Question

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.4 µm ✕ 1.4 µm in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 86.0 mV.
a) Determine the amount of current that flows through this portion of the membrane
b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do no change
increase by factor of 2
decrease by factor of 8
decrease by factor of 2
decrease by a factor of 4
increase by factor of 4

Answer 1

Answer:

Part a: The current is 1.49x10⁻¹² A

Part b: The current is decreased by a factor of 4.

Explanation:

Part a

The area is given as

A =(1.4*10⁻⁶)² =1.96*10⁻¹² m²

The resistance is given as where resistivity of the membrane material is 1.30 x 10⁷ ohms*m

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(1.96*10⁻¹²)

R=4.97x10¹⁰ ohms

So the resistance is 4.97x10¹⁰ ohms.

I=V/R =86.0x10⁻³/5.77x10¹⁰

I=1.49x10^⁻¹² A

So the current is 1.49x10⁻¹² A

b)

S=So/2=1.4/2 =0.7μm

A=(0.7*10^-6)^2=4.9*10⁻¹³ m²

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(4.9*10⁻¹³)

R=1.98x10¹¹ ohms

So the resistance is 1.98x10¹¹ ohms.

I=V/R =86.0x10⁻³/1.98x10¹¹

I2=4.52x10^⁻¹³ A

So the ratio of the new current vs the old current is as

I2/I=4.52x10^⁻¹³/1.49x10⁻¹²=0.25

So the current is decreased by a factor of 4.