Mathematical Physiology, Lecture 1: Enzyme kinetics and the law of mass action. 4th year lecture.

What this episode is about

This lecture focuses on enzyme kinetics, explaining the role of enzymes as catalysts that facilitate the conversion of substrates into products without being consumed in the reaction. It emphasizes the importance of understanding enzyme functions in various biological processes, including digestion and DNA replication, and introduces the mathematical representation of enzyme reactions based on the law of mass action.

Key takeaways

• Enzymes: nature's catalysts—turning substrates into products without breaking a sweat or, well, themselves!
• Reaction rates are like a fine dance; temperature, size, and concentration lead the choreography.
• The law of mass action: a reminder that chemical reactions can feel democratic—more reactants, more products!
• Math and biology unite! Differential equations are the secret love language of enzyme kinetics.
• Kinetics: where chemistry meets speed dating—finding out how quickly A and B can make C.

What is discussed, and when

1. 0:00 – 5:00: Enzyme Kinetics
2. 5:00 – 10:00: Law of Mass Action
3. 10:00 – 15:00: Law of Mass Action, Enzyme Kinetics
4. 25:00 – 30:00: K2
5. 35:00 – 40:00: Enzyme Kinetics
6. 40:00 – 45:00: Michaelis Constant

Notable quotes from "Mathematical Physiology, Lecture 1: Enzyme kinetics and the law of mass action. 4th year lecture."

• Ask any questions if you have while I teach the course. I hope that this can be as interactive as it can be. So the first chapter is on enzyme kinetics.

  about Enzyme Kinetics · at 0:15

• Will write here PS1 question two so that you know to come back to this point okay this has a name this is called the law of mass action the law of mass action and here I have made the following.

  about Law of Mass Action · at 6:26

• You're going to have some enzyme. Again the letter denotes the concentration of this enzyme. And then I will specify that in the forward direction when these two react together I'm going to have a reaction rate of K1.

  about K1 · at 8:43

• Basically I will add k2 now. So K2 remember is this.

  about K2 · at 26:05

• And then when you end up with this basically you have this new constant. This K constant is now defined as the sum of these two rates divided by K1. And this again has a name. This is called the megais.

  about Michaelis Constant · at 41:56

Who and what is mentioned in this episode?

Concepts & topics

• Enzyme Kinetics — Concept · Core topic · 4 mentions · first at 0:20
  Central theme of the episode, discussing the mathematical modeling of enzyme reactions.
• Law of Mass Action — Concept · Key argument · 4 mentions · first at 6:33
  Fundamental principle used to derive the differential equations describing enzyme reactions.
• Michaelis Constant — Concept · Framework · 1 mention · first at 42:01
  A constant that characterizes the enzyme-substrate interaction in the Michaelis-Menten kinetics.
• K1 — Concept · Key argument · 5 mentions
  Represents the forward reaction rate in the enzyme kinetics model.
• K-1 — Concept · Key argument · 4 mentions
  Represents the backward reaction rate in the enzyme kinetics model.
  So this is a one-way interaction. So you just have that the C breaks with this rate into E and P. But I have that this intermediate step, the complex can also be broken down to…
• K2 — Concept · Key argument · 2 mentions · first at 26:05
  Represents the reaction rate for the complex breaking down into enzyme and product.

Full transcript with timestamps

0:12 I will get into the math now. So please ask any questions if you have while I teach the course. I hope that this can be as interactive as it can be. So the first chapter is on enzyme kinetics. And you must also already have some access to lecture notes that are typed. So there you will always find more details than the ones I will be doing in class. So if you want to do some extra reading maybe it's a good time for you to start like looking through that as well as I go. But everything that I will be covering in class is basically what you need. So if I don't cover something in class from the actual type set nodes, it means it won't be covered. Okay.

0:58 All right., So, the, first, one, is, on, enzyme kinetics. So if this is the first time you are taking a course in physiology, I first want to give you some background on this. So first of all, what is an enzyme? What does it do? Why is it important for us to study it? So enzymes are catalysts and what they do is that they help convert other molecules which are known as substrates.

1:51 into products. And importantly, so they do this basically. You have some substrate, you have some enzyme, it comes, it binds on it, and then it helps it produce these other molecules that are called the products. But the most important thing for you to remember is that enzymes don't get used up in the process. So you have an enzyme at the beginning. It binds to the substrate and you get also a product plus the enzyme. Okay. So I will make this point down. So but are not used up in the reaction themselves.

