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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So today

the basic idea is to try to understand

the toggle switch. How such a thing can be made,

why it represents a memory module.

But then we'll

relatively quickly get into two themes

that are going to be useful throughout

the rest of the semester. First is these

dimensionless equations that often pop up in the

analysis of these gene circuits. And it's absolutely

essential that you understand how to get to the

dimensionless equations, and also what these

parameters end up meaning. And then we'll, at the end,

talk about stability analysis. How is it that you can determine

whether a particular set of interacting pieces in

a cell or in an ecosystem whatnot– once you get it in

the form of an equation, how's is it you can determine whether

a particular fixed point or a particular

location is going to be stable to perturbations? This is going to

be useful for us to determine where

the gene network is going to kind of move towards. Also it'll be useful for us

to determine whether a gene network is going to oscillate.

And later, it'll be relevant

in the context of predator prey oscillations, and

a bunch of things. So can somebody

maybe explain briefly the idea of the toggle switch? Yes, please. AUDIENCE: You have these two

genes that repress expression to each other. So when one of them is

large, the other one is not expressed. [INAUDIBLE]. PROFESSOR: Perfect. So you'll have two genes that

are going to mutually repress each other. So in this case, each of them

will be a transcription factor of some sort that will

bind to the other promoter and repress it.

So there are various

levels of abstraction that we might use to

describe such things. So we might, for example,

just say A repressing B, B repressing A. Of course

when you write it that way, it doesn't have to be in the

context of a gene network. This A's and B's could

just be chemicals, they could be species

eating each other. It could be almost anything. Now in this framework,

to get a basic sense of why this thing might have

two alternative stable states, is that often we like to take

the Boolean approximation. This thing will not always

work, but it's a useful thing to do to just kind of first get

a sense of what might possibly be happening.

So we might say, 0 corresponds

to some sort of low. 1 might correspond to high. And of course, these

things have to be put it in quotes, because we haven't

specified what we mean by this. But it's useful

to just make sure that we're all thinking

about the same things. And then in the

context of A and B, we can just say, well, there's

a number of different states it could possibly be in. And you can ask whether this

assignment of logic values keeps everybody happy. And so you might ask, well, is

0, 0 a mutually happy state? And of course then

you have to say well, if you're in

the 0 state, you're not repressing the

other guy, but maybe 0 is sort of your

equilibrium anyway.

So what we have to do is we

have to make the assumption that when you're not being

repressed, in that case the promoter will be

actively making that protein. So then you'll go to some

sort of high or 1 state. So in that case, you say, if you

start out with both repressed, maybe both of them

should start trying to increase their levels. So this, in some ways,

is not a stable state. Similarly here, this is

also not a stable state. Because in this

case, they're both going to be trying to

repress one another. So then they'll both

start coming down and then the situation may

resolve into one of these two. So this is just where

either A or B is on, and repressing the other one. And so for example, in

context of the repressilator, on Thursday, this

is just a useful way to start imagining how

this A repressing B, repressing C repressing

A– how such a loop can lead to oscillations. This kind of analysis

does not at all prove that there are

oscillations in any given manifestation of this thing.

But it's useful

to just make sure that you're roughly getting the

idea of what the system might be doing. Now in many cases, we'll want to

be a little bit more explicit, and draw the gene

network in more detail. And there are multiple

manifestations of the Thomas Switch. There are many of them

that have been made. So the important thing

is not too necessarily keep track of exactly

what the components are, but in one case for example,

we might have A corresponding to something here. It's coming back and

repressing again. Expression of this

B. And this might all be on one piece of DNA. Whereas this B here will

come back and repress A. Now one thing that is– and

I just want to mention here, they also have a GFP. And this actually

is a case where those two can be expressed

off of a single promoter. So this is often

done in bacteria where there's a

single promoter– so RNA polymerase actually will

transcribe both of these genes. This repressor B, as well

as this fluorescent protein.

So there are going to be

alternative loading sites for the ribosome in that case. And eukaryotes typically

do not do this. Yeah? AUDIENCE: [INAUDIBLE]. PROFESSOR: That's right. So– AUDIENCE: And it will

transcribe everything until– PROFESSOR: Exactly. So here comes– so an

RNA polymerase down here made this whole thing. And now you might have two

separate locations where the ribosome loads and

makes this protein B. And then a different

ribosome would make this GFP. AUDIENCE: And when does it stop? It just keeps going down? PROFESSOR: Yes. So there's a

termination sequence. AUDIENCE: No, no, but

it would B and then GFP, and then it would

just keep going? PROFESSOR: No. The ribosome is told to

basically start here and end here. So then it just

makes the B protein. And then another

ribosome binds here. AUDIENCE: If you had more

proteins on the same strand after GFP– PROFESSOR: And when

you say protein, you're referring

to the ribosome.

[INTERPOSING VOICES] AUDIENCE: I mean G's right? PROFESSOR: Oh! OK, you're saying if

there were another gene? AUDIENCE: No, if you

have more genes coded, if you're coding for

more proteins after GFP– if you have more genes on a [INTERPOSING VOICES] –it will just keep going

for an arbitrarily long– [INAUDIBLE] PROFESSOR: Arbitrary is

always a dangerous word. But they can be more than two. And actually at the biophysics

retreat that some of you guys were at just last two days,

there was a great talk by Gene-Wei Li, who's going

to be a new incoming biology faculty member. And he was talking about

the FO F1 ATP synthase. So it's the thing

responsible for making ATP. He analyzes process work. And there are many sub units. So there are half a dozen or so. And so it's a very

long transcript. And then what he showed

is that actually you have different

rates of synthesis of the different genes

on this one transcript. And in some cases you want

actually more copies of one of the subunits

than another one. And so then actually

if the final protein, if it needs 12 of these,

only one of these, then actually you can make 12

times as much of this, because you just

have more translation here than you did here.

