WEBVTT
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So, from where we left, which is that effective
diffusion in the fungal pellet and if we go
00:23.440 --> 00:28.710
to the slide, now, what we see is that, we
wrote this equation, if you remember, which
00:28.710 --> 00:31.610
is the diffusion equation. So, on your left
hand side is the diffusion of oxygen, and
00:31.610 --> 00:37.670
on your right hand side is the reaction that
occurs because of the growth of the fungus,
00:37.670 --> 00:41.379
fungal pellet. So, it is, the fungal pellet
essentially has to grow. What was the major
00:41.379 --> 00:45.480
assumption out here? The major assumption
was upto the steady state approximation. So,
00:45.480 --> 00:50.159
the idea was that, though the fungal pellet
is actually growing in radius, but, the rate
00:50.159 --> 00:56.100
of growth of the radius is much smaller than
the rate of diffusion of oxygen, as a result.
00:56.100 --> 01:00.479
And, how did we conclude this? We concluded
this through time scale analysis. We measured
01:00.479 --> 01:05.210
what is the time scale of the growth of the
radius of the fungus, and we measured, what
01:05.210 --> 01:09.159
is the time scale for diffusion of oxygen
in the fungal pellet. And, what we found was
01:09.159 --> 01:14.440
that, the time scale for the growth of the
radius is much, much larger than the time
01:14.440 --> 01:19.360
scale for the growth of the fungal, of the
diffusion of the oxygen; or, in other words,
01:19.360 --> 01:25.970
the diffusion of oxygen is a much faster process,
as compared to the actual growth of the radius
01:25.970 --> 01:29.720
of the fungal pellet.
So, as the result of which, we can write the
01:29.720 --> 01:34.890
pseudo, steady state approximation, or in
other words, we can assume that, the diffusion
01:34.890 --> 01:40.140
of oxygen is the steady state process, as
compared to the growth of the radius, fine.
01:40.140 --> 01:45.270
So, based on that, we write this equation
out here, and what you see over here is that,
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the temporal term in oxygen concentration
is missing, because of the pseudo steady state
01:50.750 --> 01:57.750
approximation, fine. So, this is the radial
diffusion, excuse me; this is the radial diffusion,
01:57.930 --> 02:03.560
and this is the reaction term and this is
written in dimensionless coordinate; all of
02:03.560 --> 02:07.690
it is written in dimensionless coordinate;
and, I am not redefining these dimensionless
02:07.690 --> 02:13.129
coordinates. The reason I am not redefining
is that, we already did this same, that immobilized
02:13.129 --> 02:18.709
enzyme case, you know, the last example we
did in, in the immobilized enzyme, already
02:18.709 --> 02:21.440
had the same variables and same dimensionless
coordinates.
02:21.440 --> 02:28.440
So, you can go back and look up; so, I am
not redefining it. Now, the assumption here
02:28.989 --> 02:34.530
is that, C oxygen is much, much greater than
K M, which is a very reasonable assumption,
02:34.530 --> 02:38.689
because, if you want to grow something, you
know, fungus is growing for example, oxygen
02:38.689 --> 02:45.650
is typically in abundance, as compared to
the k m. So, that is a very straightforward
02:45.650 --> 02:51.930
assumption. Now, when you make that assumption,
what you find is that, this, this equation,
02:51.930 --> 02:58.079
which is a Michaelis-Menten kinetics, reduces
to a first order kinetics over here; is that
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clear? So, the Michaelis-Menten kinetics over
here, reduces to a zeroth order, sorry, reduces
03:03.290 --> 03:08.189
to a zeroth order kinetics over here, because,
in the, when C O is much larger than K M,
03:08.189 --> 03:13.680
then, what happens? C O in the numerator and
the denominator cancel out, and this is what
03:13.680 --> 03:15.889
you have.
So, this is, this is what we got over here,
03:15.889 --> 03:22.889
and you know S 0 is all of this combined,
K M and beta and so on, combined. So, R square,
03:23.950 --> 03:30.950
r is a radius of the mold, of the fungal mold,
at that point of time; is that very clear?
03:31.139 --> 03:35.989
R, ideally, R is a function of time; it is
increasing with time, but, since we are doing
03:35.989 --> 03:40.840
a pseudo steady state approximation, so, R
over here is the radius of the fungal, at
03:40.840 --> 03:46.529
that point of time, fine. So, we write this
equation; then, this is a very straightforward
03:46.529 --> 03:50.980
zeroth order, on the right hand side, and
you can integrate it straight away, and what
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are my boundary conditions? Some are written
on the screen, which is that, the oxygen concentration
03:57.069 --> 04:04.069
at the outside radius r equals 1, or r equals
big r, whatever it is, is 1 and d d d d r
04:05.819 --> 04:09.049
of C O 0 is 0, ok.
04:09.049 --> 04:16.049
So, let me show you this by assuming this
assumption. So, this is my cylindrical mold;
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my reaction equation over here is 1 over r
d d r of r; nu max I used, instead of r max;
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the reason being that, r, now, we are using
for the overall radius of the mold; that is
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why, I changed the…D effective; S 0 is that
K M over beta, that you, K M times beta, sorry.
04:44.500 --> 04:51.500
So, this is my equation. The boundary condition
inside is…So, this is my r; this is r equals
04:54.690 --> 05:01.690
0; this is r equals 1. So, r equals 0, boundary
condition is symmetry, right. So, equals 0,
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r equals 1; I am saying that, oxygen is present
in a certain concentration. So, C O 2 equals
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1; 1 is the dimensionless coordinate, which
means that, actual value of concentration
05:17.710 --> 05:24.440
is something, like some value C, C O 0, or
something like that. So, C O 2 is equal to
05:24.440 --> 05:31.440
C O 0 in dimensionless; dimensional, when
you turn into dimensionless, this is what
05:35.160 --> 05:42.160
you get, right. So, this is a boundary conditions
we have. We can straightaway integrate this.
05:53.550 --> 05:58.810
This is a straight forward integration and
we can get the profile of concentration within
05:58.810 --> 06:04.930
the mold. So, you know, so, why are we doing
this and the reason would be obvious, you
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know, some of, one of you, I think, asked
this question the other day, and the reason
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would be obvious, once we come up with the
solution.
06:12.570 --> 06:19.570
So, look at the solution. So, this is how
it goes as, you know, it goes quadratically;
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essentially, as you see, it increases with
radius, right. It increases quadratically
06:27.000 --> 06:34.000
with radius. So, it is some constant plus
another constant times r square; is that clear.
