1. Lecture 3: Resemblance Between Relatives

2. Heritability Estimates of VA require known collections of relatives
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

3. Ancestral relatives e.g., parent and offspring
Collateral relatives, e.g. sibs

4. Half-sibs
Full-sibs

5. 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

6. 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.
Both alleles IBD
One allele IBD
No alleles IBD

7. Regressions and ANOVA Parent-offspring regression Sibs
Single parent vs. midparent
Parent-offspring covariance is a intraclass (between class) covariance
Sibs
Covariances between sibs is an interclass (within class) covariance

8. 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)

9. Situation 1 Situation 2 Var(B) = 0 Var(B) = 2.5 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

10. Parent-offspring genetic covariance
Cov(Gp, Go) --- Parents and offspring share
EXACTLY one allele IBD
Denote this common allele by A1
IBD allele
Non-IBD alleles

11. ) All red covariance terms are zero.
• 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.

12. { 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

13. Half-sibs The half-sibs share no alleles IBD
• occurs with probability 1/2
The half-sibs share one allele IBD
• occurs with probability 1/2
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

14. Full-sibs Paternal allele IBD [ Prob = 1/2 ]
Maternal allele IBD [ Prob = 1/2 ]
-> Prob(both alleles IBD) = 1/2*1/2 = 1/4
Paternal allele not IBD [ Prob = 1/2 ]
Maternal allele not IBD [ Prob = 1/2 ]
-> Prob(zero alleles IBD) = 1/2*1/2 = 1/4
Each sib gets
exact one allele
from each parent
Prob(exactly one allele IBD) = 1/2
= 1- Prob(0 IBD) - Prob(2 IBD)

15. 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)
Cov(Full-sibs) = Var(A)/2 + Var(D)/4

16. 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) +

17. Components of the Environmental Variance
Total environmental value
Common environmental value experienced
by all members of a family, e.g., shared
maternal effects
E = Ec + Es
Specific environmental value,
any unique environmental effects
experienced by the individual
VE = VEc + VEs
The Environmental variance can thus be written in terms of variance components as
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.

18. 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

19. 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

20. 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
Draw A1
IBD
Draw A2 fx
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

21. Qop = Parent & Offspringfp
Offspring inbred
Parent inbred
Mother
Offspring fo
Parental allele
1/2 = Prob random offspring allele from father.
Prob = qmf = fo that this allele is IBD to mother giving
a contribution of fo/2
qmf

22. Full sibs (x and y) from parents m and f
Q = 1/8 + 1/8 = 1/4
Q =(2+fm+ff)/8
Q =(2+fm+ff +4 Q mf)/8
1/2
(1+ff)/2
1/2
(1+fm)/2
Q mf
Q mf (1/2)(1/2)
Q mf
Q mf /4 m f m f m f
(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)

23. 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

24. Computing qxy -- chain counting
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
Coefficient of coancestry of i
Paths through a single common
ancestor (i) to both x and y
Paths through two remove ancestors (j and k)

25. Compute q for Royal Duke of Gloster and Princess Royal
Lord Raglan
Champion of
England
The Czar
Mistletoe
Mimulus
Grand Duke
of Gloster
Roan
Gauntlet
Royal Duke
Carmine
Princess
Royal
Duchess of
Gloster, 9th
Path through
Lord Raglan, n = 7
Contribution = (1/2)(1/2)6
Royal Duke
of Gloster
Princess
Royal
Lord Raglan
Qii = 1/2
q = (1/2) 7
+ (1/2) 4
+ (1/2) 4
Lord Raglan
Princess
Royal
Royal Duke
of Gloster
Path through Lord
Raglan, n = 7
Contribution = (1/2)(1/2)6
Qii = 1/2
+ (1/2) 7 = 0.141
Champion of
England
Royal Duke
of Gloster
Princess
Royal
Path through Champion
of England, n = 4
Contribution = (1/2)(1/2)3
Qii = 1/2
Path through Champion
of England, n = 4
Contribution = (1/2)(1/2)3
Champion of
England
Royal Duke
of Gloster
Princess
Royal
Qii = 1/2
Compute q for Royal Duke of Gloster and Princess Royal
Princess
Royal
Royal Duke
of Gloster
f for Roan Gauntlet
= 0.141

26. Dxy, The Coefficient of Fraternity
Dxy = Prob(both alleles in x & y IBD)
qmxfy
qfxmy
qmxmy
qfxfy x y fx fy mx my
Dxy = qmxmyqfxfy
+ qmxfyqfxmy

27. Examples of Dxy Dxy = qmxmyqfxfy + qmxfyqfxmy Dxy = qmmqff + qmf2
(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

28. Dxy = qmxmy(1/4) + (1/4) qfxmy Dxy = qmxmy(1/4) + qmxfyqfxmy
Lord Raglan
Champion of
England
The Czar
Mistletoe
Mimulus
Grand Duke
of Gloster
Roan
Gauntlet
Royal Duke
Carmine
Princess
Royal
Duchess of
Gloster, 9th my fy
Champion of
England
Carmine
q = 1/4
q = 1/4 mx fx
Mimulus
Grand Duke
of Gloster
q = (1/2)5
q = (1/2)5
Royal Duke
of Gloster
What is D for Royal Duke of Gloster and Princess Royal
Princess
Royal y x
Dxy = qmxmy(1/4) + (1/4) qfxmy
Dxy = qmxmy(1/4) + qmxfyqfxmy
Dxy = (1/2)5(1/4) + (1/4) qfxmy
Dxy = qmxmyqfxfy + qmxfyqfxmy
Dxy = (1/2)5(1/4) + (1/4) (1/2)5 = (1/2)6

29. General Resemblance between relativesD D D