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Quantitative genetics

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Title: Quantitative genetics


1
Quantitative genetics
  • Many traits that are important in agriculture,
    biology and biomedicine are continuous in their
    phenotypes. For example,
  • Crop Yield
  • Stemwood Volume
  • Plant Disease Resistances
  • Body Weight in Animals
  • Fat Content of Meat
  • Time to First Flower
  • IQ
  • Blood Pressure

2
The following image demonstrates the variation
for flower diameter, number of flower parts and
the color of the flower Gaillaridia pilchella
(McClean 1997). Each trait is controlled by a
number of genes each interacting with each other
and an array of environmental factors.
3
  • Number of Genes Number of Genotypes
  • 1 3
  • 2 9
  • 5 243
  • 10 59,049

4
Consider two genes, A with two alleles A and a,
and B with two alleles B and b.- Each of the
alleles will be assigned metric values- We give
the A allele 4 units and the a allele 2 units-
At the other locus, the B allele will be given 2
units and the b allele 1 unit
  • Genotype Ratio Metric value
  • AABB 1 12
  • AABb 2 11
  • AAbb 1 10
  • AaBB 2 10
  • AaBb 4 9
  • Aabb 2 8
  • aaBB 1 8
  • aaBb 2 7
  • aabb 1 6

5
A grapical format is used to present the above
results
6
Normal distribution of a quantitative trait may
be due to
  • Many genes
  • Environmental effects
  • The traditional view polygenes each with small
    effect and being sensitive to environments
  • The new view A few major gene and many
    polygenes (oligogenic control), interacting with
    environments

7
Traditional quantitative genetics research
Variance component partitioning
  • The phenotypic variance of a quantitative trait
    can be partitioned into genetic and environmental
    variance components.
  • To understand the inheritance of the trait, we
    need to estimate the relative contribution of
    these two components.
  • We define the proportion of the genetic variance
    to the total phenotypic variance as the
    heritability (H2).
  • - If H2 1.0, then the trait is 100 controlled
    by genetics
  • - If H2 0, then the trait is purely affected
    by environmental factors.

8
  • Fisher (1918) proposed a theory for partitioning
    genetic variance into additive, dominant and
    epistatic components
  • Cockerham (1954) explained these genetic variance
    components in terms of experimental variances
    (from ANOVA), which makes it possible to estimate
    additive and dominant components (but not the
    epistatic component)
  • I proposed a clonal design to estimate additive,
    dominant and part-of-epistatic variance
    components
  • Wu, R., 1996 Detecting epistatic genetic
    variance with a clonally replicated design
    Models for low- vs. high-order nonallelic
    interaction. Theoretical and Applied Genetics 93
    102-109.

9
Genetic Parameters Means and (Co)variances
  • One-gene model
  • Genotype aa Aa AA
  • Genotypic value G0 G1 G2
  • Net genotypic value -a
    0 d
    a

  • origin(G0G1)/2
  • a additive genotypic value
  • d dominant genotypic value
  • Environmental deviation E0 E1 E2
  • Phenotype or
  • Phenotypic value Y0G0E0 Y1G1E1 Y2G2E2
  • Genotype frequency P0 P1 P2
  • at HWE q2 2pq p2
  • Deviation from population mean ? -a - ? d -
    ? a - ?
  • -2pa(q-p)d (q-p)a(q-p)d
    2qa(q-p)d

10
  • Population mean ? q2(-a) 2pqd p2a
    (p-q)a2pqd
  • Genetic variance ?2g q2(-2p?-2p2d)2
    2pq(q-p)?2pqd2 p2(2q?-2q2d)2
  • 2pq?2 (2pqd)2
  • ?2a (or VA) ?2d (or VD)
  • Additive genetic variance, Dominant genetic
    variance,
  • depending on both on a and d depending only on
    d
  • Phenotypic variance ?2P q2Y02 2pqY12 p2Y22
    (q2Y0 2pqY1 p2Y2)2
  • Define
  • H2 ?2g /?2P as the broad-sense heritability
  • h2 ?2a / ?2P as the narrow-sense heritability
  • These two heritabilities are important in
    understanding the relative contribution of
    genetic and environmental factors to the overall
    phenotypic variance.

