mapping course-lecture 1

8/14/2019 Mapping Course-Lecture 1

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Mapping Course 2007

Project funded by USDA-CSREESProject funded by USDA-CSREESJorge DubcovskyJorge Dubcovsky


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Before Mapping

6B 7A

7B6A5D5B5A4D4A3D3B3A2D2B1D1B1A

1BL1BS

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 1B

Ditelosomic lines: missing 1 arm. The name indicates the arm PRESENT.Example: microsatellitelocated on chromosome arm 1BL

1BL2 1BL1 1BL6

C-1BL31BL3

1BL3-0.85-100

1BL2-0.69-0.85

1BL1-0.47-0.69

1BL6-0.32-0.47

C-1BL6-0.32

Deletion bin map of chromosome 1B

Deletion stocks: missing 1 segment. The name indicates the segmentABSENT.

Example: microsatellite located on bin 1BL1.

Previous mapping: check chromosome location in GrainGeneshttp://wheat.pw.usda.gov/GG2/index.shtml
http://wheat.pw.usda.gov/GG2/index.shtmlhttp://wheat.pw.usda.gov/GG2/index.shtml


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Mapping populations

Temporary populations Immortal populations


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Crossovers

Meiotic chromosomes Meiotic products

Meiosis with

no crossover

between loci

1 and 2.

Meiosis with

1 crossover

between loci

1 and 2.

Parental A

Parental A

Parental B

Parental B

Parental ARecombinant

Parental B

Recombinant

A1 A2

A1 A2

A1 A2

A1 A2

A1 A2A1 A2

A1 A2 A1 B2

B1 A2

B1 B2

B1 B2

B1 B2

B1 B2 B1 B2

B1 B2

B1 B2

Linked loci can be separated

by a physical exchange of

chromosome parts during

meiosis.

This process, called crossover

generates recombinant

haploid genotypes that differfrom the haploid parental

genotypes.

There is an average of 1crossover per short arm and

2 crossovers per long arm.

Therefore the average

chromosome length is 150

cM


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Recombination frequency as estimate of distance

L1 L2 L3 L4

Parental A1 A2= 7

Parental B1 B2= 7

Recombinant A1 B2= 5

Recombinant B1 A2= 5

Rf= 10/24= 42%

Parental A3 A4= 10

Parental B3 B4= 10

Recombinant A3 B4= 2

Recombinant B3 A4= 2

Rf= 4/24= 16%

There is a rough proportionality

between the distance betweenlinked loci and the chance of a

crossover between non-sister

chromatids between loci.

By determining the frequency

of recombinants, it is possible to

obtain a measure of the mapdistance between the genes.

A recombinant frequency of

0.01 is the distance between loci

for which one product of

meiosis out of 100 is

recombinant


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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.


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Recombination fraction and CentiMorgans

Meiotic chromosome

Parental AA1 A2

A1 A2

B1 B2

B1 B2

A1 A2

A1 A2

B1 B2

B1 B2

Meiotic product

Parental A

Parental B

Parental B

When two crossovers are produced between two markers, no recombination is detected unlessan 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*[ log [(1+2r)/(1-2r)]

Recombination fraction Kosambi cM

Rf= 0.01 1.0 cMRf= 0.02 2.0 cMRf= 0.10 10.1 cMRf= 0.20 21.2 cM

Rf= 0.49 114.9 cM


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Three point analysis

If the distance between three linked markers is known

A--------5cM---------B

A-----3cM-----C

C------------8cM---------------B

We can determine the likelihood of each order:

C-----3cM-----A--------5cM---------B Likely

A-----3cM-----C---2cM---B Unlikely

A--------5cM---------B----------8cM-------------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


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Expected Genotypic Classes

1:1:1:11:1:1:1DH

1:1:1:11:1:1:1RIL or RSL

1:2:1:2:4:2:1:2:19:3:3:1F2

1:1:1:11:1:1:1BC1F

2

Codominant

markers

Dominant markersPopulation

A1 A2

A1 A2

B1

B2

B1 B2Expected Genotypic Classes


10/21

Significance of linkage

Example of observed and expected segregation values.

