ADH.H+ Reoxidation during Anaerobic Conditions

Posted by biochemistryquestions on October 10, 2008
NAD
  (CM-15) During strenuous exercise, the NADH formed in the glyceraldehyde 3-phosphate dehydrogenase reaction in skeletal muscle must be reoxidized to NAD+ if glycolysis is going to continue.  The most important reaction involved in the reoxidation of NADH in anaerobic conditions is: a)     Dihydroxyacetone phosphate to  glycerol 3-phosphate

b)     Glucose 6-phosphate  to  fructose 6-phosphate

c)      Glucose 6 (P) to Phosphogluconate

d)     Isocitrate  to a-ketoglutarate

e)     oxalacetate to  malate

f)       malate to oxalacetate

g)     pyruvate to  lactate
 

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This entry was posted on October 10, 2008 at 12:03 am and is filed under Bioenergetics (Q),Carbohydrate Metabolism (Q). Tagged: Anaerobiosis, biochemistry, Carbohydrate metabolism,NADH. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
6 Responses to “NADH.H+ Reoxidation during Anaerobic Conditions”TAREK said

1. March 7, 2009 at 9:17 pm

why NAD & NADH+H+ are written as it, and what is the different?
ReplyBiochemistryquestions said

• March 7, 2009 at 10:44 pm

The right form of writing NAD in the oxidized form is NAD+ since this chemical structure has a positive charge (like the N in NH4+); in a reduction process,
NAD+ + H2 —> A Hydrogen with two electrons are bound to the NAD+, so it becomes NADH
As you see, the only “part” of the H2 that results bound to the NAD+ is H: and that is why we should write the reduced form as NADH
but since a proton H+ of the original H2 is released to the environment, then it is used to write the reduced form of NAD+ as NADH. H+, indicating also the proton (H+) that has been released.
Thanks for your question.
ReplyMatthew said

1. April 8, 2009 at 3:29 pm

But hang on…..
in all those reactions assuming only 1 glucose molecule is used therefor only 12 hydrogen atoms are available for the whole reaction. If this system of NADH + H+ is used, I find a total of 22 hydrogen atoms would be required. What am I missing? where are the extra H+ ions coming from?
Ie. Glycolysis: Glucose ==> 2x Pyruvate 4H removed (2x NADH + H+)
Link Reaction: 2x Pyruvate ==> 2x Acetyl CoA 4H removed (2x NADH + H+)
Krebs Cycle:2x Oxaloacetate + Acetyl ==> Citrate ==>…==> Oxaloacetate(2x (3x NADH + H+)
+ (FADH2) = 18H)
18H + 2H + 2H = 22H
Please explain where the 10 extra H+ ions are coming from cos this is a nightmare. The entire reaction is limited to 12H as far as I can see as no more are added anywhere.
ReplyBiochemistryquestions said

1. April 9, 2009 at 7:34 pm

Hi, Matthew!
Thanks for sharing your doubts with me.
(I assume that you are very young since your call it a nightmare…happily you have not found real nightmares yet!)
Lets try to answer you:
Let’s balance some global equations:
This is the very global equation:
C6 H12 O6 (glucose) + 6 O2 —–> 6 CO2 + 6 H2O (you can see 12 Hydrogens at the left and 12 at the right)
Let’s see if it is true, buy analizing two big steps:
Aerobic glycolysis:
C6 H12 O6 +O2——-> 2 (C3 H4 O3) (pyruvic) + 2 H2O
From Pyruvic to CO2 -includes pyruvic decarboxylation and krebs cyle-(you can balance and check)
2 (C3 H4 O3)pyruvic + 5 O2 —> 6 CO2 + 4 H2O
(If you add both reactions you obtain the more global reaction written above)
So, you can see that there are 12 hydrogens at the left and 12 hydrogens at the right.
This 12 hydrogens are included in 6 molecules of water, since they go, from the reduced cofactors formed in these processes, to the respiratory chain.
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