Hello. Next set of lectures that we're going to cover in this course, are grouped under the overall heading Introduction to Dynamical Systems. And there are four lectures to this section here on dynamical systems. Just to give you a brief introduction of what we're going to discuss. We we've previously taught you about how to use the Matlab computing environment to perform scientific computations. Now we're going to actually proceed to deriving systems of ordinary differential equations describing biochemical signaling networks. And we're going to show you some of the tools you can use to analyze the results of these, the simulations that you perform on these models. And these are grouped under the overall heading of Dynamical Systems Tools. And this is part one of a, of a four part series on dynamical systems. So the topics that we're going to cover in this, part one of this lecture, are are generally modelling using ODEs. ODEs in this case stands for Ordinary Differential Equations. What we're going to discuss here is the law of mass action, the Michaelis-Menten approximation for enzyme-catalyzed reactions. And how you can convert from a diagram to a system of equations that describe the behavior of our system. [SOUND] So first, let's consider the example that we showed before in the Introduction to MATLAB lectures. We considered bi binding of ligand to a receptor. So you can have this scheme like this. Ligand plus receptor, yields ligand receptor complex with sum association rate constant, k plus. And then, ligand receptor complex can dissociate to free ligand plus free receptor, with a dissociation rate constant, k minus. And what we considered before was the steady state solution to this. The bound receptor can be calculated with this equation here, that we discussed previously. And where the kd is defined as a ratio of the two rate consonants. And these are the results that you get ligand receptor, as a function of free ligand for several different values of the kd. Where L in this case refers to free ligand or for free receptor. Our tot is the total amount of receptor. Ligand receptor, is ligand receptor complex and we thought of us for a different values of ligand and different values of kd. So now we want to ask ourselves this, what about non-steady state solutions, right. This is ligand receptor complex as a function of y in a steady state but what about the dynamics as the ligand receptor complex is forming. So how can we think about the non-steady-state solutions to this? So if we have ligand binding to a receptor, more generally, this reaction implies the following set of equations. We have a single ordinary differential equation here for free ligand with respective to them. This is a derivative of ligand with expected time. This equation is the derivative of free receptor with respect to time. And this is the equation for Ligand receptor complex, with respect to time. And, someone who's expert in this can look at this chemical scheme right here. And can say immediately, well, these would be the three differential equations that we described, in this system. But that might not be obvious to a non expert. So that's what we want to discuss next is where do these particular equations come from. Why are the 3 differential equations why do the 3 differential equations of this form rather than some other form? So the first topic we need to discuss. In order to determine you know, what would be the form of the differential equations that would describe the behavior of a system is a law of mass action. And this can be defined as follows. The rate of an elementary, elementary reaction, in other words, a reaction that proceeds through only one transition state, or one mechanistic step, Is proportional to the product of the concentrations of the participating molecules. This definition came from Wikipedia. This sounds like a mouthful, but if you, if you look at it carefully you can actually see where, how, how it can get very simplified. This part here. The rate, is proportional to the product of the concentrations of the participating molecules. In other words if you know all the molecules that participate in the reaction, all you have to do is mon, multiply them together and then you'll get something that's proportional to the rate. So let's consider a, single, very, very simple reaction, where you can have some chemicals species A. That can get converted into subspecies B, with some rate constant k1 plus. Well, the forward rate the rate at which A get converted into B in units of concentration over time is like the the log mass action says. It's proportional to the product of all the participating molecules, and in this case there is only one participating molecule which is A. So we can calculate this forward rate as this rate constant, k1 plus times concentration of a. And this makes intuitive sense when we think about it. The more A that we have, the greater the rate at which it's going to get converted to B. And if we didn't have any a at all, if a was zero, then it would be impossible. To have A being converted into B, so it makes sense that this forward rate gets bigger as A gets bigger. What if we had a slightly more complex reaction where A plus B came together to form C. Well, in this case, the forward rate would be this rate constant K2 plus. Times A, times B. Again, like the law of mass action states, it's proportional to the product of the concentrations of the participating molecules. So we have two participating molecules. We have to multiply both of them together. And we can see something immediately from considering these two relatively simple reactions is that we know that the overall rate. Has to end up in the same units, units of concentration per unit time. So, this product k1 plus times A has to be in the same units as this other product here, k2 plus times A times B. So that tells us that our two rate constants, k1 plus. And k2 plus are going to have different units from one another. And if we understand the law of mass action this will allow us to write down equations for many biochemical reactions that we will encounter in practice. What about a slightly more complex biochemical reaction? Specifically, what