Question

The average molar mass of one base pair of nucleotides in DNA is approximately 600g/mol. The spacing between successive base pairs is about 0.34nm, and a complete turn in the helical structure of DNA occurs about everynm. If a DNA molecule has a molar mass of$\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$g/mol, approximately how many complete turns exist in the DNA a-helix structure?

Short answer

There are$7.50\times {10}^5$helix turns in the DNA molecule

Step-by-step solution

 The given data

$$\begin{gathered} \mathrm{M}(\mathrm{Base} \mathrm{Pair})=600\mathrm{g}/\mathrm{mol} \\ \mathrm{M}\left(\mathrm{DNA}\right)=4.5\times {10}^9\mathrm{g}/\mathrm{mol} \\ \mathrm{turn}=3.4\mathrm{nm} \\ \mathrm{L}(\mathrm{x}-\mathrm{x})=0.34\mathrm{nm} \\ \mathrm{n}\left(\mathrm{turns}\right)=? \end{gathered}$$

First, we will determine how many base pairs we have in DNA molecule (N(BP))

$\begin{array}{rcl}\mathrm{N}\left(\mathrm{BP}\right)& =& \frac{\mathrm{M}\left(\mathrm{DNA}\right)}{\mathrm{M}\left(\mathrm{BP}\right)}\\ & =& \frac{4.5\times {10}^9\mathrm{g}/\mathrm{mol}}{600\mathrm{g}/\mathrm{mol}}\\ & =& 7.50\times {10}^6\\ & & \end{array}$

Now, we will calculate how many base pairs are per turn in \alphaα helix (N(turn)).

$\begin{array}{rcl}\mathrm{N}\left(\mathrm{turn}\right)& =& \frac{\mathrm{L}\left(\mathrm{turn}\right)}{\mathrm{L}(\mathrm{x}-\mathrm{x})}\\ & =& \frac{3.4\mathrm{nm}}{0.34\mathrm{nm}}\\ & =& 10\\ & & \end{array}$

To calculate number of \alphaα helix turns (n(turns)) we will divide the number of base pairs by the number of base pairs per turn.

role="math" localid="1663894601363" $\begin{array}{rcl}\mathrm{n}\left(\mathrm{turns}\right)& =& \frac{\mathrm{N}\left(\mathrm{BP}\right)}{\mathrm{N}\left(\mathrm{turns}\right)}\\ & =& \frac{7.50\times {10}^6}{10}\\ & =& 7.50\times {10}^5\end{array}$

Hence, there are $7.50\times {10}^5\alpha$helix turns in the DNA molecule.