Flow of electrons in Z scheme

·         The Z-scheme describes the complete route by which electron flow from  H2O to NADP+, according to the equation:  2H₂O + 2NADP⁺  + 8 photons               O₂ + 2 NADPH + 2 H⁺

·         For every two photons absorbed (one by each photosystem) one electron is transferred from H2O to NADP+, a total eight photons must be absorbed, 4 by each PS.

·         Excitation of P680 in PSII produces P680*, an excellent electron donor that transfers an e to pheophytin, giving it (-ve) charge.

·         With the loss of its e ,P680* is transformed into a radical cation P680+, pheo- very rapidly passes  its extra electron to a protein bound plastoquinone, PQ_A which in turn passes its e to another more loosely bound plastoquinone PQ_B.

·         When PQ_B has acquired 2 e in two such transfer from PQ_A  and two photons from solvent water, it is in its fully reduced quinole form PQ_BH₂. The overall reaction in PSII is:

·         4 P680 + 4 H⁺ + 2 PQ_B + 4 photons                      4 P680* + 2 PQ_BH₂

·         Eventually the e in PQ_BH₂  passes through the cytochrome b6f complex. The e initially removed from P680 are replaced with an e from oxiadtion of water.

·         The excited reaction center P700* loses e to an acceptor A0 again excitation results in charge separation at the photochemical reaction center. P700+ is a strong oxidizing agent, which quickly acquires an e from plastocyanin, a soluble Cu-containing e transfer protein.

·         A0- is a strong reducing agent that passes its e through a chain that leads to NADP+.

·         First phylloquinone (A1) accepts an e and passes it to an Fe-S protein three Fe-S center in PSI.

·         From here, e moves to ferredoxin (fd).

·         The fourth e carrier in chain is the flavoprotein ferredoxin : NADP+oxidoreductase which transfer e from reduced ferredoxin to NADP+.

·         2 Fd(red) + 2 H+ + NADP+                 2 Fd (ox) + NADPH + H+