Topics in Neurophysics

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1 Topics in Neurophysics Alex Loebel, Martin Stemmler and Anderas Herz Exercise 2 Solution (1) The Hodgkin Huxley Model The goal of this exercise is to simulate the action potential according to the model proposed by Alan Lloyd Hodgkin and Andrew Huxley in This model describes the ionic mechanisms underlying the initiation and propagation of action potentials in the squid giant axon. It is worth noting that, to date, this is the only mathematical theory for which a Nobel Prize in Physiology or Medicine has been awarded. Hodgkin and Huxley received this prize in As introduced in the lecture, the main dynamical equation of the model is: c dv Ie = im (1) dt A m + with: i m = g L (V E ) + g L K n 4 (V E K ) + g Na m 3 h (V E Na ) (2) V stands for the voltage, with units of mv; C m stands for the capacitance per unit area, with units of µf/cm 2 ; i m is the current flowing through a unit membrane area, with units of pa/cm 2 ; I e is the external current input into the cell, also with units of pa; A is the cell membrane area, in cm 2 ; E L, E K, and E Na are the reversal potentials of the leakage, potassium and sodium currents, in mv; g L, g K, and gna are the maximum conductances per unit area for the leakage, potassium and sodium, in ms/cm 2 ; and finally, n, m and h are the gating variables, which determines the effective current through the different channels. The equations that describe the dynamics of the different gating variables, n, m and h will be (to save space, and as the equations has the same form for different variables, x stand for either n, m or h ): dx τx = x (V) x (3) dt 1 τ x(v) = (4) α (V) + β (V) x x αx(v) x (V) = (5) α (V) + β (V) x x

2 The exact form of α x (V) and β x (V) are different for the different gating variables, and (for the squid giant axon) they read as follow: 0.01(V + 55) α n( V) =, 0.1(V + 55) 1 e 0.1(V + 40) α m( V) =, 0.1(V + 40) 1 e β (V) = 0.125e n m (V + 65) (V + 65) β (V) = 4e (6) α 0.05(V + 65) m(v) = 0.07e, β h( V) = 0.1(V + 35 ) 1+ e 1 Finally, the parameter values that we will use are: E L = -54.4mV, E K = -77mV, E Na = 50, c m = 1µF/cm 2, g L = 0.3mS/cm 2, g K = 36mS/cm 2, g Na = 120mS/cm 2. Use a membrane area of A = 1cm 2 to scale the external current I e. (a) Plot the different α x (V), β x (V), τ x (V) and x (V) ( x stands for the n, m or h gating variables), for values of V between -100mV and +20mV. Note that the functions describing α x (V) and β x (V) for m and h are not the same. Compare your results to Figs. (1, 2). Fig. 1 only shows α n (V), β n (V), τ n (V) and n (V), so you need to complete the overall picture. Explain the different roles of the gating variables in allowing currents to flow in or out of the cell during the different stages of the action potential. Solution: The first cell of the attached script calculates the different variables as a function of the voltage, and plots the required graphs: (i) From the graph of n (V) it is observed that the opening probability of n, the gating variable of the potassium channel, increases with the voltage. In addition, the higher the voltage, the faster are the changes in n, as observed in the graph of τ n (V). (ii) The graphs of m (V) has similar properties than that of n (V), albeit with a shift along the voltage, that is, it takes higher voltage values for the m gating variable to be activated. Notably, this dependency of m on the voltage is an important factor in determining the voltage values above which an action potential can occur. Furthermore, the dynamics of m is the fastest among the gating variables, as observed from the graph of τ m (V), and changes in m occurs an order of magnitude faster than for the other gating variables. (iii) Finally, the properties of h (V) are a mirror image of m (V): the h gating variable is inactivated at higher voltage values. The h gating variable is also the most sluggish, with τ h (V) having the largest values at sub-threshold membrane potential values. (iv) Hence, although there is very little overlap of voltage values in which both gating variables of the sodium channel are open, the difference in the dynamics of m and h means that for a short time window, as the voltage values increases from resting potential, the value of m is high enough and the value of h has not yet decreased enough such that sodium current flows into the cell, depolarize the membrane potential, which lead to higher m values and stronger sodium current. This positive feedback in the sodium current leads in turn to the upward phase of the action potential.