11
What is ? a(q-p)d?
  • It is the average effect due to the substitution
    of gene from one allele (A say) to the other (a).
  • Event A a contains two possibilities
  • From Aa to aa From AA to Aa
  • Frequency q p
  • Value change d-(-a) a-d
  • ? qd-(-a)p(a-d)
  • a(q-p)d

12
Midparent-offspring correlation
  • __________________________________________________
    __________________
  • Progeny
  • Genotype Freq. of Midparent AA Aa aa Mean
    value
  • of parents matings value a d -a of progeny
  • __________________________________________________
    __________________
  • AA AA p4 a 1 - - a
  • AA Aa 4p3q ½(ad) ½ ½ - ½(ad)
  • AA aa 2p2q2 0 - 1 - d
  • Aa Aa 4p2q2 d ¼ ½ ¼ ½d
  • Aa aa 4pq3 ½(-ad) - ½ ½ ½(-ad)
  • aa aa q4 -a - - 1 -a
  • ________________________________________________

13
  • Covariance between midparent and offspring
  • Cov(OP)
  • E(OP) E(O)E(P)
  • p4a a 4p3q ½(ad) ½(ad) q4 (-a)(-a)
    (p-q)a2pqd2
  • pq?2
  • ½?2a
  •  
  • The regression of offspring on midparent values
    is
  • b Cov(OP)/?2(P)
  • ½?2a / ½?2P
  • ?2a /?2P
  • h2
  • where ?2(P)½?2P is the variance of midparent
    value.

14
  • IMPORTANT
  • The regression of offspring on midparent values
    can be used to measure the heritability!
  • This is a fundamental contribution by R. A.
    Fisher.

15
You can derive other relationships
  • Degree of relationship Covariance
  • __________________________________________________
    __
  • Offspring and one parent Cov(OP) ?2a/2
  • Half siblings Cov(FS) ?2a/4
  • Full siblings Cov(FS) ?2a/2 ?2a/4
  • Monozygotic twins Cov(MT) ?2a ?2d
  • Nephew and uncle Cov(NU) ?2a/4
  • First cousins Cov(FC) ?2a /8
  • Double first cousins Cov(DFC) ?2a/4 ?2d/16
  • Offspring and midparent Cov(O) ?2a/2
  • __________________________________________________
    __
  •  

16
Cockerhams experimental and mating designs
  • By estimating the covariances between relatives,
    we can estimate the additive (or mixed additive
    and dominant) variance and, therefore, the
    heritability.
  • Next, I will introduce mating and experimental
    designs used to estimate the covariances between
    relatives.

17
Mating design
  • Mating design is used to generate genetic
    pedigrees, genetic information and materials that
    can be used in a breeding program
  • Mating design provides genetic materials, whereas
    experimental design is utilized to obtain and
    analyze the data from these materials

18
Objectives of mating designs
  • Provide information for evaluating parents
  • 2) Provide estimates of genetic parameters
  • 3) Provide estimates of genetic gains
  • 4) Provide a base population for selection

19
Commonly used mating designs
  • 1) Open-pollinated
  • 2) Polycross
  • 3) Single-pair mating
  • 4) Nested mating
  • 5) Factorial mating tester design
  • 6) Diallel mating (full, half, partial
    disconnected)
  •  

20
Nested mating (NC Design I)
  • Each of male parents is mated to a subset of
    different female parents

21
  • Cov(HSM)1/4VA
  • V(female/male) Cov(FS) Cov(HSM)
  • 1/2VA1/4VD 1/4VA
  • 1/4VA 1/4VD
  •  
  • - Provide information for parents and full-sib
    families
  • - Provide estimates of both additive and
    dominance effects
  • - Provide estimates of genetic gains from both
    VA and VD
  • - Not efficient for selection
  • - Low cost for controlled mating

22
Example Date structure for NC Design I
  • Sample Male Female Full-sib family Individual Phen
    otype
  • 1 1 A 1 1 y1A1
  • 2 1 A 1 2 y1A2
  • 3 1 B 2 1 y1B1
  • 4 1 B 2 2 y1B2
  • 5 1 C 3 1 y1C2
  • 6 1 C 3 2 y1C2
  • 7 2 D 4 1 y2D1
  • 8 2 D 4 2 y2D2
  • 9 2 E 5 1 y2E1
  • 10 2 E 5 2 y2E2
  • 11 2 F 6 1 y2F1
  • 12 2 F 6 2 y2F2
  • 13 3 G 7 1 y3G1
  • 14 3 G 7 2 y3G2
  • 15 3 H 8 1 y3H1
  • 16 3 H 8 2 y3H2
  • 17 3 I 9 1 y3I1
  • 18 3 I 9 2 y3I2

