0:20

Last time we talked about bistability in general, we introduced a

Â concept and we talked some of the qualitative requirements for bistability.

Â Now, we want to get more quantitative.

Â In particular, we want to predict, we want to, and we want to ask

Â the question, how can one predict whether bistability will actually be present?

Â And to get start to address this question,

Â we're going to consider a simple, one dimensional example.

Â 0:43

We're going to introduce a concept of rate-balance plot,

Â which is very helpful for understanding these one dimensional examples.

Â And this is going to lead to a

Â conclusion that ultrasensitive positive feedback can create bistability.

Â Some of these terms may not be clear right

Â now, but they'll be defined as we go through this.

Â 1:11

Remember the Michaelisâ€“Menten equation we talked about earlier.

Â That's why we call this a Michaelian system.

Â When you, you have a protein a that can either

Â be exist in either this state over here on the

Â left which is the unphosphorylated state or this state over

Â here a star which is the phosphorylated form of A.

Â And let say a moves from unphosphorylated to phosphorylated with some [UNKNOWN].

Â That we call k+, and then moved backwards with some rate constant we call k-.

Â And

Â 1:40

if we don't include production of synthesis of A, or degradation of

Â A, in this system, then the total amount of A is constant.

Â In other words, the unphosphorylated plus the phosphorylated is going to

Â be equal to some constant number that we call A total.

Â 1:57

And we want to solve for what is A* become in this steady

Â state, and in this case we can write down a differential equation that looks

Â like this, and in our previous lectures of dynamical systems we talked about how

Â the law of mass action allows us

Â to write down this particular differential equation.

Â 2:14

And in a study state that means that there is no change in A*with respect to time so

Â we can set this equal to 0, and using some simple algebra we can solve for A* as

Â a function of A total and a very constant k plus and k minus.

Â Or, we can rearrange this to say A star is a fraction of A total so now

Â A* is going to be a number that goes

Â between 0 and 1 equals this expression right here.

Â 3:06

which is the forward rate constant k plus, times A total minus

Â A* so this a total minus A* is equal to our concentration of unphosphorylated A.

Â And then we can write our backward rate as equal to the negative the rate

Â constant that moves us from right to left, k minus, times our concentration of A*.

Â 3:41

Well in this case it's because we can very easily

Â plot what the rates are as a function of A*.

Â And here again, we've normalized A*.

Â To the total amount of A.

Â So that A* will always go between 0 and 1.

Â So if we look at these 2 equations, and we say.

Â What do we, what do they look like when they're, when they're graphed?

Â Well, they're both linear, right?

Â In both cases, you have, the forward rate depends linearly on A*.

Â And backward rate depends linearly on A*, but we notice that as A*

Â gets bigger, the backward rate gets bigger, and the forward rate gets smaller.

Â So, one of these decreases and one of these increases and I should know that

Â this analysis here we are going to show in the next several slides is based closely.

Â In a very nice review article written by Ferrell and Xiong,

Â that was published in the journal, Chaos, in the year 2001.

Â One other thing we note about this, is that when we reach the

Â steady state, that occurs when the forward rate is equal to the backward rate.

Â So, if we look at where these 2 curves intersect.

Â Where the backward rate curve intersects the forward rate curve.

Â This tells us where we're going to be at steady state.

Â This is where the forward rate and

Â the backward rate are balanced from one another.

Â And this concept of steady state is going to be

Â important in the in the analysis we're going to show later.

Â 5:17

In those examples, we plotted a derivative on the

Â y axis, in that case it was a [UNKNOWN]

Â of a voltage with respect to time, and then

Â on the x axis we plotted that variable itself.

Â We plotted voltage on the x axis.

Â And a derivative of voltage on the y axis.

Â We could do the same kind of thing here, again remember this

Â is our scheme the derivative of A* with respect to time, is

Â equal to the forward ray minus the backward ray and what we

Â discussed before in the last slide was forward ray minus the backward ray.

Â is equal to 0.

Â So if we plotted this, we'd plot for rate,

Â forward rate minus backward rate, we would get a curve

Â that looked like this, where at very low levels

Â of of A the forward rate exceeds the backward rate.

Â And therefore the derivative is positive.

Â At very high levels of A*, the backward rate

Â exceeds the forward rate, and therefore the derivative is negative.

Â 6:13

Why do we plot it in, in this way?

Â Well, remember how we analyzed the phase line plots before.

Â This is our steady state, where the forward

Â rate minus the backward rate is equal to 0.

Â So where your derivative line crosses your variable.

Â That's where you have a By definition that's where you have a steady state.

Â And the reason it's helpful to plot things this way is you

Â can then analyze what happens when you deviate from that steady state.

