A cardiac cell can be satisfactorily represented by the shape of a brick with ed

Question

A cardiac cell can be satisfactorily represented by the shape of a brick with edges defined by a length, width and depth. The cell's surface area can be approximated by that of the four long sides excluding the ends (width x depth). Assume such a cell with length 90 mu m, width 5 mu m and depth 8 mu m. At rest the membrane resistivity is 20, 000 Ohm cm^2 and capacitance is 1.1 mu F/cm^2. Suppose an amount of charge is instantly moved across the membrane and changes the transmembrane voltage by 20 mV. How much charge was moved? If the cell membrane functions as passive (parallel resistor-capacitor combination) what is the time constant of the cell? Suppose the membrane potential is raised to 20 mV and the cell is assumed to remain passive (i.e. no change in cell resistance and capacitance), how long will it take for the 20 mV initial rise to decay to 1 mV?

Explanation / Answer

a)

area of the cell

A=(90*10-6)(5*10-6)=4.5*10-10 m2

Capacitance

C=(4.5*10-10)(1.1*10-6/10-4)=4.95*10-12 F

Charge

Q=CV=(4.95*10-12)(20*10-3)=9.9*10-14 C

b)

Resistance

R=pL/A=(20000*10-4)*(8*10-6)/(4.5*10-10)=35555.6 ohms

Time constant

T=RC=35555.6*4.95*10-12

T=1.76*10-7 s

c)

In a RC circuit voltage as a function of time is

V=Voe-t/T

1=20e-t/(1.76*10^-7)

Ln(0.05)=-t/1.76*10-7

t=5.27*10-7 s

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