Spreadsheet Exercises in Ecology and Evolution

Answers to Exercise 8 Logistic Population Models 1. Inspect your graph of Nt against time. You should see the following:...

Author: Therese M. Donovan | Charles Welden

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Answers to Exercise 8 Logistic Population Models 1. Inspect your graph of Nt against time. You should see the following: • Population size increases slowly at first, then accelerates (the curve gets steeper), then decelerates (the curve gets less steep), and finally stabilizes. A curve of this shape is said to be sigmoid, and is typical of logistic population growth. A geometrically or exponentially growing population grows at an ever-increasing rate and does not stabilize. • The size of the stable population is the carrying capacity, K, which is 50 in this example. 2. Inspect your graph of per capita birth and death rates against population size (this applies only if you used the version of the model with explicit birth and death rates). You should see the following: • Per capita birth rate decreases as Nt increases. •Per capita death rate increases as Nt increases. This is the meaning of density dependence: per capita birth and death rates change as Nt changes. Per capita birth and death rates become equal when Nt reaches K. (This was demonstrated algebraically in the Introduction to this exercise.) 3. Inspect your graph of ∆Nt and ∆Nt/Nt, or dN/dt and dN/dt/N, against Nt. You should see the following: • ∆Nt or dN/dt starts low, rises to a maximum when Nt ≈ K/2, and then declines to 0 when Nt reaches K. This confirms what you saw in the graph of Nt against time: The population grows fastest when its size is half its carrying capacity. This property of logistic models will be important in our harvesting model. • ∆Nt/Nt or dN/dt/N starts high, declines linearly as Nt increases, and reaches 0 when Nt reaches K. These two features differ from the exponential model, in which ∆Nt increased linearly with Nt and ∆Nt/Nt was constant. 4. If you extrapolate the line for ∆Nt/Nt until it crosses both axes of the graph, its y-intercept is R and its x-intercept is K. If you extrapolate the line for dN/dt/N, its y-intercept is r and its xintercept is K. You can use this method of analysis to estimate R (or r) and K for real populations from periodic population counts or estimates, even if you know nothing about per capita birth and death rates. You will use this analysis to answer Question 11. 5. Change the initial population size to 100 by entering that value into cell B6. Look at your graph of Nt against time. In the discrete-time models, you should see the population start out at 100, fall below the carrying capacity, and then increase back to the carrying capacity of 50.

Exercise 8

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Try other values for initial population size and see what happens. If you make the initial population too large, Nt may go negative in the discrete-time models, which makes no sense. If you used the model with explicit birth and death rates, look at your graph of per capita birth and death rates against Nt. It should resemble the graph below.

Per capita birth and death rates

Logistic Model, Explicit b and d 1.20 1.00 0.80 Birth rate Death rate

0.60 0.40 0.20 0.00 0

50

100

150

Population size (Nt)

6. Carrying capacity is a stable equilibrium. Two lines of evidence support this: • Population size moves toward K from below (N0 = 1) and from above (N0 = 100). • The lines for per capita births and deaths (graph above) cross at Nt = K. When Nt < K, per capita births are greater than per capita deaths, so the population will grow. When Nt > K, per capita births are less than per capita deaths, so the population will shrink, returning toward K. At Nt = K, per capita births and deaths are equal and the population stays at equilibrium. 7. You can use the spreadsheet to answer this question only if you built the version with explicit birth and death rates. To simplify matters, restore N0 to 1.00, and set b = 2.00 and d = 1.00. R will become 1.00. Try making b′ = 0.04 and d′ = 0.05. This represents a scenario in which per capita birth rate increases with increasing population size, while per capita death rate increases because of resource limitation. Your graph of Nt should be a sigmoid curve stabilizing at Nt = 100 = K. Your graph of per capita births and deaths should show both increasing, but because per capita deaths increase more steeply, the two lines still cross, producing a stable equilibrium at Nt = 100. Exercise 8

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Exercise 8

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Try making b' = –0.02 and d' = –0.01. This represents a scenario in which per capita birth rate decreases with increasing population size because of resource limitation, while per capita death rate decreases. Your graph of Nt should be a sigmoid curve stabilizing at Nt = 100 = K. Your graph of per capita births and deaths should show both decreasing, but because per capita births decrease more steeply, the two lines still cross, producing a stable equilibrium at Nt = 100. The important point here is that the population will have a stable equilibrium size if the following conditions are met: • Per capita births > per capita deaths when Nt is small. • Per capita births < per capita deaths when Nt is large. You can use the spreadsheet to answer Questions 8 and 9 only if you built the version with explicit birth and death rates.

8. Set b = 1.50 and d = 1.00. Set b' = 0 and d' = 0 and examine your graphs and spreadsheet. Then set b' = 0.001 and d' = 0.001 and examine your graphs and spreadsheet. In both cases, you should see that K becomes undefined—that is, the spreadsheet will show an error message in the cell for K. (If your graph of ∆Nt/Nt shows nonsensical ups and downs, make sure the minimum of the left-hand y-axis is set to zero.) You should also see that the population grows at an increasing rate. Does it grow geometrically (exponentially)? You can test this by changing the y-axis of the graph of Nt against time to a logarithmic scale. It should become a straight line. Also examine the graph of ∆Nt/Nt and its column in your spreadsheet. All this should convince you that the population grows exponentially. In both scenarios, even though per capita birth and death rates are changing, the difference between them is constant, as you can see by examining the graph of per capita birth and death rates (they will be parallel lines). The result is exponential population growth. Try other values of b' and d', keeping them equal to each other. 9. Set b = 1.25 and d = 1.00. Set b' = 0.005 and d' = –0.001. Examine the graph of per capita birth and death rates. You should see per capita births increasing linearly, and per capita deaths decreasing linearly, with increasing Nt. If your graph of Nt against time has a linear y-axis, you will see what looks like an exponential curve. Try changing the y-axis to a logarithmic scale. You should see that the line still curves upward. What does this mean? It means the population is growing faster than exponentially. Examine the graph of ∆Nt/Nt and its column in the spreadsheet. You should see that the per capita rate of population growth increases as the population grows, in contrast to an

Exercise 8

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exponentially growing population, in which ∆Nt/Nt is constant. Something like this may have occurred in human populations during demographic transitions.

10. You can use the spreadsheet to answer Question 10 only if you built one of the discrete-time models. • If you built the model with explicit birth and death rates, set b = 2.00, d = 1.00, b' = –0.001, and d' = 0.001. • If you built the model with explicit carrying capacity, set N0 = 1.00, R = 1.00, and K = 500. • If you built the model with explicit birth and death rates, increase b by small increments, keeping the other parameters unchanged. •If you built the model with explicit carrying capacity, increase R by small increments, keeping K unchanged. Your graph of ∆Nt will be easier to read if you remove the line connecting the data points. Double-click on one of the data points. In the resulting dialog box, select the Patterns tab. In the left-hand box, for Line, choose None. You should observe a variety of behaviors. The values given here are a few interesting examples. Try these, and experiment with others as well. We give values of b for the model with explicit birth and death rates, and values of R for the model with explicit carrying capacity. • b = 2.00, R = 1.00: Nt shows smooth sigmoid growth, stabilizing at K • b = 2.50, R = 1.50: Nt overshoots K slightly, but soon stabilizes at K. • b = 2.75, R = 1.75: overshoot, followed by damped oscillations (decreasing amplitude), eventually stabilizing at K. • b = 3.00, R = 2.00: overshoot, followed by persistent oscillations. Careful examination of the spreadsheet column of Nt will reveal that these oscillations are also damped, but it will take much longer for Nt to stabilize at K. • b = 3.10, R = 2.10: oscillations of very slowly increasing amplitude. • b = 3.25, R = 2.25: essentially stable oscillations around K. • b = 3.50, R = 2.50: an interesting pattern, called a 2-point limit cycle, meaning oscillations of two amplitudes. A large oscillation is followed by a small one, which is followed by a large one, and so on. Not only 2-point cycles, but 4-, 8-, and 16-point cycles exist—see if you can find them. • b = 3.75, R = 2.75: Chaos! Nt changes radically, and with no apparent pattern, from each time to the next. If you change the initial population size, you will see a completely different sequence of Nt values. Examine your graph of ∆Nt and ∆Nt/Nt against Nt. You should see that even in the most chaotic scenarios, ∆Nt still forms a smooth parabola with its peak at Nt = K/2, and ∆Nt/Nt still Exercise 8

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forms a straight line with its y-intercept at R. This shows the order underlying the chaos of Nt. In other words, the values of Nt are not truly random, and we can recover an interpretable pattern from them with the proper analysis.

11. Recall from Exercise 7 that we calculated dN/dt/N of a continuous-time exponential model from population sizes by the formula dN/dt/N = r = ln(Nt+1 /Nt). We can perform this calculation on human population estimates for 1963 to 2000 (available from the U.S. Census Bureau Web site, http://www.census.gov/) to estimate dN/dt/N for each time interval. This is one way to set up the spreadsheet: 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 Exercise 8

A Date 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000

B Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

C Nt 3,205,706,699 3,276,816,764 3,345,837,853 3,416,065,246 3,485,807,350 3,557,675,690 3,632,341,351 3,707,610,112 3,785,190,759 3,862,197,286 3,938,708,588 4,014,598,416 4,088,224,047 4,160,391,803 4,232,928,595 4,305,403,287 4,380,776,827 4,456,705,217 4,532,964,932 4,613,401,886 4,693,932,150 4,773,566,805 4,854,602,890 4,937,607,708 5,023,570,176 5,110,153,261 5,196,333,209 5,283,755,345 5,366,938,089 5,449,663,819 5,531,001,812 5,610,978,348 5,690,865,776 5,768,612,284 5,846,804,802 5,924,574,901 6,002,509,427 6,080,141,683

D dN/dt/N 0.02194 0.020845 0.020772 0.02021 0.020408 0.02077 0.02051 0.020709 0.02014 0.019617 0.019084 0.018173 0.017499 0.017285 0.016977 0.017355 0.017184 0.016966 0.017589 0.017305 0.016823 0.016834 0.016954 0.01726 0.017089 0.016724 0.016684 0.01562 0.015296 0.014815 0.014356 0.014137 0.013569 0.013464 0.013214 0.013069 0.01285 Page 6 of 6

In this case, the formula in cell D6 is =LN(C7/C6), and this formula is pasted down the column to cell D42. As you did with your continuous-time logistic model, you can now graph dN/dt/N against Nt. If you determine the linear trendline for the resulting graph, you can find the overall r as the y-intercept and K as the x-intercept. To add a linear trendline to the graph, click on the graph, then open Chart | Add Trendline . Choose a Linear trendline and click on the Options tab. From the options, select “Display equation on chart” and “Display R-squared value on chart.” The resulting graph should look like the one below. Human Population 1963 to 2000: Logistic?

