CHAPTER 23 Quantitative Genetics
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Transcript CHAPTER 23 Quantitative Genetics
Peter J. Russell
CHAPTER 23
Quantitative Genetics
edited by Yue-Wen Wang Ph. D.
Dept. of Agronomy,台大農藝系
NTU
遺傳學 601 20000
Chapter 23 slide 1
1. Traits with a few distinct phenotypes are
discontinuous traits. There is usually a simple
relationship between the genes responsible and
formation of the phenotype.
2. Often, penetrance, expressivity, pleiotropy,
epistasis and environmental factors are involved
in producing a continuous distribution of
phenotypes (continuous traits). Quantitative
genetics is used to characterize continuous traits.
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Chapter 23 slide 2
Fig. 23.1 Discontinuous distribution of shell color in the snail Cepaea nemoralus from
a population in England
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 3
Fig. 23.2 Distribution of birth weight of babies (males + females) born to teenagers in
Portland, Oregon, in 1992
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 4
The Nature of Continuous Traits
1. Statistical study of continuous traits began with human traits such as
height, weight and mental traits, even before Mendel’s principles were
understood.
2. Galton and Pearson (late 1800s) showed that these traits are statistically
linked between parents and offspring, but could not determine the mode
of transmission.
3. Johannsen (1903) showed that continuous variation in bean seed weight
is partly genetic and partly environmental.
4. Nilsson-Ehle proposed that continuous variation in wheat results from
multiple genes segregating by Mendelian principles.
5. Fisher demonstrated that mathematical models of Mendelian and
population genetics also apply to traits controlled by multiple gene loci.
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Chapter 23 slide 5
Why Some Traits Have Continuous Phenotypes
1. Multiple traits arise in several ways:
a. When a trait is influenced by many loci (polygenic), a range of
phenotypes results from the numerous genotypes. For example:
i. When a single locus with two alleles determines a trait, there are
three possible genotypes (AA, Aa and aa).
ii. With two loci, each with two alleles, there are nine possible
genotypes.
iii. In general, the number of genotypes is 3n, where n is the number
of loci with two alleles. If the number of alleles is higher, the
number of genotypes will be even larger.
iv. If every genotype in a polygenic trait produces a different
phenotype and the differences between phenotypes are slight, the
trait appears to be continuous.
v. More often, several genotypes produce the same phenotype for a
polygenic trait. Reasons for this include:
(1) Dominance, producing the same phenotype in both
heterozygous and homozygous dominant individuals.
(2) Epistasis, resulting from control of the expression of other
loci.
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Chapter 23 slide 6
2. A range of phenotypes is also produced when environmental
factors affect the trait. Each genotype will have a range of
possible phenotypes, the norm of reaction.
3. Most traits are influenced by both multiple genotypes and
environmental factors, and are thus multifactorial. The rules
of transmission genetics and gene function still apply.
Understanding the role of each gene is difficult, and so
quantitative genetics is employed to analyze continuous
traits.
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Chapter 23 slide 7
Questions Studied in Quantitative Genetics
1. What role is played by genetics, and what role by environment?
2. How many genes are involved in producing phenotypes of the
trait?
3. Do some genes play a major role in determining phenotype,
while others modify it only slightly, or are the contributions equal?
4. Do the alleles interact with each other to produce additive
effects?
5. What changes occur when there is selection for a phenotype,
and do other traits also change?
6. What method of selecting and mating individuals will produce
desired phenotypes in the progeny?
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Chapter 23 slide 8
Statistical Tools
1. Genes are always expressed in an environmental context,
and without genes there would be nothing to express. The
nature vs. nurture question provides an opportunity to
examine the relative contributions of both.
2. How much of a variation in phenotype (VP) is due to
genetic variation (VG) and how much to environmental
variation (VE)? This can be expressed: VP = VG + VE.
3. To work this equation, variation must be measured and
then partitioned into genetic and environmental
components.
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Chapter 23 slide 9
Samples and Populations
1. It is difficult to collect data for each individual in a
large population.
