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Question #45007

Difficulty: 8/10 | 🟢 Free

The Question

A synthetic nucleobase analog, XX, undergoes a keto-enol tautomeric equilibrium in an aqueous buffer at 298 K298\text{ K}:

Xketo⇌XenolX_{keto} \rightleftharpoons X_{enol}

The standard Gibbs free energy change for this process is ΔG∘=+22.8 kJ mol−1\Delta G^\circ = +22.8\text{ kJ mol}^{-1}. The XenolX_{enol} tautomer possesses a hydrogen-bonding donor-acceptor profile identical to Adenine and can theoretically form stable hydrogen bonds with Thymine (T). However, high-fidelity in vitro kinetic assays reveal that the probability of DNA polymerase incorporating XX opposite T is ≈10−8\approx 10^{-8}.

Given R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} and e−9.2≈10−4e^{-9.2} \approx 10^{-4}, determine the approximate thermodynamic mole fraction of XenolX_{enol} in solution and the primary biochemical mechanism explaining why the incorporation rate is significantly lower than this mole fraction.

Options

  • A.

    Mole fraction ≈10−4\approx 10^{-4}; DNA polymerase employs an induced-fit geometric selection, where altered spatial alignment of the sugar-phosphate backbone in the XenolX_{enol}-T pair prevents the active site from adopting the catalytic conformation.

  • B.

    Mole fraction ≈10−4\approx 10^{-4}; The enzyme operates via a rigid lock-and-key mechanism, and the hydrophobic active site thermodynamically shifts the tautomeric equilibrium entirely back to the XketoX_{keto} form prior to binding.

  • C.

    Mole fraction ≈10−8\approx 10^{-8}; The high activation energy of tautomerization kinetically traps the analog in the XketoX_{keto} form, making the effective concentration of the H-bonding compatible tautomer 10−810^{-8}.

  • D.

    Mole fraction ≈10−8\approx 10^{-8}; Chemical compatibility (H-bonding) strictly dictates enzyme activity, meaning the thermodynamic mole fraction must exactly equal the 10−810^{-8} incorporation probability.

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Enzyme Specificity 'Lock and Key'|Nitrogenous Base Keto-Enol Tautomerism|Base Pairing Stability

Why This Question Is Tricky

Common Mistakes

This question tests the 'Hidden Variable' of geometric selection in enzyme specificity, contrasting it with pure thermodynamic and chemical (H-bonding) compatibility.

Students often fall into the trap of assuming that if a minor tautomer can form the correct hydrogen bonds (chemical compatibility), the enzyme will incorporate it at a rate equal to its thermodynamic mole fraction.

By forcing the calculation of the mole fraction (10−410^{-4}), the student must reconcile the 10410^4-fold discrepancy with the observed rate (10−810^{-8}).

The correct logical path requires recognizing that DNA polymerase does not use a simplistic 'lock and key' model based solely on H-bonds, but an 'induced fit' mechanism that strictly enforces the spatial geometry of the sugar-phosphate backbone, rejecting tautomers that cause steric clashes.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptEnzyme Specificity 'Lock and Key'
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree