Bioenergetics: ΔG in a metabolic pathway
Textbooks say “ATP is the energy currency of the cell.” The more precise statement is: ATP hydrolysis has a large negative ΔG under physiological conditions, and living systems couple this favourable ΔG to reactions that would not run on their own. But the whole framework rests on one arithmetic fact — the ΔG of a coupled reaction is the sum of the ΔGs of the constituent steps. This guide walks through what ΔG really means in a metabolic context, why ΔG°’ and ΔG differ by orders of magnitude, and how a pathway with two “unfavourable” steps can still spontaneously flow forward.
Prerequisites: Gibbs free energy in three intuitions and physiological buffers. The former gives you ΔG; the latter tells you why pH 7 is a special reference state.
Three ΔGs and why they disagree
Biochemistry uses three closely related but distinct free energies:
ΔG° — the standard free energy change at 298 K, 1 atm, and 1 M concentrations of every species. This is what you find in a physical-chemistry textbook.
ΔG°’ — the biochemical standard free energy change: 298 K, 1 atm, 1 M concentrations of everything except H⁺ (which is set to 10⁻⁷ M, i.e., pH 7). Water activity is fixed at 1. Divalent cations (Mg²⁺) are typically at their physiological concentration. This is the reference used in biochemistry tables (Lehninger 7e, Berg 9e).
ΔG — the actual free energy change under cellular conditions. Concentrations are whatever they are in the cell (typically μM to mM, not 1 M) and pH is 7.4 in cytoplasm.
The relationship:
where Q is the mass-action ratio [products]/[reactants] under actual cellular conditions.
The difference between ΔG°’ and ΔG is often huge. ATP hydrolysis has ΔG°’ = −30.5 kJ/mol (Lehninger 7e §13.2), but in a resting cell where [ATP] ≈ 5 mM, [ADP] ≈ 0.5 mM, and [P_i] ≈ 5 mM, the mass-action correction RT ln Q pushes ΔG to about −50 kJ/mol. In working muscle, the cellular ΔG for ATP hydrolysis can reach −60 kJ/mol. The tabulated ΔG°’ is a starting point; ΔG is the physiologically real value.
ATP hydrolysis: why it works
The reaction is:
ΔG°’ = −30.5 kJ/mol at 298 K, pH 7. Three factors combine to make this favourable (Berg 9e §15.3):
- Electrostatic repulsion. ATP⁴⁻ has four negative charges packed close together on the phosphate chain. Cleaving off a terminal phosphate relieves electrostatic strain.
- Resonance stabilisation. Inorganic phosphate P_i has three resonance structures at physiological pH; ATP’s terminal phosphate has fewer options because of the anhydride linkage. Products are entropically more stable.
- Hydration. P_i is more heavily hydrated than the terminal phosphate of ATP; the solvation change contributes to the favourable ΔG.
Note that “high-energy bond” is a misnomer. The P–O bond in ATP is not particularly weak; the products of hydrolysis are more stable than the reactants for the three reasons above. The bond does not “store energy”; the disequilibrium between ATP and ADP+P_i does.
Worked example — the cellular ΔG of ATP hydrolysis
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Under cellular conditions, ATP hydrolysis is 60% more exergonic than the tabulated ΔG°' suggests. The reason: [ATP] is ~10× [ADP], and [P_i] is comparable to [ATP], so Q ≈ 10⁻³ and the RT ln Q correction is a large negative number. Cells maintain this disequilibrium via a continuous flux of ATP synthesis (mitochondrial oxidative phosphorylation), which is what makes ATP hydrolysis a useful energy source in the first place.
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Coupling — how “unfavourable” reactions run
The first step of glycolysis is:
Positive ΔG°’. It should not run. But in cells, the reaction is catalysed by hexokinase and coupled to ATP hydrolysis:
The −30.5 kJ/mol of ATP hydrolysis pays for the +13.8 kJ/mol of glucose phosphorylation, leaving a net −16.7 kJ/mol. This is not a mystery: enzymes lower the activation barrier; the thermodynamics is set by the coupled ΔG.
The formal rule: ΔG is a state function, so ΔG_total = ΣΔG_individual for any sequence of steps. As long as the sum is negative, the pathway is spontaneous — even if some individual steps are positive.
