Chem*4520 Metabolic Processes

Fall Semester 2000

Modified Sept 2000

schematic view of the enzyme citrate synthase,
with bound acetyl-CoA analog in green

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Lecture 2:

Wed Sept 13

Free energy changes in the glycolysis pathway:
the aldolase step, the phosphoglycerate kinase step and the pyruvate kinase step.

Voet Chapter 16, pp. 445-446, 459-460, 463-464, 470-472;
Mathews, Van Holde Chapter 13, pp.450-458,460;
Stryer, Chapter 19, pp. 485-491.

The Glycolysis pathway: from glucose to pyruvate or lactate

Glucose is a widely distributed hexose sugar, found either as the monosaccharide, or combined with frucose in sucrose or in polymer form such as starch. Almost all organisms possess some kind of glucose breakdown pathway, and in animals, these pathways are adapted to provide the best means for rapid release of energy.

Glycolysis is the name given to the sequence of reactions which couple partial breakdown of glucose to the three-carbon acid pyruvate, linking this process to synthesis of ATP by condensing Pi onto ADP. Pyruvate can then be consumed in other pathways.

Glycolysis can be divided into three distinct stages:
Set-up Activation of glucose in preparation for breakdown steps; some ATP is used
Mid-phase Hexose is split into two trioses, and oxidized to a carboxylate so that some of the oxidation energy is trapped as bond energy in 1,3-bisphosphoglycerate.
Pay-off Recovery of bond energy to drive ATP formation. More ATP is made (4 ATP per glucose) than was consumed at the start (2ATP per glucose).


Energetics of glycolysis

Several experiments have been done to measure the in vivo concentrations of glycolysis intermediates so that actual can be calculated. The most complete data comes from a study on erythrocytes (see Mathews p. 460). Voet (p. 472) also reports values for muscle tissue which are roughly similar.
substratecofactors, enzymes




glucose, 5 mM; ATP as donor
hexokinase, glucokinase-16.5 kJ/mol-33 kJ/mol
glucose-6-P, 83 然;
glucose-6-P isomerase+2.0 kJ/mol-2 kJ/mol
fructose-6-P, 14 然;ATP as donor
phosphofructokinase-13.7 kJ/mol-26 kJ/mol
fructose-1,6-bisP, 35 然;
aldolase*+22.0 kJ/mol-1 kJ/mol
dihydroxyacetone-3-P, 150 然;
triose phosphate isomerase+5.6 kJ/mol+0.7 kJ/mol
glyceraldehyde-3-P, 20 然;NAD+ as e- acceptor
Pi as phosphate donor
glyceraldehyde-3-P dehydrogenase+6.3 kJ/mol+0.6 kJ/mol
1,3-bisphosphoglycerate, <1 然;ADP as acceptor
phosphoglycerate kinase-18.8 kJ/mol-1 kJ/mol
3-phosphoglycerate, 118 然;
phosphoglyceromutase+4.8 kJ/mol+0.8 kJ/mol
2-phosphoglycerate, 30 然;
enolase+3.2 kJ/mol+1 kJ/mol
phosphoenolpyruvate, 23 然;ADP as acceptor
pyruvate kinase-31.6 kJ/mol-17 kJ/mol
pyruvate, 51 然;NADH as e- donor
lactate dehydrogenase-26.2 kJ/mol-0.8 kJ/mol
lactate, 2.9 mM;

calculations use [ATP] = 1.0 mM, [ADP] = 0.1mM, [Pi] = 1.0 mM and [NADH]/[NAD+] = 0.002.

Note *: Aldolase gives one mole each of dihydroxyacetone-3-P and glyceraldehyde-3-P as products. Triose phosphate isomerase converts the molecule of dihydroxyaceton-3-P into glyceraldehyde-3-P so that all subsequent steps react with a stoichiometry of two moles per mole of glucose. Although there are only two phosphorylation steps, this stoichiometry gives rise to production of 4 ATP.


Glycolysis is internally balanced for oxidation and reduction when extended to the lactate dehydrogenase step. The formula of glucose, C6H12O6 is exactly double that of protonated lactic acid C3H6O3, so one can write a balanced equation for the hypothetical reaction:
C6H12O62 C3H5O3-++ 2 H+
glucose lactate
-917.2
kJ/mol
2 x -516.6
kJ/mol
2 x -39.9
kJ/mol
f

= 2 x (-516.6 - 39.9) -(-917.2) kJ/mol = -195.8 kJ/mol

This represents the cumulative standard free energy change for anaerobic glycolysis, and it drives the reversed hydrolysis of 2 ATP:

2 ADP + 2 Pi 2 ATP = 2 x +30.3 kJ/mol = + 60.6 kJ

The net = -195.8 + 60.6 = -135.2 kJ / mol, which represents the net chemical "driving force" behind glycolysis. Part of this excess energy drives ATP synthesis so that there is a large excess of [ATP] versus [ADP], (actual for ATP formation at the concentrations given above is about 53 kJ/mol). The remaining excess is what it takes to ensure that the reactions proceed in the required direction.


Significance of the data

Most of the values fall close to zero (± 1 kJ/mol) implying that these reactions are very close to equilibrium. They should all be slightly negative; since no step is greater than +1 kJ/mol, this can safely be put down to small errors in concentrations or values of . All of these reactions are easily reversed by a small change in the relationship of substrate and product concentrations.