And then it's great,

because then you have all the right

ratios, all the components to make the protein. So you can actually have

additional regulation even at that stage. It's possible not everybody

followed that discussion, and my apologies. But feel free to just

erase it from your brain if you're too confused. But what you need to keep track

of here is the level of GFP is going to be perhaps

proportional to the level of B. Because they're being

expressed at the same time. Now in order for this thing

to be a memory module, you also want to be

able to reset the state. So if you were in

this state, you'd like to be able to somehow

get it to move to this state instead. Does anybody remember

what the inputs were in the context of

the sample toggle switch, that it was in that review? AUDIENCE: [INAUDIBLE].

PROFESSOR: Right. OK, so there are multiple

versions of toggle switch. And indeed in one case this was

just a small molecule, IPTG, and that's because this was

in that case, the lac I. And this then represses

the repression. So in many contexts

this class, you'll have to remember that a minus,

minus is equal to a plus. And in different toggle

switches you indeed have different ways of

inhibiting this repression. So this could be another

small molecule, ATCN. In the example that

they had in this review, it was actually

heat that did this.

But that's just because this

transcription factor was a temperature sensitive mutant. So above some temperature,

it could no longer repress. So for example, if

you start out in high GFP– so in this case,

where they– all right. If the cells are in

high GFP, and you want to switch it into

the alternative state, what stimulus do you

want to apply here? So I'll give you a guess. It's going to be

either heat or IPTG. I just want to make sure that

we can all read these diagrams.

This is to switch

from the high GFP, and we want to go

to the low GFP. Give you 15 seconds to just

try to read off this diagram. Do you need more time? No. Ready. Three, two, one. So we have a majority are

B. So a majority of people are saying, well, in this

case, you have a lot of GFP. Means you have a

lot of this protein B. In this case, the lac I. And we don't have very

much of this other protein.

So that means that if we

want to switch the state, and get a lot of A, we have

to stop this repression. So we have to add IPTG. Any questions about what we

mean by the various symbols up on this board? So after we had IPTG, then

indeed the GFP should go down. How long is it going to

take for it to go down? Does this switch go

down immediately? After you add the

IPTG, how long do you think it's going to take

for this repressor, lac I to fall off of that promoter? Do you think it's going

to be seconds or hours? It's actually

seconds, and that's because this IPTG rapidly

can go across the membrane, it'll rapidly bind to

the inhibitor lac I, and then lac I will fall off.

So in this case, lac I is

actually still present, so it's going to take hours,

actually for lac I to go away. But it takes seconds for the

lac I to become inactive, ineffective. So this is the separation

of time scales idea. But how long is it going to

take for the concentration of– well, how long is it going

to take for GFP to go away maybe? Is that going to be

seconds or hours? AUDIENCE: [INAUDIBLE]. PROFESSOR: Yeah, In this

environment the bacterium– they might be dividing

every half hour. That's the characteristic

time– if these are stable, that would be the characteristic

time scale for example, for GFP to go down. And of course, that's

even after it's inhibited. After you stop making GFP. In this case, once you

had the IPTG, the lac I is going to fall

off of this promoter, and then we have

to first make A.

And only after we make A

will we start repressing expression of the B and GFP. That's the process that

you expect to take, hours, based on the generation times. But this is a memory

module because after we've added the IPTG, now we can, in

principle, take the IPTG away, and it'll stay low. So in principle,

this IPTG signal can be a transient signal.

And the cells will remember

that they encountered IPTG. That's what makes this a memory. This input can be transient. Transient signal is remembered. Now a big part of why this

why this paper was important is because this toggle

switch was constructed out of components that previously

in principle, had not even ever seen each other before. So maybe that lac I and this

promoter had been put together, but this whole system,

this whole gene network was composed of

individual components that they were synthetic. And were put together

because they thought, oh this should kind of work. And it's led to, I think

a real flowering of, this intersection between

modeling and experiment. This thing was built based

on a model that told them that maybe if we do

this, these things they multiplarize in order

to repress and so forth.

So there was a real sense in

which the modeling– and then we're going to talk more

about the modeling– on how it was essential to get an

idea of how we should construct this thing, and

to guide our work. Because if any of you

do work in the lab, you'll know that

things are hard. And often components

don't behave the way you think they

should and so forth. Now modeling can't save

you from all that pain, but at least it can guide

you in the right direction so that it limits the number

of things you have to try. Are there any questions about

this element, before we switch over to some of

the dimensionless equations that we're

going to be using? So the reading for

today was composed of a review that

these two pieces, and one of which I think

might be hard for some people, and the other one might

be hard for other people.

For those of you with maybe more

limited experimental experience in biology, maybe the

review was a challenge to try to understand

all of the nomenclature and the words, whereas

those of you that have not played with differential

equations as much recently, may have found the reading

on modeling the toggle switch and stability analysis

to be more challenging. We will, in some

cases, do this where we have two different hopefully

shorter kinds readings. And hopefully they're

not both hard for you, because then you'll spend

a lot of time reading. But then it will

be good for you. So don't run away quite yet. But from my standpoint,

it's essential that you develop intuition

behind these ideas. So once you have this equation

that somebody gives you, you have to be

able to figure out what assumptions have they made,

and what should the behavior be roughly, before you go

off and you do simulations and full mathematical analysis.

