WEBVTT
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welcome to lecture number five of this
n p t e l online certification course on bioreactors
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in the last lecture we saw that the two major
products from a bioreactor could be the cells
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themselves are the products that are produced
by the cell we said that time we will refined
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it a little better in this lecture we also
saw the various models for the variation
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of specific growth rate with substrate
concentration with product concentration and
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essentially the models for the growth
kinetics and we also saw the a model the
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luedeking piret model for the product formation
kinetics then we saw a few concepts such as
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the yield coefficient the maintenance and
that is were we finished up the [la/last]
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last in the last lecture
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let us begin this lecture with enzymes we
said we will slightly improve the outcomes
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from the bioreactor cells themselves and the
products that are made earlier we said that
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the products that we are made by the cells
it could also be the products that result
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from an enzymes mediated reaction all of you
would have heard of enzymes enzymes usually
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proteins which are produced by microorganisms
in bioreactors are used extensively in the
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industry and that is the interest in the
enzymes themselves then the enzymes could
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mediate products from let say from a to be
conversion and so on so forth the b could
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be the preferred product over a and the transformation
through enzymes the bio transformation through
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enzymes could be carried out in a bioreactor
but before we look at that let us look at
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enzymes themselves these are produced by
microorganisms and are extensively used in
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the industry i have given a few examples here
and this use has been for a few decades
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now
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the listing is as enzyme industry and the
enzymes used they are heavily used in the
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detergent industry especially enzymes such
as proteases and amylases are used in the
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detergent industry to improve the cleaning
ability of the detergents in the baking industry
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amylases are used in the brewing industry
amylases glucanases and proteases are used
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in the dairy industry rennin lipases lactases
in the starch industry amylases glucose isomerase
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remember we said that starch can be broken
down to glucose for the fermentation amylases
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are the once that could carried that out textile
industry amylase leather industry trypsin
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proteases and cocktails of enzymes could be
used mixtures of enzymes could be used in
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the leather industry
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pharmaceutical trypsin and that is just
examples of the use of enzymes in the industry
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as was mentioned previously enzymes can be
used in the bioreactors as the agents that
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cause different products to form and there
is a special name for that bioreactors such
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bioreactors are called enzyme bioreactors
examples here penicillin acylase for six amino
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penicillanic acid or six a p a production
from penicillin g penicillin g is converted
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to six a p a through penicillin acylase
six a p a is widely used and glucose isomerase
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is used to produce sweet invert sugar um this
is also used heavily in the industry in the
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food industry industry
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and as before we need to have good understanding
of the kinetics of the enzyme reaction earlier
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we looked at the kinetics the rates of
growth in the rates of products formation
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here we need to have a good understanding
of the kinetics or the rates of enzymes reaction
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and they become essential in the design and
operation of enzyme bioreactors so we will
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look at enzyme kinetics predominantly in this
particular lecture let us make things as simple
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as possible let us see how an enzyme reacts
to form the products or how the product is
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made in the presence of an enzymes how the
enzymes mediates the process how does it get
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involved in the process one of the simplest
mechanisms that have wide acceptance for simple
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enzymatic reactions this is follows if e is
the enzyme which converts substrate s to product
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p the mechanism that is considered here is
the enzyme reacts with the substrate in
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reversible fraction the rate of the forward
reaction is k one the backward reaction here
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is k two this is the reversible reaction to
form an enzyme substrate complex and this
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reaction is fast there is also an equilibrium
reaction due to the k one is rather high
