0:07

Hello.

Â Next set of lectures that we're going to cover in this

Â course, are grouped under the overall heading Introduction to Dynamical Systems.

Â And there are four lectures to this section here on dynamical systems.

Â Just to give you a brief introduction of what we're going to discuss.

Â We we've previously taught you about how to

Â use the Matlab computing environment to perform scientific computations.

Â Now we're going to actually proceed to deriving

Â systems of ordinary differential equations describing biochemical signaling networks.

Â And we're going to show you some of the tools you can use to

Â analyze the results of these, the

Â simulations that you perform on these models.

Â And these are grouped under the overall heading of Dynamical Systems Tools.

Â And this is part one of a, of a four part series on dynamical systems.

Â 1:09

What we're going to discuss here is the law

Â of mass action, the

Â Michaelis-Menten approximation for enzyme-catalyzed reactions.

Â And how you can convert from a diagram to a

Â system of equations that describe the behavior of our system.

Â [SOUND]

Â So first, let's consider the example that we

Â showed before in the Introduction to MATLAB lectures.

Â We considered bi binding of ligand to a receptor.

Â So you can have this scheme like this.

Â Ligand plus receptor, yields ligand receptor complex

Â with sum association rate constant, k plus.

Â And then, ligand receptor complex can dissociate to free ligand

Â plus free receptor, with a dissociation rate constant, k minus.

Â 1:53

And what we considered before was the steady state solution to this.

Â The bound receptor can be calculated with

Â this equation here, that we discussed previously.

Â And where the kd is defined as a ratio of the two rate consonants.

Â And these are the results that you get ligand receptor, as

Â a function of free ligand for several different values of the kd.

Â 2:13

Where L in this case refers to free ligand or for free receptor.

Â Our tot is the total amount of receptor.

Â Ligand receptor, is ligand receptor complex and we thought of us

Â for a different values of ligand and different values of kd.

Â So now we want to ask ourselves this, what about non-steady state solutions, right.

Â This is ligand receptor complex as a function of y in a steady

Â state but what about the dynamics as the ligand receptor complex is forming.

Â 2:48

more generally, this reaction implies the following set of equations.

Â We have a single ordinary differential equation

Â here for free ligand with respective to them.

Â This is a derivative of ligand with expected time.

Â This equation is the derivative of free receptor with respect to time.

Â And this is the equation for Ligand receptor complex, with respect to time.

Â And, someone who's expert in this can look at this chemical scheme right here.

Â And can say immediately, well, these would be the

Â three differential equations that we described, in this system.

Â But that might not be obvious to a non expert.

Â So that's what we want to discuss next

Â is where do these particular equations come from.

Â Why are the 3 differential equations why do the 3

Â differential equations of this form rather than some other form?

Â 3:46

And this can be defined as follows.

Â The rate of an elementary, elementary reaction, in

Â other words, a reaction that proceeds through only one

Â transition state, or one mechanistic step, Is proportional to

Â the product of the concentrations of the participating molecules.

Â This definition came from Wikipedia.

Â 4:12

The rate, is proportional to the product

Â of the concentrations of the participating molecules.

Â In other words if you know all the

Â molecules that participate in the reaction, all you have

Â to do is mon, multiply them together and

Â then you'll get something that's proportional to the rate.

Â 4:27

So let's consider a, single, very, very simple

Â reaction, where you can have some chemicals species A.

Â That can get converted into subspecies B, with some rate constant k1 plus.

Â Well, the forward rate the rate at which A get converted into B in units

Â of concentration over time is like the the log mass action says.

Â It's proportional to the product of all the participating molecules, and

Â in this case there is only one participating molecule which is A.

Â So we can calculate this forward rate as

Â this rate constant, k1 plus times concentration of a.

Â 5:08

The more A that we have, the greater the

Â rate at which it's going to get converted to B.

Â And if we didn't have any a at all, if a was zero, then it would be impossible.

Â To have A being converted into B, so it makes

Â sense that this forward rate gets bigger as A gets bigger.

