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Biophysics of Neural Computation : Introduction to Neuroinformatics Prof. Rodney Douglas, Kevan Martin, Hans Scherberger, Matthew Cook Ass. Frederic Zubler [email protected] http://www.ini.uzh.ch/∼fred/bnc.html WS 2008-2009 Exercise 15 : Cable Equation The membrane potential V (x, t) is determined by solving the following partial differential equation (cable equation) : τ where: ∂v ∂t =λ 2 ∂2v ∂x2 − v + rm ie τ = (cm rm ) sets the scale for the temporal variation in the membrane potential λ = [(arm )/(2rL )]1/2 sets the scale for the spatial variation a = radius of the axon (= 2 µm) v = V − Vrest rm = specific membrane resistance (= 1 MOhm mm2) rL = longitudinal resistance (= 1kOhm mm) ie = the current injected into a cell We now assume an infinite cable and inject a constant current ie locally at x = 0. If we wait for the system to reach its steady state ( so that ∂v ∂t = 0 ), and with some boundary conditions, the cable equation has the solution: v(x) = ie Rλ 2 e−|x|/λ , where Rλ = rL λ πa2 1. The ratio of the membrane potential at the injection site to the magnitude of the injected current is called the input resistance of the cable. Can you express this value with the help of the last formula ? 2. We consider an infinitely long axon, which can be assumed passive, apart from two nodes where there are sodium channels. These nodes are at x = 0 and x = 2λ. Let the resting potential be -70mV and the threshold for action potential generation at the nodes be -50mV. Artificially we generate a prolonged "action potential" at x = 0 (sufficiently long to assume constant current injection) of peak voltage 30mV. Will this procedure trigger an action-potential at the second node? 3. If we double the radius of the axon, does this change the answer to question 2 ? 4. In a second axon extra membrane tightly surrounds the axon between the nodes. As a consequence the membrane resistance increases by a factor of 4, while the capacitance decreases by a factor of 4. In all other respects the axon is similar to the first one, with radius of 2 microns. Again, does this change the answer to question 2 ? Exercise 16 : Power-law for axon diameter at branch point Studying the resistivity of dendrites, Rall came up with the now famous 3/2 power law (Rall 1959), that says that in dendrites, the diameter of a parent segment, d0 , is related to the diameters of its daughter segments, d1 and d2 , as (d0 )e = (d1 )e + (d2 )e , where the branch power parameter e is equal to 3/2 (cf Fig 1). In axons, experimental data could be consistent with a similar law. But what would be the physiological justification for the existance of such a power law? With Chklovskii and Stepanyants (BMC, 2003), we reckon that : 1. Thicker axons conduct action potential faster → it’s better to have a big diameter d. 2. Thick axons are costly, because they require large amount of cytoplasm, and occupy space → it’s better to have a small diameter d. These two requirements are conflicting. To tackle this problem, we try to minimize C, the combined cost of conductance delay T and volume V : C = aT + bV where: a and b are unknown constant coefficients V = πLd2 /4 is the volume of an axon branch T = L/(kd) is the time delay (L = length, k = constant factor for non myelinated axons) 1. Find the axon diameter that minimizes the cost C (Hint: set ∂C/∂d = 0 and solve for d). 2. The parameters a and b are still unknown, but we can use the same method to make a prediction about the relationship among branch parameters. If the cost function for a bifurcation consisting of three segments is : C = a1 (T0 + T1 ) + a2 (T0 + T2 ) + b(V0 + V1 + V2 ) , with Vi = πLd2i /4 the volume of the segment number i Ti = L/(kdi ) the time delay along the ith segment a1 , a2 the relative costs of conduction for delays for synapses on the daughter branches. We can group together the terms corresponding to the same segment: C = [(a1 + a2 )T0 + bV0 ] + [a1 T1 + bV1 ] + [a2 T2 + bV2 ] What are the segment diameters d1 , d2 and d3 which minimize the cost C ? (Hint : each term in the latest equation depends on the diameter of only one segment. Set ∂C/∂di = 0 for i = 1, 2, 3). 3. Do these diameters satisfy the branching law (d0 )e = (d1 )e + (d2 )e ? What is the value of e ? 4. Can you think of any other branching structure in our body ? Figure 1: Diameters at a branching point Exercise 17 : Review on Action Potential Assume a cell whose membrane is permeable to calcium and potassium. It shows a resting potential (RP) and an action potential (AP) as indicated in the Fig 2. If one varies the concentration of extracellular potassium [K+ ]out only, or the concentration of extracellular Calcium [Ca++ ]out only, one observes the changes showed in Fig 3. Assume that the membrane is normally permeable only to K+ ions, Ca++ ions and water. Answer the following questions: 1. Which of these is true in the resting state (P stands for permeability): PK > PCa , PK = PCa or PK < PCa . Explain why. 2. Same question for the peak of the AP. Figure 2: An Action potential Figure 3: Effect of single ion extracellular concentration changes 3. Compare [K+ ]in to [K+ ]out . Compare also [Ca++ ]in with [Ca++ ]out . In which direction do more ions diffuse to? What is the resulting current flow? Discuss how potassium and calcium currents relate to each other. 