outline: 1) basics 2) means and values (ch 7): summary 3) variance (ch 8): summary 4) resemblance...
TRANSCRIPT
Outline:
1) Basics
2) Means and Values (Ch 7): summary
3) Variance (Ch 8): summary
4) Resemblance between relatives
5) Homework (8.3)
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
Value components Variance components
P Phenotypic value VP Phenotypic varianceValue 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 = Σxi2/N). Given that we work with frequencies, V = Σxi
2fi
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 = Σxi2/N). Given that we work with frequencies, V = Σxi
2fi
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 = Σxi2/N). Given that we work with frequencies, V = Σxi
2fi
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 = Σxi2/N). Given that we work with frequencies, V = Σxi
2fi
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:
- 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 coefficient
t = s2B
/ (s2B + s2
W)
- 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 = s2B
/ (s2B + s2
W)
- 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 s2P → bOP = covOP / s2
P
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
= (½SAi2 + ½SAiDi) ) / N
= ½SAi2/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
= (½SAi2 + ½SAiDi) ) / N
= ½SAi2/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 offspring
P & P’ = genotypic values of the two parents
mid-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 offspring
P & P’ = genotypic values of the two parents
mid-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
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
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