WEBVTT
00:18.140 --> 00:22.539
In the last lecture, we started off with microbial
growth models and we will continue from where
00:22.539 --> 00:27.119
we left; a little bit may be, earlier. So,
we talked about the process of growth and
00:27.119 --> 00:30.750
what are the nutrients that are necessary,
what are the chemical reactions that happened
00:30.750 --> 00:31.640
during growth.
00:31.640 --> 00:36.260
And then, we started to look at the different
kinds of growth models that is there. So,
00:36.260 --> 00:41.079
if you look at that screen here, so, we started
off with the Malthusian model, which simply
00:41.079 --> 00:46.460
said that, the r x is d x d t. And then, we
went on to look at the Logistics model, which
00:46.460 --> 00:53.460
tried to propose the dependent, second order
dependence of, of of, on the reaction rate
00:53.550 --> 01:00.550
of the substrate and, and then, we got d x
d t is k x times 1 minus beta x.
01:02.379 --> 01:07.780
And then, the next model was the most important
model we looked at, which was the Monod growth
01:07.780 --> 01:14.780
model. Just check; so, I think something skipped
here. So, we, we had the Monod model to start
01:17.610 --> 01:24.610
with, which was mu times S over k S plus S.
And then, we modified it using the Monod growth
01:24.740 --> 01:29.049
model, which we called the modified Monod
model; we added the K S plus S 0 term in the
01:29.049 --> 01:32.860
denominator. And then, we had the Konak model.
As you can see over here, what happens in
01:32.860 --> 01:37.470
the Konak model is, that Konak model is for
the limit of p going to 2, for the case of
01:37.470 --> 01:42.790
p going to 2, goes to the Monod model, right.
So, this is where we stopped, if I remember.
01:42.790 --> 01:48.170
Now, where we will start today is the value.
So, if you are doing a Monod model for example,
01:48.170 --> 01:54.329
you have these K S, the rate constant and
the mu max, maximum rate. And so, we, what
01:54.329 --> 01:59.909
we going to start with today, are the values
of mu max and K S, for various organisms and
01:59.909 --> 02:06.439
substrates, fine. So, the first one, we look
at is the Escherichia coli, e coli that is,
02:06.439 --> 02:10.360
and these are growth temperatures that you
see over here, or the optimum temperature;
02:10.360 --> 02:14.220
that, as I told you before, that there is
a range within which, which these microorganisms
02:14.220 --> 02:19.900
can grow; but then, within that range also,
there is a temperature that is an ideal, or
02:19.900 --> 02:26.239
optimum temperature. As, as we discussed before,
what happens here is that, for an ideal, or
02:26.239 --> 02:30.569
optimum temperature, and I talked about this
point, you know, tropical or subtropical countries,
02:30.569 --> 02:34.360
you have the growth of these bacteria are
a lot more than colder countries; the reason
02:34.360 --> 02:39.310
being that, the optimal temperature, as we
can see over here, is ideally suited for a
02:39.310 --> 02:46.310
tropical or subtropical climate. So, excuse
me. So, e coli, for example, it can grow under
02:46.620 --> 02:51.659
different kinds of nutrients. So, depending
on the nutrients that it is growing in, the
02:51.659 --> 02:54.480
mu max, and the K S are obviously going to
be different, obviously; very straightforward;
02:54.480 --> 02:59.530
because, you know, if you change the substrate,
the growth rate is going to change.
02:59.530 --> 03:04.670
So, if you look here for glucose, for example,
the, if the limiting reactant nutrient is
03:04.670 --> 03:08.800
glucose for example, if you are using the
glucose to grow the bacteria, to grow the
03:08.800 --> 03:15.739
microorganism, then, mu max is 0.8 to 0.14
and the K S in between 2 to 4. If you are
03:15.739 --> 03:22.390
using glycerol to grow e coli, then, the mu
max is 0.87; K S is 2. If you are using lactose,
03:22.390 --> 03:28.379
then, it is 0.8 and 20. So, K S is much high,
larger over here. And, sacromyces cerevisiae
03:28.379 --> 03:33.440
which is yeast, you know, if you are growing,
if you are using glucose to grow that and
03:33.440 --> 03:39.159
the mu max is 0.5 to 0.6 and K S is 25; and,
this Candida tropicalis and this, this if
03:39.159 --> 03:46.159
you are growing with glucose, the mu max is
0.5 and K S varies between 25 and 75. And,
03:48.750 --> 03:54.230
Klebsiella aerogenes, if you are using the
glycerol, then, mu max is 0.85 and K S is
03:54.230 --> 04:01.230
9. So, what you find is that, the mu max,
more or less, you know, is independent, irrespective
04:03.670 --> 04:06.989
of whatever bacteria you are growing, or microorganism
you are growing, and what is the substrate
04:06.989 --> 04:10.000
you are using, more or less remains around
the same number, whereas, K S kind of changes;
04:10.000 --> 04:15.409
but again, the range of K S for many of these
is around 20, and some of these in the 10;
04:15.409 --> 04:18.970
but again, you know the order of, what I am
trying to say here is, the order of magnitude
04:18.970 --> 04:23.090
remains more or less the same. So, the order
of magnitude does not change too much; it
04:23.090 --> 04:27.890
still is in the range of 1 to 100, that order.
04:27.890 --> 04:31.060
And then, the values of activation energy
and we had discussed, if you remember, we
04:31.060 --> 04:35.089
had discussed this in the, when we were doing
enzymes, the dependence of temperature. So,
04:35.089 --> 04:39.210
whenever there is a dependence on temperature,
there is a effect of activation energy and
04:39.210 --> 04:44.690
that is also related to the question that,
why a certain temperature is optimal, or good
04:44.690 --> 04:49.099
to grow certain microorganism and why other
temperatures are not good, right. And then,
04:49.099 --> 04:52.450
you know, the question of tropical and subtropical
climates and other things arrived; because,
04:52.450 --> 04:56.339
if it is, for example, if the, these are the
temperature ranges and if the temperatures
04:56.339 --> 05:03.200
are too low, then, this is an activation energy.
Now, the, the activation energy is required,
05:03.200 --> 05:07.880
but the total amount of heat that is generated,
or required, is dependent on temperature.
05:07.880 --> 05:11.950
So, if the temperature is very low, for example,
that amount of heat would not be generated,
05:11.950 --> 05:18.849
or, or required to do that. So, that is why,
this gives you the range for the e S for the
05:18.849 --> 05:24.210
different bacteria. So, most of them as you
see, are growing between 20 to 37; this is
05:24.210 --> 05:29.289
20 to 40; but essentially, that is the growth
range. Then, that as I said, again, is a very
05:29.289 --> 05:35.649
ideal climate, you know, the subtropical climate
and these numbers are around 13 and 14. This
05:35.649 --> 05:41.360
psychrophilic pseudomonad, the temperature
range is lower here and the, it grows in colder,
05:41.360 --> 05:46.640
you know, cold temperature and the e a for
that is, obviously, larger, right; because,
05:46.640 --> 05:52.719
its temperature, activation temperature is
low, lower, then, e a has to be larger.