06:36.140 --> 06:42.650
So, it is one constant, some a, plus b times
r square. So, it increases quadratically with
06:42.650 --> 06:48.970
the radius. Now, what you know, what you know
is that, the concentration at the outside,
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at this interface between the fungus and the
air is a constant, which means, in other words,
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it decreases quadratically, as you go inside
the fungus; is that clear. So, what will happen
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is, just try to, you know, invasion this.
As you, as you keep increasing the size of
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the mold, what will happen? The gradient will
get steeper and steeper, right, because the
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concentration that you provide at the outside,
that is the interface between the mold and
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the, and air is a constant. So, as the radius
keeps increasing, the gradient will get steeper
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and steeper, fine; and there, will come a
point in time…
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So, this is a steady, pseudo steady, under
a pseudo steady state approximation, we are
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making this; but remember, this r is actually
a function of time. So, when you make a pseudo
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steady state approximation, how do you do
that? how do you enforce that? So, first you
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solve the steady… So, if I ask you to solve
a problem using pseudo steady state approximation,
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so, what is the idea? You decouple the state,
the two, two phenomena. How do you decouple
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the two phenomena? By first showing that,
the time scales of these two phenomena are
08:01.520 --> 08:05.330
different. So, the time scale which is much
smaller is under steady state. So, you write
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a steady state equation like, just like we
did, and solve for it; then, you solve the
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unsteady state equation and get how radius
varies with time. For example, here, if you
08:15.120 --> 08:22.120
have your d r d t, if I remember, here d r
d t equals K, fine.
08:26.470 --> 08:33.470
So, if this is what you have, then, you got
your x as a function like this, and you can
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also integrate d r d t equals K over here,
right away, and get how radius varies with
08:39.639 --> 08:44.050
time. So, radius here, would vary in some,
you know, linearly, for example, here it is
08:44.050 --> 08:51.010
varying, some K t, fine. So, what you do here
is that, after we have solved for the steady
08:51.010 --> 08:57.760
state, you impose that temporal dependence
of radius on this equation; or, in other words,
08:57.760 --> 09:02.200
if radius, for example, varies linearly with
time over here, or, radius varies exponentially
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with time over here, then, you go and put
that into this equation; is that clear? Let
09:07.260 --> 09:09.060
me write, if it is not clear.
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So, your C 0, the solution that you get here
is, C 0 minus R square nu max, 1 minus r square.
09:23.020 --> 09:30.020
Now, if R equals some R 0 at t equal 0 time,
let us say, exponential; we are just assuming;
09:30.300 --> 09:37.300
exponential minus, you know, K t, something
like this, sorry; it should be plus K t, in
09:37.590 --> 09:44.590
this case, if it is increasing with time.
So, what you do is, you put…So, this is
09:45.500 --> 09:50.480
your, essentially your R, at any point of
time. So, you put that over here, and what
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you get is, 1 minus R 0 exponential K t, fine.
So, if this exponential K t is some linear
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dependence and that will come in. So, this
is a way to, is that, is that clear? This
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is the way to solve the pseudo approximation,
when you have a pseudo steady, steady state,
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or there is the decoupling of state, two states,
then, this is the way to solve it. First,
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go and solve for the steady state, and then,
impose; then, solve for the unsteady state
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and impose that unsteady state condition on
the steady state solution; is that clear?
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This is the way to do it, fine.
Now, excuse me. So, what you find over here,
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what you find is that, if just by looking
at this equation that, as the radius keeps
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increasing, the concentration gradient becomes
steeper. Now, the concentration outside is
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a constant, remember; if the concentration
outside will not a constant, if you are increasing
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it, then, no problem; but, because the concentration
outside the mold...So, concentration here
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is a constant, is a constant and this radius
is keeps increasing, what will happen? There,
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there will come a point in time, when the
center of the mold would stop getting any
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oxygen at all; is that clear? There will come
a point in time, when the center of the mold
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will stop getting any oxygen at all, and that
means that, the center of the mold can grow,
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no longer grow; and, that is my critical radius.
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So, let me show you the definition of that.
So, at critical radius, the critical radius
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is defined as that radius, when the center
of the mold is depleted of oxygen. And, if
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the center of the mold is depleted of oxygen,
then, the growth stops; because, this thing
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has to push from outside, you know; so, this
is a, this is a radius that we want to evaluate;
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and how would you evaluate that, from this
expression? We put r equals, this r equals
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what? What, how do we get this R critical?
.
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C. concentration of.
Yes, but, there, that is fine. This side is
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0; but, there is a variable out here; critical
radius is a constant and r over here, is a
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variable. So, C O 2 equals 0, I agree; what
else?
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Sir, that will happen at this center.
Yes; so?
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That will also have to be 0.
Yes. So, this r has to be 0; this r, remember
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big r, is the external radius; this radius
at r equals 0. Thus C O 2 has to be 0. So,
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you put 0 over here, 0 over here and you will
simply get the critical radius right away,
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as square root of 6 S 0 D effective over nu
max. So, I hope this answers your question
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from last time that, fungus cannot grow, grow
indefinitely and there will come a point,
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when it will stop growing and that is the
time it will try to divide itself, so that,
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it can continue to grow. So, this is the limit
of the colony. So, it is a colony, that kind
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of, it keeps expanding, but, it does not expand
indefinitely; it expands only up till the
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time, that the radius equals the critical
radius; so, which is given over here, square
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root of 6 S 0 D effective over nu max.
So, once this critical radius is, is attained,
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then, the funguses stop growing and it will,
maybe divide and allow for future growth;
13:45.839 --> 13:52.839
is it clear? So, what we looked at is, you
know, we are looking at different, different
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things, at the same time. So, first, we looked
at the normal growth kinetics, and you know,
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how the kinetics changes, and what is the
dynamics of the kinetics, and how different
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models had been proposed, to sort of quantify
these dynamics, and what we figured is this,
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that, of all the kinetics model, kinetic models
that have been proposed, it the Monod model,
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or the modified Monod model, which are the
most useful models.
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Then, we looked at the effective mass transfer,
gave you a couple of examples to look at the
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effect of mass transfer, and just as we had
seen that, in the case of, case of a immobilized
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enzymes, mass transfer effects kinetics; or,
in other words, kind of decelerates kinetics.
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Similarly, if mass transfer effects here are
going to decelerate, or slow down the growth
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process; but, there is no running away from
it, because, for the growth process, just
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as in the enzymatic process, you need the
substrate; you need to provide the oxygen,
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and the nitrogen, and the hydrogen, and the
carbon, just as we showed in the first, I
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think, the first slide of this whole chapter.
You need all of these substrates to be available
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to the, to the cell and without this, there
is no way you can have cell growth.
