So, from where we left, which is that effective diffusion in the fungal pellet and if we go
to the slide, now, what we see is that, we wrote this equation, if you remember, which
is the diffusion equation. So, on your left hand side is the diffusion of oxygen, and
on your right hand side is the reaction that occurs because of the growth of the fungus,
fungal pellet. So, it is, the fungal pellet essentially has to grow. What was the major
assumption out here? The major assumption was upto the steady state approximation. So,
the idea was that, though the fungal pellet is actually growing in radius, but, the rate
of growth of the radius is much smaller than the rate of diffusion of oxygen, as a result.
And, how did we conclude this? We concluded this through time scale analysis. We measured
what is the time scale of the growth of the radius of the fungus, and we measured, what
is the time scale for diffusion of oxygen in the fungal pellet. And, what we found was
that, the time scale for the growth of the radius is much, much larger than the time
scale for the growth of the fungal, of the diffusion of the oxygen; or, in other words,
the diffusion of oxygen is a much faster process, as compared to the actual growth of the radius
of the fungal pellet. So, as the result of which, we can write the
pseudo, steady state approximation, or in other words, we can assume that, the diffusion
of oxygen is the steady state process, as compared to the growth of the radius, fine.
So, based on that, we write this equation out here, and what you see over here is that,
the temporal term in oxygen concentration is missing, because of the pseudo steady state
approximation, fine. So, this is the radial diffusion, excuse me; this is the radial diffusion,
and this is the reaction term and this is written in dimensionless coordinate; all of
it is written in dimensionless coordinate; and, I am not redefining these dimensionless
coordinates. The reason I am not redefining is that, we already did this same, that immobilized
enzyme case, you know, the last example we did in, in the immobilized enzyme, already
had the same variables and same dimensionless coordinates.
So, you can go back and look up; so, I am not redefining it. Now, the assumption here
is that, C oxygen is much, much greater than K M, which is a very reasonable assumption,
because, if you want to grow something, you know, fungus is growing for example, oxygen
is typically in abundance, as compared to the k m. So, that is a very straightforward
assumption. Now, when you make that assumption, what you find is that, this, this equation,
which is a Michaelis-Menten kinetics, reduces to a first order kinetics over here; is that
clear? So, the Michaelis-Menten kinetics over here, reduces to a zeroth order, sorry, reduces
to a zeroth order kinetics over here, because, in the, when C O is much larger than K M,
then, what happens? C O in the numerator and the denominator cancel out, and this is what
you have. So, this is, this is what we got over here,
and you know S 0 is all of this combined, K M and beta and so on, combined. So, R square,
r is a radius of the mold, of the fungal mold, at that point of time; is that very clear?
R, ideally, R is a function of time; it is increasing with time, but, since we are doing
a pseudo steady state approximation, so, R over here is the radius of the fungal, at
that point of time, fine. So, we write this equation; then, this is a very straightforward
zeroth order, on the right hand side, and you can integrate it straight away, and what
are my boundary conditions? Some are written on the screen, which is that, the oxygen concentration
at the outside radius r equals 1, or r equals big r, whatever it is, is 1 and d d d d r
of C O 0 is 0, ok.
So, let me show you this by assuming this assumption. So, this is my cylindrical mold;
my reaction equation over here is 1 over r d d r of r; nu max I used, instead of r max;
the reason being that, r, now, we are using for the overall radius of the mold; that is
why, I changed the…D effective; S 0 is that K M over beta, that you, K M times beta, sorry.
So, this is my equation. The boundary condition inside is…So, this is my r; this is r equals
0; this is r equals 1. So, r equals 0, boundary condition is symmetry, right. So, equals 0,
r equals 1; I am saying that, oxygen is present in a certain concentration. So, C O 2 equals
1; 1 is the dimensionless coordinate, which means that, actual value of concentration
is something, like some value C, C O 0, or something like that. So, C O 2 is equal to
C O 0 in dimensionless; dimensional, when you turn into dimensionless, this is what
you get, right. So, this is a boundary conditions we have. We can straightaway integrate this.
This is a straight forward integration and we can get the profile of concentration within
the mold. So, you know, so, why are we doing this and the reason would be obvious, you
know, some of, one of you, I think, asked this question the other day, and the reason
would be obvious, once we come up with the solution.
So, look at the solution. So, this is how it goes as, you know, it goes quadratically;
essentially, as you see, it increases with radius, right. It increases quadratically
with radius. So, it is some constant plus another constant times r square; is that clear.
