PowerPoint Presentation Materials
to accompany
Genetics: Analysis and Principles
Robert J. Brooker
CHAPTER 24
QUANTITATIVE GENETICS
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24.1 QUANTITATIVE TRAITS

A quantitative trait is any trait that varies measurably in a
given species
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24-3
Quantitative Traits Exhibit a
Continuum of Phenotypic Variation

Quantitative traits show a continuum of
variation within a group of individuals

Quantitative traits do not naturally fall into a
small number of discrete categories
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24-5
Quantitative traits can be described by using a
frequency distribution
The trait is divided arbitrarily into a number of
discrete phenotypic categories
Figure 24.1 Normal distribution of a quantitative trait
24-7
Statistical Methods Evaluate A
Frequency Distribution

Mean:
SX
X =
N

where X is the mean
S X is the sum of all values in the group
N is the number of individuals in the group
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24-8

A more generalized form of this equation is used
S fi Xi
X =
N

where
X is the mean
S fiXi is the sum of all values in the group; each value is
multiplied by its frequency in the group
N is the number of individuals in the group

E.g., A bushel of corn had ears with the following lengths:
15, 14, 13, 14, 15, 16, 16, 17, 15 and 15
4(15) + 2(14) + 13 + 2(16) + 17
X =
10
X = 15 cm
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24-9
In genetics we are often interested in the amount of
phenotypic variation that exists in a group
 A common way to evaluate variation within a population is
a statistic called variance

S fi (X – X)2
VX =
N–1

where, VX is the variance
X – X is the difference between
each value and the mean
N is the number of observations
Degrees of
freedom

E.g., Lengths of ears of corn
4(15 - 15)2 + 2(14 - 15)2 + (13 - 15)2 + 2(16 - 15)2 + (17 VX = 15)2
10 – 1
0+2+4+2+4
VX =
9
12 cm2
VX =
9
VX = 1.33 cm2
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24-10

Variances are very important in the analysis of quantitative
traits because they are additive under certain conditions

The variances of genetic and environmental factors that contribute to a
quantitative trait can be added to predict the total variance for that trait

To gain an intuitive grasp of variation, we can take the square
root of the variance
 This statistic is called the standard variation

Again using the same example in corn
SD = √ Vx
SD = √ 1.33 = 1.15 cm
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24-11
~68%
~95%
~99.7%
Less than 0.3% of the individuals will have values that are
more or less than three standard deviations from the mean
Figure 24.2 The relationship between the standard deviation
and the proportions of individuals in a normal distribution
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24-12
Table 24-2
Copyright © 2006 Pearson Prentice Hall, Inc.
Some Statistical Methods Compare
Two Variables to Each Other

In many biological problems, it is useful to compare
two different variables

For example, we may wish to

1. Compare the occurrence of two phenotypic traits


2. Compare between traits and environmental factors


Do obese animals have larger hearts?
Is heavy body weight more prevalent in colder climates?
3. Compare between traits and genetic relationships

Do tall parents tend to produce tall offspring?
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24-13


To gain insight into such questions, a statistic known
as correlation is often applied
To calculate this statistic, we need first to determine
the covariance

Describes the degree of variation between two variables
within a group
CoV(X,Y) =

S [ (X – X) (Y – Y) ]
N–1
where

X are the values for one variable and X is the mean value in the group

Y are values for another variable and Y is the mean value in that group

N is the total number of pairs of observations
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24-14

As an example, let’s consider the weight of cattle at 5
years of age between parents and offspring


A farmer might be interested in this relationship to
determine if genetic variation plays a role in cattle weight
The data on the next slide show the 5-year weights
for 10 different pairs of cows and their “daughters”
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24-15
Mother’s
weight (kg)
Offspring’s
weight (kg)
570
568
– 26
– 30
780
572
560
– 24
– 38
912
599
642
3
44
132
602
580
6
– 18
– 108
631
586
35
– 12
– 420
603
642
7
44
308
599
632
3
34
102
625
580
29
– 18
– 522
584
605
– 12
7
– 84
575
585
– 21
– 13
273
X = 596
Y = 598
SDX = 21.1
SDY = 30.5
CoV(X,Y)
(X – X) (Y – Y)
(X – X) (Y – Y)
S = 1,373
S [ (X – X) (Y – Y) ]
1,373
=
=
= 152.6
N–1
10 – 1
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24-16

The strength of association between two variables
can be evaluated by the correlation coefficient (r)
r(X,Y) =