2. Sampling of a subset is an alternative method.
a. The sample must be large enough to minimize chance
differences between the sample and the population.
b. The sample must be a random subset of the population.
3. Birth weight in humans is an example (Figure 23.2).
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Chapter 23 slide 10
Distributions
1. Phenotypes are not easily grouped into classes when a continuous range
occurs. Instead, a frequency distribution is commonly used, showing
the proportion of individuals that fall within a range of phenotypes.
2. In a frequency distribution, the classes consist of specified ranges of the
phenotype, and the number of individuals in each class is counted.
a. An example is Johannsen’s study of seed weight in the dwarf bean
(Phaseolus vulgaris).
b. A histogram is used to show the distribution of individuals into
phenotypic classes.
c. Tracing the outline of the histogram gives a curve characteristic of the
frequency distribution.
3. Continuous traits often show a bell-shaped curve (normal distribution),
due to influences of multiple genes and environmental factors.
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Chapter 23 slide 11
Fig. 23.3 Frequency histogram for bean weight in Phaseolus vulgaris
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Chapter 23 slide 12
The Mean
1. Frequency distribution of a phenotypic trait can be summarized with
two statistics, the mean and the variance.
2. The mean (average) represents the center of the phenotype distribution,
and is calculated simply by adding all individual measurements and
then dividing by the number of measurements added.
3. An example is body length of spotted salamanders (Table 23.2).
4. Another example is East’s study of tobacco flower length in genetic
crosses.
a. He crossed a short-flowered strain (mean length of 40.4 mm) with a
long-flowered strain (mean length of 93.1 mm).
b. The F1 progeny (173 plants) had a mean flower length of 63.5 mm.
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Chapter 23 slide 13
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Chapter 23 slide 14
The Variance and the Standard Deviation
1. Variance is the measure of how much the individual measurements
spread out around the mean (how variable they are).
a. Two sets of measurements may have the same mean, but different variances
(Figure 23.4).
b. The variance (s2) is the average squared deviation from the mean. To
calculate s2:
i. Subtract the mean from each individual measurement.
ii. Square the difference for each.
iii. Add the squared values.
iv. Divide by the number of original measurements minus 1 (n - 1).
c. Standard deviation is used more often than variance, because it shares the
same units as the original measurements (rather than units2 as in variance).
Standard deviation is the square root of the variance.
d. Table 23.2 shows sample calculations for variance and standard deviation.
i. A broad curve indicates a large variability in the measurements and a
large standard deviation.
ii. A narrow curve implies little variability
and a遺傳學
small601
standard
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20000 deviation.
Chapter 23 slide 15
Fig. 23.4 Graphs showing three distributions with the same mean but different
variances
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 16
e. When mean and standard deviation are known, a theoretical normal
distribution is specified. Its shape is shown in Figure 23.5. In a
theoretical normal distribution:
i. One standard deviation above or below the mean (± 1s) includes
66% of the individual observations.
ii. Two standard deviations (± 2s) includes 95% of the individual
values.
iii. Three standard deviations (± 3s) includes > 99% of the individual
values.
f. Analysis of variance is a statistical technique used to help partition
variance into components.
2. Variance and standard deviation provide information about the
phenotypes of a group. In the tobacco flower example:
a. The original cross of short-flowered with long-flowered produced an F1
with a mean flower length of 63.5 mm, intermediate to the parents.
b. The F2 had a mean of 68.8 mm, very similar to the F1. But the F2 had a
variance of 42.2 mm2, while F1 variance was only 8.6 mm2, indicating
that more phenotypes occur among the F2 than among the F1.
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Chapter 23 slide 17
Fig. 23.5 Normal distribution curve showing proportions of the data in the
distribution that are included within certain multiples of standard deviation
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 18
Correlation
1. Traits in individuals are often correlated, due to the pleiotropic effects
of genes and environmental factors. Height and weight, for example,
are aspects of a more general trait size.
a. When traits are correlated, change in one is associated with change in the
other (e.g., human leg and arm length are usually correlated).
i. Correlation coefficient measures the strength of association between
two variables in the same individual or experimental unit. In the arm
and leg example, x = arm length, and y = leg length:
(1) Obtain the covariance of x and y (the variance shared by both
traits) by taking the deviation from the mean for each.
