WEBVTT
00:17.390 --> 00:22.289
Welcome to the second lecture in the series
of biochemical engineering. We will start
00:22.289 --> 00:26.500
off today from where we left and as a title
of the lecture suggests, this is going to
00:26.500 --> 00:31.119
be concentrated to a kinetics of enzyme enzymatic
reactions.
00:31.119 --> 00:38.119
So, we will just quick wrap up from where
we left. In the last class so the enzyme essentially
00:39.530 --> 00:45.620
as I said is catalyst which does not participate
in the reaction but, accelerates the reaction
00:45.620 --> 00:52.010
by reducing the threshold activation energy.
So, if we are to denote the reaction of S
00:52.010 --> 00:57.350
being substrate going to P by reversible reaction,
S reversible going to P; then in the presence
00:57.350 --> 01:03.219
of the enzyme it turns out that E plus S goes
to P plus E where E is enzyme that reacts
01:03.219 --> 01:07.690
to the substrate to generate the product P
and the enzyme is regenerated back.
01:07.690 --> 01:13.310
So, this is the basics basic of this basics
of the kinetics. But, now what we are going
01:13.310 --> 01:17.070
to do it in today’s classes; there will
be little deeper in to the kinetics and try
01:17.070 --> 01:23.470
and understand how this really happens. How
the enzyme catalyzes or accelerates the reaction
01:23.470 --> 01:27.930
and things like that?
And as I said, the the two point here to note
01:27.930 --> 01:32.799
as the the enzyme is not consumed in the reaction
a very important point note to actually, it
01:32.799 --> 01:38.200
only accelerates the reaction and it does
not alter the final equilibrium state. So,
01:38.200 --> 01:42.130
both of these are important points which will
come into play while we do our calculation.
01:42.130 --> 01:47.500
Today, I would like you do have your exercise
books and pens ready because we are going
01:47.500 --> 01:50.479
to do some small calculation as we go along.
01:50.479 --> 01:57.479
So, just quick wrap up you know, from yesterday
the RS, the reaction rate for the Michaelis-Menten
01:58.259 --> 02:03.060
kinetics which is most popular kinetics use
in my all of biochemical engineering and much
02:03.060 --> 02:07.659
of this course that we are going to do is,
going to be concentrated to towards Michaelis-Menten
02:07.659 --> 02:12.790
kinetics is given by the reaction rate is
given by R max which is the constant and the
02:12.790 --> 02:19.790
maximum reaction rate possible time C S divided
over K M plus C S. Now, a C S being the concentration
02:20.489 --> 02:27.480
of the substrate, K M is called the Michaelis
constant and it is its value is evaluated
02:27.480 --> 02:34.480
by finding out what is the R? What is the
C S at which R S goes to half of R max? right
02:35.030 --> 02:41.129
So, let us lets lets go little beyond this
and let us try and understand how this reaction
02:41.129 --> 02:48.129
kinetics comes out of the reaction between
the enzyme and the substrate right. So, this
02:50.540 --> 02:53.599
we cannot except this reaction as it has been
given us.
02:53.599 --> 02:59.019
So, what we will today is derivation of the
Michaelis-Menten kinetics and I would like
02:59.019 --> 03:03.040
you to participate in that as I said by you
know going through the derivations with me
03:03.040 --> 03:09.470
while we go through this lecture. So, the
first in so if I have to thing I am to ask
03:09.470 --> 03:16.470
you to split up the last last reaction that
we have which is E plus S go going to P plus
03:17.709 --> 03:22.360
E intuitively
So, there will be lets us think of it is a
03:22.360 --> 03:29.360
two step process. So, the first step in this
process is going to be the enzyme E reacts
03:29.390 --> 03:36.390
with substrate S to form a complex E S. So,
I want you to write down in your copies very
03:36.430 --> 03:43.430
quickly what that reaction is going to be.
It is a very straightforward thing just as
03:44.250 --> 03:49.959
the enzyme E reacts with the substrate from
the complex E S.
03:49.959 --> 03:56.959
Because if you remember when we did the thermodynamics
of it, here in the presence of the enzyme,
03:57.629 --> 04:02.610
a complex is formed which is a E B plus. So,
here we not dealing with two substrates this
04:02.610 --> 04:06.920
diagram is for two substrate A plus B. But,
now we are dealing with only single substrate.
04:06.920 --> 04:11.829
There is a result which what will happen in
S is going react with the enzyme to form a
04:11.829 --> 04:17.060
complex A C right which will then go to the
product. This this has to be clear. So, this
04:17.060 --> 04:23.440
is the mechanism of the Michaelis-Menten kinetics
right. That you first form a complex the enzyme,
04:23.440 --> 04:28.360
first reacts with the substrate to form a
complex and then from that complex you form
04:28.360 --> 04:29.479
get to the product right.
04:29.479 --> 04:36.479
So, now let us get your pen down to the paper
and turn and write this reaction. Done? So,
04:40.550 --> 04:44.760
this is the reaction as it turns out. So,
which is that? E is the enzyme reacts with
04:44.760 --> 04:51.760
the substrate reversibly excuse me reversibly
to form the complex E S and the forward rate
04:53.349 --> 04:59.760
constant here I put as k 1 and the backward
rate constant I put as k minus 1. Important
04:59.760 --> 05:04.940
to note it what is given in parenthesis is
that this complex E S that is formed is a
05:04.940 --> 05:11.820
unstable complex that is short-lived. It does
not live for very longtime. It degenerates
05:11.820 --> 05:17.339
or degrades or you know change it over to
something else and so the next step and this
05:17.339 --> 05:22.510
is the intermediate species right. So, the
next step is this. In this second step, this
05:22.510 --> 05:27.279
unstable complex that is E S it is which is
short-lived complex dissociates to form the
05:27.279 --> 05:34.279
enzyme and the product so write that reaction
down. Fine. So, what is it Sahil?
05:39.479 --> 05:43.520
E S gets P plus C
What kind reversible or?
05:43.520 --> 05:44.479
Reversible
Huh
05:44.479 --> 05:47.520
Reversible
So, all of you wrote reversible? Actually,
05:47.520 --> 05:52.909
it is irreversible. As I said that in the
class that all your reactions are you know
05:52.909 --> 05:58.969
typically reversible but, in this reaction
you have to have it irreversible otherwise
05:58.969 --> 06:04.469
not enough product are going to be formed.
And if you look at the overall reaction, the
06:04.469 --> 06:09.460
overall reaction is actually reversible.
So, it is obvious that this reaction also
06:09.460 --> 06:14.599
has to be reversible. But, we take it as a
irreversible reaction because the backward
06:14.599 --> 06:19.560
reaction is very slow typically and this it
turns out this is rate limiting step. right
06:19.560 --> 06:24.649
I hope all of you are are aware of the concept
of rate limiting step, which is that if you
06:24.649 --> 06:29.959
have multiple reactions is the rate network
of reaction then the rate of final product
06:29.959 --> 06:34.760
formation is governed or dictated by the rate
of the slow step right.
06:34.760 --> 06:39.740
So, I give this example in class you know
that for example, if you walking somewhere
06:39.740 --> 06:44.580
with your little sister and she walks much
slower than you then the speed at which you
06:44.580 --> 06:48.510
can walk is governed or dictated at the speed
at which she can walk can and not so speed
06:48.510 --> 06:53.550
at which you can walk. So, you know this kind
of examples will probably impend this concept
06:53.550 --> 06:56.719
of rate limiting step and this is very important
concept in all of chemical engineering.
