Redox Energy Calculation:
E°'NAD+ = - 0.13 V (NAD+ + H+ + 2e– NADH)
E°'O2 = 0.815 V (1/2 O2 + 2H+ + 2e– H2O)
E°' = E°'e-acceptor - E°'e-donor
E°' = 0.815 V - (- 0.13 V) = 1.130 V = 1.130 J/C
G°' = -nFE°' = - (2 mol e-)(96,485 C/mol e-)(1.130 J/C) = 218,056 J/mol = 218.06 J/mol
Assume that the hydrolysis energy of ATP = 30.0 kJ/mol
Efficiency = (actual/possible) x 100%
actual = 2.5 (current accepted value)
possible = 218.6 / 32.0 = 6.83
Efficiency = (2.5 / 6.83) x 100% = 36.6%
(assuming 3.0 ATP/NADH, Efficiency = (3.0 / 6.83) x 100% = 43.9%)
Consider the relative reduction pattern of the ETS in the presence of NADH (highly reduced) oxygen (highly oxidized) and low [ADP].
What would a plot of reduction vs. ETS components (e.g. NADH, FADH2, CoQH2, cyt C, etc.) look like?
Click here for a Simulated BAR GRAPH
- Due to tight coupling of ETS to ATP production, the low [ADP] will slow electron transport at each complex where proton pumping occurs (Complexes I, III, and IV).
- Thus the componenets before each step will be more reduced than those after:
- NADH and FADH2 will be highly reduced
- CoQ, cyt b and c1 will be intermediate
- cyt c, cyt a, a3 will be highly oxidized.
Calculate the P/O ratio (ATPs produced for each oxygen atom consumed) for the aerobic catabolism of glucose in mammalian fast twitch muscle. Show all work in a table such as we have used in class.
Do this calculation again for glucose in insect flight muscle.
Why should this difference exist? (That is, why should evolution favor this lower value of energy for insect flight muscle?) - Because of the high rate of glycolysis in active flight muscle, it is critical to keep glycolysis flux at a high rate. It is thus worth while for the insect to lose some ATP in order to insure a high [NAD+]/[NADH]ratio to insure the Glyceraldehyde-3-P DH reaction does not slow the overall glycolytic flux.
©R A Paselk
Last modified 19 April 2011