Area under P(r) curve

Interactive and automated data processing tools (PRIMUS, GNOM, AUTORG).
Scattering from simple bodies (BODIES), peak analysis (PEAK), data plotting (SASPLOT) etc.
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emartin1
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Area under P(r) curve

#1 Post by emartin1 » 2014.05.30 18:16

Perhaps this is a naive question, but I am curious about the integrated area under P(r) curves. I was examining data of several constructs of a protein that have different oligomeric states. When looking at the P(r) curves it was clear that they did not have the same area. I had never really paid much attention to this before, but it is obvious that, they do not normalize to 1 and are quite different for different data sets.

Could someone shed some light on this for me, what in the data determines the area under the distribution?

Thanks,
Erik

Alex
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Re: Area under P(r) curve

#2 Post by Alex » 2014.05.30 18:46

they do not normalize to 1 and are quite different for different data sets.
You should normalize them to unity yourself if you want to compare them. It is not normalized by default.
Best, Alex

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Re: Area under P(r) curve

#3 Post by emartin1 » 2014.05.30 19:03

Thank you for the quick reply!

This makes sense. I was just curious if there was something inherent within the raw data that would result in a lower or higher non-normalized distribution. Or perhaps this is just arbitrary?

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Re: Area under P(r) curve

#4 Post by Alex » 2014.05.30 23:02

if there was something inherent within the raw data that would ..
As far as I know - no, not really.
Or perhaps this is just arbitrary?
I think it is arbitrary - it has something to do with gamma function.
A.

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AL
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I(0) and Rg from p(r)

#5 Post by AL » 2014.06.05 09:57

Since

I(s) = 4Π ∫p(r)sin(sr)/sr dr

the integrated area under the p(r) function is directly related to the forward scattering I(0):

I(0) = 4Π ∫p(r) dr

and less directly to Rg:

Rg2 = 2Π ∫r2p(r) dr / I(0)

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Re: Area under P(r) curve

#6 Post by Alex » 2014.06.05 13:09

Al, you mean this is after the normalization to unity, right?

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Re: I(0) and Rg from p(r)

#7 Post by AL » 2014.06.05 13:39

No, this is how to get the actual I(0) from the p(r). The first formula is just the Fourier transform p(r) → I(s), if you replace s with zero you get I(0).

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