Ecology Lab Report Essay

An organism has several ways to avoid being prated upon. One way to avoid this is to practice crptis. Crypis is the action of organisms avoiding predation by blending in with their backgrounds and matching the color pattern of a bark, twigs or leaves. Palatable animals often utilize this strategy as well. Another type of defense is aposematism or warning coloratio. Organisms that produce noxious chemicals or accumalate them from food plants, advertise the fact that they are harmful with conspicous color patterns. Mimicry is the resemblence of an organism toward some other organism or an object inthe enviroment, evolved to decieve predators or prey into confuding the organism with that which it mimics. The prey involved within this experiment utilized mimisry as their defense stategy.

Batesia mimicry, Mullerian mimicry, and aggressive mimicry are all various forms of mimicry. Batesian mimicry is a resemblance of an unpalatable species (model) by an edible species (mimic) to decieve predators. Mullerian minicry is a mutual resemblance of two or more conspicuousyly marked upalatable species to enhance predator avoidance. Aggressive mimicry is a tactic that enables predators to avoid being detected by their prey or even to attract prey. The type of mimicry that was utilized within this experiment is Batesian mimicry.

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The purpose of thsi experiment is to observe predators or in this case birds, and determine whether or not mimicry developed based on the fact that the birds learned about what was really going on. Mimicry doesn’t always develop,. The predator must be smartiin order to realize that some are tasteful, while others are distasteful. Our predators can determine this by realizing that a color strategy is being put to use. The null hypothesis for this experiment is that if the don’t eat or remove most of the red larvae which happens to be 100%palatable, by the end of the week, then they did not realize that mimicry has occured. These hypothesis are based on the assumption that the birds realized that the red larvae are more tasteful , while the blue and lime larvae were often distasteful. The focus of this experiment is to find out whether or not mimicry developed and whether the birds either learnt from the effects of mimicry or caught on to the color pattern.
Materials and Methods
The procedure that was utilize from this experiment was very time consuming and involved an assortment of small procedures. The first step was to decide upon the different colors of the food, and which speculate on whch colors would be most efficient for the experiment. In our case we choose red,which represented a kind of dangerous color, blue which was still dangerous but a little more attractive than red and lime which seemed like a nice friendly color. After than the ratios between the models were agreed upon based upon what we thought the colors would represent toward the birds. The pattern that was used is pattern A. The red color had 100% mimicry, (palatable) and 0% for models (unpalatable). The blue color had 75% mimicry (palatable) and 25% for models (unpalatable). The lime had 25% mimicry and 75% for models (unpalatable). After that decision making the class broke up into different groups to facilitate the larvae making process. The class divided into mixers, cutters,those that would make the charts fort what food would be place outside the food net and various other job titles. The mixers had to mix lard with 1.66g as much flour, quinine sulfate, to give the unpalatable taste, and food coloring in order to produce the larvae. After all the mixing was done it was then put into the refrigerator. Shortly after, the different lard combinations began to be placed out on wax paper awaiting the cutters. The cutters began cutting the lard combinations into reasonable sizes for the predators to feed upon, about 1.5cm. The palatable and unpalatable were distiguished by the by the addition of quinine sulfate to the unpalatable portion. It was paramount that the students handeling the unplatable, didn’t deal with the palatable larvae for this could some how utler the results if there were more unpalatable ones than there were supposed to be. The rest of the class was to create random arrays to which the food was to be placed. These arrays are specific intructions as to where each larvae was to be placed. For example blue unpalatable would be next to the lime palatable. By this idea, each day would have a different deployment and retrieval data. Every student was then assigned to a specific time slot, so they could either deploy the larae, or retrive the remaining data, or both deployment and retrieval on the same day but diffrenet times. The procedure of this experiment was veru time consuming, but the team effort in the class realy simplified the procedure in numerous ways.

