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<!DOCTYPE html>
<html>
<head>
<title>Gradient Free Energy</title>
<style type="text/css">
body {
background-color: #FFFFFF;
font-family: Verdana, sans-serif;
font-size: 12 px
}
</style>
<link rel="stylesheet" type="text/css" href="styles.css" />
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<script type="text/javascript" src="functions.js"></script>
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<body>
<div style="display:none">
$
\newcommand{\comb}[1]{{#1}^{\mathrm{comb}}}
\newcommand{\conc}[1]{[\mathrm{#1}]}
\newcommand{\conceq}[1]{[\mathrm{#1}]^{\mathrm{eq}}}
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\newcommand{\eq}[1]{{#1}^{\mathrm{eq}}}
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%\newcommand{\ztot}{Z^{\mathrm{tot}}}
$
</div>
<h2>
The Free Energy in a Concentration Gradient
</h2>
One key way the cell stores <a href="javascript:changeTo('freeGradient','free')">free energy</a> is by having different concentrations of molecules in different "compartments" - e.g., extra-cellular vs. intracellular or in an organelle compared to cytoplasm.
Here, we will study the simplest example of such a <i>gradient</i>, differing concentrations of <a href="javascript:changeTo('freeGradient','idealGas')">ideal gas</a> molecules across an idealized rigid membrane.
Elsewhere, we consider <a href="javascript:changeTo('freeGradient','ionsNernst')">a simple model of ion concentration gradient across a membrane</a>.
<h3>
Two ideal gases separated by a barrier
</h3>
<p style="text-align:center"><img src="images/ideal-two-compartments.gif" /></p>
Analyzing the model depicted above will enable us to understand free energy storage in a concentration gradient, but the basic ideas generalize to <a href="javascript:changeTo('freeGradient','carriers')">activated carriers</a> as well.
<p>
To be precise, the model consists of $N$ non-interacting atoms in the volume $V$ maintained at constant temperature $T$.
Beyond <a href="javascript:changeTo('freeGradient','idealGas')">the simple ideal gas studied elsewhere</a>, our present system is divided into two compartments by a rigid "membrane," with $V_1$ the volume of the inner compartment and $V_2$ the outer volume such that $V_1 + V_2 = V$.
Similarly, there are $N_1$ atoms in the inner compartment and $N_2$ outside, with $N_1 + N_2 = N$.
Although particles could pass through the channel shown in the figure, we assume it is closed so that $N_1$ and $N_2$ are constant.
</p>
<h3>
A quick mass-action analysis
</h3>
<p style="text-align:center"><img src="images/ideal-two-compartments-leak.gif" /></p>
We can derive the key result for this system very quickly using a <a href="javascript:changeTo('freeGradient','massAction')">mass action</a> "thought experiment".
Our simple kinetic analysis will provide a key reference when we delve into some specific <a href="javascript:changeTo('freeGradient','ionsNernst')">limitations of the mass-action picture in the context of ionic gradients</a>
<p>
Instead of our original system, we consider the leaky cell or organelle shown above.
The leak should be considered a simple hole (unlike a channel, which is expected to interact strongly with molecules passing through it).
Hence the inside-to-outside rate constant $k_{12}$ for the leak/hole must be equal to the outside-to-inside rate constant $k_{21}$.
After all, if the hole is large enough the "atoms" will not interact with the membrane at all - or if they do, the effects should be symmetric.
</p>
<p>
In <a href="javascript:changeTo('freeGradient','equil')">equilibrium</a>, we know that the total number of events in each direction will match: $\conc{1} \, k_{12} = \conc{2} \, k_{21}$.
Cancelling the equal rates on both sides of this relation, we find that the <b>equilibrium concentrations must be equal inside and outside</b>:
\begin{equation}
\label{eqleak}
\conceq{1} = \conceq{2}.
\end{equation}
</p>
Deriving this result thermodynamically in a careful way requires more effort (see below).
<p>
This equilbrium result for the leaky condition actually tell us something about the channel rates.
Because the channel is a passive element that uses no energy, it cannot change the (equal) equilibrium concentrations just derived.
Hence, applying the principle of <a href="javascript:changeTo('freeGradient','equil')">detailed balance</a> to the channel (which implies the flows through the channel must be equal and opposite) we see that the <i>channel rate constants must be equal in both directions</i>.
This perhaps obvious result only holds for a channel separating two systems with no driving force or external field applied - a condition which breaks down in the case of <a href="javascript:changeTo('freeGradient','ionsNernst')">trans-membrane ionic gradients</a>.
</p>
<h3>
Deriving the free energy
</h3>
Because we have non-interacting ("ideal") particles which cannot exchange across the membrane, the total free energy for the combined systems $\comb{F}$ is simply the sum of the two free energies calculated independently for the two systems.
