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Chapter 13: Problem 9

The enzyme fumarase catalyzes the stereospecific addition of water to fumarate to form \(\mathrm{L}\) -malate. A standard \(0.150 \mu \mathrm{M}\) solution of fumarase has a rate of reaction of \(2.00 \mu \mathrm{M} \min ^{-1}\) under conditions in which the substrate's concentration is significantly greater than \(K_{m}\). The rate of reaction for a sample under identical condition is \(1.15 \mu \mathrm{M} \mathrm{min}^{-1}\). What is the concentration of fumarase in the sample?

Short Answer

Expert verified
The fumarase concentration in the sample is 0.08625 渭M.

Step by step solution

01

Understand the Relationship

We know from enzyme kinetics that, under conditions where the substrate concentration is much greater than the Michaelis constant (\( K_m \)), the reaction rate is directly proportional to the enzyme concentration. Thus, if the rate of reaction changes, it is because the enzyme concentration has changed.
02

Set up the Proportion

The initial standard reaction with a fumarase concentration of \( 0.150 \mu M \) has a rate of \( 2.00 \mu M \text{ min}^{-1} \). The sample's rate is \( 1.15 \mu M \text{ min}^{-1} \). We set up the proportion based on these values: \[\frac{[E_{sample}]}{[E_{standard}]} = \frac{r_{sample}}{r_{standard}} = \frac{1.15}{2.00}\]
03

Solve for Sample Enzyme Concentration

Now we solve for the enzyme concentration in the sample \( [E_{sample}] \):\[[E_{sample}] = [E_{standard}] \times \frac{1.15}{2.00}\]\[[E_{sample}] = 0.150 \mu M \times \frac{1.15}{2.00}\]\[[E_{sample}] = 0.08625 \mu M\]
04

Verify Solution

Check the calculations and ensure the numbers are accurate by verifying that the ratio between the rates and the concentrations match.With the given numbers:\[\frac{0.08625}{0.150} =\frac{1.15}{2.00} \approx 0.575\] The calculation confirms that the concentration of fumarase in the sample is consistent with the rate.

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Key Concepts

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

Michaelis-Menten constant
In enzyme kinetics, the Michaelis-Menten constant, represented as \(K_m\), is a crucial figure in understanding how enzymes interact with substrates. Essentially, \(K_m\) offers insight into the affinity that an enzyme has for its substrate. A low \(K_m\) indicates that the enzyme binds tightly to its substrate, even at low substrate concentrations, while a high \(K_m\) suggests the opposite.
Here's how this plays into enzyme kinetics:
  • When the substrate concentration is much higher than \(K_m\), the enzyme operates at nearly its maximum speed, referred to as \(V_{max}\).
  • In such scenarios, changes in reaction rate are affected by enzyme concentration, rather than changes in substrate concentration.
Understanding \(K_m\) helps to predict how an enzyme behaves in different substrate concentration environments, making it a valuable tool for both academic and industrial purposes.
Enzyme concentration
Enzyme concentration is a key variable affecting the rate of an enzymatic reaction. Simply put, the more enzyme molecules available, the greater the number of reactions that can occur in a given time frame. This holds true under the condition where substrate concentration is saturating, i.e., much greater than \(K_m\).
Here's why enzyme concentration matters:
  • It determines the maximum rate of product formation. The reaction rate is directly proportional to the enzyme concentration in situations where excess substrate is present.
  • Changing enzyme concentration is a common way to regulate enzyme activity in biological systems.
In practical applications, understanding how enzyme concentration influences reaction rates can help in designing processes that maximize efficiency, such as in pharmaceutical production or biotechnological applications.
Reaction rate
The reaction rate tells us how quickly a chemical reaction occurs. In the context of enzyme kinetics, this refers to how fast the substrate is turned into a product by an enzyme. The reaction rate is influenced by several factors including enzyme and substrate concentrations, temperature, pH, and enzyme characteristics.
Let's explore these factors:
  • In environments where substrate concentration is much higher than \(K_m\), the reaction rate is primarily influenced by enzyme concentration.
  • Temperature and pH can affect enzyme activity, often increasing the rate up to an optimal point before denaturing the enzyme.
An accurate understanding of reaction rates helps in predicting how systems respond to changes and in optimizing conditions for biological reactions, whether in a laboratory setting or within living organisms.

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Most popular questions from this chapter

To study the effect of an enzyme inhibitor \(V_{\max }\) and \(K_{m}\) are measured for several concentrations of inhibitor. As the concentration of the inhibitor increases \(V_{\max }\) remains essentially constant, but the value of \(K_{m}\) increases. Which mechanism for enzyme inhibition is in effect?

The following data were collected for a reaction known to be pseudofirst order in analyte, \(A\), during the time in which the reaction is monitored. $$ \begin{array}{cc} \text { time }(s) & {[A]_{t}(\mathrm{mM})} \\ \hline 2 & 1.36 \\ 4 & 1.24 \\ 6 & 1.12 \\ 8 & 1.02 \\ 10 & 0.924 \\ 12 & 0.838 \\ 14 & 0.760 \\ 16 & 0.690 \\ 18 & 0.626 \\ 20 & 0.568 \end{array} $$ What is the rate constant and the initial concentration of analyte in the sample?

The enzyme urease catalyzes the hydrolysis of urea. The rate of this reaction is determined for a series of solutions in which the concentration of urea is changed while maintaining a fixed urease concentration of \(5.0 \mu \mathrm{M}\). The following data are obtained. $$ \begin{array}{cc} \text { [urea }](\mu \mathrm{M}) & \text { rate }\left(\mu \mathrm{M} \mathrm{s}^{-1}\right) \\ \hline 0.100 & 6.25 \\ 0.200 & 12.5 \\ 0.300 & 18.8 \\ 0.400 & 25.0 \\ 0.500 & 31.2 \\ 0.600 & 37.5 \\ 0.700 & 43.7 \\ 0.800 & 50.0 \\ 0.900 & 56.2 \\ 1.00 & 62.5 \end{array} $$ Determine the values of \(V_{\max }, k_{2}\), and \(K_{m}\) for urease.

The concentration of chloride in seawater is determined by a flow injection analysis. The analysis of a set of calibration standards gives the following results. $$ \begin{array}{cccc} {\left[\mathrm{Cl}^{-}\right](\mathrm{ppm})} & \text { absorbance } & {\left[\mathrm{Cl}^{-}\right](\mathrm{ppm})} & \text { absorbance } \\ \hline 5.00 & 0.057 & 40.00 & 0.478 \\ 10.00 & 0.099 & 50.00 & 0.594 \\ 20.00 & 0.230 & 75.00 & 0.840 \\ 30.00 & 0.354 & & \end{array} $$ A 1.00-mL sample of seawater is placed in a 500 -mL volumetric flask and diluted to volume with distilled water. When injected into the flow injection analyzer an absorbance of 0.317 is measured. What is the concentration of \(\mathrm{Cl}^{-}\) in the sample?

The steady state activity for \({ }^{14} \mathrm{C}\) in a sample is 13 cpm per gram of carbon. If counting is limited to \(1 \mathrm{hr}\), what mass of carbon is needed to give a percent relative standard deviation of \(1 \%\) for the sample's activity? How long must we monitor the radioactive decay from a 0.50 -g sample of carbon to give a percent relative standard deviation of \(1.0 \%\) for the activity?
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