Presentation on theme: "Forward Genetics Phenotype (Function) Genetics Gene A Gene B Gene C Proteins A B C P."— Presentation transcript:
Forward Genetics Phenotype (Function) Genetics Gene A Gene B Gene C Proteins A B C P
Survey In classical or forward genetics, we commonly use chemicals or radiation to generate mutations in model organisms such as yeast, worm, fly or mouse. In doing so, we typically try to generate mutations in A: somatic cells B: germline cells (gametes, sperm or oocytes) C: not sure
Mutagenesis-sex chromosome Oocyte gametes X XY 50% XX meiosis female Sperm YX 50% XY Male fertilization Oocyte X 2n XX 50%
Mutagenesis-sex chromosome Oocyte Sperm gametes x YX XX XY meiosis female Male fertilization Oocyte X 2n Mutagen xX xY heterozygotemutant
Mutagenesis-autosome Oocyte Sperm gametes A AA 50% AA 50% AA 50% A meiosis female Male fertilization Oocyte A 2n x y
Mutagenesis-autosome Oocyte gametes A 50% AA 50% A meiosis female fertilization Oocyte A 2n AA 50% X-ray a Aa some Heterzygous mutant Sperm AA A Male x y
Your opinions When you mutagenize the gametes of P0 animals, you usually do not get homozygous mutants in the F1 generation. A: Yes B: No C: not sure. Do you usually get homozygous mutants in the F2? A: Yes B: No C: not sure.
F1 screens vs. F2 screens ++++ Po WT m+m+ X ++++ F1 mutant ++++ Po WT m+m+ X ++++ F1 WTX m+m+ mmmm F2 mutant But, how do we get the same m in two F1 and let them mate?
Balancer chromosomes Chromosomes that suppress crossover. Homozygous of the chromosome is either lethal or with a visible phenotype. Usually contain inversions and translocations AB BA AB BA Neither can be paired with WT wt balancer Resulting abnormal chromosomes
Traditional F2 screen in fly X-ray X TM * * X F1 * * X * * F3 homozygotes
F2 screens in the worm Sperm Oocyte gametes X X 100% XX 100% Self-progeny XX meiosis Hermaphrodite fertilization x xX Po F1
Worm F2 screen xXxX F1 Sperm Oocyte gametes X X meiosis X x xx F2 Homozygous mutant 1/4
summary ++++ Po WT m+m+ X ++++ F1 WTX ++++ mmmm F3 mutant m+m+ F1 WTX m+m+ Fly, mouse, … ++++ Po WT m+m+ F1 WT mmmm F2 mutant worm mutagen
Basic mutagenesis in the worm Treat with 50mM EMS for 4 hrs Several L4 or young adult worms/plate. Multiple large plates P0P0 F 1 eggs F1F1 Remove P 0 parents after laying ~100 eggs Recovering for a few hours each F1 carries two mutagenized chromosomes m/+ heterozygotes for each mutation F2 eggs Remove F1 worms after laying ~ 2000 eggs for 20 hrs F2 Worms with m/m genotype for each mutation are mixed with m/+ and +/+ animals mutant worms (m/m) are individually picked on to a fresh small plate. Total genomes screened = 2X # of F1 animals x # of F1 plates
Question If you plan to mutagenize and screen for a mutation in a tumor suppressor gene that may leads to tumorigenesis, would you do F1 screen or F2 screen? A: F1 B: F2
Genetic and physical map rme unc-5skn-1 him-3 unc-77 fem-1 sup-41 mor-2 sup-23 evl-7 Genetic map map unit Cloned genes Non-cloned genes mDf4 nDf41 rme-2 unc-5 fem-1skn-1 Physical map cosmids YACs Named genes eP14 nhr-48 Genetic map and physical map will completely unified when every gene has been mutated. A: yes. B: No. C: not sure.
A few Dumpy (Dup) dpy X A few wild type Uncoordinate (Unc) unc X dpy + Wild type Pick several male progeny Segregate no dumpy progeny, discard If the two genes are linked unc unc + + dpy Segregate dumpy progeny, continue mapping with the plates dpy + unc + ; unc + ; Pick several wild-type hermaphrodites and place each to one pate If the two genes are unlinked 50% Figure Linkage analysis between two mutations. +; dpy+; + unc; + + dpy+ unc +
Wild type 9/16 genotypes unc + dpy + ; dpy + ; unc + ; 4/16 2/16 + ; 1/16 Unc 3/16 unc dpy + ; 2/16 unc + ; 1/16 Dpy 3/16 unc + dpy ; 2/16 + dpy ; 1/16 Dpy-and-Unc 1/16 unc dpy ; 2/16 Situation 1. The two genes are unlinked unc + dpy + ; Phenotypes of worms In the plate A single plate Pick 20 Unc animals If ~ 2/3 of these Unc worm segregate Dpy-and-Unc animals, non-linkage between the two genes is deduced.
