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

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

January 15, 2020 2 By Jose Scott


[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.