Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2.

Mapping a disease locus

Fig. 11.A

A1 D

A2 d

A1 d

A1 d

A2 d

A1

A2


Fig. 11.A

A1 D

A2 d

A1 d

A1 d

A1 D

A1

A2


Fig. 11.A

A1 d

A1 d

A2 D

A1 D

A2 d(sperm)

A1

A2

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

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

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)


0.5(total # meioses)

k = 1 recomb, n = 7 non-recomb.

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

More realistic situation: in dad, phase of alleles unknown

A1 d

A1 d

A1 D

A2 d

A1

A2

or

A1 d

A2 D

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

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)

Modern genetic scans

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

(single family)



Age of onset in breast cancerage of onset

Coins

Odds = P(your flips | r)

P(your flips | r = 0.5)

r = intrinsic probability of coming up heads (bias)


0.5(total # flips)

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)

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

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.

Simulation/theory

Expect 0.09 of a locus to reach LOD=3 by chance.

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.

Candidate gene approachHypothesize that causal variant will be in known pigment gene or regulator. NOT randomly chosen markers genome-wide.

Candidate gene approach

Red progeny have RFLP pattern like

red parent

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

Qualitative but polygenic

Fig. 3.12

Two loci.

Need one dominant allele at each locus to get phenotype.

“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

More generally (one locus):

AA x BB

AB (F1)

AB x AB

AAABBABB

(F2)

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

1 locus, complete dominance

75% 25%

ABBAAA

BB

Dominance is a kind of epistasis: nonadditive

A real example



(F2’s)

CC x SS

CS

CS x CS

CCSSCSSC

(F2’s)

Quantitative trait linkage test



(F2’s)

Not counting recombinants.Statistical test for goodness of fit.

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

>1 locus controlling trait



(One mouse family)

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”

Heritability in exptal organisms

Genetic variance = total var - “environmental var”

Heritability H2 =

e

t

g = t

- e

g/t

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

Linkage mapping (quantitative)



intolerant tolerant

Transgenic test



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.

Fine-mapping: new markersPosition of true causal variant

Increased marker density



Two loci, incomplete dominance

0.5 1 1.5 2

2-locus interaction



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”

NO progeny as extreme as diploid hybrid





Three mutant genes

From pathogenic strainFrom pathogenic strain

From pathogenic strain

Alleles from the same strain at

different genes/loci can have different

effects.

Linked mutations of opposite effect

Path

Lab

Very unlikely

Fine-mapping



Inject into golden larvae



Golden uninjected

WT uninjected

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.

Correlates with human differences



AA

AG

GG

Allele is rare

Perhaps explains phenotypic variation among people of African ancestry

Thr

Thr, Ala

Ala

Association mapping (qualitative)

Fig. 11.26

Blue alleles at markers are on the same haplotype as

the M allele of the disease

locus

Association scan, qualitative

osteoarthritis controls

C’s 141 797

G’s 47 433

2 test

Fine-mapping-lo

g(2

p-v

alue

)

rs377472

Beginnings of molecular confirmation

coding polymorphisms

Association scan, quantitative

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

diabetes control

Gm 23 270

no Gm 1343 3284

Association mapping causal loci

“Gm is protective against diabetes?”

Association and admixture

these are all the Caucasians…

Association and admixture

Cases

Controls

=

=

Don’t believe any one locus is causative!

Genotyping by array

Fig. 11.8

Coding sequence array

Fig. 1.13

Marker is linked to polymorphism in expression regulation cascade

ORFTFTF

G

kinaseTF

G

G

Marker is linked to polymorphism in expression regulation cascade

ORFTFTF

G

kinaseTF

mRNA level shows linkage to locus of polymorphic regulator(s).

Clinical applications



Colored curves = fat mass at different body locations

Association of human transcripts



linkage (families)

assoc (unrelated)

Protein inheritance



PSI+ phenotypes





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Mapping a disease locus Fig. 11.A A1D A2d A1d d A2d A1 A2.

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