### How flexible is NEURON ?

Posted:

**Fri Aug 02, 2013 6:53 am**Hi,

I have been trying to build new channels for an STG model, with functions that obey this equation :

I = g m^p h^q (v-e)

The activation and inactivation variables (m, h) are described by these equations :

mtau(v) m' = minf(v) - m

htau(v) h' = hinf(v) - h'

Now the voltage dependencies for m and h are given , i.e xtau and xinf,

however the alpha and beta values are not given, they're incorporated in the xtau and xinf. I'm asking this because in every example for constructing new channels in the book values are used for alpha and beta for the function block to slove derivative states.

My question is this: Can NEURON "figure out" the alpha and beta values? If, for example, I used this solution for the Na channel :

I = g m^3 h (v-e)

m = minf(v) - mtau(v) m'

h = hinf(v) - htau(v) h'

minf(v) = malpha(v)/(malpha(v)+mbeta(v))

mtau(v) = 1/(malpha(v)+mbeta(v))

hinf(v) = halpha(v)/(halpha(v)+hbeta(v))

htau(v) = 1/(halpha(v)+hbeta(v))

m' = malpha(v) (1-m) - mbeta(v) m

h' = halpha(v) (1-h) - hbeta(v)h

where

minf = 1/(1+exp(-(v+24.7)/5.29))

mtau = 1.32-(1.26/(1+exp(-(v+120)/25)))

hinf = 1/(1+exp((v+48.9)/5.18))

htau = 0.67/(1+exp(-(v+62.9)/10))

I know this looks messy, I'm a beginner and I couldn't find a solution for this.

My question again, can NEURON derive the values of alpha and beta through these relations, to solve the derivative state, in order to solve the breakpoint equation?

If yes, how can I achieve that in the NMODL language?

(note : i'm using data from this paper

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1941697/

)

Thanks,

Ahmad.

I have been trying to build new channels for an STG model, with functions that obey this equation :

I = g m^p h^q (v-e)

The activation and inactivation variables (m, h) are described by these equations :

mtau(v) m' = minf(v) - m

htau(v) h' = hinf(v) - h'

Now the voltage dependencies for m and h are given , i.e xtau and xinf,

however the alpha and beta values are not given, they're incorporated in the xtau and xinf. I'm asking this because in every example for constructing new channels in the book values are used for alpha and beta for the function block to slove derivative states.

My question is this: Can NEURON "figure out" the alpha and beta values? If, for example, I used this solution for the Na channel :

I = g m^3 h (v-e)

m = minf(v) - mtau(v) m'

h = hinf(v) - htau(v) h'

minf(v) = malpha(v)/(malpha(v)+mbeta(v))

mtau(v) = 1/(malpha(v)+mbeta(v))

hinf(v) = halpha(v)/(halpha(v)+hbeta(v))

htau(v) = 1/(halpha(v)+hbeta(v))

m' = malpha(v) (1-m) - mbeta(v) m

h' = halpha(v) (1-h) - hbeta(v)h

where

minf = 1/(1+exp(-(v+24.7)/5.29))

mtau = 1.32-(1.26/(1+exp(-(v+120)/25)))

hinf = 1/(1+exp((v+48.9)/5.18))

htau = 0.67/(1+exp(-(v+62.9)/10))

I know this looks messy, I'm a beginner and I couldn't find a solution for this.

My question again, can NEURON derive the values of alpha and beta through these relations, to solve the derivative state, in order to solve the breakpoint equation?

If yes, how can I achieve that in the NMODL language?

(note : i'm using data from this paper

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1941697/

)

Thanks,

Ahmad.