JEE Advanced · Chemistry · 29. Biomolecules
For a double strand DNA, one strand is given below:

The amount of energy required to split the double strand DNA into two single strands is _______ kcal \(\mathrm{mol}^{-1}.\)
[Given: Average energy per \(\mathrm{H}\)-bond for A-T base pair \(=1.0 \mathrm{kcal} \mathrm{mol}^{-1}, \mathrm{G}-\mathrm{C}\) base pair \(=1.5 \mathrm{kcal}\) \(\mathrm{mol}^{-1}\), and A-U base pair \(=1.25 \mathrm{kcal} \mathrm{mol}^{-1}.\) Ignore electrostatic repulsion between the phosphate groups.]
- A 30
- B 54
- C 19
- D 41
Answer & Solution
Correct Answer
(D) 41
Step-by-step Solution
Detailed explanation
\(\mathrm{A}=\mathrm{T} \quad 2 \mathrm{H}\)-bond
\(\mathrm{G} \equiv \mathrm{C} \quad 3 \mathrm{H}\)-bond
Number of \(\mathrm{A}=\mathrm{T}\) pair \(=7\)
Number of \(\mathrm{G} \equiv \mathrm{C}\) pair \(=6\)
Number of H-bond involve in \(\mathrm{A}=\mathrm{T}=7 \times 2=14\)
Number of \(\mathrm{H}\)-bond involve in \(\mathrm{G} \equiv \mathrm{C}=6 \times 3=18\)
Energy required for \(\mathrm{A}=\mathrm{T}=14 \times 1=14\)
Energy required for \(\mathrm{G} \equiv \mathrm{C}=18 \times 1.5=27\)
Total energy required \(14+27=41\)
\(\mathrm{G} \equiv \mathrm{C} \quad 3 \mathrm{H}\)-bond
Number of \(\mathrm{A}=\mathrm{T}\) pair \(=7\)
Number of \(\mathrm{G} \equiv \mathrm{C}\) pair \(=6\)
Number of H-bond involve in \(\mathrm{A}=\mathrm{T}=7 \times 2=14\)
Number of \(\mathrm{H}\)-bond involve in \(\mathrm{G} \equiv \mathrm{C}=6 \times 3=18\)
Energy required for \(\mathrm{A}=\mathrm{T}=14 \times 1=14\)
Energy required for \(\mathrm{G} \equiv \mathrm{C}=18 \times 1.5=27\)
Total energy required \(14+27=41\)
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