"subberry" <zhangpy at gmail.com> wrote:
> It's so kind of you to give so detailed answers.
>> "the borderline was around 600V for the high voltage of
> the detector"
>> Yes, i recorded the HT. Is that say whet the HT above 600V, the data si
> not creditable? My spectrometers is Jasco 810.
Yes, approximately at 600V. You should see it easily in the data, too.
Especially the buffer blank will start to scatter wildly at wavelengths
where the HT is too high - there's simply not enough light to be able to
calculate the difference between the absorbance of right- and left
circular polarized light.
> "For a monomeric protein, there's no reason to assume that it unfolds
> slower, or not at all, if the protein concentration is increased."
>> Is that say a monomeric protein must be unfolded at all? And what is
> the relationship between this and protein concentration? How about
> inclusion bosy?
Let's first assume that the protein does *not* aggregate. For the
reaction between folded F and unfolded U, you have:
F <==> U
and the equilibrium constant is
[F]
K = ---
[U]
(or upside down, as you like it). Let's define the total concentration
c_0 = [F] + [U] => [U] = c_0 - [F]
now:
[F]
K = ---------
c_0 - [F]
solving this for \alpha = [F] / c_0, the fraction of native molecules,
gives
\alpha c_0 \alpha
K = ---------------- = --------
c_0 - \alpha c_0 1-\alpha
K
\alpha = ---
1+K
All this is of course dependent on the guanidine concentration, but at a
given [GdnHCl], \alpha, the fraction of native molecules, is independent
of protein concentration. On the other hand if you have a homodimer:
F_2 <==> 2 U
[F_2]
K = -----
[U]^2
(where [U]^2 means [U] squared, F_2 means F index 2). Now the total concentration is
c_0 = [F_2] + [U] (in terms of monomers)
and
2 [F_2]
\alpha = ------- or 2 [F_2] = \alpha c_0
c_0
substituting this into the equation for K, we get
1/2 \alpha c_0 1/2 \alpha c_0
K = ---------------------- = ------------------
(c_0 - \alpha c_0)^2 c_0^2 (1-\alpha)^2
Even without explicitly solving this for \alpha, you can see that c_0
doesn't cancel out, and that the fraction of native molecules will
depend on c_0.
You can also explain it with the Le Chatelier principle ("principle of
least force" or similar): If you increase the number of particles, the
equilibrium will be shifted towards the side that has a lower number of
particles, thus escaping the "force" of more particles.
So this means: If you have a monomeric protein, and you know that it
unfolded at 5M GdnHCl at one particular protein concentration, you know
that it is also unfolded at 5M at other protein concentrations.
Therefore, if you prepare a concentrated unfolded stock solution, you
just take the conditions you already know about.
On the other hand, if the protein is a dimer (or more complicated
oligomer), you have to check whether it is really unfolded
*at*exactly*the*concentration*used*. Of course once you've calculated K
(and are sure about the stoichiometry), you can simply calculate
\alpha.
Now, when it comes to aggregates, things get even more complicated.
Usually aggregates dissolve readily in 6M GdnHCl, but some don't, or
rather take very long. You should be able to monitor that by light
scattering. There are some proteins that tend to aggregate when you
store them for long, or when they are frozen and thawed repeatedly, but
when you denature them and renature them again, the renatured solution
is native, active, fine :-).
> "light scattering is a good method for that"
>> what type spectrometers can do this? what i have is Zeta Potential from
> brookhaven, is it okay?
I must admit that I never used light scattering intensively myself. For
a first check, even measuring the amount of scattered light by looking
at the apparent "absorbance" at wavelengths above 300nm might be
sufficient. But when you have a real light scattering instrument that
even gives you an estimate of the size of the particles, you're lucky.
Regards, Frank
--
In der Zeit, in der man einen defekten Riegel derartig seziert hat, dass man
verlässlich bestimmte Speicherbereich als "heile" garantieren(!) kann, kann
man auch Pfandflaschen im Stadtpark sammeln, dass reicht dann für einen
doppelt so großen neuen Riegel. ;) [Jörg Rossdeutscher in d-u-g at l.d.o]
