Outline:

1) Basics2) Means and Values (Ch 7): summary3) Variance (Ch 8): summary4) Resemblance between relatives5) Homework (8.3)

Values & means: summary(Falconer & Mackay: chapter 7)

Sanja FranicVU University Amsterdam 2011

Genotype

A1A1 A1A2 A2A2

Genotypic value (gi) a d -a

Genotype frequency(fi) p2 2pq q2

Mean genotypic value (μgi =

gifi)p2a 2pqd -q2a

Mean gen. value across genotypes: μG = ∑μgi = a(p – q) +

2dpq

Frequencies of genotypes produced Mean values of genotypes

produced

(μGj)

Average effect of allele

(αj = μGj - μG)A1A1 A1A2 A2A2

Parental gametes A1 p q pa + qd α1 = q[a + d(q – p)]

A2 p q pd – qa α2 = – p[a + d(q – p)]

Average effect of allele substitution: α = α1 - α2 = a + d(q – p)

Breeding value (Ai) 2α1 = 2qα α1 + α2 = (q – p)α 2α2 = –2pα E[A] = ∑Aifi = 2p2qα + 2pq(q – p)α – 2pq2α = 2pqα(p + q – p – q) = 0

Genotypic value (Gi = μgi - μG) 2q(α – qd) (q – p)α + 2pqd –2p(α + pd) E[G] = ∑Gifi = 2p2q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq2(α + pd) = 0

Dominance deviation (Di = Gi –

Ai)

–2q2d 2pqd –2p2d E[D] = ∑Difi = –2p2q2d + 4p2q2d – 2p2q2d = 0

Summary table: P = G + E = A + D + I+

E

Variance: summary(Falconer & Mackay: chapter 8)

Value components Variance components

P Phenotypic value VP Phenotypic variance Value decomposition: P = G + E G Genotypic value VG Genotypic variance = A + D + I + E

A Breeding value VA Additive genetic variance

D Dominance deviation VD Dominance genetic variance

Variance decomposition: VP = VG + VE

I Interaction deviation VI Interaction variance = VA + VD + VI + VE*

E Environmental deviation VE Environmental variance

*the more general expression is VP = VG + VE + 2covGE + VGE, where covGE is the covariance between genotypic values and the environmental deviations, and VGE the variance due to interaction between genotypes and the environment.

VA and VD: 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(xi- μx)2/N, μx = 0, thus V = Σxi

2/N). Given that we work with frequencies, V = Σxi2fi

Genotype

A1A1 A1A2 A2A2

Genotypic value (gi) a d -a

Genotype frequency(fi) p2 2pq q2

Mean genotypic value (μgi =

gifi)p2a 2pqd -q2a

Mean gen. value across genotypes: μG = ∑μgi = a(p – q) +

2dpq

Breeding value (Ai) 2α1 = 2qα α1 + α2 = (q – p)α 2α2 = –2pα E[A] = ∑Aifi = 2p2qα + 2pq(q – p)α – 2pq2α = 2pqα(p + q – p – q) = 0

Genotypic value (Gi = μgi - μG) 2q(α – qd) (q – p)α + 2pqd –2p(α + pd) E[G] = ∑Gifi = 2p2q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq2(α + pd) = 0

Dominance deviation (Di = Gi

– Ai)

–2q2d 2pqd –2p2d E[D] = ∑Difi = –2p2q2d + 4p2q2d – 2p2q2d = 0

VA and VD: 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(xi- μx)2/N, μx = 0, thus V = Σxi

2/N). Given that we work with frequencies, V = Σxi2fi

Genotype

A1A1 A1A2 A2A2

Genotypic value (gi) a d -a

Genotype frequency(fi) p2 2pq q2

Mean genotypic value (μgi =

gifi)p2a 2pqd -q2a

Mean gen. value across genotypes: μG = ∑μgi = a(p – q) +

2dpq

Breeding value (Ai) 2α1 = 2qα α1 + α2 = (q – p)α 2α2 = –2pα E[A] = ∑Aifi = 2p2qα + 2pq(q – p)α – 2pq2α = 2pqα(p + q – p – q) = 0

