[lkirk]

Can anyone explain the free energy change in step 7 of glycolysis, so I can see why it is not exergonic and irreversible like steps, 1, 3 and 10?  For example, in step 1:

(one of the highly exergonic--not reversible steps), the negative net free energy change is -4,000 cal/mol.

     glucose + Pi --> G6P [G = +3,300 cal/mol]

     ATP --> ADP + Pi [G = -7,300 cal/mol]

     Therefore, in this coupled reaction, G = -4,000 cal/mol

I cannot produce the above for step 7 to show me that the reaction is not exergonic.

This is my guess, but I know I am missing something:

     ADP +  Pi --> ATP [G = +7,300 cal/mol]

     1,3-biphosphoglycerate + H2O (hydrolysis) --> 3-phosphoglycerate [G = -11,800 cal/mol]

     Therefore, in this coupled reaction, G = -4,500 cal/mol    ?

[lkirk]

Ok, I believe I answered my own question.  I was thinking of the reaction in the context of standard conditions, and of course, the reaction is occuring under cellular conditions.  While step 7 would be highly exergonic under standard conditions, it is not under physiologic conditions.

[Yggdrasil]

Exactly the correct answer.  The deltaG is much less exergonic under physiological conditions because many of the previous reactions are unfavorable and are driven forward by a low concentration of 1,3-bPG.