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 |
If the concentration of substrate in an enzymatic reaction is set to the Km of the...
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?
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 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
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...
The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. Graph a Lineweaver-Burk plot. What are the apparent values of vmax and km for each experiment? what is the inhibition mechanism If the concentration of inhibitor is 0.5 mM, what is the value of K1?
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)
Consider a description of an enzymatic reaction pathway that begins with the binding of substrate S to enzyme E and ends with the release of product P from the enzyme. E+S →ES → EP E+P Under many circumstances, KM = [E] [S] / [ES] What proportion of enzyme molecules are bound to substrate when [S] = KM? Why? Recall that when [S] = KM, the reaction rate is Vmax/2. Does your answer to Part A make sense in light of...
Km is: A. the concentration of an inhibitor needed to give 1/2 Vmax B. a measure of the rate of product formation C. related to the affinity of the enzyme for its substrate D. a measure of the activation energy of the enzyme-catalyzed reaction E. the velocity equal to one-half of the maximum 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
1.5 1.9 2.0 2.1 22 2.3 24 25 26 2.7 28 3) (a) An enzyme is used to convert a substrate at a temperature of 25°C. The Michaelis constant of this reaction is 0.042 mol dm3. The velocity of the reaction is 2.45 x 104 mol dm s when the substrate concentration is 0.890 mol dm3. Find the maximum velocity of this reaction? hint: vk2Er +161 and vmax (b) Plot v - vs -[S] for a standard enzymatic reaction that...