Current Dipole in Laminar Neocortex (Lee et al. 2013)

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Accession:151685
Laminar neocortical model in NEURON/Python, adapted from Jones et al 2009. https://bitbucket.org/jonescompneurolab/corticaldipole
Reference:
1 . Lee S, Jones SR (2013) Distinguishing mechanisms of gamma frequency oscillations in human current source signals using a computational model of a laminar neocortical network. Front Hum Neurosci 7:869 [PubMed]
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Model Information (Click on a link to find other models with that property)
Model Type: Realistic Network;
Brain Region(s)/Organism: Neocortex;
Cell Type(s):
Channel(s): I Na,t; I K; I M; I Calcium; I h; I T low threshold; I K,Ca;
Gap Junctions:
Receptor(s): GabaA; GabaB; AMPA; NMDA;
Gene(s):
Transmitter(s):
Simulation Environment: NEURON (web link to model); Python (web link to model); NEURON; Python;
Model Concept(s): Magnetoencephalography; Temporal Pattern Generation; Activity Patterns; Gamma oscillations; Oscillations; Current Dipole; Touch;
Implementer(s): Lee, Shane [shane_lee at brown.edu];
Search NeuronDB for information about:  GabaA; GabaB; AMPA; NMDA; I Na,t; I T low threshold; I K; I M; I h; I K,Ca; I Calcium;
TITLE Calcium low threshold T type current for RD Traub, J Neurophysiol 89:909-921, 2003

COMMENT
    Implemented by Maciej Lazarewicz 2003 (mlazarew@seas.upenn.edu)
ENDCOMMENT

INDEPENDENT { t FROM 0 TO 1 WITH 1 (ms) }

UNITS {
    (mV) = (millivolt)
    (mA) = (milliamp)
}

NEURON {
    SUFFIX cat
    NONSPECIFIC_CURRENT i   : not causing [Ca2+] influx
    RANGE gbar, i
}

PARAMETER {
    gbar = 0.0  (mho/cm2)
    v eca       (mV)
}

ASSIGNED {
    i           (mA/cm2)
    minf hinf   (1)
    mtau htau   (ms)
}

STATE {
    m h
}

BREAKPOINT {
    SOLVE states METHOD cnexp
    i = gbar * m * m * h * ( v - 125 )
}

INITIAL {
    settables(v)
    m  = minf
    h  = hinf
    m  = 0
}

DERIVATIVE states {
    settables(v)
    m' = ( minf - m ) / mtau
    h' = ( hinf - h ) / htau
}

UNITSOFF
PROCEDURE settables(v) {
    TABLE minf, mtau, hinf, htau FROM -120 TO 40 WITH 641

    minf  = 1 / (1 + exp(( -v - 56 ) / 6.2))
    mtau  = 0.204 + 0.333 / (exp(( v + 15.8) / 18.2) + exp((-v - 131) / 16.7))
    hinf  = 1 / (1 + exp((v + 80) / 4))

    if (v < -81) {
        htau  = 0.333 * exp((v + 466 ) / 66.6)
    } else {
        htau  = 9.32 + 0.333 * exp((-v - 21) / 10.5)
    }
}
UNITSON