Hey guys,
so I'm new here and start with my first question: I thought copy number variants (CNV) will be rounded (after log2) to 1,2,3... which tells me, how many extra-copies of a gene there are. Today someone asked me to give the exact number, because it's important if it's 0.5 or 0.7 etc. Searching the Internet I found this paper where it says:
"Hemizygous deletion refers to the loss of one of the alleles, whereas homozygous (biallelic) deletion refers to the loss of both alleles identified by allele-specific analysis in the clinical samples. In cell lines, homozygous deletions were identified by CNAG2.0 with a copy number of the probes equal to zero in default settings. The absolute log2 ratios of one- and two-copy losses varied depending on the ploidy levels of the cell lines but can be clearly distinguished according to the criteria described previously [24]. For example, in near-triploid cell lines, we used log2 ratios <-0.2 with an average of ∼-0.25 in the 10-SNP genomic smoothed data to define one-copy loss, whereas we used log2 ratios <-0.5 with an average of ∼-0.7 to define twocopy loss and log2 ratios <-0.9 with an average of ∼-1.8 to define three-copy (complete) loss.We used the same method to define gains. For example, in near-triploid cell lines, we used log2 ratios >0.2 with an average of ∼0.24 in the 10-SNP genomic smoothed data to define one extra copy gain, whereas we used log2 ratios >0.4 with an average of ∼0.6 to define two-extra-copy gain and >0.8 with an average of ∼1 to define three-extra-copy gain or more. Where the gains were three or more extra copies, we define these gains as amplifications. We used both quantitative polymerase chain reaction (PCR) and FISH data to validate and refine the criteria for the copy number estimations."
So I still don't understand whats going on here. What exactly tells me a CNV of -1.8, 0.24(...) and why do they define 0.6 as 2 (...) extra copies? And can I compare those Microarray CNVs with CNVs of WES?
Thank you in advance