06:56.719 --> 07:01.060
So, whenever you have a network of reactions
you have to go down and figure out which is
07:01.060 --> 07:04.529
the, what is the rate limiting step. And in
this case this is the rate limiting step.
07:04.529 --> 07:08.899
Why are we so worried about what is the rate
limiting step? Because the rate of formation
07:08.899 --> 07:15.899
of product is going to be dependent on the
rate limiting step so, this is the rate of
07:15.990 --> 07:22.020
disappearance of the substrate.
So, if you look at reaction two over here,
07:22.020 --> 07:29.020
the substrate is a P S in reaction two from
the reversible reaction. right Is that clear?
07:29.380 --> 07:34.070
So, these are the two terms that come in from
equation two. Look at equation. If there is
07:34.070 --> 07:38.700
forward and backward reaction and all are
written in equation four is simply the rate
07:38.700 --> 07:43.880
of so the disappearance of the product d C
S d t S, the rate of disappearance due to
07:43.880 --> 07:48.459
the forward reaction minus the the rate of
disappearance due to the forward reaction
07:48.459 --> 07:53.719
plus rate of formation due to the backward
reaction. Fine, now can you write what would
07:53.719 --> 08:00.719
be the rate of product formation? Product
is here and product is formed from equation
08:01.740 --> 08:08.740
three. right This should be very straightforward.
What is that?
08:12.729 --> 08:17.700
constant
YeahWe will come to that little bit. So, what
08:17.700 --> 08:24.700
I now want you to look at is this is the intermediate
species right which is unstable and short-lived.
08:28.490 --> 08:32.240
What I now want you to look at is what is
balance for this intermediate species, that
08:32.240 --> 08:36.839
we wrote a balance for the for the substrate.
We can write balance for the product which
08:36.839 --> 08:40.330
you already have written. What is a balance
for the intermediate species?
08:40.330 --> 08:43.899
And I want you to write that down is very
straightforward because intermediate species
08:43.899 --> 08:49.530
is part of two reactions; one of them reversible
one of them irreversible. So, there will be
08:49.530 --> 08:56.530
just an additional term there. So, just use
equation two and three to write what is your
08:57.490 --> 09:04.490
d d C E S d t. Fine? So, this is I hope what
you got right. One for the forward reaction,
09:20.360 --> 09:26.070
one for the backward reaction of equation
two and one for the reversible reaction of
09:26.070 --> 09:32.520
equation three.
So, I want to go back a little to this now
09:32.520 --> 09:37.720
one the important things in all of this biochemistry
and biochemical engineering is a fact that
09:37.720 --> 09:42.270
there are some constrains on the system. What
are these constraints? It is the mass balance
09:42.270 --> 09:46.990
constraint or in other words when you talking
of an enzyme, these enzymes as you know participate
09:46.990 --> 09:51.700
in the reaction do do not participate in the
overall reaction, participate in the intermediate
09:51.700 --> 09:53.990
steps of the reaction but, not in the overall
reaction.
09:53.990 --> 09:58.460
So, at the end of the day the enzyme, the
amount of enzyme remain unaltered and this
09:58.460 --> 10:03.520
is a very important thing physiologically.
Why is that? Because these enzymes are typically
10:03.520 --> 10:09.840
available in trace quantities in the human
body so you cannot afford to waste them. So,
10:09.840 --> 10:16.650
one of this this physical constraint boiled
down or you know give rise. What is a mathematical
10:16.650 --> 10:21.500
constraint on the amount of enzyme that is
present in the system and if you look at this
10:21.500 --> 10:25.890
and we done this before I know if you look
at this what is the amount of enzyme that
10:25.890 --> 10:32.890
is present in the system? What are the different
forms in which the enzyme is present?
10:34.720 --> 10:41.080
Absolutely So the free free enzyme and in
the enzyme in the complex form. So, if you
10:41.080 --> 10:47.460
look at this, so C look at equation six. C
E O or C E not, is a total amount of enzyme
10:47.460 --> 10:52.540
present before the reaction starts. That equals
the amount of free enzyme and the enzyme in
10:52.540 --> 10:55.820
complex form form I am sorry in here I wrote
it in the other way.
10:55.820 --> 11:02.420
So I should have written C not equal C E plus
C A C S right because here free enzymes come
11:02.420 --> 11:07.150
first. So, the free enzyme plus so which is
C E and the enzyme in the complex form which
11:07.150 --> 11:13.570
is C E S right. So this is this is this my
equation. I want you to make a note of that.
11:13.570 --> 11:17.550
This is known as the constraint equation.
So, whenever this is very simple system that
11:17.550 --> 11:21.780
we are trying to derive and we will try and
derive much more complex systems later on
11:21.780 --> 11:25.000
in this course and in the you know and in
the assignments and test probably give you
11:25.000 --> 11:30.060
lot more complicated system to solve. One
of the things that is important and to make
11:30.060 --> 11:34.150
a note and always remember and and put it
in to practice this is the fact that you have
11:34.150 --> 11:40.260
a constraint equations so these systems are
all biological systems are governed or dictated
11:40.260 --> 11:45.610
by certain constraints and to figure out what
that constraint constraint is. So, this is
11:45.610 --> 11:47.020
known as a constraint equation.
11:47.020 --> 11:51.680
So, the rate of product formation is something
that you have already written. Now, I give
11:51.680 --> 11:58.330
you few minutes, couple of minutes. What to
do is that this rate of product formation
11:58.330 --> 12:03.350
is written in terms of C E S the complex?
Write the concentration of the complex. What
12:03.350 --> 12:07.110
is the problem with the concentration of the
complex? It is an unstable compound as I said
12:07.110 --> 12:12.360
intermediate unstable compound and it is not
measureable, value of that is not measurable.
12:12.360 --> 12:18.779
So, what we have to do is we have to calculate
this the value of this concentration intermediates,
12:18.779 --> 12:24.160
of the intermediate species in terms of parameters
and variables that we can measure right and
12:24.160 --> 12:27.250
then we have to put it back in to the rate
of product formation to be able to get what
12:27.250 --> 12:31.840
the final rate of product formation is right.
So, this is what I want you to do quickly
12:31.840 --> 12:38.840
in the next couple of minutes is calculate
what is C E S? So, to help you I will just
12:43.460 --> 12:50.460
give you some hint. So, what do you think
Pallavi, would be would be the first thing
13:00.320 --> 13:05.950
that you need to do here.
We will be using the constraint equation
13:05.950 --> 13:10.910
Constraint comes later before that what? What
is the
13:10.910 --> 13:13.390
Yeah
intermediate is shortly so
13:13.390 --> 13:20.390
So yeah right
So that d C S by d t equation if we have
13:23.420 --> 13:25.050
Is going to be
zero
13:25.050 --> 13:28.550
That is going to be 0. So, this is very good.
So, that is the key assumption that we have
13:28.550 --> 13:32.400
to use. So, look at equation number eight
I put up there. So, the quasi state, this
13:32.400 --> 13:37.620
is known quasi-steady state assumption. There
is a complex is very short-lived, it breaks
13:37.620 --> 13:43.560
down very quickly and as a result its rate
of accumulation is 0. So, d d t of C E S is
13:43.560 --> 13:49.290
0 so what you need to do is look at equation
five, where you where you give the rate of
13:49.290 --> 13:54.140
formation of this d d t of C E S right put
that equation to 0.
13:54.140 --> 14:00.920
So, that is what you need to do and then use
the constraint. So, couple these two to now
14:00.920 --> 14:07.100
write C E S. Express C S is very straightforward.