3 (v) At the same time, two things occur that gives end to the rise in membrane potential: as the voltage increases, the value of the h gating variable is decreasing more rapidly, which in turn shuts down the sodium current; and the value of the n gating variable is increasing more rapidly, leading to an increasing potassium current to flow into the cell, which in turn lead to a decrease in the membrane potential. Once the voltage returns to values near the resting potential, the n gating variable slowly inactivate, and the h gating variable slowly de-inactivate, completing the cycle of an action-potential. Figure 1: The forward and backward transition rates, the steady state and the time constant of the dynamics of the potassium gating variable, as a function of the voltage. Figure 2: The steady state and the time constant of the dynamics of the sodium and potassium gating variables, as a function of the voltage. (b) Simulate the complete model for 20ms with the following initial conditions: V(t=0) = - 70mV, n(0) = 0, m(0) = 0, h(0) = 1, and with no external input, i.e., I e = 0. You should get something similar to Fig. (3). What do you make from the behavior of the different gating variables? Do they behave as you expected from their dependency on V (Figs. 1-2)? Solution: The second cell of the attached script simulates the HH model for the given initial conditions and with no external input. One action potential is observed, and the

4 dependencies of the gating variables on the voltage, as shown in Figure (3), are reproduced. Figure 3: The action potential. The membrane voltage, the current across the membrane, and the values of the different gating variables during an action potential are shown from top to bottom. (c) Consider now external inputs that range between I e = 0: 1: 100. For each input run a simulation that lasts 600ms, where the external input is introduced at t = 50ms. Use the same initial conditions as in the previous question. As the input increases in amplitude, you should see more and more spikes emitted by the model neuron (e.g., check figure 4). Plot the firing-rate vs. input amplitude graph. You should get something like in Fig. 4. (Note: neurons that have a similar rate-input relation, i.e., that as the amplitude increases they suddenly start emitting spikes at a finite frequency, are called Type - II neurons). As in the previous question, your model should emit one spike before the external input is turned on. Why does this happen? Can you get rid of it? Can you have more than one spike before the input is turned on? Solution: The third cell of the attached script simulates the HH model for different external inputs, and it reproduces Figure (4). As for the appearance of the single spike at the beginning of each simulation, it is due to the choice of the initial conditions of the variables. In particular, their values are not the

5 stationary values for V = -70mV (the chosen initial state of the membrane potential). That is, the system is not in a steady state, and the ensued spike is a manifestation of the dynamics of the system away from the chosen initial condition. With h(t = 0) = 1; with m rapidly increasing from its initial zero value due to its short time constant τ m ; and with n(t = 0) = 0, there is enough sodium current to depolarize the cell to lead to the positive feedback dynamics of the sodium channels and to the action potential. This dynamics of the sodium channels is supported by the fact that there is initially no potassium current to negate the depolarization effects of the sodium current, as the potassium channels are all closed. Finally, the reason only one spike is observed, as due to the fact that once the action potential is over, the values of all variables relax into their stationary solution of the resting potential, which is a stable state. That is, the system will stay at the resting potential unless perturbed. Figure 4: Left: examples of the model output for different amplitude of the external input, which is initiated at t=50ms. The input was (from top left to bottom right) I e = 0, 4, 6 and 8. Right: firing rate of the HH model neuron for different amplitudes of the external input. Notice how the firing rate starts from a finite value, a feature that characterizes Type - II neurons. (d) Now, plot the individual currents i L = g L (V E L ), i K = gk n 4 (V E K ), and i Na = gna m 3 h (V E Na ) from one spike to the next. Show these currents together in a single plot. Compute the ionic load (the total Na + charge) accumulating during one interspike interval. You may ignore the contribution from i L as being small. This is the load the ATP pumps need to remove from the cell's interior after a spike. Solution: Interestingly (e) What happens to the height of the action potentials, along the spike train response, as the external input increases in amplitude? Why do you think this happens? Solution: Interestingly, as the firing rate increases with the external input, the height of the action potentials decreases. The reason for that is the following: as τ m is short for all the relevant V values, the m gating variable has a similar dynamics till very high firing

