Question

A single ion channel is selectively permeable to K+ and has a resistance of 1.0 GΩ. During an experiment the channel is open for approximately 1.0 ms while the voltage across the channel is maintained at +80.0 mV with a patch clamp. How many K+ ions travel through the channel?

Answer 1

Answer:

5.0 × 10⁵

Explanation:

Quantity of charge Q = It

V= IR

I = V / R

Q = Vt/R

Q = ne

ne = Vt/R

n, number of K⁺ ions = Vt / Re

V = 80 × 10 ⁻³

t = 1 × 10⁻³ s

R = 1  × 10⁹ Ω

e = 1.6 × 10⁻¹⁶C

n, number of K⁺ ions = (80 × 10 ⁻³ V ×  1 × 10⁻³ s ) / (1  × 10⁹ Ω ×  1.6 × 10⁻¹⁹C) = 5.0 × 10⁵

Answer 2

Based on the resistance of the single ion channel and the amount of time it was open, the number of K+ ions that travel through are 500,000 ions.

How many K+ ions travel through?

First find the current:

= Voltage / Resistance

= (80 x 10⁻³) / (1 x 10⁹)

= 8 x 10⁻⁹ A

Then find the charge:

= Current x time

= 8 x 10⁻⁹ x 1 x 10 ⁻³

= 8 x 10 ⁻¹⁴

The number of K+ ions that escaped is:

= Charge / Charge of K+ ions

= (8 x 10 ⁻¹⁴) / (1.6 x 10⁻¹⁹)

= 500,000 ions

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