# Succinic Acid Efflux Energy Coupling

### From bio-physics-wiki

Under natural conditions succinic acid transport is coupled to proton efflux. Dependent on the structure of the transporter, the number of the protons transported out of the cell together with succinic acid can vary. If a complex of succinic acid and one proton is carried through the channel the net charge of the complex is zero and no additional work due to the electric potential has to be invested. However, if two protons form a complex with succinic acid, the net charge of the complex is positive. In this situation additional energy associciated with the moving charge in the electric potential is involved. In the article on Cells and Membrane Potentials - Equivalent Circuits we found that in equilibrium conditions the **Nernst equation**

\begin{align} \Delta \psi = \frac{2.3 RT}{z_1 F} log_{10} \frac{[{suc^-}^{(OUT)}]}{[{suc^-}^{(IN)}]}- \frac{2.3 RT}{z_2 F} \Delta pH \end{align} is valid. The reported values$^{1)}$ for the membrane potentials $\Delta \psi$ in Saccharomyces cerevisiae are very diverse and range form $-50mV$ in the resting state and for low pH ($pH=4.5$) up to $-206mV$ for the metabolising state and high pH ($pH=7.5$). Assuming a value of $\Delta \psi$, we can plot the ratio $\frac{[{suc^-}^{(OUT)}]}{[{suc^-}^{(IN)}]}$ for different delta $pH$'s. To this end, we rearrange the above equation. For the moment we assume that $z_1=1$ and we call $z_2$ simply $z$. \begin{align} \frac{2.3 RT}{z F} \Delta pH + \Delta \psi = \frac{2.3 RT}{F} log_{10} \frac{[{suc^-}^{(OUT)}]}{[{suc^-}^{(IN)}]}\\ \Delta pH + \frac{z F}{2.3 RT} \Delta \psi = z \, log_{10}\frac{[{suc^-}^{(OUT)}]}{[{suc^-}^{(IN)}]}\\ \left( \frac{[{suc^-}^{(OUT)}]}{[{suc^-}^{(IN)}]} \right)^{z}=e^{\Delta pH + \frac{z F}{2.3 RT} \Delta \psi}= C \cdot e^{\Delta pH} \end{align}

If we now plot $\frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}$ for different energy couplings $z$ \begin{align} \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]} = \left( e^{-\Delta pH - \frac{z F}{2.3 RT} \Delta \psi}\right)^{1/z} \end{align} and $\Delta \psi=-60mV$ we get

Assume for the moment, that the succinic acid ratio (e.g. $\frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}=1$, green line) is constant. If one varies the energy couplings different values of delta $pH$'s can be overcome. For highly charged complexes $z=3$ the $pH$ than can be overcome in equilibrium is about three times smaller than for $z=1$. On the other hand if we imagine a vertical line at e.g. $pH=3$, we see that for different energy couplings different equilibrium ratios of succinic acid inside and outside are possible.

Moreover, if one imagines a vertical line for a certain $pH$ the succinic acid ratio must be much lower for high $z$ values than for low $z$ values, which means a larger chemical potential due to succinic acid osmosis is required for large charges.

1) Borst-Pauwels GW. - Ion transport in yeast. Biochim Biophys Acta. 1981