Per capita change in population size

0.025

0.02

0.015

0.01

y = -3E-12x + 0.0305 R2 = 0.9356

0.005

0 3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

Population size (billions)

The R2 here refers to the coefficient of determination, not to the geometric growth factor. Such a high value of R2 indicates that the trendline fits the data very well. The equation of the line is y = 0.0305 – 3(10–12)x, where y = dN/dt/N and x = Nt. From this we can see that the y-intercept is 0.0305, which we can take as an estimate of r. A little algebra shows that the x-intercept is 10,166,666,000, which we can take as an estimate of K. If we use these values in the continuous-time logistic model, with an initial population size of 3,205,706,699 (the population size in 1963), we get excellent agreement between predicted and observed population sizes through the year 2000. Exercise 8

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The Census Bureau projects a population of 9.1 billion in the year 2050. Extrapolating beyond the year 2000, our model reaches that population about 9 years later—not bad for an estimate based solely on population sizes. Our model predicts a population of 10 billion in the year 2122, and reaches K at 10,166,666,000 in the year 2756. Remember, however, that projections are not guarantees. Future human population may behave quite differently, and such prognostications are an active area of research.

Exercise 8

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Answers to Exercise 9 Island Biogeography 1. Note in Figure 2 that as species richness increases, immigration rate decreases, and extinction rate increases. At some value of species richness, the two lines cross, indicating that immigration and extinction rates are equal. At that point, species richness is at equilibrium. In this figure, equilibrium species richness is a little less than 400 species. 2. It is stable, as you can see from examination of Figure 2. When species richness is less than the equilibrium value, the immigration rate is higher than the extinction rate, and so species richness will increase. If species richness exceeds the equilibrium value, the extinction rate is greater than the immigration rate, and species richness will decrease. Thus, even if species richness deviates from the equilibrium value, it tends to return to it. 3. It is a dynamic equilibrium. Figure 2 shows that when species richness is at equilibrium, immigration and extinction are both still greater than zero: thus even though overall species richness has reached equilibrium, individual species are still going extinct and new species are still arriving. In this example, the rates of immigration and extinction are equal at about 20 (e.g., in units of species per year). This is the turnover rate. 4. Examine Figure 4, the graph of immigration rates for islands at three distances from the mainland. You should see that the immigration line for the near island is the steepest and the line for the far island is the shallowest. Consequently, the immigration line for the near island crosses the extinction line farthest to the right, and that for the far island, farthest to the left. If you drop a vertical line from each of the three crossing points to the horizontal axis, you will see that the near island has the highest equilibrium species richness (about 550), and the far island has the lowest (about 225). 5. Again examine Figure 4. This time, draw a horizontal line from each crossing point to the vertical axis. You should see that the near island also has the highest turnover rate (about 30 species per time unit), and the far island has the lowest (about 12 species per time unit). 6. Examine Figure 5, the graph of extinction rates for islands of three different areas. You should see that the extinction line for the small island is the steepest and the line for the large island is the shallowest. Consequently, the extinction line for the small island crosses the immigration line farthest to the left, and that for the large island, farthest to the right. If you drop a vertical line from each of the three crossing points to the horizontal axis, you will see that the large island has the highest equilibrium species richness (about 525), and the small island has the lowest (about 250).

Answers: Exercise 9

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7. Again examine Figure 5. This time, draw a horizontal line from each crossing point to the vertical axis. You should see that the large island has the lowest turnover rate (about 15 species per time unit), and the small island has the highest (about 25 species per time unit). 8.

OPTIONAL:

If you made the optional graph described in Step 12 on p. 132, draw vertical and horizontal lines from the nine crossing points to the axes. You should see that the highest species richness occurs on a large island near the mainland: the lowest on a small island far from the mainland. The highest turnover rate occurs on small island near the mainland: the lowest on a large island far from the mainland.

9. As you can see from Figure 7, species accumulate at a decreasing rate, eventually stabilizing at an equilibrium value of species richness. 10. You should also see from Figure 7 that the immigration rate declines with time and increasing species richness, but does not reach zero. Likewise, the extinction rate increases with time and increasing species richness, but levels off. At some value of species richness, immigration and extinction rates become equal, and species richness stabilizes at its maximum value. You can think of this as roughly analogous to changing per capita birth and death rates and stabilizing population size in the logistic population model. Note again that species continue to turn over after equilibrium species richness has been reached. From Figure 7, you should be able to see that the turnover rate in this example is about 20 species per time unit. 11. The fact that the species-area curves are nearly straight on a log-log plot indicates a power relationship. This is also shown in the equation of equilibrium species richness, in which area is raised to a power. 12. The species-area curve for near islands has the shallowest slope; the curve for far islands has the steepest. Islands near the mainland are well stocked with species regardless of area because immigration is relatively easy, and so even small islands have high species richness. Because no island can have more species than the mainland pool, there is little room for increasing species richness with increasing area—hence the shallow species-area curve. Islands far from the mainland are species-poor regardless of area because immigration is so difficult. However, as island area increases, the extinction rate decreases, so species richness increases rapidly in comparison to near islands. Species-area curves are used to estimate the rate of species loss as habitats become increasingly fragmented, effectively producing smaller islands of suitable habitat for endangered species. These curves predict that as habitats islands become smaller, species will be lost at an increasing rate (see Figure 10).

Answers: Exercise 9

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Answers to Exercise 10 Life Tables, Survivorship Curves, and Population Growth

1. We plot survivorship curves on semi-log graphs because lx is a proportion: the proportion of the original cohort surviving to age x. The distance between points on a logarithmic axis reflects their proportional relationship, and so a logarithmic scale is appropriate. This kind of graph also makes clear important differences between the three types of survivorship curve. Note that on the linear graph, type II and type III curves have qualitatively similar shapes, whereas on the semi-log graph they look quite different. 2. The keys to interpreting the shapes of survivorship curves are to look at their slopes compared with the graphs of age-specific survival (gx ). The type I curve begins with a shallow slope, indicating low mortality among the young—that is, a high proportion of individuals of each age survive to the next age throughout the early part of life. The curve steepens at older ages, indicating increased mortality in old age. A type I curve tells us that most individuals in this population survive a long time and die at old ages. The graph of gx supports this interpretation: survival is high in the young, and drops off with age. Inspect columns E and H to see these patterns numerically. The type III curve reflects the opposite pattern of survival and mortality. The curve begins with a steep slope, indicating low survival (and high mortality) among the young. The shallow slope in middle and old ages indicates that most of the individuals that survive their youth survive to old age. The graph of gx shows the same thing—survival among the young is very low, and increases with age (until the oldest age). Inspect columns F and I to see these patterns numerically. The type II curve indicates a constant rate of survival across all ages (until the last). Note that a straight line on a semi-log plot indicates change by a constant proportion. The graph of gx (a horizontal line) also shows that survival is the same for all ages (except the oldest). Inspect columns G and J to see these patterns numerically. 3. The type I curve of life expectancy is probably not surprising, as it shows that the expected number of years of life remaining decreases with increasing age. The type II curve may strike you as somewhat surprising. It indicates that an individual can expect to live about 2 more years, regardless of age, up to about 6 years of age. After this, life expectancy decreases, to 1 year at 10 years of age. How can this be? A type II survivorship curve occurs when the same proportion of survivors die at each age (in other words, when the risk of death is constant for all ages). In this circumstance, life expectancy is the reciprocal of risk of death. If you look at the spreadsheet, you will see that half of the survivors die at each age. The reciprocal of _

Answers: Exercise 10

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is 2—the life expectancy.


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The type III curve is probably the most unexpected. It shows life expectancy increasing with age up to 3 years of age, after which it declines. How can an older individual expect to live longer than a younger one? Because of high juvenile mortality. A newborn individual is unlikely survive its first year. However, individuals that do survive can expect to live a bit more than 3 additional years. Those that survive to 2 years of age can expect to live almost another 6 years. This last pattern of survival and mortality brings up a common misinterpretation of demographic data. You may have heard that people in some region or society have high infant and juvenile mortality and short life expectancy. Sometimes people conclude from this that everyone in the society dies young and there are no old people. However, the type III pattern of survival and life expectancy should show you that this is not exactly the case. Once a person in reaches young adulthood, he or she may live many years more, and the oldest members of the society may be nearly as old as the oldest individuals in any other. It is true, however, that there will be proportionately fewer old people. 4. You should see that Dall Moutain Sheep have a type I survivorship curve, Song Thrushes a type II, and barnacles a type III. None of these will be as neat as the hypothetical examples given in the procedures, but they come close. Other populations may display a mixture of types. Human populations, especially in developing countries, often have high infant and juvenile mortality followed by low mortality until old age. Such curves have features of type III and type I survivorship. 5. Notice that the Sx values given in Figure 8 on p. 148 result in a type I survivorship curve. With the bx values given, the net reproductive rate (R0) is 1. The estimated and corrected values of r are 0. This indicates that the population is stable, and the members are exactly replacing themselves. If you shift reproduction earlier in the life cycle, by changing b1 to 4.0 and all other bx values to 0.0, you will see that R0 and r increase. This occurs for two reasons. First, because reproduction occurs sooner, offspring appear sooner and begin reproducing sooner. Second, because survival is always a decreasing function of age, there are more individuals alive at the earlier age, and the same per capita fertility means more total offspring. If you shift reproduction later in the life cycle, by changing b3 to 4.0 and all other bx values to 0, you will see R0 and r decrease. This is the mirror image of shifting reproduction to younger ages. That is, reproduction occurs later, so offspring appear later and begin reproducing later. Furthermore, there are fewer survivors to reproduce at the later age.


