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← JEE Advanced 2024 · Paper 2
ChemistryInteger+4Question 44 of 51

JEE Advanced 2024 · Paper 2, Chemistry question 10

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

5′-A G T C A C G T A A G T C-3′5'\text{-A G T C A C G T A A G T C-}3'

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

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

Answer and worked solution

Official answer: 41

Base pairs. The strand AGTCACGTAAGTC\text{AGTCACGTAAGTC} has A = 4, T = 3, G = 3 and C = 3. The complementary strand pairs every base:

  • A-T pairs: 4+3=74+3=7, with 2 H-bonds each, giving 1414 H-bonds;
  • G-C pairs: 3+3=63+3=6, with 3 H-bonds each, giving 1818 H-bonds.

(DNA has no A-U pairs.)

Energy.
14×1.0+18×1.5=14+27=41 kcal mol−1.14\times1.0+18\times1.5=14+27=41\ \mathrm{kcal\,mol^{-1}}.

Topics:Nucleic acidsHydrogen bonding
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