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QJEEAdv24Chdae8: JEE Advanced 2024 Chemistry - Class 12 JEEAdv24Chdae8 Solution: For a double strand DNA, one strand is given below: The amount…

For a double strand DNA, one strand is given below:

JEE Advanced 2024 Paper 2 Online Chemistry - Biomolecules Question 1 English

The amount of energy required to split the double strand DNA into two single strands is _______ kcal mol⁡−1\operatorname{mol}^{-1} .

[Given: Average energy per H-bond for A-T base pair =1.0 kcal mol−1=1.0 ~\mathrm{kcal}~ \mathrm{mol}^{-1} , G-C base pair =1.5 kcal=1.5 ~\mathrm{kcal} mol−1\mathrm{mol}^{-1} , and A-U base pair =1.25 kcal mol−1=1.25 ~\mathrm{kcal} ~\mathrm{mol}^{-1} . Ignore electrostatic repulsion between the phosphate groups.]

Step-by-Step Verified Solution

JEE Advanced 2024 Paper 2 Online Chemistry - Biomolecules Question 1 English Explanation

A=T⇒2H\mathrm{A}=\mathrm{T} \quad \Rightarrow 2 \mathrm{H} -bond

G≡C⇒3H\mathrm{G} \equiv \mathrm{C} \quad \Rightarrow 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\mathrm{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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