Transcript Correlated Characters

Correlated characters
Sanja Franic
VU University Amsterdam 2008
•
Relationship between 2 metric characters whose values are correlated in the individuals of
a population
•
Relationship between 2 metric characters whose values are correlated in the individuals of
a population
•
Why are correlated characters important?
•
Effects of pleiotropy in quantitative genetics
– Pleiotropy – gene affects 2 or more characters
– (e.g. genes that increase growth rate increase both height and weight)
•
Selection – how will the improvement in one character cause simultaneous changes in
other characters?
•
Relationship between 2 metric characters whose values are correlated in the individuals of
a population
•
Why are correlated characters important?
•
Effects of pleiotropy in quantitative genetics
– Pleiotropy – gene affects 2 or more characters
– (e.g. genes that increase growth rate increase both height and weight)
•
Selection – how will the improvement in one character cause simultaneous changes in
other characters?
•
Causes of correlation:
•
Genetic
– mainly pleiotropy
– but some genes may cause +r, while some cause –r, so overall effect not always
detectable
•
Environmental
– two characters influenced by the same differences in the environment
•
•
We can only observe the phenotypic correlation
How to decompose it into genetic and environmental causal components?
•
•
We can only observe the phenotypic correlation
How to decompose it into genetic and environmental causal components?
rP XY 
covPXY
 PX  PY
covPXY  rP XY  PX  PY
covPXY  covAXY  covE XY
rP XY  PX  PY  rAXY  AX  AY  rE XY  EX  EY
 A  h P
(phenotypic correlation)
(phenotypic covariance)
(phenotypic covariance
expressed in terms of A and E)
(substitution gives)
(because σ2P= σ2A+ σ2E
σP= σA+ σE
σP=hσP+eσP)
 E  e P
rP XY  PX  PY  rAXY hX  PX hY  PY  rE XY e X  PX eY  PY
rP XY  rAXY hX hY  rE XY e X eY
(substitution gives)
(phenotypic correlation
expressed in terms of A and E)
•
•
We can only observe the phenotypic correlation
How to decompose it into genetic and environmental causal components?
rP XY 
covPXY
 PX  PY
covPXY  rP XY  PX  PY
covPXY  covAXY  covE XY
rP XY  PX  PY  rAXY  AX  AY  rE XY  EX  EY
 A  h P
(phenotypic correlation)
(phenotypic covariance)
(phenotypic covariance
expressed in terms of A and E)
(substitution gives)
(because σ2P= σ2A+ σ2E
σP= σA+ σE
σP=hσP+eσP)
 E  e P
rP XY  PX  PY  rAXY hX  PX hY  PY  rE XY e X  PX eY  PY
rP XY  rAXY hX hY  rE XY e X eY
(substitution gives)
(phenotypic correlation
expressed in terms of A and E)
Estimation of the genetic correlation
•
Analogous to estimation of heritabilities, but instead of ANOVA we use an ANCOVA
Estimation of the genetic correlation
•
Analogous to estimation of heritabilities, but instead of ANOVA we use an ANCOVA
Half-sib families
• Design: a number of sires each mated to several dames (random mating)
• A number of offspring from each dam are measured
Estimation of the genetic correlation
•
Analogous to estimation of heritabilities, but instead of ANOVA we use an ANCOVA
Half-sib families
• Design: a number of sires each mated to several dames (random mating)
• A number of offspring from each dam are measured
s=number of sires
d=number of dames per sire
k=number of offspring per dam
Estimation of the genetic correlation
•
Analogous to estimation of heritabilities, but instead of ANOVA we use an ANCOVA
Half-sib families
• Design: a number of sires each mated to several dames (random mating)
• A number of offspring from each dam are measured
s=number of sires
d=number of dames per sire
k=number of offspring per dam
 2 P   2 S   2 D   2W
between-sire
between-dam
within-sire
within-progeny
observational components
Estimation of the genetic correlation
•
Analogous to estimation of heritabilities, but instead of ANOVA we use an ANCOVA
Half-sib families
• Design: a number of sires each mated to several dames (random mating)
• A number of offspring from each dam are measured
s=number of sires
d=number of dames per sire
k=number of offspring per dam
 2 P   2 S   2 D   2W
between-sire
between-dam
within-sire
within-progeny
A
D
E
observational components
causal components
σ2S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ VA
σ2S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ VA
σ2W  VT = VBG + VWG
VWG = VT – VBG
VBG = covFS
covFS = ½ VA + ¼ VD
σ2W = VWG = VT - ½ VA - ¼ VD
= VA + VD +VE - ½ VA - ¼ VD
= ½ VA + ¾ VD + VEW
σ2S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ VA
σ2W  VT = VBG + VWG
VWG = VT – VBG
VBG = covFS
covFS = ½ VA + ¼ VD
σ2W = VWG = VT - ½ VA - ¼ VD
= VA + VD +VE - ½ VA - ¼ VD
= ½ VA + ¾ VD + VEW
σ2D = σ2T-σ2S -σ2W
= VA + VD +VE - ¼ VA - ½ VA – ¾ VD - VEW
= ¼ VA + ¼ VD + VEC
(VE = VEC +VEW)
σ2S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ VA
σ2W  VT = VBG + VWG
VWG = VT – VBG
VBG = covFS
covFS = ½ VA + ¼ VD
σ2W = VWG = VT - ½ VA - ¼ VD
= VA + VD +VE - ½ VA - ¼ VD
= ½ VA + ¾ VD + VEW
σ2D = σ2T-σ2S -σ2W
= VA + VD +VE - ¼ VA - ½ VA – ¾ VD - VEW
= ¼ VA + ¼ VD + VEC
(VE = VEC +VEW)
•
In partitioning the covariance, instead of starting from individual values we start from the
product of the values of the 2 characters
covS = ¼ covA
rA 
•
•
•
covXY
 XY

