Example lab exercises in evolutionary biology
Jonathan M. Brown
Biology Department
Grinnell College
Grinnell, IA 50112 USA
brownj@grinnell.edu
Website: http://www.grinnell.edu/individuals/brownj/authentic
These labs were created for use in introductory and advanced undergraduate courses at
Grinnell. While I authored all of the labs here, my colleagues Vince Eckhart, Kathryn
Jacobson and Susan Kolbe provided critical input into revising them. I owe them great
thanks. Labs may be freely edited and reproduced for educational purposes only.
Proper acknowledgement and an email notification of your use are greatly appreciated.
Contents:
1. Heritability and plasticity in fungi
1.1. Sources of phenotypic variation (introductory)
2
1.2. Phenotypic plasticity (advanced)
11
2. Natural Selection on Gall Flies
16
3. Sexual selection in natural populations
28
1
Sources of Phenotypic Variation
Darwin identified three necessary and sufficient conditions for natural selection: (1)
individuals within populations must vary in traits; (2) variation in traits must be
associated with differences in survivorship and reproduction; and (3) the variation in the
traits must be heritable, i.e., parents and offspring must resemble each other in these
traits. We now know that heritable differences among individuals are due to differences
in the information passed from parents to offspring via molecules of DNA. However, it
is important not to forget that the features of organisms result from developmental
processes that are influenced by environmental conditions as well as genes. It is perhaps
most accurate to describe a phenotype (the physical expression of a trait) as a product of
the interaction between a set of genes and environments. Thus if we observe a population
of variable individuals, it is not immediately obvious how much of this variation is due to
variation in genes among individuals versus variation in the environments these
individuals experienced during development. As Darwin’s third condition suggests, the
ability of natural selection to cause evolutionary change depends on the extent to which
phenotypic variation is due to genetic variation among individuals.
The fact that phenotypic variation can have both genetic and environmental sources
generates two complications for biologists in understanding evolutionary change. First,
not all phenotypic change over generations is evolutionary. For example, human height
has increased dramatically in developed countries over the last 100 years, not because
taller individuals have higher survivorship and reproductive opportunities, but because
childhood nutrition has improved in these countries. Should levels of nutrition decline,
average height of populations would decrease. Second, consistent differences between
individuals in survivorship and reproduction may not lead to phenotypic change over
generations. For example, many of you performed a nutrition experiment in BIO 135 in
which you varied levels of fertilizer and noted the growth responses of Brassica plants.
Let’s say that you decide to become an entrepreneur and start the Better Brassica Seed
Co. You collect the seeds from the largest plants in your fertilizer experiment in an
2
attempt to select for larger and larger Brassica plants, but not having had BIO 136 yet,
you do not realize that the differences in your original population were not heritable.
They were primarily (or perhaps completely) due to environmental effects associated
with different levels of fertilizer.
Because of these complications, biologists need to be able to quantify the contribution of
genes and environment to variation among individuals, to assess the potential for traits to
respond to selection. The proportion of phenotypic variation in some trait that is due to
genetic differences among individuals is called heritability, a statistic that can range
continuously from 0 (in which case genetic differences do not contribute to phenotypic
variation) to 1 (in which case genetic differences account for all of the phenotypic
variation among individuals).
Measuring heritability -- Evolutionary biologists, breeders, and statisticians have
developed techniques to quantify the sources of phenotypic variation among individuals.
This field is known as quantitative genetics, since most of the traits of interest vary
continuously --rather than discretely -- and may thus be measured on a quantitative scale.
In this exercise, you'll use a simple method for estimating heritability, which is to (1)
replicate genotypes (i.e., make multiple copies of each of several genotypes), (2) raise
individuals in a common environment (a so-called "common garden"), and (3) measure
how much individuals of the SAME genotype vary (which estimates the contribution of
environmental differences to phenotype variation) compared to the differences
BETWEEN genotypes (which estimates genetic contributions to phenotypic variation).
This method is easiest for organisms that reproduce asexually (“cloning”) most or all of
the time, because it is easy to replicate genotypes in such organisms. [Somewhat more
complicated experimental designs are necessary for organisms that reproduce solely by
sexual means, because they have genetically variable offspring.] Note that while we are
placing all individuals in a common environment, we can in principle NEVER guarantee
that each individual will receive EXACTLY the same environmental conditions. This
means that, when designing a common garden experiment, positions of individuals in the
3
environment should be randomized. It is the relative effects of those, uncontrolled,
environmental variables on phenotype to effects of different genes that we are trying to
measure.
The statistical measure used to describe how much a group varies in a trait is called the
variance, which is equal to the average of the squared deviations from the mean of the
group, or
N
 (X  X)
i
2 
2
i1
N
where X1 is the first measurement, X2 the second, etc., X is the mean value of the group,
and N the total number of measurements.
[Note: The above equation calculates the variance (2) for an entire population. If you
are estimating the variance based on a sample of the population (s2), replace N with (N-1)
in the denominator of the above equation.] You may also recognize that the square root
of the variance is equal to the standard deviation ( or s), a statistic you should be
familiar with from BIO 135.
