Genomics 2011 lecture 2
- 1. C. elegans Genetics C.elegans has 2 sexes, self fertilizing hermaphrodites and males. Sex determined chromosomally - XX-hermaphrodite, X-male. Diploid for 5 autosomes. Standard classical genetic techniques can be applied. Life cycle – Zygote to adult ~3 days. Grow on petri dish – they eat bacteria. Can store them frozen in liquid nitrogen indefinately. Why might the hermaphrodite sex be useful for genetics?
- 2. Chromosome I Genetic mapping. Left arm m.u. bli-3 m.u. = map unit. -15 egl-30 Genetic mapping – recombination. mab-20 -10 1 m.u. is 1% recombination per meiosis. -5 fog-1 unc-73 unc-57 Central 0 dpy-5 dpy-14 cluster fer-1 5 lin-11 unc-29 unc-75 Parent Recombinant 10 unc-101 15 20 glp-4 fog-1 + fog-1 + 25 unc-54 glp-4 + + glp-4 Right arm
- 3. We want to understand how life works – at the molecular level. We had mutant genes with informative phenotypes. The mutated genes were mapped onto linkage groups – chromosomes. What kinds of proteins do these genes encode and how do these proteins function? In 1983, identifying the molecular sequence of a gene defined by mutation was a complicated and time consuming business, even in the worm. If we only new the sequence of the genome!
- 4. As the term applies to recombinant DNA, what is a clone? Starting with DNA extracted from any organism, Vector How can you take that and get one single fragment into a vector and grow billions of copies of that single “cloned” molecule? Cloned DNA insert
- 5. C. elegans Genome Project unc-101 unc-75 unc-54 unc-73 mab-20 lin-11 dpy-5 glp-4 fog-1 egl-30 fer-1 Mutants - function bli-3 Genetic map 25 10 15 20 0 5 -15 -10 -5 Chromosomes AACGTTCCACG....... DNA sequence – genes Cloned DNA and proteins fragments Identify DNA sequences corresponding to genes defined by mutation.
- 6. If you wanted to clone sections of chromosomes for sequencing, how many copies of each chromosome would you start with? DNA Of the order of millions – millions of copies of each chromosome
- 7. Purified genomic DNA Fragment the chromosomal DNA – either restriction enzyme or mechanical shear.
- 8. Cloning methods used by the C. elegans genome project Cosmid clones – ~ 40 Kb insert size – Genomic Library. Cosmid cloning vector Linearised cosmid vector Random fragments of genomic DNA – Drug resistance marker E. coli origin of replication millions of them. cos site Useful restriction sites DNA Ligase Long concatenates of cosmid vectors interspaced with random fragments of genomic DNA.
- 9. Mixed population “inserts” In vitro lambda packaging extracts Lambda Terminase Other phage proteins COS sites in cosmid vector E. coli Critical step Phage “transfects” single cosmid into an E. coli cell.
- 10. CLONING This is a clone Cells are plated onto medium with antibiotic selection. Cells grown up to form bacterial colonies. Insert X Each colony is derived from a single transfected cell. Each colony is a clonal population. E. coli - clonal population with a single cosmid clone – single genomic DNA fragment. Billions of copies of one cloned insert. Freeze it for storage. Purify cosmid DNA. Sequence the insert. Solid medium on plates Liquid culture Sub-clone fragments etc.
- 11. Started with many millions of different fragments of chromosomal DNA in one tube. End up with potentially millions of CLONED fragments, each in a different E.coli colony – or culture.
- 12. We have got as far as random cloned fragments of genomic DNA. What next? Average cosmid insert size – 40 Kb C.elegans genome ~100.3 Mb = 100,300 Kb 100,300/40 = 2,507.5 i.e. ~2,500 cosmid clones could contain the entire C. elegans genome – but WOULD they?
- 13. In principle, 2500 cosmid clones could contain all the DNA of the C. elegans genome. Why not just start sequencing ~2500 clones picked at random? Imagine this: I give you a large and awkwardly shaped dice with 2500 faces, with a single number on each face, the numbers 1-2500. Roll the dice and write down the number on top. Repeat this – again and again and……. How many times would you have to roll the dice so that every face of the dice would have been on top at least once? ~ 4x2500 will give ~95% probability of any one side or DNA fragment, appearing. ~10x2500 raises probability to ~99%
- 14. The Golden Path What if you could identify clones that overlapped slightly with ones another? How can we get these clones? Cloned DNA fragments – moderate overlaps. With this approach you could sequence the entire genome by sequencing less than 5000 cosmid clones (2x2500)
- 15. Cosmid fingerprinting 1. Restriction digest of cosmid DNA. 2. Separate fragments according to size by gel electrophoresis. 3. Digitise the ladder of different sized DNA fragments obtained. Multiple common fragments – clones probably overlap. C. elegans genome project, ~17,000 cosmid clones fingerprinted. A B C Assembled into “contigs” – overlapping clones. “Contig” ~17,000 random cosmid clones A Fingerprinting ~700 contigs B C D C.elegans genome 100 Mb ~2,500 cosmid clones
- 16. 700 contigs. What is the minimum number of contigs the C. elegans genome could be contained in? Or – how would we know when we had succeeded in joining all the contigs? A method of filling the gaps – joining the contigs – was needed.
