Transcript A 2

Reminder - Means, Variances and
Covariances
  E ( X )   xi f xi 
i
Var ( X )  E ( X   )
2
  xi    f xi 
2
i
Cov( X , Y )  E  X   X Y  Y 
 x    y    f x , y 
i
i
X
i
Y
i
i
Covariance Algebra
Cov(aX , Y )  aCov( X , Y )
Cov( X  Y , Z )  Cov( X , Z ) Cov(Y , Z )
Var ( X  Y )  Var ( X )Var (Y )  2Cov( X , Y )
Covariance and Correlation
Correlation is covariance scaled to range [-1,1].
rX ,Y
Cov X , Y 

Var  X Var Y 
For two traits with the same variance:
Cov(X1,X2) = r12 Var(X)
Phenotypes, Genotypes and environment
 A phenotype (P) is composed of genotypic values (G) and
environmental deviations (E):
P=G+E
 Whether we focus on mean, variance, or covariance, inference
always comes from the measurement of the phenotype
 A distinction will be made:


V will be used to indicate inferred components of
variance
s2 will be used to indicate observational components of
variance
 Mean genotypic value is equal to the mean phenotypic value
 Genotypic values are expressed as deviations from the mid-
homozygote point
E(Gj) = EPj )= Pj
Genotypic values
 Consider two alleles, A1 and A2, at a single locus.
 The two homozygous classes, A1A1 and A2A2, are assigned
genotypic values +a and -a, respectively.
 Assume that the A1 allele increases the value of a phenotype
while the A2 allele decreases the value.
 The heterozygous class, A1A2, is assigned a genotypic value of
d
 Zero is midpoint between the two genotypic values of A1A1 and
A2A2; d is measured as a deviation from this midpoint
Genotype
A2A2
-a
Genotypic value
A1A2
0
d
A1A1
+a
Properties of the Genotypic Values and
Environmental Deviations
 The mean of environmental deviations is zero
Ej = Pj – Gj , ,(E = 0)
 The correlation between genotypic values
and environmental deviations for a population
of subjects is zero (rGE = .00)
Elements of a population mean
Genotype
Frequency
Value
Freq x Value
A1A1
p2
+a
p2a
A1A2
2pq
d
2pqd
-a
2
A2A2
q
2
-q a
Sum = a(p - q) + 2dpq
Population mean
P = G = SGkpk
 Multiply frequency by genotypic value and
sum
 Recall that p2 - q2 = (p + q)(p - q) = p - q
P = a(p - q) + 2dpq
Additive model
 Assume aA and aB correspond to A1A1 and
B1B1
 A1A1B1B1 = aA + aB
 So that
P = Sa(p - q) + 2Sdpq
Average effect for an allele (a)
 Population properties vis a vis family structure
 Transmission from parent to offspring; parents pass
on genes and not genotypes
 Average effect of a particular gene (allele) is the
mean deviation from the population mean of
individuals which received that gene from one parent
(assuming the gene transmitted from the other parent
having come at random from the population)
Average effect for an allele (a)
Gamete
A1
A2
Frequ. & Values
A1A1
& +a
A1A2
&d
p
q
p
G
Minus pop.
mean
a
pa + qd
-[a(p-q) +
2dpq]
q[(a + d(q-p)]
-qa + pd
-[a(p-q) +
2dpq]
-p[a+d(q-p)]
A2A2
& -a
q
Average effect for an allele (a)
 Thus, the average effect for each allele also can be
calculated for A1 and A2 in the following manner
(Falconer, 1989):
a1 = pa + qd - [a(p - q) + 2dpq] and
a2 = -p[a + d(q - p)]
Average effect of a gene substitution
 Assume two alleles at a locus
 Select A2 genes at random from population; p
in A1A2 and q in A2A2
 A1A2 to A1A1 corresponds to a change of d to
+a, i.e., (a - d); A2A2 to A1A2 corresponds to a
change of -a to d, (d + a)
 On average, p(a - d) plus q(d + a) or
a = a + d(q - p)
When gene frequency is greater a is greater
q = 0.10
q = 0.40
a1 =
+0.24
+1.44
a2 =
-2.16
-2.16
a = a1 - a2
2.40
3.60
Breeding Value (A)
 The average effects of the parents’ genes determine the mean




