Session 2: Physiology of excitable cells

Physiology of Excitable cells Fanny Casado, Ph.D. [email protected]

ING338: Human Physiol for Engineers. 2017-1

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Feher, J. - Quantitative Human Physiology: An introduction – Fig. 2.8.1. 2012, Elsevier

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Notas de la presentación

Transduction of mechanical, electrical and chemical signals Transduction of signals: Some kinds of sensory cells can transduce mechanical stimuli to electrical signals which can be conveyed along their surface for rapid spatial relay of the signal. At the end of the cell, the electrical signal is transduced to a chemical signal to convey the signal across the gap between the cells. The postsynaptic cell transduces this chemical signal back to an electrical signal

Learning objective Understand the equilibrium potential

arising from the balance between electrical force and diffusion as modeled by the Goldman-Hodgkin-Katz equation.

Explain the physical and chemical bases for the cellular membrane resting potential and the action potential.


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Solvent drag: External forces can alter the diffusive flux


A generic force, f:

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An unknown force produces a flux:


When both diffusion and an external force work:

If the force is electrical:

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The Electrochemical Potential and Free Energy

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J:flux D:difussion coefficient C: concentration K: Boltzman’s constant = R/No Z: valence of the particle (+/- integer) e: charge of the electron E:Electric field (electric force per unit charge)


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The electrochemical potential

F=Nof

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The trident is the electrical potential energy (volts) R=8.314 J mol-1 K-1 F’:faraday=9.649x104 C mol-1

The electrochemical potential is the Gibbs Free Energy per mole when pressure and temperature are constant during diffusion


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Coupling of spontaneous process with non-spontaneous processes


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Fig. 1.7.1 Coupling of spontaneous process with non-spontaneous processes

Fig. 1.7.2 Hydrolysis of ATP


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Activity Heart cells contain a Na-Ca exchanger protein in their cellular membrane with a stoichiometry of exchanging 3 molecules of sodium per molecule of calcium. A. Calculate the free energy for the reaction Ca2+

outside Ca2+inside

Given that the free energy of transport of Ca2+ across the sarcolemma of the heart cell can be calculated from the following conditions during the rest phase of the heart beat: [Ca2+]outside = 1.2 x 10-3 M; [Ca2+]inside = 0.1 x 10-8 M; Em = -0.085 V B. Calculate the free energy for the reaction Na+

outside Na2+inside

Given the following conditions during the rest phase of the heart: [Na+]outside = 145 x 10-3 M; [Na+]inside = 14 x 10-3 M; Em = -0.085 V What is the overall free energy change per mole of exchange?


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A. The free energy of transport of Ca2+ across the sarcolemma of the heart cell can be calculated from the following conditions during the rest phase of the heart beat: [Ca2+]o = 1.2 x 10-3 M [Ca2+]i = 0.1 x 10-8 M Em = -0.085 volts Calculate the free energy for the reaction Caout Y Cain . Recall that R= 8.314 joul mol-1°K-1, T = 310 °K, T = 9.649 x 104 coulombs mol-1. Remember that Ca2+ has two electrical charges per atom. The free energy change per mole for the reaction Caout Y Cain is given by Which can be simplified to Inserting in the values, we calculate ):Cao6Cai = 8.314 joules mol-1 °K-1 x 310 °K x ln [0.1 x 10-8 M /1.2 x 10-3 M]+ 2 x 96490 coulombs mol-1 x (-0.085 volts) = -36.08 kj mol-1 - 16.4 kj mol-1 = -52.48 kj mol-1 B. Calculate the free energy for the reaction Nain Y Naout for the following conditions during the rest phase of the heart: [Na+]o = 145 x 10-3 M [Na+]i = 14 x 10-3 M Em = -0.085 volts The free energy change per mole of Na+ for the reaction Nain Y Naout is calculated by a process entirely analogous to that performed in part A. The results give Which is simplified to Inserting the appropriate numbers here we get ):Nai6Nao = 8.314 joules mol-1 °K-1 x 310 °K x ln [145 x 10-3 M /14 x 10-3 M] + 1 x 96490 coulombs mol-1 x (0.085 volts) = 6.02 kj mol-1 + 8.2 kj mol-1 = 14.22 kj mol-1 C. Which way does the Na-Ca exchange proceed at rest? The Na-Ca exchanger exchanges 3 Na+ for each Ca2+. The overall free energy change per mole of exchange is given as Which is 3 x 14.22 kj mol-1 - 52.48 kj mol-1 = -9.82 kj mol-1 Since this is negative, the reaction proceeds in the direction noted: Ca2+ enters the cell and Na+ leaves it.

Light Spectrophotometry


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The absorbance is defined as: Awavelength = log ( I0 / I ) I0: incident light intensity, I : transmitted light intensity.

Beer-Lambert Law: Awavelength = λ C d A: absorbance, λ: molar extinction coefficient (M-1), C: concentration, d :pathlength.


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Fig. 1.PS2.1. Light path in a single-beam spectrophotometer. The view is from above. Light from a source is collimited (making a narrow beam) and passed through a monochromator that selects a narrow band of wavelength of light to be passed through the sample. A photomultiplier tube (PMT) detects the light and measures its intensity. Comparison of this intensity, I, to the intensity when the sample is missing, I0, allows calculation of the absorbance. Absorbance is recorded with time or as a function of wavelength.

