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Answers to Problem Set 3
Lecture 18 (PDH and TCA cycle)
1. Catalytic coenzymes (TPP, lipoic acid, and FAD) are modified but regenerated in each
reaction cycle. Thus, they can play a role in the processing of many molecules of
pyruvate. Stoichiometric coenzymes (coenyme A and NAD+) are used in only one
reaction because they are the components of products of the reaction.
2. (a) After one round of the citric acid cycle, the label emerges in C-2 and C-3 of
oxaloacetate. (b) The label emerges in CO2 in the formation of acetyl CoA from
pyruvate. (c) After one round of the citric acid cycle, the label emerges in C-1 and C-4
of oxaloacetate. (d and e) Same fate as that in part a.
3. (a) The steady-state concentrations of the products are low compared with those of
the substrates. (b) The ratio of malate to oxaloacetate must be greater than 1.57 x 104 for
oxaloacetate to be formed.
GG19-10. [isocitrate]/[citrate] = 0.1. When [isocitrate] = 0.03 mM, [citrate] = 0.3 mM.
Lecture 19 (Electron transport system)
1. a. 4; b. 3; c. 1; d. 5; e. 2.
2. Rotenone: NADH and NADH-Q oxidoreductase will be reduced. The remainder will
be oxidized. Antimycin A: NADH, NADH-Q oxidoreductase and coenzyme Q will be
reduced. The remainder will be oxidized. Cyanide: All will be reduced.
GG20-1. [FAD] + 2cyt c(Fe2+) + 2H+ <==> [FADH2] + 2 cyt c(Fe3+)
The e- acceptor is FAD; the e- donor is cyt c(Fe2+) ΔEo’ = -0,219 – (0.254) = -0.473 ΔGo’ = -nFΔEo’ = -2(96.5 kJ/V-mol)(-0.473 V)
This reaction does not go (forward). Turned around (FADH2 –e- -> cyt c) it would be ΔGo’ = -91.3 kJ/mol and would go.
GG20-3. NAD+ + 2H+ + 2e- <==> NADH + H+
= -0.32 + (8.31)(298)/(2)(96,500) ln([NAD+][H+]2/[NADH][H+])
Because [NAD+]/[NADH] = 1 and [H+]2/[H+] = [H+], We can rewrite the formula above to: E = = -0.32 + RT/nF ln([H+] = -0.32 – 2.31(0.0128)(ΔpH) ΔpH = 1 At pH 8, ΔpH = 1 and E = - 0.35 V
GG20-8. NADH + H+ + 2UQ(ox) <==> NAD+ + 2UQH(red)
a. ΔEo’ = Eo’(acceptor: UQ) - Eo’(donor: NADH)
ΔGo’ = -nFΔEo’ = -2(96,400)( 0.038)
= exp10((73,300)/(2.3)(8.31)(298)) = 7.4 x 1012
c. Assume that the energy available = (-73.3)(0.75) = -54,975 kJ/mol
so the energy needed to form ATP can be as much as +54,975
To form ATP, ADP + Pi --> ATP + H2O, ΔGo’ = +30.5 kJ/mol;
ΔGo’ – RT ln ([Pi]) + RT ln ([ATP]/[ADP])
54,975 = 30,500 –(8.31)(298) ln (0.001) + (8.31)(298) ln ([ATP]/[ADP]) ln ([ATP]/[ADP]) = 2.976; [ATP]/[ADP] = 19.6
GG20-9. succinate + ½ O2 <==> fumarate + H2O
a. ΔEo’ (succinate -> fumarate) = -0.031 V ΔEo’ (1/2 O2 -> H2O) = 0.816 V ΔEo’ (combined) = 0.785 ΔGo’ = -nFΔEo’ = -2(96.400)(0.785) = -151 kJ/mol b. Keq = exp10(-ΔGo’/RT) = exp10(151,000/2.3(8.3)(298)) = 3.24 x 1026
Lecture 20 (ATP synthesis)
1. a. 12.5; b. 14; c. 32; d. 13.5; e. 30; f. 16.