CoV(X,Y)
SDXSDY
This value ranges between +1 and -1 and indicates
how two factors can vary in relation to each other

If r > 0


If r = 0


As one factor increases, the other factor will increase with it
The two factors are not related
If r < 0

As one factor increases, the other factor will decrease
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24-17
r(X,Y) =

SDXSDY
152.6
=
= 0.237
(21.1)(30.5)
This is a positive correlation between the 5-year weights of
mother and offspring


CoV(X,Y)
This suggests that heavy mothers have heavy offspring and lighter
mothers have lighter offspring
However, now we must evaluate whether the r value
represents a true association between the variables
or whether it is simply due to chance
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24-18

To accomplish this, we can test the hypothesis that
there is no real correlation (i.e., the null hypothesis)


The r value differs from zero only as a matter of chance
Like the chi square value, the significance of the
correlation coefficient is directly related to


Sample size
Degrees of freedom


In this case, df = N – 2
 Which is one less than the degrees of freedom of variance
Table 24.2 shows the relationship between the r
values and df at the 5% and 1% significance levels
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24-19
24-20

Consider the correlation we have just calculated for
5-year weights of mother and offspring
 r = 0.237 and N = 10
Therefore, df = 10 – 2 = 8

According to Table 24.2, it is fairly likely that this
value could have occurred as a matter of random
sampling error


Therefore, we cannot reject the null hypothesis
We cannot conclude that the positive correlation is due
to a true association between the weights of mother and
offspring
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24-21

If a correlation of 0.237 was observed for N = 1000



The value would be significant at the 1% level
Therefore we can reject the null hypothesis
We can conclude that there is a real association
between the weights of mothers and their offspring
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24-22
24.2 POLYGENIC INHERITANCE

Most quantitative traits are polygenic and exhibit a
continuum of phenotypic variation

Polygenic inheritance refers to the transmission of
traits that are governed by two or more genes

The locations on chromosomes that affect the
outcome of quantitative traits are called quantitative
trait loci (QTLs)

QTLs may contain many genes

Some or all of which may affect quantitative traits
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24-24
Polygenic Inheritance and
Environmental Factors

The first demonstration that continuous variation is
related to polygenic inheritance occurred in 1909

The Swedish geneticist Herman Nilsson-Ehle studied the
inheritance of red pigment in the hull of wheat Triticum
aestivum
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24-25

Nilsson-Ehle performed the following cross



P
F1
F2
True-breeding red X true-breeding white
Intermediate red
Great variation in redness:



As shown in Figure 24.3b, Nilsson-Ehle discovered that
the colors fell into a 1:4:6:4:1 ratio
He concluded that this species is diploid for two different
genes that control hull color


White, light red, intermediate red, medium red, dark red
Each gene exists in two alleles: red or white
He hypothesized that these two loci must contribute
additively to the color of hull
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24-26
Figure 24.3
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24-27


Many polygenic traits are difficult or impossible
to categorize into several discrete genotypic
categories
This is especially true when


1. The number of genes controlling the trait
increases
2. The influence of the environment increases
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24-28

Figure 24.4 illustrates how genotypes and phenotypes may
overlap for polygenic traits

Here, the environment (soil, sunlight, etc.) may affect the phenotypic
outcome of a trait in a plant, namely seed weight
Separate, well-defined categories
One gene
Figure 24.4
More phenotypic variation
in each category
Effect of environment
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24-29
Effect of # of genes
• If there is no
environmental effect,
the number of
contributing genes
can be estimated by
either:
– the number of
phenotypic classes
or
– by the fraction of the
total population of an
extreme phenotype
class
Nearly all individuals fall into a
single category that
corresponds to their genotype
Different genotypes
have regions of
overlapping phenotypes
Three genes
Figure 24.4
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24-30
Phenotypic expression of some
quantitative traits display dependence on
threshold effects
Experiment 24A: Polygenic Inheritance
Explains DDT Resistance in Drosophila

In 1957, James Crow conducted one of the earliest
studies to link quantitative to discontinuous genes


Crow, who was interested in evolution, spent time studying
insecticide resistance in Drosophila
He noted:


“Insecticide resistance is an example of evolutionary change, the
insecticide acting as a powerful selective sieve for concentrating
resistant mutants that were present in low frequencies in the
population.”
Crow’s aim was to determine the genetic basis for
insecticide resistance in Drosophila melanogaster
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24-31

