Anjan Kr Dasgupta
Lecture 3
Anjan Kr Dasgupta
Lecture 3
Let us recall the linear phenomenological equation as we discussed before
\[J_i=\Sigma L_{ij} X_j\]
Few important attributes of the above are:
- J and X are fluxes and forces (efffects and causes)
- In chemical systems X is sign changed free energy
- In transport or diffusion systems X is electrochemical potential gradient
In oxidative phosphorylation we have
- Flux \[J_P\] the rate of ATP synthesis
- Flux \[J_o\] the rate of oxygen uptake
\[ J_o= L_{OO}(-\Delta G_o)+ L_{OP}(-\Delta G_P)\] \[ J_P= L_{PO}(-\Delta G_o)+ L_{PP}(-\Delta G_P)\] We know that oxidative process and phosphorylation process are indeed coupled.But how can we have in absence of any common high energy intermediate, \[L_{OP} \ne 0,L_{PO}\ne 0\]Secondly, will \[L_{OP}=L_{PO}\] be satisfied? If so, it will be a proof for validity of Onsager reciprocal relation in biological systems.
\[oxidant + e^{-} \rightarrow reductant \]
Oxidant | Reductant | n | E´0 (V) |
---|---|---|---|
Succinate + CO2 | alpha-Ketoglutarate | 2 | - 0.67 |
Acetate | Acetaldehyde | 2 | - 0.60 |
Ferredoxin (ox) | Ferredoxin (red) | 1 | - 0.43 |
2 H+ | H2 | 2 | - 0.42 |
NAD+ | NADH + H+ | 2 | - 0.32 |
NADP+ | NADPH + H+ | 2 | - 0.32 |
Oxidant | Reductant | n | E´0 (V) |
---|---|---|---|
Lipoate (oxidized) | Lipoate (reduced) | 2 | - 0.29 |
Glutathione (ox) | Glutathione (red) | 2 | - 0.23 |
FAD | FADH2 | 2 | - 0.22 |
Acetaldehyde | Ethanol | 2 | - 0.20 |
Pyruvate | Lactate | 2 | - 0.19 |
Fumarate | Succinate | 2 | 0.03 |
Oxidant | Reductant | n | E´0 (V) |
---|---|---|---|
Cytochrome b (+3) | Cytochrome b (+2) | 1 | 0.07 |
Dehydroascorbate | Ascorbate | 2 | 0.08 |
Ubiquinone (ox) | Ubiquinone (red) | 2 | 0.10 |
Cytochrome c (+3) | Cytochrome c (+2) | 1 | 0.22 |
Fe (+3) | Fe (+2) | 1 | 0.77 |
½ O2+ 2 H+ | H2O | 2 | 0.82 |
The relation used to translate redox potential to free energy is: \[\Delta G^{o'} = nF E_o'\] - n is the number of electrons transferred
- F is a proportionality constant called the faraday
- F=23.06 kcal mol-1 V-1
- F= 96.48 kJ mol-1 V-1
- Redox potential is in volts, free energy is in kilocalories or kilojoules per mole.
scale=0.1
Oxidant | Reductant | n | E´0 (V) |
---|---|---|---|
NAD+ | NADH + H+ | 2 | - 0.32 |
Pyruvate | Lactate | 2 | - 0.19 |
For the reaction involving conversion of the nicotinamide adenine dinucleotide from its its reduced to oxidized form: \[NADH \rightarrow NAD+ H^+ + 2e^- (\Delta G_1)\] the free energy is given by
## Free energy in KJ /mole = -61.75
For the pyruvate \[pyruvate + 2H^+ + 2e^- \rightarrow Lactate (\Delta G_2)\]
## Free energy in KJ /mole = 36.66
\[ \Delta G_1 + \Delta G_2\] (KJ/Mole=)
## [1] -25.0848
2nd law demands \[ J1X1+J2X2>0 \] If J1X1 <0 , then '1' is driven and 2 is dricing reaction and vice versa. - The driving force in this reaction is given by oxidation of NADH - The driven reaction in this case is the pyruvate to lactate conversion ## A few Notes on Fluxes and Forces
One can see that in this case \[\Delta G_1 <0 , \Delta G2>0 \] implying \[X_1>0 , X_2<0 \]
In the phenomenological context we can define the two fluxes as J1 and J2 (assuming both are positive).
The non-equilibrium efficiency of the process is given by : \[ \eta = -\frac{J_1 (-\Delta G_1 )}{J_2 (-\Delta G_2)}\] One can easiliy prove that second law of thermodynamics implies that \[ \eta \le 1\].
This difference drives Electron Transport Through the Chain and Favors the Formation of a Proton Gradient.In this case let us take the two redox couples namely oxygen /water and NADH/NAD having potentials -0.32 and +0.82 V. The overall all direction will be towards the terminal acceptor oxygen and the magic number 1.14 V is obtained. ## Data Bases and links
The basic assumption with which Mitchell started was the importance of proton gradient .Under typical conditions for the inner mitochondrial membrane we may assume that the outside environment is 1.4 pH unit lower and if the membrane potential is of the order of 140 mV the energy carried by the protons will be of the order of 21KJ/mole