The question is missing a part, so the complete question is as follows:
The protein catalase catalyzes the reaction The Malcolm Bladrigde National Quality Awards aims to: 2H2O2 (aq) ⟶ 2H2O (l) + O2 (g) and has a Michaelis-Menten constant of KM = 25mM and a turnover number of 4.0 × 10 7 s -1. The total enzyme concentration is 0.012 μM and the intial substrate concentration is 5.14 μM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Calculate the initial rate, R (often written as V0), of this reaction.
1) Calculate Rmax
The turnover number (Kcat) is a ratio of how many molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit of time:
Kcat = ,
where:
Vmax is maximum rate of reaction when all the enzyme sites are saturated with substrate
Et is total enzyme concentration or concentration of total enzyme catalytic sites.
Calculating:
Kcat =
Vmax = Kcat · Et
Vmax = 4× · 1.2 ×
Vmax = 4.8 × M
2) Calculate the initial rate of this reaction (R):
The Michaelis-Menten equation studies the dynamics of an enzymatic reaction. This model can explain how an enzyme enhances the rate of a reaction and how the reaction rate depends on the concentration of the enzyme and its substrate. The equation is:
V0 = , where:
[S] is the substrate's concentration
KM is the Michaelis-Menten constant
Substituting [S] = 5.14 × , KM = 2.5 ×
and Vmax = 4.8 ×
, the result is V0 = 0.478 M.
The answers are Vmax = 4.8 × M and V0 = 0.478 M.