In an experiment, $DNA$ is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distancebetween two consecutive base increases. from $0.34- 0.44\,nm$ calculate the length of $DNA$ double helix $($ which has $2 \times 10^9\,bP)$ in the presence of saturating amount of this compound.
In an experiment, $DNA$ is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distancebetween two consecutive base increases. from $0.34- 0.44\,nm$ calculate the length of $DNA$ double helix $($ which has $2 \times 10^9\,bP)$ in the presence of saturating amount of this compound.
$2 \times 10^{9} \times 0.44 \times 10^{-9} / \mathrm{bp}=0.88 \mathrm{~m}$
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- [AIIMS 1981]
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- [AIIMS 2017]
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