Osmosis and the Cell Membrane

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Imagine a cell membrane that separates two regions IN and OUT. The membrane shall contain a transporter that allows only specific molecules to pass. To be explicit assume a Symporter, that simultaneously exports succinic acid ($suc^-$, shown in green) and $n$ protons $n H^+$, shown in red out of the cell, additionally there will also be a solvent.

Symporter.jpg

Let's assume that the succinic acid concentration inside the cell is high compared to the outside and that the pH outside is lower $pH=3,5$ (which means high proton concentration $10^{-3.5}M$) than inside $pH=6$. In this situation the solutes tend to mix with the solvent. This phenomenon is known as osmosis and is a direct consequence of the second law of thermodynamics. Since molecules can pass only in one direction (from inside to outside) succinic acid is pulled out of the cell to mix with the protons. In the mixed state the entropy of the closed system (inside + outside) is larger than in the unmixed state, the mixed state is much more probable so the system will evolve to that state and reach equilibrium.

The Gibbs-Energy of the total (closed) system is given by the Gibbs-Energy inside and the Gibbs-Energy outside \begin{align} dG=dG^{(IN)}+dG^{(OUT)} \end{align} With the Fundamental Equation we get \begin{align} dG&= -S \, dT + dp \, V + \sum_j \mu_j^{(IN)} dN_j+ \sum_j \mu_j^{(OUT)} dN_j \end{align} with the chemical potentials $\mu$ and the number of solvent molecules $dN$. For only one molecular species (solute and solvent) the equation becomes \begin{align} dG&= -S \, dT + dp \, V + \mu^{(IN)} dN+ \mu^{(OUT)} dN \end{align} and the chemical potential is simply \begin{align} \left( \frac{dG(T,p,N)}{dN} \right)_{p,T}&= \mu = \mu^{(IN)}+ \mu^{(OUT)} \end{align} and reaches equilibrium when $\mu=0$ \begin{align} \mu^{(IN)}=- \mu^{(OUT)} \end{align} Away from equilibrium ($dG$ is negative), the reaction will proceed spontaneous towards equilibrium.

Gibbs-Energy Reaction
$dG<0$ spontaneous
$dG>0$ non-spontaneous
$dG=0$ equilibrium

When $dG$ is negative \begin{align} 0 &>\mu^{(IN)} + \mu^{(OUT)}\\ -\mu^{(OUT)} &> \mu^{(IN)} \end{align} This means that if $|\mu^{(OUT)}| > |\mu^{(IN)}|$ the reaction will proceed spontaneously. For more species this equation generalises to \begin{align} \sum_i |\mu_i^{(OUT)}| > \sum_j|\mu_j^{(IN)}| \tag{1} \end{align}

In another article we derived an explicit expression for the chemical potential. By means of Statistical Mechanics we were able to calculate the Microcanonical Ensemble and hence the entropy, which - inserted in the Gibbs-Energy - gives the chemical potential. We also need to take the solvent into account, so there are three species that we have to mind. If we calculate with concentrations this already implies the solvent (just like in the the derivation of the chemical potential, where we switched to concentrations in the end). For the inside potential we have

\begin{align} \sum_i \mu_i^{(IN)} = \sum_i \mu_{i0}^{(IN)} + k_B T \ln c_i^{(IN)}={\mu_{suc^-0}}^{(IN)} + k_B T \ln [{suc^-}^{(IN)}] +\mu_{H^+0}^{(IN)} + k_B T \ln [{H^+}^{(IN)}] \end{align} and for the outside potential \begin{align} \sum_i \mu_j^{(OUT)} = \sum_j \mu_{j0}^{(OUT)} + k_B T \ln c_j^{(OUT)}=\mu_{suc^-0}^{(OUT)} + k_B T \ln [{suc^-}^{(OUT)}] +\mu_{H^+0}^{(OUT)} + k_B T \ln [{H^+}^{(OUT)}] \end{align}

In biological thermodynamics the potential difference $\Delta \mu$ is defined as the difference between the chemical potential inside and outside the cell membrane \begin{align} \Delta \mu &= \sum_j \mu_j^{(IN)} -\sum_i \mu_i^{(OUT)} \tag{2} \\ \end{align} This is a purely conventional matter, since we could also define $\Delta \mu$ as $\sum_i \mu_i^{(OUT)}-\sum_j \mu_j^{(IN)}$. However, in biology the definition (2) has a tradition, stemming from the study of proton gradients, which are maintained by the electron transport chain. The electron transport chain is part of the photosynthesis machinary and pumps protons through the cell membrane. This process requires energy, which for example comes from the harvesting of light. Since protons are pumped out e.g. the mitochondria or the chloroplast, there are more protons outside than inside. The resulting force due to the gradient points therefore from outside to inside. This force is called proton-motive force (PMF) and drives the ATP production via the ATP Synthase.


