Hodgkin-Huxley model simulator

Transcription

1 University of Tartu Faculty of Mathematics and Computer Science Institute of Computer Science Hodgkin-Huxley model simulator MTAT Introduction to Computational Neuroscience Katrin Valdson Kristiina Pokk Marit Asula Tartu 2015

2 Contents Introduction Part I: Equilibrium Potential Part II: Membrane Potential Part III: The Action Potential Part IV: The Fast Sodium Channel Drugs TTX, TEA and Pronase Step Current Input References 1

3 Introduction The Hodgkin-Huxley model is a mathematical model of action potential in a neuron devised by Alan Hodgkin and Andrew Huxley that is a basis for understanding and modelling neural excitability. Their influential work was published in a series of articles in 1952 (with assistance of Bernard Katz) and the authors were awarded with a Nobel Prize in Physiology and Medicine (shared with John Eccles) in (Nelson & Rinzel, 2003). In 1939 Hodgkin and Huxley started working together in Plymouth on nerve conduction using squid giant axon. They were able to record the potential difference across the nerve member using fine capillary electrode inserted in the nerve fibre, thus for the first time record an intracellular action potential. Their experiments showed that large action potential overshoots the apparent zero potential. The collaboration of Hodgkin and Huxley was interrupted by World War II. (Schwiening, 2012) In late 1940s they continued with their work. Voltage clamp technique enabled them to directly record the ionic currents flowing across the axonal membrane without change in membrane potential and to investigate voltage kinetics in the ion channels. The mathematical model, published in 1952, incorporated four currents - capacitance, K +, Na +, leak using three voltage-and time-dependent variables (K +: n as activation variable ; Na + : m as activation variable and h as inactivation variable). The Hodgkin-Huxley equation: I = C dv m dt + gkn 4 (V V K ) + g m3 Na h(v V Na ) + g l (V V l ) predicted with remarkable accuracy the time course of an action potential, making the Hodgkin-Huxley model a significant landmark in neuronal modelling. (Schwiening, 2012) 2

4 Part I: Equilibrium Potential Question 1. Calculate the effect of halving the external sodium concentration. Remember, we're assuming that only sodium channels are present. You will first need to calculate RT/zF at 6.3o, which you can do from the information provided in the Membrane window. Then, confirm your answer in the simulator by changing the concentration in the Membrane window and looking at the new value of passive Vr. (You can type new values into any of the boxes in the Membrane window, or use the < or > buttons to change the values.) Na out Na in Answer 1. Used the Nernst equation V Na = R T z F ln. 1 1 R = J.K.mol T = = K z = F = C.mol Na out = 440 Na in = 50 V Na = ln = l n8.8 = mv With halved external sodium concentration: V Na = ln = l n4.4 = mv Confirmation with the simulator is shown in figure 1. 3

5 Figure 1. Halved external sodium concentration. Question 2. Calculate the [Na+] external concentration required to achieve Confirm your answer using the simulator. = 55.5 mv. V r Answer 2. Deriving the formula for calculation: V Na = R T z F ( ln(na out) ln(na in )) V Na = R T z F ln(na out) R T z F ln(na in ) R T z F ln(na out ) = V Na + R T z F ln(na in ) V ln(na out ) = Na R T + ln(na in ) Na out = e z F V Na R T +ln(na in) z F Calculation: 55.5 Na out = e ln50 = 501 Confirmation with the simulator is shown in figure 2. 4

6 Figure 2. [Na+] external concentration required to achieve = 55.5 mv V r Question 3. Suppose we double the temperature from 6.3o C to 12.6o C. What is the new value of Vr? Explain why Vr doesn't double. Answer 3. V Na = ln = l n8.8 = mv V R does not double because the temperature is in K not Celsius in the formula. Since K = C elsius, multiplying the Celsius degrees by 2 does not multiply K by 2: C elsius = / 2 ( C elsius) Thus, the temperature value in K changes only by a little. Part II: Membrane Potential Question 1. Write down the parallel conductance equation for the resting potential as a function of equilibrium potentials and conductances. Given the equilibrium potentials shown in the 5

