An enzyme ATPase has a molecular weight of $5 \times 10^{4}$ daltons, a $K_{M}$ value of $10^{-4} M$, and a $k_{2}$ value of $k_{2}=10^{4}$ molecules ATP/min molecule enzyme at $37^{\circ} \mathrm{C}$. The reaction catalyzed is the following:
$$
\mathrm{ATP} \stackrel{\text { ATPase }}{\longrightarrow} \mathrm{ADP}+\mathrm{P}_{i}
$$
which can also be represented as
$$
\mathrm{E}+\mathrm{S} \underset{k_{1}}{\stackrel{A_{1}}{\longrightarrow} \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P}
$$
where $S$ is ATP. The enzyme at this temperature is unstable. The enzyme inactivation kinetics are first order:
$$
\mathrm{E}=\mathrm{E}_{0} e^{-k_{\alpha} r}
$$
where $\mathrm{E}_{0}$ is the initial enzyme concentration and $k_{d}=0.1 \mathrm{~min}^{-1}$. In an experiment with a partially pure enzyme preparation, $10 \mu \mathrm{g}$ of total crude protein (containing enzyme) is added to a $1 \mathrm{ml}$ reaction mixture containing $0.02 \mathrm{M}$ ATP and incubated at $37^{\circ} \mathrm{C}$. After 12 hours the reaction ends (i.e., $t \rightarrow \infty$ ) and the inorganic phosphate $\left(\mathrm{P}_{i}\right)$ concentration is found to be $0.002 M$, which was initially zero. What fraction of the crude protein preparation was the enzyme? Hint: Since $[\mathrm{S}]>>K_{m}$, the reaction rate can be represented by
$$
\frac{d(\mathrm{P})}{d t}=k_{2}[\mathrm{E}]
$$