2:41 Is it visible to the people at the back? Okay perfect. Okay, so some applications of this are in the digestive system. You can think about it. Basically, enzymes act on molecules to give you glucose that you use for energy. They can also be used in DNA replication or in liver enzymes to destroy basically toxins or other chemicals. So now how do we write this reaction mathematically? So if I tell you to consider two chemicals A and B. So consider chemicals A and B reacting. So they collide with each other to form some chemical which I will denote by C.

3:36 And I have to also specify the rate at which this reaction happens. So it will be with some rate k. So very simply I can write this as a plus b. So this this a letter basically denotes a concentration of this chemical a and then I have a concentration of a chemical b. They react together under a rate k and they produce c. Is it clear so far? Okay. So when I have this basically I will also say here that the rate will depend on the size of your molecule. It will depend on the temperature in which this reaction is taking place and also on the shapes and so on. So now you can write this as a differential equation for the process and you say the rate at which this chemical is being produced. So dt is proportional to the product of the concentrations of the two and the rate of proportionality is this k. So I'm gonna, write, down, K, A, C, A A, B., So, this, is what I have.

4:54 So if I asked you, let's instead imagine I had another example and I gave you that I had AB but with an extra molecu an a larger concentration of A reacting again under the same rate to give me product C. Does anyone have any idea of how I can rewrite this using this relationship that I gave here? So now you have double the concentration basically of a.

5:24 Yes., Yeah., Yeah., Exact., Is, that, what you're going to say? Perfect. Exactly. So you can write basically this is going to be a squared. If for someone is not that clear and you're like ah how can I know? You can just imagine doing a plus a right? So you have a * a * b. So it gives you this is the same thing. Okay great.

5:48 Worried, about, the, reaction, volume, or uh, no, because, it's, taken, into, the concentration, right? So yeah, you can have like more complicated things of course like dependence on the space and so on. But for the purposes of this class for now, you don't need to worry about that.

6:05 Yes. Is, are, there, any, constraints, of, K?, Is, it Yes., It's, not, going, to, be, a, constant., is more complicated than this

6:13 or, is, it, is, it, less, than, one, or, is, it like it's not a probability it's just the rate

6:18 yeah, it's, just, the, rate, there, are, no restriction okay, so, actually, you, will, be, asked, more on this on problem sheet one so maybe I will write here PS1 question two so that you know to come back to this point okay this has a name this is called the law of mass action the law of mass action and here I have made the following assumption. So we assume that the mixture is well steered. And what this actually means is that the concentrations of these reactants A and B that I talked about the concentrations are uniform of you can say of reactants are uniform in space.

7:33 Okay, congratulations. This section 1.1. The next thing is section 1.2. By the way, this love mass action will keep coming up. So, it's important we understand it. The next thing that I want to talk about is this Miguel mentioned kinetics. So for now this won't mean anything to you but we will derive it together and it will make sense I hope. So now I will give you a more general case of an enzyme reaction and how it looks like and then using the law of mass action we will be able to derive some differential equations and then not only we will be able to write down this model that describes the reaction but we will also focus on how to solve it and in general this class is on developing the model and also solving it.

8:20 So an enzyme reaction in general will look as follows. So I will give you some substrate with concentration S and then with this you're going to have some enzyme. Again the letter denotes the concentration of this enzyme. And then I will specify that in the forward direction when these two react together I'm going to have a reaction rate of K1. And when the two react they give me some product which for now it's an intermediate step which is called a complex. So I get some C and then the C can also break down now with a rate of K2 into the enzyme because I told you it doesn't get used up in the whole process plus the final product P.

9:24 So this is a one-way interaction. So you just have that the C breaks with this rate into E and P. But I have that this intermediate step, the complex can also be broken down to produce the substrate and the enzyme but with a different rate. So it's also a backward process. So here I will also have K minus one. So to help you a bit with your notes, let me denote what these different things are. So this is what I call the substrate. This is the enzyme. Then this I will write here that is a reversible reversible reaction.