23
Estimates by statistical software
  • VTotal 40
  • VFS Cov(FS) 10
  • VM Cov(HSM) 4
  • VE VTotal VFS 40 10 30
  • V(female/male) Cov(FS) Cov(HSM)
  • 10 4 6
  • VA 4Cov(HSM) 4 4 16 h2 16/40
    0.x
  • V(female/male) 1/4VA 1/4VD 4 1/4VD 6
  • VD 8, VG VA VD 16 6 22
  • H2 22/40 0.x

24
Factorial mating (NC Design II)
  • Each member of a group of males is mated to each
    member of group of females

25
  • Cov(HSM) 1/4 VA
  • Cov(HSF) 1/4 VA
  •  V(female ? male) Cov(FS)Cov(HSM)Cov(HSF)
  • 1/4 VD
  •  
  • - Provide good information for parents and
    full-sib families
  • - Provide estimates of both additive and
    dominance effects
  • - Provide estimates of genetic gains from both
    VA and VD
  • - Limited selection intensity
  • - High cost

26
Tester mating design (Factorial)
  • Each parent in a population is mated to each
    member of the testers that are chosen for a
    particular reason

27
  • Cov(HSM)1/4VA
  • Cov(HSF)1/4VA
  • V(female ? male) Cov(FS)COV(HSM)-COV(HSF)
  • 1/4VD
  •  
  • - Provide good information for parents and
    full-sib families
  • - Provide estimates of both additive and
    dominance effects
  • - Provide estimates of genetic gains from both
    VA and VD
  • - Limited selection intensity
  • - High cost

28
Diallel mating design
  • Full diallel each parent is mated with every
    other parent in the population, including selfs
    and reciprocal
  •  

29
  • Half diallel each parent is mated with every
    other parent in the population, excluding selfs
    and reciprocal

30
  • Partial Diallel selected subsets of full
    diallels
  •  

31
  • Disconnected half diallel selected subsets of
    full diallels

32
  • Diallel analysis
  •  
  • Cov(HS) 1/4VA
  • Cov(FS) 1/2VA 1/4VD
  • Cov(FS) Cov(FS) 2Cov(HS) 1/4VD
  •  
  • - Provide good evaluation of parents and
    full-sib families
  • - Provide estimates of both additive and
    dominance effects
  • - Provide estimates of genetic gains from both
    VA and VD
  • - High cost

33
Genomic Imprinting or parent-of-origin effectThe
same allele is expressed differently, depending
on its parental origin
  • Consider a gene A with two alleles A (in a
    frequency p) and a (in a frequency q)
  • Genotype Frequency Value
  • AA p2 a Average effect
  • Aa pq di No imprinting ? a
    d(q-p)
  • aA qp d-i Imprinting ?M a
    i d(q-p) A ? a
  • aa q2 -a ?P a i d(q-p)
    A ? a
  • Mean a(p-q)2pqd
  • No imprinting ?g2 2pq?2 (2pqd)2
  • Imprinting ?gi2 2pq?2 (2pqd)2 2pqi2
  • Imprinting leads to increased genetic variance
    for a quantitative trait and, therefore, is
    evolutionarily favorable.

34
Genomic Imprinting
The callipygous animals 1 and 3 compared to
normal animals 2 and 4 (Cockett et al. Science
273 236-238, 1996)
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Predicting Response to Selection
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Population Mean, Xp - phenotypic mean of the
animals or plants of interest and expressed in
measurable units. Selection Mean, Xs - phenotypic
mean of those animals or plants chosen to be
parents for the next generation and expressed in
measurable units. Selection Differential, SD -
difference between the phenotypic means of the
entire population and its selected mean.
44
Genetic Gain the amount that the phenotypic
mean in the next generation change by selection.
- that change can be or -
45
Selection Differential
G h2 SD
46
How to Calculate Genetic Gain
M2 M h2 (M1 - M) M2 resulting mean
phenotype M mean of parental population M1
mean of selected population h2 heritability of
the trait ? M2 - M h2 (M1
- M) ? G h2 SD (SD/?p)h2?p ih2?p i
selection intensity h2 narrow-sense
heritability ?p standard phenotypic deviation
47
  • Factors that influence
  • the Genetic Gain
  • Magnitude of selection differential
  • Selection intensity
  • Broad-sense heritability heritability
  • Phenotypic variation