Genotypic value (Gi = μgi - μG) 2q(α – qd) (q – p)α + 2pqd –2p(α + pd) E[G] = ∑Gifi = 2p2q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq2(α + pd) = 0

Dominance deviation (Di = Gi

– Ai)

–2q2d 2pqd –2p2d E[D] = ∑Difi = –2p2q2d + 4p2q2d – 2p2q2d = 0

VA = 4p2q2α2 + 2pqα2(q – p)2 + 4p2q2α2

= 2pqα2(2pq + q2 – 2pq + p2 + 2pq) = 2pqα2(q2 + 2pq + p2) = 2pqα2(p + q)2

= 2pqα2

= 2pq[a + d(q – p)]2

VA and VD: 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(xi- μx)2/N, μx = 0, thus V = Σxi

2/N). Given that we work with frequencies, V = Σxi2fi

Genotype

A1A1 A1A2 A2A2

Genotypic value (gi) a d -a

Genotype frequency(fi) p2 2pq q2

Mean genotypic value (μgi =

gifi)p2a 2pqd -q2a

Mean gen. value across genotypes: μG = ∑μgi = a(p – q) +

2dpq

Breeding value (Ai) 2α1 = 2qα α1 + α2 = (q – p)α 2α2 = –2pα E[A] = ∑Aifi = 2p2qα + 2pq(q – p)α – 2pq2α = 2pqα(p + q – p – q) = 0

Genotypic value (Gi = μgi - μG) 2q(α – qd) (q – p)α + 2pqd –2p(α + pd) E[G] = ∑Gifi = 2p2q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq2(α + pd) = 0

Dominance deviation (Di = Gi

– Ai)

–2q2d 2pqd –2p2d E[D] = ∑Difi = –2p2q2d + 4p2q2d – 2p2q2d = 0

VA = 4p2q2α2 + 2pqα2(q – p)2 + 4p2q2α2

= 2pqα2(2pq + q2 – 2pq + p2 + 2pq) = 2pqα2(q2 + 2pq + p2) = 2pqα2(p + q)2

= 2pqα2

= 2pq[a + d(q – p)]2

VD = 4q4d2p2 + 8p3q3d2 + 4p4d2q2

= 4q2p2d2(q2 + 2pq + p2) = 4q2p2d2(p + q)2

= 4q2p2d2

= (2pqd)2

VA and VD: 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(xi- μx)2/N, μx = 0, thus V = Σxi

2/N). Given that we work with frequencies, V = Σxi2fi

Genotype

A1A1 A1A2 A2A2

Genotypic value (gi) a d -a

Genotype frequency(fi) p2 2pq q2

Mean genotypic value (μgi =

gifi)p2a 2pqd -q2a

Mean gen. value across genotypes: μG = ∑μgi = a(p – q) +

2dpq

Breeding value (Ai) 2α1 = 2qα α1 + α2 = (q – p)α 2α2 = –2pα E[A] = ∑Aifi = 2p2qα + 2pq(q – p)α – 2pq2α = 2pqα(p + q – p – q) = 0

Genotypic value (Gi = μgi - μG) 2q(α – qd) (q – p)α + 2pqd –2p(α + pd) E[G] = ∑Gifi = 2p2q(α – qd) + 2pq[(q – p)α + 2pqd] – 2pq2(α + pd) = 0

Dominance deviation (Di = Gi

– Ai)