genotypic value of its progeny
Average effect can not be measured (gene substitution), while
breeding value can
Breeding value: Value of individual compared to mean value of
its progeny
Mate with a number of random partners; breeding value equals
twice the mean deviation of the progeny from the population
mean (provides only half the genes)
Breeding value is interpretable only when we know in which
population the individual is to be mated
Breeding Value
 Genotype: Breeding value
A1A1:
A1A2:
A2A2:
2a1 = 2qa
a1 + a2 = (q - p)a
2a2 = -2pa
 Mean breeding value under HWC equilibrium
is zero
2p2qa + 2pq(q - p)a - 2q2pa
2pqa(p + q - p - q) = 0
which equals...
Dominance deviation
 Breeding values are referred to as “additive
genotype”; variation due to additive effects of genes
 A symbolizes the breeding value of an individual
 Proportion of s2P attributable to s2A is called
heritability (h2)
G=A+D
 Statistically speaking, within-locus interaction
 Non-additive, within-locus effect
 A parent can not individually transmit dominance
effects; it requires the gametic contribution of both
parents
Genotypic values, breeding values, and dominance deviation
2qa
+a
d
0
(q - p)a
}a
0
-2pa
-a
Genotypic values
Breeding values
A2A2
q2
A1A2
2pq
A1A1
p2
Genotypic values, breeding values, and dominance deviation
 Regression of genotypic value on gene dosage yields the
genotypic values predicted by gene dosage


average effect of an allele
that which “breeds true”
 If there is dominance, this prediction of genotypic values from
gene dosage will be slightly off

dominance is deviation from the regression line
Epitasis - Separate analysis
 locus A shows an association with the trait
 locus B appears unrelated
AA
Aa
Locus A
aa
BB
Bb
Locus B
bb
Epitasis - Joint analysis
 locus B modifies the effects of locus A
AA
Aa
aa
BB
Bb
bb
Genotypic Means
Locus B
BB
Bb
bb
AA
AABB
AABb
Aabb
AA
Locus A
Aa
AaBB
AaBb
Aabb
Aa
aa
aaBB
aaBb
aabb
aa
BB
Bb
bb

Partitioning of effects


Locus A
M

P
Locus B
M
P
4 main effects
M
P
M
P
Additive
effects
6 twoway interactions
M 
M

P
M 
P
M 
P
P
Additive-additive
epistasis
4 threeway interactions
M 
P

M
M 
P

P
M 
M 
P

M 
P
P
Additivedominance
epistasis
1 fourway interaction
M 
P

M

P
Dominancedominance
epistasis
Two loci
AA
BB m + aA + aB + aa
Bb
m + aA + dB + ad
bb
m + aA – aB – aa
Aa
aa
m + dA + aB + da
m – aA + aB – aa
m + dA + dB + dd
m – aA + dB – ad
m + dA – aB – da
m – aA – aB + aa
Covariance matrix
Sib 1
Sib 1 s2A + s2D + s2S + s2N
Sib 2 s2A + zs2D + s2S
Sib 2
s2A + zs2D + s2S
s2A + s2D + s2S + s2N
Sib 1
Sib 1 s2A + s2D + s2S + s2N
Sib 2
½s2A + ¼s2D + s2S
Sib 2 ½s2A + ¼s2D + s2S
s2A + s2D + s2S + s2N
Detecting epistasis
 The test for epistasis is based on the
difference in fit between


- a model with single locus effects and
epistatic effects and
- a model with only single locus effects,
 Enables us to investigate the power of the
variance components method to detect
epistasis
True Model
A
a
Y
Assumed Model
B
A
b
a*

