Integration, Inhibition and Neuromodulation

A neuron can receive an inhibitory input that moves its membrane potential by zero millivolts and still stop it from firing. To see why, stop thinking of synaptic inputs as voltages that add and start thinking of them as conductances that compete. That picture explains how inputs sum, how inhibition works unseen, why one GABA input can help or hinder, and how neuromodulators change the rules.

The idea: inputs add as conductances, in time and in space

An excitatory postsynaptic potential (EPSP) is a postsynaptic potential that makes the cell more likely to fire an action potential. An inhibitory postsynaptic potential (IPSP) makes it less likely. Both are defined by their effect on firing, not by which way the trace moves. See synaptic transmission and the resting membrane.

Each fast synapse opens channels, which adds a conductance (g, in nanosiemens, nS) with its own reversal potential (E, the voltage at which current through those channels is zero). The current it passes is

I_syn = g_syn × (Vm − E_syn)

The term (Vm − E_syn) is the driving force: how far the membrane sits from that channel’s reversal potential. For AMPA-type glutamate receptors E is near 0 mV, so at rest the current depolarizes. The GABA_A receptor is a chloride channel (it also passes some bicarbonate), so E_GABA lies near the chloride equilibrium potential, which the Nernst equation gives from the chloride concentrations (derived in transport across membranes).

In a compact cell modelled as isopotential (the same voltage everywhere), the steady voltage under several constant conductances is a weighted average of their reversal potentials:

V_ss = (gL × EL + gE × EE + gI × EI) / (gL + gE + gI)

τ_eff = C / (gL + gE + gI)

Here gL and EL are the leak conductance and its reversal potential, C is the membrane capacitance, and τ is the membrane time constant, the time a passive membrane takes to get 63% of the way to a new steady voltage (input resistance times capacitance). Every open conductance pulls the voltage toward its own E.

Temporal summation is adding inputs that arrive at different times: a second EPSP that lands before the first has decayed rides on its tail, so a longer τ gives more overlap. Spatial summation is adding inputs at different sites. In a long passive dendrite at steady state, a voltage decays with distance as V(x) = V0 × e^(−x/λ), where the length constant λ is the distance over which it falls to 37%. Suppose, with illustrative numbers, that λ = 500 µm and a synapse sits 250 µm from the soma: e^(−0.5) = 0.61, so 61% of its steady depolarization reaches the soma. A brief EPSP loses more, because capacitance also filters it.

Inhibition you cannot see: shunting

All values here are illustrative. Treat the cell as isopotential and passive, with C = 200 pF, gL = 10 nS, EL = −65 mV and a fixed spike threshold of −50 mV. Real thresholds shift with channel state (the action potential).

Worked problem.

Step 1: the resting cell. Input resistance R = 1/gL = 1/(10 × 10^−9 S) = 100 MΩ. Time constant τ = C/gL = (200 × 10^−12 F)/(10 × 10^−9 S) = 20 ms.

Step 2: sustained excitation. Turn on gE = 5 nS with EE = 0 mV. V_ss = (10 × −65 + 5 × 0)/(10 + 5) = −650/15 = −43.3 mV, above threshold. The voltage approaches it with τ_eff = 200/15 = 13.3 ms, so it crosses −50 mV when −50 = −43.3 − 21.7 × e^(−t/13.3). Then e^(−t/13.3) = 6.67/21.67 = 0.308, and t = 13.3 × ln(3.25) = 15.7 ms.

Step 3: add a shunt. Add an inhibitory conductance gI = 10 nS whose reversal potential equals rest, EI = −65 mV. V_ss = (−650 + 0 − 650)/(10 + 5 + 10) = −1300/25 = −52.0 mV. That is below threshold: no spike. τ_eff is now 200/25 = 8 ms.

Step 4: the shunt alone. With gE = 0, V_ss = (−650 − 650)/20 = −65 mV. The trace does not move, yet in step 3 it vetoed the spike.

This is shunting inhibition: an inhibitory conductance with a reversal potential near rest lowers the input resistance, so a given excitatory current produces a smaller voltage change. Think of excitation as water poured into a leaky bucket and the shunt as extra holes: the resting level is unchanged, but the same pouring no longer fills it as high.

Step 5: temporal summation, with and without the shunt. Now give the cell two brief excitatory charge pulses of 0.4 pC, 10 ms apart. Each jumps the voltage by q/C = 0.4 pC / 200 pF = 2 mV, then decays. Without the shunt (τ = 20 ms), just after the second pulse the depolarization is 2 + 2 × e^(−10/20) = 2 + 1.21 = 3.21 mV. With the shunt on (total conductance 20 nS, τ = 10 ms), it is 2 + 2 × e^(−10/10) = 2 + 0.74 = 2.74 mV. The shunt did not shrink each jump, which charge and capacitance set. It shortened the time constant, so less of the first EPSP was left for the second to ride on.