Activity: ATP concentration You need to make 250 mL of a stock solution of 0.1 M Na2 ATP. Its formula weight is 605.2 g mol-1. How much Na2ATP should you weigh out? From this stock solution, you take 25 µL and dilute to 100 mL. What absorbance at 259 nm do you expect of the final diluted solution, if you made it up correctly,? Note that the molar extinction coefficient of ATP at a wavelength of incident light of 259 nm λ259 = 15.4 x 103 M-1cm-1.


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Amount of ATP = C x V = 0.1 mol L-1 x 0.25 L = 0.025 mol 0.025 mol x 605.2 g mol-1 = 15.13 g Dilution gives C1 x V1 = C2 x V2 ; 0.025 mL x 0.1 M = 100 mL x CATP CATP = 0.025 mL x 0.1 M / 100 mL = 25 x 10-6 M A = lambda C l = 15.4 x 103 M -1 x 25 x 10-6 M = 0.385

Activity: ATP is hydrolyzed by ATPases The progress of the reaction can be followed spectrophotometrically by the change in absorbance of NADH. What is the activity of the Ca-ATPase in units of µ mol min-1 mg-1 if the progress of the reaction was followed at A340? λ340 NAD+ is negligible λ340 NADH = 6.2 x 103 M-1cm-1 [Ca-ATPase] = 0.22 mg mL-1 A340 = 0.65 @ t=0 min A340 = 0.455 @ t=2.0 min.


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The decrease in [NADH] is given as ( 0.65 - 0.455)/ 6.2 x 103 M-1 = 0.0315 x 10-3 M = 0.0315 x 10-6 mol mL-1, but this decrease occurs in 2 min. So the rate of decrease is 0.0157 x 10-6 mol mL-1 min-1. The Ca-ATPase activity is obtained by dividing this rate of decrease in NADH by the amount of protein, which was 0.22 mg mL-1. So we have 0.0157 x 10-6 mol min-1 mL-1 / 0.22 mg mL-1 = 0.0715 x 10-6 mol min-1 mg-1


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The Origin of the Resting Membrane Potential: Forces generating the electrolyte equilibrium


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Chapter 3.1 THE ORIGIN OF THE RESTING MEMBRANE POTENTIAL Fig. 3.1.1 Forces generating the Na equilibrium 1) An imbalance of charged ions across the membrane generates the resting polarization (i.e., the resting potential). The charge difference is localized to the plasma membrane. The membrane itself acts as a capacitor that builds up excess charged ions. 2) Each bulk solution: extracellular and intracellular is approximately net neutral, (i.e., has an equal balance of + and – ions). The exception is the relatively few unbalanced charges held on/near the membrane. 3) The amount of excess ions creating the charge difference on the plasma membrane is very small, relatively speaking. For example, the -70 mV in charge may be generated by a deficit of < 0.001% of total potassium ions inside the cell. This is why, when ion channels open and ions move in or out during an action potential, the overall ionic concentrations inside and outside the cell do not change (for all purposes, are immeasurable).


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The equilibrium potential arises when the flux between the

electrical force and diffusion is

balanced

Nernst equation


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Activity: Sodium’s equilibrium potential Given that the concentration outside the cell of Na+ is 145 mM and the concentration inside is 12 mM, calculate the equilibrium potential for Na+. Note that the potential inside will be positive, as it should be to impede further influx of a positive ion.

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Given that the initial concentration of Na+ is 145 mM and the final concentration is 12 mM, calculate the equilibrium potential for Na+. Note that the potential inside is positive, as it should be to impede further influx of a positive ion. Equation 3.1.8


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Activity: Potassium’s equilibrium potential Given that the concentration outside of the cell of K+ is 4 mM and the concentration inside is 155 mM, calculate the equilibrium potential for K+. Note that the potential inside will be strongly negative inside.

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Equation 3.1.9


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Activity: Chloride’s equilibrium potential Given that the concentration outside of the cell of Cl- is 100 mM and the concentration inside is 5 mM, calculate the equilibrium potential for Cl-. Note that since the ion has a negative charge, z=-1.

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Equation 3.1.10

BUT the membrane of muscle cells is permeable to Na, K, and Cl at once!


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Fig. 3.1.2. Equilibrium potentials for Na, K, and Cl in a muscle cell

The Goldman-Hodgkin-Katz current equation It elates the current carried by each ion to

its concentration on both sides of the membrane and to the membrane potential


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At steady-state, the resting membrane potential is:

GHK equation

ACTION POTENTIAL


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• Electrically excitable cells produce a difference of potential across their cell membranes

• An action potential is defined as a non-linear amplification of a depolarization due to positive and negative feedback from voltage gated ion channels

• Action potentials do not spread smoothly down a myelinated axon (which is the majority of axons in the brain). Instead only a small region of the axon not covered by the myelin stealth undergoes an action potential.

• The resulting depolarization spreads to the next region (node of Ranvier) where another axon potential is elicited, and so on. This is called saltatory conduction.

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The difference in membrane potential is accompanied by a flow of current longitudinally through the intracellular and extracellular media, as well as current flow across the cell membrane along its entire length. The spatial coupling of current flow through neighboring segments of the membrane results in the propagation of the action potential along the cell surface.

Location of the motor neuron in the spinal cord


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Fig. 3.2.1.

Parts of the motor neuron


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Fig. 3.2.2.

Action potential


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Session 2: Physiology of excitable cells

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Transcript of Session 2: Physiology of excitable cells