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GENETIC EPIDEMIOLOGY
Institute of Clinical Medicine
Po-Hsiu Kuo – 04/16/2008
Outline
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Overview of gene mapping
Allele frequency
Quantitative genetics
Genetic component and Heritability
Hardy-Weinberg Equilibrium
Linkage Disequilibrium
Gene mapping
LOCALIZE and then IDENTIFY a locus that regulates a trait
Linkage
analysis
Association
analysis
Allele frequency and genotype frequency
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
Allele (gene)
Sample
Allele
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30
60
10
100
A1A1
A1A2
A2A2
Total
A1
60
60
0
120
A2
0
60
20
80
Genotype frequency?
0.3, 0.6, 0.1
Allele frequency f(A1) ? 120/200 = 0.6
Allele frequency f(A2) ? 80/200 = 0.4
Factors influence on allele frequency
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Population size
Reproduce rate and Survival rate
Immigration and mutation
Pattern of assortative mating
Quantitative genetics
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Study of continuous traits (such as height or weight) and its
underlying mechanisms
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Combined effect of the many underlying genes results in a continuous
distribution of phenotypic values
Quantitative genetics is not limited to continuous traits, but to
all traits that are determined by many genes
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Continuous traits are quantitative traits with a continuous phenotypic
range. They are usually polygenic, and may also have a significant
environmental influence
Traits with ordinal numbers, such as number of bristles on a fruit fly.
These traits can be either treated as approximately continuous traits or
as threshold traits
Some qualitative traits can be treated as if they have an underlying
quantitative basis, expressed as a threshold trait (or multiple thresholds)
Population level
Genotype frequencies (Random mating)
Allele 1
Allele 2
A (p)
a (q)
A (p)
AA (p2) Aa (pq)
a (q)
aA (qp)
aa (q2)
Hardy-Weinberg Equilibrium frequencies
P (AA) = p2
P (Aa) = 2pq
P (aa) = q2
p2 + 2pq + q2 = 1
Phenotype level
Quantitative traits
g==-1
g==0
.128205
.072
Fraction
AA
g==-1
.128205
.128205
g==-1
g==0
g==1
g==0
-3.90647
Fraction
.128205
0
Fraction
Fraction
Aa
0
g==1
.128205
0
0
g==1
-3.90647
-3.90647
.128205
-3.90647
2.7156
2.7156
qt
Histograms by g
aa
0
-3.90647
2.7156
qt
e.g. cholesterol levels
0
-3.90647
0
-3.90647
2.7156
qt
Histograms by g
2.7156
qt
Histograms by g
Components of Phenotypic Values
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Phenotype (P) = Genotype (G) + Environment (E)
Considering variances (Var), this becomes:
Var(P) = Var(G) + Var(E) + 2 Cov(G,E)
 G = A (additive genotypic value) + D (dominance deviation)
 A: breeding value (i.e. the sum of the average effects of the
two alleles)
 D: intra-allelic interaction
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In planned experiments, we can often take Cov(G,E) = 0
Heritability
Heritability

Proportion of phenotypic variation in a population
that is attributable to genetic variation among
individuals
 Variation
among individuals may be due to genetic
and/or environmental factors
 Heritability analyses estimate the relative contributions
of differences in genetic and non-genetic factors to the
total phenotypic variance in a population.
Resemblance between relatives
VA
VD
VEC
Parents & offspring (PO)
1/2
0
VEC(PO)
Full - siblings (FS)
1/2
1/4
VEC(FS)
Half - siblings (HS)
1/4
0
VEC(HS)
Fraternal twins (DE)
½
1/4
VEC(DE)
Identical twins (ME)
1
1
VEC(ME)
Phenotypic covariance between
rMZ = rDZ = 1
rMZ = 1, rDZ = 0.5
E
E
e
rMZ = 1, rDZ = ^
C
c
C
A
a
A
Q
Q
q
q
Twin 1
mole count
a
Twin 2
mole
count
c
e
Twin study
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One of behavioral genetics study design
Highlighting the role of environmental and genetic causes on
behavior
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Shared environmental influences common to members of family —class,
parenting styles, education etc
Shared genes, inherited from parents
Compares the similarity between twins (MZ vs. DZ)
Modern history of the twin study derives from Sir Francis
Galton's pioneering use of twins to study the role of genes and
environment on human development and behavior
The Law of Hardy’s
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
Assumption
 Radom
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mating, no selection, no mutation, no immigrant
Question: What’s the allele frequency between
generations?