Quantitative geneticists use the variance rather than the standard deviation to describe
variability because variances are additive. This means that for a population of
individuals divided into groups, the total variance will be equal to the between-group
variance plus the (average) within-group variance. The ability to partition the total
variance into within- and between-group components can be very useful. If the groups
are genetically identical individuals (clones), then the within-group variance is a measure
of the effect of environmental variation and between-group variance a measure of the role
of genotypic differences on phenotypic variation. The following example illustrates the
property of additivity of variances.
4
Imagine that you've raised five clones each of three genotypes of goldenrod under the
same conditions and measured their heights (in cm). The raw data are:
Genotype 1
Genotype 2
Genotype 3
-------------
--------------
--------------
14
12
11
14
12
12
15
11
13
16
10
14
16
10
15
Mean
15
11
13
Variance
0.8
0.8
2
[Here’s the variance calculated for genotype 1:
N
 (X  X)
i
 
2
i1
N
2

(14  15)2  (14  15)2  (15  15)2  (16  15)2  (16  15) 2 4
  0.8
5
5
If we consider all the clones of all the genotypes as our population, the mean is 13 and
the variance is 3.87. (Do these yourself to make sure you understand the calculations)
The latter quantity is referred to as the total phenotypic variance (P2).
The property of additivity means that the total phenotypic variance of the population
should be equal to the sum of the within-genotype and between-genotype variances:
(1) The average within-genotype, or environmental, variance (E2) is
E2 =
(0.8  0.8  2)
= 1.2
3
5
[Note: This calculation only works when you have equal sample sizes for each
genotype!].
(2) The between-genotype variance (G2) is just the variance of the genotype means or
G2 =
(15  13)2  (11 13)2  (13  13) 2
= 2.67
3
where 13 is the mean of genotype means (i.e., the mean of 15, 11 and 13).
(3) The total phenotypic variance (P2) should be equal to the sum of these two
components
P2 = E2 + G2 = 1.2 + 2.67 = 3.87
VOILA!
All that’s left to do now is to calculate the heritability, denoted by the symbol h2, which is
the proportion of the variance that is due to genetic differences among individuals:
 G 2 2.67
h  2 
 0.69
P
3.87
2
On the next page is a worksheet to help you with these calculations, as well as to help
you understand WHY they measure the degree of genetic determination.
6
Heritability Worksheet
Genotype 1 Genotype 2 Genotype 3 Genotype 4 Genotype 5
Clone A
Clone B
Phenotypic Varianc e ( V
P
)
Clone C
Clone D
Clone E
Clone F
Genotype Means
Variance among Genotype
Means ( Genetic variance -VG)
Within-Genotype
variances
Mean of within-genotype
variances (Environmental
Variance -- V E )
P henotypic variance
(V P = V G + V E )
Heritability (
h2 = V
G /V P
)
Note: On this worksheet V=2
7
[Special Note: The measure of heritability we have calculated here is often called
heritability in the broad sense or degree of genetic determination. If you move on to
advanced courses in evolution, you'll learn that the ability of selection to act on
phenotypic variation and produce evolutionary change actually depends on a portion of
the genetic variance called the additive genetic variance (A2). The ratio A2/P2 is called
heritability in the narrow sense. How much of the genetic variance is additive depends
on the importance of dominance, gene interactions and gene-environment interactions.
For this course, however, you needn’t concern yourself with this distinction (unless you
want to!).]
You and your partner will be assigned one of three fungal species in which to measure
the heritabilities of several traits. As you calculate these heritabilities, you are identifying
whether populations contain the raw material necessary for evolutionary change via
natural selection. Other things being equal, the potential rate of evolutionary change in a
trait is proportional to its heritability. While you read the following descriptions consider
what hypotheses you might make about the heritabilities of different traits. On what
basis can you make such hypotheses?
Schizophyllum commune
Many fungi are wood decomposers, and their fruiting bodies (e.g. mushrooms) can be
found growing on logs or branches on the forest floor. Most fruiting bodies are
ephemeral, lasting only a few days while environmental conditions (moisture and
temperature) are appropriate for spore dispersal. (Mushrooms are analogous to flowers in
that they produce dispersal propagules (spores versus seeds). The fruiting bodies of
Schizophyllum commune, however, are capable of folding up and withstanding periods of
seasonal desiccation, and reviving and re-opening when moistened. Mushrooms of this
species can thus form in early spring and continue to produce and disseminate spores
during moist periods throughout the summer and fall. Spores form over the entire surface
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of the gills, on the underside of the fruiting body. Hence, the larger the surface area of the
fruiting body, the greater the volume of spores produced.
Schizophyllum commune has a world-wide distribution, and can decompose many
different types of wood including oak and pine, at temperatures ranging from 12 - 35 C.
The mycelium penetrates wet wood, using cellulose and lignin from plant cell walls as a
carbon source.
It is relatively easy to obtain living material of Schizophyllum for laboratory studies by
placing a small piece of a fruiting body on nutrient media in a petri dish. These cultures
grow luxuriously at 20-25 C, forming a colony of mycelium which grows outward from
the point of inoculation, using the rich supply of nutrients in the complex malt media.
