Transcript CHAPTER 14 Quantitative Genetics

Peter J. Russell
A molecular Approach 2nd Edition
CHAPTER 14
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. 14.1 Discontinuous distribution of shell color in the snail Cepaea nemoralus from
a population in England
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Chapter 23 slide 3
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 4
The Inheritance 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 they could not determine the mode
of transmission.
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Chapter 23 slide 5
Polygene Hypothesis for Quantitative
Inheritance
Animation: Polygene hypothesis for wheat kernel color
1. Johannsen (1903) showed that continuous variation in bean seed weight is
multifactorial (partly genetic and partly environmental).
2. Nilsson-Ehle (1909) showed that a trait may be controlled by many genes, the
polygene (multiple-gene) hypothesis for quantitative inheritance. He studied
kernel color in wheat, crossing true-breeding red kernel wheat with truebreeding white (Table 14.1):
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Chapter 23 slide 6
a. The F1 were all the same intermediate color between red and white, so
incomplete dominance was a possibility.
b. But 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
reds;1 white).
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. Alleles R (red) and C (crimson) result in red pigment, while r and c
do not produce pigment.
ii. In the cross RRCC (dark red)´rrcc (white), the F1 are all RrCc
(intermediate red).
iii. Interbreeding the F1 produces an F2 with genotypes distributed as in
a typical dihybrid cross.
iv. 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 7
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 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 14.2).
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Chapter 23 slide 10
Fig. 14.2 Distribution of birth weight of babies (males + females) born to teenagers in
Portland, Oregon, in 1992
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Chapter 23 slide 11
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 (Table 14.2) 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(Figure 14.3).
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 12
Fig. 14.3 Frequency histogram for bean weight in Phaseolus vulgaris
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Chapter 23 slide 13
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 East’s study of tobacco flower length in
genetic crosses.
a. He crossed a short-flowered strain (mean length of 40.4mm) with a
long-flowered strain (mean length of 93.1mm).
b. The F1 progeny (173 plants) had a mean flower length of 63.5mm.
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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 14.4)
i. A broad curve shows high variability and a large variance.
ii. A narrow curve indicates little variability and a small variance.
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 units squared as in variance).
Standard deviation is the square root of the台大農藝系
variance.遺傳學 601 20000 Chapter 23 slide 15
Fig. 14.4 Graphs showing three distributions with the same mean but different
variances
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Chapter 23 slide 16
d. When mean and standard deviation are known, a theoretical
normal distribution is specified. Its shape is shown in Figure
14.5. 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) include 95% of the individual
values.
iii. Three standard deviations (±3s) include>99% of the
individual values.
e. An example is body length of spotted salamanders (Table 14.3).
f. Analysis of variance is a statistical technique used to help
partition variance into components.
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Chapter 23 slide 17
Fig. 14.5 Normal distribution curve showing proportions of the data in the
distribution that are included within certain multiples of standard deviation
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 18
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Chapter 23 slide 19
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 longflowered produced an F1 with a mean flower length of
63.5mm, intermediate to the parents.
b. The F2 had a mean of 68.8mm, very similar to the F1.
But the F2 had a variance of 42.2mm2, while F1
variance was only 8.6mm2, indicating that more
phenotypes occur among the F2 than among the F1.
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Chapter 23 slide 20
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 14.4).
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Chapter 23 slide 21
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Chapter 23 slide 22
ii.Correlation coefficient is a standardized measure of
covariance that can range from -1 to +1 (Figure 14.6).
(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 23
Fig. 14.6 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 24
Regression
1. Regression analysis is used to determine the precise relationship
between variables (Figure 14.7).
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. (Figure 14.8)
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 25
Fig. 14.7 Regression of sons’ height on fathers’ height
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 26
Fig. 14.8 Regression lines with different slopes
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Chapter 23 slide 27
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 such as genetics or the environment
by partitioning the variance.
2. An example is selective breeding to increase milk
production in dairy cattle.
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Chapter 23 slide 28
Quantitative Genetic Analysis
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 29
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 14.9).
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 30
Fig. 14.9 Inheritance of ear length in corn
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 31
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.
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Chapter 23 slide 32
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. To
assess heritability, we must:
a. Measure the variation in the trait.
b. Partition the variance into components attributable
to different causes.
.
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Chapter 23 slide 33
Components of the Phenotypic Variance
1. Phenotypic variance (VP) is the measure of all variability observed for a
trait.(Figure 14.10)
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
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Chapter 23 slide 34
if genes and environment were acting independently.
