Why do 60S & 40S subunits make an 80S (not 100S) ribosome and, similarly, 50S & 30S make 70S? 60+40 is not equal to 80, nor is 50+30 equal to 70, so why are the subunits of the 80S and 70S ribosomes named so weirdly? I couldn't come up with any other mathematical explanation.


3 Answers 3


That's not a simple mathematical addition!

If those units were mass, for instance, you could (and should) add them: a 60g Lego piece combined with a 40g Lego piece will, of course, form a combo that has 100g.

However, those numbers are Svedberg units, which is...

a non-metric unit for sedimentation rate [...] The Svedberg coefficient is a nonlinear function. A particle's mass, density, and shape will determine its S value. (emphasis mine)

Funnily enough, most of my students thought that that S stands for Sedimentation. However, it's a tribute to Theodor Svedberg, winner of the Nobel Prize for Chemistry in 1926.

Thus, in a oversimplified explanation, the prokaryotic ribosome has two sub-units. The large sub-unit sediments at 50s, the small sub-unit sediments at 30s, but the two together (that is, the whole ribosome) sediments at 70s, not 80s.

The same way an eukaryotic ribosome has a large sub-unit that sediments at 60s, a small one that sediments at 40s, but the whole structure sediments at 80s, not 100s.

Also, the rRNAs that constitute the sub-units have their own sedimentation rates (in svedberg units) as well:

enter image description here

For a detailed explanation with all math you need, the other answers here and here explain it beautifully.


When a complex mixture of particles undergoes ultracentrifugation, they separate based on their shape and mass due to the force applied by the centrifuge and the counteracting frictional force of the solvent. You can read more about this procedure here. S stands for Svedberg, which is a measurement of the sedimentation rate of a particle. This is given by the formula $s=\frac{v}{a}$ where $s$ is the sedimentation coefficient and $v$ is the velocity a particle moves when an acceleration, $a$, is applied.

One svedberg is defined as exactly 10−13 s. Essentially the sedimentation coefficient serves to normalize the sedimentation rate of a particle by the acceleration applied to it. The resulting value is no longer dependent on the acceleration, but depends only on the properties of the particle and the medium in which it is suspended. Sedimentation coefficients quoted in literature usually pertain to sedimentation in water at 20°C.

A 1S particle, for example, would move at 10-13 m/s when an acceleration of 1 g is applied (ignoring diffusion).

What's important to realize is that the sedimentation rate of a particle depends not only on its mass but also on its shape (among other things) since its cross-sectional area determines the effective frictional force it experiences.

When this was done with E. coli extracts, peaks were seen at 32S, 51S, 70S and 100S, depending on Mg2+ concentration. The researchers concluded that the 32S and 51S peaks were components of the 70S peak and that the 100S peak was a dimer of two 70S particles. They also determined that the mass of the 50S particle was about double the mass of the 30S particle which, together, added up to the mass of the 70S particle.

  • $\begingroup$ Svedberg unit means 10^-13 sec.... So that means 70s will take 70×10-13 seconds to settle down? $\endgroup$
    – roxaite
    Jun 6, 2017 at 17:28
  • $\begingroup$ @roxaite I've edited my post to answer your comment. $\endgroup$
    – canadianer
    Jun 6, 2017 at 17:43

The other two answers give nice detail, so I want to give a bit more mathematical answer here.

First, the S you are talking about is Svedberg units (of sedimentation coefficient, named after Swedish chemist Theodor Svedberg), used to characterize the behavior of a particle in sedimentation process, especially centrifugation. One svedberg corresponds to exactly 10-13 seconds (see other answers for more details).

Now, lets begin with equations. Sedimentation coefficient is written in equation as:

$s = \frac{v_t}{a}$

where $v_t$ is the terminal velocity. Terminal velocity of a particle in a given medium is constant because the force of gravitation (or centrifugation) is cancelled by the viscous force of the medium. In such case, terminal velocity is calculated as ratio of centrifugal force to the viscous resistance (from Stoke's law) and its equation becomes:

$v_t = \frac{mr\omega^2}{6\pi\eta r_0}$

Also, centrifugal acceleration:

$a = r\omega^2$

Putting the value of $v_t$ and $a$ in first equation:

$s = \frac{v_t}{r\omega^2} = \frac{m}{6\pi\eta r_0}$

As you see, although sedimentation coefficient is the ratio of terminal velocity to centrifugal acceleration, it does not depend on either of them! Now, all other things are constant, and mass would add linearly. So why does the value of $s$ decrease? It is because of non-linear addition of $r_0$ (radius). See the diagram below:

70s and 80s ribosome shape

It is clearly visible now that the subunits don't just mix into each other (like two oil droplets joining to form one bigger droplet), the small subunit fits on the large subunit (real shape does vary, but roughly the shape is like this). This makes the average value of $r_0$ of ribosome larger than the sum of average value of $r_0$ of subunits i.e. $r_0 (small) + r_0 (large) < r_0 (ribosome)$. This makes the overall sedimentation rate a bit lesser than expected (remember that $s \propto \frac{1}{r_0}$). I hope this makes the concept more clear.


  • $\begingroup$ You are too polite, mate: the other answer does, but my answer don't provide any detail! It's a very general, oversimplified explanation! $\endgroup$
    – user24284
    Jun 7, 2017 at 5:52
  • 4
    $\begingroup$ @GerardoFurtado I think all 3 answers complement each other quite nicely; it's always good to have different explanations. $\endgroup$
    – canadianer
    Jun 7, 2017 at 6:12
  • 1
    $\begingroup$ @gerardofurtado you mean I need to incorporate more details in my answer? If I do, who would want to scroll down and see other answers? :P $\endgroup$ Jun 7, 2017 at 6:25
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    $\begingroup$ @canadianer I was about to add technical details to my answer when you posted yours, that explains the issue perfectly. So, I just left mine as it was. $\endgroup$
    – user24284
    Jun 7, 2017 at 6:42
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    $\begingroup$ @gerardofurtado I've added all the details I could. What if people say "wow, these 3 answers explain everything so beautifully...here, have my upvote, all of you"! ;) $\endgroup$ Jun 7, 2017 at 7:04

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