1. Outline: 1) Basics 2) Means and Values (Ch 7): summary 3) Variance (Ch 8): summary 4) Resemblance between relatives 5) Homework (8.3)

2. Values & means: summary (Falconer & Mackay: chapter 7) Sanja Franic VU University Amsterdam 2011

3. Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Genotypic value (g i )ad-a Genotype frequency(f i )p2p2 2pq q 2 Mean genotypic value (μ gi = g i f i )p2ap2a2pqd-q 2 aMean gen. value across genotypes: μ G = ∑μ gi = a(p – q) + 2dpq Frequencies of genotypes produced Mean values of genotypes produced (μ Gj ) Average effect of allele (α j = μ Gj - μ G ) A1A1A1A1 A1A2A1A2 A2A2A2A2 Parental gametes A1A1 pqpa + qd α 1 = q[a + d(q – p)] A2A2 pqpd – qa α 2 = – p[a + d(q – p)] Average effect of allele substitution: α = α 1 - α 2 = a + d(q – p) Breeding value (A i ) 2α 1 = 2qαα 1 + α 2 = (q – p)α 2α 2 = –2pα E[A] = ∑A i f i = 2p 2 qα + 2pq(q – p)α – 2pq 2 α = 2pqα(p + q – p – q) = 0 Genotypic value (G i = μ gi - μ G ) 2q(α – qd)(q – p)α + 2pqd –2p(α + pd) E[G] = ∑G i f i = 2p 2 q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq 2 (α + pd) = 0 Dominance deviation (D i = G i – A i )–2q 2 d2pqd –2p 2 dE[D] = ∑D i f i = –2p 2 q 2 d + 4p 2 q 2 d – 2p 2 q 2 d = 0 Summary table: P = G + E = A + D + I+ E

4. Variance: summary (Falconer & Mackay: chapter 8)

5. Value componentsVariance components PPhenotypic valueVPVP Phenotypic variance Value decomposition: P = G + E GGenotypic valueVGVG Genotypic variance = A + D + I + E ABreeding valueVAVA Additive genetic variance DDominance deviationVDVD Dominance genetic varianceVariance decomposition: V P = V G + V E IInteraction deviationVIVI Interaction variance = V A + V D + V I + V E * EEnvironmental deviationVEVE Environmental variance *the more general expression is V P = V G + V E + 2cov GE + V GE, where cov GE is the covariance between genotypic values and the environmental deviations, and V GE the variance due to interaction between genotypes and the environment.

6. V A and V D : obtained by squaring the breeding values and dominance deviations, respectively, multiplying by the genotype frequency, and summing over the three genotypes. * Note: As the means of A, G, and D are all 0, no correction for the mean is needed and their variance is obtained simply as the mean of squared values (i.e., in V = Σ(x i - μ x ) 2 /N, μ x = 0, thus V = Σx i 2 /N). Given that we work with frequencies, V = Σx i 2 f i Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Genotypic value (g i )ad-a Genotype frequency(f i )p2p2 2pq q 2 Mean genotypic value (μ gi = g i f i )p2ap2a2pqd-q 2 aMean gen. value across genotypes: μ G = ∑μ gi = a(p – q) + 2dpq Breeding value (A i ) 2α 1 = 2qαα 1 + α 2 = (q – p)α 2α 2 = –2pα E[A] = ∑A i f i = 2p 2 qα + 2pq(q – p)α – 2pq 2 α = 2pqα(p + q – p – q) = 0 Genotypic value (G i = μ gi - μ G ) 2q(α – qd)(q – p)α + 2pqd –2p(α + pd) E[G] = ∑G i f i = 2p 2 q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq 2 (α + pd) = 0 Dominance deviation (D i = G i – A i )–2q 2 d2pqd –2p 2 dE[D] = ∑D i f i = –2p 2 q 2 d + 4p 2 q 2 d – 2p 2 q 2 d = 0

