Steady-State Conditions - Online Tutor, Practice Problems & Exam Prep

In enzyme-catalyzed reactions, the steady state occurs when the concentration of the enzyme-substrate complex remains constant, indicating that the rate of formation equals the rate of dissociation. This is expressed as $v=v+v-+v2$. The Michaelis constant (K_m) can be derived under these conditions, highlighting its significance in enzyme kinetics. Understanding these concepts is crucial for grasping enzyme behavior and kinetics in biochemical processes.

Steady State Conditions

In this video, we're going to begin our discussion on the steady state conditions. So what's really important for you guys to know is that during a typical enzyme-catalyzed reaction, the concentration of the enzyme-substrate complex quickly builds up and reaches a constant value or a stable value. And so when the concentration of the enzyme-substrate complex reaches this constant or the stable value, we refer to this as steady state. And so steady state is an incredibly important assumption that biochemists make when they're studying enzyme-catalyzed reactions. As we move forward in our course, we're going to talk more details about exactly what happens during the steady state conditions that biochemists tend to assume while they're studying enzyme kinetics.

Now, there's also another period during an enzyme-catalyzed reaction referred to as the pre-steady state. And so pre just means before. And so the pre-steady state describes the conditions that exist before the concentration of the enzyme-substrate complex reaches this stable or this constant value that we refer to as steady state. In our next lesson video, we're going to talk more details about the pre-steady state conditions, and then after that, we'll talk about the steady state conditions. So I'll see you guys in our next video.

Steady-State Conditions

So in our last lesson video, we said that the pre-steady state period exists before the steady state period is reached. And so in this video, we're going to talk about the pre-steady state period of an enzyme-catalyzed reaction. What's important to know is that when biochemists are studying enzyme-catalyzed reactions within a laboratory setting, they only add free enzyme and free substrate to the reaction mixture, but they do not add any enzyme-substrate complex or any product. This means that initially, at the very beginning of an enzyme-catalyzed reaction, the following four conditions exist just a few microseconds into the reaction. The first is that the free substrate concentration is really high, and so is the free enzyme concentration. It is relatively high, not quite as high as the free substrate concentration. We already know from our previous lesson videos that the free substrate concentration is much higher than the free enzyme concentration, which is why I'm going to only draw one arrow here. But still, relative to other periods during the reaction, the free enzyme concentration initially at the very beginning is relatively high. And again, since we do not add any enzyme-substrate complex or any product initially to the reaction, the concentration of enzyme-substrate complex initially at the very beginning is going to be really, really low, essentially 0, and so will be the concentration of the product.

What's important to know is that as the enzyme-catalyzed reaction begins and proceeds during this pre-steady state period, the concentrations of all four of these substances that we mentioned above are actually going to change in opposite directions, which means that if the concentrations start off really high, they're going to decrease and go lower over time. And if the concentrations start really, really low, then they're going to increase over time. And that's exactly what we see in our plot down below, which is showing us the pre-steady state period. Notice the plot has concentration on the y-axis and the time as the reaction proceeds on the x-axis. The colors of the curve correspond with the colors that we see of the substances in our reaction up above. This blue curve here represents the concentration of the substrate, which we can see initially starts really high but decreases over time, and so does the enzyme substrate complex, which starts relatively high and decreases over time. Whereas the red curve here, which corresponds to the concentration of enzyme-substrate complex, starts really low and increases over time, and the concentration of the product also starts really low and increases over time.

During this pre-steady state period of an enzyme-catalyzed reaction, the concentration of the enzyme-substrate complex is going to continuously change. In fact, it's going to continuously increase as we see in our plot down below. It continues to increase until a period called steady state is reached, and we're going to talk about the steady state period in our next lesson video, so I'll see you guys there.

Steady-State Conditions

Alright. So now that we talked about the pre steady state conditions in our last lesson video, in this video, we're going to talk about the steady state conditions. And so steady state is literally just referring to a specific period of time during an enzyme catalyzed reaction where the concentration of enzyme substrate complex stays exactly the same. And so if the concentration of enzyme substrate complex stays exactly the same, then, of course, what this means is that the rate or the velocity of the enzyme substrate complex association must be exactly equal to the rate or the velocity of the enzyme substrate complex dissociation.

And so, in other words, to say the same exact thing right here, we can say that if the concentration of enzyme substrate complex remains constant or remains exactly the same, then, of course, what this means is that the rate of the enzyme substrate complex association or \( v_1 \) for that matter is going to be exactly equal to the rate of the enzyme substrate complex dissociation or \( v_{-1} + v_2 \). Now just as a reminder down below over here, recall that at the very, very beginning of an enzyme catalyzed reaction, the enzyme substrate complex here in the middle can only associate or form in one way via this forward reaction right here. And so the rate or the velocity of this forward reaction that allows the enzyme substrate complex to associate is going to be \( v_1 \).

