ChemWiki: The Dynamic Chemistry E-textbook > Biological Chemistry > Enzymes > Enzymatic Kinetics > Michaelis-Menten Kinetics

Two 20^{th} century scientists, Leonor Michaelis and Maud Leonora Menten, proposed the model known as Michaelis-Menten Kinetics to account for enzymatic dynamics. The model serves to explain how an enzyme can cause kinetic rate enhancement of a reaction and explains how reaction rates depends on the concentration of enzyme and substrate.

The general reaction scheme of an enzyme-catalyzed reaction is as follows:

\[ E + S \xrightarrow[ ]{k_1}[ ES ] \xrightarrow[ ] {K_2} E + P \]

The enzyme interacts with the substrate by binding to its active site to form the enzyme-substrate complex, ES. That reaction is followed by the decomposition of ES to regenerate the free enzyme, E, and the new product, P. For more general information about enzyme-catalyzed reactions, please refer to GENERAL PRINCIPLES [of Enzymes].

To begin our discussion of enzyme kinetics, let's define the number of moles of product (P) formed per time as *V*. The variable, *V,* is also referred to as the rate of catalysis of an enzyme. For different enzymes, *V* varies with the concentration of the substrate, S. At low S, *V* is linearly proportional to S, but when S is high relative to the amount of total enzyme, *V* is independent of S. Concentrations is important in determining the initial rate of an enzyme-catalyzed reaction. A more thorough explanation of enzyme rates can be found here: Definition of Reaction Rate.

To understand Michaelis-Menten Kinetics, we will use the general enzyme reaction scheme shown below, which includes the back reactions in addition the the forward reactions:

\[ E + S \xrightarrow[ ]{k_1}[ ES ] \xrightarrow[ ] {K_2} E + P \]

\[ E + S \xleftarrow[ ]{k_3}[ ES ] \xleftarrow[ ] {K_4} E + P \]

The table below defines each of the rate constants in the above scheme.

Rate Constant | Reaction |

k_{1} | The binding of the enzyme to the substrate forming the enzyme substrate complex. |

_{k2} | Catalytic rate; the catalysis reaction producing the final reaction product and regenerating the free enzyme. This is the rate limiting step. |

_{k3} | The dissociation of the enzyme-substrate complex to free enzyme and substrate . |

_{k4} | The reverse reaction of catalysis. |

Table 1

\( E + S \xrightarrow[ ]{k_1} ES \) \( v_o = k_1[E][S] \)

\( ES \xrightarrow[ ]{k_2} E + S \) \( v_o = k_2[ES] \)

\( ES \xrightarrow[ ]{k_3} E + P\) \( v_o = k_3[ES] \)

\( E + P \xrightarrow[ ]{k_4} ES \) \( v_o = k_4[E][P]=0 \)

The ES complex is formed by combining enzyme E with substrate S at rate constant k_{1}. The ES complex can either dissociate to form E_{F }(free enzyme) and S, or form product P at rate constant k_{2} and k_{3}, respectively. The velocity equation can be derived in either of the 2 methods that follow:

E, S, and the ES complex can equilibrate very rapidly. The instantaneous velocity is the catalytic rate that is equal to the product of ES concentration and k_{2} the catalytic rate constant.

\[ v_o = k_2[E - S] \]

The total enzyme concentration (E_{T}) is equal to the concentration of free enzyme E (E_{F}) plus the concentration of the bound enzyme in ES complex:

\[ [E]_T = [E_F] + [ES] \]

\[ K_s = \dfrac{k_2}{k_1} = \dfrac{[E][S]}{[ES]} \]

\[K_s \dfrac{([E_o] - [ES])[S]}{[ES]}\]

\[[ES] = \dfrac{[E_o][S]}{K_s + [S]}\]

\[v_o = \left(\dfrac{dP}{dt}\right)_o = k_3[ES]\]

\[v_o = \left(\dfrac{dP}{dt} \right)_o = \dfrac{k_3[Eo][S]}{Ks + [S]}\]

At high substrate concentrations, \([S] >> K_s\) we get:

\( v_o = \left(\dfrac{dP}{dt}\right)_o = k_3[E_o] = V_{max} \) **(1)**

*Main Article:* Steady-State Approximation

Figure 1

The figure above shows the relatively low and constant concentration of the enzyme-substrate complex due to the complex's slow formation and rapid consumption. Note the falling substrate concentration and the rising product concentration.

