Two questions

1. is product P or substrate S favored at equilibrium?

2. how fast is the product formed (or how fast does the substrate

Question 1: in this case, the product is favored (free energy is negative) at equilibrium, but we do not know how quickly the equilibrium will be established.

Question 2: rates are determined by the free energy of activation

The rate (abbreviated as shown below by the mathematical symbol dP/dt) is proportional to the substrate concentration [S]. The rate is set equal to a proportionality constant, called the first-order rate constant k, times the concentration of S:

The first order rate constant k is equal to a formula that contains the free energy of activation. If the free energy of activation decreases ("hill" gets smaller), the rate increases and vice versa.

Enzymes exhibit extraordinary specificity and rate enhancements. Consider the following cartoon version of an enzymatic reaction.

First key notion: enzyme stabilizes the transition state (shown above as the species in the brackets), not just the substrate.

Second key notion: enzyme achieves stabilization of transition state through multiple weak interactions. Approximately at 6kJ decrease in free energy of activation translates into a ten-fold rate increase; therefore, a 60kJ decrease prompted by multiple small interactions would translate into a 10¹⁰ rate enhancement.

Third key notion: same weak interactions that confer rate enhancement also confer specificity.

Fourth key notion: enzymes use properly positioned functional groups to aid bond cleavage/formation (i.e., covalent interactions)

How are enzymes studies?

Michaelis-Menten

For a reaction in which an enzyme E converts substrate S into product P, we can write the following expression:

Assumptions: one substrate leads to one product; ignore the reverse reaction in which E and P reform ES; assume that the conversion of ES to EP does not need to be analyzed as a separate step.

How will the rate of product formation vary as we increase the substrate concentration? The following graph shows the result of adding increasing concentrations of substrate S to an enzyme E and then immediately measuring the rate of product formation. You will note that the rate increases as the concentration of S increases but then begins to "level off".

A detailed analysis of the equation shown above using the steady state approximation, ultimately leads to the following equation:

How do we extract these terms from our rate data and what do they mean?

1. v_max is fairly obvious: the maximum possible velocity or rate

2. K_m can be extracted from the rate data. It's the same as the substrate concentration at half-maximal velocity

What can we learn from these values?

1. The Michaelis Menten constant K_m is used as a measure of substrate affinity for E or as an approximate dissociation constant for ES. The smaller the value of K_m, the tighter the binding of the substrate S. Note that K_m is often in the range of cellular concentrations of S so that E can respond to changes in substrate concentration.

2. The maximum velocity v_max is equal to the rate constant k_cat times the total enzyme concentration. k_cat is often called the "turnover number" (reactions per sec). When we compare these k_cat values for specific enzymes with the same reaction under non-enzymatic conditions k_n, we get a clear picture of the phenomenal rate enhancement produced by enzymes. That is, the non-enzymatic reaction is much slower; the ratio of k_cat / k_n is 10⁷ to 10¹⁷!

3. The ratio of k_cat /K_m is referred to as the "specificity constant". Values for this ratio often approach the diffusion-controlled limit indicating the extraordinary efficiency of many enzymes in literally converting every S to P that shows up at its doorstep.

Types

Irreversible: covalent modification of E

Reversible: non-covalent