Introduction to Bioinformatics 1
Introduction to Bioinformatics. LECTURE 8: Whole genome comparisons * Chapter 8: Welcome to the Hotel Chlamydia 2
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 8. 1 Uninvited guests * Symbionts: organisms that live together in a beneficial relation * E. coli and Human: receives nutrients, gives vitamin K * Numerous examples in Nature: flowers and bees, tickbird and rhinoceros, pea aphid and Buchnera, mitochondria and Eukaryotes 3
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 8. 1 Uninvited guests * Some symbionts have moved permanently into the cells of the host * They have become entirely dependent on the host to provide them with nutrients, oxygen, specific proteins … * In the process they have lost many genes necessary to produce such products themselves * As a result, intracellular obligate symbionts have the smallest genomes – both in total size as in number of genes 4
Introduction to Bioinformatics 8. 1 – UNINVITED GUESTS Chlamydia trachomatis * Chlamydia trachomatis is an intracellular symbiont that gives no benefit to the host : it is a parasite * C. trachomatis is the most common sexually transmitted disease with +/- 3 M new infections per annum in the USA * It has lost the ability to produce many biochemical products and must live in specific cells in the human (hence the characterisation of: obligate endo-symbiont) 5
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS Chlamydia trachomatis 6
Introduction to Bioinformatics 8. 1 – UNINVITED GUESTS Chlamydia pneunomia * Chlamydia pneumonia is a related bacterial parasite of the human respiratory tract : it causes pneumonia and bronchitis * Like C. trachomatis it has a very small genome ~ 1 Mb 7
Human respiratory tract Chlamydia pneunomia 8
Introduction to Bioinformatics 8. 1 – UNINVITED GUESTS Chlamydia pneunomia Phylogenetic analysis of the parasitic lifestyle of Chlamydia shows that it dates back to 700 Myrs with the emergence of Eukaryotes * This is the same date as the pure symbiontic lifestyle of mitochondria 9
Introduction to Bioinformatics 8. 1 – UNINVITED GUESTS Whole genome comparisons * In this lecture we study the problems involved with the comparisons of entire genomes * Because of its very small genome Chlamydia are a perfect case study * Moreover, Chlamydia shows a high conservation of the order of the genes and virtually no horizontal gene transfer. 10
Introduction to Bioinformatics 8. 1 – UNINVITED GUESTS Hotel Chlamydia ‘Hotel’ Chlamydia is not so much that we are a living hotel for the Chlamydia (which is true), but that genes are guest in hotel Chlamydia – guests that come in, reshuffle rooms, move out, pass along … 11
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 8. 2 Patterns of genome evolution * Genome comparison looks at the differences between the entire set of genes between two organisms * This provides insight in evolution and function of genes * Single nucleotide polymorphisms form the bulk of the genetic variability * Also rearrangement and shuffling of genes: inversion, duplication, translocation 12
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 8. 2 Patterns of genome evolution * Often translocation between two organisms: horizontal gene transfer * Some 20% of E. coli‘s genes derive from horizontal transfer * Chromosomes can break apart and or stick together * Whole genomes can be duplicated → polyploid individuals * This is the basis for new functions as these extra genes are free to evolve 13
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 14
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 8. 3 Beanbag genomics * Genome = beanbag of genes + junk-DNA * Comparison of whole genomes is more than comparing individual genes * inversions, transpositions, duplications, deletions, chromosomal rearrangements * Therefore an alignment of entire genomes will not work 15
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Basic mechanisms of gene evolution 16
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8. 3 - BEANBAG GENOMICS : Comparison of two genomes gene evolution an alignment of entire genome will not work ! GAC ACTTTTTGG GGG TATATA CATGTAGTAC AAATAATCG AACCCCCG inversion duplication transposition deletion Therefore we have to break the problem in smaller pieces and build it back up … … with multiple single-gene analysis 18
8. 3 - BEANBAG GENOMICS : Comparison of two chromosomes chromosome evolution GAC ACTTTTTGG GGG TATATA CATGTAGTAC AAATAATCG AACCCCCG AAATAAT CATGTAGTAC GGG TATATA GAC ACTTTTTGG CG Splitting into new chromosomes Reshuffling of genes over the chromosomes 19
8. 3 - BEANBAG GENOMICS : Comparison of two genomes * STEP 1: Find which genes are present in both * STEP 2: use ORF-finder with threshold of 100 codons * EXAMPLE: Chlamydia trachomatis and C. pneumonia : --- Organism ----- size (nt) --- ORFs --C. trachomatis 1 042 519 916 C. pneumonia 1 229 853 1048 E. coli 5000 20
8. 3 - BEANBAG GENOMICS : Comparison of two genomes --- Organism ----- size (nt) --- ORFs --C. trachomatis 1 042 519 916 C. pneumonia 1 229 853 1048 E. coli 5000 * Intracellular symbionts like CT and CP have lost many genes: they parasite on their host and ‘steal’ the gene-products * CT lives in urinary tracts and CP lives in respiratory tracts * The differences in their genomes tells us something about the function of their retained genes * What are suitable algorithmic methods for comparing lost 21 and gained genes in Chlamydia?