So, it is one constant, some a, plus b times r square. So, it increases quadratically with
the radius. Now, what you know, what you know is that, the concentration at the outside,
at this interface between the fungus and the air is a constant, which means, in other words,
it decreases quadratically, as you go inside the fungus; is that clear. So, what will happen
is, just try to, you know, invasion this. As you, as you keep increasing the size of
the mold, what will happen? The gradient will get steeper and steeper, right, because the
concentration that you provide at the outside, that is the interface between the mold and
the, and air is a constant. So, as the radius keeps increasing, the gradient will get steeper
and steeper, fine; and there, will come a point in time…
So, this is a steady, pseudo steady, under a pseudo steady state approximation, we are
making this; but remember, this r is actually a function of time. So, when you make a pseudo
steady state approximation, how do you do that? how do you enforce that? So, first you
solve the steady… So, if I ask you to solve a problem using pseudo steady state approximation,
so, what is the idea? You decouple the state, the two, two phenomena. How do you decouple
the two phenomena? By first showing that, the time scales of these two phenomena are
different. So, the time scale which is much smaller is under steady state. So, you write
a steady state equation like, just like we did, and solve for it; then, you solve the
unsteady state equation and get how radius varies with time. For example, here, if you
have your d r d t, if I remember, here d r d t equals K, fine.
So, if this is what you have, then, you got your x as a function like this, and you can
also integrate d r d t equals K over here, right away, and get how radius varies with
time. So, radius here, would vary in some, you know, linearly, for example, here it is
varying, some K t, fine. So, what you do here is that, after we have solved for the steady
state, you impose that temporal dependence of radius on this equation; or, in other words,
if radius, for example, varies linearly with time over here, or, radius varies exponentially
with time over here, then, you go and put that into this equation; is that clear? Let
me write, if it is not clear.
So, your C 0, the solution that you get here is, C 0 minus R square nu max, 1 minus r square.
Now, if R equals some R 0 at t equal 0 time, let us say, exponential; we are just assuming;
exponential minus, you know, K t, something like this, sorry; it should be plus K t, in
this case, if it is increasing with time. So, what you do is, you put…So, this is
your, essentially your R, at any point of time. So, you put that over here, and what
you get is, 1 minus R 0 exponential K t, fine. So, if this exponential K t is some linear
dependence and that will come in. So, this is a way to, is that, is that clear? This
is the way to solve the pseudo approximation, when you have a pseudo steady, steady state,
or there is the decoupling of state, two states, then, this is the way to solve it. First,
go and solve for the steady state, and then, impose; then, solve for the unsteady state
and impose that unsteady state condition on the steady state solution; is that clear?
This is the way to do it, fine. Now, excuse me. So, what you find over here,
what you find is that, if just by looking at this equation that, as the radius keeps
increasing, the concentration gradient becomes steeper. Now, the concentration outside is
a constant, remember; if the concentration outside will not a constant, if you are increasing
it, then, no problem; but, because the concentration outside the mold...So, concentration here
is a constant, is a constant and this radius is keeps increasing, what will happen? There,
there will come a point in time, when the center of the mold would stop getting any
oxygen at all; is that clear? There will come a point in time, when the center of the mold
will stop getting any oxygen at all, and that means that, the center of the mold can grow,
no longer grow; and, that is my critical radius.
So, let me show you the definition of that. So, at critical radius, the critical radius
is defined as that radius, when the center of the mold is depleted of oxygen. And, if
the center of the mold is depleted of oxygen, then, the growth stops; because, this thing
has to push from outside, you know; so, this is a, this is a radius that we want to evaluate;
and how would you evaluate that, from this expression? We put r equals, this r equals
what? What, how do we get this R critical? .
C. concentration of. Yes, but, there, that is fine. This side is
0; but, there is a variable out here; critical radius is a constant and r over here, is a
variable. So, C O 2 equals 0, I agree; what else?
Sir, that will happen at this center. Yes; so?
