Michaelis-Menten Equation - Online Tutor, Practice Problems & Exam Prep

The Michaelis-Menten equation describes the initial reaction rates (V₀) of enzyme-catalyzed reactions, relating V₀ to substrate concentration, maximum reaction velocity (V_max), and the Michaelis constant (K_m). This relationship is represented by a rectangular hyperbola, where V₀ is plotted against substrate concentration. The equation can be expressed as V₀=V_maxK_m+[S]. Understanding this equation is crucial for analyzing enzyme kinetics.

Michaelis-Menten Equation

Alright. So now that we've covered the Michaelis-Menten assumptions, in this video, we're going to talk about the Michaelis-Menten equation. We commonly abbreviate it as just the MM equation in our lesson. The Michaelis-Menten equation is a mathematical description of the initial reaction rates, or the \( V_0 \), of an enzyme-catalyzed reaction. That should be no surprise to us, as we know from our previous lesson videos that biochemists focus mainly on the initial reaction rates, or the \( V_0 \), of enzyme-catalyzed reactions. Now, we're talking about an equation that allows us to calculate these initial reaction rates. The Michaelis-Menten equation mathematically relates the initial reaction velocity \( V_0 \) to the substrate concentration and does that via the theoretical maximum reaction velocity \( V_{\text{max}} \) and the Michaelis constant \( K_M \). It relates all of these enzyme kinetics variables, which we've already talked about in our previous lesson videos, and we'll see it demonstrated in our example on the left-hand side.

Now, if we take a quick look at our example below, notice the enzyme kinetics plot, which is also commonly known as a Michaelis-Menten kinetics plot. We've seen this plot many times before in our previous lessons. On the y-axis, we have the initial reaction rate or \( V_0 \), and on the x-axis, we have the substrate concentration. The curve typically seen on these plots is called a rectangular hyperbola. The Michaelis-Menten equation is actually describing this shape. Just like a line has an equation \( y = mx + b \), a rectangular hyperbola also has an equation.

The equation for a rectangular hyperbola is shown here:
y = ax b + x
The Michaelis-Menten equation is simply substituted with enzyme kinetics variables into this equation. We plug in \( V_{\text{max}} \) for \( a \) and \( K_M \) for \( b \) in the equation. Here is how it appears when substituted correctly:
V_0 = V_{\text{max}} ⋅ S K_M + S
Moving forward in our course, we'll use this equation to solve for the initial reaction velocity and other variables as applicable. This concludes our lesson on the introduction to the Michaelis-Menten equation. In our next video, I'll show you an example of how to utilize this equation. I'll see you in that video.

Michaelis-Menten Equation Example 1

All right. So now that we formally introduced the Michaelis-Menten equation in our last lesson video, in this video, I'm going to show you guys how to use the Michaelis-Menten equation in a 2-part example problem. And so this example problem wants us to consider the following enzyme kinetics data down below for the enzyme catalyzed reaction shown right here, where we have substrate A being converted into product B. And so looking at this enzyme kinetics data down below, notice in the first column on the left-hand side, what we have is the concentration of our substrate A in units of micromoles. And in the second column on the right, what we have is the \( v_0 \) or the initial reaction velocity of the enzyme catalyzed reaction in units of micromoles per minute. And so looking at part A, it's asking us what is the Michaelis constant or the Km of the enzyme. And so what we need to recall from our previous lesson videos is that the Km can be defined as, notice that the Km can be defined as the exact substrate concentration when the initial reaction velocity is exactly equal to half of the \( V_{max} \). And so notice that we are given the initial reaction velocity in our data over here, but we're not directly given the \( V_{max} \). And so what we need to do is use this data on the right to help us; it starts really, really low and it steadily increases, and then the increases get greater and greater to a point where the concentrations are so great that the initial reaction velocity doesn't even change. And so this is suggesting that the substrate concentrations are so high that it's completely saturating the enzyme. And recall from our previous lesson videos that when an enzyme is completely saturated with substrate, the initial reaction velocity can approach the \( V_{max} \). And so what we're saying is that these initial reaction velocities of 80 must be approaching and getting really close to the \( V_{max} \). And so we can pretty much say that our \( V_{max} \) is equal to a value of 80 micromoles per minute. And so now that we have our \( V_{max} \), if we want to get half the \( V_{max} \), then all we need to do is take half of the value of the \( V_{max} \), and half of our \( V_{max} \), which is 80, is going to be 40, of course. And notice that 40 actually shows up here in our data. And so, this is a velocity that's equivalent to half of the \( V_{max} \). And notice that the substrate concentration, when the initial reaction velocity is half the \( V_{max} \), is going to be the Km. And so this substrate concentration of 50 micromoles corresponds with half of the \( V_{max} \). And so we can say that this substrate concentration of 50 micromolar is our Km, and that is the answer to part A.