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Similarity on a genomic scale Similarity between pairs of genes informs about: * Blocks of conserved gene order * Changes in size of gene families * Nucleotide substitutions between orthologous genes 22
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Central idea for genomic comparison Define the similarity scores between genomes as: * Nucleotide sequences of all genes found in both genomes * Fill out a matrix with alignment scores between each possible pair of sequences * For the Chlamydiae CT and CP this is a 1048 x 916 matrix * Use Needleman-Wunsch or BLAST to compute similarity scores 23
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Identifying orthologous and paralogous genes * Use genome similarity matrix to distinguish between paralogs and orthologs * remember: homologs are genes that have a common ancestor, orthologs arise as homologs evolve in sisterspecies; paralogs arise from duplication and subsequent specialisation * Result of evolution of homologs and paralogs: no one-toone relationship, but (many/one)-to-(many/one) 24
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Reciprocal similarity * Recognition of orthologs: Best Reciprocal similarity Hits (BRHs) * A pair of ORFs is a BRH if it is the best match between the two genomes (using alignment scores) * possible: ORFs without BRH * possible: ORF with ortholog in other species and a paralog in the same species. 25
8. 3 - BEANBAG GENOMICS : Comparison of two genomes EXAMPLE 8. 1: Homology in Chlamydia * Similarity matrix → BRH → orthologs * With threshold = 100 codons we find 1964 ORFs (CT: 916 and CP: 1048) * Among these 1964 ORFs are 728 ortholog pairs * Also 126 pairs of paralogs (CT: 56, CP: 70) * These paralogs are more similar to each other than to orthologs → result of duplication after the species split * The remaining 13% (=253 ORFs) perhaps older paralogs that have been lost in the other species due to specialisation 26
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Identifying gene families * Defining a gene family is a tricky thing; at some high level all genes are ‘family’ * The very first DNA-based organism, ancestral to all present living beings, had a set of genes. * All (? ) genes have derived from these ancestral genes through duplication and subsequent specialization. * Genes that cooperate tend to move close together (Dawkins: like rowers in a rowing boat) 27
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Identifying gene families * Practical solution: only consider genes that are > 50% similar; they are ‘closely’ related and probably have a similar ‘function’ * Method for finding ‘similar’ genes: clustering * Draw-back: all clustering methods have some degree of arbitrariness * Cluster both genomes simultaneously, then count #genes in each cluster (=gene family). 28
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Identify gene families with Hierarchical Clustering * input: genome similarity matrix d * method: cluster d with NJ- or UPGMA-algorithm to group the genes in families * This is called Hierarchical Clustering (HC) 29
Hierarchical Clustering on both Chlamidiae Application of HC on Chlamydia reveals a large number of small gene families and a small number of large families 30
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Similar function of gene families Largest gene families in C. trachomatis and C. pneumonia: ---- CT ---- CP ------ Function --------------12 12 ABC transporters 6 15 G family outer membrane protein 9 10 Function not known ABC transporters are transmembrane proteins with binding sites on both sites : major role in transport in/out the cell They are very old: they are (near) identical in all organisms 31
ABC transport ATP-binding cassette ABC transporter · Hydrophobicity 32
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Schematic of the E. coli vitamin B 12 importer system. 33
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Snapshots of pore formation in the bilayer, with an applied field of 0. 5 V/nm in the presence of 1 M Na. Cl 34
8. 3 - BEANBAG GENOMICS : Comparison of two genomes Alternative approaches to finding orthologs * Clustering of genes in families has some arbitrariness * Ortholog genes are separated by a speciation event * Thus, a phylogenetic tree is also a useful metaphor * A phylogenetic tree is a better representation, but it is less amenable to an automated analysis 35
Introduction to Bioinformatics LECTURE 8: WHOLE GENOME COMPARISONS 8. 4 Synteny * In Section 8. 3 the emphasis was on analysis of genes, here the emphasis is on chromosomes * syn- = together, tenia = ribbon, band, * synteny : the relative ordering of genes on the same chromosomes 36
