# Calculating the molartity 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 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 $4 \times 10^6$ 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?

My working:

The volume comes to be $2 \times 10^{-5}$ liter. Number of moles come to be $6.7 \times 10 ^{-18}$. Dividing number of moles by volume my answer comes $3.3 \times 10^{-13}$ but the answer given in my book is $3.3 \times 10^{-9}$.

• 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 $1\:\mu\text{m}$, the volume can be derived in terms of liters remembering the linear relation between cubic meters: $$V_{\text{cell}}=\frac{4}{3}\pi\left(0.5\times10^{-6}\:\text{m}\right)^3=\frac{4}{3}\pi\left(0.5\times10^{-5}\:\text{dm}\right)^3=5\times 10^{-16}\:\text{L}$$ inside this volume, the organism contains a certain number of base pairs; so the total number of moles 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 acqueous 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 number of moles: $$n_{\text{tot bp}}=\frac{N_{\text{bp}}}{N_{\text{A}}}=\frac{4\times 10^{6}\:\text{molecs}}{6\times10^{23}\:\text{molecs mol}^{-1}}=6.7\times10^{-18}\:\text{mol}$$ Then, by the definition of molarity, dividing the number of moles contained inside the cell by it's volume gives a decent number, for a cell: $$M_{\text{DNA}}=\frac{n_{\text{tot bp}}}{V_{\text{cell}}}=\frac{6.7\times10^{-18}\:\text{mol}}{5\times 10^{-16}\:\text{L}}=1.34\times 10^{-2}\:\text{M}$$ I think that the wrong result is due to the volume, because to me it seems rather surprising that one sphere of $1\:\mu\text{m}$ in diameter has: $$2\times 10^{-5}\:\text{L}=2\times 10^{-2}\:\text{ml}=20\:\mu\text{l}=20\:\text{mm}^3\neq 5\times10^{-8}\:\text{mm}^3$$ of occupied volume. I suspect that something went wrong with the conversions, because I don't see (for now) any errors in my derivation.
Did you calculate the volume using $V = \frac{4}{3} \cdot 3 \cdot (5 \cdot 10^{-7}\;m)^3\,$? Then your volume is in $m^3$, not in liter.