Re: [Gusdev-gusdev] RAD3 Schema Question

[Angel P. <an...@pc...>]

Paul Boutros wrote:

>Hi again all,
>
>One more question about the RAD3 schema, if I may:
>
>The BioMaterialImp table has three main views:
>- BioSource	==> raw sample
>- BioSample	==> treated sample
>- LabeledExtract	==> labeled sample
>
>How can I relate the bio_material_id of a BioSource with those of its 
>BioSamples and LabeledExtracts?  The treatment table appears to detail the 
>treatments performed on a BioSource, but not what the resulting BioSample is 
>called.
>  
>
It follows the same mechanism as does the MAGE model. Namely, a 
Treatment will have a set of BioMaterialMeasurements, which in turn 
point back to BioMaterialImp.
Thus for a BioSample that was a result of a pooling of a couple of  
BioSources and subsequently labeled, you would have:

BioSource (mouse 1)  id= b1
BioSource  (mouse 2) id =b2
BioSample (sample1 ) id= b3
Treatement (pool) id = t1  bio_material_id = b3
BioMaterialMeasurement id = bm1   treatment_id = t1, bio_material_id = b1
BioMaterialMeasurement id = bm2   treatment_id = t1, bio_material_id = b2
LabeledExtract (lex 1) id = b4
Treatment (label) id = t2 bio_material_id = b4
BioMaterialMeasurement id = bm3   treatment_id = t2, bio_material_id = b3

And don't forget about the FK's to Protocol and LabelMethod for that 
appropriate BioMaterials.

I'll leave your examples as homework for yourself ;) see what you can do 
using the above shorthand and I'll tell you if it is correct.

Angel


>Let me specify the biological case, in case I'm being unclear.  We have animals 
>(BioSources) that are sacrificed, and the resulting tissues handled in 
>different ways (BioSamples), then labeled (LabeledExtracts) and run on arrays.  
>I'm struggling to see how to relate a particular BioSource to all the 
>BioSamples that have been derived from it.  The same idea could correspond to, 
>for instance, taking different tissues from the same animal.
>
>Thanks in advance for any suggestions/explanations -- very much appreciated.
>Paul
>
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