1
$\begingroup$

I'm interested in the diffusivity of macromolecules inside the mammalian retina, relative to the vitreous.

For example is the diffusion coefficient for a macromolecule (protein, 50-150 kDa for example) 10 or 100 times smaller in the retina than the vitreous?

The diffusion coefficient for such a macromolecule is about 10^(-6) cm^2/sec in the vitreous - roughly that of a protein in water. I can't seem to find any information on how much smaller it would be in the retina, a tightly packed tissue.

$\endgroup$
2
$\begingroup$

Unfortunately, I don't have a good, simple answer for you. The reason is that the answer may depend significantly on details of the macromolecule more than its size - like how strongly it binds to the environment.

In addition, within an environment that is more elastic than viscous, and with a very complex shape, you might expect to get various sorts of anomalous diffusion, rather than a single diffusion coefficient.

Transport in tightly confined biological spaces is complex, and still pretty controversial. A paper I found useful as a simple introduction to some of this: Morphogen Transport - http://dev.biologists.org/content/140/8/1621

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! This is useful information. It's interesting that this is a controversial topic! I'm building a diffusion based model you see, so for my purposes I require some numerical estimate to compare between the two areas. The macromolecules I'm interested are antibodies and fragment antibodies, which bind to VEGF, all these molecules are in the range of 50-150 kDa, and do not bind (at least for my purposes within the model) to any structures within the vitreous or retina. $\endgroup$ – Freeman Jul 12 '17 at 8:54
  • $\begingroup$ It's an interesting issue! I recommend seeing what experimental data you can find for any sort of transport properties, but be prepared for there to be a significant spread (don't stop after getting the first estimate). $\endgroup$ – AJK Jul 12 '17 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.