Transcript Quantitative traits

Chapter 6
Quantitative Genetics
数量(性状)遗传
Quantitative Genetics and Polygenic Traits
Phenotypic
variation in
quantitative
traits often
approximates
a bell-shaped
curve: the
normal
distribution!
Example:
distribution of
height in
humans
1914 Class of the Connecticut Agricultural
College (fig. 13.17 from text)
Quantitative Inheritance
Analysis of Polygenic Traits
Heritability (遗传力\遗传率)
Mapping Quantitative Trait
6.1 Quantitative Inheritance
The Multiple-Factor Hypothesis
Additive Alleles: The Basis of
Continuous Variation
Calculating the Number of Genes
The Significance of Polygenic
Inheritance
6.2 Analysis of Polygenic Traits
The Mean
Variance
Standard Deviation
Standard Error of the Mean
Analysis of a Quantitative Character
6.3 Heritability
Broad-Sense Heritability
Narrow-Sense Heritability
Artificial Selection
Twin Studies in Humans
6.4 Mapping Quantitative
Trait Loci
• Trait that exhibit quantitative phenotypic variation
are often under the genetic control of alleles whose
influence is additive(累加/加性) in nature, resulting
in continuous variation.
• Such trait can be analyzed and characterized using
statistical methods, which also allow the assessment
of the relative importance of genetic factors during
phenotypic expression.
• Such calculation establish the heritability of traits in
a population. In humans, twin studies provide a
similar, but less precise, estimate.
• Recently developed techniques allow the localization
within the genome of loci contributing to
quantitative traits.
Previous chapters:
Up to now, we have focused on genetics of:
Qualitative traits – genetic variants fall into discrete,
easily detectable classes.
phenotypic variation classified into distinct traits,
e.g.
1. Seed shape in peas (round or wrinkled) ,pea
plant tall and dwarf ,
2. Eye color in fruit fly Drosophila (red or white)
3. Blood types in humans (A, B, AB, or O)
4. squash shape spherical, disc-shaped and
elongated,
These phenotypes exemplifies discontinuous
variation
F
F
P
F1
F2
H
质量性状遗传
图 6-
数量性状遗传
质量性状遗传和数量性状遗传的区别
But what about:
Quantitative traits – phenotypic variation
continuous, and individuals do not fall into
discrete classes.
Example: Height in humans
Many other traits in a population
demonstrate fairly more variation and are
not easily categorized into distinct classes.
For Example:
Some Plant Disease Resistances
Weight Gain in Animals ,
Fat Content of Meat
IQ ,
Learning Ability ,
Blood Pressure
Quantitative Characteristics
Definitions
In natural populations, variation in a
character takes the form of a continuous
phenotypic range rather than discrete
phenotypic classes.
In other words, the variation is
quantitative, not qualitative .
Mendelian genetic analysis is extremely
difficult to apply to such continuous
phenotypic distributions, so statistical
techniques are employed instead.
Quantitative Characteristics
Many traits are influenced by the combined
action of many genes and are characterized
by continuous variation. These are called
polygenic traits.
Continuously variable characteristics that are
both polygenic and influenced by
environmental factors are called
multifactorial traits. Examples of quantitative
characteristics are height, intelligence & hair
color.
Discontinuous traits – qualitative traits with only
a few possible phenotypes that fall into discrete
classes; phenotype is controlled by one or only
a few genes (ex.: tall or short pea plants; red,
pink or white snapdragon flowers)
质量性状——变异不连续的
Continuous traits - Quantitative trait do not fall
into discrete classes; a segregating population
will show a continuous distribution of
phenotypes
- more common term for continuous trait; a trait
that has a quantitative value (yield, IQ)
数量性状——变异连续的
Quantitative Genetics - the field of genetics that
studies quantitative traits
Threshold(阈值)model
1.基本物质为呈连续分布的
数量性状，而表型性状则
为不连续分布的质性状。
2.基本物质处于某一特定范
围内，表现为正常，如果
超出某一阈值，表型就不
正常，如血压，血糖含量，
抗病力等。
3.基本物质受多基因控制，
但性状的改变仅发生在基
本物质达到或者超过某一
阈值时才发生。所以多基
因控制的性状，也可以表
现为非此即彼，全或无的
表型。
Quantitative (discontinuous)traits are
controlled by multiple genes, each
segregating according to Mendel's
laws.
These traits can also be affected by
the environment to varying degrees.
1、Quantitative Inheritance
A major task of quantitative
genetics is to determine the ways
in which genes interact with the
environment to contribute to the
formation of a given quantitative
trait distribution.
At the beginning of the 20th century, geneticists
noted that many characters in different species
had similar patterns of inheritance,
such as height and stature in humans,
seed size in the broad bean,
grain color in wheat, and
kernel number and ear length in corn.
In each case, offspring in the succeeding
generation seemed to be a blend of their parents'
characteristics.
1. What is the genetic and
2.
3.
4.
5.
environmental contribution to the
phenotype?
How many genes influence the trait?
Are the contributions of the genes
equal?
How do alleles at different loci
interact: additively? epistatically?
How rapid will the trait change
under selection?
What is the genetic basis of variation
in quantitative traits?
Characteristics of Quantitative Traits
1. Polygenic – affected by genetic variation at
many different gene loci.
2. Phenotypic effects of allelic substitution are
usually small and additive.

