Re: Why when more peroxide is added to more catalase, is there a greater reacti

Tony,
    You didn’t describe how you were monitoring the catalase reaction, so 
I’m going to assume that you were judging the rate of the reaction by 
observing the formation of bubbles of oxygen gas.  Catalase decomposes 
hydrogen peroxide into water and oxygen.
     Like any enzymatic reaction, the rate is described by a formula called 
the Michaelis-Menten equation.  The rate of the catalase reaction with 
hydrogen peroxide would be given by this equation:

reaction rate = 

    (Vmax) x [Concentration of hydrogen peroxide] x  [Enzyme concentration]
    ------------------------------------------------------------------------
     Km + [Concentration of hydrogen peroxide]

Vmax is the maximum reaction rate, and is a constant. 
Km is the Michaelis-Menten constant, and for catalase, Km = 25mM (25 
millimoles/liter).  This number gives a measure of how well an enzyme binds 
its substrate.

The Michaelis-Menten equation reflects the fact that the reaction rate for 
an enzyme-catalyzed reaction exhibits saturation.   That is, the rate 
reaches a maximum reaction rate (Vmax), and stays at that maximum rate 
regardless of how much additional hydrogen peroxide you add – since the 
concentration of hydrogen peroxide appears in both the numerator and the 
denominator.  

A good way to think about this is to remember that the enzyme is a catalyst, 
and is present in small concentrations.  The concentration of the catalase 
is much less than that of the hydrogen peroxide.  Once you have enough 
hydrogen peroxide present so that all of the enzyme molecules are rapidly 
binding hydrogen peroxide and decomposing it, adding more hydrogen peroxide 
will not make the reaction go any faster.  This may have been what your 
teacher meant when she told you that having more peroxide wouldn’t make the 
reaction go faster.  However, the rate is also dependent on the 
concentration of enzyme present.  In your experiment, you used 3 mL of 
catalase in 125 mL total volume, instead of 1 mL of catalase in 100 mL total 
volume.  So, under your conditions the rate of the reaction WILL be higher, 
because the enzyme concentration is higher.  In addition, since you have 
much more peroxide in your reaction mixture that you were supposed to, much 
more oxygen gas will be produced.