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Cellular Neuroscience (207) Ian Parker

Cellular Neuroscience (207) Ian Parker. Lecture # 1 - Enough (but not too much!) electronics to call yourself a cellular neurophysiologist. http://parkerlab.bio.uci.edu. Ohm’s Law. Current I. V = IR. resistor R. battery V. V (Volts) - electrical driving force (water pressure)

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Cellular Neuroscience (207) Ian Parker

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  1. Cellular Neuroscience (207)Ian Parker Lecture # 1 - Enough (but not too much!) electronics to call yourself a cellular neurophysiologist http://parkerlab.bio.uci.edu

  2. Ohm’s Law Current I V = IR resistor R battery V V (Volts) - electrical driving force(water pressure) [voltage, potential, potential difference, p.d. are all synonyms] I (Amperes) - electrical current flow(gallons per minute) R (Ohms) - resistance(how narrow the pipe is) R = V/I so, if V = 1 volt for R = 1 W I = 1A for R = 1 k W I = 10^-3 A (1 mA)

  3. Charge Charge = amount of electricity(number of electrons : gallons of water) = current * time 1 A * 1 sec = 1 Coulomb (C) [How many electrons are there in a Coulomb??]

  4. Resistors in series and parallel I R1 Total R = R1 + R2 I = V / (R1 + R2) V R2 1/ total R = 1/R + 1/ R2 R2 R1 I1 I2 I = I1 + I2

  5. Conductance Conductance is the reciprocal of resistance (i.e. how easily something conducts electricity) Conductance (G) = 1/R Unit : Siemen (S) = 1/ 1W total conductance G = G1 + G2 From Ohms law I = V / R So I = V * G Itotal = V * (G1 + G2) G2 G1 I1 I2 I = I1 + I2

  6. Voltage dividers E - V * R2 /(R1 + R2) R1 V R2 [ If V = 1 V, R1 = 9 kW and R2 = 1 kW whatis E? : what current flows through R1?] E

  7. Capacitance Capacitor - two conductors separated by an insulating gap (dielectric) e.g. 2 metal plates separated by an air gap Capacitance (C) increases as; 1. The area of the plates is increased 2. The separation between the plates is decreased 3. The dielectric constant of the insulator is increased Capacitors store electricity, but cannot pass a steady current Unit : Farad (F) 1 F = capacitor that will store 1 Coulomb when connected to 1 V Charge (q) stored on a capacitor = C * V

  8. RC (resistor/capacitor) circuits 1. Low-pass RC circuit switch R E V C • Voltage rises exponentially from zero to V with time constant t • t is time for change to 1/e of final voltage ( e = 2.71828…) • t (sec) = R (W) * C (F) • [what is t if R = 1 MW, C = 1 mF?] Switch closed

  9. The effect of a low-pass circuit is to pass steady or slowly changing signals while filtering out rapidly changing signals brief change in voltage longer change in voltage

  10. RC (resistor/capacitor circuits) 2. High-pass RC circuit switch E C R V • Output voltage instantly rises to match input voltage, then decays exponentially. • Time constant of decay • = RC Effect is to block rapidly-changing voltages (capacitor is an insulator), but pass rapidly changing signals

  11. What does all this mean for a NEURON? The cell membrane (lipid bilayer) acts as a very good insulator, but has high capacitance. Specific membrane resistance Resistance of 1 cm2 of membrane (Rm) 1 cm Rm of a lipid bilayer >106W cm2 But membrane channels can greatly increase the membrane conductance

  12. Specific membrane capacitance extracellular fluid membrane intracellular fluid The insulating cell membrane (dielectric) separates two good conductors (the fluids outside and inside the cell), thus forming a capacitor. Because the membrane is so thin (ca. 7.5 nm), the membrane acts as a very good capacitor. Specific capacitance (capacitance of 1 cm2 of membrane : Cm) Cm ~ 1 mF cm-2 for cell membranes

  13. Input resistance of a cell Record voltage (V) Inject current (I) Input resistance Rin = V/I cell Rin decreases with increasing size of cell (increasing membrane area) Rin increases with increasing specific membrane resistance [If I = 10 nA and V = 5 mV, what is Rm ???]

  14. A neuron as an RC circuit Record voltage (V) Inject current (I) I cell inside Rm Cm outside Voltage changes exponentially with time constant tm

  15. tm = Rm * Cm • Sotm will be longer if Rm is high • “ “ “ “ “ and if Cm is high • We can directly measure Rm and tm • so we can calculate Cm = tm / Rm • Given that Cm ~ 1 mF cm2,we can thencalculate the membrane area of the cell

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