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November 16
Remaining deadlines: Research paper Friday Dec 4
Three point cross lab report Dec 2 or 3
Exam Dec 14 at 12:00 noon
Forensics lab results
Genetic identification of the gene responsible for cystic fibrosis
Tuesday’s lab class pMCT118 alleles
VNTR polymorphisms 16 bp repeats 14-40 repeats
29 known alleles ranging in size from 369-801 bp)
1857 bp
1058 bp
929 bp
Common alleles?
383 bp
121 bp
Primer-dimer
Wednesday’s lab class pMCT118 alleles
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
½X½
531 435 heterozygous ½ X ½
643 435 heterozygous ½ X ½
643 531 heterozygous ½ X ½
Mapping of the gene responsible for cystic fibrosis
1. Identification of polymorphic DNA markers linked to disease
2. Location of DNA on Chromosome
3. Determination of region in which polymorphic markers are
tightly linked – no recombinants
4. 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
White et al 1985 Nature show that MET DNA is polymorphic when cut
with Taq1 and polymorphism segregates as a Mendelian trait
White et al Nature 1985 Met polymorphism associated with cystic fibrosis
Most families are single backcrosses
with respect to the Met locus
LOD score at recombination value 0
In most families, affected offspring have
same genotype for Met locus
and cf locus suggesting close linkage
LOD score is used to determine if two traits are linked
in human pedigrees
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.
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 accepted
Linkage distance is based on the linkage distance that
gives the maximum value for the data.
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.
Tsui et al 1985 Science
identify another polymorphic DNA marker
D0CRI-917 Partial restriction digest fragment 17.5 kb
(polymorphic HindIII site)
Tsiu et al 1985 Science
RFLP analysis shows affected children have
common alleles for DNA marker and cystic fibrosis
HincII digested DNA
HindIII digested DNA
Cystic fibrosis linked to D0CRI-917 lambda phage clone insert 18 kb
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.
Kerem et al 1989
Table of RFLP markers associated with
cf by identification in family studies
Specific 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
How do we go from a list of linked markers
to a map of the chromosome?
Use end probes and fingerprinting to generate contigs
Fig. 10.8
Fig. 10.11
Combination of mapped polymorphic sequences and
genomic DNA clones enables reconstruction of
chromosome sequence
STS are polymorphic DNA sequences
BACS are cloning vectors with genomic DNA inserts
Kerem et al 1989 Science. Chromosome alignment of CF region.
Gaps between contigs filled in with jumping technology.
DNA jumping
Collins 1987 Science
Rommens et al
Science 1989
put together DNA
contig of cystic
fibrosis region
defined by DNA
markers.
Restriction Map
of CF region
Vertical bars represent
mRNA identified as
cDNA clones
Several transcribed regions were found in the 500 kb segment of chromosome 7
Located between polymorphic sequences associated with a recombination event
Between the marker and cf in family studies
Having the physical map of the entire region made it possible to identify
many 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 be
demonstrated to be homozygous in affected individuals
and not in healthy individuals.
Riordan et al Science 1989 Candidate gene expressed in tissues affected by CF
Riordan et al. Science 1989
All carriers in a family had ΔF deletion
Quinton 1983 CF defects in
chloride permiability of sweat
glands from CF patients
Riordan et al 1989
Science predicted
structure of CF
protein - chloride ion
channel transporter
Nature review 2009 Helen Pearson, editor
The following slides are for information only and will not be discussed in lecture.
There will not be exam questions about these slides.
LOD nomenclature
Slides from Greg Copenhaver
• 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
Example linkage of marker and trait
θ
Z
θ
Σ(Z)
0.001
-6.0
0.05
28.2
0.01
-3.0
0.1
31.2
0.05
-1.1
0.15
30.4
0.1
-0.4
0.2
27.8
0.2
0.1
0.25
24.0
0.3
0.2
0.3
19.4
0.4
0.1
04
9.0
Zmax = maximum likelihood score (MLS)
Inheritance Probabilities
Assuming a doubly heterozygous parent:
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
Sibship Probabilities
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
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, q
q = 0.1 means 10 cm (10% recombinants)
q = 0.25 means 25 cm (25% recombinants)
Linkage of ABO & NPS1
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
BO
OO
BO
BO
BO
BO
OO
OO
OO
BO
What must the
NPS1 genotype
be?
