plot:

inhibitor concentration (mol/L)

2.0

inhibition mechanism

This simulation plots rate data for Michaelis-Menten enzyme kinetics for four types of inhibition. For competitive inhibition, an inhibitor forms an inactive complex with the enzyme. For uncompetitive inhibition, an inhibitor forms an inactive complex with the enzyme-substrate complex. For mixed inhibition, the inhibitor forms both types of inactive complexes. For self-inhibition, the substrate itself inhibits the reaction by forming an inactive complex with the enzyme-substrate complex.

Select "Michaelis-Menten" to plot the rate of substrate consumption versus substrate concentration. Select "Lineweaver-Burke" to plot -1/rate versus 1/(substrate concentration) so as to obtain a straight line (except for the self-inhibited case). The slope and intercept of this line are related to \(K_M\), \(K_I\), and \(V_\text{max}\). Use the slider to change the inhibitor concentration, which changes the slope and/or the y intercept, depending on which type of inhibition is selected. When self-inhibited is selected, the substrate concentration is the inhibitor concentration, so the slider is hidden.

\(k\) values come from respective reaction coefficients

\([E]\) represents the concentration of free enzyme

\([S]\) represents the concentration of substrate

\([E \cdot S]\) represents the concentration of an enzyme-substrate complex

\([E \cdot I]\) represents the concentration of an enzyme-inhibitor complex

\([E \cdot I \cdot S]\) represents the concentration of an enzyme-inhibitor-substrate complex

\([E_t]\) is the total enzyme concentration \([E]+[E \cdot S]+[E \cdot I]\)

$$V_\text{max}=k_3 [E_t] ,$$ $$K_M=\frac{k_2+k_3}{k_1} ,$$ $$K_I = \frac{k_5}{k_4} ,$$ The Lineweaver-Burke method in a method of linearizing experimental data. The slope and y intercept from experimental data can be used to determine \(V_\text{max}\), \(K_M\), and \(K_I\) from plotted data.

The line is given by $$y=mx+b$$ where \(m\) is the slope, \(b\) is the y intercept, and \(c\) is the x intercept (the value of x when y=0).

The inhibitor competes with the substrate for the enzyme to form an inactive complex. $$(1) E+S \rightarrow E \cdot S$$ $$(2) E \cdot S \rightarrow E + S$$ $$(3) E \cdot S \rightarrow P + E$$ $$(4) I + E \rightarrow E \cdot I (\text{inactive})$$ $$(5) E \cdot I \rightarrow E + I$$ $$-r_s=r_p=\frac{V_\text{max}[S]}{[S]+K_M(1+\frac{[I]}{K_I})} ,$$ $$-\frac{1}{r_s}=\frac{1}{[S]}\frac{K_M}{V_\text{max}}(1+\frac{[I]}{K_I})+\frac{1}{V_\text{max}},$$ $$m=\frac{K_M}{V_\text{max}}(1+\frac{[I]}{K_I}) ,$$ $$b=\frac{1}{V_\text{max}},$$ $$c=\frac{1}{K_M(1+\frac{[I]}{K_I})}.$$

The inhibitor binds to the enzyme-substrate complex forming an inhibitor-enzyme-substrate complex. $$(1) E+S \rightarrow E \cdot S$$ $$(2) E \cdot S \rightarrow E + S$$ $$(3) E \cdot S \rightarrow P + E$$ $$(4) I + E \cdot S \rightarrow I \cdot E \cdot S (\text{inactive})$$ $$(5) I \cdot E \cdot S \rightarrow I + E \cdot S$$ $$-r_s=r_p=\frac{V_\text{max}[S]}{K_M+[S](1+\frac{[I]}{K_I})} ,$$ $$-\frac{1}{r_s}=\frac{1}{[S]}\frac{K_M}{V_\text{max}}+\frac{1}{V_\text{max}}(1+\frac{[I]}{K_I}),$$ $$m=\frac{K_M}{V_\text{max}} ,$$ $$b=\frac{1}{V_\text{max}}(1+\frac{[I]}{K_I}),$$ $$c=\frac{1}{K_M}(1+\frac{[I]}{K_I}).$$

Mixed inhibition is most commonly seen for enzymes with two or more substrates. The inhibitor can bind to the enzyme or enzyme-substrate complex. The substrate can also bind to the inhibitor-enzyme complex. $$(1) E+S ⇌ E \cdot S$$ $$(2) E + I ⇌ I \cdot E (\text{inactive})$$ $$(3) E \cdot S \rightarrow P + E$$ $$(4) I + E \cdot S ⇌ I \cdot E \cdot S (\text{inactive})$$ $$(5) S + I\cdot E ⇌ I \cdot E \cdot S (\text{inactive})$$ $$-r_s=r_p=\frac{V_\text{max}[S]}{(K_M+[S])(1+\frac{[I]}{K_I})} ,$$ $$-\frac{1}{r_s}=\frac{1}{[S]}\frac{K_M}{V_\text{max}}(1+\frac{[I]}{K_I})+\frac{1}{V_\text{max}}(1+\frac{[I]}{K_I}),$$ $$m=\frac{K_M}{V_\text{max}}(1+\frac{[I]}{K_I}) ,$$ $$b=\frac{1}{V_\text{max}}(1+\frac{[I]}{K_I}),$$ $$c=K_M.$$

In the case of uncompetitive inhibition, the inhibitor ties up the enzyme-substrate complex to form an inactive substrate-enzyme-substrate complex. $$(1) E+S \rightarrow E \cdot S$$ $$(2) E \cdot S \rightarrow E + S$$ $$(3) E \cdot S \rightarrow P + E$$ $$(4) S + E \cdot S \rightarrow S \cdot E \cdot S (\text{inactive})$$ $$(5) S \cdot E \cdot S \rightarrow S + E \cdot S$$ $$-r_s=r_p=\frac{V_\text{max}[S]}{K_M+[S]+\frac{[S]^2}{K_I}} ,$$ $$-\frac{1}{r_s}=\frac{1}{[S]}\frac{K_M}{V_\text{max}}+\frac{1}{V_\text{max}}(1+\frac{[S]}{K_I}).$$ This plot is not linear because the y intercept depends on the substrate concentration \([S]\).

**Reference**

[1] H. S. Fogler, *Essentials of Chemical Reaction Engineering*, 1st ed., Upper Saddle
River, NJ:
Pearson Higher Education, 2010 pp. 364-370.

This simulation was created in the Department of Chemical and Biological Engineering, at University of Colorado Boulder for LearnChemE.com by Nicholas Larson under the direction of Professor John L. Falconer and was converted to HTML5 by Drew Smith. This simulation was prepared with financial support from the National Science Foundation. Address any questions or comments to learncheme@gmail.com. All of our simulations are open source, and are available on our LearnChemE Github repository.

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