Office hours 3-4pm Wednesdays 304A Stanley Hall
description
Transcript of Office hours 3-4pm Wednesdays 304A Stanley Hall
Office hours3-4pm Wednesdays304A Stanley Hall
Simulation/theory
Expect 0.09 of a locus to reach LOD=3 by chance.
Not 1-locus dominant, or 1-locus incomplete dominance, or…
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SWR parentC3H parent
(F2’s)
Doesn’t look like this…
Quantitative trait linkage test
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(F2’s)
Not counting recombinants.Statistical test for goodness of fit.
>1 locus controlling trait
QuickTime™ and aTIFF (Uncompressed) decompressor
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(One mouse family)
Just reporting significance of goodness of fit.QuickTime™ and a
TIFF (Uncompressed) decompressorare needed to see this picture.
What if…
C3Hparent
F2’s, C/C atmarker
F2’s, C/S atmarker
F2’s, S/S at
marker
SWRparent
Magnitude of spread within group has not changed.Locus effect is weaker.
Correct interpretation:
C3Hparent
F2’s, C/C atmarker
F2’s, C/S atmarker
F2’s, S/S at
marker
SWRparent
Difference between S and C at this locus has a causal role in blood pressure variation, but effect is modest.
Correct interpretation:
C3Hparent
F2’s, C/C atmarker
F2’s, C/S atmarker
F2’s, S/S at
marker
SWRparent
Difference between S and C at this locus has a causal role in blood pressure variation, but effect is modest.
“Effect of having an S allele”
Correct interpretation:
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”
Complex traits(one family, mouse model)
Complex traits(one family, mouse model)
Just reporting significance of goodness of fit.QuickTime™ and a
TIFF (Uncompressed) decompressorare needed to see this picture.
Complex traitsGenetic differences at both loci affect the trait
(one family, mouse model)
Complex traitsEach locus responsible for half?
(one family, mouse model)
Complex traitsEach locus responsible for half? Depends on the model.
(one family, mouse model)
Complex traits(one family, mouse model)
Complex traits(one family, mouse model)
Each locus responsible for a third?
Complex traits
If 5 loci, each responsible for a fifth? 10 loci? …
The more loci, the smaller the effects and the harder to detect.
Complex traits
Genetic complexity is the rule; simple 1- or 2-locus models are the exception
One common result of a linkage study is no significant linkage anywhere.
We haven’t talked about humans lately…
With model organisms, can always study a single cross/family with lots of progeny, so better statistical power to detect weak loci.
And less chance of locus heterogeneity.
Distributions
x
N
= mean
Distributions
Genetically identical
Genetically different
Heritability in exptal organisms
Genetically identical
Genetically different
Which population has the bigger variance?A. RedB. Green
Heritability in exptal organisms
Genetically identical
Genetically different
Why is the green curve taller?A. There are more mice in the green populationB. More mice in the green population have high blood pressureC. Fewer differences between mice in the green populationD. Less environmental error/noise in the green population
Heritability in exptal organisms
Genetically identical
Genetically different
Why is the green curve taller?A. There are more mice in the green populationB. More mice in the green population have high blood pressureC. Fewer differences between mice in the green populationD. Less environmental error/noise in the green population
(each curve adds to 100%)
Heritability in exptal organisms
Green = genetically identical, red = genetically different
Trait 1 Trait 2
Heritability in exptal organisms
(blood pressure) (blood cholesterol)
Green = genetically identical, red = genetically different
Trait 1 Trait 2
Which trait is more likely to be controlled by polymorphisms between the mice in the red population?
A. Trait 1B. Trait 2
Heritability in exptal organisms
Heritability in exptal organisms
Genetic variance = total var - “environmental var”
e
t
g = t
- e
Heritability in exptal organisms
Genetic variance = total var - “environmental var”
“How much of the trait difference between genetically different individuals is due to polymorphisms?”
e
t
g = t
- e
Heritability in exptal organisms
Genetic variance = total var - “environmental var”
Heritability H2 =
e
t
g = t
- e
g/t
Green = genetically identical, red = genetically different
Trait 1 Trait 2
Which trait has a higher heritability?A. Trait 1B. Trait 2
Heritability in exptal organisms
Fig. 21.13
???
What is (the square of) this quantity?A. Environmental varianceB. Total varianceC. Genetic varianceD. Population variance
Why h2?
“Are DNA differences controlling my trait?”
Otherwise, why bother with genetic mapping?
Heritability in humans: MZ twins
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Mean over all = z
Heritability in humans: MZ twins
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Mean each pair = zi
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
Total mean sq = (zij - z)2
T
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2
N
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
“Environment” alone
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2
N
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
Between pairs mean sq = (zi - z)2
N-1
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2
N
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
“Environment” and genetics
Between pairs mean sq = (zi - z)2
N-1
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2
N
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
Between pairs mean sq = (zi - z)2
N-1
= b2 = g
2 + e2
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2
N
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
NBetween pairs mean sq = (zi - z)2
N-1
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2 = w2
= b2
Heritability in humans: MZ twins
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Each individual = zij
Mean each pair = zi
Between pairs mean sq = (zi - z)2
N-1
Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2 = w2
= t2
= b2
Analysis of variance (ANOVA)
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Each individual = zij
Mean each pair = zi
Between pairs mean sq = (zi - z)2
N-1
= w2
= t2 =b
2 + w2Total mean sq = (zij - z)2
T
Within pairs mean sq = (zij - zi)2
= b2
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
Heritability in humans: MZ twins
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h2 = b
2w2
t2
The fraction of the total variance that is attributable to differences between pairs (i.e. is genetic).
Another approach: MZ and DZ
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Within pairs mean sq = (zij - zi)2
N
= w2
Another approach: MZ and DZ
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Within pairs mean sq = (zij - zi)2
N
= w2
h2 = w2(DZ)w
2 (MZ)]
Another approach: MZ and DZ
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Within pairs mean sq = (zij - zi)2
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= w2
h2 = w2(DZ)w
2 (MZ)]
environment only
Another approach: MZ and DZ
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Within pairs mean sq = (zij - zi)2
N
= w2
h2 = w2(DZ)w
2 (MZ)]
Genetic + environment
Another approach: MZ and DZ
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Within pairs mean sq = (zij - zi)2
N
= w2
h2 = w2(DZ)w
2 (MZ)]
DZ twins are half as dissimilar as two unrelated people
Heritability in humans: MZ and DZ
Heritability in humans: MZ and DZ
(A more sophisticated model-fitting
method)
Adoptee studies
Biological parent # in sample% adopted sons
alcoholic
Alcoholic mother 89 39.4
Alcoholic father 42 28.6
Non-alcoholic mother 723 13.6
Non-alcoholic father 1029 15.5
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/abs
trac
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RE
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Rates of alcoholism in adopted males
A qualitative argument for genetic contribution