38.2Example of F2, two locus L1 L2

Genotype: AA : AH : AB : HA : HH : HB : BA : BH : BB

Observed: 5.0 : 2.0 : 0:0 : 0:0 :11.0 : 0:0 : 0:0 : 0:0 : 6.0

Expected: 1.5 : 3.0 : 1.5 : 3.0 : 6.0 : 3.0 : 1:5 : 3.0 : 1.5

0.44Genotype: AA : AB : BA : BB

Observed: 6.00 : 5.00 : 7.00 : 7.00

Expected : 6.25 : 6.25 : 6.25 : 6.25

Example of RSL, two locus L1

L2 2

The df of the 2 is the number of classes minus 1.

Look for the critical value in a 2 table: critical value =0.05, 23df

= 7.81

Calculate 2=[(O-E)2/E]=(0.252+1.252+0.752+0.752)/6.25= 0.44

Since 0.44


11/21

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 B B B A B B - A A A A A A B A B B B A B A B B B B

L2 A A B B B B B A B B - B B B A A A B A A A B A A A A B

#CO 0 0 0 0 0 1 0 1 1 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 0

There are 12 recombiantion events in 25 scored pairs

RF= 12/25= 0.48 Mapmaker calculates it as 0.48


12/21

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 A1

A2

A x B B

A x B1

B2

B B B

Score: A H1 H2 H B B

Example 2: H1H2 genotype derived from recombinant chromosomesA A

1x B

2B B B

B B1

xA2

A A A

Score: H H1

H2

H H H

Therefore, in an F2counting the recombinants is just an approximation.

Mapmaker uses Maximum Likelihood to estimate the distances more precisely:

Example 3 estimating RF in an F2:

L3 A H A H B H H B A H B H A A H H B B A A H H H B

L4 H H A H B H H B A H B H A A H H B B A B H H H B#CO 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0

RF 3/48 0.0625 Mapmaker calculates it more precisely as 0.065


13/21

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

(ormost 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 abc 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.


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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 apopulation derived from a single meiosis using Haldane and Waddington (1931)formulas.

2r R R= ------- or r = -------

1+2r 2-2R

Some examples

R r R/r

0.01 0.005 1.98

0.1 0.056 1.80

0.2 0.125 1.60

0.3 0.214 1.40

0.4 0.333 1.20

0.5 0.500 1.00

When linkages are tight, the

recombination frequency

among RILs is twice the

conventional F2 rate.


15/21

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!


16/21

Map comparisons

Compare order with previous maps.http://maswheat.ucdavis.edu/

Use our site in collaboratorsarea: user: cap05, password:Genomic$

Map visualization tooluser: wheatcap, password:

jagger

We can enter your map there!

Japanese site for theWheat Gene Catalogue:http://map.lab.nig.ac.jp:8080/cmap/
http://maswheat.ucdavis.edu/http://map.lab.nig.ac.jp:8080/cmap/http://map.lab.nig.ac.jp:8080/cmap/http://maswheat.ucdavis.edu/


17/21

Answers to exercises

Exercise 1: You have assigned four loci to chromosome 1A of wheat using nullitetrasomics. Using fifty doublehaploids from a hybrid between two varieties A and B you recover the following number of recombinant and parentalchromosomes:

Locus 1- Locus 2: AA:25 AB:0 BA:1 BB:24 -> 1/50=0.02 = 2%Locus 1- Locus 3: AA:23 AB:3 BA:4 BB:20 -> 7/50=0.14 = 14%Locus 1- Locus 4: AA:25 AB:1 BA:1 BB:23 -> 2/50=0.04 = 4%Locus 2- Locus 3: AA:21 AB:3 BA:5 BB:21 -> 8/50=0.16 = 16%Locus 2- Locus 4: AA:24 AB:2 BA:1 BB:23 -> 3/50=0.06 = 6%Locus 3- Locus 4: AA:23 AB:2 BA:3 BB:22 -> 5/50=0.10 = 10%

Construct a map of these four loci.AnswerL2- (2%) - L1- (4%) - L4 (10%) - L3

Exercise 2: Chi Square for Double haploids, RSLs or BackcrossesIn similar population of 100 double haploids you mapped two loci: 1 and 5. You checked parental and recombinantchromosomesL1 L5 Observed: AA:28 AB:22 BA:24 BB:26L1 L5 Expected: AA:25 AB:25 BA:25 BB:25|O-E| 3 3 1 1

2=[(O-E)2/E]=(32+32+12+12)/25= 20/25=0.8 Critical 2 = 7.81 No significant linkage


18/21


Exercise 3: 2 for the F2 population

5+2+11+6=24 individuals and 16 classesExpected per class: 24 individuals / 16 classes = 1.5 per class