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If you spread reproduction out over a longer time span by changing b1, b2, b3, and b4 to 1.0 and leaving b0 at 0.0, you will see that the resulting values of R0 and r are slightly larger than the original scenario (with b2 = 4.0). However, the increase is much smaller than the increase that occurred when you shifted reproduction entirely to age 1. The reason is that some reproduction occurs earlier, which increases R0 and r. However, much of the reproduction occurs later, decreasing R0 and r. The net result is a slight increase. The values given here are only examples; you can experiment with other numbers. Another interesting thing to try is to figure out how many offspring an individual must produce to keep the population stable under various fertility schedules. This number of offspring is called replacement fertility—another statistic you may have heard. It is often said that the replacement fertility for human females is about 2.1. You may ask yourself, how does that depend on the ages at which women bear children and the number they bear at each age? 6. Try changing the Sx values given to those given in the table below, observing the effects on R0 and r. Table 1. Hypothetical survivorship schedules for investigating the effect of survivorship on population growth

Age (x)

0 1 2 3 4

1000 500 250 125 0

Sx

Age (x)

0 1 2 3 4

1000 250 125 100 0

Sx

Notice that the first set of Sx values in the table gives a type II survivorship curve, and the second set gives a type III curve. If you now try the various fertility schedules described in the answer to Question 1, you will see somewhat different effects, depending on the survivorship schedule. Shifting reproduction earlier produces the greatest increase in R0 and r in the type III survivorship schedule. This occurs because type III survivorship has the highest mortality in young ages, and so reproducing earlier means there are many more survivors to reproduce. Shifting reproduction later decreases R0 and r to in all three types of survivorship schedules, but most noticeably in type II. In the type I schedule, there is little mortality in middle ages, so delaying reproduction has little effect unless it is delayed Answers: Exercise 10

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to the very end of the life-span. In the type III schedule, most of the mortality has already


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occurred by the time reproduction starts, so delaying reproduction slightly has little effect. In the type II schedule however, the proportional mortality is constant across ages, so delaying reproduction has a greater effect. Spreading reproduction out over a longer time span decreases R0 and r in both type II and type III survivorship schedules, and the effect is greater in type III. This results from the lower number of survivors in the later reproductive ages. In type II and type III, this effect of delayed reproduction outweighs the effect or earlier reproduction, and results in a net decrease in R0 and r. As you did in analyzing the effects of changing fertility schedules, you may also ask yourself the effects of different survival schedules on replacement fertility.


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Answers to Exercise 11 Age-Structured Matrix Models 1. All age classes are increasing geometrically over time. Although λ is difficult to determine by examining the graph, it must be greater than 1 in order for geometric increases to occur. 2. At year 19 the population’s finite rate of increase stabilizes at 1.180768. This is the asymptotic growth rate. Signs of stabilization appear as early as year 7. This stabilized growth is readily apparent by examining the semi-log graph, where the projection lines for each age class become parallel. At that point, individuals in age class 1, followed by age classes 2, 3, and 4, dominate the population. 3. The stable age distribution is H I J K Stable Age Distribution 10 1 2 3 4 11 12 0.493786 0.33455 0.14166719 0.0299947

In other words, when the population has reached a stable distribution, 49% of the population consists of individuals from age class 1, 33% of the population consists of individuals from age class 2, 14% of the population consists of individuals from age class 3, and 3% of the population consists of individuals from age class 4. Note that these values sum to 100%. This is called the stable age vector. The formulae used in these calculations included •H13 =B37/$F$37 •I13 =C37/$F$37

•J13 =D37/$F$37 •K13 =E37/$F$37

4. You should see that the initial population vector has no influence on the stable age distribution or λ once the stable age distribution has been attained. However, the initial dynamics (before the distribution stabilizes) are strongly affected by the initial population vector. With an initial distribution of 75, 1, 1, and 1 individuals in age classes 1 through 4, respectively, λ in year 1 is 0.827; the population decreased by approximately 17%. (Recall that with the initial distribution of 45 individuals in age class 1, 18 individuals in age class 2, 11 individuals in age class 3, and 4 individuals in age class 4, λ in year 1 was 1.116.) Early irregularities in the age structure and growth rate are instabilities reflecting initial departure from a stable age distribution; the irregularities are not due to stochasticity in fertility or survival rates because these rates remained fixed in the Leslie matrix. How far the age distribution is from a stable distribution has important implications for population management because, initially, the composition of individuals affects whether the population will increase, decrease, or remain stable, given the set of parameters entered into the Leslie matrix.


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Projection of Population Size for Age Structured Populations: Semi-Log Scale

Log number of individuals

10000 1000

Age Class 1

100

Age Class 2 Age Class 3

10

Age Class 4

1 0.1

Total Pop 0

10

20

30

Time

Graph for Question 4

5. The assumptions of the matrix model are the same as those for exponential growth, except that we have accounted for age structure. In other words, once we account for age, the population’s finite rate of increase can be calculated, and the population will increase geometrically, decrease geometrically, or remain stable over time. As with the exponential growth model, this matrix model assumes that resources are unlimited. For this model, we have also assumed that individuals give birth the moment they enter a new age class, and that population censuses occur immediately after birth. We have also assumed that, within an age class, birth and death rates are homogeneous. 6. The life cycle diagram would be adjusted as follows:

1

2

3

4

B

C

D

1

Age class 2 3

The adjusted Leslie matrix has the form A 3 4 5 6 7 8


A =

0 0.8 0 0

1 0 0.5 0

1.5 0 0 0.25

E

4 1.2 0 0 0.25

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For this matrix, λ stabilizes at 1.188 compared to 1.180. The stable age distribution has the following proportions:

11 12

H 1 0.49218

I 2 0.33131

J 3 0.13938385

K 4 0.03713087

Thus, there are slightly more individuals in age class 4 (3.71%) than previously (2.99%), and slightly fewer individuals in age classes 1, 2, and 3 than previously.


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Answers to Exercise 12 Stage-Structured Matrix Models

1. The assumptions of the matrix model are the same as those for exponential growth, except that age structure has been accounted for. In other words, once we account for age, the population’s finite rate of increase can be calculated, and the population will increase geometrically, decrease geometrically, or remain stable over time. As with the exponential growth model, the matrix model assumes that resources are unlimited. For this model, we have also assumed that individuals give birth the moment they enter a new age class and that population censuses occur immediately after birth. 2. At year 60, the population’s finite rate of increase, λ, is 1. This stabilized growth is readily apparent by examining the semi-log graph, where the projection lines for each stage class become parallel. The horizontal nature of the lines indicates that the population is neither increasing nor decreasing over time; it remains constant. At stable distribution, the population is dominated by small juveniles, then hatchlings, large juveniles, subadults, and adults. For all of these age classes, λ = 1.

Population Growth of Five Classes of Sea Turtles Hatchlings

S. juvs

L. juvs

Subadults

Adults

Total

Number of individuals

10000

1000

100

10

1 0

10

20

30

40

50

60

70

80

Year


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3. When the population reaches a stable distribution, it consists of 27.6% hatchlings, 62.7% small juveniles, 8.7% large juveniles, 0.64% subadults, and 0.397% adults. This vector of abundances is called a right eigenvector for the L matrix. 4. You should see that the initial population vector has no influence on the stable stage distribution, or on λ once the stable age distribution has been attained. However, the initial dynamics (λt, pre-stable distribution) are strongly affected by the initial population vector. With an initial distribution of 75 hatchlings, 1 small juvenile, 1 large juvenile, 1 subadult, and 1 adult, λ in year 1 is 1.53, or a 53% increase over time. (Recall that the value for year 1 had previously been 1.16.). Early irregularities in the stage structure and growth rate are instabilities reflecting initial departure from a stable stage distribution. The irregularities are not due to stochasticity in fertility or survival rates, because these rates remained fixed in the Lefkovitch matrix. How far the stage distribution differs from a stable distribution has important population management implications because, initially, the composition of individuals affects whether the population will increase, decrease, or remain stable, given the set of parameters entered into the Lefkovitch matrix. 5. The population reaches a stable distribution at approximately year 13. At t his time it is composed of approximately 29.1% hatchlings, 59.3% small juveniles, 10.4% large juveniles, 0.85% subadults, and 0.37% adults. Each of these stages increases by 3% per year, and their proportions remain constant over time. Thus, TEDS have increased the growth rate by 3% per year.

Population Growth of Five Classes of Sea Turtles Hatchlings

S. juvs

L. juvs

Subadults

Adults

Total

10000000


1000000 100000 10000 1000 100 10 1 0

10

20

30

40

50

60

70

80

Year


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6. In order to increase λ to 1.03, and keeping the original P values in the L matrix, fertilities of subadults and adults would need to increased by a factor of around 3.5. We obtained this number by setting up spreadsheet headings as shown below. Cell G4 is a factor to be multiplied by the original fertility value. Cell E4 has the formula =4.665*G4, and cell F4 has the formula =61.896*G4. Enter values in cell G4 until the stabilized λ = 1.03. (You can also use Excel’s Solver function to determine this answer.)

3 4 5 6 7 8

B

C

D

E

F (h ) 0 0.675 0 0 0

F (sj ) 0 0.703 0.047 0 0

F (lj ) 0 0 0.657 0.019 0

F (sa ) 16.3275 0 0 0.682 0.061

F

G

F (a ) Factor 216.64 3.5 0 0 0 0.8091

7. N/A


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Answers to Exercise 13 Reproductive Value: Matrix Models 1. A graph of the various inoculates shows that age class 1 produces the highest population size over time.

