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

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    $\begingroup$ Surely as a polymer of indeterminate length, the molar concentration of DNA is an entirely meaningless phrase? $\endgroup$ – Aesin Jan 18 '14 at 20:50
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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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  • $\begingroup$ I found the mistake in my volume of 2X10^-5 L. $\endgroup$ – biogirl Jan 20 '14 at 5:09
  • $\begingroup$ Although I cannot find any mistake in your calculation, your answer does not match the one given in my book ! $\endgroup$ – biogirl Jan 20 '14 at 5:10
  • $\begingroup$ @biogirl My suggestion is to tell your professor about this mistake, show the calculation proposed and let him give you the correct advice. Maybe you should take the problem backwards with the given result: since the volume calculation is correct, there should be something missed with the moles. I'll try by myself and edit my answer if any breakthrough comes up. $\endgroup$ – TheVal Jan 20 '14 at 11:08
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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}

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