122 Chapter 4. Random walks, friction, and diffusion[[Student version, December 8, 2002]]
Your Turn 4e
a. Show that the units ofPsare the same as those of velocity.
b. Using this simplified model of the cell membrane, show thatPsis given byD/Ltimes the
fractionαof the membrane area covered by pores.Example Think of a cell as a spherical bag of radiusR=10μm,bounded by a membrane that
passes alcohol with permeation constantPs=20μms−^1 .Question: If initially the
alcohol concentration iscoutoutside the cell andcin(0) inside, how does the interior
concentrationcinchange with time?
Solution: The outside world is so immense, and the permeation rate so slow,
that the concentration outside is essentially always the same. The concentration
inside is related to the numberN(t)ofmolecules inside bycin(t)=N(t)/V,where
V=4πR^3 /3isthe volume of the cell. According to Equation 4.20, the outward flux
through the membrane is thenjs=−Ps(cout−cin(t))≡−Ps×∆c(t). Note thatjs
can be negative: Alcohol will move inward if there’s more outside than inside.
LetA=4πR^2 bethe area of the cell. From the definition of flux (Section 4.4.2),N
changes at the rate dN/dt=−Ajs.Remembering thatcin=N/V,wefind that the
concentration jump ∆cobeys the equation−
d(∆c)
dt=
(APs
V)
∆c. relaxation of a concentration jump (4.21)But this is an easy differential equation: Its solution is ∆c(t)=∆c(0)e−t/τ,where
τ=V/(APs)isthedecay constantfor the concentration difference. Putting in the
given numbers shows thatτ ≈ 0. 2 s. Finally, to answer the question we needcin,
which we write in terms of the given quantities ascin(t)=cout−(
cout−cin(0))
e−t/τ.
Wesay that an initial concentration jumprelaxesexponentially to its equilibrium value. In one
second, the concentration difference drops to about e−^5 =0.7% of its initial value. A smaller cell
would have a bigger surface-to-volume ratio, so it would eliminate the concentration difference even
faster.
The rather literal model for permeability via membrane pores, as sketched above, is certainly
oversimplified. Other processes also contribute to permeation. For example, a molecule candissolve
in the membrane material from one side,diffuseto the other side, then leave the membrane. Even
artificial membranes, with no pores at all, will pass some solutes in this way. Here too, a Fick-type
law, Equation 4.20, will hold; after all,somesort of random walk is still carrying molecules across
the membrane.
Since artificial bilayers are quite reproducible in the laboratory, we should be able to test the
dissolve→diffuse→undissolve mechanism of permeation by checking some simple quantitative de-
duction from the model. Figure 4.13 shows the result of such an experiment by A. Finkelstein and
coauthors, who measured the permeation constants for sixteen small molecules. To understand
these data, first imagine a simpler situation, a container with a layer of oil floating on a layer of
water. If we introduce some sugar, stir well, and wait, eventually we will find that almost, but not
all, of the sugar is in the water. The ratio of the concentration of sugar in the water to that in the
oil is called thepartition coefficientB;itcharacterizes the degree to which sugar molecules prefer
one environment to another. We will investigate the reasons for this preference in Chapter 7; for