That will also have to be 0. Yes. So, this r has to be 0; this r, remember
big r, is the external radius; this radius at r equals 0. Thus C O 2 has to be 0. So,
you put 0 over here, 0 over here and you will simply get the critical radius right away,
as square root of 6 S 0 D effective over nu max. So, I hope this answers your question
from last time that, fungus cannot grow, grow indefinitely and there will come a point,
when it will stop growing and that is the time it will try to divide itself, so that,
it can continue to grow. So, this is the limit of the colony. So, it is a colony, that kind
of, it keeps expanding, but, it does not expand indefinitely; it expands only up till the
time, that the radius equals the critical radius; so, which is given over here, square
root of 6 S 0 D effective over nu max. So, once this critical radius is, is attained,
then, the funguses stop growing and it will, maybe divide and allow for future growth;
is it clear? So, what we looked at is, you know, we are looking at different, different
things, at the same time. So, first, we looked at the normal growth kinetics, and you know,
how the kinetics changes, and what is the dynamics of the kinetics, and how different
models had been proposed, to sort of quantify these dynamics, and what we figured is this,
that, of all the kinetics model, kinetic models that have been proposed, it the Monod model,
or the modified Monod model, which are the most useful models.
Then, we looked at the effective mass transfer, gave you a couple of examples to look at the
effect of mass transfer, and just as we had seen that, in the case of, case of a immobilized
enzymes, mass transfer effects kinetics; or, in other words, kind of decelerates kinetics.
Similarly, if mass transfer effects here are going to decelerate, or slow down the growth
process; but, there is no running away from it, because, for the growth process, just
as in the enzymatic process, you need the substrate; you need to provide the oxygen,
and the nitrogen, and the hydrogen, and the carbon, just as we showed in the first, I
think, the first slide of this whole chapter. You need all of these substrates to be available
to the, to the cell and without this, there is no way you can have cell growth.
Now, as soon as you need all of these substrates to be available to the cell, you immediately
need, you know, mass transfer effect, would, all the mass transfer limitations would automatically
come in. So, there is no running away from it. So, we looked at some of these and with
the effect of, with the, for the case of normal growth of single cells and the growth of fungus.
So, what we said was that, the major difference between the growth of cells, cell colony,
and the growth of fungus is that, a cell, cell in, when a cell colony grows in general,
it is like a cell is surrounded by the fluid and therefore, by the substrate, and therefore,
it has direct effect, direct contact with the substrate; whereas, a fungus would always
grow as a pellet, or a mold, you know, as, as the, as a lump of colony and therefore,
the cells inside the fungus do not have any contact with the nutrients.
So, the only way the nutrients can come in, is through the interface, from the outside,
it is slowly diffuse it and this is the basic difference. And therefore, we did, looked
at the mass transfer effect for a single cell and then, we looked at the mass transfer effect
for the fungus mold. So, we are at this point. Now, what we are going to do for the rest
of the lecture today, is look at some other effects, and those effects are the effects
of multiple substrates and the effects of inhibition, which means that, if you have
more than one substrate, right. So, you can provide substrates in different forms, and
both can lead to cell growth; then, what happens, right. So, for example, you can give your,
your sugar, you know, both in terms of sucrose and glucose, you know, maybe two different
form. So, fructose, sucrose or glucose, you know,
two or three different forms. Now, all of these substrates, they have different kinetics,
remember; each of them will have different kinetic, but they, all of them lead to growth
of the cell, fine. So, how do we account for that, a; b is that, what would be the effective
inhibition, just like we did in the enzyme case. So, there is, just as you have substrate,
there could be inhibitory things that are there, a and b is that, when you give too
much of substrate, that, one of the thing that we talked about, I think, earlier, you
can have, what is known as substrate inhibition. So, how do these affect the cell growth process
and we are just going to look at that. I hope we can finish this today. Then, we can start
a new chapter; but anyway, we are not going to hurry through it; we are going to do it
very slowly.
The first thing we will look at, is the effect of multiple substrates. So, multiple substrate
means that, you have cells which are exposed to two different substrates, and both these
substrates will lead to the product formation, that is daughter cells, but through different
kinetic routes. So, this is, what you see on the screen is, x is your cell, starting
cell, and S 1 is substrate number one and you get the daughter cell x, 2 x; but, through
the formation of the intermediate compound, intermediate cell x prime, which is one route,
with a set of reaction coefficients k 1, k minus 1 and k 3; and then, you have another
substrate, that is, they are parallelly in the system. So, you have, might have, you
know, given two substrates to the system. So, that gives again, the daughter cells,
2 x, but through different route, x double prime; and the reaction rate constants are
k 2, k minus 2 and k 4. Now, I talked about the yield yesterday, if you remember, in the
last class, I talked about yield and so, the yield is the amount of daughter cells produced
per, per mole of the substrates, ok. So, for case one, what is your yield? One
daughter cell is produced; from two, two cells, you get, one cell, you get two cells, which
means, one, effectively one extra cell has been produced for a one mole of the substrate.