Log number of individuals

Age class 1

Age class 2

Age class 3

Age class 4

1000000 100000 10000 1000 100 10 1 0

1

2

3

4

5

6

7

8

9

10

Year

2. In this population, age class 1 has the highest reproductive value, followed by age class 2. Age classes 3 and 4 have low reproductive values. If populations are to be kept at a high level, the older age classes should be harvested because removing these individuals will have little impact on future population growth. In contrast, if you were trying to introduce an endangered species to a new habitat, you would want to introduce the younger individuals to the area, since this group will reproduce and produce the greatest increase in population growth quickly. I

2 3 4 5 6 7 8

K L M Reproductive value Inoculate method Transpose method Age class Final pop size RV RV Standardized 1 1.64938E+19 1 0.54594587 1 2 1.18204E+19 0.71665214 0.39125327 0.716652136 3 1.89731E+18 0.11503129 0.06280086 0.11503129 4 0 0 0 0


J

Page 1 of 2


Page 2 of 2

3. Remember that type I survival curves indicate high survival of young right until old age, when mortality rates suddenly become high due to senescence. A Leslie matrix of type I survivorship might look something like this:

B 5 6 7 8

C 0.0 1.0 0.0 0.0

D 0.3 0.0 0.8 0.0

E 6.0 0.0 0.0 0.8

0.0 0.0 0.0 0.0

With this life history schedule, age class 3 has the highest reproductive value. This occurs because its current contribution to the population’s growth is high compared to the other age classes, even though its future contribution (in age class 4) is 0. Be careful to make sure the population has reached a stable equilibrium before computing the reproductive values. We have assumed that the population reaches a stable distribution within 50 years, and hence used the proportions of each age class at year 50 to compute reproductive values. However, depending on the Leslie matrix entries, this might not be the case, and you may need to project your population’s growth well into the future. In some cases, the population may never stabilize. 4. A Type III survival curve indicates very high mortality in the young, such as occurs in spawning fish or among tadpoles. A Leslie matrix might look like this:

3 4 5 6 7 8

B 1 0 0 0 0

C D Leslie Matrix 2 3 30 100 0 0 0 0 0 1

E 4 0 0 0 0

With this life history schedule, adults are by far the most “valuable” for population growth because the mortality rates are so high for the younger age classes. Adults have made it past the critical mortality periods and have achieved the ability to reproduce with high fertility rates.


Page 3 of 2

Answers to Exercise 14 Reproductive Value: Life Table Approach 1–3. Your reproductive values should be as follows: J 2 3 4 5 6 7 8

K

L

M

Reproductive value distribution rx

e /l x 1 2.7166524 11.80832 102.65313

e

-rx

lxbx 0 0.7362002 0.2540581 0.0097415

-ry

e lyby 0.9999999 0.9999999 0.2637997 0.0097415

vx 0.9999999 0.7166521 0.1150313 0

In this population, age 0 has the highest reproductive value, followed by age 1. Ages 2 and 3 have low reproductive values. You might think that if populations are to be kept at a high level, the older age classes should be harvested because the removal of these individuals will have little impact on future population growth. In contrast, if you were trying to introduce an endangered species to a new habitat, you would want to introduce the younger ages to the area, since this group will reproduce and produce the greatest increase in population growth quickly. You would be right if in fact the numbers of individuals in each age class was approximately the same (i.e., the stable age distribution consists of roughly the same proportion of individuals from all age classes). However, if there individuals with the highest reproductive value in the population also makes up a very small proportion of the stable population, simply targeting individuals with the highest reproductive value may not be beneficial. See Exercise 15, “Sensitivity and Elasticity Analysis,” to learn more about this topic. 4. For a population with a Type III survival curve and a single reproductive bout at the end of life, the reproductive values are as follows: M 3 4 5 6 7 8

vx 0.9999993 2.3207934 4.3088684 0

Three-year-olds are “worth” over 4 times as much in terms of future population growth because younger individuals have a very high mortality rate.


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5. In contrast to Question 4, a Type I survival schedule produces the following reproductive values: J 2 3 4 5 6 7

K

L

M

Reproductive value distribution e rx /l x 1 1.5559809 3.0142525 6.2409854

e -rx l x b x 0 0 0.6635144 0.3364853

e -ry l y b y 0.9999998 0.9999998 0.9999998 0.3364853

vx 0.9999998 1.5559805 1.0142517 0

In this population, 1-year-olds have the highest reproductive value. Since mortality is low in the early age classes, 1-year-olds are just about to enter their reproductive life, and have their entire reproductive life ahead of them. Two-year-olds already have started reproducing, so in terms of future offspring, they have less to contribute to the population than the 1-year-olds. Three-year-olds can reproduce, but since they all die before they reach the age of 4, their reproductive value is forced to 0 because age class y(x + 1) in the reproductive value calculations is equal to 0.


Page 2 of 2

Answers to Exercise 15 Sensitivity and Elasticity Analysis 1. Overall, sea turtle population growth, l, is much more sensitive to small changes in survival values than fertility values. The model was most sensitive to the transition period from large juveniles to small adults, where a very small increase in Plj,sa will cause a change in l by a factor of 0.29 when all of the other elements in the L matrix are held constant. That is, increasing Plj,sa by 0.01 will increase l by (0.01)(0.29) = 0.0029.

10 11 12 13 14 15 16

R

S

T

U

V

W

F (sa ) 0.0018 0.0025 0.0132 0.2044 0.7682

F (a ) 0.0008 0.0011 0.0056 0.0875 0.3289

Sensitivity matrix Hatchlings Small juveniles Large juveniles Subadults Adults

F (h ) 0.0579 0.0816 0.4318 6.6950 25.1585

F (sj ) 0.1573 0.2217 1.1725 18.1795 68.3155

F (lj ) 0.0251 0.0354 0.1871 2.9005 10.8996

2. Elasticity analysis estimates the effect of a proportional change in the vital rates on population growth. In essence, elasticities are proportional sensitivities, scaled so that they are dimen-sionless. This allows you to directly compare survival and reproductive matrix elements. Inspection of the elasticity matrix shows that, by far, adult survival (the probability that an adult will remain an adult in the next time step) is the most important matrix element in terms of l. The elasticity for adult survival 0.279, indicates that a 1% increase in adult survival will cause 0.279 % increase in l.

R 19 20 21 22 23 24 25


S

T

U

V

W

Elasticity matrix F (h ) F (sj ) F (lj ) F (sa ) F (a ) Hatchlings 0 0 0 0.008669156 0.049244572 Small juveniles 0.057913728 0.16378255 0 0 0 Large juveniles 0 0.057913728 0.129163877 0 0 Subadults 0 0 0.057913728 0.146512753 0 Adults 0 0 0 0.049244572 0.279641337

Page 1 of 2

3. Sensitivity and elasticity calculations depend on the parameters entered in the original matrix, and how well those parameters have been estimated. A change in one parameter will change the sensitivity and elasticity computations (try it for yourself by changing some original matrix entries). These parameters are likely to vary over time and are often difficult to estimate. Be careful not to over-interpret your results. Your results basically tell you what would happen if the matrix parameters are in fact unbiased, and if you are able to hold other elements in the matrix constant to examine the effect of each element.


Page 2 of 2

Answers to Exercise 16 Metapopulation Dynamics 1. In our simulation, the population never reaches an equilibrium point, calculated as f = 0.75, where the fraction of patches remains constant over time.

Fraction occupied

Fraction of Patches Occupied over Time 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

2

4

6

8

10

12

Year

This is likely due to the relatively small size of the metapopulation modeled (25 patches). In Exercise 27 on demographic stochasticity, you will learn that random fluctuations in population dynamics occur when population size is small. The same idea holds for metapopulations. When the number of patches in the system is small, there are chance variations in extinction and colonization for each patch, creating unpredictable variation in the system. Thus, even when I = E, the population may not reach a stable equilibrium. 2. Your table should have the following values, which give the probability of regional persistence:

8 9 10 11 12 13 14 15


J

K

Pe 0 0.2 0.4 0.6 0.8 1

1 1 0.8 0.6 0.4 0.2 0

L M Number of patches 2 4 1 1 0.96 0.998 0.84 0.974 0.64 0.87 0.36 0.59 0 0

N 8 1 0.999 0.999 0.983 0.83 0

Page 1 of 4

A graph of the results shows that as the number of patches in the system increases, the probability of regional persistence increases dramatically. Thus, the probability that all patches will simultaneously go extinct decreases considerably as the number of patches increases. This illustrates an important point: the existence of multiple patches “spreads the risk” of extinction (Gotelli 2001).

Probability of regional persistence (Px )

Effect of Number of Populations on Probability of Regional Persistence under Varying Local Extinction Probabilities 1.2 1 1 patch

0.8

2 patches

0.6

4 patches

0.4

8 patches

0.2 0 0

0.2

0.4

0.6

0.8

1

Probability of local extinction

3. Your table should have the following values, which estimate f at equilibrium:

18 19 20 21 22 23 24 25


J

K

L

M

Pe 0 0.2 0.4 0.6 0.8 1

pi = 0 1 0 0 0 0 0

p i = 0.2 1 0.5 0.33 0.25 0.2 0.167

p i = 0.4 1 0.667 0.5 0.4 0.33 0.285

N

O

P

p i = 0.6 1 0.75 0.6 0.5 0.43 0.375

p i = 0.8 1 0.8 0.67 0.57 0.5 0.44

pi = 1 1 0.83 0.71 0.625 0.55 0.5

Pi

Page 2 of 4

Graphically, the relationship between pe and pi is:

Fraction of Patches Occupied at Equilibrium, i , under Varying p e and p i 1.2 pi = 0

1

pi = 0.2

f

0.8

pi = 0.4

0.6

pi = 0.6

0.4

pi = 0.8

0.2

pi = 1

0 0

0.2

0.4

0.6

0.8

1

pe

4. With pe = 1, and pi = 0.9, the metapopulation persists over time with an equilibrium f = 0.47. This persistence occurs because colonization of vacant patches is independent of how many patches in the system are occupied. In other words, even if all patches are extinct, each has a 90% chance of being colonized in the next time step. This sort of metapopulation system is called a propagule rain metapopulation, or an island-mainland system (Gotelli 2001), because the colonists originate from someplace other than the 25-patch system. 5. The simplest way to let pi increase as the number of patches increase is to write an equation so that the colonization rate is directly related to the proportion of patches that are occupied. Thus, if a large fraction of patches are occupied, the colonization rate will be high, and if there is small fraction of patches occupied, the colonization rate will be low. Graphically, the relationship between the colonization rates and fraction of patches occupied is shown on the next page. We entered =E35 in cell E9 to reflect internal colonization. When the simulation is run under these conditions, the population persists and approaches an equilibrium fraction of patches. (Again, note the stochastic nature of the metapopulation, due to the low number of patches in the system). The equilibrium fraction of patches occupied in an internal colonization model can be calculated as p fˆ = 1 − e i See Gotelli (2001) for further information.


