1. Introduction to Basic and Quantitative Genetics

2. Darwin & Mendel Darwin (1859) Origin of Species
Instant Classic, major immediate impact
Problem: Model of Inheritance
Darwin assumed Blending inheritance
Offspring = average of both parents
zo = (zm + zf)/2
Fleming Jenkin (1867) pointed out problem
Var(zo) = Var[(zm + zf)/2] = (1/2) Var(parents)
Hence, under blending inheritance, half the variation is removed each generation and this must somehow be replenished by mutation.

3. Mendel Mendel (1865), Experiments in Plant Hybridization
No impact, paper essentially ignored
Ironically, Darwin had an apparently unread copy in his library
Why ignored? Perhaps too mathematical for 19th century biologists
The rediscovery in 1900 (by three independent groups)
Mendel’s key idea: Genes are discrete particles passed on intact from parent to offspring

4. Mendel’s experiments with the Garden Pea
7 traits examined

5. Mendel crossed a pure-breeding yellow pea line
with a pure-breeding green line.
Let P1 denote the pure-breeding yellow (parental line 1)
P2 the pure-breed green (parental line 2)
The F1, or first filial, generation is the cross of
P1 x P2 (yellow x green).
All resulting F1 were yellow
The F2, or second filial, generation is a cross of two F1’s
In F2, 1/4 are green, 3/4 are yellow
This outbreak of variation blows the theory of blending
inheritance right out of the water.

6. Mendel also observed that the P1, F1 and F2 Yellow
lines behaved differently when crossed to pure green
P1 yellow x P2 (pure green) --> all yellow
F1 yellow x P2 (pure green) --> 1/2 yellow, 1/2 green
F2 yellow x P2 (pure green) --> 2/3 yellow, 1/3 green

7. Mendel’s explanation
Genes are discrete particles, with each parent passing
one copy to its offspring.
Let an allele be a particular copy of a gene. In Diploids,
each parent carries two alleles for every gene
Pure Yellow parents have two Y (or yellow) alleles
We can thus write their genotype as YY
Likewise, pure green parents have two g (or green) alleles
Their genotype is thus gg
Since there are lots of genes, we refer to a particular gene
by given names, say the pea-color gene (or locus)

8. Each parent contributes one of its two alleles (at
random) to its offspring
Hence, a YY parent always contributes a Y, while
a gg parent always contributes a g
An individual carrying only one type of an allele
(e.g. yy or gg) is said to be a homozygote
In the F1, YY x gg --> all individuals are Yg
An individual carrying two types of alleles is
said to be a heterozygote.

9. The phenotype of an individual is the trait value we
observe
For this particular gene, the map from genotype to
phenotype is as follows:
YY --> yellow
Yg --> yellow
gg --> green
Since the Yg heterozygote has the same phenotypic
value as the YY homozygote, we say (equivalently)
Y is dominant to g, or
g is recessive to Y

10. Explaining the crosses
F1 x F1 -> Yg x Yg
Prob(YY) = yellow(dad)*yellow(mom) = (1/2)*(1/2)
Prob(gg) = green(dad)*green(mom) = (1/2)*(1/2)
Prob(Yg) = 1-Pr(YY) - Pr(gg) = 1/2
Prob(Yg) = yellow(dad)*green(mom) + green(dad)*yellow(mom)
Hence, Prob(Yellow phenotype) = Pr(YY) + Pr(Yg) = 3/4
Prob(green phenotype) = Pr(gg) = 1/4

11. Dealing with two (or more) genes
For his 7 traits, Mendel observed Independent Assortment
The genotype at one locus is independent of the second
RR, Rr - round seeds, rr - wrinkled seeds
Pure round, green (RRgg) x pure wrinkled yellow (rrYY)
F1 --> RrYg = round, yellow
What about the F2?