Now, moving on to part B, it's asking what is the value of the \( v_0 \) or the initial reaction velocity when the concentration of substrate A is equal to 43 micromolar. And so in order to solve this, we're going to need to utilize our Michaelis-Menten equation or the equation, which is provided down below in this box here. And so, we want to solve for the initial reaction velocity; all we need to do is plug in all of these other values here and calculate it using our calculator. So let's go on and do that. So we have the initial reaction velocity \( v_0 \) is equal to the \( V_{max} \) here, and the \( V_{max} \) we previously identified as being 80 micromoles per minute, and so we're going to multiply this by the concentration of our substrate, which is given to us as 43 micromolar. So we'll go ahead and multiply this. And then this is added to the substrate concentration again, which we know is 43 micromolar. And so all we need to do is plug all of this into our calculator. So if you do \( 80 \times 43 \), and take that answer and divide it by \( 50 + 43 \), you'll get the answer, which is approximately equal to 36.99, which rounds off to about 37, and the units of the initial reaction velocity are going to equal the units of the initial reaction velocity that we have in our chart, so it's going to be micromoles per minute. And so this here is the answer we are looking for, so 37 micromoles per minute. And so we can say that when the concentration of our substrate is equal to 43 micromolar, the initial reaction velocity will be 37 micromoles per minute. And so we could have kind of checked to look at our data over here to see if our answer makes sense. And so notice that a concentration of 43 micromolar is just below the concentration of 50 micromolar, just a little bit smaller. And so we expect that the velocity that corresponds with 43 is just going to be a little bit smaller than the velocity that corresponds with 50. And so when we see that this 40 here, we can see that our velocity is in the right ballpark. So we know that our answer is in the right ballpark. And so, also what this example shows us is that as long as we have 3 out of the 4 total variables that are present in our Michaelis-Menton equation, then we can solve for that 4th and missing variable. And so that shows us the power of the Michaelis-Menten equation, and we'll be able to continue to utilize it as we move forward in our next video. So I'll see you guys there.

A) Suppose the [S] = 10 K_m. Use the Michaelis-Menten equation to determine what percentage of the V_max will be equal to the value of V ₀.

B) Now suppose the [S] = 20 K_m. Use the Michaelis-Menten equation to determine what percentage of the V_max will be equal to the value of V ₀. What conclusion can be made from these calculations?

