Following a previous post, in which I have detailed the reactions of glycolysis (preparatory phase), here are the reactions of the second part, the payoff phase.
This reaction is catalyzed by glyceraldehyde-3-phosphate dehydrogenase and it is a double reaction. The substrate will be oxidized (the carbonyl group is a carboxylic acid), and then will be added to a carboxylic group phosphorylates newly formed through a phosphoester linkage. Therefore, the reaction produces NADH (oxidation of substrate) and consumes inorganic phosphate (Pi). This addition of the phosphoryl group will give the product (1,3-bisphosphoglycerate) a high potential for transfer of phosphoryl group, ie, the molecule will become chemically unstable, because it is a small molecule (only have 3 carbons) but with many electronegative atoms (electrostatic repulsion will be many). This instability will be useful to understand what goes on in the following reaction...
This is the first reaction of glycolysis that will lead to the synthesis of ATP and requires the presence of Mg2+. It is catalyzed by the enzyme phosphoglycerate kinase (the name derives from the enzyme reverse reaction, which occurs during photosynthetic fixation of CO2). I've mentioned in the previous reaction that formed a compound chemically unstable. This instability will cause the 1,3-bisphosphoglycerate to display a tendency to lose one of its phosphoryl groups (leading to 3-phosphoglycerate), releasing a large amount of energy (remember that a molecule is much more stable and have less internal energy…). Therefore, if it releases the phosphoryl group, it becomes more stable, because it loses energy. This energy will be used to engage a reaction to the second reaction (which requires energy to occur): the synthesis of ATP! In fact, the phosphoryl group is added to the released ADP, resulting in ATP. As the phosphoryl group is derived from a substrate, the synthesis of ATP is called phosphorylation level of the substrate. However, the question that probably some will be thinking is: "So it was better to add once the phosphoryl to ADP in the previous reaction? Why do we bother to first transfer it to a interemediário, if we in the following reaction remove? " The answer to this question relates to the thermodynamics of the reactions involved. That is, the energy released in the oxidation reaction in the existing 6 is not as high as it is necessary to spend in the synthesis of ATP. Therefore, there are only two chances ... Or do not use that energy from oxidation, but this is a waste, or retains a part of the energy released in the form of chemical energy, the 1,3-bisphosphoglycerate (this is what happens!). So in the following reaction we'll have enough energy to synthesize ATP. Together, the steps 6 and 7 show the role of phosphate groups in the conservation of metabolic energy, which I mentioned in post about the reactions of the preparatory phase of glycolysis.
The enzyme that catalyzes this reaction is the phosphoglycerate mutase and is dependent on Mg2+. What will make this enzyme is to transfer the phosphate group from position 3 to position 2 of the substrate. In reality, this transfer is not direct ... The active enzyme has a phosphate group, which transfers to the substrate, leading to an intermediary known as 2,3-bisphosphoglycerate. This molecule is an important modulator of the affinity of hemoglobin for O2! Thereafter, the 2,3-bisphosphoglycerate yields a phosphate group to the enzyme, converting it into 2-phosphoglycerate and regenerating the active form of the enzyme.
This reaction is catalyzed by enolase and is an example of a reaction of elimination. What this enzyme will do is remove a water molecule of the substrate, creating a double bond. As the double bonds are electron-rich regions, its presence in the vicinity of the phosphate group will make your product, phosphoenolpyruvate, again presents a high potential for transfer of phosphoryl group. This reaction requires the presence of Mg2+.
We came to the last reaction of glycolysis ... In this reaction, catalyzed by pyruvate kinase (the third regulatory enzyme of glycolysis!), the phosphoenolpyruvate will lose the phosphoryl group, releasing a large amount of energy. Again, this energy will be partially preserved in the form of an ATP molecule. This reaction is irreversible and requires the presence of Mg2+ (or Mn2+) and K+.
Main bibliographic sources:
- Quintas A, Freire AP, Halpern MJ, Bioquímica - Organização Molecular da Vida, Lidel
- Nelson DL, Cox MM, Lehninger - Principles of Biochemistry, WH Freeman Publishers