Enzyme Inhibition Kinetics: Problem Set#5 Question #5
Recommended problems in Campbell, Chapter 5: #25, 26.
This page simulates the outcomes of initial velocity measurements on the enzyme, invertase (sucrase). Invertase hydrolyzes the glycosidic bond in the disaccharide, sucrose. The reaction catalyzed is:
Sucrose --> Glucose + Fructose
Classically, the reaction rate was measured by following the change in optical rotation of plane-polarized light. When sucrose is hydrolyzed the sign of the rotation is inverted; hence the common name, "invertase". Invertase is inhibited by both of its products. One of them is a competitive inhibitor; the other is a noncompetitive inhibitor.
Background on enzyme inhibition kinetics can be found in the Lecture 15 notes.
- You have a 10.0 mg/ml solution of invertase from carrot root. This enzyme is an octamer; the Mr = 456,000 Da. The inhibitor used in this question is fructose.
You can add any concentration of sucrose and fructose to the reaction tube.
- a) Use the resulting values of vo to calculate 1/vo and 1/[S];
- b) Plot the data (both without and with added fructose) on a single double reciprocal plot;
- c) Determine KM's and Vmax's;
- d) Replot the slopes and/or intercepts vs. [fructose] for each concentration of the inhibitor used;
- e) Calculate KI from the replot(s);
- f) Submit your data, calculations, and the graphs that yield KM, Vmax, and KI.
(Each calculated vo value has a small "experimental error" added to it.)
Use the steps described on Topic #5: Enzyme Inhibition Calculations to solve this problem.
Hint: For your submitted graph, ignore the data <10% and >90% Vmax when drawing a line. These values will have more relative error and they are not necessary to determine an accurate slope and intercept on the reciprocal plot.
Print an Answer Form for submitting your solution.
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