A student asks: "Light has already lifted the electron at photosystem II…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A student asks: "Light has already lifted the electron at photosystem II. Why does a second photosystem lift it again before it reaches NADP+?" (a) What happens to the energy of the electron on the way from PS II to PS I? (b) What does the electron need to reduce NADP+? (c) How many photons does each electron absorb on the whole way from water to NADPH?
Need a reminder? The key ideaShow
Non-cyclic: water, PS II (P680), acceptor, PQ, cytochrome b6-f, PC, PS I (P700), acceptor, NADP+ to NADPH. It gives ATP, NADPH and oxygen. Each electron is lifted twice, so 2 photons. Two water give 4 electrons, so one oxygen and 2 NADPH. Cyclic: PS I only, the electron returns to PS I, only ATP, in the stroma lamellae, in far-red light beyond 680 nm.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The loss. On the way down the chain the electron gives up energy. This energy builds the proton gradient. The electron arrives at PS I with little energy left.
Step 2. The need. To reduce NADP+ the electron must be at a much higher energy level, with a greater redox potential. PS I lifts it there.
Step 3. The photons. One photon at PS II and one at PS I make photons for each electron.
Answer: (a) it loses energy, which builds the proton gradient; (b) a lift to a higher energy level; (c) photons.
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