nanoHUB-U Principles of Nanobiosensors/Homework 3: Sensitivity of Potentiometric Sensors Solutions
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[Slide 1] Ok, welcome to the homework session. Now this week we have discussed potentiometric sensors and in the beginning of course the the potentiometric sensors look very simple the biomolecules which have certain amount of the charge be it a DNA or a protein the land on they the transistors surface or potentiometric sensor surface changes the potential and you measure the changes in the current for example in the MOSFET or the change in the voltage for other types of potentiometric sensors but we soon found out that life is a little bit more complicated than that partly because of this-- sort of the dual role of salt and the salt of course was necessary otherwise the biomolecules will not bind I've told several times now and that's also true for antibody and protein and then there was but at the same time the salt essentially took away a significant amount of charge or potential away from the biomolecules and the MOSFET didn't have much left giving the logarithmic dependence as I explained for so salt was one the other one was the pH you know I mentioned that you could say that okay I'm not going to I'm going to be very careful and will not have any pH in the solution well you cannot have that because water by definition has a potential of hydrogen there's the pH of 7 which is about ten to the power of 7 moles they are ionized automatically and the salt and the pH sort of made things complicated but we analyzed it very simply as you saw and then we had a set of-- Set of discussions or a series of discussions on on the ways to beat this sensitivity limit so what I'd like to do in this a homework is to sort of amplify certain points a little bit more now the number of formulas in this set are there are quite a few homeworks and if you go through them systematically you know some them less difficult on the more difficult and then there are these biosensor lab-based simulation if you go through them systematically actually you should have a very clear understanding about potentiometric sensors, right? So let me walk you through some of the solutions which I think I didn't cover in the original lectures that much, I wanted to point out a few things about
[Slide 2] In problem 3.1 I asked you to see whether you can solve this following problem assume you have a the nanosphere sensor that has a certain amount of charge now this could be a protein for example in the salt solution but lets consider that you have a little sphere and a certain amount of charge is sitting on top of it so the biomolecules have landed on the sensors-- spherical sensor surface and the total amount of charge that you have on the-- on the sensor surface the spherical surface is Q. The important point I wanted you to-- see was that this potential that goes into the solution drops very very rapidly exponentially rapidly so the interaction is very short range that is the electrostatic field doesn't keep going if you had a sphere-- in empty space the electric field would keep going and it will gradually decrease as 1 over r but here the important point is that it decreases much much faster and I wanted to show you how that comes about if you know how to do the 3D of course you'd be able to do the 2D and 1D. So let's see how it works
[Slide 3] So by the way why we are so interested this short range because what happens assume that you have a spherical sensors and all the red molecules are sort of the biomolecules that have covered the surface and you'd have a certain amount of charge and we want to know that how does the potential decay away from this spherical sensor 00:04:56,740 --> 00:05:01,770 Now the important point is what we're going to show that potential is very strong very close by so when the molecules, other molecules come in salt molecules come in it attracts them very quickly very short distance away from this sphere but if you go little bit away you know look at this sort of isolated atom the isolated atom you wouldn't know that there is a sensor sitting here so this strong local interaction is something that I wanted to emphasize in this homework Alright, so the basic idea is that we have a certain amount of charge sitting on the spherical sensor, red spherical sensor immersed in the salt and of course if this is a positive charge the chlorine will-- if it's sodium chloride for example chlorine is negative and the chlorine atoms will come close by and the sodium atoms to be pushed away so therefore there'd be a small region where there will be more negative charge than the positive charge in the solution so that this positive charge within this sensor is fully compensated within a distance LD, and you may remember this D stands for Debye within this very short distance so let's see how that solution gives you this particular potential profile
[Slide 4] now I have already solved this type of problem in the class so therefore we'll just copy but I want to emphasize where you have to make a modification now when you solve this type of problem the presence of salt and charges we said that we have to solve something called a Poisson Equation and Poisson equation you have gradient squared psi has to do with the double layer charges with how many sodium atoms you have and solve for n minuses whatever number of chlorine atoms you have at a given location, r and we saw that if you have a boundary condition that at this point close to the surface, remember it's a planar-- planar sensor this surface the potential is psi naught then far out of course the effect of this charge completely be screened out by the salt and so therefore far out nobody would know that there was a positive charge here, positive blue charge here so the potential at x equals infinity will be 0 because at that point there is no effect of this charge now I already discussed that if you know the potential psi the electron-- the chlorine density would go up exponentially and the sodium ion density will go down exponentially depending