How Does a Nerve Cell Membrane Model Circuit Behave?

[muscles00gt]

Sorry if this is in the wrong area, but I'm having a bit of difficulty on one of my homework problems.

A simple model of the membrane of a nerve cell is shown in the figure at the right (seen below). It consists of 2 batteries (ion pumps) with voltages V1=100mV and V2=50mV. The resistance to flow across the membrane is represented by two resistors with resistances R1=10KΩ and R2=90KΩ. The variability is represented by a switch, SW1. Four points are labeled by the letters a-d. Point b represents outside of membrane and point d inside the membrane.

Picture of circuit:

A) What is the voltage difference across the membrane when the switch is open?

B) What is the current flowing around the loop when the switch is closed?

C) What is the voltage drop across the resistor R1 when switch is open? closed?

D) What is the voltage drop across the resistor R2 when switch is open? closed?

E) What is the voltage difference across the membrane when switch is closed?

F) If the locations of resistance R1 and R2 were reversed, would the voltages across the cell membranes be different?

 

[denverdoc]

so what are your thoughts abouit this, and what do you know?

One battery is probably the Potassium gradient (have you learned about Nernst potentials?) while the other is tied to the Na gradient. the switch is the fact that the sodium channels are normally closed, but if raised to a certain potential within a certain time window, will open (opposite the convention for switch terminology) and the transmembrane potential shoots up from say a -70mV resting state to +40mV

 

[m2003]

I would be happy to help you with your homework problem! Here are the answers to the questions:

A) When the switch is open, no current is flowing through the circuit, so the voltage difference across the membrane is 0V.

B) When the switch is closed, the current flowing around the loop is V1/R1 + V2/R2 = (100mV/10KΩ) + (50mV/90KΩ) = 0.01mA + 0.000556mA = 0.010556mA.

C) When the switch is open, no current is flowing through R1, so the voltage drop across it is 0V. When the switch is closed, the voltage drop across R1 is (0.01mA)(10KΩ) = 0.1V.

D) When the switch is open, no current is flowing through R2, so the voltage drop across it is 0V. When the switch is closed, the voltage drop across R2 is (0.000556mA)(90KΩ) = 0.05V.

E) When the switch is closed, the voltage difference across the membrane is V1 - V2 = 100mV - 50mV = 50mV.

F) If the locations of resistance R1 and R2 were reversed, the voltages across the cell membranes would be different because the current flows in different paths in the circuit. This would result in a different voltage drop across each resistor and therefore a different voltage difference across the membrane.