Final Exam KEY
Multiple choice. Follow the instructions to choose either the single best answer or all correct answers. Be careful--even if a statement is true, it may not be a correct answer to the question. Each question is worth 4 points.
1. Why is the assimilation efficiency generally higher for carnivores than it is for herbivores? CIRCLE ALL CORRECT ANSWERS
a. Carnivores must eat more to maintain their body temperatureb. Carnivores are generally larger than herbivores
c. Carnivores are less abundant than herbivores
d. Carnivores' food is more similar to their own bodies than herbivores' food
2. Which of the following is/are true about mycorrhizae? CIRCLE ALL CORRECT ANSWERS
a. mycorrhizae fix nitrogen for plantsb. mycorrhizae vastly increase surface area for nutrient uptake
c. almost all plants associate with mycorrhizae
d. mycorrhizae are bacteria that associate with plant roots
3. Why might one expect a greater number of trophic levels with an increase in total solar energy input? CIRCLE ALL CORRECT ANSWERS
a. Higher temperatures allow less energy expenditure on maintaining body temperature.b. Higher temperatures allow endotherms to have more trophic levels than ectotherms.
c. More light allows greater photosynthesis, allowing a greater 1o producer biomass.
d. Higher temperatures increase respiratory rates in ectotherms.
4. In which of the following plant communities would you expect species richness to be highest? CIRCLE THE SINGLE BEST ANSWER
a. a community with both high fertility soil and high frequency of disturbanceb. a community with both low fertility soil and low frequency of disturbance
c. a community with both moderate fertility soil and moderate frequency of disturbance
d. a community with high fertility soil and low frequency of disturbance
e. a community with low fertility soil and high frequency of disturbance
5. Which of the following is/are true about the nitrogen cycle? CIRCLE ALL CORRECT ANSWERS
a. Nitrogen fixation is an energy releasing processb. Less ammonia is converted to nitrate under acidic soil conditions
c. Only prokaryotes fix nitrogen
d. Most plants pick up ammonia nitrogen (NH4) and convert it to nitrate nitrogen (NO3) before incorporating it into proteins.
6. Which of the following have been postulated to produce a humped ( /\ ) relationship? (CIRCLE ALL CORRECT ANSWERS)
a. species richness vs. soil fertilityb. species diversity vs. disturbance frequency
c. number of visits/flower vs. number of flowers in a patch
d. the number of successful seedlings vs. distance from the parent tree in the tropics
7. Low population sizes result in lowered genetic diversity. In which of the following ways is low genetic diversity a problem for species long-term survival? (CIRCLE ALL CORRECT ANSWERS)
a. greater risk of inbreeding
b. more rapid fixation of alleles
c. lower ability of populations to adapt to environmental changes
d. lower resistance to diseases and parasites
(a is an optional answer: low genetic diversity does increase the chances of inbreeding because there is a greater likelihood that a randomly selected partner will share the same alleles even if the population is large. However, we didn't talk about that aspect)
8. Because of your allergies, your doctor has suggested you move to a warm and very dry climate. Since you're independently wealthy, you can choose anywhere in the world. Which of the following would have a suitable climate for you? (CIRCLE ALL CORRECT ANSWERS)
a. Macapa, Brazil (0o latitude, east coast of South America)b. The Canary Islands (30o N latitude, just off west coast of Africa)
c. Hong Kong (22o N latitude, on southeast coast of China)
d. Copiapo, Chile (27o S latitude, just west of the Andes mountains)
9. You have collected data on the number of birds that visit each of a series of feeders. The feeders have either thistle seeds or millet, and one of each type of feeder is placed at ten environmentally different locations (see diagram). What type of test would likely be best for these data? CIRCLE THE SINGLE BEST ANSWER
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a. two-sample t-test b. two-way ANOVA c. Fisher's PLSD d. paired t-test e. ANCOVA |
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10. We considered several hypotheses to explain the tropics to poles latitudinal gradient in species diversity. Pick ONE of the following hypotheses, briefly explain it, and evaluate it using data presented in class or in your textbook
Hypothesis: Time OR Productivity OR Animal Pollinators
(circle one)
Explanation (6 pts):
Time: Because the tropics have been undisturbed for a much longer period than the temperate zone, the communities have had a longer time to accumulate species through migration and speciation than in the temperate zone.Productivity: The tropics have high evapotranspiration and a long growing season, which some ecologists have argued allows high diversity.
Animal Pollinators: Because there is high humidity and little wind in the tropics (air is ascending), wind pollination is not very effective. Therefore, plants rely on animal pollinators more heavily than in the temperate zone. Animal pollinators can be agents of speciation if there is some mechanism by which they maintain reproductive isolation between segments of a population. Thus, the more animal pollination, the higher the chance of speciation.