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Now, as soon as you need all of these substrates
to be available to the cell, you immediately
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need, you know, mass transfer effect, would,
all the mass transfer limitations would automatically
15:11.880 --> 15:16.910
come in. So, there is no running away from
it. So, we looked at some of these and with
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the effect of, with the, for the case of normal
growth of single cells and the growth of fungus.
15:23.430 --> 15:28.740
So, what we said was that, the major difference
between the growth of cells, cell colony,
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and the growth of fungus is that, a cell,
cell in, when a cell colony grows in general,
15:33.230 --> 15:37.630
it is like a cell is surrounded by the fluid
and therefore, by the substrate, and therefore,
15:37.630 --> 15:41.589
it has direct effect, direct contact with
the substrate; whereas, a fungus would always
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grow as a pellet, or a mold, you know, as,
as the, as a lump of colony and therefore,
15:47.110 --> 15:51.209
the cells inside the fungus do not have any
contact with the nutrients.
15:51.209 --> 15:55.839
So, the only way the nutrients can come in,
is through the interface, from the outside,
15:55.839 --> 15:58.890
it is slowly diffuse it and this is the basic
difference. And therefore, we did, looked
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at the mass transfer effect for a single cell
and then, we looked at the mass transfer effect
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for the fungus mold. So, we are at this point.
Now, what we are going to do for the rest
16:06.790 --> 16:11.269
of the lecture today, is look at some other
effects, and those effects are the effects
16:11.269 --> 16:16.700
of multiple substrates and the effects of
inhibition, which means that, if you have
16:16.700 --> 16:21.120
more than one substrate, right. So, you can
provide substrates in different forms, and
16:21.120 --> 16:27.160
both can lead to cell growth; then, what happens,
right. So, for example, you can give your,
16:27.160 --> 16:32.750
your sugar, you know, both in terms of sucrose
and glucose, you know, maybe two different
16:32.750 --> 16:35.029
form.
So, fructose, sucrose or glucose, you know,
16:35.029 --> 16:39.160
two or three different forms. Now, all of
these substrates, they have different kinetics,
16:39.160 --> 16:42.670
remember; each of them will have different
kinetic, but they, all of them lead to growth
16:42.670 --> 16:48.970
of the cell, fine. So, how do we account for
that, a; b is that, what would be the effective
16:48.970 --> 16:53.540
inhibition, just like we did in the enzyme
case. So, there is, just as you have substrate,
16:53.540 --> 16:59.140
there could be inhibitory things that are
there, a and b is that, when you give too
16:59.140 --> 17:02.630
much of substrate, that, one of the thing
that we talked about, I think, earlier, you
17:02.630 --> 17:06.819
can have, what is known as substrate inhibition.
So, how do these affect the cell growth process
17:06.819 --> 17:11.260
and we are just going to look at that. I hope
we can finish this today. Then, we can start
17:11.260 --> 17:15.360
a new chapter; but anyway, we are not going
to hurry through it; we are going to do it
17:15.360 --> 17:15.610
very slowly.
17:15.539 --> 17:22.539
The first thing we will look at, is the effect
of multiple substrates. So, multiple substrate
17:23.120 --> 17:28.270
means that, you have cells which are exposed
to two different substrates, and both these
17:28.270 --> 17:34.460
substrates will lead to the product formation,
that is daughter cells, but through different
17:34.460 --> 17:41.460
kinetic routes. So, this is, what you see
on the screen is, x is your cell, starting
17:42.420 --> 17:49.420
cell, and S 1 is substrate number one and
you get the daughter cell x, 2 x; but, through
17:52.010 --> 17:57.870
the formation of the intermediate compound,
intermediate cell x prime, which is one route,
17:57.870 --> 18:04.870
with a set of reaction coefficients k 1, k
minus 1 and k 3; and then, you have another
18:05.530 --> 18:09.170
substrate, that is, they are parallelly in
the system. So, you have, might have, you
18:09.170 --> 18:16.170
know, given two substrates to the system.
So, that gives again, the daughter cells,
18:16.429 --> 18:21.100
2 x, but through different route, x double
prime; and the reaction rate constants are
18:21.100 --> 18:26.380
k 2, k minus 2 and k 4. Now, I talked about
the yield yesterday, if you remember, in the
18:26.380 --> 18:33.380
last class, I talked about yield and so, the
yield is the amount of daughter cells produced
18:33.490 --> 18:39.260
per, per mole of the substrates, ok.
So, for case one, what is your yield? One
18:39.260 --> 18:43.500
daughter cell is produced; from two, two cells,
you get, one cell, you get two cells, which
18:43.500 --> 18:48.290
means, one, effectively one extra cell has
been produced for a one mole of the substrate.
18:48.290 --> 18:55.290
So, the yield of x over S 1 is 1 over a 1;
and for case two, the yield of x over S 2,
18:56.710 --> 19:03.620
that is, one daughter cell has been produced
for a 2 moles of S 2. So, yield is 1 over
19:03.620 --> 19:08.510
a 2, fine. Now, how do we find…So, what
we are trying to find is, what the, how would
19:08.510 --> 19:12.450
we find the overall growth rate of the system?
So, what is the goal, that has to be very
19:12.450 --> 19:16.350
clearly defined. So, we did…What did we
do? We have done only one single substrate
19:16.350 --> 19:21.760
before, and we have found the growth rate;
so, which is d x d t equals nu, nu, nu times
19:21.760 --> 19:28.260
x, where nu is given by the Monod, Monod kinetics,
S 0 over K plus S 0, some r max S 0, double
19:28.260 --> 19:31.450
K plus S 0.
So, here we are trying to find out, what would
19:31.450 --> 19:36.580
be the form of the growth, overall growth
of the system, because, what we are in, essentially
19:36.580 --> 19:40.090
interested at the end of the day is the overall
growth. I start with 5 million cells; how
19:40.090 --> 19:47.090
many cells do I get after 24 hours or so,
that is what I am interested; excuse me. So,
19:47.890 --> 19:54.890
the balance equations, excuse me, the balance
equations over here, are this. So, what I
19:57.820 --> 20:02.150
am writing is, I am writing a balance for
the intermediate species; the del x t, del
20:02.150 --> 20:09.150
x prime del t and as you see over here, the
first term is k 1 x S 1 which is the term,
20:11.620 --> 20:16.410
because of this forward reaction over here.
See the, where the arrow is. So, this forward
20:16.410 --> 20:22.330
reaction over here, we this term. The second
term is minus k minus 1 x prime, which is
20:22.330 --> 20:27.320
the result of the backward reaction over here.
The first reaction, the backward reaction
20:27.320 --> 20:34.320
you had; and the third term over here, is
minus k 3 x prime, which is the result of
20:35.010 --> 20:40.470
the formation of the product, of the daughter
cells. Is it clear, these three terms? It
20:40.470 --> 20:45.740
is very straightforward; and then, the same
thing is repeated for x double prime, ok.
