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Correlated characters Sanja Franic VU University Amsterdam 2008.

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Presentation on theme: "Correlated characters Sanja Franic VU University Amsterdam 2008."— Presentation transcript:

1 Correlated characters Sanja Franic VU University Amsterdam 2008

2 Relationship between 2 metric characters whose values are correlated in the individuals of a population

3 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?

4 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

5 We can only observe the phenotypic correlation How to decompose it into genetic and environmental causal components?

6 We can only observe the phenotypic correlation How to decompose it into genetic and environmental causal components? (phenotypic correlation) (phenotypic covariance) (phenotypic covariance expressed in terms of A and E) (substitution gives) (because σ 2 P = σ 2 A + σ 2 E σ P = σ A + σ E σ P =hσ P +eσ P ) (substitution gives) (phenotypic correlation expressed in terms of A and E)

7 We can only observe the phenotypic correlation How to decompose it into genetic and environmental causal components? (phenotypic correlation) (phenotypic covariance) (phenotypic covariance expressed in terms of A and E) (substitution gives) (because σ 2 P = σ 2 A + σ 2 E σ P = σ A + σ E σ P =hσ P +eσ P ) (substitution gives) (phenotypic correlation expressed in terms of A and E)

8 Estimation of the genetic correlation Analogous to estimation of heritabilities, but instead of ANOVA we use an ANCOVA

9 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

10 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

11 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 observational components between-sirebetween-dam within-sire within-progeny

12 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 observational components causal components between-sirebetween-dam within-sire within-progeny ADE

13 σ 2 S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ V A

14 σ 2 W  V T = V BG + V WG V WG = V T – V BG V BG = cov FS cov FS = ½ V A + ¼ V D σ 2 W = V WG = V T - ½ V A - ¼ V D = V A + V D +V E - ½ V A - ¼ V D = ½ V A + ¾ V D + V EW

15 σ 2 S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ V A σ 2 W  V T = V BG + V WG V WG = V T – V BG V BG = cov FS cov FS = ½ V A + ¼ V D σ 2 W = V WG = V T - ½ V A - ¼ V D = V A + V D +V E - ½ V A - ¼ V D = ½ V A + ¾ V D + V EW σ 2 D = σ 2 T -σ 2 S -σ 2 W = V A + V D +V E - ¼ V A - ½ V A – ¾ V D - V EW = ¼ V A + ¼ V D + V EC (V E = V EC +V EW )

16 σ 2 S = variance between means of half-sib families (phenotypic covariance of half-sibs) = ¼ V A σ 2 W  V T = V BG + V WG V WG = V T – V BG V BG = cov FS cov FS = ½ V A + ¼ V D σ 2 W = V WG = V T - ½ V A - ¼ V D = V A + V D +V E - ½ V A - ¼ V D = ½ V A + ¾ V D + V EW σ 2 D = σ 2 T -σ 2 S -σ 2 W = V A + V D +V E - ¼ V A - ½ V A – ¾ V D - V EW = ¼ V A + ¼ V D + V EC (V E = V EC +V EW ) In partitioning the covariance, instead of starting from individual values we start from the product of the values of the 2 characters cov S = ¼ cov A

17 var SX = ¼ σ 2 AX var SY = ¼ σ 2 AY

18 cov S = ¼ cov A var SX = ¼ σ 2 AX var SY = ¼ σ 2 AY 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 cov A

19 cov S = ¼ cov A var SX = ¼ σ 2 AX var SY = ¼ σ 2 AY 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 cov A

20 Correlated response to selection If we select for X, what will be the change in Y?

21 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

22 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

23 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 because:

24 [11.4]

25 [11.4]

26 [11.4]

27 [11.4]

28 [11.4]

29 [11.4] Coheritability

30 [11.4] Coheritability Heritability [11.3]

31 Questions?


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