Lecture #3 - C. Elegans Flashcards

Question

Hermphriditc C. elegans

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Supresor screen = looking for mutation that improve the starting phenotype - Preferom the supressor screen using mutant worms

Answer 8

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

Answer 9

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)

Answer 10

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

Answer 11

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

Answer 12

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)

Answer 13

1. What phenotypes are you looking for? --> Looking for vulvaless phenotype 2. 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 3. Is this a screen or selection --> Screen because you need to look at every worm 4. 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

Answer 14

To do complementation testing in C.elegans – cross hermaphridites and males

Answer 15

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 )

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

SO cool to get alleles in the same gene with 3 phenotypes --> BECAUSE it tells you that the gene is a master regulator of development - Know the gene is a master regulator of development because it can be mutated to flip the switch either way (flip the vulval cells on or off) Because the null of the gene is lethal = gene is likely doing more than vulval development

Answer 21

let-60 is a swicth whose activity (high or low) determines the choice between induced and uninduced fates

Answer 22

Start with e67/e67 --> mutegnize with EMS --> F1 has e67/e67me20 (hypermorph/null) --> lays eggs --> find e20 (get e67/e67 ' e67/e20 ; and e20/20 that are dead) - ASK RIANA - Look for dominant supressors --> get mutatons that would be lethal as homozygous

Answer 23

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

Answer 24

Overall – Order genes in a pathway using epistasis

Answer 25

Epistasis Analysis – Uses mutations that have opposite pheotypes and combine them to make a double mutant and ask what the phenotype of the double mutant is - Epistatic anlaysis = tells you the order of two genes in a pathway - Can ONLY do epistasis by combining mutants that have different phenotypes

Answer 26

In developmental pathway --> epistatic gene is the downstream gene = the phenotype of the double mutant will be the phenotype of the downstream gene

Answer 27

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

Answer 28

KNOW because the null alleles have opposute phenotypes IF null alleles have opposite phenotypes = then they antagonize each other (means one suppresses the other) Example – lin15 LOF = Muv ; lin3 LOF = Vul --> oppostes null = antagonize ecah other

Answer 29

1. Make doubles between mutants with different phenotypes (make double mutants) 2. 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 3. If the phenotypes you are scoring are terminal phenotypes THEN the epistastic gene is the downstream gene 4. IF the phenotype is an “intermoediate” phenotypes then the epistatic gene is the upstream gene 5. If the genes have the same loss of function phenotypes = have arrow (-->) between them 6. If the genes have opposite loss of funtion phenotypes THEN put a negative bar between them

Answer 30

If the phenotype of he double mutant matches one of the single mutants THEN you have epistasis - Epistasis has a clear dominance of 1 phenotype over the other - Epistastic gene = gene whose phenotype is seen in the double mutant IF you make a single mutant and the double mutant has a phenotype that is different from either of the single mutant then you don’t have epistasis - Example – double mutants has a WT phenotype of a lethal phenotypes

Answer 31

Intermediate phenotypes = a phenotype that is normally tansient/consumed in WT (Ex. Biochemcial pathway)

Answer 32

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

Answer 33

Called heterochromatic genes because it cases cells to act inappropriately - Causes teh cells to behave as if they are in a different developmental state)

Answer 34

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)

Answer 35

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

Answer 36

lin4 was the first miRNA that was discovered - Lin14 is the target of lin4 Can’t know lin4 is a miRNA from epistatic analysis

Answer 37

Epistasis can imply a sungle pathway and gene order /regulatory relationship between genes BUT not a specific mechanism - This is a god thing about epistasis because that means we can use epistasis anylysis to study many diverse biochemical pathways

Answer 38

To discover – cloned lin4 and lin14 - They did may tests where they positioanlly clone din lin4 --> found that lin4 has no ORF but there is small RNA there that can be found on a nortehrn blot - Found an alele of lin4 = prevents the stability of the RNA (of lin14 RNA?) - Gary ruvkin cloned lin14 --> found 21 bass of lin4 are complementary to the 3’UTR of lin14 - Lin14 has an ORF Hard to clone lin4 because they were looking for a protein coding gene but then they had to conclude that a mutation prevented the expression of a small 21 base RNA Later Ruvkin and ambros were able to show that lin4 comes on in developmnet when lin14 expression goes down

Answer 39

Cloning lin4 and lin14 showed: 1. Lin4 is a small 21 base RNA 2. Lin14 3’ UTR contains repeats that are complenetary to lin4 3. lin4 RNA levels inversely correlates with in14 protein levels

Answer 40

Big deal because no one thought that a 21 base RNA could do anything They were able to find these things because of genetics --> because genetics can be unbiased (Ex. EMS makes unbiased mutations) When they found that lin4 and lin14 have homology people thought that this would be specific to C. elegans --> BUT THEN Ruvkin found let7 (another miRNA) and showed that let7 has a human homolog - NOW we know that there are mamy miRNA in all genomes - Challenge with miRNA = hard to find the targets of miRNA

Answer 41

Drawback - gene affected is not known Issue - You can characterize the mutants BUT you will also need to isolate the gene In C. elegans going from mutant t gene is often the botteneck (hard to identify the gene)

