Cellular Neurobiology BIPN 140 Fall 2016 Problem Set #2

Transcription

1 Cellular Neurobiology BIPN 140 Fall 2016 Problem Set #2 1. (PeiXi) You are performing research on a novel ion channel and want to learn some of its characteristics. a) When you conducted voltage clamp experiments in which you set the voltage of the membrane at different values, you observed the following currents in single channel recordings. a) What is the value of the reversal potential of this channel? What is likely to be the ion to which this channel is selectively permeable? How would you confirm this ion selectivity? The value of the reversal potential can be obtained by determining the membrane potential at which there is no current flow. In this case, Erev is equal to +60 mv, which is close to the equilibrium potential of Na+ ions. To confirm this tentative conclusion, administer TTX (blocker of voltage-dependent sodium channels) and measure the current using single-channel recording to see if any channel activity remains. If little to no current is measured at +20 mv, +40 mv, +80 mv and +100 mv (or any clamped voltages), then the channel is permeable to sodium ions. b) You would then like to measure the currents flowing through this single ion channel in response to the effects of a drug taken up by the cell. What method would you use to do so? Excised inside-out patch recording or simply cell-attached patch clamp recording. c) If a channel opens twice for 50 ms in a period of 2-minutes, and the ratio of the currents recorded by conducting both whole-cell and single-channel recordings at + 40 mv is 1.5. What is the total number of these channels present in the membrane? Popen = Σ to / total time = (0.050s x 2) s/ 120 s = 1/1200 = I/i = 1.5 I = N i Popen => I/i = N Popen => N = 1.5 / = 1800 ion channels 2. (Stephanie) a) You are trying to record from a single channel using a cell-attached recording and obtain the following result: 1

2 a) Are you successfully recording current through a single channel? No, because channels open in an all-or-none manner. The presence of three different values for current (0, 2 pa, and 4 pa) suggests that you actually have two or more channels in this patch, and each channel is opening in an all-or-none fashion, with two channels open simultaneously at the highest point. b) Would using the equation Popen =Σ topen / total time work with this recording? Why or why not? No, it would not, because you are recording from two or more channels; using this equation would greatly overestimate the probability that a single channel is open. c) If the voltage were clamped at -40 mv for this experiment, which of the following ionic species could this channel be permeable to, Na+, K+, or Cl-? Remember that the upward direction is defined as outward current. This could be K+ or Cl-. K+ + would flow out of the cell at this voltage, which we would observe as an outward current. Cl- would actually flow into the cell, but because it is an anion this would also be observed as an outward current. 3. (Stephanie) You discover a mutation in the gene encoding voltage-gated sodium channels in a family prone to epilepsy. This mutation is located in the protein domain responsible for inactivation following depolarization and abolishes the function of this domain. a) Briefly explain how this mutation would prolong depolarization. When inactivation is removed, the inward sodium current will be sustained. This mutation will increase the effective sodium channel conductance (GNa) since inactivation is affected, and the sodium current is proportional to the sodium conductance. Since the inward sodium current depolarizes the cell, this sustained current would cause a longer depolarization. b) With this mutation, would you expect the overshoot to actually reach the reversal potential of sodium? Why or why not? No, it would not. Even without inactivation, there is still a significant potassium current building up during the overshoot as voltage-gated potassium channels open. Additionally, the sodium current will still decrease because INa = GNa (Vm-ENa), and as the voltage ascends to the peak of the overshoot, (Vm-ENa) becomes smaller. c) In a neuron expressing only this mutated voltage-gated sodium channel (and not the wild-type channel) graph the sodium membrane current over time when the cell is voltage clamped at 0 mv. Explain any differences between this curve and the same curve for a neuron with only the wild-type channel. Which drug might be helpful for isolating this mutated current, and why? 2

3 We can compare the above graph to figure 3.5 in the text, where t=0 is the time at which the membrane potential is clamped at V=0 mv and the membrane potential is at the resting potential before this time. Even though it is clamped at 0 mv, the sodium current decreases after time for the wild-type channel because the inactivation gate closes. This effect is not seen in the mutant channel. TEA would likely be helpful for isolating the sodium current because it blocks voltage-gated potassium channels, so the readout of current will be entirely from sodium channels alone (in a real experiment, remember that TEA may not completely block all K+ channels). 4. (PeiXi) You voltage-clamp a squid giant axon at +5 mv and at +55 mv, and measure the currents. You observe transient inward as well as delayed outward currents generated at both voltages. a) However, you notice that these currents do not completely resemble each other (they are not superimposable). What would they look like when plotted together with respect to time? b) What are the ions carrying each current? Why is there a difference between the sizes of each current? The ions carrying the inward current are Na+ ions, while the ions carrying the outward current are K+ ions. The ion conductances at the two clamp potentials are different because of the nature of the ion channels opening or closing in response to the two membrane potentials. We remember that Iion = Gion (Vm-Eion) to guide our thinking. At Vm = +5 mv, there is a large driving force for sodium ions to enter through sodium channels, and a strong likelihood for Na+ conductance to activate at this depolarizing membrane voltage; therefore, the inward current is larger than when Vm = +55 mv. At Vm =+55 mv, a membrane potential close to ENa, the current carried by sodium ions is smaller because the driving force is reduced and many of the sodium ion channels have become inactivated at this point. For the outward current carried by K+ ions, both GK and (Vm-EK) increase with increased depolarization, there is no inactivation of GK, and so the current gets larger as Vc goes from +5 to +55 mv. 3