Page 3 of 4

Internal Colonization Metapopulation Model

Colonization rate

1.2 1 0.8 0.6 0.4 0.2

1

0. 92

0. 84

0. 76

0. 68

0. 6

0. 52

0. 44

0. 36

0. 28

0. 2

0. 12

0. 04

0

Proportion of patches occupied

6. The rescue effect can be incorporated into the model by entering =1-E35 in cell E6. This formula establishes a proportional relationship between the extinction probability of a patch, pi, and the fraction of patches currently occupied. Given the starting conditions of the metapopulation, the population dynamics proceed roughly as follows:

3 4 5 6 7 8 9 10 11 12

A B C D Model parameters: x = number of patches in system n = number of years under consideration p e = probability of local extinction 1 - p e = probability of local persistence p n = probability of continued local persistence p i = probability of local colonization P x = probability of regional extinction 1 - P x = probability of regional persistence f = equilibrium number of patches occupied

E 25 10 0.36 0.64 0.01152922 0.6 8.0828E-12 1 0.625

The equilibrium fraction of patches for a rescue effect metapopulation model can be calculated as p fˆ = i e See Gotelli (2001) for additional information.


Page 4 of 4

Answers to Exercise 17 Source-Sink Dynamics 1. The source reaches an equilibrium number at year 3, when the carrying capacity (K) has been reached. The source population remains at K because excess individuals are forced to emigrate to the sink habitat. The sink does not reach equilibrium within 20 years and shows increased growth over time. This is because the source continues to produce an excess of emigrants—more than are needed to sustain the sink over time. As a result, the entire population grows. Number of individuals in the source, sink, and total population

Number of Individuals

140.0 120.0 100.0

Souce

80.0

Sink

60.0

Total

40.0 20.0

20

18

16

14

12

10

8

6

4

2

0

0.0

Year

Because the global (source + sink) population is increasing over time, and because the source has a fixed carrying capacity, the proportion of the total population in the sink increases over time, while the proportion of the total population in the source decreases over time. Note that, for the current model inputs, that less than 25% of the population resides in source habitats, yet this is sufficient to maintain numbers in the entire source-sink system. This is an important consideration for populations in heterogeneous habitats: the numbers of individuals in each habitat type determines the overall growth rate. 2. With a 100-year projection (graph on next page), the source sink system reaches equilibrium at 125 total individuals. The source has an equilibrium number of 25 individuals, as before. The sink reaches an equilibrium number of 100 individuals. Thus, the numbers of individuals in each habitat type, as well as the whole population, stabilize, because of dispersal between sources and sinks.


Page 1 of 3

Number of Individuals in the Source, Sink, and Total for 100- Year Projection


140.0 120.0 100.0 Souce

80.0

Sink

60.0

Total

40.0 20.0

98

91

84

77

70

63

56

49

42

35

28

21

14

7

0

0.0 Year

3. When the source is extirpated (locally extinct), the sink population indeed declines to extinction, as does the entire population.

100.0 80.0 Source

60.0

Sink

40.0

Total

20.0

20

18

16

14

12

10

8

6

4

2

0.0

0


Number of Individuals in the Source, Sink, and Total Population

Year


Page 2 of 3

4. Increasing and decreasing the numbers of individuals in the source population has little effect on the long-term equilibrium values of the system. However, it does affect the short-term growth in both the source and the sink, as well as the proportion of the total population occupying the various habitat types. Decreasing the number of individuals in the source to 5 causes a delay in the emigration of individuals to the sink because the source takes longer to reach its carrying capacity. Number of Individuals in the Source, Sink, and Total Population


140.0 120.0 100.0

Source

80.0

Sink

60.0

Total

40.0 20.0

20

18

16

14

12

10

8

6

4

2

0

0.0

Year

Increasing the number of individuals in the source causes earlier emigration from the source to the sink because the carrying capacity of the source is reached earlier in time. Increasing the survival rate of the source (lower the death rate) makes the source “stronger.” Although its equilibrium numbers remain unchanged, the result is a greater capacity to sustain the sink, and a larger equilibrium value for the entire source-sink system. For example, decreasing d to 0.1 increases the equilibrium number of individuals in the system to 150 individuals (compared to 125 individuals with d = 0.2). Increasing or decreasing K affects not only the equilibrium number of individuals in the source, but also the equilibrium number of individuals in the sink and in the whole system. An increase in K will increase both the sink and system equilibrium values because more individuals in the population are located within “prime” habitats for breeding. Similarly, a decrease in the K will decrease the equilibrium values. Enter a variety of values for K and examine your 100 year projection graphs.


Page 3 of 3

Answers to Exercise 18 Population Estimation and Mark-Recapture Techniques 1. If M = 20 and C is low, when you press F9 several times you should see that the LincolnPetersen index is poorly estimated and quite variable. Often (but not always) the index becomes somewhat more stable as C increases to 30 or 40. In some cases, the LincolnPetersen index will come close to estimating the true population size of 100. In other cases, as shown in the graph below, the index will be off, even after 100 individuals are recaptured. This question should impress on you the need to compute confidence intervals.

Lincoln-Petersen index

300 250 200 150 100 50 0 0

20

40

60

80

100

120

Sample size (C )

2. You should see that as the proportion of individuals in the population that are marked increases, the Lincoln-Petersen index tends to stabilize around the true population size more quickly. One of our results with 70 marked individuals looked like this:

Lincoln-Petersen index

120 100 80 60 40 20 0 0

20

40

60

80

100

120

Sample size ( C )


Page 1 of 3

When 100 individuals are marked, the Lincoln-Petersen index is 100, the true population size, because all individuals in the population are marked. The index is 100 no matter how many individuals are recaptured in the second sampling bout. 3. The relationship between C and the Lincoln-Petersen index, as developed so far, is sequentially autocorrelated—a value for a specific sample size is correlated with its values in other sample sizes. In other words, sample 99 takes advantage of the sampling up through sample 98; sample 98 takes advantage of the sampling up through sample 97, and so on. Because of this autocorrelation, this method does not reveal the true relationship between C and the Lincoln-Petersen index. Hence, we need to push ahead and determine the relationship when the Lincoln-Petersen index is calculated independently for a variety of levels of C. 4. Note first that the Lincoln-Petersen index returns only a handful of different estimates under these circumstances; if we mark 20 individuals and recapture 20 individuals, 4 of which are marked, the estimate will always be 84, regardless of the total population size. If 3 marked individuals are recaptured, the estimate will be 105. Under these circumstances, there is no way to get an estimate between 84 and 105, so if you require more sensitive measurements than this (i.e., if you are expecting small fluctuations in population size), you will need to arrange to mark and recapture more individuals when you design the experiment. Even if you are anticipating large fluctuations in population size, the Monte Carlo simulation shows that if the population is indeed around 100, the Lincoln Petersen estimate will regulary return estimates from 46.7 to 210 without any change in the actual population size. Most likely you will want more reliable estimates than this. M and C do not seem to have equal influence on the range of estimates returned. When we tried M = 50 and C = 20, we found that 95% of the estimates fell between 70 and 150, but with M = 20 and C = 50, the range was still 60 to 204—something to bear in mind during the design phase of your study. Getting meaningful results in this case may actually involve marking and resampling more than 50% of the population. 5. Suprisingly, the violations do not appear to drastically alter the range of estimates returned 95% of the time when M = 50 and C = 30. The mean Lincoln-Petersen index did not appear to differ significantly either (100.47 versus 99.27 by our results). This lack of difference can be explained by the fact that marked and unmarked individuals are equally likely to leave the population (or evade capture), so the ratio of marked to unmarked individuals remaining does not change much. However, the Monte Carlo simulations of the violations appeared to Answers: Exercise 18

Page 2 of 3

increase the total range of Lincoln-Petersen estimates and produced a skewed distribution. Our results are shown on the following page; yours will be a bit different. Under different levels of C and M, the violations may be more severe.


Page 3 of 3

32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48

M N M = 50, C = 30, no violations Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count Largest(25) Smallest(25)

100.4789 0.614051 96.875 103.3333 19.418 377.0586 2.572796 1.171355 156.8452 64.58333 221.4286 100478.9 1000 140.9091 70.45455

O

P Q M = 50, C = 30, violations Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count Largest(25) Smallest(25)

99.26785 0.607237 96.875 103.3333 19.20251 368.7364 7.518532 1.753205 193.75 64.58333 258.3333 99267.85 1000 140.9091 70.45455

6. There are many ways to modify the model; here is one suggestion using nested functions. Set cell E6 to 1 since the population is closed. Set cell E7 to 1 since this cell represents the probability that both marked and unmarked individuals become trap shy. Add a new probability in cell E8, which will represent the probability that only marked individuals become trap shy. Insert new cells in column E (shift the current cells to the right). Use a nested IF, AND, and RAND formulae in new cell E11: =IF(AND(D11="m",RAND() 0.4, the probability of extinction was 1, indicating that 100% of our trial simulations went extinct within 100 years. Probabililty of Extinction for Lambda = 1, Varying Standard Deviations Probabillity of extinction

1.2 1 0.8 0.6 0.4 0.2 0 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1

Standard deviation of lambda (s )

5. For a fixed parameters, l = 1, s = 0.25, risk of extinction decreases as the initial population size increases. This relationship is not linear. The extinction probability changes more slowly when the population size was 500 or greater in our model (your results may be different).