–2q2d 2pqd –2p2d E[D] = ∑Difi = –2p2q2d + 4p2q2d – 2p2q2d = 0

VA = 4p2q2α2 + 2pqα2(q – p)2 + 4p2q2α2

= 2pqα2(2pq + q2 – 2pq + p2 + 2pq) = 2pqα2(q2 + 2pq + p2) = 2pqα2(p + q)2

= 2pqα2

= 2pq[a + d(q – p)]2

VD = 4q4d2p2 + 8p3q3d2 + 4p4d2q2

= 4q2p2d2(q2 + 2pq + p2) = 4q2p2d2(p + q)2

= 4q2p2d2

= (2pqd)2

G = A + D, VG = VA + VD + 2covAD

covAD = -4p2q3αd + 4p2q2(q – p)αd + 4p3q2αd = 4p2q2αd(- q + q – p + p) = 0

VG = VA + VD = 2pq[a + d(q – p)]2 + (2pqd)2

Resemblance between relatives

(Falconer & Mackay: chapter 9)

- 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 (VA)

- last chapter: causal components of phenotypic variance (V: VE, VG, VA, VD, VI)- observational components of phenptypic variance: s2

- 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 (VA)

- last chapter: causal components of phenotypic variance (V: VE, VG, VA, VD, VI)- observational components of phenptypic variance: s2

- 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 coefficientt = s2

B / (s2

B + s2W)

- between group variance (s2B) = 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)

- 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 (VA)

- last chapter: causal components of phenotypic variance (V: VE, VG, VA, VD, VI)- observational components of phenptypic variance: s2

- 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 = s2

B / (s2

B + s2W)

- between group variance (s2B) = 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: covOP = bOP s2

P → bOP = covOP / s2P

OPbOP

s2P

- 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

- 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)

- 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

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

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: *

covOP = ( S ½Ai (Ai + Di) ) / N = (½S Ai (Ai + Di) ) / N = (½SAi

2 + ½SAiDi) ) / N = ½SAi

2/N + ½SAiDi/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

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: *

covOP = ( S ½Ai (Ai + Di) ) / N = (½S Ai (Ai + Di) ) / N = (½SAi

2 + ½SAiDi) ) / N = ½SAi

2/N + ½SAiDi/N

where i (i = 1, …, N) denotes parent-offspring pair.

½SAi2/N = ½VA

½SAiDi/N = ½covAD covAD = 0 (from Ch 8)

Therefore: covOP = ½VA → 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

Offspring and one parent

Another way of deriving the covariance:

A2A2

0 d + a- aGenotypic value

Genotype A1A2 A1A1

Genotype frequency

2pq

p2q2

Parents

Genotype A1A1 A1A2 A2A2

Freq p2 2pq q2

Genotypic value 2q(a - qd) (q-p)a + 2qpd

-2p(a + pd)

Offspring Mean genot. value

qa ½ (q - p)a -pa

Genotypic values (a, d, -a) expressed as deviations from the population mean

Offspring and one parent

Another way of deriving the covariance:

Parents

Genotype A1A1 A1A2 A2A2

Freq p2 2pq q2

Genotypic value 2q(a - qd) (q-p)a + 2qpd

-2p(a + pd)

Offspring Mean genot. value

qa ½ (q - p)a -pa

A2A2

0 d + a- aGenotypic value

Genotype A1A2 A1A1

Genotypic values (a, d, -a) expressed as deviations from the population mean

m

a – m= a – [(p – q) + 2dpq]= 2q(a - qd)

Offspring and one parent

Another way of deriving the covariance:

Parents

Genotype A1A1 A1A2 A2A2

Freq p2 2pq q2

Genotypic value 2q(a - qd) (q-p)a + 2qpd

-2p(a + pd)

Offspring Mean genot. value

qa ½ (q - p)a -pa

mean cross-product of these

Offspring and one parent

Another way of deriving the covariance:

covOP = 2q2p2a(a – qd) + pqa(q – p) [(q – p)a + 2qpd] + 2p2q2a(a + pd) = pqa2

= ½VA,

since VA = 2pqa2.

covOP = bOP * VP

bOP = covOP / VP

= ½VA / VP

Parents

Genotype A1A1 A1A2 A2A2

Freq p2 2pq q2

Genotypic value 2q(a - qd) (q-p)a + 2qpd

-2p(a + pd)