Each allelic substitution results in an incremental
change in overall phenotype.
3. Phenotypic variation in quantitative traits
usually influenced by environmental variation
as well as by genetic variation.
The Multiple Factor Hypothesis
Polygeic Traits
tabaco plants in a
cultivated field or
wild at the side of
the road are not
neatly sorted into
categories of "tall"
and "short,"
(Figure 6-1).
The issue of whether continuous variation could be
accounted for in Mendelian terms caused
considerable controversy in the early 1900s.
William Bateson and Gudny Yule, who adhered to the
Mendelian explanation of inheritance, suggested that
a large number of factors or genes could account for
the observed patterns.
This proposal, called the multiple-factor hypothesis,
implied that many factors or genes contribute to the
phenotype in a cumulative or quantitative way.
However, other geneticists argued that Mendel's unit
factors could not account for the blending of parental
phenotypes characteristic of these patterns of
inheritance and were thus skeptical of these ideas.
NB - no nice 3:1,
9:3:3:1 ratios!
This is the normal
type of result for
most traits.
Due to:1. Multiple genes
each having some
cumulative effect on
the phenotype (QTL’S)
plus
2. Environmentally
caused variation
By 1920, the
conclusions of
several critical sets of
experiments largely
resolved the
controversy and
demonstrated that
Mendelian factors
could account for
continuous variation.
In one experiment,
Edward M. East
crossed two strains
of the tobacco plant
Nicotiana longiflora.
Additive alleles:
The Basis of Continuous variation
The multiple-factor hypothesis, suggested
by the observations of East and others.
embodies the following major points:
l. Characters that exhibit continuous variation
can usually be quantified by measuring,
weighing, counting, and so on
2. Two or more pairs of genes, located
throughout the genome, account for the
hereditary influence on the phenotype in an
additive way. Because many genes can be
involved, inheritance of this type is often
called polygenic.
3. Each gene locus may be occupied by either
an additive allele, which contributes a set
amount to the phenotype,or by a non
additive allele, which does not contribute
quantitatively to the phenotype.
4. The total effect on the phenotype of each
additive allele, while small, is approximately
equivalent to all other additive alleles at
other gene sites.
5. Together the genes controlling a single
character produce substantial phenotypic
variation.
6. Analysis of polygenic traits requires the
study of large numbers of progeny from a
population of organisms.
These points center around the
concept that additive alleles at
numerous loci control quantitative
traits. To illustrate this, let's examine
Herman Nilsson-Ehle's experiments
involving grain color in wheat
performed In one set of experiments,
wheat with red grain was crossed to
wheat with white grain (Figure 6-3).
Nilsson-Ehle’s crosses
demonstrated that the
difference between the
inheritance of genes
influencing quantitative
characteristics and the
inheritance of genes
influencing discontinuous
characteristics is the number
of loci that determine the
characteristic. Both
quantitative (continuous) and
discontinuous traits are
inherited according to
Mendel’s laws.
P
F1
F2
Three genes act additively to determine
seed color.
The dominant allele of each gene adds an
equal amount of redness; the recessive
allele adds no color to the seed.
A adds 1 unit of redness, a does not.
B adds 1 unit of redness, b does not.
C adds 1 unit of redness, c does not.
As the number of
polymorphic loci increases,
phenotypic variation
approaches a continuous
normal distribution!
As the number of loci
affecting the trait increases,
the # phenotypic categories
increases.
Number of phenotypic
categories =
(# gene pairs × 2) +1
Connecting the points of a
frequency distribution
creates a bell-shaped curve
called a normal distribution.
Proportion of Individuals
2. Additional phenotypic variation introduced by
environmental variation will further blur distinctions
between genotypic classes!
0.18
Ten
0.16 Polymorphic
0.14 Loci
0.12
0.10
0.08
0.06
0.04
0.02
0.00
-10 -8
-6
-4
-2
0
+2 +4 +6
+8 +10
Relative Height (Deviation from Mean)
基因 变异 表型 受环
控制 分布 分布 境影
响
数量 多基 正态 连续 大
性状 因
分布
遗传
规律
性状 研究对
特点 象
非孟
德尔
遗传
易度 群体
量
质量 单基 二项 分散 小
性状 因
分布
孟德
尔遗
传
不易 个体和
度量 群体
Calculating the Number of Genes
When additive effects control polygenic traits, it is of
interest to determine the number of genes involved.
If the ratio (proportion) of F2 individuals resembling
either of the two most extreme phenotypes (the
parental phenotypes) can be determined, then the
number of gene pairs involved (n) can be calculated
using the following simple formula:
1/4n=ratio of F2 individuals expressing
either extreme phenotype
Table 6.1 lists the ratios and number of F2 phenotypic classes
produced in crosses involving up to five gene pairs.
For low numbers of gene pairs, it is sometimes easier to use the (2n
+ 1) rule. When n = 2. 2n + 1 = 5. That is, each phenotypic
category can have 4. 3, 2, 1, or 0 additive alleles. When n = 3. 3n +
1 =7, and each phenotypic category can have 6, 5, 4, 3, 2, 1, or 0
additive alleles, and so on.
Calculation of Ratio of F2 with a
Given Genotype
