nanoHUB-U Principles of Nanobiosensors/Lecture 3.8: Amperometric Sensors - Glucose Sensors II
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[Slide 1] Welcome back. So we are -- if you remember we are now talking about amperometric sensors. We had five lectures on potentiometric sensors. We got worried about screening, and so we are now using amperometric sensors, and already in the last lecture we saw that the response is linear. The current response is linear, and -- which is very good. Which didn't happen for the potentiometric sensors. Now I did not complete the study about the amperometric sensors. I just told you half of it. Now the other half.
[Slide 2] So I will briefly give you an analysis of the enzyme-assisted reaction on the platinum surface because that gave us the selectivity, and is really a very important point of glucosensors that can detect -- allows very selective detection of the biomolecules. But what, you would like to emphasize that we will then again have to come back about this diffusion-limited response. And I'll explain what I mean by that. And then I -- then we conclude. By the way, so there will be a big formula for the diffusion bottleneck, but since we have discussed them in detail from the Lecture 4 through 12, so I will just summarize many of the results. That's why you spend so much time in sort of establishing the formula in a clean way. So the derivation will be in an appendix. Send me a note if you have any questions about the derivation stuff.
[Slide 3] So we are talking about the glucosensors, and very briefly you may remember that we have glucose coming in. There is an oxidase -- glucose oxidase which is helping the reaction between glucose and oxygen. Gluconic acid is coming out. Hydrogen peroxide is going to the sensor itself and catalyzing a reaction, putting a -- in response to this 350 millivolt electrolysis of this hydrogen peroxide, giving us this current. And we discussed this part in the first lecture, that why this current is proportional to hydrogen peroxide. You remember, right? The Butler-Volmer equation that allowed us this -- and had this connection between current and the concentration of hydrogen peroxide. Now let's focus on this side now because we didn't talk about this in our last lecture.
[Slide 4] So it turns out that -- or so therefore, in terms of broad scheme of things we are -- now we focused on this part in the last lecture, and we are coming back and want to discuss this enzyme-assisted reaction. This is a very important reaction, this type of reaction. And so therefore, I thought that it would be important that we understand this because we'll -- later on when we talk about selectivity we'll return to this point again.
[Slide 5] So this is how it works. Assume that the glucose is coming in. We'll call that a substrate. That's from the biochemistry literature. Or S, or simply the target molecule because that's what we are after. Now remember that this molecule on its own, if it simply landed on the platinum surface nothing would happen because platinum surface wouldn't recognize glucose. Instead what you need to do is this glucose oxidase will sort of hold the molecule in place and allow the reaction to occur so that hydrogen peroxide comes out. Hydrogen peroxide will be our product, and E is the enzyme. So if you wanted to describe this chemical process, how would you do that? You'd say that my enzyme reacts with my target molecule to produce a temporary compound. This is the role of a catalyst, right? It just sort of holds it together in a particular shape so that a certain reaction can proceed. Now after this temporary product ES has formed it can go in one of the two ways. It can go downhill to really create the product, which is the hydrogen peroxide, leaving the enzyme behind so that it can participate in a second reaction if itself doesn't get consumed. Or what might happen that it may not be able to go forward, that it essentially goes backward and falls apart, the enzyme. And the target essentially turns to the original value. Now how do you write an equation for this? Well, this is for college-level -- differential equation course. It should be very easy to write. You see, if you wanted to know how much target molecule do you have left? So you would say that, okay, that depends on how many has been consumed by the enzyme itself, E multiplied by S. That's the product, so more you have E the more you have S, the faster it will go forward. But of course part of it can fall apart. This kr is giving me the reverse reaction. And if you also have a straight concentration or the target concentration is relatively high, you can almost -- this forward and reverse are almost close to each other. So you can set this approximately equal to 0. We'll do it in a second. Now what about the fate of the enzyme? Well, many things are pulling enzyme in different directions. One is that every time a target molecule binds to it the free enzyme is gone. And so I have this with a minus sign the first term, and every time essentially it either falls apart, goes back proportional to kr, or goes forward successfully. In both cases I have my enzyme back, here E or the E over there. So that would be added to it. And if I have the catalyst, which is the forward part, then I would also have the E back. So that's why these two terms are positive. And