题目
The Km value of a certain enzyme is known to be 25 , (mmol/L). To achieve 50% of the maximum reaction velocity of the enzyme catalyzed reaction, the concentration of the substrate should beA. 12.5 , (mmol/L)B. 25 , (mmol/L)C. 37.5 , (mmol/L)D. 50 , (mmol/L)
The Km value of a certain enzyme is known to be $25 \, \text{mmol/L}$. To achieve $50\%$ of the maximum reaction velocity of the enzyme catalyzed reaction, the concentration of the substrate should be
A. $12.5 \, \text{mmol/L}$
B. $25 \, \text{mmol/L}$
C. $37.5 \, \text{mmol/L}$
D. $50 \, \text{mmol/L}$
题目解答
答案
根据米氏方程,当 $ v = \frac{1}{2} V_{max} $ 时,有:
\[
\frac{1}{2} = \frac{[S]}{K_m + [S]}
\]
解得 $ [S] = K_m $。
题目中Km = 25 mmol/L,故 $ [S] = 25 $ mmol/L。
答案:B. 25 mmol/L