Biology · Introductory biology · Worked example

Calculate an equilibrium potential

A mammalian neuron holds about $135 mM K^{+}$ and $10 mM {Na}^{+}$ inside, with about $3 mM K^{+}$ and $150 mM {Na}^{+}$ outside. At $37 ° C$ , find the equilibrium potentials of $K^{+}$ and ${Na}^{+}$ . How do they compare with a resting potential of $- 70 mV$ and a peak of $+ 30 mV$ ?

Find the constant

$R T / F$ is the thermal voltage, with $R = 8.314 J / (mol \cdot K)$ , $T = 310 K$ and $F = 96, 485 C / mol$ . It comes out in volts.

\frac{8.314 (310)}{96485} \approx 0.0267

Convert to base-10 logs

Why multiply by $2.303$ ? The Nernst equation uses the natural $log$ , and $ln x = 2.303 {log}_{10} x$ . In millivolts, $0.0267 V$ is $26.7 mV$ .

2.303 (26.7) \approx 61.5

Find $E_{K}$

$K^{+}$ has $z = + 1$ . The ratio is outside over inside, $3 / 135$ , which is less than $1$ , so the $log$ is negative.

61.5 {log}_{10} (\frac{3}{135}) \approx - 101.7

Find $E_{Na}$

${Na}^{+}$ also has $z = + 1$ , and it is $15$ times more concentrated outside, so its $E$ is positive.

61.5 {log}_{10} (\frac{150}{10}) \approx 72.3

Compare with rest and the peak

Rest, $- 70 mV$ , lies $31.7 mV$ above $E_{K}$ , because a little ${Na}^{+}$ leaks in. The peak, $+ 30 mV$ , stops $42.3 mV$ short of $E_{Na}$ , because ${Na}^{+}$ channels inactivate and $K^{+}$ channels open before the voltage gets there.

\begin{array}{l} - 70 - (- 101.7) = 31.7 \\ 72.3 - 30 = 42.3 \end{array}

Result

$E_{K} \approx - 101.7 mV$ and $E_{Na} \approx + 72.3 mV$ . Rest at $- 70 mV$ lies between them, much nearer $E_{K}$ , and the peak at $+ 30 mV$ stops well short of $E_{Na}$ .

Your turn

1. One like the example

During spreading depolarization in the brain, $K^{+}$ outside neurons can rise to about $50 mM$ while $K^{+}$ inside falls, by one estimate, to about $112 mM$ . What is $E_{K}$ then?

Show the answer and explanation

About $- 21.5 mV$ .

$61.5 {log}_{10} (50 / 112)$ $\approx 61.5 \times (- 0.350)$ $\approx - 21.5 mV$ . With $E_{K}$ that close to zero, the membrane depolarizes far past threshold, the ${Na}^{+}$ channels inactivate, and the neurons cannot fire.

61.5 {log}_{10} (\frac{50}{112}) \approx - 21.5

2. A variation

${Ca}^{2 +}$ is about $10, 000$ times more concentrated outside a cell than inside. What is its equilibrium potential at $37 ° C$ ?

Show the answer and explanation

About $+ 123 mV$ .

${Ca}^{2 +}$ has $z = + 2$ , so the constant is $61.5 / 2 = 30.75 mV$ . Then $30.75 \times {log}_{10} 10, 000$ $= 30.75 \times 4$ $= 123 mV$ . Forgetting the charge gives twice that, $246 mV$ .

\frac{61.5}{2} {log}_{10} (10000) = 123

3. Somewhere new

Hodgkin and Huxley studied squid axons at $6.3 ° C$ , or $279.45 K$ . What is the Nernst constant $2.303 R T / F$ at that temperature, in volts?

Show the answer and explanation

About $0.0555 V$ , or $55.5 mV$ .

$2.303$ $\times 8.314$ $\times 279.45 / 96, 485$ $\approx 0.0555 V$ . The constant is proportional to the absolute temperature, so a cold squid’s is about a tenth smaller than the $61.5 mV$ of a warm mammal.

\frac{8.314 (279.45)}{96485} (2.303) \approx 0.0555

\begin{array}{l} \frac{8.314 (279.45)}{96485} (2.303) \\ \approx 0.0555 \end{array}

Keep exploring

Trace the equilibrium potentials in Math starts from $8.314 (310) / 96485$ $\approx 0.0267$ . Write $2.303 (26.7) \approx 61.5$ under it, and each row is checked.

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Find the constant

Convert to base-10 logs

Find EKEK​

Find ENaENa​

Compare with rest and the peak

Result

Your turn

1. One like the example

About −21.5 mV−21.5 mV.

2. A variation

About +123 mV+123 mV.

3. Somewhere new

About 0.0555 V0.0555 V, or 55.5 mV55.5 mV.

Keep exploring

Study it, then try it

Find $E_{K}$

Find $E_{Na}$

About $- 21.5 mV$ .

About $+ 123 mV$ .

About $0.0555 V$ , or $55.5 mV$ .