Example


The proton-motive arises from the potential difference $\Delta \mu_{H^+}$ of protons across the membrane. For the difference in potential $\Delta \mu$ we need to know ${H^+}^{(IN)}$ and ${H^+}^{(OUT)}$, then we can calculate \begin{align} \Delta \mu_{H^+} &= \sum_j \mu_j^{(IN)} -\sum_i \mu_i^{(OUT)} \\ \Delta \mu_{H^+} &= k_B T \ln \frac{[{H^+}^{(IN)}]}{[{H^+}^{(OUT)}]}\\ \Delta \mu_{H^+} &= 2.3 k_B T \log_{10} \frac{[{H^+}^{(IN)}]}{[{H^+}^{(OUT)}]}\\ \Delta \mu_{H^+} &= 2.3 k_B T \Delta pH\\ \tag{3} \end{align} The proton-motive force is defined as \begin{align} \Delta p = \frac{\Delta G_{H^+}}{F} \end{align} where $F=e \cdot N_A$, the Faraday constant $F$ is the product of electric charge $e$ and Avogadro constant $N_A$. On the other hand the Gas constant $R=k_B \cdot N_A$ with the Boltzmann constant $k_B$. Knowing these relations for $R$ and $F$ and hence $F/e=R/k_B$ we can go on and find the potential per $z$ fold electron charge $e$ \begin{align} \frac{k_B \cdot T}{z \cdot e} \ln \frac{[{H^+}^{(IN)}]}{[{H^+}^{(OUT)}]}\\=\frac{2.3 R \cdot T}{z F}\log_{10} \frac{[{H^+}^{(IN)}]}{[{H^+}^{(OUT)}]}\\ \end{align} which is known as the electrode potential. The proton electrode potential (chemical potential per $z$ fold electron charge $e$) can be modified to \begin{align} E_{H^+} &= \frac{2.3 R \cdot T}{z F} \Delta pH\\ \tag{4} \end{align}

and the explicit form for the proton-motive force is \begin{align} \Delta p = \frac{\Delta G_{H^+}}{F} &= \frac{2.3 R \cdot T}{zF} \Delta pH\\ \tag{5} \end{align} Many biology and chemistry textbooks that do not elaborate on the thermodynamics and the origin of this equation (compare Rob Philipps with Nicholls and Harold), don't discriminate between $\Delta G$ and $\Delta \mu$ and define the PMF as $\frac{\Delta \mu_{H^+}}{F}$, which is wrong by the factor $N_A$. As you can see, the form is equivalent, with the definition of the electrode potential.


Example


Succinic acid transport is like the proton transport driven by the osmotic pressure due to the chemical potential difference. Let's make the chemical potential balance \begin{align} \Delta \mu &= \sum_j \mu_j^{(IN)} -\sum_i \mu_i^{(OUT)} \\ \Delta \mu &=k_B T \ln \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]} + k_B T \ln \frac{[{H^+}^{(IN)}]}{[{H^+}^{(OUT)}]}\\ \Delta \mu &=k_B T \ln \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]} + 2.3 k_B T \Delta pH\\ \tag{6} \end{align}

Notice, that the sign in this calculation is only valid under the assumption (1). Let's think about that for a moment. If the concentration of a solute is higher outside than inside, than osmotic pressure from the outside to the inside will push molecules in. In other words the molecules follow the chemical potential gradient and the potential is higher outside than inside. If we assume $[{suc^-}^{(IN)}]<[{suc^-}^{(OUT)}]$ for equation $k_B T \ln \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}$, we find \begin{align} k_B T \ln \underbrace{\frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}}_{<1} < 0 \Rightarrow \Delta \mu < 0 \end{align} so indeed the transport would proceed spontaneously under this condition. However, for $suc^-$, assumption (1) is not valid and we have to change the sign from $+ k_B T \ln \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}$ to $- k_B T \ln \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}$ (or equivalently, change the labels IN and OUT). Physically this means molecules have to flow in the other direction (from outside to inside).

Finally we find the chemical potential due to osmosis for the succinic acid - proton Symporter \begin{align} \Delta \mu &=-k_B T \ln \frac{[{suc^-}^{(IN)}]}{[{suc^-}^{(OUT)}]}+ 2.3 k_B T \Delta pH \end{align}

Moreover, what we did not include yet is the fact, that the different levels of charge on both sides of the membrane give rise to an electric potential in which the succinic acid moves through the transporter. This electric potential can either add to, or counteract the chemical potential due to osmosis. The calculations above would be correct, if the $n=1$ and with each $suc^-$ molecule one proton $H^+$ (together they form a neutral complex) is transported. Charged molecules or complexes in contrary are affected by the electric potential. In general a charged particle is subject to the Lorentz force. The influence of the electric potential on ions is discussed in the article Cells and Membrane Potentials - Equivalent Circuits.



Furhter Reading:

  • Rob Phillips - Physical Biology of the Cell [1]
  • David G. Nicholls and Stuart J. Ferguson - Bioenergetics
  • Franklin M. Harold - The Vital Force: A study of Bioenergetics