Â 5:24

What if we had a slightly more complex reaction

Â where A plus B came together to form C.

Â Well, in this case, the forward rate would be this rate constant K2 plus.

Â Times A, times B.

Â Again, like the law of mass action states, it's proportional

Â to the product of the concentrations of the participating molecules.

Â So we have two participating molecules.

Â We have to multiply both of them together.

Â 5:47

And we can see something immediately from considering these two

Â relatively simple reactions is that we know that the overall rate.

Â Has to end up in the same units, units of concentration per unit time.

Â So, this product k1 plus times A has to be in the same

Â 6:37

And frequently when people write down diagrams they, they use shorthand.

Â They don't necessarily write down all of the steps that

Â are involved in the reaction, but write down something like this.

Â Substrate, gets converted to product, and often they'll draw an

Â enzyme and they'll draw an arrow, pointing to the other arrow.

Â And what that means is that this enzyme, will catalyze

Â its reaction, of taking Substrate, and converting it into Product.

Â 7:02

And, we're going to write down some equations now where,

Â capital S is going to, refer to free substrate,

Â capital E to free enzyme, capital E with a

Â subscript TOT, for total enzyme and capital P, for product.

Â 7:18

So this shorthand that you see here

Â implies the following more complete reaction scheme.

Â Where enzyme plus substrate come together to form an enzyme substrate complex, and

Â then this enzyme substrate complex gets

Â converted into product plus free enzyme again.

Â 7:35

And what people generally assume when they're, when they're writing

Â down systems of ordinary differential

Â equations for biochemical signaling networks.

Â They often assume Michaelis-Menten kinetics.

Â What that means is we can write down an equation here this is an approximation.

Â But it's frequently a relatively good approximation

Â where we can say, the velocity of

Â the reaction, in other words, how much

Â product you're producing with respect to time.

Â So the differential equation for product is equal to the maximum velocity times the

Â substrate concentration plus, sorry, divided by the

Â substrate concentration plus some term k m.

Â Where the variable definitions that we see in this equation are as follows.

Â V max is defined as this reaction rate k2, times the total amount

Â of enzyme, and this constant here, KM, is defined by our three

Â elementary array constants, k minus 1, k2, and k plus 1.

Â 8:34

So when people, write down a, a scheme that looks like

Â this, substrate getting converted into product with, catalyzed by an enzyme.

Â This is the more complete scheme, that, that is implied by this.

Â But frequently then we'll use a, sort of short

Â cut in order to, In order to simplify the equation.

Â And, this shortcut is what's known as

Â the Michaelis-Menten equation, which has this form here.

Â So, if we're going to write down

Â equations for an enzyme-catalyzed reaction, and we're

Â going to use the Michaelis-Menten approximation, what are

Â these differential equations going to look like?

Â 9:23

Remember, this is the scheme that we're looking at.

Â Substrate gets converted to product, and that conversion is catalyzed by

Â the enzyme, and the Abbreviations we're going to use are listed here.

Â 9:46

The derivative of, the change in substrate with respect to

Â time is equal to minus k cat times enzyme concentration.

Â Times substrate divided by substrate plus the Km.

Â And then the change in product with respect to

Â time, it's just the negative of this term here.

Â K cat times enzyme, time substrate

Â concentration, over substrate concentration plus Km.

Â We've introduced a new term here, k cat.

Â k cat refers to the catalytic constant of the enzyme.

Â And this was called k2 in the previous slide, so it, it means the same thing.

Â 10:21

But the overall point in this case, is that we can look at

Â a scheme like this, substrate getting

Â converted to product catalyzed by an enzyme.

Â And if we assume that the Michaelis-Menten approximation is going to hold,

Â then we know what the differential equations are that describe this system.

Â 10:39

So now let's look at a more complicated biochemical reaction.

Â And work our way through how we can write

Â down the system of ODEs to describe that system.

Â So the example we're going to use

Â is a generic three component repressive network.

Â And this is a model that can be display

Â oscillation but for some values of the relevant parameters.