Y

a* is the apparent co-efficient
a* will deviate from a to the extent that A and B are correlated
Phenotypic variance
 Again, assume
P=G+E
 Thus differences in phenotypes, measured as variance and
symbolized as VP, can be decomposed into both genetic and
environmental variation, VG and VE, respectively.
VP = VG + VE
 VG is comprised of three kinds of distinct variance: additive (VA),
dominant (VD), and epistatic (VI).
VP = (VA + VD + VI ) + VE
Analysis of variance
Variance
Symbol
Value
Phenotypic
VP
Phenotypic value
Genotypic
VG
Genotypic value
Additive
VA
Breeding value
Dominance
VD
Dominance deviation
Epistasis
VI
Epistatic deviation
Environment
VE
Environmental
deviation
Additive (VA) and dominance variance (VD)
 The covariance between breeding values and
dominance deviations equals zero so that
VG = VA + VD + VI
VA = 2pq[a + d(q - p)]2
VD = d2(4q4p2 + 8p3q3 + 4p4q2) = (2pqd)2
Additive and dominance variance
 If d = 0, then VA = 2pqa2, where q is the recessive
allele
 If d = a, then VA = 8pq3a2
 If p = q = .50 (e.g., cross of two inbred strains)


VA = 1/2a2
VD = 1/4d2
 In general, genes at intermediate frequency
contribute more variance than high or low
frequencies
Epistatic variance (VI)
 Epistatic variance beyond three or more loci
do not contribute substantially to total
variance
 Three types of two-factor interactions
(breeding values by dominance deviations)
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

additive x additive (VAA)
additive x dominance (VAD)
dominance x dominance (VDD)
Environmental variance
 Special environmental variance (VEs)


within-individual component
temporary or localized circumstance
 General environmental variance (VEg)


between-individual component
permanent or non-localized circumstances
 Ratio of between-individual to total phenotypic is an
intraclass correlation (r)
Summary of variance partitioning
Data needed
Partition made
Ratio Estimated
Resemblance
between relatives
(VA):(VNA+VEg+VEs)
Heritability, VA / VP
Genetically uniform
group
(VA+VNA): (VEg+VEs) Degree of genetic
= (VG):(VE)
determination, VG/VP
Multiple
measurements
(VG+VEg):VEs
All three
VA:VNA:VEg:VEs
Repeatability,
(VG+VEg)/VP
Components of variance - Summary
Phenotypic Variance
Environmental
Genetic
GxE interaction
and correlation
Components of variance - Summary
Phenotypic Variance
Environmental
Genetic
GxE interaction
and correlation
Additive
Dominance
Epistasis
Components of variance - Summary
Phenotypic Variance
Environmental
Additive
Genetic GxE interaction
and correlation
Dominance
Quantitative trait loci
Epistasis
Resemblance of relatives
 Degree of relative resemblance is a function of
additive variance, i.e, breeding values
 The proportionate amount of additive variance is an
estimate of heritability (VA / VP)
 Intraclass correlation coefficient

t = s2B / s2B + s2W
Between and within full-sibships, for example
Resemblance of relatives
bOP = CovOP / s2P
 New property of the population is covariance of
related individuals
Cross-Products of Deviations for
Pairs of Relatives
AA
Aa
aa
AA
Aa
aa
(a-m)2
(a-m)(d-m) (d-m)2
(a-m)(-a-m) (-a-m)(d-m) (-a-m)2
The covariance between relatives of a certain class is the
weighted average of these cross-products, where each
cross-product is weighted by its frequency in that class.
Offspring and one parent
 Individual genotypic values and those of their
offspring produced by random mating
 When expressed as normal deviations, the mean
value of the offspring is 1/2 the breeding value of the
parent
 Covariance between individual’s genotypic value (G)
with 1/2 its breeding value (A)
Covariance for Parent-offspring
(P-O)
AA
Aa
aa
AA
p3
p2q
0
Aa
aa
pq
pq2
q3
Covariance = (a-m)2p3 + (d-m)2pq + (-a-m)2q3
+ (a-m)(d-m)2p2q + (-a-m)(d-m)2pq2
= pq[a+(q-p)d]2
= VA / 2
Offspring and one parent
 G = A + D so that covariance is between (A + D) and
1/2A; sum of cross products equal
S1/2A(A + D) = 1/2SA2 + 1/2SAD
 CovOP = (1/2SA2 + 1/2SAD) / # of parents
 Recall that CovAD = 0
 CovOP = 1/2VA (i.e., 1/2 the variance of breeding
values)
Offspring and one parent: Effects of a single locus
Parents
Offspring
Genotype
Frequency
Genotypic value
Mean genotypic
value
A1A1
p2
2q(a - qd)
qa
A1A2
2pq
1/2(q - p)a
A2A2
q2
(q - p)a +
2pqd
-2p(a + pd)
-pa
Offspring and one parent
 Mean genotypic values of the offspring are 1/2A of
the parents
 Mean cross product equals Frequency X Genotypic
value of the parent X Mean genotypic value of the
offspring
 CovOP =
pqa2(p2+2pq+q2)+2p2q2ad(-q+q-p+p)=pqa2=1/2VA
 (Note: VA = 2pqa2)
Covariance of MZ Twins
AA
Aa
aa
AA
p2
0
0
Aa
aa
2pq
0
q2
Covariance = (a-m)2p2 + (d-m)22pq + (-a-m)2q2
= 2pq[a+(q-p)d]2 + (2pqd)2
= VA + VD
Twins
 Dizygotic twins are fulls sibs and their genetic
covariance is that of full sibs
 Monozygotic twins have identical genotypes,
i.e., no genetic variance within pairs so that
Cov(MZ) = VG
Covariance for Unrelated Pairs (U)
AA
p4
2p3q
p2q2
AA
Aa
aa
Aa
aa
4p2q2
2pq3
q4
Covariance = (a-m)2p4 + (d-m)24p2q2 + (-a-m)2q4
+ (a-m)(d-m)4p3q + (-a-m)(d-m)4pq3
+ (a-m)(-a-m)2p2q2
=0
Resemblance in general
 Let r be the fraction of VA and u the fraction of VD so




