Variability exists in all natural populations. For a wide variety of reasons, some phenotypes (visible characters) or genotypes exploit the environment more efficiently than others do. These phenotypes leave proportionately more offspring than their counterparts. If this phenotypic characteristic is heritable, offspring will resemble their parents, and the population will eventually consist mostly of individuals of the successful phenotypes. This in turn will alter the allele frequencies for that trait within the population. This weeding out of the less fit phenotypes is the process of natural selection. The accompanying change in the allele frequencies within the population is evolution. The features that ultimately come to characterize the species are then viewed as adaptations to the environment.
Since the key to evolutionary success is the production of fertile offspring, it requires numerous generations to demonstrate change in allele frequencies through time. A model system then may provide a simulation of the process.
In this model system, a plastic fork/spoon and a cup represent a predator. Two seeds represent different phenotypes within a prey species. Picking up seeds with a fork from the tabletop and putting them in the cup can simulate a feeding frenzy. Seeds that remain after the frenzy are the survivors and are the ones that can reproduce to reestablish the population. [An alternative model is to use the thumb and index or middle finger on one hand to pick up the seeds and deposit them in the cup].
Procedure:
- Work in groups of two or three.
- To start, count out 200 of each seed type (chick pea and green split pea), mix them and spread them evenly on the entire table clear of everything except the seeds.
- Assign roles to each person. You will need two predators and one timer. Rotate the tasks for each “feeding frenzy”.
- To create a “feeding frenzy”, the two predators should pick up as many seeds as possible with the fork/spoon in a 15 second interval and place them in a cup (the timer should indicate when to start and when to stop). These seeds in the cup represent the prey that have been eaten by predators and thus removed from the population. Record the number of each type of seed eaten (i.e. the number of each type of seed in the cup) under “number eaten”.
- Count the left over survivors of each prey phenotype (each type of seed) and record this in the chart provided under “survivors”.
- It is the survivors that breed according to their survivorship and reestablish the population at the equilibrium population size of 400. To determine the number of each type of seed to start with in your next generation:
- Calculate the “proportion of surviving population” by dividing the number of each type of survivor by the total number of individuals in the surviving population (i.e. the total number of seeds left on the table).
- Multiply this by 400 to determine how many of that type of seed you will need to start your next generation.
An example of this is shown below.
Prey | Starting Population | Number Eaten | Survivors | Proportion Surviving Population | Adjusted Population |
Green Split Pea | 200 | 20 | 180 | 0.56 x 400= 224 | |
Corn | 200 | 60 | 140 | 0.44 x 400=176 |
- The adjusted population then serves as the starting population for the next generation and next 15 second feeding frenzy. Repeat the above procedure for ten generations.
Merely observing changes through time are insufficient evidence that change has occurred. We need to provide statistical evidence. Therefore, we will use the Chi-Square test (explanation and examples below) to determine if a significant change has occurred.
Observations:
Table #1: Surviving peas after 15 seconds of “feeding frenzy” by fork/spoon predators for 4 generations.
Generation | Number of prey | Initial Population | Number of prey eaten | Number of survivors | Proportion of surviving population
(# of type of prey/total # of survivors) |
Adjusted Population
(proportion of surviving population x400) |
Split Pea | 200 | 13 | 187 | 187/368=0.50 | 0.50 x 400=200 | |
Chick Pea | 200 | 19 | 181 | 181/368=0.49 | 0.49x 400=196 | |
TOTAL | 400 | 368 | ||||
Split Pea | 203 | 15 | 188 | 188/362=0.51 | 0.51 x 400= 207 | |
Chick Pea | 197 | 23 | 174 | 174/362= 0.48 | 0.48 x 400= 192 | |
TOTAL | 400 | 362 | ||||
Split Pea | 208 | 15 | 193 | 193/362=0.53 | 0.53 x 400= 213 | |
Chick Pea | 192 | 23 | 169 | 169/362=0.46 | 0.46 x 400=186 | |
TOTAL | 400 | 362 | ||||
Split Pea | 213 | 23 | 190 | 190 /359=0.52 | 0.52 x 400=211 | |
Chick Pea | 187 | 18 | 169 | 169/359=0.47 | 0.47x 400=188 | |
TOTAL | 400 | 359 | ||||
Split Pea | 212 | 19 | 193 | 193/354=0.54 | 0.54 x 400= 218 | |
Chick Pea | 188 | 27 | 161 | 161/354=0.45 | 0.45x 400=181 | |
TOTAL | 400 | 354 |
Assume that the green Split Pea is the homozygous recessive phenotype in this population (aa)
Data Analysis:
Determining Allele frequencies – Hardy-Weinberg Equilibrium
A population is in genetic equilibrium when allele frequencies in the gene pool remain constant across generations.
