densitometry on western blots

Tom Anderson via methods%40net.bio.net (by ucgatan At ucl.ac.uk)
Fri Nov 24 13:26:24 EST 2006


On Fri, 24 Nov 2006 martinhoehne At gmx.net wrote:

> Thank you for your response. A few more questions ...
>
> > >1) As I understand it the bands have to be non-saturated. Does this
> > >really mean that as soon as I have pixel values of 255 (on a 8-bit
> > >image) I can forget about doing a densitometric analysis?
> >
> > No, it does not. As long as there is a calibration curve and your
> > signal is within reasonable dynamic response range, you are fine. If
> > you don't have response calibrated, going to 24 bits or whatever will
> > change nothing at all.
>
> Reasonable dynamic range - is the what one of the other posters said
> just a few saturated pixels?

Yes.

Bear in mind also that the film itself can become saturated - you know how
when you have a really bright band, the film gets a short of shiny
appearance? If that happens, you're stuffed. I suspect the film's response
is nonlinear even below actual saturation, but the serial dilution
experiment will tell you this. Another way to do it would be to expose the
same blot for increasing lengths of time, and compare the scans of the
different films; that eliminates error from lane-to-lane variation in
loading, but introduces error from film-to-film variation in exposure
time. If you do serial exposures of a serial dilution, though, you have
enough redundancy to correct for both, but that would be an impressively
anal thing to do, or

> I read somewhere that saturation does not matter too much since the
> bands are not only getting darker but also the area covered by the band
> is getting larger, and this can be used as a measurement as well. But
> this seems to be an exotic (if not wrong) statement to me.

In principle, yes, although i've never heard of anyone doing that.

> > In order to deal quantitatively with non-linear processes in the
> > course western/densitometry, you *have to have* a calibration curve
> > that tells you how your signal changes quantitatively with changing
> > input. If this can be done with purified protein, you then have a
> > chance to express results in true protein quantity. In most cases the
> > relative numbers are perfectly acceptable too: Use your pre-PI
> > material and do 1:2 or so serial dilutions that cover your entire
> > range of response. When done in parallel with the sample (same gel,
> > same development, etc) this guarantees that you can express your
> > results quantitatively, in relative units as % of total.
>
> would it be ok to run the lysates on a second gel in parallel? The gel
> with the IPs is already full.

I'd say so. You should expose both on the same film, though.

> Excuse me if the question sounds stupid, but how exactly do I proceed
> then? Please correct me if I´m wrong.
>
> Let´s assume I load 1:2 serial dilutions of the lysates and stain for
> protein B.
> - I do a densitometric analysis (I use the gel analysis tool of
> ImageJ).
> - I plot in a graph the amount of input (e.g. 0,125x; 0,25x; 0,5x; 1x)
> versus the value obtained from ImageJ for the given band.
> - I do the densitometric analysis for the precipitates (for both
> proteins, A and B). For each band I take the value I obtained from
> ImageJ and look up in my graph to what amount of protein this value
> corresponds. Let´s say for protein A_wt it´s 0,8x and the
> coprecipitated protein B in this lane it´s 0,6x. And for protein
> A_mut1 it´s 0,7x and for the coprecipitated protein it´s 0,9x.
> - I calculate (0,6 / 0,8) / (0,6 / 0,8) =1 and (0,9 / 0,7) / (0,6 /
> 0,8)=1,7
>
> Can I then state that protein A_mut1 precipitates 1,7x more protein B
> as protein A_wt does?

With some caveats. Essentially, you're taking the simple equilibrium
binding reaction:

A + B <=> AB

And its corresponding equation:

Keq = [AB] / [A][B]

And putting your measurement of the amount of complex,as measured by IP,
in as [AB] and the amount of your protein from the loading control in as
[A], and calculating Keq/[B]. Now, problems with this:

- This equation only applies if the reaction is as described, ie one to
one stoichiometry, no cooperativity, no funny business.

- Your loading control measures the total amount of protein, but [A] isn't
the total, it's the amount of free protein; this is therefore only valid
if the vast majority of your protein is free, rather than complexed.
the concentration of the proteins.

- You aren't measuring and using [B], so you're calculating Keq/[B] rather
than Keq; that's okay if [B] is constant, but might not be otherwise.

In practice, the latter two conditions require that both proteins are
present in substantial excess over the complex, which in turn requires
that the affinity of the interaction is fairly low. For example, you
could probably measure a kinase binding to a scaffold like this, but i
think you'd be on thin ice measuring an antibody binding to an antigen.

Disclaimer: i don't do this stuff for a living, and it's years since i
studied it as a student, so i could have this wrong. Does this make sense
to anyone else?

> Is it ok to scan the blot with an ordinary scanner using 300dpi
> greyscale or do I have to opt for more dpi or even have to use a
> transillumination scanner?

It'll work; whether it'll work well is another question. Your calibration
will tell you how well, though, and as long as you work in the range you
find to be linear, you'll be okay.

tom

-- 
Tom Anderson, MRC Laboratory for Molecular Cell Biology, UCL, London WC1E 6BT
(t) +44 (20) 76797264   (f) +44 (20) 76797805   (e) thomas.anderson At ucl.ac.uk


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