And then also recall that the enzyme substrate complex here in the middle can actually dissociate in 2 different ways. It can dissociate backwards here to form the free substrate and the free enzyme, and the enzyme substrate complex could also dissociate forward to form the free enzyme and the free product. And so, really, the rate or the velocity of the enzyme substrate complex dissociation is gonna be the sum of the backwards rate, or \( v_{-1} \), plus the rate of this forward dissociation, \( v_2 \). And so really this equation that we see right here is the assumption that we can make under steady state conditions.

And so in our next lesson video, we'll see that this assumption here in this equation is very important and that's because we can actually use this equation and this assumption here, to derive the Michaelis constant kilometers. And we'll be able to talk about in our next lesson video, how to rearrange this equation to derive the kilometers in our next lesson video. But for now, all I want you guys to know is that this equation is the most important assumption of steady state conditions. And again, it's important because we can derive the Michaelis constant kilometers.

Now even though we have not yet talked about the Michaelis-Menten enzyme kinetics equation, we are going to talk about this equation later in our course. And so what's important to note now is that this Michaelis-Menten Enzyme Kinetics equation that we'll talk about later is actually derived under these steady state conditions that we're talking about here. And so, that's another reason why steady state conditions and this equation is so important.

And so really notice down below over here, we're just reminding you guys of the important assumption of steady state conditions. And that is that the velocity of the association of the enzyme substrate complex or \( v_1 \) is going to be exactly equal to the velocity of the dissociation of the enzyme substrate complex, which is gonna be \( v_{-1} + v_2 \).

Now, notice over here we're showing you this graph where we have on the y-axis and the time as the reaction progresses on the x-axis. And notice that in the light blue background over here at the very of the reaction, what we have are the pre steady state conditions, which we already talked about in our last lesson video. Now notice over here in the yellow background, what we have are the steady state conditions, which again is just where the concentration of the enzyme substrate complex stays exactly the same. And so notice that in this yellow region, the concentration of enzyme substrate complex here in red stays exactly the same. And, of course, this is going to mean that the \( v_1 \) is going to equal to the \( v_{-1} + v_2 \), during this steady state period.

Now notice that in this yellow region, it is possible for the substrate concentration to change and it is possible for the product concentration to change. And again, steady state conditions is really just applying to the enzyme substrate complex. But you'll also notice that the concentration of free enzyme stays the same. But again, steady state conditions is more so referring to the concentration of enzyme substrate complex.

Now, last but not least, I want you guys to notice that we have this third star over here. And recall that the thirds, the stars are going to be important later in our course when we're talking about the assumptions that we need, to use the Michaelis-Menten Enzyme Kinetics Equation when we cover it later in our course. But for now, the main takeaways here of the steady state conditions is that the enzyme substrate complex stays the same and that allows us to use this equation, and that equation allows us to derive the kilometers and the Michaelis Menten Enzyme Kinetics Equation. And so that concludes this video here and we'll be able to get some practice in our next video. So, I'll see you guys there.

True or false: A reaction system at steady-state must also be at equilibrium.

