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Basics of enzyme kinetics graphs

AP.BIO:
ENE‑1 (EU)
,
ENE‑1.G (LO)
,
ENE‑1.G.2 (EK)
,
ENE‑1.G.4 (EK)
How to read enzyme kinetics graphs (and how they're made). Km and Vmax. Competitive and noncompetitive inhibitors.

Introduction

Let’s imagine that you’re in the market for a sports car. What might you want to know about your various options (Ferrari, Porsche, Jaguar, etc.) to decide which one is best? One obvious factor would be how fast the car can go when you floor it. But you might also want more fine-grained information on car’s performance, such as how quickly it can accelerate from 0 to 60 mph. In other words, instead of just knowing its maximum speed, you’d also want to know the kinetics of how the car reaches that speed.
Biochemists tend to feel similarly about the enzymes they study. They want to know as much as possible about an enzyme’s effects on reaction rate, not just how fast the enzyme can go in a flat-out scenario.
As a matter of fact, you can tell a remarkable amount about how an enzyme works, and about how it interacts with other molecules such as inhibitors, simply by measuring how quickly it catalyzes a reaction under a series of different conditions. The information from these experiments is often presented in the form of graphs, so we’ll spend a little time here discussing how the graphs are made (and how to read them to get the most out of them).

Basic enzyme kinetics graphs

Graphs like the one shown below (graphing reaction rate as a function of substrate concentration) are often used to display information about enzyme kinetics. They provide a lot of useful information, but they can also be pretty confusing the first time you see them. Here, we’ll walk step by step through the process of making, and interpreting, one of these graphs.
Enzyme kinetics graph showing rate of reaction as a function of substrate concentration.
Image modified from "Enzymes: Figure 3," by OpenStax College, Biology (CC BY 3.0).
Imagine that you have your favorite enzyme in a test tube, and you want to know more about how it behaves under different conditions. So, you run a series of trials in which you take different concentrations of substrate - say, 0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, and 1.0 M - and find the rate of reaction (that is, how fast your substrate is turned into product) when you add enzyme in each case. Of course, you have to be careful to add the same concentration of enzyme to each reaction, so that you are comparing apples to apples.
How do you determine the rate of reaction? Well, what you actually want is the initial rate of reaction, when you’ve just combined the enzyme and substrate and the enzyme is catalyzing the reaction as fast as it can at that particular substrate concentration (because the reaction rate will eventually slow to zero as the substrate is used up). So, you would measure the amount of product made per unit time right at the beginning of the reaction, when the product concentration is increasing linearly. This value, the amount of product produced per unit time at the start of the reaction, is called the initial velocity, or V, start subscript, 0, end subscript, for that concentration.
Now, let’s say you’ve found your V, start subscript, 0, end subscript values for all your concentrations of interest. You can then plot each substrate concentration its V, start subscript, 0, end subscript as an (X, Y) pair. Once you’ve plotted all your (X, Y) pairs for different concentrations, you can connect the dots with a best-fit curve to get a graph. For many types of enzymes, the graph you get will resemble the purple line shown above: the V, start subscript, 0, end subscript values will increase rapidly at low substrate concentrations, then level off to a flat plateau at high substrate concentrations.
Enzyme kinetics graph showing rate of reaction as a function of substrate concentration, with Vmax (maximum velocity) and Km (substrate concentration giving reaction rate of 1/2 Vmax) marked.
Image modified from "Enzymes: Figure 3," by OpenStax College, Biology (CC BY 3.0).
This plateau occurs because the enzyme is saturated, meaning that all available enzyme molecules are already tied up processing substrates. Any additional substrate molecules will simply have to wait around until another enzyme becomes available, so the rate of reaction (amount of product produced per unit time) is limited by the concentration of enzyme. This maximum rate of reaction is characteristic of a particular enzyme at a particular concentration and is known as the maximum velocity, or V, start subscript, m, a, x, end subscript. V, start subscript, m, a, x, end subscript is the Y-value (initial rate of reaction value) at which the graph above plateaus.
The substrate concentration that gives you a rate that is halfway to V, start subscript, m, a, x, end subscript is called the K, start subscript, m, end subscript, and is a useful measure of how quickly reaction rate increases with substrate concentration. K, start subscript, m, end subscript is also a measure of an enzyme's affinity for (tendency to bind to) its substrate. A lower K, start subscript, m, end subscript corresponds to a higher affinity for the substrate, while a higher K, start subscript, m, end subscript corresponds to a lower affinity for the substrate. Unlike V, start subscript, m, a, x, end subscript, which depends on enzyme concentration, K, start subscript, m, end subscript is always the same for a particular enzyme characterizing a given reaction (although the "apparent," or experimentally measured, K, start subscript, m, end subscript can be altered by inhibitors, as discussed below).

