Caenorhabditis elegans

Sydney Brenner ( ) 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 The Genetics of Caenorhabditis Elegans, 1973

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

C. Elegans lifecycle

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.

A plate with C. Elegans

Refresher: diploid cells C. Elegans is a diploid organism with 6 pairs of chromosomes IIIIIIIVVX 5 autosomes1 sex chromosome A pair of homologous chromosomes IIIIIIIVVX zygote Regular cellGametes IIIIIIIVVX One from mommy: ovum One from daddy: sperm mitosys

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

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 (A+a)(A+a) = AA + aa + 2Aa 2(A+a)(A+a) = 2AA + 2aa + 4Aa 4(A+a)(A+a) = 4AA + 4aa + 8Aa -a-- -A-- Consider a pair of chromosomes heterozygous in trait A/a: -a-- -A-- -a-- -A-- -a-- -A-- -a-- x = eggovum

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

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

Big picture 1.Induce a mutation in one of the chromosomes 2.Create a line homologous in this mutation (aa) 3.Isolate and study several different phenotypes 4.Create a genetic map of the worm 5.See why C. Elegans is a great model organism “Dumpy” worm a-- Suppose “a” is a recessive mutation: -a--

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

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

Mutation phenotypes

Blistered phenotype on a plate Blistered worm

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

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

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?

Genetic complementation Given 10 independent mutants with blistered phenotype: –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 -a-- ---? 10 plates with homologous mutants with the same phenotype:

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

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

Next step: linkage groups 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: IIIIIIIVVX -a-- --b- ---c --b- ---c -a-- --b- ---c 10 plates w/ blistered homologs 3 plates w/ blistered homologs, corresponding to 3 different genes cis-trans

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: ---- u--d cis Chromosome I is mutation-free Chromosome II carried both u & d u d trans 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.

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

Meiosis in trans heterozygote u d u-----u--d Consider a pair of homologous chromosomes: What are the possible gamete configurations? 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:

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

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

Locating linkage groups 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 = 0.25 – making it a very sensitive test! Pick out all dumpy worms and count how many are also uncoordinated: P(dpy) = (1-P) 2 + 2P(1-P) + P 2 = 1 – 2P + P 2 +2P – 2P 2 + P 2 = 1 P(unc+dpy) = P 2 Ratio = P(unc+dpy)/P(dpy) = P 2 u d Consider self-progeny of trans heterosyzote: u d

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

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

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

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

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

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

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

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

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

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

Some of the sources A couple of intro genetics textbooks