7. V A and V D : obtained by squaring the breeding values and dominance deviations, respectively, multiplying by the genotype frequency, and summing over the three genotypes. * Note: As the means of A, G, and D are all 0, no correction for the mean is needed and their variance is obtained simply as the mean of squared values (i.e., in V = Σ(x i - μ x ) 2 /N, μ x = 0, thus V = Σx i 2 /N). Given that we work with frequencies, V = Σx i 2 f i Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Genotypic value (g i )ad-a Genotype frequency(f i )p2p2 2pq q 2 Mean genotypic value (μ gi = g i f i )p2ap2a2pqd-q 2 aMean gen. value across genotypes: μ G = ∑μ gi = a(p – q) + 2dpq Breeding value (A i ) 2α 1 = 2qαα 1 + α 2 = (q – p)α 2α 2 = –2pα E[A] = ∑A i f i = 2p 2 qα + 2pq(q – p)α – 2pq 2 α = 2pqα(p + q – p – q) = 0 Genotypic value (G i = μ gi - μ G ) 2q(α – qd)(q – p)α + 2pqd –2p(α + pd) E[G] = ∑G i f i = 2p 2 q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq 2 (α + pd) = 0 Dominance deviation (D i = G i – A i )–2q 2 d2pqd –2p 2 dE[D] = ∑D i f i = –2p 2 q 2 d + 4p 2 q 2 d – 2p 2 q 2 d = 0 V A = 4p 2 q 2 α 2 + 2pq α 2 (q – p) 2 + 4p 2 q 2 α 2 = 2pq α 2 (2pq + q 2 – 2pq + p 2 + 2pq) = 2pq α 2 (q 2 + 2pq + p 2 ) = 2pq α 2 (p + q) 2 = 2pq α 2 = 2pq[a + d(q – p)] 2

8. V A and V D : obtained by squaring the breeding values and dominance deviations, respectively, multiplying by the genotype frequency, and summing over the three genotypes. * Note: As the means of A, G, and D are all 0, no correction for the mean is needed and their variance is obtained simply as the mean of squared values (i.e., in V = Σ(x i - μ x ) 2 /N, μ x = 0, thus V = Σx i 2 /N). Given that we work with frequencies, V = Σx i 2 f i Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Genotypic value (g i )ad-a Genotype frequency(f i )p2p2 2pq q 2 Mean genotypic value (μ gi = g i f i )p2ap2a2pqd-q 2 aMean gen. value across genotypes: μ G = ∑μ gi = a(p – q) + 2dpq Breeding value (A i ) 2α 1 = 2qαα 1 + α 2 = (q – p)α 2α 2 = –2pα E[A] = ∑A i f i = 2p 2 qα + 2pq(q – p)α – 2pq 2 α = 2pqα(p + q – p – q) = 0 Genotypic value (G i = μ gi - μ G ) 2q(α – qd)(q – p)α + 2pqd –2p(α + pd) E[G] = ∑G i f i = 2p 2 q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq 2 (α + pd) = 0 Dominance deviation (D i = G i – A i )–2q 2 d2pqd –2p 2 dE[D] = ∑D i f i = –2p 2 q 2 d + 4p 2 q 2 d – 2p 2 q 2 d = 0 V A = 4p 2 q 2 α 2 + 2pq α 2 (q – p) 2 + 4p 2 q 2 α 2 = 2pq α 2 (2pq + q 2 – 2pq + p 2 + 2pq) = 2pq α 2 (q 2 + 2pq + p 2 ) = 2pq α 2 (p + q) 2 = 2pq α 2 = 2pq[a + d(q – p)] 2 V D = 4q 4 d 2 p 2 + 8p 3 q 3 d 2 + 4p 4 d 2 q 2 = 4q 2 p 2 d 2 (q 2 + 2pq + p 2 ) = 4q 2 p 2 d 2 (p + q) 2 = 4q 2 p 2 d 2 = (2pqd) 2