on the potential so what you did, if you remember from the lecture that we inserted these two things and e to the power plus 6 and e to the power minus 6 the difference of that became sine hyperbole Now one assumption we made was that when psi is very small if the amount of charge is very small-- size is very small in that case I can make an assumption-- I can make an approximation expand the sine hyperbole, in the first order and essentially write the whole thing as psi divided by LD squared and LD squared had this complicated expression you can see-- it's not very complicated this is this constant divided by the other constant associated with psi when you-- expanded to the first order alright and then of course we solved this first order differential equation to get the solution-- the final solution and we did see that the potential drops very, very fast away from the sensor surface, exponentially fast from the sensor surface okay so this is easy let's see what you need to do-- if we--
[Slide 5] are thinking about in in 3D, remember it's a spherical sensor not a planar sensor so if you were paying attention in your college calculus courses then you might remember that the gradient squared in 3D goes as 1 over r 00:09:43,149 --> 00:09:46,660 and r psi second derivative with respect to r it is no longer just secondary derivitive psi divided by dx squared naught-- it is slightly more complicated in spherical coordinate but of course the right hand side depends only on every point so therefore the right hand side will not change but the left hand side would change a little bit now we have to solve these equations, this differential equation subjected the boundary condition just like before if you're far out the charge has already been screened by the salt and so therefore the potential will be 0 but the important thing is that at any radius a-- at any radius a, the Gauss Law says that the electric field at this point is simply 4 pi epsilon a squared, a is the radius and since the electric field is the negative radian of the potential, so therefore now I have both a boundary-- this is second order equation I need two boundary conditions so this is the one boundary condition and this is the second boundary condition things that fully specified
[Slide 6] and the way then one would one would-- solve this equation this is the full solution this is a trial solution you can try it out so one thing you can see that solves in general that differential equation in the cylindrical coordinate except when r is equal to infinity you see the potential has to be 0 now this term at r infinity equals infinity will decay but this term will blow up this first term because when you put r equals infinity exponential of infinity this term will blow up of course the solution cannot be physical in that case so therefore we must have A1 equal to 0, so which is good. that means this will-- this first term will dropout now how do I get the A2 well in order to get the A2, I'll have to use the second boundary condition take a derivative take a derivative and equate it to the total charge from the last slide and if you solve that you will get an expression for A2, this is how it works you see if you take a derivative of the equation there will be a psi here when you take derivative with respect to r and d psi dr multiplied by r, that on the one side and then from this side if you take a derivative you pick up a 1 over LD this psi then you can pull to the other side because you already know that expression and once you have done that you will see that at r equals A that is why I want a evaluate the gradient at r equals A, then correspondingly will lead to this particular-- this particular expression and the bottom line here is that the constant A2 therefore will have this-- these are all sets of constants you know A is the size of the molecule and correspondingly you will have an expression for A2 that you can plug it back in in the original solution now if you do that
[Slide 7] then that gives you the full solution so the bottom line is you see it's good to have some practice in solving this type of Poisson equation in different configuration because one of the things as I said the salt screening is very important and-- in potentiometric sensors and so this is one example-- why don't you try the cylindrical sensors and see whether you can calculate out this type of response in that case also okay one thing you see if the-- size of the spherical sensor was very very small in that case A would go to zero and when A goes to 0 of course this will go to 0 this will give me one and it will have e to the power r over LD and when r goes 0 very close by not too much salt around in that case you can see Q divide by 4 pi epsilon r, that will be sort of the charge associated--potential associated with the point charge, something that you'll learn in the college physics anyway this so the only thing that salt did was to add an exponential term other than that everything is exactly the same as you have seen in your previous courses so there's nothing surprising about about how this thing work okay so the bottom line is that this is this short interaction is very important for life in general because anytime you have charge you want it to locally interact very strongly with the neighbors bind things with the neighbor's properly but a little bit away the effect that this charge, it should not interfere with other things so therefore biomolecules in the solution salt really plays a very important role it not only screens away-- screening away a bad thing for potentiometric sensors granted but you see screening away also keeps neighbors fully separated so that they can go on with their own life in biology every-- every protein can do their thing, a neighbor can do their thing and if they are about one or two nanometers away they will not know even that the other is present so it's as if they're in their own world so this is this reason this exponential decay is the reason which allows them to stay in their separate world although they are physically in the same box.
