An electron starts in a water molecule and ends in NADPH…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
An electron starts in a water molecule and ends in NADPH. (a) Put these in order: photosystem I, plastocyanin, water, cytochrome b6-f complex, NADP+, plastoquinone, photosystem II. (b) At which two points is the electron lifted by light? (c) Between which two points does it lose energy on the way down, and where does the cell use that energy?
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 order. Water, photosystem II, plastoquinone, cytochrome b6-f complex, plastocyanin, photosystem I, NADP+.
Step 2. The lifts. Light lifts the electron at photosystem II and again at photosystem I.
Step 3. The energy. On the way down from photosystem II to photosystem I, along the plastoquinone, the cytochrome b6-f complex and plastocyanin, the electron loses energy. The energy builds the proton gradient, which makes ATP.
Answer: (a) water, PS II, PQ, cyt b6-f, PC, PS I, NADP+; (b) at PS II and PS I; (c) between PS II and PS I, and the energy builds the gradient that makes ATP.
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