 So, it is this entropy that we will be concerning ourselves with today. So, here is a very simple calculation to get started. So, here is RNA polymerase this example that we were talking about today. This RNA polymerase binds on the DNA it translates along this template DNA stand and it produces the same RNA. So, if you again if I am just concerned about the binding of this RNA polymerase to the DNA, I can approximate. So, I create a cartoon in my mind of this process that this DNA has certain regions which are binding regions for the RNA polymerase right. So, given by this dark red regions. So, if I think of this DNA as a one-dimensional lattice there are certain sites where the RNA polymerase can come and bind. This is true not only for RNA polymerase, what could any DNA binding protein? So, it is a generic model for a generic lattice model for a DNA protein complex ok. So, an RNA polymerase or some protein can come and bind on certain specific sites and maybe you want to calculate what is the probability of occupancy. So, what fraction of these binding sites on the DNA will be occupied by an RNA polymerase? We will do that, but let us first do a simpler problem that what is the let us say I have some. So, remember the free energy is E minus T s. So, let me for the time being first consider just the entropy part, I will come back to the energy part later on. Let us say I have this one-dimensional DNA. This is of course, an approximation in that I am forgetting about the sequence specificity and so on. I am just saying that certain sites can bind my RNA polymerase. So, let us say I have n sites that can that are binding sites and I have n p of these RNA polymerases. What would be the what would be the number of microstates? So, in how many ways can I arrange this n p molecules among these n available binding sites? Yes. Why a permutation? So, for example, almost correct. So, let us say I have an RNA polymerase, I put the first RNA polymerase somewhere here, I put the second one over there 1, 2, I put the third one here right. This is indistinguishable from the fact that if I have put the first one here and maybe the second one there right because these RNA polymerases it is difficult to distinguish which each individual RNA polymerase. All you can see is whether in site is occupied by an RNA polymerase or not, which means that this p is not the right thing, but it becomes a C right. And then the entropy is of course, Kb log of omega this clear right. If you have n sites in how many ways can you choose to fill n p of them? What is this entropy? Anyone? This is simple two state problem right. If you remember your classical statistical mechanics, you have some filled sites, you have some empty sites, what is the corresponding entropy? Or even if you do not remember the expression, how does this entropy behave? As let us say a function of n, n p. If I were to plot this entropy as a function of n p, so I can calculate this of course, I can do Kb log of n factorial by n p factorial n minus n p factorial right. You use the Sirling's approximation, you assume that n is large, n p is large and so on and you can reduce this hopefully to a recognizable form. But this is fairly elementary two state system in stat mech. Anyone remembers how the entropy looks like, how the entropy behaves? Maximum at the center, whereas some Shogatha is saying like this, it will fall like something like this ok. So, how that they tell me why a maximum at the center when n p is n by 2 ok. Shogatha, why will it decrease? Let us say I had these winding sites and I wanted to I had only one RNA polymerase right. So, this one polymerase I can place in any one of these n ways. So, my each of them is a configuration. So, my omega is n right. Say if I had 2, so this is omega 1 number of states say if I had omega 2, the first one I could have placed in n ways, the next one I could have placed in n minus 1 ways or something like n square right. So, if I take the log of this which one is going to be larger right. So, similarly if you had many many many many RNA polymerases, let us say you had equal number of RNA polymerases that you had sites, you again come back to very few options because if you had equal exactly equally where only one way which is everything is filled ok. So, this is the right this is the right behavior of entropy, it is very low when you have no molecules, it is very low when you have very high molecules, it is exactly it is maximum exactly at n p equal to n by 2. Now, anyone does that make you recall the formula for the entropy you know where it is. So, this is the this is what this is the micro state sort of counting which we have done. So, you should again go back and recall. So, if you do this Stirling's approximation on this, what you will get is something like n kb right log of n p by n n p by n log of n p by n plus n minus n p by n log of n minus n p. Does this make sense? Actually you might as well just spend time. This is log of n factorial by n p factorial n minus n p factorial which is n log n minus n and that is this minus n p log n p plus n p and then minus n minus n p log n minus n p plus n minus n p right this Stirling's approximation. So, this n will cancel with this n p and n minus n p. So, what you are left with is n log n minus n p log n p minus n minus n p log of n minus n p right and then you just massage it. So, you can break this n up into n p plus n minus n p right log n and then the remaining two terms you keep the same and then if you divide this. So, you take n p common then it becomes log n minus log n p. So, this is n p log n p by n and then this n minus n p common. So, plus n minus n p again log of n minus n p by n and then if you take an overall n outside then you divide by n throughout and this is a standard two state entropy right. So, if in some sense you call this the linear density of