Download Linkage, Crossing Over, and Chromosome Mapping

Linkage, Crossing Over, and Chromosome Mapping
Chapters 5 and 6; 13&14
I. Linkage
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Linkage = genes on same chromosome
Linked genes violate Independent Assortment
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Dihybrid should produce 4 gametic types
Linked genes tend to follow Segregation
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Hybrid should produce 2 gametic types
I. Linkage
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Discovered by William Bateson and R.C. Punnett in early 1900’s
Bateson coined term “genetics”
I. Linkage
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Bateson and Punnett studied sweet pea
Flower color gene (P = purple, p = red)
Pollen grain shape (L = long, l = round)
Created dihybrid and made dihybrid cross
I. Sweet Pea
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Parentals in excess
Recombinants rare
Proposed physical coupling and repulsion of alleles
I. Linkage in Drosophila
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T.H. Morgan used test cross and demonstrated “coupling” and “repulsion”
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Suggested coupling and repulsion due to linkage on homologous chromosomes
Coupled arrangements due to linkage
Repulsion due to crossing over
Coupled (or cis) arrangement
II. Crossing Over
II A. Phenomenon
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Pachynema of Prophase I
Homologues paired via synaptonemal complex
As chromosomes shorten in diplonema, chiasmata become visible
Male Drosophila lack crossing over entirely
In humans, crossover rates are 75-100% higher in females than in males!
II B. Effect of Crossing Over
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If parentals cis (coupling), then crossovers show repulsion (or trans) arrangment
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Result of crossover is to generate non-parental or recombinant arrangements of alleles
II C. Testing for Linkage
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Compare rates of gametic types to those predicted by independent assortment
If rates deviate, then linkage supported
If rates do not deviate, then linkage not supported
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Use test crosses, as independent assortment predicts equal frequencies of alternatives
Monohybrid test cross  1:1
Dihybrid test cross  1:1:1:1
Trihybrid test cross  1:1:1:1:1:1:1:1
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Number of alternatives 2n (why?)
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Dihybrid creates 4 gametic types
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2 parental
2 recombinants
If independent, in 1:1:1:1 ratio
Testcross of Dihybrid
Testcross of Dihybrid
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Recombinants are 50% of progeny
Recombination frequency of 50%
II C. Testing for Linkage
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If genes are linked, and no crossing over occurs, then only get parentals
If genes are linked, and get a single crossover, get two parentals and two recombinants
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If linked, recombinants should be rare and only produced by crossovers
Each parental type more than ¼ of progeny
Each recombinant less than ¼ of progeny
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Cis/trans and coupling/repulsion
pr
vg
pr+
vg+
Cis arrangements: both alleles have either wild type or mutant or dominant or
recessive
Also called coupling
Cis/trans and coupling/repulsion
pr
vg+
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pr+
vg
trans arrangements: each chromosome has one wild type and one mutant (or one
dominant and one recessive)
Also called repulsion
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pr • vg (red eye, normal wing) = 157 recombinant (cis, coupled)
pr • vg (purple eye, vestigial) = 146 recombinant (cis, coupled)
pr+ • vg (red eye, vestigial)
= 965 parental (trans, repulsion)
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pr • vg (purple eye, normal wing)
=1067 parental (trans, repulsion)
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II C. Testcross progeny:
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If independent, expect 1:1:1:1 ratio
χ2 test for linkage:
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Null hypothesis is results are not different from (are consistent with) predictions from Independent
Assortment model.
Thus, if we REJECT model of Independent Assortment, we conclude Linkage.
χ2 = 806.15, 3df, P < 0.01
Reject null hypothesis, reject model of Independent Assortment, conclude Linkage
II D. Map Units
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1 map unit = 1% recombination
Map Distance = % recombination =
(157 + 146)/2335 • 100 = 13.0 mu = 13.0 cM
13.0
Map:
pr
vg
II E. Cytological Proof of Crossing Over
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Curt Stern (Drosophila)
Harriet Creighton and Barbara McClintock (Corn, 1931)
II E. Cytological Proof of Crossing Over
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Two loci
Seed color
C = colored; c = colorless
Endosperm Wx = waxy, wx = starchy
Chromosome with C and wx with knob at “C” end and longer piece at “wx” end
A trans arrangement
(note: text says C and Wx, but original 1931 paper says C and wx, as shown below
from original paper)
Recombinant genotypes (cis) were associated with recombinant chromosomes
II F. Multiple Crossovers & Maximum Recombination Rate
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Maximum detectable recombination rate due to crossovers is 50%
In dihybrid, # parentals = # recombinants
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That is, linked genes may act independent
But why is maximum detectable crossover rate 50%?
