 We talk about biochemical and environmental reaction engineering and let us quickly look at what is this? We look at respiration, nitrogen, fixation, photosynthesis, say nitrification, denitrification. Now, all these reactions are well known to us and they are all regulated by microorganisms and enzymes are most important part of the work that the microorganisms do. So, what we will do now is quickly look at what do enzymes do and how the enzyme kinetics looks like. So, first thing we want to do is enzyme kinetics. Now, it is all known to us that enzymes are proteins. So, enzymes are proteins, they are proteins that is one, two their active sites, active sites they arise from the way they fold in space. The active sites arise from three dimensional configuration. Third thing is this configuration that these enzymes are able to achieve is when you add certain cofactors. Cofactors, cofactors enable the enzymes, enzymes exhibit the 3D, 3D structure. These are the three important features that enzymes are able to do. So, what is done in the enzyme literature is that if you have an enzyme which has its active form, it forms a complex and this complex reacts to form enzyme product plus the enzyme. Now, enzymes are products of biological activity, but it is now known that enzymes can be extracted purified and used outside the living systems that we are able to do in many cases. Now, if I call this as k 1, call this as k minus 1, I call this as k 2. You can write suppose we perform this reaction in an equipment which has substrate S naught, it has an enzyme E naught. Then you can write your material balance for the substrate as R times S times V. This is something that we all have written. Similarly, you can write a material balance for the enzyme substrate complex. This material balance for the substrate complex is the material balance for the enzyme substrate complex. So, let me write all these functions properly. R S, please note R S, R S rate of formation rate which is gets consumed then R S is I will write minus of R S k 1 E times S minus k minus of 1 E S. R S, R S is k 1 times E times S minus of. Now, R E S that means rate at which the enzyme substrate complex is getting formed. I will write here. So, it is k 1 E times S minus of k minus of 1 E S minus of k 2 times E S. Please notice here rate of formation of E S complex is k 1 times E times S minus this, minus this. Now, what is frequently done in the enzyme literature is that I have try to V here. I forgot try to V here. Now, this rate of formation of the E S complex is assumed to be this is rate of that means the E S complex assumed stationary or in other words E S remains constant. This does not change or E S equal to 0 is an assumption which is called as quasi steady state approximation. On other words what we are saying is that as the enzyme reaction proceeds the complex does not change in concentration remains reasonably constant. So, that R E S can be assumed to be 0. It is an assumption that the enzyme literature makes right through for a very very long time. In biological process it seems to be a good assumption, but in commercial synthetic processes it applies under certain conditions. The condition is the condition under which it applies is E naught by S naught should be small. If this assumption is good then this approximation is considered to be very good or in other words if you want to use this approximation we have to see that this ratio is quite low. Now, setting this as 0 that means 0 equal to k 1 E times S minus k minus 1 E S minus k 2 E S equal to 0 gives you E S k minus 1 plus k 2 equal to k 1. What is E? E is simply E plus E S though I will write this as E naught minus of E S into S. So, I will do a further simplification k minus 1 plus k 2 plus k 1 S equal to k 1 E naught. Is it all right? I have just taken this on to the other side E naught is the good question good question E naught. See we start this enzyme reaction with certain amount of initial enzyme in the process. So, it is the total amount of enzyme that you added to the process for initiating this enzyme reaction. So, you get E S let us go forward E S equal to k 1 E naught S divided by k minus of 1 plus k 2 plus k 1 S. Now, our enzyme reaction is what E plus S equal to E S and then E S giving you products plus enzyme. Therefore, the rate at which rate of formation of product is k 2 times E S. This is the rate at which product is formed is it all right. So, we have got E S here. So, I can get rate of formation of products let me write here R p is k 2 times E S. Therefore, R p equal to k 2 and E S is E S is this one k 1 E naught I will write it again. So, it is k 1 E naught S divided by k minus 1 plus k 2 plus k 1 S. So, I can simplify this little k 2 E naught S divided by k minus 1 plus k 2 divided by k 1 plus S is it all right. So, all I have done is I have just divided numerator and denominator by k 1. So, I have got S here k minus 1 plus k 2 divided by k 1 let us just recall what is k minus 1 k minus 1. Let me quickly write it again E plus S equal to E S this is 1 and this is minus 1. So, E S going to products plus enzyme this is 2. Therefore, k minus 1 plus k 2 divided by k 1 this term is often called as k m. This is the nomenclature enzyme let us k 2 E naught is often called as V m. Therefore, R p is denoted like this this is the way in which you will find in literature around the world the enzyme reaction kinetics is represented like this where V m is k 2 E naught and then k m is k minus 1 plus k 2 k minus 1 plus k 2 you can see here k minus 1 plus k 2. Now, if you want to find out the parameters of the enzyme reaction typically what people do is this make a plot. So, I will write this again R p equal to V m S divided by k m plus S. So, when you do an inverse transformation 1 by R p equal to k m by V m S plus 1 by V m this is of course, we have done this many times in the past. So, this is often called as line weaver bulk plot L B plot this is the way in which it is described. So, you get straight line like this your points like this. So, this intercept is 1 by V m and this slope is k m by V m. On other words an enzyme reaction data if you plot in this form 1 by R p versus 1 by S slope and intercept gives you all the parameters. Now, what is important to recognize here is that whenever you do an inverse transformation let me write this once again 1 by R p equal to k m by V m S divided by 1 by V m. So, when we do an inverse transformation what happens is that we do measurements in R p therefore, R p when we measure there is some standard value say R p bar plus an error. When we do an inverse transformation this error is no longer having the same distribution or same normal kind of distribution. So, inverse transformations have this problem that the error distribution get affected. Therefore, estimation of parameters by doing an inverse transformation by itself is not a good thing. So, what is generally