Risk of Extinction for a Population with Mean Lambda = 1, Standard Deviation = 0.25

Probability of extinction

0.5 0.4 0.3 0.2 0.1 0 100

200

300

400

500

600

700

800

900

1000

Initial population size


Page 2 of 2

Answers to Exercise 29 Harvest Models Fixed-Quota versus Fixed-Effort: First Questions QUESTIONS 1. What are the effects of different values of R and K on catches and population sizes in the two harvest models? 2. What are the effects of different values of Q on catches and population size in the population harvested under a fixed quota scheme? 3. What are the effects of different values of E on catches and population size in the population harvested under a fixed quota scheme? 4. How does the maximum sustainable yield (MSY) relate to R and/or K? 5. Is there any difference between fixed-quota and fixed-effort harvesting in terms of risk of driving the harvested population to extinction? ANSWERS 1. Change R by entering different values into cell B6. Change K by entering different values into cell B7. Your changes will be automatically echoed in the models. After each change, examine your spreadsheet, your graph of population size and harvest, and your graphs of recruitment curve and quota or effort line. Try changing K from 2000 to 4000, leaving R at 0.50. You should see that the catch remains unchanged at 250 individuals per time period in the fixed quota scheme. This is hardly surprising, because that's the nature of a fixed quota—it does not change. The catch in the fixed effort scheme starts at 1000 and declines to an equilibrium of 500, which is twice the catch in the fixed effort harvest with the original carrying capacity of 2000. Your graph of the recruitment curve and fixed quota now shows the quota line (horizontal line) crossing the recruitment curve at two points, in contrast to just touching the recruitment curve at its peak with the original value of K. If you drop a vertical line from the right-hand crossing point to the X-axis, you should see that this is the equilibrium population size. Your graph of the recruitment curve and fixed effort line still shows the effort line crossing the recruitment curve at its peak, although that peak has shifted to a higher value of N. Reducing K to 1000 has more drastic effects. The population harvested by a fixed quota goes extinct, and the harvest goes to zero. The fixed effort population declines to an equilibrium at 250 individuals, and catch declines from 250 individuals per time period to 125. Your graph of the recruitment curve and fixed quota now shows the quota line (horizontal line) lying entirely above the recruitment curve. This means that the quota is larger than recruitment regardless of population size, and harvesters will remove more individuals from the population than are added to it. So it is not surprising that the population goes extinct. Your graph of the recruitment curve and fixed effort line still shows the effort line crossing the recruitment curve at its peak, although that peak has shifted to a lower value of N. Restore K to 2000, and change R to 1.00. The population harvest according to a fixed quota now stabilizes at about 1707 individuals, a larger value than when R was 0.50. The harvest, of course, does not change. The fixed-effort population stabilizes at 1500 individuals, and the harvest stabilizes at 375 individuals per time period, more than with the original value of R = 0.50.

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Your graph of the recruitment curve and fixed quota now shows the quota line (horizontal line) crossing the recruitment curve at two points again. As before, if you drop a vertical line from the right-hand crossing point to the X-axis, you should see that this is the equilibrium population size. Your graph of the recruitment curve and fixed effort line now shows the effort line crossing the recruitment curve to the right of its peak. Now change R to 0.40. The fixed-quota population goes extinct, and its harvest goes to zero. The fixed-effort population stabilizes at 500 individuals, and its harvest stabilizes at about 187 individuals per time period. Your graph of the recruitment curve and fixed quota again shows the quota line (horizontal line) lying entirely above the recruitment curve. This means that the quota is larger than recruitment regardless of population size, and harvesters will remove more individuals from the population than are added to it. So it is not surprising that the population goes extinct. Your graph of the recruitment curve and fixed effort line now shows the effort line crossing the recruitment curve to the left of its peak. Finally, change R to 0.20. Again, the fixed-quota population goes extinct. The fixed effort population is still declining at the end of 50 time units, and seems headed for extinction also. Its harvest is likewise declining toward zero. Your graph of the recruitment curve and fixed quota again shows the quota line (horizontal line) lying entirely above the recruitment curve, now by a wider margin. This means that the quota is larger than recruitment regardless of population size, and harvesters will remove more individuals from the population than are added to it. So it is not surprising that the population goes extinct. Your graph of the recruitment curve and fixed effort line now shows the effort line lying entirely above the recruitment curve. As in the fixed quota model, the harvest is now greater than recruitment regardless of population size. And likewise, the population goes extinct. 2. Restore R to 0.50 and K to 2000. Change Q from 250 to 300. You should see that the quota line once again lies entirely above the recruitment curve, and the population goes extinct. Change Q to 200. Now the quota line crosses the recruitment curve at two points, and the population stabilizes at the right-hand crossing point. The original value of Q = 250 was the maximum sustainable yield, as you can confirm by setting Q to values closer and closer to 250, above and below. You should see that any value of Q greater than 250 will drive the population to extinction. Any value of Q less than 250 will result in a larger equilibrium population, but will of course result in a smaller harvest. 3. Change E from 0.25 to 0.20. The effort line now crosses the recruitment curve to the right of its peak, and the population stabilizes at the crossing point. This results in a larger equilibrium population and a smaller harvest. Change E to 0.30. The effort line now crosses the recruitment curve to the left of its peak, and the population stabilizes at the crossing point. This results in a smaller equilibrium population and a smaller harvest. Change E to 0.50. The effort line now touches the recruitment curve at N=0 and lies above it everywhere else. The population and harvest are both still declining at the end of 50 time units, heading for extinction. The original of E = 0.25 gave the maximum sustainable yield of 250 individuals per time period, as you can confirm by plugging in values of E above and below 0.25.

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Note that the MSY is the same in both models. This must be the case, because the MSY is a property of the population, not of the harvesting scheme. The harvesting schemes are just two ways of attempting to harvest the MSY. 4. Examine your graphs of recruitment against population size (either model). You should see that recruitment is greatest when N ˜ K/2. As we said earlier, a sustainable harvest cannot exceed recruitment. Therefore, the maximum sustainable harvest, or MSY, must equal the maximum recruitment. Changing values of R and K will have shown you that both of these parameters affect the recruitment curve, so MSY must be related to both R and K. We will derive the exact relationship in the next section. 5. The two harvest schemes have rather different implications for the risk of driving the harvested population to extinction. You should have noticed as you were changing values of R, K, Q, and E, that any decrease of R or K from the original values, and any increase of Q, drove the fixedquota population to extinction. However, the fixed-effort population persisted under most of the same values of R and K, and only the highest values of E drove it to extinction. So it would seem that a fixed effort harvest is less likely to drive the harvested population to extinction. We will investigate this in more detail in the last part of the exercise. Fixed-Quota versus Fixed-Effort: More Questions QUESTIONS 6. What happens to a population harvested at the MSY under a fixed quota, if it starts out at its carrying capacity? 7. What happens to a population harvested at the MSY under a fixed quota, if it starts out at less than half its carrying capacity? 8. What happens to a population harvested at the MSY under a fixed effort strategy, if it starts out at its carrying capacity? 9. What happens to a population harvested at the MSY under a fixed effort strategy, if it starts out at less than half its carrying capacity? 10. You can calculate exact values for the fixed quota and the fixed effort that produce the maximum sustainable yield because you know the exact values of R and K for the harvested populations. In the real world, however, managers must estimate these values, and estimates are always subject to error. What happens to each of the harvested populations if you overestimate K? In other words, what happens if K is actually smaller than you think it is? 11. What happens if you overestimate R? 12. What happens if you set the quota or the effort too high? 13. Do you see another reason why setting a fixed-quota at the maximum sustainable yield is a risky strategy? 14. Is the fixed-effort harvest less risky? 15. What happens if managers decide to build a safety margin into a fixed-quota harvest? 16. In answering Questions 6–15, you will have seen two ways in which a fixed-effort harvest is less risky than a fixed-quota harvest. A fixed-effort harvest is not completely without risk, however. What happens to a population harvested under a fixed-effort strategy, if the effort is set too high? How does a high effort increase the risk of extinction in a fixed-effort harvest? ANSWERS

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6. Return R to 0.50, K to 2000, Q to 250, and E to 0.25. Examine your graph of Nt and harvest versus time for the fixed-quota population. You should see the population decline from K to K/2, and stay there. Thus, taking the maximum sustainable yield from a population by fixed quota means reducing it to half its carrying capacity. 7. Set N0 for the fixed quota population to a value slightly less than K/2. Overwrite the formula in cell D12 with the value of your choice. You should see the population decline to zero. This happens because the harvest line is above the recruitment curve for all values of N < K/2. In other words, if the population falls to less than half its carrying capacity, the fixed quota harvest will remove more individuals from the population than are recruited into it, and the population will shrink. Thus, if for any reason (e.g., bad weather or poor food supply), the population declines below K/2, a fixed quota harvest will drive it to extinction. If the population has already been reduced to K/2 by harvesting at the MSY, falling below K/2 is very likely to occur sooner or later, given the normal annual variability in most populations. Incidentally, this also answers a question posed in the Introduction, about whether the equilibrium at N=K/2 under a fixed quota harvest is a stable equilibrium or an unstable one. The answer is: neither - it is a metastable equilibrium. If N departs from K/2 to a higher value, it will return to K/2, as it would for a stable equilibrium. However, if N departs from K/2 to a lower value, it will continue to move away from K/2, as it would for an unstable equilibrium. N will continue to shrink until it reaches zero (which is a stable equilibrium, although undesirable in most cases.) This is a mathematical way of expressing the riskiness of fixed-quota harvesting. 8. Examine your graph of Nt and harvest versus time for the fixed-effort population. You should see the population decline from K to K/2, and stay there. Thus, taking the maximum sustainable yield from a population by fixed effort also means reducing it to half its carrying capacity. 9. Set N0 for the fixed quota population to values between K/2 and zero. Overwrite the formula in cell G12 with the value of your choice. As long as N0 > 0, you should see the population increase to K/2, and stay there. This happens because the harvest line is below the recruitment curve for all values of N < K/2. In other words, if the population falls to less than half its carrying capacity, the fixed effort harvest will remove fewer individuals from the population than are recruited into it, and the population will grow. This is one way in which a fixed effort harvest is less risky than a fixed quota harvest. If the population has already been reduced to K/2 by harvesting at the MSY, falling below K/2 will result in a harvest that is not only smaller than the MSY but, more importantly, smaller than recruitment, which allows the population to recover (at least to K/2). Incidentally, this demonstrates that the MSY is a stable equilibrium under fixed-effort harvesting. If N departs from K/2 in either direction, it will return to K/2. Again, this is a mathematical expression of the lower risk associated with fixed-effort harvesting. 10. Restore the formula =B7 to cells D12 and G12. Leave Q and E unchanged, and adjust K in cell B7 downward. Do you see the logic of this? Try several values of K, changing it slightly each time. Examine all your graphs after each change. You should see that any overestimate of K results in extinction of the fixed quota population. The fixed effort population declines, but eventually stabilizes without going extinct.