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equilibrium reaction and then this enzyme
substrate complex reacts further converts
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further two enzyme and product this is
through the rate constant here is k three
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this is the overall mechanism of simple enzyme
mediated reaction
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if this is the case and let us say this is
occurring inside a closed vessel a batch reactor
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which we take as a system if we write a mass
balance on the product we are going to focus
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on the product and write a mass balance for
the product that gives the rate of generation
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of the product that is given here minus the
rate of consumption remember this is a batch
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reaction and therefore the input and output
terms are zero the input and output rates
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are zero so the other the those two terms
disappear and we have only the generation
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and consumption terms associated with the
product so the rate of generation minus the
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rate of consumption of the product equals
the rate of accumulation of the product all
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this in mass terms
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there is no if we assume that there is no
consumption of the product then the rate of
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generation of the product equals the rate
of accumulation of the product here rate of
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generation rate of accumulation and we know
from chemical kinetics that the rate on a
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per unit volume bases the volumetric rate
as we call is r g p per unit volume ok this
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is the normal way in which we write our rates
on a per unit volume bases this rate is on
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a mass bases mass per time this is mass per
volume per time this is what you would
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have learnt in your high school high secondary
school and so on the rate on a per unit volume
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bases is k three into the concentration of
the enzyme substrate complex thing there that
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is the earlier equation so to get the value
of interest to us which is mass per time we
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need to multiply both sides by the volume
of the system because this is on a volumetric
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basis we are looking for a mass basis concentration
is nothing but mass per volume if we multiply
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that then we get rate on a mass basis the
rate of product generation on a mass basis
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let us call it r p for simplicity equals
k three into the concentration of the enzyme
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substrate complex into the volume v
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now let us write a mass balance on the enzyme
in the vessel earlier we wrote a mass balance
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the product now let us write a mass balance
on the enzyme if we do that at any time of
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interest the enzyme is either free or bound
to the enzyme substrate complex as we had
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viewed in our mechanism therefore the mass
of the total enzyme e t is the concentration
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of the total enzyme times the volume the mass
of the total enzyme is the mass of the free
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enzyme plus the mass of the enzyme that is
bound to the enzyme substrate complex right
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therefore the mass of the free enzyme is concentration
of the free enzyme to the volume the same
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volume is this plus the mass of the enzyme
that is associated with the enzyme substrate
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complex is the concentration of the enzyme
substrate complex times the volume so this
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is just the mass balance this is what can
be equated and the concentrations cannot
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since the concentrations cannot be equated
directly but the masses can be equated now
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since all the volumes are the same we get
equality in concentrations the concentration
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of the total enzyme equals the concentration
of the free enzyme plus the concentration
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of the bound enzyme or bound as enzyme substrate
complex therefore this is just transportation
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of this equation the concentration of the
enzyme substrate complex is total concentration
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of the enzyme minus the concentration of the
free enzyme now if we plot the concentrations
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of let's say the substrate the product and
the enzyme substrate concentration verses
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time as the reaction proceeds it will be something
like this the substrate is going to react
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to give the product therefore that goes down
the product is going to get formed therefore
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that keeps going up and the enzyme substrate
complex behave something with this at very
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small times there is an increase and then
there is no change in the concentration of
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the enzyme substrate complex ok so apart from
very short times here early times the concentration
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of the enzyme substrate complex does not change
therefore if you are interested in this region
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this range of times then we could say that
the accumulation rate of the enzyme substrate
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complex is zero because if it is accumulating