if you have a biochemical reaction that's catalyzed, by an enzyme. And frequently when people write down diagrams they, they use shorthand. They don't necessarily write down all of the steps that are involved in the reaction, but write down something like this. Substrate, gets converted to product, and often they'll draw an enzyme and they'll draw an arrow, pointing to the other arrow. And what that means is that this enzyme, will catalyze its reaction, of taking Substrate, and converting it into Product. And, we're going to write down some equations now where, capital S is going to, refer to free substrate, capital E to free enzyme, capital E with a subscript TOT, for total enzyme and capital P, for product. So this shorthand that you see here implies the following more complete reaction scheme. Where enzyme plus substrate come together to form an enzyme substrate complex, and then this enzyme substrate complex gets converted into product plus free enzyme again. And what people generally assume when they're, when they're writing down systems of ordinary differential equations for biochemical signaling networks. They often assume Michaelis-Menten kinetics. What that means is we can write down an equation here this is an approximation. But it's frequently a relatively good approximation where we can say, the velocity of the reaction, in other words, how much product you're producing with respect to time. So the differential equation for product is equal to the maximum velocity times the substrate concentration plus, sorry, divided by the substrate concentration plus some term k m. Where the variable definitions that we see in this equation are as follows. V max is defined as this reaction rate k2, times the total amount of enzyme, and this constant here, KM, is defined by our three elementary array constants, k minus 1, k2, and k plus 1. So when people, write down a, a scheme that looks like this, substrate getting converted into product with, catalyzed by an enzyme. This is the more complete scheme, that, that is implied by this. But frequently then we'll use a, sort of short cut in order to, In order to simplify the equation. And, this shortcut is what's known as the Michaelis-Menten equation, which has this form here. So, if we're going to write down equations for an enzyme-catalyzed reaction, and we're going to use the Michaelis-Menten approximation, what are these differential equations going to look like? So, now let's look at what, differential equations we'll have for an enzyme-catalyzed reaction, when we assume Michaelis-Menten kinetics. Remember, this is the scheme that we're looking at. Substrate gets converted to product, and that conversion is catalyzed by the enzyme, and the Abbreviations we're going to use are listed here. If we assume Michaelis-Menten kinetics for this reaction, the relevant ordinary differential equations become the following. The derivative of, the change in substrate with respect to time is equal to minus k cat times enzyme concentration. Times substrate divided by substrate plus the Km. And then the change in product with respect to time, it's just the negative of this term here. K cat times enzyme, time substrate concentration, over substrate concentration plus Km. We've introduced a new term here, k cat. k cat refers to the catalytic constant of the enzyme. And this was called k2 in the previous slide, so it, it means the same thing. But the overall point in this case, is that we can look at a scheme like this, substrate getting converted to product catalyzed by an enzyme. And if we assume that the Michaelis-Menten approximation is going to hold, then we know what the differential equations are that describe this system. So now let's look at a more complicated biochemical reaction. And work our way through how we can write down the system of ODEs to describe that system. So the example we're going to use is a generic three component repressive network. And this is a model that can be display oscillation but for some values of the relevant parameters. And this was described in a really nice review article by Alice McGillner, published in Developmental Cell, several years ago. And this has, this contains three proteins A, B, and C that can either be phosphorylated. So that's this form with a dot dash p on it. Or it can be dephosphorylated, which is a form here. So you have protein A in the unphosphorylated form over here, and then protein A in the phosphorylated form over here. And it's similarly with B and similarly with C. And this becomes a relatively complicated system because of the regulation that occurs here. A can catalyze dephosphorylation of B, so when A goes up it's going to make it's going to catalyze pushing B from the, from the right to the left, the dephosphorylation reaction. Similarly, B is going to catalyze the dephosphorylation of C. But then C is going to catalyze the dephosphorylation of A. And not when When Alex McGillan wrote this review article he based this off of oscillatory biochemical circuit and soil bacteria. And so the original reference that he he took this scheme from is listed here down at the bottom. So we can look at this relatively complex scheme from this review article and we can say the full set of equations that underly this scheme. Are these equations here and this looks pretty complicated. This looks like a mess. But we want to go through this step by step in order to demystify it for you so that, it makes sense to you. And that when you're confronted with something like this in the future, you might be able to write down the relevant equations. So the two questions we need to address right away are. Why do we only have three differential equations rather than six? If you look at this scheme here you can see that there's one, two, thee unphosphorylated proteins. And three phosphorylated proteins. But I said that we could describe this system with three differential equations over here. So why do we only have three rather than six? And then if you look carefully at the equations over here you can see that there's a lot of commonalities in this structure. And it's written in a way to accentuate that where you can see lots of common themes if you look at the