nanoHUB-U Principles of Nanobiosensors/Lecture 4.2: Selectivity of Nanobiosensors Physics of Sequential Adsorption
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[Slide 1] Welcome back. We're talking about the selectivity of nanobiosensors. In the last lecture I discussed one of the ways that incorrect binding between two DNA molecules can give you a false positive. And the main point of that lecture was that if you bind things correctly then it has a certain amount of energy that you need in order to break it. Incorrectly bound ones have a little less energy required to detach them, therefore if you can increase the difference by designing the DNA correctly, then by using the temperature as a probe, you can wash away all the incorrectly bound ones. Let me tell you about the second sensor selectivity problem that is again, fundamental, and
[Slide 2] we have to sort of work with it. Recall that we had this target molecule that we really want to capture, the red molecule, this is a correct conjugation, the first one, red molecule has been collected or conjugated by the correct sensor, which is the signal we desired, the red is a signal that we desire, it turns out that if you have a lot of blue parasitic molecules that incorrectly binds to the sensor then that gives me a false positive and this is something we discussed in the last lecture. Today I want to focus on the black, the black term and the black terms are the parasitic molecules which sort of sit on the sensor but doesn't conjugate with the target molecule or receptor molecule at all. But it is sitting there physically close to the sensors surface and therefore will induce a false signal in the sensor; something that we have to avoid. So what we would like to see is where are these empty spaces where the molecules can land. And again the goal is to reduce beta. Remember in beta I have this number of components, parasitic components and so we want to reduce this as much as possible. The first term, antiprime, we could reduce by enhancing the temperature and washing it away. Now let's talk about this black term, it is called angometry, how we can reduce it so that we can reduce beta and the noise term associated
[Slide 3] with it. So today I'll focus on the empty spaces. Let's say you have a sensor and this is your view and all those little circles you see everywhere, the little ones, those are the receptor molecules and the target molecule. And let's say they have been captured leaving behind a lot of empty spaces where,this is the top view so where in these empty spaces a parasitic molecule can just sit on top and if you have a biosensor underneath, the biosensor may think that it has caught a molecule, like a potentiometric biosensor and it may think that it has caught a molecule, actually this is not the right molecule to begin with, which is a concern. So what we like to see how much empty space you have and couldn't you simply eliminate the empty space by packing things densely? It turns out that you cannot. It turns out that no matter what you do you cannot cover more than approximately half of the total surface. The other half remains empty. That was a big surprise. Let me explain why that thing comes in. And by the way this is a game you can play with somebody as I will explain also in a second and convince yourself. So today I'll
[Slide 4] be talking about as a selectivity problem arising from random sequential absorption. So the molecules are coming down randomly one after another and sequentially getting absorbed on the sensor surface; the physics of it. And then I will derive a corresponding limit and show you some results
[Slide 5] before I conclude. Think about a periodic array. It arranges nicely. We all know that if the total amount of fraction of area covered is essentially pi r squared, the area of a blue circle. Divided by the diameter of 2r so the box that contains the circle is 4r squared. You divide pi over 4 is 78 percent. 78 percent if things are nicely arranged, 78 percent of the area is covered, 22 percent of the area is empty spaces, which is bad because other molecules can come in and sit and so therefore give a false signal. Of course you can immediately say no no, there is a way to do better. How do I do better? Well if I put it in as sort of a hexagonal lattice then I know I can calculate 90 percent of that approximately that area is covered; two dimensional area, we are looking at the top view of a sensor, sensor surface and these blue are the receptor molecules and there is about 10 percent of the area left. This would be very good if we could do that, arrange things like this it would be very good because there would be very little empty spaces where a parasitic molecule could sit and cause trouble. It turns out that if you throw things one at a time like a child putting pennies onto a two dimensional surface it turns out that if they sort of get stuck, this is the glue, they get stuck and then it'll turn out that if you try to bring anymore you cannot put it in because it doesn't fit and eventually in a steady state you will only have slightly more than half you can fill. No matter what configuration you try, no matter how you throw it, every time it will be a different configuration because it is a random process but at the end of the day when you are done in a large enough circle you can only cover about half. The remaining half will remain empty and susceptible to parasitic absorption. This would be a big problem.
