BIOL 101L: Principles of Biology Lab
Artificial Selection
Two weeks ago, you were introduced to methods for sampling and describing
populations. Specifically, you measured the cold tolerance of fruit flies sampled from a
large population. First, you measured ten flies and selected the three flies that tolerated
cold the best (i.e., they recovered from freezing the fastest). Then, you measured ten
more flies and selected the three flies that tolerated cold the worst (i.e., they recovered
from freezing the slowest). Finally, you measured ten flies and selected three of them
randomly with respect to their cold tolerance. You probably did not realize it at the time,
but you were performing artificial selection. Your experiment was analogous to the
process used by breeders of dogs, horses, cattle, crops, and other organisms that have
been domesticated by humans. Usually, such breeders select for changes in a trait in one
particular direction (e.g., bigger or smaller). In your case, you selected flies for changes
in both directions and randomly selected flies as a control group. Because each lab
section performed the same experiment, these selection regimes were replicated
throughout the week. This experimental design will enable us to observe evolution in
action.
This week, you will quantify the evolutionary response to selection by measuring the cold
tolerance of the offspring of your selected flies. You lab instructor will give you a Petri
dish containing these offspring. You will likely have fewer flies to measure than you did
on the first week because (i) the selected populations were initially small and (ii) only
two weeks have passed since selection occurred. Don’t worry about your sample size. We
will pool data from all of the lab groups to increase the power of our experiment. By the
end of the period, you should accomplish the following:
1) Measure the cold tolerance of offspring from your three selected populations
2) Graph the data in Excel to visualize the variation within and among populations.
3) Compare the cold tolerance among these populations using simple statistical
analyses.
4) Determine whether heritable variation in cold tolerance existed in the original
population.
Evolutionary Response to Selection
Evidence for natural selection abounds, and humans have shrewdly manipulated this
process to alter the traits of organisms in desirable directions. When humans play the role
of the selective agent, biologists refer to the process as artificial selection. Recall that
Darwin himself dabbled in the artificial selection of pigeons to convince himself of the
power of natural selection.
Whether selection stems from the stresses of nature or the whims of a human, one should
expect to observe an evolutionary response. The magnitude of this response (R) depends
on two factors: the intensity of selection (i) and the heritability of a trait (h2):
R = i·h2
The intensity of selection equals the difference between the mean of the trait in the
original population and the selected sample of the population (see Fig. 1 below).
Nevertheless, only genes can be passed on from parents to offspring. Consequently,
selection only leads to change in the trait if the variation among individuals results from
genetic factors. The heritability of a trait varies from 0 to 1; a value of zero means that
the trait does not depend on genetic factors, whereas a value of one means that the trait
depends entirely on genetic factors. If one performs artificial selection, the magnitude of
the response can be used to calculate the heritability of the selected trait. For example, if
no response occurs, one can conclude that the variation in the trait was not heritable.
Figure 1. The intensity of selection equals the difference between the mean phenotypes
of two groups: the original population and the selected population. The plot on the left
shows a linear relationship between a trait’s values and the fitness of the organism. This
relationship causes a difference in the mean and variance of the trait when measured in
the original and selected populations. The plot on the right highlights the difference
between the means for these two populations (i).
Methods for Data Collection
1. Obtain a sample of flies from your lab instructor. These flies are the offspring of your
selected flies from the first lab (Sampling Populations). Depending on the number of flies
that have developed, you may receive one to three groups of flies. Regardless of how
many flies you receive, you will pool your data with everyone else in your lab section
before beginning your analyses.
2. Measure the cold tolerance of your flies as you did in the first week of lab. Recall that
cold tolerance was defined as the time required to recover from a brief exposure to 0°C.
If you forget how you measured this trait, see the handout for the pervious lab (Sampling
Populations).
3. Import you data into Excel. Do not bother to calculate means and variances yet.
Instead, give your data to your lab instructor by using a flash drive or by e-mailing the
data. Your lab instructor will then pool the data for your lab section and give you the
pooled data set.
