caenorhabditis elegans anton kapliy february 17, 2009
Post on 13-Jan-2016
227 Views
Preview:
TRANSCRIPT
Caenorhabditis elegans
Anton KapliyFebruary 17, 2009
Sydney Brenner (1927 - )
•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
HandlingIsolated from soil(see first picture on the right)
Cultures reside on small platesCovered w/ E.Coli lawn that provides nutrition for the worms
Preparation of monoxenic culturesGerms & worms are killed with a chemical, but eggs survive
Long storage via freezingEarly 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 cellsC. Elegans is a diploid organism with 6 pairs of chromosomes
I II III IV V X
5 autosomes
1 sex chromosome
A pair of homologous chromosomes
I II III IV V X
zygote
Regular cellGametes
I II III IV V X
One from mommy: ovum
One from daddy: sperm
mitosys
XO sex-determination systemSex is determined by the number of X chromosomes:
Ovum Sperm Ovum Sperm
XX XO
Note that the ovum always contains an X chromosome, but the sperm may or may not.
A small twist: HermaphroditesAn XX worm produces both ovum and spermThus, 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 + 2Aa2(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---A--
-A---a--
x =
egg ovum
Hermaphrodites
Figure A:Arrows point to head, tail, and vulva
Figure B:Anus
Figure D:An egg leaving the vulva
MalesIn 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 phenotypes4. Create a genetic map of the worm5. See why C. Elegans is a great model organism
“Dumpy” worm
-----a--Suppose “a” is a recessive mutation:
-a---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------
--------
-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-- --
---- 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-
---- a----- --
Progeny male always gets its X chromosome from mother
Progeny female gets one X chromosome from each parent
= =
IV X IV 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---a--
-a---a--
---?---?
10 plates with homologous mutants with the same phenotype:
Complementation & cis-trans test
-a---a--
Cross the homologous mutant with wild-type males & examine progeny
CASE I: same gene CASE II: different genes
1st homologous mutant (male)
2nd homologous mutant (female)
+
Progeny has different phenotypes!
=
IV IV
+
=
-a---a--
-a---a--
-a---a--
+
=
---b---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---a--
Generic cis-trans test requires that mutant males mate with hermaphrodites.But: mutated males with many phenotypes (e.g., uncoordinated) can’t mate!
1st homologous mutant (female)
Wild-type male
+
=
+
=
--------
-a------
Progeny males (wild phenotype!)
+
=
-a---a--
-a---a--
2nd homologous mutant (female)
+
-----a--
or
-a---a--
+
=
--------
-a------
+
=
---b---b
-a-----b
-------b
or
=Examine presence of mutated phenotype in progeny males!
mutant wild wild wild
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 seesix linkage groups,mapping to 6 chromosomes:
I II III IV V X
-a---a--
-a---a--
-a---a--
-a---a--
--b---b-
--b---b-
--b---b-
---c---c
--b---b-
---c---c
-a---a--
--b---b-
---c---c
10 plates w/ blistered homologs3 plates w/ blistered homologs, corresponding to 3 different genes
cis-trans
Aside: Cis and trans configurationsConsider 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-freeChromosome II carried both u & d
u------d
trans
Chromosome I is carries uChromosome 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 heterozygoteu---u---
---d---d
Start with two homozygous mutants:
u---u---
--------
Wild ♂Mutant1 ♀
+ = u-------
Baby ♂
+ ---d---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 wild Discard
Use
Meiosis in trans heterozygote
u------d
---du--- ----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
---du--- ----u--d
u------d
u---
---d
u--d
----
Sperms:
Ovums: u---u---
---du---
u--du---
----u---
u------d
---d---d
u--d---d
-------d
u---u--d
---du--d
u--du--d
----u--d
u-------
---d----
u--d----
--------
1-P
1-P
P
P
P = probability of crossover
Progeny in trans heterozygote (II)u------d
---du--- ----u--d
u------d
u---
---d
u--d
----
Sperms:
Ovums: Unc
Wild
Unc
Wild
Wild
Dpy
Dpy
Wild
Unc
Dpy
UncDpy
Wild
Wild
Wild
Wild
Wild
1-P
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 0.52 = 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) + P2 = 1 – 2P + P2 +2P – 2P2 + P2 = 1P(unc+dpy) = P2
Ratio = P(unc+dpy)/P(dpy) = P2
u------d
Consider self-progeny of trans heterosyzote:
u- ---- -d
Results: classification of linkage groups
1st chromosome
2nd chromosome
Etc…
Results: mapping of mutantsWe 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 networkTheir interconnections have been completely mapped out
Gene silencing via RNA interferenceInject 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 textbookshttp://www.wormbook.org/chapters/www_nematodeisolation/nematodeisolation.html http://www.wooster.edu/biology/wmorgan/bio306/C.elegans_Week3_Directions.htmlhttp://www.sanger.ac.uk/Projects/C_elegans/ http://www.wormbook.org/chapters/www_dominantmutations/dominantmutations.htmlhttp://www.wormatlas.org/handbook/anatomyintro/anatomyintro.htm http://www.wormclassroom.org/ge.htmlhttp://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
top related