•For example, you are asked what proportion of the F2
generation of a cross between a dark red and a whiteseeded wheat plant would have two dominant alleles.
•Aassuming 2 loci, this would include the following
phenotypes:
–AaBb (1/2 x 1/2 = 1/4 or 4/16)AAbb (1/4 x 1/4 = 1/16)
Isolate each gene and determine the proportion of each
class (product law), then add the three together (sum
law). Final answer = 6/16 aaBB (1/4 x 1/4 = 1/16)
Corn Height
Model: minimal height = 32”
4 dominant alleles interact equally and additively
A adds 8” to height, a does not
B adds 8” to height, b does not
C adds 8” to height, c does not
D adds 8” to height, d does not
e.g. AABbCcDd = 32 + (5 × 8) = 72” tall
Strain 1 AABBccdd (64”)
P Strain 2 aabbCCDD (64”)
[32 + (4x8)]
F1 AaBbCcDd (64”)
F2 a continuous distribution from
aabbccdd (32”) to AABBCCDD (96”)
16 × 16 Punnett square
Height in Corn
80
Frequency (out of 256)
70
60
50
40
30
20
10
0
32
40
48
56
64
72
80
88
96
Height (inches)
The number of genes involved is n, where the ratio of F2
individuals expressing either extreme phenotype is 1/4n.
In this case 1 out of 256 is minimum height, therefore the
number of genes involved is 4 (44=256).
The significance of
Polygenic Inheritance
Polygenic inheritance Is a significant
concept because it appears to serve as the
genetic basis for a vast number of traits
involved in animal breeding and agriculture.
For example.
height, weight, and physical stature in animals,
Size and grain yield in crops.,beef and milk
production in cattle, and
egg production in chickens are all thought to
be under polygenic control.
In most cases, it is important to
note that the genotype, which is
fixed at fertilization, establishes the
potential range within which a
Particular phenotype falls.
However, environmental factors
determine how much of the
potential will be realized. ln the
crosses described thus far we have
assumed an optimal environment,
which minimizes variation from
external sources
Homozygous parental pops.
selected to have very different
phenotypes - still show some
environmental variation
F1 - intermediate - shows some
variability (environmental)
F2 - much greater variability mean is intermediate in length
variation here genetic + envir.
Plants selected from different
parts of F2 produce F3 with
corresponding phenotypes proves F2 phenotype was partly
genetically based.
Typical inheritance where
phenotypes show
continuous range.
2.Analysis of polygenic Traits
Statistical analysis series three purposes:
l. Data can be mathematically reduced to
provide a descriptive summary of the
sample.
2. Data from a small but random sample can
be used to infer information about groups
larger than those from which the original
data were obtained (statistical inference)
3 Two or more sets of experimental data can
be compared to determine whether they
represent significantly different populations
of measurements.
•Additional phenotypic variation introduced by
environmental variation will further blur distinctions
between genotypic classes!
•Several statistical methods are useful in the analysis of
traits that exhibit a normal distribution, including the mean,
variance, standard deviation, and standard error of the
mean
Proportion of Individuals
The Mean and Variance
Means of Polygenic Traits
Population mean (µ) – estimated by
sample mean
Sample mean (X) = ∑ Xi
n
Sum of sample scores
Number in sample
Population variance (σ2)- estimated by
sample variance (s2)
Variance of Polygenic
Traits
Sample variance (s2)- an “average”
squared deviation from the mean
S2 =
∑ (Xi – X)2
n-1
Standard deviation of a population (σ)estimated by
sample standard deviation (s)
S = √ s2
Take a look at the simple example
X1
X2 X-X1 X-X2 ∑(X –X )2 ∑(X –X)2
3
4
-3
-2
9
4
4
5
-2 -1
4
1
6
6
0
0
0
0
8
7
2
1
4
1
9
8
3
2
9
4
X
6
6
∑(X –X )
0
0
∑(X –X )2
24
8
Analysis of Polygenic
Traits
Standard error of the mean (sx)measures variations in sample mean for
similar samples from same population
Sx =
s
√n
Frequency
Features of a normal
distribution
Analysis of Quantitative
Traits
We know that genotype, and the
environment a particular genotype inhabits,
contribute to the overall variation observed
in a quantitative trait.
So, how can we distinguish genetic from
environmental components of variation?
Question: For a quantitative trait, how much
of the total phenotypic variation is
attributable to genetic variation, and how
much is attributable to environmental
variation?
Phenotypic variation (VP) can be partitioned
into its genetic (VG) and environmental (VE)
components:
VP = VG + VE
Heritability
The heritability of a quantitative trait is the
proportion of the total phenotypic variation (VP)
that is attributable to genetic variation (VG ).
Given VP = VG + VE , then…
Heritability (h2) = VG / VP = VG / (VG + VE)
Heritability is a proportion and varies between 0
and 1.
If h2 = 0, none of the phenotypic variation is
attributable to underlying genetic variation.
If h2 = 1, all of the phenotypic variation is
attributable to underlying genetic variation.
Analysis of Quantitative
Traits
VP=VG+VE =VA+VD +VE
VG=VA+VD
If,
VP = VA + VD + VE + V(GXE)
Total observed
variation
Genetic
Environmental
Analysis of Quantitative
Traits
To distinguish genetic from environmental
components of variation, we need:

Genetically heterogenous population where:


VP = VG + VE
Genetically identical population where:
VG = 0
 So, VP = VE

Analysis of Quantitative
Traits
We can cross 2 highly inbred parents to
achieve the requirements needed to
distinguish genetic and environmental
variation
VG=VA+VD+VI
VP=VA+VD+VI+VE+VGE
S
VP1  VP 2
2
VE 
 Se 
2
VE=(VP1+VP2+VF1)/3
2
p1
S
2
2
P2
Analysis of Quantitative
Traits
Example: Inbred corn lines:
P1: A1A1B1B1C1C1 x A2A2B2B2C2C2
VG = 0,
So VP = VE
F1:
A1A2B1B2C1C2
F2:
segregates into many genotypes
VP =VG + VE
Analysis of Quantitative
Traits
VP (F2) – VP (F1) = VG
So, if F1 variance for plant height = 1.360
and F2 variance for plant height = 2.193
VG for plant height = (2.193-1.360) = 0.833
Heritability
(broad-sense)
Heritability (broad-sense) is the proportion of a
population’s phenotypic variance that is
attributable to genetic differences.
Heritability
(narrow sense)
Heritability (narrow sense) is the proportion of
a population’s phenotypic variance that is
attributable to additive genetic variance as
opposed to dominance genetic variance
(interaction between alleles at the same locus).
Additive genetic variance responds to
selection.
Broad-Sense Heritability
Broad-sense heritability (H2)- is the ratio of
genetic variance (VG) to total phenotypic
variance (VP)
So, for corn plant height: H2 = VG/VP
= 0.833 / 2.193
= 0.38 = 38%
This means that 38% of the observed
phenotypic variation for plant height is under
genetic control
(x