10:15 This thing is called the complex and this final is the product. So it's a basically I should write here is a two-step process. Okay. So if we actually want to think about this problem, you will see that if I want to simplify this, I can think about it as the substrate producing the P with some new rate R right? with some intermediate things that I don't know about. So the overall reaction the overall reaction is going to be S producing P with some rate R. So you could model I will write here you could model the single reaction

11:33 and capture all the intermediate steps. with this overall reaction. This has a name. This is what is known as the heel equation and we'll also come up later today again. So now I will use this over overall rate R and I'm going to write it as a function of the substrate. So I have some initial rate R0 the substrate S raised to some power n and then k n + sn and these are some constants. We will come back to this equation so don't worry if it doesn't make sense for now. Okay, so from this you see that now that I wrote R, even though this looks really simple that the substrate breaks down into a product with this rate, this is not a constant rate is something that's complicated and it depends on my substrate and the overall reaction.

12:59 Okay, so now that you have this and you also learned just now the law of mass action, we will be able to derive together what the system of differential equations is that describes this simple process over here. So from this you should expect to have one, two, three four differential equations, right? I should be able to tell you how the substrate, the enzyme, the complex and the product change with time. And for that we will use this notion of law of mass action. So I will use it on this board over here so that you have it in front of you. So it's better instead of doing that thing with the hel equation to use the law of mass action and I will go in the order that I have them there. So first the substrate, then the enzyme, the complex, and then the product. So let's write an equation down for ds dt.

14:06 So you see here, so do you expect the s to increase with c or to decrease? And with what rate?

14:21 Yeah., So, what, should, I, write, down, there? What do you think? K1us K perfect, because, S, gets, converted, to, C. So it gets less and less right with this rate but I also know that C can break down with this rate to give me S. So in this direction it should increase it right. Exactly. Nice. So it's exactly what your classmate said. But notice that when I do one of these like when when basically S and E react together with K1 I get C. Right? So in one direction you have to be careful that DDS is reacting with E. In the other direction it does not right. So I should get a product somewhere. So in this direction with the K1 I should get just that with plus because C gives me S. But in the other direction I should have a minus and then it's the product of S with E.

15:20 Does that make sense? Okay, great. So then I go to the next thing. So it's D E DT. Okay, now that we have this one, what do you think this should be? So E in that direction decreases, right, to give me C, but it reacts with S. So what should be the first term?

15:49 Yeah,, exactly., So, it's, going, to, be, minus K1 SE and then I get E produced with these rates for C because notice that you have the enzyme also on the very right hand side and then let me since you understood how to do this I will write the other two down. So the complex basically increases when S and E react together and otherwise decreases with these two rates. And then finally for the product the product one is the simplest one. It basically increases with rate K2 when you have C being produced. So we have these equations. So I will name them one, two, three and four because I will be using these equations now. Were you able to follow?

16:44 Okay great. All right, so my question now is can we actually simplify this? It looks kind of complicated. We have four cobalt ODs. So we want to simplify this. The answer is yes. Otherwise I wouldn't have asked. So yes and the reason is that if you see P basically do you see P anywhere else apart from here? No right it's decoupbled. So basically once you manage to solve everything else the C S and E you will be able to plug it in here and solve for P as well. So you can simplify it a bit. So I will say here that the product only appears in four.

17:35 So it decouples. So basically you just integrate one two and three and then once you have it you can get the p. Now I will also point out the fact that if I add equations now two and three you will notice that hopefully you see this that dt plus d c dt gives you how much these are opposite signs right and then this is the same so it's just zero so if I wrote this out it will be in dt of v plus C. So what does it imply about E and C? It's a constant, right? It doesn't change with time. So I know that E plus C is equal to a constant. And I'll just name it, let's see, E is zero.

18:34 Why did I name it is zero? Because basically at the initial time, if you look at that reaction that I have over here, you don't have any of the complex at time zero. So you only have the enzyme but it also tells us that the en enzyme concentration is being conserved during the process. This is what this says. Okay, great. So once we have this then I can write that the system now reduces to this reduces the system to only two ODS and I read them out for you. These are going to be the rate of change of the substrate which is going to be K1 with a negative 1 C minus K1 S the initial enzyme now. So I'm substituting for the E minus C.

