Calculate kcat in units of s-1 for the polyphenol oxidase, where Vmax for the reaction with the dopamine substrate is 0.36 U·min–1·mL–1 (U stands for enzyme units). The enzyme concentration is 2.0 × 10–2 U·mL–1
. Entry field with incorrect answer 0.18 s-1 the tolerance is +/-2% Your answer is incorrect. Try again.
The polyphenol oxidase also acts on catechol as a substrate, with a Vmax of 0.39 U·min–1·mL–1. Assuming the same enzyme concentration, what is the kcat in s-1 for the oxidase-catalyzed catechol reaction?
Entry field with incorrect answer 19.5 s-1 the tolerance is +/-2%
The Correct Answer and Explanation is :
To calculate the kcat (turnover number) for the polyphenol oxidase, we use the following formula:
[
k_{cat} = \frac{V_{max}}{[E]}
]
Where:
- kcat is the turnover number in units of s⁻¹ (per second),
- Vmax is the maximum velocity of the enzyme-catalyzed reaction, in units of U·min⁻¹·mL⁻¹ (enzyme units per minute per milliliter),
- [E] is the concentration of the enzyme, in units of U·mL⁻¹ (enzyme units per milliliter).
For the dopamine reaction:
- Vmax (dopamine) = 0.36 U·min⁻¹·mL⁻¹,
- [E] = 2.0 × 10⁻² U·mL⁻¹.
We need to convert the units of Vmax from U·min⁻¹·mL⁻¹ to U·s⁻¹·mL⁻¹ by dividing by 60 (since there are 60 seconds in a minute):
[
V_{max} (dopamine) = \frac{0.36}{60} \, \text{U·s}^{-1}·\text{mL}^{-1} = 0.006 \, \text{U·s}^{-1}·\text{mL}^{-1}
]
Now, we can calculate kcat (dopamine):
[
k_{cat} (dopamine) = \frac{0.006}{2.0 \times 10^{-2}} \, \text{s}^{-1} = 0.30 \, \text{s}^{-1}
]
This value of kcat (dopamine) is 0.30 s⁻¹, not 0.18 s⁻¹.
For the catechol reaction:
- Vmax (catechol) = 0.39 U·min⁻¹·mL⁻¹,
- [E] = 2.0 × 10⁻² U·mL⁻¹.
First, we convert the Vmax (catechol) from U·min⁻¹·mL⁻¹ to U·s⁻¹·mL⁻¹:
[
V_{max} (catechol) = \frac{0.39}{60} \, \text{U·s}^{-1}·\text{mL}^{-1} = 0.0065 \, \text{U·s}^{-1}·\text{mL}^{-1}
]
Now, we can calculate kcat (catechol):
[
k_{cat} (catechol) = \frac{0.0065}{2.0 \times 10^{-2}} \, \text{s}^{-1} = 0.325 \, \text{s}^{-1}
]
So, the correct value for kcat (catechol) is 0.325 s⁻¹, and not 19.5 s⁻¹.
Conclusion: The calculations for kcat were done by converting the units of Vmax from per minute to per second and dividing by the enzyme concentration to obtain the turnover number.