A given double-stranded DNA molecule is 1,00, | vedclass

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(B) The distance between two consecutive base pairs in a $DNA$ helix is $0.34 \ nm$ $(0.34 	imes 10^{-9} \ m)$.Given that the total number of base pa

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$3.4 	imes 10^4 \ nm$

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(B) The distance between two consecutive base pairs in a $DNA$ helix is $0.34 \ nm$ $(0.34 	imes 10^{-9} \ m)$.Given that the total number of base pairs is $1,00,000$ ($10^5$ base pairs).The total length of the $DNA$ molecule is calculated as: $	ext{Total length} = 	ext{Number of base pairs} 	imes 	ext{Distance between two base pairs}$.$	ext{Total length} = 1,00,000 	imes 0.34 \ nm = 34,000 \ nm$.This can be expressed in scientific notation as $3.4 	imes 10^4 \ nm$.

List-$I$	List-$II$
$a$. Bacteriophage $\phi \times 174$	$i$. $48502$ base pairs
$b$. Bacteriophage lambda	$ii$. $5386$ nucleotides
$c$. Escherichia coli	$iii$. $3.3 \times 10^9$ base pairs
$d$. Haploid content of human $DNA$	$iv$. $4.6 \times 10^6$ base pairs

$A$ given double-stranded $DNA$ molecule is $1,00,000$ base pairs long. The length of this $DNA$ molecule will be:

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