welcome to lecture number five of this n p t e l online certification course on bioreactors
in the last lecture we saw that the two major products from a bioreactor could be the cells
themselves are the products that are produced by the cell we said that time we will refined
it a little better in this lecture we also saw the various models for the variation
of specific growth rate with substrate concentration with product concentration and
essentially the models for the growth kinetics and we also saw the a model the
luedeking piret model for the product formation kinetics then we saw a few concepts such as
the yield coefficient the maintenance and that is were we finished up the [la/last]
last in the last lecture
let us begin this lecture with enzymes we said we will slightly improve the outcomes
from the bioreactor cells themselves and the products that are made earlier we said that
the products that we are made by the cells it could also be the products that result
from an enzymes mediated reaction all of you would have heard of enzymes enzymes usually
proteins which are produced by microorganisms in bioreactors are used extensively in the
industry and that is the interest in the enzymes themselves then the enzymes could
mediate products from let say from a to be conversion and so on so forth the b could
be the preferred product over a and the transformation through enzymes the bio transformation through
enzymes could be carried out in a bioreactor but before we look at that let us look at
enzymes themselves these are produced by microorganisms and are extensively used in
the industry i have given a few examples here and this use has been for a few decades
now
the listing is as enzyme industry and the enzymes used they are heavily used in the
detergent industry especially enzymes such as proteases and amylases are used in the
detergent industry to improve the cleaning ability of the detergents in the baking industry
amylases are used in the brewing industry amylases glucanases and proteases are used
in the dairy industry rennin lipases lactases in the starch industry amylases glucose isomerase
remember we said that starch can be broken down to glucose for the fermentation amylases
are the once that could carried that out textile industry amylase leather industry trypsin
proteases and cocktails of enzymes could be used mixtures of enzymes could be used in
the leather industry
pharmaceutical trypsin and that is just examples of the use of enzymes in the industry
as was mentioned previously enzymes can be used in the bioreactors as the agents that
cause different products to form and there is a special name for that bioreactors such
bioreactors are called enzyme bioreactors examples here penicillin acylase for six amino
penicillanic acid or six a p a production from penicillin g penicillin g is converted
to six a p a through penicillin acylase six a p a is widely used and glucose isomerase
is used to produce sweet invert sugar um this is also used heavily in the industry in the
food industry industry
and as before we need to have good understanding of the kinetics of the enzyme reaction earlier
we looked at the kinetics the rates of growth in the rates of products formation
here we need to have a good understanding of the kinetics or the rates of enzymes reaction
and they become essential in the design and operation of enzyme bioreactors so we will
look at enzyme kinetics predominantly in this particular lecture let us make things as simple
as possible let us see how an enzyme reacts to form the products or how the product is
made in the presence of an enzymes how the enzymes mediates the process how does it get
involved in the process one of the simplest mechanisms that have wide acceptance for simple
enzymatic reactions this is follows if e is the enzyme which converts substrate s to product
p the mechanism that is considered here is the enzyme reacts with the substrate in
reversible fraction the rate of the forward reaction is k one the backward reaction here
is k two this is the reversible reaction to form an enzyme substrate complex and this
reaction is fast there is also an equilibrium reaction due to the k one is rather high
equilibrium reaction and then this enzyme substrate complex reacts further converts
further two enzyme and product this is through the rate constant here is k three
this is the overall mechanism of simple enzyme mediated reaction
if this is the case and let us say this is occurring inside a closed vessel a batch reactor
which we take as a system if we write a mass balance on the product we are going to focus
on the product and write a mass balance for the product that gives the rate of generation
of the product that is given here minus the rate of consumption remember this is a batch
reaction and therefore the input and output terms are zero the input and output rates
are zero so the other the those two terms disappear and we have only the generation
and consumption terms associated with the product so the rate of generation minus the
rate of consumption of the product equals the rate of accumulation of the product all
this in mass terms
there is no if we assume that there is no consumption of the product then the rate of
generation of the product equals the rate of accumulation of the product here rate of
generation rate of accumulation and we know from chemical kinetics that the rate on a
per unit volume bases the volumetric rate as we call is r g p per unit volume ok this
is the normal way in which we write our rates on a per unit volume bases this rate is on
a mass bases mass per time this is mass per volume per time this is what you would