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11. Restore the value of K in cell B7 to 2000. Leave Q and E unchanged, and reduce R in small increments, as you did with K above. Examine all your graphs after each change. You should see the same results as when you overestimated K. 12. Return R and K to their original values (0.50 and 2000), and increase Q and E slightly. You should see that the fixed quota population goes extinct, but the fixed effort population persists, albeit at a smaller equilibrium size. 13. A fixed-quota MSY harvest is risky because even slightly overestimating K or R, or setting Q even a little too high, drives the harvested population to extinction. 14. A fixed-effort MSY harvest is less risky because overestimating R or K, or setting E too high (within limits), does not drive the population to extinction, although it may result in a smaller surviving population and a smaller harvest. To drive the population to extinction, you must set E = R. 15. To simulate a safety margin, set a fixed quota less than the MSY value. Examine your graph of the recruitment curve with the fixed-quota line superimposed. Notice that the fixed-quota line now crosses the recruitment curve at two points (N = N* 1 and N = N* 2 , see graph below). Both points are equilibria, where harvest equals recruitment, and are therefore sustainable yields. Both points represent lower harvests than the MSY, so they are not maximum sustainable yields. Set the initial population size of the fixed-quota population to values greater than N* 2 , between N* 2 and N* 1 , and less than N* 1 . What happens to the population size in each case? For N0 values between K and N* 2 , and for values between N* 2 and N* 1 , N returns to N* 2 . For N0 values less than N* 1 , N goes to zero. Thus, reducing the quota below the MSY reduces the risk of extinction associated with a fixedquota harvest. Under normal circumstances, N will remain stable at N* 2 . If chance variability reduces N below N* 2 , the population will recover to N* 2 , unless the reduction is so severe that N falls below N* 1 . By setting a low quota, managers can increase the spread between N* 1 and N* 2 , and reduce the risk of extinction. But doing so also reduces the allowed harvest, and it may be politically difficult to set quota low enough to ensure the harvested population's survival.

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Figure 1. Fixed quota harvest with a safety margin.

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16. Incrementally increase E, leaving other parameters unchanged. Examine your graphs of harvest and population size, and of effort and recruitment. As you increase E, you should see the effort line become steeper and steeper, and thus closer and closer to the recruitment curve on the left hand side. As you increase E, you should also see the equilibrium population size get smaller and smaller. Both factors increase the risk of extinction. Because the effort line is close to the recruitment line, anything that reduces recruitment may cause recruitment to fall below harvest, and lead to population decline, perhaps to zero. Likewise, a small population is at greater risk of extinction from random variation in N. Annual Variation QUESTIONS 17. Does the addition of stochastic variation to the model increase the risk of extinction for a population harvested at the MSY under a fixed quota? 18. Does the addition of stochastic variation to the model increase the risk of extinction for a population harvested at the MSY by fixed effort? 19. Does reducing a fixed quota below the MSY (i.e., building in a safety margin) prevent extinction of a stochastically-varying population?

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20. Does the addition of stochastic variation to the model increase the risk of extinction for a population harvested above the optimal effort (i.e. the effort that yields the MSY)? ANSWERS 17. Examine your graph of population size and harvest for the fixed-quota population. Press the recalculation key. You should see all the population values and graphs change, as the spreadsheet recalculates the random values and resulting population sizes. Repeat this recalculation many times, while watching your graph of population size and harvest. You should see that with the quota set at the MSY, and R varying by 0.5, the population often goes extinct within the 50 time-units shown. Recall that the MSY was a sustainable harvest in the deterministic model. Plainly, variation increases the risk of extinction. You can quantify this increased risk, if you wish, by keeping a tally of the total number of trials (recalculations) and the proportion of trials in which the population goes extinct.

18. Repeat the recalculation experiment described above for the fixed effort population. You should see that the fixed-effort population rarely, if ever, goes extinct with these parameter values. 19. Simulate a safety margin by changing the quota from 250 to 200 in cell F9. Repeat the recalculation experiment, while observing your graph of population size and harvest. You should see that, although the risk of extinction is reduced compared to harvesting at the MSY, the population still goes extinct fairly often. Even a quota as low as 100 produces occasional extinctions. 20. Simulate harvesting at greater than optimal effort by changing the effort from 0.25 to 0.40 in cell I9. Repeat the recalculation experiment, while observing your graph of population size and harvest. You should see that the fixed-effort population often becomes quite small and may be still declining after 50 time units. Whether or not the population goes extinct in the model, a very small population is plainly at greater risk of extinction than a large one. This scenario represents a serious overestimate of the MSY. Overestimating the optimal effort to this degree in the deterministic model of fixed effort did not produce extinction, but with stochastic variation, it can. Thus, stochastic variation also increases the risk of extinction under a fixed-effort harvest, but to a lesser degree than under a fixed quota harvest.. Try smaller values of E, representing better estimates of the MSY. You should see the risk of extinction decrease with smaller efforts. The Bonus Analysis (Allee effect) is left for the reader to work out.

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Answers to Exercise 30 Landscape Ecology: Quantifying Landscape Patterns

1. Changing the definition of a patch (as in Figure 2) alters many of the patch level metrics computed for a given landscape. For instance, if you consider a patch to be a cell plus any adjacent and diagonal cells (such as Species 2 in Figure 2), the number of patches in the landscape will decrease. The definition of a patch should center on the species of interest, and how those species perceive changes in habitat. For example, as you move from class to class, subtle changes in patch composition (from grass to weeds to bare ground) might affect how you “disperse” between buildings. However, a beetle crossing the very same area is likely to perceive and react to these same habitats differently. Thus, for a given landscape structure and composition, the metrics will be quite different depending on the patch “rule” used in analysis.

2. The categories that define each pixel, as well as the grain size of each pixel, are critical in landscape quantification. For example, imagine you have an aerial photograph of your school and the surrounding areas, and are given the task of quantifying the landscape pattern. You would have to decide first what resolution should be evaluated (do you assign habitats based on a 50 m2) or some other resolution. The resolution will critically impact how you classify each pixel’s habitat type. For example, if you choose a 50 m2 resolution, and a pixel in your photo consists of both forest and field habitat, how will you determine whether to call that pixel a “forest” or a “field”? As resolution increases so that more detail can be quantified (e.g., pixel size is 20 m2), you might be able to classify the “forest” and “field” more accurately. However, there are trade-offs. If you have limited funding for analysis, it is unlikely that you can map in detail a very large area. Decisions on the appropriate scale will need to be considered in light of the study objectives.

Exercise 30

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3. As the extent changes, the presence and relative proportions of the two land cover types change. Many other metrics vary as well as a function of extent. For example, in Landscape a, the patch in the upper right-hand corner consists of 2 cells (using the restricted rule for patch identification). Increasing the extent of the landscape, we now see that the patch consists of 3 cells. The smaller the extent of a map, the more serious the problem of artificial truncation of patches by the map boundary, resulting in biased measurements of patch size, shape, and complexity (Turner et al. 2001).

0 1 1 0 1 0 0 1 0 1 1

Exercise 30

Landscape a : Extent = 81 cells 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 1 0 0 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 1 1 1

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Landscape b : Extent = 121 cells 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 1 0 0 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 1 1 1 1 1 1 0 0 0

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4. One way to aggregate the smaller pixels “up” so that the landscape consists of only 9 units is to assign each block a habitat type based on the proportion of pixels that consist of each habitat type. For example, the upper-left cell of the right-hand landscape would be classified as a shaded habitat type because over 50% of the habitat is shaded. The bottom-left block of cells would be classified as the nonshaded habitat type because over 50% of its habitat is nonshaded. The re-sampled landscape is very different from the landscape on the left in terms of landscape metrics. Thus, grain size is a significant consideration in how landscapes are quantified. n =1

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Exercise 30

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Answers to Exercise 31 Edge Effects and Ecological Traps 1. The total population increases exponentially and does not reach K within 50 years. This positive growth occurs because there is a significant amount of core habitat, and 80% of the juveniles attempt to breed in the core habitat where birth and survival rates are high. Equilibrium is not reached within 50 years because neither habitat type reaches K. At first, the population size decreases in the edge habitat, and slightly increases in the core, indicating that the core is functioning as a source habitat and the edge is functioning as a sink. By Year 9 the edge population begins to increase.This is because at this point, there are enough juveniles produced in the system so that the 20% preferring edge habitat more than offsets the mortality of adults in the edge habitat, causing the population in the edge to increase.

Population Size with a Post-Breeding Census


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2. When the selection for core habitat is 0.7, the entire system declines to extinction.

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This indicates that enough of the new adults actively seek edge for breeding, which places enough individuals in the edge habitat so that the whole system does not produce enough young to sustain the population. In this case, the sink edge habitat functions as a trap and “drains” the population away from core habitat, leading to extirpation.


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3. When b = 2.0 the population goes extinct. When b = 2.5 the population increases exponentially in the 50 year period. When b = 2.75 the population reaches K. With a substantial number of young being produced in the core, and 50% of these individuals seeking breeding territories in the edge, K is reached first in the edge habitat because it is less abundant. The new adults seeking breeding territories in the edge that are unsuccessful then attempt to breed in the core, pushing the core population towards K very quickly.

