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.
This can't be calculated unless km is not given. Because km is required just to find out the substrate concentration.
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. Km= 15.42M
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
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
For an enzyme that follows the Michaelis-Menten kinetic, what substrate concentrations (relative to Km) are needed for the speed of the reaction to be 0.12 there vmax 0.25 there vmax 0.5 there vmax 0.9 there vmax.