Schizophyllum cultures will even form fruiting bodies in the petri dish, although the
environmental conditions (in the petri dish) required for fruiting depends on the
genotype. Some genotypes will fruit soon after inoculation on to the media, whereas
others produce an extensive mycelium before fruiting. To avoid exposure to spores we
will grow our cultures in the dark. Clones of genotypes are easily made on new media
and maintained at cold temperatures for long-term storage.
*** Insert info on other fungal species
Week 1
Five genotypes of Schizophyllum commune were collected from a forest in Vermont and
maintained in this manner. These genotypes were collected from the same site, but from
different substrates, over a period of three years. With the help of your partners, set up
the common garden experiment by inoculating 6 plates for each genotype. How will you
randomize these plates to ensure that differences between genotypes are not due to
uncontrolled environmental variation?
Week 2
Sometime during the first week, measure the following traits for each clone:
9

submerged colony radius

aerial colony radius
Calculate heritabilities for these two traits.
Sometime later (but before the colonies grow to the edge of the plates!) measure:
submerged colony radius
10
Phenotypic Plasticity
In this lab, we will explore the connections between the ideas of heritability and geneenvironment (GxE) interaction. The fact that environmental, as well as genetic variation,
contributes to phenotypic variation has more interesting consequences for evolutionary
change than you most likely have been exposed to in BIO 136 or elsewhere. These
complications turn around two basic ideas:
(1) Since environmental conditions vary among populations of species, genotypes
will express different phenotypes in different populations (i.e., they will show
phenotypic plasticity). The range of phenotypes expressed by a genotype across
environments is called its norm of reaction.
(2) Genotypes may not show the same response to a change in environmental
conditions, a condition known as gene-environment interaction. As we will see,
this has importance consequences when thinking about how selection in different
populations may lead to differentiation among those populations.
I. Measuring heritability in a clonal species
Read the attached handout from BIO136, which explains the basic concept of measuring
heritability in a single population using a “common garden” experiment. [Note the
heritability worksheet at the end of the lab, which you will need to use in analyzing your
own experiment. You can easily adapt this worksheet to an Excel spreadsheet and let the
computer do all the calculations for you. See me for tips on this, if you’re not familiar
with using formulas in Excel.]
II. Heritability and analysis of variance
Although students in BIO136 didn’t know this, they were performing a statistical analysis
called analysis of variance (usually called ANOVA) when calculating heritability.
ANOVA is a commonly used statistical tool for understanding how factors influence
variation in some measured variable. For example, someone who did a replicated
experiment to determine the influence of different levels of fertilizer on growth of a plant
would use ANOVA to test whether the different levels had a significant effect of growth.
This is done by comparing the mean growth for the different levels of fertilizer (called
the main effect) with the variation among replicates within a single level of fertilizer (the
error effect).
Read the attached handout on ANOVA (also available on a website for access to internal
links-- see the class page for a link to this useful stats resource).
(1) In the 136 experiment the “main effect” was “genotype”, while the “error” is
equivalent to the effects of environmental variation within the common garden.
11
(2) The statistical significance of an ANOVA tells you whether variation among
genotypes is significant, given variation within genotypes (i.e., environmental variation).
This is equivalent to testing whether the heritability value obtained is significantly
different from 0.
(3) It is possible to do experiments in which two factors are simultaneously varied and
the effects of each evaluated -- these are “multi-factor” ANOVAs. If we expand our
common garden experiment to measure heritability by replicating it under two different
environmental conditions which we control, e.g. temperature, our ANOVA would have
two main effects, genotype and temperature environment. The error effect in this
analysis refers to the environmental variation among replicates with the same genotype
and temperature.
(4) When multiple factors are tested in an experiment, it is possible that the effects of one
factor may depend on the condition at another factor, a so-called “interaction” effect in
the ANOVA. In our experiment, this is equivalent to asking whether genotypes show
different responses to changes in the environment, illustrating gene-environment
interactions.
III. Plasticity and norms of reaction in fungi
The experiment you will be running over the next two weeks will consist of an
investigation of norms of reaction in multiple genotypes of one of three species of fungi.
You and your partners will be responsible for designing, setting up, taking data and
analyzing the experiment. You may want to do a little research on your organism before
setting up your experiment. Run your proposal for an experiment by me before starting
off on it.
Since science is rarely done 3 hours a week on Monday afternoons, you may have to
make plans to take data at other times. Coordination among the members of the team is
crucial! Because of this, I won’t expect you to be in the lab the entire time during the
next two weeks, but I’d like a progress report each week on Monday afternoon, and will
of course be available for troubleshooting, and advice.
Some things to consider:
1. Temperature and light are likely to be important environmental influences on growth
and reproduction in these species. I will let you know the options for temperatures.
Your group should decide on what environmental condition you want to vary (temp
in dark, temp in light, or light/dark at one temp).
2. Don’t forget to randomize the positions of genotypes and replicates within each
environment!