So a lot of what we're

going to do in this class, and in particular

during the lectures, is to try to work

on that intuition. And the first thing

you need to do is make sure that you know if

you don't understand something. Sometimes things

look really simple, especially these

dimensionless equations. Part of what's attractive about

them is that they are simpler. But the connection

to experiments can be quite challenging. And we'll kind of

see some of this. So these dimensionless

equations– dimensionless

equations– I think that they're both good and bad. The good is that

you can figure out what are the essential

features of the model. So you can focus your attention

on the essential mathematical features. The disadvantage is

that you might not even know which of the parameters

change when you do something experimentally.

So the problem is

that the connection to the biology or experiments

is obscured in some cases. Connection to experiments. And as an example of

this, what we want to do is look at these equations

for the toggle switch. I know that you guys

just did reading on how to get to these

final pair of equations. It's important to remember that

in the context of the paper, they had a model. Here are the

dimensionless equations that describe our system. And we use them to

design the toggle switch. But just from that,

you don't necessarily realize what's been done. So we want to make

sure we understand it. So the equations that we

want to be comfortable with are the following. So it's u dot du dt. Now here they use alpha

as the rate of expression. So beware, in the

past, we sometimes used alpha as a

rate of degradation. So fair warning. Alpha 1 1 plus that's v beta.

v to the beta minus u v dot is equal to– here is an

alpha 2 divided by 1 plus u to the gamma minus v.

Now in many, many cases, we're going to

get– equations that look like this are going to

pop up time and time again over the rest of this

semester, and you just have to be intimately

familiar with them.

So first of all,

can somebody say why this might be capturing the

dynamics of a toggle switch? I can say something. Yes, please. AUDIENCE: There's some more

of each u or [INAUDIBLE]. PROFESSOR: That's right. So the more v you have,

the less production you're going to have of

u, and vice versa. Now in many cases,

beta and gamma are going to be something

that's larger than 1. This is capturing some

element of cooperativity in the repression on each side. So this thing is indeed just a u

repressing a v, and vice versa. Now these things are

wonderfully simple equations. See that there are four

parameters that are completely specifying the dynamics.

Now you'll notice that

the world is always to be more complicated

than the four parameters. Now there are two ways in which

these complications went away. One is that we are

modeling a simple– it's a simple model

of a complex system. But the other is that

even the simple model has been simplified by going

to this dimensionless version. So I want to make sure that

you understood the reading to the point where at

least you understand what units of concentrations,

times, and so forth are in this model. So first I want to ask about the

effective lifetimes of proteins u and v. Of u verses v. So

we'll maybe call this tau u and tau v. And I want

to know which one– and how these things are

related to each other. Tau u greater than [INAUDIBLE]. DK is again, don't know. Now of course in principle,

the lifetimes of u and v can be anything. The question is, once we've

written down that equation, have we actually said

anything about that? Have we already made

in assumption or not? Do you need more time? I see a fair number of

quizzical faces, which may mean that even with extra

time, the quizzical faces would not go away.

Question. Yes? AUDIENCE: Are you asking

about the lifetime of individual [INAUDIBLE]? PROFESSOR: OK. All right. Yeah. So when I say

effective lifetime, that's because if

it's a stable protein, then the lifetime of

that individual protein is maybe infinite, but you get

an effective lifetime because of this dilution effect. So when I say effective

lifetime in general, in this class, what

I'm referring to is the sum of two effects of

dilution to the cell growth, as well as actual degradation. AUDIENCE: And you're

saying [INAUDIBLE]. PROFESSOR: No. No, I'm saying given that the

Collins Lab– in their paper, wrote down those

set of equations. I'm asking have they

already specified anything about the effective lifetimes

of these two proteins? AUDIENCE: So in those

equations, [INAUDIBLE]. Your question asks in units

of that dimensionless time? PROFESSOR: Yeah, I

mean we can compare two times in some

dimensionless– yeah, we're going to also

talk about that. Maybe I should have

done that first. But this is a nice

question because it's highlighting that you

get a pair of equations, they look obvious, but then

some of those basic things about the system are

some how not quite clear.

So I'm going talk about in

this units of whatever time is being– however

time– we're going to discuss how time is being

measured in a moment as well. But however time

is being measured, is there some

relationship here or not? Let's go ahead and vote. And it's fine, if

you really do not know what I'm talking about,

you can say E and that's fine. Let's see where we are. Ready, three, two, one. So I would say that it's

split between B's and D's. And that's great,

because that means we have something to talk about. There's broad agreement

that it's one of these two. So turn your neighbor. You should be able

to find somebody that disagrees with you. If you can't find anybody

that disagrees with you, you could think

about how parameters are going to change as you

vary other experimental things.

[CLASSROOM CHATTER] PROFESSOR: Why don't we

go ahead and reconvene? I just want to see where we are. So let's get our cards ready. And we're still working on this

one, so you can ignore this. Is it B or D? Ready? Three, two, one. So I'd say there's some

migration towards B, but not 100% still. So I'm going to

side with this here. Can somebody volunteer

why they're saying that? AUDIENCE: Well, if I remember

correctly then they [INAUDIBLE] time by multiplying time

by degradation rate.

