TITLE sodium membrane channels for STh
COMMENT
Sodium from pyramidal, Traub 91. He based them on Sah (1988) data,
which were at 22degC, but he scaled them so that they "were fast
enough"? The Q10 measured from Sah (at least for the peak
conductance) was 1.5
How the q10 works: There is a q10 for the rates (alpha and beta's)
called Q10 and a Q10 for the maximum conductance called gmaxQ10. The
q10s should have been measured at specific temperatures temp1 and
temp2 (that are 10degC apart). Ideally, as Q10 is temperature
dependant, we should know these two temperatures. We are going to
follow the more formal Arrhenius derived Q10 approach. The
temperature at which this channel's kinetics were recorded is tempb
(base temperature). What we then need to calculate is the desired
rate scale for now working at temperature celsius (rate_k). This is
given by the empirical Arrhenius equation, using the Q10.
ENDCOMMENT
UNITS {
(mV) = (millivolt)
(mA) = (milliamp)
}
INDEPENDENT {t FROM 0 TO 1 WITH 1 (ms)}
NEURON {
SUFFIX Na
USEION na READ nai,ena WRITE ina
RANGE gna, m, h
GLOBAL rest,activate_Q10,Q10,gmaxQ10,rate_k,gmax_k,temp1,temp2,tempb
}
PARAMETER {
v (mV)
dt (ms)
gna = 1.0e7 (mho/cm2)
rest = 60.0 (mV) : for conversion from Traub WAS 60
ena
nai
celsius
activate_Q10 = 1
Q10 = 1.980105147e+00
gmaxQ10 = 1.980105147e+00
temp1 = 19.0 (degC)
temp2 = 29.0 (degC)
tempb = 23.0 (degC)
}
STATE {
m h
}
ASSIGNED {
ina (mA/cm2)
alpham (/ms)
betam (/ms)
alphah (/ms)
betah (/ms)
rate_k
gmax_k
}
BREAKPOINT {
SOLVE states METHOD cnexp
ina = (gna*gmax_k)*m*m*h*(vena)
}
UNITSOFF
INITIAL {
LOCAL ktemp,ktempb,ktemp1,ktemp2
if (activate_Q10>0) {
ktemp = celsius+273.0
ktempb = tempb+273.0
ktemp1 = temp1+273.0
ktemp2 = temp2+273.0
rate_k = exp( log(Q10)*((1/ktempb)(1/ktemp))/((1/ktemp1)(1/ktemp2)) )
gmax_k = exp( log(gmaxQ10)*((1/ktempb)(1/ktemp))/((1/ktemp1)(1/ktemp2)) )
}else{
: Note, its not 1.0, as we have rescaled the kinetics
: (reverting the scaleing Traub did), the original is
: acheived using this rate
rate_k = 1.60
gmax_k = 1.60
}
settables(v)
m = alpham/(alpham+betam)
h = alphah/(alphah+betah)
}
DERIVATIVE states {
settables(v) :Computes state variables at the current v and dt.
m' = alpham * (1m)  betam * m
h' = alphah * (1h)  betah * h
}
PROCEDURE settables(v) { :Computes rate and other constants at current v.
:Call once from HOC to initialize inf at resting v.
LOCAL vadj
TABLE alpham, betam, alphah, betah DEPEND rest,celsius FROM 100 TO 100 WITH 400
vadj = v  rest
:"m" sodium activation system
alpham = rate_k * 0.2 * vtrap((13.1vadj),4.0)
betam = rate_k * 0.175 * vtrap((vadj40.1),1.0)
:"h" sodium inactivation system
alphah = rate_k * 0.08 * exp((17.0vadj)/18.0)
betah = rate_k * 2.5 / (exp((40.0vadj)/5.0) + 1)
}
FUNCTION vtrap(x,y) { :Traps for 0 in denominator of rate eqns.
if (fabs(x/y) < 1e6) {
vtrap = y*(1  x/y/2)
}else{
vtrap = x/(exp(x/y)  1)
}
}
UNITSON