12
1. Your group may use up to 100 plates for your experiment. The growth medium is
called CYM (complete yeast medium), the recipe for which is below:
0.50g MgS04-7H2O
0.46g KH2PO4
1.00g K2HPO4
2.00g peptone
20.00g dextrose
2.00g yeast extract
15.00 g agar
in: 1L distilled water

Transfer small plugs from the stock plate to a new plate -- these are best taken from
the edge of the growing mycelium. Use sterile technique when transferring plugs
from the stocks to each of your plates. Place the plug in the center of the plate, taking
care not to drop mycelium on any other part of the plate. Mark the initial position of
the plug on the bottom of the plate. Place plates upside-down and inside one of the
sealable tupperware containers (this reduces contamination and keeps the relative
humidity from fluctuating).
Analyzing your Fungal Experiment
I will ask you to do four types of analyses of your fungal experiments:
1. Heritability calculated within each environment. For each trait you measure
(remember that the same feature measured at different times can be considered a
different trait), calculate a heritability value within each environment separately using
the matrix approach described in class.
2. Do an analysis of variance (ANOVA) to test for significant effects of Genotype,
Temperature (or Light) and their interaction. Remember the latter is a measure of the
significance of Gene-Environment interaction.
Set up your data sheet in Minitab in the following way:
Genotype
Clone
Temp
Trait1
Trait 2
etc.
1
1
..,
1
2
1
2
18
18
23
19
6
8
..
..
6
1
18
18
30
26
4
8
..
..
etc.
13
After all the data are entered, choose “Balanced ANOVA” from the “ANOVA” submenu
of the Stats menu. In the “Responses” box choose all the traits you want to analyze. In
the the “Model” box type “Genotype Temp Genotype*Temp” -- the latter is asking for
the interaction effect in addition to the main effects. In the “Random” box type
“Genotype” -- you didn’t set levels of genotype yourself (as you did temperature), and
the assumptions of the tests of significance are different with such random factors.
Here’s what the output should look like:
Analysis of Variance (Balanced Designs)
Factor
Type Levels Values
Genotype random
3
4
Temp
fixed
5
18
7
24
88
30
37
42
Analysis of Variance for DiaWk1
Source
Genotype
Temp
Genotype*Temp
Error
Total
DF
2
4
8
30
44
SS
384.53
6325.56
2525.91
522.00
9758.00
MS
192.27
1581.39
315.74
17.40
F
0.61
5.01
18.15
P
0.567
0.026
0.000
This is analysis of Montagnea arenaria (a fungus from the Namib desert). The trait is
colony diameter after 1 week of growth. Three clones of each of three genotypes were
grown at 5 temperatures. Note that genotype is not significant here, but temperature and
the interaction effect are. How would you interpret these results?
3. Plot norms of reaction for each trait. Here’s an example of the above data:
14
Norms of reaction for 1st week growth
Colony Diameter (mm)
60
50
40
Gen4
30
Gen7
Gen88
20
10
0
10
20
30
40
50
Temperature
NOTE: I didn’t put standard error bars on the means in this figure (with only 3
reps/genotypes they are big), but you should on your figures.
4. One of the interesting implications of crossing norms of reaction is that a trait
measured in two different environments may show negative genetic correlations -- this
is related to the idea raised by Gupta and Lewontin that the phenotypic rank order of
genotypes can be different in different environments. To do such an analysis, pair up
average phenotypic values in two environments from each genotype and do a correlation
analysis. With only 3 replicates/genotype in the above data, this is a VERY weak
analysis statististically, but the correlation coefficient between mean phenotype at 24 and
37  is -0.96. You may have more power with a greater number of genotypes to consider.
15
Natural Selection on Gall Flies
Note: This lab was originally conceived by Warren Abrahamson and Art Weis, and has
been informally distributed across the US for many years. The ideas draw heavily on
their extensive publications on the evolutionary ecology of this system, which has been
compiled in a recent book (Abrahamson, W. and A. Weis. 1997. Evolutionary ecology
across three trophic levels: goldenrods, gallmakers and their natural enemies. Princeton
UP. 456 pp.) The text of this lab handout, however, was written by J. Brown.
Introduction
The determination of how natural selection acts in contemporary populations constitutes
an important link between the studies of ecology and evolution. While we have referred
to natural selection as a “force” that causes populations to evolve, it is perhaps more
properly considered as an outcome of an interaction between phenotypic variation in a
population and the current environment that population experiences (where the
environment is broadly construed to include abiotic and biotic factors). This interaction
leads to consistent differences in survivorship and/or reproduction between phenotypic
variants, one of the criteria for natural selection to operate (see the introduction to Lab 2).
Understanding how biotic interactions and/or the physical environment create selection
may provide a clue as to how the current characteristics of a population have been
molded through evolution. Through such studies, we come to appreciate the link
between ecological interactions and their evolutionary effects.
If we constrain our study of selection to differences in viability (as we will in this lab),
we are looking for significant associations of phenotypic variants with the probability of
surviving. Phenotypic variation in populations often takes the familiar form of the “bell
curve,” defined mathematically as the normal distribution. The effects of selection can
be seen by comparing the distribution of phenotypes in the population before and after
selection acts (i.e., before and after individuals die); specifically, we can compare the
means and variances of the distributions and look for three types of effects of biased
survivorship associated with different phenotypes:
16
(1) Directional selection -- The population of survivors can have a higher or lower
mean value for the characteristic than the population before selection acted. If
individuals with larger values of the trait survived with higher probability, and
therefore the mean after selection is greater than the mean before selection, we say
that “upward” directional selection has occurred. “Downward” directional selection
has occurred when smaller individuals survive with higher probability.
