nanoHUB-U Principles of Nanobiosensors/Lecture 2.8: First Passage and Narrow Escape Time
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[Slide 1] Welcome back and this is sort of the last of the lectures on the diffusion problem and in the next lecture we'll wrap things up. So far you have seen that we have understood a great deal about biosensors. What's special about biosensors at nanoscale simply by thinking about nanoscale as a geometrical object, how particles come to and they come and get attached to the sensor surface. Regardless of the method of transduction that has already told us what are the fundamental limit of sensing is. For example, if you're not willing to wait beyond a certain point then that already determines the minimum value of the analyte concentration that you can reliably report. Now we learned a lot of things from biology. I often made in the analogy from biology from the jellyfish example and others for example, about the droplet. Can we now based on the understanding of the sensor teach something to the biologist? That is something the goal of this particular lecture is and in the next lecture I'll wrap things up. The two important things I want to discuss today, one is called the first passage time and another would be called the narrow escape time. Both have significant biological implications.
[Slide 2] So the first question I'll be asking is that if you have a blind molecule within a cell, then how does it know where the exit door is? Think about a let's say a crowded movie theater and the lights have been turned off, it's very crowded and the question is there are a few doors. Assume that everybody cannot see and the question is how long will it take if you are in a given seat to exit by random walking through one of the doors? That's the question we're asking. And this would be a problem relevant either of the first passage time. You want to simply get out and you want to know on average how long it takes and I'll give you some examples. There will be more examples in your homework and so therefore you'll get a feel very quickly. The narrow escape time is a relatively similar concept. I will discuss that in the beginning of next lecture and then I'll conclude.
[Slide 3] When you think about a cell in biology one thing obviously, first thing that probably strikes most people is the complexity of the cell itself. But for the time being we're not going to focus on the complexity but rather I want to draw your attention to the holes that we have within the nucleus and correspondingly on the cell surface there are these holes which allows it to communicate with outside wall. So proteins from inside, they will randomly walk around in this environment and eventually get out through one of the doors. Other elements, proteins and other things that are created are those can also get out from one of the cell walls, for example. And the question is how long does it take? This is a very fundamentally important question for biophysical activity and we'll see that there's a very simple answer to that question based on the understanding of nanosensors that we have so far. Assume this is a physicist's cell that is very simple. We have simplified the problem and had made it an oval and we are asking a question that if a protein starts from a given location X naught location away from one of the doors and let's assume for the time being that we have one or two doors, then the question is how long does it take given the separation? Now of course it will not go out directly the black line, the blue line, the green and the red are that if you put four molecules each are going to take slightly different path and what we are asking that the first time that it sort of gets out then you say times T1, T2, T3 and T4 and you take the average, hence the name means first passage time. First passage mean it is sort of getting out and never coming back. That's the passage aspect of it. Somewhat even simpler electrical engineer cell would be something like this, a box, a couple of exit doors with a flux I2 and I1, the rate at which proteins are getting out. Total integrated volume of the protein is integrated over the volume is Rho dv we'll call that Q. Now remember this Q is not the same as the flow rate. This is by definition sort of integrated charge within the system. It's just an integrated number that we are referring to.
[Slide 4] So here is the difference between the problems that we have talked before and the problem that we are discussing now. When we talked about the settling time how long it takes for the biomolecules to swim around and then get collected by the sensor surface we simply asked this question that how long does it take for NS number of nanomolecules, let's say 5 or ten regardless of where this start on average how long it takes to be collected by the sensor surface? First passage time is slightly different. It asks a slightly related but slightly different question. It says that if a biomolecule is a distance R from one of the sensors, from one of the exit doors then how long does it take on average for this particle to get out through this door? Here there's no chance that anything will come back and also we are not talking about average time from all over the space but a fixed position within the sensor itself, within the cell itself. Now the narrow escape time is almost related but slightly different. Narrow escape time may involve if opening the door is particularly narrow that means only flux which comes vertically would be able to exit. If it comes at an angle then it will sort of bounce off and so narrow escape time is more difficult than the first passage time. Both are biologically relevant and therefore has great implications.
[Slide 5] So let me tell you how the first passage time works. It is very simple you will see. The diffusion equivalent capacitance will give a simple one line answer to this very complex problem.