[Slide 6] Where does this 54 percent number come from? Well even before that let me tell you that not only is this 54 percent but it will take forever for this to want to reach the 54 percent. Why is that? So what I have plotted here is a log of the fraction field, since it is a fraction I put a negative sign to say that this is the fraction of the molecule so this is like 10 to the power minus 9, a very low fraction and zero is 100 percent. Ten to the .1, ten to the .0 is 100 percent. And it turns out that at low concentration, let's say you have one nanomolar, in low concentration, even if you had relatively high reaction dominated, it will take many hours for the surface to reach 54 percent. If it were sort of dominated by diffusion, the molecules, the blue molecules coming from the top and slowly landing on the sensors surface diffusing through the molecule, diffusing through the fluid, then it turns out it will take even longer to reach this 54 percent. So reaching this 54 percent requires a long incubation time and if you do not wait until that time there will be a lot of empty spaces where the red parasitic molecules can sit. This is something we do not want. So this is something we want to avoid. So let me tell you the physics of this 54 percent, why so low? Why can't I make it 90 percent? By the way if you usea very high density then you can reach the 54 percent relatively quickly, so the capture probes that you have often you will see in the recipe to use the high concentration and incubate overnight, before you do the detection. This is why. So let me.
[Slide 7]
[Slide 8] Let me explain how the process works. Assume that you have a sensor surface like a car parking lot. Each has a slight slot and the cars that are the exact same size as the parking lot and so you can see as the molecules are coming down they will sit on this surface even if it is not successful, if the same molecules come here it will try again-- diffuse out and it will try again but you can see eventually the surface will be fully covered, 100 percent covered, no problem. It turns out that this is a fully ordered structure, 100 percent coverage is possible. If you can ensure that individual sites are exactly equal to the biomolecules. But biomolecules are relatively larger compared to individual atoms on the sensor surface so therefore a more realistic situation I will explain in the next slide. But in this particular one we can simply compute the rate at which the molecules are arriving. Let's assume this is a sticky surface so nothing gets out once it comes down, it gets stuck so this only has a forward association, N naught is the number of parking spots, so N naught N is the number of parking spots available at a particular time. And you can see the fraction that is available at a given time, will decay exponentially as a fraction of time. At T equals infinity it will be 100 percent covered, which is good as you might have expected. And of course that kf depends on the diffusion as we have already discussed the relation between arrival time and the fractal dimension so you can put it in and you can solve the problem. So if you are willing to wait long enough then everything will be good, in this particular problem 100 percent coverage is possible. If you are not willing to wait long enough or you accidentally if you don't wait long enough, you will see only a fraction of the surface will be covered and you will be in trouble because in places where things are not covered, parasitic molecules will come in and produce spurious signal.
[Slide 9] Now let's think about slightly real, more realistic situation. The molecules are larger than the lattice. The lattice is atomic lattice, two atomic lattices wide; the molecule is slightly bigger than the atomic lattice. Then of course, a completely different thing happens. Because you see it is true that sometimes they will be able to come early, they will be able to sit. But you see once these have, sort of these 4 molecules have landed, no molecule can land in here anymore because it doesn't fit. The car is bigger than the parking spot and so therefore this will not be able to fit so this space will be wasted. And similarly if it comes down here, although there is a space here, it will reflect back, diffuse around and then eventually it can come back. But there will be many places where there will be this gap and it will not be filled. Correspondingly, parasitic molecules can come in and cause us trouble. So let's do a calculation associated with how molecules can land on the surface and what fraction of them can remain, what is the ultimate coverage that can be covered in this condition, remember I want to explain where the 54 percent number came from, so that is what we are trying to get to. And instead of drawing this picture over and over again I will be drawing a slightly simpler picture. You will see the two blue molecules connected by the deep blue connectors that essentially represent this pair of molecules. And every time every once in a while you will see these empty whites, the empty whites are the places where the molecule cannot see, these are individual spaces and so there are a bunch of molecules which have sat in their respective places, leaving behind a bunch of white spaces and