9. V A and V D : obtained by squaring the breeding values and dominance deviations, respectively, multiplying by the genotype frequency, and summing over the three genotypes. * Note: As the means of A, G, and D are all 0, no correction for the mean is needed and their variance is obtained simply as the mean of squared values (i.e., in V = Σ(x i - μ x ) 2 /N, μ x = 0, thus V = Σx i 2 /N). Given that we work with frequencies, V = Σx i 2 f i Genotype A1A1A1A1 A1A2A1A2 A2A2A2A2 Genotypic value (g i )ad-a Genotype frequency(f i )p2p2 2pq q 2 Mean genotypic value (μ gi = g i f i )p2ap2a2pqd-q 2 aMean gen. value across genotypes: μ G = ∑μ gi = a(p – q) + 2dpq Breeding value (A i ) 2α 1 = 2qαα 1 + α 2 = (q – p)α 2α 2 = –2pα E[A] = ∑A i f i = 2p 2 qα + 2pq(q – p)α – 2pq 2 α = 2pqα(p + q – p – q) = 0 Genotypic value (G i = μ gi - μ G ) 2q(α – qd)(q – p)α + 2pqd –2p(α + pd) E[G] = ∑G i f i = 2p 2 q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq 2 (α + pd) = 0 Dominance deviation (D i = G i – A i )–2q 2 d2pqd –2p 2 dE[D] = ∑D i f i = –2p 2 q 2 d + 4p 2 q 2 d – 2p 2 q 2 d = 0 V A = 4p 2 q 2 α 2 + 2pq α 2 (q – p) 2 + 4p 2 q 2 α 2 = 2pq α 2 (2pq + q 2 – 2pq + p 2 + 2pq) = 2pq α 2 (q 2 + 2pq + p 2 ) = 2pq α 2 (p + q) 2 = 2pq α 2 = 2pq[a + d(q – p)] 2 V D = 4q 4 d 2 p 2 + 8p 3 q 3 d 2 + 4p 4 d 2 q 2 = 4q 2 p 2 d 2 (q 2 + 2pq + p 2 ) = 4q 2 p 2 d 2 (p + q) 2 = 4q 2 p 2 d 2 = (2pqd) 2 G = A + D, V G = V A + V D + 2cov AD cov AD = -4p 2 q 3 α d + 4p 2 q 2 (q – p) α d + 4p 3 q 2 α d = 4p 2 q 2 α d(- q + q – p + p) = 0 V G = V A + V D = 2pq[a + d(q – p)] 2 + (2pqd) 2

10. Resemblance between relatives (Falconer & Mackay: chapter 9)

11. - resemblance between relatives: - one of the basic genetic phenomena displayed by metric traits - easy to determine by simple measurements of the trait - provides the means of estimating the amount of additive genetic variance (V A ) - last chapter: causal components of phenotypic variance (V: V E, V G, V A, V D, V I ) - observational components of phenptypic variance:  2

12. - resemblance between relatives: - one of the basic genetic phenomena displayed by metric traits - easy to determine by simple measurements of the trait - provides the means of estimating the amount of additive genetic variance (V A ) - last chapter: causal components of phenotypic variance (V: V E, V G, V A, V D, V I ) - observational components of phenptypic variance:  2 - e.g., grouping of individuals into families of full sibs: - ANOVA: we can partition the total variation into between and within group variance - these components can be used to estimate the covariation between full sibs, as the intraclass correlation coefficient t =  2 B / (  2 B +  2 W ) - between group variance (  2 B ) = covariance of the members of the group → the variance between groups (families) of full sibs = covariance between full sibs - this can be explained in great detail in the next lecture (on heritability)

13. - resemblance between relatives: - one of the basic genetic phenomena displayed by metric traits - easy to determine by simple measurements of the trait - provides the means of estimating the amount of additive genetic variance (V A ) - last chapter: causal components of phenotypic variance (V: V E, V G, V A, V D, V I ) - observational components of phenptypic variance:  2 - e.g., grouping of individuals into families of full sibs: - ANOVA: we can partition the total variation into between and within group variance - these components can be used to estimate the covariation between full sibs, as the intraclass correlation coefficient: t =  2 B / (  2 B +  2 W ) - between group variance (  2 B ) = covariance of the members of the group → the variance between groups (families) of full sibs = covariance between full sibs - this can be explained in great detail in the next lecture (on heritability) - offspring and parents: - the grouping is in pairs: parent (or mean parent) and offspring (or mean offspring) - the intraclass correlation coefficient (ICC) therefore not necessary (in addition: the phenotypic variance is often not the same in parents and offspring – the sums-of-squares approach is therefore inadequate) - instead, the covariance of offspring with parents is calculated in the standard way (from the sum of cross-products), and standardized as in regression: cov OP = b OP  2 P → b OP = cov OP /  2 P OP b OP 2P2P