II F 1. Single Crossovers
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If every cell undergoing meiosis had SCO, half of chromatids recombinant, half
parentals
Thus, 2 of 4 chromatids recombinant, = 50% recombination = 50 mu = 50 cM
II F 2. Double Crossovers
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2-strand DCO yields 0 recombinants
3-strand DCO yields 2 recombinants
4-strand DCO yields 4 recombinants
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Average of the three alternatives (0/4; 2/4; 4/4) yields 2/4 recombinant strands
2/4 = 50% recombination
II F Multiple Crossovers & Maximum Recombination Rate
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Thus, genes linked but far apart on a chromosome may act independent (50%
recombination)
II G. Additive Map Units and Linkage Maps
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By using multiple loci, can build linkage maps with additive map distances
Additive distances can be greater than 50 mu
Human Chromosome #1
Additive distance of 356 cM
Additive distances (cM) and cytological distances may differ in distances
Same in linear order
What accounts for the differences in distances?
III. Three-point Cross
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2-strand DCO masks crossover by generating parental arrangment
Note arrangement of outer loci (v and cv)
Parentals in trans arrangement
Create 2-stand DCO
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Note that outer loci are still trans (parentals)
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If add a third locus, then 2-strand DCO distinguishable
III. Method of Analysis
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For trihybrid, 23 = 8 different gametic types
If independent, the 8 classes occur in a 1:1:1:1:1:1:1:1 ratio in the gametes and in the
progeny of a testcross
If linked, the ratio will be violated
III A. Method 1
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Arrange 8 classes as reciprocal classes
Identify parental classes by finding the most common reciprocal class
Identify DCO by finding the least common reciprocal class
Identify linear order of the three loci by comparing the allelic arrangements to the
DCO
Calculate the map distances
Calculate Interference
Linear order of loci
• What change to parentals will result in DCO?
• Switch each locus and see which matches DCO
• That switch represents the middle locus
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Which choice matches actual DCO?
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Actual DCO is
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Actual DCO is A b C and a B c
Flipping “B” gives DCO arrangement
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Actual DCO is A b C and a B c
Flipping “B” gives DCO arrangement
Therefore, “B” is in middle
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Calculate A-B distance
A-B = (SCO for A-B + DCO) / Total
(130 + 150 + 6 + 4) / 2000 * 100 = 14.5 mu
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Calculate B-C distance
B-C = (SCO for B-C + DCO) / Total
(65 + 65 + 6 + 4) / 2000 * 100 = 7.0 mu
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A b C and a B c
Additive distance A-C = (A-B) + (B-C)
=14.5 + 7.0 = 21.5 cM
What is the A-C map distance if the DCO are “ignored”?
That is, treat A-C as a two-point cross
MD = (SCO A-B + SCO B-C)/2000 * 100
MD = (130 + 150 + 65 + 65)/2000 * 100
MD = 20.5 cM
Why is 20.5 less than the additive 21.5?
