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:
$$ \text{Luciferyl adenylate} + \ce{O2 -&gt;} \text{Oxyluciferin} + \text{AMP} +h\nu $$

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
$$ \ce{2CO2 + 3H2O -&gt; C2H5OH + 3O2} $$
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.