(2) Take the product of each pair of x and y values.
(3) Add all products together.
(4) Divide the sum by n - 1 to give the covariance of x and y where
n= the number of xy pairs.
(5) The correlation coefficient (r) is then obtained by dividing the
covariance by the product of the standard deviations of x and y
(Table 23.3).
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Chapter 23 slide 19
台大農藝系 遺傳學 601 20000
Chapter 23 slide 20
ii.Correlation coefficient is a standardized measure of
covariance that can range from -1 to +1 (Figure 23.6).
(1) A positive correlation coefficient means that an
increase in one variable is associated with an increase in
the other variable.
(2) A negative correlation coefficient means that an
increase in one variable is associated with a decrease in
the other variable.
(3) A correlation coefficient near 0 indicates a weak
relationship between the variables.
b. A correlation between variables does not imply a cause-effect
relationship.
c. Correlation is not the same thing as identity. Variables may be
highly correlated yet have very different values.
iActivity: Your Fate in Your Hands?
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Chapter 23 slide 21
Fig. 23.6 Scatter diagrams showing the correlation of x and y variables
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 22
Regression
1. Regression analysis is used to determine the precise relationship
between variables (Figure 23.7).
a. A graph is plotted for the individual data points, one on the x axis and
the other on the y. The regression is the line that best fits the points (the
squared vertical distance from the points to the regression line is
minimized).
b. The regression line can be represented with the equation y = a + bx,
where:
i. x and y are values of the two variables.
ii. b is the slope (regression coefficient).
iii. a is the y intercept (the expected value of y when x is 0).
c. The slope shows how much of an increase in the y variable is associated
with a unit increase in the x variable.
2. Regression analysis is a common method for measuring the extent to
which variation in a trait is genetically determined.
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Chapter 23 slide 23
Fig. 23.7 Regression of sons’ height on fathers’ height
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 24
Fig. 23.8 Regression lines with different slopes
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 23 slide 25
Analysis of Variance
1. Analysis of variance (ANOVA) determines if differences in means are
significant, and divides the variance into components.
a. It can tell whether a variation between two groups is likely to be due to
chance, rather than to a true difference.
b. ANOVA can also determine how much of a difference is due to a factor
like genetics or the environment, by partitioning the variance.
2. Details of the calculations are beyond the scope of this text.
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Chapter 23 slide 26
Polygenic Inheritance
1. In the experiments described here, it was not clear at first
how these traits were inherited, only that their pattern of
inheritance differed from that seen with discontinuous
traits.
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Chapter 23 slide 27
Inheritance of Ear Length in Corn
1. Emerson and East (1913) experimented with two pure-breeding strains of corn.
a. Each strain shows little variation in ear length.
i. The Black Mexican sweet corn variety has short ears (mean length 6.63 cm)
with a standard deviation (s) of 0.816.
ii. Tom Thumb popcorn has long ears (mean length 16.80 cm), and s = 1.887.
b. The two strains were crossed, and the F1 plants interbred (Figure 23.9).
i. The mean ear length in the F1 is 12.12 cm, approximately intermediate, and s =
1.519.
ii. Since both parents were true-breeding, all F1 plants should have the same
heterozygous genotype, and any variation in length would be due to
environmental factors.
iii. The mean ear length of the F2 is 12.89 cm, very similar to the F1, but in the F2,
s = 2.252, reflecting its greater variability.
iv. It is expected that the environment would have the same effect on the F2 that it
had on the P and F1 plants, but it would not be expected to have more effect.
v. The increased variability in the F2 most likely results from its greater genetic
variation.