[Slide 6] So the idea is this as I mentioned that previously we solve these two problems or these two equations. You remember the diffusion equation? And you remember perhaps the capture equation, capture and release equation and we simply asked how long does it take to capture NS particles regardless of the initial placement of the atoms, or initial placement of the biomolecules. And of course we solved these two equations you remember that Heisenberg uncertainty principle like analog of that rule 00:08:07,106 --> 00:08:10,586 that you had that allowed us to solve this problem easily. Now the problem that we'll be asking here or the problem that we'll be solving here is slightly different, yes? The molecule will still diffuse. I have the diffusion equation but what we'll be focusing on is the total number of particles that exist, dNTdt that depends on how many coming in and how many getting out through the door. Now we do not have anybody coming in. We are only looking at how many are getting out so FN will be zero and this rate of loss, NT our tau will be the rate of loss of this particle because nothing is coming back. There's no term like this and there's no finite capacity like this also because it can continuously get out. There's no receptor limitation either. This problem and this problem is not strange that they'll be related but we'll see what the relation is and how long does it take on average for this molecule to escape.
[Slide 7] Let's do a very quick calculation so that we can reformulate the diffusion problem. So what we wish to do is instead of looking at this diffusion problem point by point we want to integrate the whole density within the biomolecule density within the cell itself so therefore we integrate over the volume and call that dQ/dt. If you integrate over the diffusion equation then that will be equal to the flux that is getting out and the integral across the surface will essentially be equal to I1 plus I2, I1 plus I2 and so therefore you can immediately say the total amount of particle loss, the total particle loss from this box is directly proportional to the flux that is getting out. Now this is not rocket science. You remember that if you have a bucket with two holes in it the rate at which the water gets lost is equal to the flux through those two holes. That is essentially the essence of this equation and therefore if I know what the total "Charges" or "total number" of particles are at a given point and if I know that two fluxes that are coming out of those two holes then I can capture the first passage time or the mean first passage time. So that will be my goal.
[Slide 8] Let's consider even a simpler case. Let's say I have a single hole, so in that case I don't even have I2. Life would be simpler. So in this case the loss of particles from this box is simply given by I don't have anything coming in so it's I out is simply I1 and so the tau will be equal to the total number of particles that I have which is proportional to the volume multiplied by rho naught. Rho naught is density and divide by I1. Now I don't know I1 so how am I going to calculate it? It turns out that calculating I1 is not a problem. We have done it many times. You see if I have a spherical opening, hemispherical opening I1 is simply CD,ss rho naught minus rho S. Rho naught is the density here and rho S is the density at the exit point and let's say the density is very low because everything is sort of getting out and so therefore I1 is CD,ss for this hemispherical sensor or hemispherical exit door which is something we already know but in general you see the time mean first versus time is simply given by if you rearrange this, let's put this I1 in this expression and calculate tau. If we do so, we'll see that if we just knew the diffusion equivalent capacitance. The same diffusion equivalent capacitance that we had been talking about then this very complex problem of protein escape through a cell door essentially becomes an easy problem to solve. It is simply inverse of CD,ss. Does this result make any sense? I mean it's so simple you might think that you might be skeptical that it works, so let's work out my one example in some detail so that you feel comfortable. You'll be doing more on the homework and that will make things even clearer.
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[Slide 10] Consider I have a cell which has a-- its a one-dimensional cell meaning that there's an escape door here and the cell extends all the way to the other side at the length let's say L from the escape door all the way to the back of the cell. That distance is L. Assume that a protein is trying to escape from a position X. Now of course the protein will do a random walk, it doesn't know where the door is and even if it knew it's blind. There's no way to sort of guide itself to the protein itself or the door itself so it will do some random walk. Some will go back. The other one will go a little bit back and then it may go forward and so on and so forth and everybody will have a slightly different time. The question we are asking on average if you take ten thousand of those rate particles on average what will be the time for this to escape? Now our theory so far said that all I have to do is to calculate I1, the flux that is coming out from here to here. Now we have already solved this problem, right? We know that in a parallel plate capacitant configuration the steady state flux escape is simply CD,ss multiplied by rho naught. Rho naught is the height of this density. This is the rho naught is the bulk density and what, because this is the only way it can sort of get out there's no escape door in the back. What is the total number of particles, "Q" in the system? Well, this is simply area under this rectangle and this triangle. This is the triangle. This is the rectangle. Now why is it a rectangle? Why doesn't it sort of diffuse in this direction also? The reason is there's no escape door so there is no net flux. Had there been a slope in this line we'd have expected the flux to get away but remember there's no escape door so the slope must be equal to zero so there's no diffusion flux hence the rectangle. So I know Q and I know I1. Let's put them together to calculate tau. Once I have done that I essentially take these two quantities, divide it up, looks a little messy you can rearrange things a little bit but it immediately tells you that if you start from a given location X, on average this is the time you would need in order to get out. Of course if you had the diffusion was faster, if it was light molecule it will get out faster. That makes sense. If your X is farther and farther out from the exit door then of course it will also take longer. That also makes sense. Now this is more or less an exact result to the 0th order. Does my CD,ss base approximate formula give me the same result?
