November 16 Remaining deadlines: Research paper Friday Dec 4 Three point cross lab report Dec 2 or 3...
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Transcript of November 16 Remaining deadlines: Research paper Friday Dec 4 Three point cross lab report Dec 2 or 3...
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November 16
Remaining deadlines: Research paper Friday Dec 4Three point cross lab report Dec 2 or 3Exam Dec 14 at 12:00 noon
Forensics lab resultsGenetic identification of the gene responsible for cystic fibrosis
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1857 bp
1058 bp929 bp
383 bp
121 bp
Tuesday’s lab class pMCT118 allelesVNTR polymorphisms 16 bp repeats 14-40 repeats29 known alleles ranging in size from 369-801 bp)
Common alleles?
Primer-dimer
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Wednesday’s lab class pMCT118 alleles
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Forensics lab:
Are there some especially common alleles?Rare alleles can be more useful for forensics. The type of alleles that are rare will be different in isolated populations.
Question:The sizes of the PCR products of the pMTC118 locus in one family were 531 and 643 bp in the mother and 435 and 531 bp in the father. What are all possible fingerprints their children could have and what is the probability of any child getting each combination of alleles?
531 homozygous ½ X ½ 531 435 heterozygous ½ X ½643 435 heterozygous ½ X ½ 643 531 heterozygous ½ X ½
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Mapping of the gene responsible for cystic fibrosis
1. Identification of polymorphic DNA markers linked to disease2. Location of DNA on Chromosome3. Determination of region in which polymorphic markers are
tightly linked – no recombinants4. Contig assembly and sequence analysis of region 5. Compare polymorphisms in candidate gene between normal
and disease chromosomes to establish all affected family members have mutation
6. Test expression of gene, in expected tissues?7. Identify potential function of protein and explain its role in
disease
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White et al 1985 Nature show that MET DNA is polymorphic when cut with Taq1 and polymorphism segregates as a Mendelian trait
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White et al Nature 1985 Met polymorphism associated with cystic fibrosis
LOD score at recombination value 0Most families are single backcrosseswith respect to the Met locus
In most families, affected offspring have same genotype for Met locus and cf locus suggesting close linkage
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Odds of linkage is:
(Probability gene and marker are linked at a certain map distance) divided by (Probability they are unlinked).
Maximum likelihood odds of linkage; Change estimated linkage distance (θ) to get the best Odds of linkage score for the data.
LOD is the log10 of the Odds of linkage score
LOD is used so that information from separate families, in which parental allele combinations are distinct, can be combined.
LOD score is used to determine if two traits are linkedin human pedigrees
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Combine odds of linkage for many families:
p1(L)/p1(NL) x p2(L)/p2(NL) xp3(L)/p3(NL)
In practice we combine the log of odds:
LOD1 + LOD2 + LOD3.
Continue until LOD > 3.0 before linkage is acceptedLinkage distance is based on the linkage distance that gives the maximum value for the data.
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If genes and markers are unlinked, the odds of linkage will be <1.0 in some families and the final LOD score will be negative (<0).
Therefore, as you add more families the LOD score will only increase if the data from the majority of families supports linkage.
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D0CRI-917
(polymorphic HindIII site)
Partial restriction digest fragment 17.5 kb
Tsui et al 1985 Science identify another polymorphic DNA marker
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Cystic fibrosis linked to D0CRI-917 lambda phage clone insert 18 kb
Tsiu et al 1985 Science RFLP analysis shows affected children have common alleles for DNA marker and cystic fibrosis
HindIII digested DNAHincII digested DNA
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Knowlton et al 1985 Nature
Marker linked to cf in family studies by Tsui et al shown to be on Chromosome 7 fragment by hybridization to somatic hybrid lines. Other markers potentially more closely linked can be identified using these hybrid lines.
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Kerem et al 1989
Table of RFLP markers associated with cf by identification in family studiesSpecific alleles of these markers were found to be linked to disease in families with genetic recombination in chromosome 7, the chromosome that carried the disease.
