6.021/Notes/2006-09-13

Diffusion

Membrane is like oil, cytoplasm and outside bath is like water
Some solutes like oil, some like water
Find relative solubilities of solute n in oil and water
Partition coefficient $k_{oil:water}=\frac{c_n^{oil}}{c_n^{water}}$ (at equilibrium)

Difficult to solve analytically but numerically easy
From point of view of membrane, both inside and outside baths are constant
If wait long enough (reach equilibrium), the concentration will become flat in membrane
But short term, will be a straight line
How long to straight line?
- Membrane width d
- Can estimate it. For half of particles to cross membrane is $t = d 2 / D$ but this is overestimate as don't need that many particles to cross membrane. For midway is $t = d 2 / (4 D)$
- Exact solution: $\tau_{ss}=\frac{d^2}{\pi^2 D}$ (steady state time constant for membrane)

$\phi_n(t)=P_n(c_n^i(t)-c_n^o(t)), P_n=\frac{D_nk_n}{d}$
$P n$ : permeability of membrane to solute n
concentration in cell changes: 2 compartment diffusion
assume volumes constant, baths are well-stirred, membrane is thin (ignore solute in membrane), and membranes always in steady state
$c_n^i(t)V_i + c_n^o(t)V_o = N_n$ (total amount of solute is conserved)
$\phi_n(t)=P_n(c_n^i(t)-c_n^o(t))$
Solution to equations: $c_n^i(t)=c_n^i(\infty)+(c_n^i(0)-c_n^i(\infty))e^{-t/\tau_{eq}}$
$c_n^i(\infty) = \frac{N_n}{V_i+V_o}, \tau_{eq}=\frac{1}{AP(1/V_i+1/V_o)}$

2 time constants
$\tau_{ss}=\frac{d^2}{\pi^2 D}, \tau_{eq}=\frac{1}{AP(1/V_i+1/V_o)}$
For cell $V_o\gg V_i$ so $\tau_{eq}=\frac{V_i}{AP}$
Assume spherical cell, $r = 10μm, d = 10nm, k = 1$
$\frac{\tau_{eq}}{\tau_{ss}}=\frac{V_i}{AP}\frac{\pi^2D}{d^2} \approx 3\frac{r}{d} \approx 10^3$
Assumption that $\tau_{eq} \gg \tau_{ss}$ is ok