At what range of substrate concentration will an enzyme with a $$k_{cat}$$ of $$30 s{^-1}$$ and  a $$K_m$$ 0.005 M show one-quarter of its maximum rate?​

Correct option is C

Let's solve this based on the Michaelis-Menten equation:

​$$v = \frac{V_{\text{max}} \cdot [S]}{K_m + [S]}$$

Given:

• We want one-quarter of the maximum rate => v=Vmax4v = \frac{V_{\text{max}}}{4}v = V(max)4\frac{V_(max)}{4}4V(​max)​​

K_m = 0.005 M

Step-by-step Solution:

$$\frac{V_{\text{max}}}{4} = \frac{V_{\text{max}} \cdot [S]}{K_m + [S]}$$

​Cancel out V_max from both sides: 

$$\frac{1}{4} = \frac{[S]}{K_m + [S]}$$

​Cross-multiplying:

K_m + [S] = 4[S]

K_m = 4[S] - [S] = 3[S] 

[S] = $$\frac{K_m}{3}$$ =$$\frac{0.005}{3}$$

= 1.6667 × 10^-3 M

So the substrate concentration must be close to 1.67 × 10^-3 M