Question

Match the following.\[ \begin{array}{|c|l|c|l|} \hline & \textbf{List-I} & & \textbf{List-II} \\ \hline A & \text{Apoenzyme} & I & \text{Non-protein Part} \\ B & \text{Coenzyme} & II & \text{Protein Part} \\ C & \text{Holoenzyme} & III & \text{Enzyme + Cofactor} \\ D & \text{Active Site} & IV & \text{Substrate-binding Region} \\ \hline \end{array} \]

The correct answer is

• A. A -- II, B -- I, C -- III, D -- IV
• B. A -- I, B -- II, C -- IV, D -- III
• C. A -- III, B -- IV, C -- I, D -- II
• D. A -- IV, B -- III, C -- II, D -- I

Hint:

Remember: \[ \boxed{ \begin{aligned} \text{Apoenzyme} &= \text{Protein part} \text{Coenzyme} &= \text{Organic cofactor} \text{Holoenzyme} &= \text{Apoenzyme + Cofactor} \text{Active site} &= \text{Site where substrate binds} \end{aligned} } \]

Answer 1

The Correct Option is A

Step 1: Recall the definitions. \[ \boxed{ \begin{aligned} \text{Apoenzyme} &\rightarrow \text{Protein part of an enzyme} \text{Coenzyme} &\rightarrow \text{Organic non-protein cofactor} \text{Holoenzyme} &\rightarrow \text{Apoenzyme + Cofactor} \text{Active site} &\rightarrow \text{Substrate-binding region} \end{aligned} } \]

Step 2: Match List-I with List-II. Thus, \[ \boxed{ A-\mathrm{II},\; B-\mathrm{I},\; C-\mathrm{III},\; D-\mathrm{IV} } \] Therefore, \[ \boxed{(A)} \] is the correct answer.