(2) Stabilizing selection -- The population of survivors can have a reduced
variance of the characteristic compared to the original population, if individuals
with extreme phenotypes have higher rates of mortality than individuals with
intermediate phenotypes.
(3) Disruptive selection -- The population of survivors can have a higher variance
compared to the original population, if individuals with intermediate phenotypes
have higher rates of mortality than individuals with extreme phenotypes.
It’s important to note that selection can affect both the mean and the variance of
populations, i.e., both directional and stabilizing(or disruptive) selection can occur.
Comparing selection events -- Our primary goals in studying selection in natural
populations is to compare the strength of selection on different traits, different species
and between events on the same traits at different times or place. From the first two we
can potentially learn why some traits or species evolve and others do not; from the latter,
we learn how consistent natural selection is, i.e., whether we should expect traits to
evolve in certain directions over long periods of time and whether we should expect
different populations to evolve in different directions. Can you think of why this might be
important?
17
But how can we compare selection on two traits or over two events? Let’s just consider
directional selection: It might seem logical to compare traits or event for how much the
mean changes, but there are two problems with this approach:
(1) Scale bias -- Let’s say you are measuring the effect selection on body mass in
your favorite insect and you obtain an answer of 10.2 mg for the strength of
selection (i.e., zAfter  zBefore , where z refers to the mean of the trait z). If I were to
repeat your study and measure mass in grams (rather than milligrams), I would get
an answer of 0.0102. It would nice if our measure of selection was independent of
the units used. The problem gets even worse if you wanted to compare selection on
body size in elephants vs. flies, for example. Comparing the differences in means
between species, even if measured in the same units, would not be very illuminating.
(2) Variation bias -- Consider Figure 6.1, which illustrates the phenotypic
distributions of two populations that have the same mean values for the trait being
measured both before and after selection.
Figure 6.1 -- Two populations experiencing selection such that the mean value of body
size before and after selection are identical.
If you were to use the difference in mean values before and after selection (2 cm) as
the “strength” of selection, you would conclude that selection acted in the same way
18
in both populations. However, there is a real sense in which selection is more
“intense” in population B than in population A: For selection to change the mean
value by 2 cm in population A, a relatively small proportion of the population has to
die; however, for population B to move in mean 2 cm upwards would require
MOST individuals in the population to die.
Both these problems can be avoided by standardizing the measurement of change in the
mean before and after selection by dividing by a measure of the original amount of
variation, the standard deviation (s). This value, called the intensity of selection has the
following formula:
i
z After  zBefore
sBefore
Natural History of the Solidago-Eurosta System
Galls are growth deformities induced in certain plants by various insects. These
interactions are frequently species-specific, with a particular species of insect inducing
galls in a specific tissue of one species of plant. Galls are used by the insects that induce
them as sites for larval development and as food. Characteristics of the gall are often
under the influence of both the insect that provides the stimulus for gall formation and the
plant producing the gall. Therefore, some features of gall morphology may evolve in
response to selection on the gall-forming insect. Previous work on this insect-plant
system has shown that gall diameter is a heritable character of the insect, as well as the
plant. In this laboratory an analysis of gall diameter will be used to determine whether
there is selection on the gall-inducing insect for gall size.
Solidago gigantea, or Late Goldenrod, is a common perennial of the eastern and
midwestern United States that is frequently parasitized by the gall fly, Eurosta
solidaginis. In the spring, adult female gall flies lay a single egg in each of many
19
terminal buds of developing goldenrod shoots. The fly larva tunnels into the stem just
below the apical meristem, where it secretes compounds believed to be similar to normal
plant growth substances. As a result the plant undergoes abnormally high rates of cell
division in the area occupied by the larva, resulting in the formation of a spherical gall.
Gall fly larvae feed off the plant tissue, growing to full size by early Fall, overwintering
in the gall, and pupating in the Spring. After metamorphosis is completed in May, the
adult emerges from the gall to seek a mate. [Note that a related species, Solidago
altissima, is also attacked by a separate, reproductively isolated, host race of this fly
species. The natural history of this interaction is almost identical to that between the fly
and S. gigantea, and has received a greater amount of study.]
Sources of Eurosta Mortality - Mortality of fly larvae within galls may result from a
number of different causes, including interactions with predators or other herbivores of
the goldenrod. The following lists the major, diagnosable causes of mortality in this
system:
(1) Parasitoid wasps --- Parasitoids are insects that lay their eggs on or in a host, but
whose effect is to kill the host (unlike a true parasite). The wasp Eurytoma gigantea is
such a species -- a female wasp inserts its eggs into the central chamber of goldenrod
galls. The resulting wasp larva eats the fly larva, and then switches to a vegetarian diet,
eating gall tissue the rest of the growing season. Flies in smaller galls may be more
susceptible to attack by this parasitoid wasp, since wasps can attack only those fly larvae
that are within reach of the wasp's ovipositor. If this is the case, then attack by wasps
may be a factor causing directional selection on the size of galls induced by the flies.
