Biochemistry

Lecture 14: Enzyme Kinetics

Assigned reading in Campbell: Chapter 5.1-5.6

Take the Review Quiz on Lecture 14 concepts.

Substrate Saturation Kinetics shows various graphs used to analyze kinetic data.

Campbell's text is excellent on all of the assigned sections for today's lecture. These notes develop the same concepts in a slightly different order and add some material in the Transition State Theory and Catalytic Efficiency and the Specificity sections below.

For the simple enzyme-catalyzed reaction:

S <==> P

S is substrate;

P is product.

The enzyme forms a complex with the substrate, in much the same way as a protein-ligand complex, and performs some chemical reaction/transformation of the bound substrate. The resultant product is released.

Important Features of Enzyme Catalysis.

1. Enzymes increase the rate of reactions.
2. Enzymes do not change the equilibrium point of reactions.
3. Enzymes are specific for their substrates:
   • Geometric complementarity;
   • Energetic complementarity.
4. Enzymes are regulated:
   • at genetic level (transcription, translation);
   • by concentration of substrate and product;
   • allosterically.

Background Kinetics

1. Order of Reactions:

A-> P       rate ~ [A]¹       1^st order

A + A-> P       rate ~ [A]²       2^nd order

A+B->P       rate ~ [A][B]       2^nd order

2. Velocity of a reaction:

v = -d[A]/dt = d[P]/dt

= k[A][B] (for a second order reaction)

3. Transition State Theory:

Is there a physical limit to the rate of a chemical reaction? There is indeed such an upper limit; and transition state theory provides an estimate of the rate constant. As the reaction progresses, the substrates are converted to products. At some point along the reaction progress curve, the chemical species reach a high energy configuration called the transition state (Campbell, Fig. 5.1). This state is labeled by the symbol X^‡.
Transition state theory states that the rate of reaction is directly proportional to the concentration of the transition state:

v = k'[X^‡]

The key assumption made in transition state theory is that the substrates and products are in equilibrium with the transition state. We can then write an equilibrium constant, e.g. for the forward reaction:

K^‡ = [X^‡]/[A][B]

Therefore the velocity of the reaction is: v = k'K^‡[A][B]. Comparing this equation to that for a second order reaction gives: k = k' K^‡.

k^' is the rate at which the transition state will decay to products. This is given by:

Using the relationship between free energy and the equilibrium constant gives the following:

k = (k_BT/h)exp(-DG^‡/RT)

From the above equation it is clear how enzymes increase the rate of the reaction - they must do so by lowering the energy of the transition state in the enzyme substrate complex. This is shown by the free energy diagrams below:

Note that enzymes bind to both the substrates and products and stabilize the transition state. The transition state of the enzyme substrate complex is stabilized in two ways:

• First, the enzyme transition-state complex is stabilized by direct interactions between the enzyme and the transition state. This reduces the free energy of the transition state.
• Second, the enzyme may provide a number of functional groups to aid in catalysis. Since these groups are positioned in well defined locations, the entropy of bringing these groups into the catalytic site is substantially less than having these functional groups freely diffuse in solution.

Steady State Enzyme Kinetics:

• Steady-state reactions are easy to measure because the rate of the reaction is constant for relatively long periods of time.
• Steady-state rates are those which are most relevant to metabolic levels.
• The analysis of steady state kinetics can only provide limited information on the kinetic mechanism.

The simplest reaction scheme is:

where,
k₁ is the forward rate constant for substrate binding
k_-1 is the reverse rate constant for substrate binding.
k₂ is the catalytic rate constant (containing terms related to the transition state).
The ES complex is also called the "Michaelis complex".

The velocity of the reaction is:

The change in [ES] as a function of time is:

During the steady state: d[ES]/dt = 0

The goal is to relate this equation to readily measurable experimental parameters, such as:

• The total amount of enzyme: E_T = [E] + [ES]
• The concentration of substrate: [S]
• The measured steady state velocity (v = k₂ [ES])

We do not have a suitable way to measure [E], so the total enzyme concentration will be used in its place:

[E] = E_T - [ES]

[ES](k_-1 + k_-2) = k₁[S](ET -[ES])

[ES](k_-1 + k₂) = k₁ E_T[ES] - k₁[ES][S]

[ES](k_-1 + k₂ + k[S]) = k₁ E_T[S]

[ES] = k₁E_T[S]/(k_-1 + k₂ + k₁ [S])

The velocity of the reaction:

v = k₂[ES]

= k₁k₂E_T[S]/{k_-1 + k₂ + k₁[S]}

= k₂E_T[S]/{(k_-1 + k₂)/ k₁+ [S]}

Defining a few terms:

The V_max or maximal velocity: V_max = k₂E_T

The K_M or Michaelis constant: K_M = (k_-1 + k₂)/ k₁

The use of these definitions gives the Michaelis-Menten Equation

The class handout, Substrate Saturation Kinetics shows a plot of reaction velocity versus [S].
Also shown is a graph of the data in a double reciprocal plot, otherwise known as a "Lineweaver-Burk Plot".

Catalytic Efficiency and the Specificity of Enzyme Reactions: k_cat/K_M

The catalytic constant of an enzyme is defined as:

k_cat = V_max/E_T

In the case of the simple example discussed above, k_cat = k₂.
(The units of k_cat are time^-1; typically, sec^-1. Thus, if k_cat = 1000 sec^-1, the enzyme can convert 1000 molecules of substrate into product each second at saturating [S].)

Treating the enzyme catalyzed reaction as a simple second order reaction:

v = k[E][S]

Compare this equation to the Michaelis-Menten equation above. At very low substrate concentrations (E_T =[E]), the Michaelis-Menten equation becomes:

k_cat/K_M is the second order rate constant for this reaction. It represents the binding and catalysis of the enzyme and the free substrate.

Using the definitions of K_M:

If the chemical reaction is rate-limiting (k_-1>>k₂) then: k_cat/K_M = k₁k₂/k_-1 , and the observed reaction rate depends both on how well the substrate binds as well as the rate of catalysis.

If the chemical reaction is not rate-limiting (k_-1<<k₂) then: k_cat/K_M = k₁ and the rate is only limited by how fast the substrate can diffuse into the active site.

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