Michaelis-Menten vs. Lineweaver-Burk Plots - Online Tutor, Practice Problems & Exam Prep

The Michaelis-Menten and Lineweaver-Burk plots are essential for analyzing enzyme kinetics. The Michaelis-Menten plot approximates V_max and K_m but can yield inaccuracies due to limited data points contributing to the plateau. In contrast, the Lineweaver-Burk plot, a double reciprocal plot, provides a linear representation where all data points contribute to determining V_max and K_m, enhancing accuracy. This graphical method is crucial for understanding enzyme behavior and optimizing reactions in biochemical pathways.

Michaelis-Menten vs. Lineweaver-Burk Plots

In this video, we're going to do a little compare and contrast between the Michaelis-Menten plot and the Lineweaver-Burk plot. It's important to note that if multiple enzyme kinetic variables are missing or just completely unknown, then we cannot use the equation form of either the Michaelis-Menten equation or the Lineweaver-Burk equation. However, we can still determine these enzyme kinetic variables such as the vmax and the Michaelis constant KM, graphically. We can determine them graphically via plotting enzyme kinetic experimental data onto a Michaelis-Menten plot or a Lineweaver-Burk plot. As we'll see below in our image, we'll see that a Lineweaver-Burk plot actually provides some graphical advantages over the Michaelis-Menten plot.

If we take enzyme kinetic experimental data and graphically plot that data onto a Michaelis-Menten plot, we can only approximate the values of vmax and KM. However, if we take the same exact enzyme kinetic experimental data and graphically plot that data onto a Lineweaver-Burk plot, we can improve the accuracy of the estimations of both vmax and KM. Let's take a look down below at our image to clear up some of this idea.

Notice on the left here, we have our Michaelis-Menten plot, and on the right here, we have our Lineweaver-Burk plot. Starting with the Michaelis-Menten plot, notice we have the same plot that we've seen so many times before in our previous lesson videos. On the y-axis, we have the initial reaction velocity. The vmax is represented as a horizontal asymptote on this plot and the Michaelis constant KM is the exact substrate concentration that allows for the initial reaction velocity to be equivalent to exactly half of the vmax or vmax2.

Now, if we were to take enzyme kinetics experimental data, let's say we had five experiments. Suppose the first experiment was at low substrate concentration, and the initial reaction velocity data point was right here. We continue with higher substrate concentrations to plot additional points. From these five experimental data points, notice that they do not form a perfect curve. However, if we were able to get a curve of best fit here, then we would see the typical rectangular hyperbola curve. However, of these five experiments, only two are contributing to the horizontal plateau, which is the region where the initial reaction velocity approaches the vmax. This results in an inaccurate measurement of vmax.

On the other hand, if we look at our Lineweaver-Burk plot on the left, recall that it is also known as a double reciprocal plot because the y-axis is the reciprocal of the initial reaction velocity, and the x-axis is the reciprocal of the substrate concentration. By plotting reciprocals, we form a line of best fit. All five of these experimental data points contribute to this line, enhancing the accuracy of both the y-intercept, which is associated with vmax, and the x-intercept, which is associated with KM. Thus, the Lineweaver-Burk plot helps us get a more accurate measurement of both the vmax and the KM in comparison to the Michaelis-Menten plot.

That's the main takeaway of this video, and we'll be able to apply the concepts that we've learned here in our practice problem. I'll see you there.

Why is it preferable to use a Lineweaver-Burk over a Michaelis-Menten plot when studying enzyme kinetics?

• A. To directly visualize K_m & V_max on the plot.
• B. To plot kinetic data as a hyperbolic curve instead of a line.
• C. To obtain a more accurate measure of the V₀.
• D. To remove terms that cannot be calculated in a typical enzyme kinetics experiment.
• E. To get more accurate estimates of K_m & V_max.

You measure V₀ of an enzyme at 6 different [S] & plot the data on a Lineweaver-Burk plot. You then determine the line of best fit to the data to visualize the x & y intercepts. Calculate the V_max & K_m of the enzyme. Hint: pay close attention to the indicated units.