Â And this was described in a really nice review article

Â by Alice McGillner, published in Developmental Cell, several years ago.

Â 11:10

And this has, this contains three proteins A,

Â B, and C that can either be phosphorylated.

Â So that's this form with a dot dash p on it.

Â Or it can be dephosphorylated, which is a form here.

Â So you have protein A in the unphosphorylated form over

Â here, and then protein A in the phosphorylated form over here.

Â And it's similarly with B and similarly with C.

Â And this becomes a relatively complicated system

Â because of the regulation that occurs here.

Â A can catalyze dephosphorylation of B, so when A goes up it's going to make it's

Â going to catalyze pushing B from the, from

Â the right to the left, the dephosphorylation reaction.

Â Similarly, B is going to catalyze the dephosphorylation of C.

Â But then C is going to catalyze the dephosphorylation of A.

Â 11:57

And not when When Alex McGillan wrote this review article

Â he based this off of oscillatory biochemical circuit and soil bacteria.

Â And so the original reference that he he took

Â this scheme from is listed here down at the bottom.

Â 12:15

So we can look at this relatively complex scheme from this review article

Â and we can say the full set of equations that underly this scheme.

Â Are these equations here and this looks pretty complicated.

Â This looks like a mess.

Â But we want to go through this step by step in order

Â to demystify it for you so that, it makes sense to you.

Â And that when you're confronted with something like this in the

Â future, you might be able to write down the relevant equations.

Â So the two questions we need to address right away are.

Â Why do we only have three differential equations rather than six?

Â If you look at this scheme here you

Â can see that there's one, two, thee unphosphorylated proteins.

Â And three phosphorylated proteins.

Â But I said that we could describe

Â this system with three differential equations over here.

Â So why do we only have three rather than six?

Â And then if you look carefully at the equations over here

Â you can see that there's a lot of commonalities in this structure.

Â 13:07

And it's written in a way to accentuate that where you can see lots of common

Â themes if you look at the equation, differential

Â equation for A, or for B, or for C.

Â And why do these three equations share this characteristic structure?

Â So that's what we want to go through next.

Â Let's consider the equations that we get with this three component reaction.

Â And as a reminder, generic three component repressive network

Â 14:04

Each term in here is a Michaelis-Menton equation.

Â The other way we can sort of think how we ended up

Â with these particular differential equations for

Â A and phosphorylated A Here's to say,

Â what are the reactions that can make the concentration of A decrease,

Â and what are the concentrations that can make the concentration of A increase.

Â So if we look, if we just look at A right here, when

Â A has phosphorylated, then you're going to decrease

Â the amount of non-phosphorylated A you have.

Â And then, when phosphorylated A gets dephosphorylated, then you're

Â going to increase the amount of unphosphorylated A you have.

Â So, you're going to have one term here, which, increases the concentration of A.

Â One positive term, and this one represents the dephosphorylation

Â reaction, and then you're going to have one, negative term here.

Â And this negative term, represents the, the phosphorylation reaction.

Â 14:56

Furthermore, as I just mentioned.

Â Each of these terms is a Michaelis-Menten equation, right?

Â So when A gets converted from the B

Â phosphorylated form, to the physphorylated form, what's the substrate?

Â The substrate in this case is A.

Â So that's why this equation here will have this

Â form, where A is in the numerator, that would be

Â substituting for substrate in the numerator, And in the, in

Â the denominator you have A plus some kn term here.

Â Right?

Â In the Michaelis-Menten equations you have substrate plus kn in the denominator.

Â Now, we can't just use the term kn because we have

Â many different reactions going on here, and then it would be ambiguous.

Â Which reaction are you talking about?

Â So we've employed a, a different syntax here,

Â to label our different to label our different reactions.

Â We have a big K, to indicate that it's a Michaelis constant.

Â And then this little k here, it tells us that it's a kinase reaction.

Â So this phosphor layered, phosphorylation reaction

Â occurs, through the action of a kinase.

Â And then this dephosphorylation reaction occurs is catalyzed by a phosphatase.