So, the yield of x over S 1 is 1 over a 1; and for case two, the yield of x over S 2,
that is, one daughter cell has been produced for a 2 moles of S 2. So, yield is 1 over
a 2, fine. Now, how do we find…So, what we are trying to find is, what the, how would
we find the overall growth rate of the system? So, what is the goal, that has to be very
clearly defined. So, we did…What did we do? We have done only one single substrate
before, and we have found the growth rate; so, which is d x d t equals nu, nu, nu times
x, where nu is given by the Monod, Monod kinetics, S 0 over K plus S 0, some r max S 0, double
K plus S 0. So, here we are trying to find out, what would
be the form of the growth, overall growth of the system, because, what we are in, essentially
interested at the end of the day is the overall growth. I start with 5 million cells; how
many cells do I get after 24 hours or so, that is what I am interested; excuse me. So,
the balance equations, excuse me, the balance equations over here, are this. So, what I
am writing is, I am writing a balance for the intermediate species; the del x t, del
x prime del t and as you see over here, the first term is k 1 x S 1 which is the term,
because of this forward reaction over here. See the, where the arrow is. So, this forward
reaction over here, we this term. The second term is minus k minus 1 x prime, which is
the result of the backward reaction over here. The first reaction, the backward reaction
you had; and the third term over here, is minus k 3 x prime, which is the result of
the formation of the product, of the daughter cells. Is it clear, these three terms? It
is very straightforward; and then, the same thing is repeated for x double prime, ok.
So, k 2 x times S 2 minus k minus 2 x double prime minus k 4 x double prime; is it clear
to all? So, is there a question?. No; power, no, no, no, no; yes; that, I will
explain that. See, this is not an elementary reaction. See, when do you have the power?
Let us, let me explain that, if there is any doubt.
A plus B giving C; then, rate of formation of, of C is K C A C B; if it is an elementary
reaction; A plus 2 A, 2 B giving C, the rate is, could be given as K C A C B square, only
if, if and only if, it is an elementary reaction; is that clear? If it is not an elementary
reaction, there is, nobody said that, that is a, that is a rate. So, it could be this.
This one? No; no, no, A plus 2 B equals to C.
Yes, elementary reaction means, the definition of the elementary reaction is that, the rate
of the reaction is written exactly as the same way in which the molar concentration,
composition of the reactants react; that is the definition. See this reaction for example;
it is defined, an elementary reaction is, will be defined only if…So, if this, this
molar concentration is 1 is to 2 mole, it reacts in 1 is to 2 mole ratio and the reaction
rate is also given as C A C B square; but, it is not necessary that, exactly the, the
mole ratio in which they react, will be the ratio in which the, the rate of the reaction
is measured; because there, see that, the elementary reaction means that, there are
no other effects, apart from the molecular collision between the, between the two molecules
that govern the rate of the reaction. Do you understand what I am saying?
So, if there are no other effects…So, it is…See, this is very unlikely that, this
will be an elementary reaction. Why, because, the probability of 2 molecules of B and 1
molecules of A, coming and colliding and forming 1 molecule of C, is very, very small. So,
typically, a elementary reactions are of the, or either first order, or second order. So,
you know, when we talk about first order reaction, so, it cannot be an elementary reaction; because,
1 molecule cannot combine with your, itself; there has to be a molecular collision. So,
what is the elementary reaction mean that, the way 2 molecules collide with each other
is exactly the way the reaction is represented, as also, when the reaction rate is written.
So, do not assume that, whenever you write a...So, this is the molar ratio; this is its
stoichiometry. So, do not confuse stoichiometry with the
power of the reaction. The stoichiometry would be same as the power of the reaction only,
and only if, it is an elementary reaction; elementary reaction means that, all other
effects are ignored. So, it is directly that, the 2 molecules are combining and that is
forming the reaction; but, in the most cases, it is not like that. You know that, there
are complex steps out there; for example, this step that I have written over here, as
I, as we do in the enzyme, they are not necessarily elementary reactions, which means that, it
is not that this step alone is occurring; there are multiple steps that are, that are
probably occurring over here, and they are summarized as these reversible reactions.
And, elementary reaction means that, one single step, single step reaction and in that ratio.
