Lecture #3 - C. Elegans Flashcards
Epistasis
Epistasis – suppression of the effect of a gene by a non-allelic gene
Epistasis Anlysis (Overall)
Epistasis analysis = allows us to identify the order of genes in a pathway
To do an epistasis analysis = need to examine pairs of mutations in different genes with opposing or disict phenotypes
Epistasis Anlysis - Example (uracil biosynthesis pathway)
Uracil biosynthesis pathway in yeast:
- In WT organsims –> Enzyme A – converts orotidine 5’-phosphate to uridine 5’-phosphate AND Enzyme B - converts uridine 5’-phosphate to uracil
IF enzyme A is deleted (dA) –> have an increase in orotidine 5’-phosphate concetration
IF enzyme B is deleted (dB) –> have an increase in uridine 5’ phosphate concentration
IF BOTH enzyme B and enzyme A are deleted (cross these two mutants so that both enzyme A and B are deleted) –> only observe an increase in orotidine 5’ phosphate
- See the same phenotype that you see when you only delete enzyme A
How do you know epistasis is happening in Epistasis Analysis
Because the double mutant (deleate A and B) phenotype gives the same phenotype as just deleteding A –> we can conlcude that enzyme A is epistastic to enzyme B
- Enzyme A masks the effect of enzyme B
Double mutant A and B phenotypes is the same as just Mutant A phenotyp = enzyme A is epistatsic to Enzyme B AND that enzyme A is upstream in the pathway
Epistastic Gene in Intermediate phenoypes
In intermediate phenotypes –> the epistatic gene is upstream
- Double mutant has the same phenotype as enzyme A = enzyme A is the epistatic gene = enzyme A is upstream
Intermediate phenotypes
Looking at a pathway where each step is using the product of the previous step as its substrate so removing and enzyme in this process will result in the buildup of that enzyme’s substate
- Enzyme substarte is the intermediate
- If one step fails the prevous intermediate acucmulates
Example - Uracil pathway
Final output in Intermdiate phenotypes
Example - Uracil Pathway
Mutations in enzymes along this pathway will likely all produce a similar change in the final output of the pathway not particularly informative phenotype on its own)
- Looking at the final output of the pathway in either single mutant is not informative because will aways see reduced uracil = not information BUT by looking a the intermediates we can gather more information
Terminal Phenotypes
Don’t always have easily detectable intermediates BUT instead produce clearly opposite final outputs when their components are mutated
- Pathways where we focus on the final output = terminal phenotypes
- Generally terminal phenotypes = controls an ongoing process that has a default output in the absence of a single BUT once the signal is prescent the output chnages
Example – transcription of a specific set of genes –> if the pathway is off then the genes are not transcribed BUT when you turn the pathway on then the genes turn on
Terminal Phenotypes - Example #1
Example Terminal phenotype pathway – Wnt signling –> Wnt signaling controls transcription of a set of target genes
- When the Wnt ligand is absent (default stae) –> the genes are not expressed (default state is off)
- When Wnt ligand is present –> activate transcription (Wnt present = on)
Have 2 clearly defined outputs (transcription is either on or off) –> Use this binary phenotype as a readout for how particular mutations affect the function of the pathway
How does the Wnt pathway actially work to control gene expression
No Wnt ligand –> Transcrtion factor (Beta-catenin) is inhibited by the destruction complex –> Wihout beta catenin transcrtion of this set of target genes is turned off
- Inhibition is show by a barred arrow
Have Wnt ligand –> promotes the activit if Frizzled protein –> active frizzed can actiavte Disheveled –> Dishelvels destabilizes and inhibits the destruction complex –> No destrcution complex –> Beta-catenin can enter the nuclear and promote transcription
Epistasis in Wnt pthway
Have 2 muatnts:
1. A deletion in disheveled –> FIND that deleted dishevled results in transcription always being off (even in the prescence of the Wnt Ligand)
- Makes sense because dishevelved normally inhibits the destruction complex (if teh destruction complex is never inhibited then Beta-catenin can never function and transcrtion will always be off)
- A deletion in the destructruction complex –> Transcrtion is alwats on even without the Wnt Ligand
- Makes sense because if the destruction complex is never functional then beta-catenin will never be inhibited
Phenotype of the double mutant –> transcription is always on
Based on this = know that the destruction complex is epistatic to deshevaled (because the double mutant phenotype is the same as the phenotype of the destruction complex mutant) –> means that the destruction complex is downstream of disheveled
