Friday, March 28, 2014

Reversibility and Product Inhibition

 Originally Posted on  by hsauro

It is common in biochemistry to categorize reactions as either reversible or irreversible. What do we mean by these terms? Let’s look at a simple reaction such as:

  \[A \rightarrow B\]

where A is transformed to B. The rate of reaction for this transformation can be assumed to be first-order, that is:

  \[v = k_1 A\]

Mathematically this is an irreversible reaction because the reaction only goes from reactant A to product B. We can easily make the reaction reversible by adding the reverse reaction:

  \[A \rightleftharpoons B\]

If we again assume first-order kinetics for the reverse process then the reaction rate is now given by:

  \[v = k_1 A - k_2 B\]

Depending on the ratio of k_1 to k_2, if the concentration of B is high enough the reaction rate, v will actually go negative. This is the main characteristic of a reversible reaction, the potential for the forward reaction rate to go negative. If k_2 were very small relative to k_1 we might find that the concentration of B required to make the reaction go in reverse is unrealistically high. For example, if k_1 = 1 and k_2 = 0.0001, and the concentration of A was 1 mM, then the concentration of B required to reverse the reaction would be 10 M. In a biological cell such a concentration would be highly unlikely. In this situation we would probably assume for all practical purposes that our reaction is irreversible.

Another way to look at this is to recall that the ratio of the forward to backward rate constant (k_1/k_2) is equal to the equilibrium constant, K_{eq} = B/A. In our example the equilibrium constant was 10,000. One may think that such equilibrium constants are quite rare but as Cornish-Bowden and Cardenas reported, the equilibrium constant for pyruvate kinase is of the order of 100,000.

An important question is how should we represent reactions that have high equilibrium constants in our computer models? Do we simply use irreversible rate laws to model such reactions? The short answer to this is probably no.

What we often forget is that the product of an enzyme catalyzed reaction can reversible bind to the enzyme and when it does it will compete with the substrate. This means that the reaction rate is likely to decrease as product accumulates simply due to the competitive effect of product. This effect is called product inhibition. For a Michaelis-Menten like rate law it is easy to adjust the equation to take into account product inhibition, we just treat the product as a competitive inhibitor:

  \[v = \frac{V_m S}{S + K_m \left( 1 + \frac{P}{K_p} \right)}\]

where V_m is the maximal rate, K_m the substrate concentration with no product that gives half the maximal rate and K_p is the dissociation constant for the product. Depending on the value for K_p it is quite possible for the product to greatly affect the forward rate. The reaction is irreversible but the forward reaction can be reduced via product inhibition.