welcome to lecture eighteen the n p t e l online certification course on bioreactors
we will begin module five today which is the last module of this course the title of this
module is the cell view of a bioreactor we have always maintained that the cells are
the actual factories that produce the product in the bioreactor they could be products themselves
now if you recall module two the title was two major outcomes of bioreactor the cells
themselves or what is known as the biomass or the products that are made by cells or
through enzyme reactions so we are going to look closer in to what is happening or look
in to what is happening inside the cells in a bioreactor and how we could use some
of those two make bioreactors perform better is a just to repeat that the cells are the
actual factories in a bioreactor an any significant improvement needs to happen at the cell level
because that's the more fundamental level understanding anything from a fundamental
level is always much better for guaranteed improvement and that is what we are trying
to do here
we will look at some approaches that target the cells in the bioreactor infect i have
a think for this we will first look at the cell has a black box you know these documentary
has specs black box what is happening what we could do then we will open an in a window
into that black box at through the indicators of status of the cell the status is quote
unquote there are many statuses that are talked about we will look at some aspects of that
then we look inside through the window maybe from other directions also vary maybe
we will open up some doors also and look inside and an example of that is metabolic flux analysis
and then try to modify things according to our needs such as d n a modification use
of recombinant d n a technologies and so on so for we will this barely touch upon them
because that's an entire field and itself recombinant d n a technology i will just
talk about it's relevants to improving the operation or the outcomes from a bioreactor
so that's going to be the theme of a this particular module first the bioreaction stoichiometry
which we will cover in this lecture then a let us look at a problem then in the next
lecture we will solve the problem and then go forward bioreaction stoichiometry the term
stoichiometry it is use heavily in chemistry it is also used in engineering in engineering
it actually denotes the material and energy balances or things that are based on the
mask conservation principle and the energy conservation principle that is what is called
stoichiometry in some fuzzy fashion in engineering you have books on stoichiometry
similarly bioreaction stoichiometry refers to material and an energy balance calculations
that involve biosystems bioreaction stoichiometry helps us make educated guesses when the complete
data is unavailable that's how we are going to look at it we are we are just started looking
into the cell at the cell level we had let to this in our earlier discussions
but this is where we are actually getting to look in to this cell and the first approaches
that the cell is a black box and what we can gather from whatever we observed it as a black
box but with a thinking that there are things going on inside the cell
a very useful concept for the use of stoichiometry is what is called the degree of reductance
this is a formulation which makes bioreaction stoichiometry calculations easier and it is
denoted by the capital greek symbol gamma as us given here you see here this is the
degree of reductance we have degree of reductance given for each substance as
a first approximation it can be understood as the number of electrons that a substance
can donate so it's reasonably fundamental fundamental concept it works with the electron
electrons are conserved and so you can work you can do things there so number of
it is understood as the number of electrons that is substance can donate this is only
the first approximation this does not always valid ok so it is more of a number there is
assign to a certain molecule we we will see this thus the number of electrons that oxygen
can donate is minus four ok and o two is minus four o is minus two carbon is plus four
and nitrogen is minus three you know it is a number of electrons at a substance can donate
and let us complete this and then come back to this particular aspect gamma of phosphorous
is plus five gamma of sulphur is plus six gamma of c o two this is plus four c this
is minus four o two so we can just add the degree of reductances and arrive at the degree
of reductance of a compound that is made up of these elements so this is zero and similarly
h two o would would turn out to be zero here because hydrogen is one and oxygen is minus
two you have two of hydrogens and one of oxygen that turns out to be zero
so the basic things are o c n p and s and from that you get get oxygen nitrogen c o
two h two o oh sorry oxygen c o two h two o and so on and so for ok so the negative
value actually implies that the substance can accept electrons instead of donating them
and this is understood as a number of electrons at a substance can donate that is the degree
of reductance let us see what we can do with this concept
before that a as was a briefly mentioned earlier the degree of reductance of any substance
can be obtained by the algebraic addition of the degree of reductance degrees of reductance
of it constituent elements of it's constituent elements for example even for a complex
sub molecule such as a substrate you know we had earlier looked at o two h two
o c o two even if you have something like a carbohydrates here c h a o b the degree
of reductance of the substrate is the degree of reductance of it's constituent elements
added together therefore this as one carbon four into one the degree of reductance of
carbon is four therefore four into one plus the degree of reductance of hydrogen is one
one in it has a atoms of hydrogen so one into a and the degree of reductance of o is minus
two and the number of atoms is b so minus two into b so that will turn out to be
whatever it is four plus a minus two b and you can also think of a degree of reductance
for cells it is indicated as gamma x and let us say that the cellular composition people
have determined cellular compositions there are tables in one of your text books felina
analysis that which is from another paper actually which gives you the compositions
of the elemental compositions of various cells and you will find this to be a approximately
ok the representative formula chemical formula for cells is c h a n b and o c predominantly
so the degree of reductance of cells or gamma x would be four into one for carbon plus a
into one for sorry one into a for hydrogen plus minus three for n into b and plus
minus two for o into c so this would be the degree of reductance of the cells
this gamma x has been found to be about four point two for a wide variety of cells and
here in lies the strength of this method you know this is been estimated in similarly the
parameters that we are going to look at chiefly the chief parameters have been estimated for
a wide variety of substances these happened to fall within a very narrow range and therefore
if you take the average value of that range as an as a representative value we can do
first level calculations very easily and get insights into what is happening which can
intern be use to improve the bioreactive performance therefore in the absence of actual data gamma
