BIO 356 - All Labs and Lab Homeworks

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BIO 356 All Labs & Homeworks SBU Spring 2013 N/A

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Lab #1 Exercise 1.1 1963: 10,000 blue whales

50,000 = 10,000(1.1)𝑡 50,000 10,000

????: 50,000 blue whales

5 = (1.1)𝑡

10,000

= 10,000 (1.1)𝑡

ln(5) = 𝑡 ∗ ln⁡(1.1)

⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡50,000 = 10,000(1.02)𝑡 50,000 10,000

Exercise 1.2

=

10,000 (1.02)𝑡 10,000

𝑡=

5 = (1.02)𝑡

ln(5) ln⁡(1.1)

𝒕~𝟏𝟔. 𝟖𝟗⁡𝒚𝒓𝒔 𝑡=

ln(5) ln⁡(1.02)

ln(5) = 𝑡 ∗ ln⁡(1.02) 𝒕~𝟖𝟏. 𝟐𝟕⁡𝒚𝒓𝒔

Table 1.3

Year (t)

1800 1850 1870 1890 1910 1930 1950 1970 1975 1980 1985 1990 1995

Population [N(t)]

0.91 1.13 1.30 1.49 1.70 2.02 2.51 3.62 3.97 4.41 4.84 5.29 5.75

Time Interval (T)

Previous Population [N(t-T)]

-----

----50 20 20 20 20 20 20 5 5 5 5 5

0.91 1.13 1.30 1.49 1.70 2.02 2.51 3.62 3.97 4.41 4.84 5.29

Growth Rate in T years [N(t)]/[N(t-T)] ----1.24176 1.15044 1.14615 1.14094 1.18824 1.24257 1.44223 1.09669 1.11083 1.09751 1.09298 1.08696

Annual Growth Rate [N(t)]/[N(tT)]^(1/T) ----1.00434 1.00703 1.00684 1.00661 1.00866 1.01092 1.01848 1.01863 1.02124 1.01878 1.01794 1.01682


Growth Rate vs. Year

Growth Rate (%)

1.024 1.022 1.02 1.018 1.016 1.014 1.012 1.01 1.008 1.006 1.004 1.002 1800

1900

2000

2100

Time (Years)

Table 1.4

**Although growth rate continually decreases, the number of people added continues to increase**

Year

Population Size (in Billions) [N] 1975 1985 1995

3.97 4.84 5.75

96,715,000

264,973⁡

Time Completed:

Annual Growth Rate (R)

1.01863 1.01878 1.01682

Annual number of people added to the population [N*(1-R)] 73,961,100 90,895,200 96,715,000

⁡𝑝𝑒𝑜𝑝𝑙𝑒 1⁡𝑦𝑒𝑎𝑟 𝒑𝒆𝒐𝒑𝒍𝒆 ∗ ~𝟐𝟔𝟒, 𝟗𝟕𝟑⁡ 𝑦𝑒𝑎𝑟 365⁡𝑑𝑎𝑦𝑠 𝒅𝒂𝒚

𝑝𝑒𝑜𝑝𝑙𝑒 𝑑𝑎𝑦

11,041⁡

𝑝𝑒𝑜𝑝𝑙𝑒

11,041⁡


ℎ𝑜𝑢𝑟

ℎ𝑜𝑢𝑟

1⁡𝑑𝑎𝑦

∗ 24⁡ℎ𝑜𝑢𝑟𝑠 ~𝟏𝟏, 𝟎𝟒𝟏⁡ 1⁡ℎ𝑜𝑢𝑟

𝒑𝒆𝒐𝒑𝒍𝒆 𝒉𝒐𝒖𝒓

𝒑𝒆𝒐𝒑𝒍𝒆

∗ 60⁡𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ~𝟏𝟖𝟒⁡ 𝒎𝒊𝒏𝒖𝒕𝒆 1⁡ℎ𝑜𝑢𝑟


∗ 2⁡ℎ𝑎𝑙𝑓−ℎ𝑜𝑢𝑟𝑠 ~𝟓, 𝟓𝟐𝟏⁡ 𝒕𝒊𝒎𝒆⁡𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒆𝒅

30 minutes Exercise 1.3 Year

1995

Table 1.5 Fecundity (f)


0.02730

1.01682

10-Year Growth Rate (R10)

Population at Population at the beginning the end of of the 10the 10-Year Year interval interval (in (in billion) billion) 1.18152 5,750,000,000 6,793,740,000


2005 2015 2025 2035

0.02309 0.01889 0.01468 0.01048

1.01261 1.00841 1.00420 1.00000

1.13350 1.08736 1.04280 1.00000

6,793,740,000 7,700,704,290 8,373,437,817 8,731,820,955

7,700,704,290 8,373,437,817 8,731,820,955 8,731,820,955

1.00000 + 4𝑥 = 1.01682

𝑅 =𝑓+𝑠

4𝑥

1.01682 = 0.02730 + 𝑠

4

=

0.01682 4

𝑥 = 0.00420

𝑠 = 0.98952 ∆𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝑃𝑜𝑝𝑓 − 𝑃𝑜𝑝𝑖

∆𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 8,731,820,955 − 5,750,000,000 = 2,981,820,955⁡⁡⁡⁡⁡⁡⁡⁡ If the fecundity were to decrease continually over an 80 year period as opposed to a 40 year period, the final population would be larger than it is as described in the preceding table. Because the decrease in fecundity is more gradual in the 80 year declination than the 40 year declination, there is more time to accrue a more sizable population. Although the R-values continually decrease, there are more time steps that compound the geometric progression of the population’s growth, leading to an overall increase in the population.

Homework #1

1980: 450 2007: 2,300

2,300 = 450(𝑅)27

5. 1̅ = 𝑅 27

27 𝑅 = √5. 1̅

2,300

27

𝑅~1.06229

450

450

= 450 (𝑅)27

√5. 1̅ = 27√𝑅 27

R=? This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

Table 1.3 Year (t)

Population [N(t)]

Time Interval (T)

Previous Population [N(t-T)]

-----

-----

1800 1850 1870 1890 1910 1930 1950 1970 1975 1980 1985 1990 1995

0.91 1.13 1.30 1.49 1.70 2.02 2.51 3.62 3.97 4.41 4.84 5.29 5.75

Year

Population Size (in Billions) [N]

50 20 20 20 20 20 20 5 5 5 5 5

Growth Rate in T years [N(t)]/[N(t-T)] ----1.24176 1.15044 1.14615 1.14094 1.18824 1.24257 1.44223 1.09669 1.11083 1.09751 1.09298 1.08696

0.91 1.13 1.30 1.49 1.70 2.02 2.51 3.62 3.97 4.41 4.84 5.29

Annual Growth Rate [N(t)]/[N(tT)]^(1/T) ----1.00434 1.00703 1.00684 1.00661 1.00866 1.01092 1.01848 1.01863 1.02124 1.01878 1.01794 1.01682

Table 1.4

1975 1985 1995 2000 2005 2010


3.97 4.84 5.75 6.01 6.50 6.86

1.01863 1.01878 1.01682 1.00888 1.01580 1.01084

Annual number of people added to the population [N*(1-R)] 73,961,100 90,895,200 96,715,000 53,394,109 102,693,439 74,358,132

Time Completed: 30 minutes (2010, 1995) 𝑁(𝑡) = 𝑁0 (𝑅)𝑡

𝑁(𝑡) = 𝑁0 (𝑅)𝑡

𝑁(𝑡) = 𝑁0 (𝑅)𝑡

𝑁(2000) = 𝑁(1995)(𝑅)5

𝑁(2005) = 𝑁(2000)(𝑅)5

𝑁(2010) = 𝑁(2005)(𝑅)5

6.01 = 5.75(𝑅)5

6.50 = 6.01(𝑅)5

6.86 = 6.01(𝑅)5

𝑹~𝟏. 𝟎𝟎𝟖𝟖𝟖

𝑹~𝟏. 𝟎𝟏𝟓𝟖𝟎

𝑹~𝟏. 𝟎𝟏𝟎𝟖𝟒


2.C: Even though the growth rate (R) is continually shrinking, the population size will continue to increase because R>1. Due to this fact, the geometric progression of the population’s growth every year [N*(R-1)] will be larger than the last. Essentially, the compounding nature of the population’s growth supersedes the fact that the growth rate is dwindling. In addition, the Rvalue is a relativistic measurement, whereas the increase in population is a more concrete, absolute measurement, giving it a more fundamental influence over the statistics that were collected. 2.D: 74,358,132 203,721⁡