Answer 42

Overall - outcross mutant against a polymorphic strain + use whole genome sequencing to simultaneously map and identify the mutations - Can go from mutant to gene using sequencing analsysis Issue – can't JUST sequence the mutant because mutant has many other mutations generated by EMS in addition to the mutation in the gene of interest - EMS makes 100 polymorphism and you wouldn’t know which is the causative mutation

Answer 43

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

Answer 44

Issue – now you have a region BUT the region can have a couple of EMS mutations in that area --> NOW use CRISPR - Use CRISPR if have 2 mutants that mapped to teh same region - When do this – add oligio with the EMS mutaton Went from 100 EMS mutants to 2 --> NOW you ca receate the mutaion in the WT N2 bckground using CRIPSR --> look to see if get teh Vul phenotype

Answer 45

1.Use recombination mapping with known marker strains 2. Sequencing cosmids 3. By whole genome sequencing

Answer 46

Once you identify the mutants regulating vulval development they need to be mapped back to a particular gene Past – Use recombination mapping with known marker strains - Allows genes to be placed in relative locations on chromosomes - Done by mapping the genomic region responsible for the phenotype with respect to known phenotypic markers

Answer 47

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

Answer 48

Cosmid = large chuck of genome DNA spanning the area of the identified locus

Answer 49

NOW – mapping and cloning of a gene can be accomplished simultaneously using whole genome sequencing and the Hawiian strain Mutant of interest in N2 bristol background is crossed to a WT hawaiian worm --> F1 cross progeny are allowed to self --> F2 are screened for those displaying the mutant phenotype --> F2s that have the mutant phenotype can further self --> Pool the progeny and send for whole genome sequencing - The sequencing data can be combed for regions that are depleted for Hawaiian marker SNPs - In the sequence – look at the ratios of Hwaiian to N2 SNPs across the genome

Answer 50

Hawaiian strain of C.elegans = highly polymoprhic strain - Hawaiian strain = has a SNP relative to N2 at a frequency of 1 SNP per 1000 base pairs SNPs = can be used as markers (replaces the phenotypic markers used before) to trace which genomic regions are being inherited from which parents

Answer 51

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

Answer 52

Reverse genetics = want to examine the effects of a known gene - Can mine the genome using reverse genetics - Use reverse genetics to go form gene to mutant Compared to EMS = identifies new genes (forward genetics)

Answer 53

1. RNAi 2. CRIPSR

Answer 54

We know that C. elegans has 19,980 genes 50% of C.elegans genes have a human homolog 3575 have a human disease associated

Answer 55

Now we know the whole genome of C.elegans --> NOW we can target each gene specifically for mutations using CRISPR or RNAi

Answer 56

RNAi – discovered in C.elegans WHEN you feed dsRNA against a particular gene --> will knockdown the gene - Can screen for any phenotype you are interested in (screen the whole genome of C. elegans) Example - IF you give adult hermphdrite dsRNA against a gene needed for embryogensis then all of the progeny would be dead

Answer 57

1. Use a pre-selected group of genes (Ex. Look for all kinases) 2. Genes can have homology 3. Genes can be expressed in a specific tissue Can also taregt all annotated genes BUT Disadvantage of Reverse genetics = can’t discover new genes this way

Answer 58

RNAi = uses endogenous cellular machinery to target and knock down an mRNA transcript that is homologous to a double stranded RNA trigger that was introduced from an exogenous source - Can target almost any gene that we know the sequence of In C.elegans - dsRNA trigger can be provided by feeding the C.elegans E.coli that is expressing the trigger Knockdown by RNAi occurs over 12-24 hours period + can be inherited by the next generation of worms

Answer 59

RNAi = can be used to do a screen in C.elegans --> make mutants and sequence them Companies have generated libraries of RNAi constructs that target almost any gene in C.elegans genome --> Allows researchers to run large scale screens where they knockdown a list of genes of interest and look at the phenotypes that result from the knockdown - STILL reverse genetics because we know the genes being targeted in advance

Answer 60

YES Might not know about alternatively splived isoforms or viruses inserted into genes or genes in minorities that is not in original DNA sequence or lncRNAs or alternative ORFs (ex. Don't start with AUG) ALSO might not know small ORFs - Example – 5 coding amino acids and a stop codon - Small ORFs = used to be ignored by genome projects Shows there is still parts of the genome that we don’t know about Genetics will still identify any gene as long as you go through a sufficient number (of mutants?) to identify small genes - Genetics can odentofy what you don’t even know exists

Answer 61

YES - when we sequence RNA we might miss low abudnce RNA

Answer 62

1. Production of specific point mutations 2. Knock ins 3. Deletions (frameshifts) 4. Tagging of endogenous genes with flourscent laels

Answer 63

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

Answer 64

RNAi is NOT only in worms – functions across a broad range of different species to knockdown RNA transcripts

Answer 65

Endogenous – from within the cell Exogenous - Comes from outside the cell There are endogenous and exogenous pathways to regulate mRNA stability using small RNAs - Endogenous – piRNAs (PIWI intrecatig RNAs) + miRNAs - Exogenous – siRNA (short interfering RNAs) --> function in RNAi

