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Caenorhabditis elegans

Caenorhabditis elegans. Anton Kapliy February 17, 2009. Sydney Brenner (1927 - ). The Genetics of Caenorhabditis Elegans , 1973. South African biologist (originally chemist) D.Phil from Oxford Extensive work in molecular biology Nobel Prize in 2002

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Caenorhabditis elegans

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  1. Caenorhabditis elegans Anton Kapliy February 17, 2009

  2. Sydney Brenner (1927 - ) The Genetics of Caenorhabditis Elegans, 1973 • South African biologist (originally chemist) • D.Phil from Oxford • Extensive work in molecular biology • Nobel Prize in 2002 Established C. Elegans as a model organism to study genetics and cell development. In his honor, another worm was named C. Brenneri

  3. Meet C. Elegans Small nematode worm (roundworm) Natural habitat: soil Length: ~1 mm Food: E.Coli Life cycle: ~3 days Cellular structure: ~1000 eukaryotic cells; ~300 neurons First multi-cellular organism to have its genome sequenced

  4. C. Elegans lifecycle

  5. Handling Isolated from soil (see first picture on the right) Cultures reside on small plates Covered w/ E.Coli lawn that provides nutrition for the worms Preparation of monoxenic cultures Germs & worms are killed with a chemical, but eggs survive Long storage via freezing Early larvae survive freezing for weeks Individual worms can be examined Lifted with paper strips and studied under the microscope.

  6. A plate with C. Elegans

  7. I II III IV V X I I II II III III IV IV V V X X Refresher: diploid cells Gametes Regular cell C. Elegans is a diploid organism with 6 pairs of chromosomes One from mommy: ovum A pair of homologous chromosomes zygote mitosys One from daddy: sperm 5 autosomes 1 sex chromosome

  8. XO sex-determination system Ovum Sperm Ovum Sperm Sex is determined by the number of X chromosomes: Note that the ovum always contains an X chromosome, but the sperm may or may not. XX XO

  9. (A+a)(A+a) = AA + aa + 2Aa 2(A+a)(A+a) = 2AA + 2aa + 4Aa 4(A+a)(A+a) = 4AA + 4aa + 8Aa A small twist: Hermaphrodites An XX worm produces both ovum and sperm Thus, it can self-fertilize to produce progeny In the wild, self-fertilizing hermaphrodites tend to homozygosity: homologous chromosomes contain identical alleles egg ovum -a-- -a-- -a-- -A-- Consider a pair of chromosomes heterozygous in trait A/a: -a-- -A-- -a-- -A-- x = -A-- -A-- -A-- -a--

  10. Hermaphrodites Figure A: Arrows point to head, tail, and vulva Figure B: Anus Figure D: An egg leaving the vulva

  11. Males In the progeny of self-fertilizing hermaphrodites, there is an occasional male due to nondisjunction (<0.1%) This fan-shaped tail is the male’s reproductive organ. It also allows to distinguish males on the plate. Males can mate with hermaphrodites, and their sperm has advantage over hermaphrodite’s own sperm.

  12. Big picture • Induce a mutation in one of the chromosomes • Create a line homologous in this mutation (aa) • Isolate and study several different phenotypes • Create a genetic map of the worm • See why C. Elegans is a great model organism ---- -a-- Suppose “a” is a recessive mutation: “Dumpy” worm -a-- -a--

  13. Inducing mutations Sexual timeline of a hermaphrodite worm: Sperms Zygote Ovums Since sperms are already produced, only ovums contribute a mutated chromosome Introduce a powerful mutagen: EMS - ethane methyl sulfonate

  14. Properties of mutations • EMS works at DNA level by producing point mutations: G/C > A/T • EMS is a very powerful mutagen: mutation rate = ~5x10-4 per gene per generation With ~100 identifiable genes, this means 1 in 20 worms mutate • Most mutations in C. Elegans are recessive In this discussion, I will ignore dominant/semidominant

  15. Mutation phenotypes

  16. Blistered phenotype on a plate Blistered worm

  17. Site of mutation Isolating recessive alleles ---- ---- P 1. Start with wild type hermaphrodite -a-- ---- 2. Induce mutation in the ovum -a-- ---- -a-- ---- F1 3. Let the baby self-fertilize -a-- -a-- -a-- ---- ---- -a-- ---- ---- F2 4. Examine the progeny ¼ mutant phenotype ¾ wild phenotype 5. Pick out homologous mutants