𝑝𝑒𝑜𝑝𝑙𝑒 𝑦𝑒𝑎𝑟

𝑝𝑒𝑜𝑝𝑙𝑒 𝑑𝑎𝑦

8,488⁡


8,488⁡


ℎ𝑜𝑢𝑟

ℎ𝑜𝑢𝑟

1⁡𝑦𝑒𝑎𝑟

∗ 365⁡𝑑𝑎𝑦𝑠 ~𝟐𝟎𝟑, 𝟕𝟐𝟏⁡ 1⁡𝑑𝑎𝑦

∗ 24⁡ℎ𝑜𝑢𝑟𝑠 ~𝟖, 𝟒𝟖𝟖⁡ 1⁡ℎ𝑜𝑢𝑟

𝒑𝒆𝒐𝒑𝒍𝒆 𝒅𝒂𝒚

𝒑𝒆𝒐𝒑𝒍𝒆 𝒉𝒐𝒖𝒓


∗ 60⁡𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ~𝟏𝟒𝟏⁡ 𝒎𝒊𝒏𝒖𝒕𝒆 1⁡ℎ𝑜𝑢𝑟


∗ 2⁡ℎ𝑎𝑙𝑓−ℎ𝑜𝑢𝑟𝑠 ~𝟒, 𝟐𝟒𝟒⁡ 𝒕𝒊𝒎𝒆⁡𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒆𝒅

Table 1.5 Year

1995 2005 2015 2025 2035

Fecundity (f)


10-Year Growth Rate (R10)

0.02730 0.02309 0.01889 0.01468 0.01048

1.01682 1.01261 1.00841 1.00420 1.00000

1.18152 1.13350 1.08736 1.04280 1.00000

Population at the beginning of the 10Year interval (in billion) 5,750,000,000 6,793,740,000 7,700,704,290 8,373,437,817 8,731,820,955

Population at the end of the 10-Year interval (in billion) 6,793,740,000 7,700,704,290 8,373,437,817 8,731,820,955 8,731,820,955

3. A:


B: The 1995-2005 projection in Table 1.5 is relatively accurate, although slightly overestimated. The over-calculation is made clear in the 2005-2015 bracket, which dictates that within two years of the current date (despite the decreasing growth rate), the population will shoot up an additional 700 Million people. This error is due to the oversimplified nature of this chart, as several important variables (such as survivability) are neglected or considered constants. 4. Week 1: 6 snails Week 2: 60 snails

60 = 6(𝑅)2

10 = 𝑅 2

𝑅 = √10

6∗10

√10 = √𝑅 2

𝑅~3.16228

6

𝑁(10) = 6(𝑅)10

𝑁(10) = 6 ∗ 105

𝑁(10) = 6 ∗ (10.5 )10

𝑵(𝟏𝟎) = 𝟔𝟎𝟎, 𝟎𝟎𝟎⁡𝒔𝒏𝒂𝒊𝒍𝒔

%∆=

𝑁𝑜𝑣𝑟 (10)−𝑁(10)

%∆=

354,294−600,000

6∗9

600,000

6

𝑁(10)

%∆= 𝟒𝟎. 𝟗𝟓%⁡𝒍𝒆𝒔𝒔

%∆=

𝑁𝑢𝑛𝑑 (10)−𝑁(10)

%∆=

6

= 6 (𝑅)2

54 = 6(𝑅)2 6

= 6 (𝑅)2

9 = 𝑅2

𝑅 = √9

√9 = √𝑅 2

𝑅=3

𝑁𝑜𝑣𝑟 (10) = 6(𝑅)10

𝑁𝑜𝑣𝑟 (10) = 6 ∗ 59,049

𝑁𝑜𝑣𝑟 (10) = 6(3)10

𝑵𝒐𝒗𝒓 (𝟏𝟎) = 𝟑𝟓𝟒, 𝟐𝟗𝟒⁡𝒔𝒏𝒂𝒊𝒍𝒔

66 = 6(𝑅)2

11 = 𝑅 2

𝑅 = √11

966,306−600,000

6∗11

600,000

6

√11 = √𝑅 2

𝑅~3.31662

𝑁(10)

%∆= 𝟔𝟏. 𝟎𝟓%⁡𝒎𝒐𝒓𝒆

6

= 6 (𝑅)2

𝑁𝑢𝑛𝑑 (10) = 6(𝑅)10

𝑁𝑢𝑛𝑑 (10) = 6 ∗ 161,051

⁡𝑁𝑢𝑛𝑑 (10) = 6(11.5 )10

𝑵𝒖𝒏𝒅 (𝟏𝟎) = 𝟗𝟔𝟔, 𝟑𝟎𝟔⁡𝒔𝒏𝒂𝒊𝒍𝒔


5. A: Life expectancy and longevity are inversely proportional. While the exact proportionality differs by country and region, the longer the population lives, the less fertile it seems to be. Over time, fertility decreases while longevity increases. B: Both fecundity and fertility refer to the propensity/ ability to produce viable offspring. While fertility is strictly biological and hormonal perspective on reproductive ability, fecundity is more of a socio-political viewpoint. Abiotic factors limit not the biological ability to be fertile, but rather the organism’s focus, turning away from reproduction and toward other factors (such as enough food, proper shelter, or optimal temperature). Survival rates and life expectancy also deal with the same topic of life, but approach the idea in different ways. Life expectancy is how long an organism will live based upon the biotic and abiotic factors of its environment. Survival rates, however, deal with the probability of remaining alive in a certain habitat. If in an environment the survival rates for a group of organisms is the same, but one individual has a genetic variation that confers better defense against predators, that organisms life expectancy will be longer despite the same survival rates as its peers. Furthermore, the dichotomy between the 2 terms relates to a question of “if” (survival rates) and a question of “to what extent” (life expectancy).


Lab #2 Exercise 2.1

Bold Italics

Plus 1 (+1) Minus 1 (-1)

Oxen Number

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

Random Random Number 1 Number 2 (Fecundity (Survivability Check) Check) 0.18987 0.76982 0.86879 0.66965 0.00311 0.56135 0.28415 0.19026 0.60143 0.41116 0.23347 0.48411 0.03642 0.56234 0.60778 0.13405 0.61075 0.22032 0.64483 0.13234 0.05310 0.61525 0.17882 0.81667 0.63560 0.77641 0.44961 0.60393 0.28041 0.25283 0.64348 0.38131 0.95248 0.26458 0.58714 0.66393 0.88535 0.01913 0.26695 0.87845 0.43678 0.17713 0.09522 0.38212 0.62118 0.70191 0.09546 0.07640 0.61561 0.57725 0.42844 0.72485 0.78639 0.12758 0.93093 0.33435 0.03296 0.92517 0.58906 0.20929 0.49616 0.51858 Final Population Size (N)=37

F=0.227 (less) S=0.921 (more)

Oxen Number

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



Oxen Number

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


Oxen Number

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



Exercise 2.2 **Only ±3 oxen between simulations, results precise and clustered** Oxen Number

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

Random Random Number 1 Number 2 (Fecundity (Survivability Check) Check) 0.06279 0.46944 0.76153 0.45395 0.01457 0.25379 0.59864 0.15684 0.81774 0.22411 0.77611 0.44202 0.77611 0.88644 Trajectory 1 (Deterministic) The trajectory shows a maximum population of 163 musk oxen. 0.02405 0.70979 0.28824 0.98621 0.03909 0.04074 0.59588 0.65882 0.42476 0.83511 0.66241 0.28823 0.17190 0.15227 0.06089 0.87041 0.66345 0.67240 0.76907 0.92859 0.43813 0.45198 0.63463 0.27708 0.19712 0.89524 0.91955 0.18064 Trajectory 12 (Stochastic) The trajectory is more widespread than the 0.78845 0.61314 deterministic plot, although the abundance average was closely 0.28298 0.70625 aligned with that of the aforementioned trajectory. The differences 0.15321 0.09208 in the graphs are due to the fact that multiple new variables were introduced in the stochastic model, and these variables were 0.81192 0.31409 applied on an individualistic basis, not a holistic way in the first plot. 0.46803 0.34397 The final populations from the stochastic simulation range from 66 0.95918 0.98470 to 335 musk oxen. 0.96832 0.95959 0.49785 0.42909 0.18644 0.23727 0.85189 0.17877 Final Population Size (N)=35


Population Minima 35 46 32 33 33 38 40 36 37 32

Exercise 2.3

33 35 33 45 34 39 36 34 40 40

Trajectory 3 Based upon the graph and the numerical analysis, the probability of declining to a threshold of 31 is 0.1600.