Offspring Mean genot. value

qa ½ (q - p)a -pa

mean cross-product of these

OPbOP

VP

Offspring and mid-parent

- mid-parent = mean of the two parents

O = mean genotypic value of the offspringP & P’ = genotypic values of the two parentsmid-parent genotypic value: P = ½(P + P’)

covOP = SOP/N = S½(P + P’)O / N = (½SPO + ½SP’O) / N = ½SPO/N + ½SP’O/N = ½covOP + ½covOP’

If the P and P’ have the same variance, then covOP = covOP’, so:

covOP = covOP = ½VA → Therefore the covariance is the same as in the case of offspring and one parent

Offspring and mid-parent

- mid-parent = mean of the two parents

O = mean genotypic value of the offspringP & P’ = genotypic values of the two parentsmid-parent genotypic value: P = ½(P + P’)

covOP = SOP/N = S½(P + P’)O / N = (½SPO + ½SP’O) / N = ½SPO/N + ½SP’O/N = ½covOP + ½covOP’

If the P and P’ have the same variance, then covOP = covOP’, so:

covOP = covOP = ½VA → 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, VP = ½VP.

bOP = covOP / ½VP

= ½VA / ½VP = VA/VP

OPbOP

½VP

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?):

covHS = V½A = ¼VA

If you wanted me to derive it:

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?):

covHS = V½A = ¼VA

- 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 = ¼VA/VP

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’)

var½(A + A’) = ¼(VA + VA’) = ½VA if the additive genetic variance is equal in the two sexes.

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’)

var½(A + A’) = ¼(VA + VA’) = ½VA if the additive genetic variance is equal in the two sexes.

- dominance variance:- let parents have genotypes A1A2 and A3A4

- then there are 4 genotypes in the progeny: A1A3, A1A4, A2A3, A2A4, 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 D2

- for other pairs, this cross-product is 0- therefore, on average, the (mean) cross-product is ¼SD2 / N, which equals ¼VD

- total genetic covariance of full sibs is therefore:covFS = ½VA + ¼VD

Full sibs

- the correlation of full sibs is:

t = (½VA + ¼VD) / VP

Twins

- dizygotic (DZ) twins are related as full sibs; their genetic covariance is that of full sibs:

covDZ = ½VA + ¼VD

- monozygotic (MZ) twins have identical genotypes, therefore:

covMZ = VG

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 VEc: common environment- the remainder of environmental variance (VEw) arises from causes of difference that are unconnected to whether the individuals are related or not- therefore, VEw 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:

VE = VEc + VEw,

where the VEc component contributes to the covariance of related invididuals.

Summary

Relationship Genetic covariance

Offspring and one parent covOP = ½VA

Offspring and mid-parent covOP= ½VA if variance is equal in two sexes

Half sibs covHS = ¼VA

Full sibs covFS = ½VA + ¼VD

DZ twins covDZ = ½VA + ¼VD

MZ twins covMZ = VG

Summary

Relationship Covariance

Offspring and one parent covOP = ½VA

Offspring and mid-parent covOP= ½VA if variance is equal in two sexes

Half sibs covHS = ¼VA

Full sibs covFS = ½VA + ¼VD + VEc

DZ twins covDZ = ½VA + ¼VD + VEc

MZ twins covMZ = VG + VEc

Summary

Relationship Genetic covariance Standardized genetic covariance

Offspring and one parent covOP = ½VA ½VA / VP

Offspring and mid-parent covOP= ½VA VA / VP

Half sibs covHS = ¼VA ¼VA / VP

Full sibs covFS = ½VA + ¼VD (½VA + ¼VD) / VP

DZ twins covDZ = ½VA + ¼VD (½VA + ¼VD) / VP

MZ twins covMZ = VG VG / VP

Applications in structural equation modeling

Applications in structural equation modeling

Homework

9.3

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