1 Balanced Translocation detected by FISH. 2 Red- Chrom. 5 probe Green- Chrom. 8 probe.

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Balanced Translocation detected by FISH

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Red-Chrom. 5probe

Green-Chrom. 8probe

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2D Protein Gels

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MS-peptide size signature: match to all predicted proteins

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1. Follow the mutation

2. Follow which regions of DNA are co-inherited (linked)

Positional Cloning by Recombination Mapping

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1. Follow the mutation

To determine disease genepresence or absence (genotype) from phenotype you must first establish

Dominant / recessiveAurosomal / sex-linked


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SINGLE GENE DEFECTS

Modes of Inheritance

To deduce who (likely) has one or two copies of mutant gene

Affected FemaleUnaffected Male

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AUTOSOMAL DOMINANT

+/+ D/+

D/++/+

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RECESSIVEAUTOSOMAL X-LINKED

RECESSIVE

a/+

a/a

a/+x/++/Y

x/+

x/Y

+/Y

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2. Follow which DNAs are co-inherited (linked)

Use DNA sequences that differamong individuals within a family-Polymorphisms.


T G

C A

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VNTR / STRP DETECTION

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A1A1

A3

A1

A3

A4

A2

A1

A2

A2 A4A4 A3

A3

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X

X

X

2 3

Parent

Gamete

Child

A1

A2

B1

B2

C1

C2

B1

B1

C1

C1

A2

A2

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Recombination Mapping

Measures distance between 2 sites on a chromosome according to frequency of recombination

Distance between 2 DNA markersor

Distance between a “disease gene”and a DNA marker

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No fixed proportional

Conversion between

Genetic distance (cM)

and

Physical distance (kb, Mb)

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FAMILY A

A1 D

A2 +

NR NR NR NR NR RD D D D+ +

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FAMILY B

A1 D

A2 +

NR NR NR NR NR R

A1 +

A2 D

R R R R R NR

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INFORMATIVE MEIOSIS

Ideally:- unambiguous inheritance of mutation and markers

(requires heterozygosity for each in parent)

knowledge of which alleles linked in parent (phase)

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Assign numbers to results of linkage analysis

to deal with non-ideal meioses

to sum data from many meioses in a family

to sum data from several families

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Likelihood of R

Likelihood of NR

If linked and RF = If unlinked:-

1 -

1/2

1/2

Family A has 1 recombinant and 5 Non-RecombinantsLikelihood, given linkage of

L ( )

= . (1- ) 5

L (1/2) = (1/2) 6

Z = Lod = log { L ( ) / L (1/2)}

Or given unlinked:-

Z 0.58 0.62 0.51 0.3 0

0.1 0.2 0.3 0.4 0.5

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Z = 3

Lod

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FAMILY B

A1 D

A2 +

NR NR NR NR NR R

A1 +

A2 D

R R R R R NR

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Family B:- Disease gene may be linked to A1 or A2

Consider equally likely

50% chance Family B has 1 R and 5 NR

50% chance Family B has 5 R and 1 NR

L ( )L (1/2) = (1/2) 6

Z = Lod = log { L ( ) / L (1/2)}Z 0.28 0.32 0.22 0.08 0

0.1 0.2 0.3 0.4 0.5

= . (1- ) 51/2 { } + . (1- ) 5

1/2 { }

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Z 0.28 0.32 0.22 0.08 0

0.1 0.2 0.3 0.4 0.5

Phase unknown

Z 0.58 0.62 0.51 0.3 0

0.1 0.2 0.3 0.4 0.5

Phase known

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Z = Z1 + Z2 + Z3 + Z4 +…..

Z = Z(A) + Z(B) + Z(C) + Z(D) + ….

For family “A” with meioses 1, 2, 3, 4 …..

For multiple families, “A”, “B”, “C”, “D”…..

Assumption: same gene responsible for disease in all families

Problem: locus heterogeneity

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Z = 3

Lod

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LINKAGE DISEQUILIBRIUM

Manygenerations

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PCR test DNA segments

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Testing for specific mutations

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ARMS 3’ mis-match of primer

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OLA

Aa BB CC DD Ee FF Gg HH II JJ AA BB CC Dd Ee FF GG HH II Jj

AA BB CC Dd Ee FF Gg HH II JJ

a B C D e/E F G H I J

A B C d E/e F G H I JA B C D E/e F g H I J

A B C D e/E F G H I j

Mother Father

Son/Daughter

Family Trio SNP genotypes reveal haplotypes

Deduced haplotypes- ignoring recombination

Creation of variant sequencesRearrangement of sequence variants by recombination

First, consider just the creation of variant sequences within a short stretch of DNA where there is no significant rearrangementdue to recombination (an assumption that turns out to be valid)

ABCDEFGHIJKLMNOPQRST

AbCDEFGHIJKLMNOPQRST

ABCDEFgHIJKLMNOPQRST

AbCDEFGHIJKLMNOPqRST

aBCDEFgHIJKLMNOPQRST

AbCDEFGHIJkLMNOPqRST

AbCDEFGhIJkLMNOPqRST

ABCDEfGHIJKLMNOPQRST

aBCDEFgHIJKLMNOPQrST

aBCDEFgHIJKLMnOPQrST

bbqbqkbqkh

ggagargarn

f

History

ABCDEFGHIJKLMNOPQRST

AbCDEFGHIJKLMNOPQRST

ABCDEFgHIJKLMNOPQRST

AbCDEFGHIJKLMNOPqRST

aBCDEFgHIJKLMNOPQRST

AbCDEFGHIJkLMNOPqRST

AbCDEFGhIJkLMNOPqRST

ABCDEfGHIJKLMNOPQRST

aBCDEFgHIJKLMNOPQrST

aBCDEFgHIJKLMnOPQrST

bbqbqkbqkh

ggagargarn

f

Retention & amplification of only a few haplotypes

For any short region of DNA typically only 4-6 haplotypesare found in a sampling of present day humans (of the manymillions that must have existed in at least one copy en route).These local haplotypes provide some information about ancestry.

Now consider how the major haplotypes of each short region ofDNA are associated with neighboring haplotypes to see whererecombination events took place.

aBCDEFgHIJKLMnOPQrSTUVwXyZ







High LD regions?








High LD segment High LD segment

Recombination hot-spot

85% of genome made up of 5-20kb high LD blocks

Only 4-5 different major haplotypes per block in the world!

Haplotype blocks

100 kb

1 2 3 4 5 6 7 8

Disease No disease/2,000 /3,000Minor allele frequency

SNP-2a 93 130SNP-2b 21 27SNP-3a 140 62SNP-3b 24 35SNP-3c 140 260SNP-3d 87 120……. …. ….……. …. ….……. …. ..

1 Balanced Translocation detected by FISH. 2 Red- Chrom. 5 probe Green- Chrom. 8 probe.

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