The Hardy-Weinberg principle states that a gene pool will be in genetic equilibrium under the following conditions:
• the population is very large
• individuals in the population mate randomly
• there is no migration into or out of the population
• natural selection does not act on any specific genotypes
• males and females have the same allele frequencies [vs. individuals are diploid and reproduce sexually]
• no mutations occur
Another way to look at it is that allele frequencies in a population will remain constant over time (from generation to generation) unless there are evolutionary pressures to change them. This means, that if the environment is constant, we expect allele frequencies to be constant as well. Some evolutionary pressures include genetic drift, gene flow, natural selection, sexual selection, bottleneck effect, etc.
Let “A” = p and “a” = q. Thus, allele frequencies in a given population can be expressed as p + q = 1
Let “AA” = p2 Let “Aa” = 2pq Let “aa” = q2
Thus, genotype frequencies in a population can be expressed as p2 + 2pq + q2 = 1
(note: This is (p+q)2 = 12)
If we count or sample the number of each phenotype in a population, we can calculate allele frequencies.
Sample problem:
In humans, the Rh factor genetic information is inherited from our parents, but it is inherited independently of the ABO blood type alleles. In humans, Rh+ individuals have the Rh antigen on their red blood cells, while Rh- individuals do not. There are two different alleles for the Rh factor known as Rh+ and rh. Assume that a dominant gene Rh produces the Rh+ phenotype, and that the recessive allele produces the Rh- phenotype.
In a population that is in Hardy-Weinberg equilibrium, if 163 out of 200 individuals are Rh+, calculate the frequencies of both alleles.
If 163 individuals are Rh+ (dominant), that means that p2 + 2pq = 163/200. This doesn’t really help us.
It is easier to start with the recessive condition.
200-163 = 37 individuals are Rh-. This means that:
q2 =
q2 = 0.185
q =
q = 0.43
Since we know p + q = 1,
p = 1 – 0.43
p = 0.57
Therefore, the frequency of the Rh – (recessive) allele is 0.43 and the frequency of the Rh+ (dominant) allele is 0.57
Are these new frequencies different enough to say evolution has happened? This is where the Chi-Square test comes in. We will do a sample calculation together.
Natural Selection Lab Report
THE REPORT: Should include the following sections:
- Research Question
- clearly stated (easier to write when you have your conclusion)
- Data Analysis (T/I)
- Table (s) showing summary of your data
- frequencies (allele and genotype)
- Sample calculation of the frequency of each allele for all 4 generations (you will have to use Hardy-Weinberg here)
- Sample calculation of the frequency of each genotype for all 10 generations.
- Graph
- Showing allele frequencies over the 4 generations
- Scatterplot with points connected
- Make sure axes are labelled properly (including units where necessary)
- Both alleles should be graphed on the same axes
- Give your graph a “footer” – a detailed description that goes at the bottom of the graph
- Showing allele frequencies over the 4 generations
- Calculation of Chi-Square test where your original population is your expected value (E) and your final population is your observed (O).
- Table (s) showing summary of your data
- Conclusion
- A statement answering your research question (Is evolution occurring?)
- Discussion (A)
- A paragraph or two justifying your conclusion using the data and scientific context
- Include the results of your data
- A scientific context – a discussion of natural selection and how it was at work in this simulation. (For example, does this model support the theory of natural selection of Darwin’s survival of the fittest? Explain)
- What are the weaknesses of this experiment (based on the procedure you were given)?
- Some things you should think about are:
- What was missing in this simulation? What factors in nature were not accounted for or included that could impact results?
- How could you change or add to the lab in order to add in some of those missing factors or make it better?
- Some things you should think about are:
- A paragraph or two justifying your conclusion using the data and scientific context
Overall, For Communication (C)
- Use appropriate scientific terminology
- Follow appropriate conventions (headers on tables (e.g., Table 1: detailed description of what is found in the table) and footers on graphs, Figure 1: detailed description of what is found in the graph)
- Write in plain language. Don’t try and make it “fancy”. I should be able to clearly understand what you are saying after reading it once. (It is a good idea to have someone who is not familiar with the lab read it over for you to check for this!)