2. a. The P:O ratio is equal to the product of (H+/2e-) and (P/H+). Note that the P:O ratio is identical with the P:2e- ratio. (Note also that the P/H+ ratio depends on the number of subunits in the ATP synthase collar, and this may vary among organisms.)
H+ energy gradient problem As described in lecture, the free energy of a reaction involving electrically charged materials can be described by the following formula:
ΔG = RT ln ([product]/[reactant]) + z F ΔEo’ (z = charge)
This formula can apply to the movement of H+ (z = 1) across a membrane:
ΔG = -2.3 RT ΔpH + z F ΔEo’ ΔpH = pHout – pHin (H+ moving from inside to outside the membrane: thus pHout is the product, pHin is the reactant)
a) Show that a difference of 1 pH unit is equivalent to a 59 mV difference.
-2.3(8.31)(298)(-1 [low pH out]) ~ (1)(96,485)(ΔEo’) 5695.7 ~ 96,487 ΔEo’ ΔEo’ = 0.059 V [positive voltage outside]
b) With a difference across the membrane of 1 pH unit and 59 mV (0.059 V) per proton, how many protons are needed to force synthesis of one ATP: ADP + Pi → ATP + H2O? (Assume standard conditions for ATP, ADP, and Pi concentrations: ΔG’ = ΔGo’)
ΔG = -2.3 RT ΔpH + z F ΔEo’ = -2.3(8.31)(298)(-1) + (n)(96,485)(0.059) = 30,500
c) Alternatively, what voltage difference would be needed if one proton were to force the synthesis of one ATP? (Assume the same pH difference and standard conditions as above.)
ΔG = -2.3 RT ΔpH + z F ΔEo’ = -2.3(8.31)(298)(-1) + (1)(96,485)(x) = 30,500
3. The available energy from the translocation of two, three, and four protons is -38.5, -
57.7, and -77.4 kJ/mol, respectively. The free energy consumed in synthesizing a mole
of ATP under standard conditions is 30.5 kJ. Hence, the residual free energy of -8.1, -
27.2, and -46.7 kJ/mol can drive the synthesis of ATP until the [ATP]/[ADP][Pi] ratio is
26.2, 6.5 x 104, and 1.6 x 108, respectively. Suspensions of isolated mitochondria
synthesize ATP until this ratio is greater than 104, which shows that the number of
protons translocated per ATP synthesized is at least three.
4. 12/3 = 4; 14/3 = 4.7 Lecture 21 (Photosynthesis)
1. CO2 is the acceptor. H2O is the donor. Light energy powers e- flow.
GG21-1. Given: by absorption of 1 photon @ 700 nm, photosystem II ΔEo’ = 1 V
Absorbed energy = hν = (hc/λ) N = 171,139 J/mol
2. We need to factor in the NADPH because it is an energy-rich molecule. Recall that NADH is worth 2.5 ATP if oxidized by the electron-transport chain. 12 NADPH = 30 ATP. 18 ATP are used directly, and so the equivalent of 48 molecules of ATP are required for the synthesis of glucose. GG21-5.
Given: non-cyclic H+ pumping: 3 H+/e- = 3 H+/2 hν
Efficiencies: non-cyclic: 3 ATP/14 H+ (3 H+/2 hν) = 0.32 ATP/hν
cyclic: 3 ATP/14 H+ (2 H+/1 hν) = 0.43 ATP/hν
GG21-8. Assume 12 c-subunits means 12 H+ are needed to drive one turn of the c-subunit rotor and the synthesis of 3 ATP by the CF1 part of the ATP synthase. If the R.
