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Cellular Neuroscience (207) Ian Parker Lecture # 1 - Enough (but not too much!) electronics to call yourself a cellular neurophysiologist

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Presentation on theme: "Cellular Neuroscience (207) Ian Parker Lecture # 1 - Enough (but not too much!) electronics to call yourself a cellular neurophysiologist"— Presentation transcript:

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 battery V Current I resistor R V = IR 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  I = 1A for R = 1 k  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 R1 R2 V I Total R = R1 + R2 I = V / (R1 + R2) R1 R2 I = I1 + I2 I1 I2 1/ total R = 1/R + 1/ R2

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

6 Voltage dividers R1 R2 V E E - V * R2 /(R1 + R2) [ If V = 1 V, R1 = 9 k  and R2 = 1 k  what  is E? : what current flows through R1?]

7 Capacitance Capacitor - two conductors separated by an insulating gap (dielectric) 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 e.g. 2 metal plates separated by an air gap 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 V E C R switch Switch closed Voltage rises exponentially from zero to V with time constant   is time for change to 1/e  of final voltage ( e = 2.71828…)  (sec) = R (  ) * C (F)  [what is  if R = 1 M , C = 1  F?]

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 V C switch E R 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 1 cm Resistance of 1 cm 2 of membrane (R m ) R m of a lipid bilayer >10 6  cm 2 But membrane channels can greatly increase the membrane conductance

12 Specific membrane capacitance membrane extracellular fluid 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 cm 2 of membrane : C m ) C m ~ 1  F cm -2 for cell membranes

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

14 A neuron as an RC circuit Inject current (I) Record voltage (V) cell RmRm CmCm inside outside I Voltage changes exponentially with time constant  m

15  m = R m * C m So  m will be longer if R m is high “ “ “ “ “ and if C m is high We can directly measure R m and  m so we can calculate C m =  m / R m Given that C m ~ 1  F cm 2, we can then calculate the membrane area of the cell


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