20:45.740 --> 20:52.740
So, k 2 x times S 2 minus k minus 2 x double
prime minus k 4 x double prime; is it clear
20:55.530 --> 21:02.530
to all? So, is there a question?.
No; power, no, no, no, no; yes; that, I will
21:06.410 --> 21:13.410
explain that. See, this is not an elementary
reaction. See, when do you have the power?
21:13.440 --> 21:18.380
Let us, let me explain that, if there is any
doubt.
21:18.380 --> 21:25.380
A plus B giving C; then, rate of formation
of, of C is K C A C B; if it is an elementary
21:32.690 --> 21:39.690
reaction; A plus 2 A, 2 B giving C, the rate
is, could be given as K C A C B square, only
21:44.549 --> 21:51.549
if, if and only if, it is an elementary reaction;
is that clear? If it is not an elementary
21:51.610 --> 21:54.540
reaction, there is, nobody said that, that
is a, that is a rate. So, it could be this.
21:54.540 --> 22:01.540
This one?
No; no, no, A plus 2 B equals to C.
22:01.990 --> 22:07.030
Yes, elementary reaction means, the definition
of the elementary reaction is that, the rate
22:07.030 --> 22:13.650
of the reaction is written exactly as the
same way in which the molar concentration,
22:13.650 --> 22:20.650
composition of the reactants react; that is
the definition. See this reaction for example;
22:20.710 --> 22:25.549
it is defined, an elementary reaction is,
will be defined only if…So, if this, this
22:25.549 --> 22:31.010
molar concentration is 1 is to 2 mole, it
reacts in 1 is to 2 mole ratio and the reaction
22:31.010 --> 22:38.010
rate is also given as C A C B square; but,
it is not necessary that, exactly the, the
22:39.330 --> 22:43.830
mole ratio in which they react, will be the
ratio in which the, the rate of the reaction
22:43.830 --> 22:46.730
is measured; because there, see that, the
elementary reaction means that, there are
22:46.730 --> 22:52.260
no other effects, apart from the molecular
collision between the, between the two molecules
22:52.260 --> 22:56.080
that govern the rate of the reaction. Do you
understand what I am saying?
22:56.080 --> 23:00.299
So, if there are no other effects…So, it
is…See, this is very unlikely that, this
23:00.299 --> 23:05.240
will be an elementary reaction. Why, because,
the probability of 2 molecules of B and 1
23:05.240 --> 23:10.110
molecules of A, coming and colliding and forming
1 molecule of C, is very, very small. So,
23:10.110 --> 23:14.980
typically, a elementary reactions are of the,
or either first order, or second order. So,
23:14.980 --> 23:19.220
you know, when we talk about first order reaction,
so, it cannot be an elementary reaction; because,
23:19.220 --> 23:23.640
1 molecule cannot combine with your, itself;
there has to be a molecular collision. So,
23:23.640 --> 23:27.870
what is the elementary reaction mean that,
the way 2 molecules collide with each other
23:27.870 --> 23:32.990
is exactly the way the reaction is represented,
as also, when the reaction rate is written.
23:32.990 --> 23:36.870
So, do not assume that, whenever you write
a...So, this is the molar ratio; this is its
23:36.870 --> 23:40.360
stoichiometry.
So, do not confuse stoichiometry with the
23:40.360 --> 23:45.059
power of the reaction. The stoichiometry would
be same as the power of the reaction only,
23:45.059 --> 23:48.730
and only if, it is an elementary reaction;
elementary reaction means that, all other
23:48.730 --> 23:53.929
effects are ignored. So, it is directly that,
the 2 molecules are combining and that is
23:53.929 --> 23:57.280
forming the reaction; but, in the most cases,
it is not like that. You know that, there
23:57.280 --> 24:01.190
are complex steps out there; for example,
this step that I have written over here, as
24:01.190 --> 24:06.299
I, as we do in the enzyme, they are not necessarily
elementary reactions, which means that, it
24:06.299 --> 24:10.280
is not that this step alone is occurring;
there are multiple steps that are, that are
24:10.280 --> 24:15.510
probably occurring over here, and they are
summarized as these reversible reactions.
24:15.510 --> 24:20.520
And, elementary reaction means that, one single
step, single step reaction and in that ratio.
24:20.520 --> 24:23.720
So, the molar ratio, the stoichiometric ratio
that is there; so, they combine, these 2 molecules,
24:23.720 --> 24:28.500
it combine and form. So, this is something
very fundamental; you should not make a mistake
24:28.500 --> 24:32.169
about it.
So, if I want to say, tell you that, this
24:32.169 --> 24:36.340
is the rate of the reaction, then, I will
say that, this is the reaction and which reaction,
24:36.340 --> 24:40.270
one is an elementary reaction, which means
that, in this case, you will raise to the,
24:40.270 --> 24:45.980
raise this to the power a 1; and, but then,
elementary reaction, if a 1 is not an integer,
24:45.980 --> 24:50.299
then, it cannot be an elementary reaction
also; do you understand what I am trying to
24:50.299 --> 24:55.850
say? If a, a is say 0.75, it cannot be a elementary
reaction, because 0.75 of a molecule cannot
24:55.850 --> 25:02.140
combine with, you know, another molecule.
So, for elementary reaction a 1 has to be
25:02.140 --> 25:08.230
a number, typically, 1 or at best 2; even
2 is hard, but, at best 2; but, not more than
25:08.230 --> 25:13.039
2; because, 3 molecule, the probability of
3 molecules combining with each other goes
25:13.039 --> 25:17.309
down. You know, if you have read some of mechanics,
you will understand that, the probability
25:17.309 --> 25:24.000
goes down exponentially. So, anyhow, now,
so, this is the reaction rate and…
25:24.000 --> 25:30.150
So, the next step, what we do is, we write
the constraint equation like we did before,
25:30.150 --> 25:35.789
which is the total amount of cells present
in the system. So, x t is the total amount
25:35.789 --> 25:39.860
of cells present in the system. So, the total
amount of cells present in the system is either
25:39.860 --> 25:45.260
in the form of x, which is a daughter or the
mother cell, or in the form of any of the
25:45.260 --> 25:52.260
intermediate species x prime or x double prime.
So, the, this, this is my constraint equation.
25:53.309 --> 25:57.960
Next, what I do is, use a pseudo steady state
approximation, the quasi steady state approximation
25:57.960 --> 26:04.960
like we did before; that del del t of x prime
equals 0 and del del t of x double prime equals
26:05.660 --> 26:12.660
0, fine. Now, let me explain this a little
bit, from what is written in the last line.