COVxy=
i
 x )( y i  y )
n 1
为了便于计算，上式可写成
1
 xi yi  n ( xi yi）
COVxy=
n 1
相关系数γ等于 xy 的协方差除以二者标准差的乘积
COVxy
γ=
SxSy
体 长
（ mm
）xi
72.00
62.00
86.00
76.00
64.00
82.00
71.00
96.00
87.00
103.00
86.00
74.00
xi =

959.00
xi  x
-7.92
-17.92
6.08
-3.92
-15.92
2.08
-8.92
16.08
7.08
23.08
6.08
-5.92
表 6-6 虎螈体长和头宽的相关性
头 宽 y  y ( y  y) 2
( xi  x) 2
i
i
(mm ）
yi
62.67
17.00
-0.75 0.56
321.01
14.00
-3.75 14.06
37.01
20.00
2.25
5.06
15.34
14.00
-3.75 14.06
253.34
15.00
-2.75 7.56
4.34
20.00
2.25
5.06
79.51
15.00
-2.57 7.56
258.67
21.00
3.25
10.56
50.17
19.00
1.25
1.56
532.84
23.00
5.25
27.56
37.01
18.00
0.25
0.06
35.01
17.00
-0.75 0.56
2
( xi  x)
yi
 ( y i  y) 2


=1686.92
=213.00
=94.25
xi yi
1224
868
1720
1064
960
1640
1065
2016
1653
23.69
1548
1258
x y
i
i
= 17385
x   xi / n  959 / 12  79.92
y   y i / n  213 / 12  17.75
S 2x   ( xi  x) 2 /( n  1)  1686.92 / 11  153.35
S x  S 2x  153.35  12.38
S 2y   ( y i  y ) 2 /( n  1)  94.25 / 11  8.75
S y  S 2y  2.93
协
方
差
1
x
y

 i i n ( xi yi）
COVxy=
=
n 1
959  213
17385 
12
 32.97
12  1
相 关 系 数
γ =
COVxy/SxSy=32.97/（12.38× 2.93 ）
=0.91
Narrow-sense
heritability
Narrow-sense heritability (h2) is the
ration of additive genetic variance (VA)
to total phenotypic variance (VP)
Only h2 (not H2) will predict changes in
population mean in response to
selection
广义遗传力（broad-sense heritability）
和狭义遗传力（narrow-sense heritability）
VG
H 
VP
2
VA
h 
VP
2
aa
m
Aa AA
d
-a
a
图6-6 表示AA,Aa,aa不同表型计量的模式图。
AA－aa =2a, m为AA和aa的平均值，杂合体
Aa和均值之间的数量差异为d。d/a为显性程度。
表 6-8 F2 的均值和方差计算
fi
xI
基因型
AA
1/4
a
Aa
1/2
d
aa
1/4
-a
1
合计
xi  d


f i xi 
表 6-9 B1 的平均数和遗传方差的计算
fi
XI
fiXi
AA
1/2
a
1/2 a
Aa
1/2
d
1/2 d
xi  a  d
1
合计