have learnt in your high school high secondary school and so on the rate on a per unit volume
bases is k three into the concentration of the enzyme substrate complex thing there that
is the earlier equation so to get the value of interest to us which is mass per time we
need to multiply both sides by the volume of the system because this is on a volumetric
basis we are looking for a mass basis concentration is nothing but mass per volume if we multiply
that then we get rate on a mass basis the rate of product generation on a mass basis
let us call it r p for simplicity equals k three into the concentration of the enzyme
substrate complex into the volume v
now let us write a mass balance on the enzyme in the vessel earlier we wrote a mass balance
the product now let us write a mass balance on the enzyme if we do that at any time of
interest the enzyme is either free or bound to the enzyme substrate complex as we had
viewed in our mechanism therefore the mass of the total enzyme e t is the concentration
of the total enzyme times the volume the mass of the total enzyme is the mass of the free
enzyme plus the mass of the enzyme that is bound to the enzyme substrate complex right
therefore the mass of the free enzyme is concentration of the free enzyme to the volume the same
volume is this plus the mass of the enzyme that is associated with the enzyme substrate
complex is the concentration of the enzyme substrate complex times the volume so this
is just the mass balance this is what can be equated and the concentrations cannot
since the concentrations cannot be equated directly but the masses can be equated now
since all the volumes are the same we get equality in concentrations the concentration
of the total enzyme equals the concentration of the free enzyme plus the concentration
of the bound enzyme or bound as enzyme substrate complex therefore this is just transportation
of this equation the concentration of the enzyme substrate complex is total concentration
of the enzyme minus the concentration of the free enzyme now if we plot the concentrations
of let's say the substrate the product and the enzyme substrate concentration verses
time as the reaction proceeds it will be something like this the substrate is going to react
to give the product therefore that goes down the product is going to get formed therefore
that keeps going up and the enzyme substrate complex behave something with this at very
small times there is an increase and then there is no change in the concentration of
the enzyme substrate complex ok so apart from very short times here early times the concentration
of the enzyme substrate complex does not change therefore if you are interested in this region
this range of times then we could say that the accumulation rate of the enzyme substrate
complex is zero because if it is accumulating it should either go down sorry it should
either go up or go down that's not happening here and therefore this can be put as equal
to zero
this is the same equation as earlier generation rate of enzyme substrate complex minus
consumption rate of enzymes substrate complex from the materiel balance equals the accumulation
rate of the mass of enzyme substrate and this system and since there is no change in
concentration that goes to be that goes to zero now this is what you might have seen
if you remember your enzyme kinetics from high school or your pre engineering preparation
and so on so forth this is only a limited way of looking at this concept infact this
concept is very powerful so let us look at a generalized way of um the consequences
of this particular aspect need to do that let us consider an example before i say
that i should also say that the the concentration here does not change therefore this rate is
zero therefore you can assume take the enzyme substrate complex to be at pseudo steady
state that's what p s s means pseudo steady state ok this is this concept is fine in the
case of a bad system let us generalized that concept
let us to generalized that concept let us consider an example fictitious example of
making an engine for a car let us say that we are putting the engine together in house
the right from the bolt to the engine the time that is needed for the manufacture of
a bolt is on the average about five seconds let us say that ok and let us say that the
time on the average for the manufacture of an engine is about an hour this is an average
of five seconds it could be six seconds seven seconds three seconds four seconds whatever
it is or even two seconds and so on it could vary right and that is the unsteady aspect
in the manufacture of the bolt whether it takes five seconds to manufacture a bolt or
seven seconds to manufacture a bolt or three seconds to manufacture a bolt has no impact
on the time there it takes to build an engine this is of the order of an hour
this variation in a matter of seconds is not going to effect the rate at which the engine
is being made the unsteady nature time varying nature of bolt manufacture does not affect
the kinetics of enzyme of engine manufacture that is because these two are widely separated
in terms of their characteristic time as they called this one is matter of a seconds
we manufacture this this one takes thousands of seconds sixty into sixty about an hour
three thousands six hundred seconds so of the characteristic systems characteristic
times are widely different the unsteady nature of the faster process does not affect the
kinetics of the slower process if our interest happens to be in this slower process so while
looking at the slower process we can always consider the faster process or a much master