Why is the double mutant always on in Wnt pathway
No disheved then the Detsuction complex theraotically would always be on BUT this does not matter in the double mutant because the destruction complex is not functioning in the first place –> THEREFORE the double mutant phenotye would be the same as in teh detsruction complex mutant –> transcripton would always be on even in teh asnsence of teh Wnt ligand
Epistatic gene in terminal phenotypes
For terminal phenotypes – the epistasic gene is going to the be downstream gene
What is needed for good epistasis
Good epistasis analysis = requires opposite phenotypes
Caveats to epistasis anlysis
Occurs when making a double mutant and the phenotype of the double mutant doesn’t replicate the phenotype of ether individual mutations –> prevents us from being able to order the genes in a pathway
Caveats to epistasis anlysis - example #1
Example – Genes controling flow color
One mutation might give white flowers ; different mutation gives red flow BUT the double mutant is pink
It would be hard to conclude what is going on from this information BUT you could guess that if the genes are involved in the same pathway then it more complicated than a simple linear relationship OR that the two genes might not be in the same pathway at all and both indviually influencing phenotype
Caveats to epistasis anlysis - example #2
Have 2 genes controlling the same process that each produce the same terminal phenotype where the process is turned off BUT the double mutant phenotype is lethal
Normally we like to have distict phenotypes to examine epistasis BUT the fact that the double mutant is lethal does provide some information
- HERE the evidence points towards the genes participating in parallel pathways rather than a single pathway where both pathways individually are not required for life BUT if both are inhibited then we have synthetic lethality
Genetic Advantages of C. elegans
Genetic Advantages – makes them a good system in which we can study the function of genes and genetic pathways
- Short life cycle and fast generation time – worms can be quickly propagated in the lab
- Can be grown in bulk
- Adults are primarily self-fertilzing hermphtrdtes = further simplifies propegation
- Large brood size
- Can screen so many mutants that you can find small genes
Propogeation of C. elegans
C. elgans = Easy to grow and propegate in the lab
Generation time is 73 hr from zygote to adults
Get progeny in 3.5 days –> helps generate many mutants
- WHEN use EMS on parents (mutagenize the mother) –> get 250 progeny where each have a different mutations in thw body (because mutated the germ cells in the mother)
1 adult hermaphrdite makes 250 offspring = allows them to be grown in bulk
Advantages of C. elegans - Complexity Vs. Ease of use
C. elgans = Multicellular organism with clearly defined organ systems and structures (Has typical body plan)
C. elegans are transparent –> allows for live imaging (can see every cell)
Has 959 somatic cells
Basic genetics of C. elegans
6 chromsomes in total – 5 autosomes (labed I-V) ; 1 sex chrosomes (X)
- Hermaphirditic worms have XX
- Male worms = have 1 X chrosomes (X0)
Most commonly used lab strain = N2 bristol strain (often refered to as N2/WT)
C. elegans nomenclature
Genes = lowercase and italicised (three or four letters long plus a hype and a number)
- Ex. alg-1
Alleles = indicated in parentheses after a gene name
- Ex. alg-a(gk214)
Proteins = All uppercase and NOT italicisezed
- Ex. ALG-1
Phenotypes have only their first letter capitlized
- Ex. Muv
Hermphdrites Sex
C. elegans are primarily self fertile hermphrites –> produce sperm AND oocytes
- Self feryalization generates more hemaphritic worms BUT 1 in every 1,000 fertalizations = an X chromosome has a non-disjunction event that produces a male worm with only 1 X chromosme
Male C. elegans
Males demonstrate several structural differences compared to hermphirdiates
- Males only produce sperm
- Males can mate (can cross) with hermaphridites
Crosses –> Because males worms have1 X chrosmomes –> when a male mates with a hermaphridte teh resulting progencey are 50% hermaphridte and 50% amle
Hermphriditc C. elegans
Hermphridutes – modified females that genrtate oocytes AND sperm
Hermphrdites self fertilize –> producses mostly herphridites (rarely makes males)
During selfing –> Sperm and oocyte will fertilzie inside the hermphridite (essicially genrates 250 new hermphridtes)
- When hermaphridates self = produce ~100% hermpahrdtes
- Hermphridites can’t mate with other hemphridtes they can only mate with males
Cell lineage mapping
The fate of every cell in C.elegan division is known
- All cell divisions that give rise to somatic cells in the adult have been worked out by microscopy (watching embryoes divide)
