​$$\text{An enzyme has a } K_m \text{ of } 5 \times 10^{-5}\ \text{M and a } V_{\max} \text{ of } 100\ \mu\text{moles} \cdot \text{lit}^{-1} \cdot \text{min}^{-1} \ (\text{where } K_m \text{ is the Michaelis constant and } V_{\max} \text{ is the maximal velocity}).\\\text{What is the velocity in the presence of } 1 \times 10^{-4}\ \text{M substrate and } 2 \times 10^{-4}\ \text{M competitive inhibitor,}\\\text{given that the } K_i \text{ for the inhibitor is } 2 \times 10^{-4}\ \text{M}?$$​​

Correct option is B

Explanation-

To solve this, we use the Michaelis-Menten equation with competitive inhibition:

$$v = \frac{V_{max} \cdot [S]}{K_m \left(1 + \frac{[I]}{K_i}\right) + [S]}$$

Given-

$$K_m = 5 \times 10^{-5} \, \text{M} \\V_{max} = 100 \, \mu\text{mol} \cdot \text{L}^{-1} \cdot \text{min}^{-1}$$

​​$$\textbf{Step-by-Step:}\text{1. Calculate the inhibition factor:} \\1 + \frac{[I]}{K_i} = 1 + \frac{2 \times 10^{-4}}{2 \times 10^{-4}} = 1 + 1 = \boxed{2}\\[10pt]\text{2. Modified } K_m \text{ (apparent):} \\K_m^{\text{app}} = K_m \cdot \left(1 + \frac{[I]}{K_i}\right) = 5 \times 10^{-5} \cdot 2 = \boxed{1 \times 10^{-4} \, \text{M}}\\[10pt]\text{3. Apply to Michaelis-Menten equation:} \\v = \frac{V_{\text{max}} \cdot [S]}{K_m^{\text{app}} + [S]} = \frac{100 \cdot 1 \times 10^{-4}}{1 \times 10^{-4} + 1 \times 10^{-4}} = \frac{100 \cdot 1 \times 10^{-4}}{2 \times 10^{-4}} = \frac{100}{2} = \boxed{50}\\[10pt]\textcolor{black}{\boxed{\text{Final Answer:}}} \quad \boxed{50 \, \mu\text{mol} \cdot \text{L}^{-1} \cdot \text{min}^{-1}}$$

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