Answer 66

Strat - have a long ddRNA molecule or short hairpin RNA (shRNA) - dsrNA or shRNA need to have homology to a particular mRNA transcript that we want to target for silencing - Long dsRNA/shRNA = processed and trigger RNAi response Introduce the long dsRNA or shRNA to a cell --> After added to the cell --> RNA triggers are processed by DICER --> After DICER --> siRNAs are recognzied by RISC (RNA induced silencing complex) --> RISC seperates the siRNA duplex and hodls onto 1 strand from the dsRNA (holds onto the antisense/guide strand) --> RISC uses the strand as a guide to recognize and bind with perfect complementary to a region of mRNA that were are targeting --> perfect complementarity licenses the RISC to cleave the target transcript --> eventually leads to the degredation of that transcript and THEREFORE reduced expression of that particular gene product

Answer 67

Different model organisms use different delivery mechanism fr the dsRNA trigger: In cell culture – plasmids encoding the dsRNA trigger sequence are transfected into cells --> the trigger can then be expressed from the plasmid and processed In C. Elegans – dsRNA can be intrduced through feeding the worms E.coli that express the dsRNA - RNAi response can then be inherited in subsequent generations of worms Flies (transgenic integration) – shRNA sequences have been integrated into the genome under the control of different promoter elements - Allows for unique control over where and when the shRNA is expressed

Answer 68

DICER = endogenous enzyme --> processes the dsRNA or shRNA - DICER chops the dsRNA or shRNA int 21-23 bp siRNA DICER = also processes the endogenous small RNAs (like miRNA and piRNA)

Answer 69

1. When using RNAi you can never assume that you are getting 100% knockdown - Some fraction of the transcripts are likely going to escape degradation - Any stable residua protein from that transcript will still be present in your cells - Only way to look at a true knockout phenotype is to make a mutant in which you delete the gene that you are interested in 2. It is possible that you dsRNA trigger might target more than 1 transcript = results in off target effects that can confound your results

Answer 70

Sequencing costs have sharply decreased since the first draft of the human genome project in 2001 - First human genome cost $100 million to produce - NOW - $1000 to sequence a human genome Have large drop off in cost around 2008 --> mostly due to the introduction of next generation sequencing technologies

Answer 71

Sanger = first sequencing technology Overall - Sanger sequencing works by incorporating nucleotides that stop chain elongation into a growing nucleotide chain - Chain terminating nucleotide = ddNTPs Past – did this in 4 separate reactions where ddNTPs is added to 1 of 4 aliqouts of a sample at low concentration along with polmerase and normal dNTPs --> DdNTPs rendomly incorporate into the growing chain --> stops the chain growth at a particular length --> Once the reaction is done you run the products side by side on a massive gel --> radioactivley labeled ddNTPs allows the stalled products can be visualized - Because the recation was divide din 4 parts the terminal nucleotde and length could be recored across each well --> information from each well could be combined to generate a sequence

Answer 72

Sanger is still used today for low throughout-put applications BUT in a modified form NOW use flourecstley labeled ddNTPs and capilaries + laser imagers + computers - Capillaries + lasers + computers = replaced the massive gels and manual counting of bands

Answer 73

Shotgun sequencing --> large chunks of DNA were fragmented into smaller fragments which could then be cloned into a vector --> vectors could be transformed into bacteria and amplified - Vectors were small enough so that the original sanger sequencing methods could process them and the sequencing could then be aligned and assembled into a draft genome The first whole genomes were sequenced using Shotgun sequencing

Answer 74

NOW – biological amplification uses PCR instead of vectors - PCR Amplification = can be done using emulsion PCR or bridge amplification (more common) For Amplification – DNA is fragmented and adapters are ligated onto the ends of the fragments to amplify the fragments for library generations - Adpaters allow for the used of a single primer pair (Primer is attached to beads or a flow chamber)

Answer 75

Actual sequencing method varies depending on the company whose technology you ate using Best methods detect fluorescent changes as bases are incorporated into growing chains and can process this data en mass to determine the order of bases in a sequence Platform used = affects the read length + run time + the volume of data generated

Answer 76

Raw data needs to be aligned back to a reference genome and converted to a format that humans can process - People have developed algorithms and pipelines to streamline this process When doing assembly – it is important to know what your read depth is

Answer 77

Read depth = coverage depth --> BOTH how many times a given point in the genome is represented in your data - Want 25-40x read depth IF a given point is only represented once (and that 1 time is not correct) THEN you won't know there is an error IF each point is represnted multiple times = errors can be drowned out

Answer 78

Oxford nanopore = long read sequecning - Generate long continuous reads BUT at the cost of read accuracy Process - Nanopore works by circularizig DNA and pulling a single strand of DNA through a protein pore embedded in a membrane --> As the DNA passes through the pore, it generates an electrical signal that fluctuates depending on the base that is passing through --> These signals can then be decoded by an algorithm and converted to sequence - Electrical signals ca be used to determine each base in the chain

Lecture #3 - C. Elegans Flashcards

(108 cards)