finally the product. Well, product simply depends on at the rate at which this is getting dissociated. And that depends directly proportional to the number of joint molecules you have. It cannot go directly from the target to the product without the intermediate step, assisted by the enzyme. Can you solve these equations? Shouldn't be too worried; it's relatively simple. You see, what you have to do is realize that these two are approximately equal to zero. That's an assumption. If you don't like the assumption put it in the computer; it will be done. Just simple equations, but let's try it with some approximation. Let's assume that it's approximately equal to 0. So therefore, you can take this equal to this, and so therefore, in for KS we can substitute with the enzyme multiplied by the target product here. No problem. Then you realize that the enzyme catalyst must be conserved. It is either in the bound form, ES, or it is free. Doesn't matter. Whatever number of enzyme you started with, let's say you started with 100; 70 may be bound, 30 may be free. But it will always have to be 100 because otherwise it cannot -- enzymes are not consumed. And so you can immediately see that ES 0, from here you can replace it with the -- with this particular reaction which is set to 0. And so that will connect E naught with E. And if you have E naught equal to E then we are done.
[Slide 6] Because then it will immediately give us the solution product which is directly proportional to the number of enzyme you have. More enzyme you have faster the reaction. And equal to the analyte concentration; that also makes sense. And so the rate at which the reaction -- the products will form is given by this particular reaction. And in this particular case you can see that the maximum is directly proportional to the maximum velocity or maximum rate of formation of the product and is directly proportional, then, through that to the number of enzymes you have. So if you increase the enzyme by a factor of 5 presumably your reaction rate will be faster. Also if you increase the analyte concentration to a higher value that will also increase the rate. And this km essentially said how bad is the reverse reaction compared to the forward reaction? We want faster forward reactions so that reaction -- the whole process can go fast. And so therefore, smaller the value of km better off we are because then v will be equal to -- approximately equal to v max, because you see the two Ss will cancel, and we'll have maximum product formation if the reactions are essentially one-directional. And so therefore, you can -- for any particular system you can start with some analyte, which is S. The enzyme is the blue curve. Start with a certain amount of enzyme. Part of it goes and -- free enzyme goes away because they are bound as ES, so there is a dip. But eventually when the reaction is done everybody comes back. So the blue at this end and the blue on the other end are exactly the same. And you can see the reaction by products are gradually coming up. And that one is essentially presenting the depletion of the analyte is being reflected in the gradual growth of the product. So essentially the assisted reaction is all there is to it.
[Slide 7] And so one important thing, by the way, in this particular case is look, if you didn't have any enzyme glucose is not going to become hydrogen peroxide. This is the sort of the gatekeeper. And this gatekeeper is very important, otherwise that is where the selectivity comes from. Now this equation is often written in a slightly different form. The -- different people use different forms to do this. So there is this form which is line Linweavver Burk form, which is that simply inverted. If you invert it 1 over v and then it essentially becomes a linear equation. So if you plot 1 over S, inverse of the analyte concentration, and 1 over the rate, right, then it will be a linear plot. The slope will give you the -- these two constants, km over v max. And the intercept will give you v max. And so therefore, you can get both these constants if you simply do the experiment for a number of different concentrations. Once it is done then -- so this is very low concentration; 1 over S is sort of infinity. So your first very low concentration point is here. As your concentration of the analyte is increasing -- glucose concentration is increasing, the -- gradually going to the right, this height gives you the maximum velocity of the intercept, and the slope gives you through this the value for km. Now you could be telling me, "How do I measure the velocity of a product? Do I have to go in then follow every atoms on their back?" Fortunately not because that would be very painful. What you have to do is to realize that if you multiply with q on both sides then this is -- what you have is really a current, 1 over SS is the steady-state current. The number of electrons coming out through the electrode is a fair measure of how fast the reactions are proceeding. And therefore, from this plot, I over 1SS, the steady-state current, as a function of the analyte concentration, you can get the same curve. And so therefore, this is a relatively easy experiment to do. And here are, for example, some experimental values. You can see that most of the time turns out to be pretty good approximation, this linear plot of 1 over v as 1 over S.
