The total number of hydrogen bonds in a DNA d | vedclass

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(A) Two nucleic acid chains are wound about each other and held together by $H$-bonds between pairs of bases.Adenine $(A)$ forms two hydrogen bonds wi

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$33$

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(A) Two nucleic acid chains are wound about each other and held together by $H$-bonds between pairs of bases.Adenine $(A)$ forms two hydrogen bonds with thymine $(T)$,and guanine $(G)$ forms three hydrogen bonds with cytosine $(C)$.The given sequence is $5^{\prime}-G-G-C-A-A-A-T-C-G-G-C-T-A-3^{\prime}$.Counting the bases in the sequence:$G$: $5$ bases$C$: $3$ bases$A$: $4$ bases$T$: $1$ baseWait,let us re-count the sequence $G-G-C-A-A-A-T-C-G-G-C-T-A$:$G$: $1, 2, 9, 10$ ($4$ bases)$C$: $3, 8, 11$ ($3$ bases)$A$: $4, 5, 6, 13$ ($4$ bases)$T$: $7, 12$ ($2$ bases)Total $G-C$ pairs = $4$ (since $G$ pairs with $C$,we take the count of $G$ or $C$ pairs,here $4$ $G$ and $3$ $C$ is not possible in a single strand,let's re-examine the sequence: $G, G, C, A, A, A, T, C, G, G, C, T, A$. Total bases = $13$.$G-C$ pairs: $G$ $(4)$ and $C$ $(3)$. This implies the complementary strand has $3$ $G$ and $4$ $C$. Total $G-C$ pairs = $7$.$A-T$ pairs: $A$ $(4)$ and $T$ $(2)$. This implies the complementary strand has $2$ $A$ and $4$ $T$. Total $A-T$ pairs = $6$.Total $H$-bonds = $(7 	imes 3) + (6 	imes 2) = 21 + 12 = 33$.

The total number of hydrogen bonds in a $DNA$ double helix strand,where one strand has the following sequence of bases,is $...........$
$5^{\prime}-G-G-C-A-A-A-T-C-G-G-C-T-A-3^{\prime}$

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The total number of hydrogen bonds in a $DNA$ double helix strand,where one strand has the following sequence of bases,is $...........$$5^{\prime}-G-G-C-A-A-A-T-C-G-G-C-T-A-3^{\prime}$