Download Example 18-5 Resistance of a Potassium Channel

 Example 18-5 Resistance of a Potassium Channel
Using an instrument called a patch clamp, scientists are able to control the electrical potential difference across a tiny
patch of cell membrane and measure the current through individual potassium channels. In one experiment a voltage of 0.120 V was applied across a patch of membrane, and as a result K+ ions carried 6.60 pA of current (1 pA =
1 picoampere = 10212 A). What is the resistance of this K+ channel?
Set Up
Whether the mobile charges are negative
electrons or positive ions, we can use Equation 18-9 to relate potential difference
(voltage), current, and resistance.
Relationship among potential
difference, current, and
resistance:
V = iR
(18-9)
V=0
K+
membrane
K+
V = 0.120 V
Solve
Rewrite Equation 18-9 to solve for the
­resistance.
From Equation 18-9,
R =
V
i
The current is i = 6.60 pA = 6.60 * 10212 A, so
R =
Reflect
0.120 V
= 1.82 * 1010 
6.60 * 10-12 A
This is an immensely high resistance compared to the values found in circuits like that shown in Figure 18-6. Yet it is very
characteristic of the channels in cell membranes, which typically have resistances in the range from 109  to 1011 .
A potassium channel is only about 1029 m wide—so small that K+ ions must pass through it in single file—so it’s not
surprising that the resistance is so high.
Although the current through a single potassium channel is very small, the electric field required to produce that
current is tremendous. The thickness of the cell membrane is only about 7.5 nm = 7.5 * 1029 m. Can you use Equation 18-5 to show that a potential difference of 0.120 V across this distance corresponds to an electric field magnitude of
1.6 * 107 V>m? (By comparison, the electric field inside the wires of a flashlight is only about 10 V>m.)