The materials for this experiment were also numerous. This experiment required water proof paper to write down data from the retrieval and deployment, in case if it rained and a thermometers to moniter the experiment. A feeding tray was used to deploy the food outside in the bird feeder. A large mixing bowl was used for mixeing the lard and flour, while meduim size bowls were used for mixing quinine sulfate and the required colors. Cookie pressors were utilzed for seperating the paltable from the unpalatable and making them ooze out into single rows in oerder to be measured by 15 cm rulers into eatable pieces and then cut with razor blades. Weighing trays spoon and top loading balances were used to make and mesure the larvae. 2kg/5lb of white flour along with 1kg/2lb of lard were the main components of the larvae, in addition to quinine sulfate and food colorig. The larvae were placed into little plastic petri dishes by the specific intructions laid out by the arrays. The feeing array charts were arranged on the waterproof paper and marked with marking pens. Disposable gloves, wax paper, and masking tape assisted the making process of making the food easier. A refrigerator was utalized for the storage of prepared petri dishes with larvae. The chi-sqare analysis and graphs, will help analyze how many larvae were removed or remained at the end of each deployment/retrieval period.
The results of this kind of experiment are typically used to either prove a null hypothesis (H) or approve an alternative hypothesis (H). The chi-square analysis is an effective means to prove or disprove hypothesis. The total # of prey removed for day 1 was 3. The # of red larvae deployed was 64. The observed removed from the red larvae was 2. The expected removed was 64*3/200 =.96. This was calculated frm taking the total amount deployed times the total amount of larvae that was observed as being removed, divided by the total amount of larvae deployed in that day. THe chi-square analysis for the red larvaewas (2-.96)^2/.96 = 1.1. For day one, the blue larvae deployed was 70., while the observed removed was 0. The expected removed was 70*3/200=1.05. The chi-square analysis for the blue was(0-1.05)^2/1.05= .0024. For day one, the lime larvae was 66, while the observed removed was 0. THe expected removed was 66*3/200= .99. The chi-square analysis for the green larvae was (0-.99)^2/.99=.99. In order to find the chi-square analysis for the total # of prey by color (combining AM and PM data) on day #1, you would simply add up all the chi-square numbers calculated and compare it to the (x^2 0.05, 2=5.991). In our case we added up 1.1+ .0024 + .99=2.0924.
The results for eight, the red larvae deloped was , while the obsereved removed was 19. The expected removed were 75*19/200=7.1. The chi square analysis for the red larvae deployed was (19-7.1)^2/7.1=19.9. For day 8 the blue larvae developed was 70, while the observed removed was 0. The expected removed were 70*19/200=6.7. THe chi-square analysis for the blue larvae was (0-6.7)^2/6.7=6.7. For day 8 of the lime larvae, deployed was 55, while the observed removed was 0. The expected removed were 55*19/200=5.2. THe chi square analyis for the lime was (0-5.2)^2/5.2=5.2. When all of these are added up, the total chi square analysis was 19.9+6.7+5.2=31.8
For the time period in which the number of prey was removed, you calculated the chi anaylsis for the AM and period PM of they eight. For the time perid of AM we would take the number, of larvae deployed, times the number removed divided by the total number deployed for the AM time period. Then we would add them up to get the chi analysis. For the expected removed for red larvae, the calculated was 36*9/100=3.24. For the expected removed number of blue, we calculated 35*9/100=3.15. For expected removed number of lime larvae, we calculated 26*9/100=2.34. To calculate the chi analysis for red larvae we would take the number removed minus the expected , then square it and divide by the number expected. That is (9-3.24)^2/3.24= 10.24. To find the chi analysis for the blue larvae you would calculate, (0-3.15)^2/3.15=3.15. To calculate the chi square analysis for the lime larvae you would calculate (0-2.43)^2/2.34=2.34. When all of these are added up 10.24+3.15+2.43=15.82
For the time period of PM, wwe would also take the number of larvae deployed, time the number removed removed divived by the total number deployed for the PM time period. Then we would add it up to find the chi analysis for the PM period. For the expected removed for red larvae, the calculated was 36(10)/100=3.6. For the expected removed for blue 35(10)/100=3.5. For the expected removed red larvae, the calculated was 29(10)/100=2.9. To calculate the chi analysis was the red larvae, you take the number removed minus the expected, then square it and divide by the numberexpected. That is (10-3.6)^2/100=11.37. To find the chi analysis for the blue larvae, you would calculate, (0-3.5)^2/3.5=3.5. To find the chi analysis for the lime larvae, the calculate was (0-2.9)^2/2.9. When all of these are added up 11.37+3.5+2.9=17.8.