of of x equals four point two is a good first approximation
let us look at the application of this the couple of applications and i will assign
a problem let us consider at typical cell and product formation this is a stoichiometric
equation that it that represents the cell and product formation starting with a substrate
carbohydrate c h m o l and ammonium both these are substrates this is a nitrogen source here
so carbohydrate c h m o l plus a this is a stoichiometry coefficient of n of n h three
plus b of o two oxygen is of course required gives you a certain y x which is a stoichiometric
coefficient of cells i think ya this is a substrate this is a cells c h p o n n n
q and plus y p which is stoichiometric coefficient of the product c h r o s and t it's a product
plus water c molecules of water and d stoichiometry coefficient of c o two
so this equation stoichiometric equation represents the cell and the product formation in reasonably
general format we will choose this has the equation for our analysis of a few things
that we will do in this lecture electrons we know are conserved right so by the prince
by that principle the ya the number of electrons need to be conserved that's what
we just mentioned so if you count the number of electrons in this side that must equal
the number of electrons on this side how do we count the number of electrons we have degree
of reductance ok that's how the degree of reductance so gamma of s the substrate plus
the gamma of ammonia into a because that's a stoichiometry coefficient here a times minus
three for n plus three for hydrogen plus b times minus four i am writing this is as an
equation therefore i am bringing all these to the other side minus y x gamma x minus
y p gamma p minus c times h two o was zero so that is zero minus d times c o two gamma
of c o two that is zero that equals zero so ammonia drops out water drops out c o two
drops out and we have the remaining things therefore this b you know a is already dropped
out c is dropped out d as dropped out b is the one that remains b is nothing but one
fourth of by appropriately transposing this equation all these would become pluses so
this would become minus one and this is multiplied by a factor of of four minus four so
that when taken to the other side would be gamma s minus y x gamma x minus y p gamma
p divided by four so that is b
so you could write this as multiply again four b by gamma s plus y x gamma y by gamma
s right we are what we are doing is we have multiplied this by four and then we are dividing
throughout by gamma s so four b by gamma s plus y x s by x gamma x by gamma s plus y
p gamma p by gamma s that equals one ok this is a powerful expression this indicates the
fraction of the available electrons from the substrate that has been transferred to oxygen
because b is a stoichiometric coefficient of oxygen four b is the number electrons that
is associated within so on so four b by gamma s gives us the fraction of available electrons
transferred to oxygen gamma sorry y x gamma s by gamma s gives us a fraction of available
electrons transferred to cells similarly y p gamma p by gamma s gives us the fraction
of available electrons transferred to the product all these add up to one these are
all fraction so they should add up to one there are some special symbols that we use
for each of these the fraction of available electrons transferred to oxygen is represented
by epsilon the fraction of available electrons transferred to cells is represented by eta
the fraction of available electrons transferred to the product is psi so epsilon plus eta
plus psi is one so this is something that we can use now using these let me show you
how to estimate the heat that is generated from the cells in the bioreactor during growth
it is known you know i said the power of this method comes because the rules of thumb
values are known right so it is known that twenty seven kilo calories of heat are generated
per mole of available electrons transferred to oxygen you know that epsilon the fraction
of available electrons transferred to oxygen for every mole of electron transfer to oxygen
twenty seven kilo calories of heat are generated this is known in our example the equivalents
of available electrons transferred to oxygen is four b therefore the heat generated is
twenty seven times four b kilo calories from an another angle if the rate of oxygen consumption
say q o two in moles per gram cell is measured using a d o probe you know if you process
the d o data you can get the oxygen consumption rate the rate at which heat is generated can
be estimated then it becomes powerful ok once we know how much of heat is metabolically
generated we can apriori save what kind of a cooling make an a cooling design we
need to look at you know that's rather powerful this will help us design cooling coils for
obtaining the desired rate of cooling if it's a if it's cooling it's ok you 17.00)
can use a coil because the temperature is not going to go down too much is not going
to effect the cells were as heating heating coils are not really used use the
jacketed system
similarly a jacketed system it can also be use for cooling but cooling coils are also
used so the rate at which electrons is transferred to oxygen is four times q o two x into v
you know the we each oxygen corresponds to four electrons this is the oxygen up take
rate in moles per gram cell we need to multiplied by the cell concentration to get it an appropriate
unit's and then multiplied by the bioreactor volume to get it in the unit's that we need
so this is the rate at which electrons are transferred to oxygen number of electrons
per unit time right so that is the case the rate of heat generation for each mole it is
twenty seven times four b so for in this case the for each mole sorry for each mole it
is twenty seven kilo calories so in this case it is twenty seven times four q o two x v
this gives you the rate of heat generation and estimate of the heat rate of heat generation
which can be helpful for us to make many decisions so this is based at what is happening at the
cell level
now let me assign this problem and we can finish up this lecture here when we come
back we will solve the problem in the next lecture the practice problem five point one
reads us let us consider the same stoichiometry is earlier is reasonably general as long as
ammonia reuses and nitrogen source and some carbohydrate is uses the carbon source and
it's aerobic this is reasonably general so c h m o l plus a n h three plus b o two giving
you y x c h p o n n q plus y p c h r o s n t plus c h two o plus d c o two estimate the
cell oxygen yield coefficient y x slash o defined is the amount of cells produce to
the per amount of oxygen consumed and this we need to find out in terms of the relevant
ones among gamma x gamma s gamma p or epsilon eta new right in terms of these if we find
we have estimates of all these gamma x we know is four point two gamma s gamma p
can be easily we calculated epsilon eta new values some estimates some ranges are known
and therefore we can estimate the cell oxygen yield coefficient and while doing this assume
that the biomass fraction of carbon in the cells is point five so we can say that
the the cells are growing aerobically so you get a fraction of point five it's anaerobically
it's around point one or so ok then see you in the next lecture when we solve this problem
and then we go a head
thank you