48
Knowing the Selection Differential, and the
response to selection, an estimate of the traits
heritability can be calculated G / SD Realized
Heritability
49
Realized heritability can also be calculated
as M2 M h2 (M1 - M) rearranged,
(M2 - M) (M1 - M)
h2
50
  • Maximizing Genetic Gain
  • Examples

51
N48, Population Mean 109.7
52
Goal Improve the Mean Select those in red, N
6, Mean of Selected 119.5 SD 9.8 G h2 SD
0.7 x 9.8 6.86
53
Goal Reduce the Mean Select those in blue, N
8, Mean of Selected 100.4
54
Nature 432, 630 - 635 (02 December 2004)The
role of barren stalk1 in the architecture of maize
  • ANDREA GALLAVOTTI1,2, QIONG ZHAO3,
    JUNKO KYOZUKA4, ROBERT B. MEELEY5,
    MATTHEW K. RITTER1,, JOHN F. DOEBLEY3,
    M. ENRICO PÈ2 ROBERT J. SCHMIDT1
  • 1 Section of Cell and Developmental Biology,
    University of California, San Diego, La Jolla,
    California 92093-0116, USA2 Dipartimento di
    Scienze Biomolecolari e Biotecnologie, Università
    degli Studi di Milano, 20133 Milan,
    Italy3 Laboratory of Genetics, University of
    Wisconsin, Madison, Wisconsin 53706,
    USA4 Graduate School of Agriculture and Life
    Science, The University of Tokyo, Tokyo 113-8657,
    Japan5 Crop Genetics Research, Pioneer-A DuPont
    Company, Johnston, Iowa 50131, USA Present
    address Biological Sciences Department,
    California Polytechnic State University, San Luis
    Obispo, California 93407, USA

55
Mapping Quantitative Trait Loci (QTL) in the F2
hybrids between maize and teosinte
56
Maize Teosinte tb-1/tb-1 mutant maize
57
Effects of ba1 mutations on maize development
Mutant Wild type No tassel
Tassel
58
Data format for a backcross
  • Sample Height Marker 1 Marker 2 QTL
  • (cm, y)
  • 1 184 Mm (1) Nn (1) ?
  • 2 185 Mm (1) Nn (1) ?
  • 3 180 Mm (1) Nn (1) ?
  • 4 182 Mm (1) nn (0) ?
  • 5 167 mm (0) nn (0) ?
  • 6 169 mm (0) nn (0) ?
  • 7 165 mm (0) nn (0) ?
  • 8 166 mm (0) Nn (1) ?

59
  • Heights classified by markers (say marker 1)
  • Marker Sample Sample Sample
  • group size mean variance
  • Mm n1 4 m1182.75 s21
  • mm n0 4 m0166.75 s20

60
The hypothesis for the association between the
marker and QTL
  • H0 m1 m0
  • H1 m1 ? m0
  • Calculate the test statistic
  • t (m1m0)/?s2(1/n11/n0),
  • where s2 (n1-1)s21(n0-1)s20/(n1n02)
  • Compare t with the critical value
    tdfn1n2-2(0.05) from the t-table.
  • If t gt tdfn1n2-2(0.05), we reject H0 at the
    significance level 0.05 ? there is a QTL
  • If t lt tdfn1n2-2(0.05), we accept H0 at the
    significance level 0.05 ? there is no QTL

61
Why can the t-test probe a QTL?
  • Assume a backcross with two genes, one marker
    (alleles M and m) and one QTL (allele Q and q).
  • These two genes are linked with the recombination
    fraction of r.
  • MmQq Mmqq mmQq mmqq
  • Frequency (1-r)/2 r/2 r/2 (1-r)/2
  • Mean effect ma m ma m
  • Mean of marker genotype Mm
  • m1 (1-r)/2 (ma) r/2 m m (1-r)a
  • Mean of marker genotype mm
  • m0 r/2 (ma) (1-r)/2 m m ra
  • The difference
  • m1 m0 m (1-r)a m ra (1-2r)a

