Question

if you laid out all of the dna in one of your cells end - to - end, it would be 2 meters (m) long. but an average - sized nucleus in your body is just 10 micrometers long. a micrometer (symbolized \\( \mu m \\)), is one millionth of a meter, or \\( 1 \times 10^{-6} m \\). which of the following represents \\( 10 \mu m \\)?
\\( \bigcirc 0.01 m \\)
\\( \bigcirc 0.00001 m \\)
\\( \bigcirc 0.000001 m \\)
\\( \bigcirc 0.00000001 m \\)
if an average nucleus is \\( 0.00001 m \\) long and all of the dna is \\( 2.0 m \\) long, how many times does the dna have to be compacted to fit inside? (remember that it is compacted by winding around specialized dna - binding proteins.)
\\( \bigcirc 2.0 m\times0.00001 m = 0.00002 \\)
\\( \bigcirc 2.0 m/0.00001 m = 200,000 \\)
\\( \bigcirc 0.00001 m/2.0 m = 0.000005 \\)
in addition to the nucleus, there is a second organelle that is unique to eukaryotes: the mitochondrion (mi - toe - kon - dree - un; plural: mitochondria). like the nucleus, each mitochondrion is surrounded by two membranes. but in mitochondria, the inner membrane forms a series of tubes and sacs.

Explanation:

First question:

Step1: Convert micrometers to meters

Given \(1\ \mu m = 1\times10^{-6}\ m\). To convert \(10\ \mu m\) to meters, we use the formula \(x\ m=x\times10^{-6}\ m\) (where \(x = 10\) for \(10\ \mu m\)).
So, \(10\ \mu m=10\times1\times10^{-6}\ m = 1\times10^{-5}\ m=0.00001\ m\)

Second question:

Step1: Calculate the compaction factor

The compaction factor \(n\) is given by the formula \(n=\frac{\text{Length of DNA}}{\text{Length of nucleus}}\)
We know the length of DNA \(L_{DNA}=2.0\ m\) and the length of the nucleus \(L_{nucleus}=0.00001\ m\)
Substitute the values into the formula: \(n=\frac{2.0\ m}{0.00001\ m}\)

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Answer:

First question: \(0.00001\ m\) (the second option). Second question: \(2.0\ m/0.00001\ m = 200,000\) (the second option)