Three leaves are tested. Leaf 1 is lit with white light…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Three leaves are tested. Leaf 1 is lit with white light. Leaf 2 is lit only with far-red light, of wavelength beyond the limit of P680. Leaf 3 has plenty of light, but all its NADP+ has been used up. (a) Which path is used in leaf 1, and what does it give? (b) What does leaf 2 give? (c) In leaf 3, which path can still run, and why?
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
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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. Leaf 1. With white light both photosystems work. The path is non-cyclic, and it gives ATP, NADPH and oxygen.
Step 2. Leaf 2. Far-red light excites only PS I, so the path is cyclic. It gives ATP only.
Step 3. Leaf 3. The non-cyclic path ends at NADP+, so it stops. Cyclic flow does not need NADP+, because the electron goes back to PS I. ATP can still be made.
Answer: (a) non-cyclic, giving ATP, NADPH and oxygen; (b) ATP only; (c) the cyclic path, because it does not need NADP+.
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