Real cells keep the logic but not always the arithmetic. In layer V pyramidal neurons from juvenile rat somatosensory cortex, Ulrich (2003) used a dynamic clamp (a feedback system that injects the current a chosen conductance would pass) to add a shunt. It divided the subthreshold voltage responses but subtracted from the firing frequency. Do not assume either rule elsewhere.

Where and when inhibition lands

An IPSP can depolarize. If E_GABA lies above rest but below threshold, opening GABA_A channels moves the voltage up toward E_GABA while also holding it below threshold. Gulledge and Stuart (2003) studied this in layer 5 pyramidal neurons in somatosensory cortex slices from 3- to 4-week-old rats. They used gramicidin perforated-patch recording, which leaves the cell’s own chloride in place because gramicidin pores do not pass chloride. GABA responses depolarized the cells from rest and, combined with proximal excitation, could help trigger spikes, even though E_GABA was below threshold. Timing mattered: somatic GABA inhibited excitation that arrived at the same time but facilitated excitation that arrived later. A likely reading is that the conductance had closed while some depolarization remained. This does not make GABA excitatory in adult cortex in general; in that preparation, its sign depended on timing and site. The larger depolarizing GABA responses of immature neurons, which come from higher internal chloride, belong to development: see CNS synapses.

The chloride gradient is not fixed either. Each time GABA_A channels let chloride in, internal chloride rises a little and E_GABA creeps upward. In branched-dendrite computer models, Currin and Raimondo (2022) found that this activity-dependent chloride loading weakened inhibition, and that with chloride dynamics included the most effective placement of inhibitory synapses lay close to, but not at, a shared branch junction. These are model predictions, not measurements.

Dendrites that do not add linearly

The conductance equation already predicts one departure from simple addition. In the made-up cell above, one 5 nS excitatory input gives −43.3 mV, a 21.7 mV depolarization. Two such inputs together (10 nS) give −650/20 = −32.5 mV, a 32.5 mV depolarization rather than 43.3 mV. As Vm approaches EE the driving force shrinks, so each added conductance pushes less. Inputs sharing one small patch of membrane show this sublinear summation most.

Thin dendrites can also do the opposite. Polsky, Mel and Schiller (2004) used confocal imaging to find thin dendrites of rat neocortical pyramidal neurons in slices and stimulated inputs at two sites on them. Nearby inputs on the same thin branch summed sigmoidally: as the combined input grew, the response rose slowly, then steeply, then levelled off. Inputs that were widely separated, or on different branches, summed linearly. In these cells, location set the combination rule. The steep phase is generally attributed to local regenerative events in the branch, in which NMDA receptors are thought to take part; their voltage-dependent block is in ion channels as receptors.

Modulators change the rules

Fast transmitters open ion channels for milliseconds. Neuromodulators act mostly through G-protein-coupled receptors and change how a neuron responds to input, often by altering its ion channels.

Dopamine on D1 receptors. In D1 receptor-expressing striatal projection neurons, the D1 receptor couples through Gs/olf to adenylyl cyclase, which raises cAMP and activates protein kinase A, PKA (GPCRs and G proteins, cAMP and PKA). Lahiri and Bevan (2020) stimulated dopamine axons optogenetically in slices of dorsolateral striatum from adult mice of both sexes and recorded identified D1 neurons by perforated patch, with GABA, glutamate and acetylcholine receptors blocked. Firing to the same injected current rose within hundreds of milliseconds (a median onset of 458 ms by their criterion) and stayed raised for at least 10 minutes after one bout of stimulation. A D1 antagonist blocked the effect, and their results suggest that the persistence required PKA. Conventional whole-cell recording showed little or no modulation, possibly because it dialyzes the cell interior.

Acetylcholine on muscarinic receptors. KCNQ2/KCNQ3 channels carry the M-current, a slowly activating potassium current that opposes repetitive firing, named for its suppression by muscarinic agonists. Gq-coupled muscarinic receptors activate phospholipase C, which hydrolyzes the membrane lipid PIP2 into IP3 and DAG (PLC, IP3, DAG and PKC). Which event closes the channel? Suh, Inoue, Meyer and Hille (2006) depleted PIP2 directly, without PLC, in tsA-201 cells expressing KCNQ2/KCNQ3, and the current fell: loss of PIP2 was enough. Closing KCNQ channels increases excitability, so the neuron fires more to the same input.

Cascade timing is in the shape of a signal; lasting changes lead to synaptic plasticity, and dopamine returns in the basal ganglia and motivation and reward.

Design problem

In a made-up perforated-patch recording from a D1 striatal projection neuron in a mouse brain slice, with fast synaptic receptors blocked, a 200 pA, 1 s current step gives 8 spikes. After optogenetic stimulation of dopamine axons, the same step gives 14 spikes, and still 12 at 10 minutes. With a D1 antagonist in the bath, the counts are 8 and then 8. With a PKA inhibitor, they are 13 at first and back to 8 by 10 minutes. What do you conclude, and what control is missing?