 recap of the assignment process in the case of nucleic acids and by and large we do this by the nosy experiment which we discussed last time in greater detail and here is just to show that once more this is a two dimensional nosy spectrum of a particular olivore nucleotide whose sequence is given here on the top. So this has three thymines here and are very well characterized by the presence of the methyl groups in the base. So here you have a methyl group in the thymine base and that is very distinctly seen at the top here. So that actually fixes the positions of the thymine residues. So and as you know this one is the area which corresponds to the base protons and the base protons the NOEs are there to the sugar rings these are the H1 prime protons of the sugar ring and then you have these are to the H3 primes and these are to the H4 primes 5 primes and then these are the 2 prime 2 double prime area from the base protons H8 and H6 to the base protons that is indicated here H8, H6, H2 but the NOEs come from the H8 and H6 you do not see so much from the H2 protons all of these are base protons and then you have here very characteristically these are the thymines. So therefore these fix these are the NOEs from the methyl of the thymines to their own H6 protons there are three of them here you can see them very strongly here these three and therefore this fixes the positions of the thymines here. So everything what you do should be consistent with that that actually provides a very good starting point here because you see I will illustrate this to you in this particular sequence here we saw this in the last time I will go over it again you start with the T at the 5 prime end then the T at the 3 prime end. So what we have seen earlier was that you will see from the base proton NOEs to the sugar protons of the same nucleotide and to the sugar protons of the previous nucleotide that is at the 5 prime end. So therefore this T does not have anyone on the top there. Therefore this thymine base will have only one connectivity from its base proton. Secondly we also know where are the thymines the thymine positions are fixed here in this particular sequence there are two thymines and the NOEs from the methyl to the base proton fix the two thymines to be here. So therefore one of the thymine is here other thymine is here. Now if you look at these two thymines one of them shows a sequential connectivity here to the 5 prime end therefore obviously this must correspond to this whereas this one does not show any sequential peak therefore this must correspond to this. Now you can start from either end of these ones and you can walk and ultimately you have to reach this place therefore you can see from T8 self you see to the next one G7 from G7 you go to its own base from the G7 you go to the G6 this is the sequential peak then it will go to its own base then it is G6 here G6 you go to it is the sequential one you go to G5 then from G5 you go to the C4 now that is here there is a sequential this is the self from C4 you go to the G3 and then that is the sequential connection here to this center here now you notice that this peak consists of two peaks here. So there is one sequential which is connecting the self peak is here and its sequential is also in this area and this connects to the G2 here from G2 you have you go to T1 and you reach here therefore you see the sequential walk is easily done in this entire sequence. Once you have this one prime identified you will also identify the one prime two prime the two prime two double prime and the two prime three prime peaks from the NOEs which are occurring in this area. So you have this one one prime are identified here you see NOEs to the two primes in the two double primes of the individual nucleotides and also to the three primes here and then to the four primes in the five primes here therefore once the one primes are identified you connect them to the one two prime and two double prime these are also seen in the COSY spectrum one prime two prime peaks will also be seen in the COSY spectrum therefore once you have the one prime identified you will have the two prime two double primes identified there you will also see the two prime two double prime cross peaks here and this we are clearly distinctly seen at a different place and then you will have from the one prime you will have the three prime four prime peaks which are present here in this area. So the whole spectrum therefore can be analyzed by once you get these one prime assignments here and you also have this base to base assignments here in this area. So having done these assignments now you have the full assignments of the entire nucleotide now you have to go to the structure. The structure as we said consists of the backbone structure and then the sugar ring structure. The sugar ring we have seen earlier is called the sugar puckering it is described by the pseudo rotation angle P and we have all the sugar puckers are classified into two categories north and the south which we described earlier the north corresponds to the C3 prime endo geometry which is the center of the