With a AGO'of -16.7 kJ/mol, the reaction catalyzed by hexokinase is considered to be at equilibrium substrate and product concentration dependent O freely reversible metabolically irreversible none of the above
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- Energetics of the Hexokinase Reaction The standard-state free energy change. Gfor the hexokinase reaction, is — 1 6.7 kJ/mol. Use the values in Table I to calculate the value of Gfor this reaction in the erythrocyte at 37°C.Examine the ActiveModel for alcohol dehydrogenase and describe the structure and function of the catalytic zinc center.Distinguishing the Mechanisms of Class I and Class I Aldolases Fructose bisphosphate aldolase in animal muscle is a class 1 aldolase, which forms a Schiff base intermediate between substrate (for example. fructose-1, 6-bisphosphate or dihydroxyacetone phosphate) and a lysine at the active site (see Figure I8.12). The chemical evidence for this intermediate conies from studies with aldolase and the reducing agent sodium borohydride, NaBH4. Incubation of the enzyme with dihydroxyacetone phosphate and NaBH4 inactivates the enzyme. Interestingly, no inactivation is observed if NabH4 is added to the enzyme in the absence of substrate. Write a mechanism that explains these observations and provides evidence for the formation of a Schiff base intermediate in the aldolase reaction.
- With a ∆G°´ of -16.7 kJ/mol, the reaction catalyzed by hexokinase is considered to be _____. at equilibrium substrate and product concentration dependent freely reversible metabolically irreversible none of the abovewhen saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.Proline racemase catalyzes the conversion between L-proline and D-proline. The Km and kcat for this reaction are 0.15 M and 550/sec respectively. If the enzyme concentration is 1.45 X 10-5 mmole/ml what is the Vmax of this reaction?
- An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)If a 0.1 M solution of glucose 1- phosphate at 25 °C is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Calculate Keq and ΔG′° for this reaction.
- The mechanism involved in the reaction catalyzed by phosphoglyceromutase is known to involve a phosphorylatedenzyme intermediate. If 3-phosphoglycerate is radioactively labeledwith 32P, the product of the reaction, 2-phosphoglycerate, does nothave any radioactive label. Design a mechanism to explain these facts.An uncompetitive inhibitor interacts with the enzyme•substrate complex to form a ternarycomplex (ES•I). This equilibrium reaction can be described as follows:ES + I ⇌ ESIModify the simplified kinetic scheme you drew for E + S ⇌ E + P to include this equilibriumexpressionAn enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?