26:22.490 --> 26:29.490
So, d x d t. So, x, my x equals, x t equals,
x plus x prime plus x double prime. So, d
26:32.470 --> 26:39.470
x d t would be d x t d t, minus d x prime
and d t minus, d x double prime d t, right;
26:45.350 --> 26:52.350
but, these are 0, because of pseudo steady
state approximation. So, you get this as d
26:53.520 --> 27:00.520
x t d t, fine; you get this as d x t d t.
Now, this equals k 3 x prime plus k 4 x double
27:09.590 --> 27:16.590
prime. Can anybody tell me, why is that? d
x t d t equals k 3 x prime plus k 4 x double
27:20.390 --> 27:24.100
prime?.
Right. So, x t is what? x t is the overall
27:24.100 --> 27:30.990
amount of cells that are present. Now, see,
the only difference here from what we did
27:30.990 --> 27:36.490
in the enzyme case, if I go, if I am allowed
to go back to this slide. So, look at this.
27:36.490 --> 27:42.210
So, this is the reaction that is occurring
because of the formation of the daughter cells
27:42.210 --> 27:47.730
that are occurring through the, from x prime
or x double prime to 2 x. So, what is the
27:47.730 --> 27:51.250
major difference from what we are doing here,
from what we are doing in the constraint?
27:51.250 --> 27:56.500
This x t is not a constant; that is the only
difference. In the enzymes case and all the
27:56.500 --> 28:00.370
previous cases we had done, this x t was a
constant, right. But here, the x t is not
28:00.370 --> 28:04.850
a constant. So, that is a major difference
that, that is there; why, because, so, there
28:04.850 --> 28:09.549
is a overall growth, growth of the cells.
And, how is this overall growth quantify,
28:09.549 --> 28:13.770
if I come here. So, what is the overall growth?
If you look at this system, for example, the
28:13.770 --> 28:20.000
overall growth is, 1 cell, 1 x, 1 cell, 1
daughter cell grows from this, ok.
28:20.000 --> 28:27.000
So, the total growth for the system d x t
d t, d d t of x t equals, so, k 3 times x
28:28.390 --> 28:35.390
prime plus k 4 times x double prime. So, one
daughter cell growth, each of these reactions;
28:35.640 --> 28:40.570
first reaction, one daughter cell growth,
and second reaction, one daughter cell growth;
28:40.570 --> 28:46.200
is it clear to everybody? This is a little
tricky part. So, this is what we have. So,
28:46.200 --> 28:53.200
I hope you agree with me. Now, d x prime d
t equals 0. So, x prime equals, if you go
28:57.880 --> 29:02.990
back, you know, to your notes and check the,
I, I can go back to the slide in a minute,
29:02.990 --> 29:09.990
but, just let me write this; this is what
I get here, if you go back to your notes and
29:10.370 --> 29:13.590
the pseudo steady state; this is a pseudo
steady state approximation; pseudo steady
29:13.590 --> 29:20.590
state approximation; this is what we meant.
So, you will see that, d x prime d t equals
29:25.000 --> 29:32.000
0 would give you this and if you have a doubt,
I will go back to it in the, here. So, look
29:33.900 --> 29:40.900
at this. So, what you have is k 1 x S 1 minus
k minus 1 x prime minus k 3 x x prime, ok.
29:41.890 --> 29:46.620
So, from here, you can, if this side, the
left hand side is 0, you can straightaway
29:46.620 --> 29:53.620
express your x prime as a function of x and
S 1; clear to all of you? So, the same thing
29:54.250 --> 30:00.150
you can do over here, with x double prime.
If your d x, d d t of x double prime is 0,
30:00.150 --> 30:05.539
then, you can express your x double prime
as a function of x and S 2; yes, S 2, right.
30:05.539 --> 30:07.669
So, this is what we do over here.
30:07.669 --> 30:14.669
So, x, x, d d t of x prime is written as 0,
and x prime is given by this number, and x
30:15.150 --> 30:19.890
double prime is written as this number. So,
why we are doing it because, we intend to
30:19.890 --> 30:25.059
substitute these two into our, our constraint
equation, just like we had done before. So,
30:25.059 --> 30:29.179
when we substitute these two into our constraint,
our constraint equation is here at the top
30:29.179 --> 30:33.950
and what we do is, we substitute these two
into our constraint equation. So, this is
30:33.950 --> 30:38.620
what I get. So, x t equals x times…So, x
is the first term, then, x prime here. So,
30:38.620 --> 30:44.330
you take x out of the parameters, it is common,
out of the parameter; so, you get this and
30:44.330 --> 30:51.330
get this. But, so, only thing that you have
to keep in mind again that, x t is not a constant
31:05.650 --> 31:09.169
over here; it is varying with time. So, that
is the major difference between what we are
31:09.169 --> 31:16.169
doing now, to what we did in the previous
constraint equation. Now, we can take a derivative
31:17.039 --> 31:24.039
of this; I will, I will continue, as soon
as you finish writing. Shall I go ahead?
31:28.820 --> 31:31.390
Now, we take the derivatives. So, d x this,
this was already something that I have written
31:31.390 --> 31:36.700
d x d t, d d t of x t is k 3 x prime plus
k 4 x double prime. So, you do the same thing
31:36.700 --> 31:41.190
again; you replace your x prime by what you
got from the pseudo steady state approximation;
31:41.190 --> 31:44.799
you replace your x double prime by what you
got from the pseudo steady state approximation,
31:44.799 --> 31:51.799
fine. Just write the first step, then, I will
show the, the next step; just write the first
32:13.159 --> 32:17.159
line, then, I will show you the, you know,
how to get the second line. You probably know
32:17.159 --> 32:22.110
that already, but, I will still show you.
32:22.110 --> 32:29.110
So, d x d t equals k 3 x prime plus k 4 x
double prime. So, this, you replace; as soon
32:29.320 --> 32:36.320
as you replace, you get x times this number
k 3, fine. Now, what you do is, let us go
32:47.260 --> 32:52.950
to this slide. So, here, you can have the…Here,
the last line, you have the, you have the
32:52.950 --> 32:59.830
relationship between x t and x. So, because,
why I am doing this because, we, we, we want
32:59.830 --> 33:04.630
to find everything, in terms of the total
amount of cell that is present in the system.
