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 HypothesisAdditive Alleles: The Basis of Continuous Variation

Calculating the Number of Genes The Significance of Polygenic

Inheritance6.2 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 HeritabilityArtificial SelectionTwin Studies in Humans6.4 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 数量性状遗传 图 6- 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 humansMany 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

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 Definitions

Quantitative CharacteristicsMany 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 )model1.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 environmental contribution to the phenotype?

2. How many genes influence the trait?

3. Are the contributions of the genes equal?

4. How do alleles at different loci interact: additively? epistatically?

5. 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).

P

F1

F2

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.

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.

2. Additional phenotypic variation introduced by environmental variation will further blur distinctions

between genotypic classes!

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18

-10 -8 -6 -4 -2 0 +2 +4 +6 +8 +10

Relative Height (Deviation from Mean)

TenPolymorphic

Loci

Pro

port

ion

of I

ndiv

idua

ls

基因控制

变异分布

表型分布

受环境影响

遗传规律

性状特点

研究对象

数量性状

多基因

正态分布

连续 大 非孟德尔遗传

易度量

群体

质量性状

单基因

二项分布

分散 小 孟德尔遗传

不易度量

个体和群体

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 white-seeded 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 additivelyA adds 8” to height, a does notB adds 8” to height, b does notC adds 8” to height, c does notD adds 8” to height, d does note.g. AABbCcDd = 32 + (5 × 8) = 72” tall

Strain 1 AABBccdd (64”)Strain 2 aabbCCDD (64”)

F1 AaBbCcDd (64”)

F2 a continuous distribution fromaabbccdd (32”) to AABBCCDD (96”)16 × 16 Punnett square

[32 + (4x8)]P

Height in Corn

0

10

20

30

40

50

60

70

80

32 40 48 56 64 72 80 88 96

Height (inches)

Fre

qu

en

cy (

ou

t o

f 25

6)

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).

Height in Corn

0

10

20

30

40

50

60

70

80

32 40 48 56 64 72 80 88 96

Height (inches)

Fre

qu

en

cy (

ou

t o

f 25

6)

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

Plants selected from different parts of F2 produce F3 with corresponding phenotypes - proves F2 phenotype was partly genetically based.

F2 - much greater variability - mean is intermediate in length

variation here genetic + envir.

F1 - intermediate - shows some variability (environmental)

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

Population variance (σ2)- estimated by sample variance (s2)

n Number in sample

Sum of sample scores

Variance of Polygenic Traits

Sample variance (s2)- an “average” squared deviation from the mean

Standard deviation of a population (σ)- estimated by sample standard deviation (s)

S2 = ∑ (Xi – X)2

n - 1

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

Features of a normal distribution

Fre

quen

cy

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

HeritabilityThe 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 observedvariation

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 = 0So, VP = VE

Analysis of Quantitative Traits

We can cross 2 highly inbred parents to achieve the requirements needed to distinguish genetic and environmental variation

VP=VA+VD+VI+VE+VGE

VG=VA+VD+VI

22

22

21221 Pp

ePP

E

SSS

VVV

VE=(VP1+VP2+VF1)/3

Analysis of Quantitative Traits

Example: Inbred corn lines:

P1: A1A1B1B1C1C1 x A2A2B2B2C2C2

F1: A1A2B1B2C1C2

F2: segregates into many genotypes

VG = 0,So VP = VE

VP =VG + VE

Analysis of Quantitative Traits

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

VP (F2) – VP (F1) = VG

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 HeritabilityBroad-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

COVxy=1

))((

n

yyxx ii

为了便于计算，上式可写成

COVxy=1

(1

n

yxn

yx iiii ）

γ相关系数等于xy的协方差除以二者标准差的乘积

γ=SxSy

COVxy

表 6-6 虎螈体长和头宽的相关性体 长（ mm）x i

xx i 2)( xx i 头 宽(mm ）yi

yy i 2)( yy i x i yi

72.00 -7.92 62.67 17.00 -0.75 0.56 122462.00 -17.92 321.01 14.00 -3.75 14.06 86886.00 6.08 37.01 20.00 2.25 5.06 172076.00 -3.92 15.34 14.00 -3.75 14.06 106464.00 -15.92 253.34 15.00 -2.75 7.56 96082.00 2.08 4.34 20.00 2.25 5.06 164071.00 -8.92 79.51 15.00 -2.57 7.56 106596.00 16.08 258.67 21.00 3.25 10.56 201687.00 7.08 50.17 19.00 1.25 1.56 1653103.00 23.08 532.84 23.00 5.25 27.56 23.6986.00 6.08 37.01 18.00 0.25 0.06 154874.00 -5.92 35.01 17.00 -0.75 0.56 1258

ix =

959.00

2)( xx i

=1686.92 iy=213.00

2)( yy i

=94.25 ii yx= 17385

92.7912/959/ nxx i

75.1712/213/ nyy i

35.15311/92.1686)1/()(S 22x nxx i

38.1235.153SS 2xx

75.811/25.94)1/()(S 22y nyy i

93.2SS 2yy

协 方 差

C O V x y =1

(1

n

yxn

yx iiii ）=

97.32112

12

21395917385

相 关 系 数 γ =C O V x y / S x S y = 3 2 . 9 7 / （ 1 2 . 3 8 × 2 . 9 3 ）= 0 . 9 1

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）

P

G2

V

VH

P

A2

V

Vh

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 F 2 的 均 值 和 方 差 计 算

基 因 型 f i x I f i x i f i x i2

A A 1 / 4 a 1 / 4 a 1 / 4 a 2 A a 1 / 2 d 1 / 2 a 1 / 4 d 2 a a 1 / 4 - a - 1 / 4 a 1 / 4 a 2 合 计 1 dx i dxf ii 2

