Genetic clustering of depressed patients and normal controls based on single-nucleotide variant proportion

This manuscript version is made available under the CC-BY-NC-ND 4.0 license http://creativecommons.org/licenses/by-nc-nd/4.0/ This author accepted manuscript is made available following 12 month embargo from date of publication (Nov 2017) in accordance with the publisher’s archiving policyBackground Genetic components play important roles in the susceptibility to major depressive disorder (MDD). The rapid development of sequencing technologies is allowing scientists to contribute new ideas for personalized medicine; thus, it is essential to design non-invasive genetic tests on sequencing data, which can help physicians diagnose and differentiate depressed patients and healthy individuals. Methods We have recently proposed a genetic concept involving single-nucleotide variant proportion (SNVP) in genes to study MDD. Using this approach, we investigated combinations of distance metrics and hierarchical clustering criteria for genetic clustering of depressed patients and ethnically matched controls. Results We analysed clustering results of 25 human subjects based on their SNVPs in 46 newly discovered candidate genes. Conclusions According to our findings, we recommend Canberra metric with Ward's method to be used in hierarchical clustering of depressed and normal individuals. Futures studies are needed to advance this line of research validating our approach in larger datasets, those may also be allow the investigation of MDD subtypes. Limitations High quality sequencing costs limited our ability to obtain larger datasets

A simple proof of the triangle inequality for the NTV metric

Dress A, Lokot T. A simple proof of the triangle inequality for the NTV metric. Applied Mathematics Letters. 2003;16(6):809-813.We study a certain metric d on R-n that was recently introduced by Nieto, Torres and Vazquez-Trasande [1], give a simple proof for the triangle inequality, and describe when exactly d(p, q) = d(p, r) + d(r, q) holds for some p, q, r is an element of R-n. Remarkably, one has d(p, q) = 2 for some p, q is an element of R-n if and only if 0 = p + q not equal p holds, in which case, one has also d(p, q) = d(p, r) + d(r, q) for all r is an element of R. (C) 2003 Elsevier Science Ltd. All rights reserved