 Shunting inhibition

Shunting inhibition is suppression of another input's effect through increased synaptic conductance: opening additional ion channels lowers effective electrical resistance, reducing the voltage response to a given added electric current. When the added reversal potential equals the resting membrane potential, there can be no voltage change at rest and still be inhibition of a simultaneous input. See 1.4 Limitations of the Leaky Integrate-and-Fire Model.

Use the passive synaptic potential model below with leak $g_L=10\mathrm{nS}$, $E_L=-65\mathrm{mV}$ and capacitance $C=200\mathrm{pF}$. A sustained inward test electric current of $I_{\mathrm{ext}}=100\mathrm{pA}$ alone gives a steady voltage increase of $I_{\mathrm{ext}}/g_L=10\mathrm{mV}$, so $V_{\infty }=-55\mathrm{mV}$. Effective electrical resistance is $100\mathrm{M\Omega }$ and the time constant is $20\mathrm{ms}$.

Now hold a shunting conductance $g_i=20\mathrm{nS}$ open with $E_i=E_L=-65\mathrm{mV}$. With no test electric current, rest stays at $-65\mathrm{mV}$. With the same $100\mathrm{pA}$ test electric current, total conductance becomes $30\mathrm{nS}$, so the steady voltage increase is only $100/30\simeq 3.333\mathrm{mV}$: $V_{\infty }\simeq -61.667\mathrm{mV}$. Effective electrical resistance is $33.333\mathrm{M\Omega }$ and the time constant is $200/30\simeq 6.667\mathrm{ms}$. A faster response can therefore be smaller. At the original rest the shunting electric current is zero; at $V=-55\mathrm{mV}$ it is $20(-65+55)=-200\mathrm{pA}$, opposing the voltage excursion.

The factor of three reduction applies to this constant-electric current, constant-conductance example. A finite pulse, conductance-driven input or spatially extended dendrite need not have the same reduction. The interesting feature is conditional action: the shunt can leave a quiet neuron's voltage unchanged yet reshape its response when another input arrives. A model that represents all inhibition as a fixed outward electric current misses this behavior.