Macroergic bond does not exist
Jiří Wilhelm
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.
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.
For the equation:
we can express the rate from left to right:
v1 = k1 *[A]*[B]
and from right to left:
v2 = k2 * [C]*[D]
Under equilibrium v1 = v2 and thus:
k1*[A]*[B] = k2*[C]*[D]
we can express k1/k2 :
As the ratio of two constants is a constant, we label it K
and call equilibrium constant:
For the description of changes of energy in chemical
reactions we use Gibbs function G, which can be
calculated under equilibrium at sdandard conditions:
ΔG0 = - 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:
ΔG = ΔG0 + 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
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
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 + H2O  ADP + Pi
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.
The amount of energy released during hydrolysis of
ATP can be measured in a calorimeter:
ΔG0 = -7,5 kcal/mol
From this value we can calculate the equilibrium
concentrations:
[ATP]
_________________
-6
-1
= 4 * 10 M
[ADP] [Pi]
Under intracellular concentration of phosphate of 10 mM:
[ATP] / [ADP] = 4*10-8
However, physiologic ratio of [ATP] /[ADP] in cell is 4 – 6.
It means, that cell maintains the ratio of [ATP] / [ADP]
hundredmilion times greater than corresponds to the
equilibrium ratio
Let us consider reaction:
 C+D
A+B 
[1]
and we know that for the direction from left to right is
ΔG1 > 0, thus it does not proceed in this direction.
When we couple this reaction with ATP hydrolysis:
[2]
ATP + H2O  ADP + Pi
we obtain reaction:
A + B + ATP + H2O  C + D + ADP + Pi
[3]
For equation [3] we get:
ΔG3 = ΔG1 + ΔG2
= ΔG1 + ΔG20 + RT ln K2 * 10-8
= ΔG1 – 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.
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?
Even greater problem makes myokinase reaction:
 ATP + AMP
ADP + ADP 
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.
Daniel E. Atkinson: Cellular Energy Metabolism and its
Regulation.
Academic Press 1977