Answer to Question #99292 in Molecular Biology for amir

Question

Rhodobacter lauracita is a newly discovered bacterium and is a manganesebased
photoautotrophy. Light provides the energy necessary to drive the
production of organic carbon. (Be careful, this question set is asking about
assimilation, rather than dissimilation reactions)
a. How much free energy in kJ/mol would be required to form one mole of
glucose (C6H12O6) if reduced manganese (Mn2+) served as the electron donor
for R. lauracita? Assume that Mn2+ is oxidized to the solid MnO2 (activity =
1), pH = 7, and that Mn2+ and glucose are present at 10-4 M. The partial
pressure of carbon dioxide is 10-3.5 atm. Give the answer in kJ per mole of
glucose produced.
E0’ (V)
1/2 MnO2(s) + 2 H+ + e- = 1/2 Mn2+ + H2O +0.58
1/4 CO2(g) + H+ + e- = 1/24 C6H12O6 + 1/4 H2O -0.43
Fe(OH)3(am) + 3 H+ + e- = Fe2+ + 3 H 2O +0.06

Answer 1

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