So, the molar ratio, the stoichiometric ratio that is there; so, they combine, these 2 molecules,
it combine and form. So, this is something very fundamental; you should not make a mistake
about it. So, if I want to say, tell you that, this
is the rate of the reaction, then, I will say that, this is the reaction and which reaction,
one is an elementary reaction, which means that, in this case, you will raise to the,
raise this to the power a 1; and, but then, elementary reaction, if a 1 is not an integer,
then, it cannot be an elementary reaction also; do you understand what I am trying to
say? If a, a is say 0.75, it cannot be a elementary reaction, because 0.75 of a molecule cannot
combine with, you know, another molecule. So, for elementary reaction a 1 has to be
a number, typically, 1 or at best 2; even 2 is hard, but, at best 2; but, not more than
2; because, 3 molecule, the probability of 3 molecules combining with each other goes
down. You know, if you have read some of mechanics, you will understand that, the probability
goes down exponentially. So, anyhow, now, so, this is the reaction rate and…
So, the next step, what we do is, we write the constraint equation like we did before,
which is the total amount of cells present in the system. So, x t is the total amount
of cells present in the system. So, the total amount of cells present in the system is either
in the form of x, which is a daughter or the mother cell, or in the form of any of the
intermediate species x prime or x double prime. So, the, this, this is my constraint equation.
Next, what I do is, use a pseudo steady state approximation, the quasi steady state approximation
like we did before; that del del t of x prime equals 0 and del del t of x double prime equals
0, fine. Now, let me explain this a little bit, from what is written in the last line.
So, d x d t. So, x, my x equals, x t equals, x plus x prime plus x double prime. So, d
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;
but, these are 0, because of pseudo steady state approximation. So, you get this as d
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
prime. Can anybody tell me, why is that? d x t d t equals k 3 x prime plus k 4 x double
prime?. Right. So, x t is what? x t is the overall
amount of cells that are present. Now, see, the only difference here from what we did
in the enzyme case, if I go, if I am allowed to go back to this slide. So, look at this.
So, this is the reaction that is occurring because of the formation of the daughter cells
that are occurring through the, from x prime or x double prime to 2 x. So, what is the
major difference from what we are doing here, from what we are doing in the constraint?
This x t is not a constant; that is the only difference. In the enzymes case and all the
previous cases we had done, this x t was a constant, right. But here, the x t is not
a constant. So, that is a major difference that, that is there; why, because, so, there
is a overall growth, growth of the cells. And, how is this overall growth quantify,
if I come here. So, what is the overall growth? If you look at this system, for example, the
overall growth is, 1 cell, 1 x, 1 cell, 1 daughter cell grows from this, ok.
So, the total growth for the system d x t d t, d d t of x t equals, so, k 3 times x
prime plus k 4 times x double prime. So, one daughter cell growth, each of these reactions;
first reaction, one daughter cell growth, and second reaction, one daughter cell growth;
is it clear to everybody? This is a little tricky part. So, this is what we have. So,
I hope you agree with me. Now, d x prime d t equals 0. So, x prime equals, if you go
back, you know, to your notes and check the, I, I can go back to the slide in a minute,
but, just let me write this; this is what I get here, if you go back to your notes and
the pseudo steady state; this is a pseudo steady state approximation; pseudo steady
state approximation; this is what we meant. So, you will see that, d x prime d t equals
0 would give you this and if you have a doubt, I will go back to it in the, here. So, look
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.
So, from here, you can, if this side, the left hand side is 0, you can straightaway
express your x prime as a function of x and S 1; clear to all of you? So, the same thing
you can do over here, with x double prime. If your d x, d d t of x double prime is 0,
then, you can express your x double prime as a function of x and S 2; yes, S 2, right.
So, this is what we do over here.
So, x, x, d d t of x prime is written as 0, and x prime is given by this number, and x
double prime is written as this number. So, why we are doing it because, we intend to
substitute these two into our, our constraint equation, just like we had done before. So,
when we substitute these two into our constraint, our constraint equation is here at the top
and what we do is, we substitute these two into our constraint equation. So, this is
what I get. So, x t equals x times…So, x is the first term, then, x prime here. So,
you take x out of the parameters, it is common, out of the parameter; so, you get this and
get this. But, so, only thing that you have to keep in mind again that, x t is not a constant
over here; it is varying with time. So, that is the major difference between what we are
doing now, to what we did in the previous constraint equation. Now, we can take a derivative
of this; I will, I will continue, as soon as you finish writing. Shall I go ahead?