[Slide 8] So let me give you an example. So this is taken from a recent paper. So what you have is the molecules are being -- the electrodes are being prepared. So this is, by the way, the working electrode. And once you have sort of created the basic working electrode structure, look at this enzyme. So enzymes are being decorated. This will be the gatekeeper, the E naught, right -- the E naught value. So if you have 10 to the power 10 per centimeter squared here then that will be the value of your E naught. That will facilitate this reaction. And when the glucose comes in, then of course the reaction starts, and the current will flow. Remember the counter electrode and other things, I have not really shown them here because that's sort of a given that there has to be a second electrode. or even a reference electrode. But I'm just focusing on the electrode -- working electrode itself. And once again you can see if you go from -- every time you change the concentration look at the current. Remember, I multiplied q multiplied by v. The velocity is multiplied by q in order to give me the current. And therefore, the current is increasing. And in principle when you plot the current as a function of concentration you find very good correlation, and you can also plot 1 over I which is the current -- inverse of the current with the inverse of this concentration to get the constant that you need. Okay, so this is very good. So we understand how a glucose sensor work, and all it took us was probably 40 minutes. Last class 30 minutes and maybe 15 minutes now.
[Slide 9] This is good because the concentration was high. Do you remember my block test? If not, go back to the -- maybe the first or second lecture. And I showed my block test for a very good reason. I wanted you to remember that in glucose concentration was on the millimolar range. And so therefore, I didn't have to worry about any diffusion. The molecules are all present. The single drop of blood immediately gave the glucose sensor the response. It was reaction-limited. And I told you about reaction limit. But most of the time, like nitric oxide or DDT in water, in that case the solution will be very small. And then the -- sort of the ugly head of diffusion limits will come back in and change everything that we know. And that's what I want to tell you next in the next few minutes.
[Slide 10] So this is the basic reaction. I assume that the glucose is plentiful, being catalyzed, producing hydrogen peroxide. Two places diffusion limit could come. One is glucose is very small, and so therefore, it takes a certain amount before it comes here because it's being consumed fast-- by the electrodes. Or this diffusion process of the hydrogen peroxide itself could be slow, so there is one diffusion limit on this direction. That is something we knew. But the problem is that there is another diffusion limit, that this hydrogen needs to get out quick because if the hydrogen sort of hangs around close to the electrode then it will start the reverse process. And if it starts the reverse process it will diminish my current because this sort of parasitic reverse current which I don't want -- I want R to O. I don't want O to R, the reverse current, to go. And the linearity will be compromised, you remember. So let's take a look at how the diffusion limit could come into play.
[Slide 11] On the very left what it is showing here is a singular electrode -- working electrode. That's -- stands for WE. We are not showing the auxiliary electrode and reference electrode, assuming they are all there. Now once this reaction starts occurring this could be either the glucose or hydrogen peroxide, in this particular case hydrogen peroxide let's say. Once it starts occurring then they'll be converted to O, and the concentration of R will gradually go down. You know this diffusion limit, and so therefore, the concentration, if you looked at it as a function of time you'll initially see the blue curve. Initially it will be flat, of course. And then after a little bit as the R is getting converted to O the concentration will begin to drop. Little bit later it'll drop further. And similarly the concentration of O will gradually, gradually build up. And then there will be a dynamic balance between forward and reverse reaction. And at exactly -- if we've reached an equilibrium the current will be 0 independent of this concentration. Therefore, we cannot then detect anything. So therefore, diffusion limit is very important, as I will show you now.