Many mutations were known in this species, and
thus these can serve as genetic markers

The general strategy in identifying QTLs is this:



Cross two strains that differ in genetic markers and in the
quantitative trait of interest
Backcross the F1 offspring to the parental strains
This produces a population of F2 offspring that differ with
regard to their combinations of parental chromosomes


Most individuals will contain a few chromosomes from one
parental strain and the rest from the other strain
The genetic markers on the chromosome provide a way
to determine which chromosome came from which parent
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24-32



At the start of the experiment,
it is not known if alleles
affecting DDT resistance are
on this chromosome
Figure 24.5

M = minute bristles
m = normal bristles
R = rough eye
r = smooth eyes
The use of gene markers to map a QTL affecting DDT resistance
24-33
Possible chromosome 3
combinations in F2 offspring
(in the absence of crossing over)
The backcross offspring can have both copies of chromosome 3 from the
DDT-resistant strain, both from the sensitive strain, or one of each
This can be discerned by the phenotypes of the F2 offspring
Figure 24.5
The use of gene markers to map a QTL affecting DDT resistance
24-34
The Hypothesis

DDT resistance is a polygenic trait
Testing the Hypothesis

Refer to Figure 24.6
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24-35
Figure 24.6
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24-36
Figure 24.6
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24-37
The Data
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24-38
Interpreting the Data

The data shows that each copy of the chromosomes
1 (X), 2, and 3 confers some insecticide resistance



Maximal resistance is obtained when chromosomes from
only the insecticide-resistance parental strain are present
DDT resistance was less than maximal, even when only a
single chromosome was derived from the sensitive strain
These results are consistent with the hypothesis

Resistance to DDT is a polygenic trait involving multiple
genes on chromosomes 1 (X), 2 and 3
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24-39
24.3 HERITABILITY

All traits of biological organisms are influenced by
genetics and the environment


A geneticist, however, can never actually determine
the relative amount of a quantitative trait that is
controlled by



This is particularly true with quantitative traits
1. Genetics
2. Environment
Instead, the focus is on how variation, both genetic
and environmental, will affect the phenotypic results
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24-45
Yield (bushels/acre)
Genotype
Year
Roughrider
Seward
Agassiz
1986
47.9
55.9
47.5
1987
63.8
72.5
59.5
1988
23.1
25.7
28.4
1989
61.6
66.5
60.5
1990
0.0
0.0
0.0
1991
60.3
71.0
55.4
1992
46.6
49.0
41.5
1993
58.2
62.9
48.8
1994
41.7
53.2
39.8
1995
53.1
65.1
53.
24.3 HERITABILITY

Heritability is the amount of phenotypic variation
within a group of individuals that is due to genetics

If all the phenotypic variation in a group was due to
genetic factors


If all the phenotypic variation was due to environmental
factors


Heritability would have a value of 1
Heritability would have a value of 0
In most cases, the heritability value lies between these
two extremes
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24-46
Phenotypic Variance

To study quantitative trait variation, we first have to
divide it into genetic and environmental components

We then have to assume the following

1. Genetic and environmental factors are the only two
components that determine a trait

2. These factors are independent of one another
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24-47

VT = VG + VE

VT is the total variance




VG is the relative amount of variance due to genetic factors
VE is the relative amount of variance due to environmental
factors
If VG is very high and VE is very low


The amount of variation at the phenotypic level
Genetics is more important in promoting variation
If VG is very low and VE is very high

The environment causes much of the phenotypic variation
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24-48

Consider a domesticated or experimental animal

VG and VE can be determined by comparing the variation in
traits between genetically-identical and -disparate groups

For example, geneticists have developed genetically
homogeneous strains of mice through inbreeding


In such an inbred strain of mice, VG = 0


These mice are monomorphic
 They are all homozygous for the same allele of a given gene
Therefore, all phenotypic variation is due to VE
The variance of a homogeneous population with regard to
weight, for example, can be compared to the variance of a
heterogeneous population
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24-49

The two populations could be raised under the same
environmental conditions and the weights measured

Let’s suppose we obtain the following results



In the case of the homogeneous mice, VT = VE (VG = 0)


VT = 0.30 sq oz for the group of genetically homogeneous mice
VT = 0.52 sq oz for the group of genetically heterogeneous mice
Therefore, VE = 0.30 sq oz
The VE of the heterogeneous population should also be
equal to 0.30 sq oz

