Biochemistry I, Spring Term
Lecture 28
April 1, 2013
Lecture 28: Biochemical Energetics
1. Phosphate Hydrolysis:
NH2
O
O
O
O P O P O P
O
O
O
NH2
N
N
N
N
O
O
OH OH
ATP
(adenosine triphospate)
O
O
P O
O
CH2
O
OH
OH
OH
OH
glucose-6-P
N
N
O
O
O P O
O
+
O
O P O P
O
O
N
O
Pi
(inorganic
phosphate)
ADP
O
O
Pi
(inorganic
phosphate)
O
OH OH
H
O P O
O
N
H2C
O
+
OH
OH
OH
OH
glucose
2. Electrons (Redox Chemistry)
An example of a redox reaction involving inorganic metals:
Fe3+ + Cu+  Fe2+ + Cu2+
This reaction can be broken down into two balanced half reactions:
Fe3++e-  Fe2+
Cu+  Cu2++ eIn this case the iron is reduced from the 3+ state to the 2+ state while the copper is oxidized from the 1+ state to the
2+ state.
In biological oxidations and reductions the electrons are often carried by protons (i.e. hydrogen atom). Thus a
general rule for oxidation-reduction are:
Loss of electrons or hydrogen atom (H+ + e-) = oxidation
Gain of electrons or hydrogen atom = reduction
There are several common acronyms to help you remember the above, here are two:
LEO GER - lose electrons oxidation, gain electron reduction.
OIL RIG - oxidation involves loss, reduction involves gain.
In the case of biological systems the redox state of an organic
molecule may be difficult to discern. The following shows
the oxidation (loss of hydrogen atoms) of three functional
groups. Note that all of these reactions involve a two electron
transfer.
H
H
H R2
R1
R2
R1
H
H
alkane
2H+ + 2e-
+
2H+ + 2e-
ketone
H
OH
O + H2O
aldehyde
O
carboxylic acid
H
H
H
H
H
1
+
O
H
alcohol
Example: Is the conversion of ethene to ethanol a redox reaction?
2H+ + 2e-
alkene
HO
Balancing organic redox reactions. It can be difficult to
determine whether a change in an organic compound is an
oxidation or not. The two steps for balancing an organic
redox reaction are:
i) If oxygen is needed to balance the reaction, use H2O.
ii) Use H+, or H+ + e-, or e- to balance hydrogen atoms
and/or charge.
A redox reaction has occurred if electrons are consumed or
released.
+
H
H
O
H H
H
H
Biochemistry I, Spring Term
Lecture 28
April 1, 2013
Organic Electron Carriers: Two compounds are widely used in biological reductions and oxidations. These are
nicotinamide adenine dinucleotide (NAD) and flavin adenine dinucleotide (FAD). If these compounds are
involved it’s a good bet that a 2 electron oxidation/reduction has occurred.

NAD is loosely bound to enzymes and can thus
be considered a true substrate (only the functional
group that undergoes a redox change is shown).
H
H
COHN 2
H
+
H
N
COHN 2
N
H
H

NADH (Reduced)
(Oxidized)
O
FAD is tightly bound to enzymes and is thus
considered a cofactor. (only the functional group
that undergoes a redox change is shown).
N
N
H
R
R
NAD+
H
H
N
N
O
H
H
N
O
R
N
N
N
R
H
FADH2 (Reduced)
FAD (Oxidized)
3. Non-Equilibrium Thermodynamics:
ΔGo, the standard energy, is the energy released when one mole of reactants are converted to products.
[ BEQ ]
 Bo   oA  G o   RT ln K EQ   RT ln
[ AEQ]
The standard free energy difference between products and reactants, ΔGo, is insufficient to describe energy
changes in metabolism because the concentrations of intermediates in a biochemical pathway are seldom at their
equilibrium concentrations. The Gibbs free energy, ΔG is used instead.
Each species in a reaction (e.g. A and B) has an intrinsic free energy associated with it, called the chemical
potential:
 A   oA  RT ln[ A]
ΔG, Gibbs Free Energy, is the difference in chemical potential, ΔG = Δµ = µB - µA.
Assuming the usual direction for the reaction of AB, where A is the reactant and B is the product.
G   B   A  (  Bo  RT ln[ B])  (  Ao  RT ln[ A])
chemical potential
(energy)
of the system (A+B)
 (  Bo   Ao )  RT ln[ B ]  RT ln[ A]
[ B]
[ A]
 ΔG is a measure of the potential energy of the system, or its ability to do
work (release energy).
 ΔG indicates the spontaneous direction of the reaction.
ΔG = 0 the reaction is at equilibrium, the difference in the potential energies
of A and B is zero, no energy can be gained converting A to B or B to A.
ΔG < 0 the reaction is spontaneous, and energy will be released as the system
approaches equilibrium, the potential of A is larger than B, A→B is spon.
