Transcript linkage map

Human
Molecular Genetics
Institute of Medical Genetics
Yaoqin Gong
2003
Human Gene Mapping




Introduction
Common methods
Strategies
Examples
Introduction




Gene mapping: determine the gene
location on specific chromosomal
region
Genetic map (linkage map) - a map of
genetic loci based on recombination
frequencies.
Physical map - a map of the physical
distance between genetic loci measured
in base pairs.
Historically, genetic maps have been
made first, and subsequently correlated
with physical maps.
Why map a disease gene ?

Can lead to a genetic test
 provide more info to people at risk

First step in cloning the gene
 Tells you sequence of the protein it
encodes, which may give clues to the
pathology and etiology of the disease
 May enable protein therapy (e.g.
Factor VIII) or gene therapy
Things human geneticists
can’t do
 Establish
true-breeding lab
strains
 Perform testcrosses or
backcrosses
 Score lots of progeny from the
same mating
Human Gene Mapping


Gene mapping before 1967 was limited to the X
chromosome
First autosomal gene mapped was Duffy blood
group locus
Heterochromatic region
on chromosome 1
Human Gene Mapping
I
2
1
II
a/b
b/b
a/b
a/b
b/b
a/b
a/b
b/b
a/b
a/b
b/b
III
IV
a/a
b/b
5
a/a
a/b
a/b
a/b
a/a
a/b
Solid symbols indicate the presence of the heterochomatin
Duffy blood group locus is on chromosome 1
b/b
Methods used in human gene
mapping





In Situ Hybridization
Somatic Cell Hybrids
Chromosomal abnormalities
Dosage effect analysis
Linkage analysis
Fluorescent in situ hybridization
If a gene is cloned, it can be directly
mapped to a chromosomal locus
 Requirements : Karyotype & Labeled
Gene probe
 fluorescent spots appear in the same
location on homologous
chromosomes

Fluoresence In Situ Hybridization
FISH
Somatic cell hybrid mapping



Based on ability to fuse somatic cells of
different species, e.g., human and rodent
Somatic cell hybrids typically retain only
small number of human chromosomes
Isolate stable partial hybrids using
selectable markers
Human-rodent cell hybrids

Basic discovery: cells of different species
can be fused if brought close together (e.g.
with Sendai virus).

Nuclear fusion may follow.

Over many cell cycles, rodent cells will lose
some human chromosomes at random
before becoming stable.
Human-rodent cell hybrids

Rodent cells are chosen that are deficient
for enzymes of interest.

Those that carry a human chromosome that
bears the human homolog of that gene will
be complemented.

Identify human chromosome by process of
elimination.
Somatic
cell
hybrid
mapping
Somatic Hybrid Mapping
Probe1
Probe2
Probe3
1
-
Chromosome
2
3
4
+
+
+
+
-
5
+
Probe1 -- maps to chromosome 2
Probe2 -- maps to chromosomes 3
Probe3 -- maps to chromosome 2, 3, and 5, --possible paralogs,
pseudogene, or low-copy repeat
Somatic Hybrid Mapping
EXP
+
+
WIL1
WIL6
WIL7
wil14
SIR3
…
% discord
1
+
+
2
+
+
+
3
+
+
+
4
+
+
5
+
+
+
6
+
+
+
7
+
+
+
8
+
+
+
+
-
9 10 11 12 13
- - - - - + + - - + + - +
- + - + + + + + +
14 15 16
+ - + - + - + + - - +
17 18 19 20 21 22
+ - - - + + - + + + + + - - + + - - - - + + + + + +
X
+
0 32 17 24 31 21 21 31 21 24 30 21 21 28 14 24 21 28 17 34 41 21
27
+
+
+
Radiation Hybrid Mapping

Fragment human chromosomes with X-rays.

Allow uptake of chromosome fragments into rodent cells.
Most will be incorporated into rodent genome, but are still
recognizable by their banding.

Find complemented cells and correlate with human
chromosome fragments that are present.