Because both populations were raised under the same
environmental conditions
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24-50




VT = VG + VE
0.52 = VG + 0.30
VG = 0.22 sq oz
The results tell us that the phenotypic variance in
the genetically heterogeneous population is due to

In part, the environment


0.30 sq oz
In part, genetic variation in alleles that affect body weight

0.22 sq oz
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24-51
Broad versus Narrow Sense
Heritability

Heritability is the proportion of the phenotypic
variance that is attributable to genetic variation



HB2 = VG/VT
Where, HB2 is the heritability in the broad sense
VG is the variance due to genetics
VT is the total phenotypic variance
HB2 is called the broad sense heritability

It accounts for all genetic variation that may affect the phenotype
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24-52

The total genetic variance, VG, is composed of three
different genetic categories

VG = VA + VD + VI

VA is the variance due to additive alleles
VD is the variance due to alleles that follow a
dominant/recessive pattern of inheritance
VI is the variance due to genes that interact in an epistatic
manner



The heritability of a trait due to the additive effects of
alleles is called the narrow sense heritability

hN2 = VA/VT
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24-53

A common strategy to estimate narrow sense heritability:
 1. Measure a quantitative trait among groups of
genetically related individuals
 2. Use data to compute a correlation between the
individuals
 3. Calculate narrow sense heritability as
2
 hN = robs/rexp


robs is the observed phenotypic correlation between related
individuals
rexp is the expected correlation based on the known genetic
relationship

Note: For siblings, rexp = 0.5



For identical twins, rexp = 1.0
For parent-offspring relationship, rexp = 0.5
For uncle-niece relationship, rexp = 0.25
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24-54
Experiment 24B: Heritability of
Human Fingerprints

Fingerprints are inherited as a quantitative trait

Francis Galton was the first to study fingerprint
patterns

However, Kristine Bonnevie made the trait more
amenable to genetic studies

She developed a method for counting the number of ridges
within a human fingerprint
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24-55

Human fingerprints can be categorized as having an


Arch, loop or whorl
The primary difference among these is the presence of the
number of triple junctions, each known as a triradius
Figure 24.8
Human fingerprints and the ridge count method of Bonnevie
24-56

Bonnevie conducted a study on a small population


She found that ridge count correlations were relatively high in
genetically related individuals
Sarah Holt carried out a more exhaustive study of ridge
counts in a British population

In groups of 825 males or 825 females, the ridge count on all 10
fingers varied from 0 to 300


Mean value ~ 145 for males (S.D. = 51.1)
Mean value ~ 127 for females (S.D. = 52.5)
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24-57
The Hypothesis

Dermal ridge count has a genetic component

This experiment aims to determine the contribution of
genetics in the variation of dermal ridge counts
Testing the Hypothesis

Refer to Figure 24.9
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24-58
Figure 24.9
24-59
Interpreting the Data
Type of
Relationship
Number of
Pairs Examined
Correlation
Coefficient
Parent-child
810
0.48 ± 0.04*
0.96
Parentparent
200
0.05 ± 0.07
__#
Siblingsibling
642
0.50 ± 0.04
1.00
Identical
twins
80
0.95 ± 0.01
0.95
Fraternal
twins
92
0.49 ± 0.08
0.98
*± Standard error of the mean
# We cannot calculate a heritability value
because rexp is not known
Heritability
robs/rexp
Genetically unrelated
individuals have a
negligible correlation
for this trait
Genetically related
individuals have an
average heritability
value of 0.97
Fraternal and identical twins have
substantially different correlation coefficients
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24-61
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24-62
Heritability Values Are Relevant

Heritability is a widely misunderstood concept


It describes the amount of phenotypic variation due to
genetic variation for a particular population raised in a
particular environment
The terms variation, particular population and
particular environment cannot be overemphasized 


For example, the heritability for milk production may be
 0.35 in one cattle population and 0.1 in another
In addition, a heritability value of 1.0 only means that the amount of
variation within this group is due to genetics
 The environment may be quite important
 It is just not causing much variation within this very group
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24-63
Heritability Values Are Relevant

The role of genetics and environment in human
intelligence, is a topic that has been hotly debated

As a trait, intelligence is difficult to define or to measure


Although performance on IQ tests is a method of assessing
intelligence
Various studies have attempted to estimate a heritability for
IQ testing ability


The values have ranged from 0.3 to 0.8
0.6 is fairly common among many studies
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24-64
Heritability Values Are Relevant

