In oxygenic photosynthesis (as in green plants and cyanobacteria), water is split at Photosystem II. What, specifically, does this photolysis of H2O supply to the photosynthetic apparatus?
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Aelectrons needed to reduce P680 (i.e., to re-reduce oxidized P680+ in Photosystem II)
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BO2 needed for the dark reactions (Calvin cycle)
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Celectrons needed to reduce NADH
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Delectrons needed for cyclic photophosphorylation
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Eprotons needed exclusively for mitochondrial ATP synthesis
Answer
Correct Answer: electrons needed to reduce P680 (i.e., to re-reduce oxidized P680+ in Photosystem II)
Explanation
Introduction / Context:In oxygenic photosynthesis, water serves as the ultimate electron donor. When light energizes the reaction center chlorophyll P680 in Photosystem II (PSII), it ejects an electron and becomes oxidized to P680+. The system must quickly replace that electron to keep electron flow and ATP/NADPH production moving. This question tests whether you know exactly what the photolysis of water provides and where it fits in the Z-scheme of photosynthesis.
Given Data / Assumptions:
- Oxygenic photosynthesis employs two photosystems in series (PSII then PSI).
- Light excites P680 at PSII; the excited electron is passed down an electron transport chain.
- Water is split by the oxygen-evolving complex (OEC) associated with PSII.
- The Calvin cycle (dark reactions) uses ATP and NADPH, not O2 as a reactant.
Concept / Approach:PSII oxidizes water via the OEC (manganese cluster), generating electrons, protons (contributing to the thylakoid proton gradient), and molecular oxygen as a by-product. The critical role of water here is to supply electrons that re-reduce the strong oxidant P680+ back to P680, allowing continuous photo-oxidation cycles and downstream reduction of plastoquinone, cytochrome b6f, plastocyanin, and PSI.
Step-by-Step Solution:
Light excites P680 → P680* → electron transferred to primary acceptor.P680* becomes P680+ (a very strong oxidant) and must be reduced.Water is split at the OEC: 2 H2O → O2 + 4 H+ + 4 e-.These electrons reduce P680+ back to P680, sustaining electron flow.Verification / Alternative check:O2 evolution strictly correlates with PSII activity and water splitting; inhibitors of PSII halt oxygen production and electron donation to PSI, demonstrating the role of water as the source of electrons for P680+ reduction.
Why Other Options Are Wrong:
- O2 for dark reactions: the Calvin cycle does not require O2; it fixes CO2 using ATP and NADPH.
- Electrons to reduce NADH: photosynthesis reduces NADP+ to NADPH, not NADH.
- Electrons for cyclic photophosphorylation: cyclic flow occurs around PSI and does not require water splitting.
- Protons exclusively for mitochondria: the protons contribute to the thylakoid proton motive force, not mitochondrial ATP synthesis.
Common Pitfalls:Confusing NADPH with NADH and thinking oxygen produced is immediately “used” in the Calvin cycle. Oxygen is a by-product released to the atmosphere; the Calvin cycle needs ATP and NADPH, not O2.
Final Answer:electrons needed to reduce P680 (i.e., to re-reduce oxidized P680+ in Photosystem II).