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Chapter 23 slide 28
Fig. 23.9 Inheritance of ear length in corn
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 29
2. Aside from the environmental influence, four observations emerge that
apply generally to similar quantitative-inheritance studies:
a. The F1 will have a mean value for the trait intermediate between the
means of the two true-breeding parental lines.
b. The mean value in the F2 is about the same as that for the F1.
c. F2 shows more variability around the mean than the F1 does.
d. Extreme values for the trait in the F2 extend farther into the parental
range than the extreme values for the F1.
3. The data are not consistent with a single Mendelian locus, because the
discrete classes expected do not occur.
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Chapter 23 slide 30
Polygene Hypothesis for Quantitative
Inheritance
Animation: Polygenic Hypothesis for Wheat Kernel Color
1. The simplest explanation for the corn ear length data is the polygene
(multiple-gene) hypothesis for quantitative inheritance, which says that
these traits are controlled by many genes.
2. Nilsson-Ehle (1909) studied kernel color in wheat. He crossed truebreeding red kernel wheat with true-breeding white.
a. The F1 were all the same intermediate color between red and white.
Incomplete dominance was a possibility.
b. Intercross of F1 gave an F2 with four discrete shades of red
(ranging from parental red to very light) plus white, in a ratio
of 1 : 4 : 6 : 4 : 1 (15 red : 1white). 1⁄16 of the F2 are parental red, and
1⁄16 are parental white.
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Chapter 23 slide 31
c. A 15:1 pattern is typical of a trait that results from the interactions of the
products of two pairs of alleles. Both genes affect the same trait, and so are
duplicate genes.
i. In the case of wheat, there appear to be two pairs of alleles that
segregate independently. Both control red pigment.
ii. Alleles R (red) and C (crimson) result in red pigment, while r and c do
not produce pigment.
iii. In the cross RRCC (dark red) 3 rrcc (white), the F1 are all RrCc
(intermediate red).
iv. Interbreeding the F1 produces an F2 with genotypes distributed as in a
typical dihybrid cross.
v. The numbers in the phenotypic ratio are the same as the coefficients in
the binomial expansion of (a + b)4.
vi. The range of phenotypes results from incomplete dominance of the
alleles, with each copy of R and C acting as a contributing allele to
produce more red pigment. The r and c alleles are noncontributing
alleles.
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Chapter 23 slide 32
d. Some F2 populations show only three phenotypic classes (3 red : 1 white),
while others show a ratio of 63 red : 1 white, with many shades of red.
i. The 3:1 ratio is consistent with a single-gene system with two
contributing alleles.
ii. The 63:1 ratio is consistent with a polygene series with six
contributing alleles, (a + b)6.
3. This multiple-gene hypothesis has been applied to other traits, including
corn ear length. It proposes that some attributes of quantitative
inheritance result from the action and segregation of a number of allelic
pairs (polygenes), each making a small but additive contribution to the
phenotype.
4. Quantitative trait inheritance appears to be complex, however, and
molecular aspects are often not yet well understood.
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Chapter 23 slide 33
Quantitative Trait Loci
1. The application of statistical analysis to polygenic inheritance has been
powerful, but it is limited by lack of specific knowledge of the genes
involved.
a. Traditional methods are of only limited use, due to environmental
effects and the action of other genes.
b. Genomic segments correlated with phenotypic variation, quantitative
trait loci (QTLs), may be identified.
i. Typically, inbred lines with different phenotypes (homozygotes for
different alleles) are crossed, producing an F1 that is heterozygous
at most loci.
ii. Crossing the F1 either to parental lines or itself will increase
phenotypic variation as segregation is increased.
iii. The F2 is analyzed for marker genotypes that correlate with
phenotypic variation. The number of markers available and the
tightness of linkage between the QTL and the marker are factors in
this analysis.
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Chapter 23 slide 34
2. The number of genes involved in quantitative traits, the size of their
effects and their location on the genome are being investigated using
this technique.
a. Identification of QTLs responsible for differences between species
have emerged from this work.
b. In plants, important species differences include the suites of traits that
attract pollinators (color, shape, odor, nectar rewards). Monkeyflowers
are an example, with pollinators ranging from hummingbirds to bees,
and corresponding changes in traits (Figure 23.11).
i. Mimulus cardinalis has red color, deep nectar tubes with abundant
nectar and reflexed petals. It is a hummingbird pollinated species.
ii. M. lewisii has pink flowers, broad petals and little nectar reward. It
is a bee pollinated species.
iii.Crosses between these species show that many genes are involved,
some with large effects, and many of the QTLs are physically close
in the genome, indicating genetic correlation.
