Biometrical
Genetics
QIMR workshop 2013
Slides from Lindon Eaves
Biometrical Genetics from a BG point
of view
• Focus is on individual differences and variation
in a population
• Why are people different
• Why are family members similar
Environment
Genotype
r
G
h
Measured variable
E
Latent variables
e
P
Phenotype
The Basic Model
Phenotype=Genotype+Environment
P=G+E {+f(G,E)}
f(G,E) = Genotype-environment
interaction and correlation
GENES (G)
• Contribution (“Heritability”)
• Type of Action (“Additive”, “Dominant”,
Epistatic”)
• Number, location and function
Basic Model for Effects of a Single Gene on a Quantitative Trait
Think about a variant with two alleles A and a…
Decreasing
Mid-homozygote
Dominance
deviation
- Homozygous effect
Increasing
+ Homozygous effect
Fisher (1918): Basic Ideas
• Continuous variation caused by lots of genes
(“polygenic inheritance”)
• Each gene followed Mendel’s laws
• Environment smoothed out genetic differences
• Genes may show different degrees of “dominance”
• Genes may have many forms (“mutliple alleles”)
• Mating may not be random (“assortative mating”)
• Showed that correlations obtained by e.g. Pearson
and Lee were explained well by polygenic inheritance
Mendelian Basis of Continuous Variation?
Experimental Breeding Experiments
a.D is trib u tio n o f s c o re s p ro d u c e d b y tw o g e n e s
b.
T h e "s m o o th in g " e ffe c t o f th e e n v iro n m e n t
(N = 1 0 0 0 s u b je c ts )
(N = 1 0 0 0 s u b je c ts , 2 g e n e m o d e l)
0 .4
0 .4
0 .3
0 .3
0 .2
0 .2
0 .1
0 .1
0 .0
0 .0
0
1
2
3
4
Y1
5
-2 .5
-1 .5
-0 .5
0 .5
2 .5
S1
c.
C o n tin u o u s d is trib u tio n o f p o lyg e n ic tra it
(1 0 0 g e n e s w ith s m a ll c u m u la tiv e e ffe c ts )
0 .0 6
0 .0 4
0 .0 2
0 .0 0
75
1 .5
79
83
87
91
95
99
Y1
103
107
111
115
119
123
3 .5
4 .5
5 .5
6 .5
Environment “E”
• Contribution (“1-heritability”)
• Type (Shared by family, unique to individual,
remote, proximal,short-, long-term)
• Non-genetic inheritance
• Identification
Interactions and Correlations f(G,E)
• Mating system, population structure
• GxE interaction
• Multiple variables: Genetic and Environmental
Correlation
• Direction of Causation and Causal networks
• G x E interaction
• G – E correlation
• Remembering, Forgetting, Development (GxAge,
G x Time etc.)
Francis Galton (1822-1911)
1869: Hereditary Genius
1883: Inquiries into Human Faculty and its Development
1884-5: Anthropometic Laboratory at “National Health Exhibition”
Hereditary Genius (1869, p 317)
Galton’s Anthropometric Laboratory:
Karl Pearson (1857-1936)
1903: On the Laws of Inheritance in Man: I Physical Characteristics (with Alice Lee)
1904: II Mental and Moral Characteristics
1914: The Life, Letters and Labours of Francis Galton
Pearson and Lee’s diagram for measurement of “span” (finger-tip to finger-tip
distance)
From Pearson and Lee (1903) p.378
From Pearson and Lee (1903) p.378
From Pearson and Lee (1903) p.387
From Pearson and Lee (1903) p. 373
Modern Data
The Virginia 30,000
(N=29691)
The Australia 22,000
(N=20480)
ANZUS 50K: Extended Kinships of Twins
Parents of Twins
Siblings of Twins
Spouses of Twins
Twins
Offspring of Twins
© Lindon Eaves, 2009
Overall sample sizes
Relationship
Parent-offspring
Siblings
Spouses
DZ Twins
MZ Twins
# of pairs
25018
18697
8287
5120
4623
Nuclear Family Correlations for Stature
(Virginia 30,000 and OZ 22,000)
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
US
Australia
© Lindon Eaves, 2009
Nuclear Family Correlations for Liberalism/Conservatism
(Virginia 30,000 and Australia 22,000)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
US
Australia
© Lindon Eaves, 2009
The (Really!) BIG
Problem
Families are a mixture of
genetic and social
factors
Galton’s Solution:
Twins
(Though Augustine may
have got there first –
5th cent.)
One (ideal) solution
Twins separated at
birth
But separated MZs are rare
An easier alternative:
Identical and non-identical
twins reared together:
Galton (Again!)