Evaluation (4 pts):
Time: There is some evidence in support of the Time hypothesis. The fossil record shows increasing diversity over time in the absence of large scale disturbances. Also, Great Slave Lake in Canada (origin after glaciation) has many fewer species than Lake Baikal in Russia (pre-dates glaciation). No data refute this hypothesis, but the above is not a lot of support.Productivity: Two main issues here. First, the experimental evidence shows that adding nutrients to increase productivity usually decreases diversity. Second, the tropics actually have low soil nutrients (although moderate nutrient levels if you consider all that is stored in the biomass). These two issues combined have suggested a modified version of the productivity hypothesis (which could have been stated above): because the tropics have high energy input but relatively low nutrients, it puts the tropics at the peak of the productivity vs. diversity curve.
Animal Pollinators: A tiny bit of data has been collected showing that grass family plants, which are wind pollinated in the temperate zone, are animal pollinated in the tropics. There is a reasonable body of evidence that animal pollinators can be a mechanism for reproductive isolation in other systems. Not a lot of evidence, but an intriguing theory.
11. Rainforests in the tropics (let's assume at the equator) and rainforests in the temperate zone (let's assume at 45o N latitude on the west coast of North America) are similar in many ways, but also differ in many ways. Briefly explain how these two forest types would be expected to be similar or different in:
a. annual pattern of rainfall? (include a phrase or diagram describing the rainfall pattern(s) of the two sites) (4 pts)Tropical rainforests at the equator would have two rainy seasons and two dry seasons a year.Temperate rainforests at 45 o latitude on the west coast of NA would have a single rainy season in the winter.
b. species richness? (4 pts)
The tropical rainforest would have a much higher species diversity than the temperate rainforest.c. size of the nutrient pool stored in the soil? (4 pts)
The temperate rainforest would have a much higher pool of nutrients stored in the soil.d. abundance of carbon-based vs. nitrogen-based defenses in canopy trees? (4 pts)
In both types of rainforest the canopy trees would have ample light for carbon-fixation. Therefore, the real difference would be in the nitrogen availability, which would be lower in the tropics. Therefore, the tropcial trees would be expected to have a higher proportion of carbon-based defenses than the temperate trees.
12. Consider our discussion of removing the Snake River dams.
a. What are two likely benefits of dam removal? (4 pts)Presumably, dam removal would help fish successfully travel both upstream to spawn and downstream to the sea. In addititon, one could cite increased area for spawning, less variability in flow volume, more opportunities for some kinds of recreation (river rafting, kayaking), and also a boon for the Washington city at the terminus of navigability of the Columbia/Snake river system.
b. What are the costs? Please state one economic cost, one ecological risk, and one social cost. (6 pts)
Economic costs: dam removal itself, loss of irrigation water, loss of hydropower...
Ecological risk: the pollutants that have accumulated in the sediment behind the dams might contaminate the habitats downstream and the gouging effects of the water release could damage sensitive spawning grounds.
Social costs: displacement of farm workers and their families, disruption of economic base for many communities, changes in the kinds of recreation opportunities...
13. In successional sequences on soil embankments from road cuts through forests, annuals are typically followed by short-lived perennials, which are typically followed by shrubs (e.g., Oregon grape, Salal).
a. Is this an example of primary or secondary succession? Why? (4 pts)Secondary succession because soil is present
To assess which mechanism of succession was involved in the change from perennials to shrubs, an experiment compared the success of Oregon grape and salal where the perennials had been removed with their success when the perennials were present. The data are shown below:
b. Which mechanism of succession is supported by the salal data? (4 pts)Inhibition mechanism
Which mechanism of succession is supported by the Oregon grape data? (4 pts)
Tolerance mechanism
c. Should the answers to the two questions in part b be the same? Why or why not? (4 pts)Not necessarily. Each species combination is an independent interaction.
14. Consider the following map showing islands A, B, C, and D. According to MacArthur and Wilson's Equilibrium Model of Island Biogeography,
a. which island would be expected to have the highest species richness? (Circle one) (3 pts)
A . . . . .B . . . . .C . . . . .D b. which island would be expected to have the highest turnover rate? (Circle one) (3 pts)
A . . . . .B . . . . .C . . . . .D c. Which of these predictions is better supported by data from real islands, the species richness or the turnover rate prediction? (2 pts)
species richness prediction
15. The following diagram illustrates the ZNGIs (zero net growth isoclines) for two species, A and B.
a. What is the effect of species B on species A? How do you know? (3 pts)
Because the slope of ZNGIA is positive, species B has a positive effect on A. The more B around, the more A can be there at equilibrium.
b. What is the effect of species A on species B? How do you know? (3 pts)
Because the slope of ZNGIB is negative, species A has a negative effect on B. The more A around, the less B can be there at equilibrium.
c. Using arrows for each species and for the joint trajectory, show the trajectory of the populations at the point labeled X. (3 pts)
d. Is the intersection point a stable or unstable equilibrium point? Why do you think so? (4 pts)
The joint trajectory will cycle around the intersection point clockwise. Because the intersection is not a right angle, the trajectory will spiral in toward the intersection, and therefore, be stable. (However, an answer of stable cycling or even an outward spiral is acceptable if you explain how you arrived at that conclusion.)