The rates of formation and breakdown of the E - S complex are given in terms of known quantities:

- The rate of formation of E-S = \(k_1[E][S]\)

(with the assumption that [P] =0)

- The rate of breakdown of E-S = \(k_2 [ES] + k_3[ES] = (K_2 + K_3) [ES] \)

At steady state,

\[ \dfrac{d[ES]}{dt} = k_1[E][S] +k_2[ES] + k_3 [ES] =0\]

Therefore, rate of formation of E-S is equal to the rate of breakdown of E-S

So,

\[k_1[E][S] = (k_2 + k_3)[ES]\]

Dividing through by \(k_1\):

\[ [E] [S] = \dfrac{(k_2 + k_3)}{k_1} [E-S]\]

Substituting \(\dfrac{(k_2 + k_3)}{k_1}\) with k_{M}:

\([E] [S] = K_M [ES]\)

\[ k_M = \dfrac{breakdown [ES]}{formation [ES]}\]

\(K_m\) implies that half of the active sites on the enzymes are filled. Different enzymes have different \(K_m\) values. They typically range from 10^{-1} to 10^{-7} M. The factors that affect \(K_m\) are:

- pH
- temperature
- ionic strengths
- the nature of the substrate

Substituting [E_{F}] with [E_{T}]-[ES]: E_{T} = [ES] + [E_{F}]

([E_{T}] - [ES]) [S] = k_{M} [ES]

[E_{T}] [S] -[ES][S] = k_{M} [ES]

[E_{T}] [S] = [ES] [S] + k_{M} [ES]

[E_{T}] [S] = [ES] ([S] + k_{M})

Solving for [ES]:

[ES] = \(\dfrac{([E_T] [S])}{([S] + k_M)}\)

The rate equation from the rate limiting step is:

_{Vo} = \(\dfrac{dP}{dt}\) = k_{2}[ES]

Multiplying both sides of the equation by k_{2}:

\[ k_2 [ES] = k_2 (\dfrac{([E_T][S])}{(K_M + [S])}\]

\[ V_o = k_2 (\dfrac{([E_T][S])}{(K_M + [S])}\]

When S>>K_{M,} v_{o} is approximately equal to k_{2}[E_{T}]. When the [S] great, most of the enzyme is found in the bound state ([ES]) and V_{o} = V_{max}

We can then substitue k_{2}[E_{T}] with V_{max}to get the **Michaelis Menten Kinetic Equation:**

v_{o} =\(\dfrac{(v_{max}[S])}{(k_M+ [S])}\)

When \([S] << K_m\),

\(v = \dfrac{V_{max}[S]}{K_m}\)

This means that the rate and the substrate concentration are directly proportional to each other. The reaction is first-order kinetics.

When \([S] >> K_m\),

\[v = V_{max}\]

This means that the rate is equal to the maximum velocity and is independent of the substrate concentration. The reaction is zero-order kinetics.

Figure 2

Then, at

\(v = \dfrac{V_{max}}{2}\), \(K_m = [S]\)

\[v = \dfrac{V_{max}}{2} = \dfrac{V_{max}[S]}{K_m + [S]}\]

Therefore, \(K_m\) is equal to the concentration of the substrate when the rate is half of the maximum velocity. From the Michaelis Menten Kinetic equation, we have many different ways to find \(K_m\) and \(V_{max}\) such as the Lineweaver-Burk plot, Hanes-Woolf plot, and Eadie-Hofstee plot, etc.

For example, by taking the reciprocal of the Michaelis Menten Kinetics Equation, we can obtain the Lineweaver-Burk double reciprocal plot:

\[ v_o= \dfrac{(V_{max}[S])}{(K_M +[S])} \]

\[ \dfrac{1}{v} = \dfrac{(k_M + [S])}{v_{max} [S]} \]

\[ \dfrac{1}{v} = \left( \dfrac{K_m}{V_{max}} \right) \left(\dfrac{1}{[S]} \right) + \dfrac{1}{V_{max}} \]

Apply this to equation for a straight line \( y = mx + b \) and we have:

\[ y = \dfrac{1}{v} \]

\[ x = \dfrac{1}{[S]} \]

\[ m = {slope} = \dfrac{K_m}{V_{max}} \]

\[ b = {y-intercept} = \dfrac{1}{V_{max}} \]

When we plot \(y = \dfrac{1}{v} \) versus \( x = \dfrac{1}{[S]} \), we obtain a straight line.

\[ x-intercept = \dfrac{-1}{K_m} \]

\[ y-intercept = \dfrac{1}{V_{max}} \]

\[ slope = \dfrac{K_m}{V_{max}} \]

Figure 3

Another way to calculate these values (k_{M}, V_{max}) and represent enzyme kinetics:

\[ V_o = \dfrac{(V_{max}[S])}{(K_M +[S])} \]

v_{o} (k_{M} +[S]) = v_{max}[S]

v_{o} k_{M} + v_{o}[S] = v_{max} [S]

v_{o} [S] = -vo k_{M} = vmax [S]

Dividing through by [S] \( v_o = -k_m \dfrac{v_o}{[S]} + v_{max} \)

Figure 4

- Chang, Raymond.
*Physical Chemistry for the Biosciences*. Sansalito, CA: University Science, 2005. Page 363-371. - Atkins, Peter and de Paula, Julio.
*Physical Chemistry for the Life Sciences*. New York, NY: W. H. Freeman and Company, 2006. Page 309-313. - Stryer, Lubert.
*Biochemistry (Third Edition)*. New York, NY: W.H. Freeman and Company, 1988. Page 187-191.

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