14. - the phenotypic covariance is composed of the causal components of variance (V) discussed in Ch 8, but in proportions differing according to the sort of relationship → by finding out how the causal components contribute to the covariance, we will see how the observed covariance can be used to estimate the causal components

15. - the phenotypic covariance is composed of the causal components of variance (V) discussed in Ch 8, but in proportions differing according to the sort of relationship → by finding out how the causal components contribute to the covariance, we will see how the observed covariance can be used to estimate the causal components - for the time being, we will focus on genetic covariance between relatives (i.e., will not consider the non-genetic covariance) - this means we are considering the covariance between the genotypic values (G) of individuals - assumptions: - Hardy-Weinberg equilibrim - random mating with respect to the trait in question - no epistasis (these assumptions can be tested, and the effects can be explicitly modeled if the assumptions do not hold)

16. - the phenotypic covariance is composed of the causal components of variance (V) discussed in Ch 8, but in proportions differing according to the sort of relationship → by finding out how the causal components contribute to the covariance, we will see how the observed covariance can be used to estimate the causal components - for the time being, we will focus on genetic covariance between relatives (i.e., will not consider the non-genetic covariance) - this means we are considering the covariance between the genotypic values (G) of individuals - assumptions: - Hardy-Weinberg equilibrim - random mating with respect to the trait in question - no epistasis (these assumptions can be tested, and the effects can be explicitly modeled if the assumptions do not hold) - we will consider 4 types of relationships: - parent-offspring - half sibs - full sibs - twins

17. Offspring and one parent - the covariance of genotypic values of individuals with the mean genotypic value of their offspring (under random mating) - if values are expressed as deviations from the population mean, then the mean value of the offspring is by definition half the breeding value of the parent (Ch 7) - therefore: the covariance in question is the covariance between the genotypic value of an individual with half its breeding value - i.e., the covariance between G and ½A

18. Offspring and one parent - the covariance of genotypic values of individuals with the mean genotypic value of their offspring (under random mating) - if values are expressed as deviations from the population mean, then the mean value of the offspring is by definition half the breeding value of the parent (Ch 7) - therefore: the covariance in question is the covariance between the genotypic value of an individual with half its breeding value - i.e., the covariance between G and ½A G = A + D, therefore we are looking at the covariance of A + D with ½A: * cov OP = (  ½A i (A i + D i ) ) / N = (½  A i (A i + D i ) ) / N = (½  A i 2 + ½  A i D i ) ) / N = ½  A i 2 /N + ½  A i D i /N where i (i = 1, …, N) denotes parent-offspring pair. * Note: the variables do not need centering, as they are already expressed as deviations from the population mean

19. Offspring and one parent - the covariance of genotypic values of individuals with the mean genotypic value of their offspring (under random mating) - if values are expressed as deviations from the population mean, then the mean value of the offspring is by definition half the breeding value of the parent (Ch 7) - therefore: the covariance in question is the covariance between the genotypic value of an individual with half its breeding value - i.e., the covariance between G and ½A G = A + D, therefore we are looking at the covariance of A + D with ½A: * cov OP = (  ½A i (A i + D i ) ) / N = (½  A i (A i + D i ) ) / N = (½  A i 2 + ½  A i D i ) ) / N = ½  A i 2 /N + ½  A i D i /N where i (i = 1, …, N) denotes parent-offspring pair. ½  A i 2 /N = ½V A ½  A i D i /N = ½cov AD cov AD = 0 (from Ch 8) Therefore: cov OP = ½V A → The genetic covariance between parent and offspring is half the additive genetic variance of the parents. * Note: the variables do not need centering, as they are already expressed as deviations from the population mean

20. Offspring and one parent Another way of deriving the covariance: A2A2A2A2 0d+ a- a Genotypic value Genotype A1A2A1A2 A1A1A1A1 Genotype frequency2pqp2p2 q2q2 Parents GenotypeA1A1A1A1 A1A2A1A2 A2A2A2A2 Freqp2p2 2pqq2q2 Genotypic value 2q(  - qd)(q-p)  + 2qpd-2p(  + pd) OffspringMean genot. value qq ½ (q - p)  -p  Genotypic values (a, d, -a) expressed as deviations from the population mean

