Hydrogen-bonding penalty: approximate free-energy cost of leaving a potential H-bond unsatisfied inside a DNA duplex Choose the closest value for the unfavorable free energy.

Introduction / Context:
When macromolecules fold or associate, buried polar groups prefer to form hydrogen bonds. Leaving such a group unsatisfied is energetically costly and can destabilize structures like DNA duplexes and protein interiors.

Given Data / Assumptions:

• We are estimating an unfavorable free-energy penalty for an unsatisfied hydrogen-bonding opportunity after helix formation.
• Typical hydrogen bond strengths in aqueous systems are on the order of 10–30 kJ/mol, depending on geometry and environment.
• Values are approximate and context dependent.

Concept / Approach:
An unsatisfied polar group buried from solvent cannot form compensating interactions with water. The penalty is roughly comparable to the stabilization ordinarily conferred by forming a good hydrogen bond. A middle-of-the-range estimate around 20 kJ/mol is commonly cited for this penalty in biophysical discussions.

Step-by-Step Solution:
Recall typical H-bond free energies: approximately 8–25 kJ/mol in water; stronger in well-organized, low-dielectric environments.Associate penalty magnitude with loss of a stabilizing H-bond: pick a representative mid-value.Select 20 kJ/mol as the closest standard estimate among the options.

Verification / Alternative check:
Protein design and nucleic-acid thermodynamic models penalize buried unsatisfied donors/acceptors by values near tens of kJ/mol, aligning with ≈20 kJ/mol as a rule-of-thumb.

Why Other Options Are Wrong:
10 kJ/mol may underestimate typical penalties for buried unsatisfied H-bonds; 25–30 kJ/mol can occur but are often higher than the commonly quoted average; 5 kJ/mol is too small.

Common Pitfalls:
Confusing enthalpy with free energy and ignoring environmental modulation (dielectric constant, geometry, cooperativity).

Final Answer:
20 kJ/mol.