# 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 10μm
• 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 = d2 / D but this is overestimate as don't need that many particles to cross membrane. For midway is t = d2 / (4D)
• Exact solution: $\tau_{ss}=\frac{d^2}{\pi^2 D}$ (steady state time constant for membrane)

### Solute enters cell

• $\phi_n(t)=P_n(c_n^i(t)-c_n^o(t)), P_n=\frac{D_nk_n}{d}$
• Pn: 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 = 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