And they do that with

the same degradation rate for both [INAUDIBLE]. PROFESSOR: Right. So the answer here is yeah– so

the way that we got to this non dimensional time, that we're

going to discuss in a moment, is multiplying by– or divided

by some degradation rate. And you remember the derivation. You remember that

it was [INAUDIBLE]. But in many cases you don't

get to read the derivation before you have to answer it. In many cases you just

get these equations, and you have to figure out

what the authors have assumed. So I think your answer

is very much correct, but it might be–

I'm glad that you're using the pre-class reading,

but at the same time we have to be able to

answer this question just from these equations.

Because it is contained there. Yeah? AUDIENCE: So if you remove

production [INAUDIBLE]. PROFESSOR: Perfect. AUDIENCE: If you

solve that equation, it's exponential to E,

and the [INAUDIBLE]. PROFESSOR: Yes. That's right. So the statement

here is that it's nice to just imagine that

we shut down production. So these first terms go away. All right now we just have

u dot is equal to minus u. So the concentration of u and

B will both fall exponentially. And they're going to fall the

same rate in whatever units of time– and we'll

discuss this in a moment– but however time

is being measured, they're going to

fall at the same rate because it's whatever

appears in front of these two that determines that rate.

So we've already assumed

that the effective lifetime of u and v are the

same once we've written down these equations. Now that doesn't

have to be true. That means that if

experimentally, you want to make a toggle

switch with two proteins, with different st

abilities, then you can't use these equations. You have to add a delta on

one of the two equations to capture that dynamic that

they have different lifetimes.

So these equations are

simple because we've already combine things, but they've

also already assumed some things to make it look more

simple and more symmetric. So for example,

they're allowing for different effective

cooperativities, beta gamma, for the two repressors. They didn't have to do that

if they didn't want to. If they wanted to,

they could have just had beta in both equations. And that case they

would be assuming that the effective

cooperativity of the repression is the same for the two. So depending on

what you write down, you're making different

assumptions about the system. And so you have to be able

to look at these equations and figure out what

assumptions have been made. Yes, question. AUDIENCE: I'm having

trouble seeing what the first term is doing. So the second one

is [INAUDIBLE]. PROFESSOR: So broadly, the

first term in these equations is the production

rate of the protein. And the second term is some

sort of effective degradation, but it could be due to dilution.

The nice thing about just

ignoring the production term is that it's a way of focusing

only on the effective lifetime portion. This effective lifetime

is it's captured here, irrespective of whether

there's production or not. Of course the

production is going to affect what the equilibrium

is that we go to and so forth, but remember for example, that

this effective lifetime that's set for time scale both to

come up to some equilibrium, as well as come down

to some equilibrium. That's telling us about

this effective lifetime is essential to tell

us about the rate that concentration is going

to change within the cell, whether you're going up or down.

Did that answer your

question a little bit? Yes? AUDIENCE: That

last thing you said is not quite true because– PROFESSOR: Because

of the toggle. I agree. What I was really referring to

there was if we get rid of one, and we're just talking

about where we just manually set the production

rate to one thing or another. The actual dynamics of this are

more complicated, certainly. So is everybody

happy with this idea that just by writing down

those equations, we've made some assumptions constraining

relationships between u and v in some ways, but

not in other ways. Yeah? So we've kind of already danced

around this other question, but I want to make sure

that we address it head on. We want to know what

is this unit of time. If I say oh, at time T

[INAUDIBLE] 1 versus time [INAUDIBLE] to 2– so if delta

t– what is 1 in this case? Are we referring to one second? CGS. Or is it one hour

corresponding to another unit? Cell generation time, effective

lifetime, or don't know? Ten seconds, since we've

already kind of said this, but it's important

enough to make sure.

Ready? No? Yes? I'll give you another ten

seconds, just to make sure. Do you need more time? Let's vote. Ready? Three, two, one. OK. A majority of the group

is agreeing in this case it's the effective lifetime. And it's not the cell

generation time necessarily because if we actually

have a degradation tag on this protein, if it's

not a stable protein then the effective lifetime

is going to be shorter than the cell generation time.

So we've normalized

time by dividing out that– whatever that delta thing

would have been on the right. Now this is great because it

makes the equations simple. But remember what that

means is that if we change the degradation rate,

then all of a sudden it's not obvious

what's going to happen. Experimentally

we're always allowed to affect the degradation rate. But the question is, which

parameter or parameters change if you add

a degradation tag? So if you increase

the degradation rate. That's what I want to do. We'll do that in

just moment there. I have another question

that I wanted to do. Do you guys remember

the equations up there? Maybe I'll come over

to the other side just so that I– it's useful

to be able to stare at those equations as we discuss. So the question is,

what is the unit of concentration of protein u? And of course, these

are dimensionless. What I really mean is

what does u equal to 1 mean in real units? I'll just write that. What does u equal to 1 mean? What does u equal

to 1 correspond to? In something that

is recognizable to an experimentalist in the lab? You guys understand what I mean? So if in this model I

say, u is equal to 1, or u is equal to 10.

What do those numbers–

what do they mean? So are there any questions

about my question first of all? Yes? AUDIENCE: Are we [INAUDIBLE]

solely on seeing those equations? Or based on the

reading of [INAUDIBLE]? PROFESSOR: I would say

that the reading actually has a somewhat more

complicated model, and involves multiple steps. So this is really just

looking at those equations, given that we can

see that there is some statement about how u and

v are repressing each other.