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Chapter 23 slide 35
Fig. 23.11 QTL maps for 12 floral traits in monkeyflowers
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 36
Heritability
1. Heritability is the proportion of a population’s phenotypic variation
attributable to genetic factors.
2. Continuous traits are often determined by multiple genes and by
environmental factors.
a. Many ecologically important traits (e.g., body size, fecundity, development
rate) are polygenic, and understanding them will contribute to knowledge
of how natural populations evolve.
b. Understanding the roles of genetics and environment in human health (e.g.,
blood pressure, birth weight) will lead to better health care.
c. Human social behaviors (e.g., alcoholism, criminality) may also have
genetic components, and scientific information has the potential to be
misused in constructing social policy.
3. Heritability can be divided into two types, broad-sense and narrowsense. To assess heritability:
a. Measure the variation in the trait.
台大農藝系 遺傳學
601 20000causes.
Chapter 23 slide 37
b. Partition the variance into components attributable
to different
Components of the Phenotypic Variance
1. Phenotypic variance (VP) is the measure of all variability observed for a
trait.(Figure 23.12)
a. The portion of phenotypic variance caused by genetic factors is the genetic
variance (VG).
b. Nongenetic sources of variation (e.g., temperature, nutrition, parental care)
constitute environmental variance (VE).
c. The relationship is VP = VG + VE.
2. The partitioning of phenotypic variance is complex, and VG and VE
may covary, so that their sum is not the same as VP, and another term,
COVG,E, is needed.
a. For example, milk production in cows is influenced by both genetics and
nutrition.
b. If a farmer provides progeny of good milking cows with less food than
progeny of poor milking cows, a covariance between genes and
environment is produced.
c. In that case, the variance in milk production increases beyond that expected
台大農藝系 遺傳學 601 20000
Chapter 23 slide 38
if genes and environment were acting independently.
Fig. 23.12 Hypothetical example of the effects of genes and environments on plant
height
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 39
3. Knowledge that there is a genetic component to a trait does
not allow accurate predictions about offspring. Another
analysis method for phenotypic variance, genotype-byenvironment (G X E) interaction, is needed.
a. G X E variance exists when the relative effects of the genotypes
differ among environments
b. An example is temperature affecting plant height:
i. In a cold environment, height of genotype AA plants averages
40 cm, while those with genotype Aa are 35 cm.
ii. In a warm climate, genotype AA is 50 cm tall, while genotype
Aa is now 60 cm.
iii. Both genotypes grow taller in warm temperatures (an
environmental effect). There is also a genetic effect, but it
depends on the environment. Genetic-environmental interaction
is represented by VGx E.
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Chapter 23 slide 40
4. Phenotypic variance is represented by the equation:
VP = VG + VE + 2COVG,E + VG x E
a. Phenotypic variation arises from:
i. Genetic variation (VG).
ii. Environmental variation (VE).
iii. Genetic-environmental covariation (COVG,E).
iv. Genetic-environmental interaction (VG x E).
b. Relative contributions of each factor depend on the genetic
composition of the population, specifics of the environment and
the manner in which the genes interact with the environment.
