SME
Principles of Biology I
Metabolism
Use this for “Glutamate + NH₃ → glutamine has ΔG = +14.2 kJ/mol. Does it run when one ATP is hydrolyzed with it?”
Diagram
ΔG₁: Glu + NH₃ + ATP → Gln + ATP, ++14.2 kJ/mol
ΔG(ATP): Gln + ATP → Gln + ADP + Pᵢ, −30.5 kJ/mol
ΔG = ΔG₁ + ΔG(ATP)
ΔG = +14.2 + (−30.5) = −16.3 kJ/mol
Down the ladder: ΔG < 0, so it runs forward on its own (exergonic).
This is an example. Type your own values to start a new problem.
ΔG (kJ/mol)
=
ΔG₁ (kJ/mol)
+
n
×
(−30.5)
ΔG(ATP)
Free energy from ATP
calculated · kJ/mol
Formulas
ΔG(ATP)
=
n
×
(−30.5)
−30.5 = 1 × (−30.5)
ΔG = ΔG₁ + ΔG(ATP)
−16.3 = 14.2 + (−30.5)
Assumptions
•
Free energies add when the reactions share an intermediate; a total ΔG below 0 runs forward on its own.
•
Each ATP hydrolyzed to ADP + Pᵢ gives ΔG = −30.5 kJ/mol.
•
A ΔG above 0 is uphill: alone, that reaction does not run forward.
Step-by-step
WE KNOW
ΔG₁ = +14.2 kJ/mol, n = 1
FIND
free energy from ATP (ΔG(ATP)), total free energy change (ΔG)
Step 1 · Find free energy from ATP (ΔG(ATP))
Use ΔG(ATP) = n × (−30.5)
Each ATP gives −30.5 kJ/mol.
ΔG(ATP) = 1 × (−30.5)
ΔG(ATP) = −30.5 kJ/mol
Step 2 · Find total free energy change (ΔG)
Use ΔG = ΔG₁ + ΔG(ATP)
The coupled reactions’ free energies add.
ΔG = 14.2 + (−30.5)
ΔG = −16.3 kJ/mol → Below 0: it runs forward on its own.
Check
Put the numbers back in. Both sides should match.
−30.5 = 1 × (−30.5) ✓
−16.3 = 14.2 + (−30.5) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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