cytochrome bc1 complex drives the translocation of 2 H+/e-, then 2 hν gives 4 H+, and 6 hν gives 12 H+ and thus 3 ATP (thus 2 hν yield 1 ATP.) GG21-10. RuBP + CO2 + H2O --> 2 3-PGA
2 3-PGA + 2 ATP --> 2 1,3-bisPGA + 2 ADP 2 1,3-bisPGA + 2 NADPH + 2 H+ --> 2 G3P + 2 NADP+ + 2Pi 2 G3P + 2 H2O --> glucose + 2 Pi RuBP + CO2 + 2 ATP + 2 NADPH + 3 H2O + 2 H+ -->
6 CO2 --> C6H12O6 + 6 H2O Added: 6 CO2 + 12 H2O --> C6H12O6 + 6 O2 + 6 H2O Thus, water is taken up in light reactions and produced in dark reaction.
Lecture 22 (Gluconeogenesis)
1. In glycolysis, the formation of pyruvate and ATP by pyruvate kinase is irreversible.
This step is bypassed by two reactions in gluconeogenesis: (1) the formation of
oxaloacetate from pyruvate and CO2 by pyruvate carboxylase and (2) the formation of
phosphoenolpyruvate from oxaloacetate and GTP by phosphoenolpyruvate carboxykinase. The formation of fructose-1,6-bisphosphate by phosphofructokinase is bypassed by fructose-1,6-bisphosphatase in gluconeogenesis, which catalyzes the conversion of fructose-1,6-bisphosphate into fructose 6-phosphate. Finally, the hexokinase-catalyzed formation of glucose-6-phosphate in glycolysis is bypassed by glucose-6-phosphatase, but only in the liver. 2. For synthesis of one molecule of glucose from two molecules of pyruvate: 6 NTP ((4 ATP, 2 GTP); 2 NADH 3. NTP molecules for synthesis of glucose from a. glucose-6-phosphate: none.
c. two molecules of oxaloacetate: 4 (2 ATP, 2 GTP). d. two molecules of dihydroxyacetone phosphate: none. 4. (a) glycolysis: 2, 3, 6, 9; (b) gluconeogenesis: 1, 4, 5, 7, 8. 5. a. What was the rationale for comparing the activities of these two enzymes? If both enzyme operated simultaneously, the two reactions would take place and the net results would be simply, ATP + H2O --> ADP + Pi. The energy of ATP hydrolysis would be
released as heat.
b. Do these results support the notion that bumblebees use futile cycles to generate heat?
Not really. For the cycle to generate heat, both enzymes must be functional at the same
time in the same cell.
c. In which species might futile cycling take place? The species B. terrestris
might show some futile cycling because both enzymes are active to a
d. Do these results prove that futile cycling does not (or does) participate in heat
generation? No. These results simply suggest that simultaneous activity of
phosphofructokinase and fuctose-16-bisphophatase is unlikely to be employed to
generate heat in the species shown. Lecture 23 (Fatty acid catabolism)
1. b. fatty acid in the cytoplasm
c. activation of fatty acid by joining to CoA
a. reaction with carnitine
g. acyl-CoA in mitochondrion
h. FAD-linked oxidation d. hydration e. NAD+-linked oxidation f. thiolysis 2. Palmitic acid yields 106 molecules of ATP. Palmitoleic acid has a double bond between carbons C-9 and C-10. When palmitoleic acid is processed in ß oxidation, one of the oxidation steps (to introduce a double bond before the addition of water) will not take place, because a double bond already exists. Thus, FADH2 will not be generated,