done is that these k m by V m that comes from this is taken as the initial estimate to do a non-linear search. Which means you should use these values of k m and V m that comes out of this linearization and you should do a non-linear search. What is meant by non-linear search? That means these parameters are now occurring in a non-linear way you should do a non-linear regression to find out V m and k m using this as the initial approximation. Is it clear? So, k m and V m coming out of linearized estimation to be used only as an initial approximation for a non-linear search. So, the non-linear search is the best way to determine the parameters and you should not give too much confidence to the k m and V m that comes out of the linear estimations. L B means this way of representing the enzyme data is called as line weaver Burke plot. This is the name this comes out of those people who did this many years ago line weaver Burke. Line weaver I can write line weaver Burke. Now, in pharmacokinetics as you all know that R p equal to V m s by k m plus s. So, what is normally done is that we want to control certain processes in say a medicine that we all take. The object is to administer this medicine. So, there are particular pathway is stopped or it enhanced activated whatever. So, lot of material that is available which in which this enzyme you can inhibit this enzyme by putting an appropriate chemical. So, when you inhibit this enzyme what happens is that this enzyme goes and then blocks some active sites. So, when you put your substrate this substrates compete for the same active site. So, that your reaction now takes place because only as part of the active site is now available to the enzyme is that clear. So, you can have what is called as competitive inhibition. What is called as competitive inhibition in which the enzyme goes and blocks the active site. Now, we have talked about enzymes we have talked about what enzymes are we said enzymes are proteins. We also talked about a very simple form of enzyme kinetics which you call as Michaelis-Menton kinetics. And we said we are talking about Michaelis-Menton kinetics that the basic formulation of Michaelis-Menton kinetics is that enzyme reacts with substrate forms complex with substrate to form an ES complex. And ES complex decomposes to give you product an enzyme this is the basic formulation of Michaelis-Menton kinetics. Now, over the years people have understood that enzymes do get inhibited due to variety of reasons. And so and in form of kinetics in medical treatments and all that we often look at use of inhibitors to manage certain reactions. Therefore, if i is an inhibitor which complexes with the enzyme. So, that it forms an EI complex on other words what we are saying is that you could have a situation in an enzyme process that the substrate complexes with the enzyme to form the enzyme substrate complex. And an inhibitor complexes with the enzyme to form an inhibitor enzyme complex. So, that the active sites on the enzyme is partly consumed by the inhibitor. So, that the amount of sites available for the substrate is less this could happen. So, that the ES which is decomposing to form product an enzyme there is less of ES complex that is available and to that extent the reaction rates are affected. Now, the context is something like this that in enzyme literature it is tremendous amount of material on how an inhibitor performs on an enzyme. So, the use of inhibitors for managing enzyme reactions are very common and that is that is one of the reasons why we are looking at this kind of models. Now, I mean the basic procedure for understanding the enzyme kinetics we already done. So, you write the complexes RES complex R EI complex. What is RES complex that means the rate at which the enzyme substrate complex is formed. We can see here ES is formed from reaction 1. So, it is k 1 e times s you can see here ES is formed from reaction k 1 e times s it is consumed in reaction minus 1. Therefore, it is minus k k k minus 1 ES at the same time you find that ES is consumed in reaction 3 therefore, minus k 3 ES. On other words the rate of formation of ES we have taken into account the reactions in which ES is formed that reactions in which ES is consumed. So, all that is written here that is our equation 1. Now, if you look at equation this reaction E plus i equal to EI you notice that R EI complex R EI complex gets formed from k 2 k 2 k 2 e times i is the formation of EI complex and minus of k minus 2 EI is the consumption of EI complex that is reaction 2. Now, we also know that whatever enzyme we start with then that enzyme is partly free partly complex by the ES complex partly complex by the EI complex. So, that this is the material balance for the enzyme of course, we can put you know multiply throughout by V throughout by V to show that that is the total amount of materials that are inside the equipment with the batch equipment and so on. Now, the rate at which substrate s minus of R S that means the rate at which substrate s is consumed s is consumed in what are the reactions in which s is consumed the reactions which n says is consumed k 1 e times s you can see here k 1 e times s is consumed in reaction k 1 e times s. Now, s is also consumed in which other reaction s is consumed k 1 e s reaction 1 it is consumed and then the reaction minus 1 it is formed s is formed. So, k 1 e times s minus k minus 1 e s is the rate of formation of the R S complex formation of substrate. The rate of formation of product is simply what k 3 times e s that is rate of formation of product. Now, something that we like you to sort of recognize I have not derived it here, but it is so elementary so I thought it is not so essential that when you put the e s complex equal to 0 that means we say that this is stationary this is equal to 0 this is equal to 0 when we say that R e s is stationary what is meant by R e s is stationary when you say R e s is stationary what is meant what is meant is that if you make a plot of e s versus time e s sort of it goes through like e s what we are saying is that we are actually looking at the process during this in this phase where e s is not changing basically what we are saying is that e s is not changing e s is constant that is the meaning of R e s equal to 0 the rate at which e s is changing is not important it is called quasi steady state approximation. Now, when you put R e s equal to 0 what it means is that that reaction 1 this equation 1 k e s plus k minus 1 all that this is 0. Now, if you look at reaction 1 and reaction 3 and equation 1 equation 3 and equation 5 and 4 you will realize that under the quasi steady state approximation R p equal to minus of R s or rate of formation of product is equal to the rate of consumption of substrate and that makes sense also that is the but what happens here is that the presence of e i the inhibitor slightly introduces an element of of