Population Size (NC)

Number of runs that reached a size =NC

32 33 34 35 36 37 38 39 40 45 46

2 4 2 2 2 1 1 1 3 1 1

Cumulative number of runs that reached a size ≤ NC 2 6 8 10 12 13 14 15 18 19 20

Probability of decline to NC 0.10 0.30 0.40 0.50 0.60 0.65 0.70 0.75 0.90 0.95 1.00

Probability of Decline

Probability of Decline vs. Population 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 30

35

40

45

50

Threshold (Number of oxen) This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

Exercise 2.4

Initial Abundance

31

Standard Model Growth Survivability St. Rate Deviation of Growth Rate 1.1480 0.921 0.075

Trajectory 4 At a threshold level of 150 musk oxen, the probability of “explosion” is 0.5500

Parameter Initial Abundance Growth Rate Survivability St. Deviation of Growth Rate

Parameter Initial Abundance Growth Rate Survivability St. Deviation of Growth Rate

High Value (↑ 10%)

With High Value 0.76 1.00 0.56 0.57

Low Value (↓ 10%)

34 1.2628 1.0000 0.0825

With Low Value 0.46 *Experiment Fails* 0/undefined 0.55 0.64

28 1.0332 0.8289 0.0675

Difference 0.30 1.00 0.01 -0.07

Conclusion: Increasing the initial abundance of oxen and the growth rate of the population were both positively correlated with generating a larger population. Based upon the geometric progression of population growth, a larger principle (initial abundance) and compounding rate (growth rate) would substantially increase the population. The survivability (survival rate) decreased by only one (1) percentage point, so further testing is necessary to determine its relationship to this population’s growth. In this trial, the change in survival rates is determined to be negligible. However, the standard deviation of the growth rate proved to be negatively correlated with population growth. A smaller st. dev. means that the growth rate stays relatively unreactive to other variables, meaning that it is less prone to rapid vacillations that could potentially diminish the population’s growth. The evidence is clear that growth rate (R) is by far the most important, as a 10% decrease in value effectively fails the experiment; it will take much longer than 12 time steps to achieve the required threshold. The contrapositive of this fact is that a 10% increase mathematically yields a “perfect” This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

probability, meaning that the oxen will almost certainly recover to the ideal threshold within the allotted time. The implication of these data shows that the growth rate is extremely sensitive, and if humans intervene to save a species, they must be careful not to upset the local ecology in the process. In terms of creating simulations, a ±10% tolerance is too lax for such an influential variable.

Homework #2

1. The population simulations were rather precise, as the populations ranged from 35 to 38 oxen. This range of ±3 is determined by the growth rate, which itself is a composite of the survival rate and the fecundity rate. Because the fecundity and survival rates are applied per capita in a stochastic model, each trial will yield a different final population. Demographic stochasticity is the variation in population growth rates caused by random changes in individuals due to the individualistic application of survival and fecundity rates. By nature, demographic stochasticity does not have a “larger” or “smaller” effect on a population’s growth due to size, but rather produces a broader trajectory over which the population can progress. While there may be fewer possible trajectories for a smaller population than a larger one, and that may be considered a “larger” effect by comparison, the fact remains that the distribution of potential trajectories is relatively even across all population simulations. 2. The interesting point of comparison between the two models is that the trajectory for the deterministic model is almost exactly the same as the average trajectory for the stochastic model. Because the per capita application of fecundity and survival rates for the stochastic model produced a distribution, it would make sense that a “neutral” selection technique like a deterministic model would be aligned with the average value. In contrast, the stochastic model is broader in scope and encompasses 100 models at once, while the deterministic model has only one. (Deterministic, Stochastic)


3.

Based upon the graph and the numerical analysis, the probability of declining beneath a threshold of 31 is 0.1600.

Probability of Decline

Probability of Decline vs. Population 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 30

35

40

45

50

Threshold (Number of oxen)

The trajectory that I generated was not as smooth as Figure 2.7a because my graph utilized only 20 trials. The Figure in the textbook utilized 1,000 trials, providing a smoother, more uniform line to connect the points. According to the curve, the population is expected to be at 33 individuals at least 80% of the time.

4. Parameter Initial Abundance Growth Rate Survivability St. Deviation of Growth Rate

With High Value 0.76 1.00 0.56 0.57

With Low Value 0.46 *Experiment Fails* 0/undefined 0.55 0.64

Difference 0.30 1.00 0.01 -0.07


Conclusion: Increasing the initial abundance of oxen and the growth rate of the population were both positively correlated with generating a larger population. Based upon the geometric progression of population growth, a larger principle (initial abundance) and compounding rate (growth rate) would substantially increase the population. The survivability (survival rate) decreased by only one (1) percentage point, so further testing is necessary to determine its relationship to this population’s growth. In this trial, the change in survival rates is determined to be negligible. However, the standard deviation of the growth rate proved to be negatively correlated with population growth. A smaller st. dev. means that the growth rate stays relatively unreactive to other variables, meaning that it is less prone to rapid vacillations that could potentially diminish the population’s growth. The evidence is clear that growth rate (R) is by far the most important, as a 10% decrease in value effectively fails the experiment; it will take much longer than 12 time steps to achieve the required threshold. The contrapositive of this fact is that a 10% increase mathematically yields a “perfect” probability, meaning that the oxen will almost certainly reach the ideal threshold within the allotted time. The implication of these data shows that the growth rate is extremely sensitive, and if humans intervene to save a species, they must be careful not to upset the local ecology in the process. In terms of creating simulations, a ±10% tolerance is too lax for such an influential variable. The student’s values do not exactly match mine, but they have the same pattern; the 10% decrease and 10% increase yield almost the same value. The survival rates behaved the same way in both of our experiments. The reason that increasing the survival rate does not correlate with a larger population is because the growth rate is a fixed constant. If the R-value remains the same while the survival rate increases, the fecundity must decrease accordingly. It is this decrease in fecundity that negates the potential for the population to grow substantially. Because the R-value relies on a summation of 2 different measurements, survivability has an optimal “mixture” in the growth rate calculation that, if undershot or overshot, will greatly diminish its momentum in the final population calculation. 5. 𝑷(𝒂𝒍𝒍 𝒅𝒊𝒆) = (𝟏 − 𝒔)𝑵

𝑷(𝒂𝒍𝒍 𝒅𝒊𝒆) = (𝟏 − 𝟎. 𝟒𝟎)𝟏𝟖

𝑷(𝒂𝒍𝒍 𝒅𝒊𝒆) = (𝟎. 𝟎𝟎𝟎𝟏𝟎𝟐)

𝑷(𝒂𝒍𝒍 𝒅𝒊𝒆) = (𝟏 − 𝒔)𝑵 𝑷(𝒂𝒍𝒍 𝒅𝒊𝒆) = (𝟏 − 𝟎. 𝟏𝟏)𝟏𝟖 𝑷(𝒂𝒍𝒍 𝒅𝒊𝒆) = 𝟎. 𝟏𝟐𝟐𝟕𝟓


Lab #3 Exercise 3.1

Trajectory 1 for initial abundance=2, R=2.0000, and K=550 Trajectory 2 for initial abundance=800, R=2.0000, and K=550 individuals. The section of the graph after time step 12 is very individuals. The trajectory plummets rapidly, then plateaus at similar to that of Figure 3.8, as a relatively stable plateau is the carrying capacity of 550. reached in both cases.

