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I am slightly confused about the role of $\ce{H2O}$ lysis in PII. I know that this releases $\ce{H+}$ ions, which I think are important because they drive the production of $\ce{ATP}$ through F-Atpase across the thylakoid membrane. But I don't understand how else this is important? With my current understanding, I would think that it is perfectly sufficient for PII to receive photons to excite an electron in the primary pigment reaction centre, which goes along the electron transport chain to PI. Is the water lysis there to replace electrons lost by photosystem 2? If so, how is it controlled that the correct number of $\ce{H2O}$ molecules are lysed so that excess electrons aren't generated?

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Generation of ATP via photons is the primary aim of the photosynthesis pathway, which is accomplished by the generation of a chemiosmotic gradient of $\ce{H+}$, which then drives the ATPase pump. All the other enzymatic setups are designed to just let this process run in the most efficient way.

I would think that it is perfectly sufficient for PII to receive photons to excite an electron in the primary pigment reaction centre

While you are partially correct in assuming this, since the ETC also participates in creation of a chemiosmotic gradient (Plastoquinone), such a mecahinsm would not be able to run repeatedly. Without the replenishment of the electrons of the PSII, it will not be able to initiate another round of electron transport. Hence, the oxidation of water is crucial in providing PSII the electrons it passed onto plastoquinone, so as to allow another round of photon-absorption and continuation of the photosynthetic process.

how is it controlled that the correct number of H2O molecules are lysed so that excess electrons aren't generated?

This is achieved by a complicated interplay of the enzymatic components within the PSII. In summary, utilizing the four stable oxidation states of manganese core in the redox center ($\ce{Mn}^{(2-5)+}$, four separate states of the center ($S1\rightarrow 4$) are formed. These states cyclically shunt one electron during each step, until a water molecule is bound to $S4$ which is then directly converted to $\ce{O2}$ by $P680+$, formed in PSII due to photon absorption, which is a strong reducing agent. Since the final step occurs only with water molecule bound to $S4$, which in turn is formed only by the cyclical transition between the various states of the redox center with systematic sequential loss of electrons, the overall process is very tightly constrained to extract only the limited number of electrons preventing any troublesome redox intermediates.

See here for some basic treatment of photosynthesis.

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  • $\begingroup$ Thank you for your answer. Your answer and some additional reading allowed me to understand the role of water lysis! However there is one question that has come up for me which I cannot seem to find the answer to: so quinone B, after accepting electrons and hydrogen, transfers electrons to cytochrome b6f as plastoquinone/quinol. These electron are accepted by cytochrome b6f, to re-form quinone B, which is not lipid soluble. How is the quinone B replaced in PSII? It cannot diffuse back in the phospholipid bilayer as it came? $\endgroup$ – Meep Nov 22 '16 at 12:03

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