So we're saying that there is

some function that describes– it's a phenomenological

function– describes the input output relationship in terms of

some concentration of u leads to some repression

of v and vice versa. From that actually, we should

be able to say something about what we've already

assumed in these equations. Do you need more time? Yeah, question. AUDIENCE: So the k of u promoter

is– that's defined as the– PROFESSOR: That's the binding of

v to the u promoter– the piece of DNA in front of the u gene. Do we need more time, or

shall we give it a go? Ready? Three, two, one.

All right. So we got some A, B, C, D's. No E's. At least it means that your

neighbor has an opinion. He or she cannot say that– Turn

to your neighbor and discuss. [CLASSROOM CHATTER] Did you guys all decide

you're comfortable? Why don't we go

ahead and reconvene. It seems like it may be we're

coming to consensus here. Ready? Three, two, one. So now we got a clear majority

agreeing that it should be D. So what happened was that over

here in the original equations we had a real concentration

of u divided by some k. And that was the k for

repressing v. Because v is this guy that is over here. And what we did is we then

just turned u divided by this k for binding this v

promoter into just u.

So in particular, when

that concentration of u is equal to its

associated k, that corresponds to half repression. Same thing here. One way to think about this is

just that when u is equal to 1, we're getting half of

maximum possible repression. Similarly, that's what

v equal to 1 means. U and v, do they have to

have the same– I mean, does u equal to 1

and v equal to 1 mean the same thing

in terms of the number of the proteins in the cell? No. Not necessarily. So we've allowed

for the possibility that those things are measured

in different real units. But in both cases,

it's telling us about how the strength

of repression. And u and v equal to 1 tells

us about that crossing point where the other promoter

is half repressed.

Are there any questions

about what happened there? This is especially confusing

because it doesn't enter into the equations at all. Things just went away. But you know that this is

the dimensionless versions of the equations

because we have u here, and we're adding 1 to it. Are we allowed to add

things with different units? No. Never. What that means is that–

since they're being added, that means that we

already know that we made– this is the dimensionless

version of the equation. And it actually then

immediately tells us what u equal to 1 means. So now what we want

to do is we want to make sure that our intuition

on this is tip top shape. In particular, we

want to know if I want to change the

dynamics of the system– so let's say that we go

and we spend lots of time calculating the fixed

point stability, and we do everything

on these equations. And then we know

what we need to do is we need to change

some parameter in order to get say, a toggle switch.

We need to know how we do that. We need to know how the

parameters in real life, or even in the context

of the model, how is it that the parameters you can

actually change experimentally, how is it that the affect or not

the parameters in this model? So the question is, let's say

we increase the degradation rate of these two

transcription factors u and v, which of the parameters

are going to change? So I'll give you 30

seconds to think about what should be happening here. Do you need more time? Let's see where we are. Ready? Three, two, one. And this is the

possibility where you can put up two things. Remember our

fabulous card system? So I think most people are

saying it's going to be A and B are going to change.

So beta and gamma are capturing

how cooperative that transition is. And the cooperativity

is not affected by the questions of the exact

concentration, or time scale and so forth. Because that has to do

with the molecular nature of the interactions

with the promoter. So in some ways beta gamma

are the simplest things in this system. Now the question is, do

alpha 1– do they go up or do they go down? Now that I've told

you that they change. If degradation rate goes

up, alpha 1 and alpha 2, do they go up or

do they go down? I'll give you 10 seconds

to think about it. Do you need more time? Let's see it. Ready? Three, two, one. It's a majority are

saying it's going go down. Can somebody offer up

an intuitive explanation for why this might be? [INAUDIBLE] AUDIENCE: The

degradation rate goes up, it means the times

scale goes down. As you produce a fixed

number of per unit time, the unit time gets

smaller and produces less. PROFESSOR: Right, so if the

degradation rate goes up, that's kind of reducing

this unit of time. So you just are not going

to make as much protein in that unit of time.

The way that I like

to think about this maybe is that if the

degradation rate goes up, then the real concentration

of the protein should go down. And that means

that the repression should be less effective. And then wait, is

this helping me? No, now that I'm saying

this– I don't like to think about it that way. [INTERPOSING VOICES] That's right. Yeah, so– that's right. You should decrease

the sort of– So if the degradation rate

goes up, you decrease the real concentration,

which it means indeed that you are at steady state. Say a less effective repressor. That means the

concentration goes down in these units of how

repressive are you. I think the way of thinking

about it was correct, just the words were not. You can also then go– and it's

useful in the context of when you actually go and you do

the math of removing all these parameters, just to

make sure that– because it's easy to do this thing where

you just divide everything out, and you're happy, and

whistling and so forth.

But at the end of the

day, you really just have no sense of what happened. Of how the parameters

that come out of the model are affected by the real

things that you can change. And in some ways what's

funny about these equations is that essentially

everything is in the alphas. Because beta and gamma, that's

this cooperativity parameter. That's what it is. That's all it is. So that means that everything

else ends up in the alphas. So the strength of the

promoter, the concentration you need to repress,

the lifetime, everything rolls up in these alphas. And this is the beauty of

the dimensionless equations. Is it's telling you that you

don't have all these different, separate knobs. You can't change one

and change another, and go into some funny regime. Because it's all rolled into

one fundamental parameter, these alphas. That's telling you that

once you understand how these equations behave,

then you in principle understand everything that could possibly

happen in that simple model.