equation, differential equation for A, or for B, or for C. And why do these three equations share this characteristic structure? So that's what we want to go through next. Let's consider the equations that we get with this three component reaction. And as a reminder, generic three component repressive network is summarized by this, by this scheme over here. where A B and C can be phosphorylated or de-phosphorylated, and A B and C influence the reactions of the of the other species. So let's write down our equations and we're just going to get for A and phosphorylated form of A. and the differential equations for A, and phosphorylated A are as follows here. What do we notice about these equations? Well, each one of the is a Michaelis-Menton equation. Each term in here is a Michaelis-Menton equation. The other way we can sort of think how we ended up with these particular differential equations for A and phosphorylated A Here's to say, what are the reactions that can make the concentration of A decrease, and what are the concentrations that can make the concentration of A increase. So if we look, if we just look at A right here, when A has phosphorylated, then you're going to decrease the amount of non-phosphorylated A you have. And then, when phosphorylated A gets dephosphorylated, then you're going to increase the amount of unphosphorylated A you have. So, you're going to have one term here, which, increases the concentration of A. One positive term, and this one represents the dephosphorylation reaction, and then you're going to have one, negative term here. And this negative term, represents the, the phosphorylation reaction. Furthermore, as I just mentioned. Each of these terms is a Michaelis-Menten equation, right? So when A gets converted from the B phosphorylated form, to the physphorylated form, what's the substrate? The substrate in this case is A. So that's why this equation here will have this form, where A is in the numerator, that would be substituting for substrate in the numerator, And in the, in the denominator you have A plus some kn term here. Right? In the Michaelis-Menten equations you have substrate plus kn in the denominator. Now, we can't just use the term kn because we have many different reactions going on here, and then it would be ambiguous. Which reaction are you talking about? So we've employed a, a different syntax here, to label our different to label our different reactions. We have a big K, to indicate that it's a Michaelis constant. And then this little k here, it tells us that it's a kinase reaction. So this phosphor layered, phosphorylation reaction occurs, through the action of a kinase. And then this dephosphorylation reaction occurs is catalyzed by a phosphatase. So that's why we have this, this big K, because it's Micalah's constant. Little k for kinase. And one because there's, there's three kinase reactions going on here, one for A, one for B, and one for C. So, this is a, so what we can see here is that in this particular term, you know, Capital A here, here represents the substrate. Capital C represents the enzyme concentration. Remember that the total enzyme concentration is important, in determining the, the, Maximal rate in a Michaelis-Menton equation. And then, this term in the denominator here is the Km for this reaction. And then every other term in, in these two differential equations we can understand the same way. This is a Michaelis-Menton equation for the dephosphorylation reaction where the substrate in this case is phosphorylated A. And this Michaelis constant denominator here, the little p is indicate that it's a phosphatase here. And then this k, k1, and this kp1 are, are the are the rate constants, the k cats for the for the corresponding enzymes. Now, if we look carefully at these two equations we notice something really important. Notice that the change in a with respect to time, is a negative of the change in a, phosphorylated with respect to time. So dA, dt equals the negative of d, AP, dt. We can also notice that from looking at the scheme here, right? The only way that phosphor layered A goes up is if A goes down. And conversely, the only way that A goes up is if A phosphor layer goes down. And that's expressed mathematically here by the fact that one right hand side here is the negative of the other right hand side. And that simplifies our life. That's, I mean, this mathematically equivalent to saying that If, if a goes up then a phosphorylated must go down. Or in other words, the sum of A and phosphorylated A, must equal some constant a total and this a total is not going to change in this case. And because these two differential equations are just the negative of one another, we don't need to keep track of both of them. We can illuminate one of these differential equations. So we can get rid of our equation for phosphorylated A and we can just substitute in there phosphorylated A is always equal to a total minus A, and that comes directly just from rearranging this equation here. So now we, this is how we end up with our overall differential equation for A. Is what we wrote down here at the top but we've just substituted for A, for phosphorylated A, we substituted in A total minus A. And this is how we get our overall differential equation for A, we can look at our differential equations for B and C exactly the same way. And that's how we can end up with a complete set of three differential equations, that we had previously. [BLANK_AUDIO] Now there's an important point to to make here which is that when you have a, a scheme like the one that we had on the last slide that explicitly states which reactions are catalyzed by A, B and C. Then it's relatively straightforward to write down a set of ODE's describing the system. But sometimes when you look at a cartoon, that's in a paper, that that shows a mechanism that illustrates how the the author of the paper, feels that the system works. Sometimes they use a shorthand that's even more extreme. And this is an example here. This is this is a mutagen activated protein kinase pathway where you have Raf going to MEK going to ERK. But this doesn't actually mean that Raf gets converted into MEK. And then MEK gets converted into