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
If the enzyme concentration is 3.8 X 10-8, what concentration of substrate would generate a velocity...
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.
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
1. Michaelis and Menten examined how the velocity of enzyme catalyzed reactions change with substrate concentration. Which of the following is (are) common to all enzyme catalyzed reactions? Velocity is insensitive to changes in [substrate] at all substrate concentrations. Km is the [substrate] required to reach 50% of Vmax. Velocity is responsive to changes in [substrate] when the Km > [substrate]. Velocity is insensitive to [substrate] when [substrate] is much greater than Km. Velocity reaches 90% of Vmax when [substrate]...
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration will be required to obtain 55% of Vmax for this enzyme? (10 pts)
112 marks] 3. The relationship between initial velocity (V.) and substrate concentration of most of the enzyme- catalized reactions are explained by Michaelis-Menten equation. IMPORTANT: Show the calculations and indicate the units for all your answers. a. For an enzyme which follows the Michaelis-Menten enzyme kinetics, Km is 50 mmol L. Calculate the substrate concentration required to obtain the initial velocity (V.) equivalent to 90% of the maximum velocity (Vmax). b. The Vmax of the above reaction is 250 mmol...
What is substrate concentration, expressed as a multiple of Km, when an enzyme reaction is observed to have an initial rate Vo = 0.75 Vmax. Select one: O a. [S] = 0.33 x km O b. [S] = 0.25 km O c. [S] = 0.75 x Km O d. [S] = 0.3 x km O e. [S] = 0.5 x km Check Next page ime Jump to...
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 ?
4. An enzyme hydrolyzed a substrate concentration of 0.03mmol/L, the initial velocity was 0.5 X 10-3 mmol/L.min' and the maximum velocity was 4.5 x 10-3 mmol/L.min.l. Calculate the Km value. 5. Urease hydrolyzed urea at [s]=0.03mmol/L with a km of 0.06 mmol/L. The initial velocity observed was 1.5X10-3 mmol/L.min-1 Calculate the maximum velocity of the enzyme reaction.
What is the velocity of a Michaelis-Menten enzyme reaction (in terms of vmax) when the concentration of substrate is 4 times the value of KM? Show your work.
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