C16H20N3SCl + O ⇆ C16H18N3SCl + H2O
(leuco-methylene blue) (methylene blue)
Briefly—MH2 + O ⇆ M + H2O
To reduce methylene blue we can add the two hydrogen atoms directly from nascent hydrogen formed in the solution or we can split up water by a catalyzer in the presence of some substance, which will take up the oxygen of water, thus:
NaH2PO2 + H2O + Pd = NaH2PO3 + H2 + Pd
(Sodium hypophosphite) (Sodium phosphite)
This reaction occurs in presence of finely divided palladium. Methylene blue can be reduced by the H2 and the hypophosphite oxidized.
Since oxyluciferin can be reduced by palladium and sodium hypophosphite (Harvey, 1918), it is probable that we can write the equation for reduction of oxyluciferin and oxidation of luciferin in a similar manner to that of methylene blue:
Luciferin + O ⇆ Oxyluciferin + H2O
Briefly—LH2 + O ⇆ L + H2O.
Just as in the case of methylene blue the reaction proceeds in the right hand direction spontaneously if the
pressure of O is sufficiently high. If luciferase is also present we have luminescence.
LH2 + O + luciferase ⇆ L + H2O + luciferase (luminescence)
The reaction proceeds in the left hand direction under low oxygen pressure, in the presence of nascent hydrogen or with some catalyzer which is able to split water, transferring the H2 to oxyluciferin and the O to an acceptor (A). NaH2PO2 plays the part of the acceptor.