Calculating the molarity of DNA in a cell


> 
> In the following questions use a value of 3 for $\pi$, $6 \times 10^{23}$ for Avogadro’s number and $660$ for the molecular weight of $\pu{1 bp}$ of DNA. The volume of a sphere of radius $r$ is $4/3\,πr^3$. A bacterium has a single copy of a $\pu{4 \times 10^6 bp}$ circular genomic DNA.
> 
> 
> If the diameter of this spherical cell is 1 micrometer, what would be the molar concentration of DNA in this cell?
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> 
> 


The volume comes to be $\pu{2 \times 10^{-5} L}$.
Amount of substance is $\pu{6.7 \times 10^-18 mol}$.
Dividing this by the volume, my answer comes $\pu{3.3 \times 10^{-13} M}$, but the answer given in my book is $\pu{3.3 \times 10^{-9} M}$.




There is one particle of dsDNA in the cell. Divide by $N\_\mathrm{A}$ to get the amount of substance, and divide by the volume to get the concentration matching the concentration given in the answer.


\begin{align}
 V\_{\mathrm{cell}} 
 &= 4 \cdot {(\pu{0.5 μm})}^{3}\\[0.5ex]
 &= \pu{0.5 μm3}\\[0.5ex]
 &= \pu{5E-16 L}\\[3.5ex]
 N\_{\mathrm{A}}
 &= \pu{6E23 1//mol}\\[3.5ex]
 n\_{\mathrm{DNA}} 
 &= \frac{1}{N\_{\mathrm{A}}}\\[0.5ex]
 &= \pu{1.7E-24 mol}\\[3.5ex]
 c\_{\mathrm{DNA}}
 &= \frac{n\_{\mathrm{DNA}}}{V\_{\mathrm{cell}}}\\[1.5ex]
 &= \frac{\pu{1.7E-24 mol}}{\pu{5E-16 L}}\\[1.5ex]
 &= \pu{3E-9 mol//L}
\end{align}


 2

If the stated bacterium's cell has a diameter of $\pu{1 \mu m}$, the volume can be derived in terms of liters remembering the linear relation between cubic meters:
$$
 V\_\text{cell}
 =\frac{4}{3}\pi \left(\pu{0.5\times10^-6 m}\right)^3
 =\frac{4}{3}\pi \left(\pu{0.5\times10^-5 dm}\right)^3
 =\pu{5\times 10^-16 L}
$$
Inside this volume, the organism contains a certain number of base pairs; so the total amount of substance has to be known.


From Avogadro's Number, a mole is defined to be that portion (number of particles) of every substance in a defined physical phase; since this is an aqueous solution, the volume, as well as the molecular weight is meaningless: if one mole is defined by a certain number of particles, a different number of particles defines a different amount of substance:
$$
 n\_{\text{tot bp}}
 =\frac{N\_{\text{bp}}}{N\_{\text{A}}}
 =\frac{\pu{4\times 10^6 molecules}}{\pu{6\times 10^23 molecules mol-1}}
 =\pu{6.7\times10^-18 mol}
$$
Then, by the definition of molarity, dividing the amount of substance contained inside the cell by its volume gives a decent number, for a cell:
$$
 M\_\text{DNA}
 =\frac{n\_\text{tot bp}}{V\_\text{cell}} 
 =\frac{\pu{6.7 \times 10^-18 mol}}{\pu{5 \times 10^-16 L}}
 =\pu{1.34 \times 10^-2 M}
$$
I think that the wrong result is due to the volume, because to me it seems rather surprising that one sphere of $\pu{1 \mu m}$ in diameter has:
$$
 \pu{2\times 10^-5 L}
 =\pu{2\times 10^-2 mL}
 =\pu{20 \mu L}
 =\pu{20 mm3} 
 \neq \pu{5 \times 10^-8 mm3}
$$
of occupied volume. I suspect that something went wrong with the conversions, because I don't see (for now) any errors in my derivation.


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