Question: Analysis of BAF from SNP array data
gravatar for Roman Hillje
4.3 years ago by
Roman Hillje40
Milan, Italy
Roman Hillje40 wrote:

Hi guys, I have a question about the BAF pattern I see when looking at SNP array data. I recently found this paper from 2012 that discussed a linear relationship between tumor cellularity and a shift in the BAF pattern. Here is the link: Please have a look at this image from the paper. BAF shift I understand that, in a sample of only healthy cells, the BAF plot shows signals at 0, 1, and 0.5. I also understand that for the pure tumor cell line, in the region of deletion, there is only 0 and 1. But, what I can't make any sense of, is the way the BAF pattern splits in between. If there are healthy cells in the sample, where did their signal at 0.5 go?

Can anybody explain this?

snp baf • 2.1k views
ADD COMMENTlink modified 3.0 years ago by alexander.kononov10 • written 4.3 years ago by Roman Hillje40
gravatar for alexander.kononov
3.0 years ago by
alexander.kononov10 wrote:

Look!. Let's imagine that there is tumor with 50% of normal and 50% of aberrant cells with duplication in target region. The reference genome have only "A" nucleotides. In normal cell have also "G" nucleotides as alternative allele. In this case normal cell have different positions a long this region: reference homozygote - A A , alternative homozygote - G G, heterozygote - A G and heterozygote - G A. First variants of position was reflected by points in 0, second by points in 1, third and fourth by 0.5 (i mean on the plot).

And now we add the cells with duplication. There are next variant: homozygotes position - A A A and G G G, heterozygotes - A G G and G A A.

In total there are 50% normal and 50% tumor cells in our samples. In first case we have AA from normal and AAA from tumor cells. B-allele is absent. there are points in 0. Second case: GGG and GG, 100% of B-allele, points is of 1.

Position of third case have A G from normal and A G G from tumor cells. There are 3 G and 2 A, B-allele frequency is 3/5 = 0.6.

Position of fourth case have G A and G A A. There are 2 G and 3 A, B-allele frequency is 2/5 = 0.4 Now we have positions which are reflected in 1 , 0 , 0.6 and 0.4 bands on the plot.

ADD COMMENTlink modified 3.0 years ago • written 3.0 years ago by alexander.kononov10
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