The hexokinase can be inhibited by a non competitive inhibitor. On your enzyme show where the non competitive inhibitor would bind What is the AG for this reaction? (note the AG for the conversation of G to G6P is +3.0kcal/Mol).
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- The hexokinase can be inhibited by a non competitive inhibitor. On your enzyme show where the non competitive inhibitor would bind
-
What is the AG for this reaction? (note the AG for the conversation of G to G6P is +3.0kcal/Mol). Explain.
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- 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 | - |Handwritten Identify the molecule names, enzyme name, enzyme classification and change in reaction(for glycolysis pathway)16. The overall reaction for the glycolysis reaction is C6H₁2O6(aq) + 2NAD+ (aq) + 2ADP³(aq) + 2HPO(aq) + 2H₂O(1) 2CH3COCO₂ (aq) + 2NADH(aq) + 2ATP4 (aq) + 2H3O+ (aq). What is A,G at chemical equilibrium?
- The glucose/glucose-6-phosphate substrate cycle involves distinct reactions of glycolysis and gluconeogenesis that interconvert these two metabolites. Assume that under physiological conditions, [ATP] = [ADP]; [P;] = 1 mM. Consider the glycolytic reaction catalyzed by hexokinase: ATP + glucose ADP + glucose-6-phosphate AG = - 16.7 kJ/mol (a) Calculate the equilibrium constant (K) for this reaction at 298°K, and from that, calculate the maximum [glucose-6-phosphate]/ Iglucose] ratio that would exist under conditions where the reaction is still thermodynamically favorable. (b) Reversal of this interconversion in gluconeogenesis is catalyzed by glucose-6-phosphatase: glucose-6-phosphate + H20 = glucose + P AG" = -13.8 kJ/molPyruvate carboxylase catalyzes the first step of gluconeogenesis. ATP + HCO3─ + pyruvate → oxaloacetate + ADP + Pi ∆G0’ = ─2.1 kJ mol-1 a) Calculate ΔG for this reaction under the following physiological conditions: 370C, pH 7 [pyruvate] = [HCO3─] = 4.0 mM [oxaloacetate]= 2.0 mM [ATP] = 3.5 mM [Pi] = 5.0 mM [ADP] = 1.8 mMThe degradation of glycogen is catalyzed by the enzyme phosphorylase and has AGO" equal to +3.1 kJ · mol1. The equation for this reaction is shown below. glycogen (n residues) + P;→ glycogen (n-1 residues) + G1P What is the ratio of [P;]/[G1P] under standard conditions? Use 2 significant figures. [P;] : [G1P] = i :1 What is the value of AG under cellular conditions when the [P;/[G1P] ratio is 50/1? Use 2 significant figures. AG = i kJ. mol-1
- The inilial reactions in the biosynthesis of the amino acid aspartate at 298 K are: Carbamoyphoephate Cartamoyi + phosphate ADP + phoaphate - ATP AO=-12,300 calmol AG-T800 calimol Which of the following statements is gorrect? A) The energy which is released fron the conversion of 1 mole of carbamoyphosphate to carbamoyl+ phosphate (P) in reaction I is sufficient to drive the synthesis of 1 mole of ATP in reaction II. B) AG for the reaction Carbamoyphosphate + Carbamoyl + phosphate is +123 kcalimol. c) AHP for reaction I cannot be detemined trom the information given. D) Al of the above. E) None of the above.Draw the products of the reaction of xylulose-5-phosphate and erythrose-4-phosphate catalyzed by transketolase in the pentose phosphate pathway. Provide the structure in the protonation state found in physiological conditions. H H H OH FO HO-H H-OH H OPO3²- Q transketolase Draw glyceraldehyde-3- phosphate H H- H H H O OH OH OPO3²- Draw fructose-6- phosphate Q I IIntramitochondrial 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]
- Choose the enzyme and cofactors involved in the reaction of fructose-1,6-bisphosphate to fructose-6-phosphate in gluconeogenesis. H H-TOPO Fo HOTH HDH HOH MOTOPOP K E H-TOH H OPO Problem 3 of 17 A B Q D E fructose-1,6-bisphosphatase phosphoglucose isomerase glucose-6-phosphatase phosphoglycerate kinase hexokinaseConsider the mechanism of enolase, as indicated below. Which of the following correctly describes the roles of the Mg2+ as illustrated in the figure? (This is a multi- select question). Mg2+ Mg2 Enolase PO3- OH -C-C-H H OH HO H-N-H Lys 345 Glu211 2-Phosphoglycerate bound to enzyme Mg2+ Mg2 PO3- OH C-C-H OH HO H H-N*-H Lys 345 O Glu211 Enolic intermediate HOH PO3- H Phosphoenolpyruvate The metal ion (Mg2+) is helping to stabilize the extra negative charge that developed on the carboxyl group in the enolic intermediate. The metal ion (Mg2+) is serving as a general base, removing a proton in order to improve the quality of the nucleophile. The metal ion (Mg2+) is assisting in the oxidation of the carboxyl carbon through metal ion catalysis. The metal ion (Mg2+) is helping to orient the substrate properly in the active site. The metal ion (Mg2+) is accepting a proton in order to improve the quality of the leaving group.The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1