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
- for 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 transformationsConsider 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/reasoning
- When 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 enzymeWhen k2 >> k-1, KM approximates the affinity of the enzyme•substrate complex.The (circle one) higher / lower the KM, the more tightly the enzyme binds to its substrate