Now, we take the derivatives. So, d x this, this was already something that I have written
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
again; you replace your x prime by what you got from the pseudo steady state approximation;
you replace your x double prime by what you got from the pseudo steady state approximation,
fine. Just write the first step, then, I will show the, the next step; just write the first
line, then, I will show you the, you know, how to get the second line. You probably know
that already, but, I will still show you.
So, d x d t equals k 3 x prime plus k 4 x double prime. So, this, you replace; as soon
as you replace, you get x times this number k 3, fine. Now, what you do is, let us go
to this slide. So, here, you can have the…Here, the last line, you have the, you have the
relationship between x t and x. So, because, why I am doing this because, we, we, we want
to find everything, in terms of the total amount of cell that is present in the system.
So, because it does not matter to us, which one is the, is a pseudo steady state, which
is the daughter, which is the, because, what you can measure essentially at the end of
the day is, a total amount of cells; you know, you, under microscope, you cannot differentiate
that, this is in the intermediate species, this is the final and this is the initial
species. So, what you can measure, is the total amount of cell. So, that is why, all
our calculations, we want to do it in terms of the total amount of cell. So, x t over
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
quantity is a little too big, so, let me call it, this B or something. So, if x is x t times
B inverse, then, you can put it over here, into this equation, this whole thing. So,
x t is… So, then, you can get it in that form and let us call this, this thing A; this,
this term is A; this term is B; then, what you have is…So, what I did was, I just said,
this term is A; this term is B.
And then, what you have is d d, d x d t equals x t times A times B inverse, fine. Now, what
is my mu? mu is the specific growth rate, fine. How is it defined? 1 over x t times
d x t d t; take the first order rate constant. So, from this, what we will get from this
equation, from here? So, what I get, this equals A times B inverse. So, I got my specific
growth rate.
Now, the specific growth rate for the Monod model, if you remember, were some S, you know,
K S times r x r max S times K over K plus S, right; that was the Monod growth model,
mu, right. So, now, you can…So, this is what you get. So, the exactly the, you know,
once we do the, the steps that we did, did in, on paper, this is what you get and mu
is your, mu, a nu rate, specific growth rate. So, once you do the simplification, which
we are not doing because of lack of time…So, this is your mu; this inverse time this. And,
there is, if you multiply both of these, and you know, you can do a little bit of simplification,
then, you see, find out that, mu comes out to be in a very nice compact form, and what
is interesting over here is that, the, it is interesting to note is that, the substrates
influence each other. So, the specific growth rate of the two substrates
put together, is not equal to the specific growth rate of substrate one, and the, plus
the specific growth rate of substrate two. They are not independent; the substrate, the,
the, the two growths are not independent of each other, and what you find over here is
that, the, this is specific growth rate of substrate one; but, it is being influenced;
in the presence of substrate two, is, there is an influence out here, alpha 2 S 2. Similarly,
the specific growth rate of substrate two is being influenced by this specific growth
rate of substrate one and that is influenced out here in terms of alpha 1 s 1. And, alpha
1 and alpha 2 are constants, obviously. So, again, when you come to alpha 1 and alpha
2, interestingly, we will see that, alpha, alpha 2 and alpha 1, both of them involve
all the rate constants, all the rate constants of the reactions.
So, k 1, k minus 1, k 3, k 2, k minus 2, k 4. So, all the six rate constants of the two
reactions are involved in alpha 1 and alpha 2. And not surprisingly, what you find that,
mu max 1 equals k 3 and mu max 2 equals k 4. Why is that not surprising, because, the
maximum rate possible is the one, you know, in the absence of everything, just the straight
forward production rate. And, why I am saying that is interesting, is because, you know,
you, we did not presume anything; we just went ahead and did the calculation, and when
you got the final results, you can make some physical sense out of those results. So, k
1 is the specific, you know, the rate constants in, involved in the reaction one; k 2 are
the rate constants involved in reaction two. And, as I said that, the mu max, maximum growth
rate of reaction one is k 3; the maximum growth rate of reaction two is k 4, fine.
So, this is an important result and, and you know, you might want to at least remember,
if not remember it completely, at least, have a sense of what is going on. And, this is
not difficult to remember actually, because, this is a Monod growth kinetics and you just
add one term in the denominator and this is a Monod growth kinetics, add one term in the
denominator; and what are the difficult to remember, might be these constants. So, those
might be difficult to remember. So, the next thing we do is, the effect of inhibition.
So, here, first thing we looked at, is the effect of multiple substrates; but the next
thing that we do, is the effective inhibition.