[Slide 11] The answer is yes. Let's calculate. So tau over V the CD,ss is my formula. All I have to do is calculate CD,ss. Now you remember just from before that NT also there's two statements. NT is proportional to the volume multiplied by rho naught when the cell is big and the escape time is given by inverse of CD,ss. The first statement you can check out because NT is actually equal to the triangle plus the rectangle and if you can rearrange it and if L is large then you can see this term will drop out and if the term drops out, it becomes Rho naught L, L is like the volumes 1D and rho naught that rho naught. So in D, the particle number for relatively large L is given by this formula, approximate formula. We didn't do too much wrong in making that approximation. What about the second one, this formula? I already know CD,ss. D over X. Remember it's parallel plate capacitor so therefore the inverse of it will be X over D. Does it make sense? Is this answer correct? That's an approximate answer. It turns out that from the previous slide if you got the exact result you would have found you could have rewritten this form and when L is reasonably large compared to x this term will drop out and the two results approximate and exact will be exactly the same. And this proves the point, you see the exact result has a correction linear to X so therefore it bends over but so long as you're close to the exit door the approximate and the exact one essentially gives you almost identical result. That's very surprising that such a simple thing would give you such a profound result.
[Slide 12] Let's make life more complicated because after all whoever has seen a one dimensional cell? I mean that's one the electrical engineer would love but or a biologist would certainly frown upon. Let's consider a slightly more complicated cell still very approximate. Here is the exit door, the sphere, the green sphere and the protein begins to a journey at a distance R from the exit door. Now of course the protein will not go directly from this point to the exit door but rather it will diffuse around them. Every particle that you put in, every protein that you put in is going to the exit door at a slightly different time with a slightly different path. So therefore all I have to do in order to solve this problem is to calculate that diffusion equivalent capacitance between this sphere and another sphere which contains the rate point at a distance R because if I could do that then essentially I'll be able to calculate how long does it take for the system for this protein to escape? How would I know that how would the affect that the volume will come in? Because after all it will take longer if it had bigger volume to sort of bounce around. Well we will see that is the total number of particles within the system that NT rho naught will, I'm sorry, rho naught multiplied by V that will contain the information. And just for simplicity we can also solve an exact problem of two concentric spheres where a particle or a protein is sort of released from here. We're asking the question how long it takes for it to be captured by the sensor surface. This problem can be exactly solved. This problem cannot be solved exactly but approximately solved but if the results are very close it gives us confidence that we are not too far away from truth.
[Slide 13] All right, so we know what the CD,ss is of two concentric spheres, one with A, one over A and the other hypothetical sphere where the protein started at 1 over r naught and so therefore our approximate formula say that tau over V is inverse of CD,ss and tau over V therefore I will just invert the formula and this V will contain the cell size. If you have a bigger cell size you'll see tau will be V multiply this so it will take longer for the molecules to get out. If your door is very narrow it will take longer. If your diffusion coefficient is small big protein trying to get out that will take longer and if you are far out that also takes longer. So therefore overall, overall this gives you the approximate formula. Here I could solve this problem exactly and you will do that in the homework. B is the outer radius of the "Cell" and once you have solved it you will have two terms when b is relatively big. The cell is relatively big and the protein is trying to escape from the nearest door. In that case you can make the b grow large. These terms drops out. The exact and approximate results essentially just match on top of each other. A very simple answer but it tells you something profound already about proteins, size of proteins, the size of the door and volume of the cell, how long that it takes for the molecules to get out of these doors. So I will stop at this point and we'll take a five minute break. After the break, we will start discussing the narrow escape problem, narrow escape time which is a related concept but it's slightly different and once we have done that we'll conclude this set of lectures with a summary.