Before beginning the tutorial on this reaction revisit oxidation/reduction.This series of reacions continues the story of how the previously 'lysed' glucose remnants are further broken down. We begin with the two molecules of PGAL which were produced at the end of glycolysis I. Click View Model (above) and print out the glycolysis II reactions to follow during this tutorial. You'll notice a '2X' on each reaction arrow as a reminder that each stem in this sequence occures twice per original glucose molecule.
Before beginning the tutorial on this reaction revisit oxidation/reduction.
The reactant is PGAL. In the illustration it is shown losing a hydride ion (curved arrow). Several things are going:
Oxidation is the loss of electrons (remember OIL) from the reactant (PGAL). Reduction is the gain of electrons (RIG) by a reducing agent. In this case the reducing agent is the transporter NAD+. Two electrons are lost as a hydride ion. There is a positive charge remaining on the atom where electrons were removed.
The enzyme responsible is PGAL dehydrogenase indicating that PGAL has lost hydrogen. As indicated in a previous tutorial ( NAD+/NADH ) a second hydrogen in the form of an hydrogen ion (H+) is often lost from the same molecule. However, in the current reaction, the hydrogen ion is supplied by a water molecule (red). The reduced form of the transporter is NADH.
The hydroxyl ion (OH-) (red) from the water molecule is combined with a new hydrogen ion (red) that comes from monohydrogen phosphate (HPO4-2); this generates a new water molecule (note arrows). The remainder of this molecule is the negatively charged phosphate ion (PO4 -3) (blue). This phosphate ion will attach to the positive charge (teal) on the PGAL intermediate (note arrow) forming the product 1,3-bisphosphoglycerate. (The final product has a double-bonded oxygen on #1 with an hydroxyl on an adjacent carbon, #2; this is classified as an aldol.) Its prominent feature is the presence of a phosphate group at each end. Verify that the molecule's name indicates this as well as the carbons to which they are attached. Of future interest is the fact that phosphorylation using an inorganic phosphate adds a fourth oxygen instead of three as when ATP is the source.
This is our first encounter with the formation of an ATP; it is referred to as 'substrate-level
phosphorylation' because the source of ATP's third phosphate group is the reactant (substrate).
When the reactant loses the phosphate group (blue) an oxygen (red) remains behind. The phosphate group
(-PO3-2) will be picked up by ADP that loses a hydrogen ion (H+) to provide
a bonding site. This hydrogen ion becomes part of the hydrogen ion pool. The enzyme that catalyzes this
reaction is phosphoglycerate kinase. Remember that a 'kinase' indicates ATP is involved in the
reaction.
As shown at the left, the reactant, 3-phosphoglycerate, will undergo an internal rearrangement without the lose or gain of any
atoms. The product formed will be an 'isomer' of the reactant; they will both have the
same molecular formulas but different molecular structures. The product is 2-phosphoglycerate
because the #2 carbon now possesses the phosphate group. The red atoms show the before and after
locations of the atoms involved.
The enzyme that catalyzes this reaction is phosphoglycerate mutase because it has 'mutated' the reactant.
The product of the previous reaction, 2-phosphoglycerate, is dehydrated to form phosphoenolpyruvate (PEP).
The atoms lost as a water molecule are shown in teal within the reactant. Compare the bonds between
carbons #2 and #3 in the reactant and the product; the single bond has become a double bond.
The enzyme catalyzing this reaction is enolase.
The remaining phosphate will finally be removed leaving behind pyruvate. As in step 7 the acceptor of
this phosphate is ADP to form an ATP. Note that the oxygen (teal) that was binding the phosphate group to the reactant
is left behind and becomes double-bonded in the product. The third hydrogen (blue) seen on carbon #3 of pyruvate was
donated by the ADP molecule providing binding room for its new phosphate group. The enzyme for this reaction
is pyruvate kinase ... remember that a kinase involves ATP.
After each PGAL contributes to the reduction of two (2) molecules of NAD+, each of the two (2) three-carbon molecules receives another phosphate from inorganic phosphates. These are then removed forming two (2) ATP molecules. Shortly thereafter, the remaining phosphate groups are removed to form two (2) more ATP molecules. The end molecules are two (2) three-carbon pyruvate ions.
Continue to the intermediate step and Krebs' cycle.