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Lecture 3: Resemblance Between Relatives. Heritability. Central concept in quantitative genetics Proportion of variation due to additive genetic values (Breeding values) h 2 = V A /V P Phenotypes (and hence V P ) can be directly measured Breeding values (and hence V A ) must be estimated

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heritability
Heritability
  • Central concept in quantitative genetics
  • Proportion of variation due to additive genetic values (Breeding values)
    • h2 = VA/VP
    • Phenotypes (and hence VP) can be directly measured
    • Breeding values (and hence VA ) must be estimated
  • Estimates of VA require known collections of relatives
slide3

Ancestral relatives e.g., parent and offspring

Collateral relatives, e.g. sibs

slide4

Half-sibs

Full-sibs

key observations
Key observations
  • The amount of phenotypic resemblance among relatives for the trait provides an indication of the amount of genetic variation for the trait.
  • If trait variation has a significant genetic basis, the closer the relatives, the more similar their appearance
genetic covariance between relatives

One allele IBD

Both alleles IBD

No alleles IBD

Genetic Covariance between relatives

Sharing alleles means having alleles that are identical by descent (IBD): both copies of

can be traced back to a single copy in a

recent common ancestor.

Genetic covariances arise because two related

individuals are more likely to share alleles than

are two unrelated individuals.

regressions and anova
Regressions and ANOVA
  • Parent-offspring regression
    • Single parent vs. midparent
    • Parent-offspring covariance is a intraclass (between class) covariance
  • Sibs
    • Covariances between sibs is an interclass (within class) covariance
anova
ANOVA
  • Two key ANOVA identities
    • Total variance = between-group variance + within-group variance
      • Var(T) = Var(B) + Var(W)
    • Variance(between groups) = covariance (within groups)
    • Intraclass correlation, t = Var(B)/Var(T)
slide9

Situation 1

Situation 2

Var(B) = 0

Var(W) = 2.7

Var(T) = 2.7

Var(B) = 2.5

Var(W) = 0.2

Var(T) = 2.7

t = 0

t = 2.5/2.7 = 0.93

parent offspring genetic covariance

IBD allele

Non-IBD alleles

Parent-offspring genetic covariance

Cov(Gp, Go) --- Parents and offspring share

EXACTLY one allele IBD

Denote this common allele by A1

slide11

• By construction, a and D are uncorrelated

• By construction, a from non-IBD alleles are

uncorrelated

• By construction, D values are uncorrelated unless

both alleles are IBD

)

All red covariance terms are zero.

slide12

{

Hence, relatives sharing one allele IBD have a

genetic covariance of Var(A)/2

The resulting parent-offspring genetic covariance

becomes Cov(Gp,Go) = Var(A)/2

half sibs

The half-sibs share one allele IBD

• occurs with probability 1/2

The half-sibs share no alleles IBD

• occurs with probability 1/2

Half-sibs

Each sib gets exactly one allele from common father,

different alleles from the different mothers

Hence, the genetic covariance of half-sibs is just

(1/2)Var(A)/2 = Var(A)/4

full sibs

Paternal allele not IBD [ Prob = 1/2 ]

Maternal allele not IBD [ Prob = 1/2 ]

-> Prob(zero alleles IBD) = 1/2*1/2 = 1/4

Paternal allele IBD [ Prob = 1/2 ]

Maternal allele IBD [ Prob = 1/2 ]

-> Prob(both alleles IBD) = 1/2*1/2 = 1/4

Full-sibs

Each sib gets

exact one allele

from each parent

Prob(exactly one allele IBD) = 1/2

= 1- Prob(0 IBD) - Prob(2 IBD)

slide15

Cov(Full-sibs) = Var(A)/2 + Var(D)/4

Resulting Genetic Covariance between full-sibs

If both alleles IBD, cov(Gsib1, Gsib2 )=

cov(ax+ay+dxy, ax+ay+dxy )

= cov(ax+ay, ax+ay) + cov(dxy, dxy) = Var(A) + Var(D)

genetic covariances for general relatives
Genetic Covariances for General Relatives

Let r = (1/2)Prob(1 allele IBD) + Prob(2 alleles IBD)

Let u = Prob(both alleles IBD)

General genetic covariance between relatives

Cov(G) = rVar(A) + uVar(D)

When epistasis is present, additional terms appear

r2Var(AA) + ruVar(AD) + u2Var(DD) + r3Var(AAA) +

components of the environmental variance

Total environmental value

Common environmental value experienced

by all members of a family, e.g., shared

maternal effects

Specific environmental value,

any unique environmental effects

experienced by the individual

The Environmental variance can thus be writtenin terms of variance components as

VE = VEc + VEs

Components of the Environmental Variance

E = Ec + Es

One can decompose the environment further, if

desired. For example, plant breeders have terms

for the location variance, the year variance, and the

location x year variance.

slide18

Shared Environmental Effects contribute

to the phenotypic covariances of relatives

Cov(P1,P2) = Cov(G1+E1,G2+E2)

= Cov(G1,G2) + Cov(E1,E2)

Shared environmental values are expected

when sibs share the same mom, so that

Cov(Full sibs) and Cov(Maternal half-sibs)

not only contain a genetic covariance, but

an environmental covariance as well, VEc

coefficients of coancestry
Coefficients of Coancestry

Suppose we pick a single allele each at random from

two relatives. The probability that these are IBD is

called Q, the coefficient of coancestry

Qxy denotes the coefficient for relatives x and y

Consider an offspring z from a (hypothetical) cross

of x and y. Qxy = fz, the inbreeding coefficient of z

q xx the coancestry of an individual with itself
Qxx : The Coancestry of an individual with itself

Self x, what is the inbreeding coefficient of its offspring?