it should either go down sorry it should
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either go up or go down that's not happening
here and therefore this can be put as equal
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to zero
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this is the same equation as earlier generation
rate of enzyme substrate complex minus
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consumption rate of enzymes substrate complex
from the materiel balance equals the accumulation
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rate of the mass of enzyme substrate and this
system and since there is no change in
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concentration that goes to be that goes to
zero now this is what you might have seen
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if you remember your enzyme kinetics from
high school or your pre engineering preparation
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and so on so forth this is only a limited
way of looking at this concept infact this
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concept is very powerful so let us look
at a generalized way of um the consequences
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of this particular aspect need to do that
let us consider an example before i say
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that i should also say that the the concentration
here does not change therefore this rate is
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zero therefore you can assume take the
enzyme substrate complex to be at pseudo steady
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state that's what p s s means pseudo steady
state ok this is this concept is fine in the
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case of a bad system let us generalized that
concept
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let us to generalized that concept let us
consider an example fictitious example of
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making an engine for a car let us say that
we are putting the engine together in house
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the right from the bolt to the engine the
time that is needed for the manufacture of
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a bolt is on the average about five seconds
let us say that ok and let us say that the
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time on the average for the manufacture of
an engine is about an hour this is an average
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of five seconds it could be six seconds seven
seconds three seconds four seconds whatever
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it is or even two seconds and so on it could
vary right and that is the unsteady aspect
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in the manufacture of the bolt whether it
takes five seconds to manufacture a bolt or
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seven seconds to manufacture a bolt or three
seconds to manufacture a bolt has no impact
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on the time there it takes to build an engine
this is of the order of an hour
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this variation in a matter of seconds is not
going to effect the rate at which the engine
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is being made the unsteady nature time varying
nature of bolt manufacture does not affect
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the kinetics of enzyme of engine manufacture
that is because these two are widely separated
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in terms of their characteristic time as they
called this one is matter of a seconds
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we manufacture this this one takes thousands
of seconds sixty into sixty about an hour
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three thousands six hundred seconds so of
the characteristic systems characteristic
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times are widely different the unsteady nature
of the faster process does not affect the
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kinetics of the slower process if our interest
happens to be in this slower process so while
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looking at the slower process we can always
consider the faster process or a much master
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process to be at pseudo steady state in actuality
it could vary but that variation does not
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affect the process the slower process so the
faster process in comparison can be taken
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to be a pseudo study state and this is a concept
that has wide applications you can also look
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at at this way the enzyme substrate complex
formation variation is very fast compare
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to the times over which seeds a product gets
formed and therefore the enzyme substrate
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complex could be assumed to be at pseudo study
state compare to the product
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so whenever we use the pseudo study states
assumption we need two processes which are
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widely separated in terms of the characteristic
types we will visit this later but for now
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which have we understand that's fine it's
not going to effect the understanding of the
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derivation of the kinetic equation ok let's
go back to the this we had enzyme plus substrate
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giving you reversibly and enzyme substrate
complex and irreversibly and enzyme and the
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product the rate of generation of the enzyme
substrate complex we going to focus here is
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nothing but k one into the concentrations
of enzyme and substrate into the volume this
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is on a this is on a volumetric basis this
is on a mass basis therefore we need to multiplied
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by the volume and the rate of consumption
of the enzyme substrate complexes you see
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here it is consumed through the reverse reaction
with the rate constant k two and this reaction
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with the rate constant k three the consumption