But that's what's

wonderful about the dimensionless equations. But the problem is

that you sometimes lose track of how it's related

to the real experimental things. So I would say that I very

much like these dimensionless equations. They clarify things for you. But you have to spend

the time to make sure that you understand where

all of reality went.

Because it all ends up in

this mathematical equation, and that simplifies things. You're not floating in a

sea of symbols anymore, but it's really easy to

lose track of the connection to real measurements. And that's why we

want to do modeling so we can make that connection. So do the dimensionless

equations, but make sure that you play with them

a bit so you know what changes when you change what. And we'll actually see a

bit later, some other cases where it's actually quite tricky

to figure out what's happening. On exams I always want to

just ask a equation like this, if this parameter goes off–

and then the TAs always say, it's too hard of an question. I feel like it's like the most

basic thing you would want from an equation like this. Is that if you increase

the strength of this– But it actually is– it's

surprisingly difficult. So maybe this year

I'll convince the TAs that it's an OK question. Are there any questions

about where we are right now? So what I want to do

for the last half hour is talk about

stability analysis.

The first context in

which stability comes up in this class is indeed

in this toggle switch. But it's not– I would say–

the most satisfying application of it in some ways. Just because you end up with

equations that all you can do is plot them. And it's useful to be

able to recapitulate, to understand how the figures

from that paper come about. But at the same time

it's not always– after you find the solution,

you don't feel so happy about it either. Because in terms of complexity

as a function of time, things always start out simple. And then in the course

of the calculations things get complicated.

And the problems

that are fun to solve are the cases where

it comes simple again. This one– it never

quite converges. I do want to talk about

the stability analysis so that you can understand

the calculation that was in the notes. But also so that you can

just get some more intuition about some of the other

problems that we're going to be solving

in the next few weeks. Just to make sure that we're all

talking about the same thing, it's useful to start

by just making sure that in a one

dimensional problem, we understand what

we mean by stability. We're going to be fast. X equals 0 is stable

if and only if what? I'll give you 10 seconds. Ready? Three, two, one. All right.

So this is always–

So there's actually a fair number of answers here. And this is tricky because

the temptation is always to jump into the two

dimensional stability analysis, or the n dimensional

stability analysis. And the thing is that

we have to make sure that we are completely

comfortable with a one– talked about one dimension before you

talk about multiple dimensions, because then everything is lost. I'm not going to have

you guys discuss, the but it's going to

be C. Many people are saying A or other things. There is a context in

which this guy comes in. But let's just make sure. So x equal to 0. Now first of all, is this

thing– is x equals 0, is it always a fixed

point of the system? Yes. So fixed point means that

if you go right there, then in principle you

don't move off it.

So it doesn't say anything

about whether it's stable or unstable. But if x equals zero, then

indeed x dot is equal to 0. So that's a fixed point. But if you have positive x–

in order for that to be stable, you have to have a

negative change in x. So if you talk about the

behavior as a function of time, this function of x here. If you start out above 0,

the definition of stable is if you go a little bit

away, you should come back.

And that means that x

dot has to be negative. So if positive x– and

similarly if x is negative, you want it to be

stable, then you need the x dot to be positive. So this is– A less than 0. Now there is a context in

which a stability condition looks something a little

bit more like this. And can somebody say when

it is that you get something to look– condition around 1? Yes? AUDIENCE: Discreet. PROFESSOR: Yeah, in

discrete maps then indeed the condition for stability

looks something like this.

If you have something that

looks the x of t plus 1, is equal to axt, then

if the condition for x equal to 0 being stable is for

indeed a to be less than 1. Or it's in this case it's really

the magnitude of a being less than 1. Because in that case

you might get hopping. But then the condition

is around the 1 thing, whereas here 0

[INAUDIBLE] indeed. Solution is different– it's

going to go exponentially to– so x is a function

of time is just going to be some x naught

e to the– and is it at, or a is less than 0. It's just at, right? And in this case, we

just have one dimension.

So the eigenvector

[INAUDIBLE] is really just x. There's just one eigenvalue,

and a is indeed the eigenvalue. So the stability for x

equals 0 to be stable is that all– and in

general, for n dimensions, is that you need all the

eigenvalues to be less than 0. And in this case, a is

indeed the eigenvalue. And you need that

to be less than 0.

So if you're not

comfortable with this, or you got something

other than C, then I think it's essential

that you take the time now to go over all of these ideas. First in one dimension. Make sure you're all

comfortable with that. But then to look again

at these two dimensional, high dimensional

stability analyses. Because if these things

you're finding tricky just because it's been

a while since you looked at this stuff, that's fine. But if you don't spend

the time to iron out now, then everything gets

much more painful later. I highly recommend Strogatz's

book on dynamical systems. It's a beautiful

book with just as clear as a textbook could be. So that book is available at

various libraries and reading rooms and so forth. So it's a great reference. Now we want to do is

generalize this idea to think about in n dimensions. So now what we have a vector x.

And there's going to be some

matrix A that– oh, and x dot. So the change in

this vector x is going to be described by

a linear set of equations, so that they can be

specified by some matrix. To determine the

stability, we're going to then look at the

eigenvalues of this matrix. Now in many cases in this

class, what we're going to do is we're going to find that

some set of non-linear equations has a fixed point

somewhere, and then we're going to linearize

around that fixed point to convert into a linear

problem that looks like this. But for now what

we're going to do– the general statement

is for n n dimensions.

