Interpreting ASCAT CNV output

Question

[philomath_physicus]

Hello,

I have two queries w.r.t ASCAT CNV data:

1. How does one go about getting the total copy number for a given locus or chromosome from the ASCAT CNV.output results. What is the optimal way to filter the dataset to reduce the noise?

2. I would like to compare my dataset with some TCGA results for my cancer and to this end - I would like to get the ASCAT CNV data in .seg format to do some concordance analysis.

Thank you for your help! PP

Answer 1

[zizigolu]

ASCAT gives a .csv output in with you have allele specific copy number I mean the number of copies for minor and major allele for tumour and matched normal sample separately By adding miner and major allele column for each position of genome for tumour you would get the total copy number of the locus

For second question you need to convert ASCAT output to segment file for which you divide total copy number of each locus for each sample by the ploidy of that sample which gives segment mean of that locus after taking log2 However you still need to count the number of probs for each segment which is a bit complex but I cab help if you want

After having your segment mean you can compare that with TCGA but I suggest you to first run GISTIC on your segment file and compare the output with TCGA GISTIC output because GISTIC says about significantly amplified or deleted segment which reduce noise

Another way for reducing noise is ignoring segment means by a threshold

[Thind amarinder]

@A I am using a similar approach, but I used Sequenza tool. While preparing gistic2 input, I defined Seg.CN = log2(2 × depth.ratio) – 1. Do you think, I should divide the total copy number of each locus for each sample by the ploidy of that sample?

https://crazyhottommy.blogspot.com/2017/11/run-gistic2-with-sequenza-segmentation.html?showComment=1591775787493#c757830296052537465

[zizigolu]

Sorry I don't know about sequenza I devide by ploidy because SCAT estimates the average ploidy for sample so I can not simply assume the ploidy as 2 I just know that the sequenza output is perfect for clonality analysis

[binbinZhao]

Hi, I want to prepare the input for GISTIC2 analysis, according to your above comment, Seg.CN (of the Segmentation File) = log2((nMajor+nMinor)/ploidy of the sample). Do I understand correctly? What do you mean still need to count number of probs for each segment? Thanks.