reaction rate decrease which we will come to shortly. So, what we are saying now what we are saying now is that R e s equal to 0 R e a equal to 0 and then the the material balance for the enzyme is given by equation 3 and under the quasi steady state approximation R s R p equal to minus of R s which you can derive yourselves not a problem with these we can now go on and determine what is the kinetics for the case of competitive inhibition. So, what I have done what I have done is that I have just rewritten equation 1 equation 1 I have written it like this I have taken the other side and similarly I have written equation 2 slightly differently. So, the idea of writing equation 1 and 2 equation 1 and 2 what I have done is that I have taken this put as 0 I have taken it to one side I have taken this to one side. So, it sort of this gives certain advantages in terms of simplifications. So, when I write it in this form equation 1 and 2 as 6 and 7 I can divide 6 by 7 e cancels off you see if I divide 6 by e cancels off. So, that I get directly from here the e a complex that means from equation 6 and 7 I am able to tell that what is the amount of e a complex formed. So, you can see very clearly that e i equal to k minus of 1 by k 3 k minus of 2 this this is equation I can call it equation 8 if you like. So, what we have done what we have done is that equation using equation 1 and 2 and rearranging we have eliminated e and then got an expression for the e a complex. That means e i now depends on k minus 1 k 3 k 1 s k 2 k m and i n e n and so on. Now, so the advantage seems is that. So, I have just what I have done here is that I have done e a complex I have just got the e a complex here and the equation 8 then I have substituted for e i I have substituted for e i from here. So, what I have done here is that I have substituted for e i from equation 8. So, what do we get what do we get is that I have substituted for e i here and then simplified. So, what do we get we get let me go through this once again just for all. So, what do we have what do we have here is that r e s equal to 0 k 1 e s k minus 1 e s k 3 e s. Now, this e we can eliminate from equation 3 what is e equal to e naught minus e s minus e i. So, what I have done is what I have done is the following from equation 6 and 7 just now we talked about from equation 6 and 7. So, equation 6 and 7 we found out what is e i now we know e i now if you go to this equation 3 equation 3 now you want to eliminate this e here in equation 1 can be written as e naught minus e s minus e i. So, what we have done is that we have substituted in equation 1 the replaces e from equation 3 and then substituted for e i we have already calculated and the whole thing simplifies simplifies very nicely. It simplifies something like this it simplifies as that it simplifies like this. So, that what we get is that the e s that means you can see here e s appears everywhere e s appears. So, that we can then determine what we have done is that we have determine what is the amount of e s complex that exists in the process in the presence of the inhibitor. So, you can see here this equation 9 I call this is equation 9 we find the equation 9 tells us quantitatively how the quantity e s complex is related to system parameters. So, that is what has been done. So, we are able to determine from equation 1 2 up to 9 that e s complex e s complex in the case of this kind of competitive inhibition is given by this expression. Notice here that in this expression in this expression it involves i the amount of inhibitor we have added and also involves k i the ionization constant for the inhibitor. How do you define ionization constant? We define ionization constant as k minus 2 by k 2. What is k minus 2 and k 2? What is k minus 2 and k 2? This is k minus 2 this is the ionization of e i is called minus 2 and the reaction e plus i is called k 2. So, ionization constant is k minus 2 by k 2 and with that simplification with that formulation with that we find that e s complex can be given by this expression. Now, this from elementary algebra is involved. So, I could have derived everything, but I thought it is not necessary. So, elementary that I did not think it is important to spend time on this. So, I have got the e s complex here and then r p is defined as k 3 times what is r p rate of formation k is given as k 3 times e s. So, e s is given by this therefore the rate at which the enzyme reaction would occur is given by this expression. What we have got here? What we have got here is the rate of formation of product is given by this expression. Now, what we have got here? This k 3 e naught is represented as v m and then this term k 1 k minus 1 by k 3 by k 1 is denoted as k m this is k m this is denoted as k m. So, the rate of formation of product is given as v m s divided by s plus k m multiplied by a term which is determined by the inhibitor. Let us just go through this once again just for the sake of understanding the basic approach of determining the chemical kinetics. Let me do this once again just to emphasize to you this is very elementary there is nothing much in here very elementary algebra is involved I did not do it because it is not necessary. What did we do? We said this is the case of competitive inhibition. What is competitive inhibition? Competitive inhibition is an instance where the inhibitor attaches itself to the enzyme. So, blocking some of the active sites because it blocks some of the active sites the amount of e s complex that is formed is less than in the case where there is no inhibitor. Now, what have we done? We have written the rate of formation of e s complex e i complex and a material balance for the enzyme. We also write the rate of formation of this rate of formation of substrate and rate of formation for the product. We said under the quasi steady state approximation where v s is 0 we can show I am not shown this very elementary that R s equal to R p equal to minus of R s. On other words whatever product whatever substrate is consumed that much product is formed under the quasi steady state approximation. We also said that this approximation is valid if e naught by s naught is small that means the amount of enzyme to the amount of substrate is small. Now, having said this we said that we can now by looking at reactions 2 and 3 we can make some rearrangements and find out what is the amount of e i complex that is formed. Then we said that the amount of e i complex that is formed can be shown to be equal to equation 8 it is something very elementary. Now, then we said that it is now possible to eliminate this equation 1 e can be replaced as e naught minus of e s minus of e i therefore, you can substitute for e is replaced by e naught minus e s minus of e i. Then both e i can be given in terms of e s therefore, we can get an equation for e s that is what we did. We said now e s can be said to be equal to. So, we