[**NOTE: This handout does not describe a second parasitoid species, Eurytoma
obtusiventris, which can be a very significant source of mortality in some populations, as
it is extremely rare in populations west of the Great Lakes. It is extremely abundant in
the eastern U.S. See the Solidago-Eurosta website for more info.]
(2) Bird predators -- During the winter, downy woodpeckers (Picoides pubescens) and
black-capped chickadees (Parus atricapillus) also prey on the gall fly larvae. These birds
20
peck through the tissue of the gall and extract the soft-bodied fly larva. These birds are
visual predators and thus larvae living in galls more easily seen by birds may suffer
higher rates of mortality. If gall size is a determinant of which fly larvae are attacked by
birds, then predation by birds will cause directional selection on the size of galls induced
by flies.
(3) Other herbivores -- The stems and galls of the goldenrods are attacked by a large
number of herbivorous insects. One common herbivore often found in the galls is
Mordellistena unicolor, a beetle species that lays its eggs on the surface of the gall early
in the summer. When many larvae burrow into the gall tissues they often cause the death
of the fly larva and may consume it. If gall size is a determinant of which fly larvae are
killed by this herbivore, beetle attack will be a cause of selection on the size of galls
induced by the flies.
(4) Plant interactions -- Plants may have mechanisms to resist herbivory, in some cases
causing the death of the herbivore. This may be an explanation of the phenomenon of
Early Larval Death for Eurosta flies -- the gall continues to form although the fly larva
has died early during gall formation. This is a common cause of mortality for flies on S.
gigantea, and often leads to smaller than average gall sizes due to the early death of the
gall inducer. Since the gall size is not indicative of a fly phenotype in this case (the fly
hasn’t been present to stimulate gall growth throughout the entire growing period), we
will remove these galls from the analysis before estimating selection on gall size.
By collecting galls, measuring the phenotypic distribution of gall sizes and determining
how the distributions change after mortality agents act, you will be able to estimate the
strength and form of selection on this phenotype for these populations. You will also be
able to look at the rates of mortality due to different agents, which will help you interpret
why selection has acted in the way it has. Finally, you’ll be able to compare your data to
the data from last year’s class and the literature to determine how consistent selection has
been over time or place. This may be important in understanding why populations have
the phenotypic traits they do.
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Methods
Sampling Galls (Week 1) -- When collecting a sample of individuals from a population,
it is important to consider carefully how the methods used to choose measured
individuals may bias the results. Sampling is a complicated area of ecology, with
different techniques used in different situations. Ideally, we want to choose individuals
from a population randomly, although sometimes this is not practical. The technique
described below does not produce a truly random sample of the population of galls, but
should guard against systematic biases in the sample (can you think of biases that might
be introduced by other methods of sampling?):
1. Lay out a 30m measuring tape along one edge of the population and determine the
end-points of belt transects along this tape using a random number table.
2. Run a belt into each population perpendicular to the end-point line using the long
measuring tapes.
3. Collect all galls within 0.5 meter of the measuring tape. Make sure you do not miss
small galls!
Data Collection (Week 2):
1.
Measure the diameter at the widest point of each gall by fitting it into the metric
template -- find the smallest hole that the gall will fit through.
2.
In order to determine the fate of the fly that induced the gall, carefully cut open the
gall with the pruners and examine the contents.
a.
If a cream-colored, fat larva or a tan-colored fly pupal case is present in the
gall, that fly has clearly survived all the mortality agents discussed above and
will almost certainly survive to emerge later in the spring. All other galls
were induced by a fly that did not survive.
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b.
After examining each gall, categorize the fate of the fly that induced the gall.
Examples of galls in each class will be available in lab. For each gall you will
record the diameter and the fate on the data sheet and in the class computer
file.
Data Analysis -- Your instructor will inform you how to access the data from your class
and previous years. The data are organized into two columns, “Fate” and Gall size”.
First, organize the data so you can use EXCEL to do all the number crunching for you:
3. Sort the data by gall fate by highlighting all the cells in both columns, choosing Sort
from the Data menu, and indicating that you want to sort the cells by the “Fate”
column.
4. Divide the data into separate columns on Sheet 2. Start by transferring the list of gall
sizes for all galls EXCEPT those that died due to Early Larval Death into the first
column, “Pre-selection.” Then highlight all the gall sizes associated with
“Survivors”, copy them, and paste them into the “Survivors” column on Sheet 2. Do
the same with gall sizes associated with mortality due to parasitoid wasps, birds,
beetles, and unknown insects.
5. Calculate the rates of mortality for different agents (% killed by each agent), and the
rate of survivorship for flies (% survived), using the number of “pre-selection” galls
as the denominator for these rates. [You may also wish to calculate the rate of early
larval death, as it may prove interesting in interpreting year-to-year and/or site-to-site
variation in overall fly survivorship.]
6. Calculate averages, variances, and standard errors for each column. [In a cell at the
bottom of the column type “=Average(“, then highlight the column of data, type “)”
and press <Return>. Type “=Var(“ to get the variance. The S.E. can be calculated
from the variance and the sample size.]
7. Determine whether the mean size of galls attacked by the each of the mortality agents
is significantly different from the mean size of “Pre-selection” galls, using a t-test.
