Only section 6-31 1.252.55.020.01/Velocity (sec/M)36 x 10³20 x 10³12 x 1036 x 10³6-31 You do an enzyme kinetic experiment and calculate a Vmax of 100μmol of product per minute. If each assay used 0.1 mL of anenzyme solution that had a concentration of 02 mg/mL, what wouldbe the turnover number if the enzyme had a molecular weight of128000 g/mole?6-32 The enzyme D-amino acid oxidate has a very high turnover numberbecause the D-amino acids are potentially toxic. The Ky for thenzyme is in the range of 1 to 2 mM for the aromatic amino aciand in the range of 15 to 20 mM for such amino acids as serinealanine and the acidic amino acids. Which of these acmino acienzyme?are the preferred substrates for the

Turnover number is the number of substrate molecules transformed into product per minute by one…

Answered: 6-31 You do an enzyme kinetic…

Biochemistry

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Chapter1: Biochemistry: An Evolving Science

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Problem 1P

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Why can’t an enzyme have a kcat/KM value greater than 109 M−1 ∙ s−1?
Consider the following chemical equation whose delta(G) = 9kcal/mol: AC + BD ---> AB + CD what are the reactants and what are the products is this reaction spontaneous? How do you know? Is energy released or consumed by this reaction? If an enzyme, which catalyzes this reaction is added, what will happen to delta (G) If this reaction is coupled to another reaction, whose delta(G) is -12 kcal/mol, what will be the net delta(G) value? will the overall reaction be spontaneous
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What is the likely keq for an enzyme that has a deltaG =-18.3kj/mol?
The turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?
The conversion of glucose-1-phosphate to glucose-6-phosphate by the enzyme phosphoglucomutase has a △G°' of -7.6 kJ/mol. Calculate the equilibrium constant for this reaction at 298 K and a pH of 7. (R = 8.315 J/K-mol) A. 0.003 B. 0.047 C. 1.00 D. 21
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
Most of the enzyme reactions followed the mathematical kinetic plots suggested by Michaelis-Menten plots: Give the Michaelis-Menten equation of an enzyme reaction and draw the Michaelis-Menten plot of [S] versus V0.
The phosphoryl group transfer potentials for glucose-1-phosphate and glucose-6-phosphate are 20.9 kJ/mol and 13.8 kJ/mol, respectively. (a) What is the equilibrium constant for the reaction shown below at 25 °C? (b) If a mixture was prepared containing 1 m glucose-6-phosphate and 1 x 10-3 M glucose-1-phosphate, what would be the thermodynamically favored direction for the reaction?
Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
Consider the following reaction: Glucose-1-phosphate → Glucose-6-phosphate ΔG° = −7.1 kJ/mol What is the equilibrium constant for this reaction at 25oC?

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6-31 You do an enzyme kinetic experiment and calculate a Vmax of 100 If each assay used 0.1 mL of an μmol of product per minute. enzyme solution that had a concentration of 02 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128000 g/mole?