4. When the cell is perturbed mechanically, the membrane transiently hyperpolarizes. What permeabililty change(s) might be responsible? Explain. What ionic currents typically lead to hyperpolarization at physiological concentrations? Exercise 18 : Equivalent circuit for a Synapse Fig 4 depicts a simplified neuron, represented by an equivalent circuit including a fast chemical synapse. Vrest is the resting potential of the cell, R is the input resistance and Vm the membrane potential. By Ohm’s law you obtain the postsynaptic current due to a single synapse (left hand side of the diagram): Isyn = gsyn (t) (Vm (t) − Esyn ) Note, that the synaptic conductance is a function of time, which depends on the opening of the synaptic ionic channels. The complete equation for the above circuit (by Kirchhoff’s law) yields: 0 = Isyn + Irest = gsyn (t) (Vm (t) − Esyn ) + (Vm (t) − Vrest ) /R Assume gsyn to be switched "on" at t = 0, so that potential arrives at the synapse) and is switched "off" from the synaptic cleft) such that: 0 gsyn (t) = 1nS 0 Assume Vrest = -70mV and R = 2GOhm current may pass through (when a presynaptic action at t = 1ms (when all neurotransmitter has been cleared for t < 0, for 0 < t 6 1ms, for t > 1ms 1. Let Esyn = 10mV. Is this a excitatory or inhibitory synapse? Calculate Isyn and the amplitude of the postsynaptic potential (i.e. Vm for 0ms < t < 1ms). Biophysics of Neural Computation, Introduction to Neuroinformatics (Winter 2006 2. Let Esyn = -90mV. Is this a excitatory or inhibitory synapse? Calculate Isyn and the amplitude of the postsynaptic potential. 3. Let Esyn = -70 mV. Calculate Isyn and the amplitude of the postsynaptic potential. Calculate gtotal (total Question membrane conductance). What effect1 will this synapse have on the responsiveness of the membrane? Is this a excitatory or inhibitory synapse? The figure below depicts a simplified neuron, represented by an equivalent circuit i 4. What circuit element can you add to make the circuit more realistic (hint: think of the time-course of the synapse. responses)? Figure 4: An equivalent circuit including a fast chemical synapse Vrest is the resting potential of the cell, R is the input resistance and Vm the membrane po By Ohm’s law you obtain the postsynaptic current due to a single synapse (left hand sid Isyn = gsyn (t) * (Vm (t) - Esyn ) Exercise 19 : Pre- and post- synaptic recordings. Question 2 in Fig 5 you see four traces of recordings from a presynaptic cell and its postsynaptic cell. Voltage and current In the figure you see four traces of recordings from aofpresynaptic cell and its postsynaptic cel Note, thatbelow the synaptic conductance is a function time, which depends on the open are plotted on the same time scale. Identify which trace is andchannels. current areThe plotted on the same time scale. Identify trace complete equation for the abovewhich circuit (byis Kirchhoff’s law) yields: 1. the postsynaptic current 1) the postsynaptic current = Isyn + Irest = gsyn (t) * (Vm (t) - Esyn ) + (Vm (t) - Vrest ) / R 2. the presynaptic action potential2) 0 the presynaptic action potential 3. the postsynaptic potential 3) the postsynaptic potential 4) gthe Ca2+-current Assume to be switched ‘on’ at t=0, so that current may pass through (when a pre 4. the presynaptic Ca2+-current synpresynaptic how to yourand conclusion. arrives atyou thecame synapse) is switched ‘off’ at t = 1ms (when all neurotransmitter h Explain how you came to Explain, your conclusion. synaptic cleft) such that: 0 for t < 0 gsyn (t) = 1nS for 0 < t <= 1ms 0 for t > 1ms Assume Vrest = -70mV and R = 2GOhm (a) Let Esyn = 10mV. Is this a excitatory or inhibitory synapse? Calcula postsynaptic potential (i.e. Vm for 0ms < t < 1ms) and Isyn . (b) Let5: ESome -90mV. ... Is this a excitatory or inhibitory synapse? Calcula syn =recordings Figure Question 3 postsynaptic potential and Isyn . You are designing an experiment test Calculate the probabilistic nature neurotransmitter You know, (c) Let Esyn = -70 tomV. Isyn and theofamplitude of therelease. postsynaptic po the probability of (total releasemembrane is small (e.g.conductance). if you lower extracellular Ca++will in your and the gtotal What effect this preparation) synapse have on nt synaptic vesicles available for release is large (as at the neuromuscular junction) then the expected n membrane? Is this a excitatory or inhibitory synapse? vesicles released following a presynaptic action potential follows a Possion distribution. Therefore if Exercise 20 : Quantal release I You are designing an experiment to test the probabilistic nature of neurotransmitter release. You know that when the probability of release is small (e.g. if you lower extracellular Ca++ in your preparation) and the number of synaptic vesicles available for release is large (as at the neuromuscular junction) then the expected number of vesicles released following a presynaptic action potential follows a Possion distribution. Therefore if m is the mean number of vesicles released per trial, then the probability p of observing a particular number x x (x = 0, 1, 2, 3 ...) of vesicles released in a given trial is: px = mx! e−m You stimulate a nerve 500 times. Assume that each trial is independent of the other trials performed. Given that the number of quanta released in one trial follows Possion statistics and the mean quantal content is 5. 