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Enzymes bind to substrates with a specific affinity, called Question Blank 1 of 20. For example, hexokinase binds to ATP (Km = 0.4 mM) and glucose (Km = 50 M), and thus it binds to Question Blank 2 of 20 at a lower concentration and much more tightly. Catalyzed reactions are dependent upon substrate concentration to a point, and the maximum rate of catalysis where substrate concentrations only have a minimal effect is called Question Blank 3 of 20. Both Km and Vmax can be affected by inhibitors.

          Enzymes bind to substrates with a specific affinity, called Question Blank 1 of 20. For example, hexokinase binds to ATP (Km = 0.4 mM) and glucose (Km = 50 M), and thus it binds to Question Blank 2 of 20 at a lower concentration and much more tightly. Catalyzed reactions are dependent upon substrate concentration to a point, and the maximum rate of catalysis where substrate concentrations only have a minimal effect is called Question Blank 3 of 20. Both Km and Vmax can be affected by inhibitors.
        
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Biology for AP Courses
Biology for AP Courses
Julianne Zedalis, John Eggebrecht
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Enzymes bind to substrates with a specific affinity, called Question Blank 1 of 20. For example, hexokinase binds to ATP (Km = 0.4 mM) and glucose (Km = 50 M), and thus it binds to Question Blank 2 of 20 at a lower concentration and much more tightly. Catalyzed reactions are dependent upon substrate concentration to a point, and the maximum rate of catalysis where substrate concentrations only have a minimal effect is called Question Blank 3 of 20. Both Km and Vmax can be affected by inhibitors.
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Transcript

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00:01 30 nanogram enzyme present in 1 ml of solution.
00:17 Now 30 nanogram that equals 30 multiplied by 10 to the power minus 9 gram.
00:25 Now molar heat that will be weight over molecular weight divided by volume in liter.
00:46 Now 1 ml that equals 10 to the power minus 3 liter.
00:51 So here weight will be 30 nanograms 30 multiplied by 10 to the power minus 9 gram divided by molecular weight that will be given 1, 0, 0, 0, 0 gram per mole divided by volume that will be 10 to the power minus 3 liter.
01:23 Now kcat that will be equal vmax over enzyme concentration.
01:34 Vmax will be given 0 .14 and enzyme concentration that will be 3 into 10 to the power minus 10 molar.
01:42 So that will be 3 into 10 to the power minus 10 molar...
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