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Wize University Biochemistry Textbook > Enzyme Catalysis

Enzyme Kinetics

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Enzymes Kinetics

  • Kinetics: the study of enzymatic reactions
  • Investigates the maximum rate of reaction as well as substrate and/or inhibitor specificity
  • Enzymes lower the energy barrier a reaction must overcome
  • The enzyme does not contribute the the reaction (is in native state at the end of reaction)
  • It does not alter the equilibrium of the reaction
  • Rate of reaction (V): The quantity of substrate that disappears in a unit of time
  • Initial rate of reaction (V0): Rate when substrate concentration is constant
  • Maximum rate of reaction (Vmax): Maximum rate of reaction when the enzyme is saturated by substrate
V0=Δ[S]ΔtV_0=\frac{Δ[S]}{Δt}




https://commons.wikimedia.org/wiki/File:Carbonic_anhydrase_reaction_in_tissue.svg. Fvasconcellos. This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.
https://commons.wikimedia.org/wiki/File:Enzyme_accelerating_reaction.jpg. Kristie.leong. This file is licensed under the Creative Commons Attribution 3.0 Unported license.
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Michaelis-Menten Equation

  • A way of expressing the relationship between initial velocity, maximum velocity, and initial substrate concentration

V0=Vmax⁡⋅[S]Km+[S]=kcat⋅[E][S]Km+[S]V_0=V_{\max}\cdot\frac{\left[S\right]}{K_m+\left[S\right]}=k_{cat}\cdot\frac{\left[E\right]\left[S\right]}{K_m+\left[S\right]}

https://commons.wikimedia.org/wiki/File:Michaelis-Menten_saturation_curve_of_an_enzyme_reaction_LARGE.svg. U+003F. This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication.

Michaelis-Menten constant (Km): The [S] at which V0=1/2 Vmax
  • Indicates how [S] affects enzyme function
Km=k2+k−1k1K_m=\frac{k_2+k_{-1}}{k_1}

Wize Tip
The lower the Km value, the more efficient the enzyme is.

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Turnover Number (kcat): The maximum theoretical reaction rate for a single saturated enzyme
  • eg. if kcat=9 s-1, then for every 1 second, 1 enzyme can process 9 substrate molecules
kcat=Vmax⁡[E]k_{cat}=\frac{V_{\max}}{\left[E\right]}

Wize Tip
The higher the kcat value, the more powerful the enzyme is.


Specificity Constant (kcat/Km): The ratio of the turnover number and the Michaelis-Menten constant.
  • Used to compare enzymes
  • The higher the ration the more efficient and more powerful (overall better) the enzyme

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Lineweaver-Burke Plot

  • The double reciprocal of the Michaelis-Menten equation
  • Allows for a more precise identification of Vmax and Km

1V0=KmVmax⁡⋅1[S]+1Vmax⁡\frac{1}{V_0}=\frac{K_m}{V_{\max}}\cdot\frac{1}{\left[S\right]}+\frac{1}{V_{\max}}

https://commons.wikimedia.org/wiki/File:Lineweaver-Burke_plot.PNG. GFDL. This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.
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Select the Option that lists the proper labels from A -> E for the following plot:


a) 1[S]\frac{1}{\left[S\right]} , 1V0\frac{1}{V_0}, 1Vmax⁡\frac{1}{V_{\max}}, 1Km\frac{1}{K_m} , KmVmax⁡\frac{K_m}{V_{\max}}

b) 1V0\frac{1}{V_0} , 1[S]\frac{1}{\left[S\right]} , 1Vmax⁡\frac{1}{V_{\max}} , 1Km\frac{1}{K_m} , KmVmax⁡\frac{K_m}{V_{\max}}

c) 1[S]\frac{1}{\left[S\right]}, 1V0\frac{1}{V_0}, 1Km\frac{1}{K_m} ,−1Vmax⁡-\frac{1}{V_{\max}} , Vmax⁡Km\frac{V_{\max}}{K_m}

d) 1V0\frac{1}{V_0} ,1[S]\frac{1}{\left[S\right]} , 1Vmax⁡\frac{1}{V_{\max}} ,−1Km-\frac{1}{K_m} , KmVmax⁡\frac{K_m}{V_{\max}}

d) Is the correct order of labels
Which Statement is TRUE about enzyme kinetics?
You are studying an enzyme and have calculated that the Michaelis-Menten constant is 2M. You also know that when the substrate concentration is 4M, the maximum velocity of the reaction is 6M/min. What is the initial velocity of this reaction?