north region and this happens in the A type DNA helix or in the RNAs and in the B type double helix you have the C2 prime endo geometry these are characteristically shown here what how do they look like the sugar pucker is in this in this way and then you have the glycoxidic torsion angle here which determines how the base is oriented with respect to the sugar ring there. So now we have to determine the sugar geometry we have there is only one variable here that is a pseudo rotation angle and how do we determine the sugar geometry what are our observables from the NMR point of view and that is the sugar ring as you know has many protons and all these protons are coupled J coupled. So you have this one prime on every position you have the protons here one prime two prime two double prime then you have the three prime the four prime and the five prime it goes like that. So far the sugar ring is concerned we have these couplings which are present here one bound three bound couplings and those ones are indicated here. So you will have the ribose ring we will have this coupling constants here I will show you the coupling constant a little bit more detail and this is once again the pseudo rotation circle. So you have the in the southern part this is the so called C2 prime endo region and this is the northern part this is the C3 prime endo region all other intermediate puckers which occur in this area. Now how do we characterize these sugar rings what are our observables the observables are the coupling constants the coupling constants are indicated here. Now the coupling constants depend on the sugar geometry what is plotted here how many coupling constants are there we have one prime to two double prime then one prime to two prime two prime to three prime two double prime to three prime and three prime to four prime. So we have five different coupling constants five different coupling constants which can be used to characterize the sugar ring and these ones depend on the confirmation of the sugar ring and that you can see from here these are plotted here the coupling constants vary with P in this manner. This big one here this one this is the one prime two prime coupling this peaks here at this about 10 hertz is the coupling constant here for the C2 prime endo geometry. For the C2 prime endo geometry this one prime two prime coupling is 10 hertz. So and we can write here some of these things so for the C2 prime endo for the C2 prime endo geometry you have the one prime two prime is approximately 10 hertz and you can look at the other ones the one prime two double prime one prime two double prime is approximately you can see which one is that that is the one which is below there so that is approximately 6 hertz that is here. So in this area this is 10 hertz this is 6 hertz and this one is my two prime three prime what is two prime two prime three prime this is the third one and that is approximately seven hertz okay so that is about 7 to 8 hertz 7 to 8 hertz and what about two double prime three prime two double prime three prime is this one okay we are talking here. So this is about 2 hertz or less than 2 hertz 1 to 2 hertz and 3 prime 4 prime 0 okay and these are very characteristic coupling constants and they will show up in the cross peak fine structures in this we are going to discuss. Now what happens in the case of C3 prime endo let me write here the C3 prime C3 prime endo geometry so you can see what are the coupling constants C3 prime endo means we are here we are here so here 1 prime 2 prime is 0 okay and all others are relatively larger all others are greater than others all others greater than 6 hertz in this area you can see this one is very small this is the C3 prime endo area the 6 hertz and this is about 7 hertz then you have 10 hertz among these the strongest ones are the 3 prime 4 prime this is the 3 prime 4 prime this is the highest value for 10 hertz and then the 2 double prime 3 prime is also around 8 to 9 hertz so that is this one here. So therefore these are very distinct coupling constants which will allow you to distinguish between the sugar geometries okay now how do they show up in the sugar geometry so we will have to calculate the sugar geometry fine structure for this therefore we will try and calculate that in the cosy we will have to use the cosy spectrum okay so fine structures in the cross peaks of the sugar ring. First of all let me draw the schematic of the cosy spectrum what all you are going to get so let us say I have this this is the C3 prime endo let us say this is why where are the 1 primes let us say this is the H1 prime then I have 2 prime 2 double prime here then I have the 3 prime here then these are the 4 primes 5 prime 5 prime so we will not go so up to the 4 prime one can go so what do you expect here from the 1 prime this is the diagonal okay so you have the 1 prime to 2 prime to double prime so you will let us say this is my I will draw that here 2 prime 2 double prime there are 2 here then I will have the 3 prime and this is the 4 prime here and then I have the 1 prime here so this is the characteristic now what cross