Question

If only one water molecule were oxidized by PSII, how many photons of light would have to be absorbed and transformed into chemical energy? How many NADPHs would be produced at the other end of the light reaction, and where would the electrons and protons for the last step in the light reaction come from? Explain.

In the same situation (one water oxidized), how many protons would accumulate, and where and how would they accumulate in support of making a proton gradient? Explain.

Answer 1

In case of oxidation of only one water molecule by the PSII two photons of light would have to be absorbed and transformed into chemical energy. On the other hand, if two water molecules are needed to be oxidized for the production of one NADPH molecule on the stroma side of the PSI.

Remember that the photosynthesis follows the Z scheme of electron travel from the PSII to PSI so the electrons that were generated from the oxidation of the water molecule at the PSI will go to PSII and the photons will come from the sunlight in order to generate NADPH.

The oxidation of water results into splitting of the H2O into 2H+ and O, since it is a light reaction and photons cause continuous oxidation of water molecule at PSII a lot of H+ will be produced and thus the concentration of H+ towards the internal side of the thylakoid membrane will be more this will thus drive the ATP synthase which transfers H+ down its concentration gradient and generate ATP from ADP, this is how the ATP will be produced.

Source http://en.wikipedia.org/wiki/Image:Z-scheme.png

Figure source (http://cnx.org/contents/[email protected])

The figure indicates the chemiosmosis phenomenon where the H+ ions generated at the PSII are crossing from the ATP synthase protein embedded at the thylakoid membrane to generate ATP towards the stroma side of the chloroplast.