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Question #45004

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The Question

An enzyme operates optimally at $pH = 7.5$ and $298\text{ K}$ , utilizing a highly specific "Lock-and-Key" mechanism. Its active site features two critical residues for substrate binding and catalysis: a Histidine ( $pK_{aR} = 6.0$ ) acting as a general base, and an Aspartic acid ( $pK_{aR} = 3.9$ ) acting as an electrostatic anchor for the cationic substrate. When the reaction buffer is abruptly adjusted to $pH = 4.8$ , the catalytic rate drops by over $95\%$ . However, Circular Dichroism (CD) spectroscopy reveals that the enzyme's overall $\alpha$ -helical and $\beta$ -sheet content remains identical to that at $pH = 7.5$ . Which of the following best explains the sudden loss of enzymatic activity?

Options

A.
Protonation of the Histidine imidazole ring disrupts the active site's electrostatic complementarity and general base capability, without causing macroscopic denaturation.
B.
The enzyme undergoes global denaturation due to the disruption of internal salt bridges, destroying the rigid macroscopic geometry.
C.
Protonation of the Aspartic acid carboxylate group eliminates the essential electrostatic interaction required to anchor the cationic substrate.
D.
The peptide backbone undergoes a localized zwitterionic shift, causing the active site to collapse via acid-catalyzed hydrolysis.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Enzyme Specificity 'Lock and Key'|Amino Acid Zwitterion pH Sensitivity|Protein Denaturation Specificity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Chemical Exception' that loss of enzyme activity does not strictly imply macroscopic protein denaturation.

Students often fall into the trap of assuming the 'Lock-and-Key' model is purely about rigid macroscopic shape (Option B).

The hidden variable is the local pH-dependent ionization of specific amino acid side chains.

By providing CD spectroscopy data, the problem forces the student to rule out global denaturation and instead apply the Henderson-Hasselbalch relationship to individual $pK_a$ values, realizing that Histidine protonates at pH 4.8 while Aspartate remains deprotonated.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	Enzyme Specificity 'Lock and Key'
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free