Note that if a mean size of gall for a mortality agent is LESS THAN that of preselection galls, that agent imposes directional selection favoring LARGER galls.
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8. Determine whether the mean gall size for Survivors is significantly different from the
mean for pre-selection galls, using the t-test. This will tell you whether the total
effect of all mortality agents has resulted in significant directional selection. If it
has, calculate the intensity of directional selection for gall size in this population.
7. Are changes in variance significant?
How would you determine whether stabilizing or disruptive selection is occurring?
Remember that we want to know whether the phenotypic variance has increased or
decreased after selection has acted. But just as it is legitimate to ask “how much different
does it have to be to be REALLY different” when we compare two means, we need a
statistical test to determine how much of a difference in variances is significant (i.e., not
likely due to chance). This test is called the F-test.
Here’s the problem: Imagine you have two populations that have the same variance,
but you estimate that variance by taking a sample of each. If you take a small number of
individuals from each population, it’s quite likely by chance that your estimates of the
variances would be different. The more you sample, the more likely the two variances
should be equal. The statistical test uses this principle by asking you to express the
difference between the variances as a ratio (larger variance divided by smaller variance)
called the F-ratio. The null hypothesis is that the variances are equal, i.e., that the ratio is
equal to 1. The F-test asks whether ratio you got could happen easily by chance (given
how many individuals you sample) when the true ratio is 1; if the probability of getting
this ratio by chance is less than 5%, then we reject the null hypothesis of equal variances
and say that the variances are significantly different.
To test whether there is significant stabilizing or disruptive selection , calculate the ratio
of pre- and post-selection variances (larger variance over smaller variance) and compare
that ratio to the appropriate critical value in the table below. [Remember that d.f. = n-1
and use the largest value in the table below that is less than your actual d.f.] If the ratio is
24
higher than the critical value, you can reject the null hypothesis of equal variances with
95% confidence.
Critical values of F (p = 0.05)
d.f. numerator
20
d.f. denominator
30
60
120
Infinity
20
2.12
2.04
1.95
1.9
1.84
30
1.93
1.84
1.74
1.68
1.62
60
1.75
1.65
1.53
1.47
1.40
120
1.66
1.55
1.43
1.35
1.25
Infinity
1.57
1.46
1.32
1.22
1.00
9. Data sets from the last four years are available (note that the class samples two
different sites at CERA, “Plantation” and “Lake”). Choose at least one other data set and
compare your results to it. Repeat your calculations for the other data set(s) and compare
the results.
Questions for consideration in your papers:
1. What do your analyses suggest about the evolution of gall size? Can you think of
possible constraints on adaptation by the gallmaker to such selection?
2. Is natural selection consistent from year-to-year or place-to-place? Why or why
not? What are the ramifications of this for understanding the evolution of the
flies?
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26
27
Sexual selection in natural populations
Darwin described sexual selection as occurring due to “the advantages that certain
individuals have over others of the same sex and species, in exclusive relation to
reproduction.” Studying sexual selection in natural populations entails the demonstration
of phenotypic variation in heritable characters that is significantly associated with
reproductive success either through competition with others of the same sex
(“intrasexual” selection) or attraction to the opposite sex (“intersexual” selection). The
study of sexual selection is important not only to demonstrate its ability to create
intersexual differences; it has also been implicated in the processes that produce
reproductive isolation during speciation.
One can easily measure the potential intensity of sexual selection in natural populations
of organisms that have prolonged copulatory periods by comparing characteristics of
individuals found mating with characteristics of solitary individuals. Local species of
insects that exhibit these features include members of the coenagrionid damselfly genus
Enallagma , commonly known as “bluets,” and the acantharid beetle genus
Chauliognathus, commonly known as the goldenrod soldier beetles. We will split into
two teams to capture an adequate sample of solitary males and pairs of these species. We
will then return to the lab to measure phenotypic differences between successful and
unsuccessful males. In our analyses, we will again have to pay attention to potential
correlations between characters in our interpretation of the mechanism of sexual
selection.
Let us assume, for the moment only, that differences we may measure between
successful and unsuccessful males are due to female choice, rather than male-male
competition. One set of theories of sexual selection contends that females should make
choice based on variation in certain traits because these traits are good indicators of the
genetic quality of a male. For example, Hamilton and Zuk (1982) proposed that females
choose male characters that are good indicators of low parasite load, since resistance to
parasitism is a important and heritable determinant of fitness in many organisms. Several
recent studies of vertebrate and invertebrate species have indicated that developmental
stability may also be such a target of sexual selection. The most common measure of
stability is fluctuating assymetry (FA), defined as “the random deviation from bilateral
symmetry in a morphological trait for which differences between the right and left sides
have a mean of zero and are normally distributed” (Watson and Thornhill, 1994). Under
this model of sexual selection, males that are mated should be more symmetrical than
those that do not obtain mates.
We will use a computer image analysis system to gather morphometric data on the
individuals we collect. We will compare both absolute characters of males and their
symmetry to explore how sexual selection may be acting in these populations. Finally,
we will compare the morphology of mated pairs, to see whether there is any evidence of
assortative mating (“like mating with like”), an important component of many models
of speciation.