In this paper the linkage of these markers with disease is validated
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Use end probes and fingerprinting to generate contigs
How do we go from a list of linked markers to a map of the chromosome?
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Fig. 10.8
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Fig. 10.11
Combination of mapped polymorphic sequences and genomic DNA clones enables reconstruction of chromosome sequence
STS are polymorphic DNA sequencesBACS are cloning vectors with genomic DNA inserts
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Kerem et al 1989 Science. Chromosome alignment of CF region.Gaps between contigs filled in with jumping technology.
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DNA jumpingCollins 1987 Science
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Restriction Map of CF region
Rommens et al Science 1989 put together DNA contig of cystic fibrosis region defined by DNA markers.
Vertical bars represent mRNA identified as cDNA clones
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Several transcribed regions were found in the 500 kb segment of chromosome 7 Located between polymorphic sequences associated with a recombination eventBetween the marker and cf in family studies
Having the physical map of the entire region made it possible to identifymany transcribed sequences.
Riordan et al 1989 and Kerem et al 1989 found a transcript for an ion channel encoded in the region.That transcript was expressed in lung tissue.Many affected family members had a small deletion in the coding region of this transcript that would lead to deletion of a single amino acid.
That polymorphism was the only candidate gene in the region that could bedemonstrated to be homozygous in affected individuals and not in healthy individuals.
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Riordan et al Science 1989 Candidate gene expressed in tissues affected by CF
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Riordan et al. Science 1989
All carriers in a family had ΔF deletion
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Riordan et al 1989 Science predicted structure of CF protein - chloride ion channel transporter
Quinton 1983 CF defects in chloride permiability of sweat glands from CF patients
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Nature review 2009 Helen Pearson, editor
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• θ = recombinant fraction
• L(θ) = likelihood of linkage at 0 > θ < 0.5
• L(0.5) = likelihood of independent assortment
• Log[L(θ)/L(0.5)] = log-of-odds ratio = LOD = Z
• LOD scores > 3 indicate linkage
• LOD scores < -2 indicate non-linkage
LOD nomenclatureSlides from Greg Copenhaver
The following slides are for information only and will not be discussed in lecture.There will not be exam questions about these slides.
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θ 0.001 0.01 0.05 0.1 0.2 0.3 0.4
Z -6.0 -3.0 -1.1 -0.4 0.1 0.2 0.1
θ 0.05 0.1 0.15 0.2 0.25 0.3 04
Σ(Z) 28.2 31.2 30.4 27.8 24.0 19.4 9.0
Example linkage of marker and trait
Zmax = maximum likelihood score (MLS)
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Aa•Bb
The probability of progeny inheriting either non-recombinant chromosome from that parent is:
½ (1-θ) + ½ (1-θ) = 1- θ
The probability of inheriting a recombinant chromosome is:
½(θ) + ½(θ) = θ
Assuming a doubly heterozygous parent:
Inheritance Probabilities
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Aa•Bb
The probability of 5 progeny inheriting a non-recombinant chromosome is:
(1-θ)5
The probability of 4 progeny inheriting a recombinant chromosome is:
(θ)4
The probability of 9 progeny with 5 inheriting a non-recombinant chromosome and 4 inheriting a recombinant is:
(1-θ)5(θ)4
Sibship Probabilities
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Calculating lod (logarithm of odds) Scores
lod = logarithm of odds = Z
Z = log( )probability of pedigree if linked probability of pedigree if not linked
calculated for different linkage values, = 0.1 means 10 cm (10% recombinants)
= 0.25 means 25 cm (25% recombinants)
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Is blood type (alleles O and B) linked to Nail-patella syndrome(alleles N and D where D is autosomal dominant mutant allele with full penetrance – shown as affected)?
Linkage of ABO & NPS1
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OO BO
BO BO BO BO BO BOOO OO OO OO OO
What must theNPS1 genotypebe?
Linkage of ABO & NPS1
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OO BO
BO BO BO BO BO BOOO OO OO OO OO
What must theNPS1 genotypebe?