- 17. YACs – Yeast Artificial Chromosomes DNA inserts of ~100 kb – 2 Mb. Grown in yeast. Clonal growth of yeast colonies, much like cosmids in E. coli. YAC DNA separated by pulsed-field gel electrophoresis. C. elegans genome is ~100 Mb. Cosmid clones – approximately 40 kb inserts. YAC clones – select average 500 kb inserts. ~2500 cosmid clones would permit 1x coverage of the genome. ~200 YAC clones would permit 1x coverage of the genome.
- 18. ~17,000 fingerprinted cosmid clones – ~700 unlinked contigs. Cosmid clone contigs ? ? 6 Chromosomes AACGTTCCACG....... unc-101 unc-75 unc-54 unc-73 mab-20 lin-11 dpy-5 glp-4 fog-1 egl-30 fer-1 bli-3 Genetic map 15 10 20 25 0 5 -15 -10 -5
- 19. Joining up the contigs Contig X Contig Y YAC clone ~700 contigs – grids of representative cosmid clones. • Large YAC clones (> 1Mb). • Purify YAC DNA – (PFGE). • Radio-label YAC DNA. • Hybridise to cosmid grid. • Expose to X-ray film. Linked cosmid clones
- 20. unc-101 unc-75 unc-54 unc-73 mab-20 lin-11 dpy-5 glp-4 fog-1 egl-30 fer-1 bli-3 Genetic map 10 15 20 25 0 5 -15 -10 -5 A physical map of the genome - the “Golden Path” – chromosomes represented in ordered overlapping clones or “clone contigs”. YACs Cosmids The Sequence of The Genome
- 21. Sequencing the C. elegans Genome Individual cosmid clone. Randomly fragmented and shotgun cloned into sequencing vectors. Generally smaller insert size is best for primary sequence determination – 2-10 Kb. Sequence of cosmid or YAC etc, determined and compiled in silico. Finishing – directed cloning to fill in any gaps. Check for overlap of sequence with overlapping cosmids.
- 22. Gaps between cosmid contigs ~20% of genome. Most of these gaps were not random. They contained regions that could not be cloned in cosmids. YAC clones covering most of the gaps. YAC DNA shotgun cloned into M13 or plasmid vectors. Most of the DNA contained in these awkward regions was successfully sub-cloned into small insert size vectors, and sequenced. The sequence as published in December 1998 was generated from: 2527 cosmids, 257 YACs, 113 fosmids, 44 PCR products.
- 23. C. elegans cosmid K06A5, 24323 bp. Flat sequence file –3955 bp shown. >CEK06A5 acaagagagggcgcctcggccgtatgttgaatgggagatcgatggaaccgagacaacgagaaaaggaatagagacggagaaagagagagagagcgcgcgttgttggaaggatg aaaaagaaaaaagacatgagctgcttcacaagagcttggcgaaagcaaagggcaaagtgttgacagcttagtggtggtagttggatcttctctcctcgttctctgctcacaac tcgtctatcactcatatcacatttatttcccaatatcattttaacaacatcttccgatgcatgttcgtcaatattgcgcaaccactttgcaatattgtcaaaacttttcgcat ttgtgatatcgtaaaccagcataattcccattgctccgcggtaatatgatgttgtgattgtgtggaatcgttcttgtccagctgtgtcccagatttgtaatttaatctttttt ccttttaattcgatagttttaattttgaagtcgattcctgaatgaaaaaagaaaattattttgaaatcactagattctgaataaaaactaaccaatagttgagatgaatgtgg tgttaaaggcatcatccgaaaatctgtacagaatgcaagtttttccaactcctgagtcgcctattagcagcaatttgaagagcatgtcatacggtcggcgagccatttttctt ctgaaatgagaaaaagttgagaactaaagttgcacaaaagtaagagaaaagcacttgagtcatggcaaatagaacgaacactttgagatttcgaagaagttatcaagagttga caattggaagatatttggaagaactttctaatttttttctagttttccaaaattaggtttttgtcataaaatgttgtcaaagaaaaaacaggacaaaatagttaattgttgtt tccattataacaaaaaaaaatttgaacggagctattaacgcgtgcatgcgcaaatcacatcgattagctgtttctgggaaattctcgggaaaaggtgaacagcagctgctggc ttcctctgcgggtcacgaaaacacaaagagatcattataattgttatttggaaaggaagcgaatctaaaacgggtacaggtggacgtttattgatcgaaagtgctttttattt gaaattgaatggtgaactttgcaattttgtaatgcaaagtacgttatcagatggcatgagatgtgtgaagtgataaggaataaaatgtgaacgacatgttcaagaaactgtga