MCB 102University of California, Berkeley July 24, 2009Isabelle Philipp Online Document

Problem Set 5

——–Answer Key

1. Match the K′eq values with the appropriate ∆Go′

values

K′eq ∆Go′

(kJ/mol)a) 1 i) 28.53b) 10−5 ii) –11.42c) 104 iii) 5.69d) 102 iv) 0e) 10−1 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 + H2O proceeds with an overall ∆Go’of –2863 kJ/mol. When this process occurs in a typical cell, 32 moles of ATP areproduced from ADP + Pi. Assuming that cellular ∆G for ATP→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 ofATP?

Energy released: –2863 kJ/mol glucose.Energy conserved: 41.7 kJ/mol ATP x 32 moles →. ATP/mol glucose = 1334.4 kJ/mol glucose.

Fraction conserved: 1334.4kJ/mol2863kJ/mol

x 100 = 46.6%

3. Oxidative Phosphorylation

Normally ATP synthesis is tightly coupled to electron transfer through the electrontransport chain. Under these conditions, the ratio of ATP produced per atom ofoxygen consumed (P/O ratio) is about 2.5. Predict the effect of a low concentrationof an uncoupling agent on the P/O ratio.

Uncouplers depolarize the inner mitochondrial membrane, and ATP synthesis is com-promised. The electron transport chain remains unaffected, so protons are still pumpedby complexes I, III and IV. Protons pumped from the matrix to the intermembranespace are rapidly returned to the matrix by the membrane-soluble uncoupler, so energythat would normally be used to activate ATP synthase is ultimately expended as heat.

2 Online Document: Problem Set 5 ——– Answer Key

Low concentrations of uncouplers dissipate some of the proton gradient necessary forATP synthesis. Overall, more protons have to be transferred across the membrane foreach ATP synthesized. This requires more NADH per ATP synthesized, so the P/Oratio 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-electronprocess for each complex

(c) calculate how many moles of protons can be translocated across the inner mito-chondrial membrane if translocation of 1 mole requires 23 kJ.

Complex I: NADH is oxidized in a 2 electron process, and UQ is reduced, also ina 2 electron process. The standard reduction potentials are:NAD+ + 2H+ + 2e− ↔ NADH + H+ Eo’ = -0.32 VUQ + 2H+ + 2e− ↔ UQH2 Eo’ = 0.045 V

(a) redox potential for complex I Eo’ = 0.365 V

(b) ∆Go’ = -nF∆Eo’ so for a 2 electron process ∆Go’ = -70.4 kJ/mol

(c) 3.1 moles of protons can be translocated

Complex II: succinate is oxidized in a 2 electron process, and UQ is reduced, alsoin a 2 electron process. The standard reduction potentials are:fumarate + 2H+ + 2e− ↔ succinate Eo’ = 0.031 VUQ + 2H+ + 2e− ↔ UQH2 Eo’ = 0.045 V

(a) redox potential for complex II ∆Eo’ = 0.014 V

(b) ∆Go’ = -nF∆Eo’ so for a 2 electron process ∆Go’ = -2.7 kJ/mol

(c) no protons are translocated by complex II

Complex III : UQH2 is oxidized in a 2 electron process. Cytochromec1 is reducedin two 1 electron processes. The standard reduction potentials are:UQ + 2H+ + 2e− ↔ UQH2 Eo’ = 0.045 Vcytc1 (Fe3+) + e− ↔ cytc1 (Fe2+) Eo’ = 0.22 V

(a) redox potential for the net equation of Complex III is ∆Eo’ = 0.175 V

(b) ∆Go’ = -nF∆Eo’ 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 isreduced in a 2 electron process. The standard reduction potentials are:cytc (Fe3+) + e− ↔ cytc (Fe2+) Eo’ = 0.254 V0.5 O2 + 2H+ + 2e− ↔ H2O Eo’ = 0.816 V

Online Document : Problem Set 5 ——– Answer Key 3

(a) redox potential for complex IV ∆Eo’ = 0.562V

(b) ∆Go’ = -nF∆Eo’ so for a 2 electron process ∆Go’= -108.4 kJ/mol

(c) 4.7 moles of protons can be translocated

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 themethyl 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 thantwo or three well–chosen sentences.

Malonyl CoA carbons provide all of the carbons added to a growing fatty acid chainexcept the first two. The first two are the methyl end of the fatty acid, and theycome from the initial acetate group. The carboxyl carbons from successive malonylCoA additions provide carbons 1 (COO) and further odd-numbered carbons of thechain, except carbon 15 (which is from acetate); the methyl carbons provide the even-numbered 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 Acetyl-Co and enzyme that uses the co-factor, biotin; (ATP REQ. STEP STEP no.2) A ketobond must be reduced by NADPH (2.5 ATP); (ATP REQ. STEP STEP no.3) A C=Cdouble bond must be reduced by NADPH (2.5 ATP). NADPH is used over NADH, sinceNADH levels are low in the cytosol, where synthesis occurs. 4-Phosphopantethenic acidfrom Acyl Carrier Protein (ACP) is a thiol-ended tether that is used in the transferand catalysis reactions, and from the thiol end, malonyl groups can be transferred tothe growing acyl chain in each step. Eight condensations are required, which is 8 x 6= 48 ATP.

