NATURAL SELECTION: A SIMULATION GAME
Introduction
Natural selection is one of the primary driving forces behind evolution. Natural selection
leads to directed changes in the frequencies of allele, due to differences among individuals in
their ability to obtain the resources necessary to survive and reproduce. In today’s lab we want to
simulate natural selection in action, using a fairly simple, and hopefully fun, procedure with
some basis in ecology.
Suppose that we have populations of two different species, a predator and a prey. The
prey species are represented by (almost completely) defenseless beans, and the predator species
are represented by voracious Homo sapiens. For the beans, we will focus on a single trait, bean
color, which we’ll pretend is determined by two codominant alleles, C and C (codominance
occurs when neither allele is dominant over the other). We will also focus on a single predator
trait, the food gathering organ, that we’ll also pretend is determined by two codominant alleles,
M and M. The chart below specifies what allele combination (that is, genotype) leads to what
color or organ (that is, phenotype):
Prey (Beans)
Alleles
CC
CC
CC
Predator (Humans)
Alleles
Food Gathering Organ
MM
Spoon
Fork
MM
Tongue Depressor
MM
Color
Red
Mottled
Black
Methods
To start out with, we assume that allele frequencies are 50/50 for both predator and prey
populations. We can then use the Hardy-Weinberg equation (see 23-129 in your text for more
details) to figure out the trait (phenotype) frequencies:
p 2  2 pq  q 2  1 ,
where p = frequency of the C (or M) allele and q = frequency of the C´ or M´ allele. If p = 0.5
and q = 0.5, then our trait frequencies will be 0.25 (or p2), 0.5 (or 2pq), and 0.25 (or q2). Thus, in
the prey population we will start out with equal numbers of red and black beans, but twice as
many mottled beans. For the predators, we have equal numbers of spoons and tongue depressors,
with twice as many forks. The initial number of each bean color will be: 300 red, 300 black, and
600 mottled. The initial number of food gathering organs will depend on the class size.
Now comes the fun part! We’ll begin by assigning each predator a food gathering organ
and a cup (which represents your stomach). The class will then go outside to a designated
foraging area, where you’ll gather in a large circle facing away from the middle. The lab
instructor will then randomly distribute the prey (that is, toss them around the foraging area), get
out of the way, and shout GO!! Once you hear this, begin foraging…BUT with the following
rules:
1. You may collect prey only with your foraging organ.
2. You must keep your stomach (cup) off the ground at all times.
3. You may not upset another predator’s cup.
4. You may collect any prey another predator has dropped.
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We will forage this way for one minute, after which time will be called. We’ll then count up the
number of each type of bean collected, and from that figure out the frequencies of the
“survivors” (those beans still on the ground). Using these frequencies, we’ll then calculate the
composition of the next generation of beans, assuming random mating.
For the predators, we’ll arbitrarily assume a 50% survival rate. Thus, the 50% of the
predators which gather the most prey will survive to pass on their food-gathering organ alleles,
while the other 50% of the predators will be eliminated along with their pathetic alleles. We will
then count the collecting organs of the survivors, determine the allele frequencies, and use these
to determine the composition of the offspring (just as we did above with the beans). Finally,
we’ll allow the “deceased” predators to re-animate with new foraging organs, and repeat this
process 1 or 2 more times.
Lab Write-Up
For this week’s lab, you are to answer the following questions, and turn them in no later
than the beginning of the next lab period.
1. What are the allele frequencies and phenotype frequencies of the prey population during
each generation? What are the allele frequencies and phenotype frequencies of the
predator population during each generation? NOTE: You should calculate these during
class.
2. How did allele frequencies change over time in the prey population? How did allele
frequencies change over time in the predator population? How and why might these
observed changes have arisen? HINT: Consider the foraging organs, the beans, and the
habitat in answering this last question.
3. Briefly describe a real-world example that would resemble the results of this experiment.
In other words, can you come up with a “real” example where prey organisms would
evolve ways to avoid getting eaten and predator organisms would evolve ways to better
catch their prey?
4. EXTRA CREDIT: (For this question, use only the results from the first selection trial)
(A) Calculate the selection coefficient s for the prey and the predator populations. (B)
Use this to predict the change in p (p) that should arise, again for both the prey and
predator populations. (C) Finally, about how long (that is, how many generations) will it
take for the “better” allele in each population to reach fixation? NOTE: See pages 4 & 5
for details on how to calculate these variables.