- All mutant phenptypes can be desirbed using names of cells and how they misbehave in the mutants
Knowledge = allowed for the study of genetic mechanism that govern cell fate specification
How do we use this cell mapping infomration to study genetics
Use foward genetics –> search for mutants in which normal cell fate patterns are disruspted
Process – Mutegnze the parental Hermphrdites (P0) using EMS –> plate the F1 genertion that has mutations individuals and allow F1 to self –> produces the Wt heterozygous mutant AND homozygous mutants in F2 (F2 with have hetrosyzgous WT/mutant AND homosygous mutants)
- Dominant mutations can be identified in the F1 generation
- Recessive mutations = detected in the F2 generation
Mutogenizing the parental hermaphridtes
When mutagenize paternal hermphridites –> causes mutations in the germline of the parental –> mutations in the germ line of P0 can be passed on to the F1 generation –> plate the F1 generation individually to isolate independent mutations and allow the F1 to self
Worm Vulva
WT worms = have a single vulva
Vulva connects the internal germline to teh outside world –> used to deposit embroys onto a plate
- Used to lay embryoes
Mutations in specific genes = can lead worms to produce more than 1 vulva structure (Multiple vulvas – Muv OR no Vulvals -Vul)
- Vulvaless worms = can’t deposit any embryos –> eventually the embryos will hatch inside of the hermaphridte –> produces a bag of worms
Identofying the genes required for proper vulva formation in worms
To find genes involved in vulva formation – carry out EMS screen on Muv worms
- Looking for additional regulators of cell fate –> do supressor screen
Process - Mutogenize Multivulval worms with EMS –> worms in the F1 and F2 generation were scanned for those that only had 1 Vulva –> any worms in the F1 or F2 generation that no longer had multiple vulvas were considered to have acquired a revertant mutations which were a supressor of the multivulva phenoytype
- Mutations in womrs that only ahve 1 vulva in F1/F2 = revertants
- Worms with 1 vulva in F1/F2 = carrier new mutations that are supressors of the Muv phenotype
Supressor screen Goal
Supresor screen = looking for mutation that improve the starting phenotype
- Preferom the supressor screen using mutant worms
Vulva Cell Fate determination
Looking at which cell or cells are repsonsible for proper formation of the Vulva?
Image – shows how the Vulva is made
- Can look at the 6 cells that make the vulva
- Cells divide to make an opening in the skin
- Different cells adopt different fates
Fates of cells:
1. Primary fate – when P6p divides to make 8 cells
2. Secondary fate – P5p and P7p devide to make 7 cells
3. Tertiary fate – P3p, 4p, P8p – divide only once and then self with nieghboring skin
4. Anchor cell in Gonad = Directley above P6p
Is there signalling between difefrent cells to position the vulva
Want to look at whether there is signaling between different cells to position the vulva –> FOUND anchor cells signal to cells to make the vulva
Expeirment done - Killed the anchor cells early in development using a laser –> found all of the cells adopt a tertiary fate where they fuse with the skin and don’t make a Vulva
When you preserve the anchor cell and kill all of the other cells in the gonad –> can still make vulvaes (get primary, secondary, and tertiary fates)
Overall – shows that the anchor cells signal to the other cells to induce vulval cell fates (anchor cell tells the other cells what to do)
Image of vulvaless phenotype
WT = see eggs are laid on the plate
Vulvaless mutant –> keeps all of the embryoes inside of the mother
- Empbyros will contonie to develop and hatch as larvea in the mother –> the larvea will eat the inside of the moth –> END = have a bag of worms phenotype –> Eventially the worms will come out of the mother
Why is worm in a bag a good phenotype
Worm in a bag = good phenotype for geneticists because it is easy to screen for AND the worms are fertile/viable and can propagate
- The mother only dies AFTER she generates many mutants –> can then porpegate the mutants
End - Easy to screen for –> can study what mutation causes the phenotype
Deisgning a screen to idetofy the anchor cell signal (you have WT worms + NNGM plates + OP50 + EMS)
Goal – screen for mutants that disrupts the signaling pathway
Process – Take WT worms –> grow on NNGMp plates with OP50 –> mutogenize WT worms with EMS (mutagenize P0) –> self F1 –> look for mutant worms in F2 –> propagate the mutant on its own as a clone –> Sequence
- Have homosyzgous the worms in F2 –> takes m/m (homozygous mutant) hermphridite and plate alone –> all the progeney will be homozygous for the mutants (Shows an advantage of selfing)
Deisgning a screen to idetofy the anchor cell signal (you have WT worms + NNGM plates + OP50 + EMS) - QUESTIONS ON SLIDE
What phenotypes are you looking for?
