 Now, let us come to this problem that I want to discuss today which is the cell signaling problem. So, if I just to remind you. So, we discuss this sort of last class. So, here was this problem that I have some sort of a chemical which the cell is going to sense ok. And it is trying to if it is an attractant it is going to try move towards that chemical if it is a repellent it is going to try to move away from that chemical and so on. So, we saw this H pylori movies which had we had this H pylori movies where we saw it moving either towards urea or away from H Cl and so on. So, this was the picture that I have this cell the cell has certain chemo receptors on its surface which are proteins which can detect and recognize the chemical. What I want to know is that if I have some chemical source far away at what rate is the cell going to detect these chemicals with the help of these chemo receptors. So, that is that is the basic problem that I will try to answer. So, let me write down the proper question. So, what I want to ask is that what is the rate what is the rate of rate at which signaling molecules which are like your urea or the H Cl in the case of pylori make their way to the surface make their way the surface of the cell or in other words let me say I want to calculate something which is like this rate. So, dn dt so, the number of signaling molecules coming to the surface of the cell per unit time that is what I want to ask at this as far as the cell signaling problems. So, we will make a number of assumptions and I will say. So, your cell could be whatever arbitrary complicated shape right for example, E coli is like this ferrocylinder we saw if you take an animal cell that is soft. So, it can take variety of shapes, but what will solve this for is this model cell which is a spherical cell spherical cell yes. So, let me write down my assumptions. So, I assume that the cell is a perfect sphere the first step to doing a physics modeling of anything is that assume it to be a perfect sphere, but it is not a very hard constraint to relax. Of course, if you take arbitrary shapes writing down analytic solutions becomes difficult, but you can always solve it numerically following the same technique. If you have some other shape for example, well defined shape with some symmetry for example, a cylindrical cell or something like that then you can again do it analytically the maths will just become a little more complicated. So, I will do it for the simplest case which is a perfect sphere and let me say it is a perfect sphere of some radius of some radius. So, that is my model of the cell let us say it has some m number of receptors m number of receptors on its surface which are distributed homogeneously uniformly on the surface of the cell. So, often we will see you will see that for example, even in this cartoon of the E. Coli the chemoreceptors are clustered at the ends of the cell as far as this calculation goes I will not take anything like that into account I will see that the receptors are sort of spread uniformly over the surface and there are m of them. So, because it is a perfect sphere and the receptors are spread uniformly on the surface let me make another assumption that the problem is very good. So, the concentration is spherically symmetrical spherically symmetric. So, the concentration of signaling molecules is spherically symmetric which means that if I write this concentration in general it is a function of the vector r right it might depend on which direction you are looking at and so on, but I will just say that if I make this spherical symmetric approximation it is just a distance away from the cell that is the only thing that matters. And I will assume that you have a source of signaling molecule somewhere which was your pipette in these pylori experiments far away from the cell. So, I will assume a far field concentration I will assume a far field concentration which is C. So, as r tends to infinity I will say that my concentration goes to C naught. So, this is then my thing that I have a cell over here of radius a far away from that at infinity I maintain a concentration of C naught using some apparatus that I have and if everything is spherically symmetric look what I want to solve for is the concentration profile in all of this space in between ok. So, I want to solve for the concentration profile and because I have made the assumption that it is spherically symmetric it only depends on the scalar r not the vector r and this. So, the signaling molecules are being maintained at a constant concentration here they are going to diffuse right they are going to perform a random walk through the space until they hit the surface of the cell and the surface of the cell you have these receptors which are going to absorb the signaling molecule and then respond to that by moving in some particular direction ok. So, the setup is clear. So, because they are diffusing I will solve the diffusion equation for