Chemistry Problem Set 5

Problem Set 5 ——– Answer Key 0 0 1. Match the Keq values with the appropriate ∆Go values 0 a) b) c) d) e) Keq 1 −5 10 104 102 10−1 0 ∆Go (kJ/mol) i) 28.53 ii) –11.42 iii) 5.69 iv) 0 v) –22.84 a) 0; b) 28.53; c) – 22.84; d) – 11.42; e) 5.69 2. Energy conservation The complete combustion of glucose to CO2 + H2 O proceeds with an overall ∆Go ’ of –2863 kJ/mol. When this process occurs in a typical cell, 32 moles of ATP are produced from ADP + Pi . Assuming that cellular ∆G for AT P →ADP hydrolysis is –41.7 kJ/mol and that ∆G∼ =∆Go ’ for glucose oxidation under intercellular (in vivo) conditions, what fraction of the potential energy of glucose is conserved in the form of ATP? Energy released: –2863 kJ/mol glucose. Energy conserved: 41.7 kJ/mol ATP x 32 moles → . ATP/mol glucose = 1334.4 kJ/mol glucose. x 100 = 46.6% Fraction conserved: 1334.4kJ/mol 2863kJ/mol 3. Oxidative Phosphorylation Normally ATP synthesis is tightly coupled to electron transfer through the electron transport chain. Under these conditions, the ratio of ATP produced per atom of oxygen consumed (P/O ratio) is about 2.5. Predict the effect of a low concentration of an uncoupling agent on the P/O ratio. Uncouplers depolarize the inner mitochondrial membrane, and ATP synthesis is compromised. The electron transport chain remains unaffected, so protons are still pumped by complexes I, III and IV. Protons pumped from the matrix to the intermembrane space are rapidly returned to the matrix by the membrane-soluble uncoupler, so energy that would normally be used to activate ATP synthase is ultimately expended as heat. Low concentrations of uncouplers dissipate some of the proton gradient necessary for ATP synthesis. Overall, more protons have to be transferred across the membrane for each ATP synthesized. This requires more NADH per ATP synthesized, so the P/O ratio will decrease. 4. Oxidative Phosphorylation – Electron Transport Chain 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-electron process for each complex (c) calculate how many moles of protons can be translocated across the inner mitochondrial 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 2 electron process. The standard reduction potentials are: NAD+ + 2H+ + 2e− ↔ NADH + H+ E o ’ = -0.32 V + − UQ + 2H + 2e ↔ UQH2 E o ’ = 0.045 V (a) redox potential for complex I (b) ∆Go ’ = -nF ∆E o ’ so for a 2 electron process (c) 3.1 moles of protons can be translocated E o ’ = 0.365 V ∆Go ’ = -70.4 kJ/mol Complex II: succinate is oxidized in a 2 electron process, and UQ is reduced, also in a 2 electron process. The standard reduction potentials are: fumarate + 2H+ + 2e− ↔ succinate E o ’ = 0.031 V UQ + 2H+ + 2e− ↔ UQH2 E o ’ = 0.045 V (a) redox potential for complex II (b) ∆Go ’ = -nF ∆E o ’ so for a 2 electron process (c) no protons are translocated by complex II ∆E o ’ = 0.014 V ∆Go ’ = -2.7 kJ/mol Complex III : UQH2 is oxidized in a 2 electron process. Cytochromec1 is reduced in two 1 electron processes. The standard reduction potentials are: UQ + 2H+ + 2e− ↔ UQH2 E o ’ = 0.045 V + − + cytc1 (Fe3 ) + e ↔ cytc1 (Fe2 ) E o ’ = 0.22 V (a) redox potential for the net equation of Complex III is ∆E o ’ = 0.175 V (b) ∆Go ’ = -nF ∆E o ’ so for a 2 electron process ∆Go ’ = -33.8 kJ/mol (c) 1.5 moles of protons can be translocated Complex IV: cytochrome c is oxidized in two 1 electron processes, and 0.5 O2 is reduced in a 2 electron process. The standard reduction potentials are: cytc (Fe3+ ) + e− ↔ cytc (Fe2+ ) E o ’ = 0.254 V 0.5 O2 + 2H+ + 2e− ↔ H2 O E o ’ = 