1 222

2

1

2

1daxf ii

表 6 - 9 B 1 的 平 均 数 和 遗 传 方 差 的 计 算 f i X I f i X i f i X i

2 A A 1 / 2 a 1 / 2 a 1 / 2 a 2 A a 1 / 2 d 1 / 2 d 1 / 2 d 2 合 计 1 dax i )(

2

1daxf ii )(

2

1 222 daxf ii

表 6 - 1 0 B 2 的 均 值 和 遗 传 方 差 的 计 算 f i x i f i x i f i x i

2 A a 1 / 2 d 1 / 2 d 1 / 2 d 2 a a 1 / 2 - a - 1 / 2 a 1 / 2 a 2 合 计 1 adx i )(

2

1adxf ii )(

2

1 222 daxf ii

ΣX n n

(ΣX)2

n n (ΣX)2

n V F2 = Σfx 2- 1 1 1 2 2 4

1 1 n n

2 4 若有几对基因 2 4

VA= a12+a2

2+ +an2 VD= d1

2+d22+ +dn

2

( )2Σx2-nS2=

ΣX2-S2=

S2 = ΣX2-

(Σfx)2

n

= a2+ d2– d2

a2+= d2 a2+= d2

在多对基因时

1 1 2 4 1 1 2 3 1/2VA+1/4VD 1/2VA

1/2VA+1/4VD+VE 1/2VA+1/4VD+VE

(ΣX)2 1 1 1 n 2 4 4 (ΣX)2 1 1 1 n 2 4 4

回交一代的平均方差 =1/2 （ S12+S2

2） =1/4 a2+1/4 d2

=1/4VA+1/4VD+VE

VF2= VA + VD + VE

VE= (VP1+VP2) or VE= (VP1+VP2+ VF1)

H2= h2 =

S12= ΣX2― = (a2+d2) ― (a+d)2 = (a-d)2 = ¼(a2-2ad+d2)

S22= ΣX2― = (a2+d2)― (d-a)2 = (a+d)2 =

¼(a2+2ad+d2)

将 A a 与 A A 回 交 ， 回 交 1 代 B 1 的 频 度 和 观 察 值 如 表 6 - 9 所 示 ， 根 据 表 6 - 9 ， B 1 差 异 部 分 的 均 值

)(2

11 daxfB ii

B 1 的 方 差 22222

221 )(

4

1)(

4

1)(

2

1)(dadada

n

xxS i

i

将 F 1 A a 与 a a 回 交 ， 回 交 一 代 B 2 的 频 度 和 观 察 值 如 表 所 示 。 B 2 差 异 部 分 的 均 值

)(2

12 adxfB ii

B 2 的 方 差 22222

222 )(

4

1)(

4

1)(

2

1)(adadda

n

xxS i

i

回 交 一 代 的 平 均 方 差 为 2222

21 4

1

4

1)(

2

1daSS

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

2

21222

2)](

2

1[2

21

F

BBF

PP

a

B

A

V

VVV

V

a

V

S

V

Vh

1 2

1 1 1 1 2 4 4 4 1 4 1 2 1 1 2 4 1 2 VF2

若 VF2 ― （ VB1+VB2 ）

= ( VA+ VD+VE) ― ( VA + VD+VE)

= VA

h2=VA

VA+ VD + VE

2 × [ VF2 ― （ VB1+VB

2 ） ] h2 =

（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 总双生子对数

×100%

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

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

H2=

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 improvementin closed populations

improvement is gradual but consistent .

Response to selection

R = h2 x S

where

R is the responseh2 is the heritabilityS is the selection differential .

Response to selection

average daily gain in pigs. h2 = .4 herd mean = 1.6 lb/dayselected 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. h2 = .5

all boars = 1.0 in all gilts = 1.2 inselect boars = .8 in select gilts = 1.1 in

selection differential

males .8 - 1.0 = -.2 infemales 1.1 - 1.2 = -.1 inoverall (-.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 P

where

R is the responsei is intensity or

standardized selection differentialh2 is the heritabilityP is the phenotypic standard deviation .

Standardized Selection Differentials

Proportion saved i.70 .50

.50 .80

.25 1.27

.10 1.76

.05 2.06

.01 2.67

.001 3.37

weaning weight in beef cattleh2 = .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

responseR = i x h2 x P = 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 P .

Heritability and Selection Response

M2= M + h2 (M1-M)

h2 = M2 - M M1 - M

Response is also related to heritability

R = h2SWhere, (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

Correlation Heritability (H2)

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

Control Strain

F1 Hybrids

Resistant Strain

Example of Polygenic Inheritance: Insecticide Resistance in Drosophila

Resistant Strain

F1 Hybrids

Control Strain

Nilsson-Ehle’s Model

Genotype

“Redness” score Color

AABBCC 6 Dark red

aabbcc 0 No red (white)

AaBbCc 3 Medium red

AABbcc 3 Medium red