05:52.719 --> 05:57.440
And, you know, certain, there are certain
bacteria, or virus, for example, these, the
05:57.440 --> 06:01.839
bacteria that we are used to, at some point
of time, including this time, that is, the
06:01.839 --> 06:06.729
common cold virus. So, it is a, it is, you
know, so, for example, it show your mom will
06:06.729 --> 06:10.250
say that you caught a cold. What do you mean
by you caught a cold? You did not catch a
06:10.250 --> 06:15.469
cold; cold is not a thing that you can catch;
it is the virus that you catch. So, the reason
06:15.469 --> 06:19.300
people say this so easily that you caught
a cold, then, it, it is kind of, becomes a
06:19.300 --> 06:25.240
part of our everyday linguistics; but essentially,
we are talking about viruses. When people
06:25.240 --> 06:28.930
say that you caught a cold, what they mean
is, you caught the common cold virus. So,
06:28.930 --> 06:32.140
you know, if you expand the statement that,
that is what it means, you caught the common
06:32.140 --> 06:38.779
cold virus. So, it is one of the rare viruses
which actually can exist in a cold, colder
06:38.779 --> 06:42.979
climate; in a, lot more than the summer climates.
So, you would not have the common cold virus
06:42.979 --> 06:47.529
even in summer or high temperature season,
whereas you would have this especially during
06:47.529 --> 06:52.099
the cold temperature. So, this is one of those,
so, just, you know, the, so, which grows in
06:52.099 --> 06:56.289
2 to 12, then, common cold virus actually
grows in temperature this and much lower than
06:56.289 --> 06:56.539
this.
06:56.349 --> 07:03.349
Now, what we are going look at is essentially,
the major topic of today, is the effect of
07:05.510 --> 07:10.260
mass transfer and growth. Now, if you remember,
this is exactly how we did the enzyme. So,
07:10.260 --> 07:14.510
we look, looked at kinetics of the problem.
So, we looked at the effect, the reaction
07:14.510 --> 07:18.070
aspect of it and then, we looked in immobilized
enzyme, enzyme we looked at the transport
07:18.070 --> 07:22.219
aspect, it was the mass transfer aspect. And
then, what you did, much of the first half
07:22.219 --> 07:29.219
of the semester was, the combined effect of
mass transfer and reaction, which was, I think,
07:29.740 --> 07:34.430
sort of something new, you know; I am not
sure, exactly what you did in the last, previous
07:34.430 --> 07:41.430
courses, but, sort of something new, as compared
to what you might have done in previous courses.
07:41.490 --> 07:45.500
You would have done, I mean, there are courses
where would have done mass transfer, or transport
07:45.500 --> 07:49.990
and then, there are courses where you have
done reaction, but what, but, in real life,
07:49.990 --> 07:54.390
what happens is, these two things, kind of
compete with each other. So, reactions and
07:54.390 --> 08:00.039
mass transfer, either occur in series, or
occur parallelly, or simultaneously.
08:00.039 --> 08:04.909
So, then, we, if you remember, what we did
in the last part of the semester was, we looked
08:04.909 --> 08:11.019
at these several, if, cases, where reaction
follows mass transfer and there are cases,
08:11.019 --> 08:15.029
where reaction occur simultaneously with mass
transfer. And, the same thing is actually
08:15.029 --> 08:19.649
applicable to growth model, the reason being
that, growth is essentially a reactive process.
08:19.649 --> 08:23.950
It is a reactive uptake process. And, if you
remember, you know, just because, in a few
08:23.950 --> 08:30.950
days, where I can go back and show you the
reactions over here. Some reason, I think,
08:37.180 --> 08:44.180
something happened. Anyway, so, so the, if
you remember, I do not have the reactions
08:48.050 --> 08:52.880
with me right now, but, if you remember, the
reaction that we wrote was, the reaction of
08:52.880 --> 08:59.880
glucose, or some kind of carbohydrate and
nitrogen with the cell, which resulted in
08:59.910 --> 09:06.280
an uptake of carbon, hydrogen, oxygen and
nitrogen to the cell, right. And, one of the
09:06.280 --> 09:10.640
confusion that you might have, see for example,
what we are doing in the model are, we are
09:10.640 --> 09:15.840
using 2 symbols, x and S. So, what does the
x stands for? x stands for the cell and S
09:15.840 --> 09:20.070
stands for the substrate. So, please make
a note of that, you know; do not, do not forget
09:20.070 --> 09:23.970
this. So, S stands for the substrate.
So, S is the substrate as a whole; so, it
09:23.970 --> 09:28.320
could be the nitrogen; it could be the oxygen;
it could be the carbon, or the hydrogen; but,
09:28.320 --> 09:33.760
substrate as a whole, in general and x stands
for the, stands for the cell. So, essentially,
09:33.760 --> 09:37.240
there is an reactive uptake of these substrates
by the cell and the cell grows as a result
09:37.240 --> 09:42.120
of, result of it. So, there is a reaction
aspect of it, but, there is a mass transfer
09:42.120 --> 09:47.270
aspect of it to, mass transfer aspect to that
as well. And, why is that because, the substrate
09:47.270 --> 09:51.730
essentially has to overcome some mass transfer
limitations, or transport limitations, to
09:51.730 --> 09:56.280
go to the point, where the cell can actually
uptake it through a reactive process, exactly
09:56.280 --> 10:01.870
like the enzyme thing, that immobilized enzyme
thing that we did, and we will do some, few
10:01.870 --> 10:07.150
cases today, or may be tomorrow as well.
So, this is an example that we are looking
10:07.150 --> 10:11.700
at here. This is the cell; this inner, inner
circle that you see over here is the cell
10:11.700 --> 10:18.700
and the outer circle that you see over here
is the liquid film, or the boundary layer.
10:19.430 --> 10:23.740
So, so, what happens essentially that, you,
this is, this is cell and it is the cell is
10:23.740 --> 10:28.290
growing in a medium, right. I discussed this
in the last lecture, all of it. The cell is
10:28.290 --> 10:32.790
growing in the medium and you might have in
the medium, may not have in the medium; you
10:32.790 --> 10:38.500
might have, you can use a plate inside the
incubator to kind of, shake it. So, whether
10:38.500 --> 10:44.730
you have in the medium or not, there is always
going to be a liquid layer, you know, a stagnant
10:44.730 --> 10:48.130
layer; it is not really a boundary layer,
but, it is a stagnant layer, which is similar
10:48.130 --> 10:52.830
to a boundary layer; boundary layer is for
a moving system, but, you know, even if you
10:52.830 --> 10:56.950
have or not ((stirring)); so, what happens,
because the cell is solid, there is always
10:56.950 --> 11:01.730
going to be a stagnant layer of liquid around
the cell, right; is it clear to everybody,
11:01.730 --> 11:06.980
or is that something that you are agree with?
So, then, this layer, liquid layer, say, has
11:06.980 --> 11:11.010
a thickness of d, or something, whatever,
I might say.
11:11.010 --> 11:13.460
Now, what happens as a result of the presence
of this liquid layer is that, the substrate
11:13.460 --> 11:19.980
has a concentration S 0 outside and the concentration
at the interface of the liquid layer and the
11:19.980 --> 11:23.420
cell is obviously, going to be lower than
the concentration outside, which means that
11:23.420 --> 11:30.380
there is a concentration gradient in between
the bulk and this one. Now, unlike in the
11:30.380 --> 11:35.620
case of the enzyme, we may not actually want
to go and solve for the concentration profile
11:35.620 --> 11:40.050
of, in the, in this liquid layer, stagnant
cell; the reason being, the stagnant cells
11:40.050 --> 11:45.460
are typically very thin, and it is probably
of not much practical use, to actually go
11:45.460 --> 11:49.570
and solve for the concentration profile in
this.
11:49.570 --> 11:53.650
So, what we do, we simply use the concept
of the mass transport coefficient. So, what
11:53.650 --> 12:00.650
you see over on, here on the screen, k L times
a prime x over rho cell times S 0 minus S.
12:00.700 --> 12:07.700
Let us, let me explain each little thing.
So, k l times a, a prime is, the k l is probably
12:08.490 --> 12:15.490
the mass transfer, mass transfer coefficient
you times this area and then, x over rho cell
12:16.740 --> 12:22.370
is the volumetric increase of the, you know,
increase of the cell volume, volumetric increase.
12:22.370 --> 12:29.370
So, we divide it, say x s by weight and you
divide it by the density. So, you get the
12:29.779 --> 12:34.500
volume of the cell. So, this is, sorry, this
is, k l times a prime would then be for unit
12:34.500 --> 12:38.620
volume. The unit of this would be for unit
volume. So, this would be volume. So, k l
12:38.620 --> 12:43.690
times a prime times x, x over rho cell would
be the, for, for a particular given volume,
12:43.690 --> 12:49.160
so, this would be the mass transfer coefficient
times, the difference, of course, S 0 minus
12:49.160 --> 12:54.260
S. So, this is the amount of substrate transport.
Why is it proportional to x? Why is that proportional
12:54.260 --> 13:00.490
to x, because, the more is the, more is the
volume of the cell, more is a total transport
13:00.490 --> 13:05.330
going to be, right; and, the volume of the
cell is proportional to its weight; is it
13:05.330 --> 13:09.610
clear? The more is the volume of the cell,
the more is the total transport going to be.
13:09.610 --> 13:15.090
Is it something you agree with?
So, next is the...So, this is the rate of
13:15.090 --> 13:18.610
substrate transport and then, what would be,
what, what is that we, we are going to balance
13:18.610 --> 13:22.600
it with what is going to happen. So, at the
interface, for example, the rate of substrate
13:22.600 --> 13:27.110
transport at the interface. So, at the interface,
for example, if there are no internal mass
13:27.110 --> 13:30.930
transfer gradient inside the cell, it is very
likely that, there are internal mass transfer
13:30.930 --> 13:34.430
gradient inside the cell, and you know, we
did this in another course, at some point
13:34.430 --> 13:40.990
of time; some of you did that course, ignore
that; what happens, we solve for the concentration
13:40.990 --> 13:46.270
profile within the cytoplasm; you can, you
can even solve for concentration profiles
13:46.270 --> 13:51.180
of certain materials of substrate within the
cytoplasm, if possible. But here, we assume
13:51.180 --> 13:55.700
the cell to be very small and we assume that,
the reaction within the cell is more or less
13:55.700 --> 14:00.900
uniform; we do not assume a mass transfer
gradient within the cell, which is not an
14:00.900 --> 14:05.480
unreasonable assumption, let me tell you.
It is possible to go, see, you decide what
14:05.480 --> 14:08.779
is the level of details you want to go into.
So, you can look at, for example, if you want
14:08.779 --> 14:13.980
to, this model that we are solving, if you
want to solve a really detailed model, what
14:13.980 --> 14:19.430
would you do? You would take the liquid film
and then, solve for the profile within the
14:19.430 --> 14:23.960
liquid film; solve for the profile within
the cell, and then balance all of that. What
14:23.960 --> 14:27.600
we are doing instead is that, we not solving
for any of the profile even within the cell,
14:27.600 --> 14:31.839
or the profile within the liquid film; we
are simply saying, the rate of substrate transport
14:31.839 --> 14:37.500
is mass transfer coefficient times volume
of the cell times S 0 minus S; and, for the
14:37.500 --> 14:43.250
cell, what is now, what is the uptake. So,
the uptake is due to the reaction in the cell,
14:43.250 --> 14:47.580
right.
So, what would the uptake be? We already did
14:47.580 --> 14:54.580
that, you know, the growth. What is a uptake?
We did the growth model. So, if you are using
14:56.550 --> 15:03.550
a Monod model, for example here, what would
be the uptake? The uptake is, you know, what
15:05.800 --> 15:12.800
I gave here. So, which is S times x divided
by K S times K S plus S times q max. So, q
15:13.750 --> 15:19.860
times x is exactly the Monod growth model,
if you remember. So, I used mu; instead of
15:19.860 --> 15:24.460
mu I use q, over here, and I will come to
the reason of it,in a moment; you will see
15:24.460 --> 15:30.430
why I used that. But, the same thing exactly,
so, q times x. So, q is the growth rate per
15:30.430 --> 15:35.690
unit weight of the cell, fine. Growth rate
for unit cell times number of cell again,
15:35.690 --> 15:40.430
because more is the amount of cell, you know,
more is a growth; is it clear? Linearly dependent
15:40.430 --> 15:46.380
on growth, cell. So, q. So, q is given as
q max S over K S plus S, Monod growth model
15:46.380 --> 15:51.550
times x. So, what happens is that, at steady
state, these two should balance each other.
15:51.550 --> 15:57.660
So, these are, this is my basic equation that
you have, q max S X divided by K S plus S
15:57.660 --> 16:04.660
equals k L a prime X over rho cell times S
minus S 0.
16:14.260 --> 16:21.260
Then, so, this is a basic equation. What it
entails is that, the rate of substrate transport
16:30.649 --> 16:37.649
at steady state equals the rate of reactive
uptake by the cell. Now, so, here, q is just
16:38.360 --> 16:45.360
as which mentioned here; q is q max S over
K S plus S. So, what we do here is that, we
16:46.670 --> 16:53.670
expressed S as a function of q. So, we expressed
S as a function of q. Why do we do that? I
17:00.760 --> 17:06.669
will come to that; it is written here. So,
once you do that, we express S; so, S becomes
17:06.669 --> 17:13.409
q times K S over q max minus q. Once you do
that, what you can do is, look at this equation;
17:13.409 --> 17:18.439
you can go and put back your S here in terms
of all of the variables; then, this becomes
17:18.439 --> 17:25.439
in equation in S 0 and you can solve for S
0 straightaway. And, this becomes a linear
17:25.790 --> 17:30.800
equation in S 0. You can solve for S 0 straightaway,
and this is what you get for S 0. So, S 0
17:30.800 --> 17:37.800
is now expressed in terms of different transport
properties and q. So, q is a variable remember,
17:37.990 --> 17:44.990
but everything else, is a constant out here.
We will see in a minute, why I am trying to
17:45.530 --> 17:52.530
do that, you know.
Now, what I do is, now, this new thing comes
17:54.740 --> 18:01.740
in over here, is that, I defined the yield
of x per unit S. See, here, when you were
18:02.240 --> 18:08.730
writing a balance, let us see, when you are
writing a balance, your are writing a balance
18:08.730 --> 18:13.540
for the substrate; remember, you are not writing
a balance for the cell; you are writing a
18:13.540 --> 18:18.010
balance for the substrate; that is why, I
used q over here. Now, if I am doing it, if
18:18.010 --> 18:22.770
I am want to convert it into the cell, the
yield of cell, so, when you did the, that
18:22.770 --> 18:27.260
is a only little bit different between what
we did to the Monod growth model and here.
18:27.260 --> 18:32.710
Monod growth model was quantifying the amount
of cell growth. Therefore, it was given as
18:32.710 --> 18:39.710
mu times S, mu times x; whereas, this one
quantifies the amount of substrate uptake
18:40.120 --> 18:45.600
and Y is the one that gives the yield of x
per unit S. What does it mean that, see, there,
18:45.600 --> 18:49.840
it is not necessary that, there has to be
a one is to one relationship between the uptake
18:49.840 --> 18:53.450
of the substrate and the production of the
cell. It is not necessary that, one mole of
18:53.450 --> 18:56.570
the substrate should produce one mole of,
of the cell.