Â So, what, what happens if we had a, a sudden

Â decrease from a steady state level to a lower level here?

Â 6:44

Well, at this level, the derivative is

Â positive the production exceed the the degradation.

Â The for grade exceeds the backward rate.

Â Therefore we're going to have an increase in a [UNKNOWN]

Â and it's going to push us back towards a steady state.

Â Conversely if we go, go from a steady state to

Â some level up here, here the overall derivative is negative.

Â The backward rate exceeds the forward rate, and

Â it's going to push us back to the left.

Â Again, back to the stay state.

Â So, we can analyze this system and say, intuitively, we can look at this

Â and determine that the fixed point in this case is going to be stable.

Â [BLANK_AUDIO]

Â What we've seen so far is all relatively simple.

Â We're going to have a constant forward rate, rate constant k plus.

Â And a constant backward rate constant, k minus.

Â Now let's make it a little more interesting.

Â Lets assume that the forward rate is a function of some stimulus.

Â Where k plus is not just a constant,

Â but it's constant, which we're going to call k,

Â but with the symbol plus, rather than spelled

Â out plus, times the value of some stimulus, S.

Â So, for low values of the stimulus, we don't have very much production of A*.

Â We have relatively low rate, but then for higher values

Â of the stimulus up here, we have much greater rates.

Â Now we can look at where the small stimulus concentration, where the small

Â stimulus curve intersects with the backward rate, and then the

Â medium values of the stimulus intersects with the backward rate, and

Â there where the high value of stimulus intersects with the backward rate.

Â 8:28

By doing this, by plotting forward rate for

Â many different values of the stimulus, we can use

Â these curves to then derive how stimulus strength is

Â going to affect your state state value of A*.

Â Normalized to a total.

Â What I mean by this is, consider the family of curves here.

Â For one value of stimulus, you have one steady state.

Â For another value, you get a different steady state.

Â For yet another value, you get a different steady state.

Â You can collect all these points, for all your different values of stimulus.

Â And then what you can derive from this is A* normalized A total on

Â the y axis, this is a variable that starts at zero with for the lowest

Â values of stimulus and then it goes up to 1 and you can see

Â that this value, this your output here A* normalized A total is going to increase.

Â And eventually it's going to saturate at a value

Â of one meaning that 100% of your your A,

Â 100% of that is, is phosphorylated, and this, the

Â shape of this curve will probably look familiar to you.

Â This is a hyperbola, which is analogous to what you get with a Michaelis-Menten

Â equation, and that's why we initially refer

Â to this is a simple Michaelian system.

Â [SOUND]

Â Now let's make it a little more complicated.

Â Now let's continue to make, make a

Â Michaelian system but let's include linear feedback.

Â What do I mean by that.

Â What I mean is that a 4 rate constant that

Â rate constant determines how much of your unphosphorylated come how quickly

Â your unphosphorylated A gets converted in to a phosphorylated A,

Â A* Depends not just on the stimulus, like it did before.

Â But it also depends on A* itself.

Â So when A* goes up, you see that's what this arrow represents here.

Â When A* gets bigger, it's going to make this reversion rate bigger.

Â And we can express this feedback with this term we call kf.

Â 10:20

Mathematically that's how we would express this when we have linear feedback.

Â We have K+ times stimulus plus some term kf

Â for feedback times the amount of A* that we

Â have, and again this term here, A total minus

Â A*, is how we calculate our unphosphorylated A here.

Â So this is our substrate concentration, and

Â this whole term is our overall rate constant.

Â 10:47

What do we see when we plot our rate

Â balance plots for this Michaelian system with linear feedback?

Â For very weak feedback, we can see something that looks like this, where the

Â backward rate, again, is shown in red, and the forward rate is shown in blue.

Â So if your feedback is relatively weak.

Â Then your forward rate may never exceed your backward rate,

Â and it will always fall below the backward rate constant.

Â 11:11

But if you have stronger feedback, then you can

Â get a curve that looks like this, where the

Â backward rate, and the forward rate intersect one another

Â at at 0, and then they also intersect one another.

Â Up here now we see something more complicated than what we saw,

Â then what we saw before, in our previous examples the forward rate and

Â backward rate always intersected at one point, so we could determine what the

Â unique steady state was because there was only one point where they intersected.

Â 11:36

Now we have a case where the forward

Â rate, and the backward rate intersect at 2 points.

Â We've been talking about bistability.

Â So you see 2 steady states in this

Â case, 2 locations where the 2, the curves intersect.

Â So what we have on the right looks like a bistable system, right?

Â Maybe we have one steady here and one steady state here.

Â Is this system bistable?

Â 12:51

Well if we start at 0 and we move to some

Â place a little bit greater than 0, what's going to happen?