12. Phenotype Genotype Frequency Y-R- Y-rr ggR- ggrr
Let R- denote RR and Rr. R- are round. Note in F2,
Pr(R-) = 1/2 + 1/4 = 3/4
Likewise, Y- are YY or Yg, and are yellow
Phenotype
Genotype
Frequency
Yellow, round
Y-R-
(3/4)*(3/4) = 9/16
Yellow, wrinkled
Y-rr
(3/4)*(1/4) = 3/16
Green, round
ggR-
(1/4)*(3/4) = 3/16
Green, wrinkled
ggrr
(1/4)*(1/4) = 1/16
Or a 9:3:3:1 ratio

13. Probabilities for more complex genotypes
Cross AaBBCcDD X aaBbCcDd
What is Pr(aaBBCCDD)?
Under independent assortment,
= Pr(aa)*Pr(BB)*Pr(CC)*Pr(DD)
= (1/2*1)*(1*1/2)*(1/2*1/2)*(1*1/2) = 1/25
What is Pr(AaBbCc)?
= Pr(Aa)*Pr(Bb)*Pr(Cc) = (1/2)*(1/2)*(1/2) = 1/8

14. Mendel was wrong: Linkage
Bateson and Punnet looked at
flower color: P (purple) dominant over p (red )
pollen shape: L (long) dominant over l (round)
Phenotype
Genotype
Observed
Expected
Purple long
P-L-
284
215
Purple round
P-ll 21 71
Red long
ppL-
Red round
ppll 55 24
Excess of PL, pl gametes over Pl, pL
Departure from independent assortment

15. Linkage If genes are located on different chromosomes they
(with very few exceptions) show independent assortment.
Indeed, peas have only 7 chromosomes, so was Mendel lucky
in choosing seven traits at random that happen to all
be on different chromosomes? Problem: compute this probability.
However, genes on the same chromosome, especially if
they are close to each other, tend to be passed onto
their offspring in the same configuation as on the
parental chromosomes.

16. Consider the Bateson-Punnet pea data
Let PL / pl denote that in the parent, one chromosome
carries the P and L alleles (at the flower color and
pollen shape loci, respectively), while the other
chromosome carries the p and l alleles.
Unless there is a recombination event, one of the two
parental chromosome types (PL or pl) are passed onto
the offspring. These are called the parental gametes.
However, if a recombination event occurs, a PL/pl
parent can generate Pl and pL recombinant chromosomes
to pass onto its offspring.

17. Let c denote the recombination frequency --- the
probability that a randomly-chosen gamete from the
parent is of the recombinant type (i.e., it is not a
parental gamete).
For a PL/pl parent, the gamete frequencies are
Gamete type
Frequency
Expectation under independent assortment PL
(1-c)/2
1/4 pl pL
c/2 Pl
Parental gametes in excess, as (1-c)/2 > 1/4 for c < 1/2
Recombinant gametes in deficiency, as c/2 < 1/4 for c < 1/2

18. Expected genotype frequencies under linkage
Suppose we cross PL/pl X PL/pl parents
What are the expected frequencies in their offspring?
Pr(PPLL) = Pr(PL|father)*Pr(PL|mother)
= [(1-c)/2]*[(1-c)/2] = (1-c)2/4
Likewise, Pr(ppll) = (1-c)2/4
Recall from previous data that freq(ppll) = 55/381 =0.144
Hence, (1-c)2/4 = 0.144, or c = 0.24

19. A (slightly) more complicated case
Again, assume the parents are both PL/pl.
Compute Pr(PpLl)
Two situations, as PpLl could be PL/pl or Pl/pL
Pr(PL/pl) = Pr(PL|dad)*Pr(pl|mom) + Pr(PL|mom)*Pr(pl|dad)
= [(1-c)/2]*[(1-c)/2] + [(1-c)/2]*[(1-c)/2]
Pr(Pl/pL) = Pr(Pl|dad)*Pr(pL|mom) + Pr(Pl|mom)*Pr(pl|dad)
= (c/2)*(c/2) + (c/2)*(c/2)
Thus, Pr(PpLl) = (1-c)2/2 + c2 /2

20. Generally, to compute the expected genotype
probabilities, need to consider the frequencies
of gametes produced by both parents.
Suppose dad = Pl/pL, mom = PL/pl
Pr(PPLL) = Pr(PL|dad)*Pr(PL|mom)
= [c/2]*[(1-c)/2]
Notation: when PL/pl, we say that alleles P and L
are in coupling
When parent is Pl/pL, we say that P and L are in repulsion

21. Molecular Markers You and your neighbor differ at roughly 22,000,000
nucleotides (base pairs) out of the roughly 3 billion
bp that comprises the human genome
Hence, LOTS of molecular variation to exploit
SNP -- single nucleotide polymorphism. A particular
position on the DNA (say base 123,321 on chromosome 1)
that has two different nucleotides (say G or A) segregating
STR -- simple tandem arrays. An STR locus consists of
a number of short repeats, with alleles defined by
the number of repeats. For example, you might have
6 and 4 copies of the repeat on your two chromosome 7s