(From the probability point of view embedded in the calculation of a partition function - see below - the lack interactions implies statistical independence and hence factorizability of the full-system partition function into those for the two systems.)
We have
\begin{equation}
\comb{F}(N_1, N_2) = \fidl(N_1, V_1, T) + \fidl(N_2, V_2, T) ,
\end{equation}
where $\fidl$ is defined in the <a href="javascript:changeTo('freeGradient','idealGas')">ideal gas page</a>.
We have omitted the $V_1$, $V_2$, and $T$ dependence in $\comb{F}$ because these will be held constant throughout.
<p>
Substituting in for $\fidl$, we have
\begin{align}
\comb{F}(N_1, N_2) = &
N_1 \, k_B T \ln \frac{N_1 \, \lambda^3}{V_1} + \, N_2 \, k_B T \ln \frac{N_2 \, \lambda^3}{V_2} . \\
\end{align}
Noting that $N_2 = N - N_1$, we can rewrite this further as
\begin{align}
\comb{F} (N_1, N - N_1) =
& N_1 \, k_B T \ln \frac{N_1 \, \lambda^3}{V_1} \nonumber \\
& + \, (N - N_1) \, k_B T \ln \frac{(N - N_1) \, \lambda^3}{V_2} .
\label{fcombo}
\end{align}
</p>
Eq. \eqref{fcombo} is the free energy as a function of the number of particles inside the membrane (volume $V_1$).
<p style="text-align:center"><img src="images/ideal-low-N1.gif" height="150" /> <img src="images/ideal-high-N1.gif" height="150" /></p>
<p style="text-align:center"><img src="images/ideal-Fcombo-graph.gif" height="250" /></p>
If we open the channel and allow exchange of atoms between the compartments, the value of $N_1$ can change.
<b>The probability of having $N_1$ atoms in $V_1$ is proportional to the Boltzmann factor of the free energy</b>:
\begin{equation}
\label{ftoprob}
p(N_1) \propto e^{ - \left.\comb{F}(N_1, \, N - N_1) \right/ k_B T }
\end{equation}
<p>
The most probable $N_1$ value therefore can be found by determining the minimum of $\comb{F}$.
This will represent the <a href="javascript:changeTo('freeGradient','equil')">equilibrium</a> point in the thermodynamic limit (very large $N$ - when fluctuations about the most probable $N_1$ will be very small compared to $N_1$ itself).
We set $\dee \comb{F} / \dee N_1 = 0$ in Eq. \eqref{fcombo}, then re-arrange and cancel terms to find
</p>
\begin{equation}
\label{derivzero}
0 = k_B T \ln \frac{N_1 \, \lambda^3}{V_1} - k_B T \ln \frac{(N - N_1) \, \lambda^3}{V_2}
\end{equation}
Combining the terms using the rules of logarithms, followed by exponentiation, we find that
\begin{equation}
\label{eqconc}
\mbox{Equilibrium: } \frac{N_1}{V_1} = \frac{N - N_1}{V_2} = \frac{N_2}{V_2} ,
\end{equation}
where we substitute $N_2 = N - N_1$ to obtain the last equality.
<p>
In words, Eq. \eqref{eqconc} shows that <i>the concentrations inside and outside the membrane must match in equilibrium</i>.
You probably knew that already, but we have derived it from statistical/thermodynamic principles.
</p>
<h4>
The next step: Considering ions
</h4>
<a href="javascript:changeTo('freeGradient','ionsNernst')">Ion concentration gradients</a> can also be analyzed in a similar way.
<h4>
Work that can be performed
</h4>
As we move from higher to lower free energy, the system can perform work - if it is coupled to a suitable mechanism for harvesting the work.
The <a href="javascript:changeTo('freeGradient','freeThermo')">maximum amount of work that can be extracted</a> is equal to the decrease in free energy as sketched above.
In a simple gas system, work could be extracted by placing a turbine at the "channel"/nozzle as the gas flows toward equilibrium.
Models of work extraction which are more pertinent to cell biology are discussed in the <a href="javascript:changeTo('freeGradient','transport')">transport</a> section.
<h4>
Passive Transport
</h4>
In the simplest kind of <a href="javascript:changeTo('freeGradient','passiveTransport')">passive transport</a>, molecules flow down a gradient (from high to low concentration) and that flow is not coupled other processes.
In our technical language, such a process would involve moving from a state of higher to lower free energy (see sketch above) or from lower to higher probability - see Eq. \eqref{ftoprob}.
<h3>
A deeper look at partition functions and probabilities
</h3>
A partition function $Z$ is simply a sum of Boltzmann factors for all possible states (configurations - and velocities if considered) of a system.
Because a Boltzmann factor represents a weight (an un-normalized probability), $Z$ is the sum of weights.
Many times, partition functions are easier to work with mathematically, compared to free energies.
Our system is such a case.
<p>
<h4>Factorizability</h4>
Our combined system consists of two independent sub-systems. That is, the state of one system will not affect the other.