f i xi 
表 6-10 B2 的均值和遗传方差的计算
fi
xi
fixi
Aa
1/2
d
1/2 d
aa
1/2
-a
-1/2 a
x

d

a
i
1
合计

fixi2
1/4 a2
1/4 d2
1/4 a2
fixi
1/4 a
1/2 a
-1/4 a

f i xi 
1
d
2

f i xi2 
1 2 1 2
a  d
2
2
fiXi2
1/2 a2
1/2 d2
1
(a  d )
2

f i xi2 
1 2
(a  d 2 )
2
fixi2
1/2 d2
1/2 a2
1
(d  a)
2

f i xi2 
1 2
(a  d 2 )
2
Σx2- ( ΣX )2
n
2
n
S=
n
2
V
F2
ΣX2- (ΣX)
n
S2=
n
(ΣX)2
2
S =
n
2
ΣX (Σfx)
= Σfx 2- 2 n
1 2 1
1
2
= 2 a + 2 d – 4 d2
= 12 a2+ 41 d2
VA= a12+a22+
n
若有几对基因
+an2
VD= d12+d22+
n
= 2 a2+ 4 d2
+dn2
在多对基因时
1
1
VF2= 2 VA + 4 VD + VE
1
1
VE= 2 (VP1+VP2) or VE= 3 (VP1+VP2+ VF1)
1/2VA+1/4VD
1/2VA
2
2
H = 1/2VA+1/4VD+VE
h = 1/2VA+1/4VD+VE
2
(ΣX)
S12= ΣX2― n =
1 2 2
2 (a +d ) ―
(ΣX)2
1 2 2
2
2
S2 = ΣX ―
=
(a +d )―
n
2
1
1
2
2 = ¼(a2-2ad+d2)
(a+d)
=
(a-d)
4
4
1
1
(d-a)2 =
(a+d)2 = ¼(a2+2ad+d2)
4
4
回交一代的平均方差=1/2（S12+S22）=1/4 a2+1/4 d2
=1/4VA+1/4VD+VE
将 Aa 与 AA 回交，回交 1 代 B1 的频度和观察值如表 6-9 所示，根据表 6-9，B1 差异
部分的均值
B1  
1
f i xi  ( a  d )
2
B1 的方差
S12   xi2 
( xi ) 2
n

1 2
1
1
(a  d 2 )  (a  d ) 2  (a  d ) 2
2
4
4
将 F1 Aa 与 aa 回交，回交一代 B2 的频度和观察值如表所示。B2 差异部分的均值
B 2   f i xi 
1
(d  a)
2
( xi ) 2
1 2
1
1
2
2
(
a

d
)

(
d

a
)

(d  a ) 2
B2 的方差
n
2
4
4
1 2
1 2 1 2
2
回交一代的平均方差为 ( S1  S 2 )  a  d
2
4
4
S 22   xi2 

若将 VF2 的表型方差和回交 1 代平均方差相减
VF2-1/2 (VB1+VB2)=(1/2 a2+1/4 d2)-(1/4 a2+1/4d2)=1/4a2
即使考虑环境方差的话，双方也可以减掉，并不影响
上式结果，那么狭义遗传力就可以按一式计算
1 a 2 2[VF 2
VA S
2
2
h 