You want all the eigenvalues–

lambda I in particular– to be the real part

to be less than 0. And that has to be true

for all the eigenvalues. We're going to be talking

about the conditions in a two dimensional system

where there are in principles and shortcuts,

but it's all the same thing. It's all that you need. The real part of all the

eigenvalues be less than 0 for the fixed point be stable. So for now what we're

going to do is we're just going to imagine

that we have– assume we have two dimensional problem.

This vector x, we're

going to write as x and y. So we can write the

dynamics like this. This is x. This is x dot, y dot. And this is really

the same thing as saying that x dot can be

described by some ax plus by, and y dot is some cx plus dy. And so we want to be as

comfortable as we can with all the

possible things that could happen in these two linear

and differential equations. And particularly we want to

know what the condition for 0, 0 being stable? 0, 0 is stable if and

only if– now there's a rule that you

found in your reading having to do with the

trace and the determinant. So trace is the

sum of these two. Determinant is product

minus the product here.

Now this is not

the kind of thing that you have to memorize. But it's useful to know

that there is a simple rule, because it allows you to quickly

determine whether something could possibly be stable. And you don't have

to memorize it, but you should be able to

figure it back out later. So what we have is the trace of

this thing being less than 0– so I'm going to give

you some options. Now of course, you can

always look at your notes and that is not

going to help you. You're not being graded

on your answers right now. I'm going to encourage you

to try and think about it and see if you can recapitulate

what this thing should be. So it's less than,

greater than– So for example you

can start to think about– you should be able

to write down some equation that you know is stable. And figure out, OK, what

conditions would it satisfy. That's a common, useful trick

to be able to do in life. To dredge these

things out of memory. Do you understand the question? I'm going to give you

30 seconds because it's well worth trying to figure

out what this rule should be.

So from the reading

you know there's some rule about the trace

and the determinant. And indeed they

both have to be true in order– this is an and sign. May be an and sign. Do you need more time? It's OK if you haven't actually

been able to recapitulate this. But it's useful

to think about it. Should we go ahead

and vote just to see? I'm just curious where we are. Ready? Three, two, one. So we have a lots of B's. And can somebody give me

an example of a matrix A that really ought to be stable? Yeah? AUDIENCE: Negative identity. PROFESSOR: Negative

identity, OK. So if the matrix A is

something that's minus 1 here. 0, 0 minus 1. Then x and y are uncoupled.

X decays exponentially,

y decays exponentially. Now we can just directly

say, well, this, the trace is

definitely negative, and the determinant's positive. It's 1 minus 0. So that gets us here. Because I think there are

many situations in life that are like this, where you

know that there's some rule and you can't

remember what it is. You don't need to

actually remember, once you know that

there's the rule, then you can figure out

what it had to have been. This is not a proof. The proof is only a few lines. You can do it, but the point

is that this kind of situation comes up a lot.

It's useful to just

have simple things that you know what

the answer has to be, and then that allows you to

figure out where things were. And indeed what you'll

find is that for a two dimensional system,

this condition that the trace of the

matrix is less than 0 and the determinant

is larger, that is equivalent to the

statement that the real part of both eigenvalues

is less than 0. And there's the derivation is

simple, and it's in your notes. Are there any questions

about where we are right now? So what I want to just

say a few more things about the trajectories here. Can somebody explain

the notion of what's an intuitive statement

about the eigenvectors that you get out here? Why do we like eigenvectors? What are they useful for? AUDIENCE: Decoupling the

differential equations.

PROFESSOR: Right, so

they're decoupling, and there's a sense

that– I guess the way I like to think

about this is that you have a fixed point say, like this. Now if these things

are stable, that means the arrows are

all say coming in. Are these eigenvalues–

are they purely real, or are they complex? Reminder, somebody? Real and negative. There's no imaginary component. This is a stable fixed point

with real negative eigenvalues. Now the idea of

these eigenvectors is that these are

the two directions in which if you start out on one

of them, you'll stay on them.

And in general then you

can– any other trajectory you can decompose as a

combination of the pads on the two. The position as function

of time can always described as the sum

of the eigenvectors where you grow or

you shrink along each eigenvector exponentially. Now for this thing to be

stable, all these lambda I's are going to be negative. We have this property

that– that the dynamics of this matrix kind of keep

you along the direction of the eigenvector. What this is saying is that

if you start somewhere random, that is off one

the eigenvectors, then you can decompose

the trajectory along each. But I just want to

highlight that it's often useful to draw what these

things end up looking like. Now I want to make sure

I get this one correct. I'm a little bit worried I'm

going to do something funny. So here– we're going to say

this is v1 and this is v2. So this direction

is one eigenvector, this direction is the other. So let's imagine that the

trajectories look like this.

The question is which

eigenvalue is closer to 0? Is it A eigenvector v1, or B

eigenvector– So this is really lambda 1 verses lambda 2. 1 is closer to 0 than is 2. Which of the eigenvalues

is closer to 0? I'll give you 20 seconds to

think about this because it's useful to be able to

extract these things from the trajectories. Do you need more time? And it's fine if

you don't really understand how you can get to

this question from this figure, then go ahead and flash C, D, or

E, just so I know where we are. Let's vote. Ready? Three, two, one. All right, so there's

at least a majority are saying that it's

going to be lambda 2. Can somebody offer why that is? Yes? AUDIENCE: I sort of

see it as whichever one is bigger is

going to be more effective at squeezing

things toward the origin along that axis. PROFESSOR: OK and bigger– and

you're saying more negative, in this case, you're saying? Yes, right. So yeah that's right. So there's some notion

that lambda 1 has to be more negative

than lambda 2 because you first collapse along

the direction of eigenvector 1, and then you slowly come

in along the direction of eigenvector 2.

