1. Macroergic bond does not exist Jiří Wilhelm

2. The very term chemical bond represents forces holding atoms together. In order to break it we must apply greater force than that of binding. Therefore we can not expect gain of energy during bond breaking. There are compounds releasing great amount of energy during destruction of the molecule (e.g. nitroglycerine). However, the resynthesis of the parent molecule from the products of destruction demands much higher energy than is obtained during decomposition, making these types of compounds uneconomical sources of energy.

3. Lipmann suggested in 1941 the symbol  (squiggle) for bonds that are easily broken in displacement reactions, they attach a good leaving group. In order to emphasize the importance of activation of this type in metabolic energetics, Lipmann also suggested the term high energy bond, and that term was the one that caught on. Unfortunately, it was interpreted as a macroergic bond, which indicated a special type of bond containing high amount of energy. To elucidate the problem we must refer to thermodynamics which deals with questions of energy transformation.

4. For the equation: we can express the rate from left to right: v 1 = k 1 * [A] * [B] and from right to left: v 2 = k 2 * [C] * [D]

5. Under equilibrium v 1 = v 2 and thus: k 1* [A] * [B] = k 2* [C] * [D] we can express k 1 /k 2 : As the ratio of two constants is a constant, we label it K and call equilibrium constant:

6. For the description of changes of energy in chemical reactions we use Gibbs function G, which can be calculated under equilibrium at sdandard conditions: ΔG 0 = - RT ln K Index zero indicates conditions of the standard state (concentration of reactants 1 M etc.) In the case we are dealing with the real situation in vivo, we must consider the actual concentrations of reactants and products:

7. ΔG = ΔG 0 + RT ln ________ [C] [D] [A] [B] which can be rewritten: ΔG = - RT lnK + RT lnQ ΔG = RT ln ____ Q K Q = K  ΔG = 0 Q  K  ΔG < 0 Q > K  ΔG > 0

8. What shoul be the thermodynamical properties of a fuel? For instance gasoline is: thermodynamicaly unstable – its reaction with oxygen runs quickly and equilibrium is moved completely to the right kineticaly stable - gasoline can be stored in contact with the air provided you avoid ignition

9. We know that ATP is a source of energy for the cell, it thus represents a “fuel”. Does it have the thermodynamic properties needed? Energy is released during the reaction: ATP + H 2 O  ADP + P i which is shifted markedly to the right. On the other hand, in the absence of enzymes reaction does not proceed Both conditions are thus fulfilled.

10. The amount of energy released during hydrolysis of ATP can be measured in a calorimeter: ΔG 0 = -7,5 kcal/mol From this value we can calculate the equilibrium concentrations: [ATP] _________________ = 4 * 10 -6 M -1 [ADP] [P i ] Under intracellular concentration of phosphate of 10 mM: [ATP] / [ADP] = 4*10 -8 However, physiologic ratio of [ATP] /[ADP] in cell is 4 – 6.

11. It means, that cell maintains the ratio of [ATP] / [ADP] hundredmilion times greater than corresponds to the equilibrium ratio Let us consider reaction: A + B   C + D [1] and we know that for the direction from left to right is ΔG 1 > 0, thus it does not proceed in this direction. When we couple this reaction with ATP hydrolysis: ATP + H 2 O  ADP + P i we obtain reaction: [2] A + B + ATP + H 2 O  C + D + ADP + P i [3]

12. For equation [3] we get: ΔG 3 = ΔG 1 + ΔG 2 = ΔG 1 + ΔG 2 0 + RT ln K 2 * 10 -8 = ΔG 1 – 7,5 + 1,4 (6-8) Thus, if we couple in a cell any reaction with ATP hydrolysis, the resulting ΔG will be mostly negative and the reaction will proceed. It implies that energy is not stored in some special macroergic bond, but in the ratio of ATP/ADP, which is maintained far from equilibrium.

13. Using the term macroergic bond generates a lot of obstacles. For instance in muscle we have an energetic reserve in the form of creatine phosphate, which originates from the reaction between ATP and creatine: creatine + ATP   creatine-P + ADP If the driving force for the reaction from left to right is the macroergic bond in ATP, what is the driving force of the oposite reaction under situation when ATP concentration in muscle under strenuous work decreases?

14. Even greater problem makes myokinase reaction: ADP + ADP   ATP + AMP In this case the macroergic bond originate from 2 ADP that has lost it. Myokinase reaction serves as an energy reserve in the ischemic cell unable to synthesize ATP from normal substrates. In order to generate ATP continuously, AMP must be degraded. That enables to maintain the vital ATP/ADP ratio at the expense of the size of adenine nucleotide pool.

15. Daniel E. Atkinson: Cellular Energy Metabolism and its Regulation. Academic Press 1977