the concentration of signaling molecules del C del T is equal to D Laplacian of C. So, here is my diffusion equation that I will solve. So, let me say that I want to solve it in the steady state. So, when everything has come to nothing is changing with time anymore which means that I will take del C del T is equal to 0 which means that the equation that I am looking to solve therefore, is D Laplacian of C is equal to 0 all right. So, in what coordinate system should I be solving this you can of course, solve it in any coordinate system, but the most appropriate choice given that this problem that I have defined I have a spherical cell and then there is a spherical symmetry is that I will solve it in the spherical coordinate system. So, I write down the Laplacian in the spherical coordinate system. So, what I have the is D into 1 by R square del del R of R square del C del R is equal to 0. So, here is the Laplacian in the spherical coordinates. So, this is what I am looking to solve. So, if I solve this what this means is that R square del C del R is some constant independent of R because del del R of that quantity is 0 which means that I can write down my I can write down my concentration profile as a function of the scalar distance R as sum A by R plus B right. So, if you solve this differential equation properly the solution comes out to be C of R is equal to A by R plus B there is let me put a minus sign ok. In order to determine the constants A and B what I need to do is that I need to specify the boundary conditions of the problem. Now, one boundary condition I have already talked about which is that the far field concentration is C naught that C as R goes to infinity is equal to C naught. So, that will fix one of the constants for me, but I need one more boundary condition. So, and that boundary condition will most naturally occur at the surface of the cell itself surface of the sphere. So, what I will do over here is I will make an assumption I will make an assumption that these cells are these receptors are sort of perfect receptors which means that whatever sort of signaling molecule comes to these receptors gets absorbed instantaneously and completely which means that what the other boundary condition that I will choose is that C at R is equal to A the surface of the cell is equal to 0 ok. So, this is an assumption it is called the perfect receptor assumption perfect receptor assumption and basically it means what I said that once you anything that comes to the surface of the cell gets absorbed ok. So, which means which is why that the concentration drops to 0. We will relax this in a few minutes and we will see what happens when we relax this constraint, but for the time being let me make this simple assumption that these cells are these receptors are perfect receptors ok. So, now let me I have these two boundary conditions let me now put them in this solution and see what I get for A and B. So, if I put that C at R going to infinity is C naught then what that means is that when R goes to infinity this term will drop out and therefore, what I will get is that B is equal to C naught. If I put this condition that C at R equal to A is 0 what that means is 0 is equal to minus A by R plus C naught which means that A is equal to R C naught right. So, sorry I made a mistake this is at R equal to A. So, this is A. So, this is capital A is equal to this small A times C naught. So, what I have if I substitute back this A and B into that equation is a concentration is a solution for the concentration at any R and that turns out to be C naught which is common 1 minus A by R right. So, it satisfies the two boundary conditions at R goes to infinity this comes back to C naught when R goes to A this C goes to 0 ok. So, satisfied my boundary conditions. So, this is how my concentration profile will look like given that I make this assumption of perfect absorbers. So, now from here I could ask that well now that I know the concentration profile I can ask that how many particles makes it to the surface of the cell per unit time. And to do that you can calculate first the flux the flux J of R at R equal to A the flux at the surface of the sphere the incoming flux remember the flux is nothing, but the number of particles per unit area per unit time. So, that from Fick's law is given by minus D del C del R at R equal to A right. So, if I take del C del R then I will get. So, let me see minus D into del C del R will give me 1 over R square. So, C naught by A by A square because I am calculating at R equal to A. So, this is my incoming flux the negative sign simply says that your particles that are coming into the surface and getting absorbed. So, this is my incoming flux and from there if I want to calculate the number per unit time the number of particles per unit time remember the flux is the number of particles per unit area per unit times if I want the number of particles that comes into the surface per unit time d n by d t