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Meiosis, Recombination and Mapping • • • • Meiosis Why bother to map your gene? Mapping a mutant defining a single gene Mapping a sequence Material for you: •This presentation •PDFs of important papers •Review of basic Genetics •Problem sets with answers •Definitions of IBM map words Contact me anytime! Dempsy. Maize Handbook Meiosis takes about 5 days, then there is about 8-10 more days to shed Maize meiosis Three major processes in meiotic prophase Leptotene 1. Chromosomes become visable as threads 2. The Axial Element is installed Leptotene-Zygotene Transition 1. The bouquet is formed 2. Dramtic transient change in chromatin structure 3. It all begins… Zygotene • • • Pairing Synapsis Recombination Pachytene 1. Chromosomes are completely synapsed 2. Recombination is completed 3. Interference probably works here MITOSIS METAPHASE ANAPHASE MEIOSIS METAPHASE I ANAPHASE I METAPHASE II ANAPHASE II Sister Chromatid cohesion and chiasmata holds bivalents together until the Metaphase to Anaphase to transition Diplotene- Diakinesis 1. SC falls apart 2. Chromosomes further condense 3. Homologs held together by chiasmata Homologous Chromosome Pairing Synapsis Axial element Lateral Element Synapsis starts at the chromosome ends Anderson and Stack, Chr. Res 2002 Complete Synapsis Anderson et al, Genetics 2003 Recombination 5’ 3’ Lichten Number of RAD51 foci Leptotene Zygotene 500 400 300 200 Early pachytene Diplotene Pachytene Late zygotene Mid zygotene 0 Early zygotene 100 Leptotene Distribution of RAD51 recombination protein in normal maize meiosis 600 Late pachytene Wojtek Pawlowski Early Recombination Nodules in TEM Anderson, et al Genetics 2001 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. Allers and Lichten Cell 2001 Patterns of distribution of RAD51 recombination protein in normal maize meiosis. Organisms rely on crossovers to segregate chromosomes… … but most organisms make very few crossovers per chromosome pair species S.c. N.c. C.e. A.t. D.m . M .m . H.s. Z.m. haploid 16 chromosome number (n) 7 6 5 4 20 23 10 genetic map length (m.u.) 4300 1040 300 520 285 1600 3300 1600 haploid genome size (bp) 1.2 x 107 2.7 x 107 1.0 x 108 1.0 x 108 1.4 x 108 3.0 x 109 3.3 x 109 5.4 x 109 length of DNA/ 2.8 m.u. (kb/m.u.) 26 330 200 490 1900 1000 3400 crossovers per chromosome 2-5 1 2-3 0-2* 1-2 1-7 2-4 3-8 Anne Villeneuve Inna Golubovskaya Break The Genetic Maps of Maize There 1176 genetic maps of maize. We will only use the old fashion “genetic” map and the IBM2 map Where is your gene? • • • • Relative to other genes? One the chromosome? Why do you care? How do you find out? Your gene relative to other Genes • The genetic map is based on an abstract principle: the percent recombination between two genes (or markers). • The amount of recombination is not uniform along the length of the chromosome- but that does not matter for mapping. f1 an1 rhl d8 150 170 184 197 Your gene on the chromosome Why do you care? • It makes to whole map better. • Maybe there is already a sequence mapped to your gene! • Maybe there will be soon • Maybe you think that the mutant of your gene is just like a mutant in yeast. Using the yeast sequence, get some maize GSSs, map, do they map to your gene? • Map-based cloning may (will) be possible How do you map a gene? • You need a DIFFERENCE to map- two alleles • First localize in a general region • Next fine structure mapping How do you map two genes relative to each other? y10 lg2 y10 lg2 Y10 Lg2 y10 lg2 Gametes: Parental Recombinant { { y10 lg2 y10 lg2 Y10 Lg2 y10 Lg2 Y10 lg2 Test Cross: Heterozygote cross multiply recessive (or co-dominant) tester line. The genotype of the gametes of the “tested” parent is also the phenotype of the progeny. y10 lg2 y10 lg2 Y10 Lg2 y10 lg2 Genotype Parentals Recombinants { { Phenotype Number of Progeny y10 lg2 y10 lg2 Yellow, liguleless 48 Y10 Lg2 y10 lg2 Wildtype 42 y10 Lg2 y10 lg2 Yellow 6 liguleless 4 Y10 lg2 y10 lg2 y10 lg2 y10 lg2 Y10 Lg2 y10 lg2 Number of Progeny 48 P: Recombinants X 100 = cM Total Progeny 42 6 R: 4 6+4 X 100 = 10% = 10cM 48+42+6+4 10 cM between y10 and lg2 How can you determine gene order?? Three (or more) factor cross mwy MWY MWY X mwy F1 MWY mwy X True Breeding X wild type mwy mwy Phenotype Min, white, yellow Yellow White eyes Min Yellow, white eyes White eyes, min Min, yellow Wild type True breeding miniature, white-eyed, yellow bodied fly Test cross (Genotype) mwy y++ w++ m++ yw+ wm+ my+ +++ Number 3501 25 3 1720 1734 35 6 3471 Put the 8 progeny types into 4 reciprocal classes. The number of progeny per class is a clue. P m w y 3501 Parental + + + 3471 mwy Parental: +++ m + + 1720 SR y w + 1734 Single recombinants w m + 35 SR y + + 25 Single recombinants DR w++ 3 my+ 6 Double Recombinants M w y Parental: + + + P m w y 3501 + + + 3471 m + + 1720 SR + w y 1734 w m + 35 SR + + y 25 DR +w+ 3 m+y 6 Recombinants X 100 = cM Total Progeny M w y Parental: + + + P m w y 3501 + + + 3471 m + + 1720 SR + w y 1734 w m + 35 SR + + y 25 DR +w+ 3 m+y 6 Recombinants X 100 = cM Total Progeny Recombination between m and w: (1720 + 1734 + 3 + 6) / 10495 = 33cM M w y Parental: + + + P m w y 3501 + + + 3471 m + + 1720 SR + w y 1734 w m + 35 SR + + y 25 DR +w+ 3 m+y 6 Recombinants X 100 = cM Total Progeny Recombination between m and w: (1720 + 1734 + 3 + 6) / 10495 = 33cM Recombination between w and y: (35 + 25 + 3 + 6) / 10495 = 0.7cM M w y Parental: + + + P m w y 3501 + + + 3471 m + + 1720 SR + w y 1734 w m + 35 SR + + y 25 DR +w+ 3 m+y 6 Recombinants X 100 = cM Total Progeny Recombination between m and w: (1720 + 1734 + 3 + 6) / 10495 = 33cM Recombination between w and y: (35 + 25 + 3 + 6) / 10495 = 0.7cM Recombination between m and y: (1720 + 1734 + 35 + 625) / 10495 = 33.5cM M w y Parental: + + + P m w y 3501 + + + 3471 m + + 1720 SR + w y 1734 w m + 35 SR + + y 25 DR +w+ 3 m+y 6 Recombinants X 100 = cM Total Progeny Recombination between m and w: (1720 + 1734 + 3 + 6) / 10495 = 33cM Recombination between w and y: (35 + 25 + 3 + 6) / 10495 = 0.7cM Recombination between m and y: (1720 + 1734 + 35 + 625) / 10495 = 33.5cM W is in the middle, because it took two crossovers to separate it from the other allele M w y Parental: + + + P m w y 3501 + + + 3471 m + + 1720 SR + w y 1734 w m + 35 SR + + y 25 +w+ 3 DR m+y 6 Recombinants X 100 = cM Total Progeny Recombination between m and w: (1720 + 1734 + 3 + 6) / 10495 = 33cM Recombination between w and y: (35 + 25 + 3 + 6) / 10495 = 0.7cM Recombination between m and y: (1720 + 1734 + 35 + 625) / 10495 = 33.5cM 33 cM m 0.7cM w y Crossover Interference: Crossover in one region of a chromosome reduces the probability of a second crossover nearby. If no interference: Freq. of d.c.o. involving interval x