19:33 So notice that I did E is equal to E 0 minus C to get this. And then I will also write down the equation for the complex. So for this one we have K1s it's multiplied by the difference between the initial enzyme concentration and the complex minus and I have K2 plus K1 C. So your system basically becomes this system of two ODS and I will focus on solving this now. So of course when you're doing some solution of ODS you need some initial condition right? So I have to tell you at time zero this is how much substrate you have and so on and once you have that information you can solve for the unknowns which are the substrate and the complex right now. So I will write here which can be solved subject to suitable initial conditions. So you might be given something else than the one I will give you now but it will be straightforward how to extend it to that case. So for you now I will just say that the initial substrate is s0

20:53 and that there is zero complex at time zero. Whenever we are doing some analysis of these kind of problems, it's very useful to do also a non-dimensionalization of the problem which means that you get rid of the units in the process to be able to infer something more generic. Right? So this is the next thing we have to do. So I will say here now I will focus on these two equations and we will nondimensionalize.

21:33 So this is an important thing to do the system in order to be able to analyze it. Okay. So the choice of non-dimensionalization sometimes it's not unique but for now I'll just give you uh the non-dimensionalization we will use. So I will have that the substrate can be written as the initial substrate times some other variable little less. By the way if you have a question now when I'm writing it if it's capital or not ask me and then I will have something for the complex as well. So this one will be based off on the initial enzyme concentration times a little variable C.

22:16 And finally I will also give you a new definition of time that depends on this new variable T prime divided by the rate of reaction times is zero. Okay. Now that we have specified this, I have to use these new variables now to rewrite the system in the dimensionless form. Right? So the way you do that, so you see here I have ddt for example with the original time that has units. So I want to rewrite it with those variables and end up with something that is of the form little less and t prime instead. So this is how I will do it. So I have to use the following procedure. So does note that if I want to rewrite this this is the same as having using this equation over here and this is this with little less now and similarly you can write down the DC DT this is going to be the original enzyme concentration times DC and DT and I'm going to have also the transformation of DT in order to plug it

23:31 in here. So this will become dt prime. Here I'm just doing the differential of that. And then if you want to write it as ddt you're going to end up with this instead. Okay. So taking everything into account basically this equation over here becomes

24:01 [snorts] So, I, will, take, it, that, you, trust, me, that I did the computation correctly. But if you don't agree, you can let me know. So you end up with this k minus one c. This is little c and then I have minus k1 s0 s 1 minus c. And I have some extra nodes in the write up that I will share with you later. Okay, so this is how this equation becomes. We will do exactly the same with the second equation to write in dimensionless form. But before I do that, let me simplify this one a bit. So if you see you have these extra things in front. So I want to get rid of them. I will just divide through divide through by k1 s0 to obtain and I'm going to get now d s dt is k1 this is minus one and then I have minus this just drops off so I have 1 minus Okay. So I will do some uh rearrangement here because I want to define some new constants that will become important later. But you can basically rewrite

25:26 this as -1 k1 s and then I have minus the little s and now I will introduce this lambda which is just a new parameter. So this is the same as d sdt prime. Here to introduce this new parameter I will just add and subtract a term that is the same so I don't change the equation. So basically I have I will copy this thing down k1 s0 and then let me write it also with a different color. Basically I will add k2 now. So K2 remember is this. So I have K1 as zero. Since I added it and it wasn't there, I will subtract it. And then this is minus S again, which is this one. So the reason I did this is because this will become a parameter that I call lambda.

26:33 And this will also be used throughout the model. it will be denoted as K prime. So let's write it out. So this K prime is basically this is the same denominator. So I have K minus one plus K2 divided by K1 S0 and I have defined lambda to be the thing after the minus. So it's basically K2 over K1 S0. Okay, so this is what we have for this. Now we do exactly the same for the C term. Maybe I will write it here. So again use the same procedure that I did here to rewrite that. So the DC DT equation becomes K1 S it's actually E0 and then I have DC DT prime K1 S0 and then I have the little variable so it's 1 - C minus K2 + K - one C.

27:54 All right. So I will do exactly the same thing that I did here. So notice here in order to get rid of the constant in front of this, I just divided through by it. Now I want to despite the fact that I have E0 here, I will still divide by the substrate initial concentration. And there is a reason for that. There will be a small parameter basically that multiplies my derivative there. So we now divide So again by k1 as zero like we did there. So now you will notice the k1 you see it goes away. It goes away here and then you get an extra term here and then in front of the dt I will get e 0 divided by s0.

