I am reading an overview of the CNAqc package, which defines how the algorithm computes "Variant Allele Frequency (VAF) peaks for clonal CNAs."

  • mutations present in a percentage $0<c<1$ of tumour cells, sitting on a segment $nA:nB$;
  • tumour purity $\pi$;
  • a healthy diploid normal;

Since the proportion of all reads from the tumour is $\pi(n_A+n_B)$, and from the normal is $2(1-\pi)$. Then, mutations present in $m$ copies of the tumour genome should peak at VAF value $$v_m(c)=\frac{m \pi c}{2(1−\pi)+\pi(n_A+n_B)}.$$

However, I don't understand exactly the definition $n_A:n_B$ in this context. Are $n_A:n_B$ the number of reads coming from alleles $A$ and allele $B$ in the tumor? But then, why the normal has only $n = 2$?

  • 1
    $\begingroup$ Migrated to Biology at the request of the question author $\endgroup$
    – gringer
    Jun 23 at 3:59

1 Answer 1


On a further inspection on the associated paper, $n_A$ and $n_B$ are the allele-specific copy numbers. And it definitely makes sense that we have 2 (diploid) for the normal component.


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