[Slide 8] The other thing is a notion of Debye length versus the Bjerrum length and I'll explain what that is so this was problem 3.3
[Slide 9] Now in this particular case the idea was that Bjerrum length is a distance L, lB at which the potential becomes kT, kT over q what is kT over q, kT is the thermal fluctuation in the environment that you have so you see if we have two molecules like this and the potential begins to decay in between the point where it becomes kT over q you see if you try to go a little bit farther out then the bond strength is smaller than the thermal strength, if you break the bond right and if you're closer than that then the bond stays fine so one of the consequence of the screening is this definition that how far can the two bonds be apart before they break apart, so therefore I have to equate this potential for the distance r equals lB, at which it will become kT over q now you can see that this equation will be very difficult to solve, right? nonlinear equation I know kT over q, twenty five milli electron volts in room temperature but I have a lB here, by the way LD depends on the salt concentration you can calculate but I have lB on the exponential and I have a lB sitting on the denominator so therefore only iterative solution is appropriate in the 0th order let's assumes that lB is very small and very close to a, you know it's-- very small value, let's this assume that then we'll see how it works so then what will happen that lB minus a would be approximately equal to 0 because I'm looking very close to the surface this willgive me one and then this is also in the order of one Let's assume that so then here my life is simple I have lB a here equals 2 kT over Q and then I can get a value for the lB, lB is the distance, distance over which the potential has decayed to kT over q and you can see you can easily calculate this the interesting thing is that while lB is about 56 nanometers in air why because in air you have epsilon-- epsilon relative direct constant is one so therefore, it's 56 nanometers, it goes in long-distance but in water you see epsilon r is 78 and if you have 78 in that case it is only 7 angstrom so by the time that you have traveled about 7 angstrom away from the charge, because of the screening and because of the salt concentration-- and the water polarization and dielectric constant the potential has decayed to a value which is just kT over q that's why two DNA molecules are 1.1 nanometer apart because if they were apart little bit farther you could not sort of have an interaction this screening would have completely killed it right so that's why they are a certain distance apart and this calculation essentially allows you to see that what is the binding distance between are two charges in salt solution now of course you may complain that this is not an exact calculation true, true you all you have to do it now is if you wanted to make this thing self-consistent you have to take let's say .71 nanometer put it in here and then recalculate and then once you recalculate then you can go back and forth and refine it by the time you are done you can see the value doesn't change significantly here also you can do the recalculation put it on a computer or the recalculation on your own but the value will not change significantly, in this case it will in this case because the salt is screening in air but not in water so therefore this approximation is quite reasonable
[Slide 10] Alright, now We discuss in one lectured about the origin of the charge in a biomolecule remember I said why is DNA negatively charged by now you know right that DNA is negatively charged because its back bone-- its an acid first of all and deoxynucleic acid and so therefore it has this phosphorus backbone phosphoric acid residues that's where the charge comes from now this problem really talks about pH changes and the implication because this is something many even in many, researchers essentially incorrectly the report You know capture of biomolecules where all they have done is that they have changed the pH and changed the surface, surface concentration of positive and negative charges or they changed the charge of the receptor molecule incorrectly thinking a target molecule has come, let me explain. I think-- you will understand better if I show you the picture
[Slide 12] You remember that there are this target molecule, right? This a DNA, this a polymer DNA, ok. And then I had this target I'm sorry this is receptors and the target molecule is supposed to come in and bind to it and there was salt around and the point was that initially let's say you had charge, three charges. later on you'll have six charges and from this difference the sensor would know that a target molecule has been sort of bond together now what I want to tell you next is that even if you didn't have this target molecule this blue target molecule you simply had changes in the pH in the solution the green molecule, the charge of the green molecule itself will change and as a result, somebody sitting in here on the sensor side and not paying attention what is happening to pH will incorrectly think that the blue molecule has come, blue molecule may not even be there simply because of the change in the pH the green molecule's charge will change so you will have a false positive you would say that there is some disease but there is actually not, it's just-- noise. So let me show you how that happens.
