Ecology final paper
QUESTION A
(Multiple choice question)
Which one of the following is a biotic component?
- A) Temperature
- B) Minerals
- C) Bacteria
- D) Humidity
Answer: C-Bacteria
Justification
The biotic component here is bacteria, while the alternative answer in place of bacteria is plants, or fungi. Generally, the ecosystem is made of both biotic and Abiotic factors that may either cause a decrease or increase in population of an organism or ecosystem in question. While biotic Factors are living components of an ecosystem, Abiotic factors are the non living components of an ecosystem. The biotic components include the producers that convert energy to food, consumers that depend on the producers for food and last the decomposers that breakdown chemicals into simple form that can be used. According to the choices, temperature, humidity and minerals exhibit no life hence non living things thus Abiotic. On the other hand, bacteria exhibit the ability to live hence a biotic factor under the category of decomposers. Therefore, the right choice is bacteria since it’s a living component of an ecosystem.
QUESTION B
(Fill in blank the question)
__________are organisms that live within a single year and __________are organisms that live up to two years
Answer: Annuals and biennials respectively.
Justification
The annuals exhibit only several or many generations within a year i.e. flowers and grasshoppers and biennials utilize their first year in vegetative growth where they reproduce and die I the following year i.e., onions and lettuce. Therefore, making aforementioned answers fit for the bank spaces above.
QUESTION C
(Essay question)
Recently there have been allegations of over fishing in your community pond. The community leader therefore, decides to recruit you to determine the current population size of the fish in the pond. As an ecology student, briefly describe how you could count the fish and give both the advantages and disadvantages of the method of your choice.
Answer
In order to determine the current population size of the fish in the pond, i would use the mark and recapture technique. First, I would visit the pond and capture a small portion of the fish from it. Then, I would put a harmless mark on them before they are released back to the pond. Then on the second visit I would capture the second portion and count the number of the marked fish in the sample. Thereafter, I would determine the population size by dividing the sum of marked fish by the proportion of the marked fish in the second capture sample. Since the sum of the marked fish in the second capture is proportional to the sum of the marked ones in the pond.
The advantage of this technique is that its accuracy does not depend on the size of the habitat. Furthermore, it’s non-invasive hence distance sampling is possible. Last, the method allows measuring of population Abundance. However, there are drawbacks of this method like death of the marked fish in between the sampling time. Second, change of behavior makes it harder or easier to get hold of the recapture round of sampling. Last, the accuracy of this method depends on capturing a large proportion of the population.
Justification
This method is suitable for this case since it gives an absolute estimation of the current population of the fish within a fixed unit of habitant (fish pond). The method assumes that the marked fish will not be affected by the mark and the mark does not fade, the marked fishes remain completely fixed in the population, the probability of recapturing a previously marked fish is similar to that of capturing any other fish in the pond and last, the time interval of capturing the samples is small in relation to the total time. Therefore, this makes this method the most convenient for concluding whether there was over-fishing or not in the community pond since the current population can be compared to the initial population size.
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