The more closely linked two loci are, the more likely they will
end up in the same hybrid cell.

Radiation hybrid mapping is a method for high-resolution
mapping.
Radiation hybrid mapping
Chromosome aberration Mapping


Chromosome aberrations provide
shortcuts to mapping
Deletions are particularly useful
Deletion mapping the Y
chromosome
Identify males with cytologically
aberrant Y chromosomes
 Test each male for presence or
absence of sequence known to map
to the Y chromosome
 Order the results into a conservative
map (the deletion map)

Deletion mapping the Y
chromosome
DMD: Another monument to
deletion cloning


Duchenne’s muscular dystrophy known to
be X-linked
Regional mapping accomplished by two
chromosome aberrations:
 deletion of DMD and several other
genes
 Translocation interrupting the DMD
gene
DMD: Another monument to
deletion cloning
DMD: Another monument to
deletion cloning

Patient B.B.
 Rare cytological detectable deletion
 Width of Xp21 significantly reduced
 Multiple genetic disorders
DMD
Chronic granulamatous disease (CGC)
Retinitis pigmentosa
- Deletion 10 mb in length
DMD: Another monument to
deletion cloning
Normal X
DMD
Patient B.B.
CGD
RB
Dosage Effect Analysis
SOD
1.5
:
1
Dosage Effect Analysis
Autosomal probe
X-chromosomal probe
Linkage Analysis



Principle
Human genetic maps
Mapping the disease genes
Linkage and recombination
Two classes of progeny
Parental types – progeny that result from
gametes with the same combination of
alleles as the parental gametes
Recombinant (nonparental) types - arise
from crossover between linked genes on
homologous chromosomes
(OR from independent segregation of
chromosomes in a pattern that separates
the two parental chromosomes)
Recombination Frequency
(RF)
RF =
number of recombinants
total number of progeny
x 100%
RF  50%
RF is significantly < 50%


no linkage
linkage
RF is significantly > 50%

impossible
Genetic Map Unit
One genetic map unit is the distance between genes
that gives one recombinant out of 100 meioses.
A recombination frequency of 0.01 (1%) =
1 map unit (m.u.) = 1 centiMorgan (cM)
In humans, 1 cM  1 Mb (megabase). Because many
chromosomes are > 50 Mb in size, two distant
genes on the same chromosome can behave as if
unlinked. (The maximum possible RF is 50%.)
Three-point testcross
Definition: Crossing a triple heterozygote to a
triply recessive tester
(I.e. A/a • B/b • C/c x a/a • b/b • c/c)
Purpose: To determine linkage and (if
applicable) order of genes
Note: In writing out genotypes, can omit alleles
from tester for this analysis because they will
all be “abc”
(i.e. aBC = aaBbCc)
P
aaBBCC
F1
F2:
aBC
Abc
abC
ABc
abc
ABC
aBc
AbC
Total
x
AaBbCc x
580
592
45
40
89
94
3
5
1448
AAbbcc
aabbcc
Parental Classes (highest 2 classes)
Recombinant Classes
Double Recombinants
(not always present)
Determining map distances
(A - C)
aBC
Abc
abC
ABc
abc
ABC
aBc
AbC
Total
580
592
45
40
89
94
3
5
1448
Analyze each set of 2 loci at a
time:
AC
Ac
aC
ac
94 + 5 = 99
592 + 40 = 632
580 + 45 = 625
89 + 3 = 92
RF = 99 + 92 = 13.2%
1448
(significantly < 50%  linked and
distance = 13.2 cM)
Determining Map Distances
(B - C)
aBC
Abc
abC
ABc
abc
ABC
aBc
AbC
Total
580
592
45
40
89
94
3
5
1448
Analyze each set of 2 loci at a
time:
BC
Bc
bC
bc
580 + 94 = 674
40 + 3 = 43
45 + 5 = 50
592 + 85 = 677
RF = 43 + 50 = 6.4%
1448
(significantly < 50%  linked
and distance = 6.4 cM)
Determining Map Distances
(A - B)
aBC
Abc
abC
ABc
abc
ABC
aBc
AbC
Total
580
592
45
40
89
94
3
5
1448
Analyze each set of 2 loci at a time:
AB
Ab
aB
ab
40 + 94 = 134
592 + 5 = 597
580 + 3 = 583
45 + 89 = 134
RF = 134 + 134 = 18.5%
1448
(significantly < 50%  linked and
distance = 18.5 cM)
Ordering Loci
AC = 13.2 cM
BC = 6.4 cM
AB = 18.5 cM
A
C
13.2
18.5
B
6.4
Problem: 13.2 + 6.4 = 19.6  18.5
(because double crossover class was counted as
non-recombinant when actually there were two
crossovers)
a
a
C
C
B
B
A
A
c
c
b
b
a
a
C
c
B
B
A
A
C
c
b
b
Double
Cross-Over
Recombinant
Recombinant
Human Linkage Maps