Express C E S in terms of the other variables.
14:07.100 --> 14:14.100
So, from equation five you will get C E S
equals k 1 over k minus 1 plus k 2 C S C E
14:14.170 --> 14:21.170
right right.
14:28.330 --> 14:35.330
So, I have it on this screen now that the
d d t once you put the d d t of C E S equal
14:47.790 --> 14:54.790
0. You get C S equals k 1 C S so time C E
over k minus 1 plus k 2. Clear Liza? Okay.
14:59.839 --> 15:06.839
Now, what you need to do is express use equation
nine which is equation I got here on the constraint
15:10.779 --> 15:17.490
equation over here because you have now got
C E S. So, what you need to do is substitute
15:17.490 --> 15:24.490
C S over there. You see Arush. See equation
nine. You got over here substituted back to
15:29.190 --> 15:36.190
equation six. And then you can express your
C E as a function of C E naught and C S. Why
15:48.279 --> 15:52.990
are you trying to do this Krishnapra? Why
are we trying to express C E as function of
15:52.990 --> 15:57.029
C E naught and C S?
C S
15:57.029 --> 16:04.029
I know why why
we have to find the rate of reaction
16:04.070 --> 16:07.790
I know but, why not keep C E there?
C E is not we do not know C E
16:07.790 --> 16:12.870
C E as an evolving quantity. We are not aware
of what C E is. We are aware of C S because
16:12.870 --> 16:17.690
C S is something can measure at C C S is also
dynamic quantity remember but, C S is something
16:17.690 --> 16:21.610
we can measure at any point of time. C E naught
is known quantity because this is something
16:21.610 --> 16:26.470
that we started with. So, we want to express
everything all our parameters and values in
16:26.470 --> 16:33.470
terms of C E naught and C S. You got it? What
you got? You got the same as what I have on
16:39.350 --> 16:45.170
the board? Equation ten and then C E S now
is given by so, once you got C E as a function
16:45.170 --> 16:50.330
of C E naught and C S. Then you can simply
replace that in the C E S equation and you
16:50.330 --> 16:57.230
get slightly more complicated relation but,
you get C E S as k 1 C S C E naught k minus
16:57.230 --> 17:04.230
1 plus k 2 plus k 1 C S. Fine? What shall
we do now? What is the next step?
17:04.569 --> 17:11.270
C S rate of reactions
Which one? Which reaction?
17:11.270 --> 17:14.020
This goes to k 2
17:14.020 --> 17:21.020
Very good. yeah So what we need to do now
is, figure out the what is the rate of product
17:21.159 --> 17:26.250
formation? The rate of you know reactions
so to say. So, the rate of substrate, this
17:26.250 --> 17:31.190
is the rate of substrate, the rate at which
that the substrate is taken up and actually
17:31.190 --> 17:38.190
we will see that this is this is equal to
the rate of product formation. Why is that?
17:41.250 --> 17:42.770
Which one?
17:42.770 --> 17:49.370
Yeah I know. That is not the final answer.
That is an easy answer but, I am evaluating.
17:49.370 --> 17:55.540
Question is I am evaluating the rate of substrate
production formation right but, what do I
17:55.540 --> 17:58.730
find? Actually want to evaluate. I want to
evaluate the rate of product formation but,
17:58.730 --> 18:04.770
as I said the rate of substrate that is taken
up consumption equals to the rate of product
18:04.770 --> 18:08.330
formation as intuitively it is very clear
right because S going to P.
18:08.330 --> 18:13.740
So how much equimolar things so if x amount
of S is taken up x amount of P is going to
18:13.740 --> 18:17.640
form intuitively it makes complete sense.
But, what I am trying to ask you is mathematical
18:17.640 --> 18:24.640
can you prove that to me? This is there right
in front of your eyes. You just have to tell
18:28.309 --> 18:33.700
me which equation to use.
Its overall reaction
18:33.700 --> 18:38.990
No what is that which equation to use to get
that?
18:38.990 --> 18:43.650
d plus
No straightforward equation five and equation
18:43.650 --> 18:50.650
eight. Equation five is this. The d C d t
of E S equal that that you put equal to 0
18:52.410 --> 18:57.550
then, you will find that the rate of product
formation equals to the rate of substrate
18:57.550 --> 19:04.550
consumption right. So, essentially what I
am trying to say here is that in this in this
19:05.150 --> 19:12.150
slide is that, if you look at R S R S is d
C S d t which equals minus of d C P d t. Fine?
19:15.970 --> 19:22.750
So, I want to go through this, want you to
go through this calculation quickly. You can
19:22.750 --> 19:29.750
do it this way or there is another easy way
is k C C d d t of C P is equals k 2 times
19:29.980 --> 19:34.790
k 2 times C S right.
So, you simply need to multiply this by k
19:34.790 --> 19:41.790
2. Clear pallavi? Everybody liza? This simply
multiply this by k 2 then you get the d del
19:43.850 --> 19:50.850
del t of C P equals this fine and this is
what we get here. k 2 simply multiply this
19:53.150 --> 20:00.150
this same thing multiplied by k 2. That is
my the rate of product formation equals to
20:00.620 --> 20:05.120
the negative of the rate of substrate consumption
fine good.
20:05.120 --> 20:09.720
So, now the question is, so this is what we
got. Say this is the rate of product formation
20:09.720 --> 20:15.360
or this is the rate of substrate consumption,
this is what we got. Now, my question is that,
20:15.360 --> 20:21.620
how do you recast into the Michaelis-Menten
form because what we started. These are two
20:21.620 --> 20:26.820
objectives we said for S that it is a Michaelis-Menten
form. It has been told to us that this is
20:26.820 --> 20:31.620
Michaelis-Menten form but, we do not accept
that as scientists as you know as something
20:31.620 --> 20:33.950
God given.
So, we started off with the very basic and
20:33.950 --> 20:39.530
try started to started to derive that. So,
there is no assumption or in built assumption
20:39.530 --> 20:44.620
that is Michaelis-Menten form. At any point
of time after we have got this rate expression
20:44.620 --> 20:50.500
given by equation twelve what we are trying
to figure out is this this Michaelis-Menten
20:50.500 --> 20:56.620
form? So, what we do to that the what we do
is the Michaelis-Menten form was given to
20:56.620 --> 21:03.620
us here here equation one. And what we need
to do is we need to compare this form with
21:04.429 --> 21:10.350
that form and figure out that how can be recast
this form to that form so equation twelve
21:10.350 --> 21:15.970
that I got. How can I recast equation twelve
in to a form that looks like the Michaelis-Menten
21:15.970 --> 21:16.220
form?
21:16.179 --> 21:21.980
So what I need to do so this is the Michaelis-Menten
form I need to compare this with what I got.
21:21.980 --> 21:28.980
So, if when we compared this it is very straightforward
what turns out is, k C what turns out what
21:29.179 --> 21:36.179
turns out is R max equals k 2 C naught and
K M which is known as michaelis constant is
21:36.549 --> 21:43.549
k minus 1 plus k 2 k 1 and R max is a maximum
rate. Can who will tell me will you tell me
21:44.660 --> 21:49.460
Krishnaprada, why R max is intuitively why
R is max is k 2 C naught?