  18. Autosomal vs sex-linked mutations Cross the homologous mutant with wild-type males & examine progeny CASE II: X-linked mutation CASE I: autosomal mutation X IV X IV -a-- -- -a-- -- ---- a- ---- a- Homologous mutant + + + ---- ---- -- ---- ---- -- Wild male (1 X chromosome) = = = -a-- ---- -- ---- ---- a- Progeny male always gets its X chromosome from mother ♂ ♂ -a-- -- ---- -- ---- a- ---- -- Progeny female gets one X chromosome from each parent ♀ ♀

  19. Experimental results: phenotypes As mentioned before, virtually all mutations are recessive. Located means that the mutation was mapped on one of the chromosomes Note that there are several autosomal blistered mutants. What are they?

  20. Genetic complementation -a-- -a-- -a-- -a-- 10 plates with homologous mutants with the same phenotype: • Given 10 independent mutants with blisteredphenotype: • Do we know that the same gene is responsible in each case? • Or could multiple genes cause the same phenotype? • If different genes cause the same phenotype in two mutants, they are said to show genetic complementation • To find out: use Cis-Trans test ---? ---?

  21. Complementation & cis-trans test Cross the homologous mutant with wild-type males & examine progeny CASE II: different genes CASE I: same gene IV IV -a-- -a-- -a-- -a-- 1st homologous mutant (male) + + + ---b ---b -a-- -a-- 2nd homologous mutant (female) = = = -a-- ---b -a-- -a-- Progeny has different phenotypes! Still exhibits mutation! Restored wild phenotype! Cis-trans test allows to group mutants of the same phenotype into complementation groups

  22. Cis-trans test in C.Elegans Generic cis-trans test requires that mutant males mate with hermaphrodites. But: mutated males with many phenotypes (e.g., uncoordinated) can’t mate! -a-- -a-- -a-- -a-- 1st homologous mutant (female) + + + ---- ---- ---- ---- Wild-type male = = = -a-- ---- -a-- ---- Progeny males (wild phenotype!) + + + -a-- -a-- ---b ---b 2nd homologous mutant (female) = = = -a-- -a-- ---- -a-- -a-- ---b ---- ---b Examine presence of mutated phenotype in progeny males! or or mutant wild wild wild

  23. I II III IV V X Next step: linkage groups 3 plates w/ blistered homologs, corresponding to 3 different genes 10 plates w/ blistered homologs ---c ---c --b- --b- --b- --b- -a-- -a-- -a-- -a-- cis-trans --b- --b- -a-- -a-- ---c ---c --b- --b- --b- --b- ---c ---c -a-- -a-- -a-- -a-- The cis-trans test performed on 10 independent mutations tell us how many are truly independent – that is, caused by different genes. In the example above, we reduced the problem to 3 independent mutations (genes). Next step is to determine the linkage groups of these genes. Genes in different linkage groups segregate independently (acc. to Mendel) In hindsight, we expect to see six linkage groups, mapping to 6 chromosomes:

  24. Aside: Cis and trans configurations Consider a worm that has two recessive mutations: “u” and “d”, but is exhibiting wild phenotype. There are two ways this could happen: cis trans ---- u--d u--- ---d Chromosome I is mutation-free Chromosome II carried both u & d Chromosome I is carries u Chromosome II carries d Both are wild type, since u and d are recessive! These configurations are called double heterozygotes Next slide shows how we can construct one.

  25. Constructing a trans heterozygote u--- u--- ---d ---d Start with two homozygous mutants: Two types of progeny: u--- ---d (a) Mutant1 ♀ Mutant2 ♀ Wild ♂ Baby ♂ u--- u--- ---- ---- u--- ---- ---d ---d + = + = ---- ---d (b) To filter out (b), let the progeny self-fertilize: u--- ---d u--- ---d - ¼ uu, ¼ dd, ½ wild Use + ---- ---d ---- ---d - ¼ dd, 1½ wild Discard +

