Answer to Question #99505 in Molecular Biology for amir

Question #99505
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
Expert's answer
Dear amir, your question requires a lot of work, which neither of our experts is ready to perform for free. We advise you to convert it to a fully qualified order and we will try to help you. Please click the link below to proceed: Submit order

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be the first!

Leave a comment

New on Blog