Hello. Now we're at the second part of our lectures on bistability and biochemical signaling models. Last time we talked about bistability in general, we introduced a concept and we talked some of the qualitative requirements for bistability. Now, we want to get more quantitative. In particular, we want to predict, we want to, and we want to ask the question, how can one predict whether bistability will actually be present? And to get start to address this question, we're going to consider a simple, one dimensional example. We're going to introduce a concept of rate-balance plot, which is very helpful for understanding these one dimensional examples. And this is going to lead to a conclusion that ultrasensitive positive feedback can create bistability. Some of these terms may not be clear right now, but they'll be defined as we go through this. To begin to analyze in a quantitative way whether or not bistability is present, let's consider a simple Michaelian system. Remember the Michaelis–Menten equation we talked about earlier. That's why we call this a Michaelian system. When you, you have a protein a that can either be exist in either this state over here on the left which is the unphosphorylated state or this state over here a star which is the phosphorylated form of A. And let say a moves from unphosphorylated to phosphorylated with some [UNKNOWN]. That we call k+, and then moved backwards with some rate constant we call k-. And if we don't include production of synthesis of A, or degradation of A, in this system, then the total amount of A is constant. In other words, the unphosphorylated plus the phosphorylated is going to be equal to some constant number that we call A total. And we want to solve for what is A* become in this steady state, and in this case we can write down a differential equation that looks like this, and in our previous lectures of dynamical systems we talked about how the law of mass action allows us to write down this particular differential equation. And in a study state that means that there is no change in A*with respect to time so we can set this equal to 0, and using some simple algebra we can solve for A* as a function of A total and a very constant k plus and k minus. Or, we can rearrange this to say A star is a fraction of A total so now A* is going to be a number that goes between 0 and 1 equals this expression right here. But we don't want to solve these systems by manipulating equations and doing a lot of algebra. We want to get a more intuitive understanding of how these systems work. So instead of solving equations, let's find these solutions graphically. Rather than writing down a complete differential equation like we did before, we can take this same system, and instead we can write down forward rate, which is the forward rate constant k plus, times A total minus A* so this a total minus A* is equal to our concentration of unphosphorylated A. And then we can write our backward rate as equal to the negative the rate constant that moves us from right to left, k minus, times our concentration of A*. So we just took our differential equation d A*dt and we broke it down into, what's going to make A* increase? That's our forward rate. And, what's going to make A* decrease? That's our backwards rate. Why did we do that? Well in this case it's because we can very easily plot what the rates are as a function of A*. And here again, we've normalized A*. To the total amount of A. So that A* will always go between 0 and 1. So if we look at these 2 equations, and we say. What do we, what do they look like when they're, when they're graphed? Well, they're both linear, right? In both cases, you have, the forward rate depends linearly on A*. And backward rate depends linearly on A*, but we notice that as A* gets bigger, the backward rate gets bigger, and the forward rate gets smaller. So, one of these decreases and one of these increases and I should know that this analysis here we are going to show in the next several slides is based closely. In a very nice review article written by Ferrell and Xiong, that was published in the journal, Chaos, in the year 2001. One other thing we note about this, is that when we reach the steady state, that occurs when the forward rate is equal to the backward rate. So, if we look at where these 2 curves intersect. Where the backward rate curve intersects the forward rate curve. This tells us where we're going to be at steady state. This is where the forward rate and the backward rate are balanced from one another. And this concept of steady state is going to be important in the in the analysis we're going to show later. If these rate balance plots look familiar to you they should. That's because these rate balance plots are very closely related to what we showed in some of the lectures on, on dynamical systems that are called phase line plots. In those examples, we plotted a derivative on the y axis, in that case it was a [UNKNOWN] of a voltage with respect to time, and then on the x axis we plotted that variable itself. We plotted voltage on the x axis. And a derivative of voltage on the y axis. We could do the same kind of thing here, again remember this is our scheme the derivative of A* with respect to time, is equal to the forward ray minus the backward ray and what we discussed before in the last slide was forward ray minus the backward ray. is equal to 0. So if we plotted this, we'd plot for rate, forward rate minus backward rate, we would get a curve that looked like this, where at very low levels of of A the forward rate exceeds the backward rate. And therefore the derivative is positive. At very high levels of A*, the backward rate exceeds the forward rate, and therefore the derivative is negative. Why do we plot it in, in this way? Well, remember how we analyzed the phase line plots before. This is our steady state, where the forward rate minus the backward rate is equal to 0. So where your derivative line crosses your variable. That's where you have a By definition that's