KVPY SX SB Exam 2018 Question Paper

Hint $->$ See the hydrogen bounded structure of $\mathit{A}-\mathit{T}$ (Adenine - Thymine) and - G-C(Guanine- Cytosine)
Solution :
$\Rightarrow$ No $\mathit{O}\mathit{H}$ bond between $\mathit{A}-\mathit{T}=2$
Average energy of each $\mathit{H}$ bond in $\mathit{A}-\mathit{T}$ pair $=\mathit{X}$ Kcal ${\mathit{m}\mathit{o}\mathit{l}}^{-1}$
Total no of $\mathit{A}-\mathit{T}$ pair in structure $=5$
Hence, total energy $=2\mathit{x}\times 5=10\mathit{x}$
$\to$ No of $\mathit{H}$ bond between $\mathit{G}-\mathit{C}=3$
Average energy of each H-bond in G-C pair = Y Kcal $\mathit{m}\mathit{o}{\mathit{l}}^{-1}$
Total no of $\mathit{G}-\mathit{C}$ pair in structure $=3$
Hence total energy $=3\mathit{y}\times 3=9\mathit{y}$
-> Now, the total energy required to split all double-stranded DNA into two single strands.
$=10\mathit{x}+9\mathit{y}$