Methods for Data Analysis
1. Once you have the pooled data, start by describing these data with simple statistics
such as the mean, variance, and standard deviation. Practice using the function in Excel
that you were shown in lecture (average, var, stdev). If you don’t remember how to use
these functions, try looking them up using the Help feature of Excel. You will find that
these functions make the calculations much easier than they were when you performed
them during the first lab. When you have finished these calculations, consider the
patterns in your data. Which group had the lowest mean and which group had the highest
mean? Which group had the greatest variance?
2. Use Excel to create histograms of the data for each group. Do the means a variances
that you calculated in Excel seem to correspond to the distributions of data shown in the
histograms? In other words, do both suggest similar differences among the means and
variances of the three groups?
2. Plot the data for the three groups (fast, slow, and random flies) using a standard bar
graph. Add error bars to show the standard deviations of the means. Do these means
appear different when viewed graphically? If so, do the differences among groups match
your expectations?
3. Recall that differences among groups can arise from sampling error, even when no
biological process has created real differences. This sampling error makes it impossible
to know for sure whether the mean cold tolerances of your selected and random flies
were different. Fortunately, we can use statistical tests to conclude whether the means
were probably different.
To test for whether the means were different, we can state a null hypothesis. The null
hypothesis claims that there is no real difference between the true population means
estimated by the sample means.
Write down the null hypothesis for your flies: __________________________________
_______________________________________________________________________
_______________________________________________________________________
This is not a statement that we can ever know with certainty. Instead, the best we can do
is estimate the probabilities associated with rejecting the null hypothesis when it is really
true. For example, we would commit an error if we conclude that the two generations of
flies tolerated cold differently, when in reality they had the same population mean. If we
make some assumptions about the distributions of cold tolerances (e.g., we assume
normal distributions or bell curves), we can estimate the probability of making this error.
Specifically, the P-value tells us the probability that we would record the observed
difference between sample means (or a greater difference) when there really was not
difference between the population means. Biologists, commonly accept a 5% chance of
committing this error, which means we should only conclude that the means were
different if P < 0.05.
To determine the P-value, we will use a t-test. Use the Data Analysis package of Excel to
conduct t-tests, as you did in the previous lab on natural selection (Evolution
Simulations). Note that Excel will give you two different P-values: a one-tailed value and
a two-tailed value. In this case, you can use the one-tailed P-value because you expect the
cold tolerance of each selected population to differ from the cold tolerance of the control
population in a specific way (e.g., fast flies < control flies). Conduct three t-tests to draw
conclusions about your experiment:
1) Fast selected flies vs. control flies
2) Slow selected flies vs. control flies
3) Fast selected flies vs. slow selected flies
For each test, note the following pieces of information:
1) the t-value
2) the P value
3) whether or not you should reject the null hypothesis (i.e., whether you would
conclude the population means were different)
4) the biological interpretation of this result as it pertains to your experiment
Recall from lecture that a professional biologist would usually use a different kind of
statistical test to compare means among three or more groups (an F test). Nevertheless,
these the t-tests that you performed should lead to a similar conclusion without the need
for you to learn a more advanced statistical test at this early stage in your training.
Conclusion about the Heritability of Cold Tolerance
As described above, we would ideally quantify the heritability of the selected trait from
the intensity of selection and the evolutionary response. Unfortunately, we lack the ability
to precisely control environmental factors that influence cold tolerance over the period of
three weeks. For example, we know from our previous data that the temperature of the
lab bench affects the time to recover from cold exposure (recall our discussion of the lab
data in lecture). Very likely, your lab bench was a different temperature in week one than
it was today; there’s just no practical way we could control that variable. Therefore, we
might not be able to accurately measure the evolutionary response to selection by
comparing the means of the flies between weeks. However, we can infer an evolutionary
response if the cold tolerance of the selected flies and control flies differed significantly.
Furthermore, the magnitude of the difference between the means reflects the magnitude
of the heritability. Based on your analyses, was cold tolerance a heritable trait?