Alright. So here's a 2 part practice problem that's broken up into part a up above and part b down below. Starting with part a, it says to suppose that the substrate concentration is equal to 10 times the \(k_m\). This is an interesting way to present the substrate concentration, but it's totally valid, especially since we know from our previous lesson videos that the \(k_m\) is literally a substrate concentration. What this is saying is that the substrate concentration is equal to 10 times the value of the substrate concentration that corresponds with the \(k_m\). To better understand exactly what this is saying, let's take a look at our enzyme kinetics plot that we've drawn down below. Notice that the \(k_m\) on this plot corresponds with half of the value of the \(V_{\text{max}}\). Notice that the \(k_m\) is also a substrate concentration, which is why we find it on the axis of the substrate concentration. If we have a substrate concentration that is 10 times the value of the \(k_m\) substrate concentration, that's just saying that the substrate concentration is so high that it's actually 10 times the value of this \(k_m\). We're working with a substrate concentration that's really high and that's going to correspond with our curve here where the curve is approaching the \(V_{\text{max}}\). We expect that if the substrate concentration is 10 times greater than the \(k_m\), our substrate concentration is going to be so high that the curve should be approaching the \(V_{\text{max}}\). This is important to consider as we continue to move forward in our problem. It continues to tell us to use the Michaelis Menten equation to determine the value of the initial reaction velocity or the \(v_0\). In other words, it's asking us that when the substrate concentration is 10 times greater than the \(k_m\), what percentage of the \(V_{\text{max}}\) will the initial reaction velocity be? We want our answer in this form, and we need to determine what this percentage is. We'll use the Michaelis Menten equation to solve this problem, which I've provided here. We already have the \(v_0\) all by itself, so let's go ahead and plug in values for the rest of these equations. We have the initial reaction max, and since we're not actually given a value for the \(V_{\text{max}}\) here, we'll leave it as a variable, which will be multiplied by the substrate concentration. Notice that we're given that the substrate concentration is equal to \(10 k_m\). What we can do is plug in for the substrate concentration \(10 k_m\) instead of substrate concentration. So we plug in \(10 k_m\) here, multiply it by \(10 k_m\), and this is all going to be over the actual \(k_m\) shown here, plus the concentration of the substrate again, which again is represented as \(10 k_m\) here. Now we can further simplify this expression here. So \(k_m + 10 k_m\) is just going to be \(11 k_m\) on the bottom. Let's rewrite this expression over here. On the top, we still have \(V_{\text{max}} \times 10 k_m\), and on the bottom, we have \(11 k_m\). Let's simplify this, and when we do that what we get is \(V_{\text{max}}\) all by itself is going to be multiplied by \(\dfrac{10 k_m}{11 k_m}\). Here we can see that the units of \(k_m\) are going to cancel, and we just have \(\dfrac{10}{11}\). If you take your calculator and do \(10 \div 11\), what you'll get is an answer that corresponds to \(0.91\), and of course the \(V_{\text{max}}\) is still here, so \(\times V_{\text{max}}\). Notice that we have all of these equal signs throughout here, so these are expressions that are equal to each other. We can say the initial reaction velocity here is equal to \(0.91 \times V_{\text{max}}\). Let's rewrite that down below. The initial reaction velocity is equal to \(0.91 \times V_{\text{max}}\). Now we can see that this is the decimal form of a percentage. To convert it into an actual percentage, we need to take our \(0.91\) and multiply it by \(100\% \). When we do that, what we get is \(91\%\). When the substrate concentration is equal to a value that is 10 times greater than the value of the substrate concentration, equal to the \(k_m\), the initial reaction velocity is going to be \(91\%\) of the value of the \(V_{\text{max}}\). This is the answer we were looking for. The percentage we're looking for is \(91\%\). This tells us that the initial reaction velocity is pretty close to the\(V_{\text{max}}\). It's \(91\% \) of the way to the \(V_{\text{max}}\). That's pretty close and that's exactly what we had expected earlier in our problem. Moving on to part b, it's asking us to essentially do the same thing, except this time we're supposing that the substrate concentration is equal to \(20 k_m \) this time instead of \(10 k_m\). What do you guys think the value of the percentage is going to be this time? Well, we know that it can't actually equal \(100\%\), because the initial reaction velocity can only approach the \(V_{\text{max}}\) and it can never be \(100\%\) of the \(V_{\text{max}}\). Our percentage is going to be somewhere between \(91\% \) and \(100\%\). Let's do the same thing, except using \(20 k_m\) this time. Our initial reaction velocity is going to be equal to the \(V_{\text{max}}\), rewriting this equation again. So \(V_{\text{max}} \times \) the substrate concentration, which this time is given to us as \(20 k_m\), so \( \times 20 k_m\). This is all going to be over the\( k_m\) plus, again, the concentration of our substrate, which is \(20 k_m\). We can simplify the bottom here, \(20 k_m + k_m\) is going to be \(21 k_m\). On the bottom, we have \(21 k_m\), and on the top, we have the same exact thing, we still have \(V_{\text{max}} \times 20 k_m\). What we can do is we can simplify this further, take the \(V_{\text{max}}\), and bring it out to the front of this expression here. When we do that, it's going to look like \(V_{\text{max}} \times \dfrac{20 k_m}{21 k_m}\). Here we can see that the \(k_m\)s are cancelling each other out. So all we have is \(\dfrac{20}{21}\). If you take your calculator and do \(20 \div 21\), you'll get an answer that is equal to \(0.95\), and this is still \(\times V_{\text{max}}\). Again, because we have all of these equal signs throughout here, we can say that the initial reaction velocity is equal to this value over here, and we could set that equal down below. Let's go on and do that. So we have the initial reaction velocity is equal to this expression here, \(0.95 \times V_{\text{max}}\). Again, this is the decimal form of the percentage that we're looking for. To convert it to a percentage, we have to take \(0.95\) and multiply it by \(100\% \), and that's going to be equal to \(95\%\). This \(95\%\) here is telling us that when the substrate concentration is equal to \(20\) times greater than the value of the \(k_m\), then the initial reaction velocity is going to be \(95\% \) of the \(V_{\text{max}}\). Just like we expected, this percentage here is the answer that we are looking for for part b, and it's also greater than \(91\%\), but less than \(100\%\). This is essentially the conclusion for Part b, but notice that it's also asking us what conclusion can be made from both of these calculations here. The conclusion that can be made is that no matter how many times greater the substrate concentration is than the \(k_m\), the initial reaction velocity will always be less than the \(V_{\text{max}}\), which is why we define \(V_{\text{max}}\) as the theoretical maximum reaction velocity. The initial reaction velocity will always be less than the \(V_{\text{max}}\) no matter how many times greater the substrate concentration is than the\( k_m\). That is the final conclusion that we can make from this problem. That concludes this practice problem, so I'll see you guys in our next one.

