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