VB VP
VP
2
a
1
 (VB1  VB 2 )]
2
VF 2
若 VF ―
2
1
2（VB1+VB2）
1
1
1
1
= ( 2 VA+ 4 VD+VE) ― ( 4 VA + 4 VD+VE)
=
1
4 VA
1
VA
2
h2 = 1
1
VA+ VD + VE
2
4
1
2 × [ VF ― 2（VB1+VB2）]
h2 =
VF2
2
（二）系谱分析法
判断该病是否遗传病？遗传方式？
辨别是单基因病？多基因病？染色
体病？
是否存在遗传异质性？
（三）双生子法
单卵双生（Monozygotic twin , MZ)
遗传基础相同，表型极相似
双卵双生（Dizygotic twin , DZ)
遗传基础不相同，表型有较大差异
通过比较MZ与DZ表型特征的一致性
和不一致性，估计遗传和环境因素
在表型发生中的各自作用
发病一致率（%）
双生子之一具有某种性状或疾病时，
另一个也具有此性状或疾病
同一疾病双生子对数
×100%
总双生子对数
 如MZ发病一致率 > DZ发病一致率
提示该病遗传因素具有一定影响
双生子的某些生理和病理特征的一致性
性状
MZ一致率（%） DZ一致率（%）
开始起坐
82
76
开始行走
68
31
眼色
99.6
28
血压
63
36
先天愚型
89
7
原发性癫痫
72
15
精神分裂症
80
13
十二指肠溃疡
50
14
智力低下
94
47
麻疹
95
87
H2=
CMZ –CDZ
100 –CDZ
CMZ: 同卵双生子共同发病率
CMZ：异卵双生子共同发病率
如精神分裂症25对同卵双生子中共同发病20对；
异卵双生子中共同发病2对
H2 = (80-10)/100-10=0.78
Heritability and Selection
Response
The H2 or h2 value for a trait indicates
how that trait will respond to selection
Selection response (R) is measured as
the difference between the population
mean (Xpop) and the offspring mean (Xo).
R = Xo - Xpop
Principles of Selection selection - allowing some individuals more
opportunity to reproduce that others
only means of directing genetic improvement
in closed populations
improvement is gradual but consistent
.
Response to selection
R = h2 x S
where
R is the response
h2 is the heritability
S is the selection differential .
Response to selection
average daily gain in pigs.
h2 = .4 herd mean = 1.6 lb/day
selected parents = 1.8 lb/day
selection differential
S = 1.8 - 1.6 = .2 lb/day
response
R = h2 x S
= .4 x .2
= .08 lb/day per generation
.
Response to selection
backfat thickness in pigs.
all boars = 1.0 in
select boars = .8 in
h2 = .5
all gilts = 1.2 in
select gilts = 1.1 in
selection differential
males .8 - 1.0 = -.2 in
females 1.1 - 1.2 = -.1 in
overall (-.2 + -.1) / 2 = -.15 in
response
R = h2 x S
= .5 x -.15
= -.075 in per generation
.
Response to selection
R = i x h2 x s P
where
R is the response
i is intensity or
standardized selection differential
h2 is the heritability
sP is the phenotypic standard deviation .
Standardized Selection Differentials
Proportion saved
.70
.50
.25
.10
.05
.01
.001
i
.50
.80
1.27
1.76
2.06
2.67
3.37
weaning weight in beef cattle
h2 = .30
% saved (males) = 5%
% saved (females) = 30%
standard deviation = 40 lb
selection intensity
male
5% im = 2.06
female 30% if = 1.16
overall i = ( im + if ) / 2 = (2.06 + 1.16) / 2 = 1.61
response
R = i x h 2 x sP
= 1.61 x .30 x 40 = 19.32 lb per gen
.
Effect of selection intensity
more intense selection
keeping smaller proportion
as parents
increased response
R = i x h2 x s P
.
Heritability and Selection
Response
M2= M + h2 (M1-M)
M2 - M
h2 =
M1 - M
Response is also related to heritability
R = h 2S
Where, (S) = selection differential
(Difference between parent mean XP and
overall population mean Xpop)
S = Xp - Xpop
Heritability and Selection
Response
Realized heritability can be calculated
from the selection response:
H2 = R/S
Heritability and Selection
Response
In humans, we study identical (monozygotic)
twins as sources of genetically identical
populations to study environmental and genetic
variation.
Twins usually share the same environment. It’s
more accurate to compare the correlation
between pairs of monozygotic and dizygotic
twins for the trait in question.
Twin Studies in Huamn
Table 3.6 Comparison of human twins
Heritability (H2)
Correlation
Monozygotic
twins (rM)
Dizygotic twins
(rD)
Fingerprint-ridge
count
0.96
0.47
0.98
Height
0.9
0.57
0.66
IQ
0.83
0.66
0.34
Social maturity
score
0.97
0.89
0.16
Sources of error in twin studies

GXE Interaction (Increases variance in dizygotic
twins)

Similar uterine environment (Monozygotic twins
often share membranes)

Identical twins treated more similarly by parents,
teachers

Dizygotic twins of different sexes treated more
differently than monozygotic
Mapping Quantitative
Trait loci
Because quantitative traits are influenced
by numeous genes, geneticists would like
to know where these genes are located in
the genome
Linked on a single chromosome or
Scattered throughout genome among
many chromosomes?
Example of Polygenic Inheritance:
Insecticide Resistance in Drosophila
Resistant
Strain
F1
Hybrids
Control
Strain
Example of Polygenic Inheritance:
Insecticide Resistance in Drosophila
Resistant
Strain
F1
Hybrids
Control
Strain
Nilsson-Ehle’s Model
Genotype
AABBCC
“Redness”
score
6
Color
Dark red
aabbcc
0
No red (white)
AaBbCc
3
Medium red
AABbcc
3
Medium red