33:04.630 --> 33:08.210
So, because it does not matter to us, which
one is the, is a pseudo steady state, which
33:08.210 --> 33:11.260
is the daughter, which is the, because, what
you can measure essentially at the end of
33:11.260 --> 33:15.049
the day is, a total amount of cells; you know,
you, under microscope, you cannot differentiate
33:15.049 --> 33:18.309
that, this is in the intermediate species,
this is the final and this is the initial
33:18.309 --> 33:21.630
species. So, what you can measure, is the
total amount of cell. So, that is why, all
33:21.630 --> 33:25.630
our calculations, we want to do it in terms
of the total amount of cell. So, x t over
33:25.630 --> 33:32.630
here is 1 plus k 1 S 1 k minus 1 k 3 plus
k 2 S 1 k minus 2 plus k 4, fine. So, this
33:40.779 --> 33:47.779
quantity is a little too big, so, let me call
it, this B or something. So, if x is x t times
33:52.950 --> 33:59.950
B inverse, then, you can put it over here,
into this equation, this whole thing. So,
34:07.710 --> 34:14.710
x t is… So, then, you can get it in that
form and let us call this, this thing A; this,
34:19.200 --> 34:26.200
this term is A; this term is B; then, what
you have is…So, what I did was, I just said,
34:26.909 --> 34:28.159
this term is A; this term is B.
34:28.159 --> 34:35.159
And then, what you have is d d, d x d t equals
x t times A times B inverse, fine. Now, what
34:43.740 --> 34:50.740
is my mu? mu is the specific growth rate,
fine. How is it defined? 1 over x t times
35:00.550 --> 35:06.780
d x t d t; take the first order rate constant.
So, from this, what we will get from this
35:06.780 --> 35:13.780
equation, from here? So, what I get, this
equals A times B inverse. So, I got my specific
35:15.040 --> 35:15.560
growth rate.
35:15.560 --> 35:21.280
Now, the specific growth rate for the Monod
model, if you remember, were some S, you know,
35:21.280 --> 35:28.280
K S times r x r max S times K over K plus
S, right; that was the Monod growth model,
35:30.720 --> 35:37.720
mu, right. So, now, you can…So, this is
what you get. So, the exactly the, you know,
35:40.180 --> 35:45.030
once we do the, the steps that we did, did
in, on paper, this is what you get and mu
35:45.030 --> 35:52.030
is your, mu, a nu rate, specific growth rate.
So, once you do the simplification, which
35:54.490 --> 35:58.640
we are not doing because of lack of time…So,
this is your mu; this inverse time this. And,
35:58.640 --> 36:03.070
there is, if you multiply both of these, and
you know, you can do a little bit of simplification,
36:03.070 --> 36:07.210
then, you see, find out that, mu comes out
to be in a very nice compact form, and what
36:07.210 --> 36:13.339
is interesting over here is that, the, it
is interesting to note is that, the substrates
36:13.339 --> 36:16.990
influence each other.
So, the specific growth rate of the two substrates
36:16.990 --> 36:20.480
put together, is not equal to the specific
growth rate of substrate one, and the, plus
36:20.480 --> 36:24.770
the specific growth rate of substrate two.
They are not independent; the substrate, the,
36:24.770 --> 36:29.750
the, the two growths are not independent of
each other, and what you find over here is
36:29.750 --> 36:34.540
that, the, this is specific growth rate of
substrate one; but, it is being influenced;
36:34.540 --> 36:40.200
in the presence of substrate two, is, there
is an influence out here, alpha 2 S 2. Similarly,
36:40.200 --> 36:43.560
the specific growth rate of substrate two
is being influenced by this specific growth
36:43.560 --> 36:50.150
rate of substrate one and that is influenced
out here in terms of alpha 1 s 1. And, alpha
36:50.150 --> 36:57.150
1 and alpha 2 are constants, obviously. So,
again, when you come to alpha 1 and alpha
36:57.680 --> 37:02.630
2, interestingly, we will see that, alpha,
alpha 2 and alpha 1, both of them involve
37:02.630 --> 37:06.750
all the rate constants, all the rate constants
of the reactions.
37:06.750 --> 37:13.650
So, k 1, k minus 1, k 3, k 2, k minus 2, k
4. So, all the six rate constants of the two
37:13.650 --> 37:20.650
reactions are involved in alpha 1 and alpha
2. And not surprisingly, what you find that,
37:35.660 --> 37:42.660
mu max 1 equals k 3 and mu max 2 equals k
4. Why is that not surprising, because, the
37:42.910 --> 37:47.609
maximum rate possible is the one, you know,
in the absence of everything, just the straight
37:47.609 --> 37:52.910
forward production rate. And, why I am saying
that is interesting, is because, you know,
37:52.910 --> 37:56.670
you, we did not presume anything; we just
went ahead and did the calculation, and when
37:56.670 --> 38:02.830
you got the final results, you can make some
physical sense out of those results. So, k
38:02.830 --> 38:09.830
1 is the specific, you know, the rate constants
in, involved in the reaction one; k 2 are
38:11.510 --> 38:15.900
the rate constants involved in reaction two.
And, as I said that, the mu max, maximum growth
38:15.900 --> 38:22.450
rate of reaction one is k 3; the maximum growth
rate of reaction two is k 4, fine.
38:22.450 --> 38:27.880
So, this is an important result and, and you
know, you might want to at least remember,
38:27.880 --> 38:31.109
if not remember it completely, at least, have
a sense of what is going on. And, this is
38:31.109 --> 38:35.109
not difficult to remember actually, because,
this is a Monod growth kinetics and you just
38:35.109 --> 38:38.990
add one term in the denominator and this is
a Monod growth kinetics, add one term in the
38:38.990 --> 38:42.890
denominator; and what are the difficult to
remember, might be these constants. So, those
38:42.890 --> 38:47.020
might be difficult to remember. So, the next
thing we do is, the effect of inhibition.
38:47.020 --> 38:51.670
So, here, first thing we looked at, is the
effect of multiple substrates; but the next
38:51.670 --> 38:53.290
thing that we do, is the effective inhibition.
38:53.290 --> 38:58.000
So, what is the substrate? You know, if you
have one single substrate, but, instead of
38:58.000 --> 39:04.359
the substrate, along with the substrate producing
the cells, the substrate inhibits. So, this
39:04.359 --> 39:09.500
is exactly the same, if you remember, as substrate
inhibition that we did for the case of enzyme
39:09.500 --> 39:14.609
kinetic; exactly the same. So, the kinetics
is also the same. So, in there also, we had
39:14.609 --> 39:20.200
the thing X plus S giving X S, which you know,
in, in earlier case, gives products, but,
39:20.200 --> 39:26.210
in this case, our product is the new cell.
So, X plus S giving X S gives new cell and
39:26.210 --> 39:31.930
this is the inhibition. So, X plus S, more
substrate is there, that present over there;
39:31.930 --> 39:37.890
instead of, you know, giving new cells, it
gives this complex X S 2, which does not lead
39:37.890 --> 39:42.750
to product formation. So, we have to, you
know, go through the similar set of steps
39:42.750 --> 39:49.420
that we did for inhibition in case of enzyme
kinetics, except that, in these, in this case,
39:49.420 --> 39:56.359
the difference being that, this is varying
with time. So, x t is not a constant, but,
39:56.359 --> 40:03.359
x t is varying with time, ok.
