[Strike]

Hello,

The experiment is that little paper discs are soaked in a catalase solution so each disc absorbs 1ug of the enzyme. The disc is then placed at the bottom of a beaker with a known peroxide concentration. The time it takes for the disc to reach the top of the solution is recorded.



I only know how to get the Km value was to plot 1/v vs 1/[ S] but I don't have the v values. I tried plotting the data and it does seem to max out at a 0.2 M of H2O2


(Graph starts to reach an asymptote after 0.2 M meaning Vmax is reached at 0.2 M but how do I calculate this?)


Also: Say the enzyme (is tetrameric, meaning 4 active sites) has a kcat of 40,000,000 sec-1 and the balanced equation for substrate into product is:

2H₂O₂ --> 2H₂O + O₂

The enzyme would still turnover 40 million peroxide molecules regardless of the stoichiometry and the "tetrameric enzyme with 4 active sites" right?

[Babcock_Hall]

Does the floating to the top only happen when all of the substrate is consumed?  I admit to being baffled.  Most enzyme kinetics is done using the initial rate of the reaction, before much of the substrate is consumed.  However, this problem might require the use of an integrated rate equation, which is less common in enzyme kinetics.  R. A. Alberty did some work in this area.