Q.14 Which of the following is/are CORRECT when two single complementary strands
of DNA come together to form a double helix at a given temperature?
(∆𝑆 and ∆𝐻 are changes in entropy and enthalpy of the process, respectively.)
(A) ∆𝑆 > 0
(B) ∆𝑆 < 0
(C) ∆𝐻 > 0
(D) ∆𝐻 < 0

When two single complementary DNA strands form a double helix, the process is spontaneous and exothermic, driven primarily by enthalpy despite an unfavorable entropy change. Options (B) ∆S < 0 and (D) ∆H < 0 are correct.

Option Analysis

(A) ∆S > 0 is incorrect. Helix formation orders flexible single strands into a rigid structure, reducing conformational freedom and causing negative ∆S; favorable entropy from water release partially offsets this but does not make overall ∆S positive.

(B) ∆S < 0 is correct. The net entropy decreases due to loss of rotational and translational freedom in strands, confirmed by calorimetry showing dominant -T∆S opposition to stability.

(C) ∆H > 0 is incorrect. Base stacking and hydrogen bonding release energy, yielding negative (exothermic) ∆H, as measured by ITC (-73 kcal/mol duplex at 293 K).

(D) ∆H < 0 is correct. Enthalpy-driven process from van der Waals stacking and H-bonds; ∆H becomes more negative at lower temperatures due to negative heat capacity change.

DNA double helix formation from single complementary strands involves key thermodynamic changes in entropy (∆S) and enthalpy (∆H) at a given temperature. This process, central to molecular biology and CSIR NET Life Sciences preparation, is enthalpy-driven (∆H < 0) despite negative entropy (∆S < 0).

Thermodynamic Drivers

Hydrogen bonds (A-T, G-C) and hydrophobic base stacking provide favorable enthalpy release, making ∆H negative. Single strands’ flexibility loss causes ∆S < 0, but hydration entropy gain from water release around bases supports stability (∆G < 0).

• Base stacking: Primary enthalpic contributor via van der Waals forces.

• H-bonding: Stabilizes pairs but secondary to stacking.

• Solvent effects: Counteracts ∆S loss.

CSIR NET Question Breakdown

For Q.14, correct choices are (B) ∆S < 0 and (D) ∆H < 0. ITC/DSC data confirm ∆H ≈ -73 to -420 kJ/mol (sequence-dependent), with negative ∆Cp enhancing low-temperature favorability.