Alright. So this practice problem says you measure the initial reaction velocity or the \( v_0 \) of an enzyme at 6 different substrate concentrations, and then you plot that data into the Lineweaver Burk plot shown down below. And so, notice that with the Lineweaver Burk plot, it has the reciprocal of the initial reaction velocity on the y-axis and the reciprocal of the substrate concentration on the x-axis. And so, we can tell that we've plotted 6 different experimental data points in our plot below because we have these 6 experimental data point circles in red down below. And so, the problem continues to tell us that you then determine the line of best fit to this data. So this line right here is the line of best fit that best fits the experimental data points that are shown. And, the reason that we do that is to visualize the x and the y intercept. So notice that the y intercept is equal to \( 3 \times 10^7 \), and the x intercept is equal to \( -5 \times 10^4 \). And so, it wants us to calculate the \( V_{\text{max}} \) and the \( K_m \) of the enzyme from this information, and it gives us the hint to pay close attention to the indicated units. And so, again, we know that we're going to be calculating the \( V_{\text{max}} \). And so in order to calculate the \( V_{\text{max}} \), we're going to need to pay close attention to the y intercept, and the y intercept is where our line crosses the y-axis. So that's going to be right at this particular point here, which is associated with the value \( 3 \times 10^7 \). And so, essentially what we can say is the y intercept on a Lineweaver Burk plot, we know is going to be a reciprocal just like the x-axis and the y-axis are also reciprocals. So it's going to be \( 1 \) over something, and that something is going to be associated with the y-axis, and so, because the y-axis associated with a reaction velocity, so is the y intercept. And it's actually going to be the reciprocal of the theoretical maximal reaction velocity, \( V_{\text{max}} \). And so, we just said that this value here is associated with the value \( 3 \times 10^7 \). And so, this is the value of the y-intercept, not the value of the \( V_{\text{max}} \) itself. And so, we need to set our y intercept equal to \( 3 \times 10^7 \). And then that allows us to calculate for this \( V_{\text{max}} \). So we can calculate the \( V_{\text{max}} \) by moving it up to this side by multiplying both sides of the equation by \( V_{\text{max}} \), And then after we do that, we can divide both sides of the equation by \( 3 \times 10^7 \). And when you do that, you'll get that the \( V_{\text{max}} \) is equal to \( 1 \) over \( 3 \times 10^7 \). And so if you type that into your calculator, \( 1 \) divided by \( 3 \times 10^7 \), you'll get an answer of \( 3.33 \times 10^{-8} \). And the units of the \( V_{\text{max}} \) are going to be the same as the units of the initial reaction velocity \( v_0 \), which is units of molarity per minute. So we can put that here. So what we need to notice next is that the answer here, wants us to, indicate the answer in units of nanomolar per minute, not molarity per minute. And so we need to convert the molarity into nanomolar so that they match each other. And so, in order to do that, what we can do is take our \( 3.33 \times 10^{-8} \), the same number here. And, we can say that this is in units of molarity, and we want to convert it to nanomolar. So we know that we want molarity to be here, so we can say that \( 1 \) molar, that will allow the molarity to cancel out. We can say that there are \( 10^9 \) nanomolar. And so nano here means to the \( 9^{\text{th}} \), \( 10^9 \). And so, if you take your calculators and you do \( 3.33 \times 10^{-8} \times 10^9 \), you'll get the answer of \( 33.3 \) nanomolar, per minute. And so now that we have our \( V_{\text{max}} \) in units of nanomolar per minute, we can go ahead and plug that value into our answer below. And so, the answer is going to be \( 33.3 \) nanomolar per minute for the \( V_{\text{max}} \). So now that we've done that, we can do the same with the \( K_m \), which we know is instead, associated with the x intercept instead of the y intercept. And the x intercept is going to be where our line crosses the x-axis. So it's going to be right at this particular point right here, which we can see is associated with this value \( -5 \times 10^4 \). So we can say over here that our x intercept on a Lineweaver Burk plot, just like everything else, is essentially reciprocals, so is the x intercept. It's going to be a reciprocal, so it's going to be one over something, and that something is going to be associated with what's on the x-axis, which is a substrate concentration. And we know that the \( K_m \) is an exact substrate concentration that allows the initial reaction velocity to be half of the maximum. And so, next, what we need to recall is that the x intercept always falls into this negative region here of our Lineweaver Burk plot. So notice that the values of the x intercept are always negative. And so, what that means is we need to remember to include the negative sign here with the one. And so, the x intercept on a Lineweaver Burk plot is the negative reciprocal of the \( K_m \). And, we just said that the x intercept is associated with this value, \( -5 \times 10^4 \). So we can set it equal to that. And so now we can solve for the value of our \( K_m \), our Michaelis constant. And we can do that by moving it up to this side of the equation by multiplying both sides of the equation by the \( K_m \), And then after we do that, we can divide both sides of the equation by \( -5 \times 10^4 \). When we do that, what we'll get is that the \( K_m \) all by itself is equal to \( \frac{-1}{-5 \times 10^4} \). And so if you take your calculator and you do \( -1 \) divided by \( -5 \times 10^4 \), you'll get an answer of \( 2 \times 10^{-5} \). And the units are going to be the same units that are associated with the substrate concentration on the bottom, which are just molarity. And so we can put that here as well, molarity. And so, notice that our \( K_m \), we want in the units of micromolar, not in the units of molar. And so, what we need to do is convert these units. So we can take our \( 2 \times 10^{-5} \) molar, and we can say that in one molarity, there are \( 10^6 \) micromolars. And so, that allows our units of molarity to cancel out. And so if we take our calculator and we do \( 2 \times 10^{-5} \times 10^6 \) here, we'll get an answer of \( 20 \) micromolar. And so now that our answer of micromolar, matches, the units that we need, we can say that our \( K_m \), our \( K_m \) is just \( 20 \) micromolar. And so, these here are the answers to our practice problem, and that concludes this practice, so I'll see you guys in our next video.

