[Spring2019] Question about homework 2

[Borries Demeler]

So here are some very good questions from Ariel:

>  1.  In the slides I downloaded, there are three slides with homework
>  questions, but there are more questions after these slides. I just
>  wanted to confirm that the homework questions end on the slide labeled
>  "Homework (p3 of 3)."

Sorry, yes that is confusing. But I would like you to do also the
problem on the next 2 pages.

>  2.  For the questions in part A, it does not specify what solvent the
>  measurements are taken in. Because it says in the table caption that the
>  diffusion values have been corrected to water at 20 degrees Celsius,
>  I have assumed that these are the solvent conditions (and therefore
>  the appropriate density is 1 g/mL, viscosity is 0.01 g/(cm*sec),
>  and temp is 293 K). Is this acceptable?

Yes, exactly, standard conditions refer to what one would measure at
20C (293.15K) and in distilled water. Keep in mind that density and 
viscosity of the solvent changes as a function of temperature and also
as of buffer composition. To keep things simple, scientists correct to 
"standard conditions" so they can easily compare results from different
temperatures and buffer conditions.

>  3.  Again for part A, I used the ideal gas constant 8.3145*10^5. Is
>  this the correct constant?

Stick to the cgs system, use 8.3145*10^7 erg.

>  4.  For part B, I just wanted to check my general understanding
>  because I think I'm overthinking it. Due to the relationship s =
>  (D*M*(1-vbar*rho))/RT, as the diffusion coefficient goes up, the
>  sedimentation coefficient should also go up.

Actually, that is not the case. Assuming anisotropy is constant, then 
M has to go down when D goes up, and that would make s likely go down
as well. In general, the smaller the particle is, the slower it will
sediment and the faster it will diffuse, all other things being equal
(anisotropy, buffer, temperature).

>  In addition, whenever a
>  particle increases in volume or surface area, both should increase
>  because there are more possible friction interactions with the
>  solvent. Would you say this is accurate?

No, when the particle increases in volume (but the partial specific
volume stays constant - i.e., same density) then it also increases in
molar mass. If the anisotropy doesn't change, then the sedimentation 
coefficient will also go up, but the diffusion coefficient will get
smaller, since the larger particles has more friction, so:

	D = (RT)/(Nf)

f will increase, causing D to decrease. Keep in mind that both s and D
are INVERSELY proportional to friction, while only s depends on mass.

>  5.  Specifically for part B number 3, my understanding is that the salt
>  helps neutralize the charge on the DNA backbone, making it less prone to
>  interactions with the solvent. Therefore, if there is less salt, there
>  should be more friction interactions with the solvent. Is this accurate?

Actually not:
As we saw in Marielle's paper today, when salt is decreased, the
friction of the molecule increases, and it becomes more anisotropic.
So, why is that? The charges on the backbone are all of the same
sign. Therefore, there is strong electrostatic repulsion unless the 
charges are balanced by Na+ ions from the added salt. The microscopic
effect from the bound Na+ causing more friction is negligible. But what
happens is that the negative unbalanced charges in the backbone cause a
lot of repulsion and they also cause a huge water shell to co-migrate
with the DNA molecule. The repulsion makes the DNA stretch as far as it
can, making it stiff and very extended (anisotropic). This causes it to
have a lot friction. The salt reduces the repulsion and allows the DNA
to become more flexible, allowing it to fold and overall to become more
globular and less frictional, and dragging along a smaller hydration cloud.


All good questions! Please keep them coming!

-Borries