Assume that the reaction Arabinose +Pi <> Arabinose-6-P has a delta G of +16.1 kJ/mol and the reaction XTP <> XDP + Pi has a delta G of -24.7 kJ/mol. Calculate the delta G for the combined reaction Arabinose + XTP > Arabinose-6-P + ADP, in kJ/mol to four significant figures.
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- The equilibrium constant Kc for the reaction fructose-1,6-diphosphate ⇋ glyceraldehyde-3-phosphate + dihydroxyacetone phosphate is 8.9 x 10-5 M at 250C and the behavior is assumed to be ideal. Calculate the delta G for the process Suppose that we have a mixture that is initially 0.010 M in fructose-1,6-diphosphate and 1.0 x 10-5 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is the delta GThe equilibrium constant Kc for the reaction fructose-1,6-diphosphate ⇋ glyceraldehyde-3-phosphate + dihydroxyacetone phosphate is 8.9 x 10-5 M at 250C and the behavior is assumed to be ideal. Calculate .for the process (standard state: 1M) Suppose that we have a mixture that is initially 0.010 M in fructose-1,6-diphosphate and 1.0 x 10-5 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is ?The standard state free energy of hydrolysis of acetyl phosphate isΔG° = -42.3 kJ/mol. Acetyl-P + H2O acetate + Pi Calculate the free energy change for the acetyl phosphate hydrolysis in a solution of 2 mM acetate, 2 mM phosphate and 3 nM acetyl phosphate.
- What is the likely keq for an enzyme that has a deltaG =-18.3kj/mol?Given that the AG" values for the hydrolysis of glucose 1-phosphate and glucose 6-phosphate are approximately -21 kJ/mol and –14 kJ/mol, respectively, which statement is true regarding the isomerization shown? glucose 1-phosphate → glucose 6-phosphate O The reaction can likely proceed in either direction and will therefore depend on the concentrations of glucose 1-phosphate and glucose 6-phosphate. O Only the anabolic direction will occur because the free energy change is positive. O The reaction will proceed in both directions simultaneously. This reaction can only proceed in one direction and must be the rate limiting reaction for this point in metabolism.Please note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
- The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.27. The standard-state free energy of hydrolysis for acetyl phosphate in 1,3-Bisphosphoglycerate is 1,3-Bisphosphoglycerate + H₂O 3-phosphoglycerate + Pi AGI-49.6 kJ/mol. Determine the free energy change for 1,3-Bisphosphoglycerate hydrolysis in a solution of 5.4 mM 3-phosphoglycerate, 0.0054 M inorganic phosphate, and 20 nM 1,3-Bisphosphoglycerate.The turnover number of the enzyme fumarase that catalyzes the reaction, Fumarate + H20 ===→ L-malate, is 2.5 x 103 S - l and Km = 4.0 X 10- 6 mol/L. Calculate the rate of conversion of fumarate to L-malate if the fumarase concentration is 1.0 x 1 0 - 6 mol/L and the fumarate concentration is 2.04 x 10- 4 mol/L.
- The KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affi nity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?In making wine, glucose (C6H1206) is fermented to produce ethanol (C2HSOH) and carbon dioxide (CO2) as shown in the reaction. a. If the fermentation reaction starts at 57.5 g glucose, what is the theoretical yield of ethanol (in grams)? b. If 22.0 g ethanol is produced, what is the percent yield of the reaction?A table of standard state free energies is shown below. For the reaction: Creatine phosphate + ADP----> Creatine + ATP the free energy under standard state conditions is kJ/mole. TABLE 14.1 Standard free energies of hydrolysis of some phosphorylated compounds Compound kcal mol kJ mol - 14.8 Phosphoenolpyruvate 1,3-Bisphosphoglycerate Creatine phosphate ATP (to ADP) Glucose 1-phosphate Pyrophosphate Glucose 6-phosphate -61.9 -11.8 - 49.4 - 10.3 -43.1 - 7.3 - 30.5 - 20.9 - 19.3 - 5.0 - 4.6 – 3.3 -13.8 Glycerol 3-phosphate – 2.2 - 9.2 A. +12.6 В. +73.6 С.-3.0 D. -12.6 ОЕ.-73.6