[Slide 10] I want to know how many white circles do I have, on average on this surface. How do you calculate it? It turns out there is a very simple recipe. Assume that at a given time, you have a situation where you have this long chain and only three timers have landed. Now you have how many empty spaces? You have a chain, connected chain of 7 empty spaces; we will call that 7-tuplet. How many, but this 7 contains 6-tuplex because you have 1, 2, 3, 4, 5, 6, you have 6 here, and the 6 connected chain, so this 6 connected chain in one side displaced by one makes it 7 connected chains. 7. And it will have 3 you can check that if you want to have chains of 5 connected chains of 5, random 5 then you have 3 of them. In this particular case you have one 5-tuplet, 5 empty spaces connected, 2 4-tuplet because you have 4 and 4, and so on and so forth. You can then write the equation and the next molecule comes, how the number of tuplets changes as a function of time, so this is the 7-tuplet, When a new molecule lands, you see what can happen, it can either land here on the first two, right? On one end so a 6-tuplet is destroyed because now I have only a 5-tuplet here. Or it can land on the other side also so a 7 tuplet is destroyed, simultaneously destroying a 6-tuplet because I now have this 5, and finally the molecule can land anywhere between of the 7 and you can land here, it can land here, it can land here, so the number of empty spaces, it will destroy, all of them will destroy the 6-tuplet. And so we can write a differential equation for it and then solve it as a function of time, that will give you why the coverage is never perfect. So let's do the counting, assume m is 6, so I'm trying
[Slide 11] to count the probability that the 6-tuplet will be destroyed. This is equal to the probability that a 7, 7-tuplet is destroyed and the factor of two says that I have the 7-tuplet on the two sides, the first and second set, here 2-- here is one probability and here is one probability. So this is one way the probability-- this 6-tuplet can be destroyed is if the 7-tuplet is destroyed by a pair so that explains the second term and the first term simply means the number of places this timer can sit in the 7-tuplet chain, in this chain, and so therefore it is m-1, chose 1 because you can just sit in any of the m-1 places. All of them destroys 6-tuplet, easy differential equation, the solution is known; you can just put this solution in here. And that will give you immediately, if you put it in here it can give you for Dp1, Dt, the nonprobability you have a single isolated open space will be given by this simple relationship and once you solve for it you can see that eventually at the end of the day for up to a long period of time, 1 over e squared of the white spaces will never get erased because this is at the end of the day, there will be all tuplets sitting there but this empty spaces will remain and the number is 1 over e squared. So let me explain how they fill. So they fill
[Slide 12] first of all if it were direct, if there were no space restriction, then it will fill relatively quickly and eventually saturate, but if you have this type of sequential absorption, when they are trying repeatedly to fill this space, the saturation will be very, very slow, and I'm simply calculating this probability, 1 minus this, you see how it saturates. And you can also calculate how the tuplets, triplets, and all other big spaces, how they disappear as a function of time. This is the function, at which the doublers-- singlers disappear, doublers disappear and the triplets disappear in this way. Bottom line is that in all the spaces, a fraction of the space will remain unoccupied. So everything
[Slide 13] I did was in terms of a lattice, so in terms of a continuum, then what will happen is you will see every time you have an empty space you have two lattice sides which are empty and the number of white spaces, 1 over e squared so in terms of total amount of space it took up is 2 over e squared. These are not really wasted, the solid blue ones, and that will be 1-1 e squared, the rest of them, they have each one of them has one lattice space, it takes up one lattice space, So this is the total amount of empty spaces but the real empty are this one, the ones that are associated with whites, so this 2 divided by one over e squared. If you take the divide it out you will have 23 percent and so in 1D, in 1D, in 1D, no matter what you do you cannot get more than 76 percent and in 2D any more than 54 percent. This is why. This is the statistics of random sequential absorption, fundamentally precludes us from achieving full occupancy. So it turns out that this as I mentioned previously, that
[Slide 14] in 2D, at high concentration, it treats the 54 percent relatively quickly. If you use 1 milimolar target concentration, receptor concentration, it reaches relatively quickly, if you use one micro molar, then the blue line says that it takes quite a bit and as it lowers the concentration of the target molecule you have to include more and more and spatially with the diffusion limit, getting to 54 percent saturation concentration will take a very long time. Now are any of these experimentally verified?
