# 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?

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 $$\pu{3.3 \times 10^{-9} M}$$.

• Surely as a polymer of indeterminate length, the molar concentration of DNA is an entirely meaningless phrase? – Aesin Jan 18 '14 at 20:50

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.
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}