**A Simulation of Natural Selection**: heritable variations in beetle morphology

**Introduction****:** Evolution is a process that changes the genetic makeup of a population over time. Presumably, those genetic changes are reflected in changes in phenotypic makeup (observable characteristics) of the population. This exercise will demonstrate the effect of natural selection on the frequencies of three populations of “beetles”. Natural selection, as formulated by Charles Darwin in Origin of Species (1859), is the most important cause of evolution.

An individual’s ability to reproduce depends on its ability to survive. If all gene variations conferred on every individual the same capability to survive and reproduce, then the composition of a population would never change. If a variation of a characteristic increases an individual’s ability to survive or allows it to have more offspring, that variation will be naturally selected. Darwin reasoned that the environment controlled in nature what plant and animal breeders controlled artificially. Given an overproduction of offspring, natural variation within a species and limited resources, the environment would “select” for those individuals whose traits would enable a higher chance of survival and hence more offspring in the next generation.

**Purpose:** In this activity you will use different beans to represent heritable variations in beetle morphology (size and coloration of carapace). These three beetle morphs will be studied in two different habitats.

**Procedure: **Work with a partner. Count out exactly 10 each of the three types of bean. These represent beetle type; the three types are equally common initially.

- You will choose 2 habitats and perform the procedure the same way in each one. Choose 2 from this list: sidewalk, asphalt, sand, dirt. The point is that the two substrates must be different colors. Label your data table with the type of habitat of both Habitat #1 and Habitat #2.
- Scatter your beans randomly over the area about one square meter. This will be your predator foraging area. The beans must be scattered (tossed),
dumped in a small pile.__not__ - One person will be the designated predator for the habitat. (Switch roles for the second habitat). The predators “eat” (pick up) exactly 20 beans. Remember to think like a predator. I.e. pick up what you see first as quickly as you can. Place the 20 beans picked up in the plastic bag provided. Leave the rest of the beans on the ground. They have survived the predator and will reproduce (their offspring will be represented by adding beans for each of the survivors bringing the population size back up to 30, line D).
- Count the number of each bean collected (make sure you have exactly 20) and record the numbers on line B.
- Subtract the number of each kind eaten (line B) from the number you started with (line A), to obtain the number of survivors (line C).
- Assume each survivor has two offspring. Record those values in line D. These are the numbers of each bean that need to be scattered with the survivors to bring the population back to 30. Count out and scatter the required number of beans into the same area as your P1 survivors. Now complete line E by adding lines C and D. These are your P2 or second-generation populations.
- Repeat steps 4 through 7 two more times and complete the table for Habitat #1. Remember the offspring values tell you how many beans each type need to be scattered into your predator foraging area.
- Pick up all your beans when you are finished. Repeat the entire procedure in Habitat #2.

## Results and Analysis:

## Summary of Beetle Captures

Lima | Pinto | Kidney | Total | Lima | Pinto | Kidney | Total | ||

A | P1 | 10 | 10 | 10 | 30 | 10 | 10 | 10 | 30 |

B | “eaten” | 20 | 20 | ||||||

C | “survivors” (A-B) | 10 | 10 | ||||||

D | “offspring” (2C) | 20 | 20 | ||||||

E | P2 (C+D) | 30 | 30 | ||||||

F | “eaten” | 20 | 20 | ||||||

G | “survivors” (E-F) | 10 | 10 | ||||||

H | “offspring” (2G) | 20 | 20 | ||||||

I | P3 (G+H) | 30 | 30 | ||||||

J | “eaten” | 20 | 20 | ||||||

K | “survivors” (I-J) | 10 | 10 | ||||||

L | ” offspring” (2K) | 20 | 20 | ||||||

M | P4 (K+L) | 30 | 30 | ||||||

| | | |

c^{2 }**Analysis of Beetle Captures**

Habitat | Phenotype | Expected (e) | Observed (o) | (o – e)^{2}e |

#1 | Kidney | 10 | ||

Pinto | 10 | |||

Lima | 10 | + ____________ | ||

Total | c=^{2} | . | ||

P = | . | |||

#2 | Kidney | 10 | ||

Pinto | 10 | |||

Lima | 10 | + ____________ | ||

Total | c=^{2 } | . | ||

P = |

** **You will determine whether evolution has occurred by comparing the frequencies of each bean morph as the beginning and at the end of our predation experiment. You will use a **Chi-Square Statistic (**c** ^{2}) **to determine whether or not our final frequencies are significantly different from our initial ones.

Chi square = sum of (o – e)^{ 2} o = observed count in a category

e e = expected count in that category

The Chi square value that you get is located on the Chi square table below. The degrees of freedom is one less that the number of phenotypic categories. We have 3 phenotypic categories (the three bean morphs).

**Probability of exceeding the critical value**

Degrees of Freedom | P = .99 | P = .95 | P = .75 | P = .50 | P = .25 | P = .05 | P = .01 |

1 | 0.00157 | .00393 | .102 | .455 | 1.32 | 3.84 | 6.63 |

2 | .0201 | .103 | .575 | 1.39 | 2.77 | 5.99 | 9.21 |

3 | .115 | .352 | 1.21 | 2.37 | 4.11 | 7.81 | 11.3 |

4 | .297 | .711 | 1.92 | 3.36 | 5.39 | 9.49 | 13.3 |

A probability of less than 0.05 tells you that there is less than a 5% chance that the differences between our beginning and ending counts could be due to random variation. This means that there is a 95% probability that the differences are not due to random factors but are a result of our experiment.

Any statistic can only test two hypotheses, the null hypothesis of **no difference** and the alternative hypothesis of **significant difference**. These statistical hypotheses can be summarized below:

H_{0} : (the null hypothesis): there is no difference between our beginning and ending counts.

H_{1} : (the alternative hypothesis): there is a statistically significant difference between our beginning and ending counts.

A statistical hypothesis is not the same as an experimental hypothesis: our experiment hypothesis involves the role of predation and habitat on natural selection. What hypotheses were tested in this simulation?

**Information to include in your Discussion and Conclusions**:

How and why do your results differ for the two habitats?

How can you explain your results in terms of the hypotheses?

**Lab Write-up**:

1. **TITLE**: Title of Lab and your name and date.

2. **HYPOTHESIS**: A paragraph describing the specific hypotheses tested in the simulation.

3. **RESULTS**: A paragraph describing the meaning of your Chi square statistics.

4. **DISCUSSION**: A paragraph or more discussing the relevance of your results, answering the questions above, and including relevant information to support your findings.

5. **CONCLUSION**: A well-crafted conclusion of no more that two sentences.

Make sure to type up your data table and Chi square test and include them in the paper.

This write up will be due September 23rd.