62
  • The difference of marker genotypes can reflect
    the size of the QTL,
  • This reflection is confounded by the
    recombination fraction
  • Based on the t-test, we cannot distinguish
    between the two cases,
  • - Large QTL genetic effect but loose linkage with
    the marker
  • - Small QTL effect but tight linkage with the
    marker

63
Example marker analysis for body weight in a
backcross of mice
  • __________________________________________________
    ___________________
  • Marker class 1 Marker class 0
  • ______________________ _____________________
  • Marker n1 m1 s21 n1 m1 s21 t P value
  • __________________________________________________
    ___________________________
  • 1 Hmg1-rs13 41 54.20 111.81 62 47.32 63.67 3.754
    lt0.01
  • 2 DXMit57 42 55.21 104.12 61 46.51 56.12 4.99
    lt0.01
  • 3 Rps17-rs11 43 55.30 101.98 60 46.30 54.38 5.231
    lt0.000001
  • __________________________________________________
    ___________________

64
Marker analysis for the F2
  • In the F2 there are three marker genotypes, MM,
    Mm and mm, which allow for the test of additive
    and dominant genetic effects.
  • Genotype Mean Variance
  • MM m2 s22
  • Mm m1 s21
  • mm m0 s20

65
Testing for the additive effect
  • H0 m2 m0
  • H1 m2 ? m0
  • t1 (m2m0)/?s2(1/n21/n0),
  • where s2 (n2-1)s22(n0-1)s20/(n1n02)
  • Compare it with tdfn2n0-2(0.05)

66
Testing for the dominant effect
  • H0 m1 (m2 m0)/2
  • H1 m1 ? (m2 m0)/2
  • t2 m1(m2 m0)/2/?s21/n11/(4n2)1/(4n0)
    ,
  • where s2 (n2-1)s22(n1-1)s21(n0-1)s20/(n2n1
    n03)
  • Compare it with tdfn2n1n0-3(0.05)

67
Example Marker analysis in an F2 of maize
  • __________________________________________________
    ____________________________________________
  • Marker class 2 Marker class 1 Marker class
    0 Additive Dominant
  • ____________ ______________ ______________
  • M n2 m2 s22 n1 m1 s21 n0 m0
    s20 t1 P t2 P
  • __________________________________________________
    _____________________________________________
  • 43 5.24 2.44 86 4.27 2.93 42
    3.11 2.76 6.10 lt0.001 0.38 0.70
  • 2 48 4.82 3.15 89 4.17 3.26 34
    3.54 2.84 3.28 0.001
    -0.05 0.96
  • 3 42 5.01 3.23 92 4.14 3.18 37
    3.57 2.68 3.71 0.0002
    -0.57 0.57
  • __________________________________________________
    _____________________________________________

68
Testing for the dominant effect
  • H0 m1 (m2 m0)/2
  • H1 m1 ? (m2 m0)/2
  • t2 m1(m2 m0)/2/?s21/n11/(4n2)1/(4n0)
    ,
  • where s2 (n2-1)s22(n1-1)s21(n0-1)s20/(n2n1
    n03)
  • Compare it with tdfn2n1n0-3(0.05)

69
Example Marker analysis in an F2 of maize
  • __________________________________________________
    ____________________________________________
  • Marker class 2 Marker class 1 Marker class
    0 Additive Dominant
  • ____________ ______________ ______________
  • M n2 m2 s22 n1 m1 s21 n0 m0
    s20 t1 P t2 P
  • __________________________________________________
    _____________________________________________
  • 43 5.24 2.44 86 4.27 2.93 42
    3.11 2.76 6.10 lt0.001 0.38 0.70
  • 2 48 4.82 3.15 89 4.17 3.26 34
    3.54 2.84 3.28 0.001
    -0.05 0.96
  • 3 42 5.01 3.23 92 4.14 3.18 37
    3.57 2.68 3.71 0.0002
    -0.57 0.57
  • __________________________________________________
    _____________________________________________
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