Exercises — Biological importance of Na, K, Ca, Mg

• Outside high: Na⁺ (~145 mM outside vs ~12 mM inside).
• Inside high: K⁺ (~140 mM inside vs ~4 mM outside).

Why: the Na⁺/K⁺-ATPase pump throws Na⁺ out and pulls K⁺ in. "Na-Out, K-Keep."

(a) Chlorophyll → Mg²⁺. (b) Haemoglobin → Fe²⁺. Same porphyrin frame (see Coordination Chemistry — Porphyrins), different metal.

3 Na⁺ out, 2 K⁺ in per ATP hydrolysed. Net one positive charge leaves the cell each cycle — the pump is electrogenic (it itself makes the inside slightly more negative). Remember this net charge; L4 uses it.

WHAT: plug into $E=\frac{RT}{zF}\ln \frac{[X{]}_{out}}{[X{]}_{in}}$. WHY Nernst: it is the one equation that balances the concentration push against the electrical push, giving the inside voltage at which K⁺ stops net-moving (sign convention: $E=V_{in}-V_{out}$).

WHAT (step 1): set $z=+1$ and use the prefactor $26.7\text{mV}$. WHY: K⁺ carries a single positive charge, so no halving; the prefactor was derived above. $E_K=26.7\times \ln \frac{4}{140}\text{mV}=26.7\times \ln (0.02857)\text{mV}$ WHAT (step 2): evaluate the log. WHY: the ratio (out/in) is less than 1 because K⁺ is scarcer outside, so its log is negative — this is what makes $E_K$ come out negative. $\ln (0.02857)=-3.555$. $E_K=26.7\times (-3.555)\approx -94.9\text{mV}\approx -95\text{mV}.$ Close to the resting $-70\text{mV}$ — which is why K⁺ mainly sets the resting state (see the resting-potential box above).

WHAT (step 1): insert the sodium concentrations (out on top) into the same formula. WHY out-on-top: our sign convention fixes $E=V_{in}-V_{out}$, and the derivation of Nernst puts $[X{]}_{out}/[X{]}_{in}$ inside the log precisely so the answer is the inside voltage — never flip it. $E_{Na}=26.7\times \ln \frac{145}{12}\text{mV}=26.7\times \ln (12.083)\text{mV}.$ WHAT (step 2): evaluate the log. WHY: now the ratio is greater than 1 (Na⁺ is abundant outside), so the log is positive → a positive $E_{Na}$, the mirror image of K⁺. $\ln (12.083)=2.492$. $E_{Na}=26.7\times 2.492\approx +66.5\text{mV}\approx +67\text{mV}.$ Sign is POSITIVE — opposite to K⁺. Na⁺ "wants" to rush in, so its equilibrium inside-voltage is well above the resting level. That gap is the stored energy a firing nerve unleashes.

WHAT (step 1): divide outside by inside. WHY: "how many times higher" is literally the ratio of the two concentrations — the same out/in quantity Nernst uses. $\text{Factor}=\frac{10^{-3}}{10^{-7}}=10^{-3-(-7)}=10^4=10,000\times .$ WHAT (step 2): count decades. WHY: a "decade" = one power of 10, so ${\log }_{10}(10^4)=4$ decades. This translates directly into voltage via the $61.5$-mV-per-decade shortcut later. That is 4 decades — an enormous gradient that makes Ca²⁺ a sharp ON-switch.

What the figure shows (in words): the vertical axis is membrane voltage $E=V_{in}-V_{out}$ in mV. Three horizontal lines are drawn — an orange dashed line at $E_{Na}\approx +67$ (top), a grey solid line at rest $\approx -70$ (middle), and a blue dashed line at $E_K\approx -95$ (bottom). A thick orange up-arrow runs from the rest line up toward the orange line; a thick blue down-arrow runs from the rest line down toward the blue line.

A channel lets its ion drift toward its own Nernst potential (its equilibrium voltage — the value where that ion stops net-moving).