There are three steps with strongly negative :


The results indicate that most steps are close to equilibrium, where is almost zero. Three out of the four kinase steps have large negative values, and are driven strongly left to right. These are the reactions which determine the direction of reaction in glycolysis. They would require 103 to 106-fold concentration changes to be reversed, which is unreasonable.

Good metabolic strategy

Having highly directional reactions at start and finish of a patheway is good metabolic strategy;

The hexokinase and phosphofructokinase steps drive substrates into the sequence; the pathway can continue even if substrates are significantly depleted.

Pyruvate kinase clears intermediates out of the pathway.

Implications for regulation

Large negative implies [substrate] is in excess and [product] is low relative to equilibrium values.

Catalytic activity is limited at these steps.

These enzymes are good targets for regulation and control of the pathway.

Flux = net rate of conversion in a pathway.

For example if we examine a pathway operating at a flux rate of 10 痠ol/min, the activity of the one-way step 1 should also be 10 痠ol/min.
To achieve a flux rate of 10 痠ol/min, the near-equilibrium step 2 must have much more total enzyme activity available, e.g. 100 痠ol/min forwards, 90 痠ol/min in reverse, total enzyme activity 190 units.
If activity is halved at step 1 only due to some regulatory mechanism, the flux must drop to 5 痠ol/min (97.5 痠ol/min forwards and 92.5 痠ol/min reverse for step 2).
If the overall activity at step 2 only is cut from 190 痠ol/min to 90 痠ol/min by some regulatory mechanism, a flux of 10 痠ol/min is easily maintained (50 痠ol/min forwards and 40 痠ol/min reverse for the equilibrium steps).

Regulation of glycolysis

These steps are where glycolysis is regulated, meaning that the activity is shut down if ATP is not needed.

Hexokinaseis subject to product inhibition by glucose-6-phosphate.
If glucose-6-phosphate is not being used, the enzyme is inhibited.

However, liver contains glucokinase, which is a high KM isozyme of hexokinase, and is not subject to product inhibition. In periods of low physical activity, peripheral tissues such as muscle reduce the glycolysis flux rate due to pruduct inhibition by underutilized glucose-6-phosphate. The accumulated glucose levels become high enough for liver glucokinase to continue to produce glucose-6-phosphate. Since the energy demand is low, this can then result in diversion of excess glucose to fat biosynthesis in the liver.

Phosphofructokinaseis negatively regulated by ATP and activated by 5'-AMP.
If ATP levels are high, glycolysis is not needed and phosphofructokinase activity is reduced. Citrate (-ve) and fructose-2,6-bisphosphate (+ve) are also important regulators.
Pyruvate kinase is also inhibited by high levels of ATP, reducing enzyme activity if ATP is not needed. Alanine (-ve) and fructose-2,6-bisphosphate (+ve) are also important regulators.
Regulation of phosphofructokinase and pyruvate kinase is also important in the context of gluconeogenesis, and we will return to this topic.


Thermodynamic details of specific reactions

Aldolase

The aldolase reaction has a high positive :
fru-1,6-bisP DHA-3-P +glyceraldehyde-3-P = +22.0 kJ/mol
35 然150 然20 然
The observed concentrations seem to conflict with the idea that high substrate and low product concentrations should be needed to obtain negative values of actual .

The trick is that this is a reaction with one substrate and two products, so when substrate and product concentrations are similar but very low, the reaction quotient does not cancel evenly and a negative contribution to results.


The same result is observed with two substrates and two products if one of the substrates is H2O

The opposite effect, i.e. a large +ve contribution to , is observed if a reaction has two substrates and one product.

Phosphoglycerate kinase

1,3-bisPG+ ADP 3-PG +ATP = -18.8 kJ/mol
1 然100 然118 然1 mM
This reaction has a large -ve , but near zero . This step follows a pair of reactions with +ve , and as a result the concentration of substrate, 1,3-bisphosphoglycerate, is less that 1 然. In addition the reaction transfers phosphate to ADP and the typical [ATP]/[ADP] ratio in a cell is quite high. This results in a large positve contribution to . However there is enough of an energy boost to raise the product concentration so that the remaining steps of the pathway can be maintained.

Pyruvate kinase

PEP+ ADP pyruvate + ATP = -31.6 kJ/mol
Phosphoenolpyruvate has a huge -ve for hydrolysis:

PEP + H2O pyruvate + Pi = -61.9 kJ/mol

This means that transfer to ADP to make ATP has = -31.6 kJ/mol.


If is transferred away from PEP, the logical product should be enolpyruvate, not pyruvate. Enolpyruvate then isomerizes into pyruvate. Given Keq for the isomerization, its contribution to the overall can be determined.

= - RT ln Keq
= - 2.478 ln 110000
= - 2.478 x 11.61
= - 28.8 kJ/mol

Since (overall) = -31.6 kJ/mol, the actual bond only contributes -2.8 kJ/mol in the first step of the reaction. So the glycolysis reactions end by up creating a fairly high energy phosphate ester attached to an unstable isomer of pyruvate. The enol isomerization provides most of the excess energy seen in the pyruvate kinase reaction. This is discussed in Voet, p. 464, or in Stryer, p. 504.


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