calculated that e s can be given by this representation. Now, we know that the rate of formation of product is k 3 times e s therefore, we got r p the rate of formation of product under the conditions of some competitive inhibition is given by this expression. Notice here the form of this is very very similar to the Michaelis Menten. Michaelis Menten is what Michaelis Menten equation is r p equal to v m s by s plus k m instead here instead of k m it gets multiplied by a term 1 plus i by k i or in other words what happens here is that in competitive inhibition the k m value or observed value of k m is higher than you would have gotten when you had no inhibition. So, that is the only point that you should remember essentially what it means under competitive inhibition that k m becomes higher than in the previous case when there is no competitive inhibition. We can understand this is another way which is the way in which it is available in the literature this is called as the line weaver bark plot. What is called this 1 by r p is called the line weaver line weaver bark plot line weaver bark also called as l b l b plot. In l b plot if you plot 1 by r p versus 1 by s 1 by r p plus that is what I have done here you notice here that the this it is called the same intercept as before, but the slope is higher showing that the k m value that you will get under competitive inhibition is higher than what you would get when there is no competitive inhibition. So, when you make a plot of 1 by r p you versus 1 by s you get a higher previously you had this when there was no inhibitor no inhibitor when there is inhibitor the slope becomes k m by v m times this whole thing showing all that is being said when there is a competitive inhibition the k m by v in the in the line weaver bark plot you will find that the slopes are higher than what you got before and that that represents the effect of the inhibitor. Suppose you have to find out these numbers in an experiment what you will do is that you will do an experiment without inhibitor you will do an experiment with inhibitor. So, that by from these two you can find out v m this is v m this is this gives you k m and this gives you k i is this clear when you do an experiment with inhibitor and without inhibitor without inhibitor you would get k m and v m when you put the inhibitor since you know the concentration of inhibitor you will know k i. So, you need to have to do two experiments to be able to get the values of k m v m and k i. Now, there are many types of inhibitions that are described in the literature I will quickly set out the other types of inhibition a s i plus e s this i e s e s sorry e s giving you product plus n sign. Now, please compare this with this this is e plus s is e s e plus i is e i and then e s giving you product. Now, this is what is called as this is competitive this is called uncompetitive. What is the difference between these two in competitive inhibition the enzyme active sites the inhibitor goes and sits, but in uncompetitive inhibition the the inhibitor sits on the e s complex you can see here. So, depending upon the way actually you can show I will write only the final form because it will be in this form this is how you will get please notice here in the competitive inhibition the k m gets multiplied by this term 1 plus i by k i in uncompetitive inhibition the s term in the denominator gets multiplied by that term. So, this is the you can be looking at. So, in the in the line view of birth plot if you make a plot of this r p equal to v m s divided by k m plus s multiplied by 1 plus i by k i. So, when you do 1 by r p becomes 1 by what is it v m sorry k m by v m s plus what 1 by v m multiplied by 1 plus i by k i. So, in this plot here without. So, you have slope is the same correct. So, intercept is higher. So, I should put a plot it here I think is it now. So, you have a higher intercept the intercept is previously it was now it is intercept becomes larger it is it is a larger intercept i 1 plus i by k i times v m. So, this is the intercept. So, in this case we have a larger intercept same slope while in the previous case we had in this case we had same intercept higher slope. See this is competitive correct this is competitive s or no this is competitive inhibition where the slope is. So, this is competitive inhibition you have same intercept but higher slope. Now, in uncompetitive inhibition where are we this is uncompetitive inhibition this is uncompetitive inhibition here you have same slope, but higher intercept. So, same slope slope is v m sorry k m by v m k m by k m by v m this is k m. So, it is uncompetitive slope is the same, but intercept is higher. Now, there is one more form which is also seen in the literature let me quickly write this. So, here is an instance here is an instance where enzyme combines with substrate combines with i to form i e complex combines with s to form i e s complex here combines with s to form s complex complex i i s complex. So, here is an instance where you have e plus i giving you i a competitive then i x uncompetitive here you have the other way e s and then. So, it is a combination of all this. So, when you go through this and then put this quasi-stratistic approximation your r p comes something like this v m I would not derive this it is a little messy. So, you can do it yourself. So, when you so you can see here v m s in the denominator now the i plus k a multiplies both s as well as k m. So, when you do this line view of bulk plot let me plot it here 1 by s by 1 by r p. So, you have this is no inhibitor. So, this is v m 1 by v m and this is k m by v m now when you have this kind of inhibition you will get a slope I am drawing it like this I am drawing it like this. So, this is slope here is k m by v m multiplied by 1 plus i by k i is it all right is it and this intercept is 1 by v m multiplied by 1 plus i by k i is it. So, we have 1 2 3 4 types of enzyme models we will do 1 by 1 this is an instance here when there is no inhibitor. This is an instance where there is an inhibitor competitive inhibition here this uncompetitive inhibition. So, the nomenclature is a little not convincing, but what is important to recognize is that this is competitive is here r e i equal to 0 r i e s equal to 0 and then r e s equal to 0. These are the assumptions and there is one more assumption which is implied here which is done in the enzyme literature that is the ionization constant for inhibitor attaching to i e complex and attaching to i e s complex as long as inhibitor is attaching that ionization constant does not change no this assumption is made that means i is attaching to e i whether is attaching on e s or on e i it does not matter because the sides are the same that is how those they assume that kind of inhibition constants. You can do this yourself is not very complicated. So, what I want to summarize here is that the form of this function