Problem 9
Question
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?
Step-by-Step Solution
Verified Answer
The fumarase concentration in the sample is 0.08625 μM.
1Step 1: 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.
2Step 2: 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: \[\frac{[E_{sample}]}{[E_{standard}]} = \frac{r_{sample}}{r_{standard}} = \frac{1.15}{2.00}\]
3Step 3: Solve for Sample Enzyme Concentration
Now we solve for the enzyme concentration in the sample \( [E_{sample}] \):\[[E_{sample}] = [E_{standard}] \times \frac{1.15}{2.00}\]\[[E_{sample}] = 0.150 \mu M \times \frac{1.15}{2.00}\]\[[E_{sample}] = 0.08625 \mu M\]
4Step 4: 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.
Key Concepts
Michaelis-Menten constantEnzyme concentrationReaction rate
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:
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.
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:
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.
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:
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.
Other exercises in this chapter
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