18:56.570 --> 19:00.690
So, mu is that ratio; it could be that, 2
moles of the substrate produce one moles of,
19:00.690 --> 19:06.309
of cell. So, mu, sorry, Y is the ratio that
would give that. So, mu over q. So, Y could
19:06.309 --> 19:11.679
be half; Y could be 0.3, 0.4; Y could be 2,
or 3 even, unlikely; but, you know, it could
19:11.679 --> 19:16.510
be 2, or 3, but, you understand the concept
very, very clearly; should I explain one more
19:16.510 --> 19:21.170
time? So, what happens is that, when I wrote
this equation, I used q, because q for the
19:21.170 --> 19:25.890
substrate, whereas, when you use these notations,
you have to remember. So, for the, when you
19:25.890 --> 19:31.710
use it for the cell, it is mu, and when you
use it for the substrate, it is q; and, q
19:31.710 --> 19:38.220
over, mu over q is the amount of cell produced,
amount of x produced per unit amount of S,
19:38.220 --> 19:45.220
that comes in; is that clear? So, now, I can
write the same equation in terms of mu. How
19:46.520 --> 19:53.520
do I do that? See, for example, so, q is mu
over Y, right, from this equation, here, look.
19:54.910 --> 20:01.910
So, Y is mu over q; therefore, mu, q itself
is mu over Y right. So, you substitute that.
20:02.179 --> 20:09.179
So, then, let me use the pen here.
So, S 0 here q K S, q max minus q plus. So,
20:19.910 --> 20:25.670
Y being mu over q, which means that, q itself
is mu over Y. So, S 0 then becomes, mu over
20:25.670 --> 20:32.670
Y times K S, Y is the number, remember; it
is not a variable, just a number; 1 over Y
20:39.380 --> 20:46.380
mu max. So, 1 over Y, 1 over Y cancels out
here. So, you have mu K S divided by mu max
20:59.470 --> 21:06.470
minus mu plus mu rho cell by K L prime. So,
if I go back to the screen now, so, the last
21:08.809 --> 21:15.809
thing you see is exactly what I did just now.
So, mu K S over mu max minus mu plus mu rho
21:17.400 --> 21:24.400
cell over K L a prime Y, fine.
21:26.780 --> 21:33.780
So, if you look at this equation that you
get over here, you can solve for mu in this
21:39.410 --> 21:46.410
equation. You can solve for, yes, mu in this
equation provided, because mu is a variable;
21:51.040 --> 21:56.170
now, q earlier. So, what we did was, you,
what, what did we do essentially, it was a
21:56.170 --> 22:00.170
variable; if you put your S over there, then,
it was variable in S; but, we did not want
22:00.170 --> 22:06.549
to put my, our S in that. So, we put in terms
of q. Then, we converted q to mu and then,
22:06.549 --> 22:10.990
you can solve for mu, because mu is a variable,
which is unknown over here and that could
22:10.990 --> 22:17.990
be solved for, if the determinant is greater
than the 0. So, you may want to take a minute
22:21.450 --> 22:26.440
and look at this and you, if you want, I can
show you this. Maybe, that is what I will
22:26.440 --> 22:27.510
do. So, here.
22:27.510 --> 22:34.510
So, your equation is mu K S mu max minus mu
plus mu rho cell K L a prime Y. Then, you,
22:38.540 --> 22:45.440
you know, multiply the whole thing by mu max
minus mu and so on. So, you get mu max minus
22:45.440 --> 22:52.440
mu times K L a prime Y equals mu K S K L a
prime Y plus mu rho cell mu max minus mu,
23:01.260 --> 23:07.460
right. So, this becomes a quadratic in mu.
So, this, if you take it, so, you get mu square
23:07.460 --> 23:14.460
times rho cell, plus, let us see, minus, sorry.
So, minus S 0 K L Y plus K S K L a prime Y,
23:33.830 --> 23:40.830
plus rho cell mu max times mu, plus S 0 mu
max K L a prime Y equals 0. So, this is your
23:50.170 --> 23:54.100
quadratic, and, so, all you need to do is,
look at the determinant of this. So, this
23:54.100 --> 24:00.210
is what I do over here; and, for this to have
a solution, for the equation that I just wrote,
24:00.210 --> 24:03.960
the quadratic equation, for that to have a
solution, the determinant has to be greater
24:03.960 --> 24:08.809
than 0, which is what we use. What are, why
are we trying to do this? See, what we are
24:08.809 --> 24:15.809
trying? Can you have, do you have a sense,
why we are trying to do this? You know, this
24:16.040 --> 24:20.720
looks too complicated mathematically, but
physically, what we are trying to aim at?
24:20.720 --> 24:26.020
What we are trying to find over here?
24:26.020 --> 24:30.669
Yes, minimum concentration of what?
Substrate..
24:30.669 --> 24:37.370
Yes. So, initial, S 0 is initial amount of
substrate that is, that is there, and what
24:37.370 --> 24:42.940
is my initial amount of substrate that I require
for the cell to grow; because, that is very
24:42.940 --> 24:47.000
important. We need to figure out. See, if
you, if you undernourished the cells, they
24:47.000 --> 24:52.280
are never going to grow. So, what we are going
to find out is that, what is minimum amount
24:52.280 --> 24:57.830
of substrate that is required for the cells
to grow. See, we had the equations at the,
24:57.830 --> 25:02.730
at long back, you know; if you look at what
is here on the screen, see, we have the equation
25:02.730 --> 25:06.340
right here, and what we could have done is,
straightforward, gone and solved, gone and
25:06.340 --> 25:11.960
solved for S, right. This is a quadratic equation
in S. We have it and you can go where, go
25:11.960 --> 25:16.370
and straightforward get it, solve it and get
a solution; but, we did not go that way. We
25:16.370 --> 25:19.520
went another route and the reason was, it
is more important to actually find out that,
25:19.520 --> 25:22.900
what is the minimum amount of initial substrate
that you need to nourish the cells experimentally;
25:22.900 --> 25:29.510
for experimental growth, rather than to actually
solve for what your substrate amount is. So,
25:29.510 --> 25:32.600
that is exactly what we are trying to do.
We can always solve that; that is not a big
25:32.600 --> 25:36.900
deal to solve that quadratic equation; but,
what we are trying to find out is, what is
25:36.900 --> 25:40.270
my minimum amount of substrate and how do
you get that. You get that straight from the
25:40.270 --> 25:46.290
equation, fine; because everything else is,
you see over here, mu max, rho cell, K L,
25:46.290 --> 25:51.090
Y and K S, these are all constants.
So, the only unknown over here, or thing,
25:51.090 --> 25:54.700
that you want to control, rather; so, these
are the things that you cannot control; mu
25:54.700 --> 25:58.169
max is the property of the cell; you cannot
control; rho cell is the property of the cell;
25:58.169 --> 26:02.250
you cannot control; K L again, is the property
of the cell and the substrate and substrate
26:02.250 --> 26:05.610
liquid, and the cell, you cannot control.
Y is the property of the cell and the substrate,
26:05.610 --> 26:09.140
you cannot control. So, these are the things
you cannot control; the only thing, controlling
26:09.140 --> 26:13.809
parameter that you have here, is S 0. So,
what you want do from here is that, equate
26:13.809 --> 26:20.070
this, you know; so, you put for S 0, this
equal 0 and solve for S 0, to get, what that
26:20.070 --> 26:24.750
would give you the minimum amount of substrate
that is required for the growth and nourishment,
26:24.750 --> 26:30.679
and therefore, the growth of the cell; is
it clear to all of you?
26:30.679 --> 26:37.350
So, then, once you do that, you know, so,
you can, once this condition is satisfied,
26:37.350 --> 26:44.350
then, mu, that you had over there in the equation
itself, could be written in this form. What,
26:44.830 --> 26:51.490
what I am trying to do over here is that,
what I am trying to do, do you have a sense?