• A. True
• B. False

Steady-State Conditions

Alright. So now that we've introduced steady state conditions, in this video, we're going to specifically talk about how the Michaelis constant or the km can be derived under these steady state conditions. So recall from our previous lesson videos, we already mentioned that both the Michaelis constant km, as well as the Michaelis-Menten equation, which we haven't yet talked about, but we're going to talk about in more detail later in our course. So both the km and the Michaelis-Menten equation are derived or defined under these steady state conditions, which is why steady state conditions are so important. And recall from our last lesson video, we said that steady state conditions just mean that the concentration of the enzyme-substrate complex stays exactly the same. And, of course, if the concentration of the enzyme-substrate complex stays exactly the same, that means that the rate of its formation or the rate of its association is going to be equal to the sum of the rates of the dissociation and that's exactly what we see down below in this expression right here. The velocity of the association is equal to the sum of the velocities of the dissociation of the enzyme-substrate complex. Now recall from our previous lesson videos that velocities, reaction velocities like these can actually be expressed or rewritten as rate laws. And so if we consider the rate laws for each of these velocities, then we can go ahead and, put those down below. So for this rate law for the association of the enzyme-substrate complex, recall that the rate law says that it's going to be equal to the rate constant, which would be k₁ for this association, times the concentration of the reactants. And we have 2 reactants for this association. We have the free enzyme and the free substrate. So, down below we can put the free enzyme as well as the free substrate. And so this represents the rate law for this reaction velocity. Now if we write out the rate laws for v₋₁, that's going to be, k₋₁ rate constant, times the reactant, which is going to be the enzyme-substrate complex. So we can put that in here. And then for the k₂, which is essentially going this way using this rate constant, its rate law is going to be k₂ times the enzyme-substrate complex concentration. And so here, notice that they both contain this, enzyme-substrate complex concentration. So what we can do is actually factor it out to the front, both of these out to the front. And when we do that, what we get is the concentration of the enzyme-substrate complex, times the sum of these two-rate constants. And then, of course, on the left, we still have the same exact expression. And so notice here, this is exactly where we can start to define the Michaelis constant km. And that's because notice that we have k₋₁ + k₂ here. And so all we need to do is take k₁ and divide it over under this side. So we have k₋₁ + k₂ / k₁. And so that's exactly what we see over here, is that we have k₋₁ + k₂ / k₁. And of course, this is going to be equal to the Michaelis constant km. And also notice that we have, the free enzyme times the free substrate concentration over here on the left, and then we have the enzyme-substrate complex concentration. So we can divide that over here so that it's below. And so what you end up getting is this same exact expression that we have right here, which is the free enzyme times the free substrate concentration over the concentration of the enzyme-substrate complex. So again, we already talked about how the km can be expressed in these formats in our previous lesson videos, and here, we're just showing you briefly how the Michaelis constant km can actually be derived under these steady state conditions. So it all results from making this assumption here that these velocities are going to be, equal to each other in these ways. And so this here concludes our lesson, and I'll see you guys in our next video.

Draw the curves that show the appropriate relationships between the variables in each of the plots below for a simple enzyme-catalyzed reaction that follows Michaelis-Menten kinetics.

Alright. So this practice problem wants us to draw the curves that show the appropriate relationships between the variables in each of the four plots below for a simple enzyme catalyzed reaction that follows Michaelis-Menten Kinetics. Now, we have not yet talked about Michaelis-Menten Kinetics, but we are going to later in our course. But we do know from our last lesson video on steady state conditions that Michaelis-Menten Kinetics and the equation associated with it are derived under steady state conditions. And because we're somewhat familiar with steady state conditions and we've seen a good handful of these plots before in our previous lesson videos, we should still be able to solve this problem. What we need to recall is that when biochemists are studying an enzyme-catalyzed reaction in a lab, they only add free enzyme and free substrate to the reaction mixture. They do not add any enzyme-substrate complex or any product. So at the very beginning of an enzyme-catalyzed reaction, we can say that the product concentration is going to be 0. So when we're looking at this first plot here, looking at the product concentration on the y-axis and the time as the reaction progresses on the x-axis, we know that very early on in the reaction, the product concentration is going to be 0. So, we can say that at time 0, the product concentration will be 0. So, we know we're going to start down here at this corner for this plot. And then over time, we know that the product concentration is going to increase and it's going to continuously increase until it reaches its equilibrium concentration. So at this point up here where it becomes horizontal, we know that the product has reached equilibrium. So we can write here equilibrium has been reached. And really, this is the answer to this first plot right here, this curve that we've drawn. And that's it. So we can move on to our next plot. Now, again, we know that when biochemists are studying that enzyme-catalyzed reaction, they only add the free enzyme and the free substrate, and they do not add any enzyme-substrate complex. So the enzyme-substrate complex at the very, very beginning of a reaction is going to be 0 because the reaction hasn't really had a chance to get going yet. So, we can say that on this plot here, where we have the enzyme-substrate complex concentration on the y-axis and the time as the reaction progresses on the x-axis, we know that at the very beginning of the reaction at time 0, the enzyme-substrate complex concentration will be 0. So, we know that we're also starting down here. Now, we also know that the concentration of free enzyme is going to be much smaller than the concentration of the substrate. And so what that means is that the concentration of the enzyme-substrate complex is limited by how much enzyme there is. And since there's not that much enzyme in comparison to the substrate, we know that the enzyme-substrate complex is going to have a relatively low concentration. Now, we know that from our previous lesson videos on pre-steady state conditions, before the enzyme-substrate complex gets to steady state and becomes stable, it has to increase and so it's actually going to rapidly increase. And then what's going to happen is it's going to become stable and it's going to reach steady state. And so, essentially, this is the curve that we're looking for for the enzyme-substrate complex. And so, what we can say is that this period here, as it's increasing, is the pre-steady state and then this period here where it's constant is going to be the steady state. And so that is pretty much it for this curve and we can move on to our next plot. So now down below, notice what we have is the substrate concentration on the y-axis and the time as the reaction progresses on the x-axis. And of course, we know that at the very beginning of the reaction, the substrate concentration is going to be really high because remember, the biochemist just added the substrate, so it hasn't had much of a chance to convert into the product. So the substrate concentration is still going to be really high. So we know that at the very beginning of time here, we're going to start pretty high on this axis, somewhere up here. And then over time, we know that the substrate concentration is going to decrease, not necessarily all the way down to 0 because we know that enzymes only help the substrate concentrations get to their equilibrium concentrations. And the equilibrium concentration does not necessarily mean it's going to go all the way down to 0. So we know that it's going to stabilize right at its equilibrium concentration. And so when it reaches this flat point over here we know that equilibrium has been reached just like what we saw up above with the product concentration. Just like what we saw up above here with the product concentration. Equilibrium has been reached when it gets flat. And so now, that is it for that plot. So now we can move on to our final plot here, and in this plot what we have is the initial reaction rate on the y-axis and the substrate concentration this time on the x-axis. So notice that we do not have time. All of the other plots we had time on the x-axis, whereas, in this plot here, we have the substrate concentration. So, when we're thinking of 0 here we can't be thinking at the beginning of time, so we're not going to really be thinking of it in the same way that we thought about the other three plots. So, essentially, we're going to be thinking of it in a different way because we have different axes and different variables. And so we know that this is really our typical enzyme kinetics plot and the typical curve here is for it to go up and then for it to curve across like this. Now, when it curves across at this point, the reason that it becomes horizontal is not because equilibrium has been reached because remember, equilibrium it takes some time, to reach equilibrium, but the initial velocity that we're plotting is at the very beginning of the reaction, so it's not occurring over time, so we can't say that that equilibrium has been reached over here. And so, instead, the reason that it's becoming flat is because, the substrate concentration we can say that the enzyme is saturated and the enzyme is specifically saturated with horizontal here. And so essentially, these are the four curves that we're looking to create for this problem, and that concludes this practice problem. So I'll see you guys in our next video.

