Question

. Catalase is an important enzyme that breaks down its substrate, H2O2, into water and oxygen in nearly all living cells. The enzyme has k2 = 1.1*107 sec-1 and Km = 2.0 mM. An experiment was set up under the conditions when the reaction rate is independent of [H2O2]. In this experiment, at what concentration of catalase (use SI units and scientific notation) will the reaction have a rate of 850 mM/min?

Answer 1

$V_{max}=k_2[E_T]$

$V_{max}=\frac{850}{60}×10^{-3} mols^{-1}$

$=1.1×10^7 ×[E_T]$

$E_T=1.288×10^{-9} M$