Lecture 16
Glycolysis
Glycolysis: oxidative breakdown of glucose to pyruvate with capture of some energy as
ATP, NADH (first step in respiration)
Themes:
sequential, simple enzymatic reactions
investment of energy; then recapture
control of pathway at key steps: ΔG <<0
production of useful intermediates
Glycolysis
glucose → pyruvate
Subsequent reactions
→ lactate, CO2 + ethanol, malate
→ CO2 + H2O
Compare respiration to combustion:
C6H12O6 + 6 O2 →
6 CO2 + 6 H2O
ΔGo’ = -2840 kJ/mol
In respiration, including glycolysis, some ΔGo’ (~ 35%) is retained as formation of ATP
Phases: priming (-2 ATP/glucose), payoff (+4 ATP, +2 NADH/ glucose)
Phase I: 5 reactions
1
2
3
4
glucose → G-6-P → F-6-P → F-1,6-bisP → DHAP + G-3-P
ATP → ADP
ATP → ADP
5
DHAP → G-3-P
Phase II: 5 reactions (note 2 G-3-P substrates per glucose)
NAD+ → NADH
G-3-P + Pi → 1,3-bisPGA → 3-PGA →
ADP → ATP
2-PGA → PEP →
ADP →
What happens to the pyruvate?
Anaerabic conditions: pyruvate → lactate
NADH → NAD+
pyruvate → CO2 + acetaldehyde → ethanol
NADH → NAD+
pyruvate + CO2 → OAA → malate
NADH → NAD+
Aerobic conditions: pyruvate → CO2 + H2O
Pyruvate
ATP
Reaction 1
Note the need for Mg2+
with ATP (common)
ΔGo’ =+13.8 kJ/mol
= -30.5
-16.7
ΔG’ = ΔGo’ + RTln[P]/[R]
= -33.9
Reactions with ΔG<<0 can
be regulated
Hexokinase is allosterically
inhibited by G-6-P,
Km ~ 0.1 mM ([glucose]
in blood ~ 4 mM)
Glucokinase in liver not
allosterically inhibited:
Km ~ 10 mM
Phosphorylation of glucose
traps it in the cell
Reaction 2
ΔGo’ =+1.67 kJ/mol
ΔG’ = -2.92 kJ/mol
Near equilibrium:
not regulated
Reaction 3
ΔGo’ =+16.3 kJ/mol
= -30.5
-14.2
ΔG’ = -18.89
ATP is an allosteric inhibitor
AMP reverses ATP inhibition
(ATP down 8% results in
AMP up 4-fold: p. 542)
Reaction 4
ΔGo’ =+23.8 kJ/mol
ΔG’ = -0.23 kJ/mol
Reaction 5
ΔGo’ =+7.56 kJ/mol
ΔG’ = +2.41 kJ/mol
Overall:
glucose + 2 ATP → 2 G-3-P + 2 ADP
ΔGo’ =+63.13 kJ/mol
−61.0
2.13
(glucose → 2 G-3-P)
(2 ATP → 2 ADP)
ΔG’ = -53.4 kJ/mol
(from relative concentrations in cells)
Problem
What is the ratio of [DHAP]/[G3P] in the erythrocyte, when ΔG = 2.41 ?
ΔG = ΔGo’ + RT ln ([G3P]/[DHAP])
2,410= 7,560 + (8.31)(298) ln ([G3P]/[DHAP])
ln ([G3P]/[DHAP]) = -2.1
[G3P]/[DHAP] = 0.125
[DHAP]/[G3P] = 8.0
(DHAP useful for making the glycerol in lipids)