1. How many failures of transmission do you expect to observe in the 500 trials? 2. How often do you expect two quanta to be released ? Exercise 21 : Quantal release II Fig 6 shows a histogram of end-plate potential of a cat muscular fiber. These data were recorded in 1956 by I. A. Boyd and A. R. Martin to study the release properties of the muscular end-plate. The recording were achieved under low extracellular Ca++ and high Mg++ concentrations. These special conditions permitted to reach extremely low release probabilities. This make up the end-plate potential to fluctuate according to Poisson’s law (see previous exercise). The histogram of Fig 6 corresponds to the amplitude distribution of 200 events (evoked and spontaneous). The mean amplitude of all events is 0.93 mV. 1. What do the different peaks labelled A, B, C, and D correspond to? 2. From the graph what is Q, the quantal amplitude if exactly one vesicle is released. 3. Using the result from b) and the mean amplitude of all events calculate m, the mean number of vesicles released per trial. Can you imagine another way of calculating m assuming a Poisson distribution? 4. Assume that m and Q are measured for a given experiment. In a subsequent measure in which experimental parameters have changed, m and Q are measure again and they might have changed. How can you detect an increase in the release probability? How can you detect postsynaptic modifications? 5. It is possible to fit a Gaussian to each peak of the distribution and then add them to obtain the theoretical distribution. In order to do that you need to know how many events to include under each curve. What do these number of events correspond to and how would you calculate them. Exercise 22 : Inhibitory synapses The most important sources for inhibition in the central nervous system are mediated by GABA and glycine receptors. These channels are selective for anions, such as Cl− or HCO− 3 . For most neuronal cell bodies, the equilibrium for anions is more negative then the resting potential (because Cl− is pumped out of the cell). However, in some cells, the equilibrium for ions is sometimes more positive than the resting membrane potential, and the opening of GABA or glycine channels produces a depolarization. 1. In this case, how could the effect still be inhibitory? 2. The figure 7 shows the effect of an unknown peptide that opens such a receptor-channel at different voltages (the y-axis represents the difference of current if you release the peptide). Remember that by definition a flux of positive charge going outward is a positive current. If the ions flowing through this channel (when it is open) are negative (anions), do they go outward or inward at the resting potential (-70 mV)? 3. what is the reverse potential for this channel? 4. If this peptide was designed to be an anti-epileptic drug, would it be a good candidate? Figure 6: End-plate potential of a cat muscular fiber Figure 7: Effect of our mysterious peptide at different voltages (note that the y axis shows the difference of current with the peptide) Exercise 23 : Spike time dependent plasticity It is experimentally observed that the relative timing of pre-and post-synaptic action potential can change the synaptic efficacy. Within a time window of 50ms, presynaptic spike that precede postsynaptic action potential produce LTP (strengthening of the synapse), whereas presynaptic spikes that follow postsynaptic AP produce LTD (weakening of the synapse). 1. Why is it important -for any learning rule- to have mechanisms for both the strengthening and the weakening of the synaptic weight? 2. The figure 8 shows the spike-time-dependent plasticity (STDP) modification function for a particular synapse. F (∆t) is the amount of synaptic weight modification observed after a pair of pre- and postsynaptic spikes, as function of ∆t = tpre − tpost . ( F (∆t) = A+ exp(∆t/τ + ) if ∆t < 0 A− exp(∆t/τ − ) if ∆t > 0 Estimate the value of the parameters A+ and tau for the function F of Fig 8. Figure 8: Spike-time-dependent plasticity (STDP) modification function for a particular synapse, Song & al. (2000) 3. If two neurons fire completely independently, there is on average as much occurrences where the first one spikes just before the second one does, than cases where the opposite happens. Nevertheless, a synapse between two such neurons should be weakened, which will only happen if the integral of the function F is negative, i.e. if A+ τ + < A− τ − . Draw a weight update function (similar to Fig 8) that ensures that uncorrelated pre- and postsynaptic spike produce an overall weakening of the synapse. 