Two alleles (A, a)
 F(A)
= p, f(a) = q, p+q=1
 Frequency of genotype distribution AA, Aa, aa?
Allele frequency in the next generation
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f(A) = p2 + ½ (2pq) = p(p+q) =p
f(a) = q2 + ½ (2pq) = q(p+q) =q
Therefore, under previous assumptions, allele
frequency is unchanged over generations – i.e.
reached equilibrium
Note: a German physician Wilhelm Weinberg
made the same conclusion as Hardy  so called
Hardy-Weinberg Equilibrium (HWE)
Test for Hardy-Weinberg Equilibrium
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Test for genotype frequency
Example: examining MN blood typing in 6129
Americans
MM
MN
NN
Total
1787
3707
1305
6129
Under HWE, test for χ2 = Σ(O-E)2/ E
First: f(M) = p = 0.539
f(N) = q = 0.461
Test for HWE (conti.)
P=0.539, q=0.461
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f(MM) = p2 = 0.291, f(MN) = 2pq = 0.496, f(NN)
= q2 = 0.212
MM
MN
NN
Total
Observed
1787
3707
1305
6129
Expected
0.291*(6129) 0.496*(6129) 0.212*(6129) 6129
(O-E)2/E
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χ2 = 0.00489
P(χ2,df=1 ) = 0.9
Conclusion: MN blood patterns in this sample are in
HWE
Linkage phase
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For two loci, features of being in equilibrium means
p(AB) = 2 p(A)p(B)
 Linked two loci, say A1A2/B1B2
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 Coupling
phase
 Repulsion
phase A B
1 2
A1 B1
A2 B2
A2 B1
Linkage Disequilibrium
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
Gametic Association
equilibrium, f(AB gamete) = PAB = PA× PB
 Consider 2 alleles are at each of 2 loci with random
mating, then there are ten possible zygotes
 If r is the probability that a gamete is recombinant
 In each generation, the amount of gametic
disequilibrium, measured by PAB - PA PB , is reduced by
a factor of r
 PAB‘ = PAB – r D
(where D = p11 p22 - p12 p21)
 At
 D:
a quantitative measure of the amount of linkage
disequilibrium
Linkage Disequilibrium (conti.)
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Dt=(1-r)t D0
Decay of linkage disequilibrium parameters
between two loci depends on recombination
fraction and time (generation)
Dmax = min (p1q2, p2q1), the largest positive value
When p11= p22 = 0.5 & p12= p21= 0, max D =
0.25
D´= D / Dmax, defined by Lewontin (1988)
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Falconer and Mackay (1996)
Population Admixture
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Admixture
Population 1
Population 2
m
1－m
A1
p1
P1
A2
q1
Q1
B1
p2
P2
B2
q2
Q2
A1B1
m p 1 p2
(1-m) P1 P2
A1B2
m p 1 q2
(1-m) P1 Q2
A2B1
m q 1 p2
(1-m) P2 Q1
A2B2
m q 1 q2
(1-m) Q1 Q2
Mixture proportion
Allele frequency
Gametic frequency
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D = PA1B2 PA2B1－PA1B1 PA2B2
= [m p1 q2＋(1-m) P1 Q2]‧[m q1 p2＋(1-m) P2 Q1]
－[m p1 p2＋(1-m) P1 P2]‧[m q1 q 2＋(1-m) Q1 Q2]
= m (1-m) (p1－P1) (p2－P2)
Only when p1 = P1 or p2 = P2 or (p1 = P1 and p2 = P2),
D=0