0.816 V (a) redox potential for complex IV (b) ∆Go ’ = -nF ∆E o ’ so for a 2 electron process (c) 4.7 moles of protons can be translocated ∆E o ’ = 0.562V ∆Go ’= -108.4 kJ/mol 5. Lipid Biosynthesis You have a liver homogenate that is capable of synthesizing palmitate. (a) If you supply this homogenate with malonyl-CoA that is radiolabeled at the methyl carbon, at what position(s) will the resulting palmitate be labeled? The radiolabeled positions are circled on the diagram. (b) Explain the pattern of circled carbons above. This should not take more than two or three well–chosen sentences. Malonyl CoA carbons provide all of the carbons added to a growing fatty acid chain except the first two. The first two are the methyl end of the fatty acid, and they come from the initial acetate group. The carboxyl carbons from successive malonyl CoA additions provide carbons 1 (COO) and further odd-numbered carbons of the chain, except carbon 15 (which is from acetate); the methyl carbons provide the evennumbered carbons (except carbon 16, which is from acetate). 6. Fatty Acid Synthesis How many ATP are consumed in making a C18 saturated fatty acid from acetyl-CoAs? In your answer explain what cofactors are used to synthesize fatty acids and why? (ATP REQ. STEP no.1) To make each malonyl-CoA requires 1 ATP and an AcetylCo and enzyme that uses the co-factor, biotin; (ATP REQ. STEP STEP no.2) A keto bond must be reduced by NADPH (2.5 ATP); (ATP REQ. STEP STEP no.3) A C=C double bond must be reduced by NADPH (2.5 ATP). NADPH is used over NADH, since NADH levels are low in the cytosol, where synthesis occurs. 4-Phosphopantethenic acid from Acyl Carrier Protein (ACP) is a thiol-ended tether that is used in the transfer and catalysis reactions, and from the thiol end, malonyl groups can be transferred to the growing acyl chain in each step. Eight condensations are required, which is 8 x 6 = 48 ATP. 7. ATP Synthase Mitochondrial inner membrane fragments that contain ATP synthase can be isolated. Under appropriate conditions, these membranes can be caused to form inside-out vesicles, such that the side of the membrane normally facing the mitochondrial matrix is on the outside of the vesicle (Fig. (b)). (a) Schematic mitochondrion with ATP synthases (not to scale) (b) Schematic mitochondrial vesicles with ATP synthases (not to scale) (a) The vesicles in Figure 2 are placed in a buffer at pH 7.0, which also contains ATP, ADP and Pi, at roughly equal concentrations. The pH inside the vesicles is also pH 7.0. Under these conditions, the reaction catalyzed is the hydrolysis of ATP to ADP and Pi. Explain why, in four sentences or fewer. Enzymes catalyze reactions in both directions. In absence of the proton motive force to drive ATP synthesis, the exergonic reaction of ATP hydrolysis will be favored. (b) In a separate experiment, a second set of vesicles is placed in buffer at pH 6.5 (without nucleotides or Pi) until the insides of the vesicles have reached pH 6.5. Then these vesicles are suddenly moved into a new solution. This new solution contains ATP, ADP and Pi at roughly equal concentrations, like the one in part ’a, but it is at pH 8.5. What reaction do the vesicles catalyze now? Explain, in four sentences or fewer. There will be a gradient with a higher concentration of hydrogen ions inside and a lower concentration outside. The mitochondrial membrane including the ATP synthase is inverted: this is the same gradient direction, relative to the enzyme, that is present in normal mitochondria. Thus: ADP + Pi ↔ ATP + H2 O 