polymerase molecules or whatever molecules that you are interested in if I call this quantity c. If I call this like a c then this is c log c plus 1 minus c log 1 minus c right which is what we have. So, n k b c log c plus 1 minus c log 1 minus c and that looks if you plot that entropy that will look exactly like this the maximum number of microstates that you will have available is when n p is exactly half half of n. So, c is equal to half that is the standard two state entropy. So, you can use these sort of very basic static concepts to get and handle on what are the entropy or what are the number of microstates that a binding process like this would have in a biological context. We have of course, done many simplifications we have not taken any specificities we have forgotten about the actual 3D structure of the DNA and so on, but even so even with these we will see how good these approximation these type of simple calculations do ok. I will come back to that, but let us look at for the moment related problem of this ligand receptor binding ok. Again this is generically of the same class of problems that we are thinking about it is a binding problem. So, I have two protein molecules ligand and a receptor. So, for example, this is a hemoglobin which binds this oxygen and the same group at the center binds the oxygen. So, the oxygen would be the ligand the hemoglobin protein would be the receptor. Similarly, this is a different protein molecule this is the PD 1 ligand protein this thing over here this curved protein over there is the receptor ok. So, you have a large protein generally which we call as the receptor and a smaller protein which comes and binds on it which we call the ligand. So, I have some large protein let us say like this which is my ligand and a smaller protein comes and sets on it which is my receptor ok. And then depending on whether this protein is bound or not you can perform different functions inside the cell. So, again let us say what I want to ask is that given a certain concentration of these ligands what is the probability that a receptor is bound to a ligand ok. So, let us say let me define the problem. So, let us say I have a certain concentration of ligands certain concentration of ligand molecules and I want to ask that what is the probability let us call it P bound that a receptor will have a ligand bound to a receptor has a ligand bound to it. So, what we will do is again we will make a very simple approximation we will consider this sort of a lattice model ok. So, here is my receptor molecule that is sitting over here and I have these ligands floating about in solution ok. So, we will consider a lattice model of the solution basically I will say that each microscopic volume is something let us call some V box and these ligands will occupy one of these boxes ok. Each ligand will occupy one box and then we will try to calculate the entropy and from there we will try to calculate what is the probability that this receptor will be bound to the ligand. So, let me first define again this. So, my solution for the solution I will approximate it as a lattice model. So, n boxes each of volume let us call it some V box ok. That is my model for the solution. Let us say I have l number of ligand number of ligand molecules number of ligand molecules ok and I have a single receptor molecule a single receptor molecule. So, exactly like that cartoon ok. Now of course, so when a ligand binds to the receptor it sort of lowers its energy it is energetically favorable for this ligand to come and bind to the receptor right. So, let us say a ligand in solution has some energy ligand in solution ligand in solution has some energy let me call it epsilon solution and like a bound ligand has some bound ligand has a different energy let us call it epsilon bound which is lower than this epsilon solution. So, it lowers its energy by binding to the ligand. So, that is what it would like to do energetically, but again as we discussed entropically it is going to be favorable for it not to stay bound it would like to stay in the solution. So, energetically it is favorable to be bound entropically it is favorable to be unbound and then what it will actually be or what is the probability of finding a bound the receptor with a bound ligand will depend on this competition between the energy and the entropy ok. This is clear the at least the setup is clear that ok. So, what we will first do is that we will we will calculate the partition function. Remember the partition function is the sum over all states e to the power Boltzmann factor e to the power of minus beta times the energy of that state. Now, let me say broadly there are two states one state in which the ligand is bound to a receptor and another state in which the ligand is unbound. So, let me say the energy of the bound you will have an energy of the bound state and an energy of free ligand state. So, if you have a ligand bound to the receptor ok what will be the energy of the system as a whole yes or let us say let us talk about this first if no ligand is bound to the receptor what would be the energy l into epsilon solution right all my ligands are free in solution I have l of these so l into epsilon solution. Now, if I have this bound state where one of these ligands is bound to the receptor what will be the energy of that state epsilon bound plus l minus 1 epsilon solve right one of these ligands is bound the remaining l minus 1 are in solution ok. So, at least I know these two states but there are many ways to get these states right. So, if I consider in terms of these bound state and the free state then I could say that what are the multiplicities of these states right what is omega b e to the power of minus beta e bound plus omega free e to the power of minus beta e free right where this e bound and e free we have calculated. So, what is