[Slide 12] My working electrodes are shown here in blue. That is my platinum electrode you may remember. R is like hydrogen peroxide. And my concentration is shown here in the y axis. So the green is at a given instant the concentration of hydrogen oxide, square root of DT of A. Remember, that is how far it has eaten up, that I explained before. And this is the concentration at the surface. And the oxygen is produced -- being produced, and this is walking away, so this is also increasing as square root of DT. This increasing as square root of DT. This is also increasing as square root of DT. And gradually because it cannot diffuse fast enough the concentration is building up, preventing H2O2 from going to 0. This is gradually also rising because they have to sort of balance each other. And the voltage, 350 millivolt, that will essentially govern the ratio of these two for, at a given time. Let's solve this problem. Very easy as you will see.
[Slide 13] So this problem is solved in the appendix. It simply balances three equations. And let me explain what those things are. You know the diffusion flux of H2O2 going toward the electrode, and that depends on rho R, the bulk density, and the concentration in the surface. This time you cannot assume that this concentration is 0. That's the whole point. On the other hand, there's also another diffusion found going away, the oxygen. I want it to go fast. But there is a finite limit, and that controls the second reaction. So flux going in, flux going out, and this reaction producing the current. You have to balance these three fluxes. It is essentially three lines of algebra. And once you put it in then you'll get a formula like this. Don't worry too much about the formula, the complexity of the formula, because for a given voltage applied to blue this is a constant. This is a constant. This is a constant. These three things are constants. And so therefore, look at how simple it is. All it is saying that it depends on how fast the reaction is going, 1 over k naught, and whether they can come -- diffuse fast enough. So you balance these two, so like a parallel combination. If it comes fast enough then the reaction rate says stop. It cannot accept fast enough. And if it doesn't come fast enough then the diffusion controls it. So it's essentially a parallel combination of these two multiplied by a bunch of constants. This rho R is our initial concentration. Remember the initial concentration, and Ae, larger the electrode you have the corresponding current will be larger. That's all there is to it in this equation. So I hope you will not be confused. This CDt you may have seen before. You must have seen before in this course because you spend a lot of time calculating this diffusion equivalent -- or transient diffusion equivalent capacitance. Let's see whether you remember because all I have to do -- regardless of the complexity of the electrode itself all I have to do is to find the corresponding capacitance and replace W with a 2 square root of 2Dt, or 4Dt, or 6Dt depending on the value you have, and then correspondingly calculate the whole thing. Look how simple it is regardless of how complex the electrode structure, and the reactions, and other things might be. So let's quickly take a look at the diffusion capacitance so I'll just show you a table.
[Slide 14] Let's see whether you remember. Let's start with the top line. Planar electrode, area of the planar electrode is whatever the area. 1 centimeter squared, or 1 millimeter squared, whatever the-- area electrode. What is the CDSS of two parallel plates? Epsilon naught A over D. D is a separation, or W is a separation. And D is a diffusion commission. Remember, if epsilon has to be replaced by D. And W is replaced by 2Dt -- square root of 2Dt. And you remember the cylindrical one. I don't have to tell you the spherical ones. That was -- so all I have to do in the previous expression, just look it up on the table and put it in. That will solve you -- solve working out pages and pages of algebra in the electrochemistry book if you just did this two lines of very simple calculation.
[Slide 15] There are more, of course, right? You could just look it up. Microdisk, if you have a small disk of electrodes sitting on the bottom of a sensor surface, then again we have to do the same, square root of Dt replacing W. If you have an array -- let's say you have a nanowire array, or if you have factor array. All you have to do is to look up the corresponding capacitance and replace that blue with the corresponding Dt values, and you are done.