and y = (RF x) (RF y) x y 33 cM m 0.7cM w y Exp dco=2.3% Obs dco =0.08 “coefficient of coincidence” C = obs d.c.o. / exp d.c.o. C = 1 intervals are independent C = 0 complete interference C=0.08/2.3 How can you map your MUTANT? In most cases, you must do crosses First localize it to a chromosome arm by: -Crossing to many TB or T-waxy translocation lines and scoring. This takes two (TB) to three (T-waxy) generations. Then, make crosses to close markers, and map. A possible discussion topic. And/Or -Make a mapping population, and use molecular markers to localize, and to fine map Molecular Markers for mapping: RFLPs SSRs AFLPs Indels SNPs RFLP: Restriction Fragment Length Polymorphism Allele A Allele B GCCGCATTCTATAAAG CGGCGTAAGATATTTC GCCGAATTCTATAAAG CGGCTTAAGATATTTC Detect by Southern blot: A/A A/B B/B SSR: simple sequence repeat (length polymorphism) PCR Primer (CA)n (CA)m variation between strains in number of repeats at a given locus PCR yields products of different size: Inserted Elements as Genetic Markers e.g. presence or absence of mapped transposon Product No Product SNP: single nucleotide polymorphisms Outgrowth of sequencing projects If they affect a restriction enzyme site: snip-SNP Detection of SNPs can be done without gels: highly automated/high throughput and/or highly parallel (simultaneous scoring of MANY markers) “Snip-SNP” Single nucleotide polymorphism affecting a restriction endonuclease site Step 1) PCR Step 2) cut A B Indels: Insertions- deletions Through sequencing efforts, small insertions or deletions (indels)have been discovered. These can be detected by PCR with unique primers. A B Each chromosome is divided into bins These bins are about 20 cM Bins are defined by molecular markers QuickTime™ and a TIFF (LZW) decompressor are needed to see this picture. The 90 core bin markers are at http://www.maizegdb.org/cgi-bin/bin_viewer.cgi Map your mutant relative to a molecular marker First, make a mapping population: New mutant: lg5. This mutant was found in an MTM (Mutator) population. You introgressed into B73 lg5 In a B73 background lg5 B1 B2 lg5 B3 B4 B1 B2 lg5 B3 B4 F1 x x Mo17 M1 M2 Lg5 M3 M4 M1 M2 Lg5 M3 M4 B1 B2 lg5 B3 B4 M1 M2 Lg5 M3 M4 F1 B1 B2 lg5 B3 B4 M1 M2 Lg5 M3 M4 x B1 B2 lg5 B3 B4 B1 B2 lg5 B3 B4 Mapping population This population with segregate 1:1 for lg5. You will make DNA from some number of lg plants, and an equal number of wild type plants. B73 Mo17 lg 1 lg 2 lg 3 lg 4 lg 5 lg 6 lg 7 lg 8 lg 9 lg 10 wt 1 wt 2 wt 3 wt 4 wt 5 wt 6 wt 7 wt 8 wt 9 wt10 DNA extracted Digested with EcoR1/HindIII Southern Probed with tub1 P: mutant allele of lg5 and B73 tub1 allele: R: mutant allele of lg5 and Mo17 tub1 allele: R: Wt allele of Lg5 and B73 tub1 allele: P: Wt allele of Lg5 and Mo17 tub1 allele: 5+5 20 = 50% UNLinked! 5 5 5 5 B73 Mo17 lg 1 lg 2 lg 3 lg 4 lg 5 lg 6 lg 7 lg 8 lg 9 lg 10 wt 1 wt 2 wt 3 wt 4 wt 5 wt 6 wt 7 wt 8 wt 9 wt10 DNA extracted Digested with Pst1 Southern Probed with umc128 P: mutant allele of lg5 and B73 tub1 allele: R: mutant allele of lg5 and Mo17 tub1 allele: R: Wt allele of Lg5 and B73 tub1 allele: P: Wt allele of Lg5 and Mo17 tub1 allele: 1+1 20 9 1 1 9 = 10% = 10 cM Linked! If you have mapping data, submit a maize newsletter article, and Mary Polacco will incorporate map data. Soon you can import mapping data directly to MaizeGDB, but still, write the MNL article so everyone