[Slide 13] You may remember this particular plot that as a function of pH pH going from 2 to all the way to 14 or 0 to 14 the charges of the DNA essentially kept changing the green one is when it was not charged at all at high-- low pH value means very high proton concentration essentially there is no dissociation that's the green charge and the red one, when it has one charge single charge every backbone every base pair has one electronic charge and the blue one has two electronic charges and then this final piece here is the final piece three electronic charges per base pair. So let's see. So if you start with two so somebody has changed the pH from 2 to 8 so these are the two places so now look at 2 at 2 on the green curve you will see about 60 percent of the charges there's 60 percent of the DNA base pairs will be uncharged right? That's-- what this is the green curve. So assume you have a hundred base pairs in the-- DNA 60 of them are uncharged and 40 of them a singly charged. So therefore what is the total charge for a hundred base pairs? The total charge will be at pH 2 will be 40 q that-- will be the total charge. On the other hand if you change the pH to 8 we are all talking about just a receptor molecule now there's no target molecule. Now if you are here then what you will see that at pH equals 8 about 90 percent of them are essentially singly charged ninety percent of them are doubly charged so 90 multiplied by 2 is 180 and about 5 percent are singly charged thats on this red curve what is the total charge? 180 plus 5 185q, so simply by changing the pH from 2 to 8 my charge has gone from the receptor charge has gone from 40q to 185q, so a person who is not careful will assume as if a biomolecule has been captured with the total charge of 145q. Now granted that this is an extreme example I just wanted to make a point the thing is that even if you change pH from 7 to 8 there'll be significant amount of change in the charge-- of the receptor itself and unless you are careful you will actually just report that parasitic disturbance in the fluid that has got nothing to do with the target biomolecule so this is the reason why I gave this particular homework problem.
[Slide 14] Alright, now calculating the protein charge is a little bit more complicated so but this is sort of very important because why it's more complicated because remember it's a polymer but unlike DNA which has all the same charges all negative it could be 0, 1, 2 and 3 for DNA the issue with protein is that the protein has both positive and negative charges and four of them are positive, three of them are negative and correspondingly things are little bit more-- complicated but the good news is
[Slide 15] that there are these calculators and I have given step-by-step instruction it is sort of very important that once you do it once, I mean once you have gone through this exercise you'll never have any trouble, any molecule that you have in the Protein Data Bank this-- Protein Data Bank has thousands and thousands of proteins any protein that you're interested in most likely is there and so the lots of information so it's a very good idea this PDB is the Protein Data Bank you go there work through the problem and you will have a very good understanding for
[Slide 16] any protein what the charges would be. The formula that gets evaluated in the back I discussed in the class is something like this depending on the amino acid which are negative plus amino acid which are positive you add them up and correspondingly that gives you the total Q and for every amino acid the rate constants are all given.
[Slide 17] but the point I want to make I mean that's-- those things you can look at the solution these things are all there, the point I wanted to make was this very important point where the charge becomes charge of the solution at a given pH completely disappears what it means is that the protein has lots of amino acid some of them become negative some of them become positive but when you combine them all total charge is 0 and if the total charge is 0 potentiometric sensor cannot see it. So therefore anytime you want to design a potentiometric sensor first thing you have-- especially for protein, first thing you have to do is to do this calculation, go to the Protein Data Bank go through the steps, plot this curve and once you know the pH value where the charge is appears the point of zero charge your first goal would be to move away 00:29:04,070 --> 00:29:08,820 at least one pH value away from that point, either on the positive side or the negative side you cannot stay here you will not see anything but at same time you have to be very careful don't go too far out in here because if you go too far out in here yeah you have a lot of charges but the structure of the protein will change right because the positive and negative residues will not form properly and therefore when you're trying to catch it by-- hook it by this antibody, it will not fit properly and so therefore you want to stay close by this was the purpose of this homework hopefully you have been able to work through this problem.
