Nerve cells exchange sodium (Na+) and potassium (K+) ions across their cell memb

Question

Nerve cells exchange sodium (Na+) and potassium (K+) ions across their cell membranes to maintain a voltage across the membrane. A nerve cell that is not involved in sending a nerve signal has a 70.0 mV voltage across the membrane, which is known as the “resting potential”. The inside of the cell is at a lower potential than the outside.

(a) This voltage tells us that there is charge separation so that there are charges collected on the outside of the membrane and other charges on the inside. Is the inside of the membrane positively or negatively charged?

(b) Suppose the thickness of the cell membrane is 8.00 nm (this is fairly typical). What is the strength of the E-field inside the cell membrane (you may be surprised at how large it is...).

(c) The membrane is not a perfect insulator. So some current flows through it constantly. Thus, we can think of the membrane as a (very oddly shaped) wire. For a spherical nerve cell that is 0.0600 mm in diameter what is the cross-sectional area of this “wire”?

(d) The resistivity of a cell membrane is about 3.6×107 ·m. Use this and the information from earlier in the question to find the resistance of the cell membrane and then find the current that conducts across the membrane when it is at its resting potential.

Explanation / Answer

a) In the rest state, the inside of the nerve cell membrane is negative charged with respect to the outside ( about -70 millivolts). The voltage arises from differences in concentration of the electrolyte ions K+ and Na+.

b)   strength of the E-field inside the cell membrane   = 70 * 10-3/8 * 10-9

                                                                                      = 8.75 * 106 N/C

c)   cross-sectional area of this “wire" = 3.14 * 0.03 * 10-3 * 0.03 * 10-3

                                                              = 2.826 * 10-9 m2

d)     resistance   =   3.6×107 * 8 * 10-9/2.826 * 10-9

                                        =   10.19 * 107   ohm

     => current =   70 * 10-3/ 10.19 * 107

                                     =   6.87 * 10-10 A

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