Is it because glyceraldehyde -3-phosphate (a molecule which when enter glycolysis help produce ATP through substrate level phosphorylation) can be prepared without losing an ATP through this process?

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  • $\begingroup$ I've replaced the abbreviation in your title. Avoid all but the most familiar abbreviations (such as DNA) unless you define them (and this cannot be done in titles). $\endgroup$
    – David
    Jun 25, 2016 at 18:20
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    $\begingroup$ I've modified your title again. The use of "predominate" could have been taken to imply it processed more glucose than glycolysis. However in resting erythrocytes the figure seems to be 10% PPP. Still a lot compared with many other cells, but less than glycolysis. $\endgroup$
    – David
    Jun 26, 2016 at 17:12
  • $\begingroup$ Thanks for the correction though I actually meant 'Why is the Pentose Phosphate Pathway so active in erythrocytes?' so its ok. $\endgroup$
    – Tyto alba
    Jun 26, 2016 at 20:19
  • $\begingroup$ Can you please cite a reference which says that erythrocytes use PPP relatively more than other cells? I was actually not aware of this; I don't think this some common textbook fact that I missed. $\endgroup$
    Jun 27, 2016 at 4:53

1 Answer 1


No. Your supposition is incorrect — the phosphate in glyceraldehyde 3-phosphate has to come from somewhere, and it comes from glucose 6-phosphate. The reason a second ATP is required before you get to the triose phosphate stage in glycolysis is that you are generating two molecules of triose phosphate. In the pentose phosphate pathway (energy-producing non-oxidative branch) you are not generating two molecules of triose-P from one hexose-P, you are generating two hexose-P and one trisose-P from three hexose-P, as shown in my Diagrams 2 and 3 below. (The other three carbon atoms are lost as CO2.)

A good way of approaching this question is to ask “What can be produced by the pentose phosphate pathway that cannot be produced by glycolysis?”.

Glycolysis may be summarized in this context as:

Glucose + NAD+ ➝ Pyruvate + NADH + 2ATP

But to allow glycolysis to continue the NADH is reoxidized to NAD+:

Pyruvate + NADH ➝ Lactate + NAD+

So the sole product of glycolysis for the erythrocyte is ATP (primarily for active cation transport to maintain cell shape) lactate passes into the blood for recycling.

Oxidative phase of the Pentose Phosphate Pathway

The Pentose Phosphate Pathway (the oxidative stage) is shown in Diagram 1, above, and may be summarized in the way done for glycolysis as:

Glucose + ATP + 2NADP+ ➝ Ribulose 5-P + CO2 + 2NADPH

Ribulose 5-P has two possible fates, but only one differs from glycolysis, so the two distinguishing products of the pathway are ribose and NADPH.

Ribose is important for nucleic acid synthesis, particularly in dividing cells, explaining the increased activity of the pentose phosphate pathway in those cells. This cannot be the reason for the high activity of the pathway in erythrocytes as they have no nuclei and do not divide. In fact the ribulose 5-P is fed back into glycolysis for generation of ATP as shown in diagrams 2 and 3.

NADPH, then, is the answer in this case. It is the reducing agent used in the cytoplasm for synthetic processes (unlike NADH used particularly for ATP generation in the mitochondria), and so the pentose phosphate pathway is found in cells such as adipose, mammary gland and liver, that synthesize fatty acids, and cells that synthesize steroids. However that is not there is no fatty acid or steroid synthesis in the erythrocyte.

NADPH is important in erythrocytes as it is the specific source of reducing power required to keep the molecule glutathione in a reduced state. This has an important protective role in reducing cellular molecules that have become oxidized by molecular oxygen, a problem that is more acute in erythrocytes than in perhaps any other cells as they are the carriers of molecular oxygen. The cell membrane is particularly prone to oxidative damage. You can read about this online in this chapter in Berg et al.

Conversion of pentose to hexose and triose in the Pentose Phosphate Pathway

Pentose Phosphate Pathway: Fate of the Carbon Skeleton

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    $\begingroup$ I came across this line 'This stress is most acute in red blood cells because, lacking mitochondria, they have no alternative means of generating reducing power.'. Does it mean that normal cells with mitochondria are better equipped for combating oxidative stress? Is it due to NADH and FADH2 formed in the mitochondrial matrix during Beta oxidation and Krebs cycle? Can NADH and FADH2 act as reducing agents for glutathione just as NADPH? $\endgroup$
    – Tyto alba
    Jun 25, 2016 at 20:38
  • $\begingroup$ I quoted the line from that chapter you linked. However my question still remains the same Can NADH and FADH2 act as reducing agents since they say that in absence of Mitochondria RBCs have no alternative means of generating reducing power? $\endgroup$
    – Tyto alba
    Jun 26, 2016 at 8:41
  • $\begingroup$ I'll reply in two comments because of length restriction. I don’t find the line you quote at all convincing or helpful in any way. First, if what was required were NADH, that could be generated in glycolysis as far as pyruvate, with the pyruvate going into the blood and returning to the liver. (Pyruvate can pass in and out of cells.) Second, I imagine protection against reactive oxidative species in the mitochondrion is required because the mitochondrion is the site of the electorn transport chain at which oxygen acts. Thus protection in the mitochondrion is only designed to be used there. $\endgroup$
    – David
    Jun 26, 2016 at 22:18
  • $\begingroup$ The mammalian glutathione reductase only uses NADPH. It does not use NADH or FADH<sub>2</sub>. This is entirely consistent with the differential use of these two reducing nucleotide coenzymes in mitochondrion and cytoplasm. $\endgroup$
    – David
    Jun 26, 2016 at 22:25
  • $\begingroup$ I have now updated my answer, including my own illustrations, to make it slightly more comprehensive and, I hope, of more general use. It is meant to be an introductory general answer to the question, but if anyone thinks there are defects or points that should be added, please comment. $\endgroup$
    – David
    Jun 26, 2016 at 22:27

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