So, what is the substrate? You know, if you have one single substrate, but, instead of
the substrate, along with the substrate producing the cells, the substrate inhibits. So, this
is exactly the same, if you remember, as substrate inhibition that we did for the case of enzyme
kinetic; exactly the same. So, the kinetics is also the same. So, in there also, we had
the thing X plus S giving X S, which you know, in, in earlier case, gives products, but,
in this case, our product is the new cell. So, X plus S giving X S gives new cell and
this is the inhibition. So, X plus S, more substrate is there, that present over there;
instead of, you know, giving new cells, it gives this complex X S 2, which does not lead
to product formation. So, we have to, you know, go through the similar set of steps
that we did for inhibition in case of enzyme kinetics, except that, in these, in this case,
the difference being that, this is varying with time. So, x t is not a constant, but,
x t is varying with time, ok. So, x t over, d x t, d d t of x t over here
is K times X S, right; is that clear to everybody? Why is that, because, this second reaction
is not leading to any growth of cell. So, you know…So, that is not considered over
there. So, d d t of x t is K times S. Now, K, K, K times X S, sorry. Now, from here,
from the first reaction, the reversible first reaction, what you find is, if this is assumed
to have attained equilibrium, then, X S could be written as, X times S could be written
as, K S times X S; is it clear to everybody? Very straightforward. So, we can replace my
X S that I have over here, by X times S divided by K S, fine. So, you have K over K S times
X times S, ok. Now, the other things to take care of, this
X S 2. So, my constraint equation again, is the total amount of cell that is present.
Cells are present as parent cells and daughter cells, in, as X, in terms of the complex,
intermediate complex X S, and in terms of the complex that you have is X S 2. So, the
cells are presented in these three different forms, fine. Again, I want to reiterate that,
x t is not a constant, but, it varies with time. So, d d t of x t is not 0; but, what
we do over here is that, this, from this inhibition, in inhibition reaction, the substrate inhibition
reaction, this, we again assume to have gained equilibrium, fine. We had assumed the substrate
inhibition reaction given here to have attained the equilibrium and as the result, we can
write K i as X S times X S times X S 2, clear. Then, X S, again, we can write from this equation
as, X times S over K S, clear. So, that is what we do. So, we write this X plus X S plus
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
2, I have substituted from here. So, X S 2 is, from here is, X S times S over K i, fine;
where, X S could again be substituted by, X times S over K S, right. So, I can write
this like this, and then, X, then, I can take X common, out of the whole thing, and this
is, this is the part I get. I will just show you, if there is any problem with this.
So, x t equals X plus X S plus X S 2. I am just doing away with the bracket. So, X S
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
this over here. So, what I get is, X times S square over K i K S. So, from, if I replace
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.
So, you can take it out.
So, what you, what you get over here is that, d x d t equals K S, K over K S times X times
S; the reason being that, this is the final reaction. This is the only reaction, only
thing that generates products, fine. Is that clear?
So, this is the only, this is the only step that generates product. So, you have K over
K S times X S and then, you can substitute your X, X in terms of x t that we got. So,
x in terms of x t is, we got over here; you substitute that back from here and you get
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?
That is now, my mu over here, fine. So, what you find is mu, is this. So, what is the difference?
The major difference that you find over here is, as compared to the mu that you have before,
was that, the mu before, had something which is similar to a Monod growth kinetics; but,
this is no way similar to the, close to the Monod growth kinetics. And, if you remember
what we did from last time, what, what is going to be the characteristic of the mu?
How is this? You know, how is this mu going to be different from mus’ in the last case,
graphically? Let us think graphically. So, you know, mu over S; in the Monod growth kinetics,
it is going to be something like this, right. So, Monod. So, for multiple, you will have
S 1 and S 2. So, say mu 1 is something like this; mu 2 is something like this, for multiple;
this is mu. Now, how is this.
Yes. So, it will go through a maximum and then, go down to 0. So, this one starts at
0 and saturates; this one starts from 0 and goes to 0 through a maximum. Why is that?
If you look at the expression itself, you will see that, that, it will, it will saturate
out in the absence. So, how do you obtain the maximum? So, they are simply differentiated
and do a del mu del x and get that. So, we did a del mu del x to get the maximum value,
and the maximum specific growth rate turns out to be, S critical is square root of K
i K S; square root of K i K S. It is similar, if you remember from last time, in the case
of the enzymatic growth, it is exactly similar, to what we did in the last case. when you
are done, last case. For the last case, you can do some calculations on your own also.
