The Km of an enzymatic reaction is 10^-4 M. What substrate concentration is needed in order to attain 60% if the maximum velocity of the reaction?
The Km of an enzymatic reaction is 10^-4 M. What substrate concentration is needed in order...
If the concentration of substrate in an enzymatic reaction is set to the Km of the enzyme substrate system then the velocity is equal to: 0 units / second Vmax units/second Vmax/2 units/second 1/3 Vmax units/second 2/3 Vmax units/second 100 units / second
What kind of kinetics is observed in an enzymatic reaction, under conditions where the substrate concentration is much higher with respect to KM ([S] >> KM)? Assume that a monomeric and non-allosteric enzyme is considered. A. Cooperative B. First Order C. Zero Order D. The systerm is at equilibrium and no reaction occures E. Second Order
The rate law for a certain enzymatic reaction is zero order with respect to the substrate. The rate constant for this reaction is 4.6 times 10^2 M middot s^-1. If the initial concentration of the substrate is 0.338 mol L^-1, what is the initial rate of the reaction? rate = _______ times 10 _______ M middot s^-1
The relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation a) At what substrate concentration would an enzyme with a kcat of 30.0 s-1 and a Km of 0.0050 M operate at one-quarter of its maximum rate? b) Determine the fraction of Vmax that would be obtained at the following substrate concentrations: [S]=Km/2, [S]=2Km, [S]=10Km
What would be the Km of an enzymatic reaction that showed the following kinetics? What would be the Km of an enzymatic reaction that showed the following kinetics? [substrate] mM Rxn Rate (umol/min) 8 x 10-6 80 2x 10-5 121 8 x 10-5 40 4 x 10-3 254 2 x 10-2 280 1x 10-1 279 O 8 x 10-5 8 x 10-6 1x 10-1 4x 10-3
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km of 0.0030 M have an initial velocity that is 25% of the maximum velocity? a. 0.0010M b 0.0090 M c. 0.0015 M d. 0.0060 M e.Cannot be determined
Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, PNPP, had been converted to PNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1 st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5...
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?
If the enzyme concentration is 3.8 X 10-8, what concentration of substrate would generate a velocity equal to 0.25 Vmax? Km is not needed. Km= 15.42M
3. Acid Phosphatase was assayed at [S] =2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, PNPP, had been converted to pNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5 minutes?...