal here see and this area is a as we discussed here this is the one-dimensional spectrum this belongs to the 2 prime 2 double prime and the CH3 groups see I also put the primary structure here once more to be able to identify which protons are where and then I have also shown here this H3 prime H4 prime H5 prime H5 double prime those ones appear in this area and next to that is the H1 prime and H5 this belong to the cytosine and here we have H8 H6 H2 this whole area belongs to the H8 H6 H2 these belong to the bases for the various bases okay now what does this tell you now these peaks here this peaks here this tells you about the base-base interactions so there is a base stacking and there is a base stacking you have two protons on top of each other and you will find adjacent base pairs you are showing proton-proton interactions through the nosy short distances and these ones will show you base to this region and what is this region this region is the H1 prime and H5 base to H1 prime and this is this goes to this area so this goes to the H3 prime base proton to the H3 prime correlations sequential correlations distances all of these depend upon which are the short distances here and these ones goes to the base protons to the H4 prime H5 prime H5 double prime and this goes from the base protons this area is to the 2 prime 2 double prime and the methyl so this is the very well resolved now pretty well resolved of course you also have all from the 1 prime to the 3 prime here and then 1 prime to the 4 prime 5 double primes 1 prime to the 2 prime 2 double primes you will see here then of course from the 3 prime you will see also the 2 prime 2 double prime and 4 prime 5 prime also to the 2 double prime 2 prime so and then here you have the 2 prime 2 double prime within that within the same base within the same sugar 2 prime to double prime these are short distances and therefore you will see NOE cross peaks between these protons so 2 prime 2 double prime is actually geminal so it is about 1.8 angstrom therefore you will always see this this is the very short distance so therefore you will see all of those ones in this area and you will see from the base protons the methyl are present here from the methyl you will see to the 5 prime 4 prime 1 prime and then to the other basis H8 H6 H2 so therefore the amount of information that is present here is quite substantial quite enormous therefore using this you can actually identify the individual proton signals of the individual residues so this actually made a big change here now here I show you the 2D nosy what sort of the things you will get the NOEs are dependent on the distance between the protons the distance between the protons and that is in the NOE intensity or NOE means I say in the cross peak intensity is here the various cross peaks which are present these are proportional to the inverse the sixth power of the distance between the protons therefore the distance is becomes quite important which distance is short which distance is long that becomes important in figuring out which peaks are likely to come and it turns out here I will show the distances here this is the nucleotide I this is the sugar ring of the nucleotide I and this is the sugar ring of the nucleotide I plus 1 these 2 sugar rings are indicated here this base proton this is another base proton these are in the Nt confirmation I indicated you the Nt confirmation this base proton comes closer to the oxygen this comes on the other on above the sugar ring in the sin confirmation this goes other way round this goes on that side so you will not be able to see that okay so in this confirmation you can see in the Nt confirmation you will see peaks from here to its own sugar rings own sugar protons but you will see also from the base proton one base proton to the sugar ring of the next one on the 5 prime end notice this is at the 5 prime end this base proton two short distances are there to the sugar protons of the of the nucleotide which is at the 5 prime end so this one to this you will not see from this proton this to this sugar you will not see that that is the long distance you will not be able to see those ones so first of all you will see within the nucleotide from the base proton to its own sugar rings okay and then from the base proton to the sugar ring protons of the residue adjacent to it on the 5 prime side therefore that is the important information present here if you look at this at each base if you look pick out this area here this is the base proton to the 1 prime protons this is the base to base distance the base to base protons cross peaks okay so this will tell you you can just simply walk from here one residue to another residue one nucleotide to another nucleotide following this base-base interactions base-base interactions you stepwise you go up from one base to the next to the next to the next and you can go up like that and the same thing you will figure out from this as well or this as well or this as well so you have from the base proton to the one prime of the same this of the same nucleotide and to the one prime of the