22. SNPs vs STRs SNPs Cons: Less polymorphic (at most 2 alleles)
Pros: Low mutation rates, alleles very stable
Excellent for looking at historical long-term
associations (association mapping)
STRs
Cons: High mutation rate
Pros: Very highly polymorphic
Excellent for linkage studies within an extended
Pedigree (QTL mapping in families or pedigrees)

23. Quantitative Genetics
The analysis of traits whose variation is determined by both a number of genes and environmental factors
Phenotype is highly uninformative as to
underlying genotype

24. Complex (or Quantitative) trait
No (apparent) simple Mendelian basis for variation in the trait
May be a single gene strongly influenced by environmental factors
May be the result of a number of genes of equal (or differing) effect
Most likely, a combination of both multiple genes and environmental factors
Example: Blood pressure, cholesterol levels
Known genetic and environmental risk factors
Molecular traits can also be quantitative traits
mRNA level on a microarray analysis
Protein spot volume on a 2-D gel

25. Consider a specific locus influencing the trait
For this locus, mean phenotype = 0.15, while
overall mean phenotype = 0
Phenotypic distribution of a trait

26. Basic model of Quantitative Genetics
Basic model: P = G + E
Genotypic value
Environmental value
Phenotypic value -- we will occasionally
also use z for this value
G = average phenotypic value for that genotype
if we are able to replicate it over the universe
of environmental values, G = E[P]
G x E interaction --- G values are different
across environments. Basic model now
becomes P = G + E + GE

27. Contribution of a locus to a trait
Q1Q1
Q2Q1
Q2Q2 C
C + a(1+k)
C + 2a
C + a + d
C -a
C + d
C + a
d measures dominance, with d = 0 if the heterozygote
is exactly intermediate to the two homozygotes
d = ak =G(Q1Q2 ) - [G(Q2Q2) + G(Q1Q1) ]/2
2a = G(Q2Q2) - G(Q1Q1)
k = d/a is a scaled measure of the dominance

28. Example: Apolipoprotein E & Alzheimer’s
Genotype ee Ee EE
Average age of onset
68.4
75.5
84.3
2a = G(EE) - G(ee) = > a = 7.95
ak =d = G(Ee) - [ G(EE)+G(ee)]/2 = -0.85
k = d/a = 0.10
Only small amount of dominance

29. Example: Booroola (B) gene
Genotype bb Bb BB
Average Litter size
1.48
2.17
2.66
2a = G(BB) - G(bb) = > a = 0.59
ak =d = G(Bb) - [ G(BB)+G(bb)]/2 = 0.10
k = d/a = 0.17

30. Fisher’s (1918) Decomposition of G
One of Fisher’s key insights was that the genotypic value
consists of a fraction that can be passed from parent to
offspring and a fraction that cannot.
Consider the genotypic value Gij resulting from an
AiAj individual G i j = b
Dominance deviations --- the difference (for genotype
AiAj) between the genotypic value predicted from the
two single alleles and the actual genotypic value, G i j = π + Æ π G = X i j f r e q ( Q )
Mean value, with
Average contribution to genotypic value for allele i
Since parents pass along single alleles to their
offspring, the ai (the average effect of allele i)
represent these contributions b G i j = π + Æ
The genotypic value predicted from the individual
allelic effects is thus

31. G = π + Æ ± Fisher’s decomposition is a Regression G = π + 2 Æ ( ° ) N
Residual error G i j = π + Æ
Predicted value
A notational change clearly shows this is a regression,
Independent (predictor) variable N = # of Q2 alleles
Regression slope
Intercept G i j = π + 2 Æ 1 ( ) N
Regression residual 2 Æ 1 + ( ) N = 8
>
< : f o r ; e . g , Q

32. Allele Q1 common, a2 > a1
Slope = a2 - a1
Allele Q2 common, a1 > a2
Both Q1 and Q2 frequent, a1 = a2 = 0 1 2 N G
G22
G11
G21

33. Æ = p a [ + k ( ° ) ] ± = G ° π Æ π = 2 p a ( 1 + k ) Genotype Q1Q1
Consider a diallelic locus, where p1 = freq(Q1)
Genotype
Q1Q1
Q2Q1
Q2Q2
Genotypic
value
a(1+k) 2a π G = 2 p a ( 1 + k )
Mean
Allelic effects Æ 2 = p 1 a [ + k ( ) ]
Dominance deviations i j = G π Æ