28:44 So I have just this remaining here minus and if you notice here you see this is just this constant over here right so I have k prime so this becomes k is that clear okay all right so I since I have this now just note down that this thing is a small parameter that I will call epsilon. So I will rewrite this equation basically as epsilon this indic is equal to that and you can also combine the terms a bit differently since you have c and c here to write out the equation in the following form instead. So this is the same as S minus S. Now since I have minus outside, I'm going to have a plus K prime and this is times the complex concentration in the dimensionless form.

29:53 Yes. Small., Yes,, that's, the, whole, point. Perfect. Exactly. Because basically the whole point of showing this and introducing this small parameter is the fact that in order to have this big reaction to produce the product P on that side of the board, we need very little enzyme in order to do that. It's very efficient as a method. So the enzyme you have very tiny amounts. they also get conserved during the process and you can have whatever breakdown of toxins or what not using these enzymes with small epsilon um parameters.

30:31 Yeah, thanks for the insightful question. So I will state here what your classmate asked. So where this parameter epsylon which is ep 0 divided by s0 is much smaller than one and this basically says that we have We only need a bit of enzyme. So it's an it's a it has a remarkable effectiveness. Basically this is what this says. All right. So now if I combine these two formats what I am left with is this equation for the substrate which involves again just the complex concentration of the substrate and then the simplified equation for the complex.

31:25 So overall what you need to remember is these two equations minus s + c s plus after having introduced this lambda and then with the epsilon I have this c and we will see some interesting dynamics if we set this limit to zero. Huh?

32:03 You can say here if you want. By the way, with initial again if I want to solve it, I need some initial conditions and those should be for S and C. So I should have right because I have basically s0 if you remember and then from the non-dimensionalization that I just covered I have this so this tells you that the initial little s should be one right what was the initial complex that I defined uh yeah the initial complex concentration at time zero do you remember

32:44 zero, So, little, c, at, zero, should, be, also zero. Okay. Excellent. All right. So now this thing that your classmate pointed out before that this is small. Now let's do the following thing. We can say let's look at the limit where this goes to zero. This won't go to zero but the whole thing since you multiply it will go to zero. Right? So let's see what happens in that case. So epsilon much less than one means that we can neglect the time derivative in the second equation in the C equation. Okay. So basically this means that we have a so-called quasi steady system.

34:01 So a steady system would mean that it does not depend on time. Here it sort of does. So quasi means almost almost steady system. So the reason we call it that is because S still evolves with time. You see through here but the C will not. It will be an algebraic equation. Okay. So I will say here that s evolves through a time derivative and instead the complex concentration C evolves through an algebraic equation that we will actually derive right now. So if I look at the limit of small epsilon what I have as I said is that this le that this thing goes to zero and so if I write on the right left hand side that I'm left with this so I'm trying to get an algebraic equation for c which means that this is just c is equal to s over s + k prime This is just if you remember a sum of the two reactions divided by the other reaction rate. So it's just a constant this K prime. All right. So this is an

35:28 important part of the solution. And then now I will just take this C that I solved for and I will plug it in here. That will leave me an equation for d s that just depends on the substrate concentration s. I got rid of the c right. What type of o is it? This is a first order od. What methods do you know usually to solve first order od is integrating factor right? Or separation variables. In this case it will be an integrating factor because you have the substrate concentration. Okay. So let's do it together. So I would say here if we plug in this in into the first equation d s the prime we get so I have d s dt prime minus s and then I have again plus so the c right there will be this so I have s s + k prime and then I have s + k prime minus lambda.

36:52 The trick here to simplify things you just make a common denominator. So this will give me this is little s by the way when you just have a common denominator there you end up with a simplification which I will not write down now here to not waste time but we have basically this and I hope you believe me again just just do this times that and just simplify terms to get this okay so I will denote again this equation by something so that we can refer to it. So, this thing has a name. I hope I'm not butchering the pronunciation, but I think it's malis men.

37:42 And this is a law for enzyme kinetics. Do you have any questions so far? Yes. reaction that

38:12 yes, is, of, course, dependent, on, the reaction that you have what concentrations you have initially for the substrate and the enzyme and so on but in general when you have an enzyme binding on a substrate you have very small concentrations like 10 to the minus 7 or something of enzyme compared to the substrate. So you can always make the assumption that the epsilon is really small but yeah the absolute values will change based on the reaction. Does that answer your question? Okay. [clears throat] All right. So now how can I talk about this reaction rate now? So I will ask this question.