That's saying that you have

these two exponential decays, and the directional

on the eigenvector 1 is more rapid than the

directional on eigenvector 2. I think in many of these

cases in dynamical systems, differential equations,

it's hugely valuable to be able to draw

the trajectories. So in many cases,

what we're going to do over the course

of the semester is we're going to have a simple

pair of differential equations that are going to

be two proteins, or they're going to be

rabbits and foxes or whatever. So we're going to locate

where the fixed points are. And then we're

going to figure out the stabilities and

the eigenvectors. And then we can really

understand the entire dynamics of the system without

solving the full thing, without using a computer.

But just by figuring out

where the fixed points are, and then the dynamics

around there. Then you can

basically understand all the dynamics of the system. But you have to

make sure that you develop intuition of how systems

behave near their fixed points. Now this is a case where both

of the eigenvalues were real. But of course, if you

have complex eigenvalues, if you do the

calculations, you'll see that the

eigenvalues are going to be a complex

conjugates of each other. And there you get spirals. So trajectories

might look like so. So there are two qualitatively

different ways that the fixed point can be stable.

You can have spiraling through

a state of the fixed point, or you can come in via

these straight lines. Of course, the specific

trajectories in some cases, can be curved. And so if you look at a

particular trajectory, sometimes even these can

look a little bit spirally, so be careful. Those are the two

basic ways it works. There's a simple way

of getting a sense of the dynamics of a system,

as a function of the trace and the determinant. So what we have–

something here. If you go ahead and you do

the calculation, what you find is that the two eigenvalues are

going to be described by this. And then the basic dynamics

are going to be the following. So down here, you have a case

where the eigenvalues are real, and they're of opposite sign.

And in this case, is

that stable or unstable? Unstable. So in order for it to be

stable, all the eigenvalues have to have negative

real components. So if they have opposite

sign, one of them is going to be positive. So this is all

unstable down here. Over here you have a case

where the eigenvalues are real, greater than 0. So in this case,

the trajectories are also coming out somehow. All right, here, this is the

case where they are real, but now both less than 0. So this is again, this is

like what we drew here, where everything is stable coming in. And up here is where

we get the spirals. But over here, it's the

real part is less than 0. So this is where we have

the spirals coming in, whereas over here, the

real part is greater than 0 and we have spirals coming out. Oh, I'm sorry. Uh, that would have been useful. So this is in the trace of A

and this is in the determinant.

So these are the variety

of possible outcomes when you have a two

dimensional system that's already a linear two

dimensional system. And this is indeed

consistent with it, to have a stable

fixed point, you need to have the

trace less than 0, and the determinant

greater than 0. So it's in this quadrant. Now it's– finally it's useful

to– so let's just for now, stick with the linear system. And just imagine a case

where we have in this A– so remember it's A,

B, C, and D. So we saw one way in which this

thing could be stable. Was that if b and

c were both equal to 0, so there's no

cross interaction, but a and d were both

less than 0, then it's all trivially stable. Now the question

is, if something that looks like this, a

is equal to minus 2 and d is equal to plus 1. Questions is, could

such a system be stable? So I'm seeing some nods. And on the face of this,

you say, oh, that's a little bit surprising.

Because d being plus

1, what that's saying is that y on its

own is unstable. So if you start out

with no x and no y, and you add a little bit of y,

y starts growing exponentially. So y on its own is unstable,

but x is stable its own. And the trace

being less than 0– that's saying that there's

some sense in which if y is unstable on its own,

then x has to somehow be more stable than y is unstable. Because the sum of those things

still has to be negative. Now that was necessary,

but not sufficient, in order to have stability. Because we also need to have a

condition on the determinant. Now so the trace of a

here– that's minus 1. That's less than 0. OK, that's great. Now the determinant– now

it's going to be a times d. That's minus 2. But then we also have to

say minus b times c, right? And this thing has

to be greater than 0.

So what you see here

is that b and c– they have to have opposite signs

order for this thing to work, in order for the

origin to be stable. And the product has

to somehow be strong. Now this makes sense, because

of course, if b and c, or even for matter,

just one of them were 0, then it would be impossible

to get the stability. But for example, if we

have some situation where we have some x that's inhibiting

itself– that's a here– but then y is activating itself. So y here is somehow

on its own, unstable. Then what you need is you

need maybe something that looks like this. Some cross activation

and or repression. And what's interesting is

that it doesn't matter which of the two, b or c is negative.

You can get actually, the origin

to be stable in either way. But in this situation, you need

the direction of the regulation to be in opposite directions. And if you'd like,

you could then play with– you could

think for example about the directions of these

trajectories around the origin, and so forth. But it's useful, I think, to

play with the simplest toy systems that you can imagine,

just so that you can get a sense of what are the

basic ingredients that you need in order to get stability

in something like this.

Because once you start doing

the whole linearization around the fixed point,

and then calculating traces and determinants,

you're not going to have it. You're going to lose all

your intuition about that about things at that stage. So it's useful to make

sure that you nail things down in this context. We are out of time,

so I'll let you go..