Pyrodoxal-5-phosphate (PLP) in enzymes form schiff-base to lysine residue. This can be detected by the change in the absorption from 390 nm to 420 nm (free PLP to stiff-base bound PLP).

I remember from chemistry that the pi-orbital electrons can be excited, so you get a pi-pi transition when light is absorbed.

In PLP you have the aldehyde group on top (left image) which then forms a Schiff base with a lysine residue in the enzymes active site (right image). How does this results in a absorption shift from 390 nm to 420 nm? You still have five double bonds in the molecules, which gives 10 pi-electrons as far as I can remember from my organic chemistry. The point of the Schiff base should be to lower the pKa of the C-alpha in the substrate, which replaces the enzyme bound schiff-base. I just dont see why the absorption is changed from bound to unbound form.

Does this have something to due with the free electrons pairs on the oxygen of the aldehyd group?

Left) PLP right (PLP-Schiff base to enzyme)

  • 2
    $\begingroup$ The shift is rather small in energy terms, approx 1830 cm$^{-1}$ vs energy of electronic state, approx 25600 cm$^{-1}$, and must result from the slightly different energy of the C=O vs C=N orbitals involved in the delocalisation with the ring. I think that it is rather difficult in such a complex molecule to predict which has the higher energy. $\endgroup$ – porphyrin Nov 8 '18 at 15:23

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