[RASMB] further question ln(c) vs r2

[Walter Stafford]

>Dear Walter
>
>>call it the baseline). Unless your protein has a large excluded
>>volume, is high charged and/or you are using a very low ionic
>>strength buffer; non-ideality should not be a problem.
>Actually it is fairly basic, pI of 10.66, and the buffer is only 
>20mM phosphate buffer pH3
>it has five Argines, so non ideality is quite likely, I'll have to 
>read some more on it.
>
>>	It looks as though you have a slight offset to the data which
>>is influencing the lowest concentration data.  You might consider
>>fitting this data with NONLIN both with and without a floating offset
>I'll try this.  What type of offset can occur? In the previous email 
>I meantioned that I had a negative concentration and was unsure how 
>this could occur is there likely to be any link?
>
>
>many thanks once again.
>
>heather :)

Hi Heather,

	When I read your second email I didn't realize you were using 
interference optics. With interference optics the fringe 
displacements are known only to within an additive constant (That 
means they could all negative depending on how the data were taken). 
You MUST fit with a floating additive constant (delta in NONLIN) when 
fitting to interference data. (And when doing it "manually as you 
were describing for your ln(c) ve R^2 plots, the data must be 
re-zeroed somehow). In order to do this by the ln(c) method, you must 
have run at a sufficiently  high speed that the meniscus area is 
effectively zero concentration so you have an absolute reference 
point whose concentration you know.

Your sigmas must be in the range of 2.0 to about 6 cm^-2 (sigma is 
the slope of a plot of ln(c) vs r^2/2  (not r^2)) if you are running 
at three speeds and three concentrations as is the recommended 
procedure. Only sigma's above  4.0 will give you sufficient room at 
the meniscus to nail down the offset manually.  [Sigma is also 
defined as (w^2)s/D = M(1-vp)(w^2)/RT]. You can get away with lower 
sigmas if you know your "monomer" sigma. Below a sigma of 2.0 the 
offset, pre-exponential factors and sigmas all become hopelessly 
cross-correlated such that almost any combination will fit the data 
OK: not good. There's more to this story, but this should get you 
started.

Short answer: all interference data must have a reference point 
subtracted from it to get absolute fringe displacements. NONLIN will 
do this for you if you stay above sigma=2.0,  and 3.0-4.0 is 
preferable.

Hope that helps,
Walter


-- 
Walter Stafford
mailto:stafford at bbri.org
617-658-7808