peaks I will have of course the diagonal will be present here 1 prime to 2 prime to double prime because I will have I will have 2 peaks here schematically in the cosy spectrum so 2 prime to double prime of course you will have the cross peak between these 2 prime to double prime there this will be present now 1 prime 2 prime peak will it be present in the C3 prime endo geometry for a general thing all these peaks will be there but will this 1 prime 2 prime will be present will it be present so let us look at that so 1 prime 2 prime coupling is 0 right so for the C3 prime endo geometry this peak will be absent okay 2 prime double prime will be there now all other peaks will be there now 2 prime 2 prime to 3 prime peak will be there though so 2 prime to 3 prime peak will be there okay and 3 prime to 4 prime that is here 3 prime to 4 prime will be there this is also very strong peak okay so on the other side I am not writing this symmetrically they will also be present here so I am not writing on the other side so these peaks will be present you will have 1 prime to 2 double prime and 2 prime to 3 prime this is 2 prime to the 2 double prime to 3 prime will it be present 2 double prime to 3 prime will also be present and that will also be present here 2 double prime to 3 prime. And 3 prime, 4 prime, will it be present? Yes, that will be present and that is this. This is 3 prime to 4 prime. This is the diagonal of the 3 prime. This is the diagonal of the 4 prime. And these are the diagonals of the 2 prime and 2 double, 2 double prime and 2 primes there. So, these many peak, these 4 cross peaks will be present in the case of a C-thru-primando geometry. Now, let us do that for the 2-primando geometry. It is right here. C 2 prime and 2 geometry. Now, let me do the same thing here. You have the 2 prime, 2 double prime, then you have the 4 prime, then 3 prime, then the 1 prime. So, let us draw the same thing here. 2 prime, 2 double prime, 4 prime, 3 prime and 1 prime. Now, what peaks will be present here? For the C-2-primeando geometry, go back and look here. We wrote those ones here. What are the coupling constants? 1 prime, 2 prime coupling is very large. Therefore, I will see the 1 prime, 2 prime cross peak. What I will not see? What I will not see is I am indicating below the 3 prime, 4 prime cross peak will not be seen because that coupling constant is 0. See, this coupling constant is 0. 3 prime and the 2 double prime, 3 prime is also very small, 1 to 2 hertz. Therefore, you may not see this one also. So, what you will see? You will see these 3 couplings. These 3 couplings only will be seen. So, let us draw that here. So, therefore, I will have 1 prime, 2 prime coupling. This peak will be present. 1 prime, 2 double prime also will be present. And what about 2 prime, 3 prime? 2 prime, 3 prime will be present. 2 double prime, 3 prime will not be present. And 3 prime, 4 prime also will not be present. So, therefore, I will have only 3 peaks here. On both sides, of course, correspondingly, there will be symmetrically they will be present here as well. I am not writing on this side. So, basically on one side, these are the distinctively different kinds of patterns you will get in the 2 kinds of sugar geometry. By looking at these kinds of peaks itself, you will be able to figure out whether I have the 3 prime endowed geometry or the 3 prime endowed geometry which is pure. In the case of the pure geometries. But of course, if there are equilibrium mixtures of the 2, then you may find the peaks present for the other ones also. So, this we can figure out by actually measuring the coupling constants. Now, how do we measure the coupling constants? The fine structures contain this information. All of these peaks, they have fine structures and the fine structures will have the information about the coupling constants. So, how do we measure this coupling constants? In order to do that, we must be able to understand what are the fine structures in these peaks. So, let us take a particular peak, let us say cross peak 1 prime, 2 double prime. Let me consider one particular peak 1 prime, 2 double prime for a C2 prime endowed geometry, for C2 prime endowed geometry. So, we have seen this how to calculate these fine structures, but so let us do that once more here. So, let us say you have the H1 prime chemical shift here. This is the H1 prime chemical shift, this is the H1 prime, this is the single line, this is the chemical shift here. What are the coupling constants this will have? This will have H1 prime has 1 prime, 2 prime and 1 prime, 2 double prime. These 2 couplings are there. Now, if you are looking at the 1 prime, 2 double prime cross peak, now I will have an active coupling and a passive coupling, which is the active coupling here, because I am looking at the 1 prime, 2 double prime cross peak. Therefore, this is my active coupling and this is my passive coupling. What is the consequence of this? The active coupling will lead to plus minus splitting. So, I will put this split this here, 1 plus minus, this is the 1 prime, 2 double prime coupling. So, now we notice here in