6 rates as it can follow the fast changes in the membrane potential. The dynamics of the h and n gating variables are much slower, however, and as the firing rate increases, the h gating variable remains at low values (that is, there is less recovery from inactivation); and the n gating variable remains at relatively high values. Hence, less sodium current flows into the cell and more potassium current flow out of the cell during spikes at higher firing rates, which in turn leads to the observed phenomenon of smaller spikes. (2) Some numerical experiments with the HH model. You now have a neuron under your microscope, and it is time to figure out how its different components are responsible for the specific spiking activity that you observe. (a) For the values of the initial conditions for the V, n, m and h variables given in problem 1, you saw a single spike even before the external input was introduced. You also figured out that if you change the initial conditions to the steady-state values of n, m and h for V(t = 0) = V rest, you will not observe this spike. Here, start with V(t = 0) = V rest and set the initial conditions for n, m and h to their steady-state values, and deliver a step current to the model. For what kind of step input you could get exactly one, and only one, spike? Explain the underlying mechanisms for that. (hint: sometime in order to go up, you first need to go down ). Solution: There are two step inputs with which one can get just one spike at the HH model.the first, is a short excitatory input at just the right amplitude. The second, the more interesting input, is an inhibitory input that is given for a long enough time and then is abruptly stopped. Such an inhibitory input can lead to one, and only one spike, which has the name of a rebound spike. In particular, a rebound spike occurs due to the different time constants of the m and h gating variables of the sodium channel, and their different stationary values for a given holding voltage. That is, the inhibitory input reduces the membrane potential, which pushes the value of h toward 1, and the value of m toward 0. When the input is stopped abbruptly, the voltage quickly elevates back to the resting membrane potential, and the value of m quickly increases to m (V rest ). As the time constant τ h is much slower than τ m, h remains at relatively higher values than h (V rest ). To summarize, we are now at the resting potential, but with h > h (V rest ), which allows for more sodium to flow into the cell, which lead to the depolarization of the membrane potential, and then to higher values of m, and so on, till we get a spike. Once the spike is over, all the different gating variables relax to their V rest values, and so no follow-up spike will occur without an external input (excitatory, or inhibitory). (b) What happens to the response of the neuron (for a certain amplitude of the external stimulus) when you decrease the value of E L, e.g., from E L = -54.4mV to E L = -65mV? Or to E L = -75mV? Explain what you see in the simulation, and then try to speculate what could lead to such changes in E L? Solution: To answer this problem, we need to remind ourselves what is E L. In short, E L in the HH model is the reversal potential of all the currents that are not involved in the production of the spikes. Hence, when there are no external currents and no spikes, the membrane potential will keep close to this value. A more negative value for E L means

7 that a larger depolarizing current is needed in order to induce a spike, and for a certain amplitude of an external stimulus, the firing rate will decrease when E L is lower. What could lead to such changes in E L? Changes in the ion concentration in and out of the cell; or changes in their permeability (remember what determines the reversal potential at the Goldman-Hodgkin-Katz voltage equation). (c) What happens when the dynamics of the n subunit of the potassium channel becomes slower? How can you model such a change? Support your answer with a figure. Solution: when the dynamics of the n subunit becomes slower, it means that the hyperpolarizing potassium current initiates later. The spikes therefore become wider, and the ISI (Inter-Spike-Interval) becomes longer. As a reminder, τ n (V) determines the dynamics of n and it has the following form: 1 τ n(v) = (7) α (V) + β (V) n n We can simply multiply the equation by some value a > 1 to have τ n (V) larger for each value of V. Although this will work, it is not based on any bio-physical principle. Changing either α n (V) or β n (V) would work as well. I will it for you to figure out how to do it. (d) What happens when the subunit that is responsible for the inactivation of the sodium channel doesn t work properly, and always exactly10% of the h subunits are active? How can you model that? What happens when always exactly 20% of these subunits are active? Explain your findings. Solution: first, the easiest way to model this is by simply assign the value of h(t) = 0.1 or 0.2 for all time steps of the simulation. Functionally, it means that there is always a flow of sodium into the cell which will depolarize it. In fact, for the values of h suggested here, there will be enough sodium current that will prevent the production of any spikes for any external inputs! (3) Some numerical experiments with the HH model. Set m(t) =m (V(t)) (that is, slaving the fast Na + activation variable to the voltage); and set h(t) =0.8 - n(t) (that is, combining two variables that evolve on a similar time scale), in order to get a two-variable Hodgkin-Huxley model. Note that the gating variables range between 0 and 1, so to be careful, set h(t) = 0 if (0.8 - n(t)) < 0. (a) Write down and simulate the reduced Hodgkin-Huxley model. Can you get it to spike? (b) Now examine the phase plane spanned by V(t) and n(t). Plot the solutions to dv/dt = 0 and dn/dt = 0 in this plane. These curves are known as the null-clines of the system. Note that the voltage null-cline is a function of the external current I e.

8 (c) Plot the time-dependent solution V(t), n(t) for spiking dynamics into the phase plane. Compare the orbit (the closed loop of the dynamics in the phase plane) to the null-clines for different values of I e.