To compute Qxx, denote the two alleles in x by A1 and A2

Draw A1

Draw A2

fx

IBD

Draw A1

fx

Draw A2

IBD

Hence, for a non-inbred individual, Qxx = 2/4 = 1/2

If x is inbred, fx = prob A1 and A2 IBD,

Qxx = (1+ fx)/2

q op parent offspring

fp

Mother

Offspring

fo

Parental allele

qmf

Qop = Parent & Offspring

Offspring inbred

Parent inbred

1/2 = Prob random offspring allele from father.

Prob = qmf = fo that this allele is IBD to mother giving

a contribution of fo/2

slide22

Full sibs (x and y) from parents m and f

(1+ff)/2

1/2

1/2

(1+fm)/2

Q mf

f

f

f

m

m

m

Q mf (1/2)(1/2)

Q mf /4

Q mf

Q = 1/8 + 1/8 = 1/4

Q =(2+fm+ff)/8

Q =(2+fm+ff +4 Q mf)/8

(1/2)(1/2)(1/2)

(1/2)(1/2)(1/2)

[(1 +ff )/2] (1/2)(1/2)

[(1 +fm )/2] (1/2)(1/2)

slide23

Full sibs (x and y) from parents m and f

Qxy = (2 + fm + ff + 4Qmf)/8

ff =Qsf,df

fm =Qsm,dm

Qxy = (2 + Qsm,dm + Qsf,df + 4Qmf)/8

computing q xy chain counting

)

)

(

(

Coefficient of coancestry of i

Number of individuals (including x and y) in path

Connecting x and y through i

Number of individuals, including x and y

on the path leading from two different (but related) ancestors j and k

Coefficient of coancestry of j and k

Computing qxy -- chain counting

Paths through a single common

ancestor (i) to both x and y

Paths through two remove ancestors (j and k)

slide25

Qii = 1/2

Lord Raglan

Lord Raglan

Lord Raglan

Qii = 1/2

Champion of

England

Champion of

England

Champion of

England

Qii = 1/2

Duchess of

Gloster, 9th

Mistletoe

The Czar

Qii = 1/2

Grand Duke

of Gloster

Mimulus

Carmine

Princess

Royal

Princess

Royal

Princess

Royal

Princess

Royal

Princess

Royal

Princess

Royal

Royal Duke

of Gloster

Royal Duke

of Gloster

Royal Duke

of Gloster

Royal Duke

of Gloster

Royal Duke

of Gloster

Royal Duke

of Gloster

Roan

Gauntlet

Path throughLord

Raglan, n = 7

Contribution = (1/2)(1/2)6

Path throughChampion

of England, n = 4

Contribution = (1/2)(1/2)3

Path through

Lord Raglan, n = 7

Contribution = (1/2)(1/2)6

Path throughChampion

of England, n = 4

Contribution = (1/2)(1/2)3

Compute q for Royal Duke of Gloster and Princess Royal

q = (1/2) 7

+ (1/2) 4

+ (1/2) 4

+ (1/2) 7 = 0.141

f for Roan Gauntlet

= 0.141

slide26

Dxy, The Coefficient of Fraternity

qfxmy

qfxfy

qmxmy

qmxfy

my

fy

fx

mx

x

y

Dxy = Prob(both alleles in x & y IBD)

Dxy = qmxmyqfxfy

+qmxfyqfxmy

examples of d xy
Examples of Dxy

Dxy = qmxmyqfxfy +qmxfyqfxmy

(1) x and y are full sibs:

mx = my = m, fx = fy = f

Dxy = qmmqff +qmf2

If parents unrelated, qmf = 0

Dxy = 1/4

If parents not inbred, qmm =qff = 1/2

(2) x and y are paternal half-sibs:

fx = fy = f

Dxy = qmxmyqff +qmxfqmyf

If parents unrelated, qmxf = qmyf = qmxmy = 0

Dxy = 0

slide28

Lord Raglan

Champion of

England

Champion of

England

Duchess of

Gloster, 9th

fy

Mistletoe

The Czar

q = 1/4

Grand Duke

of Gloster

Grand Duke

of Gloster

Mimulus

Mimulus

q = 1/4

Carmine

Carmine

mx

fx

q = (1/2)5

my

q = (1/2)5

Princess

Royal

Princess

Royal

Royal Duke

of Gloster

Royal Duke

of Gloster

Roan

Gauntlet

What is D for Royal Duke of Gloster and Princess Royal

y

x

Dxy = (1/2)5(1/4)+ (1/4)(1/2)5 = (1/2)6

Dxy = qmxmy(1/4)+qmxfyqfxmy

Dxy = qmxmyqfxfy +qmxfyqfxmy

Dxy = (1/2)5(1/4)+ (1/4)qfxmy

Dxy = qmxmy(1/4)+ (1/4)qfxmy