4. Such an STDP rule would tend to balance the firing rate in a neuron. As state Song & al. (2000) : "(...) a neuron can operate in two different modes with distinct spike-train statistics and input/output correlations. When excitation is strong (...) so that the mean input to the neuron would bring it well above threshold if action potentials were blocked, the neuron operates in an input-averaging or regularfiring mode. The postsynaptic action potential sequences produced in this mode are significantly more regular than the presynaptic spike trains that evoke them. The interspike intervals of the postsynaptic response depend on the total synaptic input, but the absolute timing of individual postsynaptic action potentials is fairly insensitive to presynaptic spike times. As a result, there are roughly equal numbers of presynaptic action potentials before and after each postsynaptic spike". And in this case, if the integral of F is negative, we have a weakening of the synapse. "As the excitatory synapses are weakened by STDP, the postsynaptic neuron enters a balanced mode of operation in which it generates a more irregular sequence of action potentials that are more tightly correlated with the presynaptic spikes that evoke them. The total synaptic input in the balanced mode is, on average, near or below threshold, so the postsynaptic neuron fires irregularly, primarily in response to statistical fluctuations in the total input. Because action potentials occur preferentially after a random fluctuation, there tend to be more excitatory presynaptic spikes before than after a postsynaptic response". What happends then? Exercise 24 : Dendritic computation An excitatory synapse increases locally the membrane potential when activated. This depolarization spreads passively toward the soma. An inhibitory "shunting", or "silent" synapse can fix the potential locally to the resting potential. It can be seen as a veto on any more distal synapse, whose membrane potential change will be stopped. But an inhibitory synapse has little effect in vetoing a more proximal exitatory synapse. (a) In Fig 9, which combination has a chance to elicit an action potential in the axon? i. e1 + i7 activated, all the others inactivated ii. e2 + i1 + i4 + e5 activated, all the others inactivated iii. all the synapses are activated (b) What synaptic conditions must exist for e6 to have an influence on the membrane potential at the beginning of the axon? Figure 9: Schematic dendritic tree of a cell. The excitatory synapses are e1, e2, ..., e6 , and the inhibitory ones i1, ..., i7. Figure taken from [Koch and Shepherd, 1990, in The synaptic organization of the brain , Oxford University Press]. (c) If all the inhibitory synapses have the same probability to be active, which excitatory synapse will have more influence on the membrane potential at the beginning of the axon, e2 or e5? (d) What is the number of possible combinations of activation for the synapses on this neuron? (e) If this neuron needs at least the effect of 2 excitatory synapses on the beginning of the axon to reach the firing threshold, give a possible combination that meets this condition (write the required activation condition for each synapse, ie active or not). Exercise 25 : Interaural time differences (a) Look at Fig 10. Describe the nature of the spike input from the left and right side in order to process inter-aural time differences. Is it a temporal or a rate code? Explain. (b) The neurons (A-E) serve as coincidence detectors and fire maximally when they receive the left and right sided input simultaneously. Nevertheless, the output of the neurons (A-E) has different tuning curves, due to the different lengths of the axons from the left and right sided input neurons. Describe the tuning properties of the output neurons (A-E) qualitatively. (c) Given that the speed of sound in the air is 350 m/s and the distance between the left and right ear is 15 cm (in humans), how big is the time delay between the left and right ear for a sound coming directly from the side,? How does that relate to the duration of an action potential? Exercise 26 : Reading a neuron’s specificity Fig 11 shows spike rasters and peristimulus time histograms of a parietal cortex neuron for reach (blue) and eye movements (red) in 8 center-out directions. (For the saccade task, eye position traces are shown at the bottom of each panel). (a) Describe the spiking activity of this neuron for the two behaviors, movement directions, and in within the task (baseline, cue, planning, execution phases). What is this neuron coding for? (b) What can be said about the activity variation from trial to trial? Figure 10: Coincidence detectors for sound localization. Figure 11: Raster plots and PSTH for reach (blue) and eye movements (red) in 8 center-out directions.