Both photophosphorylation and oxidative phosphorylation utilize the energy gener

Question

Both photophosphorylation and oxidative phosphorylation utilize the energy generated by a proton gradient to generate ATP as explained by the chemiosmotic hypothesis.

A)What is the chemiosmotic hypothesis?

B)In a gram negative bacterium the DpH and the Dy are .60 and 0.13V, respectively, per pair of electrons transferred through the electron transport chain. What percentage of energy is conserved in the proton gradient relative to the amount of standard free energy produced by the oxidation of NADH by O2.

C)Assuming physiological conditions in this bacterium are close to standard conditions, how many moles of ATP can be generated per pair of electrons passed down the electron transport chain?

D)Write the net reaction for photophosphorylation and calculate the percentage of energy conserved in the ATP and NADPH generated by this reaction using 8 photons with a wavelength of 450nm.

E)Calculate the DGo’ for the reduction of pyruvate to lactate by NADH.

Explanation / Answer

I would be able to answer only a and E                                                                                                                         A. The chemiosmotic hypothesis theory says that most ATP synthesis in respiring cells comes arise from electrochemical gradient across the inner membranes of mitochondria which uses the energy of NADH and FADH2 formed from the breaking down of energy-rich molecules such as glucose during glycolysis and TCA cycle.

E. The reduction of pyruvate to lactate require oxidation of NADH and reaction are as follow

Pyruvate + 2H+ + 2e' ® Lactate e0' = -0.185 volts

NAD+ + 2H+ + 2e' ® NADH + H+ e0' = -0.320 volts now the difference in reduction potentials (De0') for the two half-reactions is calculated as follows:

De0' = e0'(reduction) - e0'(oxidation)

De0' = -0.185 - (-0.320) = +0.135 volts

DG0' = -nÁ De0'

where n = the number of electrons transferred, Á is the Faraday constant (96.5 kJ/volt mol)

DG0' = -2*96.5kJ/volt mol*0.135 volt so DG0' = -26.0 kJ/mol

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