26:51.490 --> 26:55.730
So, this part, is that clear? So, this, this
is separate from what I have below, below;
26:55.730 --> 27:00.230
do not worry about this; is that, upto that,
is that clear to everybody? The S 0 thing,
27:00.230 --> 27:04.330
that what we do is, figured out, what is the
minimum amount of substrate that is necessary.
27:04.330 --> 27:08.360
Now, this is something different, I am trying
to do here. What I am trying to do here is,
27:08.360 --> 27:15.360
see, my model that I have over here, this
one, is a model for the Monod growth; but,
27:17.870 --> 27:22.240
the actual growth that is taking place, is
not the Monod growth. Why because, it is limited
27:22.240 --> 27:27.450
by mass transfer; because, it is limited by,
by transport; and therefore, the actual growth
27:27.450 --> 27:32.510
is not the same as the Monod growth. So, what
I am trying to do is, trying to find out that,
27:32.510 --> 27:37.410
what is my actual growth; and, that, with
a little bit of algebra that is here and you
27:37.410 --> 27:43.809
can have a look at that, when you go back
home is that, you convert this. So, S, you
27:43.809 --> 27:49.770
can, how, what you do is that, what is a difference.
So, if my, if I have Monod growth for example,
27:49.770 --> 27:53.820
let us see.
27:53.820 --> 28:00.820
What would be my growth, if I had Monod growth
for the cell? This is for the cell. Monod
28:01.330 --> 28:08.330
growth. You simply have mu max S times S 0
times K S plus S 0, right. And, of course,
28:15.510 --> 28:20.039
multiplied by x, that is always there. But,
so, your mu Monod, let us call this mu Monod
28:20.039 --> 28:27.039
is this, and your growth rate is d x d t would
be mu times x, fine. Now, in the presence
28:28.919 --> 28:35.070
of mass transfer, how is this going to alter?
What is the growth rate going to be, in terms
28:35.070 --> 28:40.539
of presence of mass transfer? This is going
to be mu; let us call this mu Monod. So, mu
28:40.539 --> 28:46.090
m times x. So, this is going to be mu something
else; let us call this mu m t times x. How
28:46.090 --> 28:53.090
it is going to be different, mu m t, equals,
use what; it still follows the same thing,
29:00.090 --> 29:07.090
you know. What is the mu m t? times S plus
K S plus S. We did that before, here; then
29:16.000 --> 29:23.000
look. So, this is the one. So, what is the
difference? The difference is, for pure Monod
29:24.990 --> 29:28.840
growth is, there is no mass transfer with
S 0; but, in the presence of mass transfer,
29:28.840 --> 29:32.950
you have S. And, what you need to do is, we
simply need to substitute for S. How would
29:32.950 --> 29:37.960
you substitute? If you, let us look at the
screen here. So, this is here, my equation.
29:37.960 --> 29:42.320
Now, that I have figured out, what my, so,
my major first motivation was to figure out
29:42.320 --> 29:46.220
what is the amount of, minimum amount of nutrient
that I required. Once I have figured that
29:46.220 --> 29:51.020
out, I go ahead and simply solve for the S,
which is the quadratic equation. Now, I will
29:51.020 --> 29:53.520
have two solution; one would be unfeasible
solution; one would be a feasible solution.
29:53.520 --> 29:59.789
Take that solution, and then, I can put it
back into, look, look here. So, I put it back.
29:59.789 --> 30:06.789
So, I get the solution of a, put it back over
here. So, you know. So, my, my S, that is
30:07.990 --> 30:14.990
the solution S that I get is, S 0 plus, I
am sorry; S 0 thing, that you get, yes, I
30:20.549 --> 30:23.080
think, what will happen is, it is little bit
of algebra out there.
30:23.080 --> 30:30.080
So, I am skipping the step; you can just go
and check, you will get mu max S 0, K S plus
30:31.250 --> 30:38.250
S 0 plus mu max, rho cell plus K L, a prime
into Y. So, this is what you get and so, this
30:55.710 --> 31:01.289
is what you get. So, the algebra out here
is that, you will have a denominator, you
31:01.289 --> 31:04.549
know, it will get cancelled; little bit of
algebra is out there, once you solve for the
31:04.549 --> 31:10.049
S 0. So, what you need to do is, go back,
solve for S. So, take the quadratic equation,
31:10.049 --> 31:15.100
solve for S in terms of S 0. If I need to
explain, I will do it. I think, I have it
31:15.100 --> 31:22.100
here, yes. So, this was, no, sorry, this is
not that equation. Let me write down, you
31:26.990 --> 31:29.049
know, I do not want confusion in this.
31:29.049 --> 31:36.049
So, mu max S, mu max times S times x over
Y K S plus S equals K L prime a x over rho
31:47.179 --> 31:54.179
cell S 0 minus S. So, this is your quadratic
equation that you now need to solve for S.
31:55.710 --> 32:02.710
Is it clear? Solve for S using this quadratic
equation and the solution that you get, put
32:02.929 --> 32:09.929
it in mu. So, mu equals, or mu m t let us
say, equals mu max S over K S plus S; put
32:14.289 --> 32:21.289
that, put that solution that you get, into
this; what you will get is mu max S 0 K S
32:25.850 --> 32:32.850
plus mu max rho cell K L a prime Y plus S
0. This is what you will get. So, solve for
32:43.740 --> 32:49.970
S, put this into this and after a little bit
of algebra, denominator which will get cancelled
32:49.970 --> 32:53.190
and some things will happen and this is what
you get. Now, what we are going to do is,
32:53.190 --> 32:57.549
how does this look? This looks similar to
the Monod model, if you look at it, except
32:57.549 --> 33:02.419
that, if you call this K apparent. So, if
you call this K apparent, then, this simply
33:02.419 --> 33:09.419
becomes mu max S 0 K apparent plus S 0, right.
So, then again, it looks like the Monod growth
33:17.500 --> 33:24.110
model, but with a difference.
Now, let us go to the screen. So, this is
33:24.110 --> 33:28.570
my K apparent. So, if you want to write, write
the, if you did not write what I wrote, this
33:28.570 --> 33:33.890
is what it is. So, mu m t is mu max S 0 over
K S plus S 0 plus this term, which could be
33:33.890 --> 33:38.990
written in the Monod form, mu max S 0 over
K apparent plus S 0, where K apparent is K
33:38.990 --> 33:40.140
S plus S.
33:40.140 --> 33:46.049
So, what it means is that, K S, because this
term, you know, mu max rho cell over K L Y,
33:46.049 --> 33:52.429
which is larger than 0, of course, greater
than 0, which means that, K apparent is larger
33:52.429 --> 33:58.850
than K S; which means that, the overall growth
process is slowed down; which is obvious,
33:58.850 --> 34:03.020
if you have mass transfer resistance, overall
growth processes is obviously, going to be
34:03.020 --> 34:10.020
slowed down; excuse me. So, in the absence
of mass transfer, my mu max, mu m was mu max
34:10.460 --> 34:15.079
times S 0 over K S plus S, and in the presence
of growth transfer, my mu m is going to be
34:15.079 --> 34:22.079
small, mu, mu m t is going to be smaller than
mu m; or, in other words, mu, mu, growth rate,
34:22.520 --> 34:26.109
or the specific growth rate, mu is called
the specific growth rate, the new specific
34:26.109 --> 34:33.109
growth rate is going to be smaller than the
old specific growth rate, right. Is there
34:33.129 --> 34:40.129
any question on this, anybody? So, so just,
you know, is it fine with everybody?