7 Membrane window and the conductances shown in the Channels window, calculate the resting potential V r. Answer 1. = 0 is equal to the flow of current through the passive channels and current I leak through one leak channel can be calculated from the equation: I i = g i ( V m E i ), where i denotes the given ion channel (Na, K or Cl), g i denotes the conductance per unit area, V m denotes membrane potential and E i denotes reversal potential of the i -th ion channel. In our case V m = V r, where V r denotes the resting potential. The current through all the leak channels can be calculated from the equation: I leak = 0 = g Na (V r E Na ) + g K (V r E K ) + g Cl (V r E Cl ), from where we can find V r followingly: I leak = 0 = g Na V r g Na E Na + g K V r g K E K + g Cl V r g Cl E Cl g Na V r + g K V r + g Cl V r = g Na E Na + g K EK + g Cl E Cl V r (g Na + g K + g Cl ) = g Na E Na + g K E K + g Cl E Cl g Na E Na + g K E K + g Cl ECl V r = g Na + g K + gcl ( 72.1) ( 57.2) V r = V r = 47.7 Confirmation with the simulator is shown in figure 3 (although the simulator value is slightly different). 6

8 Figure 3. The value of resting potential V r. Question 2. You can see from the Channels window that g K is micro-siemens. Calculate to two significant digits the value to which g K would have to be reduced in order to make V r be -45 mv. Verify your answer by changing the value of g K in the simulator and seeing how the value of passive V r in the Membrane window changes. Answer 2. We can deduce g K from the equation used in previous exercise: I leak = 0 = g Na (V r E Na ) + g K (V r E K ) + g Cl (V r E Cl ) g K (V r E K ) = g Na (V r E Na ) g Cl (V r E Cl ) g Na (V r E Na ) g Cl (V r E Cl ) g K = (V r E K ) ( ) 0.1 ( 45 ( 57.2)) g K = ( 45 ( 72.1)) g K = Confirmation with the simulator is shown in figures 4 and 5. 7

9 Figures 4 and 5. Changing the value of to in the channels window results in a new resting g K potential equal with mv. Question 3. We can also manipulate the resting potential V r by changing the ion concentrations. Obviously it is easier to change the concentrations outside the cell than inside. Suppose we want to raise the cell's resting potential to -39 mv, while maintaining a constant osmolarity and not disturbing the charge balance. What ion concentrations should we change, and what should their new values be? Use the simulator to find the answer empirically. Answer 3. As we wanna maintain constant osmolarity, which basically means that the number of all the ions should remain the same, and at the same time not disturb charge balance, which means that the ratio of positive ions to the negative ions should remain the same and thus the number of positive and the number of negative ions should remain the same. Therefore, we cannot manipulate negatively charged Cl ion concentrations as it would change the number of negative ions, which would disturb the charge balance or change osmolarity. The only way is to increment potassium ions by some number n and at the same time decrement sodium ions by the same number or vice versa. For example, we can take n = 37.5 and decrement sodium ions from 440 to while incrementing potassium ions from 20 to We can see from the figure 6, that such concentration change results in a resting potential mv. 8

10 Figure 6. Changing sodium and potassium external ion concentrations results in a resting potential equal with mv. Part III: The Action Potential Question 1. Why does hyperpolarization cause a spike? Answer 1. Hyperpolarization causes h (inactivation variable) to increase while conductance (gna and gk) is relatively small (decreasing during stimulus). After the stimulus m is first to respond and it starts increasing quickly to its original value. When resting potential is reached h has not had enough time to decrease to its normal value during rest. The result is that increased sodium current to the inside of the neuron causes depolarization and therefore action potential (figure 7). (Hodgkin & Huxley, 1952; Guttman & Hachmeister, 1972) 9

