Study Questions Final

Study Questions Final - BIS 102, Prof. C. Gasser 49 BioSci...

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BIS 102, Prof. C. Gasser 49 BioSci 102, Sample Study Problems for Final Exam 1. To determine the K m and k cat (= turnover number) of an enzyme, 0.1 nmole of enzyme was added to substrate at various concentrations and the appearance of product was measured. When the substrate concentration was 1 X 10 -5 M, the product formed by: 15 seconds was 0.5 μmole 30 seconds was 1.0 μmole 45 seconds was 1.5 μmole V = 2 μmol/min (or 33 nmol/sec) 1 V = 0.5 min/μmol , 1 [S] = 10 5 M -1 When the substrate concentration was 2 X 10 -5 M, the product formed by: 15 seconds was 0.75 μmole 30 seconds was 1.5 μmole 45 seconds was 2.25 μmole V = 3 μmol/min (or 50 nmol/sec) 1 V = 0.33 min/μmol , 1 [S] = 5 X 10 4 M -1 When the substrate concentration was 4 X 10 -5 M, the product formed by: 15 seconds was 1.0 μmole 30 seconds was 2.0 μmole 45 seconds was 3.0 μmole V = 4 μmol/min (or 67 nmol/sec) 1 V = 0.25 min/μmol , 1 [S] = 2.5 X 10 4 M -1 When the substrate concentration was 10 X 10 -5 M, the product formed by: 15 seconds was 1.25 μmole 30 seconds was 2.5 μmole 45 seconds was 3.75 μmole V = 5 μmol/min (or 83 nmol/sec) 1 V = 0.20 min/μmol , 1 [S] = 10 4 M -1 Using the supplied graph paper, and showing all work, calculate the K m and K cat for this enzyme.
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BIS 102, Prof. C. Gasser 50 1 V max = 0.165 sec/μmol , V max = 6.1 μmol/min (100 nmol/sec) -1 [K m ] = -5 X 10 4 M -1 , [K m ] = 2 X 10 -5 M K cat = V max [E t ] = 6.1 μmol/min 10 -4 μmol = 6.1 X 10 4 min -1 (or 1,000/sec) check V max and K m to make shure we made no error. For [S] = 10 μM V = 6.1 X 10 20 + 10 = 2.03 (very close to 2) For [S] = 100 μM V = 6.1 X 100 20 + 100 = 5.08 (very close to 5)
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BIS 102, Prof. C. Gasser 51 2. An enzyme was assayed at different concentrations of substrate in the presence and in the absence of an inhibitor. The initial velocities (μmoles X liter -1 X min -1 ) were measured at various substrate concentrations (moles/liter) were plotted according to the Lineweaver- Burk method and the following plot was obtained. Answer the follow questions: a) Label the X axis and the Y axis with the appropriate symbols and units. b) What is the K m of the enzyme for the substrate in the absence of inhibitor? -1/K m = -1000 M -1 K m = 10 -3 M c) What is the V max of the enzyme in the absence of inhibitor? 1/V max = 0.02 V max = 50 μM/min d) On the graph above, sketch the plot that would be obtained if the amount of inhibitor were decreased to about half the amount originally used. e) Write an equation which you could be used to determine the K i of the inhibitor from the above information. Km app (Km in the presence of inhibitor) = Km (1 + [I] K I )
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BIS 102, Prof. C. Gasser 52 3. The K m for an enzyme catalyzed reaction is 10 -4 M. The apparent K m for the same reaction in the presence of 9 X 10 -5 M inhibitor is 10 -3 M. a) Is the binding of substrate to enzyme greater or less in the presence of inhibitor? Why?
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Study Questions Final - BIS 102, Prof. C. Gasser 49 BioSci...

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