Which of the following statements about a V₀ vs. [S] plot for a Michaelis-Menten enzyme is false?

• A. As [S] increases, V₀ also increases.
• B. At very high [S], the curve becomes a horizontal line that intersects the y-axis at K_m.
• C. K_m is the [S] at which V₀ = ½ V_max.
• D. The shape of the curve is a hyperbola.

What is the ratio of [S] to K_m ( [S] / K_m ) when the V₀ of an enzyme-catalyzed reaction is 80% of the V_max?

• A. 1.
• B. 2.
• C. 3.
• D. 4.
• E. 5.

An enzyme-catalyzed reaction was carried out with a [substrate] initially 1000 times greater than the Km for that enzyme. After 9 minutes, 1% of the total substrate was converted into 12 μmoles of product. If in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount of product (12 μmoles) to be formed?

• A. 1.5 min.
• B. 13.5 min.
• C. 27 min.
• D. 3 min.
• E. 6 min.

An enzyme catalyzes a reaction at a velocity of 10 μmol/min when all enzyme active sites are occupied with substrate. The K_m for this substrate is 1 x 10^-5 M. Assume that Michaelis-Menten kinetics are followed, calculate the initial reaction velocity (V₀) when:

A) [S] = 1 x 10^-5 M. V₀ = ___________ 

B) [S] = 1 x 10^-2 M. V₀ = ___________

Alright. So here we have a 2 part practice problem broken up into part a on the left and part b on the right. It says that an enzyme catalyzes a reaction at a velocity of 10 micromoles per minute when all of the enzyme active sites are occupied with substrate. Now, if all of the enzyme active sites are occupied with substrate, that means that all of the enzymes are in the enzyme-substrate complex form. And when that happens, we know that the enzymes are capable of approaching their theoretical maximum velocity. So, we can say that this velocity of 10 micromoles per minute is going to be really close to and approximate the theoretical maximum velocity vmax, and that's important to note for this problem. Now moving on, it tells us that the Michaelis constant for this substrate is \(1 \times 10^{-5}\) M. It tells us to assume that the Michaelis-Menten kinetics are followed, and to calculate the initial reaction velocity or \(v_0\) when we have different substrate concentrations below. Starting with part a here, notice that the substrate concentration is equal to \(1 \times 10^{-5}\) M. This substrate concentration is exactly equal to the value for the Kilometers. When the value of the substrate concentration is exactly equal to the value of the Kilometers, we know that exactly half of all the available enzyme active sites are going to be occupied with substrate, and that means that the enzyme is going to be operating at half of its theoretical maximum rate. So, \(10 \div 2 = 5\), and we can say that the answer to part a is going to be 5 micromoles per minute. If we were to use the Michaelis-Menten equation to solve this problem, the initial reaction velocity would be equal to the vmax, which is 10 micromoles per minute. You need to make sure the units match, so converting the molarity into units of micromoles, we multiply \(1 \times 10^{-5}\) by 1,000,000, resulting in 10 micromoles over the kilometers \(1 \times 10^{-5}\). These concentrations match, simplifying the expression to: 100 20 , which simplifies to 5 micromoles per minute. For part b, the substrate concentration is different from the Kilometers, so we continue using the Michaelis-Menten equation. Convert the substrate concentration \(1 \times 10^{-2}\) to micromoles by multiplying by 1,000,000, resulting in 10,000 micromoles. Then we calculate: 10 × 10000 10 + 10000 which gives approximately 9.99 micromoles per minute. This is the answer for part b of our problem. These are the answers to our two-part problem, and that concludes this practice.

