From the statistics page, I found that the Module-M00008 is present only in bacteria and not Eukaryotes. Why so? I noticed that all the compounds necessary or the reaction are present in human body and also all the enzymes required for the reactions are present in human body too. Then why does that reaction not happen in humans?


1 Answer 1


That module represents Entner-Duodoroff pathway, which is an alternative pathway to the Embden-Mayerhof-Parnas pathway of glycolysis.

Entner-Duordoff pathway exists only in prokaryotes.

  • $\begingroup$ so does this mean that the enzymes required for this particular pathway in bacteria are not present in Eukaryotas ? $\endgroup$
    – girl101
    Commented Jul 13, 2015 at 4:33
  • $\begingroup$ @Rishika Precisely $\endgroup$
    Commented Jul 13, 2015 at 4:35
  • $\begingroup$ so if these enzymes would have been present then which pathway would have had the more probability of occurring? $\endgroup$
    – girl101
    Commented Jul 13, 2015 at 4:37
  • 2
    $\begingroup$ @Rishika, to add to this answer, there are three possibilities for discrepancy in coverage. 1) Limited information, i.e. even if the module exists in other species it has not been found (missing enzymes, proteins, etc.). 2) The information has not been added to the database (remember that this resource is manually curated). 3) It actually refers to different evolutionary paths. As WYSIWYG says, this is an alternative pathway. Either only the ancestors of prokaryotes found it or if other species found it, it did not increase fitness and was lost. You will find many more examples. $\endgroup$
    – ddiez
    Commented Jul 13, 2015 at 8:07
  • 1
    $\begingroup$ @Rishika And actually, reading the link to the Entner-Duodoroff pathway in wikipedia, it mentions that some prokaryotes use it as they lack enzymes critical for glycolysis. This suggests that even if "prokaryotes" have both glycolysis and this alternative according to the KEGG Module, maybe not the same prokaryote has both. At least it seems the case here. $\endgroup$
    – ddiez
    Commented Jul 13, 2015 at 8:12

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