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?

Biochemistry Photosynthesis and Respiration Difficulty: Easy
Choose an option
  • A
    electrons needed to reduce P680 (i.e., to re-reduce oxidized P680+ in Photosystem II)
  • B
    O2 needed for the dark reactions (Calvin cycle)
  • C
    electrons needed to reduce NADH
  • D
    electrons needed for cyclic photophosphorylation
  • E
    protons 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).

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