where you have a steady state. And the reason it's helpful to plot things this way is you can then analyze what happens when you deviate from that steady state. So, what, what happens if we had a, a sudden decrease from a steady state level to a lower level here? Well, at this level, the derivative is positive the production exceed the the degradation. The for grade exceeds the backward rate. Therefore we're going to have an increase in a [UNKNOWN] and it's going to push us back towards a steady state. Conversely if we go, go from a steady state to some level up here, here the overall derivative is negative. The backward rate exceeds the forward rate, and it's going to push us back to the left. Again, back to the stay state. So, we can analyze this system and say, intuitively, we can look at this and determine that the fixed point in this case is going to be stable. [BLANK_AUDIO] What we've seen so far is all relatively simple. We're going to have a constant forward rate, rate constant k plus. And a constant backward rate constant, k minus. Now let's make it a little more interesting. Lets assume that the forward rate is a function of some stimulus. Where k plus is not just a constant, but it's constant, which we're going to call k, but with the symbol plus, rather than spelled out plus, times the value of some stimulus, S. So, for low values of the stimulus, we don't have very much production of A*. We have relatively low rate, but then for higher values of the stimulus up here, we have much greater rates. Now we can look at where the small stimulus concentration, where the small stimulus curve intersects with the backward rate, and then the medium values of the stimulus intersects with the backward rate, and there where the high value of stimulus intersects with the backward rate. By doing this, by plotting forward rate for many different values of the stimulus, we can use these curves to then derive how stimulus strength is going to affect your state state value of A*. Normalized to a total. What I mean by this is, consider the family of curves here. For one value of stimulus, you have one steady state. For another value, you get a different steady state. For yet another value, you get a different steady state. You can collect all these points, for all your different values of stimulus. And then what you can derive from this is A* normalized A total on the y axis, this is a variable that starts at zero with for the lowest values of stimulus and then it goes up to 1 and you can see that this value, this your output here A* normalized A total is going to increase. And eventually it's going to saturate at a value of one meaning that 100% of your your A, 100% of that is, is phosphorylated, and this, the shape of this curve will probably look familiar to you. This is a hyperbola, which is analogous to what you get with a Michaelis-Menten equation, and that's why we initially refer to this is a simple Michaelian system. [SOUND] Now let's make it a little more complicated. Now let's continue to make, make a Michaelian system but let's include linear feedback. What do I mean by that. What I mean is that a 4 rate constant that rate constant determines how much of your unphosphorylated come how quickly your unphosphorylated A gets converted in to a phosphorylated A, A* Depends not just on the stimulus, like it did before. But it also depends on A* itself. So when A* goes up, you see that's what this arrow represents here. When A* gets bigger, it's going to make this reversion rate bigger. And we can express this feedback with this term we call kf. Mathematically that's how we would express this when we have linear feedback. We have K+ times stimulus plus some term kf for feedback times the amount of A* that we have, and again this term here, A total minus A*, is how we calculate our unphosphorylated A here. So this is our substrate concentration, and this whole term is our overall rate constant. What do we see when we plot our rate balance plots for this Michaelian system with linear feedback? For very weak feedback, we can see something that looks like this, where the backward rate, again, is shown in red, and the forward rate is shown in blue. So if your feedback is relatively weak. Then your forward rate may never exceed your backward rate, and it will always fall below the backward rate constant. But if you have stronger feedback, then you can get a curve that looks like this, where the backward rate, and the forward rate intersect one another at at 0, and then they also intersect one another. Up here now we see something more complicated than what we saw, then what we saw before, in our previous examples the forward rate and backward rate always intersected at one point, so we could determine what the unique steady state was because there was only one point where they intersected. Now we have a case where the forward rate, and the backward rate intersect at 2 points. We've been talking about bistability. So you see 2 steady states in this case, 2 locations where the 2, the curves intersect. So what we have on the right looks like a bistable system, right? Maybe we have one steady here and one steady state here. Is this system bistable? In fact it's not. This is, this is not yet a true bistable system, and we'll explain why next. What I just said is that this plot that we see down here is not bistable, where we have a strong linear feedback of a Michaelian system. Our forward rate constant here becomes a parabola, and our backward rate constant remains a line. And these intersect in 2 places. To see why this isn't truly bistable, let's assume that we're starting off at A* equals 0, where we don't have any of the. Substrate phosphorylated. And let's consider what happens with a tiny deviation from A* equals zero. In other words what if there was one spontaneous phosphorylation reaction? Well if we start at 0 and we move to some place a little bit greater than 0, what's going to happen? Or forward rate in this case is going to exceed the backward rate. And