As I understand it, during glycolysis, glucose is converted into fructose 1,6-bisphosphate which greatly destabilizes the molecule. This is so it can be divided into pyruvates more easily.

Now I don't understand why adding charged groups such as phosphates destabilize molecules, what actually happens to the molecule when a charged group associates with it? Why does it become destabilized?


1 Answer 1


The quick and general answer is electrostatic repulsion. Fructose-6-phosphate is already a dianion. Phosphorylating it a second time will mean adding another two negative charges to the dianion. Like charges repel each other, so the two dianionic phosphate groups are pushed apart electrostatically.

However, you cannot generalise this. Sometimes, the additional negative charge may serve to better balance charges overall. It could also be so far away from an existing negative charge that electrostatic interactions are minimal. Finally, there are many ions and positively charged organic molecules dissolved in the cell which altogether stabilise negative charges relatively with respect to a simple solution of fructose-1,6-bisphosphate. (The same can be said for the stabilisation of positive charges.)

Finally, please note that the additional negative charge does not really assist the fragmentation of fructose-1,6-bisphosphate. The reaction — a retro-aldol — is enzymatic and catalysed principally by a lysine and aspartate residue. The former is used to convert the keto functionality into an imine which has a greater tendency to form an enamine while the latter serves as a proton shuttle. The phosphate groups are merely there so the enzyme recognises the correct molecule; modifying the enzyme’s structure to allow for unphosphorylated fructose to be consumed would not alter the reaction rate significantly.

  • $\begingroup$ What actually is destabilization ? What happens to the molecule after destabilization? $\endgroup$ Sep 9, 2020 at 11:29
  • $\begingroup$ @HariharanV I didn’t use the term once, so I don’t entirely understand why you’re asking me, but: essentially, it means a molecule being somehow given more internal energy. What happens depends on the molecule and is way too broad to answer here. $\endgroup$
    – Jan
    Sep 9, 2020 at 11:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.