Bioenergetics
First 26 pages :
High-yield summary — Bioenergetics
Based on your lecture PDF, pp. 1–26. The uploaded file covers thermodynamics, reaction coupling, and the beginning of ATP hydrolysis; the oxidation–reduction section listed in the outline is not included.
MED 103-Medical Biochemistry-Bioenergetics-2026-2027.pdf
1. Core thermodynamic concepts
Bioenergetics: the quantitative study of energy transformations in living cells.
First law: energy is conserved; it can be transferred or transformed.
Second law: spontaneous processes increase the total entropy of the universe.
Cells are open systems: they exchange matter and energy with their surroundings. They can maintain internal order because the total entropy of the cell plus its surroundings increases.
Correction to slide 5: an isolated system exchanges neither matter nor energy; a closed system can exchange energy but not matter.
Quantity | Meaning | Sign to remember |
ΔG: Gibbs free-energy change | Determines whether the forward reaction is thermodynamically favorable at constant temperature and pressure | Negative = exergonic |
ΔH: enthalpy change | Heat absorbed or released at constant pressure | Negative = exothermic |
ΔS: entropy change | Change in energy dispersal/disorder | Positive = increased entropy |
\boxed{\Delta G=\Delta H-T\Delta S}
A negative ΔH and a positive ΔS both favor a negative ΔG. Temperature must be in kelvin.
2. ΔG determines reaction direction
Actual ΔG | Meaning |
ΔG < 0 | Forward reaction is spontaneous/exergonic |
ΔG > 0 | Forward reaction is endergonic; reverse direction is favorable |
ΔG = 0 | Equilibrium; no net reaction or net work available |
Spontaneous does not mean fast. A favorable reaction may proceed extremely slowly because of a high activation-energy barrier.
At equilibrium, forward and reverse reaction rates are equal. Reactant and product concentrations are constant, but not necessarily equal.
3. Standard versus actual free-energy change — the central distinction
ΔG°′ | ΔG |
Standard transformed free-energy change | Actual free-energy change |
Uses the biochemical standard state, including pH 7 | Uses the actual reactant and product concentrations |
Fixed for a given reaction under specified conditions | Changes as concentrations change |
Indicates equilibrium preference | Determines the direction under current conditions |
The prime ′ denotes the biochemical standard state. The lecture uses 25°C = 298 K.
For aA+bB\rightleftharpoons cC+dD:
Q=\frac{[C]^c[D]^d}{[A]^a[B]^b}
\boxed{\Delta G=\Delta G^{\circ\prime}+RT\ln Q}
Q uses current concentrations.
K′eq uses equilibrium concentrations.
Water is omitted from these expressions because its activity is treated as constant.
Key implication: a reaction with positive ΔG°′ can still proceed forward if its actual ΔG is negative. Removing products lowers Q, making forward conversion more favorable.
4. Relationship between ΔG°′ and equilibrium
At equilibrium, \Delta G=0 and Q=K'_{eq}:
\boxed{\Delta G^{\circ\prime}=-RT\ln K'_{eq}}
K'_{eq} | ΔG°′ | Equilibrium favors |
> 1 | Negative | Products |
= 1 | Zero | Neither side |
< 1 | Positive | Reactants |
Use R=8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}, and keep energy units consistent.
Lecture example:
\text{Glucose 1-phosphate}\rightleftharpoons\text{Glucose 6-phosphate}
At equilibrium: 1 mM glucose 1-phosphate and 19 mM glucose 6-phosphate.
K'_{eq}=19,\qquad \Delta G^{\circ\prime}=-7.3\ \mathrm{kJ/mol}
Reversing the reaction gives ΔG°′ = +7.3 kJ/mol.
5. Reaction coupling — how ATP drives unfavorable reactions
Free-energy changes add; equilibrium constants multiply.
For A\rightleftharpoons B and B\rightleftharpoons C:
\boxed{\Delta G^{\circ\prime}_{total}
=\Delta G^{\circ\prime}_1+\Delta G^{\circ\prime}_2}
\boxed{K'_{total}=K'_1K'_2}
An unfavorable reaction can proceed when it is mechanistically coupled to a sufficiently favorable reaction.
Reaction | ΔG°′ |
Glucose + Pi → glucose 6-phosphate + H₂O | +13.8 kJ/mol |
ATP + H₂O → ADP + Pi | −30.5 kJ/mol |
Net: glucose + ATP → glucose 6-phosphate + ADP | −16.7 kJ/mol |
ATP transfers its phosphoryl group to glucose. The net reaction is favorable under biochemical standard conditions.
The lecture’s equilibrium calculation gives:
K'_{total}=(3.9\times10^{-3})(2.0\times10^5)
=7.8\times10^2
6. ATP: function and favorable hydrolysis
Energy from nutrient catabolism is used to form ATP from ADP + Pi. ATP then supports:
Biosynthesis
Transport against concentration gradients
Mechanical work
\text{ATP}+\mathrm{H_2O}\rightarrow\text{ADP}+P_i
\qquad
\boxed{\Delta G^{\circ\prime}=-30.5\ \mathrm{kJ/mol}}
The slides emphasize three reasons ATP hydrolysis is favorable:
Reduced electrostatic repulsion between negatively charged phosphate groups.
Resonance stabilization of released inorganic phosphate.
Product ionization, favored in the low-H⁺ environment at pH 7.
Interpretation: the favorable energy change belongs to the whole hydrolysis reaction and its stabilized products. Breaking a bond itself requires energy.
Essential exam recall
ΔG determines direction; activation energy affects rate.
ΔG = 0 at equilibrium; ΔG°′ need not equal zero.
Negative ΔG°′ ↔ K′eq > 1.
Actual concentrations can overcome an unfavorable ΔG°′.
Coupled ΔG values add; coupled equilibrium constants multiply.
ATP → ADP + Pi: −30.5 kJ/mol under biochemical standard conditions.
ATP-driven glucose phosphorylation: −16.7 kJ/mol.