Download Lecture 1: overview of C. elegans as an experimental organism

FALL 20 C. elegans lectures
Kaveh Ashrafi
Lecture 1: overview of C. elegans as an experimental organism
Conceptual framework:
Emphasis on applying genetics to study developmental biology, neurobiology,
physiology, & behavior
Genetics learning points:
Diploid genetics
Hermaphrodite genetics
General design for mutant hunts
What do you do once you have found mutants?
SNP mapping
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REREFERENCES:
Background on all things C. elegans:
WORMBOOK
http://www.wormbook.org/
Big picture/conceptual framework:
Horvitz HR. Worms, life, and death (Nobel lecture).
Chembiochem. 2003 Aug 4;4(8):697-711.
S. Brenner. 1988. Foreword. In The nematode Caenorhabditis elegans (ed. W.B. Wood
and the Community of C. elegans Researchers pp. ix-xiii. Cold Spring Harbor Laboratory
Cold Spring Harbor, New York. (Sydney’s thoughts on why he chose C. elegans)
A field is born:
***Brenner S.
The genetics of Caenorhabditis elegans.
Genetics. 1974 May, 77(1):71-94.
Genetic screening and SNP mapping:
Jorgensen EM, Mango SE
The art and design of genetic screens: Caenorhabditis elegans.
Nat Rev Genet. 2002 May;3(5):356-69. Review.
Wicks SR, Yeh RT, Gish WR, Waterston RH, Plasterk RH.
Rapid gene mapping in Caenorhabditis elegans using a high density polymorphism map.
Nat Genet. 2001 Jun;28(2):160-4.
Williams BD.
Genetic mapping with polymorphic sequence-tagged sites.
Methods Cell Biol. 1995;48:81-96. Review.
Online resources for C. elegans researchers:
C. elegans ANATOMY
http://www.wormatlas.org/index.htm
WORMBASE: CENTRAL REPOSITORY OF C. ELEGANS GENETIC INFO
http://www.wormbase.org/
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I.
OVERVIEW
Caenorhabditis elegans
C. elegans is a metazoan (an animal!): a free-living soil nematode found on rotting
organic matter. Size: microscopic, 1-1.5mm long, 80 microns thick. C. elegans is nonpathogenic to humans but it is related to some nasty human/plant pathogens.
1. ATTRACTIONS FOR GENETIC ANALYSIS
Ease of cultivation & maintenance
Can be cultivated on solid (agar based) media or in liquid (they swim). In the laboratory,
C. elegans are fed E. coli although aexenic media (synthetic media) is now also available.
Small enough such that thousands of animals can be kept on one plate. They move
around but won’t leave the plate. Small enough such that animals can be washed off of
Petri plates, pipetted, spun into a pellet in a microfuge, yet large enough that individual
animals can be handled using a pick.
Lines can be generated from a single hermaphrodite.
For long-term storage, whole animals can be frozen in glycerol stocks and maintained at
–80 and liquid nitrogen (>30 years!!).
Short lifecycle and large brood size:
Generation time: 3.5 days at 20°C {from egg through four larval stages (L1, L2, L3, L4)
to egg-laying adults}. An egg-laying adult lays about 300 eggs during its reproductive
phase. Time from fertilization to hatching is about ~12-14 hrs at room temp. The large
brood size and fast reproductive cycle is great for genetic analysis. Adult lifespan is ~2-3
weeks.
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2. BASIC ANATOMY:
Adult hermaphrodite have 959 somatic cells.
Basic organization of the body is two concentric
tubes separated by a fluid filled space
(pseudocoelom). Shape is maintained by internal
hydrostatic fluid pressure. Outer tube is covered
by a tough, collagenous, extracellular cuticle. At
each molt, the cuticle has to be shed and remade.
Cuticle is secreted from the underlying layer of
skin (hypodermis). Outer tube also contains the
nervous system, musculature, and excretory
(kidney like) systems. The inner tube consists of
the pharynx (heart like pump used for feeding),
the intestine and, in the adult, gonad.
3. ATTRACTIONS FOR DEVELOPMENT/CELL BIOLOGY, NEUROBIOLOGY,
& GENOMICS: Stereotypical development
Animals are transparent (great for lineage analysis, direct observation of GFP reporters).
The excellent optical qualities has allowed for many processes to be observable in living,
intact animals. Combined with an invariant cell lineage the ability to observe each cell
division led to elucidation of complete cell lineage of C. elegans (>120 cell types). In
turn, this knowledge provided a framework for asking some key, fundamental questions
such as: how do cells adopt their fates? who dies/who lives? how is the timing of
divisions determined? how do cells end up in the right place? Additionally, complete
neuronal wiring diagram known allowing for detailed analysis of neuro-circuitry.