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Chapter 23 slide 41
5. Genetic variance (VG) can be subdivided into three components arising from
different gene actions and interactions between genes.
a. Some genetic variance results from average effects of the different alleles, creating
additive genetic variance (VA). Example:
i. Allele g contributes 2 cm to plant height, while G contributes 4 cm.
ii. A gg homozygote would receive 4 inches of height, a Gg heterozygote 6 inches,
and a GG homozygote 8 inches.
iii. This effect is added to the height effects produced by alleles at other loci.
iv. Nilsson-Ehle’s experiments with wheat kernel color are another example of
additive genetic variance.
b. Dominance variance (VD) occurs when one allele masks the effect of another allele
at the same locus, and prevents the effects from being strictly additive.
i. If G is a dominant allele and g a recessive one, the Gg genotype would make the
same contribution to phenotype as GG.
ii. As dominance decreases, genotypic differences become phenotypic differences,
turning dominance variance into additive genetic variance.
c. Epistatic interactions occur among alleles at different loci, adding another source of
genetic variation, the epistatic or interaction variance (VI).
d. Genetic variance is the sum of these three factors. (VG = VA + VD + VI)
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Chapter 23 slide 42
6. The environmental component of variance can also be partitioned.
a. General environmental effects (VEg) include factors like temperature and
nutrition, resulting in fairly irreversible differences among individuals.
b. Special environmental effects (VES) result in immediate changes of
phenotype (e.g., skin pigment due to sun exposure) and are often
reversible.
c. Environmental effects shared by members of a family (VECf) can be
mistaken for genetic influences, because they contribute to differences
among families (e.g., insects that sequester host plant compounds for
their own defense).
d. Maternal effects (VEm) are a common category of VECf (e.g., birth
weight and weight gain due to milk quality and volume in humans).
7. It is difficult to quantify all of these influences simultaneously, and
usually assumptions are made about some factors. In general, variation
due to nature and nurture is summarized by the equation:
VP = VA+VD+VI+VEg+VEs+VEcf+VEm+2COVG,E+VGxE
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Chapter 23 slide 43
Broad-Sense and Narrow-Sense Heritability
1. The amount of variation among individuals resulting from genetic
variance (VG) is the broad-sense heritability of a phenotype.
a. Broad-sense heritability = VG = h2B/VP (h2 is heritability, and B designates
“broad-sense”).
b. Heritability ranges from 0–1, with 0 meaning no variation from genetic
differences, and 1 meaning that all variation is genetically based.
c. Broad-based heritability:
i. Includes all types of genes and gene actions.
ii. Does not distinguish between additive, dominance and interactive
genetic variance.
iii. Assumes that interaction between genotype and environment (VG x E)
is not important.
iv. Is therefore of questionable usefulness.
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Chapter 23 slide 44
2. Additive genetic effects are more often used, because this component
allows prediction of the average phenotype of the offspring when
phenotypes of the parents are known.
a. For example, in a cross for a trait involving a single locus:
i. One parent might be 10 cm tall, with the genotype A1A1, and the
other parent 20 cm tall, with the genotype A2A2.
ii. If the alleles are additive, the F1 (A1A2) will be 15 cm tall, while if
one allele is dominant, the F1 will resemble one of the parents.
b. In the same way, epistatic genes will not always contribute to the
resemblance between parents and offspring.
3. Narrow-sense heritability is the proportion of the variance
resulting from additive genetic variance.
a. Narrow-sense heritability = VA = h2N/VP (h2 is heritability, and N
designates “narrow-sense”).
b. VA determines resemblance across generations, and responds to
selection in a predictable way.
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Chapter 23 slide 45
Understanding Heritability
1. Heritability estimates have limitations that are often ignored, leading to
misunderstanding and abuse. Important qualifications and limitations of
heritability:
a. Broad-sense heritability does not indicate the extent to which a trait is
genetic. Rather, it measures the proportion of the phenotypic variance in a
population resulting from genetic factors.
i. For example, knowledge of football requires a nervous system,
produced by genes, but differences in football knowledge are usually
not due to genetic factors, so broad-sense heritability (VG) for football
knowledge is zero.
ii. If all individuals have the same genes at loci controlling a trait (e.g.,
eye or ear number), VG = 0, even though genes are clearly involved in
producing the trait.
iii.High heritability also does not mean that environment is unimportant.
It may mean instead that environmental factors influencing the trait are
relatively uniform across the population.
b. Heritability does not indicate what proportion of an individual’s phenotype
is genetic. Heritability is a characteristic台大農藝系
of a population,
not an individual.