Polymorphic Markers
CEPH families
Linkage Maps
LOD Score
Polymorphic Markers



Types: RFLP, STR, SNP
Known location
Highly polymorphic:
Heterozygosity> 75%
CEPH Families
Human Linkage Map
LOD Score

It is the logarithm of the odds ratio

It is a statistical measure of likelihood that
two genes are linked at a particular
distance.
LOD Score

Test of linkage (H0 : =1/2 ; H1:  <1/2 )
Z= LOD score = log10 {(L  =RF)/(L =1/2)}
 Z > 3: accept linkage
 Z < -2: reject linkage
 2 < Z < 3 : uncertain (collect more data)
d
1
D d
1 2
d
1
d
1
D
2
d
1
D
2
d
1
d
1
d
1
D
2
d
1
d
1
d
1
D
2
d
1
d
1
d
1
d
2
d
1
d
1
1
1
D
1
d
1
D
2
d
1
d
1




Here: 2 recombinants out of 10 meiosis:
RF = 
Z () = N log (2) + R log  + (N-R) log (1- )
Z = 10 log (2) + 2 log (0.2) + 8 log (0.8)
= 0.837
Recall that if  = 0 and no recombinants are
observed:
Z = 10 log (2) = 3.01
Thus: the denser the map the higher the power to
detect linkage.
Mapping disease genes by
Linkage Analysis
Find a large, multigenerational,
affected family
 Test linkage of the disease to a
mapped polymorphism
 Determine the odds of obtaining
the observed pedigree assuming a
given amount of linkage, and
compare to the odds assuming no
linkage

Mapping disease genes by
Linkage Analysis
Disease locus
Dd
dd
Dd
dd
dd
Dd
Dd
Dd
dd
Marker 1
12
34
13
23
24
14
13
14
23
Marker 2
12
34
13
14
23
24
13
14
23
Marker 1 : linked to disease locus
Marker 2 : unlinked to disease locus
Mapping BDB by Linkage
Analysis
Disease locus
Dd dd Dd dd Dd Dd dd dd Dd Dd Dd dd Dd Dd dd Dd dd Dd
D9S938
21
34 23 33 23 22 14 13 24 22 22 42 23 21 31 21
14 24
D9S123
13
22 12 24 13 14 32 12 12 13 14 23 12 13 23 13
32 12
D9S938 : tightly linked to disease locus
D9S123 : observed 1 recombinant out of 16,
RF=1/16=6.25%
Summary for common methods
Method
Known
genes
Known
protein
Disease
genes
In situ hybridization
Yes
No
No
Somatic cell hybrid
Yes
Yes
No
Chromosome aberration
No
No
Yes
Dosage effect
Yes
Yes
No
Linkage analysis
Yes
Yes
Yes
Mapping strategies


Known genes
Disease genes
Mapping a Known Gene
In Situ Hybridization
NO
YES
Somatic cell hybrid mapping
YES
Specific chromosome
Linkage
Fine mapping
NO
Linkage analysis
Mapping COL9A3
by somatic cell hybrid analysis


