Question

Three ions are permeable in a cell, A, B, and C. Ea = +50 mV, Eb = -50 mV, Ec = 0 mV. 500 ion A channels are open. 10,000 ion B channels are open, and 20,000 ion C channels are open. What is the approximate membrane potential of the cell?

A) 0 to +50 mV
B) -50 mV
C) 0 to - 50 mV
D) +50 mV
E) 0

Answer 1

The approximate membrane potential of the cell is -3.33 mV.

Step-by-step explanation:

The membrane potential of the cell can be calculated by taking the weighted average of the individual ion potentials based on the number of open channels for each ion. In this case, the approximate membrane potential can be calculated as (500 * 50 + 10000 * -50 + 20000 * 0) / (500 + 10000 + 20000) = -3.33 mV. Therefore, the closest option to the approximate membrane potential of the cell is C) 0 to -50 mV.