Enzyme kinetics graphs and inhibitors

Now, what about inhibitors? We discussed two types of inhibitors, competitive and noncompetitive, in the article on enzyme regulation.
  • Competitive inhibitors impair reaction progress by binding to an enzyme, often at the active site, and preventing the real substrate from binding. At any given time, only the competitive inhibitor or the substrate can be bound to the enzyme (not both). That is, the inhibitor and substrate compete for the enzyme. Competitive inhibition acts by decreasing the number of enzyme molecules available to bind the substrate.
  • Noncompetitive inhibitors don’t prevent the substrate from binding to the enzyme. In fact, the inhibitor and substrate don't affect one another's binding to the enzyme at all. However, when the inhibitor is bound, the enzyme cannot catalyze its reaction to produce a product. Thus, noncompetitive inhibition acts by reducing the number of functional enzyme molecules that can carry out a reaction.
If we wanted to show the effects of these inhibitors on a graph like the one above, we could repeat our whole experiment two more times: once with a certain amount of competitive inhibitor added to each test reaction, and once with a certain amount of noncompetitive inhibitor added instead. We would get results as follows:
Enzyme kinetics graph showing rate of reaction as a function of substrate concentration for normal enzyme, enzyme with a competitive inhibitor, and enzyme with a noncompetitive inhibitor. For the competitive inhibitor, Vmax is the same as for the normal enzyme, but Km is larger. For the noncompetitive inhibitor, Vmax is lower than for the normal enzyme, but Km is the same.
Image modified from "Enzymes: Figure 3," by OpenStax College, Biology (CC BY 3.0).
  • With a competitive inhibitor, the reaction can eventually reach its normal V, start subscript, m, a, x, end subscript, but it takes a higher concentration of substrate to get it there. In other words, V, start subscript, m, a, x, end subscript is unchanged, but the apparent K, start subscript, m, end subscript is higher. Why must more substrate be added in order to reach V, start subscript, m, a, x, end subscript? The extra substrate makes the substrate molecules abundant enough to consistently “beat” the inhibitor molecules to the enzyme.
  • With a noncompetitive inhibitor, the reaction can never reach its normal V, start subscript, m, a, x, end subscript, regardless of how much substrate we add. A subset of the enzyme molecules will always be “poisoned” by the inhibitor, so the effective concentration of enzyme (which determines V, start subscript, m, a, x, end subscript) is reduced. However, the reaction reaches half of its new V, start subscript, m, a, x, end subscript at the same substrate concentration, so K, start subscript, m, end subscript is unchanged. The unchanged K, start subscript, m, end subscript reflects that the inhibitor doesn't affect binding of enzyme to substrate, just lowers the concentration of usable enzyme.

Michaelis-Menten and allosteric enzymes

Many enzymes act similarly to the hypothetical enzyme in the example above, producing parabolic curves when reaction rate is graphed as a function of substrate concentration. Enzymes that display this behavior can often be described by an equation relating substrate concentration, initial velocity, K, start subscript, m, end subscript, and V, start subscript, m, a, x, end subscript, known as the Michaelis-Menten equation. Enzymes whose kinetics obey this equation are called Michaelis-Menten enzymes. If you want a more detailed look at the Michaelis-Menten equation and the model underlying it, you may want to check out the Michaelis-Menten videos in the MCAT section.
Michaelis-Menten enzymes are different from allosteric enzymes (discussed in the main article on enzyme regulation). Allosteric enzymes typically have multiple active sites and often display cooperativity, meaning that the binding of a substrate at one active site increases the ability of the other active sites to bind and process substrates.
Reaction rate graphed as a function of substrate concentration for a cooperative enzyme. The curve is S-shaped (sigmoidal), with a sharp transition from low to high reaction rate over a narrow range of substrate concentrations.
Cooperative enzymes are more sensitive in their response to changes in substrate concentrations than other enzymes and display a “switch-like” transition from low to high reaction rate as substrate concentration increases. This corresponds to a velocity vs. substrate curve that is S-shaped, as shown above.

