We left off with the two 3-carbon phosphate molecules, dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. Dihydroxyacetone phosphate was converted to glyceraldehyde-3-phosphate, the molecule of choice to continue down the pathway, via triose phosphate isomerase. From here on out, every reaction we see involves two molecules of glyceraldehyde-3-phosphate: the original one produced from fructose-1,6-bisphosphate and the one converted from dihydroxyacetone phosphate. Up until this point, glycolysis has cost us energy. We’ll recoup that and net a total of two each: NADH, pyruvate, ATP, and water molecules in this second half of glycolysis. If you haven’t already, I highly suggest reading my Intro to Metabolism blog post and you can find the first half of glycolysis here.
Step 6: Glyceraldehyde-3-Phosphate to 1,3 Bisphosphoglycerate
The first step in the second half of glycolysis involves redox chemistry using NAD+ as oxidizing agent. Glyceraldehyde-3-phosphate dehydrogenase catalyzes the reaction between glyceraldehyde-3-phosphate, NAD+, and phosphate to produce 1,3 bisphosphoglycerate, NADH, and a hydrogen.
The aldehyde group of glyceraldehyde-3-phosphate is oxidized to a phosphoanhydride bond with the caboxylic acid of 1,3 bisphosphoglycerate and NAD+ is reduced to NADH in this reaction. Let’s see how this happens:
In the enzyme’s active site, a cysteine residue is present. The thiol group of cysteine is deprotonated, triggering its attack on glyceraldehyde-3-phosphate’s carbonyl group.
This creates a thiohemiacetal intermediate. This intermediate transfers a hydride to an NAD+ in the enzyme’s active site, causing its release. This hydride transfer also oxidizes our thiohemiacetal intermediate into an acyl thioester intermediate:
This is an important intermediate and demonstrates a point that I’ll get back to shortly.
A phosphate enters the enzyme’s active site and, because we have an unstable high-energy thioester, it attacks this carbon. This breaks the carbon-sulfur bond, releasing 1,3 bisphosphoglycerate and restoring the enzyme.
Going back to our high-energy thioester, I want to make a point as to why this intermediate is so important. A thioester is a high-energy bond that isn’t resonance-stabilized like a regular ester. A regular ester would form a much too low free energy state with the enzyme. The energy it would take to climb out of this “hole” so to speak would slow the reaction down greatly:
Versus what actually happens:
If you remember the section on the significance of high and low energy bonds from my introductory post, this step exemplifies why the relative energies matter.
Step 7: 1,3 Bisphosphoglycerate to 3-Phosphoglycerate
Here, we break even energy-wise (remember, we’re dealing with two molecules). This is a straightforward reaction in which phosphoglycerate kinase catalyzes substrate-level phosphorylation of ADP, producing ATP. Simply put, the phosphate on the 1 position of the molecule gets taken off, generating ATP:
This is a highly favorable reaction with a ΔG of -49.4 kJ/mol
Step 8: 3-Phosphoglycerate to 2-Phosphoglycerate
In this step, the phosphate group is being moved from the terminal hydroxide of the molecule to the secondary hydroxide – it gets moved from the 3 position to the 2 position.
Simple, yes, but why? The final product of glycolysis is pyruvate which has a carbonyl at the 2 position. If this step did not occur, we’d ultimately end up with a product that has a carbonyl at the 3 position – something that has metabolic consequences. Rather than having a ketone, we’d have an aldehyde which, by itself is toxic, and would prevent subsequent reactions. It’s good exercise to think how a misplaced carbonyl would affect the TCA cycle. There’s several “cross-road” compounds that we’ll encounter that serve important biochemical functions. When we get to the TCA cycle, we’ll see how intermediates can be drawn off (or fed into) for other purposes. As such, if these are malformed, downstream reactions will see the consequence. For example, having a carbonyl at the 2 position allows us to make alpha-amino acids – a nitrogen can be placed where the carbonyl is. If at the 3 position, we’d have a beta-amino acid – something that cannot be used to make proteins. This is an example of that chemical logic I continually keep bringing up.
Alright, let’s look at the mechanism by which phosphoglycerate mutase accomplishes this transfer.
Phosphoglycerate mutase is a phosphoenzyme – it contains a phosphate attached to a histidine residue within its active site. This is a high energy bond, much like the thioester linkage we saw earlier. Once 3-phosphoglycerate binds to the enzyme, it’s hydroxyl group at the 2 position is attacked by this high energy phosphate, giving us 2,3 bisphosphoglycerate:
2,3 bisphosphoglycerate is involved in the allosteric changes to hemoglobin, changing its oxygen-binding affinity. This is an important molecule that, if you’ve ever taken a biochemistry course, have likely discussed. Now you know where it comes from. 2,3 bisphosphoglycerate can either leave the active site of the enzyme, killing it, or rephosphorylate the histidine from the 3 position, producing 2-phosphoglycerate. If the latter happens, we’ve effectively switched the position of the phosphate group. This phosphate is not the same as the original one that entered – that one is what was used to phosphorylate the enzyme, allowing the cycle to repeat.
Step 9: 2-Phosphoglycerate to Phosphoenolpyruvate
The next step involves the enzyme enolase. Enolase makes a double bond between the terminal and 2-carbon positions of 2-phosphoglycerate through a dehydration reaction, forming the high-energy intermediate, phosphoenolpyruvate. During the catalytic process, there is a rapid equilibrium with water. A deprotonation/reprotonation takes place (which was discovered through isotopic labeling), that I haven’t shown. This, however, tells us that this enolase’s active site is not closed off to the aqueous environment.
This is a metalloenzyme that utilizes magnesium to orient 2-phoshoglycerate within its active site.
A lysine residue extracts a proton from the carbon at the second position, forming an enolate. This enolate is stabilized by magnesium and other positively charged residues in the active site of the enzyme. The hydroxyl group can be eliminated through protonation from a glutamate residue, giving us our product, phosphoenolpyruvate:
This is a highly exergonic reaction with a ΔG of -61.9 kJ/mol, making it functionally irreversible.
Step 10: Phosphoenolpyruvate to Pyruvate
This is our last step in glycolysis as well as our last energy pay-off step. Pyruvate kinase catalyzes the substrate-level phosphorylation (like we’ve seen before) of ADP to form pyruvate and ATP from phosphoenolpyruvate.
ADP attacks the phosphoenolpyruvate’s phosphate group, forming ATP and leaving enolpyruvate. This first step is endergonic – that is, it’s not favored by itself. Enolpyruvate, however, tautomerizes to the keto form, pyruvate. This is a highly exergonic transition. So when coupled together, the overall ΔG is negative. Just as the previous step, this is a functionally irreversible reaction.
That’s glycolysis – simple right? We saw two stages: a preparative stage and an energy-production stage all occurring within the cytosol of the cell. This is a fairly fast pathway – provided there’s oxidized cofactors around, when glucose is stuffed in, lots of pyruvate and ATP comes out. Here’s our net reaction:
Glucose + 2 NAD+ + 2 ADP + 2 Pi –> 2 NADH +2 pyruvate + 2 ATP +2 H20 + 4H+
What about our other sugars like fructose, galactose, and mannose? How are those broken down? There are some notable differences that I’ll cover next. Fructose metabolism in particular, is a controversial area and a very active area of research. This will transition us into regulation and flux control of glycolysis before diving into oxidative phosphorylation.