21. Offspring and one parent Another way of deriving the covariance: Parents GenotypeA1A1A1A1 A1A2A1A2 A2A2A2A2 Freqp2p2 2pqq2q2 Genotypic value 2q(  - qd)(q-p)  + 2qpd-2p(  + pd) OffspringMean genot. value qq ½ (q - p)  -p  A2A2A2A2 0d+ a- a Genotypic value Genotype A1A2A1A2 A1A1A1A1 Genotypic values (a, d, -a) expressed as deviations from the population mean  a –  = a – [(p – q) + 2dpq] = 2q(  - qd)

22. Offspring and one parent Another way of deriving the covariance: Parents GenotypeA1A1A1A1 A1A2A1A2 A2A2A2A2 Freqp2p2 2pqq2q2 Genotypic value 2q(  - qd)(q-p)  + 2qpd-2p(  + pd) OffspringMean genot. value qq ½ (q - p)  -p  mean cross-product of these

23. Offspring and one parent Another way of deriving the covariance: cov OP = 2q 2 p 2  (  – qd) + pq  (q – p) [(q – p)  + 2qpd] + 2p 2 q 2  (  + pd) = pq  2 = ½V A, since V A = 2pq  2. cov OP = b OP * V P b OP = cov OP / V P = ½V A / V P Parents GenotypeA1A1A1A1 A1A2A1A2 A2A2A2A2 Freqp2p2 2pqq2q2 Genotypic value 2q(  - qd)(q-p)  + 2qpd-2p(  + pd) OffspringMean genot. value qq ½ (q - p)  -p  mean cross-product of these OP b OP VPVP

24. Offspring and mid-parent - mid-parent = mean of the two parents O = mean genotypic value of the offspring P & P’ = genotypic values of the two parents mid-parent genotypic value: P = ½(P + P’) cov OP =  OP/N =  ½(P + P’)O / N = (½  PO + ½  P’O) / N = ½  PO/N + ½  P’O/N = ½cov OP + ½cov OP’ If the P and P’ have the same variance, then cov OP = cov OP’, so: cov OP = cov OP = ½V A → Therefore the covariance is the same as in the case of offspring and one parent

25. Offspring and mid-parent - mid-parent = mean of the two parents O = mean genotypic value of the offspring P & P’ = genotypic values of the two parents mid-parent genotypic value: P = ½(P + P’) cov OP =  OP/N =  ½(P + P’)O / N = (½  PO + ½  P’O) / N = ½  PO/N + ½  P’O/N = ½cov OP + ½cov OP’ If the P and P’ have the same variance, then cov OP = cov OP’, so: cov OP = cov OP = ½V A → Therefore the covariance is the same as in the case of offspring and one parent However, the regression coefficient is different. The variance of the mean of n variables is 1/n of variance of single variables. Therefore, V P = ½V P. b OP = cov OP / ½V P = ½V A / ½V P = V A /V P OP bOPbOP ½V P

26. Half sibs - a group of half sibs = progeny of one individual mated to a random group of the other sex, having one offspring by each mate - therefore, by definition, the mean genotypic value of a group of half sibs is half the breeding value of the common parent - the covariance of half sibs is the variance of the true values of the half-sib groups (i.e., it is the between-group variance; will be explained more in the lecture on ICC) - the true mean of each half sib group is half the breeding value of the parent - therefore, the covariance of half sibs is the variance of is half the breeding value of the parent, which is a quarter of the additive variance (trust me, or should I derive it?): cov HS = V ½A = ¼V A

27. If you wanted me to derive it:

28. Half sibs - a group of half sibs = progeny of one individual mated to a random group of the other sex, having one offspring by each mate - therefore, by definition, the mean genotypic value of a group of half sibs is half the breeding value of the common parent - the covariance of half sibs is the variance of the true values of the half-sib groups (i.e., it is the between-group variance; will be explained more in the lecture on ICC) - the true mean of each half sib group is half the breeding value of the parent - therefore, the covariance of half sibs is the variance of is half the breeding value of the parent, which is a quarter of the additive variance (trust me, or should I derive it?): cov HS = V ½A = ¼V A - degree of resemblance between half sibs is expressed as the ICC (the between group variance [i.e., the covariance] as a proportion of total variance): t = ¼V A /V P