that is nothing, but the flux at the surface times the area of the surface itself. So, if I do d C naught A into 4 pi A square. So, what I get is nothing, but 4 into pi into d into C naught into A right right. So, this is the number of part in the case where I have the simple diffusion going on and I have these perfect receptors. This is the number of particles per unit time that comes into the comes to the surface per unit time. So, d n d t is nothing, but what I did is minus j at R equal to A into the area of the surface itself. So, which is this 4 pi d C naught A. So, this is a very important limit it is called the diffusive limit it is called the diffusive limit and what it says is that you if you have sort of chemicals which are diffusing and then getting absorbed at some surface like this R equal to the spherical surface at R equal to A and if this process is happening simply by diffusion then this is the maximum number of particles that you can absorb per unit time at this surface ok. So, it provides a limiting case and it is called the diffusive limit. So, as we go along in the course you will see that this we will use this diffusive limit sort of again and again. You might say that you know we we have done this in the sort of more simplistic case which is that I have this perfect sphere and so on and most organisms will not be perfect spheres which is true, but what that means is that what will happen is that this pre factor that I have this 4 pi that will sort of change, but roughly it will be of this order. You will have some correction terms maybe which are of order 1 depending on what geometry of the cell that you choose, but roughly this is going to be your limit the fastest that you can absorb particles providing these particles are simply diffusing ok. If you have other active processes which are being driven and so on then of course, you can go faster than this, but if you are relying purely on diffusion for the signaling molecules to reach your surface of the cell this is the maximum speed that you can achieve. So, let me now just relax this constraint a bit of this perfect absorbance. So, let me say imperfect absorbance. So, it is good to have it is good to have a clear idea of what assumptions you are making whenever you are trying to model. So, that you can if you want you can systematically relax them one by one in order to go closer and closer to an experiment real experimental situation. So, I will not relax the others for now this is spherical and so on, but at least I will relax this perfect absorber approximation the imperfect absorbers. So, I will say that this receptors have some finite rate of absorption, it has some finite rate of absorption of signaling molecules which is let us say k on. So, it will absorb this signaling molecules with some rate which is characteristic of the receptors which I will call as k on and remember there are m absorbers which means that the number of molecules per unit time that gets absorbed the surface of the cell is going to be m if you have m absorbers each absorber is going to absorb with a rate which is k on and the concentration at the surface of the cell. So, you have some concentration at the surface of the cell these molecules will get absorbed at some rate in a given absorber and you have m of those absorbers. So, that is my rate of absorption. I need to calculate this of course, k on and m are some parameters which I say whatever the cell has some number of receptors it has some rate of intake and I need to solve for what is the c of a. So, now, unlike in the perfect receptor case this boundary condition no longer holds right at r equal to a this is no longer equal to c because not a these are not perfect absorbers. So, some molecules will stay there at the surface of the cell ok. So, in a situation like this how would I define my steady state intuitively before we write it down mathematically. I can say that I will reach I will reach steady state when the rate. So, here is my cell when the rate at which these refusing molecules reach the surface of the cell is exactly the equal to the rate at which this molecules get absorbed into the cell then I will reach steady state right make sense. So, the rate at which this get absorbed is this we have written m k on c of a. So, we have written m k on c of a the rate at which it reaches the surface of the cell is the diffusive flux times the area right. So, minus j of r into 4 pi r square or so, if I write in terms of the concentration j is minus del c del r. So, d del c del r 4 pi r square good. So, now, I can solve this I now have an equation for c a differential equation for c d del c del r into 4 pi r square is equal to m k on into c of a. So, I can integrate this let us see I can integrate this between a to r some arbitrary r. So, dc from c of a to c of r and then on the other side I have integral m k on c of a by 4 pi d dr by r square from again from a to r I have just written. So, dc is on one side