Problem Set 8 - *want a positive ^E so that ^G is negative...

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Problem Set 8 1. A. Redox B. isomerization C. phosphorylation 2. I. entropy increases II. phosphate groups are in resonance III. can interact w/ Mg2+ to stabilize 3. glu + atp glu-6-phos + Adp I. ^G= -4000 cal/mol Glu-6-phos fruc-6-phos II. ^G= 400 cal/mol II.i. Net REACTION= glu + atp fruc-6-phos + adp II.i.1. ^G= -3600 cal/mol A. Yes favorable B. ^G= -RT lnKeq a. Assume T=298 k b. R= 1.986 cal/mol*k c. e^(-^G/RT) = Keq just calculate it –about 420 4. ^E° 2 cytochromeCred + pyruvate + 2H+ 2 cytochromeCox + lactate -split into half reactions and add them CytochromeCreduced cytochromeCoxidized + e- 2 x ^ E° = -.254 b/c need 2 cytochrome to balance reaction Pyruvate + 2e- + 2H+ lactate 1 x ^ E° = -.185 ^G= - nFE ^ E° = -.693 N=2 b/c 2e- F=96.48 E=above answer ~~ +84.7 kJ/mol 5. FAD + 2e- + 2H+ FADH2 E° = -.219 NAD + + 2H + + 2e- NADH E° = -.320 **but you need to flip around the NAD+ equation so then the E turns positive NADH NAD + + 2H + + 2e- E° = +.320 ^ E° = + .101 so plug into bolded ^G equation above
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Unformatted text preview: ***want a positive ^E so that ^G is negative from equation Overall reaction: FAD + NADH FADH2 + NAD + B. If all at 1 M and products at 10 M FAD: ^E° + (.026/2) ln (10/1) = -.189 NADH: ^E° + (.026/2) ln (1/10) =0.035 Total: -.153 6. Enzyme- hexokinase Glu + ATP Glu-6-phos + ADP ^G= -4000 cal/mol Enzyme- phoshpfructokinase Fruc-6-Phos + ATP Fruc-1,6-bisPhos + ADP ^G= -4000 about A. Mg2+ B. Irreversible basically (need special enzymes) Glucokinase is backward for first regulation points b/c large ^G (all others about 0 so they go back and forth but this one depends on what enzyme is present so it’s a branch point) Second one (fructose) is rate limiting step 7. It inactivates cysteine b/c its sulfur (the catalytic residue) it shuts it all down the product that uses the cysteine is accumulating the step right before glyceraldehyde-3-phosphate dehydrogenase...
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Problem Set 8 - *want a positive ^E so that ^G is negative...

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