[RASMB] another vbar question

[Borries Demeler]

On Thu, Sep 28, 2017 at 01:25:11PM -0700, John Sumida wrote:
> Hi Borries,
> 
> Thank you for pointing me at the manuscript.
> 
> What would boundary data stemming from the MWL detector look like?  I am
> fascinated that the spectral shifts shown in the paper can be observed in
> real time at each radial position - if my reading is accurate?

Hello John,

The data from the MWL detector provides one complete velocity experiment
for each wavelength analyzed, in this case several hundred. Since all 
wavelengths are measured simultaneously, you can get for each radial
position (and from each scan) an entire wavelength scan for the observed
wavelength range. That allows you also to go the other direction, and 
decompose your hydrodynamic signal into multiple optical species, and 
get their relative proportions in a hydrodynamic species. In other words,
get stoichiometry from the optical separation. Here is an example:

	https://www.ncbi.nlm.nih.gov/pubmed/27977168

The details of this whole process are explained here:

	https://www.ncbi.nlm.nih.gov/pubmed/26412646

Incidentally, you can do the same type of experiments now with the
new Optima AUC, which has multi-wavelength capability. The optics
are designed somewhat differently than Helmut's design, but have the
advantage of working well in the lower UV. One drawback is the need to
accumulate wavelengths sequentially, but in UltraScan you can still
derive wavelength scans using various modeling functions available,
so in effect, you can still do the same experiments.

> Also in applying the 2DSA analysis wouldn't one still need to know what
> v-bar was for the CdSe/ZnS qdots in order to calculate the equivalent
> radius, MW etc. of the particles?  I am, perhaps, making the erroneous
> assumption that "equivalent radius" is related to the core shell diameter in
> your manuscript. 

Equivalent radius is not the core shell diameter, instead, it is
supposed to be the hydrodynamic radius. The model we show is supposed to
include the functionalization layer, and the density calculations are
an estimate and solvent contributions are certainly far from rigorous.
The assumption made here is that vbar is approximately the inverse of
the density of the overall particle (incl. bound water, ions, etc),
and of course there is some error involved. Nevertheless, the intercept
and the linear term of the molar mass fit comes out very close to the
theoretical molar mass of the magic cluster, so the density model is
probably not too far from reality. So for the 2DSA calculation we then
used a different vbar value for each s-species simulated, and that vbar
was based on this model described in the SI. Granted, this is a very rough
approximation, but it was the best we had and it compensated for the 
clear change in vbar to be expected for these particles.

Not to point you to yet another manuscript, but you can also get a pretty
good estimate for the vbar using the generalized 2DSA algorithm if you
make the assumption that the particles have a constant frictional ratio
(close unity if they are spherical):

	https://www.ncbi.nlm.nih.gov/pubmed/25010012

> Also, by "leveraging error" you are referring to the very real situation
> where the density of the q-dot is such that it would require too great a
> proportion of the density matching solvent component?

Sorry, that was not very clear. The problem is that the change in density
achieved with the heavy water is not big enough to cause a significant
buoyancy effect for these super-dense qdots, in comparison to proteins
or lipids for example, which are much closer to the maximum density 
achieved with heavy water. So when you extrapolate the observed s-differences
as a function of D2O concentration you have a very long way to go to get
to s=0. So any error in the slope gets magnified considerably since you 
are so far away from zero s. So to answer your question:

	"it would require too great a proportion of the density matching solvent component"

is true, but even if you were using 100% D2O or 100% H2O18 it would not raise
the density of your solvent sufficiently to give enough of change in s to
get a reliable.

Cheers, -b.