Interference
• If DCO are random (independent), then actual occurrence of DCO should be equal to
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joint probability of individual SCO frequencies
That is, multiply probability of SCO between middle and outer locus with probability
of SCO between middle and other outer locus
Using previous example of
A-B = 14.5 cM
B-C = 7.0 cM
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Joint Probability of simultaneous SCO between A-B and SCO between B-C =
0.145 * 0.070 = 0.01015
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About 1% chance (1.015%) of DCO if random
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However, if NOT independent, then presence of one SCO interferes/alters chance of
second SCO
Interference =
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I = 1 – ((10/2000) / 0.01015)
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I = 0.507 means obtained about half of expected values
I values typically range from 0 to 1
0 = NO interference (observed = expected)
1 = COMPLETE interference (no observed DCO)
Coefficient of coincidence is the observed/expected value
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Interference values tend to increase as distance between loci decreases
At about 10 cM between outside loci, DCO disappear (I = 1)
At about 45 cM, interference disappears (I = 0)
1 – (observed DCO/expected DCO)
= 1 – (0.005/0.01015)
= 1 – 0.493
= 0.507
III B. Method 2
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As in other method, pair up reciprocal arrangements
Identify parentals as most common class
Identify type of crossover for each remaining reciprocal class
Calculate map distances for the 3 “pairs” of loci
Outer loci will have greatest distance
Draw linkage map
Identify DCO
Calculate Interference
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v-cv = (45 + 40 + 89 + 94) / 1448
= 268/1448 = 18.5%
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cv-ct = (45 + 40 + 3 + 5) / 1448
= 93/1448 = 6.4%
ct-v = (89 + 94 + 3 + 5) / 1448
= 191/1448 = 13.2%
As v-cv (18.5%) is largest value, then v-cv are the two outside loci, with ct middle
locus
Interference value
= 1 – ((8/1448)/(0.132 * 0.064))
= 1 – (0.00552/0.0084)
= 1 – 0.663
= 0.337
IV. Mapping by Tetrad Analysis
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Ascomycete fungi contains meiotic products within ascus
Typically post-meiotic doubling (8 spores)
In some species, meiotic products not ordered within ascus
In some species (Neurospora, Sordaria), meiotic products are ordered within the ascus
Score individual asci, use counts to map genes
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With two loci, get three asci types
PD (parental ditypes): two spore phenotypes of parental type (4:4 within ascus)
NPD (nonparental ditype): two spore phenotypes of parental type (4:4 within ascus)
TT (tetratype): 4 spore phenotypes, 2 parental and 2 nonparental (2:2:2:2 within ascus)
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What causes these three asci types?
If both loci on different chromosomes
PD and NPD result from different segregation patterns at Meiosis I (Independent
Assortment)
PD should be equal to NPD
TT results from SCO between gene and centromere
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If both loci on same chromosome
PD simple segregation and 2-strand DCO
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If both loci on same chromosome
NPD from 4-strand DCO
PD should be much more common than NPD
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If both loci on same chromosome
TT results from SCO and 3-strand DCO
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Map Distance =
(½ TT + 3NPD)/Total Asci
x 100
V. Recombination Within Gene
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Up until 1950’s genes thought to be “beads” along a chromosome
“Beads” were indivisible and crossing over only occurred between beads (genes)
Seymour Benzer, using rII locus of phage T4, showed recombination occurred within
gene
We will examine evidence from the X-linked lozenge locus of Drosophila
lz locus affects eye shape and reduces eye pigments
lz mutants are recessive and many are known
Two lz mutants (lzBS and lzg)
Use flanking linked loci to identify crossover
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ct : cut wing, 7.7 cM to left of lozenge
v: vermillion eye color, 5.3 cM to right of lozenge
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Flanking loci cis in females
Production of wild-type progeny could be caused by either mutation or crossover
between the lzBS and lzg loci
If due to mutation, wild-type should have cis (parental) arrangment of ct and v
If due to crossover between the lzBS and lzg loci should have trans (recombinant)
arrangement of ct and v
From 16,000 progeny, 134 males and females with wild-eyes were found
Male wild-type progeny (w/maternal X), had cut wings and normal eye color
VI. Complementation
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Earlier we examined a technique to determine if mutants affecting the same trait were
allelic or different loci
It turns out for a particular trait there may be many mutants that alter the same
phenotypic trait
The complementation test can determine alleleism or different locus origin of
recessive mutants
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F1
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F2
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Not the ratio expected for monohybrid
Are these two white flower strains allelic or different loci?
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F1
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F2
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What does the phenotype of the F1 suggest, allelic or different loci?