that
Cov = rVA + uVD
P, Q two individual in relationship with parents A,B
and C,D and f coancestry
r = 2fPQ and u = fACfBD + fADfBC
For inbred relatives, r = 2fPQ / [(1 + FP)(1 + FQ)]1/2
Resemblance in general
 Coefficient r of the additive variance is sometimes
called the coefficient of relationship (the correlation
between the breeding values A)
 Coefficient u represents the probability of the
relatives having the same genotype through identity
by descent
 It is zero unless the related individuals have paths of
coancestry through both of their respective parents,
(e.g., full sibs and double first cousins)
Environmental covariance
 VE = VEc + VEw
 VEc; common, i.e., contributes to variance
between means of families but not the
variance within (covariance among related
individuals)
 VEw; within, i.e., arises from independent of
coefficient of relationship
 Maternal effects and competition
Phenotypic resemblance between relatives
Relatives
Covariance
Offspring and
one parent
Offspring and
mid-parent
Half sibs
1/2VA
Regression (b) or
intraclass
correlation (t)
b = 1/2(VA/VP)
1/2VA
b = VA/VP
1/4VA
t = 1/4(VA/VP)
Full sibs
1/2VA+1/4VD+VEc
t=(1/2VA+1/4VD+V
Ec)/VP
Heritability
 Regression of breeding value on phenotypic value
 Index of response to genetic selection
 Estimated with





offspring-parent regression,
sib analysis,
intra-sire regression of offspring on dam,
or combined estimates
plus other methods (Markel et al., 1995, 1999)
Heritability: Ratio of additive genetic variance to
phenotypic variance
 h2 = VA / VP
 Regression of breeding value on phenotypic value
 h2 = bAP
 rAP = bAP sP / sA = h2(1/h) = h
Heritability: Twins and human data
Between pairs, s2b
Within pairs, s2w
Identical
(MZ)
VA + VD + VEc
VEw
Dizygotic
(DZ)
1/2 VA + 1/4 VD + VEc
1/2 VA + 3/4 VD + VEw
Difference 1/2 VA + 3/4 VD
1/2 VA + 3/4 VD
Heritability: Twins and human data
Correlations
Trait
Monozygotic
Dizygotic
Height
0.93
0.48
Intelligence
0.86
0.62
Personality
0.50
0.30
Alcohol
consumption
0.64
0.27
C = B1P + B2 + B3P

P
B1
B2
C
P
B3
1.0 (MZT,DZT) or 0.0 (MZA, DZA)
1.0 (MZ) or .25 (DZ)
1.0 (MZ) or .5 (DZ)
A1
C1
a
E1
c
e
D1
d
A2
a
C2
c
E2
e
P1
P2
Twin 1
Twin 2
D2
d