For a double strand DNA,one strand is given b | vedclass

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(D) The given strand is $5'-A-G-T-C-A-C-G-T-A-A-G-T-C-3'$.The complementary strand will be $3'-T-C-A-G-T-G-C-A-T-T-C-A-G-5'$.Counting the base pairs:$

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(D) The given strand is $5'-A-G-T-C-A-C-G-T-A-A-G-T-C-3'$.The complementary strand will be $3'-T-C-A-G-T-G-C-A-T-T-C-A-G-5'$.Counting the base pairs:$A-T$ pairs: $A$ at positions $1, 5, 9, 10, 12$ (in the given strand) paired with $T$ in the complementary strand. Total $A-T$ pairs $= 5$.$G-C$ pairs: $G$ at positions $2, 7, 11$ and $C$ at positions $4, 6, 13$ (in the given strand) paired with $C$ and $G$ respectively. Total $G-C$ pairs $= 8$.Energy for $A-T$ pairs $= 5 	imes (2 \ H	ext{-bonds}) 	imes (1.0 \ kcal \ mol^{-1}) = 10 \ kcal \ mol^{-1}$.Energy for $G-C$ pairs $= 8 	imes (3 \ H	ext{-bonds}) 	imes (1.5 \ kcal \ mol^{-1}) = 36 \ kcal \ mol^{-1}$.Total energy required $= 10 + 36 = 46 \ kcal \ mol^{-1}$.Note: Based on the provided options,the calculation assumes the total number of $H$-bonds is $41$. Re-evaluating the sequence: $A(1), G(2), T(3), C(4), A(5), C(6), G(7), T(8), A(9), A(10), G(11), T(12), C(13)$. Total $A-T$ pairs $= 7$,$G-C$ pairs $= 6$. Energy $= (7 	imes 2 	imes 1.0) + (6 	imes 3 	imes 1.5) = 14 + 27 = 41 \ kcal \ mol^{-1}$.

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