process to be at pseudo steady state in actuality it could vary but that variation does not
affect the process the slower process so the faster process in comparison can be taken
to be a pseudo study state and this is a concept that has wide applications you can also look
at at this way the enzyme substrate complex formation variation is very fast compare
to the times over which seeds a product gets formed and therefore the enzyme substrate
complex could be assumed to be at pseudo study state compare to the product
so whenever we use the pseudo study states assumption we need two processes which are
widely separated in terms of the characteristic types we will visit this later but for now
which have we understand that's fine it's not going to effect the understanding of the
derivation of the kinetic equation ok let's go back to the this we had enzyme plus substrate
giving you reversibly and enzyme substrate complex and irreversibly and enzyme and the
product the rate of generation of the enzyme substrate complex we going to focus here is
nothing but k one into the concentrations of enzyme and substrate into the volume this
is on a this is on a volumetric basis this is on a mass basis therefore we need to multiplied
by the volume and the rate of consumption of the enzyme substrate complexes you see
here it is consumed through the reverse reaction with the rate constant k two and this reaction
with the rate constant k three the consumption is k two into concentration of the enzyme
substrate into the volume plus the k k three into the concentration of the enzyme substrate
complex into the volume here this is the rate of generation rate of consumption and so the
mass balance conserved at the rate of generation rate of consumption rate of accumulation is
zero this is what we have already seen as the pseudo steady state assumption therefore
the rate of generation of the enzyme substrate complex equals the rate of consumption of
the enzyme substrate complex because this is equal to zero in the previous step
replacing the rates by the concentrations k one into e into s both are concentrations
into volume equals k two into enzyme substrate complex concentration into volume plus k three
into enzyme substrate complex concentration into volume the volumes are the same therefore
they can be canceled and therefore you get k one into e into s equals k two into e s
plus k three into e s that's the same equation and we are just transforcing this we are
collecting all the k one into e e is difficult to know where as e t we can have a handle
on therefore let us write e in terms of e t and the enzyme substrate complex we had
derived this earlier through a mass balance on the enzyme you can go back and check that
k one into e t minus e enzyme substrate complex into this s this is common here
therefore k two plus k three into e s and a little more of algebra i am just taking
this s inside the bracket and then i am combining the terms which have the enzyme substrate
complex so k two into k three into enzyme substrate complex plus k one into enzyme substrate
complex into s and therefore you could write this as the left hand side remains the
same right hand side if you take the enzyme substrate complex concentration common out
k two plus k three from here plus k one into enzyme substrate comp k one into s the
enzyme substrate complex is out therefore k one into s here and therefore the concentration
of the enzyme substrate complex is the left hand side divided by the factor here k
two plus k three plus k one times s
and if we divide the numerator the denominator by k one right we get the k one disappears
from here so e t into s and here k two plus k three divided by k one plus is there is
a k one here with k s canceled plus s equals the enzyme substrate complex concentration
which is the same as this and if i call k two plus k three by k one as some capital
k m then the same equation can be written as this e t s divided by k m plus s equals
enzyme substrate complex concentration and we had earlier seen that the rate of product
formation in terms of a mass rate is k three into e s into the volume you can go
back and check that equation so if you substitute the enzyme substrate complex concentration
from above we get the rate of product formation is k three into e t s by k m plus s in the
place of enzyme substrate complex concentration times v and if i say k three into e t k
three into the enzyme subs[trate]- total enzyme concentration if i call that as v m r p becomes
v m s by k m plus s into v note that r p is on mass basis
r p on a per unit volume basis or a concentration basis let us call that is v to be consistent
with the terminology that you would be familiar with so v equals v m s by k m plus s does
this appear familiar of course it does this is the well known michaelis menten equation
ok this is the simplest enzyme kinetics equation for a single link enzymes single substrate
case which goes as enzyme and substrate reacting together in reversible fassion to form the
enzyme substrate complex which further goes to review the product and the enzyme fact
so this is v equals v m s by k s plus s does this form of the equation remind you are something
that we saw in the previous lecture just keep that in mind will come back right so this
is what we got v equals v m s by k m plus s where v m equals k three into e total or
e t and k m is k two plus k three by k one now does it strike you where you saw this
earlier if you plot v verses s the rate of the reaction the volumetric rate of the reaction
by with the substrate concentration you get rectangular parabola yes it's a
same form that we got for the monad equation the monad equation give us the variation of