11 Figure 7. Hyperpolarization causes a spike. Question 2. Why doesn't this second pulse cause a second spike? Phrase your answer in terms of gates and voltages. Answer 2. The second pulse takes place during refractory period when action potential cannot be generated. During the second stimulus m and h have not reached resting values (they are notably smaller). This means that Na gates cannot open (depolarize membrane potential). In addition potassium activation variable is still high, which makes membrane potential more negative. Figure 8. The second pulse doesn t cause a spike. 10

12 Question 3. How much time must elapse between the end of the first pulse and the beginning of the second in order for the second pulse to cause a second spike? Answer 3. There has to be at least 8 seconds between the first and the second pulse to generate the second spike. Figure 9. Two spikes are generated when the time difference between two pulses is at least 8 seconds. Figure 10. The time interval between first and second pulse is set to 8 in stimulus window. 11

13 Question 4. To the nearest 10th of a millisecond, what is the longest delay after a 1 msec 5 na positive pulse that a 1 msec -5 na pulse can block a spike? Answer 4. The longest delay after a 1 msec 5 na positive pulse that a 1 msec -5 na pulse can block a spike is 0,3 ms. Figure 11. Spike is blocked after 0.3 ms delay. 12

14 Part IV: The Fast Sodium Channel Question 1. Looking at the simulator output, at what potential does the cell settle after the spike has subsided, i.e., what is the new resting value of Answer 1. The new resting value is: V m = 6.25 mv? Click on the red line to measure the value. V m Figure 12. The new resting potential is equal with 6.25 mv. Question 2. Using the parallel conductance equation, calculate the conductance of the fast sodium channel in this new resting condition. Answer 2. We use the following formula: dv I = Cm m dt + g K n 4 ( V m E k ) + g m3 Na h (V m E Na ) + g l (V m E l ) As V m is in resting condition and doesn t change, C m dt = 0 and as we do not consider potassium channels, V E ) g K n 4 ( m k = 0 and thus we get: I = 0 = g m3 Na h (V m E Na ) + g l (V m E l ) dv m 13

15 0 = g m3 (x)na h (V m E Na ) + g Na (V m E Na ) + g K (V m E K ) + g Cl (V m E Cl ) (g Na (V m E Na ) + g K (V m E K ) + g Cl (V m E Cl )) g (x)na = m3h (V E ) m Na, where = g Na g K g Cl = = 0.1 = 6.25 V m E Na E K E Cl = 52.4 = = m = h = //Something doesn t quite add up during the calculation Figure 13. The conductance of the fast sodium channel in the new resting condition is 0.23 us. 14

16 Question 3. Stimulating the cell again (using the Stim1 button) in this condition will not cause another spike. Even if you raise the stimulus intensity to 20 na and the duration to 10 msec, the cell will not spike. (Try it. Turn off the cyan plot by selecting "-" in the pop-up menu so you can see the yellow and green plots clearly when you hit Stim1.) What is the explanation for this? Answer 3. Activation variable m and inactivation variable h are almost not affected by the stimulation, which means that they are balanced and thus spikes cannot occur. Part V: The Delayed Rectifier Question 1. Based on the voltage? V m plot (red line), what is the peak value reached by the membrane Answer 1. Peak value reached by the membrane voltage is -15,9 mv Figure 14. The peak value of membrane potential is mv. Question 2. After reaching its peak, the membrane voltage quickly declines again, even though the stimulus is still on. What is causing this? Answer 2. K +channels open (n increases), allowing K +to flow out of the neuron, making membrane potential more negative. 15