4 5. (PeiXi) a) Draw an action potential. Use an arrow to indicate where INa = -IK. The arrow should be pointed at the peak of the action potential. It could also be pointed at threshold. b) If GNa = 50GK, EK = -90 mv and ENa = +55 mv, calculate the corresponding Vm at this point. INa + IK = 0 since dv/dt=0, so INa = -IK (GNa (VM-ENa) = -(GK (Vm-EK) 50GK (Vm-55 mv) = (GK (VM+90 mv) = mv This Vm is likely from the peak of the action potential, where Vm is close to ENa. There is no change in voltage with respect to time for a brief time as the initially-small voltage-gated K+ current is balanced by the now-small voltage-gated Na+ current. 6. (Antonia) You extract an axon from an alien squid that you found floating at La Jolla Shores. First, you measure the resting membrane potential to be -70 mv by using a current clamp. After that, you inject a current that would depolarize the membrane potential of the cell to +150 mv. To be able to confirm that voltage has been reached you place a recording electrode a little further from the site of current injection. As the current starts flowing into the cell, your recording electrode reveals that Vthreshold is reached at -10 mv in 1 msec. (a) You are particularly interested in the value of tau (τ) for this axon. With the assumption that the cell is isopotential, how would you calculate tau? Vt= Vinf (1-e -t/ τ ) Vt= -10 mv-(-70 mv) = 60 mv; Vinf =150 mv-(-70 mv) = 220 mv; t=1 ms; Tau = ms (b) Proud of your catch, you show off your giant squid axon (radius = 1000 µm) to your BIPN 140 IAs. They tell you they have been secretly investigating the same alien squid for quite some time and found that Cm =10-6 F/cm2. However, they have unsuccessfully been trying to obtain its rm. You laugh at them (and wonder who allowed them to be IAs in the first place), because you have all the information you need to perform the calculation. Show your work. 4

5 Tau = Rm Cm = rm cm; Cm = 10-6 F/cm2; cm = Cm*2πa=10-6 /cm2* 2π *0.1 cm = *10-7 F/cm; rm =tau/cm = 3.140*10-3 s /6.2831*10-7 F/cm =4.99*10 3 ohm*cm. 7. (Antonia) You obtain the following plot when clamping an axon at different voltages. (a) What is the conductance of the unknown channel at Vc = -40 mv? Iion= Gion(Vc-Eion) Vc = -40 mv I = 10 ma Eion = -60 mv; Gion= 0.5 S (b) If the current measured through a single ion channel is -10 pa and its probability of being closed is 0.2, how many channels conduct current at Vc = -80 mv? I = N * i * Popen i = -10 pa I = -10 ma Popen = 1-Pclosed=0.8 N = 1.25* (Milad) You isolate a squid giant axon to measure some electrical properties of the cell. You current clamp the axon at different potentials for different durations. In the first case you stimulate the axon with a short duration, high amplitude current. In the second case, the axon is initially stimulated with a long duration, low amplitude current that is immediately followed by a high amplitude current (the same amplitude as in the first case). 5

6 If you can t figure what is going on here please refer back to your notes where Nick discussed the pre-pulse experiments of Hodgkin and Huxley and voltage-dependent Na channel inactivation. Why doesn t an action potential fire in the second case when you inject the long-duration, lowamplitude current? An action potential doesn t fire in this case because the injected current is not strong enough to raise the membrane potential to this cell s threshold potential. Even though the current is being injected for a longer period of time it still doesn t allow the cell to reach threshold. This results because voltage-gated sodium channels have been inactivated. 9. (Nick) After isolating a single myelinated axon, you made the following measurements: resting membrane potential = -80 mv, threshold potential = -50 mv, furthest internode distance at which an action potential could be generated = 3 mm. Your roommate then applied a constant stimulus voltage to depolarize the axon to +45 mv at a particular point. a) What was the length constant of the axon? V x = V 0 e -x/λ V x = 30 mv V 0 = 125 mv x = 3 mm (distance at which we know that V m will be at threshold) 30 = 125 e -3/λ λ = 2.10 mm b) After an apocalyptic disaster struck their lab, scientists from a biotech company here in La Jolla found the same axon, this time with many altered properties although the axon diameter was still 20 µm. They found that the specific membrane resistance was 100Ωcm 2 and the specific internal resistance 100 Ωcm. What was the new length constant? Λ = [Rm/2pa / Ri/pa2] 1/2 λ = [(100 Ωcm 2 /2π*0.001cm)/(100 Ωcm/ πcm 2 )] 1/2 λ = [.0005] 1/2 cm =.022 cm = 220 µm 6

7 10. (Nick) You isolate a neuron that has a resting membrane potential of -80 mv. When depolarizing the neuron by 40 mv, you observe that the action potential peaks at +50 mv. It takes roughly 1 x 10-3 sec for it to reach threshold. Measuring membrane properties, you find out that the specific membrane capacitance = 1 µf/cm 2 and the specific membrane resistance is 1000 Ωcm 2. Finally, the diameter of the axon is measured to be roughly 0.3 cm. a) What is the threshold for the neuron to fire an action potential under these circumstances? V t = V (1-e -t/t ) V t = 40 (1-e -1/1 ) V t = 40 (1-1/e) V t = 40 (0.63) V t = 25.2 mv depolarization from -80 mv. So threshold = -55 mv. b) Another group of scientists conducted an experiment using the same neuron, but observed that the time to reach threshold was shorter. Assuming the properties of the neuron stayed the same, what most likely happened to cause this? A larger stimulus was probably applied to the neuron. c) What principle from the first lecture concerning information coding applies here and in what system does it mainly take place? This deals with the concept of frequency coding, mainly seen in neurons in the peripheral nervous system. 7