when you do it on the line we were bulk plot you get a straight line from where you can get your inhibition constants. So, only thing is that this you should use it as an initial approximation for a non-linear search that is all do not use this directly for your estimation because they have lot of errors associated with it. Now, the important thing that we should do now is what happens in enzyme systems is the following. So, typically what happens is you get this e s and now frequently the substrate goes and binds. So, that your e s complex gives you p plus e now this is an instance of your substrate itself blocking the active side. So, that the rate at which reaction now takes place is affected by the fact that s is blocking the s e s complex. Now, this is not uncommon see for example, what happens is that in commercial process where we tend to use a large quantity of substrate because we want to process in large quantities then that is when this seems to happen that s goes and blocks the enzyme. Now, we want to quickly derive this because it gives some interesting relationships. So, let me write this e naught equal to e plus e s plus s e s then you have a large quantity of r e s equal to k 1 e times s minus of k minus 1 e s plus k plus minus 2 k minus 2 e s this is all right please tell me this is let me put all the nomenclature 1 minus 1 2 minus 2 and 3. So, e s k 1 e s minus k 3 e s this is correct k 1 e s k minus 1 e s k minus 2 I have to write that also minus k 2 that I have to write is it now is it all right. So, then r s e s equal to k 2 e s times s minus of k minus 2 s e s is it all right k 1 let me go through let me go through this k 1 k minus 1 k into s e s this is this is now this goes to 0 this goes to 0. So, when you set up all your equations this is not all this little bit of mess here, but I will write the final form the final form looks like this r p when you calculate it looks like k 3 e naught k 1 s divided by k 1 s divided by k 1 s divided plus k minus 1 plus k 3 divided by k 1 and then k 1 k 2 s squared divided by k minus 2 k 1 simplifies finally, simplifies as r p equal to k 3 e naught s divided by s plus k minus 1 plus k 3 divided by k 1 and then k 2 s squared divided by k minus 2 this it sort of looks like this and in the literature this term comes as s squared I have not it is not there I have not it is not there it is not there because I put k 1 on the top. Now, when I divide throughout by k 1 I knock out this I knock out this then I put k 1 here. So, this is this what I will remove this see this how it is done in the literature. So, this term is called so it becomes k 2 by k minus 2. So, final it is now let me write it in the form in which is available in the literature r p is available like this v m s divided by s plus k 2 plus s squared by k 1 where k 1 is k minus what is k 1 k minus 2 by small k 2 cut. So, and this term k minus 1 k 1 is and then other one is k 2 is k minus 1 plus k 2 plus k 2 plus k 2 plus k 2 plus k 2 plus k 2 plus k 2 plus k 3 divided by k 1 this how it is available in the literature. Now, the reason why I am doing this is the following this is an instance this is an instance of the rate function going through a maxima you will find you will find frequently in the enzyme literature that the rate function actually goes through a maxima. And this is how it is explained in the literature the maxima is because the substrate inhibits the enzyme complex and this kind of model applies and that is why this kind of functionality is seen. Now, if I ask you now what is the if I do d r p by d s equal to 0 can you quickly do this for me on this r p v m s by s plus k 2 s squared by k 1 what is d r p by d s if you put d r p by d s equal to 0 I get this happens only when s optimum becomes equal to k 1 is an interesting result in enzyme literature that this whenever you have some kind of substrate inhibition that the highest rate you obtain when the value of substrate that you use is square root of k 1 k 2 this is an interesting result where k 1 and k 2 are the two constants of the model. And square root of k 1 k 2 is the best value at which you should operate see the reason for doing the reason for doing this is the following. What happens is an enzyme tends to ionize like this please recognize this reason why this particular model of enzyme reaction has become very interesting is because enzyme tends to ionize in this form and the e minus is the active form of the enzyme. On other words if you put in so much of enzyme it is this fraction that seems to be active as far as the enzyme reaction is concerned. Now, let me write what is k 1 k 1 is e minus h plus divided by e and k 2 this first reaction second reaction is e minus 2 h plus divided by e minus 1 and h plus this is all right. So, what is k 1 and what is k 2 k 1 and k 2 are the ionization constants first ionization second ionization of the enzyme say enzymes of active sites the ionize. And it is this it is this which seems to be the active form of the enzyme and but it ionizes further this is an inactive form. So, what is the active form of the enzyme if you want to calculate using the values of k 1 and k 2. So, you want to find out what is e naught minus e naught that means, what is e minus by e naught let me quickly calculate that. So, total enzyme e naught equal to e e plus e minus plus e minus 2. So, e naught equal to from here from here what is e equal to e equal to e minus h plus divided by k 1. So, e minus h plus divided by k 1 that is the first reaction term e equal to e minus h plus by k 1 the second term I retain like that e minus third term third term e minus 2 is what k 2 e minus h is this correct what I have written. So, it will be it will be squared here all right. So, what does it become e minus 2 e minus e minus 2 becomes k 2 e minus h plus all right. So, you have e minus within brackets h plus by k 1 plus 1 plus k 2 by h plus equal to e naught is it all right h plus by k 1 plus 1 k 2 by h plus equal to e naught e minus by e naught equal to 1 divided by 1 plus h plus by k 1 plus k 2 by h plus is it all right. So, let us just see the form in which we had the previous see if you just look at this if I just divide by and this divide by s here. So, you can see here 1 plus that forms are identical h plus by k 1 you can see here s by k 1 and then k 2 by h plus k 2 by s is that clear. So, the form of the substrate inhibition that we said for enzyme kinetics and the way in which the hydrogen ion affects the e minus or the active form of the enzyme the form is identical showing that h plus actually serves as a way of inhibiting the enzyme kinetics. On other words what we are saying is that in enzyme kinetics choice of hydrogen ion concentration is crucial. So, if you choose it properly then you get very high activity if you do not you do not get good activity and what is the best choice what is the best choice of p h what is the highest value