Department of Computer Science and Center for Computational Molecular Biology, Brown University, Providence, RI, 02906, USADepartment of Computer Science and Center for Computational Molecular Biology, Brown University, Providence, RI, 02906, USA.
Department of Computer Science and Center for Computational Molecular Biology, Brown University, Providence, RI, 02906, USA.
Bioinformatics. 2014 Jun 15;30(12):i78-86. doi: 10.1093/bioinformatics/btu284.
High-throughput sequencing of tumor samples has shown that most tumors exhibit extensive intra-tumor heterogeneity, with multiple subpopulations of tumor cells containing different somatic mutations. Recent studies have quantified this intra-tumor heterogeneity by clustering mutations into subpopulations according to the observed counts of DNA sequencing reads containing the variant allele. However, these clustering approaches do not consider that the population frequencies of different tumor subpopulations are correlated by their shared ancestry in the same population of cells.
We introduce the binary tree partition (BTP), a novel combinatorial formulation of the problem of constructing the subpopulations of tumor cells from the variant allele frequencies of somatic mutations. We show that finding a BTP is an NP-complete problem; derive an approximation algorithm for an optimization version of the problem; and present a recursive algorithm to find a BTP with errors in the input. We show that the resulting algorithm outperforms existing clustering approaches on simulated and real sequencing data.
Python and MATLAB implementations of our method are available at http://compbio.cs.brown.edu/software/ .
肿瘤样本的高通量测序表明,大多数肿瘤表现出广泛的肿瘤内异质性,其中包含不同体细胞突变的多个肿瘤细胞亚群。最近的研究通过根据包含变异等位基因的 DNA 测序读数的观察计数将突变聚类到亚群中,从而量化了这种肿瘤内异质性。然而,这些聚类方法没有考虑到不同肿瘤亚群的群体频率通过它们在同一细胞群体中的共同祖先而相关。
我们引入了二叉树分区 (BTP),这是一种从体细胞突变的变异等位基因频率构建肿瘤细胞亚群的组合问题的新表述。我们表明,找到 BTP 是 NP 完全问题;为该问题的优化版本推导出一个近似算法;并提出了一种在输入中存在错误的递归算法来找到 BTP。我们表明,在模拟和真实测序数据上,所得到的算法优于现有的聚类方法。
我们方法的 Python 和 MATLAB 实现可在 http://compbio.cs.brown.edu/software/ 获得。