Length of DNA segment having 200 nitrogen bases is _______.

680 \circ A

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340 \circ A

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68000 \circ A

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68 μ m

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Solution

The correct option is B $340 \circ A$
In 1 full turn of DNA with a length of $34 A^{\circ}$ has 10 base pairs or 20 bases. Now length of a DNA segment having 200 bases is
$34 A^{\circ} - 10 b p i n d s D N A o r 20 b a s e s$

$x - 200 b a s e s i n D N A \frac{200 \times 34}{20} = 340 A^{\circ}$

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340

^{\circ} A

20

340 A^{\circ}

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