The closed loop conformation in the TIM reaction pro-
vides a striking example of the stereoelectronic control that
enzymes can exert on a reaction (Section 15-1Eb). In solu-
tion, the enediol intermediate readily breaks down with the
elimination of the phosphate at C3 to form the toxic com-
pound methylglyoxal (Fig. 17-12a). On the enzyme’s surface,
however, this reaction is prevented because the phosphate
group is held by the flexible loop in the plane of the enediol,
a position that disfavors phosphate elimination. In order for
this elimination to occur, the bond to the phosphate
group must lie, as shown in Fig. 17-12a, in the plane perpen-
dicular to that of the enediol. This is because, if the phos-
phate group were to be eliminated while this bond
was in the plane of the enediol as diagrammed in Fig. 17-
12b, the CH
2
group of the resulting enol product would be
twisted 90° out of the plane of the rest of the molecule. Such
a conformation is energetically prohibitive because it pre-
vents the formation of the enol’s double bond by eliminat-
ing the overlap between its component p orbitals. In the
mutant enzyme lacking the flexible loop, the enediol is able
to escape: ⬃85% of the enediol intermediate is released
into solution, where it rapidly decomposes to methylglyoxal
and P
i
.Thus, flexible loop closure also ensures that substrate
is efficiently transformed to product.
On the basis of the foregoing X-ray structures, it had
been widely assumed that the binding of substrate to TIM is
ligand-gated, that is, it induces loop closure. Yet, if this were
the case, the reversibility of the TIM reaction and the chem-
ical resemblance of its reactant and product (GAP and
DHAP) make it difficult to rationalize how product could
be released. However, NMR measurements by John
Williams and McDermott reveal that, in fact, loop motion
still occurs when TIM is binding either glycerol-3-phosphate
(a substrate analog) or 2-phosphoglycolate (a transition
state analog) and is sufficiently fast (on a time scale of 100
s) to account for the catalytic reaction rate (a turnover
C¬O
C¬O
time of 230 s).This is a clear example of how complemen-
tary information supplied by X-ray and NMR methods has
yielded important insights into an enzymatic mechanism
that neither technique alone could have provided.
d. TIM Is a Perfect Enzyme
TIM, as Jeremy Knowles demonstrated, has achieved
catalytic perfection in that the rate of bimolecular reaction
between enzyme and substrate is diffusion controlled; that
is, product formation occurs as rapidly as enzyme and sub-
strate can collide in solution, so that any increase in TIM’s
catalytic efficiency would not increase the reaction rate
(Section 14-2Bb). Because of the high interconversion effi-
ciency of GAP and DHAP, these two metabolites are
maintained in equilibrium: K ⫽ [GAP]/[DHAP] ⫽ 4.73 ⫻
10
⫺2
; that is, [DHAP] ⬎⬎ [GAP] at equilibrium. However,
as GAP is utilized in the succeeding reaction of the gly-
colytic pathway, more DHAP is converted to GAP, so that
these compounds maintain their equilibrium ratio. One
common pathway therefore accounts for the metabolism
of both products of the aldolase reaction.
Let us now take stock of where we are in our travels
down the glycolytic pathway. At this stage, the glucose,
which has been transformed into two GAPs, has com-
pleted the preparatory stage of glycolysis. This process has
required the expenditure of two ATPs. However, this in-
vestment has resulted in the conversion of one glucose to
two C
3
units, each of which has a phosphoryl group that,
with a little chemical artistry, can be converted to a “high-
energy” compound (Section 16-4Ba) whose free energy of
hydrolysis can be coupled to ATP synthesis. This energy
investment is doubly repaid in the final stage of glycolysis in
which the two phosphorylated C
3
units are transformed to
two pyruvates with the coupled synthesis of four ATPs per
glucose.
606 Chapter 17. Glycolysis
Figure 17-12 The spontaneous
decomposition of the enediol
intermediate in the TIM reaction to
form methylglyoxal through the
elimination of a phosphate group. (a)
This reaction can occur only when the
bond to the phosphate group lies
in a plane that is nearly perpendicular
to that of the enediol so as to permit
the formation of a double bond in the
intermediate enol product. (b) When
the bond to the phosphate group
lies in a plane that is nearly parallel to
that of the enediol, the p orbitals on the
resulting intermediate product would
be perpendicular to each other and
hence lack the overlap necessary to
form a bond, that is, a double bond.
The resulting unsatisfied bonding
capacity greatly increases the energy of
the reaction intermediate and hence
makes the reaction highly unfavorable.
C¬O
C¬O
H
HH
O
Parallel p orbitals
maximally
overlapped to
form a π bond
Perpendicular p orbitals
have no overlap; a π
bond cannot be formed
O bond to phosphate
in plane perpendicular
to plane of molecule
O bond to phosphate
in plane of molecule
Enol Methyl
glyoxal
Enediol
intermediate
(a)
(b)
P
i
P
i
CH
OHC
C
C
OPO
3
2⫺
O
CH
CH
H
OH
C
O
CH
O
C
H
3
C
H
H
H
O
CH
OHC
C
C
OPO
3
2⫺
O
CH
H
H
OHC
C
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