Diameter of DNA double helix is

A^{0}

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A^{0}

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3.4

A^{0}

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340

A^{0}

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Solution

The correct option is B 20 $A^{0}$
The B-DNA is arranged in a right-handed double helix; the helix is 20 angstrom wider and the number of base pairs per helical turn is 10.5. The plane of the base pairs in B-DNA is tilted about 6 degree with respect to the helix axis. One complete turn of the double helix is 3.4 nm long and each of which has 10.5 base pairs, the distance between two consecutive base pairs 0.34 nm. Correct answer is B.

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Similar questions

If Avogadro's number is A $_{0}$ , the number of sulphur atoms present in 200 mL of 1N H $_{2}$ SO $_{4}$ is

\frac{1}{a_{0}} + \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} + \dots + \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}} = \frac{1}{a_{0} -} \frac{a_{0} x}{a_{1} + x -} \frac{a_{1} x}{a_{2} + x -} \dots \frac{a_{n - 1} x}{a_{n} + x}

\frac{1}{a_{0}} - \frac{x}{a_{0} a_{1}} + \frac{x^{2}}{a_{0} a_{1} a_{2}} - \dots + {(- 1)}^{n} \frac{x^{n}}{a_{0} a_{1} a_{2} \dots a_{n}}

$s i n n x = \sum_{r = 0}^{n} a_{r} s i n^{r} x$ where n is an odd natural number, then

(1 + x^{2}) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . + a_{36} x^{36}

a_{0} + a_{3} + a_{6} + . . . + a_{36} = \frac{2}{3} (2^{35} + 1)

a_{0} + a_{1} + a_{2} + . . . + a_{36} = 2^{36}

a_{0} + a_{2} + a_{4} + . . . + a_{36} = 2^{35}

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