Here’s what students ask on this topic:

What is the main difference between Michaelis-Menten and Lineweaver-Burk plots?

The main difference between Michaelis-Menten and Lineweaver-Burk plots lies in their graphical representation and accuracy. The Michaelis-Menten plot is a hyperbolic curve that approximates V_max and K_m, but it can be inaccurate due to limited data points contributing to the plateau. In contrast, the Lineweaver-Burk plot is a double reciprocal plot that linearizes the data, allowing all data points to contribute to determining V_max and K_m. This results in more accurate measurements of these enzyme kinetic parameters.

Why is the Lineweaver-Burk plot considered more accurate than the Michaelis-Menten plot?

The Lineweaver-Burk plot is considered more accurate than the Michaelis-Menten plot because it uses a double reciprocal transformation, which linearizes the data. This allows all experimental data points to contribute to the determination of V_max and K_m. In contrast, the Michaelis-Menten plot often relies on fewer data points to estimate the plateau, leading to potential inaccuracies in V_max and K_m values.

How do you determine V_max and K_m from a Lineweaver-Burk plot?

To determine V_max and K_m from a Lineweaver-Burk plot, you plot the reciprocals of the initial reaction velocity (1/V₀) against the reciprocals of the substrate concentration (1/[S]). The y-intercept of the resulting straight line gives 1/V_max, and the x-intercept gives -1/K_m. From these intercepts, you can calculate V_max and K_m using the equations V_max = 1/(y-intercept) and K_m = -1/(x-intercept).

What are the limitations of the Michaelis-Menten plot?

The limitations of the Michaelis-Menten plot include its reliance on fewer data points to estimate the plateau, which can lead to inaccuracies in determining V_max and K_m. Additionally, the hyperbolic nature of the plot makes it difficult to precisely determine these parameters, especially when experimental data points do not perfectly fit the expected curve. This can result in less reliable measurements compared to the Lineweaver-Burk plot.

What are the advantages of using a Lineweaver-Burk plot in enzyme kinetics?

The advantages of using a Lineweaver-Burk plot in enzyme kinetics include its ability to linearize the data, making it easier to determine V_max and K_m accurately. By plotting the reciprocals of the initial reaction velocity and substrate concentration, all data points contribute to the line of best fit, enhancing the precision of the measurements. This method also allows for easier identification of enzyme inhibition types and more straightforward comparison of kinetic parameters.