Electron Transport Chain
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Electron Transport Chain
I am trying to study the mathmatical redox for cellular respiration C6H12O6> 6CO2 + 6H20 + energy. I have been able to get everything except the 6H20. I know it occurs in ETC from the NADH. It looks like there are 10 NADP which transfere the electrons on down to the oxidase complex where 8 e= + 2O + 8H join to form 2H2O with 4e moving up through complex. My question is that if each NADH forms 1 H2O, then there would be a formation of at least 10 H20, but the redox only allows for 6H20. How are the extra H2O handled (i.e 4 H20), accounted for as far as the redox equation is concerned? Are they ignored because at least 6 H2O were formed?
*I understand that there is also extra H20 due to FADH, ect.., but just saying if all things being equal and there were really only 10 H20 formed, how does that equal into the redox equation.
*I understand that there is also extra H20 due to FADH, ect.., but just saying if all things being equal and there were really only 10 H20 formed, how does that equal into the redox equation.
Could it be the net 4 H2O that are used in 2x Kreb's Cycle per glucose? Remember, for each loop of the Kreb's Cycle, [EDIT  inserted this] you use 1 H2O (1) to combine acetylCoA and Oxaloacetate to make Citrate and HSCoA. Then you produce 1 H2O (+1) going from Citrate to cisAconitate, and then use 1 H2O (1) to make Isocitrate. [EDIT  cut out, see below] Finally, you use one more H2O (1) going from Fumarate to Malate. Add all those together and you net 2 H2O per Kreb's cycle (i.e. per pyruvate cum acetylCoA), and that happens twice per Glucose, so total 4 H2O.
So, avg 10 produced H2O from the ETC minus 4 H2O used in the Kreb's Cycle leaves you with 6H2O, balancing out the equation.
This is only counting the electrons from NADH and FADH2 inside the mitochondria. I don't know the details of the numbers for the NADH from Glycolysis because I believe there is some kind of energy cost to transporting them into the mitochondria. I assume it all somehow balances out in the end, given the shear number of metabolic pathways all tied into this.
[EDIT  cut from above]
However, you then use another H2O (1) going from SuccinylCoA to Succinate + HSCoA.
Oops. Small mistake here. Also here, GDP + Pi > GTP produces 1 H2O (+1), so this part nets zero. Sorry about that.
[EDIT  the last one, promise ]
The Wikipedia Article on the Kreb's Cycle has a good diagram of it, as well as a nice table. It ignores the net zero at the SuccinylCoA part, so if you add up the H2O in the Products column (+1) of the table, and subtract the H2O from the Substrates column (3), you will see that you indeed get 2.
So, avg 10 produced H2O from the ETC minus 4 H2O used in the Kreb's Cycle leaves you with 6H2O, balancing out the equation.
This is only counting the electrons from NADH and FADH2 inside the mitochondria. I don't know the details of the numbers for the NADH from Glycolysis because I believe there is some kind of energy cost to transporting them into the mitochondria. I assume it all somehow balances out in the end, given the shear number of metabolic pathways all tied into this.
[EDIT  cut from above]
However, you then use another H2O (1) going from SuccinylCoA to Succinate + HSCoA.
Oops. Small mistake here. Also here, GDP + Pi > GTP produces 1 H2O (+1), so this part nets zero. Sorry about that.
[EDIT  the last one, promise ]
The Wikipedia Article on the Kreb's Cycle has a good diagram of it, as well as a nice table. It ignores the net zero at the SuccinylCoA part, so if you add up the H2O in the Products column (+1) of the table, and subtract the H2O from the Substrates column (3), you will see that you indeed get 2.
"Empathise with stupidity, and you're halfway to thinking like an idiot."  Iain M. Banks
Re: Electron Transport Chain
C6H12O6 + 6O2 > 6CO2 + 6H2O + (energy – ATP + heat)
Oxidation halfreaction
C6H12O6 > CO2
a + b + c = 0 a + b = 0
a + (+12) + (12) = 0 a + (4) = 0
a = 0 a = +4
ON of C = 0 ON of C = +4
increase of 4
Reduction halfreaction
O2 > H2O
ON of O = 0 ON of O = 2
grow less of two
Oxidation halfreaction
C6H12O6 > CO2
a + b + c = 0 a + b = 0
a + (+12) + (12) = 0 a + (4) = 0
a = 0 a = +4
ON of C = 0 ON of C = +4
increase of 4
Reduction halfreaction
O2 > H2O
ON of O = 0 ON of O = 2
grow less of two
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