of e minus by e plus you will find that the optimum e minus by e naught you will get when you choose h plus equal to root of k 1 k 2. So, this is the best the best p h of operation of an enzyme reaction is root of k 1 plus k 2. So, this is the most important message that you want to carry home from the enzyme literature now with this background in enzyme. So, what just we quickly summarize what we are saying is that the rate of enzyme reaction is this. Now, we are saying that choice of h plus choice of h plus we are saying gives you a proper reaction rate and then the effect of h plus can be accounted for by a form of this types where the active form of the enzyme is given by this relationship. So, that you choose h plus as this you get very good reaction rates. Now, what is V m we said V m is k 2 times e naught correct that is what is on V m. So, our model is e plus s is e s and then e s giving you product plus enzyme and this is k 2 correct and V m is k 2 times e naught that also we said. Now, this e naught is not what is the active what is the active is it is only this. So, to that extent the effect of the of p h would come into play is this clear to all of us. That means, we think that e naught is active working for us, but what works for you is only e minus and therefore, the effect of e minus by e naught will actually appear here. So, in here it says k 2 times e naught, but actually it is k 2 times e minus e naught. So, e minus that effect will come in the reaction or in other words V m that you will observe will be slightly less because of the fact that the effect of p h on the enzyme because of this relationship. Once again what we are saying is the rate of enzyme reaction is given by V m s by k m plus s. Now, we find in many cases the choice of p h is crucial to the rate of chemical reaction. What is the best choice best choice comes from understanding that e minus which is the active form of the enzyme is affecting the reaction rate by this relationship. The best choice of p h is given by h plus equal root of k 1 plus k 2, but whatever the p h e minus by e naught is given by this relationship. Therefore, the V m value that is appropriate to your reaction is given by naught k 2 by e naught, but k 2 e naught multiplied by y where y is e minus by e naught. And this why it takes the best value when you choose p h as root of k 1 plus k 2 is that clear. So, what we are now saying is that this whole thing is analogous to substrate kind of inhibition because it is substrate inhibition which gives us this kind of relationship. Now, having said this suppose we have a microbial reaction we have done this enzyme reaction because we really want to see how best we can go forward and understand microbial reactions. See, microbial reactions the rate at which the microorganism will work is frequently given in this form where mu is specific growth rate. And this mu is often written in this form notice this is the same form in which Michaelis Menten is written V m is maximum velocity k m is Michaelis constant. Here it is mu m by k s plus s the reason for doing this is because microbial reactions are also regulated by enzymes therefore, we think the form will not be different. Whatever form enzyme gives same kind of form we will expect in microbial reactions also what is mu m mu m is called the maximum velocity k s is called monod's constant. So, k s is called as monod's constant mu m is maximum growth rate. In enzyme literature V m is called as maximum velocity k m is called as Michaelis constant. In microbial reactions we call it as mu m as maximum growth rate and k s is monod's constant it is very analogous. Now, if you are conducting this reaction let us say we write V times d x times d t is r x in a batch vessel let us say in a batch vessel. And then we have V times some in a d x d t is r s therefore, we should have d x to d s r x divided by r s is it all right is it ok. What is the r x rate of growth of cell mass the rate of substrate consumption therefore, now this ratio r x to r s what meaning can we give r x to r s in the literature they give it a meaning of yield coefficient. This ratio is called as yield coefficient that means whatever is the substrate that is consumed how much of that substrate goes towards growth of cells. If you are spending your substrate on growing the cells this ratio is what is a yield coefficient. Now, x is growing s is getting consumed therefore, you put a negative sign other because x is growth and s is consumption. Now, you can integrate you can integrate this I will write here. So, it becomes x minus of x i equal to y times s i minus of s I will write it big and it cannot see properly I will write it again. So, d x d s equal to r x by r s integrating this you get x minus of x i equal to y times you get x minus of x i equal to y times s i minus of s this is all right yes or no I put a minus y here. So, that we get this all right now let us go back to this equation r x times v. So, we want to integrate this equation we want to integrate this. So, d x d p is r x this r x we said is mu times x. Now, mu we say sorry this is grams per liter per time second this is also grams per liter per second it is quite all right. Now, it is mu m mu m s divided by k s plus. Now, r x we said in enzyme literature to microbial literature what we are saying is that the rate at which the cells are growing is given by this representation. The rate at which cells are growing is mu specific growth rate multiplied by the cell concentration this specific growth rate this depends upon the environment in which the cell is growing that environment is described by this concentration s mu m s by k s plus s is the rate at which cell is growing to which you multiply by the cell concentration to give you. So, this is d x d t yes or no. Now, how do I now integrate this I have to replace x in terms of s I have to replace s in terms of x for which I have you have derived this relationship where we said y is called the yield coefficient this is yield coefficient. Now, this yield coefficient ask yourself when is it that the consumption of substrate to growth of the cell why should it be constant it is constant or it may not be u and i for example, beyond an age we do not grow we consume substrate, but we do not grow correct. On other words this ratio depends upon the age of the cell therefore, it may be constant it may not be constant. So, the fact that we assumed it to be a constant implies that when we apply this to a particular problem we will have to see whether to update this y as we do the integration if you know the age of the cell otherwise you have to use a mean value. So, then it applies for the entire time duration for which you are doing the integration. So, this number y is an experimental quantity. So, that will be available to you assuming that it is a constant now we can. So, let us let us replace this y I will do it in the next one. So, you