IDENTICAL TWINS
• MONOZYGOTIC: Have IDENTICAL
genes (G)
• Come from the same family (C)
• Have unique experiences during life (E)
FRATERNAL TWINS
• DIZYGOTIC: Have DIFFERENT genes
(G)
• Come from the same family (C)
• Have unique experiences during life (E)
S c a tte rp lo t fo r c o rre c te d M Z s ta tu re
13
HTDEV2
8
3
-2
r=0.924
-7
-1 2
-1 0
-5
0
5
10
HTDEV1
Data from the Virginia Twin Study of Adolescent Behavioral Development
S c a tte rp lo t fo r a g e a n d s e x c o rre c te d s ta tu re in D Z tw in s
20
HTDEV2
10
0
r=0.535
-1 0
-2 0
-1 6
-1 1
-6
-1
4
9
14
HTDEV1
Data from the Virginia Twin Study of Adolescent Behavioral Development
Genotype Frequencies
in Randomly Mating Population
“Hardy-Weinberg Equilibrium”
frequencies
What is the mean expected to be?
Note: Effects measured from mid-homozygote (“m”)
With equal allele frequencies (easier!) put u=v= ½
And the mean is expected to be….
How does A/a affect the variance?
Equal allele frequencies u=v= ½
Dominance component
Additive component
Q: What happens with lots of genes?
A: The effects of the individual genes add up.
IF… the genes are independent
(“linkage equilibrium”)
Requires random mating, complete admixture
So:
Additive Genetic Variance
Dominance Genetic Variance
Additive and Dominance Components:
Unequal allele frequencies.
Can show (see e.g. Mather, 1949)
Q: What happens when u=v?
VA
VD
Bottom line:
With unequal allele frequencies can
still separate VA and VD but their
definitions change
Plotting Effect of Allele frequency on Genetic
Variance Components (“R”)
d<-1
# Homozygous effect ("additive")
h<-1
# Heterozygous deviation ("dominance")
u<-seq(0.01,0.99,by=.01) # Vector of frequencies of increasing allele
v<-1-u
# Frequencies of decreasing allele
VA<-2*u*v*(d+(v-u)*h)^2 # Additive genetic variance
VD<-4*u*u*v*v*h*h
# Dominance genetic variance
VP<-VA+VD
# Total (genetic) variance
# Plot results
plot(u,VP,type="l",
main="VA (red) and VD (green) as function of increasing allele frequency",
xlab="Frequency of increasing allele",ylab="Variance component")
# Add line for VA
lines(u,VA,col="red")
# Add line for VD
lines(u,VD,col="green")
1.0
VA (red) and VD (green) as function of increasing allele frequency
0.2
0.4
0.6
VA
VD
0.0
Variance component
0.8
VA+VD
0.0
0.2
0.4
0.6
Frequency of increasing allele
0.8
1.0
What about the environment???
Two main sources of environment
• Individual experiences – not shared with
siblings:
VE
• “Family” environment – shared with siblings:
VC
So: the TOTAL variance
(Genes + Environment) is:
VP = VA+VD+VE+VC
“Heritability”
“Broad” heritability:
h2b=(VA+VD)/VP
Proportion of total variance explained
by genes
“Narrow” heritability:
h2n=VA/VP
Proportion of total variance explained
by additive (homozygous) genetic
effects (predicts response to selection
– Fisher, 1930)
So far: have looked at effects on
total variance…
How do VA and VD affect the
correlations between relatives?
Contribution of genes to
correlation between relatives (r):
r = C/VP
Where C=Covariance between
relative pairs
“C” depends of kind of relationship
(sibling, parent-offspring, MZ twin
etc)
But can also be expressed in terms of
VA and VD
Approach
1. For a given relationship, work out expected frequencies of
each type of pair (AA, aa etc.)
2. Write phenotypes of each type of relative
3. Compute cross-products of phenotypes of members of
type of pair
4. Each cross-product by the corresponding frequency
5. Add the result of “4” across all pair types
The answer is the covariance you want (if you have done
the algebra right!)
For equal allele frequencies….
Contribution of one gene to covariance:
Notice that terms in d2 and h2 are
separated – but their coefficients
change as a function of relationship
Can add over all genes to get
total contribution to covariance
Cov(MZ) = VA + VD
Cov(DZ) = ½VA + ¼VD
Cov(U)= 0
Can use the same approach for other
relationships
Contributions of VA and VD to covariances
between relatives (ignoring environment)
Relationship
Total variance
Sibling (DZ twin)
MZ twin
Half-sibling
First cousin
Parent-offspring
Avuncular
Grand-parent
Unrelated
VA
1
½
1
¼
1/
8
½
¼
1/
8
0
Contribution to Covariance
VD
1
¼
1
0
0
0
0
0
0
Adding effects of Environment
VP = VA + VD + VE + VC
Cov(MZ) = VA + VD + VC
Cov(DZ) = ½VA + ¼VD + VC
Cov(UT) = VC
Etc.
To get the expected correlations
Just divided expectations by expected
total variance
Results are proportional contributions
of VA, VD etc. to total variance
Practice (paper and pencil)
• Pick a “d” and “h” (e.g. d=1,h=1; d=1,h=0)
• Pick a frequency for the increasing (A) allele
(e.g. u=0.2, u=0.7)
• Work out VA and VD
• Tabulate on board
Vienna