Download Genetic Mapping

Genetic Mapping
Establishing relative positions of genes along chromosomes
using recombination frequencies
Enables location of important disease genes
Allows isolation and analysis of new genes
Essential framework for human genome project
y w
|1.5|
m
34.7
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f B
20.5
|0.3|
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Linkage Groups
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Unlinked Genes - Independent Assortment
50%
50%
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Adjacent Genes - Complete Linkage
100%
0%
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Incomplete Linkage - Recombination Possible
> 50%
< 50%
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Recombinants result from CO in heterozygote
Maximum
recombinants 50%
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Recombinants result from CO in heterozygote
50% recombinant
50% recombinant
100% recombinant
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Crossing over occurs during Meiosis I Pachynema
Drosophila - Recombination occurs in females
Complete linkage (no CO) in males
Humans -
Different rates of recombination
Males vs Females
Chromosome regions
Hot spots, Cold spots
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Genetic Mapping Principles
Recombination-based maps assume:
recombination rates

distance between genes
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Genetic Map Units Defined
# CO

# recombinants
# CO

distance between genes
Therefore,
# recombinants

distance
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Genetic Map Units Defined
Sturtevant (1913)
Distance yielding 1% recombinant testcross progeny
= 1 map unit (mu, centiMorgan, cM)
10 mu = 10%
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Calculating Map Units - Two Point Testcross
Linked mutant genes are coupled (cis-configuration)
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Calculating Map Units - Two Point Testcross
Two genes
linked
P > 50%
R < 50%
Distance between b+ b and vg + vg = 18 mu
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Calculating Map Units - Two Point Testcross
Linked mutant genes are in repulsion (trans-configuration)
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Calculating Map Units - Two Point Testcross
Linked genes are in repulsion (trans-configuration)
P > 50%
R < 50%
Distance between b+ b and vg + vg = 18 mu
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Two Point Testcross - Sample from Text
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Two Point Testcross - Sample Problem 1
Parental?
Recombinant?
Are genes linked?
If so, how many map units between them?
How are alleles arranged on F1 female’s chromosomes?
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Two Point Testcross - Sample Problem 2
Parental?
Recombinant?
Are genes linked?
If so, how many map units between them?
How are alleles arranged on F1 female’s chromosomes?
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Two Point Testcross - Sample Problem 3
Parental?
Recombinant?
Are genes linked?
If so, how many map units between them?
How are alleles arranged on F1 female’s chromosomes?
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Testcross versus F1 x F1 cross
1/2 masked
by male b+ vg +
chromosome
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Predicting Results - Two Point Testcross
If you screened 1000 testcross progeny, how many would you
expect to have normal body color and miniature wings?
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Compare Expectations to Observations - Chi Square
Degrees of freedom = # categories -1;
4 phenotypes; 3 DF
From Chi Square table, find corresponding probability
p < 0.05, differ significantly
p >0.05, observations fit expectations
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Compare Expectations to Observations - Chi Square
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Mapping Three Genes - Three Point Testcross
P vs DCO
indicates
middle
gene
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Three Point Testcross - Determining Middle Gene
Figure 15.7
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Three Point Testcross - Strategies
F1 Aa Bb Cc
X
aa bb cc
Arrange as reciprocals
A- B- Caa bb cc
(low)
(low)
A- B- cc
aa bb C-
(high)
(high)
A- bb cc
aa B- C-
(int.)
(int.)
aa B- cc
A- bb C-
(int.)
(int.)
Parental? DCO? SCO?
What is the middle gene? Arrangement in F1 heterozygote?
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Three Point Testcross - Strategies
Map distances Adjacent genes =
(SCO + DCO) x 100
total
Outside genes =
(SCOI+SCOII+2DCO) x 100
total
b - bw = ((200 + 20)/1000) x 100 =
bw - vg = ((180 + 20)/1000) x 100 =
b - vg = (((200 + 180 + 2(20))/1000) x 100 =
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Three Point Testcross - Sample from Text
Distance between
p+ p and j+ j?
j+ j and r+ r?
p+ p and r+ r?
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Three Point Testcross - Sample
Parental?
DCO?
SCO I?
SCO II?
Middle gene?
Arrangement of genes in heterozygous parent?
b
B
Map units?
Aa - Bb
Bb - Cc
Aa - Cc
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Predicting Progeny - Three Point Testcross
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Dealing with Unusual Testcross Results
No clear P or DCO
Look at two genes at a time
Aa-Bb:
P
R
Aa-Cc:
P
R
Bb-Cc:
P
R
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Genetic Mapping - Strategies
Genes separated by 50 or more map units appear unlinked.
In most cases, map units are additive.
Use shorter distance to establish complete linkage map.
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Genetic Mapping - Strategies
Studying only two genes can underestimate map distances.
Only odd # of CO counted among recombinants; even # not noticed
Fewer recombinants observed, shorter apparent map distance
Most accurate map unit calculation for distant genes:
Add shorter distances between genes
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Interference
CO in one region affects CO in adjacent region
coefficient of coincidence = cc = Observed DCO/Expected DCO
Interference = i = 1 - cc
Fewer CO than expected:
cc = 280/400
i = 1-0.7 = + 0.3
More CO than expected:
cc = 500/400
i = 1-1.25 = - 0.25
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Interference
Most organisms,
mu > 30, i = 0
Drosophila - 10 mu or less,
i = 1, no DCOs occur
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Genetic Mapping Functions
Mapping Functions - for genetic distances > ~ 7 mu,
correct for the fact that when multiple crossovers occur,
unnoticed even numbers lead to underestimate of map units
Haldane’s (assumes no interference,
d = CO frequency, e = base natural logarithms)
At 20%
recombination,
true map distance
approaches 30
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Genetic Mapping in Humans
1) Pedigree Analysis
X-linkage, Linkage to Markers (LOD scores)
Karyotype Analysis:
Chromosome changes
(deletion, rearrangement)
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Genetic Mapping in Humans
2) Molecular Approaches
DNA profiling, PCR and RE markers, in situ hybridization
Sequencing (physical mapping)
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Genetic Mapping in Humans
3) Cell Hybridization Interspecific hybrids
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Genetic Mapping in Humans
3) Cell Hybridization - Synteny testing
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Genetic Mapping in Humans
4)
LOD score analysis
Calculate likelihood of pedigree based on alternative assumption
that loci are:
linked ( = possible map distance = x) or
unlinked ( = 0.5)
Ratio of two likelihoods gives odds of linkage
Lod score = log of likelihood ratio
LOD  = x = Z x =
Log10
Odds of observed result if = x
Odds of observed result if = 0.5
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