Difficulty: Easy
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:
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:
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:
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).
Discussion & Comments