Use the information below to answer the next three questions. An enzyme can cata
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Use the information below to answer the next three questions.
An enzyme can catalyze a reaction with either one of two substrates (S1 or S2). The initial reaction rate (vo) for each of the substrates was measured (in nmoles/min) as a function of the concentration of substrate [S] (?M), and the following results were obtained:
Use the information below to answer the next three questions. An enzyme can catalyze a reaction with either one of two substrates (S1 or Se). The initial reaction rate (vo) for each of the substrates was measured (in nmoles/min) as a function of the concentration of substrate [Sl (uM), and the following results were obtained: IsI 0.0001 0.0002 0.0005 0.001 0.002 0.005 0.01 0.02 0.05 5.0 Vo (S) Vo (S2) 17 29 50 67 80 91 95 98 100 100 50 71 83 91 96 98 100 100 1 Adapted from: Chang R (2005) Physical Chemistry for the Biosciences, University Science Books, Herndon, VA. P.383.Explanation / Answer
10. d) the Km for S1 is 0.0002 and for S2 is 0.0005. Km is the value equal to substrate concentration at half of the maximum reaction velocity (Vmax). Thus other options are not valid as they represent the vecolcity not the particular substrate concentration. The Vmax is 100 for both the substrate, so Vmax will be 50. At vecolcity of 50,S has 0.0002 and S2 has 0.0005 concentrations.
11. c) S1 is a better substrate because it has lower Km. Vmax is same for both the substrate as it it the property of enzyme not substrate. Km is the constant that indicates binding of a particular substrate with enzyme.
12. d) 1/100. Intercept Y at x=0 represents the 1/Vmax in Lineweaver-Burk plot/equation, so for Vmax = 100, 1/Vmax will be 1/100.
13.
Reaction 1: K = antilog (-dG°/1.42) = antilog (10.3/1.42) = 22758309.54
Reaction 2: antilog (7.3/1.42) = 138305.67
Reaction 3: antilog (3.3/1.42) = 210.81
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