A kinetics experiment reveals that the enzyme fumarase catalyzes the dehydration of malate with a kcat of 900 s−1. If the substrate concentration is equal to the KM of the reaction, what is the reaction velocity when the enzyme concentration is 200 nM?A.3.6 × 102 µM/sB.9.0 × 101 µM/sC.4.5 × 10−3 µM/sD.2.2 × 10−4 µM/s
Question
A kinetics experiment reveals that the enzyme fumarase catalyzes the dehydration of malate with a kcat of 900 s−1. If the substrate concentration is equal to the KM of the reaction, what is the reaction velocity when the enzyme concentration is 200 nM?A.3.6 × 102 µM/sB.9.0 × 101 µM/sC.4.5 × 10−3 µM/sD.2.2 × 10−4 µM/s
Solution
The reaction velocity (v) of an enzyme-catalyzed reaction can be calculated using the Michaelis-Menten equation:
v = (kcat[E][S]) / (KM + [S])
where:
- v is the reaction velocity,
- kcat is the turnover number (the number of substrate molecules each enzyme site converts to product per unit time),
- [E] is the enzyme concentration,
- [S] is the substrate concentration, and
- KM is the Michaelis constant (the substrate concentration at which the reaction velocity is half its maximum).
In this case, we are told that the substrate concentration [S] is equal to the KM. Therefore, the equation simplifies to:
v = (kcat[E][S]) / (2[S]) v = (kcat[E]) / 2
Substituting the given values:
v = (900 s−1 * 200 nM) / 2 v = 180,000 nM/s / 2 v = 90,000 nM/s
However, the answer choices are given in µM/s. To convert from nM/s to µM/s, divide by 1,000 (since there are 1,000 nM in a µM):
v = 90,000 nM/s / 1,000 v = 90 µM/s
Therefore, the reaction velocity when the enzyme concentration is 200 nM and the substrate concentration is equal to the KM of the reaction is 90 µM/s. The correct answer is B. 9.0 × 10^1 µM/s.
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