16. Below are shown the species occupying three forest communities and the proportions of each species found in each community.
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Bigleaf maple |
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Vine maple |
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Evergreen huckleberry |
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Salal |
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Western hemlock |
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Douglas fir |
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Western red cedar |
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a. Which of these communities contains the highest species richness? (4 pts) .. .A .. . B . . .C
b. What is the diversity of Community A? (Use the Shannon-Weiner Diversity (H') Index). Show your work. (5 pts)
Comm A is actually the most diverse:H' = -[0.01)(-2)+(0.09)(-1.05)+(0.2)(-0.7)+(0.3)(-0.5)+(0.4)(-0.4)]= 0.56 (using log)
or
H' = 1.31 (using ln)
17. In the Spring 2002 issue of Conservation in Practice, there was a report that feral (escaped domestic) pigs have caused a huge decline in the population of endangered island foxes, which occur only on six of the eight Channel Islands that lie about 25 mi off the coast of California.
The pigs do not directly interact with the foxes, but instead have provided sufficient food for golden eagles, also not native to the Channel Islands, to colonize. The house-cat-sized island fox did not evolve with predatory eagles and is active in the daytime (instead of at night like most canids), so it has become easy prey for the golden eagles. The authors' data indicate that golden eagle predation is causing about 2/3 of the 90% decline in fox numbers between 1993 and 1998 on two northern Channel Islands.
In addition, the decline in the island foxes has apparently led to an increase in the nocturnal spotted skunk, which has seen a 17-fold increase as the island fox population declined. This is apparently due to foxes eating fewer mice, which are a main prey item for the skunks.
a. Fill in the boxes below to show the relationships between the pigs, foxes, golden eagles, mice, and skunks. (5 pts)
b. Label each arrow with a "+" or a "-" indicating the effect of the species at the beginning of the arrow on the species to which the arrow points. (5 pts)
In addition to removing the pigs, a component of the fox conservation plan is to re-introduce native bald eagles. The bald eagles prey almost exclusively on fish, but they are territorial and the hope is that they would eliminate the golden eagles.
c. Add a box and labeled arrows for bald eagles to the diagram. (3 pts)d. If bald eagles do eliminate the golden eagles, what would be the predicted effect on skunks? Use the signs on the arrows to show how you know. (5 pts)
bald eagles --( - ) --> golden eagles --( - )--> foxes --( - )--> mice --( + )--> skunks negative overall: (-) (-) (-) (+) = " - "
If the bald eagles do eliminate the golden eagles, that would increase the fox population, which would in turn reduce the mouse population, which would reduce the skunk population.
18. Golden paintbrush is a state and federally endangered plant species. It is difficult to establish new populations of golden paintbrush because it is a hemi-parasite (its roots attach to the roots of a host plant, from which it gains some nutrients). At Mima Mounds, one of the most common grasses is Idaho fescue, which is an acceptable host for golden paintbrush. In several areas in the reserve, there is an experiment using many 1 m2 plots to assess the effects of pre-treating the plots with vigorous raking which removed most of the dead biomass, but allowed the living biomass, both above and below-ground, to remain. This pre-treatment left a lot more bare soil than the control treatment, in which most of the ground was covered with living and dead vegetation.
Assume that for each plot, the researchers counted the number of established golden paintbrush seedlings and estimated the percent of area occupied in each plot by other vegetation. Those data are shown below:
a. Describe the relationship shown by the graph above. Be sure you assign cause and effect to the appropriate variables. (4 pts)
As occupied area in the plots increased from 0 to about 35%, the number of established golden paintbrush seedlings increased. However, as the occupied area in the plots increased still further, the number of established golden paintbrush seedlings decreased.
b. How would you statistically test to see if there really is a significant "hump" in the curve? (4 pts)
One could run a quadratic regression and check whether the x2 term was significant (just as we did in the intertidal lab).
c. Suggest a hypothesis or hypotheses for WHY the relationship looks as it does. (6 pts)
As with essentially all humped curves, there are likely to be two mechanisms, one that increases seedling establishment when occupation is low and another that decreases seedling establishment when occupation is high. Because golden paintbrush is a hemi-parasite, when occupied area is very low, it is difficult for the seedlings to find hosts. As occupied area increases, more seedlings can find hosts and the number of established seedlings increases. But because the seedlings are also photosynthesizing and collecting some nutrients on their own, high proportions of occupied area increase competition for the seedlings, and decrease their ability to establish successfully.
3 pt Bonus: What did we do in this course that you…
liked best?
liked least?
think will help you the most in the future?