Hence the probability for a configuration of the combined system is simply the <i>product</i> of the probabilities for the individual system configurations, and this also holds for the Boltzmann weights summed in Z.
</p>
If $U_1$ is the potential energy of the configuration of system 1 and $U_2$ is the energy of system 2, these two are independent, so we have
\begin{equation}
e^{-\totsub{U} / k_B T} = e^{- \left.\left( U_1 + U_2 \right) \right/ k_B T}
= e^{ - U_1 / k_B T } \, e^{ - U_2 / k_B T }
\end{equation}
We can extend this reasoning to calculate the partition function of the combined system, building on what was done for a simple <a href="javascript:changeTo('freeGradient','idealGas')">ideal gas</a>.
Denoting the configuration of the combined system by $(\rn{1}, \rn{2})$, the partition function is
\begin{align}
\label{z}
\comb{Z}\!\!&(N_1, N_2) =
\frac{\lambda^{-3N_1}}{N_1!}
\frac{\lambda^{-3N_2}}{N_2!}
\int_{V_1} d\rn{1}
\int_{V_2} d\rn{2}
\, e^{-\left. \totsub{U} \left( \rn{1}, \, \rn{2} \right) \right/ k_B T} \nonumber \\
& =
\left[ \!
\frac{\lambda^{-3N_1}}{N_1!}
\! \!\ \int_{V_1} \! d\rn{1}
\, e^{-\left. U_1 \left( \rn{1} \right) \right/ k_B T}
\! \right]
\! \!
\left[ \!
\frac{\lambda^{-3N_2}}{N_2!}
\! \! \int_{V_2} \! d\rn{2}
\, e^{-\left. U_2 \left( \rn{2} \right) \right/ k_B T}
\! \right]
\nonumber \\
& =
\frac{\lambda^{-3N_1}}{N_1!} V_1^{N_1}
\,
\frac{\lambda^{-3N_2}}{N_2!} V_2^{N_2}
= \idl{Z}(N_1, V_1) \, \idl{Z}(N_2, V_2)
\end{align}
where we evaluated the integrals in the last line, noting $U_1 = U_2 = 0$ for ideal particles, so that each atom's integration yields a factor of $V_i$ depending on which volume $i$ is occupied.
<h4>
Connecting $Z$ and $F$ to probability
</h4>
A partition function is the sum of all probability (weights) consistent with the conditions/constraints - such as constant $T$ or $V$.
This allows us to compare the (summed) probabilities consistent with different constraints, such as different $N_1$ values in $\comb{Z}$.
More specifically, recalling that $N_2 = N - N_1$, we have
\begin{equation}
\label{ztoprob}
p(N_1) \propto \comb{Z}(N_1, N - N_1) .
\end{equation}
Because the free energy is nothing other than the log of the partition function ($F = -k_B T \ln Z$ or $Z = \exp{(-F/k_BT)}$), we see that Eqs. \eqref{ztoprob} and \eqref{ftoprob} are equivalent.
<h4>
The total partition function
</h4>
Advanced readers may have considered the possibility of the <i>total</i> partition function $\totsup{Z}$, which not only sums over all configurations consistent with a given $N_1$ value - <i>but which also sums over all possible $N_1$ values from $0$ to $N$</i>.
The outcome is rather neat.
We write
\begin{align}
\label{totz}
\totsup{Z} &= \sum_{N_1=0}^N \comb{Z}(N_1, N - N_1) \nonumber \\
&= \sum_{N_1=0}^N \frac{1}{N_1!} \left( \frac{V_1}{\lambda^3} \right)^{N_1}
\frac{1}{(N-N_1)!} \left( \frac{V_2}{\lambda^3} \right)^{(N-N_1)}
\nonumber \\
&= \frac{1}{N!} \sum_{N_1=0}^N \frac{N!}{N_1! \, (N-N_1)!} \left( \frac{V_1}{\lambda^3} \right)^{N_1}
\left( \frac{V_2}{\lambda^3} \right)^{(N-N_1)}
\nonumber \\
&= \frac{1}{N!} \left( \frac{V}{\lambda^3} \right)^{N}
\end{align}
where the last line derives from recognizing the <a href="javascript:extLink('http://en.wikipedia.org/wiki/Binomial_theorem')">binomial expansion</a> for $(V_1 + V_2)^N$ implicit in the previous line.
<p>
You should recognize Eq. \eqref{totz} as the partition function for a <a href="javascript:changeTo('freeGradient','idealGas')">simple ideal gas</a> of $N$ atoms in volume $V = V_1 + V_2$.
Indeed, once the channel is open, all atoms can access both volumes and this is the correct result.
</p>
<h3>
Exercises
</h3>
<ol>
<li> Derive Eqs. \eqref{derivzero} and \eqref{eqconc}.
</ol>
</body>
</html>