Â Or forward rate in this case is going to exceed the backward rate.

Â And remember the overall rate is the forward rate minus the backward rate.

Â So if the forward rate is greater than

Â the backward rate, the overall rate is positive.

Â And if the overall rate is positive, which way is the system going to go?

Â It's going to go to the right.

Â Since the forward rate here is greater than the backward rate,

Â it's going to move to some even higher value of, of A*.

Â If it moves to some higher value of A*,

Â again the forward rate is greater than the backward rate.

Â Here again, the forward rate is greater than the backward rate.

Â 13:33

and that's going to lead to yet another increase in A.

Â Therefore, we conclude that this, the steady state and A* equal 0 is

Â unstable and that's why this [UNKNOWN] is marked with an open circle here.

Â Indicating that this is an unstable steady state.

Â 14:09

How can I make this particular steady state stable?

Â There are two ways this can be modified to make this a truly bistable system.

Â One, is what we call non-linear or ultra-sensitive

Â feedback, rather than linear feedback as we saw before.

Â And the second is if we have partial saturation of the back reaction.

Â 14:38

Again, this is our general scheme, where the forward rate constant, the rate at

Â which unphosphorylated A gets converted into phosphorylated A, or A*, depends on.

Â The stimulus strength k plus, it also depends on A* itself, but

Â if you look at the equation, you see an important difference here.

Â Now, instead of just depending on A* raised to the first power, and

Â depend, on our feedback constant kf times A star raise to the nth

Â power, and we've also normalized this A* to the nth power plus some

Â Km raised to the nth power to indicate that this this reaches a limit.

Â This saturates as A* goes up.

Â Now if we graph this particular curve, we see something that looks like this.

Â Our forward rate, instead of being a, a parabola.

Â It get's more complicated now.

Â The beginning part of this is sigmodal and then it curves back and

Â comes back down to 0, when we, when all of our A is phosphoralated.

Â What do we have now, now we have one

Â intersection of the forward rate and the backward rate.

Â A second intersection.

Â And a third intersection of the forward rate and the backward rate.

Â Now, we really have a bonafide bistable system.

Â 15:50

Moreover, we can look at these 3 steady states.

Â And by doing this rate balance plot analysis, we

Â can intuitively understand that this uppermost one is stable.

Â This lowermost one is stable and this middle one is unstable.

Â Let's think about how that works.

Â What happens if we have a deviation from this rightmost steady state over here?

Â If we deviate to the left, forward

Â rate exceeds the backward rate, therefore the incre-,

Â that we have an increase in A* and moves us back towards the steady state.

Â Conversely, if we increase from here, backward rate exceeds the forward rate.

Â That means the overall rate, is negative and

Â that pushes us back to our sustained state.

Â 16:34

Similarly, we can analyze this steady state, here,

Â in the middle and see, if we deviate

Â to the right, it's going to continue to push us to the right until we get here.

Â If we deviate to the left, it's going to

Â continue to push this down until we're here.

Â 17:08

What happens if we change this exponent n?

Â The previous example we just plotted was, I believe, for n = 4.

Â But what you can see if you plot.

Â This curve here, this forward rate for different values of n, is that the greater

Â n gets, the more curvature you see in this initial part of the curve here.

Â So for n equals 2, you see this very very subtle change.

Â It almost looks like a parameter.

Â sorry, it almost looks like a parabola in this case.

Â But it deviates a little bit.

Â the the beginning here.

Â And that allows the black curve to be a little bit

Â lower than the red curve and then to cross the red curve.

Â But for higher values of this exponent n,

Â you see much more curvature at the beginning

Â here and then for this, for a very high value here for n equals 8 you see.

Â Very steep behavior here, where there's almost no

Â change for very low values of A* and you

Â see this steep increase for medium values of

Â A*, before it reaches a beacon and turns around.

Â 18:03

So, what we can conclude from this is when you have a larger hill

Â exponent, when this value N increases, that

Â makes bistability more likely and also more robust.

Â In other words, for the black curve here.

Â This one is technically bistable, but it's barely bistable, because you, you have

Â these two intersections, these 2 steady states that are very close to one another.

Â As you increase your value of n, your stable steady state

Â and your unstable steady state become farther away from one another.

Â 18:34

Now to summarize.

Â This second lecture on bistability.

Â Rate-balance plots are very useful for assessing

Â whether bistability may occur in one-variable systems.

Â These are useful because they taught the relationship

Â between the forward rate and the backward rate.

Â And when you look at these rate-balance plots, you can intuitively understand.

Â Is the system going to move to the right?

Â Meaning we're going to increase my value of my my variable?

Â Or is it going to move to the left?

Â Meaning my variable is going to decrease.

Â