Introduction to Bioinformatics 8. 4 – SYNTENY Major mechanisms of reshuffling of synteny are: inversions and transpositions Noise on synteny is caused by: insertions, duplications, and deletions Blocks of synteny: long stretches of DNA where the relative ordering of orthologous genes is conserved Synteny allows for annotation of non-coding sequences and identification of homologous intergenetic regions 37
Introduction to Bioinformatics 8. 4 – SYNTENY 38
Introduction to Bioinformatics 8. 4 – SYNTENY Cat on Human Conserved synteny map 39
Introduction to Bioinformatics 8. 4 – SYNTENY Visualising Synteny Dot-plot: * x-axis = position on genome_1, * y-axis = position on genome_2, * For a homologous gene with genome_1 -position x, and genome_2 -position y: put a dot ‘*’ on (x, y) 40
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Introduction to Bioinformatics 8. 4 – SYNTENY Chlamidia Synteny The high level of conservation in Chlamidia is remarkable but typical for all intracellular symbionts For instance Buchnera aphidicola, a intracellular symbiont in pea aphids, has retained synteny for 50 million years. 42
Introduction to Bioinformatics 8. 4 – SYNTENY Buchneria, a endosymbiont in pea aphids, has retained synteny for 50 million years. pea aphid Buchneria aphidicola 43
Introduction to Bioinformatics 8. 4 – SYNTENY SYMBIOSIS BETWEEN PEA APHIDS AND BUCHNERA Plant sap contains little protein and aphids cannot produce ten essential amino acids The required amino acids come from their symbiotic friends, the bacterium Buchnera aphidicola. 44
Introduction to Bioinformatics 8. 4 – SYNTENY SYMBIOSIS BETWEEN PEA APHIDS AND BUCHNERA The symbionts genome reflects this biosynthetic activity. Buchnera aphidicola carries the two genes trp. EG for tryptophan synthesis. Each bacterium contains three or four plasmids that contain four tandem repeats of these genes, resulting in 12 to 16 copies of trp. EG. Thus, the symbionts supply the host with the essential amino acids and receives free nutrients and shelter. 45
Introduction to Bioinformatics 8. 4 – SYNTENY The relation between pea aphids and Buchnera aphidicola is very old …. 46
Introduction to Bioinformatics 8. 4 – SYNTENY Endosymbionts and Synteny Buchnera aphidicola has retained synteny for 50 million years. IN GENERAL: The cloistered lifestyle of endo-symbiotic organisms shields them from viruses and other bacteria that may induce gene rearrangement 47
Introduction to Bioinformatics 8. 4 – SYNTENY Homologous intergenic regions and ‘phylogenetic footprinting’ Intergenic regions are not selected for → fast evolution Non-protein coding regions of the genome that are conserved are suspicious: they may be RNA-coding or regulatory sequences Using syntenic coding regions as anchors we can find intergenic regions that are highly conserved. This is called: Genetic Footprinting 48
intergenic regions ORF CT 671 -CP 1949 ORF CT 672 CP 1950 ORF CT 673 -CP 1951 49
Introduction to Bioinformatics 8. 4 – SYNTENY A metric for the syntenic distance Two genomes can be formed by many smaller syntenic blocks rearranged by inversions or transpositions Can we define a metric for the syntenic distance of these two genomes? We are not interested in nucleotide differences but in the number of genomic rearrangements that separate the two species. METRIC = smallest number of operations (=inversion or transposition) that transform one genome into the other 50
Introduction to Bioinformatics 8. 4 – SYNTENY EXAMPLE: Sorting by reversals As an example let us consider only the case of inversions “sorting by reversals” = minimum number of inversions to transform one genome into the other Algorithm: given a permutation of N numbers find the shortest series of reversals that can sort the back into their original order 51
Introduction to Bioinformatics 8. 4 – SYNTENY EXAMPLE: Sorting by reversals 3 2 1 4 8 7 6 5 9 1 2 3 4 5 6 7 8 9 2 reversals → syntenic distance = 2 52
Introduction to Bioinformatics 8. 4 – SYNTENY Sorting by reversals NOTE: In practice we do not know the original genome, so we select either one of the two as ‘the’ standard 53
Introduction to Bioinformatics 8. 4 – SYNTENY SIMPLE REVERSAL ALGORITHM STEP 1: designate one sequence as the standard s and the other as t STEP 2: i=1, increase(i) until s(i) ≠ t(i) or i=length(t) STEP 3: j=i; increase(j) until t(j) = s(i), reverse(t(i: j) STEP 4: i=j+1; if i=length(t), stop, else goto STEP-2 54
Introduction to Bioinformatics 8. 4 – SYNTENY SIMPLE REVERSAL ALGORITHM QUESTION: Can this algorithm solve overlapping reversals? REMARK : Involving transpositions is even more complex … 55
END of LECTURE 8 56
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