# Can luciferin + ATP → luciferyl adenylate + PPi be reversed?

Reading about the light emitting reaction of luciferin + ATP, catalysed by luciferase I am confused about the following.

On this webpage the partial formula is written as:

luciferin + ATP → luciferyl adenylate + PPi

which seems to me like a one way reaction.

But on searching on Google "luciferase ATP equilibrium" I found websites like this one and this one which seem to indicate that the reaction is reversible, as there is an equilibrium.

Is it possible for this reaction to take place in reverse? And what about the link on books.google.nl showing a reversible reaction with light involved? What about that?

Thank you for clarifying.

EDIT 2014-02-25: This article: "Kinetics of inhibition of firefly luciferase by dehydroluciferyl-coenzyme A, dehydroluciferin and L-luciferin" by Luis Pinto da Silva etc (Link) shows the following reactions:

Where Luc is firefly luciferase, LH2 is luciferin, L-CoA is dehydroluciferyl-coenzyme A, L is dehydroluciferin, ATP is Adenosine triphosphate, AMP is adenosine-5-monophosphate, PPi is inorganic pyrophosphate, O2 is oxygen, CO2 is carbondioxide.

Looking at the last equation, it seems it is possible to form ATP as a product, as it is on the right side of the arrow. Remarkable, I would say. still I do not see any luciferin (LH2) on the right side of any equation, so the answer to question of this topic remains no` thus far.

In theory every chemical reaction is reversible.

The reaction that you mention does not emit any light yet, that comes later with the following reaction: $$\text{Luciferyl adenylate} + \ce{O2 ->} \text{Oxyluciferin} + \text{AMP} +h\nu$$

Since the reaction you mentioned occurs under loss of pyrophosphate ($\ce{PP_{i}}$) which is where the energy was stored beforehand, I guess it's somewhat hard to get it back.

It's almost like hoping that $$\ce{2CO2 + 3H2O -> C2H5OH + 3O2}$$ occurs under room temperature. Which it doesn't, because $\ce{CO2}$ is such a thermodynamical black hole. I think the same applies to your question.

• Simple answer. I like it. – Mike de Klerk Feb 19 '14 at 13:05

Reactions that are metabolically irreversible are usually so, because mechanisms are in place to keep the mass action ratio (Q) far away from the equilibrium constant (K). Recall that equilibrium occurs when Q = K.

The change in Gibb's free energy for standard conditions ($\Delta G°$) determines if the reaction is "inherently" thermodynamically favourable or not. But in reality the concentrations of reactants will be far away from their assumed 1 M values. Remember that the actual change in Gibb's free energy is related to $\Delta G°$ by the following expression:

$$\Delta_rG=\Delta_rG° + RT \ln Q$$

In this particular reaction, hydrolysis of ATP to release $\text{PP}_i$ makes the reaction exergonic / spontaneous ($\Delta G<0$). But in a hypothetical scenario with no substrates, and high concentrations of products (which would make $Q$ large, and thus $\Delta_r G$ positive), the reverse reaction would be spontaneous.

Now, if the reaction is practically irreversible, it's because things are being done to prevent such a situation from happening. For example, pyrophosphatase hydrolyses $\text{PP}_i$ to $2\text{P}_i$ at a high rate.

In many metabolic pathways, irreversible reactions are also catalysed by enzymes, that are inhibited by either their product, or a product downstream in the pathway, thereby inactivating the enzyme long before equilibrium is reached.