17 Question 3. When the stimulus ends, the cell does not simply return to its resting value; it undershoots it and then approaches the value from below. Why doesn't it just return to its resting value? Answer 3. Potassium channels remain open to the point when membrane potential is reaching the equilibrium potential of K + (about -70mV). Part VI: Voltage-Gated Channel Parameters Question 1. How does g_max for the fast sodium channel compare with the passive sodium conductance? Answer 1. The passive sodium conductance g max = 120. g Na = is a lot smaller than the value of Question 2. What relationship must hold between the red and blue lines for m to increase? And, what is the approximate value of Vm at which this condition occurs (if there were no other mechanisms affecting the cell)? How does this compare to the cell's normal resting potential? Answer 2. The rate at which m changes is expressed by the formula dm dt = α(1 m) β m. Because m is the fraction of channels with open activation gates and α is the rate at which segments move from closed to open, we multiply α with 1 m. β is the rate at which segments move from open to closed, thus we multiply β with m. Since we are looking at the case where m increases, we get the following inequality: dm = α (1 m ) β m > 0 dt So the relationship between α and β is the following: ( 1 m) α > m β The V m value for this condition depends on the value of m. For instance if m = 0, then for m to increase, α > 0, which applies for all values of V m. If m = 0.5, then for m to increase, α > β, which occurs at around 40 mv. And if m=1, then m can not increase further, thus the formula says β < 0, which does not happen. The normal resting potential is lower than the V m value when α > β in the case where m = 0.5. Question 3. In order to stop the cell from oscillating on its own, we can change the passive channel conductance. Hit the Run button in the main simulator window to continue the simulation. By playing with the value of g_k in the Channels window, find a value close to the 16

18 original value of 0.07 micro-siemens that prevents the cell from spiking spontaneously. The cell should of course still spike in reponse to a stimulus from the Stim1 button. Report the g_k value you find, to two significant digits. Answer 3. To stop the oscillation, it is necessary to increase the g K value. At 0.17 the spikes still do not stop, the closest value to 0.07 which made the random spiking stop was Drugs TTX, TEA and Pronase Question 1. Simulate with the "Drug window" the role of drugs TTX, TEA, and pronase on the voltage of the neuron. Describe where these drugs are found, which channel affect, and how do they affect the action potential. TTX Tetrodotoxin Tetrodotoxin or TTX is a potent neurotoxin, being about 1200 times more dangerous to humans than cyanide. It s most well-known source is the pufferfish (fugu) and the toxin was even named after it (Tetraodontidae). It has been later discovered that other animals such as blue-ringed octopuses, marine worms, star fish and some terrestrial animals: salamanders, frogs. It is likely that TTX is provided by bacteria residing in the animals that are poisonous. (Map of Life - "Tetrodotoxin", 2015) 17

19 Figure 15. (Tetrodotoxin: Mode of Action, 2015) TTX blocks nerve conduction by blocking the Na + channels, stopping the flow of sodium ions to the cell, thus preventing depolarization and the occurrence of action potential. It acts as a cork to the channel as tetrodotoxin is much larger than sodium ions. (Tetrodotoxin: Mode of Action, 2015) 18

20 Figure 15 shows two spikes occurring with a 10 msec pause. As seen on Figure % inhibition already weakens the first action potential (maximum membrane voltage is 9.41 mv) and stops the second one from occurring. With 100% inhibition no action potential occurs (maximum value of m is only 0.12). TEA Tetraethylammonium Tetraethylammonium (TEA) is an experimental drug with no approved indication or marketed formulation. TEA s mechanism of action is still being investigated, but it is known that tetraethylammonium blocks autonomic ganglia, calcium- and voltage- activated potassium channels, and nicotinic acetylcholine receptors. The most common use of tetraethylammonium presently is as a pharmacological research agent that blocks selective potassium channels. ( DrugBank - Tetraethylammonium ). Typical symptoms produced in humans include the following: dry mouth, suppression of gastric secretion, drastic reduction of gastric motility, paralysis of urinary bladder, and relief of some forms of pain. (Wikipedia - Tetraethylammonium ) 19