and palmitoleic acid will yield 1.5 fewer molecules of ATP than palmitic acid, for a total of 104.5 molecules of ATP. For heptadecanoic acid, activation fee to form acyl CoA, -2 ATP; 7 acetyl CoA, +70 ATP; 7 NADH, +17.5 ATP; 7 FADH2, +10.5 ATP; propionyl-CoA --> succinyl-CoA, -1 ATP; succinyl-CoA --> succinate, +1 ATP; succinate --> fumarate, +1.5 ATP; malate --> OAA, +2.5 ATP; total 100 ATP 3. Glucose: glycolysis, +2 ATP; 2 NADH @ 1.5 = 3 ATP; 2x4 NADH @ 2.5 = 20 ATP; 2 FADH2 @ 1.5 = 3 ATP; 2 GTP = 2 ATP; total = 30 ATP
Caprioic (hexanoic) acid: activation, -2 ATP; 2 NADH @ 2.5 = 5 ATP; 2 FADH2 @ 1.5
= 3 ATP; 3 acetyl-CoA through CAC =30 ATP; total = 36 ATP 4. Hummingbird flight:
Weight lost = 4-2.7 = 1.3 g lipid At 37 kJ/g (Table 23.1), energy used = 48.1 kJ (but this conversion number relates to dry weight of adipose tissue)
10 h across the Gulf Assume 4.81 kJ/h (above) and 0.25 l O2/h (given) Energy per O2 = 4.81/0.25 = 19.24 kJ/l O2 Compare to human 8-minute miler: 12.7 kcal/min x 4.18 J/cal = 53 kJ/min Time running on 48.1 kJ (trans-Gulf flight) = 48.1/53 = 0.9 min Miles to lose 1 lb body fat: 1 lb = 454 g => 16,798 kJ (@ 37 kJ/g)
At 53 kJ/min, losing 16,798 kJ takes 317 min
Lecture 24 (Fatty acid synthesis)
1. e. formation of malonyl ACP
d. reduction of a carbonyl
a. dehydration . c. release of a C16 fatty acid 2. All of the labeled carbon atoms will be retained. Because we need 8 acetyl CoA molecules and only 1 carbon atom is labeled in the acetyl group, we will have 8 labeled carbon atoms. The only acetyl CoA used directly will retain 3 tritium atoms. The 7 acetyl CoA molecules used to make malonyl CoA will lose 1 tritium atom on addition of the CO2 and another one at the dehydration step. Each of the 7 malonyl CoA molecules will retain 1 tritium atom. Therefore, the total retained tritium is 10 atoms. The ratio of tritium to 14C is 1.25. GG24-6.
glycerol + 2 fattyacyl-CoA + 2ATP + CTP + ethanolamine -->
phosphatidylethanolamine + 2ADP + CMP + PPi + 2CoA + Pi
ΔGo’ = - 4 high-energy bonds + combining 4 components into one
GG24-11. ATP equivalents needed to form palmitic acid from acetyl-CoA, using 3.5 ATP/NADPH:
7 acetyl-CoA + 7 ATP + 7 HCO3- --> 7 malonyl-CoA + 7 ADP + 7 Pi + 7H+ acetyl-CoA + 7 malonyl-CoA + (2x7) NADPH + 14H+
---> palmitoyl-CoA + 7 HCO3- + (2x7) NADP+ + 7 CoSH
Lecture 25 (Nitrogen metabolism)
1. The seven precursors of the carbon skeletons of the 20 amino acids: oxaloacetate,
pyruvate, ribose-5-P, phosphoenolpyruvate, erythrose-4-phosphate, α-ketoglutarate, and
GG25-1. Oxidation number of N in nitrate, nitrite, NO, N2O, and N2:
GG25-2. ATP equivalents for formation of NH4+ from NO3- and N2, using 3 ATP per NADH, NADPH, and reduced Fd:
a. NO3- --> NO2-: 1 NADH, 3 ATP NO2- --> NH4+: 6 Fd,
GG25-4. ATP equivalents for formation of urea:
carbamate + ATP --> carbamoyl-P + ADP
citruline + ATP --> argininosuccinate + AMP 2 ATP
My research interests focus on understanding, at a biochemical and structural level, the mechanisms that drive molecular machines. In particular my aim is to use cryo-EM in combination with image processing techniques to uncover conformational changes that play functional roles in protein translation (ribosome) and polyketide synthesis (PKS). Principle Investigator, CIC bioGUNE, Bilbao, Spain.
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