So, there is a mu; just do, do a del mu del x and get this. So, shall we move on?
So, the last thing that we are going to do is, in today’s lecture is, allosteric inhibition
and this is the last thing that we are going to do in this chapter also. So, in this kind
of inhibition, what happens is that, instead of stopping here…See, what is the difference
between the last case and here, is set, last time, we stopped here, X S 2 and we stopped
here. In allosteric inhibition, what happens is that, this also gives rise to daughter
cells. We do not stop here completely. We say that, this also gives a rise to, rise
to daughter cells. So, in a way, it is the combination of substrate inhibition and multiple
substrates. Come to think of it, it is a, do you agree with me? So, in a way, it is
a combination substrate inhibition. In substrate inhibition, we had everything like, we had,
but, we stopped over here, right. So, there is no daughter cells produced from the second
reaction. But, in the multiple substrate, we did not have the same substrate, or combining,
but, we had daughter cells from both the substrates. So, in a way, it is like a combination of
multiple substrates and substrate inhibition. So, I would not do all the steps for it. I
will do some steps and you can go ahead and do the rest, either here, or later, maybe
at home, as an assignment. So, the procedure is that, the same procedure is as before;
the only difference that is going to happen is that, del del t of x t is K 2 times X S
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
for the sake of algebra, times X S 2; is that clear? So, I think, there, there is nothing
hard in it. So, and then, you have to go and do the rest of the algebra, which is the x
t, you have to, you know, calculate the x t, which is the X plus X S plus X S S, as
before, and do the del del t of x t. And then, take the reversible for both cases, and this
is a final result that you will come up with for mu. mu is defined as 1 over x t del d
d t of x t and this is the final result, you will come up with. So, the difference between,
yes, the difference between last time and this time is that, last time, we had the denominator
and you had one term in the numerator; you have, you have another term in the numerator.
So, it is quadratic, both in the numerator and the denominator and when you, again, it
will have a lot more interesting kinetics. So, what I would like you to do is, probably
slide here. So, what I would like you to do is, actually go and plot this; because, this
is not necessarily going to have the same kinetics as, as that and it can even have
multiple peaks. I am not sure, but, once you differentiate it, you will be able to figure
out that, if it has multiple peaks. So, what I want you to do is, as a little assignment
is that, do the steps; steps are very straight forward; but, do the steps and check the answer.
And then, once you have got this, and I want you to plot this a and b c. The first thing
you should actually do is, go and differentiate it; find the, find the maxima and the minima;
because, once you get the maxima and the minima, you can plot it; very straight forward. Otherwise,
you know, it is like a black box, trying to plot this. So, go and find the maxima and
the minima; one of the maxima is going to be like what we had before; but, there in,
there is a quadratic term. So, it could lead to some other interesting effect.
So, I want you to see that. So, this is, this is a something, very common that happens and
this is, this is the specific growth rate in, in, in that case and we need to understand
how this works, in order to be able to figure out. So, the reason we are doing this is,
like, you know, what we are going to do in the next chapter and these, all these cases
that we did today are important, because, I am going to quiz you on, on those things,
later in the exams.
So, in the next chapter, what we are going to look at, is the process, the dynamics of
cell growth. See, we are looking at the kinetic sub cell growth. But, what happens is that,
the cell growth process happens in a, in a reactor like this, you know. So, it could
be a batch reactor; it could be a chemostat, continuous third tank reactor.
So, one of the things that we are going to look at is, starting next class, it is a new
chapter, is chemostat ,which is a C S T R, C S T R for a biological reaction. So, temperature
is controlled and you know, pH is controlled. So, pH and temperature controlled and then,
we look at the growth process inside this reactor. So, that is when all the kinetics
that we are doing in this class and the previous classes in the mass transfer effects and so
on, come into play. So, right now, we are looking at…So, this is all connected. So,
right now, we are looking at the kinetics in itself; just isolating the kinetics and
looking at it. Then, we put, check the kinetics out and put into the reactor, and look at
what the effects are, and this, these are, I am, I am assure you that, some of these
effects are going to be very interesting. So, we, we will work with, you know, some,
probably, what we will start with is, not even this, the very straight forward case
here, the Monod kinetics and then, we will go and slowly do what I will probably ask
you to do, what the effects of multiple reactions are, the effects of inhibition and the effects
of allosteric inhibition. So, we will stop here, and I will see you in the next class.
Thanks.