Aerobic Respiration problem

[gonelli]

I'm currently learning about cellular respiration in biology and have looked at the different stage involved in aerobic respiration. The book shows 12 H being left after the NADH and FADH2 have gone through the electron transport chain. So only 3 oxygen molecules are needed to make the 6 water molecules (12H + 3O2 ---> 6H2O).

But respiration has 6 oxygen molecules as an input, so i'm missing 3O2. I think that they must be used somewhere in the Krebs Cycle, but i'm not 100% sure. Can anybody confirm that or help me work it out?

[fredrik]

(*) In short

"FADH2 -> FAD" requires ½ O2

"NADH -> NAD" requires ½ O2

for details see http://en.wikipedia.org/wiki/Oxidative_phosphorylation

 

Glycolysis (glucose -> 2 pyruvate) gives you 2 NADH per glucose.

PDH conversion of (pyruvate -> acetyl-CoA) gives 2 more NADH / "glucose"

Krebs cycle gives you 3 NADH and 1 FADH2 per acetyl-CoA.

 

Total you get a "redox shift" of 10 NADH + 2 FADH2 per glucose.

 

To restore this as per the oxidative phosphorylation(*) you need 6 O2 / glucose.

 

/Fredrik

[gonelli]

I'm still a little confused. So does the 10 NADH per glucose hold a total of 10 H, and the 2 FADH2 hold total of 4 H?

[fredrik]

> So does the 10 NADH per glucose hold a total of 10 H, and the 2 FADH2 hold total of 4 H?

 

Yes you can say that, but it's hydrogen ions(H+), not atoms(H), so balancing takes more than just bundling the H's together with O's to get water.

 

In biochemistry the explicit reaction formulas are often complex and various shorthand notations are used. And to see the detailed reaction mechanism you need to go beyond the summation formulas and analyse each catalytic step involved as well as the transport of spieces between cytosol and mitochondria.

 

The two redox pairs we deal with is

 

[ce]

(1) NAD+ + H+ + 2e- <-> NADH

[/ce]

 

[ce]

(2) FAD + 2H+ + 2e- <-> FADH2

[/ce]

 

With oxygen we have

[ce]

(3) O2 + 4H+ + 4e- -> 2H2O

[/ce]

 

During glycolysis and krebs cycle, (1) and (2) moves in the forward reaction.

And you end up with 10 NADH and 2 FADH2 that needs to be put back to their oxidized states, by running (3) to the right.

 

[ce]

10 NADH + 2 FADH2 + 6 O2 + 24H+ +24e- ->

[/ce]

[ce]

-> 10 NAD+ + 2 FAD + 14H+ + + 24 e- + 6 H2O

[/ce]

 

Cleaning up

 

[ce]

10 NADH + 2 FADH2 + 6 O2 + 10H+ ->

[/ce]

[ce]

-> 10 NAD+ + 2 FAD + 6 H2O

[/ce]

 

If you wonder what's up with the 10 H+, they are also generated, 6 pcs / glucose during krebs, 4 pcs / glucose during glycolysis, basically (one per NADH).

 

/Fredrik

[gonelli]

So in your last equation you've got 6O2, giving you a total of 12 oxygen. And then the output is 6H2O, which only contains 6 oxygen. Where have the other oxygen gone to?

[fredrik]

You're correct. I noticed screwed up the formula when I wrote it because I rearranged it on the fly, I'm sorry. I found it messy to type in on board like this... but here is another one, a bit more developed. I noticed there is some annoying line limit when using the ce tags.