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4. SEX
Two sexes: males and self-fertilizing hermaphrodites (modified female: makes and
stores sperm in L4, makes oocytes later).
Males can mate with hermaphrodites (hermaphrodites do not mate with other
hermaphrodites). Self-fertilized hermaphrodites lay ~300 eggs, mated hermaphrodites
lay up to ~1000 eggs. Hermaphrodites predominate the population in the lab.
XX=hermaphrodite, XO=male (no Y chromosome). Males arise from chromosome nondisjunction (1:500 to 1:1000 chance) or through mating (1:2 chance—see below).
5 pairs of autosomes (I, II, III, IV, V), one pair of sex chromosome (XX or X0). So, this
means that the somatic tissue is diploid !! (exception: genes on X in males).
II. Genetic basics
1. NOMENCLATURE:
Gene names; three letters and a number, italicized, dpy-1, dpy-2, daf-1
Allele names in parenthesis: unc-13(e51), unc-13(e1091)
Phenotype:
Dpy (Dumpy), Unc (Uncoordinated)
Protein:
UNC-13
Transgenes: if extrachromosomal: prefix Ex, if integrated: prefix Is
stEx5[sup-7(st5) unc-22(+)]
2. GENETIC CROSS:
Remember that hermaphrodites make oocytes and sperm, males make sperm only.
Because of hermaphrodites, you must think of self progeny and cross progeny. Here is
an example: crossing an unc (uncoordinated movement) hermaphrodite with a wild type
male:
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unc-40(e271) I—recessive mutation resulting in uncoordinated movement
P0
unc-40(e271) /unc-40(e271)
X
+/+
F1
Self progeny: oocyte unc-40(e271) X sperm unc-40(e271) = unc-40(e271)/unc40(e271). Phenotype: 100% Unc. sex ratio: 100% hermaphrodites
Cross progeny oocyte unc-40(e271) X sperm “+” (wild type) = unc-40(e271)/+
Phenotype: 100% nonUnc, sex ratio: 50% hermaphrodite, 50% male
sex linkage : what if mutation is on X?
F2
From the cross, take F1 cross progeny hermaphrodite (unc-40(e271)/+), self fertilize:
F2 progeny will segregate: 3/4 WT, 1/4 Unc (3:1 phenotypic segregation), sex ratio:
100% hermaphrodites
inferring genome of mother from phenotype of progeny:
Recessive mutation:
3/4 WT, 1/4 mutant;
Dominant mutation;
3/4 mutant, 1/4 WT
Dominant mutation but recessive lethal:
1/3 WT, 2/3 mutant (why thirds?)
Counting is important! Crosses involving multiple unlinked loci: use Punnett square to
follow how genes should segregate relative to one another.
III. GENETIC SCREENS
Concept: Point of entry into a biological process. Takes advantage of what is essentially
a random process, mutagenesis, to identify specific mutants. Ideal: A simple screen
that can produce informative, tractable mutations with strong and specific
phenotypes. As you learn more about a biological process you can design more
sophisticated screens to target more specific questions. BUT is the mutant really what
you thought you were screening for? Question your assumption using secondary screens.
Forward Genetics: start with phenotype, look for mutations—you can generate mutants
by chemical mutagenesis, radiation, & less well by transposon hopping.
Reverse Genetics: knock-out or knock-down gene (RNAi), then look for phenotype.
1. MUTANT HUNT: FORWARD GENETICS
X-rays, gamma rays to cause double stranded breaks resulting in deficiencies,
inversions, duplications, translocations, single bp mutation. Efficiency is (~1:1000 to
1:2000 mutations/locus/haploid genome).
Soaking in EMS (ethylmethanesulfone) or ENU (N-nitrso N-ethylurea) to cause point
mutations, however, small rate of deletions or deficiencies also occur. Efficiency is
(~1:2000 mutations/locus/haploid genome); spontaneous rate of mutation is (1:106).
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A standard F1/F2 mutagenesis scheme:
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2. FROM MUTATION TO GENE:
Now that you’ve identified mutant strains what do you do?
--Is the phenotype reproducible? Single animals onto individual plates and observe
phenotype in progeny. Ask: what is the penetrance of the phenotype? Are there
associated phenotypes?
--Is the phenotype due to a mutation in a single locus or mutations in multiple loci?
Is the mutation recessive or dominant? Cross mutant with wild type and observe
phenotype in F1 cross progeny and ratio of F2 animals that display mutant
phenotype/wild type phenotype. How do you tell self from cross progeny especially if
mutation is dominant? Use marker mutation to follow cross
--Backcross: Cross mutant and wild type to removes unlinked mutations (50% per cross).
