​What is the fold difference between v at [S] = K_m and v at [S] = 1000K_m, where v is the initial velocity of an enzyme catalyzed reaction, [S] is substrate concentration and K_m is the Michaelis constant?​

Correct option is A

Explanation-

Given:
Two substrate concentrations:

​$$[\text{S}]_1 = K_m \quad \text{and} \quad [\text{S}]_2 = 1000 K_m$$

We need to find the fold difference between the velocities ₁ and ₂ at these two concentrations, using the Michaelis–Menten equation:​​

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

​$$\textbf{Step 1:} \text{Calculate } v_1 \text{ at } [\text{S}] = K_m$$

                                           $$v_1 = \frac{V_{\text{max}} \cdot K_m}{K_m + K_m} = \frac{V_{\text{max}} \cdot K_m}{2K_m} = \frac{V_{\text{max}}}{2}$$

$$\textbf{Step 2:} \text{Calculate } v_2 \text{ at } [\text{S}] = 1000K_m$$

                                           $$v_2 = \frac{V_{\text{max}} \cdot 1000K_m}{K_m + 1000K_m} = \frac{V_{\text{max}} \cdot 1000}{1001}$$

$$\text{Step 3: Fold difference} = \frac{v_2}{v_1} \\[10pt]$$

                                           $$\frac{v_2}{v_1} = \left( \frac{V_{\text{max}} - 1000}{1001} \middle/ \frac{V_{\text{max}}}{2} \right) = \frac{1000}{1001} \cdot 2 = \frac{2000}{1001} \approx 1.998$$

Correct answer: Option a : 1.998​​

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