 

Unless I made more mistakes it should be

 

[ce]

10 NADH + 2 FADH2 + 6 O2 + 24H+ +24e- ->

[/ce]

[ce]

-> 10 NAD+ + 2 FAD + 14H+ + 24 e- + 12 H2O

[/ce]

 

Simplifying to

[ce]

10 NADH + 2 FADH2 + 6 O2 + 10H+ ->

[/ce]

[ce]

-> 10 NAD+ + 2 FAD + 12 H2O

[/ce]

 

But

[ce]

C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O

[/ce]

 

But the apprent extra 6 water has to be consumed elsewhere. I think one need to look at all details too see the mechanisms: Water is consumed in ATP and GTP hydrolysis as well as in the krebscycle (2 or 3 or something like that), but also created during glycolysis. I don't have it all ontop of my head so I'd have to check to respond.

 

For example

 

[ce]

ATP + H2O <-> ADP + H2PO4^- + H+

[/ce]

[ce]

GTP + H2O <-> GDP + H2PO4^- + H+

[/ce]

 

But sometimes I find bio books confusing because sometimes shorthand is used. Maybe someone else has ito ntop of their head, or I'll try ro respons later.

 

/Fredrik

[fredrik]

Ok, I put some more steps down (I migth have done more mistakes though).

 

1) Glycolysis

[ce]

C6H12O6 + 2NAD+ + 2H3PO4 + 2ADP ->

[/ce]

[ce]

-> 2 pyruvic acid + 2ATP + 2H2O + 2NADH + 2H+

[/ce]

 

2) Pyruvate dehydrogenase step

[ce]

pyruvic acid + CoA-SH + NAD+ ->

[/ce]

[ce]

-> acetyl-CoA + NADH + H+ + CO2

[/ce]

 

3) Kreb cycle

[ce]

acetyl-CoA + 3 NAD+ + FAD + GDP + P + 2H2O ->

[/ce]

[ce]

CoA-SH + 3 NADH + 3H+ + FADH2 + GTP + 2CO2

[/ce]

 

Add this up, simplify, and you get

 

[ce]

(1) C6H12O6 + 10 NAD+ + 2FAD + 4P +

[/ce]

[ce]

2ADP + GDP + 2 H2O ->

[/ce]

[ce]

-> 6 CO2 + 10 NADH + 10H+ + 2FADH2 + 2 GTP + 2ATP

[/ce]

 

4) Now compare this with the oxidative posphorylation.

[ce]

10 NADH + 2 FADH2 + 6 O2 + 10H+ ->

[/ce]

[ce]

-> 10 NAD+ + 2 FAD + 12 H2O

[/ce]

 

ATP production in mitochondria is

 

2.5 ATP/NADH except for the two NADH (per glucose) who come from glycolysis and

needs to exchanged for a FADH2 inside the mitochondria.

 

1.5 ATP/FADH2

 

Check a book on the details on these steps, it's unreadable to post it here.

 

This needs that we need to balance the phosphorylation (of 10 NADH and 2 FADH2) with ATP production.

(8*2.5 + 2*1.5 + 2*1.5 = 26 ATP)

 

[ce]

26 ADP + 26 H3PO4 -> 26 ATP + 26 H2O

[/ce]

 

[ce]

10 NADH + 2 FADH2 + 6 O2 + 10H+

[/ce]

[ce]

+ 26 ADP + 26 H3PO4 ->

[/ce]

[ce]

-> 10 NAD+ + 2 FAD + 38 H2O + 26 ATP

[/ce]

 

Add this to (1)

 

[ce]

C6H12O6 + 6 O2 + 30 H3PO4 + 28ADP + 2 GDP ->

[/ce]

[ce]

-> 6 CO2 + 2 GTP + 28ATP + 36 H2O

[/ce]

 

Since GTP and ATP can be transferred to each other we can effectively write

 

[ce]

C6H12O6 + 6 O2 + 30 H3PO4 + 30ADP ->

[/ce]

[ce]

-> 6 CO2 + 30ATP + 36 H2O

[/ce]

 

[ce]

C6H12O6 + 6 O2 + (30 H3PO4 + 30ADP ) ->

[/ce]

[ce]

-> 6 CO2 + 6 H2O + (30ATP + 30 H2O)

[/ce]

 

/Fredrik