Fig. 14.10 Hypothetical example of the effects of genes and environments on plant
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 35
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 36
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 37
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 38
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 39
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 = h2B= VG /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 40
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 = h2N= VA /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 41
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 42
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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Chapter 23 slide 43
(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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Chapter 23 slide 44
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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Chapter 23 slide 45
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 14.11).
b. Slope of the parent-offspring regression line reflects the
magnitude of heritability.
i. If the slope is 0, narrow-sense heritability (h2N) is also
0.
ii. If the slope is 1, offspring have a phenotype exactly
intermediate between the two parents, and additive gene
effects account for the entire phenotype.
iii. 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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Chapter 23 slide 46
Fig. 14.11 Three hypothetical regressions of mean parental wing length on mean
offspring wing length in Drosophila
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台大農藝系 遺傳學 601 20000
Chapter 23 slide 47
c. 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.
d. 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 14.5)
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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Chapter 23 slide 48
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Chapter 23 slide 49
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.
a. Both groups are concerned with evolution (genetic change within
groups of organisms based on natural selection).
b. 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).
2. In both artificial and natural selection, genetic variation is
a key factor in determining the rate and type of evolution,
and quantitative genetics is used to:
a. Estimate the amount of genetic variation.
b. Predict the rate and magnitude of genetic change.
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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. Dog evolution under domestication (Table 14.6) is an example.
ii.Another example is selection for positive and negative phototactic
behavior in Drosophila pseudoobscura (Figure 14.12) in which the
response to selection eventually 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台大農藝系
due to genetic
遺傳學 correlations.
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Fig. 14.12 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) (Table 14.7).
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(Figure
14.13).
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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Fig. 14.13 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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Quantitative Trait Loci
1. The application of statistical analysis to polygenic inheritance has been powerful,
requires specific knowledge of the quantitative trait loci (QTLs) that affect them.
2. QTLs are identified by correlation with phenotypic differences between individuals,
and work best when analyzing a population with:
a. A detailed linkage map
b. Significant phenotypic variation.
c. Large numbers of individuals.
3. Typically, inbred lines with different phenotypes (homozygotes for different alleles)
are crossed, and then backcrossed, intercrossed, or selfed to produce a series of
recombinant inbred strains.
a. The resulting recombinant inbred strains are analyzed for marker genotypes that correlate
with phenotypic variation.
i. If the phenotype mean value is the same for all classes, the marker locus is unlinked to
a QTL.
ii.If the mean value is different for different classes, the marker locus is linked to a QTL.
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4. QTLs responsible for agronomic traits and adaptive differences
between closely related species have emerged from this work.
a. 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 14.14).
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.
b. Crosses between these species show that at least one QTL controls³25% of
phenotypic variation in pollinator attraction and efficiency traits (Figure
14.15).
c. In some cases, important phenotypic shifts correlate with mutations in
single loci.
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Fig. 14.14 Mimulus lewisii (A,C) and M. cardinalis (B, D) flowers
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Fig. 14.15 QTL maps for 12 floral traits in monkeyflowers
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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5. QTL approaches have been used extensively in corn,
tomato, and the fruitfly. Examples:
a. In corn, the tb1 (teosinte branched 1) QTL controls the number
of axillary branches, distinguishing corn from its wild relative,
teosinte. tb1 is a DNA-binding transcriptional regulator that
suppresses growth in specific tissues.
b. In tomatoes, the fw2.2 QTL controls up to 30% of the fruit
weight difference between cultivated and wild species.
i. Expressed early in fruit development, variation in fw2.2 alleles
affects timing and levels of gene expression, but not major
differences in protein sequence.
ii. The small fruit allele also causes a pleiotropic effect, as
isogenic lines homozygous for the allele produce more fruits
than do plants homozygous for the large fruit allele.
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6. QTL may sometimes be found by testing for associations
between phenotypic differences and allelic variation at
candidate loci. For example, in Drosophila a recently
discovered QTL defined for sternopleural bristle number
occurs near a known QTL for bristle number, hairy (h).
7. In humans, genes for quantitative traits cannot be found
through traditional pedigree analysis, due to the effects of
environment and other segregating genes.
a. Association studies use the occurrence of widely distributed
DNA markers in a subject population and compare them with a
random control population to detect QTL.
b. Genotype and phenotype of individuals is statistically analyzed
to correlate DNA markers with QTL.
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