# Michaelis-Menten Kinetics

By Levi Clancy for Student Reader on
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▶︎ View related▼︎ Tap to hide [E] Concentration of free enzyme. Concentration of free substrate. Concentration of enzyme-substrate complex. This is equal to [Et] [S] / (Km + [S] ). In the reaction E + S ⇔ ES ⇔ E + P, the rate of ES formation is V = k1 ( [ET - [ES] ) [S] -- the difference between [ET] and [ES] is equivalent the [E]. The rate of ES breakdown is V = k-1 [ES] + k2 [ES]. k-1 [ES] refers to the velocity of ES → E + S. k2 [ES] refers to the velocity of ES → E + P. Concentration of total enzyme, free and substrate-bound. At Vmax where the substrate saturates the enzyme, then [ET] = [ES] and [E] = 0. If the rate-determining step is the reaction rather than the enzyme-binding, then Vmax = k2 [ET] -- it changes with [ET] and by the definition of enzyme saturation, there is no [E] when Vmax is achieved so ∴ [ET] = [ES] Concentration of the product of the enzyme-catalyzed reaction. Equilibrium constant. Shortened as just K', this is equal to [products] / [reactants] which is [P] / [S]. This allows you to determine which way is the reaction proceeding. Has units time-1. A higher value means this step is favored. Has units time-1. Has units time-1. Has units time-1. kcat is the rate-determining step in a reaction. For the standard reaction used in these formulae, then k2 = kcat. However, if there are perhaps several intermediates, then the step with the highest ΔG‡ will be the kcat whether k2, k3, k4, etc. kcat = Vmax / [ET] The reaction velocity refers to how much S becomes P per unit of time. For a first order reaction where S ⇔ P, the substrate is the only factor influencing the reaction velocity: the V = k [S] where k is the rate constant. For a second order reaction E + S ⇔ E + P then V = k [E] [S] where k is the rate constant. Pronounced V-knot. This is the velocity of the reaction before all the enzyme is saturated with substrate, at which point it is represented by Vmax. At this point, assuming the enzyme concentration is fixed, the reaction rate is proportional to the substrate concentration; add more substrate and you'll get more reaction. This essentially measures how quickly the enzyme can bind the substrate under these conditions. V0 = kcat [ES] = { kcat [Et] [S] / ( Km + [S] } = { Vmax [S] / (Km + [S] ) } Pronounced V-max. At this point, addition of more substrate does not cause much difference in its rate of conversion. There is so much substrate that the enzyme is saturated. As soon as it converts one substrate molecule, it will bind another one because there are just so many in solution. This is the maximum speed of the enzyme-catalyzed reaction. Michaelis' constant, Km, has molar units. It refers to how well the enzyme binds the substrate (while kcat refers to how well it converts the substrate to product). It is equal to ( k-1 + k2 ) / ( k1 ) -- the numerator is the breakdown of ES, and the denominator is the formation of ES. Thus, Km reflects the affinity between the enzyme and substrate. More specifically, it represents the rate when every enzyme is saturated at which the ES complex becomes unbound (converts) either into E + S (K-1) or into E + P (K2). Km = [S] when V0 is ½ Vmax. In many cases, k2 << k-1 because k2 is the rate-determining step. Thus, Km may be reduced to k-1 / k1 which is the dissociation constant. However, in other cases Km remains a more complex function of all three constants.

## Michaelis-Menten Equations   The Michaelis-Menten equation assumes that the binding step is fast and the reaction step is slow.

This is a significant assumption that excludes many enzyme-catalyzed reactions. We also must assume that the rate of ES formation is equal to its breakdown before the enzyme is fully saturated with substrate.

k The rate constant k refers to the percentage of the reactant that is converted to product and is in inverse units of time (usually s-1). For example, if k = .05 s-1 for the reaction S ⇔ P then 5% of available substrate is converted to P in one second. If k = 2000 s-1 then the reaction occurs in a fraction of a second. For a first order reaction k is in inverse units of time (usually s-1) and for a second order reaction k is in inverse units time and molarity (usually s-1 M-1). The larger the value of k, then the faster the reaction. This represents the total change in energy in the reaction, the difference between the activation potential energy and the product potential energy. The ΔG'0 defines the reaction equilibrium. Correlation between ΔG'0 and [P] and [S] (the equilibrium constant) is expressed by ΔG'0 = -R T ln K'eq This represents the activation energy, the difference between the reactant potential energy and the highest potential energy peak. The ΔG‡ defines the reaction rate. The relationship between the rate constant and the ΔG‡ is represented by: k = ( k T e(-ΔG‡/RT) ) / h with k equalling Boltzmann's constant and h equalling Planck's constant and T equalling the absolute temperate 298 ° Kelvin. The relationship between the rate constant and ΔG‡ is inverse and exponential so ∴ the lower the activation energy, the faster the reaction rate. The specificity constant (aka catalytic efficiency) is kcat / Km. It has no laconic variable to represent it in equations. As it measures catalytic efficiency, it is usually just abbreviated as S.C. or C.E. in formulae. The catalytic efficiency refers to how quickly the catalyzed reaction proceeds and how well the enzyme binds the substrate. It thus refers to the reaction from E + S ⇔ E + P instead of E + S ⇔ ES ⇔ E + P. It is equivalent to kcat (measuring the rate of the reaction) divided by Km (measuring the enzyme-binding). A higher catalytic efficiency means a quicker conversion of S to P. Thus, if an enzyme has high kcat (fast reaction) and high km (poor binding) then it could be just as efficient an enzyme with low kcat (slow reaction) and low km (good binding).

## Catalytic Perfection

Catalytic efficiency between 108 and 109 M-1 s-1 is considered perfect. The enzyme binds and converts the substrate immediately upon encountering it.

Any efficiency beyond this is biologically pointless since the enzyme's performance will be capped by the rate of diffusion of the substrate to the enzyme.

## Lineweaver-Burk Plot Thus, the x-intercept (where the line hits the x-axis) is equal to 1/Km and the y-intercept is equal to 1/Vmax. The slope is Km / Vmax.

Lineweaver-Burk plots are thus useful when a few data points are available and information must be gleaned accurately.