Want to join the conversation?

  • mr pink red style avatar for user ucdeng
    "Conversely, for a competitive inhibitor, the reaction gets never reaches its normal V{max}" it's noncompetitve right?
    (16 votes)
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    • leafers tree style avatar for user emilyabrash
      Yes, you are absolutely right, I just fixed that and it should be reflected on the site soon. Thanks for your good eye!

      Also, if you find other errors in the future, please don't hesitate to report them through the "report a mistake" button.

      Thanks again!
      (5 votes)
  • blobby green style avatar for user ana.michelle.avina
    "An uncompetitive inhibitor reduces Vmax, but increases the apparent Km"... doesn't the uncompetitive inhibitor bind to the enzyme and enhancing its binding to the substrate (higher affinity means lower Km)?
    (7 votes)
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    • duskpin ultimate style avatar for user Amy
      You're right, and it should be changed in the article. The apparent Km decreases in uncompetitive inhibition because by binding to the enzyme-substrate complex, uncompetitive inhibitors are "pulling" that complex out from the reactions. This removal of substrate decreases its concentration, and allows the remaining enzyme to work better. In general, a lower Km indicates better enzyme-substrate binding.
      (5 votes)
  • leafers ultimate style avatar for user anixmc1
    Can someone please clarify why Km is always the same? If i were to add more enzyme Vmax would increase and since Km is just 1/2 of Vmax, wouldn't Km increase as well?
    (2 votes)
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  • blobby green style avatar for user vbc-vcri
    Respected sir, I have a doubt, which may be very simple also. When we are plotting Vo versus [S}, we get rectangular hyperbolic curve. The curve approaches the Vmax asympotically and it never reaches or touches it.. Then how can we fix the Vmax value? From Vmax, only, we calculate Km. I could not understand this point. Kindly explain it with an example.
    (3 votes)
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  • blobby green style avatar for user 15panjabiar1
    In regards to competitive inhibition, would it not also take a longer time to reach the Vmax?
    (2 votes)
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  • blobby green style avatar for user md.nabil176
    If an inhibitor binds to the active site, it would compete with the substrate. But if it binds permanently, like an irreversible inhibitor, wouldn't the number of enzymes available decrease, thus making it a noncompetitive inhibitor?
    (2 votes)
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  • primosaur tree style avatar for user mksureshkumar11
    I did an enzymatic reaction with a constant substrate concentration. I measured the rate of product formation at different time interval. And I plotted the rate of product formation vs time. I got the exponential curve. Can I use Michael's Kinetics to define the time at which the rate of reaction reaches V max?
    (2 votes)
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  • blobby green style avatar for user Steve Berryman
    In the article, it says that noncompetitive inhibitors will not prevent substrate from binding with the enzymes, but shouldn't that be uncompetitive inhibitors? I was under the impression that uncompetitive inhibitors bind to the ES complex, while noncompetitive inhibitors can bind to either the enzyme (and block substrate binding) or the ES complex.
    (2 votes)
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  • blobby green style avatar for user jomwit22
    In the third graph, are there any explanations why the green and the purple curve meet at the same y (Vmax at the end) at the same concentration? I thought that since noncompetitive inhibitor exist, the required substrate would have to be more than normal situation where the inhibitor doesn't exist. Meaning that I think that the Purple graph should reach Vmax at a larger substrate concentration (higher X), than that of the normal enzyme curve.

    Also the green and purple graph represents the situation with the same concentration of enzymes right?
    (2 votes)
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    • winston baby style avatar for user Ivana - Science trainee
      Exactly.

      Let me explain.
      The lines represent Vmax , but they cross each other (meet at the same point) right at the end. What does it mean? That the Vmax wasn't the same up to the very last point.
      How comes?
      Well, competitive inhibitor competes for the exactly same place like substrate do. They have same potential to bind to the enzyme. Once it binds to the enzyme in kinetics terms it is the same when substrate is bonded to enzyme.
      (1 vote)
  • hopper jumping style avatar for user cali27
    Super helpful article!!
    Why does the Vmax occur at the same determined time interval each time in a reaction?
    (2 votes)
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