remember the overall rate is the forward rate minus the backward rate. So if the forward rate is greater than the backward rate, the overall rate is positive. And if the overall rate is positive, which way is the system going to go? It's going to go to the right. Since the forward rate here is greater than the backward rate, it's going to move to some even higher value of, of A*. If it moves to some higher value of A*, again the forward rate is greater than the backward rate. Here again, the forward rate is greater than the backward rate. So, that leads to a further increase in A and that's going to lead to yet another increase in A. Therefore, we conclude that this, the steady state and A* equal 0 is unstable and that's why this [UNKNOWN] is marked with an open circle here. Indicating that this is an unstable steady state. Now the question becomes how can we make the off state stable. We've already determined that this case here where the forward, where the you have a deviation in the forward rate exceeds the backward rate. Is an unstable steady state. How can I make this particular steady state stable? There are two ways this can be modified to make this a truly bistable system. One, is what we call non-linear or ultra-sensitive feedback, rather than linear feedback as we saw before. And the second is if we have partial saturation of the back reaction. Now let's consider what we mean when we talk about ultra sensitive feedback, rather than linear feedback, like we saw before. Again, this is our general scheme, where the forward rate constant, the rate at which unphosphorylated A gets converted into phosphorylated A, or A*, depends on. The stimulus strength k plus, it also depends on A* itself, but if you look at the equation, you see an important difference here. Now, instead of just depending on A* raised to the first power, and depend, on our feedback constant kf times A star raise to the nth power, and we've also normalized this A* to the nth power plus some Km raised to the nth power to indicate that this this reaches a limit. This saturates as A* goes up. Now if we graph this particular curve, we see something that looks like this. Our forward rate, instead of being a, a parabola. It get's more complicated now. The beginning part of this is sigmodal and then it curves back and comes back down to 0, when we, when all of our A is phosphoralated. What do we have now, now we have one intersection of the forward rate and the backward rate. A second intersection. And a third intersection of the forward rate and the backward rate. Now, we really have a bonafide bistable system. Moreover, we can look at these 3 steady states. And by doing this rate balance plot analysis, we can intuitively understand that this uppermost one is stable. This lowermost one is stable and this middle one is unstable. Let's think about how that works. What happens if we have a deviation from this rightmost steady state over here? If we deviate to the left, forward rate exceeds the backward rate, therefore the incre-, that we have an increase in A* and moves us back towards the steady state. Conversely, if we increase from here, backward rate exceeds the forward rate. That means the overall rate, is negative and that pushes us back to our sustained state. Similarly, we can analyze this steady state, here, in the middle and see, if we deviate to the right, it's going to continue to push us to the right until we get here. If we deviate to the left, it's going to continue to push this down until we're here. So now we have a bona fide stable system where this lower-most one, 0, is stable, this upper most steady state is stable, and the steady state in the middle is unstable. The final thing we want to look at here is this exponent n. What happens if we change this exponent n? The previous example we just plotted was, I believe, for n = 4. But what you can see if you plot. This curve here, this forward rate for different values of n, is that the greater n gets, the more curvature you see in this initial part of the curve here. So for n equals 2, you see this very very subtle change. It almost looks like a parameter. sorry, it almost looks like a parabola in this case. But it deviates a little bit. the the beginning here. And that allows the black curve to be a little bit lower than the red curve and then to cross the red curve. But for higher values of this exponent n, you see much more curvature at the beginning here and then for this, for a very high value here for n equals 8 you see. Very steep behavior here, where there's almost no change for very low values of A* and you see this steep increase for medium values of A*, before it reaches a beacon and turns around. So, what we can conclude from this is when you have a larger hill exponent, when this value N increases, that makes bistability more likely and also more robust. In other words, for the black curve here. This one is technically bistable, but it's barely bistable, because you, you have these two intersections, these 2 steady states that are very close to one another. As you increase your value of n, your stable steady state and your unstable steady state become farther away from one another. Now to summarize. This second lecture on bistability. Rate-balance plots are very useful for assessing whether bistability may occur in one-variable systems. These are useful because they taught the relationship between the forward rate and the backward rate. And when you look at these rate-balance plots, you can intuitively understand. Is the system going to move to the right? Meaning we're going to increase my value of my my variable? Or is it going to move to the left? Meaning my variable is going to decrease. The second important point that we've learned here is that ultrasensitive positive feedback can produce bistability in a one-variable system. In subsequent lectures, we're going to look at other ways you can get bistability, and then we're going to also look at what happens in bistability in multivariable systems.