is k two into concentration of the enzyme
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substrate into the volume plus the k k three
into the concentration of the enzyme substrate
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complex into the volume here this is the rate
of generation rate of consumption and so the
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mass balance conserved at the rate of generation
rate of consumption rate of accumulation is
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zero this is what we have already seen as
the pseudo steady state assumption therefore
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the rate of generation of the enzyme substrate
complex equals the rate of consumption of
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the enzyme substrate complex because this
is equal to zero in the previous step
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replacing the rates by the concentrations
k one into e into s both are concentrations
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into volume equals k two into enzyme substrate
complex concentration into volume plus k three
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into enzyme substrate complex concentration
into volume the volumes are the same therefore
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they can be canceled and therefore you get
k one into e into s equals k two into e s
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plus k three into e s that's the same equation
and we are just transforcing this we are
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collecting all the k one into e e is difficult
to know where as e t we can have a handle
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on therefore let us write e in terms of e
t and the enzyme substrate complex we had
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derived this earlier through a mass balance
on the enzyme you can go back and check that
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k one into e t minus e enzyme substrate
complex into this s this is common here
19:42.470 --> 19:49.320
therefore k two plus k three into e s and
a little more of algebra i am just taking
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this s inside the bracket and then i am combining
the terms which have the enzyme substrate
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complex so k two into k three into enzyme
substrate complex plus k one into enzyme substrate
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complex into s and therefore you could
write this as the left hand side remains the
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same right hand side if you take the enzyme
substrate complex concentration common out
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k two plus k three from here plus k one into
enzyme substrate comp k one into s the
20:23.150 --> 20:29.441
enzyme substrate complex is out therefore
k one into s here and therefore the concentration
20:29.441 --> 20:37.429
of the enzyme substrate complex is the left
hand side divided by the factor here k
20:37.429 --> 20:40.030
two plus k three plus k one times s
20:40.030 --> 20:50.100
and if we divide the numerator the denominator
by k one right we get the k one disappears
20:50.100 --> 20:59.400
from here so e t into s and here k two plus
k three divided by k one plus is there is
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a k one here with k s canceled plus s equals
the enzyme substrate complex concentration
21:04.940 --> 21:20.890
which is the same as this and if i call
k two plus k three by k one as some capital
21:20.890 --> 21:27.900
k m then the same equation can be written
as this e t s divided by k m plus s equals
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enzyme substrate complex concentration and
we had earlier seen that the rate of product
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formation in terms of a mass rate is k
three into e s into the volume you can go
21:41.640 --> 21:48.270
back and check that equation so if you substitute
the enzyme substrate complex concentration
21:48.270 --> 21:55.820
from above we get the rate of product formation
is k three into e t s by k m plus s in the
21:55.820 --> 22:07.220
place of enzyme substrate complex concentration
times v and if i say k three into e t k
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three into the enzyme subs[trate]- total enzyme
concentration if i call that as v m r p becomes
22:14.250 --> 22:24.650
v m s by k m plus s into v note that r p is
on mass basis
22:24.650 --> 22:31.950
r p on a per unit volume basis or a concentration
basis let us call that is v to be consistent
22:31.950 --> 22:41.169
with the terminology that you would be familiar
with so v equals v m s by k m plus s does
22:41.169 --> 22:48.289
this appear familiar of course it does this
is the well known michaelis menten equation
22:48.289 --> 22:54.880
ok this is the simplest enzyme kinetics equation
for a single link enzymes single substrate
22:54.880 --> 23:01.350
case which goes as enzyme and substrate reacting
together in reversible fassion to form the
23:01.350 --> 23:07.690
enzyme substrate complex which further goes
to review the product and the enzyme fact
23:07.690 --> 23:18.120
so this is v equals v m s by k s plus s does
this form of the equation remind you are something
23:18.120 --> 23:27.419
that we saw in the previous lecture just keep
that in mind will come back right so this
23:27.419 --> 23:35.520
is what we got v equals v m s by k m plus
s where v m equals k three into e total or
23:35.520 --> 23:45.730
e t and k m is k two plus k three by k one
now does it strike you where you saw this
23:45.730 --> 23:52.850
earlier if you plot v verses s the rate of
the reaction the volumetric rate of the reaction
23:52.850 --> 24:00.879
by with the substrate concentration
you get rectangular parabola yes it's a
24:00.879 --> 24:08.620
same form that we got for the monad equation
the monad equation give us the variation of
24:08.620 --> 24:13.580
growth rate you know the culture growth rate
with substrate concentration here it is very
24:13.580 --> 24:21.610
different here it is the rate of the michaelis
menten enzyme re reaction or the enzyme
24:21.610 --> 24:26.580
reaction which takes place in the michaelis
menten form with again the substrate concentration
24:26.580 --> 24:31.450
the substrate concentration variation that
aspect is the same but this is for an enzyme
24:31.450 --> 24:37.610
reaction that is for growth kinetics ok you
must make a distinction between these two
24:37.610 --> 24:42.649
people ten to confuse one with the other so
the same kind of since is the same form of
24:42.649 --> 24:51.340
the equation the same kind of interpretations
can be drawn v m is the maximum rate and the
24:51.340 --> 24:59.230