Exercise 3.2

Trajectory 3 The trajectory with demographic stochasticity is also very similar to Figure 3.8, as there is a period of exponential growth that eventually plateaus when it reaches the carrying capacity. Initial abundance=2, R=2.0000, and K=550.

Exercise 3.3


Trajectory 4 “Scramble” The population climbs and falls rapidly, Trajectory 5 “Contest” The population grows extremely quickly, with near-vertical slopes. The graph alternates between high but unlike the “Scramble” trajectory, there is no oscillation peaks and low valleys, but most hover around the carrying above and below the carrying capacity. After reaching Day 4, capacity, zeroing in on 550. Initial abundance=2, R=20.0000, the graph plateaus at the value of 550. Initial abundance=2, K=550 R=20.0000, K=550

Trajectory 6 “Ceiling” The population expands at an even faster rate than “Contest,” but the final result is generally the same. After only Day 2, the graph reaches the carrying capacity and plateaus as expected. The transition from growth to plateau is the most abrupt in this trajectory than in “Scramble” of “Contest.” Initial abundance=2, R=20.0000, K=550.

Exercise 3.4

Trajectory 7 The points seem to oscillate around the carrying capacity, 600. This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

***COMPLETE***

Exercise 3.5 Initial Population Size 2,000 4,000 6,000 8,000 10,000 12,000 14,000

× (Natural Survival Rate) 0.5 0.5 0.5 0.5 0.5 0.5 0.5

= (Number that Survive) 1,000 2,000 3,000 4,000 5,000 6,000 7,000

-(Number to Harvest)

=(Number that Remain)

1,000 1,000 1,000 1,000 1,000 1,000 1,000

0 1,000 2,000 3,000 4,000 5,000 6,000

÷(Number at the Beginning) 2,000 4,000 6,000 8,000 10,000 12,000 14,000

=(Overall Survival Rate) 0.000 0.250 0.333 0.375 0.400 0.417 0.429

Probability of Survival vs. Original Population Size 0.5

Probability of Survival

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

2000

4000

6000

8000

10000

12000

14000

16000

Original Population (Individuals)


Initial Population Size 2,000 4,000 6,000 8,000 10,000 12,000 14,000

× (Natural Survival Rate) 0.5 0.5 0.5 0.5 0.5 0.5 0.5

= (Number that Survive) 1,000 2,000 3,000 4,000 5,000 6,000 7,000

-(Number to Harvest) 400 800 1,200 1,600 2,000 2,400 2,800

=(Number that Remain) 600 1,200 1,800 2,400 3,000 3,600 4,200

÷(Number at the Beginning) 2,000 4,000 6,000 8,000 10,000 12,000 14,000

=(Overall Survival Rate) 0.300 0.300 0.300 0.300 0.300 0.300 0.300

Probability of Survival vs. Original Population Size Probability of Survival

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

2000

4000

6000

8000

10000

12000

14000

16000

Original Population (Individuals)

The first curve is logarithmic, and has a limit of 0.500; which means that in an infinitely large population with adequate resources, only 50% of the population will live. Furthermore, a larger population is more likely to survive as compared to a small one, which makes sense with respect to group dynamics and animal culture. The second graph, however, refutes this claim by asserting that another strategy renders the probability of survival unrelated to the original population size. This graph shows a constant 0.300 rate of survival amongst individuals of all populations. According to the composite graph, smaller populations are more susceptible to the flat deduction; if a population is smaller than 5,000 individuals, it will be wiped out by the overfishing. However, this practice is not nearly as deleterious for large populations of fish, which can withstand the raw cut from their numbers. It is for this reason that smaller populations of fish survive best under the “percentage cut,” whereas larger populations are better off with a flat deduction of 1,000 fish. Failure to adhere to the proper strategy for each population size will lead to overfishing and possibly extinction.


Exercise 3.6

Population Size vs. Time

300 250

Population

200 150 100 50 0 0

2

4

6

8

10

12

Time (Years) This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

1.2

1

Growth Rate (R)

0.8 10% Decrease 0.6

10% Increase

0.4

Replacement Rate

0.2

0 0

50

100

150

200

250

300

350

Population N(t)

Homework #3 1. A: Through interpolating the graph, the carrying capacity is approximately 300 individuals. B: ***(R) is determined by interpolating from the graph at each step*** 𝑁(𝑡 + 1) = 𝑁(𝑡) ∗ 𝑅

𝑁(𝑡 + 1) = 100(2.1)

𝑵(𝒕 + 𝟏) = 𝟐𝟏𝟎 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍𝒔

𝑁(𝑡 + 1) = 𝑁(𝑡) ∗ 𝑅

𝑁(𝑡 + 1) = 450(0.6)

𝑵(𝒕 + 𝟏) = 𝟐𝟕𝟎 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍𝒔

C: The initial population of 300 individuals would remain at 300 individuals (assuming no outside interference or stochasticity) because such a population is already at the carrying capacity of the environment. Realistically, the population will fluctuate slightly above and below 300, but ideally the population will remain at its current level for an indefinite amount of time, unless the parameters of the simulation change. This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

2. A: Due to the nature of the question, it is impossible to pinpoint which strategy is the best because the population size is needed to determine the best strategy. However, if the assumption is made the fish population is equally likely to fall anywhere on the spectrum provided, the best strategy would be to take a percentage cut away from the population. Most of the populations depicted are larger than the threshold level of 5,000 fish, meaning that the majority of populations (those with numbers greater than 5,000) would produce the greatest yield if harvested by percentage. Taking a percentage of large populations (which dominate the spectrum) would ensure a large turnout, more so than if applied to small populations. B: The best strategy would be to take a fixed number away from the population. Most of the populations depicted are larger than the threshold level of 5,000 fish, meaning that the majority of populations (those with numbers greater than 5,000) would benefit most from taking a fixed number, as they will be large enough to sustain the loss in fish—a trait not present in small fish populations.


6. Density-independent growth is sensitive to abiotic factors, such as excessive heat or cold, rainfall levels, and drought. Density-independent simulations are not curtailed by populations, so an exponential pattern and lack of a carrying capacity are indicative of this type of growth. Densitydependent factors include parasitism, disease, predation, and lack of prey. Because the environment is considered finite with only adequate resources, a carrying capacity is imposed and the growth of the population is sigmoid—it contains a lag phase, log phase, and plateau phase. 1.2

1

Growth Rate (R)

0.8 10% Decrease 0.6

10% Increase

0.4

Replacement Rate

0.2

0 0

50

100

150

200

250

300

350

Population N(t)


350

300

Population N(t+1)

250

200 10% Decrease 150

10% Increase Replacement Line

100

50

0 0

50

100

150

200

250

300

350

Population N(t)


Lab #4 Exercise 4.1 𝑵(𝟓𝟎)

𝑹(𝟒𝟗) = 𝑵(𝟒𝟗)

𝑷𝟎 =

𝑵𝟎 (𝟓𝟎)

𝑷𝟏 =

𝑵𝟏 (𝟓𝟎)

𝑷𝟐 =

𝑵𝟐 (𝟓𝟎)

𝑷𝟑+ =

𝑷𝟎 = 𝟎. 𝟑𝟏𝟑𝟑𝟑𝟗

𝟐𝟑𝟓

𝑷𝟏 = 𝟎. 𝟐𝟏𝟎𝟑𝟖𝟓

𝟏𝟔𝟏

𝑷𝟐 = 𝟎. 𝟏𝟒𝟒𝟏𝟑𝟔

𝑷𝟏 = 𝟏,𝟏𝟏𝟕

𝑵(𝟓𝟎)

𝑷𝟐 = 𝟏,𝟏𝟏𝟕

𝑵(𝟓𝟎) 𝑵𝟑+ (𝟓𝟎)

𝑹(𝟒𝟗) = 𝟏. 𝟎𝟒𝟗𝟖𝟏𝟐

𝟑𝟓𝟎

𝑷𝟎 = 𝟏,𝟏𝟏𝟕

𝑵(𝟓𝟎)

𝑵(𝟓𝟎)

