Will the addition of catalase affect Keq or deltaG'? Explain your answer.
No, neither will be affected. Both Keq and deltaG' are thermodynamic quantities, describing the point of equilibrium and the amount of energy that a reaction can generate as it moves toward equilibrium. Catalysts (like enzymes) affect only the rate and mechanism of a reaction, not its equilibrium point or the net free energy change.
2. For a given set of experimental conditions (temperature, pH, salt concentration, etc) and a given concentration of catalase enzyme, how could you design an experiment to determine Km and Vmax for catalase? You may assume that you have a pure preparation of catalase and a means of measuring the rate of product formation (oxygen generation).
a) (2 pts) What sorts of assays would you need to conduct? What would you hold constant from assay to assay, and what would you vary? What would you measure?
You need to conduct assays in which you can measure the initial rate of the enzyme-catalyzed reaction, probably by measuring the rate of product formation. Each assay will need to have the same enzyme concentration and the same reaction conditions (temperature, pH, etc), but each will have a different substrate concentration. In other words...
Hold constant: enzyme concentration, temperature, pH
Vary: substrate concentration
Measure: initial rate of product formation (oxygen generation)
b) (2 pts) Draw a graph (or graphs) that depict(s) the sort of data you'd expect to collect, and the general trends you'd expect those data to exhibit. Be sure to label your axes.
answer #1 (Michaelis-Menten graph):
X axis: substrate concentrationY axis: initial rate of reaction
shape of curve: increases linearly at first, then levels off as substrate concentration increases
also an acceptable answer...(Lineweaver-Burke plot)
X axis: 1/substrate concentrationY axis: 1/initial rate
shape of curve: linear, crosses Y axis somewhere above zero (1/Vmax), crosses X axis at a number less than zero (-1/Km)
c) (1 pt) Explain how Vmax and Km could be derived from the information graphed in b.
Graph the data using a Lineweaver-Burke plot, the second version described above. The Y intercept is 1/Vmax and the X intercept is -1/Km
3. You add an inhibitor to the reactions conducted in part 2, and you observe that the presence of inhibitor affects the apparent Km of the enzyme without affecting the Vmax.
a) (2 pts) Given that this is an inhibitor, would you expect the apparent Km to be larger or smaller than the Km measured in part 2? Explain your answer.
The apparent Km will be larger than the Km without inhibitor. In other words, it will take a higher concentration of substrate to get the reaction to occur at a rate of 1/2 Vmax. That's because the substrate doesn't have unimpeded access to the enzyme--it has to compete with the inhibitor.
b) (1 pt) What type of inhibitor is this, and what would you predict is its mechanism of inhibition?
It's a competitive inhibitor, and it works by binding reversibly to the active site of the enzyme. Enzyme that is bound to inhibitor cannot bind simultaneously to substrate, so the substrate and the inhibitor compete for binding to the active site.
Maintained by: bkbaxter@lclark.edu
Updated: 6 Feb 01