Consider an enzyme with the following properties: – Km = 0.0050 M
At what substrate conc. would the enzyme operate at one-fourth of its maximum rate?
At a given substrate conc. (e.g., ½Km) what would the rate be as a fraction of Vmax?
Consider an enzyme with the following properties: – Km = 0.0050 M At what substrate conc....
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
Enzyme X has a Km for substrate A which is twice as large as its Km for substrate B. If the enzyme is added to a solution containing equal, but low concentrations of A and B and Vmax for A and B are equal: substrate A will be used at a rate equal to substrate B substrate A will be used at a rate greater than substrate substrate B will be used at a rate greater than substrate A. 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)
Estimation of Vmax and K by Inspection Although graphical methods are available for accurate determination of the Vmax and K of an enzyme-catalyzed reaction (see Box 6-1), sometimes these quantities can be quickly estimated by inspecting values of Vat increasing (Sl. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained. V (pm/min) V (m/min) 112 [S] (M) 2.5 x 10 4.0 x 100 1 x 10-5 2 x 10-5 [S(M) 4 x 10-5...
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...
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.
You are studying a new muscle enzyme, M, and need to characterize its kinetic properties. a) Experimentally, how will you determine the M’s VMAX and KM for its substrate , A? EXPLAIN in detail, including any limitations your mathematical methods may have. b) Now consider that it is known that another molecule, D, also binds to M, and interferes with the reaction that converts A to product Q, but you don't know where or how. What kind of kinetic evidence...
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
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM