1. 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

2. Tuesday’s lab class pMCT118 alleles
VNTR polymorphisms 16 bp repeats repeats
29 known alleles ranging in size from bp)
1857 bp
1058 bp
929 bp
383 bp
121 bp
Common alleles?
Primer-dimer

3. Wednesday’s lab class pMCT118 alleles

4. 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
heterozygous ½ X ½
heterozygous ½ X ½
heterozygous ½ X ½

5. Mapping of the gene responsible for cystic fibrosis
Identification of polymorphic DNA markers linked to disease
Location of DNA on Chromosome
Determination of region in which polymorphic markers are tightly linked – no recombinants
Contig assembly and sequence analysis of region
Compare polymorphisms in candidate gene between normal and disease chromosomes to establish all affected family members have mutation
Test expression of gene, in expected tissues?
Identify potential function of protein and explain its role in disease

6. White et al 1985 Nature show that MET DNA is polymorphic when cut with Taq1 and polymorphism segregates as a Mendelian trait

7. 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

8. 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.

9. 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.

10. 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.

11. identify another polymorphic DNA marker
Tsui et al 1985 Science
identify another polymorphic DNA marker
D0CRI-917
Partial restriction digest fragment 17.5 kb
(polymorphic HindIII site)

12. 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

14. 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.

15. 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

16. How do we go from a list of linked markers to a map of the chromosome?
Use end probes and fingerprinting to generate contigs

17. Fig. 10.8

18. Combination of mapped polymorphic sequences and
Fig
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

19. Kerem et al 1989 Science. Chromosome alignment of CF region.
Gaps between contigs filled in with jumping technology.

20. DNA jumping
Collins 1987 Science

21. put together DNA contig of cystic fibrosis region
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

22. 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.

23. Riordan et al Science 1989 Candidate gene expressed in tissues affected by CF

24. Riordan et al. Science 1989
All carriers in a family had ΔF deletion

25. 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

26. Nature review 2009 Helen Pearson, editor

27. LOD nomenclature Slides from Greg Copenhaver θ = recombinant fraction
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

28. Example linkage of marker and traitθ
0.001
0.01
0.05
0.1
0.2
0.3
0.4 Z
-6.0
-3.0
-1.1
-0.4
0.15
0.25 04
Σ(Z)
28.2
31.2
30.4
27.8
24.0
19.4
9.0
Zmax = maximum likelihood score (MLS)

29. 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

30. 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
(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

31. ( ) Calculating lod (logarithm of odds) Scores
lod = logarithm of odds = Z
( )
probability of pedigree if linked
Z = log
probability of pedigree if not linked
calculated for different linkage values, q
q = means 10 cm (10% recombinants)
q = 0.25 means 25 cm (25% recombinants)

32. 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)?

33. Linkage of ABO & NPS1 What must the NPS1 genotype be? OO BO BO BO BO

34. Linkage of ABO & NPS1 What must the NPS1 genotype be? OO BO ND BO BO

35. Linkage of ABO & NPS1 OO NN BO ND BO ND BO ND BO NN BO ND OO ND OO NN
Without knowing the phase we can’t identify recombinant (R) and
non-recombinant (NR) progeny….
…so we calculate both.

36. Linkage of ABO & NPS1 OO NN BD/ON Phase 1 BO ND NR R BO ND NR R BO NN

37. Linkage of ABO & NPS1 OO NN BN/OD Phase 2 BO ND NR R BO ND NR R BO NN

38. Linkage of ABO & NPS1 OO NN BN/OD Phase 2 BO ND NR R BO ND NR R BO NN
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

39. 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 =

40. 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
L(0.5) ½(1-0.5)8(0.5)3 + ½(1-0.5)3(0.5)8 = =
0.441

41. 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
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) =
Z(0.1) =
Do this for several θ values (0.001, 0.01, 0.05, 0.1, 0.2, 0.3, 0.4, 0.45) = =
0.441

42. Linkage of ABO & NPS1 θ 0.001 0.01 0.05 0.1 0.2 0.3 0.4 Z -6.0 -3.0
-1.1
-0.4
0.15
0.25 04
Σ(Z)
28.2
31.2
30.4
27.8
24.0
19.4
9.0
Zmax = maximum likelihood score (MLS)

43. Linkage of ABO & NPS1 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
0.001
0.01
0.05
0.1
0.2
0.3
0.4
Σ(Z) - -
28.2
31.2
27.8
19.4
9.0
Cumulative
Z scores from
multiple pedigrees
Below Z = -2
so not linked
at these θ
LOD score are additive!

44. 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)
L(0.1) (1-0.1)8(0.1)3
L(0.5) (1-0.5)8(0.5)3
log = =
-0.055

45. Computer-Generated lod Scores4
evidence for linkage, cM, most likely 13 cM 3 2 1
lod score
0.1
0.2
0.3
0.4
0.5 q -1
inconclusive for linkage between 3 and 11 cm or above 13 cm -2 -3
linkage excluded below -2 cm -4