all the r terms on the other side. So, dr by r square and then hopefully these are m k on c a by 4 pi. So, now, I can do this integration. So, this over here gives me c of r minus c of a this gives me minus 1 over r. So, m k on c of a by 4 pi d minus. So, let me just do the lower limit first minus m k on c of a by 4 pi d. I do this dr integration between a to r if you take the signs correctly and these are all constant m k on c a 4 pi all right. I can now apply the other boundary condition remember the other boundary condition is still valid that c at r going to infinity is c naught. So, if I put r as infinity in this equation c of infinity is c naught. So, c naught minus c of a c naught minus c of a this term will vanish if I put r equal to infinity right 1 by infinity will be 0. So, all I will be left with this term m k on c of a by 4 pi d. So, in this everything is known except for this c of a. So, I can find out the concentration on the surface of the sphere now right. So, I can solve for c of a that is going to be c naught by 1 plus m k on by 4 pi d. So, this is the concentration at the surface of the set when you have imperfect receptors which are in up taking the signaling molecules with that rate. You can check that whether this makes sense right. If this receptors were very good. So, if it sort of absorbed everything that came into it which means that this k on was very high right. So, let me try a few limiting cases. So, high k on. So, let me formally let me write m k on by 4 pi d a this object over here is much much greater than 1. If I take this limit. So, it is almost infinitely large if k on is like infinity then what will the c of a go to what will this equation reduce to 0 right which is my perfect absorber. If my k on was so large that everything got an absorbed I should get back whatever I had for the boundary condition which is c of a equal to c. So, that is good. On the other hand if these were very poor absorbers nothing got absorbed. So, very low k on right or equivalently this this object m k on by 4 pi d a was much much smaller than 1. Then what should I expect what should this c of a go to c 0 right. So, nothing gets absorbed. So, it just stays as whatever the concentration at infinity was right. So, then c of a goes to c naught. So, if this was very small then I can neglect this with respect to 1. So, it goes back to c naught ok. So, the limits work out very nice work out nicely as it should. If you are very good if you are perfect absorbers your c of a goes to 0 if you have hopeless absorbers your c of a just stays at the background concentration nothing gets absorbed ok. So, now that you know this c of a which is the concentration yes why not it can. So, it it up this you have the surface of the cell right where you have this receptor and the signaling molecule comes it can bind to this receptor and cause a cascade of changes which causes the cell to. So, but this thing is no longer free in solution ok. So, the signaling molecule is no longer free in solution it causes the chemical reaction by getting absorbed with the receptor. It can even get internalized in some cases ok what was I saying right. So, now I know this c of a which is the concentration of signaling molecules at the surface of the cell. I can just substitute that here to find out what is this new DND I know what was it in the case of the perfect absorbers. So, for this case it is then just m k on m k on c naught by 1 plus m k on by 4 pi d a right. So, for the case of imperfect receptors this is the rate at which signaling molecules get absorbed by the cell by the surface of the cell ok. The better the better the receptor is the higher k on the better this DNDT higher the number of receptors that you have higher m higher this DNDT and so on. Remember that this rate that we have calculated this DNDT that is going to be limited by this diffusion diffusive limit because this is the best that you can achieve because this is the case of perfect receptors. So, this DNDT is definitely less than this. So, what we can ask is that well if the cell sort of keeps on increasing the number of receptors that I that it has on its surface how efficient is that how better does it do. So, to do that what we will ask one question which is let us say how many receptors would this cell need. So, I will ask how many receptors would it need would it need to achieve half the diffusive limit. So, here is for the maximum speed the maximum rate at which it can take. So, I just say that well as long as I am in the order of magnitude is the same I have done reasonably well right. So, let me say that how many receptors. So, what would be the number of receptors that I need to achieve half this way. So, that is the question that I have asked. So, basically the DNDT that I am looking for the DNDT that I am looking for is half of this maximum rate. So, 4 pi DC A naught by 2 and I want to find the number of receptors m that will give me this rate. So, m k on C