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Fick's first law of diffusion

We can treat a single ion channel as a cylindrical pore through the cell membrane. The channel is of length L = 5 nm and is filled with water. The concentration differences of sodium ions across the membrane is â??c = 100 mM. This concentration difference produces a flux of ions through the channel, given by Fickâ??s Law. For small ions like sodium, the diffusion constant, D, in water is of the order of 1 µm2/ms.

a.Assuming that the channel cross-section has an area A = 1 nm2, use Fickâ??s Law to calculate the number of sodium ions that pass through the channel every second due to diffusion.

b.Figure 2 shows current recording from a single channel. Use this figure to estimate the number of ions passing through the channel per second, when it is open. Assume that each sodium ion carries an elemental charge, q.

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Fick's first law of diffusion tells us how the flux varies with the concentration gradient:

J = -D*dc/dx

Where D is a diffusion coefficient (given in the problem as 1 µm^2/ms) that represents how easily the particles in question move through the medium (or to look at it another way, how fast the sodium ions can move in the water).

Here the concentration gradient is given as well: 100 mM over 5 nm.

So we can find the flux, J, in a fairly straightforward way. The trickiest part at this point is going to be keeping the units straight. I highly recommend explicitly writing out the units at every step.

J = -D*dc/dx

J = -(1 µm^2/ms) * (100 mM / 5 nm)

Now, ...

Solution Summary

Fick's first law of diffusion is thoroughly exemplified.

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