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 $$ \frac{d(\mathrm{P})}{d t}=k_{2}[\mathrm{E}] $$

Step by step solution

01

- Calculate the total moles of product formed

The concentration of inorganic phosphate (\text{P}_{i}) is found to be 0.002 M in a 1 ml reaction mixture. To find the total moles of \text{P}_{i} formed, use: \[ \text{Moles of } \text{P}_{i} = \text{concentration} \times \text{volume} = 0.002\text{ M} \times 1 \text{ ml} = 0.002 \text{ mmol} = 2 \times 10^{-6} \text{ mol} \]

02

- Determine the total activity (enzyme turnover)

Given that the reaction rate can be represented by \( \frac{d(\text{P})}{dt}=k_{2}[\text{E}] \), and the total moles of \text{P} formed over the course of the reaction is equal to the integral of this rate over time, the total amount of product can be calculated by: \( \text{Total product} = \text{total moles of } \text{P}_{i} = 2 \times 10^{-6} \text{ mol} \). Since the enzyme decomposes over time, we calculate the integral of the remaining enzyme, learning its total contribution.

03

- Calculate the initial enzyme concentration

Given the molecular weight of ATPase is \(5 \times 10^{4} \text{ daltons} \), first convert 10 \text{ μg} of total crude protein to moles: \( \text{Moles of protein} = \frac{\text{total mass}}{\text{molecular weight}} = \frac{10 \times 10^{-6} \text{ grams}}{5 \times 10^{4} \text{ daltons}} = 2 \times 10^{-10} \text{ mol} \)

04

- Consider enzyme degradation

The enzyme degrades over time according to the inactivation kinetics, \( \text{E} = \text{E}_{0} e^{-k_{d} t} \). The total active enzyme over the time period can be calculated using an integral: \( \text{Active enzyme} = \text{E}_{0} \times \frac{1}{k_{d}} \times (1-e^{-k_{d} t}) = 2 \times 10^{-10} \text{ mol} \times \frac{1}{0.1} \text{ min} = 2 \times 10^{-9} \text{ mol} \times 600 \text{min} = 1.2 \times 10^{-6} \)

05

- Relate substrate turnover to enzyme concentration

Since \( d(\text{P})/dt=k_{2}[\text{E}] \), plugging in \( k_{2}= 10^4 \text{ mol/min} \) and using the average enzyme concentration, we get: \( [\text{E}](\text{initial})= \frac{2 \times 10^{-6} \text{ mol}}{10^4 \text{ min} } = 2 \times 10^{-10} \text{ mol}, \text{ which is comparable to the initial enzyme concentration, so the enzyme fraction must be calculated directly}. \)

06

- Compute enzyme proportion

Finally, compare ratios Percentage of mixture composed of enzyme \( \text{ fraction of weight } \) either knowing turnover or implied purely from data explicitly given:

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ATPase function

ATPase is an enzyme that catalyzes the decomposition of ATP (adenosine triphosphate) into ADP (adenosine diphosphate) and an inorganic phosphate (\text{P}_{i}). This reaction releases energy, which the cell can then use for various processes. In the given exercise, the specific reaction mediated by ATPase is expressed as follows:

\[ \text{ATP} \overset{ \text{ATPase} }{\longrightarrow} \text{ADP} + \text{P}_{i} \]

This means ATPase splits one molecule of ATP into ADP and phosphate. The energy provided by this process is crucial for numerous cellular activities—such as muscle contractions, nerve impulses, and chemical synthesis. In the context of the exercise, ATPase helps us understand enzyme function, as well as the specific kinetic properties like molecular weight, Km, and k2 values that define their activity.

Enzyme inactivation kinetics

Enzyme inactivation kinetics describe how enzymes lose activity over time due to factors such as temperature, pH, or the presence of inhibitors. In the given problem, ATPase becomes unstable at a specific temperature (37°C), leading to inactivation. The kinetic model for this inactivation is first-order, represented by the equation:

\[ \text{E} = \text{E}_{0} e^{-k_{d} t} \]

Here, \( \text{E} \) represents the active enzyme concentration at time \( t \), \( \text{E}_{0} \) is the initial enzyme concentration, and \( k_{d} \) is the degradation rate constant. For ATPase, \( k_{d} = 0.1 \text{ min}^{-1} \). This means that as time progresses, the amount of enzymatic activity decreases exponentially. These principles help us in calculating how much enzyme activity remains at any given time and assessing the enzyme's practical utility in biochemical reactions.

Michaelis-Menten equation

The Michaelis-Menten equation illustrates the relationship between the rate of an enzyme-catalyzed reaction and the concentration of substrate available. It's a fundamental framework in enzyme kinetics:

\[ v = \frac{ V_{\text{max}} [S] }{ K_{m} + [S] } \]

Here, \( v \) is the reaction rate, \( V_{\text{max}} \) is the maximum rate achieved by the system, \( [S] \) is the substrate concentration, and \( K_{m} \) is the Michaelis constant—the substrate concentration at which the reaction rate is half of \( V_{\text{max}} \). In our exercise, considering that \( [S] >> K_{m} \), simplifying the equation we find:

\[ \frac{d(\text{P})}{dt} = k_{2} [E] \]

This tells us that the reaction rate is directly proportional to the enzyme concentration \( [E] \). This direct relationship allowed us to compute the total activity through the enzyme turnover rate \( k_{2} \) given in the problem.

Enzyme activity calculation

To calculate enzyme activity accurately, we start by determining the total moles of product formed from the reaction, which in this case is inorganic phosphate (\( \text{P}_{i} \)) concentration. Using the formula:

\[ \text{Moles of } \text{P}_{i} = \text{concentration} \times \text{volume} = 0.002 \text{M} \times 1\text{ml} = 2 \times 10^{-6} \text{mol} \]

Next, we use the given enzyme turnover number \( k_{2}=10^{4} \) molecules of ATP/min per molecule of enzyme and consider the inactivation kinetics of the enzyme over time:

\[ \text{E} = \text{E}_{0}e^{-k_{d}t} \]

We calculate the total active enzyme during the reaction:

\[ \text{Active enzyme} = \text{E}_{0} \times \frac{1}{k_{d}} \times (1-e^{-k_{d} t}) = 2 \times 10^{-10} \text{mol} \times \frac{1}{0.1} \text{min} = 1.2 \times 10^{-6} \]

Finally, using the relationship \( \frac{d(\text{P})}{dt} = k_{2} [E] \), we determine the initial enzyme concentration by plugging in the values and solving. This step-by-step approach ensures precise determination of enzyme activity.