# nanoHUB-U Nanobiosensors L3.8: Sensitivity – Amperometric Sensors – Glucose Sensors II

[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.

wonderful video, If there's subtitle, maybe better

THANK u , wondefull video , I' m trying to do a circuit of glucometer , but I get trouble on the potentiel that I have to give to the Reference Electrode and the the others ….Can u help