Reading a pathway — glycolysis as an example
Glycolysis converts glucose to two pyruvate over ten steps, with net ΔG°’ ≈ −85 kJ/mol (Lehninger 7e Table 14-1). Not every step is negative:
| Step | Reaction | ΔG°’ (kJ/mol) | ΔG (kJ/mol, in cell) |
|---|---|---|---|
| 1 | Hex → G6P | −16.7 | −34 |
| 2 | G6P → F6P | +1.7 | −2.9 |
| 3 | F6P → F1,6BP | −14.2 | −19 |
| 4 | F1,6BP → 2 triose-P | +23.8 | −0.2 |
| 5 | DHAP → G3P | +7.5 | +2.4 |
| 6 | G3P → 1,3-BPG | +6.3 | −1.3 |
| 7 | 1,3-BPG → 3-PG | −18.8 | +0.1 |
| 8 | 3-PG → 2-PG | +4.4 | +0.8 |
| 9 | 2-PG → PEP | +7.5 | −0.1 |
| 10 | PEP → Pyr | −31.4 | −29.2 |
Two observations:
1. ΔG°’ can be positive while ΔG (cellular) is negative. Step 2 (phosphoglucose isomerase) has ΔG°’ = +1.7 kJ/mol but runs backward at equilibrium and forward in a cell only because F6P is drained by step 3, keeping [F6P]/[G6P] far from equilibrium.
2. Not every step has to be far from equilibrium. Steps 6, 7, 8, 9 are all “near equilibrium” (small ΔG), meaning they are freely reversible and their direction is determined by the concentration gradient rather than by the chemistry. The three irreversible regulatory steps (1, 3, 10) have highly negative ΔG under all conditions — these are the flux-control points.
The distinction matters. Cellular regulation of glycolysis operates almost exclusively on the three irreversible steps (via hexokinase, PFK-1, and pyruvate kinase). The near-equilibrium steps track wherever the metabolite concentrations push them.
Reversal and gluconeogenesis
Gluconeogenesis (making glucose from pyruvate) reverses glycolysis, but not step by step. The three irreversible glycolytic steps (1, 3, 10) have such large negative cellular ΔG that reversing them directly is thermodynamically impossible under physiological conditions. Gluconeogenesis uses different enzymes to bypass each irreversible step:
- Pyruvate → PEP is done via oxaloacetate (pyruvate carboxylase + PEPCK), using one ATP and one GTP.
- F1,6BP → F6P uses fructose 1,6-bisphosphatase (a hydrolase), not the reverse of PFK-1.
- G6P → glucose uses glucose 6-phosphatase (a hydrolase), not the reverse of hexokinase.
The net cost of running gluconeogenesis is 6 ATP-equivalents per glucose, compared to a net yield of 2 ATP per glucose from glycolysis. Both directions cannot run simultaneously (would be a futile cycle), so hormonal signals — glucagon and insulin — reciprocally regulate the two pathways at the shared irreversible steps.
Displacement from equilibrium — the ΔG shortcut
For a reaction A → B with equilibrium constant K_eq at 310 K, the ΔG in a cell is:
Interpretation: ΔG measures how far the reaction is from equilibrium. Q = K_eq gives ΔG = 0 (at equilibrium, no drive). Q < K_eq gives ΔG < 0 (product-forming). Q > K_eq gives ΔG > 0 (reactant-forming).
For a near-equilibrium step, Q ≈ K_eq and ΔG is small — the reaction can run either direction depending on which side is momentarily higher. For a far-from-equilibrium step, Q ≪ K_eq (product-forming) and ΔG is strongly negative — the reaction runs one way only until an enzyme is regulated.
Worked example — reading an equilibrium
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Even though ΔG°' for aldolase is +23.8 kJ/mol (very unfavourable), the cellular concentrations of F1,6BP, DHAP, and G3P are such that Q is close to K_eq, and ΔG under cellular conditions is nearly zero. Aldolase runs forward in glycolysis and backward in gluconeogenesis, driven purely by the small residual concentration disequilibrium at each direction of flux.
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Common pitfalls
Confusing “high-energy” with “unstable”. ATP is thermodynamically unstable relative to ADP + P_i, but kinetically stable — no non-catalysed decomposition on physiological timescales. This is why cells can store ATP at millimolar concentrations without it hydrolysing spontaneously.