So, x t over, d x t, d d t of x t over here
40:06.560 --> 40:11.960
is K times X S, right; is that clear to everybody?
Why is that, because, this second reaction
40:11.960 --> 40:16.170
is not leading to any growth of cell. So,
you know…So, that is not considered over
40:16.170 --> 40:23.170
there. So, d d t of x t is K times S. Now,
K, K, K times X S, sorry. Now, from here,
40:24.510 --> 40:29.630
from the first reaction, the reversible first
reaction, what you find is, if this is assumed
40:29.630 --> 40:34.640
to have attained equilibrium, then, X S could
be written as, X times S could be written
40:34.640 --> 40:41.640
as, K S times X S; is it clear to everybody?
Very straightforward. So, we can replace my
40:44.000 --> 40:51.000
X S that I have over here, by X times S divided
by K S, fine. So, you have K over K S times
40:51.430 --> 40:56.520
X times S, ok.
Now, the other things to take care of, this
40:56.520 --> 41:01.589
X S 2. So, my constraint equation again, is
the total amount of cell that is present.
41:01.589 --> 41:07.900
Cells are present as parent cells and daughter
cells, in, as X, in terms of the complex,
41:07.900 --> 41:14.240
intermediate complex X S, and in terms of
the complex that you have is X S 2. So, the
41:14.240 --> 41:18.540
cells are presented in these three different
forms, fine. Again, I want to reiterate that,
41:18.540 --> 41:25.540
x t is not a constant, but, it varies with
time. So, d d t of x t is not 0; but, what
41:26.630 --> 41:31.060
we do over here is that, this, from this inhibition,
in inhibition reaction, the substrate inhibition
41:31.060 --> 41:37.450
reaction, this, we again assume to have gained
equilibrium, fine. We had assumed the substrate
41:37.450 --> 41:42.170
inhibition reaction given here to have attained
the equilibrium and as the result, we can
41:42.170 --> 41:49.170
write K i as X S times X S times X S 2, clear.
Then, X S, again, we can write from this equation
41:52.140 --> 41:59.140
as, X times S over K S, clear. So, that is
what we do. So, we write this X plus X S plus
42:03.780 --> 42:10.780
X S 2. So, X plus X S, I have, I have substituted
from here. So, X times S over K S and X S
42:11.440 --> 42:18.440
2, I have substituted from here. So, X S 2
is, from here is, X S times S over K i, fine;
42:18.910 --> 42:25.910
where, X S could again be substituted by,
X times S over K S, right. So, I can write
42:27.280 --> 42:32.320
this like this, and then, X, then, I can take
X common, out of the whole thing, and this
42:32.320 --> 42:37.220
is, this is the part I get. I will just show
you, if there is any problem with this.
42:37.220 --> 42:44.220
So, x t equals X plus X S plus X S 2. I am
just doing away with the bracket. So, X S
42:45.390 --> 42:52.390
2 equals X S times S over K i. Now, X S equals,
this is from the X times S over K S and replace
42:58.820 --> 43:05.820
this over here. So, what I get is, X times
S square over K i K S. So, from, if I replace
43:08.310 --> 43:15.310
it over here, I get X plus X S X times S over
K S plus X times S square over K i K i K S.
43:22.150 --> 43:27.430
So, you can take it out.
43:27.430 --> 43:34.430
So, what you, what you get over here is that,
d x d t equals K S, K over K S times X times
43:34.750 --> 43:39.640
S; the reason being that, this is the final
reaction. This is the only reaction, only
43:39.640 --> 43:43.089
thing that generates products, fine. Is that
clear?
43:43.089 --> 43:48.280
So, this is the only, this is the only step
that generates product. So, you have K over
43:48.280 --> 43:55.280
K S times X S and then, you can substitute
your X, X in terms of x t that we got. So,
43:56.250 --> 44:01.579
x in terms of x t is, we got over here; you
substitute that back from here and you get
44:01.579 --> 44:08.579
in terms of x t. Now, my specific growth rate
is 1 over x t d d t of x t. So, what is that?
44:09.670 --> 44:16.670
That is now, my mu over here, fine. So, what
you find is mu, is this. So, what is the difference?
44:18.920 --> 44:23.849
The major difference that you find over here
is, as compared to the mu that you have before,
44:23.849 --> 44:30.109
was that, the mu before, had something which
is similar to a Monod growth kinetics; but,
44:30.109 --> 44:35.710
this is no way similar to the, close to the
Monod growth kinetics. And, if you remember
44:35.710 --> 44:40.579
what we did from last time, what, what is
going to be the characteristic of the mu?
44:40.579 --> 44:47.579
How is this? You know, how is this mu going
to be different from mus’ in the last case,
44:49.990 --> 44:56.990
graphically? Let us think graphically. So,
you know, mu over S; in the Monod growth kinetics,
44:57.970 --> 45:04.970
it is going to be something like this, right.
So, Monod. So, for multiple, you will have
45:08.630 --> 45:15.630
S 1 and S 2. So, say mu 1 is something like
this; mu 2 is something like this, for multiple;
45:19.650 --> 45:19.900
this is mu. Now, how is this.
45:19.839 --> 45:26.839
Yes. So, it will go through a maximum and
then, go down to 0. So, this one starts at
45:29.359 --> 45:36.359
0 and saturates; this one starts from 0 and
goes to 0 through a maximum. Why is that?
45:36.570 --> 45:43.430
If you look at the expression itself, you
will see that, that, it will, it will saturate
45:43.430 --> 45:50.430
out in the absence. So, how do you obtain
the maximum? So, they are simply differentiated
45:51.470 --> 45:58.470
and do a del mu del x and get that. So, we
did a del mu del x to get the maximum value,
45:58.760 --> 46:03.950
and the maximum specific growth rate turns
out to be, S critical is square root of K
46:03.950 --> 46:10.950
i K S; square root of K i K S. It is similar,
if you remember from last time, in the case
46:12.240 --> 46:19.240
of the enzymatic growth, it is exactly similar,
to what we did in the last case. when you
46:44.520 --> 46:51.520
are done, last case. For the last case, you
can do some calculations on your own also.
47:01.280 --> 47:08.280
So, there is a mu; just do, do a del mu del
x and get this. So, shall we move on?