Here’s what students ask on this topic:

What is the Michaelis-Menten equation and what does it describe?

The Michaelis-Menten equation is a mathematical description of the initial reaction rates (V₀) of enzyme-catalyzed reactions. It relates the initial reaction velocity (V₀) to the substrate concentration ([S]), the maximum reaction velocity (V_max), and the Michaelis constant (K_m). The equation is represented as:

$V_0=\frac{V_{\max }\cdot S}{K_m+S}$

This equation helps in understanding how enzymes interact with substrates and how the reaction rate changes with varying substrate concentrations.

What are the assumptions of the Michaelis-Menten equation?

The Michaelis-Menten equation is based on several key assumptions:

• Steady-State Assumption: The formation and breakdown of the enzyme-substrate complex (ES) reach a steady state, meaning the rate of formation of ES is equal to the rate of its breakdown.
• Initial Rate Assumption: The initial reaction rate (V₀) is measured when the substrate concentration is much higher than the enzyme concentration, ensuring that the product formation is negligible.
• Substrate Saturation: The substrate concentration is high enough that the enzyme is saturated, and the reaction rate approaches V_max.

These assumptions simplify the complex kinetics of enzyme reactions, making it easier to derive and use the Michaelis-Menten equation.

How do you determine Vmax and Km from a Michaelis-Menten plot?

To determine V_max and K_m from a Michaelis-Menten plot, follow these steps:

1. Plot the Data: Plot the initial reaction velocity (V₀) on the y-axis against the substrate concentration ([S]) on the x-axis.
2. Identify V_max: V_max is the asymptote of the curve, representing the maximum reaction rate when the enzyme is fully saturated with the substrate.
3. Determine K_m: K_m is the substrate concentration at which the reaction rate is half of V_max. Find the point on the x-axis where V₀ is half of V_max to determine K_m.

These values are crucial for understanding enzyme efficiency and affinity for the substrate.

What is the significance of the Michaelis constant (Km) in enzyme kinetics?

The Michaelis constant (K_m) is a key parameter in enzyme kinetics. It represents the substrate concentration at which the reaction rate is half of the maximum velocity (V_max). Mathematically, it is defined as:

$K_m=\frac{k_{\mathrm{-1}}+k_{\mathrm{cat}}}{k_1}$

where k₁ is the rate constant for the formation of the enzyme-substrate complex, k_-1 is the rate constant for its dissociation, and k_cat is the rate constant for the conversion of the substrate to product. A low K_m indicates high affinity between the enzyme and substrate, while a high K_m suggests lower affinity.

How does substrate concentration affect the initial reaction rate in the Michaelis-Menten model?

In the Michaelis-Menten model, the initial reaction rate (V₀) is influenced by the substrate concentration ([S]). At low substrate concentrations, V₀ increases linearly with [S] because there are many available enzyme active sites. As [S] increases, the rate of increase in V₀ slows down, eventually reaching a plateau at V_max. This occurs because the enzyme becomes saturated with substrate, and all active sites are occupied. The relationship between V₀ and [S] is described by the Michaelis-Menten equation:

$V_0=\frac{V_{\max }\cdot S}{K_m+S}$

This equation shows that V₀ approaches V_max as [S] becomes very large.