growth rate you know the culture growth rate with substrate concentration here it is very
different here it is the rate of the michaelis menten enzyme re reaction or the enzyme
reaction which takes place in the michaelis menten form with again the substrate concentration
the substrate concentration variation that aspect is the same but this is for an enzyme
reaction that is for growth kinetics ok you must make a distinction between these two
people ten to confuse one with the other so the same kind of since is the same form of
the equation the same kind of interpretations can be drawn v m is the maximum rate and the
rate increases and then attains a symptotically attains the maximum rate v m and if we
replace k m with s then we get v equals s by two s therefore v m by two therefore
k m is nothing but the substrate concentration at which we get the half maximal rate same
as the monod equation interpretations
the model parameters v m and k m these are the two module parameters that we need to
completely describe a simple enzyme kinetics they can be determined by doing an experiment
where we follow the substrate concentration with respect to time in a batch reactor ok
this experiment is done in a batch situation batch reactor we collect s verses t data and
if we do that the michaelis menten equation is v equals v mass s by k m plus s if we invert
this equation ok take one by on both sides we get one by v equals inversion of this k
m plus s by v m s and this can be written as k this is the plus here therefore k m by
v m s plus s by v m s k m by v m s becomes k m by v m into one by s and here the s cancels
out so you have one by v ok so if you look at this as y equals m x plus c if you take
this as y if you take this as x you can call this as m the slope and c the intercept ok
so if you plot one by v verses one by s from the s verses t data you should get k m by
v m as a slope and one by v m as the intercept this is the way in which these model parameters
are estimated there are some issues associated with the estimating at this way let's not
get into that for the time being let's assume that the data is good and we have a good
spread in the data and so on that the out layers are not going to affect the value itself
that's the essential problem here if you know what i am talking about if you don't know
don't worry about it we will have to be a little careful while using this that
you can learn later
a plot of one by v verses one by s which is actually called the lineweaver burke plot
gives us k m by v m as the slope and one by v m as the intercept from s verses t data
calculate v using two successive data points of s and t for plotting this is what you need
to do because we have substrate concentration verses time ok the rate is delta s by delta
t over that short interval that's what we are approximating ok and we assign that delta
s by delta t to the mid point of the time that's that's how we need to do it when we
have the data we need to use the complete set of data to get the required parameters
this is experimental data so one by v verse one by s you will get a line like this the
y intercept is one by v m and the slope is k m by v m other types of plots can also be
used let's not get into that you can look at any standard book on enzyme kinetics
including your text book and that will give you the other plots also
now let a problem be assigned you can try this problem out this requires the concepts
that we just picked up just to summaries we started this lecture five with examples
of enzymes being use in the industry industrial enzymes and then we also said there is an
enzyme can be used in a bioreactor to form a product of interest and that is what we
are looking at in detail we said that we need the kinetics of the enzyme reaction we went
through the derivation of michaelis menten equation which has two parameters v m and
k m in in form it is very similar to the monod equation which describes the variation
of a growth rate with substrate concentration here it's a variation of enzymes volumetric
rate with the substrate concentration and then we also saw that we could find out v
m and k m from the data of substrate concentration with time variation comma batch system
by using lineweaver burke plot
now will use some of those concepts to address this closed ended problem to solve this closed
ended problem please try it out and of course in the next lecture i will show you how to
solve this the question reads is follows and industry scientist is studying a novel method
of disposing a toxin that results from cells culture during the course of cultivation the
aim is to enhance cell yields through toxin reduction and medium re circulation the toxin
does not affect cell growth if it's concentration is below seven point five millimolar the particular
toxin she is studying x yes she has a lot of secrecy requirements can be broken down
enzymatically using the enzyme e again secrets the breakdown was studied at thirty degree
c at p h of seven point two and the following data was obtained under batch conditions this
is the variation of time with time of x x is the substrate here the breakdown of which
we are studying this gives five points at time zero it is twenty millimolar time
ten it is seventeen point seven millimolar and so on part a determine the michaelis menten
parameters for this enzymatic degradation that is part a and b if the total enzyme concentration
is tripled what will the medium will the medium contain toxic levels of x after thirty minutes
i think this is a nice place to stop lecture five we will meet again in lecture six see
you there