Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2.
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Transcript of Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2.
![Page 1: Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2.](https://reader035.fdocuments.net/reader035/viewer/2022062221/56649d2c5503460f94a026a7/html5/thumbnails/1.jpg)
Mapping a disease locus
Fig. 11.A
A1 D
A2 d
A1 d
A1 d
A2 d
A1
A2
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Mapping a disease locus
Fig. 11.A
A1 D
A2 d
A1 d
A1 d
A1 D
A1
A2
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Mapping a disease locus
Fig. 11.A
A1 d
A1 d
A2 D
A1 D
A2 d(sperm)
A1
A2
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LOD scores
Odds = P(pedigree | r)
P(pedigree | r = 0.5)
r = genetic distance between marker and disease locus
Odds = (1-r)n • rk
0.5(total # meioses)
Odds = 0.77 • 0.31
0.58
= 6.325
Data >6 times more likely under LINKED hypothesis than under UNLINKED hypothesis.
k = 1 recomb, n = 7 non-recomb.
A1
A2
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Just a point estimate
observed recombination fraction = 1/8 = 12.5 cM
Disease-causing mutation
Restriction fragment length polymorphism
True distance 30 cM
this is our observation
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LOD scoresr odds
0.1 12.244
0.2 10.737
0.3 6.325
0.4 2.867
0.5 ??
Odds = P(pedigree | r)
P(pedigree | r = 0.5)
Odds = (1-r)n • rk
0.5(total # meioses)
k = 1 recomb, n = 7 non-recomb.
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How to get an overall estimate of probability of linkage?
A. Multiply odds togetherB. Add odds togetherC. Take the largest oddsD. Take the average odds
Given r
Odds1
Given r
Odds2
Given r
Odds3
1,2 2,3
2,3 1,2 1,3
2,3 1,3
1,2 2,3
1,2 1,2 1,3
2,3 1,3
1,3 2,3
1,3 1,2 2,3
2,3 2,2 2,2
Combining families
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More realistic situation: in dad, phase of alleles unknown
A1 d
A1 d
A1 D
A2 d
A1
A2
or
A1 d
A2 D
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More realistic situation: in dad, phase of alleles unknown
Odds = 1/2[(1-r)n • rk]P(pedigree|r)
A1
A2
A1 D
A2 d
+ 1/2[(1-r)n • rk]
assume one phase for dad
7 non-recomb, 1 recomb
(k = # recomb, n = # non-recomb)
A1 d
A2 D
assume the other phase for dad
1 non-recomb, 7 recomb
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In real life this correction does matter…
best r = 0.2873best r = 0.2771
Accounting for both phasesUsing only one phase
family 1: 10 meioses, 1 (or 9) apparent recombinantsfamily 2: 10 meioses, 4 (or 6) apparent recombinantsfamily 3: 10 meioses, 3 (or 7) apparent recombinantsfamily 4: 10 meioses, 3 (or 7) apparent recombinantstotal LOD = LOD(family 1) + LOD(family 2) + LOD(family 3) + LOD(family 4)
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Modern genetic scans
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(single family)
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Age of onset in breast cancerage of onset
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Coins
Odds = P(your flips | r)
P(your flips | r = 0.5)
r = intrinsic probability of coming up heads (bias)
Odds = (1-r)n • rk
0.5(total # flips)
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Coins
3 heads2 heads1 heads
r odds
0 0
0.1 1.1664
0.2 1.6384
0.3 1.6464
0.4 1.3824
0.5 1
0.6 0.6144
0.7 0.3024
0.8 0.1024
0.9 0.0144
1 0
0 heads
r odds
0 16
0.1 10.498
0.2 6.5536
0.3 3.8416
0.4 2.0736
0.5 1
0.6 0.4096
0.7 0.1296
0.8 0.0256
0.9 0.0016
1 0
4 heads
r odds
0 0
0.1 0.0016
0.2 0.0256
0.3 0.1296
0.4 0.4096
0.5 1
0.6 2.0736
0.7 3.8416
0.8 6.5536
0.9 10.498
1 16
r odds
0 0
0.1 0.1296
0.2 0.4096
0.3 0.7056
0.4 0.9216
0.5 1
0.6 0.9216
0.7 0.7056
0.8 0.4096
0.9 0.1296
1 0
r odds
0 0
0.1 0.0144
0.2 0.1024
0.3 0.3024
0.4 0.6144
0.5 1
0.6 1.3824
0.7 1.6464
0.8 1.6384
0.9 1.1664
1 0
r = intrinsic probability of coming up heads (bias)
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Coins
By chance, can get good LOD score for just about anything.
The more students you have flipping coins, the more likely you are to see this “unlikely”
combination.
The multiple testing problem
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CoinsProbability of one student observing 0 heads and 4 tails: 1/16Estimated number of students out of 70 observing 4 tails: 70*(1/16) = 4
Probability of one student observing 1 head and 3 tails: 4/16Estimated number of students out of 70 observing 1 heads and 3 tails: 70*(4/16)= 17.5
Probability of one student observing 2 heads and 2 tails: 6/16Estimated number of students out of 70 observing 2 heads and 2 tails: 70*(6/16) = 26.3…
We would need to see >4 students get 0 heads and 4 tails before we
believe any coins are biased.
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Simulation/theory
Expect 0.09 of a locus to reach LOD=3 by chance.
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Simulation/theory
But this would change in a different organism, with different number of markers, etc.