8. Photosynthesis and Oxidative Phosphorylation List four similarities and four differences between photosynthesis and oxidative phosphorylation. SIMILARITIES: ATP synthase (Fo /F1 ATP synthase/CFo CF1 ATP synthase), chemiosmotic mechanism (electron transport coupled to proton translocation), lipophilic electron transport pool (ubiquinone/plastoquinone), dedicated organelles (mitochondria/chloroplasts), small, watersoluble 1-electron transfer proteins (cytochrome c/plastocyanin), etc. DIFFERENCES: Source of chemical reduction (light/NADH), ATP/ADP transport (passive/active), products (photosynthesis produces NADPH), coupling to TCA (photosynthesis has no analog of complex II), oxygen evolution (photosynthesis) vs. oxygen consumption (oxidative phosphorylation), etc. For your information: In the former exam no point was given for listing mitochondria and chloroplast as a difference. The question asked specifically about photosynthesis and oxidative phosphorylation. In addition, they are both dedicated organelles that are believed to have arisen from endosymbiosis. 9. Photosynthesis and Oxidative Phosphorylation Explain how photosynthesis and respiration (oxidative phosphorylation) are complementary processes (You can answer this in one sentence, dont write too much!). Carbon dioxide and water are used to synthesize carbohydrates during photosynthesis while respiration results in the conversion of carbohydrates to carbon dioxide and water. 10. Redox Potentials Consider the tables of the Standard Redox Potentials. (a) Is FAD a better oxidizer then NAD+ ? Explain your answer using numbers in the table. consider the reaction: FAD + NADH + H+ ↔ FADH2 + NAD+ ∆E o ’ = Eo ’F AD - Eo ’N ADH = –0.219 – – 0.320 = +0.101V Yes, FAD is a better oxidizer, since ∆E o ’ is positive and the oxidation of NADH will be spontaneous. 0 (b) What is the standard free energy difference, ∆Go , for the Oxidation of FADH2 with O2 , where FAD and H2 O are the final products? FADH2 + 21 O2 ↔ FAD + H2 O Eo ’O2 = 0.816V; Eo ’F ADH2 = 0.219V ∆E o ’ = 0.816 – – 0.219V = 1.035 ∆Go ’ = nF ∆E o ’ = – 2 ∗ 96.485 ∗ 1.035 ∆Go ’ = 200 kJ/mol 0 (c) The hydrolysis reaction of ATP (ATP + H2 O ↔ ADP +Pi ) has a ∆Go of 31kJ/mol. How many ATP could be produced if this oxidation of FADH2 by O2 described above in (b) were perfectly coupled to the formation of ATP from ADP and Pi . (∆Go ’ F ADH→F AD ) / (∆Go ’ AT P →ADP ) = −200kJ/mol = 6.5 ATP −31kJ/mol 11. Palmitate Synthesis Which carbon atoms in palmitate will be labeled most rapidly when a cell extract is given the following labeled substrates. Unless otherwise stated, assume all malonyl CoA is generated from the (possibly) labeled acetyl CoA. (labeled Carbon atom: 14 C) (a) 14 CH3 -(CO) SCoA + excess unlabeled malonyl CoA C16 (b) CH3 -(14 CO) SCoA + excess unlabeled malonyl CoA C15 (c) CH3 -(CO)SCoA (unlabeled) + excess O2 C-14 CH2 -(CO)SCoA C2 , C4 , C6 , C8 , C10 , C12 , C14 , but NOT C16 (d) CH-(CO) SCoA + excess O14 C-CH(CO)SCoA HERE is a typo in the question, I missed some ”H” in my Acetyl-CoA and Malonyl-CoA!!! The answer is the same, if you have different labelled substrates as needed, NONE of your C-Atoms in your fatty acid are labeled. In particular, you would not get any fatty acid. If the question would have been correct and asked you which Carbon atoms are labeled given : CH3 -(CO) SCoA + excess O14 C-CH2 -(CO)SCoA, the answer is as well : NONE, as the CO2 would be labeled here. (e) CH3 -(CO)SCoA + excess O2 C-CH2 -(14 CO)SCoA C1 , C3 , C5 , C7 , C9 , C11 , C13 , but NOT C15 (f) 14 CO2 no label