rate increases and then attains a symptotically
attains the maximum rate v m and if we
24:59.230 --> 25:08.520
replace k m with s then we get v equals
s by two s therefore v m by two therefore
25:08.520 --> 25:15.080
k m is nothing but the substrate concentration
at which we get the half maximal rate same
25:15.080 --> 25:19.890
as the monod equation interpretations
25:19.890 --> 25:28.460
the model parameters v m and k m these are
the two module parameters that we need to
25:28.460 --> 25:36.710
completely describe a simple enzyme kinetics
they can be determined by doing an experiment
25:36.710 --> 25:42.640
where we follow the substrate concentration
with respect to time in a batch reactor ok
25:42.640 --> 25:49.539
this experiment is done in a batch situation
batch reactor we collect s verses t data and
25:49.539 --> 25:57.160
if we do that the michaelis menten equation
is v equals v mass s by k m plus s if we invert
25:57.160 --> 26:04.710
this equation ok take one by on both sides
we get one by v equals inversion of this k
26:04.710 --> 26:13.029
m plus s by v m s and this can be written
as k this is the plus here therefore k m by
26:13.029 --> 26:21.200
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
26:21.200 --> 26:32.330
out so you have one by v ok so if you look
at this as y equals m x plus c if you take
26:32.330 --> 26:40.900
this as y if you take this as x you can call
this as m the slope and c the intercept ok
26:40.900 --> 26:49.909
so if you plot one by v verses one by s from
the s verses t data you should get k m by
26:49.909 --> 26:58.110
v m as a slope and one by v m as the intercept
this is the way in which these model parameters
26:58.110 --> 27:04.309
are estimated there are some issues associated
with the estimating at this way let's not
27:04.309 --> 27:09.660
get into that for the time being let's assume
that the data is good and we have a good
27:09.660 --> 27:17.080
spread in the data and so on that the out
layers are not going to affect the value itself
27:17.080 --> 27:20.470
that's the essential problem here if you know
what i am talking about if you don't know
27:20.470 --> 27:25.919
don't worry about it we will have to be
a little careful while using this that
27:25.919 --> 27:28.320
you can learn later
27:28.320 --> 27:33.409
a plot of one by v verses one by s which is
actually called the lineweaver burke plot
27:33.409 --> 27:43.450
gives us k m by v m as the slope and one
by v m as the intercept from s verses t data
27:43.450 --> 27:49.290
calculate v using two successive data points
of s and t for plotting this is what you need
27:49.290 --> 27:56.019
to do because we have substrate concentration
verses time ok the rate is delta s by delta
27:56.019 --> 28:02.020
t over that short interval that's what we
are approximating ok and we assign that delta
28:02.020 --> 28:06.080
s by delta t to the mid point of the time
that's that's how we need to do it when we
28:06.080 --> 28:11.450
have the data we need to use the complete
set of data to get the required parameters
28:11.450 --> 28:19.429
this is experimental data so one by v verse
one by s you will get a line like this the
28:19.429 --> 28:29.020
y intercept is one by v m and the slope is
k m by v m other types of plots can also be
28:29.020 --> 28:36.720
used let's not get into that you can
look at any standard book on enzyme kinetics
28:36.720 --> 28:44.519
including your text book and that will give
you the other plots also
28:44.519 --> 28:53.890
now let a problem be assigned you can try
this problem out this requires the concepts
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that we just picked up just to summaries
we started this lecture five with examples
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of enzymes being use in the industry industrial
enzymes and then we also said there is an
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enzyme can be used in a bioreactor to form
a product of interest and that is what we
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are looking at in detail we said that we need
the kinetics of the enzyme reaction we went
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through the derivation of michaelis menten
equation which has two parameters v m and
29:24.659 --> 29:32.890
k m in in form it is very similar to
the monod equation which describes the variation
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of a growth rate with substrate concentration
here it's a variation of enzymes volumetric
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rate with the substrate concentration and
then we also saw that we could find out v
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m and k m from the data of substrate concentration
with time variation comma batch system
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by using lineweaver burke plot
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now will use some of those concepts to address
this closed ended problem to solve this closed
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ended problem please try it out and of course
in the next lecture i will show you how to
30:09.630 --> 30:17.110
solve this the question reads is follows and
industry scientist is studying a novel method
30:17.110 --> 30:23.429
of disposing a toxin that results from cells
culture during the course of cultivation the
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aim is to enhance cell yields through toxin
reduction and medium re circulation the toxin
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does not affect cell growth if it's concentration
is below seven point five millimolar the particular
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toxin she is studying x yes she has a lot
of secrecy requirements can be broken down
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enzymatically using the enzyme e again secrets
the breakdown was studied at thirty degree
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c at p h of seven point two and the following
data was obtained under batch conditions this
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is the variation of time with time of x x
is the substrate here the breakdown of which
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we are studying this gives five points
at time zero it is twenty millimolar time
31:15.970 --> 31:22.280
ten it is seventeen point seven millimolar
and so on part a determine the michaelis menten
31:22.280 --> 31:30.309
parameters for this enzymatic degradation
that is part a and b if the total enzyme concentration
31:30.309 --> 31:38.419
is tripled what will the medium will the medium
contain toxic levels of x after thirty minutes
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i think this is a nice place to stop lecture
five we will meet again in lecture six see
31:46.649 --> 31:46.820
you there