47:27.700 --> 47:33.430
So, the last thing that we are going to do
is, in today’s lecture is, allosteric inhibition
47:33.430 --> 47:37.560
and this is the last thing that we are going
to do in this chapter also. So, in this kind
47:37.560 --> 47:44.560
of inhibition, what happens is that, instead
of stopping here…See, what is the difference
47:44.790 --> 47:49.890
between the last case and here, is set, last
time, we stopped here, X S 2 and we stopped
47:49.890 --> 47:54.880
here. In allosteric inhibition, what happens
is that, this also gives rise to daughter
47:54.880 --> 48:00.700
cells. We do not stop here completely. We
say that, this also gives a rise to, rise
48:00.700 --> 48:07.339
to daughter cells. So, in a way, it is the
combination of substrate inhibition and multiple
48:07.339 --> 48:12.700
substrates. Come to think of it, it is a,
do you agree with me? So, in a way, it is
48:12.700 --> 48:16.630
a combination substrate inhibition. In substrate
inhibition, we had everything like, we had,
48:16.630 --> 48:21.030
but, we stopped over here, right. So, there
is no daughter cells produced from the second
48:21.030 --> 48:25.780
reaction. But, in the multiple substrate,
we did not have the same substrate, or combining,
48:25.780 --> 48:32.260
but, we had daughter cells from both the substrates.
So, in a way, it is like a combination of
48:32.260 --> 48:37.410
multiple substrates and substrate inhibition.
So, I would not do all the steps for it. I
48:37.410 --> 48:43.380
will do some steps and you can go ahead and
do the rest, either here, or later, maybe
48:43.380 --> 48:49.760
at home, as an assignment. So, the procedure
is that, the same procedure is as before;
48:49.760 --> 48:55.380
the only difference that is going to happen
is that, del del t of x t is K 2 times X S
48:55.380 --> 49:02.380
plus this beta K 2 or X K 3, K 4 also we can
call it. But, we just call it beta K 2, just
49:05.180 --> 49:12.180
for the sake of algebra, times X S 2; is that
clear? So, I think, there, there is nothing
49:13.510 --> 49:20.510
hard in it. So, and then, you have to go and
do the rest of the algebra, which is the x
49:22.410 --> 49:29.410
t, you have to, you know, calculate the x
t, which is the X plus X S plus X S S, as
49:31.750 --> 49:38.750
before, and do the del del t of x t. And then,
take the reversible for both cases, and this
49:44.650 --> 49:50.990
is a final result that you will come up with
for mu. mu is defined as 1 over x t del d
49:50.990 --> 49:56.440
d t of x t and this is the final result, you
will come up with. So, the difference between,
49:56.440 --> 50:03.440
yes, the difference between last time and
this time is that, last time, we had the denominator
50:05.920 --> 50:09.540
and you had one term in the numerator; you
have, you have another term in the numerator.
50:09.540 --> 50:14.740
So, it is quadratic, both in the numerator
and the denominator and when you, again, it
50:14.740 --> 50:18.500
will have a lot more interesting kinetics.
So, what I would like you to do is, probably
50:18.500 --> 50:23.520
slide here. So, what I would like you to do
is, actually go and plot this; because, this
50:23.520 --> 50:28.040
is not necessarily going to have the same
kinetics as, as that and it can even have
50:28.040 --> 50:31.790
multiple peaks. I am not sure, but, once you
differentiate it, you will be able to figure
50:31.790 --> 50:34.930
out that, if it has multiple peaks. So, what
I want you to do is, as a little assignment
50:34.930 --> 50:39.970
is that, do the steps; steps are very straight
forward; but, do the steps and check the answer.
50:39.970 --> 50:45.609
And then, once you have got this, and I want
you to plot this a and b c. The first thing
50:45.609 --> 50:50.030
you should actually do is, go and differentiate
it; find the, find the maxima and the minima;
50:50.030 --> 50:53.599
because, once you get the maxima and the minima,
you can plot it; very straight forward. Otherwise,
50:53.599 --> 50:58.329
you know, it is like a black box, trying to
plot this. So, go and find the maxima and
50:58.329 --> 51:02.710
the minima; one of the maxima is going to
be like what we had before; but, there in,
51:02.710 --> 51:06.520
there is a quadratic term. So, it could lead
to some other interesting effect.
51:06.520 --> 51:11.910
So, I want you to see that. So, this is, this
is a something, very common that happens and
51:11.910 --> 51:16.500
this is, this is the specific growth rate
in, in, in that case and we need to understand
51:16.500 --> 51:21.050
how this works, in order to be able to figure
out. So, the reason we are doing this is,
51:21.050 --> 51:24.790
like, you know, what we are going to do in
the next chapter and these, all these cases
51:24.790 --> 51:28.619
that we did today are important, because,
I am going to quiz you on, on those things,
51:28.619 --> 51:29.880
later in the exams.
51:29.880 --> 51:34.849
So, in the next chapter, what we are going
to look at, is the process, the dynamics of
51:34.849 --> 51:40.910
cell growth. See, we are looking at the kinetic
sub cell growth. But, what happens is that,
51:40.910 --> 51:46.400
the cell growth process happens in a, in a
reactor like this, you know. So, it could
51:46.400 --> 51:51.099
be a batch reactor; it could be a chemostat,
continuous third tank reactor.
51:51.099 --> 51:55.410
So, one of the things that we are going to
look at is, starting next class, it is a new
51:55.410 --> 52:02.410
chapter, is chemostat ,which is a C S T R,
C S T R for a biological reaction. So, temperature
52:03.430 --> 52:10.430
is controlled and you know, pH is controlled.
So, pH and temperature controlled and then,
52:15.619 --> 52:21.150
we look at the growth process inside this
reactor. So, that is when all the kinetics
52:21.150 --> 52:25.260
that we are doing in this class and the previous
classes in the mass transfer effects and so
52:25.260 --> 52:28.700
on, come into play. So, right now, we are
looking at…So, this is all connected. So,
52:28.700 --> 52:32.980
right now, we are looking at the kinetics
in itself; just isolating the kinetics and
52:32.980 --> 52:38.130
looking at it. Then, we put, check the kinetics
out and put into the reactor, and look at
52:38.130 --> 52:42.060
what the effects are, and this, these are,
I am, I am assure you that, some of these
52:42.060 --> 52:45.460
effects are going to be very interesting.
So, we, we will work with, you know, some,
52:45.460 --> 52:51.109
probably, what we will start with is, not
even this, the very straight forward case
52:51.109 --> 52:55.869
here, the Monod kinetics and then, we will
go and slowly do what I will probably ask
52:55.869 --> 53:01.670
you to do, what the effects of multiple reactions
are, the effects of inhibition and the effects
53:01.670 --> 53:08.670
of allosteric inhibition. So, we will stop
here, and I will see you in the next class.
53:10.369 --> 53:15.319
Thanks.