a) Calculate the enzyme and specific activity of a reaction with 3 pM Hsp90 using the following information: The rate is measured in a spectrophotometer as 0.028 OD units/min in a 1 ml reaction volume. The absorbance was detected at 340nm and the extinction coefficient for NADH at this wavelength is 6200L M- 1 min-1 and the molecular mass of Hsp90 is 82.7 kDa. The rate of NADH utilisation is equivalent to the rate of ATP utilised by Hsp90. Show all your calculations and the units for your answers. b) Calculate the turnover number for the reaction described in (a) above
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- An 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?a) Calculate the enzyme and specific activity of a reaction with 3 μM Hsp90 using the following information: The rate is measured in a spectrophotometer as 0.028 OD units/min in a 1 ml reaction volume. The absorbance was detected at 340nm and the extinction coefficient for NADH at this wavelength is 6200 L M-1 min-1 and the molecular mass of Hsp90 is 82.7 kDa. The rate of NADH utilisation is equivalent to the rate of ATP utilised by Hsp90. Show all your calculations and the units for your answers. b) Calculate the turnover number for the reaction described in (a) aboveAn enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?
- Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMIf the hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase has a ΔG′∘ of −11.386 kJ/mol at 25 °C, what percentage of substrate remains once the reaction reaches equilibrium assuming no product was initially present? (Round answer to the nearest whole number)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)
- For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.Use the Michaelis-Menten equation to complete the enzyme kinetic data set, when Km is known to have a value of 1 mmol L-1For part (b) of this problem, use the following standard reduction potentials, free energies, and nonequilibrium concentrations of reactants and products: Consider the last two steps in the alcoholic fermentation of glucose by brewer’s yeast: pyruvate + NADH + 2H+ → ethanol + NAD+ + CO2 (a) Do you predict that ∆S° for this reaction is > 0 or < 0? (b) Calculate the nonequilibrium concentration of ethanol in yeast cells, if ∆G = -38.3 kJ/mol for this reaction at pH = 7.4 and 37 °C when the reactants and products are at the concentrations given above. (c) How would a drop in pH affect ∆G for the reaction described in part (b)? (d) How would an increase in intracellular CO2 levels affect ∆G for the reaction in part (b)? (e) How would an increase in intracellular CO2 levels affect ∆G°′ for the reaction in part (b)?
- Substrate 1 is plotted on the following kinetic data, the Vmax and Km for Substrate 1 are 95 uM/sec and 10 µM, respectively. Draw a curve for the data you would expect to observe for a different substrate (Substrate 2) with a Km = 30 µM (assuming that both substrates give the same Vmax), if [Enzyme] = 0.05 µM, calculate kcat for Substrate 1 and what is the catalytic efficiency in units of M-1s-1?A bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 m mol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM?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