It is important to consider what a value of 0.6 means, and
what it does not mean



It means that 60% of the variation in IQ testing ability is due to
genetic factors in a particular population in a particular environment
It does not mean that 60% of an individual’s IQ testing ability is due
to genetics and 40% is due to the environment
It cannot be overemphasized that heritability is
meaningless at the level of a single individual

Heritability is a populational value that pertains to variation
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24-65
Selective Breeding of Species Can
Alter Quantitative Traits Dramatically

Selective breeding is the modification of
phenotypes in plants and animal species of
economic importance

It is also called artificial selection


Natural selection is due to natural variation in
reproductive success
In artificial selection the breeder chooses individual with
traits that are desirable from a human perspective
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24-66

For centuries, humans have been practicing
selective breeding


The common breeds of dogs and cats have been obtained
by selective breeding strategies
Refer to Figure 24.10


The figure shows the very striking way in which selective breeding
can modify the quantitative traits in a species
Likewise, most of the food we eat has been obtained from
species modified profoundly by selective breeding


This includes products such as grains, fruit, vegetables, meat
Refer to Figure 24.11

The figure shows how six different plants were developed by selective
breeding of the wild mustard plant
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24-67
Fig. 24.10
Fig. 24.11a
Fig. 24.11b

The phenomenon that underlies selective breeding is
variation


Within a group of individuals, there may be allelic variation
that affects the outcomes of quantitative traits
The breeder chooses parents with desirable phenotypic
characteristics


Indeed, the selective breeder will often choose genetically
related individuals as the parental stock


These will pass on the advantageous alleles to their offspring
This is known as inbreeding
Figure 24.12 illustrates a common outcome when
selective breeding is performed for a quantitative trait
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24-68
Oil content
is ~ 19%


This experiment was begun in
1869! with 163 ears of corn
with an oil content of 4-6%
In each of about 80 successive
generations, corn plants were
divided into two groups


Oil content
is < 1%
Figure 24.12

1. Members with the highest
oil content
2. Members with the lowest
oil content
These groups were chosen as
the parents of the next
generation
Common results of selective breeding for a quantitative trait
24-69
Average
46 bristles

Average
40 bristles


Average
36 bristles
In this experiment flies were
selected based on their bristle
number
The starting group had an average
of 40 bristles for females and 35
bristles for males
In each of 8 successive
generations, flies were divided into
two groups


Average
30 bristles
Figure 24.12

1. Members with the highest
bristle number
2. Members with the lowest
bristle number
These groups were chosen as the
parents of the next generation
Common results of selective breeding for a quantitative trait
24-70

When comparing the curves made in Figure 24.12,
some general observations can be made

Quantitative traits are often at an intermediate value
in unselected populations


Therefore, artificial selection can increase or decrease the
magnitude of the trait
Nevertheless, there is a limit




After many generations, the population will eventually become
monomorphic for all or most of the desirable alleles in question
At this point, additional selective breeding will have no effect
At the start of the experiment the heritability for the trait is
fairly high
At the end, it is near zero
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24-71
Estimating the Offspring Phenotype
If we have determined the narrow sense heritability of a trait we can estimate
the phenotypic value for an offspring.
To = T + h2(T*-T) where
To = predicted offspring phenotype
T = population mean
h2 = narrow sense heritability
T* = midparent value [(Tf + Tm)/2]
Example:
T = 80 seeds/plant
Tf = 90 seeds/plant
Tm = 120 seeds/plant
T* = (90 +120)/2 = 105
h2 = 0.5
Then:
To = 80 + 0.5 (105-80)
To = 80 + 12.5
To = 92.5 seeds/plant

Realized heritability

The most common way to estimate narrow sense
heritability in a starting population
R
2
hN =
S

Where




Here
R = XO – X
S = XP – X

R is the response in the offspring
S is the selection differential in the parents
Where

X is the mean of the starting population
XO is the mean of the offspring

XP is the mean of the parents

So
hN2 =
XO – X
XP – X
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
24-72

Example:

An experiment is begun with a population of fruit flies in
which the average bristle number for both genders is 37.5.


The parents chosen from this population had an average bristle
number of 40
The offspring of the next generation had an average bristle number
of 38.7
hN2 =
hN2 =
XO – X
XP – X
38.7 – 37.5
40 – 37.5
=
1.2
2.5
= 0.48
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Thus, 48% of the
phenotypic
variation is due to
additive alleles
24-73