21 TEA increases the duration of the action potential (Schmidt & Stampfli, 1966) by blocking depolarization-activated delayed rectifier K+ channels in the nodal axolemma. A motor nerve terminal stimulated in the presence of TEA releases more transmitter (Katz & Miledi, 1969; Benoit & Mambrini, 1970), and may discharge repetitive action potentials (Koketsu, 1958; Payton & Shand, 1966). TEA is also known to reverse the action of drugs such as tubocurarine, a non-depolarizing blocker. TEA evokes more release of the neurotransmitter and thus it will reverse the competitive antagonistic block of any drugs belonging to the curare family. ( Amrita University - Effects of pharmacological blockers on action potential ) Figure 18. Normal action potential. Figure % TEA inhibition. 20

22 Figure % TEA inhibition. Normal action potential is displayed in figure 18, figure 19 shows that 90% TEA inhibition leads to repetitive action potentials and in figure 20 we see that no spikes occur with 100% TEA inhibition. Pronase Pronase is a group of proteolytic enzymes, meaning they break down protein molecules into peptides and then in turn into amino acids (Encyclopedia Britannica - Proteolytic enzyme ). Pronase is isolated from the extracellular fluid of Streptomyces griseus ( Wikipedia - Pronase ), a species of bacteria found in soil (Wikipedia - Streptomyces griseus ). With the condition that potassium K conductance was blocked and with the use of a voltage clamp, which helps control the conditions, it was discovered that pronase exclusively blocks the Na inactivation gate, meaning that Na + ions will keep flowing into the cell. The effect does not depend on whether the gates are currently active or not. (Scielo - The action potential: From voltage-gated conductances to molecular structures ) The result is that the action potential gets prolonged, forming a plateau of a positive voltage value for a few minutes. After that, the cell gets slowly repolarized. (C. M. Armstrong, F. Bezanilla, E. Rojas, "Destruction of Sodium Conductance Inactivation in Squid Axons Perfused with Pronase", 1973) 21

23 Action potential before and after exposure to pronase, where the first spike is without pronase and the second one with pronase, is shown in figure 21. Figure 21. Action potentials without and with pronase. 22

24 Step Current Input 23

25 References Guttman, R., & Hachmeister, L. (1972). Anode Break Excitation in Space-Clamped Squid Axons. Biophysical Journal, 12(5), Hodgkin, A. L., & Huxley, A. F. (1952). A quantitative description of membrane current and its application to conduction and excitation in nerve. The Journal of Physiology, 117(4), Nelson, M., & Rinzel, J. (2003). The Hodgkin-Huxley Model. In M. J. Bower, & D. Beeman, The Book of GENESIS. Exploring Realistic Neural Models with the GEneral NEural SImulation System (pp ). Schwiening, C. J. (2012). A brief historical perspective: Hodgkin and Huxley. The Journal of Physiology, 590 (11), Map of Life - "Tetrodotoxin". Retrieved from (Last visit: 2015, May 22) Tetrodotoxin: Mode of Action. Retrieved from (Last visit: 2015, May 22) DrugBank - Tetraethylammonium. Retrieved from (Last visit: 2015, May 24) Amrita University - Effects of pharmacological blockers on action potential. Retrieved from (Last visit: 2015, May 24) Encyclopedia Britannica - Proteolytic enzyme. Retrieved from (Last visit: 2015, May 24) Wikipedia - Pronase. Retrieved from (Last visit: 2015, May 24) Wikipedia - Streptomyces griseus. Retrieved from (Last visit: 2015, May 24) Scielo - The action potential: From voltage-gated conductances to molecular structures. Retrieved from (Last visit: 2015, May 24) 24

26 C. M. Armstrong, F. Bezanilla, E. Rojas, "Destruction of Sodium Conductance Inactivation in Squid Axons Perfused with Pronase", 1973, Retrieved from: (Last visit: 2015, May 24) Wikipedia - Tetraethylammonium. Retrieved from: (Last visit: 2015, May 25) 25