nucleotide at the 5 prime end remember not to the 5 nucleotide at the 3 prime end therefore this provides a directionality while this one does not provide directionality this base to one prime and the base to the sugar one prime two prime or three these ones will provide you directionality of to the sequential walk which you may go through in the case of the DNA. So you will see two peaks from each base proton one to this one one prime and to the one prime on the 5 prime end okay now here is just a listing of the various short distances which are present in the case of RNA and in the DNA and these are important which are the short distances these are about 3.6 angstrom 3.6 and h6 to h2 prime these are the self within the same within the same nucleotide unit these are the ones which you will actually see. So these distances you see they are all less than 5 angstroms 3.6 4 3.8 and things like that therefore you will see all these peaks and the DNA these distances are roughly similar distances which are given as less than 2 angstroms are not valid this there is some errors there in those one that is indicated here unrealistically small due to the assumption of a rigid uniform geometry and the omission of taking by hydrogen items into account from these are generated from the fiber diffraction models. So therefore the ones which are less than 2 angstroms are not really meaningful so because there is a when the means less than the means there is a steric overlap and that is not practical okay but the other distances are other valid so you will see these distances they are in the range between 3 to 5 angstroms you will see all of these distances these are the sequential distances sequential meaning from c to g g to c c to g c to g see these are all the distances between 2 nucleotide units 1 nucleotide unit to the nucleotide unit at the 5 prime n so you will see all of those distances there here is an illustration here is an illustration of a particular portion of the spectrum you but take this particular segment here t g g c g g t in fact this actually forms a cotorplex but so far as the analysis is concerned one strand analysis remains the same in either case so you see here what is present here these are the H8 or the H6 proton of the base H8 or H6 proton of the base okay and H2 proton does not produce NOEs H2 proton is far away when the H8 is close to the sugar ring the H2 proton is on the other side of the purine ring that will be further away so you won't see NOE cross peak from the H2 protons to the sugar rings you will always see from the H8 or likewise you will see from the H6 okay now you see here the you see cross peaks from but every particular cross every particular base you will have two cross peaks one to its own one to the sequential for example if this is the the self peaks are indicated by the green and the sequential peaks are indicated by the red so for example if I start here so I get here from C4 self peak its own then I go to the sequence I get a sequential peak here now from here I move horizontal to find the base which produces the self peak which produces the green peak so this is base proton of C4 this is the H6 of C4 from the H6 of C4 I find a H1 prime to its own and then the H1 prime of the residue on the 5 prime end the 5 prime end means it will be G3 it will be G3 suppose I took this this is a 5 prime end so this is G3 I will see this correlation from C4 to G3 so C4 to G3 I will get here now I go horizontal to find the base proton of G3 so I get the green signal here this is the one which is the base proton of G3 and it turns out that G3 sequential peak is also close by it is also on the same one so in overlap here because this portion of the signal is in line with this G2 where I should go from G3 I should go to G2 so therefore from G3 I go here a sequential peak I go to the G2 so go to the G2 then I have a sequence this is the self peak then I get the sequential peak here from G2 this is to the T1 from the T1 though I go horizontal again to find the self peak and then T1 now how do I know I am correct here now you see once you reach the end T1 I do not have any other residue on the other sign therefore there will be no signal below the there will be only one peak there only the self peak therefore this confirms that what we have done here is correct therefore this is the T1 base and there is nothing below here there is no sequential peaks so you can actually continue in this manner once you have that so you go from C4 you go horizontal where do I go I go to the G5 this is the sequential peak from G5 to C4 this is the sequential peak now I go from here to find a red peak here I get the G6 so from here to here I go I go to the G6 then I go to the red one here I go to the G7 then I go from G7 I go to this red one here and go to the T8 so therefore from the T8 I have T8 to G7 G7 to G6 G6 to G5 G5 to C4 C4 to G3 G3 to G2 and G2 to T1 this completes the cycle this is the beautiful demonstration of how the nosy spectrum can be used to identify the individual protons in the individual nucleotide units I think I think we can stop here