28
Hamilton, W.D. and Zuk, M. 1982. Heritable true fitness and bright birds: a role for
parasites? Science 218: 384-7.
Watson, P.J. and Thornhill, R. 1994. Fluctuating assymetry and sexual selection.
Trends in Ecology and Evolution 9:21-25.
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Measuring sexual selection
NIH Image is an image analysis program that we will use to gather morphometric data on
our insect samples. It is a free program (your tax dollars at work!), so you may do the
measuring part of the exercise on any Macintosh computer (assuming it has a screen as
big as the one the images are captured on).
I. Dissecting your organism





Set up a data sheet to keep track of each individual you measure.
Pick an individual or pair to measure and assign it a sequential code number that has
M or F as the final digit, for male and females respectively. Make a note in the data
sheet if males were paired or not (i.e., "studs" or "duds").
Weigh each individual on the analytical balance and record its mass.
For the beetles, remove the two elytra (hardened forewings) and the right and left
forelegs.
For the damselflies, remove the right and left forewings, the right and left forelegs
and the head.
II. Scanning images into the computer using NIH Image







Flip the switch on the power strip on to turn on the video camera.
Turn on the fiber optic light sources to the dissecting scope. Adjust the intensity of
light from below and above to get the best image contrast.
Start up NIH image on the computer.
Go to the “Special” menu and choose “Start Capturing”
First sharpen the image of the calibration ruler by looking through the eyepieces of
the microscope. Then adjust the focal ring on the camera tube (between the
microscope and video camera) until the image on the computer screen is sharp. From
now on the images should be in focus -- if not just use the microscope adjustments to
sharpen the computer screen image.
Capture the image by choosing “Save As” from the File menu. Find the folder for the
beetles or damselflies inside the class folder. Give the image the code number as a
name.
Put the body parts back in the tube and do another individual. If you have a pair, put
the male and female into separate tubes.
III. Measuring individual variation using NIH Image


Choose “Load Macros” from the Special menu and open “Measurement Macro” from
the Class folder.
Choose “Take Measurements” from the Special menu. Follow the directions for
naming a data file and taking measurements. For each image that you open, you will
first calibrate the image by clicking on the 10 mm increment that has been marked on
the ruler, and then clicking twice on the image for each measurement.
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



For beetles, measure right and left elytral length, right and left elytral spot height,
right tibia length, and left tibia length.
For damselflies, measure right and left wing length, right and left tibia length, and the
length of the abdomen.
If you make a mistake on a measurement, open the image again and redo the
measurements, remembering to remove the first line from the data later.
When you are done taking measurements, the macro will prompt you to save the data
to a text file.
IV. Analyzing the data


Transfer your data to Minitab. Selecting the data from the data window in NIH
Image and copying it does this most easily. Open Minitab and click the space for the
title of the first column. Choose “Paste/Insert Cells” from the Edit menu. If
something goes wrong (e.g. the computer bombs), remember you has saved the data
to a text file and can thus recover it.
Create a “Mated” column that contains a ‘0’ for unmated males and a ‘1’ for mated
males.
Looking for correlations:

Explore the relationships between measured characters, looking for evidence of
character correlations. [You may first want to create average values for the paired
variables; use the “Mathematical Expressions” command in the “Calc” menu.] You
may also want to use principle components analysis to create composite characters for
later analysis.
Looking for sexual selection -9. Convert the paired (left-right) variables into asymmetries by using the “Mathematical
Expressions” command in the “Calc” menu (Enter an empty column in the “Variable”
box at the top and enter the expression in the lower box). For each set of paired
variables, calculate the absolute asymmetry (the absolute value of the difference
between left and right) and the relative asymmetry (the absolute asymmetry divided
by the average of left and right). Which of these 2 ways of measuring asymmetry are
you going to use? Test each assymetry variable for the assumption that the mean
score is not significantly different from zero [Choose "1-sample t-test" from the
"Basic Statistics submenus of the "Stat" Menu.].
 For each non-paired variable (e.g., mass), the means of paired variables and the
asymmetry variables, test whether the mean values for mated and unmated males are
significantly different using a t-test. [Choose “2-sample t-test” from the “Basic
Statistics” submenu of the Stat menu. Indicate the variable in the “Samples” box and
the “Mated” column in the “Subscripts” box.]
Looking for assortative mating --
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
Measure females in the same way as males. Arrange the values on the spreadsheet so
that male and female measurements from a mated pair are in the same row. Then
look for significant correlations between male and female trait values for the different
characters. What is the potential significance of assortative mating?
A paper describing your results is due on September 30 by 5 PM. No rough drafts this
time, but I will be happy to discuss your results with you in advance of the deadline.
Here are some questions to consider:
1. From your data, can you conclude whether sexual selection occurred via male-male
competition or female choice? What arguments could you use to support either
mechanism? What kinds of data would provide stronger evidence?
2. If you found that characters were correlated, how do you interpret the target of sexual
selection? What studies could test your hypotheses more directly?
3. If sexual selection is occurring due to female choice, can you make conclusions as to
why certain characters are favored? What further studies could test hypotheses?
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