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The correct path of transfer of electrons in the light harvesting photosystems in vascular plants is _____________.

  • A.\(\mathrm{H_{2}O}\) → Photosystem II → Cyt \(\mathrm{b_{6}f}\) → Photosystem I → \(\mathrm{NADP^{+}}\)
  • B.\(\mathrm{H_{2}O}\) → Photosystem I → Cyt \(\mathrm{b_{6}f}\) → Photosystem II → \(\mathrm{O_{2}}\)
  • C.\(\mathrm{H_{2}O}\) → Photosystem II → Cyt \(\mathrm{b_{6}f}\) → Photosystem I → \(\mathrm{O_{2}}\)
  • D.\(\mathrm{H_{2}O}\) → Photosystem I → Cyt \(\mathrm{b_{6}f}\) → Photosystem II → \(\mathrm{NADP^{+}}\)

Answer: A

The path is H2O → Photosystem II → Cyt b6f → Photosystem I → NADP+.

In the Z-scheme, the oxygen-evolving complex of photosystem II takes electrons from water, releasing O2. Photosystem II passes them through plastoquinone to the cytochrome b6f complex, which hands them to plastocyanin. Photosystem I re-energises them and sends them via ferredoxin to ferredoxin-NADP+ reductase, which reduces NADP+ to NADPH.

Paths that start at photosystem I have the photosystems in the wrong order. Paths that end at O2 confuse the by-product of water oxidation with the final electron acceptor.

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