So in practice, everyone does their own simulation specific to their own study.
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Candidate gene approachHypothesize that causal variant will be in known pigment gene or regulator. NOT randomly chosen markers genome-wide.
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Candidate gene approach
Red progeny have RFLP pattern like
red parent
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Affected sib pair method
2,2 2,3
2,2 2,2
4,4 1,3
1,4 1,4
…
Sib pairs Observed Expected under null
Same allele
2 (1/2)*2
Different allele
0 (1/2)*2
2 = (O - E)2
E
Test for significant allele sharing.
Total # families
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Qualitative but polygenic
Fig. 3.12
Two loci.
Need one dominant allele at each locus to get phenotype.
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“A weak locus”: need lots of dataAAbb aaBB
AaBb
Flower color
inter-mate
Two loci.
Need one dominant allele at each locus to get phenotype.
AABb AaBb aaBb AaBB aaBB Aabb
Genotype at marker close to A locus
purple white
Top allele
3 1
Bottom allele
2 3
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More generally (one locus):
AA x BB
AB (F1)
AB x AB
AAABBABB
(F2)
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25% 50% 25%
“Effect of having a B”
AA
ABBA
BB
AA ABBA
BB
Effect of a B allele is the same regardless of genotype: additive
1 locus, incomplete dominance
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1 locus, complete dominance
75% 25%
ABBAAA
BB
Dominance is a kind of epistasis: nonadditive
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A real example
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(F2’s)
CC x SS
CS
CS x CS
CCSSCSSC
(F2’s)
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Quantitative trait linkage test
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(F2’s)
Not counting recombinants.Statistical test for goodness of fit.
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Locus effect vs. parents
C3Hparent
F2’s, C/C atmarker
F2’s, C/S atmarker
F2’s, S/S at
marker
SWRparent
Homozygotes do not look like
parent.What do you
infer?
A single varying locus does not explain the data
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>1 locus controlling trait
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(One mouse family)
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A weak locus
C3Hparent
F2’s, C/C atmarker
F2’s, C/S atmarker
F2’s, S/S at
marker
SWRparent
Most loci underlying human disease look like this.
“Effect of having an S allele”
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Heritability in exptal organisms
Genetic variance = total var - “environmental var”
Heritability H2 =
e
t
g = t
- e
g/t
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Heritability in humans: MZ twins
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Each individual = zij
Total mean sq = (zij - z)2
T
Mean each pair = zi
Within pairs mean sq = (zij - zi)2
NBetween pairs mean sq = (zi - z)2
N-1
= b2
= w2
= t2 h2 =
b2w
2
t2
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Linkage mapping (quantitative)
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intolerant tolerant
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Transgenic test
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Fine-mapping: new markers
Best marker
Position of true causal variant
Because you have to hunt through by hand to find the causal gene, and test experimentally. The smaller the region, the better.
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Fine-mapping: new markersPosition of true causal variant
Increased marker density
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Two loci, incomplete dominance
0.5 1 1.5 2
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2-locus interaction
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Effect of J at locus 2
Locus 2 is epistatic to locus 1: effects of locus 1 are masked in individuals with JJ or JL,LJ at locus 2
Locus 2 follows a dominance model: JJ and JL,LJ have the same phenotype, LL differs
“The dominant allele of locus 2 does the masking”
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NO progeny as extreme as diploid hybrid
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Three mutant genes
From pathogenic strainFrom pathogenic strain
From pathogenic strain
Alleles from the same strain at
different genes/loci can have different
effects.
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Linked mutations of opposite effect
Path
Lab
Very unlikely
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Fine-mapping
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Inject into golden larvae
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Golden uninjected
WT uninjected
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No truncation in humans, but…
No other species have the Thr allele: what does this mean?Could be deleterious, just an accidental mutation.Could be advantageous for some humans, no other species.
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Correlates with human differences
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AA
AG
GG
Allele is rare
Perhaps explains phenotypic variation among people of African ancestry
Thr
Thr, Ala
Ala
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Association mapping (qualitative)
Fig. 11.26
Blue alleles at markers are on the same haplotype as
the M allele of the disease
locus
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Association scan, qualitative
osteoarthritis controls
C’s 141 797
G’s 47 433
2 test
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Fine-mapping-lo
g(2
p-v
alue
)
rs377472
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Beginnings of molecular confirmation
coding polymorphisms
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Association scan, quantitative
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Association vs. linkage
Strong, easy to detect, but rare in population;may not be reflective of common disease.Also, hard to collect family data.
Common but weak effects; need 1000’s of samples to detect.If no common cause, can fail.
Unrelatedindividuals
Relatedindividuals
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diabetes control
Gm 23 270
no Gm 1343 3284
Association mapping causal loci
“Gm is protective against diabetes?”
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Association and admixture
these are all the Caucasians…
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Association and admixture
Cases
Controls
=
=
Don’t believe any one locus is causative!
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Genotyping by array
Fig. 11.8
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Coding sequence array
Fig. 1.13
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Marker is linked to polymorphism in expression regulation cascade
ORFTFTF
G
kinaseTF
G
G
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Marker is linked to polymorphism in expression regulation cascade
ORFTFTF
G
kinaseTF
mRNA level shows linkage to locus of polymorphic regulator(s).
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Clinical applications
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Colored curves = fat mass at different body locations
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Association of human transcripts
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linkage (families)
assoc (unrelated)
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Protein inheritance
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PSI+ phenotypes
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