Glycolysis

2017-11-08


Conceptual goals

  • Understand the chemical transformations by which glucose is converted to pyruvate
  • Understand how the oxidation of glucose can release energy that can be captured as ATP

Skill goals

  • Reason about metabolic energy extraction
  • Reason about metabolic pathway regulation

Are the carbons more or less oxidized than $CO_{2}$? Less

Than hexane? More

If we oxidize glucose to $CO_{2}$, we get -2850 kJ/mol

$C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O$ would release all $2850\ kJ \cdot mol^{-1}$ at once.

This needs to be captured in incremental steps

Organisms have complex pathways for dealing with sugar "metabolites"

Glycolysis is a partial oxidation of glucose to two pyruvate molecules

Glycolysis is universal and ancient.

Glycolysis occurs in 10 steps, only three of which are highly favorable

Glycolysis outline

  • Pay 2 ATP
  • Oxidize carbon and extract 4 ATP

glucose to glucose-6-phosphate

Uses hexokinase, burns one ATP

This is effectively irreversible (-16.7 kJ/mol).

Charge on G6P prevents diffusion out of cells

isomerize to fructose-6-phosphate

phosphoglucose isomerase

$\Delta G$ is -1.4 kJ/mol. Basically reversible.

General principle: intermediate steps often freely reversible.

fructose-6-phosphate to fructose-1,6-bisphosphate

phosphofructokinase, burns one ATP

$\Delta G$ is -16.2 kJ/mol. This is an irreversible step.

This enzyme is allosterically regulated by ADP and PEP (bacteria) and fructose-2,6-biosphosphate (mammals)

Glycolysis is regulated at the phosphofructokinase step.

  • Activated by excess ADP (bacteria) or excess glucose via fructose-1,6-bisphosphate (mammals)
  • Deactivated by excess product (bacteria). "Product inhibition" is critical for many pathways

Why does it make sense to regulate a step with a large drop in $\Delta G$?

This makes a "one-way" valve.

fructose-1,6-bisphosphate to dihydorxyacetone phosphate and glyceraldehyde-3-phosphate

aldolase

$\Delta G$ near 0 in vivo

We've split glucose in half at this point.

DHAP gets converted to G3P

triose phosphate isomerase

$\Delta G$ slightly positive, but G3P gets used quickly thus "pulling" reaction forward.

Now we have two copies of glyceraldehyde-3-phosphate. The remaining pathway is doubled

Big ideas (so far):

  • Metabolic pathways are usually a mixture of downhill irreversible steps and isoenergetic steps.
  • Small unfavorable steps can be overcome by downstream favorable steps that "pull" material through the pathway.
  • Irreversible steps are points of regulation because they can act as a one-way valve.
  • Excess substrates often activate pathway (ADP or F26BP, indicating excess glucose).
  • Products often inhibit a pathway via "product inhibition."

So far we've stuck two phosphates onto a sugar, rearranged it a few times, and broke it in half.

Discuss:

  • Have we changed the oxidation state of any of the carbons? No.
  • If we pulled the phosphates off could we recharge an ADP molecule? No.
    • We need to do some oxidation!

If we're going to oxidize carbon, we need to reduce something else. What will it be?

$NAD^{+}$

glyceraldehyde-3-phosphate + $NAD^{+}$ + $P_{i}$ $\rightarrow$ 1,3-bisphosphoglycerate + $NADH$ + $H^{+}$

glyceraldehyde-3-phosphate dehydrogenase

oxidizes an aldehyde and extracts a hydride to $NAD^{+}$ to make $NADH$

unfavorable (+6.7 kJ/mol)

1,3 bisphosphoglycerate + ADP $\rightarrow$ 3-phosphoglycerate + ATP

phosphoglycerate kinase

$\Delta G$ is -18.8 kJ/mol, which lets this drag previous reaction forward

This is the first energy release. 1 ATP per 3-phosphoglycerate; 2 per glucose

3-phosphoglycerate $\rightarrow$ 2-phosphoglycerate

phosphoglycerate mutase

$\Delta G$ near 0

rearrangement of phosphate

2-phosphoglycerate $\rightarrow$ phosphoenolpyruvate

enolase

$\Delta G$ near 0

basic dehydration reaction

phosphoenolpyruvate $\rightarrow$ pyruvate

pyruvate kinase

yields 1 ATP per pyruvate (2 per glucose)

highly favorable, basically irreversible

Last step is three reactions that, together, are favorable

$PEP \rightarrow enolpyruvate$ (-16 kJ/mol) and $ADP + P_{i} \rightarrow ATP$ (30.5 kJ/mol)

Not enough to pay for making ATP.

Unfavorable reaction is coupled (in the same enzyme) to a favorable reaction

$enolpyruvate \rightarrow pyruvate$ is (-41 kJ/mol)

Reaction is net favorable by $-16 + 30.5 - 41 = -26.5\ kJ \cdot mol^{-1}$

Summary

  • Each step is catalyzed by a distinct enzyme.
  • Free energy consumed/released is transferred by ATP and NADH.
  • Rate of the pathway can be controlled by altering activity of individual enzymes.
    • Key, irreversible steps are regulated.
    • Molecules like low-energy state (ADP) increase "flux"; high-energy state molecules (ATP) decrease "flux."
  • Oxidation of carbon provides the energy to produce ATP.