Major concepts:
1. What is the relationship between ΔG and ΔG°′? Between ΔG°′ and K_eq ?
2. What is the significance of ΔG°′ being additive?
3. What is the role of ATP in the cell? (Is it used for storing energy or for transferring energy?)
4. Why is ATP a "high energy" molecule? In general, what are the characteristics of high energy molecules?
5. What is the relationship between ΔG°′ and ?E°′? Why are oxidation-reduction reactions covered in biochemistry?

Core knowledge:
1. What is the equation that relates ΔG°′ to K_eq ?
2. In what ways does ATP participate in reactions?
3. What are the reactions catalyzed by nucleoside diphosphate kinase (NDP kinase) and by adenylate kinase? What reaction is catalyzed by pyrophosphatase?
4. What is the difference between H⁺, H, and H⁻ ?
     What is the half reaction for the reduction of NAD⁺?
5. What is the equation that relates ΔG°′ to ΔE°′?
6. Given two half reactions, how can you write a spontaneous whole reaction, and calculate ΔE°′ and ΔG°′ for the spontaneous reaction?

Introduction to considering energy in cells: work, energy transfer, energy storage

Thermodynamics:
A. ?G and relationship to K_eq and component concentrations
      1. ΔG°: measured at 25° C (298 K), 1 atm, initial concentrations of each component = 1.0 M
      2. ΔG°′ is ΔG measured at pH 7, [H₂O] = 55.5 M, & [Mg²⁺] = 1 mM
      3. Actual free energy change (ΔG in a cell) has different conditions, especially concentration
     4. ΔG = ΔG°′ + RT ln K_eq , so when actual ΔG = 0 (at equilibrium), ΔG°′ = − RT ln K_eq
          When K_eq > 1, ΔG°′ is < 0, and the reaction is spontaneous.
         When K_eq < 1, ΔG°′ is > 0, and the reaction proceeds in reverse.
     5. ΔG = ΔG°′ + R T ln ([products]/[reactants])
        When [reactants] increase and [product] is small, the actual ΔG in the cell may be negative,
         even if ΔG°′ is positive.
B. ΔG°′ for two reactions is additive. Significance
      1. For coupled reactions, if the overall ΔG°′ < 0, the overall reaction is spontaneous.
          example: glucose + P_i → glucose-6-P + H₂O                               ΔG°′ = 13.8 kJ/mol
               ATP + H₂O → ADP + P_i                                                        ΔG°′ = − 30.5 kJ/mol
               glucose + ATP → glucose-6-P + ADP                                        ΔG°′ = − 16.7 kJ/mol
      2. For a reversible reaction with ΔG°′ > 0 followed by a spontaneous reaction,
         removing the product of the 1^st reaction can pull the 1^st reaction.

Function of ATP
A. ATP – Why is ATP a "high energy" molecule?
     1. adjacent negative charges on ATP are relieved by hydrolysis
     2. ADP + P_i have more resonance structures available than does ATP.
     3. ATP + H₂O → H–ADP + HPO₄²⁻   → ADP + H⁺ + HPO₄²⁻    (more entropy)
      4. Products are more solvated than reactants.
      While ΔG°′ is negative and large, the reaction requires an enzyme. ΔG^† is also large.
B. ATP = donor of functional groups
     1. phosphoryl (not phosphate, because the O in R – O – PO₃²⁻ is from the alcohol)
          ATP + molecule → ADP + molecule – P (phosphorylated molecule)
      2. pyrophosphoryl:  ATP + molecule → AMP + molecule – PP
     3. AMP:  ATP + molecule → PP_i + molecule – AMP = adenylylation
         Always followed by pyrophosphatase reaction: PP_i + H₂O → 2 P_i      ΔG°′ = − 19 kJ/mol
C. Effects of group transfers involving ATP
      1. → conformation change: enzyme activity, carrier change, muscle contraction, etc.
      2. → covalent intermediates in reactions that are otherwise very unfavorable
          synthesis of energy storage molecules, for example
D. Transphosphorylations = transfer of phosphates between nucleotides, other coenzymes
     1. Nucleoside diphosphate kinase: ATP + NDP ↔ ADP + NTP                     ΔG°′ ~ 0
     2. Adenylate kinase: ATP + AMP ↔ 2 ADP                                                   ΔG°′ ~ 0
E. Other high energy compounds
     1. Other phosphorylated molecules, including creatine phosphate
     2. Thioesters:

Note: much of the time, we will ignore different ionized forms of ATP, H⁺ that is released, etc., in writing reactions (= biochemical reactions).

Oxidation-reduction reactions: more reduced molecules are higher in energy (biochemically)

A. Types of biological oxidation-reduction reactions:
      frequently related to oxidation states of carbon – see Figure 13-13.
1. directly as electrons: Fe²⁺ + Cu²⁺ → Fe³⁺ + Cu⁺ (Fe is oxidized, Cu is reduced)
2. as H atoms = H⁺ + e⁻ .
3. as hydride ions (H⁻) = H⁺ + 2 e⁻: usually in relation to NAD⁺ function
4. through combination with O₂, which accepts electrons from organic compounds

B. Reduction potential E = electron affinity; must be measured using a reference cell
1. More positive E = greater affinity for electrons.    Note: standard E°′ is comparable to ΔG°′.
2. By convention, half reactions are always written as reduction reactions.
3. To determine ΔE°′ for a spontaneous oxidation reduction reaction,
      compare E's for each half-reaction. The more positive E will get electrons (be reduced),
         and ΔE°′ for the whole reaction = E°′ (electron acceptor) − E°′ (electron donor), or
BEST METHOD:
         ΔE°′ for the whole reaction is determined by changing the sign for the half-reaction
              that is reversed (the oxidation reaction) and adding the reduction potentials.
     Example, from Table 13-7:
         ¹/₂ O₂ + 2 H⁺ + 2 e⁻ → H₂O                                              E°′ = 0.816 V
         pyruvate⁻ + 2 H⁺ + 2 e⁻ → lactate⁻                                     E°′ = − 0.185 V
      Since O₂ has the more positive E°′, reverse the other half reaction and change the sign of its E°′:
         lactate⁻ → pyruvate⁻ + 2 H⁺ + 2 e⁻                                       E°′ = 0.185 V
      Add the two half reactions and their reduction potentials together:
         ¹/₂ O₂ + lactate⁻ → H₂O + pyruvate⁻       ΔE°′ = 0.816 V + 0.185 V = 1.001 V

C. Relationship of ?E°′ to ΔG°′: ΔG°′ = − n F ΔE°′, where n = number of e⁻ transferred
?E°′ > 0 is therefore an indication of ΔG°′ < 0.
For the example used, ΔG°′ = − (2 e⁻) (96485 J/V-mol) (1.001 V) = − 193,000 J = − 193 kJ

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