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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 distance between two consecutive base pairs increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix, which has 2 x 109 bp, in the presence of saturating amount of this compound.

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Q1

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 distance between two consecutive bases increases, from 0.34 nm to 0.44 nm. Calculate the length of the DNA double helix (which has

2 \times 10^{9} b p

) in the presence of a saturating amount of this compound.

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Q2

If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of DNA double helix in haploid mammalian cell is

3.3 \times 10^{9}

bp then length of DNA is approximately

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Q3

If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is

6.6 \times 10^{9}

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Q4

If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a E.coli cell is

4.6 \times 10^{6}

bp, then the length of the DNA is approximately

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Q5

If the distance between two consecutive base pairs is $$0.34 nm$$ and the total number of base pairs of a DNA double helix in a typical mammalian cell is $$6.6 \times 10^9 bp$$, then the length of the DNA is approximately

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