left to calculate is this omega b and this omega f what is this omega b going to be or what is this omega f going to be. So, in this omega f I have these l ligands right which I have to arrange among these n available boxes of the solution. So, that is mcl right ncl and that is why and therefore, that will give me the entropy effect what would be omega b nc ncl minus 1 right one is bound. So, that does not have any freedom anymore you have the remaining l minus 1 ligands which are still in solution therefore, ncl minus 1 good. So, once I have that I have this partition function in principle. So, my partition function said is n factorial n factorial by l factorial n minus l factorial that is the free. So, then e to the power of minus beta l epsilon sol plus n factorial by l minus 1 factorial n minus l plus 1 factorial e to the power of minus beta eb minus beta into l minus 1 epsilon sol. This is my full partition function. Yes yes well not really right because that would not change between these two states between the bound state and the free state you would have some factor of course, which would encapsulate the receptor mobility, but that would that would be the same here versus here right. So, effectively it would not make a difference as far as this type this question is concerned. Ok. So, it is the same. So, this is my partition function. Now, if I am interested remember I am interested in what is the probability that the receptor has a ligand bound to it. So, if I ask what is p bound now that I know the partition function can you tell me what is the probability that the receptor has a ligand bound to it? Yes, Somya. Which one by which one? First one. It is the second one divided by the partition function. These are all the states that have a receptor bound to a ligand right. Therefore, the probability that you have a receptor bound to a ligand is this n factorial by l minus 1 factorial n minus l plus 1 factorial e to the power of minus beta epsilon v minus beta l minus 1 epsilon sol divided by that whole thing divided by z right. Now, this looks bad right, but we can make a few we can try to simplify this by using sterlings and some. So, let us say we are working in the we are working in a sort of thermodynamic limit where n is much much larger than l, but l is also much much larger than 1 ok. You have lots and lots of ligands, but equivalently you have lots of much more of free space as well right. So, n much much greater than l much much greater than 1. So, for both of these n and l terms I can use a sterling approximation. So, that is what I will do. So, let me try to use it on I have time. So, let me try to use it on the partition function. So, let me see. So, z is so that expression that I have written there. So, let me say I will take n factorial is common l minus 1 factorial is common and then n minus l factorial is common right n minus is common and then e to the power of minus beta e to the power of minus beta l minus 1 actually let me just take beta l. So, then in the first one in the first term of the partition function all I have left is a 1 over l right l into l minus 1 will make give me that l factorial. In the second term what I have is 1 by n minus l plus 1 right which into this will give me a n minus l plus 1 factorial and on top I have e to the power of minus beta epsilon b minus epsilon solve right. So, that this and this together gives me the l minus 1 epsilon solve this is the minus beta epsilon b. So, let me call this as some delta e this is the amount that you lower your free energy that you lower your energy by by by ligand when ligand binds to the receptor. So, let me call this as my delta some delta some delta. So, delta e is epsilon b minus. So, this is whatever is outside is outside n minus l plus 1 plus l into the power of beta delta and what I want is. So, over here was n factorial e to the power of minus beta l epsilon solve by l minus 1 factorial n minus l. Let me pull this one outside. So, n factorial by l minus 1 factorial n minus l plus 1 factorial then what I am left with is n by l. So, n by l minus 1 plus 1 by l plus e to the power of minus beta delta of the signs right. Why I wrote it in this form is that ultimately I want to put that z over here. So, I just want to get the pre factor is the same which is n factorial l minus 1 factorial n minus l plus 1 factorial which is n of course, I am missing this e to the power of minus beta l epsilon. So, now if I substitute if I put this back over there in this p bound expression I will get rid of this pre factor completely. So, p bound is going to be here again I will get e to the power of minus beta delta e right anything else beta delta e and on on the bottom I will get whatever is over here. So, n by l n by l minus 1 plus 1 by l plus e to the power of minus beta delta e ok. This is just I have just manipulated it I have got rid of this pre factor completely. I have got rid of this epsilon sol and epsilon b separately and just written everything as delta i the difference in pre energy difference in energies between a ligand and solution and a bound ligand. So, let me now multiply and divide by this term l over n. So, l by n e to the power of minus beta delta e 1 minus l by n into plus 1 by n plus l by n e to the power of minus beta delta ok. Now because I am working in this limit that n is much much greater than l is much much greater than 1. Let me neglect these terms in comparison to 1 right l by n is a small quantity 1 by n is an even smaller quantity. If I do this so in this limit what I get is my p bound is a much simpler expression for this p bound the probability that I will have a ligand bound to the receptor which is l by n e to the power of minus beta delta e divided by 1 plus l by n e to the power of minus beta delta. If you want to write in terms of concentration the total volume of the box is of course, n