34. A ( G ) = Æ + i j Average effects and Additive Genetic Values A = X ≥
The a values are the average effects of an allele
A key concept is the Additive Genetic Value (A) of
an individual A = n X k 1 ≥ Æ ( ) i + A ( G i j ) = Æ +
Why all the fuss over A?
Suppose father has A = 10 and mother has A = -2
for (say) blood pressure
KEY: parents only pass single alleles to their offspring.
Hence, they only pass along the A part of their genotypic
Value G
Expected blood pressure in their offspring is (10-2)/2
= 4 units above the population mean. Offspring A =
Average of parental A’s

35. æ = + Genetic Variances G = π + ( Æ ) ± 2 G A D æ ( G ) = X Æ + ± æ (j = π g + ( Æ ) æ 2 ( G ) = π g + Æ i j
As Cov(a,d) = 0 æ 2 ( G ) = n X k 1 Æ i + j
Dominance Genetic Variance
(or simply dominance variance)
Additive Genetic Variance
(or simply Additive Variance) æ 2 G = A + D

36. Key concepts (so far) ai = average effect of allele i
Property of a single allele in a particular population (depends on genetic background)
A = Additive Genetic Value (A)
A = sum (over all loci) of average effects
Fraction of G that parents pass along to their offspring
Property of an Individual in a particular population
Var(A) = additive genetic variance
Variance in additive genetic values
Property of a population
Can estimate A or Var(A) without knowing any of the underlying genetical detail (forthcoming)

37. æ = ( p a k ) æ = p a [ + k ( ° ) ] æ = E [ Æ ] X p Q1Q1 Q1Q2 Q2Q2
a(1+k) a æ 2 A = E [ Æ ] m X i 1 p
Since E[a] = 0,
Var(a) = E[(a -ma)2] = E[a2]
One locus, 2 alleles:
Dominance effects
additive variance æ 2 A = p 1 a [ + k ( ) ]
When dominance present,
asymmetric function of allele
frequencies æ 2 D = E [ ] m X i 1 j p
Equals zero if k = 0
One locus, 2 alleles:
This is a symmetric function of
allele frequencies æ 2 D = ( p 1 a k )

38. Additive variance, VA, with no dominance (k = 0)
Allele frequency, p VA

39. Complete dominance (k = 1)VA VD
Allele frequency, p

40. Epistasis æ = + 2 G A D G = π + ( Æ ) ± A D DD
Dominance value -- interaction
between the two alleles at a locus
Additive Genetic value G i j k l = π + ( Æ ) A D DD
Additive x Additive interactions --
interactions between a single allele
at one locus with a single allele at another
Additive x Dominant interactions --
interactions between an allele at one
locus with the genotype at another, e.g.
allele Ai and genotype Bkj
Dominance x dominance interaction ---
the interaction between the dominance
deviation at one locus with the dominance
deviation at another.
These components are defined to be uncorrelated,
(or orthogonal), so that æ 2 G = A + D

41. Resemblance Between Relatives

42. 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

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

44. Half-sibs
Full-sibs

45. 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

46. 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

47. Parent-offspring genetic covariance
Cov(Gp, Go) --- Parents and offspring share
EXACTLY one allele IBD
Denote this common allele by A1
IBD allele G p = A + D Æ 1 x o y
Non-IBD alleles

48. C o v ( G ; ) = Æ + D All white covariance terms are zero.p ) = Æ 1 + x D y
• 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 white covariance terms are zero.

49. C o v ( Æ ; ) = Ω i f 6 . e , n t I B D V a r A 2 V a r ( A ) = Æ + sx ; y ) = Ω i f 6 . e , n t I B D V a r A 2 V a r ( A ) = Æ 1 + 2 s o t h C v ;
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

50. 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

51. 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)

52. Resulting Genetic Covariance between full-sibsBD al l el es P
rob a
bil i ty Co n tr
but on 1/ 4 1 2 V r ( A ) /
) + Va r( D
Cov(Full-sibs) = Var(A)/2 + Var(D)/4

53. 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) +

54. 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 environmental further, if
desired. For example, plant breeders have terms
for the location variance, the year variance, and the
location x year variance.

55. 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
Cov(Full-sibs) = Var(A)/2 + Var(D)/4 + VEc