21:49.460 --> 21:56.320
It is maximum maximum rate
Yeah way is that
21:56.320 --> 22:02.640
All the enzyme
Right If all the enzyme had reacted and form
22:02.640 --> 22:07.730
the product then, the C naught see if there
is no free in enzyme left then, C naught is
22:07.730 --> 22:12.220
the total amount of enzyme that will be in
the complex form and C naught is the is it
22:12.220 --> 22:16.840
is a total catalyst so there will be no free
enzyme left in which case my R max is the
22:16.840 --> 22:18.960
max.
So, the maximum reaction rate possible this
22:18.960 --> 22:23.710
is something that you need to understand intuitively
that because you know despite all the reactions
22:23.710 --> 22:27.380
and the equations we write there might be
times when I might have to just figure out
22:27.380 --> 22:32.230
in a problem that what is a the maximum reaction
rate intuitively? So, what I am trying to
22:32.230 --> 22:35.730
promote is not just a mathematical way of
looking at it but, sometimes intuitive way
22:35.730 --> 22:41.700
of looking at it as well. So, R max is k 2
times had the total amount of enzyme reacted
22:41.700 --> 22:48.700
so that is the maximum possibility and K M
just comes out of the of the calculations
22:48.700 --> 22:53.380
right. Is it physical understanding clear
to all of you?
22:53.380 --> 23:00.380
Then move on now what will do is how to evaluate
these constants? So, these are these are the
23:01.059 --> 23:05.640
two are R max K M R now my two constants.
What have been have I been able to achieve
23:05.640 --> 23:11.540
through this in the by recasting is in the
in the Michaelis-Menten form, can you tell
23:11.540 --> 23:18.540
me that? Why did I say when I once I recast
in to the Michaelis-Menten form? I had no
23:18.820 --> 23:22.610
reason actually Michaelis-Menten they right
the form good for them but, I do not have
23:22.610 --> 23:26.570
to necessarily recast my equation in this
that platform why would I be motivated to
23:26.570 --> 23:31.420
do that? What do I gain by doing that?
We only have two
23:31.420 --> 23:33.100
Very good
calculated that why
23:33.100 --> 23:34.860
Very good not variables constants
23:34.860 --> 23:40.780
Parameters yeah very good so, what we can
manage to do or manage to achieve by recasting
23:40.780 --> 23:46.309
in the Michaelis-Menten form you see in this
first form you have one, two, three four variables,
23:46.309 --> 23:52.790
four parameter; C naught, k minus 1, k 2 and
k plus 1. When you recast in the Michaelis-Menten
23:52.790 --> 23:57.530
from, you end up with just two parameters.
Why is that good for us? Because we have to
23:57.530 --> 24:02.549
finally, at the end of the day evaluate these
parameters experimentally right. So, once
24:02.549 --> 24:07.660
we recast in the Michaelis-Menten from its
much easier to evaluate two parameter experimentally.
24:07.660 --> 24:11.600
Then four and some of you if you are doing
experiment you know how hard it is to you
24:11.600 --> 24:17.320
nourish, you probably know how hard it is
to evaluate these kinetic constants experimentally.
24:17.320 --> 24:22.100
So, if you reduce the number of kinetic constants
as much as possible it is just make your day
24:22.100 --> 24:28.130
easier right. So, what will look at now in
the next few minutes is how to evaluate these
24:28.130 --> 24:34.880
kinetic constants? So, what do you think you
know, give me a sense this is your rate equation
24:34.880 --> 24:40.630
for example, the Michaelis-Menten. What do
you think would be an easy way to measure
24:40.630 --> 24:46.980
these constants?
performance at a very constant at a very low
24:46.980 --> 24:47.380
concentration of C S
Hm
24:47.380 --> 24:48.330
perform very high of C S
So we have two rate
24:48.330 --> 24:55.330
You right that is the one way. yeah That is
a very good way actually but, that is a clever
24:59.760 --> 25:06.010
way actually of doing it. But, one things
you can do simply is measure the rate that
25:06.010 --> 25:10.230
various concentration straightforward way.
This is not a first order reaction but, I
25:10.230 --> 25:14.830
will show you how to how to convert it to
something similar to that. So, very straightforward
25:14.830 --> 25:20.530
way would be to just measure these rate rates
and various concentrations and what happens
25:20.530 --> 25:25.030
is once you measure these rates at various
concentrations you get a plot like this.
25:25.030 --> 25:29.580
So, these numbers that you see here are actually
numbers corresponding to certain experiments
25:29.580 --> 25:36.580
and R S is a rate versus C S the concentration
and this is in micromolar the concentration
25:36.880 --> 25:43.700
in the x axis and the rate in the y axis in
micromolar per second. So, what happens is
25:43.700 --> 25:48.600
that what this is what you said the R S is
R max C S over K M plus C S. You plot the
25:48.600 --> 25:53.730
whole thing because it is not a good idea
to do what you talking about is knows as asymptotic
25:53.730 --> 25:57.850
analysis. What you have on the left on the
if you see on the left case one and case two
25:57.850 --> 26:01.970
these are known as asymptotic analysis. And
I think I mentioned this yesterday’s class
26:01.970 --> 26:05.809
the asymptotic analysis is very important
in all kinds of engineering analysis. You
26:05.809 --> 26:09.830
do because you need to figure out what is
the maximum, what is the minimum and then
26:09.830 --> 26:16.419
your final solution is bounded by these two
right. So, if you look at the plot over here
26:16.419 --> 26:20.850
this is asymptotic solution.
So, case one is case two. Let us look at which
26:20.850 --> 26:26.549
C S is very, very small. C S is much much
trace quantities of C S you get and the R
26:26.549 --> 26:31.480
S you that you get the reaction rate is R
max C S over K M. I explained this yesterday
26:31.480 --> 26:36.040
but, I will I will show it one more time today.
So, if C S is very, very small then this term
26:36.040 --> 26:42.220
in here is now not present. Negligible as
compared to K M and you get R max C S over
26:42.220 --> 26:47.080
K M if C S is very, very large then K M is
negligible over here and C S in the numerator
26:47.080 --> 26:54.080
and denominator cancel out and you get simply
R R S equals R max over K M. Fine? No R x
26:54.390 --> 26:57.520
equals
26:57.520 --> 27:04.520
R max yeah right so if you plot this this
is what you get in the two asymptotic cases;
27:05.350 --> 27:11.770
one is turns out to be zeroth order case,
this one and one this is zeroth order case
27:11.770 --> 27:17.820
on the top and one is the first order case
which is near before small concentrations
27:17.820 --> 27:22.330
of C S. Now what do you do for small concentrations
you know the slope.
27:22.330 --> 27:28.260
So, you measure the slope and the slope you
know is R max over K M for large concentrations
27:28.260 --> 27:34.710
you know that, the slope is R max. R S is
R max and so straight away you can evaluate
27:34.710 --> 27:41.710
R max from the from the asymptote the saturation
limit it reaches excuse me and once R max
27:41.890 --> 27:47.820
is evaluated then, you can use your slope
to evaluate one over K M. Fine? This looks
27:47.820 --> 27:54.820
good and easy now? I am going to quiz you
little bit here. At large concentrations of
27:55.340 --> 28:01.780
C S, small and large concentration of C S
intuitively, physically tell me why does the
28:01.780 --> 28:06.980
system approach first order and zeroth order?
Let us start with the easy one which is at
28:06.980 --> 28:12.190
large concentrations of C S why does the system
approach zeroth order solution?