ΔG > 0 the reaction is not spontaneous, energy had to be added to the system
to reach this state (the reverse reaction is spontaneous). The potential of B is
 G o  RT ln
larger than A, the system will spontaneously move: B→A.
2
oA
oB
fA
1
0
0
1
fB
H
O
Biochemistry I, Spring Term
Lecture 28
Energy increases in both
directions. No change in
state of system.
oA
Case 1: At equilibrium.
G  ( oB  RT ln[ B]EQ )  ( 0A  RT ln[ A]EQ )
 ( oB  0A )  RT ln[ B ]EQ  RT ln[ A]EQ
[ BEQ ]
 G 0  RT ln
[ AEQ ]
  RT ln
[ BEQ ]
[ AEQ ]
 RT ln
[ BEQ ]
[ AEQ ]
April 1, 2013
oB
G = 0
fA
1
0
0
1
Case 2 : [A] higher than its equilibrium point.
G> 0
oB
G< 0
fA
fB
1
0
0
1
fA
fB
1
0
0
1
Case 3: [B] higher than its equilibrium point.
G  ( oB  RT ln{[ B ]EQ   })  ( 0A  RT ln{[ A]EQ   })
G  ( oB  RT ln{[ B ]EQ   })  ( 0A  RT ln{[ A]EQ   })
 ( oB  0A )  RT ln{[ B ]EQ   }  RT ln{[ A]EQ   }
 ( oB  0A )  RT ln{[ B ]EQ   }  RT ln{[ A]EQ   }
 G 0  RT ln
 RT ln
[ BEQ ]  
[ AEQ ]  
[ B ]EQ  
[ B ]EQ
 RT ln
  RT ln
[ BEQ ]
[ AEQ ]
 RT ln
[ BEQ ]  
 G 0  RT ln
[ AEQ ]  
[ A]EQ
 RT ln
[ A]EQ  
Both ln terms are <1, therefore G is <1.
[ BEQ ]  
[ AEQ ]  
[ B ]EQ  
 RT ln
[ B ]EQ
[ BEQ ]
  RT ln
[ AEQ ]
 RT ln
[ BEQ ]  
[ AEQ ]  
[ A]EQ
[ A]EQ  
Both ln terms are >1, therefore G is >1
4. Flux and Coupling of Reactions: A key feature of all biochemical pathways is that there is a constant flux
of material through the pathway. To insure a constant flux through the pathway it is necessary to insure that
the Gibbs energy of the products are lower in energy than the substrates, such that the natural flow is always
from substrates to products.
O
H
Direct Coupling: Many steps in biochemical
pathways involve reactions that are extremely
unfavorable, i.e. a large positive Go. This energy
barrier reduced by the direct coupling of the
unfavorable reaction to a favorable one, with both
reactions occurring at the same time in the active
site of one enzyme. The energetics of each
reaction are most easily seen by considering the
half-reactions. For example, the first reaction in
glycolysis, catalyzed by the enzyme hexosekinase,
converts glucose to glucose-6-phosphate. The high
energy of ATP is used to drive this unfavorable
reaction by direct coupling..
O
H2C
O
O
+
OH
OH
O
P O
O
CH2
O
OH
O P O
O
OH
OH
OH
OH
OH
NH2
NH2
N
N
O
O
O
O P O P O P
O
O
O
+
O P O
O
O
O
O P O P
O
O
N
N
O
O
ADP
Pi
OH OH
OH OH
NH2
NH2
O
N
N
H2C
O
O
OH
OH
O
O
H
OH
N
N
N
N
O
ATP
(adenosine triphospate)
OH
+
O
O
O P O P O P
O
O
O
N
N
O
O
O
O
P O
O
CH2
O
OH OH
O
+
OH
OH
N
N
O
O P O P
O
O
N
N
O
OH
OH
O
OH OH
Reactions of this type are often called “group
transfer reactions” since a group (phosphate) is transferred from one substrate to another.
Indirect Coupling: By lowering the concentration of the product, a reaction can be made spontaneous; the G
becomes < 0. This generally implies that a reaction further down the pathway has a large negative Go, such that
[B] is kept low.
G  G 0  RT ln
Example problem in indirect coupling: In the pathway: AB:
Concentration of [A] is 1 mM under normal metabolic conditions.
Standard free energy, ΔGo, of the first reaction (AB) is +5.75 kJ/mol. What is the
direction of the reaction if the concentration of [B] is 10-5 M? Assume RT=2.5 kJ/mol
(T~300K).
3
oB
spontaneous
direction A to B
At equilibrium
0
spontaneous
direction B to A
[ B]
[ A]
105
10 3
 5.75  2.5 ln 10 2
 5.75  11.51
 5.76
 5.75  RT ln
fB