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Prey (beans)
Generation 0
Predator (students)
Generation 0
p0 =______________________
q0 = _____________________
# red = ____________________
p0 =______________________
q0 = _____________________
# Spoon = ____________________
# mottled = __________________
# black = ____________________
# Fork = __________________
# TD = ____________________
# red Survivors = _______________
# mottled Survivors = ________________
# Spoon Survivors = _______________
# Fork Survivors = ________________
# black Survivors = _________________
Total Survivors = __________________
# TD Survivors = _________________
Total Survivors = __________________
Generation 1
p1 = [#Red + 0.5*#Mottled]/[Total Survivor] =
_______________
Generation 1
p1 = [#Spoon + 0.5*#Fork]/[Total Survivor] =
_______________
q1 = [#Black + 0.5*#Mottled]/[Total Survivor] =
_______________
q1 = [#TD + 0.5*#Fork]/[Total Survivor] =
_______________
# Red = p12 * 1200 = ______________
# Mottled = 2p1q1 * 1200 = _____________
# Black = q12 * 1200 = ______________
# Spoon = p12 * Class Size = ___________
# Fork = 2p1q1 * Class Size = __________
# TD = q12 * Class Size = ______________
# Red Survivors = _______________
# Mottled Survivors = ________________
# Spoon Survivors = _______________
# Fork Survivors = ________________
# Black Survivors = _________________
Total Survivors = __________________
# TD Survivors = _________________
Total Survivors = __________________
Generation 2
p2 = [#Red + 0.5*#Mottled]/[Total Survivor] =
_______________
Generation 2
p2 = [#Spoon + 0.5*#Fork]/[Total Survivor] =
_______________
q2 = [#Black + 0.5*#Mottled]/[Total Survivor] =
_______________
q2 = [#TD + 0.5*#Fork]/[Total Survivor] =
_______________
# Red = p22 * 1200 = ______________
# Mottled = 2p2q2 * 1200 = _____________
# Black = q22 * 1200 = ______________
# Spoon = p22 * Class Size = ___________
# Fork = 2p2q2 * Class Size = __________
# TD = q22 * Class Size = ______________
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Calculating Fitness and Selection Coefficients
Suppose we have the following example, starting with the predators. We begin with 24
predators in generation 0: 6 spoons (MM), 12 forks (MM), and 6 tongue depressors (or TDs;
MM). This gives initial allele frequencies of p = 0.5 and q = 0.5. After collecting beans, the 12
foragers which were least successful were killed; suppose this eliminated 1 spoon, 7 forks, and 4
TDs. There were then 5 spoons, 5 forks, and 2 TDs which survived. Now, as one measure of
absolute fitness we can use the probability of survival. We’ll call this fitness measure W, so
that:
W = (# of survivors) / (initial #)
We then have:
Wspoon = 5 / 6 = 0.833 (or, 83% of spoons survive)
Wfork = 5 / 12 = 0.417 (or, 41% of forks survive)
WTD = 2 / 6 = 0.333 (or 33% of TDs survive)
Biologists typically do not use absolute fitness; instead, we calculate something called relative
fitness. This measure, symbolized w, is given by:
wi = (absolute fitness of trait i) / (maximum absolute fitness in population)
In our predator population, the maximum absolute fitness is 0.833, for spoons, and so we have:
wspoon = 0.833 / 0.833 = 1
wfork = 0.417 / 0.833 = 0.5
wTD = 0.333 / 0.833 = 0.4
What this says is that forks have 50% of the fitness of spoons, while TDs have only 40% of a
spoon’s fitness. And now, we can use these relative fitness values to calculate the selection
coefficient s, which measures how fast selection will occur if it is directional (as is true for our
situation). In the simplest case, we have:
s = 1 – (fitness of homozygote with the lowest value);
note that this is not always the correct formula for s, but will work for us! If you recall that
homozygous means having two of the same allele, then our two homozygotes are TDs and
spoons. Since TDs have lower fitness than spoons, for our predators we have:
s = 1 – 0.4 = 0.6.
So why do we want to know this? Well…knowing the selection coefficient will let you estimate
how long it will be before the better allele (M, in this case) will completely take over the
population. This process is called fixation, and occurs when p = 1 (and thus q = 0). We do this
by calculating the rate of change of p per generation, or p:
p = [0.5*s*p*q] / [1 – (s*q)],
where p and q are the values before selection occurred. For our predators, p = 0.5, q = 0.5, and s
= 0.6, so plugging these values into the equation gives:


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p = [0.5*0.6*0.5*0.5] / [1 – (0.6*0.5)], or
p = [0.075] / [1 – 0.3], or
p = [0.075] / [0.7], or
p = 0.11
This indicates that, for each generation, p should increase by 0.11. Thus, in generation 1, p
should equal 0.61, in generation 2 it should equal 0.72, and so on. By generation 5, p should
reach fixation (p = 1) and all foraging organ alleles should be M; that is, we should have a
population full of spoons.
Just to give you a second, but shorter, example, let’s look at the prey. Here, we start out with
1200 beans (300 red, 600 pinto, and 300 white) in generation 0, and so again have p = 0.5 and q
= 0.5 initially. Suppose that after the predators have done their work, we have 203 surviving red
beans, 320 surviving pinto beans, and 133 surviving white beans. This gives absolute fitnesses
of:
Wred = 203 / 300 = 0.677 (or, 68% of red beans survive)
Wpinto = 320 / 600 = 0.533 (or, 53% of pinto beans survive)
Wwhite = 133 / 300 = 0.443 (or 44% of white beans survive)
Red beans have the highest absolute fitness, and so we have:
wred = 0.677 / 0.677 = 1
wpinto = 0.533 / 0.677 = 0.78
wwhite = 0.443 / 0.677 = 0.65
Since the homozygotes with lowest fitness are the white beans, we have for the selection
coefficient:
s = 1 – 0.65 = 0.35
Now, we can figure out the rate of change of p:
p = [0.5*0.35*0.5*0.5] / [1 – (0.35*0.5)], or
p = [0.04375] / [1 – 0.175], or
p = [0.04375] / [0.825], or
p = 0.05
Thus, p will increase by 0.05 each generation (p = 0.55 in generation 2, p = 0.6 in generation 3,
and so on), and it will be generation 10 when the prey reach fixation so that all alleles are C and
all beans are red.
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