Why can't Glycolysis go backwards? That is, what step makes the process (almost) irreversible?
ust to be annoying it actually can (however i write later today why it in principal can't). The argument for statement is the following: A reaction is given by: A + B <-> C + D \[\Delta G =\Delta G ^{\Theta}+RT*\ln(Y)\] Where Y is: \[Y=\left( \frac{ [C]*[D] }{ [A]*[B] } \right)\]
Reason I can't answer now is because I want to check a book for a special phrase.
we look upon the following situation: Delta G standard <0 RT ln(Y) > abs(Delta G standard) And we consider this true: ΔG (Q -> P) = -ΔG( P-> Q) Then the reaction towards the products would not be spontaneous.
Thanks!
Sleepyfox here is the real answer: Glucose enters cells through specific transport proteins and has one principal fate: it is phosphoylated by ATP to form glucose-6-phosphate. This step is notable for 2 reasons: 1) glucose-6-phosphate cannot pass through the membrane because it is not a substrate for the glucose transporters. 2) the addition of the phosphoryl group acts to destabilize glucose, thus facilitating its further metabolism. The transfer of the phosphoryl group from ATP to the hydroxyl group on carbon nr. 6 of glucose is catalyzed by hexokinase. and have a strong equilibrium towards the products (glucose-6-phosphate (G-6P))
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