= Nuclear pore transport balance
{title2=$R=C_n/C_c$}
Nuclear accumulation should compare <concentrations>, not just total cargo. In an original constructed example, a <cell nucleus> occupies one <volume (thermodynamics)> unit and the accessible <cytosol> occupies nine. Of $130$ tracer units, $40$ are nuclear and $90$ are cytosolic. With $n_n,n_c$ the amounts and $V_n,V_c$ the accessible <volume (thermodynamics)>, the <concentration> ratio is
$$
C_n=\frac{n_n}{V_n}=40,\qquad C_c=\frac{n_c}{V_c}=10,\qquad R=\frac{C_n}{C_c}=4.
$$
Only $40/130=4/13\approx30.8\%$ of the cargo is nuclear, yet nuclear <concentration> is four times cytosolic <concentration>. A majority outside the <cell nucleus> does not rule out nuclear enrichment.
For comparison, freely exchanging nonbinding cargo at equal <concentrations> would have $n_n=13,n_c=117$: the nuclear fraction would equal the nuclear fraction of accessible <volume (thermodynamics)>, $1/10$. These are constructed amounts, not measurements from the <nuclear pore localization experiment>. Binding, unequal recovery or cargo degradation would require additional accounting before interpreting an assay.
A minimal two-compartment <ordinary differential equation> makes the distinction precise. Assume fixed accessible <volume (thermodynamics)>, no synthesis or degradation, and effective first-order exchange rates $a,b>0$:
$$
\frac{dn_n}{dt}=a n_c-b n_n,\qquad n_c=N-n_n,\qquad \frac{dn_c}{dt}=-\frac{dn_n}{dt}.
$$
Total cargo $N=n_n+n_c$ is conserved. Setting the <derivative> to zero gives
$$
n_n^*=\frac{aN}{a+b},\qquad \frac{C_n^*}{C_c^*}=\frac{aV_c}{bV_n}.
$$
For symmetric passive exchange with conductance $P$, $a=P/V_c$ and $b=P/V_n$, so the <concentration> ratio is one even when the amounts differ. Choosing instead $a=4/9$ and $b=1$ in reciprocal model-time units gives $n_n^*=40$ for $N=130,V_n=1,V_c=9$. The exact <ordinary differential equation> solution is
$$
n_n(t)=n_n^*+\bigl(n_n(0)-n_n^*\bigr)e^{-(a+b)t}.
$$
Substitution verifies the solution, and $a+b>0$ makes departures decay. This effective model can describe an asymmetric exchange balance, but does not by itself identify a <nuclear localization sequence>, transport receptor or source of <energy>.
The usefulness of the <nuclear pore complex> is precisely this combination of separation and exchange: a <cell nucleus> can maintain a distinctive <protein> population while receiving new cargo and exporting <RNA>. The <nuclear pore localization experiment> asks what feature controls entry; the <nuclear pore transport balance> asks what an accumulation readout actually means.
Back to article page
