Mapping Course Jorge Dubcovsky. Project funded by USDA-CSREES

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1 Mapping Course 2007 Jorge Dubcovsky Project funded by USDA-CSREES

2 efore Mapping Nulli-tetrasomic lines missing 1 pair of chromosomes that is compensated with a double dose of one homoeologous chromosomes. The chromosome location is identified by lack of amplification. Example microsatellite located on chromosome 1 1A 1 1D 2 2D 3A 3 3D 4A 4D 5A 5 5D 6A 6 7A 7 Ditelosomic lines missing 1 arm. The name indicates the arm PRESENT. Example microsatellite located on chromosome arm 1L 1S 1L Deletion stocks missing 1 segment. The name indicates the segment ASENT. Example microsatellite located on bin 1L1. 1L3 1L2 1L1 1L6 C-1L3 C-1L L L L Previous mapping check chromosome location in GrainGenes http//wheat.pw.usda.gov/gg2/index.shtml 1L Deletion bin map of chromosome 1

3 Mapping populations Temporary populations Immortal populations

4 Crossovers Meiotic chromosomes Meiosis with no crossover between loci 1 and 2. Meiosis with 1 crossover between loci 1 and 2. Meiotic products Parental A Parental A Parental Parental Parental A Recombinant Recombinant Linked loci can be separated by a physical exchange of chromosome parts during meiosis. This process, called crossover generates recombinant haploid genotypes that differ from the haploid parental genotypes. Parental There is an average of 1 crossover per short arm and 2 crossovers per long arm. Therefore the average chromosome length is 150 cm

5 Recombination frequency as estimate of distance L1 L2 L3 L4 There is a rough proportionality between the distance between linked loci and the chance of a crossover between non-sister chromatids between loci. y determining the frequency of recombinants, it is possible to obtain a measure of the map distance between the genes. Parental = 7 Parental A3 A4= 10 Parental = 7 Recombinant = 5 Recombinant = 5 Parental 3 4= 10 Recombinant A3 4= 2 Recombinant 3 A4= 2 Rf= 10/24= 42% Rf= 4/24= 16% A recombinant frequency of 0.01 is the distance between loci for which one product of meiosis out of 100 is recombinant

6 Chiasma interference Recombination frequency per length unit in wheat (Lukaszewski and Curtis 1993) In wheat there is strong chiasma interference (one chiasma reduces the pb. of 2nd chiasma in its proximity). As a result of chiasma interference and the initiation of chromosome pairing from telomeres, chiasma in wheat are favored at the end of the chromosomes. Small genetic distances in the centromeric region correspond to large genetic distances in the wheat physical maps.

7 Recombination fraction and CentiMorgans Meiotic chromosome Meiotic product Parental A Parental Parental A Parental When two crossovers are produced between two markers, no recombination is detected unless an additional markers is included in the middle. Mapping functions (Haldane and Kosambi) correct the recombination fraction r for this possibility. Haldane does not consider interference and is not appropriate for wheat Kosambi considers chiasma interference cm=100*[¼ log [(1+2r)/(1-2r)] Recombination fraction Rf= 0.01 Rf= 0.02 Rf= 0.10 Rf= 0.20 Rf= 0.49 Kosambi cm 1.0 cm 2.0 cm 10.1 cm 21.2 cm cm

8 Three point analysis If the distance between three linked markers is known A cM A-----3cM-----C C cM We can determine the likelihood of each order C-----3cM-----A cM Likely A-----3cM-----C---2cM--- Unlikely A cM cM C Very unlikely The Mapmaker command THREE POINT calculates the likelihood of the three alternative orders for each three points and discards those that are very unlikely

9 Expected Genotypic Classes Expected Genotypic Classes Population Dominant markers Codominant markers C1F F RIL or RSL DH

10 Significance of linkage Example of observed and expected segregation values. χ2 Example of RSL, two locus L1 L2 Genotype Observed Expected AA A A Example of F2, two locus L1 L2 Genotype AA Observed 5.0 Expected 1.5 AH A HA HH H A H The df of the χ2 is the number of classes minus 1. Look for the critical value in a χ2 table critical value α=0.05, χ23df= 7.81 Calculate χ2=σ[(o-e)2/e]=( )/6.25= 0.44 Since 0.44<7.81, there are no significant differences with the expected frequencies, and therefore there is no significant linkage

11 Calculating recombination frequencies in DH populations In a DH, backcross, or RSL population the calculation of the recombination fractions is very simple Count the number of recombinant chromosomes and divide by the total The total is the number of chromosomes for which we have information for the two loci RSLs L1 A A A - A A A A A A A A A L2 A A A - A A A A A A A A A A #CO There are 12 recombiantion events in 25 scored pairs RF= 12/25= 0.48 Mapmaker calculates it as 0.48