have d x d by d t of what is x y times s i minus of s equal to mu m s divided by k s plus s what is y this is x i plus y times s i minus of s is it all right. So, left hand side becomes d s d t with a minus sign right hand side becomes mu m s by k s plus s and this is x i plus y times s i minus of s. Now, of course, this can be integrated in all that. So, I will write the integrated form this do not waste time on this the integrated form looks like this x y k s y s i l n x i plus x i plus x i plus x i plus x i plus x i y s i y s divided by x i equal to y k s l n s i by s equal to x i y s i mu m times d this how the integrated form looks like finally, elementary integration. So, now the context is the following if you are growing a cell if you are growing a cell correct you start with certain value for x i that means, you have a vessel in which you put your substrate and you put your cell small amount of cell x i and cell starts growing. So, when the cell starts growing what you see what you see is this this how the experiment looks like this cell what is called as batch growth curve batch growth curve generally gives you 1 2 3 or 4 distinct regions a region during which the growth is very small this is what is called as a lag phase. What is the lag phase it is a phase of growth of the cell of the cell during which you do not see much change in the cell density. So, that is the period during which the organisms present in your environment is trying to adjust itself to the environment of the growth. So, it takes some time to understand the environment to adjust and to recognize and so on once it is understood the environment starts growing fairly rapidly. So, this is what is called as the exponential growth phase exponential growth during which period the organisms grows fairly rapidly you can see here you know this region the slope is something like mu m grows very rapidly. And after a point you find that the growth starts to slow down somewhere here the growth starts to slow down. How do you explain the slowing down of growth the people explain the slowing down of growth in various ways one simple explanation is that the substrate has run out there is not enough substrate and as a result your rate function your rate function here mu m s by k s plus s s is come down with the rate is come down. So, there are various other explanations in the literature as very elementary kind of explanation is that substrate is decreasing. And therefore, you see a decrease in the rate at which the growth is taking place. And after a point you find that there is no growth you see and this is explained by saying that the substrate is exhausted there is no substrate or in other words s is 0. Therefore, there is no growth. So, a batch growth that you will see typically has one two and three phases lag phase exponential phase and a stationary phase stationary phase. Let us look at this once again this is lag exponential and then this is stationary. Now, if you go to a go to an industry you will find that the another phase which seems to be of interesting to them which is a phase what is called as death phase. Now, tell me when is that you know the user is very you know to understand this phase of the organism which process of a process industry this phase of the organism is of very great importance. You growing the organism exponential growth stationary phase and then this is the death phase. Now, if you go to a penicillin factory for example, what are their interest their interest is to make as much penicillin as possible. Now, what do they start they start with an inoculum and make that organism to grow after it grows and reaches sufficiently high density what they would do is that they would not provide a proper environment for it to grow further, but to produce penicillin. That means the conditions required to produce penicillin are different from condition required for growing the organism. So, appropriately they will provide the feed or the substrate at a rate that it is sufficient for it to make penicillin only it will not grow further. This is that means this phase of growth is useful for producing secondary metabolites. This phase is useful when you are trying to make the organism for sale say bakers yeast if you want to make you want to be in this region because you want to grow the organism because you want to sell the organism. If you want to make penicillin you want to organism has to be in this region. So, that it will produce penicillin suppose you are your interest is actually to disinfect the process see suppose the entire process of disinfection disinfection is a huge activity in fermentation industry it is in water treatment industry. Then here you are looking at this phase because you are trying to understand what is the condition required to kill the organism. So, the death phase is what would be of great concern to you. On other words what we are saying is that to biologist the lack phase the exponential phase the stationary phase the death phase all are important because different processes require different aspects of the growth to be of importance to him. Now, suppose instead of conducting this reaction in a batch equipment with S naught I conduct this reaction in a flow equipment. In the sense you make it flow upwards or downwards as far as this form is concerned makes no difference as long as you know what is see this is the excise initial cell concentration here it will be initial cell may be X naught continuously cells are coming in continuously cells are going out or in a flow process it is a in what is called as tower fermentation for example, instead of a batch fermentation continuously cells are coming in cells are growing and you get products out cells out. So, your substrate gets consumed from S naught it get becomes S to the extent it is consumed your product is produced. So, a flow process and the batch process we know the equivalent as only thing wherever T appears you provide residence time otherwise it is there is no difference is there clear. So, as far as the growth process is concerned is governed by your monod constant k s your yield coefficient y and your maximum growth rate mu m mu m k s and y are your growth parameters, but there is one fundamental difference between batch and continuous in biology. In batch process what happens is that the organisms get a lot of time to adjust to the environment while in flow process they do not get that kind of time. So, you will find biology flow processes do not work as well as in batch process. So, this is a very fundamental difference and therefore, the parameters that you determine from batch process equipment may not generally apply to a flow process. What is mu m k s and y in a batch process and the mu m k s and y in a continue may not be the same because the fact that the organisms do not like this kind of environment they like to work in the environment is there adjusted