The results for this experiment did not surprise me at all. I did not expect that on the first day, that there would be a color favor, amongst the birds. I realized that although lime might be a more pleasant color than red or blue, the birds might not find this favoritism. Although we knew that the red was the most palable, the birds would have to test the larvae first before coming to a conclusion as to which one they prefered. According to the data, I failed to reject my hypothesis because the chi sqare analysis that I calculated, 2.09, is smaller than the 5.991 which is the chi analysis given in the book. When the chi analysis is smaller, this shows that , my hypothesis was not refused and that there was no color prefrence chosen by the birds on the first day. Also through the graphs, one can also see frrom the pattern of the percentage line of removed food, vs time, that there was no larvae color preference.
By the eighth day I expected that the birds would have realize that only the red was since the most palatable larvae, especially since there was only about twenty five percent lime that was palable. However, through tha graph, One can see that toward the eight day the birds were still feeding on some of the lime larvae. This made it an oblivious indication that mimicry had occured. Also I calculated chi square value was much higher that given in the book, which is about a comparision of 31.8 vs 5.991. Since it is higher, I failed to accept my null hypothesis. This means that the birds up on till that they could still not really tell the difference, or had’t really noticed that some of the food was more palatable than others.

In addition to the eight day, I had also seperated the time periods to see if there was any effect on temperature on the out come of the feed in each the birds removed the food. After graphing the data, I failed to accept my null hypothesis. I thought that the diffent time periods of the day would affect when the birds would usually remove the larvae. I imagined that birds would come out first thig in the morning and then go bacl to their nest. However through the one could see that the results in both days were very similar. In addition when I calculated the chi square analysis for both periods, they were 17.7 for PM and 15.83 for Am. These numbers were both larger than the given value in the book of 3.84, which is why I failed to acceot my null hypothesis.
The results of this lab were sometimes a little difficult to interpret due to various mistakes. One large and costly mistake was missing data. Several groups did not report their data, which resulted in the graphs and calculations being somwhat unusual and often difficult to understand. There were also often large gaps in the graphs. The problem of negative numbers in the data threw off the graphs even more than it already was. Also, although not proven, it is possible that othe organisms or enviromental conditions could have interfiered with the data collection. The factor of the wind must be taken into account for emoving the prey. If the prey was removed, it doesn’t mean that the predators had eaten it. The predator also could it in his or her mouth and spit it right back out. This response would be very important to mimicry , because the preadator or prey is realizing what is palatable and what isn’t. The negative numbers that appeared in this experiment could have been the cause of experimental error, due to the fact that there were two extra larvae in each petri dish. There were quite a few mitakes in this data but at least we successfuly obtained viable data.

According to the graphs, the birds preyed upon the red larvae the most. The red larvea was 100% palatable and 0% unpalatable. I expected the red larvae to be preyed upon the most, but what shocked me was the amount of lime that was still being preyed on while the birds approched the eighth day. I thought that since there was only 25% of the lime larvae that was palable the would have picked it up right away. Even the blue larvae was removed about as much as s the lime was another indication that mimicry had occured. The blue was only 25% unpalatable there fore it should have yeilded much higher results than the lime. Some times as was a little confused as to whether my accepting or rejecting a null hypothesis was correct. This is because the chi square analysis isn’t a procedure thatalways ccurately reject or support the null hypothesis. The chi square analysis is most likely a test of randomness, rather than a test for supporting or rejecting the null hypothesis. This is also why we say that we fail to rejact it, because, just because we fail to reject it doesn’t mean that it is correct, therefore we cannot accept it. This is also why it is good to have graphs, in an experiment like this one, so that they can back up your results and hypothesis, because it is giving you a visual sense of what is going on.

Charlene Ngong


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