naught by 1 plus m k on 4th pi D ok. So, this question is clear I have imperfect receptors with a certain rate at which it absorbs the molecules which is k on. I ask how many of these imperfect receptors would I need on the surface of the cell to achieve this specific speed which is half of the maximum speed I could achieve. So, this is this you can solve. So, 4 pi DC A naught by 2 plus 4 pi D A. So, m k on C this I have put the this C naught and that is m k on C naught by 2 is equal to the right hand side is just m k on C and I want to solve for m. So, the solution for m is then just what m is what is m 4 pi D by k on 4 pi D A by k on right. So, this is the number this is the number of receptors that I would need provided the receptor has a rate k on at which it takes the cell has a size A the molecules have this diffusion coefficient D. This is the number of receptors that I would need in order to achieve half the diffusive limit this by itself of course, does not tell you much what is 4 pi D A by k on. So, what we will do is that we will put in some typical numbers now and see how many receptors does it translate to. So, let me see. So, let us say a typical eukaryotic cell. So, let me take A as some 10 microns right. So, my cell is a sphere of 10 microns which is roughly a cell size as we discussed. Let me take a typical diffusion constant which is again roughly let me say 100 microns square per second I will need a k on for which I use again. So, these are all order of magnitudes roughly in the same correct order of magnitude. So, 10 per micromolar per second and at some point maybe I will also need the size of a receptor. So, let me just call that something. So, typical. So, receptor is a protein molecule right. So, typically let me say a size of receptor let me call that something SR is some 10 nanometers. So, these are just order of magnitude estimates. So, typically eukaryotic cell, typical diffusion constant, typical uptake rate and some typical size. You can substitute this back here and find out how many receptors that does that translate to me. So, but before we do that. So, for it is so, this is a cell of 10 micron size right. How many receptors would you expect on the surface of this cell? Order of magnitude 100,000 million billion from our estimates of protein numbers and so on in the first class this is just a guess I mean this is 10 to the power of 4. Good we will start off with that and see. So, now, who is going to put in these numbers and tell remember you should put take everything in the right units. So, what is one molar? One molar remember is avogadro's number. So, 6 point let me just say 6 6 into 10 to the power of 23 molecules per liter. 1 liter is what is 1 liter in micron cubed is 1000 cc. So, therefore, someone 6 into 10 to the power of 23 by how many microns cubed 10 to the power of 15. So, 10 to the power of 15 per micron cube and I am talking in terms of micromolars. So, one micromolar is this into 10 to the power of minus 6. So, that is 17 15. So, 600 per micron cubed that is one micromolar. So, this is 600 per micron cubed. So, k on if I if I translate this let me do here that is getting crowded. So, k on if I translate that k on is 10 per micromolar. So, 10 by 600 into micron cubed per second right. Tell me if I am wrong if I am making any mistake anyway. So, that is the only conversion. So, then I can calculate what is the number of receptors ok. So, this is 4 pi into 100 microns square per second with diffusion into the size which is 10 microns divided by k on which is 10 by 600. So, 600 by 10 second per micron cubed right. So, the units work out micron cubed on top, micron cubed on bottom second. So, this gives me a pure number which is as it should. This is the number of receptors I am calculating. So, this is 4 pi into 6 into 10 to the power of 4. Where is 4 pi roughly 12, 12 6, let me say 75, 75 into 10 to the power of 4. So, 7.5 into 10 to the power of 5 receptors ok. So, the number of receptors that you would need to achieve half of this maximum speed maximum diffusive speed is of the order of 10 to the power of 5 receptors on the surface of this 10 micron cell. Is that a large number or is that a small number? So, how would I say that? What is it large in comparison to? Yes of course, it is a large number, but ok. So, let me say that I have this set right on which I have placed this 10 to the power of 5 receptors. Each of them occupy some area. I could ask that given I have these many receptors 10 to the power of 5 which is a very large number. What fraction of the membrane is covered by these receptors? So, what fraction of the membrane is free for other proteins? If this is taking up all my membrane area, then that is pretty bad that is pretty large right. I do not have anything left for any other proteins to do its job. So, that is one thing I can calculate. Another thing I can calculate is that well maybe what is the separation between given I have 10 to the power of 5 