this 1 prime, 2 double prime coupling is of the order of 6 to 7 hertz. Now, 1 prime, 2 prime coupling is much larger, that is about 10 hertz. So, therefore, this will further split into 2 and they will go like this. But this since it is a passive coupling, I will have this plus and plus here and minus and minus there. So, this will produce me a structure plus minus plus minus. Let us look at the 2 primes, the 2 double prime peak. Now, if we draw the 2 double prime peak here, H2 double prime, what are the coupling constants for this? It has a 2 prime, 2 double prime coupling, 2 prime, 1 prime coupling and 2 double prime, 3 prime coupling. Now, this one has 3 couplings. So, this one has 3 couplings, it has 2 double prime, 2 prime, 2 double prime, 1 prime and 2 double prime, 3 prime. But 2 double prime, 3 prime is very, very small. So, 2 double prime, 3 prime is almost close to 0. So, therefore, we will not see that. So, 2 double prime has 2 double prime, 1 prime coupling, 2 double prime, 3 prime coupling which is small for the C2 prime endo geometry and the 2 prime, 2 double prime, this is a large one. This is typically of the order of 14 hertz, 2 double, this is the geminal coupling and this is of the order of 7 hertz. Now, therefore, what is the kind of a splitting here? So, now, let us draw this pattern here. So, what is the active coupling? The active coupling is this one. This is the active coupling. So, therefore, if I want to draw first the splitting due to the active coupling, this will be plus, minus and this will be of the order of 2, this is 2 double prime, 1 prime coupling and this will be further split into 2 because of the 2 prime, 2 double prime coupling. Therefore, this will be plus, plus, minus, minus. This is 2 double prime, 2 prime coupling, 2 double prime, 2 prime coupling. So, this also will be plus, minus, plus, minus. So, what will be the fine structure? So, therefore, I will have here plus, minus, plus, minus and here also I will have plus, minus, plus, minus. So, if I multiply this, how many components I will have? I will have 16 components. This peak will have 16 components. Therefore, I want to write that here inside plus, minus, plus, minus. Then I will have minus, plus, minus, plus. This will be plus, minus, plus, minus, minus, plus, minus, plus. So, 16 components, this is the kind of a structure that will be present in the 1 prime, 2 double prime peak. So, if you look at the 1 prime, 2 prime peak, of course, then you will have to consider differently. In the case of 1 prime, 2 prime peak, what are the, you have like 2 prime, 3 prime coupling also in addition. So, go to the 1 prime, 2 prime coupling, the same geometry, C2 prime endo. Now, I want to see what will be for the 1 prime, 2 prime peak. 1 prime, 2 prime. See the 1 prime, 2 prime. Now, what are the actual coupling here? Let us split that here. 1 prime, 2 prime coupling is very large. 1 prime, 2 prime coupling, this is the active coupling now. 1 prime, 2 double prime coupling will become smaller. That is the passive coupling. So, this will be plus, minus and 1 prime, 2 double prime coupling, which is smaller. So, therefore, this will be like this. Therefore, this will be plus, plus, minus, minus. Now, let us look at this is the H1 prime. Now, let us look at the H2 prime. H2 prime, once again I will write 1 prime, 2 prime is the active coupling. This is, what are the coupling constants here? So, in this case, I will have 2 prime, 2 double prime, 2 prime, 1 prime and 2 prime, 3 prime. There are 3 couplings. And the 2 prime, 3 prime is not 0. So, therefore, let us start with this big 1, 2 prime, 2 double prime that is the big 1 there. Let us say, this is my, this is, this is plus, plus because this is the passive coupling. I am writing with the plus, passive coupling there. And now what is this one? So, this one is my 2 prime, 2 double prime and then I will have 2 prime, 1 prime. Now, this is my active coupling. Therefore, this will be plus, minus, plus, minus. This is the 2 prime, 1 prime, 2 prime, 1 prime. And 2 prime, 3 prime is smaller than 2 prime, 1 prime. So, therefore, this will be again another passive coupling. So, now we see I will write here plus, plus and this is minus, minus. This will be plus, plus and this will be minus, minus. So, what you got here? You got 8 of them, plus, plus, minus, plus, minus, plus, minus, minus. So, 8 components. Therefore, this will have how many total in the structure? There will be 32 components, 4 into 8. So, here I will, suppose I want to write here. So, this is plus, plus, minus, plus, minus, plus, minus, minus. So, and this side I have here, plus, plus, minus, minus. So, you put that here. So, plus, plus, minus, minus, plus, plus, minus, minus, minus, minus, plus, plus, plus, plus, plus, minus, minus, plus, plus, plus, plus, minus, minus, minus, plus, plus, minus, minus, plus, plus. So, you will have 32 components and all of these coupling, these will of course, will have the separation there as we indicated here. So, which is the coupling? This is the 1 prime, 2 double prime, 1 prime, 2 double prime. So, by looking at these plus, minus