densitometry on western blots

Hello,

I have some basic questions on densitometry and would be glad if anyone
can point me to a good source to learn more or even answers some of the
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?

2) I want to use the analysis for judging the binding-strength of two
proteins. I have various point mutations of protein A and want to
compare how good these mutants interact with protein B. For this I am
doing a co-IP (precipitating protein A and co-precipitating protein B).
I then measure the intensities of the bands in western blots. In order
to not being fooled by slightly varying expression levels, I divide the
value for the band of the coprecipitated protein B by the value for the
precipitated protein protein A. Is this scientifically correct? Or is
there a better/more appropriate way?

Thanks for any input,

Martin

In article <1164299514.092168.323980@f16g2000cwb.googlegroups .com>, [Only registered users see links. ] wrote:

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.


That's normalizing and is perfectly acceptable way to go. Problem is,
the pixels you derive from densitometry without any sort of calibration
curve are essentially meaningless. You might as well eyeball the blot,
say "this is less and that is more" and that will be as accurate
scientific information and your normalized pixels are.

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.

DK

In <wGk9h.197$[Only registered users see links. ]>,
DK <[Only registered users see links. ]> wrote:


That's not right. For an 8-bit image, once a pixel value is at 255,
that pixel value cannot get any higher - by definition it is already
fully saturated. Practically speaking, from an 8-bit image you get
two logs of linearity.

It is for this reason that many of the commercial systems made by Kodak,
Biorad, Alpha-Innotech etc. etc., make use of 10-bit or 12-bit CCDs.

AC
--
Email: echo 14232209320335931527179462016806497532155427589018 6P | dc

In article <E3l9h.5384$[Only registered users see links. ]>, [Only registered users see links. ] wrote:

Practically speaking, two logs is well within a range biologists
operate on Western blots. And, practically speaking, any number of
shades of gray is completely meaningless without a calibration curve.


Once again, without calibration that's like insisting on measuring protein
concentration without standards but with spectrophotometer that is
capable of accurately measuring 0.0001 absorbance unit.

Cheapest scanner and the most primitive software that's capable of
counting pixels and their intensity is all that's sufficient to get perfectly
accurate and scientifically valid quantitative westerns. No amount of
mega $$$ equipment can deliver that without calibration.

DK

<[Only registered users see links. ]> wrote in message
news:1164299514.092168.323980@f16g2000cwb.googlegr oups.com...

on any digital image system, once you reach the limit of the dynamic range
(256 for 8-bit, 65536 for 16-bit, etc) it doesn't make sense to use it.
You can get away with *a few* saturated pixels within the area you are
measuring. Usually you would use the median (or mean) of the pixels in the
area, or some other measurement, and having a few saturated pixels won't be
too bad. But you should aim to avoid any saturation if you want accurate
results.

It's one little bug of mine, when you see figures with a loading control
that's clearly saturated... and sure, they all look the same then! :-)

Jose

"DK" <[Only registered users see links. ]> wrote in message
news:gql9h.205$[Only registered users see links. ]...

Having a calibration curve is a very good thing, as long as you're not using
saturated bands, as the other poster said.

Once you reach saturation... you can't say anything about that sample, only
that it is high. If you want to do any sort of quantitative measurement,
even rough, you have to stay within the dynamic range of the capture system,
i.e: no saturation.

But I agree that an 8-bit system is enough to give you pretty good results.

Jose

In article <[Only registered users see links. ]>, "Jose de las Heras" <[Only registered users see links. ].uk> wrote:

Once you reach full saturation, you don't have calibration curve. Some
degree of saturation is not a killer though as the response is still changing
and the results can still be deduced from the curve.


Yep. Which can, of course, be done for both 8- and 12-bit image
processing with proper care.

Thank you for your response. A few more questions ...



Reasonable dynamic range - is the what one of the other posters said
just a few saturated pixels?
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.




would it be ok to run the lysates on a second gel in parallel? The gel
with the IPs is already full.
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?



And in addition some technical questions:
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?

Thanks a lot for your input,

Martin

On Fri, 24 Nov 2006 [Only registered users see links. ] wrote:


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


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


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


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?


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) [Only registered users see links. ]

In article <1164376874.555523.143430@h54g2000cwb.googlegroups .com>, [Only registered users see links. ] wrote:


Reasonable dynamic range is everything where calibration curve is
reasonably close to linearity, e.g. not completely flat and not going
up to quickly.


That's where the pain comes from. No two blots can be exactly
identical, so ideally one should have calibration points on the same
gel. That means doing many westerns if one has to analyze and
compare many points. I used to do this: on a 15-well gel, 10 points
for calibration and 3 different amounts of experimental sample
(having different amounts is essential to increase accuracy of
determination because the data will be obtained from different
parts of calibration curve). I ended up doing five westerns - four
for four samples to be compared and one final control, on which
I loaded all four samples based on previous blots in a way that
should have produced identical signal for all of them. Once this
happened, I knew I can trust the numbers and claim to be quantitative.


I think that if you assume that A_wt and A_mt behave identically in
terms of detection (which is reasonable) and determine amounts of
each from the *same* calibration (to normalize, you must express
WT and MT in the same units, and since the expression level of WT
and MT can be different, 21342660f its own lysate results in different
units) then I'd say yes, 1.7X.


I used to scan at 600 dpi with a consumer-grade HP scanner.
I don't know how much of an improvement fancier equipment
gives.

DK