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why is that 2 electrons are lost in this oxidation? doesn't H only carry one electron?
NADPH to NADP+
- Thread starter zoner
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rls303,
I think you have it right. I think veins always carry deoxy blood to the heart and arteries always carry oxy blood away from the heart. The only time you have to be careful about the terms "vein" and "artery" are when it comes to the pulmonary vein and pulmonary artery. In this case, it reverses. The pulmonary vein carries oxygenated blood to the heart (this blood has passed through the lungs where the hemoglobin in the red blood cells have picked up oxygen) and the pulmonary artery carries deoxygenated blood to the lungs in order to fill up on oxygen.
I think you have it right. I think veins always carry deoxy blood to the heart and arteries always carry oxy blood away from the heart. The only time you have to be careful about the terms "vein" and "artery" are when it comes to the pulmonary vein and pulmonary artery. In this case, it reverses. The pulmonary vein carries oxygenated blood to the heart (this blood has passed through the lungs where the hemoglobin in the red blood cells have picked up oxygen) and the pulmonary artery carries deoxygenated blood to the lungs in order to fill up on oxygen.
Because a hydride ion is transferred rather than a hydrogen atom.
NADPH --> NADP+ + H-
If you look at the charge, it makes sense that what is stripped off of NADPH has to be H-, rather than just H. Otherwise, there is no reason for the product to have a +1 charge. Structurally, it makes sense that both NADPH and NADP+ are the way they are. Looking at the link provided by tttgo, you can see that the nicotinamide ring (containing N+) is aromatic in both the oxidized (NADP+) and the reduced (NADPH) states. In NADP+, the 6 pi electrons are the 3 double bonds. In NADPH, 4 pi electrons come from the 2 double bonds, and 2 more come from the valence electrons of the N. In order to preserve aromaticity, and to not break any other general rules of chemistry, the redox reaction must be a hydride, two-electron transfer rather than a one-electron transfer. To note, you may see the reaction being written as H+ + 2e- rather than H-, but you can see that the two are equivalent.
NADPH --> NADP+ + H-
If you look at the charge, it makes sense that what is stripped off of NADPH has to be H-, rather than just H. Otherwise, there is no reason for the product to have a +1 charge. Structurally, it makes sense that both NADPH and NADP+ are the way they are. Looking at the link provided by tttgo, you can see that the nicotinamide ring (containing N+) is aromatic in both the oxidized (NADP+) and the reduced (NADPH) states. In NADP+, the 6 pi electrons are the 3 double bonds. In NADPH, 4 pi electrons come from the 2 double bonds, and 2 more come from the valence electrons of the N. In order to preserve aromaticity, and to not break any other general rules of chemistry, the redox reaction must be a hydride, two-electron transfer rather than a one-electron transfer. To note, you may see the reaction being written as H+ + 2e- rather than H-, but you can see that the two are equivalent.
hey, in my campbell textbook, under the light reactions/calvin cycle section of photosynthesis, says that water is split to provide a source of electrons and H+. NADP+ then takes these to become NADPH. When you draw a water molecule, you are transferring the O-H bond over to the NADP+ to form the C-H bond in the picture from my link above. This pushes the double bond over to negate the N's positive charge. I think this is the hydride shift that rabolisk mentioned, so it's really a H- that's being taken. hope that helps.
