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

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

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

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 ½ X ½ 531 435 heterozygous ½ X ½643 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 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

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

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

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

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.

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.

D0CRI-917

(polymorphic HindIII site)

Partial restriction digest fragment 17.5 kb

Tsui et al 1985 Science identify another polymorphic DNA marker

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

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 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

Use end probes and fingerprinting to generate contigs

How do we go from a list of linked markers to a map of the chromosome?

Fig. 10.8

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

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

DNA jumpingCollins 1987 Science

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

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.

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

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

Nature review 2009 Helen Pearson, editor

• 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.

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)

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

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

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)

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

OO BO

BO BO BO BO BO BOOO OO OO OO OO

What must theNPS1 genotypebe?


OO BO

BO BO BO BO BO BOOO OO OO OO OO

What must theNPS1 genotypebe?

ND


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.


OONN

BD/ON

BONDNRR

BONDNRR

BONNRNR

BONDNRR

BONDNRR

BONNRNR

OONDRNR

OONNNRR

OONNNRR

OONNNRR

OONNNRR

Phase 1

1


OONN

BN/OD

BONDNRR

BONDNRR

BONNRNR

BONDNRR

BONDNRR

BONNRNR

OONDRNR

OONNNRR

OONNNRR

OONNNRR

OONNNRR

Phase 2

12


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


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=


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


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


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


Zmax = maximum likelihood score (MLS)

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


LOD score are additive!Below Z = -2so not linkedat these θ

CumulativeZ scores from multiple pedigrees

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


-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

Documents

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