• $E_{Na}\approx +67\text{mV}$ sits above rest ($-70$). Open Na⁺ channels → membrane voltage is dragged upward toward $+67$ → depolarisation (the orange up-arrow).
• $E_K\approx -95\text{mV}$ sits below rest. Open K⁺ channels → voltage is dragged downward toward $-95$ → repolarisation (the blue down-arrow).

So the same mechanism (drift toward Nernst potential) gives opposite effects purely because the two equilibrium voltages sit on opposite sides of rest. That is the whole logic of a nerve spike.

Coulomb attraction between two charges is $F\propto \frac{q_1{q}_2}{r^2}$.

• A caveat on phosphate's charge (edge-case chemistry): free inorganic phosphate is a mixture whose average charge depends on pH. Near physiological $\text{pH}\approx 7.2$, it is mostly a blend of ${\text{HPO}}_4^{2-}$ and ${\text{H}}_2{\text{PO}}_4^{-}$, so its effective charge is between $-1$ and $-2$ (around $-1.5$ to $-1.8$), not a clean $-2$. In ATP the terminal phosphates carry clustered negatives that are likewise pH-dependent. The exact number shifts with pH, but the key point survives: the anionic group presents more than one unit of negative charge for a $+2$ ion to grab.
• Against a $+2$ ion the numerator $q_1{q}_2$ is roughly twice that of a $+1$ ion at the same distance → about double the pull, even after the pH correction.
• Mg²⁺ is also tiny, so $r$ is small and $\frac{1}{r^2}$ is large → the force is amplified further.

Roughly doubling the binding energy pushes the ion from "slips off in nanoseconds" (good for fast signalling) into "stays bound" (good for structure/catalysis). A factor of ~2 in the exponent of a Boltzmann-type residence time is a huge change in stickiness — that qualitative jump is why $+2$ ions build, $+1$ ions signal. See Alkaline Earth Metals (Mg, Ca).

WHAT (step 1): halve the prefactor because $z=+2$. WHY: the $z$ sits in the denominator $\frac{RT}{zF}$, so a doubly-charged ion gets exactly half the volts per decade. $\frac{RT}{2F}=\frac{26.7}{2}=13.35\text{mV}.$ WHAT (step 2): put the out/in ratio into the log. WHY: Ca²⁺ is $10^4\times$ higher outside, so out-on-top gives a large positive log — Ca²⁺ strongly "wants in." $E_{Ca}=13.35\times \ln \frac{10^{-3}}{10^{-7}}\text{mV}=13.35\times \ln (10^4)\text{mV}.$ $\ln (10^4)=4\times 2.3026=9.210$. $E_{Ca}=13.35\times 9.210\approx +122.9\text{mV}\approx +123\text{mV}.$ Comment: the ratio is a giant $10^4$, but because $z=2$ the prefactor is halved, partly taming it. Even so $E_{Ca}$ is strongly positive — Ca²⁺ is desperate to flood in, which is exactly what makes it a violent, sharp signal.

(a) Charge out $=3(+1)=+3$; charge in $=2(+1)=+2$. Net $=+3-(+2)=+1$ charge leaves the cell each cycle. (b) Removing net positive charge makes the inside more negative relative to outside (recall $E=V_{in}-V_{out}$, so losing positive charge pushes $E$ down). So beyond just building the concentration gradients, the pump is electrogenic — it directly shaves a few millivolts off the resting potential. The bulk of $-70\text{mV}$ still comes from K⁺ leak (see the resting-potential box), but the pump adds a small helping hand.

With no pump, the leaks are no longer opposed: (i) Na⁺ leaks in → $[N{a}^{+}{]}_{in}$ rises. (ii) K⁺ leaks out → the K⁺ gradient collapses (shrinks). (iii) With the K⁺ gradient shrinking, $E_K$ moves toward $0$; the membrane potential depolarises (drifts up toward 0 mV), losing the resting negativity. (iv) Extra Na⁺ trapped inside raises internal solute → water follows osmotically → the cell swells and can burst. This is why the pump burns so much ATP continuously: it's fighting a constant osmotic + electrical leak.

