SME
Biochemistry
Carbohydrates, lipids and membranes
Use this for “Ca²⁺ is 1.8 mM outside a cell and 0.0001 mM inside, and Δψ = −60 mV. Find ΔG for each mole of Ca²⁺ that enters at 37 °C.”
Diagram
V = −70 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 ≈ 3.63 mM: Ion 145 mM outside, 12 mM inside.
This is an example. Type your own values to start a new problem.
ΔG (kJ/mol)
=
R
T (K)
ln
cᵢₙ (mM)
cₒᵤₜ (mM)
+
z
F
×
(
Δψ (mV)
)
Gconc
Concentration part
calculated · kJ/mol
Gₑₗ
Charge part
calculated · kJ/mol
Formulas
Gconc
=
8.314
×
T
×
ln(cᵢₙ ÷ cₒᵤₜ)
1,000
−6.425
=
8.314
×
310.15
×
ln(12 ÷ 145)
1,000
Gₑₗ
=
z
×
96.485
×
Δψ
1,000
−6.754
=
1
×
96.485
×
−70
1,000
ΔG = Gconc + Gₑₗ
−13.18 = −6.425 + (−6.754)
Assumptions
•
The ion moves from outside the cell to inside; to move it out, swap the two concentrations and the sign of ΔG.
•
Δψ is the inside relative to the outside, in mV; z is the ion’s charge with its sign (Na⁺ +1, Ca²⁺ +2, Cl⁻ −1).
•
ΔG = RT ln(cᵢₙ ÷ cₒᵤₜ) + zFΔψ: the concentration part plus the charge part, with Δψ in volts (mV ÷ 1000), in kJ/mol.
•
R = 8.314 J/(mol·K) (÷ 1000 for kJ) and F = 96.485 kJ/(mol·V); body temperature, 37 °C, is 310.15 K.
Step-by-step
WE KNOW
cₒᵤₜ = 145 mM, cᵢₙ = 12 mM, z = +1, Δψ = −70 mV, T = 310.15 K
FIND
concentration part (Gconc), charge part (Gₑₗ), free energy per mole moved in (ΔG)
Step 1 · Find concentration part (Gconc)
Use Gconc = 8.314 × T × ln(cᵢₙ ÷ cₒᵤₜ) ÷ 1,000
Moving toward the lower concentration releases free energy: RT times ln of inside over outside, ÷ 1000 for kJ/mol.
Gconc
=
8.314
×
310.15
×
ln(12 ÷ 145)
1,000
Gconc
=
8.314
×
310.15
×
ln(0.08276)
1,000
Gconc
=
8.314
×
310.15
×
−2.492
1,000
Gconc
=
2,578.5871
×
−2.492
1,000
Gconc
=
−
6,426
1,000
Gconc = −6.425 kJ/mol
Step 2 · Find charge part (Gₑₗ)
Use Gₑₗ = z × 96.485 × Δψ ÷ 1,000
One
mole
of
the
ion
carries
z
times
F,
96.485
kJ
for
each
volt;
Δψ
in
volts
is
mV
1000
.
Gₑₗ
=
1
×
96.485
×
−70
1,000
Gₑₗ
=
−
6,753.95
1,000
Gₑₗ = −6.754 kJ/mol
Step 3 · Find free energy per mole moved in (ΔG)
Use ΔG = Gconc + Gₑₗ
The ion feels both pushes at once, so the two parts add.
ΔG = −6.425 + (−6.754)
ΔG = −13.18 kJ/mol → ΔG < 0: the ion moves in on its own, so an open channel is enough.
Check
Put the numbers back in. Both sides should match.
−6.425
=
8.314
×
310.15
×
ln(12 ÷ 145)
1,000
✓
−6.754
=
1
×
96.485
×
−70
1,000
✓
−13.18 = −6.425 + (−6.754) ✓
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Free energy of the proton gradient
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Equilibrium potential of one ion (Nernst)
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