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) / Predator (Humans)Alleles / Color / Alleles /
Food Gathering Organ
CC / Red / MM / SpoonCC¢ / Mottled / MM¢ / Fork
C¢C¢ / Black / M¢M¢ / Tongue Depressor
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:
,
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.
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 (Dp) 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.
Prey (beans) / Predator (students)Generation 0 / Generation 0
p0 =______/ p0 =______
q0 = ______/ q0 = ______
# red = ______/ # Spoon = ______
# mottled = ______/ # Fork = ______
# black = ______/ # TD = ______
# red Survivors = ______/ # Spoon Survivors = ______
# mottled Survivors = ______/ # Fork Survivors = ______
# black Survivors = ______/ # TD Survivors = ______
Total Survivors = ______/ Total Survivors = ______
Generation 1 / Generation 1
p1 = [#Red + 0.5*#Mottled]/[Total Survivor] = ______/ p1 = [#Spoon + 0.5*#Fork]/[Total Survivor] = ______
q1 = [#Black + 0.5*#Mottled]/[Total Survivor] = ______/ q1 = [#TD + 0.5*#Fork]/[Total Survivor] = ______
# Red = p12 * 1200 = ______/ # Spoon = p12 * Class Size = ______
# Mottled = 2p1q1 * 1200 = ______/ # Fork = 2p1q1 * Class Size = ______
# Black = q12 * 1200 = ______/ # TD = q12 * Class Size = ______
# Red Survivors = ______/ # Spoon Survivors = ______
# Mottled Survivors = ______/ # Fork Survivors = ______
# Black Survivors = ______/ # TD Survivors = ______
Total Survivors = ______/ Total Survivors = ______
Generation 2 / Generation 2
p2 = [#Red + 0.5*#Mottled]/[Total Survivor] = ______/ p2 = [#Spoon + 0.5*#Fork]/[Total Survivor] = ______
q2 = [#Black + 0.5*#Mottled]/[Total Survivor] = ______/ q2 = [#TD + 0.5*#Fork]/[Total Survivor] = ______
# Red = p22 * 1200 = ______/ # Spoon = p22 * Class Size = ______
# Mottled = 2p2q2 * 1200 = ______/ # Fork = 2p2q2 * Class Size = ______
# Black = q22 * 1200 = ______/ # TD = q22 * Class Size = ______
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 Dp:
Dp = [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:
Dp = [0.5*0.6*0.5*0.5] / [1 – (0.6*0.5)], or
Dp = [0.075] / [1 – 0.3], or
Dp = [0.075] / [0.7], or
Dp = 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:
Dp = [0.5*0.35*0.5*0.5] / [1 – (0.35*0.5)], or
Dp = [0.04375] / [1 – 0.175], or
Dp = [0.04375] / [0.825], or
Dp = 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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