ND
Linkage of ABO & NPS1
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OONN
BO
BOND
BOND
BONN
BOND
BOND
BONN
OOND
OONN
OONN
OONN
OONN
ND
Without knowing the phase we can’t identify recombinant (R) and non-recombinant (NR) progeny….
…so we calculate both.
Linkage of ABO & NPS1
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OONN
BD/ON
BONDNRR
BONDNRR
BONNRNR
BONDNRR
BONDNRR
BONNRNR
OONDRNR
OONNNRR
OONNNRR
OONNNRR
OONNNRR
Phase 1
1
Linkage of ABO & NPS1
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OONN
BN/OD
BONDNRR
BONDNRR
BONNRNR
BONDNRR
BONDNRR
BONNRNR
OONDRNR
OONNNRR
OONNNRR
OONNNRR
OONNNRR
Phase 2
12
Linkage of ABO & NPS1
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OONN
BN/OD
BONDNRR
BONDNRR
BONNRNR
BONDNRR
BONDNRR
BONNRNR
OONDRNR
OONNNRR
OONNNRR
OONNNRR
OONNNRR
Phase 2
12
Phase-1 L(θ) = (1-θ)8(θ)3
Phase-2 L(θ) = (1-θ)3(θ)8
L(θ) = ½(1-θ)8(θ)3 + ½(1-θ)3(θ)8
Linkage of ABO & NPS1
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Calculate likelihood for specific values of θ (e.g. 0.1)
L(0.1) ½(1-0.1)8(0.1)3 + ½(1-0.1)3(0.1)8=
Linkage of ABO & NPS1
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Calculate likelihood ratio for specific values of θ (e.g. 0.1)
L(0.1) ½(1-0.1)8(0.1)3 + ½(1-0.1)3(0.1)8
L(0.5) ½(1-0.5)8(0.5)3 + ½(1-0.5)3(0.5)8= = 0.441
Linkage of ABO & NPS1
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Calculate likelihood ratio for specific values of θ (e.g. 0.1)
L(0.1) ½(1-0.1)8(0.1)3 + ½(1-0.1)3(0.1)8
L(0.5) ½(1-0.5)8(0.5)3 + ½(1-0.5)3(0.5)8
Calculate the LOD by taking the log of the likelihood ratio
Log(0.441) = -0.356
Z(0.1) = -0.356
Do this for several θ values (0.001, 0.01, 0.05, 0.1, 0.2, 0.3, 0.4, 0.45)
= = 0.441
Linkage of ABO & NPS1
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θ 0.001 0.01 0.05 0.1 0.2 0.3 0.4
Z -6.0 -3.0 -1.1 -0.4 0.1 0.2 0.1
θ 0.05 0.1 0.15 0.2 0.25 0.3 04
Σ(Z) 28.2 31.2 30.4 27.8 24.0 19.4 9.0
Linkage of ABO & NPS1
Zmax = maximum likelihood score (MLS)
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9.019.427.831.228.2--Σ(Z)
0.40.30.20.10.050.010.001θ
0.10.20.1-0.4-1.1-3.0-6.0Z
0.40.30.20.10.050.010.001θ
Linkage of ABO & NPS1
LOD score are additive!Below Z = -2so not linkedat these θ
CumulativeZ scores from multiple pedigrees
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If the phase of the doubly heterozygous parent were knownthe calculation is siplified
Calculate log of likelihood ratio for specific values of θ (e.g. 0.1)
L(0.1) (1-0.1)8(0.1)3
L(0.5) (1-0.5)8(0.5)3 = = -0.055log
Linkage of ABO & NPS1
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-4
-3
-2
-1
0
1
2
3
4
0.1 0.2 0.3 0.4 0.5
lod
sco
re
evidence for linkage, 11-19 cM, most likely 13 cM
linkage excluded below -2 cm
Computer-Generated lod Scores
inconclusive for linkage between 3 and 11 cm or above 13 cm