遺傳學 601 20000
Chapter 23 slide 46
c.
Heritability is not fixed for a trait. It depends on the genetic makeup and
environment of a population, and so a calculation for one group may not hold
true in another. An example is human height.
i. In a population with a uniformly high quality diet, differences in height are
likely to be due to genetic factors, especially if there is high ethnic
diversity.
ii. In a population where quality of diet varies widely, height differences will
be due less to genetic factors, and relatively more to environmental
factors.
d. High heritability for a trait does not imply that a population’s differences in the
same trait are genetically determined. An example is diet in mice.
i. Genetically diverse mice were randomly split into two groups. Both groups
received the same space, water and other environmental necessities. The
only difference was diet.
(1) Mice in the group receiving nutritionally rich food grew large.
Heritability of adult body weight was calculated at 0.93.
(2) Mice in the group receiving an impoverished diet lacking calories
and essential nutrients were smaller. Heritability of adult body
weight was 0.93.
(3) Both large size and small size were calculated to be highly heritable
in these mice. This is a contradiction, since both groups are from
the same genetic stock.
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(4) Another example is the correlation between presence of books in the
home and a child’s reading skill.
e. Traits shared by members of the same family do not necessarily have high
heritability.
i. Familial traits may arise from either shared genes or shared environment.
ii. Familiality is distinct from heritability.
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How Heritability Is Calculated
1. Many of the methods used compare related and unrelated individuals, or
compare individuals with different degrees of relatedness. When
environmental conditions are identical:
a. Closely related individuals have similar phenotypes when the trait is
genetically determined.
b. Related individuals are no more similar in phenotype than unrelated
ones when the trait is environmentally determined.
2. In humans, environmental conditions are complex in structure, and
extended parental care means that it is difficult to separate the effects of
genetics from those of nurture and culture. Methods of calculating
heritability in humans include:
a. Comparison of parents and offspring.
b. Comparison of full and half siblings.
c. Comparison of identical and nonidentical twins.
d. Response-to-selection data.
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3. Heritability from parent-offspring regression is calculated by correlation
and regression of data measuring the phenotypes of parents and
offspring in a series of families.
a. The mean phenotype of the parents (mid-parent value) and mean
phenotype of offspring are plotted with each point on the graph
representing one family (Figure 23.13).
i. Random scatter across the plot indicates no relationship between the
traits of parents and offspring, and thus low heritability.
ii. Linear relationships between phenotypes of parents and offspring
indicate that heritabifity is high (unless environmental effects have
influenced the trait).
iii. Slope of the parent-offspring regression line reflects the magnitude
of heritability.
(1) If the slope is 0, narrow-sense heritability (h2N) is also 0.
(2) If the slope is 1, offspring have a phenotype exactly intermediate
between the two parents, and additive gene effects account for the
entire phenotype.
(3) If the slope is between 0 and 1, both additive genes and
nonadditive factors (dominant and epistatic genes, environment)
affect the phenotype.
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Fig. 23.13 Three hypothetical regressions of mean parental wing length on mean
offspring wing length in Drosophila
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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b. When the mean phenotype of the offspring is regressed against
the phenotype of only one parent, the narrow-sense heritability is
twice the slope, because the offspring shares only 1/2 its genes
with the parent.
c. The factor by which the slope is multiplied to obtain heritability
increases as the distance between relatives increases.
4. Heritability values have been calculated for many traits in
a variety of species and populations, using several
methods (Table 23.5).
a. Heritability estimates are not precise and may vary widely for the
same trait in the same organism.
b. Human heritability values are especially difficult, because
genetic and environmental factors are hard to separate.
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Response to Selection
1. Both plant and animal breeding and evolutionary biology are concerned
with genetic change in groups of organisms, and use the methods of
quantitative genetics to predict rate and magnitude of genetic change.
2. Natural selection is based on the idea that certain genotypes leave more
offspring than others, leading to adaptive change in the population.