Â So that's why we have this, this big K, because it's Micalah's constant.

Â Little k for kinase.

Â And one because there's, there's three kinase reactions going on

Â here, one for A, one for B, and one for C.

Â 16:17

So, this is a, so what we can see here is that

Â in this particular term, you know, Capital A here, here represents the substrate.

Â Capital C represents the enzyme concentration.

Â Remember that the total enzyme

Â concentration is important, in determining the,

Â 16:34

the, Maximal rate in a Michaelis-Menton equation.

Â And then, this term in the denominator here is the Km for this reaction.

Â And then every other term in, in these

Â two differential equations we can understand the same way.

Â This is a Michaelis-Menton equation for the dephosphorylation reaction

Â where the substrate in this case is phosphorylated A.

Â And this Michaelis constant denominator here, the little

Â p is indicate that it's a phosphatase here.

Â And then this k, k1, and this kp1 are, are the are

Â the rate constants, the k cats for the for the corresponding enzymes.

Â Now, if we look carefully at these

Â two equations we notice something really important.

Â 17:19

Notice that the change in a with respect to time, is

Â a negative of the change in a, phosphorylated with respect to time.

Â So dA, dt equals the negative of d, AP, dt.

Â 17:37

The only way that phosphor layered A goes up is if A goes down.

Â And conversely, the only way that A goes up is if A phosphor layer goes down.

Â And that's expressed mathematically here by the fact that one right

Â hand side here is the negative of the other right hand side.

Â And that simplifies our life.

Â That's, I mean, this mathematically equivalent to saying that If,

Â if a goes up then a phosphorylated must go down.

Â Or in other words, the sum of A and phosphorylated A, must equal some

Â constant a total and this a total is not going to change in this case.

Â 18:11

And because these two differential equations are just the negative of

Â one another, we don't need to keep track of both of them.

Â We can illuminate one of these differential equations.

Â So we can get rid of our equation for

Â phosphorylated A and we can just substitute in there phosphorylated

Â A is always equal to a total minus A,

Â and that comes directly just from rearranging this equation here.

Â So now we, this is how we end up with our overall differential equation for A.

Â 18:38

Is what we wrote down here at the top but we've just

Â substituted for A, for phosphorylated A, we substituted in A total minus A.

Â And this is how we get our overall differential equation for A, we

Â can look at our differential equations for B and C exactly the same way.

Â And that's how we can end up with a

Â complete set of three differential equations, that we had previously.

Â [BLANK_AUDIO]

Â Now there's an important point to to make here which

Â is that when you have a, a scheme like the

Â one that we had on the last slide that explicitly

Â states which reactions are catalyzed by A, B and C.

Â Then it's relatively straightforward to write down

Â a set of ODE's describing the system.

Â But sometimes when you look at a cartoon, that's in a paper, that that shows

Â a mechanism that illustrates how the the author

Â of the paper, feels that the system works.

Â Sometimes they use a shorthand that's even more extreme.

Â 19:38

This is this is a mutagen activated protein kinase pathway

Â where you have Raf going to MEK going to ERK.

Â But this doesn't actually mean that Raf gets converted into MEK.

Â And then MEK gets converted into ERK.

Â 19:56

What this really means is that raf catalyzes the phosphorylation of MEK.

Â And then when MEK gets double phosphorylated in

Â this case, MEK catalyzes the phosphorylation of ERK.

Â So if you had a scheme like this where

Â Raf catalyzes this reaction and MEK catalyzes this reaction.

Â Then you could write down some

Â differential equations describing how this works.

Â If you just had this cartoon over here, it's

Â says Raf and then arrow MEK and then arrow ERK.

Â You couldn't write down a set of

Â differential equations describing it in that case.

Â 20:26

So just to give you a little bit of

Â warning sometimes you can look at a A biochemical

Â reaction scheme or a cartoon that's in a paper,

Â and you can write down the differential equations immediately.

Â Other times you can't.