  26. Meiosis in trans heterozygote Consider a pair of homologous chromosomes: u--- ---d What are the possible gamete configurations? u--- ---d u--d ---- These are regular gametes, when one chromosome in the pair entirely goes to the gamete These gametes resulted from chromosomal crossover, when the pair of parental chromosomes got mixed during meiosis:

  27. Progeny in trans heterozygote (I) u--- ---d P = probability of crossover 1-P P Sperms: u--- ---d u--d ---- ---- u--- u--d ---d u--- u--- ---d u--- u--d u--- u--- ---d ---d ---d ---- ---d ---d u--d u--d u--d ---- u--d u--- ---- ---d ---- u--d ---- ---- ---- u--- u--d Ovums: u--- 1-P ---d u--- ---d u--d P ----

  28. Progeny in trans heterozygote (II) u--- ---d We observe Mendelian ratio 9:3:3:1 1-P P Sperms: u--- ---d u--d ---- Wild Dpy Unc Wild Unc Wild Dpy Wild Dpy Unc Dpy Wild Wild Wild Wild Wild Unc Ovums: u--- 1-P ---d u--- ---d u--d P ----

  29. Locating linkage groups Consider self-progeny of trans heterosyzote: u--- ---d Pick out all dumpy worms and count how many are also uncoordinated: P(dpy) = (1-P)2 + 2P(1-P) + P2 = 1 – 2P + P2 +2P – 2P2 + P2 = 1 P(unc+dpy) = P2 Ratio = P(unc+dpy)/P(dpy) = P2 u- -- -- -d Suppose genes “u” and “d” are on different chromosomes: Then, they segregate independently, with P=0.5 If genes u and d are on the same chromosome, the measured ratio will quadratically diverge from 0.52 = 0.25 – making it a very sensitive test!

  30. Results: classification of linkage groups 1st chromosome 2nd chromosome Etc…

  31. Results: mapping of mutants We can guess the order of genes on each chromosome by using P, the recombination probability, as the yard stick:

  32. Intermission • Brenner’s paper establishes C. Elegans as a perfect model organism because: • Worms are easy to handle and quick to multiply • Availability of very potent mutagen • Hermaphrodites can maintain homozygous recessive alleles • Hermaphrodites can self-fertilize even with mutations that impair movement • Rare males allow to mix genetic traits • Actually, there is a lot more that can be done with C. Elegans – the final few slides summarize some of its interesting features & recent developments

  33. Ease of observation The worm is transparent, and we can see all of its ~1000 cells in a microscope

  34. Developmental biology It is possible to trace the fate of each cell in the growing worm

  35. Complete cell lineage Constant number of cells: 959 in hermaphrodite, and 1031 in male

  36. Programmed cell death (apoptosis) 131 cells in the developing worm embryo die by apoptosis in a predetermined way

  37. First complete genetic map 100 million base pairs ~20,000 genes

  38. One of the simplest nervous systems Nervous system consists of 302 neurons that form a small-world network Their interconnections have been completely mapped out

  39. Gene silencing via RNA interference Inject double-stranded RNA Enzyme dicer breaks dsRNA into a cascade of small-interfering RNA siRNA bind to another enzyme:RNA-induced Silencing Complex RISCs silence the matching sequence in the messenger RNA

  40. Some of the sources A couple of intro genetics textbooks http://www.wormbook.org/chapters/www_nematodeisolation/nematodeisolation.html http://www.wooster.edu/biology/wmorgan/bio306/C.elegans_Week3_Directions.html http://www.sanger.ac.uk/Projects/C_elegans/ http://www.wormbook.org/chapters/www_dominantmutations/dominantmutations.html http://www.wormatlas.org/handbook/anatomyintro/anatomyintro.htm http://www.wormclassroom.org/ge.html http://www.ncbi.nlm.nih.gov/books/bv.fcgi?indexed=google&rid=ce2.section.100 http://fruitfly4.aecom.yu.edu/labmanual/16a.html http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/C/Caen.elegans.html http://en.wikipedia.org/wiki/Caenorhabditis_elegans http://en.wikipedia.org/wiki/Apoptosis http://www.bio.unc.edu/faculty/goldstein/lab/movies.html http://www.loci.wisc.edu/outreach/text/celegans.htmlhttp://www.nematodes.org/teaching/devbio3/index.shtml http://www.translational-medicine.com/content/2/1/39/figure/F1

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