ERK. What this really means is that raf catalyzes the phosphorylation of MEK. And then when MEK gets double phosphorylated in this case, MEK catalyzes the phosphorylation of ERK. So if you had a scheme like this where Raf catalyzes this reaction and MEK catalyzes this reaction. Then you could write down some differential equations describing how this works. If you just had this cartoon over here, it's says Raf and then arrow MEK and then arrow ERK. You couldn't write down a set of differential equations describing it in that case. So just to give you a little bit of warning sometimes you can look at a A biochemical reaction scheme or a cartoon that's in a paper, and you can write down the differential equations immediately. Other times you can't. So there's some times where considerable prior biological knowledge is required to, to derive equations from the cartoons that describe biological mechanisms. Now we want to summarize what we've learned in this first part of our lecture, on dynamical systems. First, biochemical signalling pathways can usually be described mathematically with systems of ordinary differential equations. Writing down the relevant ODEs is the first step in constructing a mathematical model. And third, what we've learned is that biochemical signalling ODEs are generally some combination of either the law of mass action for very simple reactions or Michaelis-Menten approximations for enzyme-catalyzed reactions. Let's conclude this lecture by considering our self-assessment question based on what we just discussed. This one is quite involved but I think it's worth it when you, when you go through it to determine what the differential equations are going to be. So, our reaction scheme is illustrated here and it's summarized in the, in the text. I'm not going to read every single word of this but I'll, I'll highlight some of the key features. You have proteins A and B here that can bind to form some complex AB and then AB can dissociate back to free A and free B. Furthermore A can be phosphorylated, so A can go from the un-phosphorylated form here to the phosphorylated form up here. And this is catalyzed by an enzyme we call C, this has a reaction rate K cap, that we call K Little k, in this case, for a kinase reacton. And then A, the phosphorylated form of A, A P up here can be dephosphorylated in this dephosphorylation reaction is catalyzed by a protein D. In this case, it's a phosphatase that catalyzes the dephosphorylation reaction. and this has a reaction rate or a K cap that we call K, K little p. furthermore we are going to define a big K subscript little k and a big K subscript little p, as the KM's for the kinase reaction, this one is catalyzed by C And the phosphor state's reaction, that's one that's catalyzed by D. And other rate constants are listed here on the scheme and so the question we want to ask is what's the correct differential equation for A. We want to derive the equation for dA, dt and it's one of these four here A, B, C, or D. So in this case I think it's probably a good idea to pause the recording, and think about it a little bit. Maybe write it down yourself. And then we'll come back and we'll explain which of these answers is correct. Okay now you have some time to think about it. The correct answer in this case is C. Now lest go through and discuss why C is the correct answer and A, B, D are not correct answers. And this is our scheme summarized here. And this is the one that we got as the correct answer for the, differential equation for A. What's the first term here? K, little k times C times A over A plus Big K, subscript, little k, with a negative in front of it. This is a decrease in a that happens when a gets phosphorylated. So, when a moves from the free, the dephosphorylated form here to the phosphorylated form, the concentration of a goes down. That's why there's a negative. And every other term here can be understood by considering this is Michaelisâ€“Menten, kinetics catalyzed by C and A in this case is a sub straight. This term here the second term is the increase in A you get due to dephosphorylation. So, this is a plus because the arrow's pointing towards A. And you can understand every term in this by considering that it's Michaelis-Menten approximation and phosphorylated A, in this case, is a substrate. The third term is a decrease in A due to binding to B. In this term here we derived using the law of mass action. And in the fourth term here is the, is an increase In a, due to dissociation of, of the AB complex. So this is a very, you know, complex equation. It looks pretty, it can look pretty daunting to see this, this whole reaction scheme. And try to figure out, what's the ODE describing, describing a? But we can understand this term by term. To see you know, what, what, which terms are going to make A go up and which are going to make A go down and what are the relevant reactions in this case. Now let's look at the choices that were not correct. Our choice B was this, da dt equals equals these terms. This is actually the ODE describing the time evolution of phosphorylated A. Where this kinase reaction is, is positive makes a phosphorylator go up. This phosphatase reaction is negative and this one doesn't include any binding terms. What about choice D, where we had dA dt equals k plus, times A times B, minus k minus, times AB complex. This was another incorrect answer. And this one is actually the ODE describing the time evolution of, of AB complex. So in this case formation A times B times k plus is going to make AB go up. And then ne k minus time AB, the dissociation reaction is going to make AB concentration go down. That's why there's a negative in front of it. And then our final choice, A. This one simply has mistakes in it. You can see that both of these depend on D. Both of them have the k and A. None of these have a phosphorylated. So this one is just kind of a mess. And we can identify this as being wrong, because, for instance, it doesn't depend on c at all. So hopefully this self-assessment question was helpful to you, and this is a way for us to practice learning how to write down ordinary differential equations when we learn, when we have reaction schemes.