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af86

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## Homework Statement

Calculate number of channels present in a lipid bilayer

q = 1.6^-19; D (diffusion constant) = 1.2E-9; C (Concentration); 4.36; k = 1.38E-23; T=298; d = 2.5E-9

Therefore G

_{1}= 1.27E-17

Channel area = 7.07E-18

measured conductance (from injecting Na+ concentration of 100mM) = 149.24uS

## Homework Equations

G1 (conductance) = q^2DC/kTd

Gm = g/[tex]\pi[/tex]r^2

## The Attempt at a Solution

Conductance of a single channel = 1.27E-12 x 7.07E-18

Number of channels = Gm/G1 ??

Calculating the number of channels is what is getting to me. I have the area of the channel, and I have a measured conductance from injecting a concentration

Thanks!