Question

If \(1000\) bp of a double-helical DNA weighs \(1\times10^{-18}\) g and the distance between two bp is \(0.34~\text{nm}\), the total amount of DNA (in mg, rounded to one decimal) required to stretch from Earth to Moon (distance \(=3.74\times10^{5}\) km) is ________________.

Hint:

Use: length \(=\) (bp count) \(\times 0.34~\text{nm}\); mass scales linearly with bp number.

Answer 1

Correct Answer: 1

Step 1: Convert length to nm: \(3.74\times10^{5}\,\text{km}=3.74\times10^{8}\,\text{m}=3.74\times10^{17}\,\text{nm}\).
Step 2: Number of bp needed: \[ N=\frac{3.74\times10^{17}}{0.34}\approx1.10\times10^{18}\ \text{bp}. \] Step 3: Mass per bp \(=\frac{1\times10^{-18}\text{ g}}{1000}=1\times10^{-21}\) g.
Total mass \(=N\times1\times10^{-21}\approx1.10\times10^{-3}\) g \(=1.1\) mg.