178. An enzyme follows Michaelis-Menten kinetics with the following parameters: Vmax= 5 mM/s and Km= 2.5
mM. The reaction velocity would be:
1. 2.5 mM/s at all substrate concentration
2. equal to Vmax at all substrate concentration
3. 1.67 mM/s at a substrate concentration of 1.25 mM
4. 5 mM/s at a substrate concentration of 2.5 mM

Understanding Michaelis-Menten Kinetics and Calculating Reaction Velocity

Michaelis-Menten kinetics are fundamental to understanding enzyme behavior and reaction rates in biochemical processes. In this article, we will explore the Michaelis-Menten equation, its components, and how to calculate the reaction velocity at various substrate concentrations.

What is Michaelis-Menten Kinetics?

Michaelis-Menten kinetics describe the relationship between the substrate concentration and the reaction velocity of an enzyme-catalyzed reaction. The equation is:

v=Vmax×[S]Km+[S]v = \frac{V_{max} \times [S]}{K_m + [S]}v=Km​+[S]Vmax​×[S]​

Where:

• v is the reaction velocity.

• Vmax is the maximum reaction velocity (mM/s).

• Km is the Michaelis constant, which is the substrate concentration at which the reaction velocity is half of Vmax.

• [S] is the substrate concentration.

The equation tells us that the reaction velocity increases with substrate concentration until it reaches Vmax, beyond which it plateaus.

Problem Breakdown

We are given the following parameters for an enzyme that follows Michaelis-Menten kinetics:

• Vmax = 5 mM/s (Maximum reaction velocity)

• Km = 2.5 mM (Michaelis constant)

We need to calculate the reaction velocity at a substrate concentration of 1.25 mM and at a substrate concentration of 2.5 mM.

Step 1: Reaction Velocity at a Substrate Concentration of 1.25 mM

Using the Michaelis-Menten equation:

v=Vmax×[S]Km+[S]v = \frac{V_{max} \times [S]}{K_m + [S]}v=Km​+[S]Vmax​×[S]​

Substitute the known values:

v=5×1.252.5+1.25v = \frac{5 \times 1.25}{2.5 + 1.25}v=2.5+1.255×1.25​ v=6.253.75v = \frac{6.25}{3.75}v=3.756.25​ $$v=1.67\text{mM/s}$$

So, at a substrate concentration of 1.25 mM, the reaction velocity is 1.67 mM/s.

Step 2: Reaction Velocity at a Substrate Concentration of 2.5 mM

Now, let’s calculate the reaction velocity when the substrate concentration is equal to Km (2.5 mM):

v=5×2.52.5+2.5v = \frac{5 \times 2.5}{2.5 + 2.5}v=2.5+2.55×2.5​ v=12.55v = \frac{12.5}{5}v=512.5​ $$v=2.5\text{mM/s}$$

So, at a substrate concentration of 2.5 mM, the reaction velocity is 2.5 mM/s.

Answer Choices Explained

1. 2.5 mM/s at all substrate concentrations: This is incorrect because the reaction velocity does not remain constant at 2.5 mM/s for all substrate concentrations. It depends on the substrate concentration and will increase until it reaches Vmax.

2. Equal to Vmax at all substrate concentrations: This is incorrect because the reaction velocity only equals Vmax when the substrate concentration is high enough to saturate the enzyme, beyond which further increases in substrate concentration will not increase the reaction velocity.

3. 1.67 mM/s at a substrate concentration of 1.25 mM: This is the correct answer, as we calculated the reaction velocity to be 1.67 mM/s at a substrate concentration of 1.25 mM.

4. 5 mM/s at a substrate concentration of 2.5 mM: This is incorrect because at a substrate concentration of 2.5 mM (which is equal to Km), the reaction velocity is half of Vmax, i.e., 2.5 mM/s, not 5 mM/s.

Conclusion

The correct answer is 3. 1.67 mM/s at a substrate concentration of 1.25 mM, based on the calculation using Michaelis-Menten kinetics.

By understanding the Michaelis-Menten equation and how substrate concentration impacts enzyme activity, you can predict the reaction velocity under different conditions and optimize enzyme-catalyzed processes.