connectivity map vehicle scan id

Question

[rv]

Hello everyone,

I have a problem in understanding the association between perturbation and ctrl array in the cmap instance file

When the chip type (array) is HT_HG-U133A, I see associations like these:

array          perturbation_scan_id          vehicle_scan_id
HT_HG-U133A    5500024053146032309538.A06    5500024053146032309538.A01
HT_HG-U133A    5500024024213121906564.C07    .H07.G08.D10.B11.A12
HT_HG-U133A    5500024051859013109328.F09    .H07.G08.E09.D10.B11.A12

The first one is clear for me.

In the second row, since the HT_HG-U133A chip is a multi-plate chip, I would have expected to find in the vehicle_scan_id a "Cxx" suffix corresponding to the "C07" in the perturbation_scan_id

As an example, if had found a ".C12" suffix enumerated in the vehicle_scan_id, I would have built the vehicle_scan_id "5500024024213121906564.C12" for the perturbation_scan_id "5500024024213121906564.C07"

But such a "Cxx" suffix is missed.

The same for the third row.

Could you, please, give me some hint?

Thanks

Answer 1

[Zhilong Jia]

.H07.G08.D10.B11.A12 means5500024024213121906564.H07

5500024024213121906564.G08 5500024024213121906564.D10 5500024024213121906564.B11 5500024024213121906564.A12

"Their full scan numbers are not listed on the instance page but can be reconstructed by appending each of the six extensions provided (eg .H01) to the twenty-two character number preceding the period from the corresponding perturbation scan number. The complete set of cmap.cel files can be downloaded in bulk from the download tab." from the help of CMap.

see figure

Update:

First, I did not process the raw data of CMap before. I think the data cmap supplied in ctrl column means using all the ctrls in this row not only Cxx one.

Here is an example:

"In the case of multiple controls per batch, we formed a more robust control by removing as an outlier the control with the highest Euclidean distance to the other controls, and then used the mean of the rest as the controls."

a ref here.

[rv]

Ok, thank you very much @Zhilong Jia.