6.021/Notes/2006-09-13

Diffusion

 * Non-linear relationship between space and time is non-intuitive

Diffusion applied to cells

 * Membrane diffusion
 * The membrane is around 10nm whereas the cell is about 10&mu;m
 * Can treat as 1D diffusion (diffusion across membrane ignoring other dimension)
 * Reference direction for flux: positive is out of cell

Dissolve and Diffuse model

 * 1) solute outside dissolves into membrane
 * 2) solute diffuses through membrane
 * 3) solute dissolves into cytoplasm
 * 4) concentration of solute in cell increases


 * Assume dissolving is faster than diffusion (assume dissolving is instant)
 * $$c_n^i$$: concentration inside of solute
 * $$c_n^o$$: concentration outside of solute

Dissolve model

 * 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)

Diffusion in membrane

 * 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 $$t=d^2/(4D)$$
 * Exact solution: $$\tau_{ss}=\frac{d^2}{\pi^2 D}$$ (steady state time constant for membrane)

Solute enters cell

 * $$\phi_n(t)=P_n(c_n^i(t)-c_n^o(t)), P_n=\frac{D_nk_n}{d}$$
 * $$P_n$$: 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: $$c_n^i(t)=c_n^i(\infty)+(c_n^i(0)-c_n^i(\infty))e^{-t/\tau_{eq}}$$
 * $$c_n^i(\infty) = \frac{N_n}{V_i+V_o}, \tau_{eq}=\frac{1}{AP(1/V_i+1/V_o)}$$

Check assumption dissolving is fast

 * 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, $$r=10\mu{\rm m}, d=10{\rm nm}, k=1$$
 * $$\frac{\tau_{eq}}{\tau_{ss}}=\frac{V_i}{AP}\frac{\pi^2D}{d^2} \approx 3\frac{r}{d} \approx 10^3$$
 * Assumption that $$\tau_{eq} \gg \tau_{ss}$$ is ok