White flower pea x
White flower pea
Purple Flower Pea
Purple:White 9:7
White flower pea x
White flower pea
Purple Flower Pea
Purple:White 9:7
VII. Mapping Human Genes
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VII A. Somatic Cell Hybrids
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Fuse human cells with rodent cells
Hybrids selectively lose human chromosomes
Form stable cell lines with different human chromosome compositions
Grow in media that requires
• TK gene product from human cells
• HGPRT from rodent cell
• Only hybrids can grow
• Correlate presence/absence of gene or gene product with presence/absence of human
chromosomes
• Test multiple cell lines
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+ means present, - means absent
Gene 1 is on which chromosome?
Gene 2 is on which chromosome?
Gene 3 is on which chromosome?
Human beta-galactosidase gene
Human Alzheimer gene – discordance data (gene and
chromosome)
VII B. Molecular Mapping
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Known Gene Product
Unknown Gene Product
LOD scores
VII B 1. Known Gene Product
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RNA or protein product known
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Use probe complementary to gene, bind probe to DNA/chromosomes
FISH (Fluorescence In-Situ Hybridization)
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Reverse transcribe mRNA to make cDNA probe labeled with fluorescent dye
If have protein, make cDNA probe that codes for protein primary structure
FISH (Fluorescence In-Situ Hybridization)
• Denature chromosomes on a slide
• Bind probe to chromosome prep
• Look for hybridization
FISH for Telomeres
FISH for Muscle Protein (Chromosome #11)
FISH of Chromosome #1
FISH Chromosome Painting
Using Probes With Cloned Libraries
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Genetic Library
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Collection of clones
Contains genetic information of an individual
Types:
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Genomic library: all genes of an individual
Chromosomal library: all genes of one chromosome
Human libraries often kept in yeast using YAC (Yeast Artificial Chromosome)
Human genomic library contained within 1,800 YAC clones
Genes found in clones by hybridization of probe to gene
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Uses complementary base pairing
Probe labeled with radioactivity
VII B 2. Unknown Gene Product
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Based on co-inheritance of gene of interest (unknown) and molecular marker
If co-inherited, then linked
Molecular markers are typically DNA sequences with no known function
Molecular Markers include:
• RFLP’s (Restriction Fragment Length Polymorphisms)
• VNTR (Variable Number of Tandem Repeats) = minisatellite DNA
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Microsatellite DNA (dinucleotide repeats)
VNTR and Microsatellite DNA also called
• STR (Short Tandem Repeats)
• SSLP (Short Sequence Length Polymorphisms)
RFLP for Homozygote
RFLP for Heterozygote
VNTR for Heterozygote
VII B 2. Unknown Gene Product
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Molecular markers are diploid (one locus), inherited in Mendelian fashion
Examine co-inheritance of molecular marker and gene of interest in pedigree
Note: 1cM is approximately 1million bp of DNA
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Which band pattern is coinherited with the dominant phenotype?
Huntington Disease Pedigree
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VII B 3. LOD Scores
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Human pedigrees typically have limited numbers of progeny
How can reliable recombination frequencies be found?
Combine data from different matings/pedigrees
Calculate LOD scores
LOD = Log of Odds
“Odds” are probabilities of two alternative events for 2 loci that generally recombine:
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Independent Assortment
Linkage with Recombination
Calculate ratio of
RF with linkage / RF with Ind. Asst.
Then calculate log of ratio
LOD of 3 or greater typically interpreted as evidence of linkage at a specific
recombination frequency
6 progeny
4 parental patterns
2 recombinant patterns
Expected Proportions of Parentals & Recombinants at
Given RF
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Independent Assortment (RF = 50) =
0.25 x 0.25 x 0.25 x 0.25 x 0.25 x 0.25 x B
= 0.00024 x B
where B = # possible birth orders for 4 parentals and 2 recombinant offspring
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For RF = 0.2
0.4 x 0.1 x 0.4 x 0.4 x 0.1 x 0.4 x B
= 0.00026 x B
where B = # possible birth orders for 4 parentals and 2 recombinant offspring
Thus ratio of the two probabilities
= (0.00026 x B) / (0.00024 x B)
= 1.08
• Hypothesis of RF = 0.2 is 1.08 times as likely as hypothesis of Independent
Assortment
Last updated 31 October 2003
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