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BME 6938 Neurodynamics Instructor: Dr Sachin S. Talathi.

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Presentation on theme: "BME 6938 Neurodynamics Instructor: Dr Sachin S. Talathi."— Presentation transcript:

1 BME 6938 Neurodynamics Instructor: Dr Sachin S. Talathi

2 Recap Ionic currents and conductance Equivalent circuit model for cell membrane Resting potential and input resistance (of passive cell) Integrate and Fire neuron model Frequency current relationship for IF model Simple extension to IF model: QIF and Exponential IF

3 Introduction to XPPAUTO A tool for simulating, animating and analyzing dynamical systems. Why XPPAUTO? What are the alternatives. Installation of XPP Creating the ODE file Running the program Solving the eq, graphing and plotting Changing the parameters and initial data The data viewer Command Summary Nonlinear ODEs (will come back to it later)

4 Example ode: Passive cell model (One dimensional linear ODE) A general two dimensional linear ODE

5 Writing the ODE file Ode file for passive cell model Open any editor and enter the following text. Save as Ex1.ode **** # Ex 1: passive membrane with constant input current delivered as step input #parameter par Rinp=10000, taum=1000, I=2, Vrest=0 par t_on=5, t_off=10 #The ode dV/dt = (Rinp*I*f(t)-1000*(V-Vrest))/taum #some initial condition V(0)=0 # Explicit function definition f(t)=heav(t_off-t)*heav(t-t_on) # track the current (auxillary variable) aux ibar=f(t)*I Done **** Note: The sentence following the # command represents comments.

6 Example 2 ode file Class Assignment: # Linear two dimensional model #parameter par a=0,b=1,c=-1,d=0 #The ode dx1/dt = a*x1+b*x2 dx2/dt=c*x1*d*x2 #some initial condition x1(0)=1,x2(0)=0 #Pre assigning the system parameters @ total=5, method=runge-kutta,bound=500,dt=0.25,dtmin=1e-12,xhi=1,xlo=-1,yhi=1,ylo=-1 Done

7 Linear cable theory for passive cells Construct models to incorporate the anatomical structure of neurons It is important to address questions such as: – What is the relative effectiveness of the synapse close to the soma as compared to the same synapse located at distal dendrites – How does dendritic branching affect integration of various inputs We will spend some time developing models to address these issues. The key assumption we will make is that the various portions of the dendritic tree are passive nerve cylinders and therefore the equations satisfied by electrical potentials are passive differential equations of linear cable theory The models here will serve as a bridge between the simple IF neuron model that we have studied thus far and the Hodgkin Huxley models that we will study later on in the course

8 Cable Theory Assumptions: Membrane parameters are linear and independent of mem. potential (passive) ; current entering the cable flows linearly (homogeneous); resistance of extracellular medium is zero (cell immersed in homogeneous isopotential medium, the reference)

9 Circuit diagram of the nerve cylinder Cable Equation: (assuming external medium is perfectly conducting)

10 Boundary & initial conditions Cable Equation With notation Eq 1 has unique solution if suitable initial data at t=0 and suitable boundary conditions at x=0 and x=l are provided Initial condition: Membrane depolarization at beginning of the expt: More often, the membrane is assumed at rest at equilibrium potential Boundary Condition: 1.Voltage Clamp: and possibly 2.Sealed end: Longitudinal current at the terminal is zero Seal at x=0 Seal at x=l This condition is also referred to as open circuit

11 Boundary conditions 3. Open end (Killed end): No voltage gradient at the ends Open at x=0 Open at x=l 4. Current injection at one end: Injected current at x=0 Injected current at x=l 5. Natural termination: The ends are sealed with membrane itself Figure shows the equivalent circuit diagram for cable that is sealed at x=0 The boundary condition is: Question: What is the boundary condition If membrane sealed at the terminal x=l?

12 Boundary Condition 6.Lumped soma: This condition assumes the soma to be a equipotential cylinder with uniform capacitance C s and resistance R s attached to nerve cylinder. Soma is typically attached to the start of nerve cylinder at x=0; with boundary condition 7.Infinite Cylinder: For semi infinit cylinder boundary condition at x=0 can be any one of the above. In addition we have

13 Dimensionless cable equation: rescaling variables Typical ConstantsSquid giant axonCat spinal motoneuron 10002500 10 -6 2x10 -6 3070 50010 (main dendrite) 1.5x10 4 8.9x10 7 15 65001000

14 Steady State We are interested in the solution to this equation when boundary conditions are given. We will consider the specific case when current applied only to boundary of the nerve cylinder General Solution:

15 Steady State Response-Current applied to terminal end Steady state current I 0 applied to terminal x=0 (a) Semi Infinite Cable: Input Resistance: Steady State Voltage at X=0 divided by the Steady State Current

16 Steady State Response-Current applied to terminal end Steady state current I 0 applied to terminal x=0 (a) Finite cable with sealed end: Input Resistance: Hint consider the general solution of the form Input resistance of a finite cylinder at X=L is greater than that of a semi infinite cylinder

17 Steady State Response-Current applied to terminal end Steady state current I 0 applied to terminal x=0 (a) Finite cable with open end: Input Resistance: Same hint: Input resistance of a finite cylinder at X=L is smaller than that of a semi infinite cylinder

18 Summary Semi-Infinite Cable Semi-Infinite Cable Finite Cable Sealed End Finite Cable Sealed End Finite Cable Open End Finite Cable Open End Finite Cable Clamped End Finite Cable Clamped End Cable TypeSolution Boundary Condition Homework problem

19 Summary

20 Transient Solution Initial Condition: Boundary Condition: Solution: Where G(x,y;t) is the Greens function corresponding to the depolarization resulting from a delta step input current delivered instantaneously at point y along the length of nerve cylinder; i.e. For the specific case of Infinite cylinder, it is easy to show that General solution for finite cable can be obtained through appropriate linear combination of G(x,y,t) so as to satisfy given boundary conditions Hint: Fourier Transform& use the formula

21 Specific Example: Transient Solution Solution to the cable equation when impulse current is applied at a given point y along a finite cable with sealed ends with initial condition V(X,0)=0 Boundary conditions: Response at x=0 to an impulse applied at t=0 at a given point y is

22 Ralls Model-Equivalent Cylinder Remember: Input resistance of a semi-infinite nerve cylinder with diameter d: Where Basic Idea: Impedence matching Assumptions: 1.The membrane properties are identical for soma and dendritic branches. 2. Membrane properties are uniform and voltage independent 3. All dendritic branches terminate at the same electrotonic length The 3/2 rule

23 Synaptic Integration-With XPPAUTO Use XPP AUTO to answer following Questions (Cable.ode) 1. Sketch the potential at the soma for the synaptic input at compartments 0, 5, 10, and 20. 1a.How do the peak amplitudes depend on distance? 1b. How about the time to peak? 1c.Does the peak appear to decay slower or faster for more distant inputs? 1d. How does the potential scale across various compartments For synaptic input at different locations on the cable Note: We have already looked at a simple version of this problem analytically

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