Given, isas ri Glucose + Pi→ Glucose-6-P ATP + H₂O → ADP + Pi Determine K 'eq for a coupled reaction Tot 24 h Glucose + ATP 18) (OUT) I MEX + H₂O K'eq = 3.9 x 10-3M-1uto beriasen K'eq=2 x 105 Marbl 93A otsdge Glucose-6-P + ADP K'eq?
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- Given the following data, calculate Keq for the denaturation reaction of the protein β-lactoglobin at 25oC: ΔH° = –88 kJ/mol ΔS° = 0.3 kJ/mol. The free energy of hydrolysis of ATP in systems free of Mg2+ is −35.7 kJ/mol. When the concentration of this ion is 5 mM, ΔG°observed is approximately −31 kJ/mol at pH 7 and 38°C. Suggest a possible reason for this effect.uizzes/67365/take Based on the image below, select the correct statements. Note: There may be more than 1 correct response. I Ribose 5-phosphate ribose phosphate pyrophosphokinase (PRPP synthetase) glutamine-PRPP amidotransferase adenylosuccinate synthetase AMP > 5-Phosphoribosylamine I adenylosuccinate PRPP lyase 9 steps Adenylosuccinate AMP IMP <-- ADP - AMP <-- GMP <-- IMP IMP dehydrogenase <- GMP - XMP ADP ATP GMP يمد XMP-glutamine amidotransferase Increased levels of ADP inhibit the production of PRPP. Increased levels of GMP inhibit the production of XMP. O Increased ADP activates PRPP synthase to increase PRPP levels. Increased IMP activates glutamine-PRPP amidotransferase to further increase IMP levels. 8 OBCUsing the ActiveModel for phosphofructokinase (Trypanosoma), describe the difference between the APO1, AP02, and holoenzyme conformations.
- Energetic of Fructose-1 ,6-bis P Hydrolysis (Integrates with Chapter 3.) The standard free energy change (G) for hydrolysis of fructose-1. 6-bisphosphate (FBP) to fructose-S-phosphate (F-6-P) and P: is -16.7 KJ/mol: FBP + H2O fructose-6-P + Pi The standard free energy change (G) for ATP hydrolysis is -30.5 KJ/mol: ATP + H2O ADP + Pj What is the standard free energy change for the phosphofructokinase reaction: ATP + fructose-6-P ADP + FBP b. What is the equilibrium constant for this reaction? c. Assuming the intracellular concentrations of [ATP] and (ADP] are maintained constant at 4 mM and 1.6 mM, respectively, in a rat liver cell, what will be the ratio of [FBP]/[fructose-6-P] when the phosphofructokinase reaction reaches equilibrium?The reaction catalyzed by phosphorylase is readily reversible in vitro. At pH 6.8, the equilibrium ratio of orthophosphate to glucose 1-phosphate is 3.6. The value of ΔG°’ for this reaction is small because a glycosidic bond is replaced by a phosphoryl ester bond that has a nearly equal transfer potential. However, phosphorolysis proceeds far in the direction of glycogen breakdown in vivo. Suggest one means by which the reaction can be made irreversible in vivo.His388 Glu357 His388 Glu357 Ring opening Proton HO HO abstraction HO но- G6P He NH- NH His388 His388 His388 Glu357 Glu357 Glu357 HO но HO но- но OH cis-enediol F6P Ring closure intermediate OH Describe the mechanism shown above for phosphoglucose isomerase. Describe the chemistry of each step • How the enzyme appears or might facilitate the chemistry How the enzyme increases the reaction rate.
- Que:- An Michaels - Menten Kenetirs eneyme obeying the following parameters K2= 1•4 x10%M's" , K,= 2:5x1o°5' gave - 1 K3 = l-5 x10s ca) So this enzyme ctalytically pufet ? Explain your heasoning . (b) What is Vmaz , ik 1-4 nmol/me and saturabing eveyrme are usecd ? substrateis this stement false? Intracellular concentrations in resting muscle are as follows: Fructose-6-phosphate (1.0 mM)Fructose-(1-6)-bisphosphate (10.0 mM)AMP (0.1 mM)ADP (0.5 mM)ATP (5.0 mM)Pi (10.0 mM)Under the above conditions the Phosphofructokinase reaction in muscle is more exergonic than under standard conditions.In muscle cells, the AG for glucose+ATP2 glucose-6-P+ADP is –33.5 kJ•mol-'. In contrast, the AG for glucose – 6 – P fructose – 6 - P is –2.5 kJ · mol-1. (a) In a physiological context, which reaction is faster? How do you know? (b) The enzyme that catalyzes glucose + ATP 2 glucose important point of regulation in glycolysis, while the enzyme that catalyzes glucose – 6 – PZ fructose – 6 – P (phosphoglucose isomerase) is not. Why is hexokinase a good step at which to regulate glycolysis relative to phosphoglucose isomerase? 6 – P + ADP (hexokinase) is an | - |
- Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate AG for ATP hydrolysis at 37 °C under these condi- tions. The energy charge is the concentration of ATP plus half the concen- tration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]Our assay had two reactions, one catalyzed by glucose oxidase, and the second by peroxidase. What must be true about the maximum possible rates (Vmax) of the two enzymes to insure that the velocity we measure corresponds to glucose oxidase activity, and not peroxidase activity? BIU SX' X. 14- A M IA rat liver (2 gms) was dissected, homogenized under optimal conditions in appropriate buffer (50 ml), pH, temperature & supplemented with protease inhibitors. The protein level in a 100 μl sample of the homogenate (H) was determined using standard procedures, & was estimated to be 2 mg/ml. Peepo is a medical student interested in purifying a protein that transports copper (Cu) at the expense of ATP hydrolysis into ADP and inorganic phosphate (Pi). Thus, prior to purification Peepo must develop an assay/test that confirms the presence of this protein. Kindly select one answer for Questions 1-10 Question -1 Compared to a control which of the following may serve as a possible assay verifying the presence of a Cu sensitive protein that hydrolyzes ATP: Add Cu to the H and measure the level of Pi Add ATP to H and measure the level of copper Vary copper level, add inorganic phosphate to H & measure residual Pi Vary Cu, add ATP to H and measure level of inorganic Pi…