𝟏,𝟏𝟏𝟕

𝑹(𝟒𝟗) = 𝟏,𝟎𝟔𝟒

𝟑𝟕𝟏

𝑷𝟑+ = 𝟏,𝟏𝟏𝟕

𝑷𝟑+ = 𝟎. 𝟑𝟑𝟐𝟏𝟒𝟎

𝑹(𝟒𝟗) = 𝟏. 𝟎𝟒𝟗𝟖𝟏𝟐

Stable Values: Age 0: 0.314

Age 1: 0.210

Age 2: 0.144

Age 3+: 0.332

𝝀 ≈ 𝟏. 𝟎𝟒𝟗𝟓 The final age distribution is almost identical to the stable distribution, and the growth rate from Year 49 to Year 50 was the same within 3 decimal places of the finite rate of increase. This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

𝑵(𝟓𝟎)

𝟗𝟔𝟓

𝑹(𝟒𝟗) = 𝑵(𝟒𝟗)

𝑷𝟎 =

𝑵𝟎 (𝟓𝟎)

𝑷𝟏 =

𝑵𝟏 (𝟓𝟎)

𝑷𝟐 =

𝑵𝟐 (𝟓𝟎)

𝑹(𝟒𝟗) = 𝟗𝟐𝟎

𝑵(𝟓𝟎)

𝑵(𝟓𝟎)

𝑷𝟑+ =

𝑷𝟎 = 𝟗𝟔𝟓

𝟑𝟎𝟑

𝑷𝟎 = 𝟎. 𝟑𝟏𝟑𝟗𝟗𝟎

𝑷𝟏 = 𝟗𝟔𝟓

𝟐𝟎𝟐

𝑷𝟏 = 𝟎. 𝟐𝟎𝟗𝟑𝟐𝟔

𝟏𝟑𝟖

𝑷𝟐 = 𝟎. 𝟏𝟒𝟑𝟎𝟎𝟓

𝑷𝟐 = 𝟗𝟔𝟓

𝑵(𝟓𝟎) 𝑵𝟑+ (𝟓𝟎) 𝑵(𝟓𝟎)

𝑹(𝟒𝟗) = 𝟏. 𝟎𝟒𝟖𝟗𝟏𝟑

𝟑𝟐𝟐

𝑷𝟑+ = 𝟗𝟔𝟓

𝑷𝟑+ = 𝟎. 𝟑𝟑𝟑𝟔𝟕𝟗

𝑹(𝟒𝟗) = 𝟏. 𝟎𝟒𝟖𝟗𝟏𝟑

Stable Values: Age 0: 0.314

Age 1: 0.210

Age 2: 0.144

Age 3+: 0.332

𝝀 ≈ 𝟏. 𝟎𝟒𝟗𝟓


The stable age values are, once again, almost superimposable with the final age values. The same can be said of the finite rate of increase and the growth rate from Year 49 to Year 50, as it is accurate to 2 decimal places. The population has approximately conformed to the stable age distribution, so it has equilibrated in the 50 time steps. The final abundance does change when the initial abundances in each age class are changed; the final population went from 1,117 individuals in the first simulation to 965 in the second simulation.

Year

Min

50

592.00

-SD

Avg

+SD

Max

766.55 1043.04 1319.53 1853.00


Year

Min

–SD

Avg

50

267.00

427.87

+SD

Max

955.86 1483.85 2712.00

The trajectory that has both demographic and environmental stochasticity was much broader in scope than was the trajectory that showcased only the demographic stochasticity. As with all simulations, the more stochastic parameters that are introduced, the more distribution the resulting trajectory will display. This explains why the minimum and maximum values of the trajectory are much more extreme than in the previous trajectory.

Factors such as food availability, water shortages, and disease would all curb the exponential growth of a population. The population hits the carrying capacity (K) and fluctuates beneath it. The final population will be approximately 150 individuals. The Leslie matrix provides a finite rate of growth of 𝝀 ≈ 𝟏. 𝟎𝟒𝟗𝟓, so the population will continue to fluctuate around the carrying capacity.

Exercise 4.2

𝝀 ≈ 𝟏. 𝟑𝟖𝟐𝟕/ 𝒅𝒆𝒄𝒂𝒅𝒆 Stage Init. distr. Stable distr. 1

0.311

0.338

2

0.243

0.229


3

0.171

0.162

4

0.115

0.113

5

0.079

0.077

6

0.051

0.051

7

0.030

0.030

The initial age distribution and stable age distribution are not equivalent, as only the oldest age groups (50-60 and 60-70) had identical values. In this population, more time steps are needed to equilibrate the age strata. Initial Population: 43,535,036 individuals Final population: 1,121,892,608 individuals (Total of 10 time steps) %∆=

𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑶𝒓𝒈𝒊𝒏𝒂𝒍 𝑽𝒂𝒍𝒖𝒆 𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝑽𝒂𝒍𝒖𝒆

%∆=

𝟏,𝟏𝟐𝟏,𝟖𝟗𝟐,𝟔𝟎𝟖−𝟒𝟑,𝟓𝟑𝟓,𝟎𝟑𝟔 𝟒𝟑,𝟓𝟑𝟓,𝟎𝟑𝟔

%∆= 𝟐, 𝟒𝟕𝟕% 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆

Overall change in Population: 𝟏, 𝟏𝟐𝟏, 𝟖𝟗𝟐, 𝟔𝟎𝟖 − 𝟒𝟑, 𝟓𝟑𝟓, 𝟎𝟑𝟔 = 𝟏, 𝟎𝟕𝟖, 𝟑𝟓𝟕, 𝟓𝟕𝟐 individuals

“Family Planning”

𝝀 ≈ 𝟏. 𝟎𝟎𝟎𝟑/ 𝒅𝒆𝒄𝒂𝒅𝒆

%𝑹𝒆𝒅𝒖𝒄𝒕𝒊𝒐𝒏 = 𝟔𝟒. 𝟏%


%∆=


%∆=

𝟔𝟏,𝟔𝟕𝟖,𝟒𝟑𝟐−𝟒𝟑,𝟓𝟑𝟓,𝟎𝟑𝟔 𝟒𝟑,𝟓𝟑𝟓,𝟎𝟑𝟔

%∆= 𝟒𝟏. 𝟔𝟖% Increase

Overall change in Population: 𝟔𝟏, 𝟔𝟕𝟖, 𝟒𝟑𝟐 − 𝟒𝟑, 𝟓𝟑𝟓, 𝟎𝟑𝟔 = 𝟏𝟖, 𝟏𝟒𝟑, 𝟑𝟗𝟔 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍𝒔 The population continued to increase, despite the finite rate of increase equaling one, because the population had not yet reached a stable age distribution. Only after this equilibrium is reached will the population’s growth plateau. “Family Planning, Pt. II” Age Class Age Class 0-10 Age Class 10-20 Age Class 20-30 Age Class 30-40 Age Class 40-50 Age Class 50-60 Age Class 60-70

𝝀 ≈ 𝟏. 𝟎𝟎𝟎𝟎/ 𝒅𝒆𝒄𝒂𝒅𝒆 Percent Reduction (%) 0.0 100.0 100.0 8.5 0.0 0.0 0.0

Because the fecundities of adolescents and 20-year-olds was reduced to zero, and those younger than age 10 are infertile, people in this population only start to reproduce in their 30’s. The 40-50 Age class is the only other class with the potential to reproduce, so children will be born to much older parents in this scenario. The fecundity reduction in this simulation directly targets the younger, more virile members of the population, while the pervious simulation targeted each age class identically.


