A canonical Watson-Crick G:C base pair is stabilized by three hydrogen bonds formed between a donor (+0.25e) and acceptor (-0.35e) atom located at a distance $$r = 2.9\,\text{\AA}.$$ Assume that the dielectric constant 'D' in the DNA core is 4, and in water is 78.5. Which one of the following options correctly represents the total difference in electrostatic interaction energy for all 3 hydrogen bonds when buried in the DNA core versus fully exposed to solvent?​
(Modified Coulomb's constant, k = 1389 kJ/mol nm $$e^{-2}$$​)

Correct option is A

Step 1: Convert distance to nm

r = 2.9 Å = 0.29 nm
​
Step 2: Calculate charge product

q₁ = +0.25e
q₂ = –0.35e
Therefore,
q₁ × q₂ = –0.0875 e²

Step 3: Electrostatic energy for one bond

Formula:
E = (k × q₁ × q₂) / (D × r)

Difference in dielectric

ΔE = (k × q₁ × q₂ / r) × (1/D_core – 1/D_water)

Substitute values:
ΔE = (1389 × (–0.0875) / 0.29) × (1/4 – 1/78.5)

Step-by-step check:

1389 × 0.0875 = 121.5375

Divide by 0.29 → 419.1

(1/4 – 1/78.5) = (0.25 – 0.01274) = 0.23726

Multiply 419.1 × 0.23726 ≈ 99.5

Apply negative sign (since q₁ × q₂ is negative) → ΔE (for one bond) ≈ –99.5 kJ/mol

For 3 bonds:

ΔE_total = 3 × (–99.5) = –298.5 kJ/mol

Final Answer:
The total electrostatic energy difference for 3 bonds ≈ –298.5 kJ/mol