28:12.190 --> 28:19.190
limited by concentration of concentration
of enzyme
28:26.260 --> 28:30.880
So R max essentially contains the enzyme concentration
the initial concentration of the enzyme. So
28:30.880 --> 28:36.860
they the react, substrate is no longer limiting
reactant. It does not matter how much substrate
28:36.860 --> 28:39.860
you are putting in because you are putting
in less amount of substrate. The substrate
28:39.860 --> 28:46.370
is no longer limiting reactant. So, as a result
so the rate is independent of the substrate.
28:46.370 --> 28:51.120
Is that clear? So, that is what zeroth order
reaction is essential you know some these
28:51.120 --> 28:55.169
questions often asked in quizzes. That is
zeroth order reaction what does zeroth order
28:55.169 --> 29:00.030
reaction essentially mean? Does it mean that
the the rate does not depend on the reaction
29:00.030 --> 29:04.179
rate at all? Then it mean it will mean that
you know it does not matter how much substrate
29:04.179 --> 29:07.580
I put in, the reaction rate is still the same
right.
29:07.580 --> 29:12.590
What is the zeroth order reaction mean? These
are like if it my mean that if there is no
29:12.590 --> 29:16.820
dependence on concentration, if I put a large
amount of substrate and small amount of substrate
29:16.820 --> 29:23.160
it makes no difference right. Is that so?
In a zeroth order reaction if the rate is
29:23.160 --> 29:29.380
independent of concentration does it mean
that it is immaterial how much material I
29:29.380 --> 29:31.770
put in?
29:31.770 --> 29:34.160
Hm
concentration
29:34.160 --> 29:41.160
No, if I am saying the reaction order is first
order itself this is zeroth order itself.
29:43.780 --> 29:50.780
Does it mean that it is independent of the
amount of substrate that I put in? It means
29:54.380 --> 29:58.010
it is the independent but can I put in a very
small amount of substrate?
29:58.010 --> 30:02.890
Hm
This way
30:02.890 --> 30:07.840
No. That is the point you know that is what
I want you to realize. It is independent but,
30:07.840 --> 30:12.730
at a large asymptotic value of the substrate
so what is a concept whole concept of the
30:12.730 --> 30:17.410
reaction mean that you are giving enough substrate
for the system to be independent of the concentration?
30:17.410 --> 30:23.260
There are very very hardly very very few systems
in the in this in the universe in reaction
30:23.260 --> 30:28.740
engineering universe where the system is actually
independent of concentration.
30:28.740 --> 30:33.480
So, if there is zeroth order reaction mean
that is independent of concentration. No,
30:33.480 --> 30:37.610
not really. It looks as if it is independent
of concentration. But, it is not real independent
30:37.610 --> 30:43.289
of concentration. What it means is that, for
large amount of substrate being present there
30:43.289 --> 30:48.020
the system behaves this is as if it independent
of concentration. Now you decrease the substrate
30:48.020 --> 30:51.950
concentration. You will figure out that it
has to be dependent and concentration and
30:51.950 --> 30:57.750
that is what happens at the at the at the
small concentration. What you find it is linearly
30:57.750 --> 31:02.070
dependence on concentration. Which mean that
its limited actually by the amount of substrate
31:02.070 --> 31:06.320
that you put in. So, this is the physicals.
You know we it is very easy to and good to
31:06.320 --> 31:10.140
do the mathematics but, at all points of time
we should have the physics of the, what really
31:10.140 --> 31:15.770
is happening at the back of our mind.
So, we will go on from here. the this is This
31:15.770 --> 31:20.159
is clear to you? So, this is how we evaluate
evaluate the R max from the top asymptote
31:20.159 --> 31:23.919
and the K M K M K M over the bottom asymptote,
from the bottom asymptote and figure it out.
31:23.919 --> 31:28.179
Let us see one way of throwing asymptotic
analysis. Can you tell me what other ways
31:28.179 --> 31:33.289
could be there apart from asymptotic analysis
if I want to use the whole whole range here.
31:33.289 --> 31:38.580
I am not using the whole range. Plot one way
R process over C
31:38.580 --> 31:45.580
Huh Very good. So, the this is known as a
Lineweaver-Burk equation. So what you do is,
31:46.740 --> 31:53.740
look at this. So, what you do is you had R
S given by that formula that we had here here
31:55.400 --> 32:01.150
here so if you take the inverse of this. Just
take the inverse of this one over R S then
32:01.150 --> 32:06.090
what will happen is 1 over C S will come in
the denominator and this is what happens K
32:06.090 --> 32:11.659
M over R max 1 over C S plus 1 over R max.
Now if look at it, what is what is what happen
32:11.659 --> 32:18.659
Lisa is that, 1 over R max 1 over R is and
you plot 1 over R S against 1 over C S. Then
32:18.950 --> 32:24.980
they linearly related right. So, then you
can use the whole range of data. Why am I,
32:24.980 --> 32:30.490
why do you think Krisnaprada did sometimes
better to use this this equation rather than
32:30.490 --> 32:34.600
the asymptotic one?
Sir error is less
32:34.600 --> 32:39.580
Hm because in in in in experiments when you
actually use experiments you are not sure
32:39.580 --> 32:44.559
if your data in the two asymptotic limits
necessarily correct right because it is and
32:44.559 --> 32:48.940
those of more prone to errors. The When the
concentration is very small of the concentration
32:48.940 --> 32:52.960
is very large, those are more prone to errors
and you actually want use your data through
32:52.960 --> 32:58.770
the entire in task over the entire spectrum
which you cannot use if the, if you are just
32:58.770 --> 33:02.720
doing an asymptotic analysis.
So, that is why this is sometimes better because
33:02.720 --> 33:06.980
in this way you can use your data over the
entire spectrum. And this is how plot looks
33:06.980 --> 33:13.980
like. This is a this is a representative plot
of the system so, if you look at this so your
33:14.480 --> 33:19.870
slope is here look at equation twelve. We
have slope is here 1 over R max sorry your
33:19.870 --> 33:25.100
your intercept is 1 over R max and your slope
is came over R max. So, once you evaluated
33:25.100 --> 33:32.100
intercept, you can directly evaluate here
slope from this. Just give you few seconds
33:32.220 --> 33:39.220
to if you want to make a note this, make a
note of this So, 1 over R is linearly related
33:44.559 --> 33:50.780
with 1 over C S. Shall we move on?
33:50.780 --> 33:55.580
So, there are other the couple of other relations
or equations that are used. So I just showed
33:55.580 --> 34:02.580
you the lineweaver-burk and the asymptotic
method and the two Scatchard plot Scatchard
34:02.690 --> 34:08.119
scatchard equation and Scatchard plot and
the Eadie-Hofstee equation and the plot, respective
34:08.119 --> 34:15.119
plot. These are also use to obtain the, these,
why are why do we go in to different forms
34:15.399 --> 34:19.530
of these plots? The reason is you know, when
you evaluate these experimentally the data
34:19.530 --> 34:24.079
may be available in different forms at different
points of time depending on what kind of experiments
34:24.079 --> 34:31.079
you are doing. So, here what you have is C
S over R S is K M K K M over R max C S over
34:31.979 --> 34:37.950
R max in the scatchard equation. So, if your
data is available as C S over R S may be your
34:37.950 --> 34:43.819
one of set of data is available C S over R
S versus C S. You know it is possible. It
34:43.819 --> 34:48.499
does not, we not aware how you did collect
the data. So, depending on how you collect
34:48.499 --> 34:52.289
so there are these several options are available
to you and depending on how you collect your
34:52.289 --> 34:59.289
data, you actually decide what form of the
plot you want to use. So, the equation thirteen
34:59.369 --> 35:02.160
uses C S over R S.
So, you measure C S over R S. That is your
35:02.160 --> 35:08.910
y axis and your x axis is C S. The equation
fourteen uses R x equals R max minus K M R
35:08.910 --> 35:13.180
S over C S. This is very straightforward.
So, your intercept is going to be R max and
35:13.180 --> 35:18.329
again you use your R S instead of C S over
R S you just use R S over C S.
35:18.329 --> 35:25.329
Now, till this point is very good and wonderful
by the thing is that all our all our analysis
35:25.349 --> 35:31.529
that we have done. We based it on an assumption
and a very important assumption which is the
35:31.529 --> 35:38.069
quasi-steady state assumption. Which is that
the C E S that you have the the intermediate
35:38.069 --> 35:43.469
species it short-lived and it accumulation
is 0 d d t f C S is zero. That was assumption
35:43.469 --> 35:48.670
we made. Now, what we have to do is, you know
as not just work as a engineer but, as a scientist
35:48.670 --> 35:55.170
we have to go back and evaluate that whether
that assumption is correct or not. So what
35:55.170 --> 36:02.170
we do about it? Sahil any thoughts?
36:06.869 --> 36:12.469
Quantitatively
Harish
36:12.469 --> 36:17.910
Experimentally
Hm well experimentally you can but,
36:17.910 --> 36:24.910
in the intermediately we do not have at different
points while conducting the experiment
36:29.729 --> 36:34.369
Yeah You can use fluorescent stuff like that.
That is the possibility. What we might do
36:34.369 --> 36:40.979
is theoretically is tha,t you know we can
actually go and not make that assumption we
36:40.979 --> 36:45.739
met that assumption and did the calculations
and we can actually go on not make that assumption
36:45.739 --> 36:50.339
and then see based on a rate constants whether
that assumption is valid or not. That is a
36:50.339 --> 36:54.489
possibility so we can do the same analysis
not making that assumption makes that calculation
36:54.489 --> 37:00.660
little complex. But, we can do that and then
compare that with the system very make that
37:00.660 --> 37:03.609
assumption.
So, that is always you know in when you validate
37:03.609 --> 37:08.410
something in a, validate a model for example,
if your lower dimensional model is an average
37:08.410 --> 37:12.219
model and you want to validate your average
model what is the procedure? What you do is
37:12.219 --> 37:15.960
you take the full three dimensional model.
Compute the solution of that. You take the
37:15.960 --> 37:19.940
average model compute the solution of that
and compare the average solution with the
37:19.940 --> 37:23.710
full solution and if that error between the
two is not high, there is obviously going
37:23.710 --> 37:26.920
to be an error because once you make an assumption
that is always going to be an error. But,
37:26.920 --> 37:31.920
if the error between these two are not very
is not very high then, you then we are say
37:31.920 --> 37:35.440
that your assumption is more or less correct.
So, that is what we are going to do.
37:35.440 --> 37:42.440
So, this is an assumption that we made the
complex formation. That is the fast step and
37:42.440 --> 37:48.739
you know it as a result of quasi-steady state
is unstable and there is no accumulation and
37:48.739 --> 37:54.019
this was an equation that was related to that.
Now, if I ask you to I we actually wrote that
37:54.019 --> 37:59.059
equations these are the equations.
So, if I ask you to go back to that, I think
37:59.059 --> 38:05.719
equation five or something if I remember.
Just give me a minute yeah equation five.
38:05.719 --> 38:10.499
So, this is my this is where we put the assumption
in this. I equate it to 0. If I do not equate
38:10.499 --> 38:16.849
this to 0 and I want to solve this in consumption
with others, we can do that one of the assumptions
38:16.849 --> 38:20.390
that we make while solve and that is exactly
what we are going to do. We are going to solve
38:20.390 --> 38:24.880
equation five. Is that clear? We are going
to solve the equation five, making some some
38:24.880 --> 38:31.880
small assumptions but, not make the big assumption
that it its accumulation is 0. yeah So, the
38:34.150 --> 38:40.680
assumption that we make is C S at this period
is equal C S naught. That is only assumption
38:40.680 --> 38:45.940
we make. What it means is that the it is at
the initial stage of reaction. As a result
38:45.940 --> 38:50.710
that would amount of substrate that is found
that is present equals the total amount of
38:50.710 --> 38:55.029
substrate that was initially present. That
is the small assumption that we make and once
38:55.029 --> 38:59.309
we make that assumption we can go ahead and
solve this.
38:59.309 --> 39:05.809
So, this is my balance equation right and
all I am doing over here so how is the different
39:05.809 --> 39:10.369
from equation fifteen is that, all I am doing
over here is I am substituting C S by C S
39:10.369 --> 39:17.369
naught right. So, you need to, who will tell
me how to solve this equation? Method, the
39:32.519 --> 39:39.519
method I am not asking for the solution but,
what will be the method? That is straightforward
39:40.349 --> 39:47.279
right. There is no method involved is d y
d t equals some constant minus a minus b y
39:47.279 --> 39:51.329
right. So, you can go ahead and solve it is
b exponential solution.
39:51.329 --> 39:55.749
So, this is what you get. Once the solve it
what is the what is assumption at equals so
39:55.749 --> 40:00.279
what is the initial condition at t equal 0,
there is no substrate no complex. That is
40:00.279 --> 40:05.259
safe fine. So, you write you can write down
equation 18 which is a solution of the system
40:05.259 --> 40:12.259
so it is an exponential solution equation
eighteen. Now, from here you have to come
40:45.430 --> 40:52.430
up with an idea that so the solution is good.
But, you to come up with an idea what to do
40:52.430 --> 40:59.430
with it, where we go from here?
To begin have a plot C S this k 1 in to k
41:05.140 --> 41:12.140
1 plus C 1 C S naught this
What do, anybody else what do you think? So,
41:18.859 --> 41:25.859
this is the first order reaction right. And
so, when you do a first order reaction what
41:27.130 --> 41:31.940
is important is the time scale for reaction.
What you trying to measure, how fast the reaction
41:31.940 --> 41:38.940
is you know, what you said you trying to measure.
How fast the reaction is? So, what how do
41:39.390 --> 41:43.450
you evaluate the time scale for reaction?
Time constant
41:43.450 --> 41:49.269
Yeah The time constant, not the time constant
but, the inverse of the time constant right.
41:49.269 --> 41:56.269
So, if you time constant which is a prefactor
of t is something you know. It is the inverse
41:57.499 --> 42:04.499
of it. Essentially, the time constant now
so what will you do from here? What you have
42:05.190 --> 42:09.420
to figure out? What the time scale for the
three actions is right or in other words what
42:09.420 --> 42:15.029
is the effective rate constant for this reaction
is? And if you look at it, what he said you
42:15.029 --> 42:19.589
know, just Krishnaprada said that is very
correct that if you look at this equation
42:19.589 --> 42:23.410
eighteen.
What is the term that matters when it comes
42:23.410 --> 42:29.180
to the dynamic? See, we are trying to understand
the fastness of the process, how fast this
42:29.180 --> 42:33.999
process is. If it is fast enough, then our
assumption is pretty good we are good with
42:33.999 --> 42:39.849
it. So, we are trying to evaluate the fastness
of the process. What is the part or the term
42:39.849 --> 42:44.690
that has to do with fastness of the process?
The term when the exponential right because
42:44.690 --> 42:51.690
that is the one that has a time unit, so you
have to take that prefactor of time out and
42:51.829 --> 42:57.170
we have to compare it with the time constant
for the other reaction.
42:57.170 --> 43:04.170
So, if you look at this, this isthe rate constant
for the first step. k 1 let us go, let me
43:07.170 --> 43:13.819
go back so k 1 yeah this one k k 1 K M plus
C S naught. This is a merit constant for the
43:13.819 --> 43:20.289
first step. Now, the second step second step
is I will go back again. my This equation
43:20.289 --> 43:25.729
three E S giving E plus P so, that is the
gain of first order reaction with respect
43:25.729 --> 43:29.319
to E S and you can evaluated its rate constant
clear.
43:29.319 --> 43:34.759
So, what we have to do is we have to compare
the rate constant or time time times time
43:34.759 --> 43:41.209
constant or rate constant whatever the time
scale of first of equation two two or reaction
43:41.209 --> 43:45.769
two with the time scale for reaction equation
three. Or in other words what it means is
43:45.769 --> 43:51.209
that equation two has to be much, much faster
than equation three. If I am talking in terms
43:51.209 --> 43:58.209
of reaction time scales, how does it translate
the reaction time scale for reaction time
43:58.309 --> 44:00.339
scale for?
Two is very less
44:00.339 --> 44:05.660
Two is very, very less as compared to reaction
time scale of equation three. Is that clear?
44:05.660 --> 44:09.969
Reaction two has to be much, much faster than
reaction three which means that the time scale
44:09.969 --> 44:16.969
for reaction two has to be much much smaller
than that of reaction three. Clear? So, we
44:17.109 --> 44:23.069
calculated our time scale for reaction one.
Now, I wanted to do is calculate the time
44:23.069 --> 44:30.069
scale for reaction two. I can help you with
the, I can you give reaction two here. So,
44:30.759 --> 44:37.759
there is equation three. What is it going
to be?
45:25.959 --> 45:32.959
Hm No idea? See this is going to be same as
you know, what you see on the board reaction
45:52.799 --> 45:59.249
twelve, equation twelve with the exception
that C S can be changed to C S naught because
45:59.249 --> 46:05.739
that that was initial idea that we can keep
it as C S naught. Because it is a pseudo pseudo
46:05.739 --> 46:10.049
first order approximation we are making over
here. Because when you couple see what we
46:10.049 --> 46:15.180
have trying to do is, we trying to decouple
the phase one from phase two. But, in reality
46:15.180 --> 46:19.930
they are coupled right unless you evaluate
what you are E S is you cannot so what should
46:19.930 --> 46:24.940
you do ideally what should you do here you
have if I ask you to solve this, we can do
46:24.940 --> 46:29.400
that is an assignment. I do not have the time
in this class do it. This is your C S C E
46:29.400 --> 46:34.440
S right. So if you are, if you write your
next equation what should it be? Next equation
46:34.440 --> 46:41.440
is E S giving E plus S E plus P right. So
your C E S is given here. You know C E S is
46:43.259 --> 46:50.259
as so I write that equation quickly. Write
the balance equation here. I have it here.
46:54.160 --> 47:01.160
So, write the balance equation for the second
reaction formation of P right.
47:02.269 --> 47:05.160
C s
C S, now we know C S. Put the value C S. Is
47:05.160 --> 47:12.160
that clear k 2 in to C S and all in to have
do is you have your C S over here and this
47:13.940 --> 47:20.549
is your C S. You have to just put the value
of C S over here. So, the rate constant for
47:20.549 --> 47:27.549
that turns out to be something like this and
for for this for the second phase k k 2 in
47:29.039 --> 47:33.910
to this part. Now, what we are doing here
this is an assumption in place it over here.
47:33.910 --> 47:40.910
Which is that the What is assumption here?
Tell me.
47:40.949 --> 47:47.949
K2? if I just multiply this by this k 2. Then
we have k 2 C in to C naught in the C S naught
47:48.219 --> 47:53.099
E naught K M plus C S naught fine. But, when
I put the rate constant over here, I am just
47:53.099 --> 47:59.660
putting k 2 C C naught K M plus C S naught
and the this part is missing over here. The
47:59.660 --> 48:03.759
time part is missing. Why am I, what is the
assumption implicit in that? There is obvious
48:03.759 --> 48:07.239
an assumption that is because that is not
the right answer. You know that right. Is
48:07.239 --> 48:14.239
that clear? The right answer is you have Del
C P del t equals k 2 in to C E C S and C S
48:16.190 --> 48:21.539
is given by that. So, your Del C P del t is
essentially k 2 in to this part time say exponential
48:21.539 --> 48:25.869
minus time.
So, the rate itself is now a function of time.
48:25.869 --> 48:29.130
You see the difference in equation seventeen,
the rate itself is not a function of time.
48:29.130 --> 48:34.309
Only after integrated, it becomes a function
of time. But, here in equation eighteen, not
48:34.309 --> 48:37.849
equation eighteen when you do the second part
the rate itself becomes a function of time
48:37.849 --> 48:43.660
right. But, I am trying to do pseudo first.
So, this is not a first order reaction. But,
48:43.660 --> 48:49.430
I am trying to do pseudo first order approximation
of that and as a result you convert this in
48:49.430 --> 48:54.609
to first order and say there is a rate reaction
rate is simply k 2 C E naught K M plus C S
48:54.609 --> 49:01.609
naught. Is that clear? Let me go through one
more time. So, this is my C E S C S naught
49:03.109 --> 49:10.109
E naught C E naught over K M plus C S naught
times this. Now, in when I write would like
49:11.059 --> 49:17.190
this this is what I get k 2 C E S you can
you can change it to C S times what I miss
49:17.190 --> 49:22.529
over here is, this times the exponential part
one minus exponential part is there. But,
49:22.529 --> 49:28.029
I neglect that exponential part. Why? Under
what circumstances do I do that?
49:28.029 --> 49:33.349
Right, very good. So, the the assumption the
whole idea or the the concept behind that
49:33.349 --> 49:39.160
we are analyzing the first part at initial
times, small times whereas we are analyzing
49:39.160 --> 49:43.089
the second part it large times. And if you
look at this, it is a little complicated.
49:43.089 --> 49:47.739
So, I am giving a little bit of time with
this. So, if you look at this so what happens
49:47.739 --> 49:54.739
is exponential minus k 1 K M plus C S naught.
This part will go to 0 because at large times
49:55.069 --> 49:59.059
this part will go to 0.
So, only the prefactor the the the part that
49:59.059 --> 50:04.569
you see over here is going to remain and that
is going to be your rate constant right. Fine?
50:04.569 --> 50:10.739
Clear to everybody? Ashwin, Liza? This is
little complicated so that is why I wanted
50:10.739 --> 50:15.199
to give you enough time for this. Now, what
we do is what we had discussed which is that
50:15.199 --> 50:20.969
step a has to be much, much faster than step
b. Or in other words two ways of putting it,
50:20.969 --> 50:25.539
the rate constant of step a has to be much,
much higher than the rate constant of step
50:25.539 --> 50:31.309
b. Or the reaction time for step a has to
be much, much smaller than the reaction time
50:31.309 --> 50:38.309
for step b. Fine? So, as I put here the rate
constant for step b has to be much much smaller
50:39.599 --> 50:44.009
than the rate constant of step a and when
you put that over here you find the rate constant
50:44.009 --> 50:49.430
for step b something we got here k 2 E C E
naught over K M plus C S naught has to be
50:49.430 --> 50:54.209
much much smaller than the rate constant for
rate step a which is k 1 times K M plus C
50:54.209 --> 51:01.209
S naught.