What generaton will you screenfor after mutagensis
Is this a screen or selection?
How will you know when to stop screening
- What phenotypes are you looking for? –> Looking for vulvaless phenotype
- What generaton will you screen for after mutagensis?
- F1 will have dominant mutations –> Since mutations are random F1 will only have 1 allele (heterozygous in F1 = only see domiannt mutatinos)
- F2 = has recessive mutations –> ¼ of the worms are homosygouze for the mutations = can screen for recessive mutations - Is this a screen or selection –> Screen because you need to look at every worm
- How will you know when to stop screening –> Do Complementation test as you go along to make sure you recover new genes –> Eventually you will keep getting alleles of the same gene
How do you do complementatino testing in C.elegans
To do complementation testing in C.elegans – cross hermaphridites and males
Results of Genetic analysis (Results of the screen)
Get vulvaless mutants and multivulval mutants
- Multivulval – phenotype where multiple cells are being turned on in vulval mode (Multiple vulva are easy to pick out in a screen because can see multiple vulval alonh the belly of the worm )
Isolating Supressors of a Multivulval mutations
NOW – doing a secondary screen to look for revertants
- Revertants = supressors of a multivulval muttaions
Process - Start with Multivulval mutant worms –> mutagenize again (Add EMS) –> look for suppressors
- F1 – has Muv/Muv and m/+
- F2 – Muv/Muv ; +/+ AND Muv/Muv ; m/+ AND Muv/Muv ; m/m
END - to find supressor mutations –> look for an egg on a plate –> KNOW that egg was derived from a WT animal that had a normal Vulva (Egg on a plate had to come from a hermaphridte that had a supressor)
- Based on the fact that ONLY WT can lay eggs on a plate
- Can select for eggs on the plate and know they came from WT
Is Isolating Supressors of a Multivulval mutations a screen or a selection
A selection
Good because selection is faster –> looking for eggs instead of looking at each worm individually for a vulva = Can easily select for mutants you want
- Can look at many more mutogenized genomes
let60
let60 locus affects vulval development
Using screens and selections in C.elegans they found many mutants –> found multiple alleles of the same complementation group where each alleles has a different phenotype
let60 gene has 3 alleles:
1. e20 –> mutant of let60 that is lethal
2. e42 –> mutant of let60 that leaves to vulvaless (all Pnps adopt tertiaty fates)
3. e67 –> mutant of let 60 that leads to multivulval (all Pnps adopt primary and secondary fates)
Shows 1 gene mutated in 3 ways leads to 3 phenotypes
Why can you have different alles at the same locus give different phenotype
Can you have different alles at the same locus give different phenotypes –> BECAUSE the different alleles have different morphs
e20 –> look at e20/e20 = e20/delation < e20/+ –> e20 is an amorph –> gives lethal phenotype
e42 –> e42/Df < e42/e42 < e42/+ < e42/+ WT Dup –> Antimorph –> reduces function
e67 –> e67/Df > e67/ + > e67/+ WT Dup –> hypermoph –> hyperactivtaing gene –> causes hyper induction of vulva
What can you conclude about let60
let-60 is a swicth whose activity (high or low) determines the choice between induced and uninduced fates
Signaling path using let60
Using screens/selection - got many genes that show the anchor cells sends signals to P6p cells (signal is sensed by EGF receptor) –> signal is tranduced through RAS
- RAS can be GTP or GDP bound –> Screens using let60 mutations idetofied mutation in RAS that showed biologists that RAS is a master regulator (switch that can on on or off)
- When let60 is on –> turns on cascade that leads to primary fate
Overall (process) - Trnascrtion factor lin15 –> Inhibits Lin 3 AND Lin 3 –> binds to let23 (EGF-like signal) –> Let23 signals is traduced by let60 (let-60 is RAS = master regulator) –> get primary cell fate
Example epistastic analysis - let60 –> can you figure out what the double mutant phenotype will be for the DEVELOPMENTAL pathway
Slide = shows all of the double mutants made + single mutant phenotypes
Start - Have many single mutatnts
Let 60 –> Vulvaless
Let-60(hypermorph) –> Multivulval
Lin-15 –> Muv
Let-23 –> Vul
Lin-3 –> Vul
Phenotypes of the double mutants:
1. Let60,lin15 –> vulveless (Lin15 supressing does not matter)
2. Let60(hyper);let23 –> multivula (Let60 is the downstream gene and it domiants and it is hypermorphic = it will make muv)
3. Let60(hyper) ; lin 3 –> MUV
4. Lin15;let23 –> Vul
5. Lin15-lin3 –> Vul
ALL TOGETHER – shows that lin15 inhibits lin 3 ; let-23 feeds into let60
- Shows that epistasis testing allows genetcist to put genes in pathways