Design primers which specifically amplify
human COL9A3, not mouse col9a3
PCR test
H
M H+M
control
Mapping COL9A3
by somatic cell hybrid analysis

Detect different cell lines
Cell lines
Human chromosomes retained
PCR result
A
1, 7, 15, 18, 21, X
-
B
3, 5, 15, 20, 21, 22
+
C
3, 8, 18, 21, 22, X
-
D
2, 7, 15, 18, 21, 22
-
E
2, 6, 14, 17, 19, 20
+
COL9A3 is on chromosome 20
Mapping COL9A3
by linkage analysis




Screening for polymorphism within COL9A3
gene
Select polymorphic markers at 10cM on
chromosome 20
Genotyping CEPH families
Linkage analysis
Mapping COL9A3
by linkage analysis

Intragenic polymorphism
Bam HI : GGATCC
GGATCC
CCTAGG
Allele 1
GAATCC
CTTAGG
Allele 2
with or without Bam HI site
RFLP
Mapping COL9A3
by linkage analysis

Detect intragenic polymorphism by
PCR-RFLP
100
Genomic DNA
GGATCC
CCTAGG
PCR
150
250bp product
Bam HI
250
150
100
22
12
11
Gel electrophoresis
Mapping COL9A3
by linkage analysis

Genotyping chromosome 20 markers
1 2 3
4
5
6
Genomic DNA
PCR
8
7
6标
5记
4
3
2
1
PCR products
Denaturing PAGE
13 24 55 36 18 57
Mapping COL9A3
by linkage analysis

Analyze CEPH families
12

Chooce the families with heterozygote
parents
Mapping COL9A3
by linkage analysis
COL9A3
12
22
12
12
22
12
22
12
22
22
D20S111
23
14
21
31
21
24
24
21
21
24
Mapping COL9A3
by linkage analysis
COL9A3
12
22
12
12
22
12
22
12
22
22
D20S115
13
24
12
14
32
14
34
12
12
34
D20S123
13
24
12
14
32
14
34
12
32
34
D20S143
13
24
12
14
32
14
34
12
32
34
D20S145
13
24
12
34
32
14
34
12
32
34
Mapping COL9A3
by linkage analysis
D20S115
13
24
12
14
32
14
34
12
12
34
COL9A3
12
22
12
12
22
12
22
12
22
22
D20S123
13
24
12
14
32
14
34
12
32
34
D20S143
13
24
12
14
32
14
34
12
32
34
D20S145
13
24
12
34
32
14
34
12
32
34
Mapping a disease gene
Collect families
Chromosomal analysis
NO
YES
Candidate chromosome
linkage
Fine mapping
Candidate genes
NO
Genome scanning
Candidate region
YES
Mutation detection
EPPK
I
1
2
II
1
2
3
4
5
6
7
9
8
III
1
2
3
4
5
6
7
8
Mapping EPPK gene
Chromosomal analysis
NO
Candidate genes
EPPK
-Candidate gene analysis
II
1
2
3
4
5
6
III
1
D17S579 5/5
D12S90 3/4
2
4/3
4/5
4/3
2/2
4/2
4/1
1/2
5/3
5/4
5/3
1/2
3/2
5/2
5/2
Linked to D17S579 which is tightly linked to KRT9 gene
EPPK
-Mutation Detection
C544T
EPPK
-Mutation Detection
BDB
BDB
D9S1820
65
36
56
46
64
34
63
24
43
46
63
56 64
44
D9S1795
52
22
22
23
52
22
52
22
22
25
52
52 25
25
D9S1842
42
33
23
32
43
33
43
33
33
34
43
43 35
35
D9S1781
54
55
45
14
51
51
55
55
55
55
55
55 51
51
D9S1815
53
34
34
52
55
35
53
36
63
65
53
54 41
51
D9S1832
72
58
28
26
72
72
75
21
15
17
75
78 86
56