Fine. So what we will do is you know, so this
51:19.539 --> 51:25.569
is you done with this. Now, you essentially
have what you have the is a relationship between
51:25.569 --> 51:31.979
your enzyme concentration and the substrate
concentration. You see what is happening?
51:31.979 --> 51:37.059
The last equations that you see essentially
see your k 1 k minus 1 K M. These are constant
51:37.059 --> 51:41.369
for the system. You have no control over these
constants right. What you have control over
51:41.369 --> 51:45.519
the substrate concentration and and the initial
enzyme concentration.
51:45.519 --> 51:51.589
So in order to make sure that Michaelis-Menten
kinetics is satisfied; you have to make sure
51:51.589 --> 51:55.180
that this relationship between the initial
enzyme concentration and in the initial substrate
51:55.180 --> 52:01.799
concentration is followed right. Or in other
words let us not say that in order to make
52:01.799 --> 52:06.019
sure and in other words if the relationship
between C E naught and C S naught, this relationship
52:06.019 --> 52:12.959
between C E naught and C S naught is maintained
then Michaelis-Menten kinetics is valid otherwise
52:12.959 --> 52:17.469
it is not valid.
So, let us not you know decide forever that
52:17.469 --> 52:22.599
Michaelis-Menten kinetics is always valid
given you know whether it the sunshines or
52:22.599 --> 52:26.549
whether it rains. Exacts not the thing so
only under certain conditions it is valid.
52:26.549 --> 52:32.069
And these are the conditions what I you know,
I think some of you might have to go. But,
52:32.069 --> 52:37.869
what I wanted to do is there is any question
on this? Let me address that if there is.
52:37.869 --> 52:44.869
So what I will quickly go to do is you know,
we just have couple of minutes more. Summarize
52:47.569 --> 52:53.630
for you what we did today very quickly and
in the next class we will start with regulation
52:53.630 --> 52:59.569
of enzyme activity and I will give the introduction
later. But, very quickly I will just summarize
52:59.569 --> 53:02.809
for you. So, this is where we started the
Michaelis-Menten kinetics.
53:02.809 --> 53:07.479
And we said that the Michaelis-Menten kinetics
is given by R max C S divided by K M plus
53:07.479 --> 53:11.539
C S. But, we do not want to take it phase
value and we want to derive it. So, what we
53:11.539 --> 53:16.670
did was we said that the enzyme reaction that
happens is the two step reaction and the first
53:16.670 --> 53:23.089
step the enzyme reacts with the substrate
to form unstable short-lived intermediate
53:23.089 --> 53:30.079
E S which has a short life and but, reduces
the threshold energy barrier. So, that the
53:30.079 --> 53:34.059
reaction happens and it accelerates it fine.
In the next step what we said is, that this
53:34.059 --> 53:38.979
complex now breaks down to form the enzyme
and the product and this reaction is more
53:38.979 --> 53:43.459
or less irreversible. In the sense that, if
and if there is a backward reaction possible
53:43.459 --> 53:48.079
it is very, very slow as compared to the forward
reaction. So, we could write the rate of disappearance
53:48.079 --> 53:53.249
of the substrate which equals which we will
see in a minute, equals the rate of formation
53:53.249 --> 53:59.180
of the products. So, this is the rate of formation
of the complex E S given by given on the screen
53:59.180 --> 54:03.949
by equation five. And our another equation
that we have is a constraint equation which
54:03.949 --> 54:09.499
says the total amount of enzyme beat in the
free form or the free form plus the form in
54:09.499 --> 54:15.329
the complex together is a constant which equals
what was what we started with right.
54:15.329 --> 54:22.329
So, once we do that, then the rate of formation
of product is d C P d t is k 2 times C E S.
54:22.890 --> 54:26.640
Now, very very important assumption that we
make which will which we test in the next
54:26.640 --> 54:32.699
few minutes is a quasi-steady state assumption
that the rate of disappearance of the complex
54:32.699 --> 54:37.180
is 0 because it is a short-lived intermediate.
So, the rate of accumulation of the complex
54:37.180 --> 54:42.459
rather is 0 because it is short-lived intermediate
compound. Now, once we make that assumption,
54:42.459 --> 54:46.589
what turns out is the equation seven, that
is the rate of formation of the product is
54:46.589 --> 54:53.589
negative of the rate of formation disappearance
of the substrate right. So this is all fine.
54:53.900 --> 54:58.819
Then, we do a little bit of algebraic manipulation.
Let us forget all of that and we get rid of
54:58.819 --> 55:04.239
C E S, the complex some concentration of the
complex and we get rid of C E E the concentration
55:04.239 --> 55:07.890
of the enzyme. Why do we do that? Because,
these are not measureable quantities. We want
55:07.890 --> 55:11.569
a solution in terms of measureable quantities
which is the concentration of the substrate
55:11.569 --> 55:16.569
and the initial concentration of the enzyme.
So, then we go through these calculations.
55:16.569 --> 55:20.809
Let us forget all these calculations and then
we come up with reaction rate which is given
55:20.809 --> 55:25.049
by equation twelve. Now, we do not know if
this reaction rate is resemble in Michaelis-Menten
55:25.049 --> 55:29.390
form. So, what we do is, we simply compared
it with a Michaelis-Menten form and figure
55:29.390 --> 55:35.039
out yes it does does look like the Michaelis-Menten
form. And then we converted in to Michaelis-Menten
55:35.039 --> 55:39.979
form. The reason we do that is we come down
from four parameters which are hard to measure
55:39.979 --> 55:42.809
to two parameters which are easier to measure
experimentally.
55:42.809 --> 55:46.640
And then we go ahead and measure it experimentally.
We should talk about four different ways of
55:46.640 --> 55:50.479
measuring it; one is the asymptotic analysis,
the second one is a linear curve where it
55:50.479 --> 55:54.279
take the inverse of that and it is easier
to measure and then there is scatchard plot
55:54.279 --> 56:00.569
and the eadie-hofstee equation and the final
thing we tested is the quasi-steady state
56:00.569 --> 56:06.099
assumption. Whether that is valid or not and
what we figured out is that it is not necessarily
56:06.099 --> 56:10.680
valid under all conditions and the way we
did it to split the entire reaction, the two
56:10.680 --> 56:14.680
reactions that happen in two parts. So, the
first reaction is supposed to be fast and
56:14.680 --> 56:18.479
the second reaction is supposed to be slower.
And during the first reaction which happens
56:18.479 --> 56:23.759
we assume that the concentration of the substrate
equals the initial concentration and taking
56:23.759 --> 56:27.890
that and taking an initial condition that
there is not complex present we integrated
56:27.890 --> 56:33.349
that we found the dynamic solution to that.
And we use that to get to the second part.
56:33.349 --> 56:36.999
And then, we compared the rate constants of
the first and the second part and we figured
56:36.999 --> 56:43.259
out that if rate the second part is much,
much slower than the first part. Then only
56:43.259 --> 56:48.660
it is possible to assume as quasi-steady state
relationship and then only can be get to Michaelis-Menten
56:48.660 --> 56:53.109
form right.
So that was and so what we figure is that
56:53.109 --> 56:57.910
yes if there is a this this possible if a
certain relationship between the initial enzyme
56:57.910 --> 57:02.219
concentration and the initial substrate concentration
is maintained. So, I guess we will stop here
57:02.219 --> 57:07.880
today and the next week we will continue with
how to regulate these enzyme activities which
57:07.880 --> 57:09.950
is a different section. So, thank you.