Exam Questions from 2008

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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 vesi-cles, such that the side of the membrane normally facing the mitochondrial matrix ison the outside of the vesicle (Fig. (b)).

(a) Schematic mitochondrionwith ATP synthases (not toscale)

(b) Schematic mitochondrialvesicles 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 alsopH 7.0. Under these conditions, the reaction catalyzed is the hydrolysis of ATPto ADP and Pi. Explain why, in four sentences or fewer.

Enzymes catalyze reactions in both directions. In absence of the proton motiveforce to drive ATP synthesis, the exergonic reaction of ATP hydrolysis will befavored.

(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 solutioncontains 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, infour sentences or fewer.

There will be a gradient with a higher concentration of hydrogen ions inside anda lower concentration outside. The mitochondrial membrane including the ATPsynthase is inverted: this is the same gradient direction, relative to the enzyme,that is present in normal mitochondria. Thus: ADP + Pi ↔ ATP + H2O

8. Photosynthesis and Oxidative Phosphorylation

List four similarities and four differences between photosynthesis and oxidative phos-phorylation.

Online Document : Problem Set 5 ——– Answer Key 5

SIMILARITIES: ATP synthase (Fo/F1 ATP synthase/CFoCF1 ATP synthase), chemios-motic mechanism (electron transport coupled to proton translocation), lipophilic elec-tron 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 (pho-tosynthesis has no analog of complex II), oxygen evolution (photosynthesis) vs. oxygenconsumption (oxidative phosphorylation), etc.

For your information: In the former exam no point was given for listing mitochondriaand chloroplast as a difference. The question asked specifically about photosynthesisand oxidative phosphorylation. In addition, they are both dedicated organelles thatare believed to have arisen from endosymbiosis.

9. Photosynthesis and Oxidative Phosphorylation

Explain how photosynthesis and respiration (oxidative phosphorylation) are comple-mentary processes (You can answer this in one sentence, dont write too much!).

Carbon dioxide and water are used to synthesize carbohydrates during photosynthesiswhile respiration results in the conversion of carbohydrates to carbon dioxide andwater.

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 thetable.

consider the reaction: FAD + NADH + H+ ↔ FADH2 + NAD+

∆Eo’ = Eo’FAD - Eo’NADH = –0.219 – – 0.320 = +0.101VYes, FAD is a better oxidizer, since ∆Eo’ is positive and the oxidation of NADHwill be spontaneous.

(b) What is the standard free energy difference, ∆Go′, for the Oxidation of FADH2

with O2, where FAD and H2O are the final products?

FADH2 + 12O2 ↔ FAD + H2O

Eo’O2 = 0.816V; Eo’FADH2 = 0.219V∆Eo’ = 0.816 – – 0.219V = 1.035∆Go’ = nF∆Eo’ = – 2 ∗ 96.485 ∗ 1.035∆Go’ = 200 kJ/mol

(c) The hydrolysis reaction of ATP (ATP + H2O ↔ ADP +Pi) has a ∆Go′of

31kJ/mol. How many ATP could be produced if this oxidation of FADH2 byO2 described above in (b) were perfectly coupled to the formation of ATP fromADP and Pi.

(∆Go’ FADH→FAD) / (∆Go’ ATP→ADP ) = −200kJ/mol−31kJ/mol

= 6.5 ATP

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11. Palmitate Synthesis

Which carbon atoms in palmitate will be labeled most rapidly when a cell extract isgiven the following labeled substrates. Unless otherwise stated, assume all malonylCoA is generated from the (possibly) labeled acetyl CoA. (labeled Carbon atom: 14C)

(a) 14CH3-(CO) SCoA + excess unlabeled malonyl CoAC16

(b) CH3-(14CO) SCoA + excess unlabeled malonyl CoA

C15

(c) CH3-(CO)SCoA (unlabeled) + excess O2C-14CH2-(CO)SCoAC2, C4, C6, C8, C10, C12, C14, but NOT C16

(d) CH-(CO) SCoA + excess O14C-CH(CO)SCoAHERE is a typo in the question, I missed some ”H” in my Acetyl-CoA andMalonyl-CoA!!!The answer is the same, if you have different labelled substrates as needed, NONEof your C-Atoms in your fatty acid are labeled. In particular, you would not getany fatty acid.If the question would have been correct and asked you which Carbon atoms arelabeled given : CH3-(CO) SCoA + excess O14C-CH2-(CO)SCoA, the answer is aswell : NONE, as the CO2 would be labeled here.

(e) CH3-(CO)SCoA + excess O2C-CH2-(14CO)SCoA

C1, C3, C5, C7, C9, C11, C13, but NOT C15

(f) 14CO2

no label