38:55 What is actually the reaction rate? Okay. What am I asking here is basically if you remember I have the product being produced as part of this reaction the P that we had written down. So the reaction rate is going to be little R. This is a definition that I'm giving you is going to be the rate of change of the product in this enzyme reaction. Okay. And this is I'm going to give it to you. is going to be minus ddt. This is because it is more the state of depletion of the substrate remember that I don't have basically if you remember I wrote that the overall reaction was this remember without the intermediate complex and so on. So I'm now saying that the substrate gets depleted with this rate to give me the product and we will be able to derive an equation for this.

40:06 So here if I use the same exact non-dimensionalization that we looked at together a bit ago, I can simplify this into the following. So this is little less right now. This is capital and I will make a note here that this is due to the non-dimensionalization.

40:32 But if you remember now I just defined this Miguel is meant law. So this gives me exactly the format of dsdt prime right. So I can take this and put it in here. So use the star from that board. So this will give me that R is basically this. Now let's substitute back S is equal to S0 S which is this non-dimensionalization and I will also give you again the definition of lambda. It's probably don't remember because it was this complicated ratio is K2 divided by K1 S0 and then we when we use these two things we basically get the following.

41:38 So I have that the R the reaction rate simplifies to the following. So there is a lot of intermediate steps that you can find in my notes. It's just a manipulation basically of um simplifying a few of the constants and then when you end up with this basically you have this new constant. This K constant is now defined as the sum of these two rates divided by K1. And this again has a name. This is called the megais uh constant.

42:31 Okay. So in general you might have asked me now but how do I know in my model what this k1 kg1 k2 constants are? how can I quantify them? So the way you do that is experimentally. So you will be given in a problem what this should be if you're asked to solve it for a specific problem. So it's actually not that straightforward to measure the constants individually. the action rates.

43:16 So what people instead do is that they say I'm going to measure the whole reactions reaction rate which is this little R instead. Okay. So instead of saying experimentally I cannot get what this K1 is or K -2 and so on, you say I'm going to start with this substrate and enzyme and I'm going to get this product and I know that the overall reaction rate is going to be this R which I can't determine. Okay. So we measure we can measure the overall reaction rate are and concentrations.

44:06 So for this of course you have to be given again an initial substrate concentration which is the S0. So if we look at the initial rate of reaction meaning I have not r of t now since this is a function of time but I have r0 the initial one. I'm going to write it as K2 E 0 S0 divided by S0 + K and then from this we have two more minutes so I can finish this. So basically what I want to do is look at 1 / this reaction rate initially. So this just becomes S0 + K divided by this. I will again do okay let me write it out for you. This is K2.

45:07 And here I will just simplify this. So you can just split this divided by that plus this divided by this. Simplify the S0 with the S0 to get basically the following. So I'm trying to write it in this particular form to point out something one over and I have S0 as well. Okay. So if I'm trying to plot a thing for this function, I can treat this now as a new variable. This is a linear equation for 1 / r0 in terms of the reciprocal of the initial substrate with a certain slope and a y intercept. Okay, so just imagine this is y is equal to mx + c. And when you plot this kind of thing, it has a name.

46:01 So this is so think about this as the y intercept and of this as the slope. These kind of plots are known as line we burk plots and I will stop here for today and we'll continue next time next time I will also cover next time meaning tomorrow uh trans ion membrane transports. So it will be again different a different application but we will see more techniques for writing down differential equations and solving them. Thanks for your attention.

Questions about this episode

What is "Mathematical Physiology, Lecture 1: Enzyme kinetics and the law of mass action. 4th year lecture." about?

This lecture focuses on enzyme kinetics, explaining the role of enzymes as catalysts that facilitate the conversion of substrates into products without being consumed in the reaction. It emphasizes the importance of understanding enzyme functions in various biological processes, including digestion and DNA replication, and introduces the mathematical representation of enzyme reactions based on…

What is mentioned in this Oxford Mathematics episode?

6 people, works and ideas are mentioned, including Enzyme Kinetics, Law of Mass Action, Michaelis Constant, K1, K-1, K2. Each one on this page links to its own page with every episode that mentions it.

How long is "Mathematical Physiology, Lecture 1: Enzyme kinetics and the law of mass action. 4th year lecture."?

The episode runs 47 min; the transcript is about 5,692 words in 53 timestamped sections.

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