can see the data! How do you map a sequence? High resolution maps: IBM2 and IBM2 neighbor map Why the huge difference? Making the Intermated B73 X MO17 mapping population Mo17 B73 X Single F2 plant was selfed X Grow up 200 plants, random mating Genotype of 5 of those 200 plants Select 100 ears, pick 5 kernals from each ear. Put in a bag, shake, plant, more random matings (2nd generation). Repeat, (repeat….) From these lines, generate Recombinant Inbred lines by repeated selfing (5x or more) x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Cool, but how do you map with them?? Date: Tue, 2 Mar 2004 13:43:57 -0600 To: [email protected] From: Ed Coe <[email protected]> Reply-To: [email protected] Subject: Want to know the map location of your gene? Want to know the map location of your gene? You can easily obtain maize map location with the Community IBM Map Data Entry (CIMDE) utility. Kits containing a 96-well plate with DNA of 94 IBM lines and the parents, with protocols, are available: http://www.maizemap.org/dna_kits.htm (alternatively, seeds of the 94-member panel of IBM stocks can be requested from the Stock Center) Data submission: http://www.maizemap.org/CIMDE/cimde.html Maps are generated and reported-back promptly after the data are submitted. The data, if you so choose, will be incorporated into the combined community map, cIBM (see MaizeGDB for recent posting of maps cIBM 1 through 10). Ed Coe Karen Cone Georgia Davis Mary Polacco $240 There are 288 lines in the IBM population. They send you DNA from 94 representatives. You use primers from your gene of interest, that amplify a different length fragment in Mo17 vs B73. Use these pimers on all 94 lines: B73 Mo17 PCR + EcoR1 B73 Mo17 Line1 Line2 Line3 Line4 Line5 Line6 Line7 Line8 Line9 Line10 EcoR1 So, you gene’s ‘code’ is: MMBBMMBBMB… Then, the MMPers figure out, with their computers, where that matches in the genome. So, you gene’s ‘code’ is: MMBBMM … Line 1 Line 2 Line 3 MMMMMB Nope! MBMBMM Nope! MMBBMM YEAH! Line 4 Line 5 Line 6 High resolution maps: IBM2 and IBM2 neighbor map Why the huge difference? IBM2 vs IBM2 neighbors On frame vs off frame Oat x maize addition line Genomic in situ hybridization of an addition line with fluorescein-labeled maize DNA. The pair of maize chromosomes are yellow-green, and the oat chromosomes are counterstained red with propidium iodide. Chromosome 1 addition line Unusual characteristics noticed in chromosome 1 OMA lines:Erect leaf blade, photoperiod neutral response, sectoring among shoots for the maize chromosome Chromosome 10 Addition Line Characteristics of chromosome 10 OMA lines: grassy type and very sterile Oat-corn chromosome 9 monosomic addition line Oat line without a corn chromosome Oat line with a corn chromosome piece transfer Selfing Oat line with a modified corn chromosome Mapping with OMA lines primer33A primer33C primer34A primer34C primer34A: will amplify ESTs AW066336 and AW066011. These two sequences are very similar, e-172 and 335/343 (97%) and may represent two alleles from one gene. primer34C: was designed to amplify TC77059. This sequence is more distantly related to AW066336, e-112, 269/289 (93%). The gene represented by ESTs AW066336 and AW066011 appears to lie on chromosome 5. The PCR primers, primer34C, detects the related sequence on chromosome 4. 3-D Deconvolution Microscopy raw data deconvolved data