1CHEM205 Problem Set 6.Oxidative Phosphorylation and PhotosynthesisQuestion 1 (6 points). Normally ATP synthesis is tightly coupled to electron transferthrough the electron transport chain. Under these conditions, the ratio of ATP producedper atom of oxygen consumed (P/O ratio) is about 2.5. Predict the effect of a lowconcentration of an uncoupling agent on the P/O ratio.Uncouplers depolarize the inner mitochondrial membrane, and ATP synthesis iscompromised. The electron transport chain remains unaffected, so protons are stillpumped by complexes I, III and IV. Protons pumped from the matrix to theintermembrane space are rapidly returned to the matrix by the membrane-solubleuncoupler, so energy that would normally be used to activate ATP synthase is ultimatelyexpended as heat. Low concentrations of uncouplers dissipate some of the protongradient necessary for ATP synthesis. Overall, more protons have to be transferredacross the membrane for each ATP synthesized. This requires more NADH per ATPsynthesized, so the P/O ratio will decrease.Question 2 (12 points). For each of the four complexes in the electron transport chain:(a) calculate the total redox potential of the complex(b) calculate the free energy available for proton translocation assuming a 2-electronprocess for each complex(c) calculate how many moles of protons can be translocated across the innermitochondrial membrane if translocation of 1 mole requires 23 kJ.Complex I: NADH is oxidized in a 2 electron process, and UQ is reduced, also in a 2electron process. The standard reduction potentials are:NAD+ + 2H+ + 2e- Æ NADH + H+E° = -0.32 VUQ + 2H+ + 2e- Æ UQH2 E° = 0.06 V(a) redox potential for complex I DE° = 0.38 V(b) DG° = -nFDE°so for a 2 electron process DG° = -73.7 kJ/mol(c) 3.19 moles of protons can be translocated2Complex II: succinate is oxidized in a 2 electron process, and UQ is reduced, also in a 2electron process. The standard reduction potentials are:fumarate + 2H+ + 2e- Æ succinate E° = 0.031 VUQ + 2H+ + 2e- Æ UQH2 E° = 0.06 V(a) redox potential for complex II DE° = 0.029 V(b) DG° = -nFDE°so for a 2 electron process DG° = -5.6 kJ/mol(c) no protons are translocated by complex IIComplex III step 1: UQH2 is oxidized in a 2 electron process. Cytochrome c is reducedand UQ is reduced to UQH. in two 1 electron processes. The standard reductionpotentials are:UQ + 2H+ + 2e- Æ UQH2 E° = 0.06 Vcyt c (Fe3+) + e- Æ cyt c (Fe2+) E° = 0.254 VUQ + H+ + e- Æ UQH. E° = 0.03 V(a) redox potential for step 1 DE° = 0.224 V(b) DG° = -nFDE°so for a 2 electron process DG° = -43.2 kJ/molComplex III step 2: UQH2 is oxidized in a 2 electron process. Cytochrome c is reducedand UQH. is reduced to UQH2 in two 1 electron processes. The standard reductionpotentials are:UQ + 2H+ + 2e- Æ UQH2 E° = 0.06 Vcyt c (Fe3+) + e- Æ cyt c (Fe2+) E° = 0.254 VUQH. + H+ + e- Æ UQH2E° = 0.19 V(a) redox potential for step 2 DE° = 0.384 V(b) DG° = -nFDE°so for a 2 electron process DG° = -74.1 kJ/molThe total free energy for steps 1 and 2 is DG° = -117.3 kJ/mol(c) 5.1 moles of protons can be translocated in the 2 steps of complex III3Complex IV: cytochrome c is oxidized in two 1 electron processes, and 0.5 O2 isreduced in a 2 electron process. The standard reduction potentials are:cyt c (Fe3+) + e- Æ cyt c (Fe2+) E° = 0.254 V0.5 O2 + 2H+ + 2e- Æ H2O E° = 0.816 V(a) redox potential for complex IV DE° = 0.308 V(b) DG° = -nFDE°so for a 2 electron process DG° = -59.4 kJ/mol(c) 2.58 moles of protons can be translocatedQuestion 3 (6 points). When the NADPH/NADP+ ratio in chloroplasts is high,photophosphorylation is predominantly cyclic.(a) is O2 evolved during cyclic photophosphorylation?(b) is NADPH produced?Briefly explain your reasoning.Oxygen is not evolved because photosystem II is not involved in cyclicphotophosphorylation. The excited electron from P700 is transferred back tocytochrome b6f via the plastoquinone pool. Transfer to plastocyanin ultimately returnsthe electron to reduce P700+. NADPH is not produced because no electrons aretransferred from ferredoxin to ferredoxin:NADP+ oxidoreductase.Question 4 (10 points). 14CO2 is administered to a green plant in a closed system. Aftera short time, the 12CO2 and 14CO2 levels in the system reach equilibrium. The plant isharvested, and the following substances are isolated: glucose, sedoheptulose-7-phosphate, ribose-5-phosphate, erythrose-4-phosphate and 3-phosphoglycerate. Drawthe structures of these five molecules and indicate which atoms will be 14C labeled.After a short time, C-3 and C-4 of glucose, C-3 C-4 and C-5 of sedoheptulose, C-1 C-2and C-3 of ribose, C-1 and C-2 of erythrose, and C-1 of phosphoglycerate. glucose sedoheptulose ribose erythrose phosphoglycerateAfter equilibrium is reached, all the carbons of all the molecules will be labeled.CH2OHOHHOOHHOHHOHHCH2O PCH2O POHHHOOHHOHHCHOHCHOCH2O POHHOHHOHHCHOCH2O POHHOHHCOO-CH2O POHH4Question 5 (6 points). A toxin is discovered that binds reversibly to Asp61 of the c (rotor)subunit in the Fo subunit of F1Fo ATP synthase.(a) What effect would such a toxin have on the P/O ratio?(b) What effect would such a toxin have on metabolic rate?(c) What effect would such a toxin have on NADPH synthesis if administered to a greenplant?Assume that the Fo and CFo subunits respond to the toxin in the same way and in allcases briefly explain your reasoning.(a) Reversible binding of the toxin to Asp61 of the c subunit will decrease the rate ofproton transport through ATP synthase, and thus will reduce the rate of ATP synthesis.The rate of proton translocation through complexes I, III, and IV will remain unchanged,as will the rate at which NADH and oxygen are consumed. The P/O ratio will decrease.(b) If cellular demand for ATP remains constant, more protons will have to betranslocated to the intermembrane space in order to compete with the toxin. This willrequire an increase in glycolysis and TCA, so metabolic rate will increase.(c) ATP synthesis by photosynthesis would be reduced by the same mechanism.NADPH synthesis would be unaffected, so NADPH would increase relative to ATPsynthesis. The cell can regulate this imbalance by using cyclic photophosphorylationwhich disables NADPH synthesis while continuing to synthesize ATP.Question 6 (6 points). An organism is discovered with a mutation in the enzyme rubisco.The mutated enzyme uses fructose-1,6-bisphosphate as its substrate in place