Here’s what students ask on this topic:

What is the steady state assumption in enzyme kinetics?

The steady state assumption in enzyme kinetics refers to a condition during an enzyme-catalyzed reaction where the concentration of the enzyme-substrate complex (ES) remains constant. This implies that the rate of formation of the ES complex is equal to the rate of its dissociation into free enzyme and product. Mathematically, this is expressed as:

$v_1=v_{-}+v_2$

where $v_1$ is the rate of ES formation, and $v_{-}$ and $v_2$ are the rates of ES dissociation. This assumption is crucial for deriving the Michaelis-Menten equation and understanding enzyme kinetics.

How is the Michaelis constant (Km) derived under steady state conditions?

The Michaelis constant (Km) is derived under steady state conditions by equating the rate of formation and dissociation of the enzyme-substrate complex (ES). The steady state assumption gives us:

$v_1=v_{-}+v_2$

Expressing these velocities as rate laws, we get:

$k_1\left[E\right]\left[S\right]=(k_{-}+k_2)\left[\mathrm{ES}\right]$

Rearranging, we get:

$K{m}_{}=\frac{k_{-}+k_2}{k_1}$

This constant is crucial for understanding enzyme affinity and catalytic efficiency.

What is the difference between pre-steady state and steady state in enzyme kinetics?

In enzyme kinetics, the pre-steady state refers to the initial phase of the reaction where the concentrations of the enzyme-substrate complex (ES) and product are changing rapidly. During this period, the ES complex is being formed and its concentration is increasing. The steady state, on the other hand, is the phase where the concentration of the ES complex remains constant, indicating that the rate of its formation is equal to the rate of its dissociation. This steady state assumption is crucial for deriving the Michaelis-Menten equation and understanding enzyme behavior in biochemical processes.

Why is the steady state assumption important in deriving the Michaelis-Menten equation?

The steady state assumption is crucial in deriving the Michaelis-Menten equation because it simplifies the complex dynamics of enzyme-catalyzed reactions. By assuming that the concentration of the enzyme-substrate complex (ES) remains constant, we can equate the rate of formation of ES to its rate of dissociation. This allows us to derive the Michaelis constant (Km) and the Michaelis-Menten equation, which describes the relationship between the reaction rate and substrate concentration. The equation is fundamental for understanding enzyme kinetics and predicting how enzymes behave under different conditions.

How does the concentration of enzyme-substrate complex change during the pre-steady state period?

During the pre-steady state period of an enzyme-catalyzed reaction, the concentration of the enzyme-substrate complex (ES) changes rapidly. Initially, the concentration of free enzyme and substrate is high, while the concentration of ES and product is very low. As the reaction proceeds, the concentration of ES increases as more enzyme and substrate molecules form the complex. This period continues until the steady state is reached, where the concentration of ES becomes constant. The pre-steady state is characterized by dynamic changes in the concentrations of all reaction components.