Introduction

An enzyme's active sites are usually composed of amino acid residues, depending on which amino acid residues are present, the specificity of the substrate can vary greatly. As we all know, depending on the pH level, the physical properties (mainly, the electric charge) of an enzyme can change. A change in the electric charge can alter the interaction between the active site amino acid residues and the incoming substrate. With that said, the substrate can bind to the active site via Hydrogen bonding or Van Der Waals Forces. Once the substrate binds to the active site it forms an enzyme-substrate complex that is then involved in further chemical reactions. 

In order for an enzyme to be active and be energetically favorable to allow a chemical reaction to proceed forward, a substrate must bind to an enzyme's "active site". An active site can be thought of as a lock and the substrate as a key, known as the lock and key theory. A key (substrate) must be inserted and turned (chemical reactions), then the lock (enzyme) will open (production of products). Note that an enzyme might have more than one active site. Another theory on the active site-substrate relationship is the induced fit theory, which is quite opposite of the lock and key theory (where the active site is seemingly inflexible). In the induced fit theory, the active site of the enzyme is very flexible, and only changes its conformation when the substrate binds to it.

Enzymes work as a catalyst by lowering the Gibbs free energy of activation of the enzyme-substrate complex. Below are two figures showing a basic enzymatic reaction with and without a catalyst.

Notice the difference of the Gibbs Energy with the presence of a catalyst; it is higher in value when a catalyst is absent. We know that Gibbs Energy can tell us whether or not a reaction is spontaneous judging from its value. In this case  ∆G­^‡,the Gibbs Free Energy, can be seen as the probability of whether or not a reaction would occur in nature, without any interference. With a lower Gibbs energy values, the probability of reaction is higher, and vice versa. Enzymes are here to increase the probability of the chemical reaction via many mechanisms, with one of the more prominent explanations being that it changes into a more favorable conformation.

Now that we have a basic understanding of enzyme kinetics, how does one determine the effiency of the enzyme? Lets start by looking at a simple enzymatic reaction:

  Fig 3

German biochemist Leonor Michaelis and Canadian biochemist Maud Menten derived an equation later known as the "Michaelis-Menten Equation", shown below.

\[ v_0 = \dfrac{v_{max}[S]}{K_M + [S]} \]

From a quick glance, this equation tells us the rate of the reaction at a given substrate concentration, given that we know V_max , which is the maximum rate the reaction can proceed at, and also K_M, the Michaelis constant. However, in a practical application of the Michaelis-Menten, V₀ is often measured, and V_max is observed when we see saturation or plateau in our data plot. Since we know the substrate concetration, K_M is usually the calculated value we are after.

For K_M we know that V₀ = V­_Max/2

V­_Max/2 = (V_max[S])/(K_M + [S])

(K_M + [S])(V­_Max/2) =  (V_max[S])

(K_M + [S]) = (V_max[S]) / (V­_Max/2)

K_M + [S] = 2[S]

K_M = [S]

The Michaelis constant can be thought of as the rate at which the substrate becomes unbound to the enzyme, which can either be in the events of substrate-enzyme complex becoming the product, or the substrate becomes unbound to the enzyme. K_M can be shown as an equation.

\[ K_M = \dfrac{k_{-1} + k_2}{k_1}  \]

Whereas k_-1 is the rate constant at which the substrate becomes unbound to the enzyme, resulting in the dissociation of the enzyme-substrate complex, K₂ is the rate constant where the substrate-enzyme complex dissappears and turns into product, and K₁ is the rate constant for the formation of the the substrate-enzyme complex formation. Therefore, we can view K_M as the rate of substrate-enzyme complex dissappearance divided by the rate of substrate-enzyme complex formation, which is the level at which half of the substrate is bound to the enzyme. K_M is a useful indicator for the presence of an inhibitor because we can look for changes in K_M and compare to our control (biological systems that we know have zero inhibitor presence). K_M is a dependant variable, and its value can change due to many reasons, including the pH level of the system, temperature, or any other condition that might affect a chemical reaction. A small K_M means that the substrate has a high affinity to the enzyme.

So in what ways can the Michaelis-Menten equation tell us about the efficiency of the enzyme? If we were to graph V₀ vs [S], we would end up with something like this.

V_max is the maximum rate the reaction can run at, regardless of [S], meaning that even if you add more substrate, the reaction would not go any faster. That is because at V_max all of the active sites on the enzyme are occupied.  After all the explanations on various forms of enzyme kinetic equations, we arrive at our conclusion of catalytic efficiency. Referring back to Fig 3, we have:

 Fig 3

Notice k₂ is an irreversible reaction as oppose to an equilibrium expression, when compared to k_-1 and k₁. K₂ here is also known as k_cat , the catalytic effiency of enzyme. From our previous discussion we learned that V₀ is the measured reaction rate, which is the product formation over time, so we can conclude that an equation would end up looking like the following:

­\[ v_0 = \dfrac{d[P]}{dt} = k_2[E]_0 \]

Where [E]₀ is the total enzyme concentration.

We also know that V­_Max is observed when all of the enzyme-substrate complex disappear and turn into products, so we can make the following assumption:

\[ V­_{Max} = k_2[E]_0 \]

and after rearrangment, we have this equation:

\[ k_{cat} = k_2 = \dfrac{V­_{Max} }{[E]_0} \]

That is the equation for calculating catalytic efficiency, to be used after we obtain data from experiments and after using the Michaelis Menten Equation. With a larger k_cat , the enzyme is efficient because less enzyme is needed.

Inside Links

•  http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Complex_Reactions/Enzymatic_Kinetics/Enzyme_Inhibition
• http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Complex_Reactions/Enzymatic_Kinetics/Michaelis-Menten_Kinetics
• http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics

Problems

1. What is the rate of product formation if k₂ is 4.3 min^-1 and [ES] is 2.3 x 10^-2 M?
2. Find the formation of ES if k₁ = 3.3 x 10³ min^-1, k_-1 = 1.1 x 10^-10 min^-1 and [E] and [S] are 1.2 x 10^-6 M and 6.3 x 10^-2 M respectively. 
3. Given that k_-1 = 8 x 10­­⁴ s^-1, k₂ = 9 x 10­­⁵ s^-1, k₁ = 7 x 10­­⁶ ­M^-1s^-1 find K_M. What does the answer tell us about the affinity of the substrate?
4. Using the K_M value from above, find V₀ if we determine experimentally the V_max value to be 1.5 x 10^-4 M min^-1 and the [S] is 5.1 x 10^-4 M. 
5. Judging from K_M from problem 1, we hypothesize that [E]₀ is half of [S], find k₂ using the V_max value given above. 

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