Situation 2. The two genes are linked on one of the six chromosomes A single plate + dpy unc + Phenotypes Wild type 1/2 Non-recombinant Genotypes + dpy unc + Dpy-and-Unc 0 unc dpy Unc 1/4 unc + Dpy 1/4 + dpy Pick 20 Unc animals linkage between the genes is indicated by segregation of no Dpy animals from all or the majority of Unc animals. majority unc + Self fertilizing The frequency of rare recombinants that segregate Dpy animals is correlated with the genetic distance between the two genes. unc dpy Rare unc ++ dpy unc + + dpy unc dpy unc + unc dpy + dpy Rare recombinants unc dpy
Figure Example of genetic three point mapping unc dpy egl mapping strain phenotype Wild type hermaphrodite genotype unc dpy egl meiosis Sperms or eggs Most of the gametes Rare gametes from a recombination event dpy egl unc unc + dpp unc + dpy Progeny from combination of the common gametes unc + dpp + egl + Progeny from combination between common gametes and recombinant gametes unc + dpp + egl dpy unc + dpp unc + + Self-fertilizing Wild type Unc and Dpy Egl Dpy non-Unc Unc non-Dpy Wild type unc egl + + egl dpy + egl + Egl Sperms or eggs
unc dpy egl to the right of the egl dpy egl unc dpy egl unc unc + dpp + egl dpy unc + dpp unc + + unc + dpp + + dpy unc + dpp unc egl + Dpy non-Unc No Unc non-Dpy Yes Dpy non-Unc Yes Unc non-Dpy No Progeny with Egl phenotype Recombination occursto the left of egl Recognizable recombinants Map position unc dpy egl a b a b = # of recombinations occurred to the left of egl # of recombinations occurred to the right of egl
egl unc X egl unc Genetic mutant derived from the strain from Bristol, England. The egl mutation is being mapped. A C. elegans strain from Hawaii. SNPs between this strain and the Bristol strain have been determined. egl unc ****** The hybrid strain. Stars indicate SNPs in the region Select Unc but non-Egl recombinants XXX egl unc ****** unc egl unc ****** unc egl unc ****** 6 Determine SNP #4 for all recombinant worms by sequencing or digestion. A B C Determine SNP #5 and #6 for those that have lost SNP#4 (worm C only) Worm C has SNP #6 but not #5: the egl gene maps to the right of SNP#5 Worms A and B have #4 SNP from the Hawaii strain Figure An example of genetic mapping using SNPs
Genetic mutation Injection of subclones sequencing mutant DNA Mapping using marker mutations SNP mapping RNAi of candidate genes Microinjection of cosmid/YAC clones Common steps involved in cloning C. elegans genes defined by mutations. What would be the flow chart for cloning in yeast? Fly? Human?
Which is the strongest evidence for claiming the cloning of the gene defined by the mutation? A. The transgene put back into the animal can rescue the mutant phenotype. B. You find a missense mutation in this gene by sequencing. C. Reducing the gene activity by RNAi mimics the mutant phenotype. D. The gene is expressed in the tissue with the mutant phenotype.
Figure Microinjection transformation in C. elegans. DNA solution is injected to the distal arms of the gonad unc-119(+) gene as a marker Select F1 transgenic animals, most are unstable F2 transgenic animals, stable lines Injection Three types of markers a strongly expressed GFP gene as a marker A dominant rol- 6 mutant gene as a marker unc-119(-) mutant Wild type Roller Wild type Green worm
Figure The prevailing model for the mechanism of RNAi dsRNA Introduced into cells Dicer Bind to Dicer-RDE- 1enzyme complex dsRNA is cut to ~22 nt siRNA Incorporated into RISC nuclease complex Unwinding siRNAs, activation of RISC AAAAAAAA 5’ Target mRNA Cleavage of target mRNA Multiple-protein components of RISC
Figure RNAi methods in C. elegans. dsRNA intestine A. Injecting ds RNA into intestine or gonad B. Soak worms with dsRNA solution Observe phenotype in progeny Soak for 24 hours Transfer to plates Feed worms the bacterial strain Observe phenotype in progeny Grow the bacterial strain containing the vector expressing dsRNA C. Feed the worms a bacterial strain that expresses dsRNA