Methods for epistasis probelms
- Make doubles between mutants with different phenotypes (make double mutants)
- If the phenotype of he double mutant matches one of the single mutants THEN you have epistasis
- Means that the genes likely function in the same pathway - If the phenotypes you are scoring are terminal phenotypes THEN the epistastic gene is the downstream gene
- IF the phenotype is an “intermoediate” phenotypes then the epistatic gene is the upstream gene
- If the genes have the same loss of function phenotypes = have arrow (–>) between them
- If the genes have opposite loss of funtion phenotypes THEN put a negative bar between them
Heterochromatic genes in C. elegans
When lookin at cell lineage in tail of the worm –> can see that the cells can divide in different ways
FOUND mutations where T cell would skip a stage and would like it is in L2 stage BUT in reality the worm was still in L1 stage
ALSO found a GOF mutation in lin14 whwre the T cell would repeat the pattern that should have only been in L1 stage
- Lin14 = swicth gene –> cause siwtch in lineage for that cell
- Lin4 gene = has 1 alelle that causes the repeat of L1 lineage
Which double mutant would you make in order to figure out the order of the genes
Answer – A –> because you need the two mutants t have different phenotypes
- Lin4 LOF repeats L1 Vs. Lin14 LOF skips to L2
- In Asnwer A = mutants have opposute phenotypes = can combine them (for epistasis analysis need to combine mutations in the double mutant that have different phenotypes)
Question – What is the order of the genes
Image – shows the phenotypes of each of the single mutants + shows that the double mutant skips L2
Answer:
Know Lin14 is downstream because the double mutant skips L2 (double mutants has the lin14 phenotye = Lin14 is the epistastic gene = lin14 is downstream)
Know there is a bar between Lin4 and lin14 because lin4 and lin14 LOF have opposite phenotypes
How do you idetify the causitive mutation out of all of the mutations made by EMS
Overall - need to shuffle the genome to select for mutants -> shows where the mutation is
Process – Cross mutants to a strain of worms that has any polymorphisms in the genome (cross mutant worm that has a bristol background wih hawaiian worm) –> get F1 –> then get F2 –> In F2 you select for mutant phenotype again (select for F2 that is homozygous for mutant –> sequence the genomes of the mutants –> only place where you see the bristol signatirs of polymorphisms is in teh region where the mutant lies (eveyrwhee else has an equal disrbution of bristol polymoprhisms and hawaiian polymoprhism)
- In F2 you select for mutant phenotype – (Example – look for Vul mutant again)
- F2 = Recombinant F2 –> have recombination in mom germline so that the blue and yellow chromosomes have recombined
Image – Blue worm = bristol –> bristol worm is being crossed with worms form hawaii that have many polymoprhisms
How do you map mutations to individual genes - sequencing cosmids
Once a general region in which the mutation likely exists is established –> worms are injected with Cosmids
- Cosmid = large chuck of genome DNA spanning the area of the identified locus
Overall - cosmid that could rescue the phenotype of interested were sequences to narrow down the sequence responsible
- By looking at where the cosmids overlap the researchers could find the gene that might cause the phenotype of interest
Example – Multivulval develoment – turned out many genes were part of EGF pathway
Anylyzing the genome after crossing with Hawian strain to map mutations to genes
When analyzing the genome - Regions that have a low percent of Hawaiian SNPs ca be compared to sequencing data from cross progeny that did not display the mutant phenotype
The causative mutation should exist in a region of the genome that homozygous for the mutant allele and therefore should not contain any hawaiian markers
CRIPSR Process
Process –> Cas9 Nuclease + gRNA + repair template ALL simultaneous injected to the germline of an adult hermaphroditic worm –> the gRNA directs the Cas9 to a site in the genome of individual germline nucleus where the Cas9 enzymes makes a dsDNA break
- Want the break to be repaired through homology direct repair according to the template that was injected
- If repaired by homology directed repair –> fraction of the progeny form the injected worm will carry the desired mutation and can be screened for the mutation through PCR
F1 and F2 are screened for individuals that carry the edit and are clones to homozygosity
- F1s that carry the mutations = singled out and allowed to self –> produces homozygous mutants for the desired edit
Fowards Vs. Reverse genetics
