Formula that I am using to calculate intercellular fluxes: $$ J = \frac{4D}{\pi d} \big([C]_1 - [C]_2\big), $$ where the flux through a hole is $d$ of a solute with a diffusion coefficient $D$ that is present in the signaling cell at concentration $[C]_1$ and in the receiving cell at $[C]_2$.
($d = \pu{4nm}$ , $D = \pu{10^{-5} cm^2/s}$, $[C]_1 - [C]_2 = \pu{100 \mu M}$)

Step 1: Standardize the units of $d$ and $D$: $$D = \pu{10^{-5} cm^2//s} = \pu{10^9 nm^2//s}$$

Step 2: Substitute values in the formula \begin{align} J &= \frac{4 \cdot \pu{10^9 nm^2}}{3.14 \cdot \pu{4 nm s}} \left(\pu{100 \mu M}\right)\\ J &= \frac{\pu{10^9 nm}}{\pu{3.14 s}}\left(\pu{10^2 \mu M}\right) \end{align}

You can compute it even further but my answer is not correct as the correct answer is $2.4 \times 10^5$ molecules per pore per second.
This problem is an example from a presentation. Here is the slide:


Kindly explain what I am doing wrong.


1 Answer 1


I believe the equation has mistakes: Why is the diameter of the hole ($d$) in the denominator? This means that the larger the hole, the smaller the flux!?

Let me try to give the whole calculation. The flux $J$ through a hole of thickness $l$ of a solute with a diffusion constant $D$ and the concentration difference $\Delta c$ is given by the Fick's law:

$$J = D\frac{\Delta c}{l}.$$

Note that the flux is a quantity that tells us what is the solute current per unit area, namely $J = \Phi/A$.

Let's assume the thickness of the pore is $l=\ce{4 nm}$, then given your values of $D=\ce{10^{-5} cm^2 s^{-1}}$ and $\Delta c=\ce{100 \mu M}$ we obtain:

$$J = \ce{2.5 \cdot 10^{-2} mol m^{-2} s^{-1}}.$$ We can convert this to a number flux by multiplying with Avogadro's constant: $$J_N = J N_{\rm A} = \ce{ 1.5 \cdot 10^{22} m^{-2} s^{-1}}.$$

To calculate the ion current through one pore, we multiply flux with the area of the pore, which is $A=\pi (d/2)^2$ for a circular hole with diameter $d$:

$$\Phi_N = J_N A = J_N \pi (d/2)^2 = \underline{\ce{1.5\cdot10^5 s^{-1}}}.$$

  • $\begingroup$ I don't see the diameter $d$ in that equation. $\endgroup$ Commented Sep 10, 2023 at 14:36
  • $\begingroup$ @Martin-マーチン, ah, I've introduced $d$ prematurely in the text. I'll fix it. $\endgroup$
    – Domen
    Commented Sep 10, 2023 at 14:51

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