separations, you can figure out what are the coupling constants in each one of those. But often you may not be able to resolve the peaks in this manner. So, what will happen? There will be cancellations in the middle, all pluses, minuses and things like that will cancel in the middle, then you may not be able to find all the peaks. So, in such a situation what do we do? When we do not have this, of course, you will have to calculate all of this. Similarly, you can calculate the fine structures for the 2 prime, 3 prime peaks and then the 3 prime, 4 prime peaks, you can calculate the fine structures for the individual sugar geometries. So, when we do that, this is an experimental spectrum of the C2 prime endo sugar geometry, which is shown here. Now, what you have here? So, you have the 1 prime, which I have already discussed with you. The 1 prime peak is present here. These are the diagonals here and then you have the 2 prime, 2 double prime cross peak present here. So, let me write here, this is the 1 prime and this is the 3 prime, this is the 2 double prime and this is the 2 prime. So, therefore, you see 1 prime to 2 double prime and 2 prime and then you see from, this is the 3 prime. So, 3 prime to 2 prime, notice 2 double prime, 3 prime peak is absent. This is what I showed you earlier, 2 double prime because that coupling constant is very, very small and then you of course, you have the 2 double prime, 2 prime peak. Symmetrically, you will also have these peaks here. Now, these ones as you can see have a fine structure. These ones have a fine structure. This is what I was discussing, explained to you just now and we will see how one can calculate these. Now, you see these are the simulations. So, since there will be cancellations in the peak intensities, you cannot do it by simply, all the components are not visible to you, all the components are not visible to you. You have an experimental spectrum, you put in these numbers, certain numbers for the particular sugar geometry. Here, what is calculated is for all the sugar geometries starting from the pseudo rotation angle, 9 degrees, 36 degrees, 72 degrees, 90 degrees and this is my C2 prime endo geometry. And we instead of see how many cross peaks are seen here, you only see 4 components. All the other components, there are lots of things which have merged, they have cancelled or merged and things like that. So, in this particular case, we see more components. If we see here, there is some cancellations here, some residual things are present here. So, when you come to this area, then you do not see these cross peaks at all, because this is the 1 prime, 2 prime peak. 1 prime, 2 prime peak, peaks around, the coupling constant peaks around 144. This is 0 here and 0 here. Therefore, you will not see the cross peaks for the C2 prime endo geometry for these values of the pseudo rotation angle. Now, therefore, these are the patterns which are calculated and simulated keeping in mind the resolutions that are there and the line weights. So, once you have those, once you get these patterns. So, experimentally what we have to do is, you will have to simulate and compare with the experimental spectra to estimate the correct values of the coupling constants. Now, this is a similar calculation, simulation for the 1 prime, 2 double prime peak. So, this is the 1 prime, 2 double prime peak here. Now, you see there are many more peaks which are visible here, because this one has only 1 prime, 2 prime coupling absent. 1 prime, 2 prime coupling is absent, but all other couplings are there. So, because of the 1 prime, 2 prime coupling is absent, you see this peak appears narrow along this axis for the various values, because at this point at 9 degree, the 1 prime, 2 prime coupling is 0. Therefore, at this value, 1 prime, 2 prime coupling is 0. Therefore, this side splitting is only 2, 1 prime, 2 double prime coupling. Therefore, this is narrow. As we keep increasing, you start getting contributions from the 1 prime, 2 prime coupling also. This actually becomes wider, you see this width becomes more, this becomes more, further more and it has the largest, you see here is the largest width here, because the 1 prime, 2 prime coupling is large and 1 prime, 2 double prime coupling is also available here. Therefore, this has the larger width along this axis. There are four components along this axis that you can see that here clearly. In fact, in this one, you can also see all the 16 components, so to say. Centrally, there are some cancellations at this point, because of the plus minus. And you keep increasing the value of the P, go over to C2 prime in the geometry and then you go over towards the end. Once again, it becomes narrow here. This is similar to this. This is again back into the North regime. North regime has the 1 prime, 2 prime coupling is going to 0. So, therefore, this is becoming a narrower one here. So, therefore, from the looking at the patterns of the cross peaks in the cozy spectra, you can actually calculate, determine whether it is the geometry is in the North domain or the South domain. So, I think we can stop here.