%∆=


%∆=

𝟓𝟒,𝟕𝟕𝟐,𝟑𝟑𝟐−𝟒𝟑,𝟓𝟑𝟓,𝟎𝟑𝟔 𝟒𝟑,𝟓𝟑𝟓,𝟎𝟑𝟔

%∆= 𝟐𝟓. 𝟖𝟏% 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆

Overall Change in Population: 11,237,296 individuals

The method of selectively decreasing the fecundities of the younger Age classes was much more effective at reducing population size than was the across-the-board decrease utilized in the first simulation. The “Family Planning” scenario would have limited success if implemented, as it would take several years, possibly even decades to decrease fecundity to the calculated level. Should the idea be implemented, it would decrease the population size, but not to the extent that was calculated. The population would continue to grow as the fecundities decreased, so the final population size would be larger than predicted by the trajectory. Both the educational and economic empowerment of women would decrease fecundity, as women would be attending school and pursuing careers as opposed to settling down and having children. Conversely, providing large stipends to older people would improve their health, and overall reproductive value. With more money and fewer medical problems, older Age classes would become more likely to reproduce, increasing the population’s fecundity. A) If there is no change in fecundity, then the industrialized value of 9.3X is applicable to this population, meaning it improved by 830%. B) If there is a change in fecundity, the factor would be calculated as follows: (Step 3) 𝑭𝒊𝒏𝒂𝒍 𝑷𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝑨 𝑰𝒏𝒅𝒖𝒔𝒕𝒓𝒚 𝑭𝒂𝒄𝒕𝒐𝒓 𝒊𝒏 𝑨

=

𝑭𝒊𝒏𝒂𝒍 𝑷𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝑩 𝑰𝒏𝒅𝒖𝒔𝒕𝒓𝒚 𝑭𝒂𝒄𝒕𝒐𝒓 𝒊𝒏 𝑩

𝟏,𝟏𝟐𝟏,𝟖𝟗𝟐,𝟔𝟎𝟖 𝟗.𝟑

=

𝟔𝟏,𝟔𝟕𝟖,𝟒𝟑𝟐 𝑿

𝑿 = 𝟎. 𝟓𝟏𝟏𝟑

After adding (1) to make the growth positive and deducting (1) to adjust to percent, the population’s energy consumption rises by 51.13%

C) If there is a change in fecundity, the factor would be calculated as follows: (Step 5) 𝑭𝒊𝒏𝒂𝒍 𝑷𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝑨 𝑰𝒏𝒅𝒖𝒔𝒕𝒓𝒚 𝑭𝒂𝒄𝒕𝒐𝒓 𝒊𝒏 𝑨

=

𝑭𝒊𝒏𝒂𝒍 𝑷𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝑪 𝑰𝒏𝒅𝒖𝒔𝒕𝒓𝒚 𝑭𝒂𝒄𝒕𝒐𝒓 𝒊𝒏 𝑪

𝟏,𝟏𝟐𝟏,𝟖𝟗𝟐,𝟔𝟎𝟖 𝟗.𝟑

=

𝟓𝟒,𝟕𝟕𝟐,𝟑𝟑𝟐 𝑿

𝑿 = 𝟎. 𝟒𝟓𝟒𝟎

After adding (1) to make the growth positive and deducting (1) to adjust to percent, the population’s energy consumption rises by 45.40%

Exercise 4.3


Survival Rate Data Average Std. Dev. of Survival Rate Survival Rate Age 0 0.4250 Age 0 0.0721 Age 1 0.1748 Age 1 0.0368 Age 2 0.0859 Age 2 0.0438 Age 3 0.0231 Age 3 0.0261 Age 4 0 Age 4 0

Fecundity Data Average St. Dev. of Fecundity Fecundity Age 0 0 Age 0 0 Age 1 1.4967 Age 1 0.5758 Age 2 5.1099 Age 2 2.3251 Age 3 11.7067 Age 3 5.6727 Age 4 11.7245 Age 4 19.4742

Exercise 4.4

Trajectory 2: Deterministic Population growth based upon 1962 data. Final population after 20 years was 14,907 individuals.

Trajectory 1: Deterministic Population growth based upon extrapolated data. Harvest Rate is equal to 0.0581.


Final Age Abundances Age 0: 5,046 individuals Age 1: 2,020 individuals Age 2: 332 individuals Age 3: 27 individuals Age 4: 1 individual

Trajectory 3: Deterministic Population growth based upon extrapolated data. Harvested amounts by age class were 125 individuals (Age 1), 20 individuals (Age 2), and 2 individuals (Age 3).

Relative Risk of Decreasing to 1,000 Individuals No Harvest: 0.1060 Proportional Harvest: 0.2240 Constant Harvest: 0.4930 Based upon the stochastic trajectories (Not Pictured) and the probabilities of decline to 1,000 individuals, the constant harvest technique would have the most detrimental effect on the population. According to the idea of Allee Effects, a low population will exhibit inverse density dependence (DECREASED growth rate) when a constant harvest is imposed on the group. A proportional harvest does not greatly upset the age strata of the population, and not harvesting at all would mean the population grows exponentially until a carrying capacity is reached. This explains why not harvesting yields the smallest probability of decline, whereas harvesting indiscriminately of This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

age, fecundity, and survival rates (constant harvest) is the most likely to cause an extreme decline in the population.

Lab #4 Homework 1.

𝑺𝟎 = 𝑺𝟏 = 𝑺𝟐 = 𝑺𝟑 =

𝑵𝟏 (𝟏𝟗𝟗𝟗) 𝑵𝟏 (𝟐𝟎𝟎𝟎) 𝑵𝟏 (𝟐𝟎𝟎𝟏) + + 𝑵𝟎 (𝟏𝟗𝟗𝟖) 𝑵𝟎 (𝟏𝟗𝟗𝟗) 𝑵𝟎 (𝟐𝟎𝟎𝟎)

𝟑 𝑵𝟐 (𝟏𝟗𝟗𝟗) 𝑵𝟐 (𝟐𝟎𝟎𝟎) 𝑵𝟐 (𝟐𝟎𝟎𝟏) + + 𝑵𝟏 (𝟏𝟗𝟗𝟖) 𝑵𝟏 (𝟏𝟗𝟗𝟗) 𝑵𝟏 (𝟐𝟎𝟎𝟎)

𝟑 𝑵𝟑 (𝟏𝟗𝟗𝟗) 𝑵𝟑 (𝟐𝟎𝟎𝟎) 𝑵𝟑 (𝟐𝟎𝟎𝟏) + + 𝑵𝟐 (𝟏𝟗𝟗𝟖) 𝑵𝟐 (𝟏𝟗𝟗𝟗) 𝑵𝟐 (𝟐𝟎𝟎𝟎)

𝟑 𝑵𝟒 (𝟏𝟗𝟗𝟗) 𝑵𝟒 (𝟐𝟎𝟎𝟎) 𝑵𝟒 (𝟐𝟎𝟎𝟏) + + 𝑵𝟑 (𝟏𝟗𝟗𝟖) 𝑵𝟑 (𝟏𝟗𝟗𝟗) 𝑵𝟑 (𝟐𝟎𝟎𝟎)

𝟑

𝑺𝟎 =

(𝟎.𝟔𝟖𝟔𝟎+𝟎.𝟔𝟏𝟓𝟒+𝟎.𝟔𝟒𝟕𝟕) 𝟑

𝑺𝟎 = 𝟎. 𝟔𝟒𝟗𝟕

𝑺𝟏 =

(𝟎.𝟔𝟖𝟒𝟐+𝟎.𝟕𝟏𝟏𝟗+𝟎.𝟔𝟔𝟎𝟕) 𝟑

𝑺𝟏 = 𝟎. 𝟔𝟖𝟓𝟔

𝑺𝟐 =

(𝟎.𝟓𝟓𝟐𝟔+𝟎.𝟓𝟖𝟗𝟕+𝟎.𝟓𝟎𝟎𝟎) 𝟑

𝑺𝟐 = 𝟎. 𝟓𝟒𝟕𝟒

(𝟎.𝟓𝟎𝟎𝟎+𝟎.𝟔𝟏𝟗𝟎+𝟎.𝟓𝟐𝟏𝟕) 𝟑

𝑺𝟑 = 𝟎. 𝟓𝟒𝟔𝟗

𝑺𝟑 =

𝑺𝟒 = 𝟎 (No 5 year olds were observed)

𝑭𝟎 = 𝟎 (As 0 year olds are not of reproductive age) 𝑭𝟏 = 𝟎 (As 1 year olds are not of reproductive age) 𝑭𝟏𝟗𝟗𝟖 = 𝑵 𝑭𝟏𝟗𝟗𝟗 =

𝑵𝟎 (𝟏𝟗𝟗𝟗)

𝟗𝟏

𝟐 (𝟏𝟗𝟗𝟖)+𝑵𝟑 (𝟏𝟗𝟗𝟖)+𝑵𝟒 (𝟏𝟗𝟗𝟖)