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4. Under the conditions given, when the birth rate in the core is 2.75 individuals per individual per year, there must be 420 hectares of core habitat to sustain the population. However, if a is less than 0.42, the population will go extinct. Thus, persistence is very sensitive to habitat selection for the high quality habitat. Population Size with a Post-Breeding Census


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Answers to Exercise 32 Triage: Prioritizing Species and Habitats

1. Sites 2 and 3 were selected to be purchased. You might have come to this same conclusion with some thought. Keeping in mind that species 2, 4, and 5, were identified as the species of highest conservation priority and the suitabilities of the different sites shown below, site 2 will benefit species 2 and 5, while site 3 will benefit species 3 and 4. Although site 4 would be valuable for both species 4 and 5, it has a low probability of development and thus is likely to be conserved to some extent even if it is not purchased. Site 1 is of value only to species 1, which is not a high-priority species.

Standardized Suitability Scores for 5 Species on 4 Sites 1 0.9

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2. It appears that the model is fairly robust to slight changes in the prioritization scheme, given the values entered in the model. This is because the decision to purchase a site not only depends on the prioritization scores, but also the probability of development. We could not find a solution that would recommend that site 4 be purchased. Even by making species 1 the highest conservation priority, site 1 is not purchased because it has little value to the other species. 3. Only when the probability of development reaches 0.8 for site 4 does the Solver select that site for purchase. At that point, site 4 is the only site that can be purchased because it is an expensive acquisition.


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1

4. In many cases, higher densities at a site indicate that the habitat there is of better quality. However, if you have completed the Source-Sink or Ecological Traps Exercise, you know that at times density can be high even though habitat quality is low. In such cases, suitability would better be estimated by measuring a demographic variable associated with fitness (birth and death rates). 5. If populations vary significantly from year to year (due to demographic or environmental stochasticity), suitability models based on a single sampling session may be misleading. If the year-to-year variation is small, single sessions may be beneficial because funding that is used for developing suitability indices may be spent on other conservation needs. If you have completed the exercise on Island Biogeography, you realize that distance to other sites, as well as size, is important in considering the location of reserves. Such information on spatial arrangements should ultimately be considered in reserve selection.


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Answers to Exercise 33 Reserve Design 1. The optimal habitat proportion is 16% habitat 1 and 84% habitat 2, producing 20 individuals of species 1 and 18 individuals of species 2, for a total abundance of 38 individuals of both species. You can’t reach your objective of 20 individuals of both species.

L 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

Answers to Exercise 33

M N Proportion of habitat Habitat 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48 0.52 0.56 0.6 0.64 0.68 0.72 0.76 0.8 0.84 0.88 0.92 0.96 1

Habitat 2 1 0.96 0.92 0.88 0.84 0.8 0.76 0.72 0.68 0.64 0.6 0.56 0.52 0.48 0.44 0.4 0.36 0.32 0.28 0.24 0.2 0.16 0.12 0.08 0.04 0

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Species 1 #NUM! 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125

Species 2 125 79.7799163 49.9546722 30.6351074 18.364629 10.7374182 6.10744449 3.36951562 1.7968344 0.9223372 0.45349632 0.21231384 0.09396082 0.03895504 0.01495855 0.00524288 0.00164527 0.00045036 0.00010367 1.9021E-05 2.56E-06 2.199E-07 9.2876E-09 1.0737E-10 5.2429E-14 0

Total #NUM! 84.7799163 59.9546722 45.6351074 38.364629 35.7374182 36.1074445 38.3695156 41.7968344 45.9223372 50.4534963 55.2123138 60.0939608 65.038955 70.0149585 75.0052429 80.0016453 85.0004504 90.0001037 95.000019 100.000003 105 110 115 120 125

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2. The optimal habitat proportion is 4% habitat 1 and 96% habitat 2, producing 5 individuals of species 1 and 79 individuals of species 2, for a total abundance of 84 individuals of both species. L 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

M N Proportion of habitat Habitat 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

0 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 0.44 0.48 0.52 0.56 0.6 0.64 0.68 0.72 0.76 0.8 0.84 0.88 0.92 0.96 1

Habitat 2 1 0.96 0.92 0.88 0.84 0.8 0.76 0.72 0.68 0.64 0.6 0.56 0.52 0.48 0.44 0.4 0.36 0.32 0.28 0.24 0.2 0.16 0.12 0.08 0.04 0

O

P Q Number of individuals

Species 1 #NUM! 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125

Species 2 125 79.7799163 49.9546722 30.6351074 18.364629 10.7374182 6.10744449 3.36951562 1.7968344 0.9223372 0.45349632 0.21231384 0.09396082 0.03895504 0.01495855 0.00524288 0.00164527 0.00045036 0.00010367 1.9021E-05 2.56E-06 2.199E-07 9.2876E-09 1.0737E-10 5.2429E-14 0

Total #NUM! 84.7799163 59.9546722 45.6351074 38.364629 35.7374182 36.1074445 38.3695156 41.7968344 45.9223372 50.4534963 55.2123138 60.0939608 65.038955 70.0149585 75.0052429 80.0016453 85.0004504 90.0001037 95.000019 100.000003 105 110 115 120 125

This habitat is different than the optimal proportion found when the objective was 20 individuals per species. Specifically, the optimal proportion includes less of habitat 1 and fewer individuals of species 1, but a much larger total abundance across both species. The total abundance is much larger in this case because species 1 achieves the objective of 5 individuals with only 1 cell of habitat, at which time species 2 has many cells of habitat at close to the maximal density. species 2 does not reach the objective of 5 individuals until 80% of the habitat is habitat 2, allowing only 25 individuals of species 1 for a total abundance of 35 individuals. Thus, given these habitat associations and objectives, the optimal habitat proportion is mostly habitat 2.


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3. As d increases, the carrying capacity of the habitat increases. As potential density increases, the range of abundance also increases for each species. Thus, fewer cells are necessary to achieve minimum abundances and higher abundances are possible. For example, here is a graph of abundances with d increased to 10.


300 250 200

Species 1

150

Species 2 Total

100 50

0.96

0.88

0.8

0.72

0.64

0.56

0.48

0.4

0.32

0.24

0.16

0.08

0

0

Proportion of habitat 1

As z increases, the critical threshold proportion increases for a species. In other words, a higher proportion of habitat is necessary for density to increase much above zero. For example, the graph below shows density as a function of proportion of habitat 1 with different values of z. Generally speaking, species with low value of d and high values of z are the limiting factors in reserve design considerations.

Density

Density as a Function of Habitat Proportion 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

z=1 z = 10

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Proportion of habitat 1


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4. Habitat configuration B maximizes total abundance across the two species. This conclusion is reached by calculating the abundance of each species and total abundance for each habitat configuration. Configuration B provides the largest abundance of species 1 that was limited by patch sizes of at least 3 cells. 5. Habitat configuration B still maximizes total abundance across the two species. The abundance of species 1 is smaller with the larger minimum patch size requirement, causing the total abundances to be smaller as well. As the patch size requirements increase, the abundances are generally smaller because fewer cells meet the patch size requirement in any given configuration.


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Answers to Exercise 34 VBA Landscape Programming

There are no specific answers to these questions.

Answers to Exercise 35 Neutral Landscape Theory and Connectivity 1. Because the landscape consists of two cover types, one cover type increases in total edge and number while the other decrease as the proportions of habitat in the landscape change. Thus, when the proportion of forest habitat is low, the proportion of non-forest habitat is high. At intermediate proportions where both habitats are approximately equal across the landscape, the total edge and total number of patches peaks for both types. As the landscape is altered to low proportions of a single habitat type, the edge is decreased. But at the same time, the edge is high for the second habitat type. 2. There are several different ways to determine when a landscape is fragmented. A conventional way is to determine the proportion of habitat when the number of patches, core area, or largest patch size suddenly changes. For example, one could argue that the graph below shows the greatest change in maximum patch size around 70% forest cover (at that point the slope of the tangent line is greatest). Another person might argue that around 50% forest cover there is a change in the relationship between n maximum patch size and proportion of forest cover. Exactly where the thresholds is not just a function of random, neutral model results, but also for the organism of interest as well as the nature of the matrix (non-forest habitat).

Maximum Patch Size

Total Core

120 100

Average

80 60 40 20 0 1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

Proportion of forest cover

3. Certainly, the small extent of our landscape (100 cells) biases our results. It would be very insightful to repeat the analysis for much larger landscapes. When the landscape is 100% forest, only 64% of the landscape is core. This is strictly a function of the boundaries of the landscape, in which all cells are considered non-core habitat. With larger maps, this bias is limited and the percolation thresholds are much closer to the 59% habitat predicted by percolation theory (Gardener et al. 1987).


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Answers to Exercise 36 Random Walks These questions have no specific answers.

Spreadsheet Exercises in Ecology and Evolution

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Ecology and Evolution in Anoxic Worlds (Oxford Series in Ecology and Evolution)

Big Questions in Ecology and Evolution

Ecology and evolution in anoxic worlds

Ecology and Evolution of Flowers

Ecology and Evolution of Flowers

Ecology and Evolution of Flowers

Ecology of social evolution

Anolis Lizards of the Caribbean: Ecology, Evolution, and Plate Tectonics (Oxford Series in Ecology and Evolution)

Ecology of Social Evolution

Reticulate Evolution and Humans: Origins and Ecology

Reticulate Evolution and Humans: Origins and Ecology

Discovering Evolutionary Ecology: Bringing Together Ecology and Evolution

Discovering evolutionary ecology: bringing together ecology and evolution

Discovering Evolutionary Ecology: Bringing Together Ecology and Evolution

Discovering Evolutionary Ecology: Bringing Together Ecology and Evolution (Oxford Biology)

Phenotypes: Their Epigenetics, Ecology and Evolution

Island Bats: Evolution, Ecology, and Conservation

Island Biogeography: Ecology, Evolution, and Conservation

Evolution and ecology : the pace of life

Insect-Fungal Associations: Ecology and Evolution

Metapopulation Biology: Ecology, Genetics, and Evolution

Plant Resins: Chemistry, Evolution, Ecology, and Ethnobotany

Ecology, Genetics and Evolution of Metapopulations

Island Biogeography Ecology, evolution, and conservation

Predatory Prokaryotes. Biology, Ecology and Evolution

Lagomorph Biology: Evolution, Ecology, and Conservation

Spreadsheet Exercises in Ecology and Evolution