Â So there's some times where considerable prior biological knowledge is required

Â to, to derive equations from the cartoons that describe biological mechanisms.

Â 21:13

And third, what we've learned is that

Â biochemical signalling ODEs are generally some combination of

Â either the law of mass action for very simple reactions or Michaelis-Menten

Â approximations for enzyme-catalyzed reactions.

Â Let's conclude this lecture by considering our

Â self-assessment question based on what we just discussed.

Â 21:43

So, our reaction scheme is illustrated here

Â and it's summarized in the, in the text.

Â I'm not going to read every single word of this

Â but I'll, I'll highlight some of the key features.

Â You have proteins A and B here that can bind to form some

Â complex AB and then AB can dissociate back to free A and free B.

Â 22:12

And this is catalyzed by an enzyme we call C, this has a reaction rate

Â K cap, that we call K Little k, in this case, for a kinase reacton.

Â And then A, the phosphorylated form of A, A P up here can be dephosphorylated

Â in this dephosphorylation reaction is catalyzed by a protein D.

Â In this case, it's a phosphatase that catalyzes the dephosphorylation reaction.

Â and this has a reaction rate or a K cap that we call K, K little p.

Â furthermore we are going to define a big K

Â 22:50

subscript little k and a big K subscript

Â little p, as the KM's for the kinase reaction,

Â this one is catalyzed by C And the

Â phosphor state's reaction, that's one that's catalyzed by D.

Â 23:02

And other rate constants are listed here on the scheme and so

Â the question we want to ask is what's the correct differential equation for A.

Â We want to derive the equation for dA, dt and it's

Â one of these four here A, B, C, or D.

Â 23:21

So in this case I think it's probably a good idea

Â to pause the recording, and think about it a little bit.

Â Maybe write it down yourself.

Â And then we'll come back and we'll explain which of these answers is correct.

Â 24:11

What's the first term here?

Â K, little k times C times A

Â over A plus Big K, subscript, little k, with a negative in front of it.

Â This is a decrease in a that happens when a gets phosphorylated.

Â So, when a moves from the free, the dephosphorylated form

Â here to the phosphorylated form, the concentration of a goes down.

Â That's why there's a negative.

Â And every other term here can

Â be understood by considering this is Michaelisâ€“Menten,

Â kinetics catalyzed by C and A in this case is a sub straight.

Â 24:52

So, this is a plus because the arrow's pointing towards A.

Â And you can understand every term in this by considering that

Â it's Michaelis-Menten approximation and phosphorylated A,

Â in this case, is a substrate.

Â 25:04

The third term is a decrease in A due to binding to B.

Â In this term here we derived using the law of mass action.

Â And in the fourth term here is the, is an

Â increase In a, due to dissociation of, of the AB complex.

Â So this is a very, you know, complex equation.

Â It looks pretty, it can look pretty

Â daunting to see this, this whole reaction scheme.

Â And try to figure out, what's the ODE describing, describing a?

Â But we can understand this term by term.

Â 25:57

Where this kinase reaction is, is positive makes a phosphorylator go up.

Â This phosphatase reaction is negative and

Â this one doesn't include any binding terms.

Â What about choice D, where we had dA dt equals

Â k plus, times A times B, minus k minus, times AB complex.

Â This was another incorrect answer.

Â And this one is actually the ODE

Â describing the time evolution of, of AB complex.

Â So in this case formation A times B times k plus is going to make AB go up.

Â And then ne k minus time AB, the dissociation

Â reaction is going to make AB concentration go down.

Â That's why there's a negative in front of it.

Â 26:40

And then our final choice, A.

Â This one simply has mistakes in it.

Â You can see that both of these depend on D.

Â Both of them have the k and A.

Â None of these have a phosphorylated.

Â So this one is just kind of a mess.

Â And we can identify this as being wrong, because,

Â for instance, it doesn't depend on c at all.

Â 27:01

So hopefully this self-assessment question was helpful to

Â you, and this is a way for us to

Â practice learning how to write down ordinary differential

Â equations when we learn, when we have reaction schemes.

Â