3x backcross=88% of unlinked mutations gone.
if recessive: re-isolate homozygous mutant based on phenotype in the F2
generation, cross this re-isolated F2 back to wild type. Repeat.
if dominant: same starting strategy as recessive mutants but problem is that even
in F2, you don’t know which animals are homozygous for the mutation. So, pick
individual F2s to separate plates, observe F3. If all F3 are mutant =>
homozygote, reinitiate backcrossing.
--How many mutations were isolated? Complementation tests.
You have isolated a number of phenotypically similar mutants from your screen.
You want to know how many different loci (genes) they represent? Are your
mutants due to different genes or multiple alleles of the same gene(s)? You can
answer this by doing complementation tests (same logic as phage/yeast).
--Mapping: What is the molecular identity of a mutation? Mapping by linkage.
The logic behind this is simple, beautiful, and very powerful: genes that are on different
chromosomes segregate independently of one another. Genes that are physically linked
to each other by the virtue of being on the same chromosome segregate with each other
unless separated from one another by recombination during meiosis. The closer two
genes are to each other on a chromosome, the more likely that they will segregate with
one another since there are less chances that a recombination event would separate these
loci from each other. So, if you have a test strain that has a set of markers (phenotypic or
molecular), you can cross this strain to your mutant strain and follow the frequencies of
segregation of various markers to determine genetic map position of mutation.
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--Single Nucleotide Polymorphism (SNP) mapping:
Consider the following scenario: you have identified a mutant strain that carries a single
recessive mutation (*). This mutant was identified in the standard laboratory N2
background strain (strain isolated in Bristol, England). The Hawaiian CB4856 strain is
also wild type but has polymorphisms relative to N2. Cross mutant (*) hermaphrodite
with Hawaiian males. Assume mutation (*) is on chromosome I. (chromosomes III, IV,
V, and X not shown for simplicity):
SNP1
I
*
*
II
X
I
SNP2
II
SNP1
F1 CROSS PROGENY:
I
*
SNP1
Phenotype: 100% wild type
50% hermaphrodites, 50% males
SNP2
II
SNP2
pick individual hermaphrodite cross progeny
onto separate plates, allow self fertilization
F2 Phenotypes: ! wild type, " mutants. 100% hermaphrodites.
Step 1: determine which chromosome carries the mutation:
Pick individual F2 progeny with mutant phenotype. What are the probabilities of seeing
SNPs on chromosomes unassociated with mutation? On chromosome associated with
mutation?
*
NON-RECOMBINANT
F1
F2
*
I
*
*
RECOMBINANT
*
F1
F2
25%
II
50%
25%
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SO, if a SNP that marks a specific chromosome is unassociated with the chromosome
that carries the mutation of interest, that SNP should appear in 75% (or more) of F2
progeny that are phenotypically mutant.
Step 2: determine region on a chromosome that the mutation:
Once you have determined which chromosome your mutation is linked to, you need to
identify the chromosomal region that carries the mutation. This can be accomplished
through reiterative rounds of SNP mapping and following recombinants. Since your
mutation needs to be homozygous to show up phenotypically, you search for regions that
are homozygous for starting strain (in which mutagenesis was performed). Example:
F1
I
snps
F2
*
a
b c d
*
*
*
*
*
*
*
*
*
*
Here is a review of basic genetic concept of recombination frequency for estimating
map positions:
Recombination Frequency: The frequency at which crossing over occurs between two
chromosomal loci. RF=100* (total # of recombinants/total # of progeny stemming from
a single cross--in this case self fertilization). Remember that the two chromosomes in
each F2 are independent and come from independent meiosis (one in sperm, one in
oocyte). Therefore each F2 “scans’ two recombination events.
Step 3: Going from SNP mapping to molecular identity:
Through reiterative rounds of testing your SNP markers, you identify the chromosomal
genomic region that carries the mutation (mapping to a chromosome is really fast & easy,
depending on availability of marker mapping down to a few map units is also relatively
easy—then things get harder, less and less chances of identifying informative
recombinants and more and more work).
-Rescue mutation by cosmid (large plasmids typically span few tens of kb of
genomic DNA) injection
-Long range PCR fragments (~10-12 kb fragments) to tile the region of interest
-Look at the region, take your best guess and sequence a couple of candidates
(you’ll have to show rescue if wt gene is reintroduced)
-RNAi candidates
Nucleic acids are introduced into C. elegans to generate transgenic animals by:
-microinjection into the gonad of hermaphrodites
-bombardment (gene gun)
bacteria/yeast like transformation doesn’t work in C. elegans
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