Question

10.5. Provide the graphs for the Lineweaver-Burk for Competitive, Non-competitive, and Un-competitive inhibition and

describe what occurs to Km and Vmax for each.

Answer 1

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.]
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.
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_0V0​V, start subscript, 0, end subscript, for that concentration.
Now, let’s say you’ve found your V_0V0​V, start subscript, 0, end subscriptvalues for all your concentrations of interest. You can then plot each substrate concentration its V_0V0​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_0V0​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.]
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_{max}Vmax​V, start subscript, m, a, x, end subscript. V_{max}Vmax​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_{max}Vmax​V, start subscript, m, a, x, end subscript is called the K_mKm​K, start subscript, m, end subscript, and is a useful measure of how quickly reaction rate increases with substrate concentration. K_mKm​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_mKm​K, start subscript, m, end subscript corresponds to a higher affinity for the substrate, while a higher K_mKm​K, start subscript, m, end subscript corresponds to a lower affinity for the substrate. Unlike V_{max}Vmax​V, start subscript, m, a, x, end subscript, which depends on enzyme concentration, K_mKm​K, start subscript, m, end subscriptis always the same for a particular enzyme characterizing a given reaction (although the "apparent," or experimentally measured, K_mKm​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.]
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_{max}Vmax​V, start subscript, m, a, x, end subscript, but it takes a higher concentration of substrate to get it there. In other words, V_{max}Vmax​V, start subscript, m, a, x, end subscript is unchanged, but the apparent K_mKm​K, start subscript, m, end subscript is higher. Why must more substrate be added in order to reach V_{max}Vmax​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_{max}Vmax​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_{max}Vmax​V, start subscript, m, a, x, end subscript) is reduced. However, the reaction reaches half of its new V_{max}Vmax​V, start subscript, m, a, x, end subscript at the same substrate concentration, so K_mKm​K, start subscript, m, end subscript is unchanged. The unchanged K_mKm​K, start subscript, m, end subscript reflects that the inhibitor doesn't affect binding of enzyme to substrate, just lowers the concentration of usable enzym