1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 :1

1.5 : 3.0 : 1.5 : 3.0 : 6.0 : 3.0 : 1.5 : 3.0 : 1.5

Genotype: A1A

2: A

1H2: A

1B2: H

1A

2: H

1H2

: H1B2

: B1A

2: B

1H2

: B1B2

Observed: 5.0 : 2.0 : 0.0 : 0.0 : 11.0 : 0.0 : 0.0 : 0.0 : 6.0

Expected: 1.5 : 3.0 : 1.5 : 3.0 : 6.0 : 3.0 : 1:5 : 3.0 : 1.5 .

|Difference| 3.5 1.0 0.5 3.0 5.0 3.0 1.5 3.0 4.5

2 =[(O-E)2/E]=38.2 Critical valueDf=8 2 =15.51

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 self50 4 0*locus1 aaaaa aaaaa bbbbb bbbbb aaaaa bbbbb aaaaa bbbbb aaaaa bbbbb

*locus2 aaaaa aaaaa bbbbb bbbbb baaaa bbbbb aaaaa bbbbb aaaaa bbbbb

*locus3 aaaaa aaaaa bbbbb bbbbb bbbbbbbbbb aaaaa bbbbb aaaaa bbbbb

*locus4 aaaaa aaaaa bbbbb bbbbb bbbbb aaaaa bbbbb bbbbb aaaaa bbbbb

L1-L2= 1/50 R= 0.02 r= 0.010

L2-L3= 4/50 R= 0.08 r= 0.043

L3-L4= 10/50 R= 0.20 r= 0.125


19/21



20/21



21/21

Answers to puzzle

Xabg601 AAAHAHAAAAAHAAAAAAHHHA AAHAAHHAAHHHAAHAAA AAHAAAAHAHHHAAAA

X X

Xabg602 AAAHAHAAAAAHAAHAAAHHHA AAHAAHHAAHHHAAHAAA AAHAAAAHHHHHAAAA

X X

XBamy AAAHAHAAAAAHAAHAAAHHHA AAHAAHHHAHHHAAHAHA AAHAAAAHHHHHAAAA

X

XksuH11 AAAHAHAAAAAHAAHAAAHHHA AAHAAHHHAHHHAAHAHA AAHAHAAHHHHHHAAA

X X X X

Xabc305 AAAHAHAAAAHHAAHAAAHHHH AAHAAHHHHHHHAAHAHA AAHHHAAHHHHHHHAA

X

Xmwg2112 AAAHAHAAAAHHAAHAAAHHHH AHHAAHHHHHHHAAHAHA AAHHHAAHHHHHHHAA

X X X X X X

Xbcd1302 AHAHAHAAAHHHAAHAAHHHHH AHHAAHHHHHHHHAHAHA AAHHHAHHHAHHHHAH

X X X X X

Xpsr164 AHAHAHAAHHHHAAHHAHHHHH AHHAHHHHHHHHHAHAHA AHHHHAHHHAHAHHAH

X X

XksuC2 AHAHAHAAHHHHHAHHAHHHHH AHHAHHHHHHHHHAHAHA AHHHHAHAHAHAHHAH

X

Xpsr1051 AHAHAHAAHHHHHAHHAHHHHH AHHAHHHHHHHHHAHAHH AHHHHAHAHAHAHHAH

X X

Xbcd1262 AHAHAHHAHHHHHAHHHHHHHH AHHAHHHHHHHHHAHAHH AHHHHAHAHAHAHHAH

X XXDhn6 AHAHAHHAHHHHHAHHHHHHAH AHHAHHHHHHHHHAHHHH AHHHHAHAHAHAHHAH

X X

Xpsr922 AHAAAHHAHHHHAAHHHHHHAH AHHAHHHHHHHHHAHHHH AHHHHAHAHAHAHHAH

XX

Xcdo669 AHAAAHHAHHHHAHAHHHHHAH AHHAHHHHHHHHHAHHHH AHHHHAHAHAHAHHAH

X X XX X X

Xwg622 AHAAA-HAHAHHAHAHHHHAAH AH-AHHHHAAHHHAHHHH AHHHHAHAHAHHHAAH

X X

Xpsr921 AHAAAAHAHAHHAHAHHHAAAH AHHAHHHHAAHHHAHHHH AHHHHAHAHAHHHAAH

mapping course-lecture 1

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