[Slide 15] This is a very strong statement, that you will have 54 percent unoccupied areas where molecules can land in. It turns out that you can indeed do so, so on the left plot we have the receptor molecules randomly decorating a planar surface let's say, and you see that at a given time, you can see that all these empty spaces where parasitic molecules can come in and corrupt your signal, you can try to fit a circle in this empty space so you can see what fraction of the space is really empty, these big red circles simply saying which fraction is really empty. And you know that at saturation, this 54 percent would be occupied by the blue and remaining will be empty. But we have not gone to full saturation yet, at full saturation the gap will be no bigger than individual biomolecules, at full saturation therefore things will be considerably better, but we are not there yet, you can see large empty spaces in various places. And you can compare with experiments for example by looking at a surface with different polyheliglychol, different target receptor molecules then you can indeed show that the experimental simulation are very close. This is indeed the physical phenomenon, that leaving behind a lot of empty space when you are decorating the surface with receptor molecules is a fundamental problem associated with biosensors.
[Slide 16] And it turns out that that is why one often puts a lot of small molecules, here, so once you have decorated with the receptor molecules you can then put lots of small molecules, like covering the whole surface with grass, so the parasitic molecules cannot sit in between and that is often done in order to improve selectivity and we use this parasitic binding. So this is one way to do this, in that case, the biomolecules actually come to this forest and the receptors cannot bind
[Slide 17] and they will be on top of this blue grass, and that way the molecule, the parasitic binding would be considerably reduced. The other technique people have used is to use a sort of flat foot, a big foot so in that case they sort of touch each other leaving behind this smaller empty space and that way also as well, one can reduce this problem with empty space and parasitic molecules binding to it. So let me conclude then about this selectivity issue.
[Slide 18] And the selectivity issue we already mentioned is a key concern for potentiometric and cantilever based sensors. Why? Because in both cases these are in one case a camera for the charge, in another case it's the camera for the mass, in that case, even when a parasitic molecule sits on the surface in the empty space where there is no receptor molecule to begin with it thinks there is a real molecule there and as soon as it thinks there is a real molecule it will immediately see it can produce this spurious signal. One thing that came to a big surprise to me is that covering a 2D surface or even a cylindrical surface is very difficult. It turns out that the coverage, if you cover them one at a time, as the molecules are diffusing and coming to the sensor surface that you cannot reach more than 54 percent, that is the fundamental mathematical limit. And so therefore, and even reaching this 54 percent takes forever because coming through diffusion and the diffusion it will try to land once, if it cannot land in the right place it will go into the fluid again and then diffuse and try to find another spot. So therefore it is like a person blindly trying to find a parking spot and it repeatedly getting hit by other preexisting cars in the parking lot. And so therefore it takes a very long time especially toward the end to as you are reaching the 54 percent limit to essentially find a place and to get toward the saturation limit. So this is why it is very important to do long incubation times, whether you do a nanosensor, nanobiosensor or some other sensor, it is very important that you incubate the surface, otherwise, if it doesn't have enough receptor molecules, then parasitic molecules will bind and you will say that I have detected a signal, actually it is false positive and should not be counted as a proper nanobiosensor technology, And the most important point of these two lectures that I want to emphasis is that selectivity is very important. It can be highly sensitive, you can detect femtomolar, you can have a very miniaturized sensor which can be implanted or embedded in various configurations. But if you cannot differentiate between the target molecule, or if the empty molecules, if you allow the empty molecules to sit without passing with a grass or peg or other molecules, then you have a technology which is not really very good and cannot be used. So I hope this is clear message that you will remember. In the next lecture I'll say that let's say you cannot solve this problem directly, this is fundamental, let's say and the 54 percent coverage or this DNA binding, you cannot really come up with a new DNA structure that can solve the problem. What do you do in that case? Are you doomed? The answer is no, when all else fails we will tag it, we will amplify it and filter it. And that is the topic of the next lecture and until the next lecture, take care.