3. Artificial selection (selective breeding) by humans mimics this process,
and can be a powerful tool for rapid change in a species (e.g., domestic
dogs).
4. In both artificial and natural selection, genetic variation is a key factor
in determining the rate and type of evolution, and so quantifying it is
important.
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Estimating the Response to Selection
1. Phenotypes will change from one generation to the next in response to
selection if the appropriate genes are present in the population. The
amount of phenotypic change in one generation is the selection
response, R.
a. An example is body size in Drosophila melanogaster.
i. A geneticist begins by weighing flies to determine the average weight
in the population (e.g., 1.3 mg).
ii. The largest flies (e.g., those ≧ 3 mg) are selected for breeding.
iii. Weights of the F1 are compared with those of the original unselected
population.
iv. If they are significantly larger, response to selection has occurred.
b. The selection response is dependent upon:
i. Narrow-sense heritability.
ii. Selection differential (s), the difference between the mean phenotype
of the selected parents and that of the unselected population. (In the
Drosophila example,s = 3.0 mg - 1.3 mg = 1.7 mg).
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c. Selection response relates to selection differential and heritability by the
“breeder’s equation,” R = h2S.
i. In the example, R (response to selection) is the difference in mean body
weight between F1 flies and the original population (R = 2.0 mg - 1.3
mg = 0.7 mg).
ii. R and S are known, so the equation can be solved for h (narrow-sense
heritability):
h2 = R/S = 0.7 mg/1.7 mg = 0.41.
iii. This is a common method for determining heritability of many traits.
d. As long as genetic variation is present, traits will respond to selection in
each generation.
i. An example is selection for positive and negative phototactic behavior
in Drosophila pseudoobscura (Figure 23.14).
ii. Eventually the response to selection decreased. Possible explanations
include:
(1) No further genetic variation for phototactic behavior exists
within the population.
(2) Some variation still exists, but genes for the selected trait have
detrimental effects on other traits due to genetic correlations.
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Fig. 23.14 Selection for phototaxis in Drosophila pseudoobscura
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Genetic Correlations
1. Phenotypes of more than one trait may be correlated, and therefore do
not vary independently (e.g., blond hair, fair skin and blue eyes).
Association is not complete, but the traits are found together more often
than chance predicts.
a. Phenotypic correlation is computed by measuring the two phenotypes
in a number of individuals, and then calculating a correlation
coefficient for the two traits.
b. Sometimes phenotypic correlation occurs because the traits are
influenced by a common set of genes (e.g., human hair, eye and skin
color).
c. A gene affecting ≧ 2 traits is pleiotropic, and the traits will be
correlated (e.g., human height and weight).
d. Environmental factors may also cause nonrandom associations between
phenotypes (e.g., fertilizing soil may cause taller plants with more
flowers).
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2. Genetic correlations may be positive or negative.
a. Positive correlation means increase in one trait correlates with an increase in
the other (e.g., body and egg weight in chickens).
b. Negative correlation means an increase in one trait correlates with a decrease in
the other (e.g., egg size and number of eggs laid). Negative correlations
represent trade-offs (genetic constraints). Examples include:
i. Garter snakes, where for unknown reasons speed in escaping predators
correlates negatively with resistance to toxic newts in the diet.
ii. Dairy cattle, where milk yield correlates negatively with butterfat content.
3. Ability to adapt to a particular environment is also often influenced by
genetic correlations among traits. An example is tadpoles, where
developmental rate and size at metamorphosis show a negative correlation.
a. Genes that accelerate development (a good thing for life in a puddle that may
dry up quickly) cause metamorphosis to occur in smaller tadpoles, producing
smaller frogs.
b. Smaller frogs have decreased survival due to water loss, predation and
difficulty in finding food, and so there are constraints on the genes that
accelerate development.
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4. Additional examples of genetic correlations
appear遺傳學
in Table
23.7.Chapter 23 slide 58
Fig. 23.15 Negative genetic correlation (r = -0.45) between speed and resistance to
tetrodotoxin in garter snakes illustrated by family means
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Chapter 23 slide 59