BIOENERGETICS
1. Define equilibrium, non-equilibrium, steady state, and free energy change.
2. Know the first and second laws of thermodynamics. Describe their importance with
respect to biological systems.
3. Distinguish G and G0 in terms of information provided about a reaction.
4. For the model reaction A + B ↔ C + D: Know the equation defining the equilibrium
constant; know the equation expressing the free energy change at given
concentrations of A, B, C, and D.
5. Understand the relationship between the standard free energy change and the
equilibrium constant.
6. Explain in terms of G values how an endergonic reaction can be coupled to an
exergonic reaction.
7. Describe how ATP serves to couple exergonic and endergonic reactions.
8. Understand why ATP, phosphoenolpyruvate, creatine phosphate, 1,3bisphosphoglycerate are considered high energy compounds.
BIOENERGETICS
1. First law of thermodynamics- law of conservation of energy
a. Total energy of a system plus its surroundings (the universe) is constant
b. Energy that leaves system equals energy that enters minus energy that remains
stored in system
energy out = energy in – energy stored
energy stored (called internal energy, E) = energy in – energy out
c. Interested in change in internal energy (E) when chemical reaction takes place
rather than actual energy values
E = Eproducts – Ereactants
d. Change in enthalpy, heat content (H), for biological reactions nearly equals E
H = Hproducts – Hreactants
2. Second law of thermodynamics- law of thermodynamic spontaneity
a. Process spontaneous only if sum of entropies (randomness, disorder) of a system
plus its surroundings (the universe) increases
b. Spontaneity refers to feasibility of reaction in particular direction, not rate of
reaction
c. For every process Ssystem + Ssurrounding > 0
3. Free energy
a. G or Gibbs free energy- measure of spontaneity for a process that only depends on
parameters within system
G = H – TS (T is temperature in Kelvin)
b. For every process G of the system must be negative
4. Using G to access feasibility of reactions under given set of conditions
a. Equilibrium constant (Keq)
A↔B
Keq = [B]eq / [A]eq
ratio of product concentrations to reactant
concentrations at equilibrium
b. Free energy is lowest at equilibrium and increases as reaction is displaced;
tendency toward equilibrium provides driving force of every chemical reaction
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c. G is means of calculating how far from equilibrium reaction lies and how much
energy is released as it proceeds (under specified conditions)
G = -RT ln([B]eq/[A]eq) + RT ln([B]specified/[A]specified)
= -RT lnKeq + RT ln[B]specified/[A]specified
(R is the gas constant = 1.987 cal/mol-K)
or more generally for aA + bB ↔ cC + dD (a molecules of reactant A, etc.)
G = -RT lnKeq + RT ln {([C]c[D]d)/([A]a[B]b)}
d. Significance of G for reaction under specified conditions
G < 0: feasible; energy produced by reaction
G > 0: not feasible; energy must be supplied to drive reaction
G = 0: at equilibrium
5. Standard free energy change
a. G0 is free energy in standard state- all reactants and products at 1 M, 25ºC (298
K); G0’ further indicates at pH 7.0
b. To get G0 equation, use G equation and plug in 1 for all concentrations
G0 = -RT lnKeq + RT ln1
= -RT ln Keq
= -0.592 ln Keq or –1.36 log Keq
(measured in kcal/mol)
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Allows G0 to be calculated directly from Keq
Inverse relationship between G0 and Keq
If Keq> 1 (at standard conditions), G0negative; if Keq< 1 G0 positive
c. Determining G at given conditions from G0
 G0 has been determined for many reactions
 To get equation relating G and G0, use G equation and replace –RT lnKeq
with G0
G = G0 + RT ln {([C]c[D]d)/([A]a[B]b)}
= G0 + 0.592 ln {([C]c[D]d)/([A]a[B]b)}
6. Steady state
a. Although reactions move toward equilibrium, most in cell do not reach it; most
compounds at steady state levels far from equilibrium
b. Reactants and products affected by other steps in metabolic pathways; continuous
uptake of energy
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c. Drive toward equilibrium used to perform useful work; cell at equilibrium would
be dead
7. Driving unfavorable reactions
b. Overall G for coupled series of reactions sum of G's of individual steps
A↔B+C
G0 = +4 kcal/mol (endergonic)
B↔D
G0 = -8 kcal/mol (exergonic)
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A↔C+D
G0 = -4 kcal/mol (exergonic)
c. Thermodynamically unfavorable reaction can be driven by favorable reaction to
which it is coupled
8. ATP is universal energy carrier
a. Nucleotide with adenine, ribose, and triphosphate units
b. Formed during oxidation of foodstuffs
c. Large free energy released from hydrolysis
ATP + H2O ↔ ADP + Pi
G0 = -7.3 kcal/mol
ATP + H2O ↔ AMP + PPi
G0 = -10.9 kcal/mol
d. Hydrolysis often used to drive unfavorable reactions
A↔B
G0 = +3 kcal/mol
A + ATP + H2O ↔ B + ADP + Pi
G0 = -4.3 kcal/mol
e. ATP-ADP cycle fundamental mode of energy exchange
9. Nature of high energy molecules
a. Large free energy difference between molecule and its hydrolysis products
b. ATP
 Strong tendency to transfer phosphoryl group to water; high phosphate
transfer potential
 Cause of big energy difference
- Negative charges on phosphates of ATP; repel each other; want to react
with water
- Hydrolysis products more stable; more resonance structures
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c. Other molecules are high energy compounds; often energy released from
hydrolysis of phosphate group, for example phosphoenolpyruvate, creatine
phosphate, 1,3-bisphosphoglycerate
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