SME
Cell & Molecular Biology · Problem type
The resting potential (Goldman)
Use this for “With these K⁺, Na⁺ and Cl⁻ levels and permeabilities 1, 0.04 and 0.45, what is the resting potential?”
Diagram
V = −67.3 mV: the inside is negative, so + charges line the outside face and − the inside.
One pair per 10 mV: 7 a side.
Dots on one scale, one dot ≈ 6.5: K⁺ 5 out, 140 in; Na⁺ 145 out, 15 in; Cl⁻ 110 out, 10 in.
This is an example. Type your own values to start a new problem.
Kₒ
K⁺ outside
example · mM
Kᵢ
K⁺ inside
example · mM
Naₒ
Na⁺ outside
example · mM
Naᵢ
Na⁺ inside
example · mM
Clₒ
Cl⁻ outside
example · mM
Clᵢ
Cl⁻ inside
example · mM
PNa
Na⁺ permeability
example
PCl
Cl⁻ permeability
example
Vₘ
Resting potential
calculated · mV
Formulas
Vₘ = 61.5 × log₁₀((Kₒ + PNa × Naₒ + PCl × Clᵢ) ÷ (Kᵢ + PNa × Naᵢ + PCl × Clₒ))
−67.3
=
61.5
×
log₁₀(
5 + 0.04 × 145 + 0.45 × 10
140 + 0.04 × 15 + 0.45 × 110
)
Assumptions
•
Permeabilities are relative to K⁺, so PK = 1; 61.5 mV is 2.303RT ÷ F at 37 °C.
•
Vₘ is the inside relative to the outside, with no net current across the membrane.
•
Chloride’s inside and outside swap places in the formula because its charge is −1.
Step-by-step
WE KNOW
Kₒ = 5 mM, Kᵢ = 140 mM, Naₒ = 145 mM, Naᵢ = 15 mM, Clₒ = 110 mM, Clᵢ = 10 mM, PNa = 0.04, PCl = 0.45
FIND
resting potential (Vₘ)
Step 1 · Find resting potential (Vₘ)
Use Vₘ = 61.5 × log₁₀((Kₒ + PNa × Naₒ + PCl × Clᵢ) ÷ (Kᵢ + PNa × Naᵢ + PCl × Clₒ))
Weight each ion by its permeability; chloride’s sides swap because it is negative.
Vₘ
=
61.5
×
log₁₀(
5 + 0.04 × 145 + 0.45 × 10
140 + 0.04 × 15 + 0.45 × 110
)
Vₘ
=
61.5
×
log₁₀(
5 + 5.8 + 4.5
140 + 0.6 + 49.5
)
Vₘ
=
61.5
×
log₁₀(
10.8 + 4.5
140.6 + 49.5
)
Vₘ
=
61.5
×
log₁₀(
15.3
190.1
)
Vₘ = 61.5 × log₁₀(0.080484)
Vₘ = 61.5 × (−1.0943)
Vₘ = −67.3 mV
Check
Put the numbers back in. Both sides should match.
−67.3
=
61.5
×
log₁₀(
5 + 0.04 × 145 + 0.45 × 10
140 + 0.04 × 15 + 0.45 × 110
)
✓
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