12 Calculating recombination frequency in F2 populations In an F2, it is not possible to determine whether the H1H2 are parental or recombinant. The following examples show two different pairs of chromosomes in an F2 individual, one parental and one recombinant. The crossovers are indicated by x. Example 1 H1H2 genotype from parental chromosomes A Score A x A x A H1 H2 H Example 2 H1H2 genotype derived from recombinant chromosomes Score A x H H1 x H2 A H A H A H Therefore, in an F2 counting the recombinants is just an approximation. Mapmaker uses Maximum Likelihood to estimate the distances more precisely Example 3 estimating RF in an F2 L3 L4 #CO A H A H H H A H H A A H H A A H H H H 0 H 0 H 0 0 A 0 H 0 0 H 0 A 0 A 0 H 0 H A 2 0 H 0 H 0 H 0 1 H 0 A 0 H 0 0 RF 3/ Mapmaker calculates it more precisely as 0.065

13 LOD scores The maximum likelihood method of estimation is a general statistical method to estimate the value of a parameter the value chosen for the parameter is that which is most probable (or most likely) given the occurrence of a certain set of observations. LOD score (Log10 of the odds ratio) is calculated as Pb. of obtaining the observed data under linkage LOD= log Pb. of obtaining the observed data under random assortment The ratio between the two probabilities is transformed to log10. The higher the value the more likely is the linkage. For example, a LOD=3 indicates that is 1000 times (103) more likely to obtain the observed data from linked markers than from unlinked ones. LOD scores are used to determine the most likely order of markers in a map. Mapmaker assigns 0 to the better order and negative LODs to alternative orders. For example, if order a b c has LOD=0 and order La Lc Lb has a LOD score of 2, that indicates that it is 100 (102) less likely to obtain the observed data from the 2nd order than from the 1st order.

14 Recombinant inbred lines Since RILs have more generations for recombination, recombination frequencies need to be adjusted. Recombination levels observed in RILs can be related to recombination in a population derived from a single meiosis using Haldane and Waddington (1931) formulas. 2r R = r R r = R or Some examples R r R/r When linkages are tight, the recombination frequency among RILs is twice the conventional F2 rate.

15 Mapping advice Take extreme care entering genotypic data. Merge linked markers into single group Review all double crossovers (if possible re-extract DNA and re run marker) Too many double crossovers for a marker indicate problems. Repeat PCR! Too many double crossovers along an individual suggest DNA problems. Recheck conflicting markers with the same tube of DNA. Try to have 1 DNA for the complete project, if not possible keep track of DNA replacements using a version number/date. Record mapping date of each markers If in doubt regarding the genotype of an individual, it is better to enter it as a missing value than to guess. Each error expands your map 2 crossovers!

16 Map comparisons Compare order with previous maps. http//maswheat.ucdavis.edu/ Use our site in collaborator s area user cap05, password Genomic$ Map visualization tool user wheatcap, password jagger We can enter your map there! Japanese site for the Wheat Gene Catalogue http//map.lab.nig.ac.jp8080/cmap/

17 Answers to exercises Exercise 1 You have assigned four loci to chromosome 1A of wheat using nullitetrasomics. Using fifty double haploids from a hybrid between two varieties A and you recover the following number of recombinant and parental chromosomes Locus 1- Locus 2 A5 A0 4 -> 1/50=0.02 = 2% Locus 1- Locus 3 A3 A3 A4 0 -> 7/50=0.14 = 14% Locus 1- Locus 4 A5 A 3 -> 2/50=0.04 = 4% Locus 2- Locus 3 A1 A3 A5 -> 8/50=0.16 = 16% Locus 2- Locus 4 A4 A 3 -> 3/50=0.06 = 6% Locus 3- Locus 4 A3 A A3 2 -> 5/50=0.10 = 10% Construct a map of these four loci. Answer L2- (2%) - L1- (4%) - L4 (10%) - L3 Exercise 2 Chi Square for Double haploids, RSLs or ackcrosses In similar population of 100 double haploids you mapped two loci 1 and 5. You checked parental and recombinant chromosomes L1 L5 Observed A8 4 6 L1 L5 Expected A5 A5 5 5 O-E χ2=σ[(o-e)2/e]=( )/25= 20/25=0.8 Critical χ2 = 7.81 No significant linkage

18 Answers to exercises Exercise 3 χ2 for the F2 population =24 individuals and 16 classes Expected per class 24 individuals / 16 classes = 1.5 per class Genotype H2 Observed Expected Difference 3.5 H H1H2 3.0 H H2 1.5 χ =Σ[(O-E) /E]=38.2 Critical value Df=8 χ = Exercise 4 Recombination fractions in RI or SSD The following is the data for MapMaker for a Recombinant Inbred Population of 50 lines and 4 loci (RI or SSD). Calculate the corrected RF r between loci 1-2, 2-3 and 3-4. Remember that in a SSD r= R/(2-2R) where R is the proportion of recombinants in the RI. data type ri self *locus1 *locus2 *locus3 *locus4 L1-L2= 1/50 L2-L3= 4/50 L3-L4= 10/50 R= 0.02 R= 0.08 R= 0.20 baaaa r= r= r= 0.125

19 Answers to exercises

20 Answers to exercises

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