[Slide 16] All right, this sounds simple, but is it too simple? Meaning, does it explain the results well? I'll show you in a second that it turns out to be very powerful, as you will see. Bottom line, let's say you are having diffusion limit meaning that the analyte molecules are coming slowly. And as soon as they are coming you are immediately gobbling up. So 1 over k is much smaller compared to this value. And so therefore, let me drop that. If I drop that then the CDt will flip up. And this therefore, the whole thing will be proportional to the diffusion capacitance -- transient diffusion capacitance. Let's see how it works. Assume on-- in here that you had no analyte in the beginning to the sensor surface. All of a sudden you put a pulse of analyte here, state pulse. Let's say you dunk your sensor into a hydrogen peroxide solution, let's say. And so that is what it means to have a state response. If you used a planar sensor, if your electrode was planar, right, it's just use a square, that is put in. Then you will see that depending on the concentration you'll have these curves. Look at this curve. The current density as a function of time, both plotted in a log, log plot, is showing a linear decrease as a function of time. And whether you do it measured, or analytical, or numerical, or from the literature, same value. You will see they're the same value here. How does it work? Why is the current decreasing? First of all, if you just went back and with your pencil you checked out how big this is versus how big the x axis is you'll immediately see that that exponent is .5. The rate is .5, going down as t to the power .5 -- .05. No, no, they're t to the power .5. That's right. Why does that happen? Recall that the CDt goes as the square root of t. Put it in here. The current should go as t to the power minus R. Log, log on both sides. The slope is indeed .5. That should make sense. What about you had a very little nanoelectrode and you put it in water rather than putting a square in the water? I know what the answer should be. This is a cylinder, right? And so therefore, what capacitance should I be using? I should be using that two concentric cylinder formula with the one virtual electrode gradually expanding. This is the formula. And once again, recall that this time dependence is very weak because it's sitting underneath the log. And so therefore, the response is more or less flat. And finally, if you had a sphere remember with a little bit longer time this term drops out 4 pi DNR, so therefore, your response is time independent. The current as a function of time is essentially a constant. Now does any of this make sense? Because it looks too simple. Is this true? Does it match with the experiment?
[Slide 17] It turns out it does because think about the spherical sensors that I just told you about, the current being more or less constant -- after initial transient the current being more or less constant. The experiments that I had been showing you and sort of has not really getting into the details too much, remember I told you about the reference electrode on the top? But look at this cube, the reactions are actually happening, and the little floating cubes -- not floating. tethered, but small cubes on the electrode surface. Not on the bottom, but on the top. And therefore, it does behave almost like a spherical sensor. And as you might expect that, therefore, you see that after initial transient the current is constant. After initial transient the current is constant. And so therefore, this indeed goes with this constancy of the current predicted by the theory. And our particular results would be very close to the experimental results.
[Slide 18] So let me conclude then, the amperometric sensors, it can be a sensitive measure, monitor of analyte density, as I explained to you that the current can be directly proportional to the analyte concentration. Not -- there's nothing -- no screening or other effects that will make it a logarithmic dependence, which is very good. And we saw that this selectivity, in addition to a sensitivity is very selective because the enzyme allows it -- enzyme makes sure that only certain reaction products are created, in this case hydrogen peroxide. And so that's the gatekeeper allowing a very selective measure of the analyte also. But depends on the value of the voltage that you apply. If you apply too high a voltage then other parasitic reactions can start, something that has to be avoided. And finally, you see although at high density -- analyte density like my blood glucose level, millimolar concentration reaction is all that matters. But if the concentration is low like nitric oxide, DDT, or some environmental product that you want to measure, in that case, however, diffusion becomes a very important thing. And because the reaction product doesn't go away too fast, it gradually builds up close to the electrode, the current is no longer directly proportional to the analyte concentration. Its value is suppressed because of the reverse reaction. And so the linearity is no longer directly linear. It becomes sublinear response. So therefore, having fast diffusion and fast reactions are both important in order to get the best response from these amperometric sensors. In the next class I will -- or in the next lecture I will tell you about how to beat the diffusion limit. Because after all, if the diffusion limits are around then you cannot get the maximum sensitivity out of the amperometric sensors. So until next time, take care.