OO OO
BO
Linkage of ABO & NPS1
BO
ND
OO
BO
BO
BO
BO
OO
OO
OO
BO
What must the
NPS1 genotype
be?
OO OO
BO
Linkage of ABO & NPS1
BO
ND
OO
NN
BO
ND
BO
ND
BO
NN
BO
ND
OO
ND
OO
NN
OO
NN
BO
ND
OO OO
NN NN
BO
NN
Without knowing the phase we can’t identify recombinant (R) and
non-recombinant (NR) progeny….
…so we calculate both.
Linkage of ABO & NPS1
BD/ON Phase 1
OO
NN
1
BO
ND
NR
R
BO
ND
NR
R
BO
NN
R
NR
BO
ND
NR
R
OO
ND
R
NR
OO
NN
NR
R
OO
NN
NR
R
BO
ND
NR
R
OO
NN
NR
R
OO
NN
NR
R
BO
NN
R
NR
Linkage of ABO & NPS1
BN/OD Phase 2
OO
NN
1
2
BO
ND
NR
R
BO
ND
NR
R
BO
NN
R
NR
BO
ND
NR
R
OO
ND
R
NR
OO
NN
NR
R
OO
NN
NR
R
BO
ND
NR
R
OO
NN
NR
R
OO
NN
NR
R
BO
NN
R
NR
Linkage of ABO & NPS1
BN/OD Phase 2
OO
NN
1
2
BO
ND
NR
R
BO
ND
NR
R
BO
NN
R
NR
BO
ND
NR
R
OO
ND
R
NR
OO
NN
NR
R
OO
NN
NR
R
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
BO
ND
NR
R
OO
NN
NR
R
OO
NN
NR
R
BO
NN
R
NR
Linkage of ABO & NPS1
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
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
= 0.441
=
8
3
3
8
L(0.5) ½(1-0.5) (0.5) + ½(1-0.5) (0.5)
Linkage of ABO & NPS1
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
= 0.441
=
8
3
3
8
L(0.5) ½(1-0.5) (0.5) + ½(1-0.5) (0.5)
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)
Linkage of ABO & NPS1
θ
Z
θ
Σ(Z)
0.001
-6.0
0.05
28.2
0.01
-3.0
0.1
31.2
0.05
-1.1
0.15
30.4
0.1
-0.4
0.2
27.8
0.2
0.1
0.25
24.0
0.3
0.2
0.3
19.4
0.4
0.1
04
9.0
Zmax = maximum likelihood score (MLS)
Linkage of ABO & NPS1
θ
Z
θ
Σ(Z)
0.001 0.01
-6.0 -3.0
0.001 0.01
-
Below Z = -2
so not linked
at these θ
0.05
-1.1
0.05
28.2
0.1
-0.4
0.1
31.2
0.2
0.1
0.2
27.8
0.3
0.2
0.3
19.4
0.4
0.1
0.4
9.0
LOD score are additive!
Cumulative
Z scores from
multiple pedigrees
Linkage of ABO & NPS1
If the phase of the doubly heterozygous parent were known
the calculation is siplified
Calculate log of likelihood ratio for specific values of θ (e.g. 0.1)
log
L(0.1) (1-0.1)8(0.1)3
=
L(0.5) (1-0.5)8(0.5)3
= -0.055
Computer-Generated lod Scores
4
evidence for linkage, 11-19 cM,
most likely 13 cM
3
lod score
2
1
0
0.1
-1
0.2
0.3
0.4
inconclusive for linkage between
3 and 11 cm or above 13 cm
-2
-3
-4
linkage excluded below -2 cm
0.5
q