# Explain estimating intercellular fluxes

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.

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 $$d$$:

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

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