# 8.4 Modeling cell assemblies  (Page 7/9)

 Page 7 / 9

## Network size and patterns

In the simulations there will be 50 E-cells and 50 I-cells. The Network will be "trained" to encode assemblies by running the weight producing algorithm on a set of patterns, each which represents a cell assembly (remember weights are only produced between the E-cells initially). There will be 8 patterns that contain 8 E-Cells in each. Each "event" or pattern will carry equal significance when the weighting algorithm runs. To help visualize the patterns look at the diagram below.

## Parameters for cells

Below are tables with parameter that are given exactly from A Lansner and E Fransen.

 Parameter E-Cell I-Cell ${V}_{leak}$ (mV) -50 -70 ${G}_{core}$ 0.04 0.0638 ${G}_{m}$ Soma $\left(\mu S\right)$ 0.0032 0.0016 ${C}_{m}$ Soma $\left(nF\right)$ 0.032 0.016 ${G}_{m}$ $\left(\mu S\right)$ Dentrites 0.0096 0.0096 ${C}_{m}$ Dentrites $\left(nF\right)$ 0.288 0.288 ${V}_{Na}$ 40 50 ${G}_{Na}$ $\left(\mu S\right)$ 1.0 1.0 ${V}_{K}$ -70 -90 ${G}_{K}$ $\left(\mu S\right)$ 0.5 1.0 ${V}_{Ca}$ 150 150 ${G}_{Ca}$ $\left(\mu S\right)$ 0 0 ${G}_{K\left(Ca\right)}$ $\left(\mu S\right)$ 0.0017 0.01 ${\rho }_{AP}$ (mV ${}^{-1}$ ms ${}^{-1}$ ) 4 0.013 ${\delta }_{AP}$ ms ${}^{-1}$ .075 .02
 m h n q p A (mV ${}^{-1}$ ms ${}^{-1}$ ) 0.2 0.08 0.02 0.08 0.7 (ms ${}^{-1}$ ) $\alpha$ B (mV) -40 -40 -15 -25 C (mV) 1 1 0.8 1 17 A (mV ${}^{-1}$ ms ${}^{-1}$ ) 0.06 0.4 0.04 0.005 0.1 (ms ${}^{-1}$ ) $\beta$ B (mV) -49 -36 -40 -20 C (mV) 20 2 0.4 20 17
 m h n q A (mV ${}^{-1}$ ms ${}^{-1}$ ) 0.2 0.08 0.02 0.08 $\alpha$ B (mV) -30 -30 -21 -15 C (mV) 1 0.2 0.2 1 A (mV ${}^{-1}$ ms ${}^{-1}$ ) 0.06 0.4 0.02 0.005 $\beta$ B (mV) -38 -26 -18 -10 C (mV) 20 0.2 0.2 20
 ${V}_{syn}$ Excitatory 0 $mV$ ${V}_{syn}$ Inhibitory -85 $mV$ $s$ variable ${G}_{syn}$ E to E (AMPA) ${w}_{ij}$ x ${w}_{E2E}$ ${G}_{syn}$ E to I (AMPA) ${w}_{ij}$ x ${w}_{E2I}$ ${G}_{syn}$ I to E ${g}_{I2E}$ ${G}_{NMDA}$ E to E ${G}_{syn}$ x ${w}_{NMDA}scalar$ ${\rho }_{NMDA}$ E to E ${G}_{syn}$ x ${\rho }_{NMDA}scalar$ ${\delta }_{NMDA}$ $m{s}^{-1}$ .02

## Remaining parameters

There are a few parameters that remain to be given values. A Lanser and Fransen do not give exact values for these unknown parameters, but instead a desired range for EPSP's (Excitatory Post Synaptic Potentials) that the parameters help determine. By experimenting with a single neuron we can find the ideal range for these unknown parameters. Hence, we can then scale the resulting Weights from the algorithm so they are mapped to the ideal range. We need an overall weighting scale constant for connections from E to E cells (AMPA), ${w}_{E2E}$ . A weighting scale constant for E to I cells, ${w}_{E2I}$ Also, a scale general conductance constant for all I to E connections, ${g}_{I2E}$ . The NMDA connections will be proportional to the AMPA weight but scaled by a constant ${w}_{NMDA}scalar$ . The influx parameter ${\rho }_{NMDA}$ for the $\left[C{a}_{NMDA}\right]$ pool is also proportional to the AMPA weight. Lastly, the binary variable $s$ needs to be given a time duration to stay active for. This time duration may be chosen different for each type of connection class as well (meaning E to E, E to I, and I to E).

Most of these unfixed variables relate to the strength of connections between cells and the EPSPs or IPSPs they create. The reason they won't be fixed for all different size simulations is that the values should vary according to size of the networks and assemblies. For instance, if we model on large networks that have large assemblies we would desire that a cell would require synaptic input from more cells in order to fire, than when the network is small. By picking these numbers accordingly we will be able to roughly determine how many cells of an assembly need to be firing in order to fully activate the assembly.

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!