Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chemistry 438 - Spring 2011

BIOCHEMISTRY REVIEW

Week 13

ETS & Oxidative Phosphorylation

Redox Energy Calculation:

E°'NAD+ = - 0.13 V (NAD+ + H+ + 2eright arrow NADH)

E°'O2 = 0.815 V (1/2 O2 + 2H+ + 2eright 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)
ATP Equivalents
(@2.5 ATP/NADH)
Oxygens consumed 
Hexokinase ADP 1 -1
-1
 0
PFK  ADP 1 -1
-1
0
Ga-3-P DH (via Malate-Asp Shuttle) NADH 2 5
6
2
PGA Kinase ATP 2 2
2
0
Pyruvate Kinase ATP 2 2
2
0
Pyruvate DH Complex

 NADH

 2

5
6
 2
Isocitrate DH NADH

2

5
6
 2
2-oxoglutarate DH Complex NADH

2

5
6
 2
Succinyl-CoA Synthetase GTP

2

2
2
 0
Succinate DH FADH2

2

3
4
 2
Malate DH NADH

2

5
6
 2

  TOTAL=

32
38
 12
P/O = ATP/O =
32/12 = 2.67
38/12 = 3.17
 

 

 

 

 

 

 

 

Do this calculation again for glucose in insect flight muscle.

 Reaction
Energy Product

factor
ATP Equivalents
(@2.5 ATP/NADH)
ATP Equivalents
(@2.5 ATP/NADH)
Oxygens consumed 
Hexokinase ADP 1 -1
-1
 0
PFK  ADP 1 -1
-1
0
Ga-3-P DH (Glycerol-P shuttle) NADH 2 3
4
2
PGA Kinase ATP 2 2
2
0
Pyruvate Kinase ATP 2 2
2
0
Pyruvate DH Complex

 NADH

 2

5
6
 2
Isocitrate DH NADH

2

5
6
 2
2-oxoglutarate DH Complex NADH

2

5
6
 2
Succinyl-CoA Synthetase GTP

2

2
2
 0
Succinate DH FADH2

2

3
4
 2
Malate DH NADH

2

5
6
 2

  TOTAL=

30
36
 12
P/O = ATP/O =
30/12 = 2.5
36/12 = 3.0
 

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.

 

 

 

Schedule

 

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©R A Paselk

Last modified 19 April 2011