times v box remember each a was. So, if I multiply and divide by v box and v box ok, this is the concentration of ligands the number per in the whole volume ok and same here v box v box and let me say see I call c naught as some concentration which is a reference concentration whatever that is and then the this concentration of ligands c is nothing, but the number of ligands in this whole solution n by v box ok. If I want to write in this concentration language I might have left it here, but if I want since I asked as a function of concentration. If I want to write it in this concentration language this is c by c naught e to the power of minus beta delta e by 1 plus c sorry c by c naught e to the power of minus beta delta. So, given a certain concentration of ligands the probability that you will have a receptor which is bound to a ligand is given by this it depends on the concentration of course, but it also depends on how much pre energy how much energy you lower by binding to the receptor molecule right. So, for example, if I were to plot it if I wanted to plot this probability the probability that the ligand will be bound this p bound as a function of the concentration of ligands ok. So, if there was no ligand of course, then of course, the probability is 0 because there is nothing to bind. So, I know it will start off from here ok. If the concentration was extremely high then what would the limiting probability be? If the c went to infinity what would the probability that the receptor would be bound be 1 right. So, I know that at very high concentrations it has to go to 1 and then you can plot this and see how this goes. So, it will go something like this. This is for example, for some given value of delta e let me call it delta e 1. If I now had a greater delta e. So, for example, if this ligand receptor binding was even stronger right it lowered its energy more by binding to the receptor. So, if I now wanted to plot it for a delta e 2 where this mod of delta e 2 is greater than mod of delta e 1 which way would the curve shift would it come below or above above right. So, if I wanted to plot it for a delta e 2 then at the same concentration of ligands you would get a higher probability that the receptor is bound ok. This this this sort of a curve has a name this is a very famous equation for binding it is called the Langmuir adsorption isotherm it is called the Langmuir adsorption isotherm adsorption isotherm. This sort of a function x by 1 plus x it is also called a hill function it is called a hill function with a hill coefficient of 1 with it I will explain what it means. So, generally you can have hill functions or functions where you will have something like this raised to the power of n ok. In this case n is 1 which means this this n is called hill coefficient which means is a hill function with hill coefficient 1. Later on as we do as we continue to do problems like this you will see that in cases where you have cooperative binding where the fact that you have bound one ligand means it is more likely that you can bind another for these multivalent receptors where you can bind multiple things you will see that you will get different values of n where. So, n is in some sense a measure of cooperativity. Anyway I will come to that later. So, that is the idea. So, we have stachmic wise we have done very simple things writing down this partition function and calculating the probability should be very easy, but at least from there we have managed to get some sort of input sorry output which tells me how this probability of binding is going to depend on the concentration of ligands right. And this is yes can only go till n of course, yeah I mean, but n has is going up to infinity remember this is the thermodynamic limit yes, but this to a certain extent that is true that is true yeah let us say whatever it will ideally depend on. So, if you look at these sort of P bound plots exactly what we were plotting right. So, this is exactly what we said this is how the P bound goes as a function of the concentration of ligands these are for different values of delta E the higher the delta E. So, remember delta E the sign will always be negative because epsilon B will be smaller than epsilon solution, but as long as the so, if the magnitude of this delta E is larger. So, green, blue, then red and that is how the curves will rise up. And in fact, this is indeed of in many binding experiments you can find curves like this. So, for example, this was a drug trial where you were trying to ascertain the efficacy of two different drugs binding to a certain receptor molecule this is in some pharmacological journal. And you will see that these drugs this this is the binding probability the number of these complexes that you form l r l is the ligand r is the receptor as. So, this is the this corresponds to the probability of binding as a function of this ligand concentration. And you will see that the curves look exactly as would be predicted by this. In fact, what you can read off to a certain extent if you assume that this sort of approach is correct from these curves you can read off what would be the value of this delta E how much would the energy be made lowered by the binding of drug A to the receptor molecule versus drug B to the receptor molecule. So, although this is a very simplistic approach we have made many many approximations even so, the overall the nature of these nature of these plots that you get from this simple calculation will often correspond to these binding binding experiments for different ligands and different receptors ok. Of course, what you will be getting out if you want to get out a delta E of these experiments you will not really get the exact delta E at a molecular level, but some effective delta E which is going to be represented within the approximations of this sort of approach.