Adding ΔG°’ values across a pathway to predict physiological direction. ΔG°’ is a reference-state quantity; cellular Q values push individual ΔGs far from ΔG°’. The direction of net flux is determined by ΔG under cellular conditions, not by ΔG°’.
Assuming every enzyme is at equilibrium. Some are (near-equilibrium enzymes in metabolic parlance); others are far from equilibrium (irreversible, flux-controlling). Regulation targets the latter class exclusively — trying to regulate a near-equilibrium enzyme has no effect on flux, only on the concentration ratio.
Ignoring compartmentalisation. Free energies depend on local concentrations. Mitochondrial-matrix pH is 7.9 (vs cytoplasmic 7.4), and [ATP]/[ADP] ratios differ between compartments. A cellular ΔG averaged over the whole cell is a fiction; real ΔGs are defined per compartment.
Practice
Problem 1 Easy The reaction glutamate + NH₃ → glutamine + H₂O has ΔG°' = +14.2 kJ/mol. In cells it runs forward because glutamine synthetase couples it to ATP hydrolysis. What is the ΔG°' of the coupled reaction?
Answer: ΔG°'(coupled) = +14.2 + (−30.5) = −16.3 kJ/mol. The coupled reaction is glutamate + NH₃ + ATP → glutamine + ADP + P_i, favourable overall. This is how nitrogen fixation into amino acids is 'paid for' by ATP hydrolysis.
Problem 2 Intermediate A reaction A ⇌ B has K_eq = 100. In a cell, [A] = 1 mM and [B] = 10 mM. What is ΔG at 310 K? Which direction does the reaction run?
Answer: ΔG = RT ln(Q/K_eq) = (8.314)(310) ln(10/100) = 2577 × ln(0.1) = −5934 J/mol = −5.9 kJ/mol. Because ΔG < 0, the reaction runs forward (A → B). Even though B/A = 10 (net product-loaded), it is still 10× less product-loaded than equilibrium would demand, so forward flux continues until Q = K_eq is reached, at which point [B]/[A] = 100.
Problem 3 Hard The 'phosphocreatine buffer' in muscle: phosphocreatine + ADP → creatine + ATP has ΔG°' = −12.5 kJ/mol. In resting muscle, [PCr] = 25 mM, [Cr] = 13 mM, [ATP] = 8.1 mM, [ADP] = 0.9 mM. Compute ΔG at 310 K and explain how this system responds to a burst of exercise.
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At rest, the phosphocreatine kinase reaction is well away from equilibrium in the forward direction (ΔG = −8.5 kJ/mol). When exercise depletes ATP, [ADP] rises sharply — even a 5-fold rise pushes Q down 5-fold and ΔG more negative, so PCr immediately dumps its phosphoryl group onto ADP to regenerate ATP. This 'buffer' can sustain maximal work for about 10 seconds; then glycolysis and oxidative phosphorylation take over. It is why sprinters run out of top-speed after ~100 m: they exhaust PCr, not ATP itself.
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Where this leads
Bioenergetics is the through-line of biochemistry:
- Oxidative phosphorylation — the electron transport chain builds a proton gradient whose free energy drives ATP synthesis via chemiosmosis (Mitchell 1961).
- Photosynthesis — light drives electron transport in the opposite direction, generating an equivalent gradient.
- Substrate cycles — futile-looking pairs of reactions (glycolysis / gluconeogenesis, fatty acid synthesis / β-oxidation) allow cells to switch metabolic direction cleanly.
- Metabolic control analysis — quantifies how sensitively pathway flux depends on each enzyme’s activity, extending the ΔG framework into a systems view.
Every one of them begins by asking: what is Q, what is K_eq, and how far is the reaction from equilibrium?
Sources
- Nelson & Cox, Lehninger Principles of Biochemistry, 7th ed. (2017), Ch. 13
- Berg, Tymoczko, Gatto & Stryer, Biochemistry, 9th ed. (2019), Ch. 15
- Voet, Voet & Pratt, Fundamentals of Biochemistry, 5th ed. (2016), Ch. 13-14
- Alberty, R. A., Thermodynamics of Biochemical Reactions (2003), Ch. 3
- MetaCyc / BRENDA — curated ΔG°' and ΔG values for metabolic reactions
- eQuilibrator (Noor et al., 2014) — computational ΔG' database at physiological conditions