34:40.299 --> 34:46.669
So, let us move on to the next part, that
we will do today is, the growth of fungi;
34:46.669 --> 34:52.710
slightly different from the growth of the
cell. The reason is, the growth of fungi is
34:52.710 --> 34:57.469
different from the growth of cell, then, you
probably know this, is that, you have seen,
34:57.469 --> 35:03.559
you know, fungi growing in, in trees, right;
you have seen how fungi grow, in trees and
35:03.559 --> 35:09.779
you know, this mushroom like structure, they
sometimes form. So, what is the major difference,
35:09.779 --> 35:15.180
that in the way the cell grows, from that
a fungi will grow. The major difference is
35:15.180 --> 35:22.180
that, the cell they grow, along each other
and with each other, but, they are separated
35:22.369 --> 35:27.289
from each other within the, in the, in the
cell growth chamber. So, if I am to draw the
35:27.289 --> 35:33.559
process, yes, so, take something like this,
and then, the cell, one cell will grow...Let
35:33.559 --> 35:35.499
us start with one cell, for example.
35:35.499 --> 35:42.489
And then, we have, 2 cell, 4, you know, 4
cell and so on. So, this, this is going to
35:42.489 --> 35:49.489
continue, let us say; but, it grows like this.
So, you have the medium out here. So, the
35:50.219 --> 35:55.229
medium is out here, all of them; that is why,
we could do a model like this, where we took
35:55.229 --> 36:01.719
the cell and we took the medium around it.
This is the cell growth, general cell, normal
36:01.719 --> 36:08.049
microbial cell growth. And, and, fungi growth
is different in the sense that, again, if
36:08.049 --> 36:14.789
you, you can start with one cell, but, they
grow in colonies; there is no medium out around;
36:14.789 --> 36:21.400
the only medium you have, is outside the colonies,
you know. So, they, they grow like this. Now,
36:21.400 --> 36:26.019
can you, this is, this is the interesting
question I want to ask you that, can you,
36:26.019 --> 36:33.019
can you tell me from, in the evolutionary
prospective that, which one is better? Which
36:33.229 --> 36:38.559
kind of growth is better? So, you have this
growth, as you know, so, there is no way,
36:38.559 --> 36:41.219
you can take this. So, if you have to take
this, you have to take the whole thing together,
36:41.219 --> 36:46.890
as a colony. So, you know, whereas, you can
isolate a single cell over here; but you cannot
36:46.890 --> 36:51.200
a isolate single cell over here; you have
to it growth like a colony. So, tumours also
36:51.200 --> 36:58.200
grow like that, in colony. So, from a evolutionary
prospective, can you tell me that, which one
36:59.789 --> 37:06.789
makes more sense? Which one is better from
an evolutionary prospective? I will give you
37:14.849 --> 37:20.219
the answer, in a way almost, that...
mass transfer rate will be there
37:20.219 --> 37:24.660
Yes. So, what is my, what is the answer to
my question? Which is better, from an evolutionary
37:24.660 --> 37:26.999
prospective? Which kind of growth?
37:26.999 --> 37:32.599
Better, right; that is correct. So, the single
cell growth is better from an evolutionary
37:32.599 --> 37:37.289
prospective than the fungal growth. The reason,
and, and you know the answer is there almost
37:37.289 --> 37:42.079
that, that you know, the cell growth includes
all kinds of cell growth, including human
37:42.079 --> 37:45.049
cell growth. So, this cell growth that we
are talking of, are of course, more evolved
37:45.049 --> 37:51.670
species. So, all kinds of cells, whereas,
fungal, fungus is a lot more primitive species
37:51.670 --> 37:55.890
and you know, much of earlier cell growth,
actually started like fungal growth. Fungal,
37:55.890 --> 38:00.900
fungus is a lot more primitive species than,
than cell, normal cell, and normal cells are
38:00.900 --> 38:07.150
the more evolved species, and, and the whole
of evolution, more or less is even put towards,
38:07.150 --> 38:12.489
getting towards the more intelligent, creating
more intelligent cells and, and species. So,
38:12.489 --> 38:17.700
that is why, human beings thinks that, they
are the most evolved because, they were the
38:17.700 --> 38:21.829
last formed, you know, the last species formed.
So, if there is another species formed after
38:21.829 --> 38:27.660
human being, they will be more evolved because,
evolution always, you know, works along the
38:27.660 --> 38:32.219
path of intelligence. So, whatever is the
more intelligent, self sustaining process,
38:32.219 --> 38:36.319
they will work towards that. So, what is the
major difference between this, this cell growth
38:36.319 --> 38:39.859
that you see over here, and the fungal growth
is, what he just, you know, ((Kaushik)) just
38:39.859 --> 38:44.640
mentioned is, this mass transfer resistance
is lot lesser here, as compared to here, because,
38:44.640 --> 38:50.170
you have fluid around it.
So, which means that, nutrients can enter
38:50.170 --> 38:55.150
each and every cell, each and every cell;
for each and every cell nutrient is made available;
38:55.150 --> 38:59.349
whereas, what happens is that, there is a
hierarchy almost here that, in a, in a fungal
38:59.349 --> 39:04.309
growth, first the outside cell, the outline
cells near the border, should have nutrients
39:04.309 --> 39:08.259
and they should grow; only after they have
eaten, they will pass on what is left, to
39:08.259 --> 39:13.299
the one that is inside. So, both mass transfer
resistance are both here and there; it is
39:13.299 --> 39:18.759
a lot more over here actually, and so, from
sustainability point of view, nutrient sustainability
39:18.759 --> 39:24.739
point of view, what happens is that, the accessibility
of nutrient to the species inside, to the
39:24.739 --> 39:30.940
cells inside, is lot less than the accessibility
of nutrient to, in, in the species outside,
39:30.940 --> 39:35.460
right. So, that is the evolutionary perspective;
from a simple, chemical engineering point
39:35.460 --> 39:39.910
of view, you, you can kind of try, sort of
trivialize it almost, and say that, the surface
39:39.910 --> 39:45.329
area for the mass transfer that is available,
is more when a single cell is there, as compared
39:45.329 --> 39:48.710
to the surface area of mass transfer, when
the colony of cell; but, I would not go to
39:48.710 --> 39:53.210
that level, because as I said, that will be
almost trivializing a process, that is as
39:53.210 --> 39:56.799
complicated as, grand and as revolutionary
as this one.
39:56.799 --> 40:01.940
So, the basic thing about fungal growth, something
that I just discussed is that, it grows as
40:01.940 --> 40:07.269
a colony, as a mold colony. ((So, this will
be here was I write)). So, it grows as the
40:07.269 --> 40:13.769
mold colony, and it shows a constant rate
of increase of radius, and, and you know,
40:13.769 --> 40:18.670
this is happens, whenever some species, anything,
grows as a colony; for example, the question
40:18.670 --> 40:25.190
I gave you in the test was about this growth
of tumour, and tumour also grows as a mold,
40:25.190 --> 40:29.489
like a mold. So, it grows together as a colony,
and therefore, this shows a constant rate
40:29.489 --> 40:35.719
of increase of the radius. So, anything that
grows as a, as a mold, mold colony, shows
40:35.719 --> 40:39.799
a constant growth of, rate of growth of the
radius; constant rate of increase of radius.
40:39.799 --> 40:46.039
The reason being that, you know, why, because,
see the, it is that, total amount of nutrient
40:46.039 --> 40:51.709
that is, that is entering the system is a
constant. So, it is not a single cell; it
40:51.709 --> 40:56.469
is, I mean, sorry, it is...So, all the cells
together, are acting like a single cell over
40:56.469 --> 41:00.549
here, right. So, that is the major difference.
