How do cellular conditions change the Gibbs free energy of a reaction?

Question

How do cellular conditions change the Gibbs free energy of a reaction? Taking glycolysis as an example, how exactly would cellular conditions affect the free energy released from this reaction?

Answer 1

To understand that, you must have a basic understanding of Gibbs-Helmholtz equation. it states-

$\Delta G'=\Delta G^{'o} +RT\,ln\,(Q) $

Here, $\Delta G'$ = biochemical free energy change & $\Delta G^{'o}$ = standard biochemical free energy change.

Suppose you have a reaction, $x+FADH_2\rightleftharpoons xH_2+FAD $

So, $Q=\dfrac{[xH_2][FAD]}{[x][FADH_2]}$

Now let's assume that $\Delta G^{'o}>0$.

Hence the spontaneity of the aforesaid reaction (in forward direction) will be dependent on the $\Delta G^{'o}$, which is a function of reaction quotient (Q) as well as temperature (T).

Now, if the body is in lower energy state, i.e. fasting; $\dfrac{[FAD]}{[FADH_2]}$ ratio will be high. So, in order to make $\Delta G'<0$, the following condition must be satisfied $-$

$RT\,ln\,(Q)+\Delta G^{'o}<0$,

or, $RT\,ln\,(Q)< -\Delta G^{'o}$

or, $RT\,ln \dfrac{[xH_2][FAD]}{[x][FADH_2]}\,< -\Delta G^{'o}$

With this reasoning, be it any metabolic pathway; we can predict its spontaneity.