The most efficient enzyme is one with O high KM and low Vmax O low KM and low Vmax O low KM and high Vmax O high KM and high Vmax
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- What would be the Vmax and KM, if you double the concentration of the enzyme Question 20 options: Vmax change and KM double Vmax double and KM double Vmax double and KM not alter Vmax not change and KM doublein the concerted model of enzyme action when the L ratio (T/R) is 10,000 the velocity vs. Substrate plot results in what type of a curve.For an enzyme that has a Km of 25 mM, Vmax of 50 mM/s and kcat of 250 s-1, how long does a single reaction take? Select one: 50 ms 250 ms 4 ms 25 s 4 s
- When substrate [S] = Km, the velocity of an enzyme catalyzed reaction is about: * 0.1 Vmax. 0.2 Vmax. 0.5 Vmax. 0.9 Vmaxfor the table below make a graph call it Factors vs Rate of Enzyme Activity rules: data points must be an x or circled dot, must be on grid papaer and half the page minimum size, the independant variable on the x axis and dependant variable on the y axis, must include titlesThe order of the data in the table is reversed. Pay attention and calculate straight. Some experiments have been done using a fixed amount of enzyme. Made in different substrate concentrations. The table below contains data. S( mM/L-min) V( mM/L-min) 2000 155 1000 150 200 120 100 100 60 80 40 64 20 40 10 20 Find Vmax and Km using all data points and at least two linear transformations
- Consider the following data set of enzymes A, B C and D which catalyze the same reaction. Enzyme A km = 5mM Vmax = 225MM/sec Enzyme B km = 50mM Vmax = 430mM/sec Enzyme C km = 15mM Vmax = 235mM/sec %3D Enzyme D km 12mM Vmax = 533mM/sec When concentrations of substrate are very low, which is the enzyme that is producing most product? MacBook Air DO F5 S0 # 12 F4 F3 %kcat is: a measure of the catalytic efficiency of the enzyme the rate constant for the reaction ES → E + P the rate constant for the reaction ES → E + S the [S] that half saturates the enzyme ½ Vmax-Inhibitor +Inhibitor [S] (mM). V&νβσπ:(μmol/sec) V0&νβσπ&ν βσπ;(μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITHOUT iinhibitor?
- An enzyme-catalyzed reaction has a KM of 5 mM and a Vmax of 60 nM/sec. What is the substrate concentration when the initial velocity of the reaction is 30 nM/sec? Show your work/reasoningWhen substrate concentration [S] << Km, the rate of an enzymatic reaction is increased by O increasing kcat/Km O increasing total enzyme concentration O increasing [S] O A and B O A, B and CIf a competitive inhibitor of an enzyme is added to the mixture of the enzyme and its substrate, the inhibitor will: O a. decrease the value of Vmax for the chemical reaction catalyzed by this enzyme O b. decrease the value of KM for the chemical reaction catalyzed by this enzyme O c. increase the value of KM for the chemical reaction catalyzed by this enzyme O d. increase the value of Vmax for the chemical reaction catalyzed by this enzyme