So, whatever nutrient is available, is available
41:00.549 --> 41:05.609
to the whole colony of cells. And then, there
is a gradient within it, and we will study,
41:05.609 --> 41:08.670
look at the gradient, how, what happens and
so on.
41:08.670 --> 41:15.670
So, so, if you have d r d t, which is a rate
of increase of the radius with time, that
41:16.719 --> 41:23.719
equals K constant and at any instant t...
K is a constant
41:25.420 --> 41:30.529
K is a constant, yes, because that is what
happens in the fungal growth; the radial...
41:30.529 --> 41:31.739
Fungi
41:31.739 --> 41:38.739
No, we are talking about single mold over
here; it is not the, a mold that is not breaking;
41:42.660 --> 41:47.499
a single colony of cells, that is growing.
See, there is a process which can break only
41:47.499 --> 41:51.950
after it has reached the threshold; it is
not that, every fungus will, will lead to
41:51.950 --> 41:56.789
another fungus, to fungus; it will reach the...What
happens is, a mold will reach a certain threshold
41:56.789 --> 42:00.809
of radius, beyond which, it cannot take itself
and then, it will break; and, I will also
42:00.809 --> 42:06.190
come to what is that threshold; I will, I
will, I will show you that. So, this is the
42:06.190 --> 42:09.660
process, where the, the fungus itself is growing
as a mold; the single fungus is growing as
42:09.660 --> 42:14.619
a mold and the rate of increase of radius
is the constant over here and the...So, we
42:14.619 --> 42:19.579
considered here, the, in this problem that
we do today, we considered a cylindrical mold.
42:19.579 --> 42:26.579
So, if you have cylindrical mold, then, the
x is pi r square over h times rho.
42:29.049 --> 42:36.049
So, this is the radius at each point of time,
which is increasing; this is the height of
42:39.880 --> 42:46.880
the mold and rho equals the density, fine.
So, this is what you have. Now, d x d t would
42:49.839 --> 42:53.369
there, therefore be, since the height is not
increasing, d x d t, it simply depends on
42:53.369 --> 42:57.799
the increase of the radius of the mold, fine.
So, this is what we do; we differentiate both
42:57.799 --> 43:04.799
sides of the problem. So, you get, you get
2 2 pi r h rho times d r d t, fine.
43:15.390 --> 43:21.039
So, what we are trying to do over here, next,
what we are trying to do is, simply express
43:21.039 --> 43:25.160
everything, in terms of x. See, when we did
the last model, we had everything in terms
43:25.160 --> 43:30.789
x. So, we want the growth rate, not in terms
of the radius, but in terms of the cell itself.
43:30.789 --> 43:36.930
So, d x d t. Then, what we do is, simply replace
your r over here, by square root of x, that
43:36.930 --> 43:41.640
you get here, from this equation, right. So,
and then, you get this; let me show it here,
43:41.640 --> 43:48.640
without any confusion. So, x is equals pi
r square h. So, d x d t is 2 pi r d r d t
43:55.180 --> 44:02.180
times h row. Now, from here, what I get is
that, r equals x over pi h rho square root
44:04.859 --> 44:11.859
of that, and I know that, d r d t equals K.
So, I put that over here. Again, d x d t equals
44:14.549 --> 44:21.549
2 pi square root of x over pi h rho, h rho
and K, which is 2 and K, right. So, then,
44:24.440 --> 44:31.440
you
44:36.339 --> 44:43.269
can simply go ahead and integrate this and
find out, how the mold changes with time.
44:43.269 --> 44:50.269
So, You assume that, at the start, the mold
had a total weight of x 0 or something, and
44:50.650 --> 44:56.599
then, you can just quickly go and integrate
it, and then, you get x as a function of time
44:56.599 --> 45:03.599
and x 0; straightforward integration. Fine,
clear?
45:25.759 --> 45:32.759
Now, what we look at, is a same problem, but,
we are now looking at spherical mold and you
45:34.150 --> 45:40.359
know, some fungi will grow cylindrical, some
are spherical. So, again, you can repeat the
45:40.359 --> 45:45.930
same steps. So, x equals 4 third pi r cubed
rho. So, d x d t would be 4 pi r square d
45:45.930 --> 45:52.479
r d t rho and then, you can replace again,
r from that, and then, you instead of half,
45:52.479 --> 45:59.479
you will get a 2 third over here, which you
can again integrate.
46:13.549 --> 46:20.549
So, integrate, this is what you get, x equals,
some gamma t times over 3 plus x 0 to the
46:46.479 --> 46:53.479
one third whole, whole cubed. So, what was
your expression for the spherical mold? It
47:03.940 --> 47:10.940
was some lambda t times something square and
this one, you get gamma t times cubed. So,
47:13.420 --> 47:17.619
which means that, the spherical mold will
grow much faster than a cylindrical mold,
47:17.619 --> 47:24.619
right; because, this is cube of time and that
has a square of time.
47:35.599 --> 47:42.599
K, got K d r d t as constant, right, to start
with; that was the assumption. It is not a
47:43.509 --> 47:50.509
strictly valid assumption, but, sort of, reasonably
valid assumption. Once you are done, just
48:07.769 --> 48:14.769
let me know. Finished?
48:15.650 --> 48:22.650
Now, comes the question that, if just like,
like we did for the, in, for the growth of
48:24.410 --> 48:29.420
the, for enzymatic reaction, then, we did
the same thing for normal cell, cellular growth;
48:29.420 --> 48:33.269
first, we looked at the, looked at the aspect
of growth, that reactive aspect, and then,
48:33.269 --> 48:38.459
we looked at a mass transfer aspect. So, same
thing, we are going to look at exactly, now,
48:38.459 --> 48:41.589
that is, look at the set of mass transfer,
that will partly answer your question about
48:41.589 --> 48:45.519
the threshold, and which is the point that,
it can keep growing. So, when it cannot keep
48:45.519 --> 48:49.819
growing any further, it will break into 2,
to reduce. So, that is again evolutionary
48:49.819 --> 48:54.319
thing, to reduce the mass, amount of mass
transfer resistance, that is offered to it.
48:54.319 --> 49:00.009
So, the, that, that is about we are going
to look at now. So, excuse me. So, this is
49:00.009 --> 49:03.979
a cylindrical, cylindrical model that we are
taking; cell cylindrical fungal palette; let
49:03.979 --> 49:08.579
me just go back. So, this is what we did for
the growth process, but, this growth process
49:08.579 --> 49:12.900
that we are talking of, is not, not a complete
growth process, the reason being that, there
49:12.900 --> 49:17.729
is mass transfer resistance, that is offered
to the substrate; so, because, the substrate
49:17.729 --> 49:21.549
can only reach the outside. So, that is the
major difference from here, as I explained,
49:21.549 --> 49:27.190
to with, major difference between this and
the single cell model. So, this is the substrate;
49:27.190 --> 49:31.759
I had referred to that as an, at oxygen, but,
it does not have to be oxygen; you know, it
49:31.759 --> 49:35.449
should be everything else, oxygen, carbon
dioxide, carbon, hydrogen, everything put
49:35.449 --> 49:39.469
together, let us say; but, let us assume it
to be oxygen.
49:39.469 --> 49:44.509
So, this is the equation in dimensionless
form and the reason I wrote it this way is
49:44.509 --> 49:49.479
that, you know, we can go back and we did
already, we did this already for the enzymes
49:49.479 --> 49:55.170
and you can go back and look at the dimensionless
form, for the immobilized enzyme; if you remember,
49:55.170 --> 50:01.589
just before the exams,. We showed how to make
a dimensionless for the immobilized enzyme
50:01.589 --> 50:07.410
and exactly the same thing and so, beta you
know. So, these are all these, same, same
50:07.410 --> 50:14.049
coefficient that we had before, and this has
to do with, yes, this has to do with the Michaelis-Menten
50:14.049 --> 50:18.699
kinetics over there. So, this is dimensionless
form, but, if you do not trust this form completely,
50:18.699 --> 50:23.079
then, we can go and derive it from what you
had, but the reason I am not repeating is
50:23.079 --> 50:28.140
because, we had already done this in the immobilized
enzyme part.
50:28.140 --> 50:33.789
So, so, 1 over r. So, r bar, this is r bar,
is the dimensionless form of r. And, how do
50:33.789 --> 50:38.979
you make it dimensionless? Just make it dimensionless
with some initial radius, because the radius
50:38.979 --> 50:45.979
is growing. So, just make it dimensionless
with an initial radius. So, d r r r bar times
50:47.819 --> 50:54.819
d d c d r times, this is the reaction. So,
what is happening is that, this is the diffusion
50:55.319 --> 51:02.289
that is occurring of the substrate, inside
the, inside the mold, and there is reaction
51:02.289 --> 51:05.420
that is occurring. Couple of things that we
need to understand this, you know, before
51:05.420 --> 51:11.219
we go further with this; one is that, as in
the cell, you know, while in the cell, we
51:11.219 --> 51:15.989
neglected the gradient, mass transfer gradient
of the substrate within the cell, if you remember,
51:15.989 --> 51:21.579
but, we cannot do it here; why is that, because,
the cell is much smaller entity.
51:21.579 --> 51:28.579
So, what we did here, when we did the cell
example was, and this is the fungal model.
51:28.789 --> 51:35.789
So, we neglected the mass transfer gradient,
neglected m t in cell; we cannot neglect m
51:40.140 --> 51:47.140
t in cell, because, the length scales are
much larger here. So, the diffusional gradients
51:50.979 --> 51:57.979
are going to be much larger over here. Second
is that, this is it, cannot neglect mass transfer
52:01.880 --> 52:08.880
resistance in fungal pellet and b, let us
go back to the screen over here, what you
52:13.809 --> 52:19.089
see is a steady state; this is the question
I have for you; what you see is the steady
52:19.089 --> 52:26.089
state equation; but, the fungal pellet is
growing. So, where is? Why is ? There should
52:26.640 --> 52:32.130
be, right; fungal pellet is growing. So, if
you have an, the whole process is steady.
52:32.130 --> 52:39.130
So, what is happening here? Why do we assume
steady state? We are running out of time today,
53:12.559 --> 53:18.119
but, I will just explain quickly, why. What
we do is, let us, let me write it here. Assume,
53:18.119 --> 53:25.079
this is a very important thing, and as we
will start tomorrow, next class with this
53:25.079 --> 53:32.079
quasi-steady state. Again, this may be a new
concept that we are, you are dealing with
53:32.359 --> 53:32.849
today.
53:32.849 --> 53:39.269
So, this is a, this is different from the
assumption of quasi-steady state in the enzymatic
53:39.269 --> 53:43.680
case; please understand. What is happening
is that, there is an unsteady process that
53:43.680 --> 53:47.479
is happening, that is the growth of the fungus;
and, there is another, probably another unsteady
53:47.479 --> 53:53.329
process, which is the diffusion of oxygen
into the cell. Now, whenever there are a couple
53:53.329 --> 53:58.019
of processes, two or three processes, and
we want to make some assumption about one
53:58.019 --> 54:03.890
of these processes, what we do; we compared
the time scales of these processes, fine.
54:03.890 --> 54:07.799
So, what are the time scales over here? How
do we compare the time scales? One is the
54:07.799 --> 54:13.279
time scale of growth, fungal growth; t g,
let us call it; the other is a time scale
54:13.279 --> 54:18.559
of diffusion of the oxygen. So, what is the
time scale for diffusion? Let us assume that,
54:18.559 --> 54:25.559
at any point of time, at a certain point of
time, at t equals t 0, R equals R 0; that
54:28.869 --> 54:35.869
is, the, at t equals t 0, R equals R 0; that
is, the radius of the fungal pellet; is some
54:38.599 --> 54:45.599
value, we need to assume some value; otherwise,
we cannot do radius of fungal pellet. So,
54:46.910 --> 54:49.309
what would be my diffusion time scale?
54:49.309 --> 54:56.309
R 0 square by d effective, fine. What would
be my growth time scale? What would be my
55:06.630 --> 55:12.099
growth? So, if the fungus is, if the fungus
is growing, in general, right, the radius
55:12.099 --> 55:16.939
of the fungus is increasing. So, what would
be my growth time scale? We had it in the
55:16.939 --> 55:23.939
last example we did; just have to use that;
what is d r d t? K. So, what would be my growth
55:30.319 --> 55:33.189
time scale?
55:33.189 --> 55:40.189
T g would be, not into, r 0 over K; r 0 over
K, fine. So, if you have to assume quasi-steady
55:55.140 --> 55:59.150
state, what is, what quasi-steady state of
what?. What does the quasi-steady state mean?
55:59.150 --> 56:03.150
Quasi-steady state, that you have assumed,
if I go, let me go back to the screen here,
56:03.150 --> 56:08.109
what it means is that, quasi-steady state
means is that, I wrote a steady state equation
56:08.109 --> 56:12.559
for the diffusion of oxygen; but, I am writing
a steady state equation for the growth of
56:12.559 --> 56:17.059
the pellet; for the growth of radius of pellet,
which means that, the diffusion of oxygen
56:17.059 --> 56:23.029
is much, much faster, than the growth of the,
the radius movement of the radius of the pellet,
56:23.029 --> 56:27.069
the radius of the pellet, the fungus radius,
and this is not true; you know, in real life,
56:27.069 --> 56:31.589
the fungus radius, it does not change every
second; it changes over a period of days;
56:31.589 --> 56:35.599
whereas, oxygen diffusion is taking place
over, over the, over time scales of minutes;
56:35.599 --> 56:36.939
is that clear to all of you?
56:36.939 --> 56:42.509
So, what I need is that, the diffusion time
scale t D, what is the relation between the
56:42.509 --> 56:47.729
diffusion time scale and the growth time scale,
and t G over here; what would be the relation?
56:47.729 --> 56:49.839
t D should be...
56:49.839 --> 56:52.309
Much much greater.
56:52.309 --> 56:59.309
Time scale, should be much, much smaller than
t G. So, t D should be much, much smaller
56:59.829 --> 57:06.059
than t g; that is my requirement; which means
that, if I put t g, t d should be much, much
57:06.059 --> 57:13.059
smaller than t g; which means, R 0 over K
should be, sorry; not R 0 over K, R 0 over
57:14.509 --> 57:20.539
D effect, R 0 square over D effective should
be much, much smaller than R 0 over K; which
57:20.539 --> 57:27.539
means that, R 0 should be, R 0 over D effective
should be much, much smaller than 1 over K;
57:32.729 --> 57:39.729
or K should be much, much greater than D effective
over R 0, fine. So, this is the criteria that
57:56.299 --> 58:00.239
we need, for quasi-steady state to be valid.
If quasi-steady state is to be valid, then,
58:00.239 --> 58:04.439
we can go ahead and write this; if quasi-steady
state is not valid, then, what you would do?
58:04.439 --> 58:11.439
Then, you have to write, d d c over d t, del
c over del t, sorry, plus 1 over r del del
58:16.519 --> 58:23.519
r of r del c o 2 del r, equals some c o 2,
c o times mu over K 1 plus beta c o. So, we
58:29.309 --> 58:33.949
will stop here today. We have run out of time
and we will continue from here in the next
58:33.949 --> 58:34.199
class.