[Slide 18] Alright finally I'll show you just one more example about how to beat the screening limit and one approach we said was this super Narst sensor. So I gave you an example, for example aluminium gallium nitride has a mobility of 2000 oxide thickness of 45, width of 1 micrometer and there is a corresponding silicon nanowire so this is the sensor and the silicon nanowire on that hand has a mobility 100 oxide thickness of let's say fifteen nanometers and width the-- 15 nanometer and I want to know what type of amplification will I get
[Slide 19] what type of amplification. So this is the silicon nitrite one, the wide one and this is the silicon nanowire the skinny one, the narrow one and we put, we always put the salt or the solution in the wide one that's very important to remember if you put it here your amplification will go the other side it will actually go, instead of amplifying, it will sort of degrade the signal. So you put, always put the solution in the big one that that's the thing to remember and what I said that, you know, the change in the current for the big one and the change in the current for the small one if you add them together the current should be 0. I mean that's how this detection scheme works and the final formula was something like that, the final gain is whatever is the pH change on this sensor would be amplified by this ratio of the capacitance and ratio of the mobility width and all. So the homework is related to that problem so let's see.
[Slide 20] So all we have to do is just put this value where is the mobility, mu1 is, I already had added there, 2000 and mu2 100 for silicon that W1 and W2, it was you know one micron which is a 1000 nanometer and 50 nanometers, so you get another ratio there length is about the same so channel length we'll not worry about it, channel length here is same and here we said the BDS their connected to the common drain so those at the same and oxide capacitance you have to be a little careful because oxcide is inversely proportional to or capacitance is inversely proportional to the oxide thickness. So when you're doing Cox1, it would be 1 over 15 and when you're doing Cox2 then this would be the corresponding 1 over what ever the thickness was. So you put it in you get about one-third. Put this whole thing in look at this value, this is 8 volt per pH. What is the normal limit if it were alone? the normal would have been 59 millivolt per pH. So you have a tremendous, tremendous gain in terms of pH sensitivity. So this is something that will be, can be very helpful if you wanted to have a super sensitive design.
[Slide 21] Finally are there was a BiosensorLab homework example there's only one reason, this is very nicely laid out and the solution has everything so you should be able to take a look I just wanted to point out one thing about this solution so this is pH, x-axis is pH and the-- y-axis is surface potential how much potential has changed when you have changed the pH a little bit for a given-- sensor? Now the important thing is that as you're going from let's say 2 to 8 remember the example that we talked about? the pH response is not linear, it begins to saturate towards the end. So what to the time I saying 59 millivolt per pH or 8 volt per pH, assuming this response is linear. Now of course if the response is not linear saying something like 59 millivolt per pH sensitivity makes no sense because that's just the slope and the slope will change depending on where you so you have to say at what pH was that slope. So the important thing is not to rely on that slope number all the time but rather do the full calculation BiosensorLab has the the ability to do the full-- full simulation. And so therefore in that case let's say you have decided to build your sensor at this point, this would be this local slope will be your pH sensitivity if we have decided to build something here then that would be your sensitivity, so the full pH response is very important don't rely always rely on this differential slope because that really doesn't give you the fully insight about how the sensor works right? It is a nonlinear device. Charge is coming in, pH is changing the surface concentration lots of equations playing playing its role so expecting that and then there'd be one number describing everything is not realistic right? But-- a tool like BiosensorLab can do the calculation for you and you'll be in good shape. Okay so this ends-- lecture--this potentiometric sensors homework associated with it now just want to make sure that this is one of the most important sensor that have been around so make sure you study this part very well because we'll come back to this again towards the end when we talk about genome sequences, right? Genome sequences that potentiometric based sensors. So if you understand this clearly all this pH response and all, then you will see that those two lectures would be crystal clear to you have no problem in understanding it all. Okay, alright I see you next time