29. Full sibs - dominance variance contributes to the covariance between full sibs (unlike the relationships considered so far) - additive variance: - full sibs share both parents; therefore, their mean genotypic value equals the mean breeding value of the two parents - therefore, the covariance is the variance of ½(A + A’) var ½(A + A’) = ¼(V A + V A’ ) = ½V A if the additive genetic variance is equal in the two sexes.

30. Full sibs - dominance variance contributes to the covariance between full sibs (unlike the relationships considered so far) - additive variance: - full sibs share both parents; therefore, their mean genotypic value equals the mean breeding value of the two parents - therefore, the covariance is the variance of ½(A + A’) var ½(A + A’) = ¼(V A + V A’ ) = ½V A if the additive genetic variance is equal in the two sexes. - dominance variance: - let parents have genotypes A 1 A 2 and A 3 A 4 - then there are 4 genotypes in the progeny: A 1 A 3, A 1 A 4, A 2 A 3, A 2 A 4, each with a frequency of ¼ - let one of the sibs have any of the genotypes - then the probability that the other sib will have the same genotype is ¼ - therefore, ¼ of full sibs have the same genotype, and consequently the same dominance deviation, D - for these pairs, the cross-product of dominance deviations is D 2 - for other pairs, this cross-product is 0 - therefore, on average, the (mean) cross-product is ¼  D 2 / N, which equals ¼V D - total genetic covariance of full sibs is therefore: cov FS = ½V A + ¼V D

31. Full sibs - the correlation of full sibs is: t = (½V A + ¼V D ) / V P

32. Twins - dizygotic (DZ) twins are related as full sibs; their genetic covariance is that of full sibs: cov DZ = ½V A + ¼V D - monozygotic (MZ) twins have identical genotypes, therefore: cov MZ = V G

33. Environmental covariance - related individuals may resemble each other for environmental reasons as well (some relatives more than others) - family members reared together share a common environment -> some environmental circumstances that cause differences between unrelated individuals are not a cause of difference between members of the same family -> there is a component of environmental variance that contributes to the variance between means of families, but not to the variance within families -> therefore, it contributes to the covariance of family members - this components is termed V Ec : common environment - the remainder of environmental variance (V Ew ) arises from causes of difference that are unconnected to whether the individuals are related or not - therefore, V Ew appears in the within-group component of variance, but does not contribute to the between-group component - the total environmental variance can therefore be partitioned as follows: V E = V Ec + V Ew, where the V Ec component contributes to the covariance of related invididuals.

34. Summary RelationshipGenetic covariance Offspring and one parentcov OP = ½V A Offspring and mid-parentcov OP = ½V A if variance is equal in two sexes Half sibscov HS = ¼V A Full sibscov FS = ½V A + ¼V D DZ twinscov DZ = ½V A + ¼V D MZ twinscov MZ = V G

35. Summary RelationshipCovariance Offspring and one parentcov OP = ½V A Offspring and mid-parentcov OP = ½V A if variance is equal in two sexes Half sibscov HS = ¼V A Full sibscov FS = ½V A + ¼V D + V Ec DZ twinscov DZ = ½V A + ¼V D + V Ec MZ twinscov MZ = V G + V Ec

36. Summary RelationshipGenetic covarianceStandardized genetic covariance Offspring and one parentcov OP = ½V A ½V A / V P Offspring and mid-parentcov OP = ½V A V A / V P Half sibscov HS = ¼V A ¼V A / V P Full sibscov FS = ½V A + ¼V D (½V A + ¼V D ) / V P DZ twinscov DZ = ½V A + ¼V D (½V A + ¼V D ) / V P MZ twinscov MZ = V G V G / V P

37. Applications in structural equation modeling

39. Homework 9.3 (you’ll need to read Ch 9 to understand the solution, please be able to explain it in class)