I was wondering where the 2 electrons and H+ go when NADPH is turned into NADP+ during the Calvin Cycle? I know that during the light dependent reaction, 2 electrons (from water) and H+ combine with NADP+ (from the Calvin Cycle) to form NADPH, but after NADPH is used and is turned into NADP+, where do those electrons and H+ go?
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1$\begingroup$ where have you looked for this answer? $\endgroup$– Vance L AlbaughMay 30, 2017 at 1:39
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$\begingroup$ @VanceLAlbaugh my biology textbook. It does not have any information on it. $\endgroup$– Serendipitous EpiphanyMay 30, 2017 at 2:09
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2$\begingroup$ This is quite basic in a photosynthesis class: they are used to reduce the 3-phosphoglycerate to glyceraldehyde-3-phosphate. $\endgroup$– user24284May 30, 2017 at 5:14
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A reasonable source of information about the Calvin Cycle is Berg et al., but for this reaction the full structures are not shown. I therefore show them below in a diagram of my own:
I have coloured the hydrogen atoms to show where they go in reducing the phosphoglycerate (acid) to an aldehyde. You can think of the two electrons being used to form the additional covalent bond.
(The hydrogen phosphate is actually in equilibrium with other phosphate species, but this is not relevant to your question and so is not shown.)
