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Next: Post synaptic behavior Up: Modeling synapses Previous: The action potential


Assuming we have built a neuron such as the one in your homework above, then we can connect these together to form a neural network. This requires a model for the synapses that connect the cells. Later on, we will create ``connectionist'' type models that mimic synapses by directly producing PSPs, but here we are interested in a more rigorous and directly biophysical approach. The synapse is remarkably complex and involves many simultaneous processes such as the production and degradation of neurotransmitter, the sequestration in the vesicles, release, binding to the postsynaptic membrane, and opening of the channels.

The basic story is that the action potential travels down the axon and terminates at many presynaptic sites invading regions called synaptic terminals. These terminals contain calcium channels which when depolarized cause (a) release of calcium; (b) calcium activates a calcium binding protein which promotes release by binding to vesicles containing the transmitter; (c) these ``docked'' vesicles release their transmitter into the synaptic cleft. This type of detail is all well and good, but if we are going to model many cells, then we want to simplify things quite a bit. In the chapter by Destexhe, Mainen, and Sejnowski, a detailed description of this model is provided as well as simulations. Click here for a copy.

We adopt here a cruder approach that can be viwed as a mean field or average over many individual synapses. Our simple model is to treat the synapse as just another ionic channel that will be modeled by Ohms law. For every synapse between two neurons, we will add a current of the form:

Isyn = gsyns(t)(V-Vsyn)


to the right-hand side of the current balance equation for voltage. (Note that the NMDA synapse discussed below is more complex since the conductance is dependent on the post-synaptic potential.) Here s(t) is the time-dependent proportion of open channels, gsyn is the maximum conductance, and Vsyn is the reversal potential for the synapse. The quantity, s(t) will always be nonnegative and is zero if the presynaptic cell is not firing. For ``excitatory'' synapses, Vsyn larger than the resting potential so that it will produce an inward current. ``Inhibitory'' synapses will be close to the reversal potential of potassium and negative relative to rest. ``Shunting '' synapses have a reversal potential close to the resting potential.
I would like you to explore a simple 5 compartment model of a dendrite that is passive and has one synapse in compartment 3. In this case, s(t) is a pulse that turns on to 1 at ton and then back to 0 at toff. The XPP file is
# pas_syn.ode 
init v1=5.08  v2=4.33  v3=3.8  v4=3.46  v5=3.3  
par c=1  i=1  gl=0.05  vsyn=0  gsyn=0.1  gc=1  
par ton=10  toff=15 
and is called pas_syn5.ode. Plot v1,v3,v5 against time for vsyn=-20mV,0mV,50mV and a variety of different times on/off and maximal conductances. Integrate the equations for 100 msec and define an appropriate window. Note the difference at the ends and at the site of the synapse. The time units are in milliseconds. What are the units of i in this equation? Note also, that the coupling conductances and the leak and the synapses are all in mS/cm2. Observe that even though compartment 1 receives input via a current injection, that the synapse that is downwind from it exerts a significant effect on the membrane potential at that point.

Now that you have explored the effects of these transient conductance changes, I return to modeling the synapse. How do we determine s(t) from the presynaptic activity? As we have noted above, the release of transmitter is quite complex. However, a good fit to the total amount of transmitter release by a single action potential is given by the expression:  
(V_{pre}) = T_{max}/(1+\exp(-(V_{pre}-V_p)/K_p))\end{displaymath} (2)
where Tmax is the maximal concentration of transmitter in the synaptic cleft and Kp,Vp determine the stiffness and threshold for the release. A good value for these constants is $V_p=2 \ mV$ and $K_p=5 \ mV.$ Typically, we will assume that 1 mM of transmitter is the maximum concentration released.

next up previous
Next: Post synaptic behavior Up: Modeling synapses Previous: The action potential
G. Bard Ermentrout