$E_{Na}\approx +67$, $E_K\approx -95$. The peak $+30\text{mV}$ is far above rest and heading toward $+67$, i.e. toward $E_{Na}$. So at the peak the membrane is dominated by open Na⁺ channels (voltage chasing $E_{Na}$). It never fully reaches $+67$ because K⁺ channels start opening and pull the other way — but the drift direction alone tells you Na⁺ rules the upstroke.

WHAT: invert Nernst for the unknown $[N{a}^{+}{]}_{in}$. Start: $E=26.7\ln \frac{145}{[N{a}^{+}{]}_{in}}$ mV. Set $E=30$: $\frac{30}{26.7}=\ln \frac{145}{[N{a}^{+}{]}_{in}}⟹1.1236=\ln \frac{145}{[N{a}^{+}{]}_{in}}.$ Exponentiate: $\frac{145}{[N{a}^{+}{]}_{in}}=e^{1.1236}=3.076$. $[N{a}^{+}{]}_{in}=\frac{145}{3.076}\approx 47.1\text{mM}.$ Interpretation: if the pump lets internal Na⁺ climb from 12 to ~47 mM, $E_{Na}$ drops from $+67$ to $+30$ — the driving force for Na⁺ entry weakens, and nerve spikes get feebler. This is why even a partial pump failure blunts excitability.

Two effects fight:

• Ca²⁺ ratio $10^4$ → $\ln =9.21$ (large).
• But Ca²⁺ prefactor is halved ($13.35$ mV) because $z=2$.

Product: $E_{Ca}\approx 123\text{mV}$ (from L3.3). Compare K⁺: ratio $\frac{4}{140}$ → $|\ln |=3.55$, prefactor $26.7$ → $|E_K|\approx 95\text{mV}$. So Ca²⁺ does end up largest in magnitude here ($123>95$), but not simply because its ratio is biggest — the halved prefactor nearly cancels the ratio advantage. The claim reaches the right answer for the wrong reason. Magnitude $=|\text{prefactor}|\times |\ln (\text{ratio})|$; you must weigh both factors, never the ratio alone.

1. K⁺ → resting. Its $E_K\approx -95$ mV lies nearest the required negative resting level, and the membrane leaks K⁺ most, so the resting blend sits close to it (see resting-potential box). A leak channel for the ion whose Nernst matches rest gives a stable baseline "for free."
2. Na⁺ → spike. Its $E_{Na}\approx +67$ mV sits on the opposite side of rest (L2.2), so opening Na⁺ channels swings the inside-voltage far and fast — perfect for a sharp, all-or-nothing upstroke (L4.3).
3. Ca²⁺ → signal. Kept $10^4\times$ lower inside (L2.3), so a tiny opening gives a huge, clean spike of concentration (not just voltage). Its $+2$ charge (L3.2) lets it also bind troponin/proteins — it does chemistry on arrival, not just carry charge.
4. Mg²⁺ → ATP. Smallest, highest charge density → grips the phosphate cluster of ATP hardest (L3.2, Coulomb), neutralising the negatives so enzymes can act. Too sticky to be a fast switch, perfectly sticky to be a permanent cofactor. See ATP and Bioenergetics and Chlorophyll and Photosynthesis.

Each assignment falls straight out of charge + size → gradient + Nernst sign. Nothing is arbitrary.

Nernst sign (out-over-in, giving $V_{in}-V_{out}$) + charge $z$ (halves the prefactor) explain every role: K⁺ negative → rest, Na⁺ positive → spike, Ca²⁺ huge ratio → signal, Mg²⁺ high charge density → grip. ::: If you can rebuild all four from those two levers, you own this topic.