proteins, what is the typical separation between two proteins and how does that compare to the size of the protein itself right. If it is of the order of the if it is the same as the size of the protein which means I would need this protein sitting next to each other in order to accommodate these many. There are some measures that yes 10 to the power of 5 is large, but does it really fill up the cell. So, that again we can do. So, let us say let me do the area calculation. I have these many receptors. So, I want to let us say I want to find this covered membrane fraction covered membrane fraction which is how much area is taken up by these proteins divided by the total area of this 10 micron cell. So, the covered membrane fraction is something 7.5 into 10 to the power of 5 into the area of the protein each individual protein I will assume as 10 nanometer square. So, 10 nanometer square and on the denominator is the surface area of the cell which is 4 pi r square. So, that is 100 micron square ok. So, what is that? It might help instead of writing 75 I had some 4 pi into 4 pi into 6 into 10 to the power of 4 then I can cancel off the 4 pi space square. And nanometer square I need to convert to micron square. So, this will be 10 to the power of minus 5 micron square. So, good. So, how much is that? So, 4 pi 4 pi cancels off 100 this becomes 100. So, roughly of the order of 10 to the power of minus 3 right 10 to the power of 2 into 10 to the power of minus 5. So, roughly of the order of 10 to the power of minus 3. So, the fraction of the true I have 10 to the power of 5 protein molecules, but the fraction of the area that it covers out of the total area that is available to the cell this is like something like 0.001. So, it is still pretty sparse and even. So, it is a lot of proteins, but it is still pretty sparsely distributed on the cell, but even with the sparse distribution I have achieved half of my maximum speed. So, you can say that well instead of 10 to the power of 5 I will put 10 to the power of 6, it does not give you an order of magnitude estimate maybe you are at 50 percent it takes you to 60 percent it is not really worth it for the cell. Cell is already doing pretty well even with this sort of sparse distribution. So, actually if you look at you can also calculate the mean separation given these many it comes to around 18 nanometers roughly between two proteins and a typical protein is square root of this. So, it is 3 nanometers size So, 3 nanometer protein separated an inters proteins separation of around 18 nanometers. So, it is by all measures on the surface of the cell it is pretty sparsely distributed. So, if you look at the effect of receptor density. So, this is some particular bacteria which I do not remember. The molecule that is getting absorbed is the virus the colophage lambda k and this is the receptor density on the surface of the bacteria ok. So, when you receptor density is small as you increase the density of course, you get a lot of improvement in the absorption, but beyond a certain point it sort of flattens out. There if you put more and more receptors on the surface of the cell it does not really make that much of a difference to the to the absorption performance of the cell itself. So, that is one lesson the cell does pretty well even with a relatively low coverage of receptors as far as this diffusion limit goes. And then like I said of course, this is majorly simplified with various approximations one can do better. For example, if you look at the distribution of chemo receptors on the surface of the E coli you will see that there are large clusters at these two ends and very few chemo receptors over here. So, if you were to model the E coli as a sphero cylinder then you could take that there is nothing on this cylindrical walls, but there is a large density on this hemisphere and on this hemisphere and then ask that you know what sort of how does that change the calculation. So, realistic to model realistic organisms the spirit is the same whatever we have done for spherical cases you take the appropriate shape you can solve it numerically or in very few cases maybe you can solve it analytically, but this is the this is the crux of it. And this like we discussed this is also true not only for the cell signaling problem, but a similar approach would also work for this diffusion and capture sort of problems. If you are talking about microtubules where these t-bulin subunits are diffusing and are getting captured at the tip of the microtubule you could do a sort of similar calculation and ask that what is the rate at which these t-bulin units subunits will arrive at the tip of this growing microtubule. So, this is these are two concrete examples that I thought I will discuss one is PrEP, the other is the cell signaling next. So, we will continue with diffusion next class and we will talk a little bit about first passage properties of diffusion. All right that is all I have for today.