covXY
varX varY
covS = ¼ covA
varSX = ¼ σ2AX
varSY = ¼ σ2AY
rA 
1 covA
4
1 covAX 1 covAY
4
4
rA 
•
•
•
covXY
 XY

covXY
varX varY
covS = ¼ covA
varSX = ¼ σ2AX
varSY = ¼ σ2AY
rA 
1 covA
4
1 covAX 1 covAY
4
4
Offspring-parent relationship
•
•
•
To estimate the heritability of one character, we compute the covariance of offspring and
parent
To estimate the genetic correlation between 2 characters we compute the “cross-variance”:
product of value of X in offspring and value of Y in parents
Cross-variance = ½ covA
rA 
•
•
•
covXY
 XY

covXY
varX varY
covS = ¼ covA
varSX = ¼ σ2AX
varSY = ¼ σ2AY
rA 
1 covA
4
1 covAX 1 covAY
4
4
Offspring-parent relationship
•
•
•
To estimate the heritability of one character, we compute the covariance of offspring and
parent
To estimate the genetic correlation between 2 characters we compute the “cross-variance”:
product of value of X in offspring and value of Y in parents
Cross-variance = ½ covA
rA 
covXY
covXX covYY
Correlated response to selection
•
If we select for X, what will be the change in Y?
Correlated response to selection
•
If we select for X, what will be the change in Y?
•
•
The response in X – the mean breeding value of the selected individuals
The consequent change in Y – regression of breeding value of Y on breeding value of X
Correlated response to selection
•
If we select for X, what will be the change in Y?
•
•
The response in X – the mean breeding value of the selected individuals
The consequent change in Y – regression of breeding value of Y on breeding value of X
b( A)YX 
covA
 2 AX
r
 AY
 AX
Correlated response to selection
•
If we select for X, what will be the change in Y?
•
•
The response in X – the mean breeding value of the selected individuals
The consequent change in Y – regression of breeding value of Y on breeding value of X
b( A)YX 
covA
 2 AX
r
 AY
 AX
because:
r
covXY
 XY
bYX 
, covXY  r X  Y
covXY

2
, covXY  bYX  2 X
X
r X  Y  bYX  2 X
b
r X  Y
 2X
r
Y
X
RX  ihX  AX
[11.4]
RX  ihX  AX
CRY  b( A)YX RX
[11.4]
[11.4]
RX  ihX  AX
CRY  b( A)YX RX
CRY  ihX  AX rA
 AY
 AX
[11.4]
RX  ihX  AX
CRY  b( A)YX RX
CRY  ihX  AX rA
CRY  ihX rA AY
 AY
 AX
[11.4]
RX  ihX  AX
CRY  b( A)YX RX
CRY  ihX  AX rA
 AY
 AX
CRY  ihX rA AY
Since  AY  hY  PY :
CRY  ihX hY rA PY
[11.4]
RX  ihX  AX
CRY  b( A)YX RX
CRY  ihX  AX rA
 AY
 AX
CRY  ihX rA AY
Since  AY  hY  PY :
CRY  ihX hY rA PY
Coheritability
[11.4]
RX  ihX  AX
CRY  b( A)YX RX
CRY  ihX  AX rA
 AY
 AX
CRY  ihX rA AY
Since  AY  hY  PY :
CRY  ihX hY rA PY
Coheritability
RX  ih 2 PX
Heritability
[11.3]
• Questions?