tttttcaataatttgtgatgaaatattttaggaacagaaatgaacatattaattgatataaaaacaataggaacactaactcataattatgataggtgaatatcaaaatgtgc tagattttttgaagttaaaaaatacatttctaatattttttcaaataataagtttcagctgaaatttcagggtgatttcagaaagctatgttttgataaattgttttgaaaat taaaagaagctacagcaaaaaaaaattaaagagaacatcgctccctcgtagtgtataatttttgattatcgaaaaaaatgagtcaatgatgaaaaggaagtcgcaatctcaaa acttcaaaaatcaaaagaagccgttgcctctgtcatcaaaaattcagaagacaaggttgttgacaagggtcaattctcagtggtggagggcattgggcgtggtgaaatttttg aaggctagtgtggttggacctctactagatagacaaaacccccgaaatagacgtttaatttgatgagatggtggagaaagaaaaggactcattctctagatgatagagagacc agagatacagacaagagagggcgcctcggccgtatgttgaatgggagatcgatggaaccgagacaacgagaaaaggaatagagacggagaaagagagagagagcgcgcgttgt tggaaggatgaaaaagaaaaaagacatgagctgcttcacaagagcttggcgaaagcaaagggcaaagtgttgacagcttagtggtggtagttggatcatgtgtttttatgttt ccggtgggagaaggttcaacaaaaaatgaaaagaaaaagttcaagcggcatgaatcattctgagtttaaaacaaaattattgcgaaaattaatattaaaaccttttcacaaaa cttcaagctaatctgttcatgaaaatttgaataatagttttttcccacctatttagaattaacttcatattaacgaaattaattaacgaatcgaaaattatgacttttcagaa tcatctgaagttttttcacattccatgctgcatggaataatttgatcctggaatcgatatgtttttatggtatactttttaaccttcaatttagctggaaaagtatggaataa ataattcccgaagctatgtacatatatgtagaattattgaatgattgtgagaacaacttgactttagcttgagtaggaatcggaatggctatcgaccgatcaacacttaggat tgtaagaatggcagtaagaatatattgaagaaagaatgtttgttcataggaagagaaagagtattgcgaaatcatcatcgcccactttagaatggacgggcggtgagcggaca tagagaattgtgaatgactaatgcttttgcagaatctagggcaaaatcgtaggaacaaacaattgtaatacggagaaaacaatcatatcgatcgatgatcatggagaaaaatg tgatttaagtgagtagacttggaaaaattaataaaagcatgaattgtcgatatttttcatttattttcattataaagctctttaaaaacaaattaaatattgagaatggcttc gaagaatattgtttcaaatatgttcaatggtgacaccttgcggataaaattaatgtaaaaatcatggaacacagattcactgatatctcattatctcaagcagtgtaattaga gattttttggaacaattattttataaaactataaataaaccgtttatactactcaaagccaaatattcaagctattaccattttttttctaactaattcttgagcaattaaag tattccccagtttttattttgcaacgactccaggcaaacacgctccgttgcacttgccgccaaggcgttgcattcaaatcagagagacatctcattccgatttctgtttttct tccaataaacggtattttatgcctaatgggtgatacggaaattgttcctcttcgagtacaaaatgtacttgatagcgaaatcattcgtctcaacttgtggtccatgaaggtaa ctgtctagtttttttaagttttcatgatttcaatatttttacagtttaacgcgaccagtttcaaactcgaaggttttgtgagaaatgaagaaggcactatgatgcagaaagtt tgttccgaatttatttgtgtaagtcgagaaacatattcgtcaacaattttcattaaatattcagagacgcttcacttctacgttgcttttcgatgtttccggacgtttcttcg acttggtcggacagattgatcgggaatatcaacaaaaaatgggaatgcctagtagaattattgatgaattttcaaatggaattcctgaaaattgggccgaccttatctattcc tgcatgtcagccaaccaaagaagcgcacttcgccctatccaacaggctccaaaagaaccaattagaactagaacagaaccaattgttacgttggcagatgaaaccgagctaac tggaggatgccagaaaaattccgaaaacgagaaagaaaggaacagacgtgagcgtgaagaacagcaaacaaaggaacgtgagagaagattagaagaagaaaaacaacgacgag atgctgaagctgaggctgaaagaaggcgaaaagaagaggaagagctggaagaagctaattacacccttcgtgctccgaaatctcagaacggcgagccaatcactccgataaga
- 24. Genome sequence of C.elegans. Sequence of entire genome. Sequence of cDNA clones. Approximately 19,500 predicted protein coding gene sequences. Large number of various kinds of functional RNAs – not discuss further. For this lecture – focus predicted proteins. Gene prediction? How? Science, December 1998.
- 25. Computer based predictions GENEFINDER Biases in coding sequence - in C. elegans non-coding is AT rich. Splice site signals, initiator methionines, termination codons. Likely exons and probable/possible splice patterns. • Evidence that a prediction is correct? • Homology with genes in other organisms – homologues. • Known protein families. •Experimental evidence.