𝑵𝟎 (𝟐𝟎𝟎𝟎) 𝑵𝟐 (𝟏𝟗𝟗𝟗)+𝑵𝟑 (𝟏𝟗𝟗𝟗)+𝑵𝟒 (𝟏𝟗𝟗𝟗)

𝑭𝟐𝟎𝟎𝟎 = 𝑵 𝑭𝒂𝒗𝒈 =

𝑵𝟎 (𝟐𝟎𝟎𝟏) (𝟐𝟎𝟎𝟎)+𝑵 𝟐 𝟑 (𝟐𝟎𝟎𝟎)+𝑵𝟒 (𝟐𝟎𝟎𝟎) 𝑭𝟏𝟗𝟗𝟖 +𝑭𝟏𝟗𝟗𝟗 +𝑭𝟐𝟎𝟎𝟎 𝟑

𝑭𝒂𝒗𝒈 =

𝑭𝟏𝟗𝟗𝟖 = (𝟑𝟖+𝟐𝟐+𝟏𝟏) = 𝟏. 𝟐𝟖𝟐 𝑭𝟏𝟗𝟗𝟗 =

𝟖𝟖 (𝟑𝟗+𝟐𝟏+𝟏𝟏)

= 𝟏. 𝟐𝟑𝟗

𝟗𝟒

𝑭𝟐𝟎𝟎𝟎 = (𝟒𝟐+𝟐𝟑+𝟏𝟑) = 𝟏. 𝟐𝟎𝟓 𝟏.𝟐𝟖𝟐+𝟏.𝟐𝟑𝟗+𝟏.𝟐𝟎𝟓 𝟑

= 𝟏. 𝟐𝟒𝟐

𝑭𝟐 = 𝑭𝟑 = 𝑭𝟒 = 𝟏. 𝟐𝟒𝟐

𝑵𝟑+ (𝟏𝟗𝟗𝟗)

𝑺𝟐+ (𝟏𝟗𝟗𝟖) = 𝑵

𝟑+ (𝟏𝟗𝟗𝟖)+𝑵𝟐 (𝟏𝟗𝟗𝟖)

𝑺𝟐+ (𝟏𝟗𝟗𝟗) = 𝑵

𝟑+ (𝟏𝟗𝟗𝟗)+𝑵𝟐 (𝟏𝟗𝟗𝟗)

𝑺𝟐+ (𝟐𝟎𝟎𝟎) = 𝑺𝟐+ =

𝑵𝟑+ (𝟐𝟎𝟎𝟎)

𝑵𝟑+ (𝟐𝟎𝟎𝟏) 𝑵𝟑+ (𝟐𝟎𝟎𝟎)+𝑵𝟐 (𝟐𝟎𝟎𝟎)

𝟑𝟐

𝑺𝟐+ (𝟏𝟗𝟗𝟖) = 𝟑𝟑+𝟑𝟖 = 𝟎. 𝟒𝟓𝟎𝟕 𝟑𝟔

𝑺𝟐+ (𝟏𝟗𝟗𝟗) = 𝟑𝟑+𝟑𝟗 = 𝟎. 𝟓𝟎𝟎𝟎 𝑺𝟐+ (𝟐𝟎𝟎𝟎) =

𝟑𝟑 𝟑𝟔+𝟒𝟐

= 𝟎. 𝟒𝟐𝟑𝟏

𝑺𝟐+ (𝟏𝟗𝟗𝟖) + 𝑺𝟐+ (𝟏𝟗𝟗𝟗) + 𝑺𝟐+ (𝟐𝟎𝟎𝟎) = 𝟎. 𝟒𝟓𝟕𝟗 𝟑


[𝑵(𝒕 + 𝟏)] = [𝑳] ∗ [𝑵(𝒕)]

𝟖𝟕 𝟔𝟏 = 𝟕𝟏

𝟗𝟒 × 𝟓𝟕 𝟕𝟎

𝑵(𝒕 + 𝟏) = 𝟖𝟕 + 𝟔𝟏 + 𝟕𝟏 = 𝟐𝟏𝟗 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍𝒔 2.


The strategy in Step 6 was “constant harvest,” which removed a fixed number of individuals from each age class. The harvest amount differed from one Age class to another, so the amounts were 125, 20, and 2 two individuals for Age classes 1, 2, and 3, respectively. These amounts stabilized the population because they were derived from the proportional harvest rate, which itself stabilized the population. No Harvest: 0.1060 Proportional Harvest: 0.2240 Constant Harvest: 0.4930

Not harvesting at all is much less likely to produce a decline to 1,000 individuals than are the proportional or constant harvest strategies. According to the data, the proportional harvest is more than twice as likely to induce a decline as not harvesting, while the constant harvest is almost five times as likely to cause a decline. The constant harvest strategy is much less safe than the proportional harvest because the latter strategy respects the age class strata by not favoring one class over another. Because it entails a percent cut, the proportional harvest strategy treats the central three age classes fairly. However, the constant harvest technique disrupts the age strata by manipulating the harvest rate from the previous exercise. Because each of the central three strata are 𝒉

treated differently (as the harvest rate involved in (𝟏−𝒉) is both directly and inversely proportional to the number harvested from each strata) the harvest is not proportional across the different age classes. It is for this reason—that the strata are treated individually and not part of a system-- that the constant harvest is much more risky than the proportional harvest.

3. A: There is no stochasticity involved in the model B: There is no density dependence and growth is exponential C: The population tends toward its stable age distribution in the time allotted Due to the fact that animals live in a density dependent, Malthusian world, and there is stochasticity in the real world, using lambda in conservation decisions would be a poor choice. The model using lambda rejects too many variables that come into play in any ecosystem.


Lab #6 Question #1 A stage structured model assumes that there is no stochasticity, no density dependence, a panmitic and isolated population, characteristics related to demographic stage, and the option is given to either progress to a new stage. The Leslie matrix, however, utilizes the age-structure model, which stipulates that minimal variation and growth between discrete time steps accurately predicts growth.

Question #2 The diagram maps onto a Leslie matrix as shown: P1 G1 0 0

F2 P2 G2 0

F3 0 P3 G3

0 0 0 P4

Question #3 Based upon matrix calculations: Unaltered matrix yields Altered matrix yields

𝝀 = 𝟎. 𝟗𝟒𝟑𝟗 𝝀 = 𝟎. 𝟗𝟓𝟑𝟎

The larger vale of 𝝀 leads to faster growth in the population of Alder when the stable stage distribution has been attained, despite the fact that some of the fecundities were decreased.

Question #4 Plans to conserve the loggerhead turtle species include protecting nest sites and implementing TEDs, or Turtle Extrication Devices, which enable sea turtles to safely escape fish trawls. Utilizing these TEDs would cause the probability of exploding past 200,000 individuals to skyrocket to around 83% over three decades, while not using the TEDs would make it all but impossible (0%) to have the turtle population grow beyond 200,000 individuals. Additionally, using TEDs yields a 0% chance of the turtles declining to under 20,000 individuals over a 30 year period, whereas the population could crash with a probability as high as 43% if no new technology was implemented. If only the nest protection strategy were employed, the fecundities of the population would have to increase by close to 300%, or quadruple, just to equal the numbers projected through the use of TEDs.


The statistics make the right course of action clear, as installing TEDs is a much more effective method to increase the number of individuals in this population of turtles.

Question #5 Sensitivity analysis highlights how each parameter in a matrix affects the population’s growth, development, and trends. There are two main types of sensitivity analysis, namely deterministic and risk sensitivity. The deterministic model tends to have difficulty explaining trends that by nature are stochastic, such as explosion or extinction curves. The risk-based model quantifies how frequently a specific event will transpire, and what impact it will have on the population. In Exercise 5.4, it was evident that the juvenile stage’s vital rates were much more sensitive than those of the adult stage, meaning that the precision of the calculations and parameters involved must be quantified in context of the stage structure of the population.


Lab #8 Homework Question #1 Background extinction rates can be defined as the “normal” amount of species that undergo extinction over a fixed period of time, as defined by the Earth’s biological timeline. It is worth noting that this extinction rate is calculated without incorporating humans’ influence on the planet or any other species. Population trends have dropped off precipitously in the past few years for many species of birds, certain fungi in Europe, and several populations of freshwater fish on the order of 20-50%. The Background rate is approximately 1/ 10^6 species each year. The current progression is around 1/10^4 species per year, rendering the extinction rate at 100 times its normal value; some estimates have put this value as high as 1,000 times the normal rate.

