I think this is more of a physics question than a biology question because the DNA might be modeled as a string (of known linear density). I attempted to solve this question, my full treatment is below. It was originally inspired by a question on S.E. Biology.

If the over-arching question is deemed off-topic for Physics S.E., my secondary question is did I apply physical principles correctly in my attempt to solve this problem?

Relevant physics concepts might be tension, tensile strength, work-energy theorem, and bond dissociation (no tags available).

Thank you for your insight! Here is my analysis:

(For those not familiar with exactly what a nucleotide is, it is the repeating unit of DNA - rungs on the ladder, so to speak.)

Say a long chain of DNA double helix is suspended from some little vise. The nucleotide at the bottom of the chain has a certain very small weight, and will probably not break the covalent bond to the nucleotide above it. The two bottom nucleotides weigh twice as much as the single bottom one, but that also probably won't break the bond above them. If you keep going up with this logic, you might reach a point where some number of nucleotides has enough weight to break the covalent bond above them, and that chain of nucleotides would fall to the floor (making a very tiny puddle).

The bonds in the sugar-phosphate backbone are: $...O-P-O-C-C-C-O-P-O...$ (just a random bit of the chain, where the carbons are 5', 4', 3' of a deoxyribose sugar). The lowest bond energy, thus the most easily broken, is $P-O$ at 335 kJ/mol (Ref). I'm taking this as this how much energy is required to break the DNA sugar-phosphate backbone (in a single place).

An experimentally derived estimate of the linear density of double stranded DNA is $2.1\cdot10^{12} \frac{g/mol}{m}$. So every meter of DNA weighs about a trillion g/mol. Does this sound like a lot? It's because that is the meter-weight of one mole of DNA helices. The linear density of one molecule of DNA helix can be found by dividing by avagadro's number: one molecule of DNA has a mass of $3.5\cdot10^{-12} g$ per meter.

Let's take our vertical strip of DNA and imagine it is divided into two segments: an upper segment held by some vise, and a lower segment which dangles freely. The two segments are connected by a covalent bond of the sugar-phosphate backbone, specifically a $P-O$ bond. The lower segment has some mass $m$, and its weight $mg$ exerts a force against the bond ($F=mg$).

The bond requires energy $E=335 kJ/mol$ to break. This energy can be provided by the lower segment doing work to break the bond. Technically this work would be done by the force of gravity on the lower segment, which releases energy if the segment is moved closer the center of the earth. But the segment will only fall toward the earth of the energy released by that process overcomes the energy barrier $E$ of the bond holding the segments together.

Work is force times distance (roughly speaking), so what distance will the lower segment move in the process of breaking the bond? I think 1 angstrom (about a bond length) may be a reasonable approximation. That is, if we stretch the $P-O$ bond (which is already around 1.5 angstroms long) by 1 angstrom, it should be reasonable to say it is broken and the atoms will not re-attract each other (see Bond Dissociation graphs for details).

So! Our energy $E = 335 kJ/mol$ to break the bond will be supplied by work $W=Fd$, performed by the force of gravity $F=mg$, which acts for a relevant distance of $d=1$ angstroms $= 10^{-10} m$:


The mass of the lower segment is $m=\rho l$, where $\rho=2.1\cdot10^{12} g/mol/m$ is the linear density of the DNA segment and $l$ is the length of the segment. The question we should ask is "how long must the lower segment be for it to have enough weight to break the bond?" Well...

$$E=mgd=(\rho l)gd$$

$$l=\frac{E}{\rho gd} \\ = \frac{335\cdot10^3 J\cdot mol^{-1}}{(2.1\cdot10^{12} g\cdot mol^{-1} m^{-1})(9.8 m\cdot s^{-2})(10^{-10} m)} \cdot \frac{kg \cdot m^2 s^{-2}}{1 J} \cdot \frac{10^3 g}{1 kg} \\ = 1.6 \cdot 10^5 m$$

So, according to my admittedly crude approach, a segment of DNA would have to be on the order of 100,000 meters long to weigh enough to break the covalent bond holding it to the rest of the chain above it. I suppose this is a testament to the powerful strength of covalent bonds (or I've made some grave error somewhere in my physics). So my calculations suggest a 6-ft segment of DNA (magically extracted and assembled from trillions of humans cell) should have no problems holding together when you grip it in a tiny vise and hold it vertically.

  • $\begingroup$ It would be much better if you can make this a self-contained post,which could include posting your answer as an answer. $\endgroup$ Apr 8, 2017 at 11:16
  • $\begingroup$ @Emilio Pisanty Will do $\endgroup$ Apr 8, 2017 at 14:51

1 Answer 1


An experimentally derived estimate of the linear density of double stranded DNA is $2.1 \times 10^{12}\frac{g/mol}{m}$.

Linear density should have units of $\text{g/m}$ (or $\text{mol/m}$ if you want linear nucleotide density instead of linear mass density). Every meter should have a specific mass or a specific count. From the abstract of your link, it seems to be $2.8 \times 10^6 \text{dalton}/\mu\text{m}$ or $3.45 \times 10^{-12}\text{g}/\text{m}$

Other sources seem to give an explicit break force for a covalent bond (C-C specifically) as around $1600\text{pN}$. You'd need $m = \frac{F}{a} = \frac{1600pN}{g} = 1.6\times10^{-7}\text{g}$ to exert that weight by hanging. That's about $35000\text{m}$. Well within the same order of magnitude as your calculation.

  • $\begingroup$ Note the figure reported in the article is 2.08 Da/um (not 2.8), giving 3.5 pg/m for the linear density as opposed to 4.6 pg/m. Would you mind providing the source for break force? $\endgroup$ Apr 8, 2017 at 21:05
  • $\begingroup$ One of my main concerns is the link between energy and force. When using a bond energy instead of directly using a break force value, is it valid to do it the way I did? energy That is, can a bond energy be converted to bream force via work, or is it not appropriate to make the connection the way I did (using W=Fd)? $\endgroup$ Apr 8, 2017 at 21:09
  • $\begingroup$ Yes, thank you on the figures. Yes, I consider your method mostly valid. The difficulty is that the force changes over the break distance. Using it to find the maximum force is likely to be close, but with quite a bit of error. $\endgroup$
    – BowlOfRed
    Apr 9, 2017 at 4:39

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