Essential Cell Biology 5th edition

Free Energy and Catalysis
95
REACTION RATES
A spontaneous reaction is not necessarily a rapid reaction:
a reaction with a negative free-energy change (ΔG ) will not
necessarily occur rapidly by itself. Consider, for example, the
combustion of glucose in oxygen:
COUPLED REACTIONS
Reactions can be “coupled” together if they share one or
more intermediates. In this case, the overall free-energy
change is simply the sum of the individual ΔG o values. A
reaction that is unfavorable (has a positive ΔG o ) can for this
reason be driven by a second, highly favorable reaction.
H
C
HO
CH 2 OH
C O OH
OH
C
H
C
C
H
+ 6O 2 6CO 2 + 6H 2 O
H OH
ΔG o = –2867 kJ/mole
SINGLE REACTION
glucose
+
fructose
NET RESULT: reaction will not occur
ATP
ADP
+
P
sucrose
ΔG o = –30.5 kJ/mole
ΔG o =
+23 kJ/mole
Even this highly favorable reaction may not occur for centuries
unless enzymes are present to speed up the process. Enzymes
are able to catalyze reactions and speed up their rate, but they
cannot change the ΔG o of a reaction.
CHEMICAL EQUILIBRIA
NET RESULT: reaction is highly favorable
COUPLED REACTIONS
glucose
P
+
+
ATP
P
glucose 1-P
glucose 1-P fructose sucrose
+
+
ADP
P
ΔG o =
23 – 30.5 =
–7.5 kJ/mole
A fixed relationship exists between the standard
free-energy change of a reaction, ΔG o , and its equilibrium
constant K. For example, the reversible reaction
Y
X
NET RESULT:
sucrose is made in a reaction driven
by the hydrolysis of ATP
will proceed until the ratio of concentrations [X]/[Y] is
equal to K (note: square brackets [ ] indicate
concentration). At this point, the free energy of the
system will have its lowest value.
HIGH-ENERGY BONDS
free
energy
of system
equilibrium
point
lowest
free
energy
One of the most common reactions in the cell is
hydrolysis, in which a covalent bond is split by adding
water.
A
B
hydrolysis
A
OH + H
B
[X]
[Y]
At 37 o C, ΔG o = –5.94 log 10 K (see text, p. 96)
K = 10 –ΔG o /5.94
For example, the reaction
CH 2 OH
CH 2 O P
O O P
O OH
glucose 1-P
glucose 6-P
has ΔG o = –7.3 kJ/mole. Therefore, its equilibrium
constant
K = 10 (7.3/5.94) = 10 (1.23) = 17
So the reaction will reach steady state when
[glucose 6-P]/[glucose 1-P] = 17
The ΔG o for this reaction is sometimes loosely termed
the “bond energy.” Compounds such as acetyl
phosphate and ATP, which have a large negative ΔG o
of hydrolysis in an aqueous solution, are said to have
“high-energy” bonds.
ΔG o
(kJ/mole)
acetyl P acetate + P –43.1
ATP ADP + P –30.5
glucose 6-P glucose + P –13.8
(Note that, for simplicity, H 2 O is omitted from the above
equations.)