Comparative Genomics

Abstract

Over the last couple of decades, numerous genetic and physical maps have been developed for a wide range of species. These data have led to the development of the field of comparative genomics, in which we analyse characteristics of whole genomes, in contrast to the analysis of single genes in comparative genetics. By comparing homologous markers between species, we can get a feel for their relative distributions on the chromosomes of the respective species. Such information also enables us to deduce segments of chromosomes where the gene content, and sometimes also the order of the genes, is similar in two or more species. More recently, DNA sequences of whole genomes have become available, enabling us to compare genomes at both the sequence and the gene levels. Furthermore, these datasets have provided us with more detailed information with which to study the processes involved in genome dynamics. The promise of the post-genome era means that we will see many fully sequenced genomes within the next decade and we should develop analytical methods that are mature when significant quantities of data become available. However, not all species will be sequenced, at least in the near future, and so we should continue to develop methods of analysis that are appropriate for both types of data. In this chapter, we give a flavour of the types of comparative genome analysis that are currently possible. We highlight particular problems such as phylogenetic inference and the use of maps to compare genomes. Finally, we look at problems that should be tackled to enable us to make the most of the emerging genomic data.

GWAS of human bitter taste perception identifies new loci and reveals additional complexity of bitter taste genetics

“…SNPs were further filtered for Hardy–Weinberg equilibrium P -value (>10 −5 ), Q -score (>0.2) and minor allele frequency (>5%). Log-transformed detection thresholds (caffeine, quinine, PROP) were normally distributed according to the Kolmogorov–Smirnov test ( 26 ), with P = 0.73, 0.22, 0.11, respectively.…”