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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Chemistry
Chapter
Biomolecules

JEE Advanced 2024 · Paper 2 · Chemistry · Biomolecules

For a double strand DNA, one strand is given below: The amount of...

For a double strand DNA, one strand is given below: For a double strand DNA, one strand is given below: The amount of energy required to split the double strand DNA into The amount of energy required to split the double strand DNA into two single strands is _______ kcal mol−1\mathrm{mol}^{-1}. [Given: Average energy per H\mathrm{H}-bond for A-T base pair =1.0kcalmol−1,G−C=1.0 \mathrm{kcal} \mathrm{mol}^{-1}, \mathrm{G}-\mathrm{C} base pair =1.5kcal=1.5 \mathrm{kcal} mol−1\mathrm{mol}^{-1}, and A-U base pair =1.25kcalmol−1=1.25 \mathrm{kcal} \mathrm{mol}^{-1}. Ignore electrostatic repulsion between the phosphate groups.]

Correct answer

0

Step-by-step solution

A=T2H\mathrm{A}=\mathrm{T} \quad 2 \mathrm{H}-bond G≡C3H\mathrm{G} \equiv \mathrm{C} \quad 3 \mathrm{H}-bond Number of A=T\mathrm{A}=\mathrm{T} pair =7=7 Number of G≡C\mathrm{G} \equiv \mathrm{C} pair =6=6 Number of H-bond involve in A=T=7×2=14\mathrm{A}=\mathrm{T}=7 \times 2=14 Number of H\mathrm{H}-bond involve in G≡C=6×3=18\mathrm{G} \equiv \mathrm{C}=6 \times 3=18 Energy required for A=T=14×1=14\mathrm{A}=\mathrm{T}=14 \times 1=14 Energy required for G≡C=18×1.5=27\mathrm{G} \equiv \mathrm{C}=18 \times 1.5=27 Total energy required 14+27=4114+27=41
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