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Question #45002

Difficulty: 8/10 | 🟢 Free

The Question

An apoenzyme EE binds a fat-soluble vitamin VV to form a holoenzyme. At 340 K340\text{ K}, the apoenzyme undergoes reversible thermal denaturation:

\ceEnative<=>Edenatured\ce{E_{native} <=> E_{denatured}}

The thermodynamic parameters for this denaturation are ΔH∘=259.0 kJ/mol\Delta H^\circ = 259.0\text{ kJ/mol} and ΔS∘=800 J/(K\cdotmol)\Delta S^\circ = 800\text{ J/(K\cdot mol)}.

The intrinsic binding constant of VV to the specific hydrophobic pocket of \ceEnative\ce{E_{native}} is KB=9.1×107 M−1K_B = 9.1 \times 10^7\text{ M}^{-1} at 340 K340\text{ K}. The denatured enzyme, lacking the specific 'lock and key' pocket, only binds VV via non-specific hydrophobic interactions with KNS=1.0×105 M−1K_{NS} = 1.0 \times 10^5\text{ M}^{-1} at 340 K340\text{ K}.

Calculate the standard Gibbs free energy of effective binding (ΔGeff∘\Delta G^\circ_{eff}) for the vitamin to the enzyme mixture at 340 K340\text{ K}.

(Given: 2.303RT=6.5 kJ/mol2.303 RT = 6.5\text{ kJ/mol} at 340 K340\text{ K}; log⁡10(9.1)=0.96\log_{10}(9.1) = 0.96; log⁡10(9.01)=0.95\log_{10}(9.01) = 0.95. The effective binding constant is defined as Keff=[\ceEVnative]+[\ceEVdenatured]([\ceEnative]+[\ceEdenatured])[\ceV]K_{eff} = \frac{[\ce{EV_{native}}] + [\ce{EV_{denatured}}]}{([\ce{E_{native}}] + [\ce{E_{denatured}}])[\ce{V}]})

Options

  • A.

    −39.0 kJ/mol-39.0\text{ kJ/mol}

  • B.

    −38.7 kJ/mol-38.7\text{ kJ/mol}

  • C.

    −32.5 kJ/mol-32.5\text{ kJ/mol}

  • D.

    −51.7 kJ/mol-51.7\text{ kJ/mol}

Check Your Answer

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

Vitamin Solubility Logic|Enzyme Specificity 'Lock and Key'|Protein Denaturation Specificity

Why This Question Is Tricky

Common Mistakes

This problem tests the interplay between 'Lock and Key' specificity and non-specific 'Hydrophobic Partitioning' under thermal stress.

At 340 K, the apoenzyme is 99% denatured (KD=100K_D = 100).

A common trap is assuming that the denatured state completely dictates the binding affinity, leading students to use only KNSK_{NS} (Option C).

However, the 1% native enzyme possesses such high specificity (KB=9.1×107K_B = 9.1 \times 10^7) that it contributes ~90% of the total binding capacity.

Another trap is ignoring the non-specific binding entirely (Option B).

The effective binding constant requires a weighted sum of both states, proving that highly specific native pockets dominate ligand binding even when the protein is overwhelmingly denatured.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptVitamin Solubility Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree