Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chemistry 438 - Spring 2013

Redox Energy Calculation:

E°'NAD+ = - 0.13 V (NAD+ + H+ + 2e– right arrow NADH)

E°'O2 = 0.815 V (1/2 O2 + 2H+ + 2e– right arrow 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?

Answer:

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.

 

 

 

 

 

 

 

 

Calculations of P/O ratios

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.

 Reaction
 Energy Product

factor
ATP Equivalents
(@2.5 ATP/NADH)
 Oxygens consumed 
Hexokinase ADP 1 -1  0
PFK  ADP 1 -1 0
Ga-3-P DH (via Malate-Asp Shuttle) NADH 2 5 2
PGA Kinase ATP 2 2 0
Pyruvate Kinase ATP 2 2 0
Pyruvate DH Complex

 NADH

 2

5
  2
Isocitrate DH NADH

2

5
  2
2-oxoglutarate DH Complex NADH

2

5
  2
Succinyl-CoA Synthetase GTP

2

2
  0
Succinate DH FADH2

2

3
  2
Malate DH NADH

2

5
  2

  TOTAL=

32
  12
P/O = ATP/O =
32/12 = 2.67
  

 

 

 

 

 

 

 

 

 

 

 

 

Do this calculation again for glucose in insect flight muscle.

 Reaction
 Energy Product

factor
ATP Equivalents
(@2.5 ATP/NADH)
 Oxygens consumed 
Hexokinase ADP 1 -1  0
PFK  ADP 1 -1 0
Ga-3-P DH (Glycerol-P shuttle) NADH 2 3 2
PGA Kinase ATP 2 2 0
Pyruvate Kinase ATP 2 2 0
Pyruvate DH Complex

 NADH

 2

5
  2
Isocitrate DH NADH

2

5
  2
2-oxoglutarate DH Complex NADH

2

5
  2
Succinyl-CoA Synthetase GTP

2

2
  0
Succinate DH FADH2

2

3
  2
Malate DH NADH

2

5
  2

  TOTAL=

30
  12
P/O = ATP/O =
30/12 = 2.5
  

 

 

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.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Calculate the P/O ratio for the aerobic catabolism of stearate (C=18). Show all work in a table such as we have used in class.

 Reaction Energy Product Factor Multiplier ATP Equiv.
ATP Equivalents
(@3 ATP/NADH)

 O's
FACoA Syn. AMP -2 1 -2 -2 0 
Flavin DH FADH2 1.5 8 12 16 8
NAD+ DH NADH 2.5 8 20 24  8

8 AcCoA's to Kreb's TCA Cycle
NADH 2.5 x 3 9 67.5 81  27
GTP 1 9 9 9  0
FADH2 1.5 9 13.5 18  9

Total=

 120 146  52
P/O = ATP/O =
 120/52 = 2.31 146/52 = 2.81  

 

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©R A Paselk

Last modified 3 April 2013