Regenie.QRS:在生物银行规模上计算高效的全基因组量子力回归
Fan Wang1, Chen Wang1, Tianying Wang2
1Department of Biostatistics, Columbia University, New York, US.
bioRxiv : the preprint server for biology
|July 14, 2025
概括
我们开发了Regenie.QRS,这是一种全基因组关联研究 (GWAS) 的新方法,可以检测复杂的基因型-表型关联. 这种方法提高了识别整个表型分布的遗传效应的能力,改进了传统的线性回归模型.
科学领域:
- 遗传学 遗传学 是一个
- 统计遗传学 统计遗传学
- 生物信息学是一种生物信息学.
背景情况:
- 基因型-表型关联往往是动态的和上下文依赖的,导致在表型的分布中产生异质的遗传效应.
- 传统的线性回归可能无法完全捕捉这些复杂的遗传效应,需要先进的分析方法.
研究的目的:
- 介绍Regenie.QRS,一个计算效率高的全基因组定量回归技术,用于分析生物库规模的全基因组关联研究 (GWAS) 数据.
- 比现有方法更有效地检测和描述异质的基因型-表型关联.
主要方法:
- 提出了一种全基因组量子回归方法,Regenie.QRS,用于大规模的GWAS数据,包括遗传结构.
- 该方法估计了多基因效应,并将它们作为非混合量子力回归模型中的偏移集成在一起.
- 通过对英国生物银行和ProgeNIA/SardiNIA数据集的模拟和应用进行验证.
主要成果:
- 与线性回归相比,模拟证实了I型错误的稳健控制和检测异质关联的功率增加.
- Regenie.QRS在边际定量回归试验中表现出更好的功率.
- 现实世界的应用突出了该方法在识别和表征异质遗传效应方面的优势,以G6PC2位点在葡萄糖调节中的作用为例.
结论:
- Regenie.QRS是一个强大而有效的工具,用于在大型GWAS中发现复杂的基因型-表型关系.
- 该方法成功地识别了基因变异,其影响在表型分布上有所不同,提供了更深入的生物学见解.
- 这些发现说明了量子回归在理解复杂特征的遗传贡献方面的有用性,例如G6PC2对低血糖的保护作用.
更多相关视频
09:23Quantification of Information Encoded by Gene Expression Levels During Lifespan Modulation Under Broad-range Dietary Restriction in C. elegans
Published on: August 16, 2017
8.2K
09:10A Fast and Quantitative Method for Post-translational Modification and Variant Enabled Mapping of Peptides to Genomes
Published on: May 22, 2018
9.3K
相关概念视频
Biostatistics: Overview
377
Biostatistics plays a crucial role in understanding and analyzing data in healthcare and biology. Biostatisticians conduct experiments, gather evidence, and draw meaningful conclusions using statistical methods and techniques. Different variables form the foundation of biostatistical analysis, allowing researchers to understand and interpret data effectively. These variables are classified into different types, each serving a specific purpose in statistical analysis.
Discrete variables are...
Discrete variables are...
377
Genome-wide Association Studies-GWAS
14.3K
Genome-wide association studies or GWAS are used to identify whether common SNPs are associated with certain diseases. Suppose specific SNPs are more frequently observed in individuals with a particular disease than those without the disease. In that case, those SNPs are said to be associated with the disease. Chi-square analysis is performed to check the probability of the allele likely to be associated with the disease.
GWAS does not require the identification of the target gene involved in...
GWAS does not require the identification of the target gene involved in...
14.3K
Genomics
37.5K
Genomics is the science of genomes: it is the study of all the genetic material of an organism. In humans, the genome consists of information carried in 23 pairs of chromosomes in the nucleus, as well as mitochondrial DNA. In genomics, both coding and non-coding DNA is sequenced and analyzed. Genomics allows a better understanding of all living things, their evolution, and their diversity. It has a myriad of uses: for example, to build phylogenetic trees, to improve productivity and...
37.5K
Quantitative Analysis
614
Quantitative analysis is a technique for measuring the amount of specific constituents in a sample. When the sample's composition is unknown, qualitative analysis is performed first to identify its components, which ensures that the correct substances are measured during the quantitative phase.
In quantitative analysis, two key measurements are made: the sample quantity and a property proportional to the amount of the analyte (the substance being analyzed). This forms the basis of the...
In quantitative analysis, two key measurements are made: the sample quantity and a property proportional to the amount of the analyte (the substance being analyzed). This forms the basis of the...
614
Quantifying and Rejecting Outliers: The Grubbs Test
2.1K
Sometimes, a data set can have a recorded numerical observation that greatly deviates from the rest of the data. Assuming that the data is normally distributed, a statistical method called the Grubbs test can be used to determine whether the observation is truly an outlier. To perform a two-tailed Grubbs test, first, calculate the absolute difference between the outlier and the mean. Then, calculate the ratio between this difference and the standard deviation of the sample. This...
2.1K