to. Now, there is an interesting feature that you will it is not common, but lot of literature that is available see what there is great interest particularly in fermentation industries to increase the throughput I mean this is there for all of us you know for a given volume can we produce more. Now, what people have done is that they have created what is called as immobilized enzyme pellets immobilized pellets what are immobilized enzyme pellets you take the sorry not immobilized micro micro pellets micro pellets. Now, this organism what they do they take a gel and then immobilize the organism inside the gel various types of gel is available in which you can immobilize the organism. So, what is it that you have now instead of having micro organisms which is suspended it is now immobilized inside a matrix like a heterogeneous catalyst something that we have seen in industry is a heterogeneous now you can put this immobilized catalyst inside here. So, what is it contain it contains your micro organism is that clear now you have your substrate coming in it may come in with some cells it may not come in with some cells that is, but all these have got x i grams of cell inside. Now, as this substrate goes through what would happen as the substrate contacts this pellet they this is get consumed correct yes or no. So, what is the concentration of cell at any point it will be what we have put in plus what comes in plus what grows do we agree x i is the cells that are put inside this per unit volume see what is called immobilized cell reactor immobilized cell reactor what they do is they take the organism and put it inside a pellet. Now, what you have is something like a heterogeneous catalyst you can pack up your reactor with these catalyst. Now, your your substrate is coming in S naught it may also bring some cells it may it may not one does not know, but if you look at any cross section of these of the reactor you have x i grams per liter of immobilized cells you have x naught coming in plus. So, much is growing because of the fact that the S naught become has become S on other words at every cross section of the equipment the concentration of cells is x i plus x naught plus what has grown because of the reaction is that right. Therefore, in our equation here d x d v equal to r x what is this r x mu times x this x previously now it has become do you see this point I am coming I am coming do we agree with this the right hand side this is mu x previously now that x is has all these extra features coming in S n o or I am talking about continuous process say it is a plug flow S l I say continuous process is it all right. Now, what is this mu S n o now what is the what is this v it is a volume when we did not have this immobilized cells here the liquid is going right through therefore, the volume that is available for reaction if it is v previously. Now, it is occupying certain amount of space is lost because of this therefore, the actual liquid volume now previously it was fully now it is v times 1 minus epsilon g times epsilon l where epsilon g is the gas what is called gas hold up because reaction rates are very high the lot of gas which is held up in the equipment epsilon l is the liquid hold up previously epsilon g was 0 epsilon l was 0 because the reaction rates were like that now because of the fact that we have put in this immobilized cells we have lost some of the reactor volume because of epsilon g and epsilon l. So, the left hand side has to be multiplied by 1 minus epsilon g and epsilon l you understand is this clear to all of us. So, now let us look back at what we have done this is what we did and for which we wrote the solution somewhere I will have it somewhere here solution is here. Please recall the solution for a batch equipment for which what did we say we said t is the time for reaction what is previously we had put x i x i is the cells in the batch equipment. Now, it is the flow equipment flow equipment the feed is coming in with x naught we are putting in so much of x i x naught is coming in so much is growing. So, the effect of all this is to increase the rate at which chemical reaction takes place correct. So, you will find that immobilized cells immobilized cells I mean a huge amount of literature exist between in 1970's and 1980's when immobilized cells was investigated as a possible way of increasing reaction rates in biological reactor huge amount of literature exist. And the only thing that we have to recognize is that whatever is our equations here you have to appropriately change these things to take into account these effects that I pointed out. But, what did turn out is the following that many of these great ideas somehow in the industry it did not work out very well let me write the final form and then we explain why. So, if you immobilized cells your answers look like this see the form of this is identical to what we have already written. So, I am not writing it again. So, this is for immobilized cell reactor that is understand quickly x naught is the continuous input of cells x i is the cells that you have put in because you have immobilized cells and put it inside the equipment. And now epsilon s and epsilon g is hold up a solids hold up of gas in the immobilized cell equipment and tau is the residence time which is v by f these are all microbial parameters. Now, the rate at which the reaction that means the extent to which we can drive the reaction really depends upon how much you lose because of this how much you gain because of this. This is the loss because you are losing volume, but this is the gain because you have put in so much of cells and you have lost so much of volume is this clear. So, the plus point of this and minus point of this you will have to see how it affects your process the experience that is reported in literature is that it is substantial it is not small. And that is how this huge amount of research got done particularly for alcohol huge amount of alcohol fermentation huge amount of research exist. Now, having said this what seem to happened is that this particular immobilized cell in which the cell is immobilized as the reaction takes place if for example, alcoholic fermentation as you all will know lot of carbon dioxide is produced. Now, this carbon dioxide which is produce inside this immobilized cell will have to come out. Now, as carbon dioxide has to come out because a huge increase in volume it is a it ruptures this immobilized cells. So, the what people found was that the stability of the immobilization was unsatisfactory. And consequently the net result that they anticipated because of immobilization did not eventuate in a process it work very well in the laboratory and all that, but in a process it did not work out very well. Therefore, immobilized cell reactor is not very popular or not very common in the process industry it work very well in the laboratory will stop there.