Homework Question #2

Using a cost-benefit analysis for this Question and its subparts will yield the best scenarios for the aforementioned species to stave off extinction. Part 1: Must not exceed $125,000 Plan F: Choosing this plan uses all of the funding, but ensures a relatively low risk (0.5) of recovering the species. Part 2: Reduce risk to 0.6 or below This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

Plan B: While this plan has a risk on the higher end of the spectrum, it is the most cost effective option that does not greatly sacrifice risk reduction. Part 3: Reduce risk to 0.5 or below Plan F: This step is where the cost benefit analysis comes into play, as making the risk reduction step more stringent causes Plan F to be the best course of action. Plan H offers the same level of risk reduction, but at a significantly larger cost. Plan 4: Must not exceed $200,000 Plan F: This also employs a use of cost benefit analysis, as the cost increase for this plan was more than counterbalanced by its risk reduction, rendering Plan B slightly less effective than this plan. Because the cost was fixed and the only other factor was risk reduction, Plan F yielded the best possible result.


Lab #9 Homework Question #1 1)

𝑯𝟎 : There is no significant difference between the fecundities of the disturbed and undisturbed populations of spotted owl. 𝑯𝑨 : There is a significant difference between the fecundities of the disturbed and undisturbed Spotted Owl populations. The environmental group would support the alternative hypothesis, as it would draw attention to the consequences of the logging company’s activities and would force them to change their policies/ ideas of expansion.

2) A Type I Error is one in which the null hypothesis is true, but rejected. For the Spotted Owl scenario described, there would be no significant difference between the fecundities, but the analysis would show that there was. This error would cause a backlash on the logging company to cease their operations in the area, so the company would want to minimize errors of this type. 3) A Type II Error is one in which the null hypothesis is false, but is accepted anyway. For the Spotted Owl scenario described, there is a significant difference between the two fecundities, but the analysis would not reflect it. The error would cause the logging company to dismiss the assertion that the owl’s fecundity was affected, and would proceed with their plans of harvesting timber. Therefore, the environmental group wants to minimize this kind of error, as it would weaken their argument and put the spotted owl population at risk. 4) The environmentalist group has an incentive in making the study as factual and thorough as possible, as they must utilize the data to refute the claims of the logging company that the harvesting will not affect the owl population’s fecundity. The data must also be accurate so that the perceived change in fecundity—a 20% decrease—is reached, which would mandate the company to seek elsewhere to harvest timber. Simply showing a decline is not sufficient, as the threshold value must hit 20% in order for the risk to be significant enough for the company to abandon the area. On the other hand, the logging company has an interest to ensure that the study is ambiguous, so that they can hide within the “gray area” and continue to harvest timber without the environmentalists opposing them. If the evidence provided by the assessment were of lower quality, it would be easier for the company to sidestep the issue and continue to harvest from the area, despite the potential consequences on the owl population. Homework Question #2


H j b e Trajectory 1 Extinction/ Decline curve for a constant harvest of 400 individuals. Total harvest amount= k 12518.81 kgs. Extinction Risk=9.2% c

Trajectory 2 Extinction/ Decline curve for a constant harvest of 250 individuals. Total harvest amount=8000 kgs. Extinction Risk~0%


Trajectory 3 Extinction/ Decline curve for a constant harvest of 350 individuals. Total harvest amount= 11146.41 kgs. Extinction Risk=2%

Trajectory 4 Extinction/ Decline curve for a constant harvest of 450 individuals. Total harvest amount= 13232.67 kgs. Extinction Risk=28.90%


Trajectory 5 Extinction/ Decline curve for a constant harvest of 550 individuals. Total harvest amount= 12338.85 kgs. Extinction Risk=81.50%

Extinction Risk (%)

Extinction Risk vs. Amount Harvested 90 80 70 60 50 40 30 20 10 0 0

100

200

300

400

500

600

Amount of cod harvested/ timestep


Trajectory A-1 Extinction/ Decline curve for a proportionate harvest of 0.2 of Stage 3 at each timestep. Total harvest amount=12749.51 kgs. Extinction Risk~0%

Proportion of Individuals Harvested

Extinction Risk (%)

Amount Harvested (kgs)

0.2 0.3 0.4 0.5 0.6

0 0 0 0 0

12749.51 14030.15 13101.05 11048.54 9225.59

Extinction Risk (%)

Extinction Risk vs. Total Harvest 90 80 70 60 50 40 30 20 10 0 -10 0

5000

10000

15000

Total Harvest (kgs) Clearly, the best course of action would be to harvest a fixed proportion of the cod population, such that fishing quotas are met but the species growth is not affected. Chapter 4’s analysis did indeed show that smaller populations will benefit most from a percent cut, which is evident in the graphs and table. This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

Homework Question #3

Probability 2-Fold Profit 3-Fold Profit 4-Fold Profit Total Loss

1st Scenario 0.2130 0.0620 0.0240 0.9390

2nd Scenario N/A 0.2140 0.0750 0.9180

3rd Scenario 0.4520 0.1670 0.0650 0.7000

Based upon the simulations, Scenario 3 (which involves improving the fecundities of the palms) is slightly better than either of the other two scenarios. While Scenario 1 produces the least favorable results, Scenario 2 actually produces the best. However, the risk of total loss in Scenario 2 is greater than 90%, while the same quantity in Scenario 3 was at a much lower value of 70%. This pushes Scenario 3 into the top spot, as a small loss in the probability of profit is more than remedied by the massive reduction in risk of the investment. Furthermore, previous labs have shown that the growth rate (and by extension, fecundity) is one of the most sensitive parameters in a simulation, which is why Scenario 3’s method of increasing fecundity has such a dramatic effect reducing the risk of declining to 1,000 individuals.


Lab Worksheet Part 1, Q1 While having access to the mortality data is imperative in constructing a trajectory for a particular population, many more parameters are necessary to determine how that population grows. Specifically, non-human mortality rates must also be analyzed so that the parameter for survivability (S) of the bird species can be determined. Furthermore, the growth rate (R) of a population is denoted as the sum of survivability and fecundity (F), meaning that the reproductive rates are integral in calculating the proper growth rate for the population. If the growth rate dips below one (1), then human induced mortality can be considered a viable excuse as to the decline of the bird population. Even with such a conclusion, the population model must incorporate (st)age classes, sex ratios, type of density dependence (positive Allee Effects or negative, exponential growth), and carrying capacity of the habitat in which the birds roost.

Part 1, Q2

Blue Jay (Cyanocitta cristata)

̂) 𝑵𝒌𝒊𝒍𝒍𝒆𝒅 = (𝑵𝒂𝒗𝒈 ) ∗ (𝒑

𝑵𝒌𝒊𝒍𝒍𝒆𝒅 = (𝟐. 𝟒 ∗ 𝟏𝟎𝟗 𝒃𝒊𝒓𝒅𝒔) ∗ (𝟎. 𝟎𝟐𝟕)

𝑵𝒌𝒊𝒍𝒍𝒆𝒅 = 𝟔𝟒, 𝟖𝟎𝟎, 𝟎𝟎𝟎 𝒃𝒊𝒓𝒅𝒔

Part 1, Q3 The blue jay (Cyanocitta cristata) is native to North America, and is extant from the East Coast to the Heartland, and from Florida up to Ontario, Canada. The blue jay is a forest bird, tending to live on the outskirts of large populations of trees, particularly oaks. They are prevalent in forests, parks, and cities, provided that they can harvest acorns. They are omnivorous, so they tend to eat insects off of trees and even the eggs of other birds. Both male and female blue jays gather twigs, rootlets, and grass with which to construct their nest. Typically, females perform the building while the males will usually collect the materials, particularly twigs, from live plants. Migration in blue jays is rather enigmatic, as some jays migrate while others do not. Additionally, those who do tend to be younger, but each year their migratory paths change. This bird is ranked at least concern (LC) by the IUCN, N5B in the United States, and is present at secure levels across all of North America. Both conservation status and susceptibility to predation by cats are linked through the blue jay’s odd migration patterns; numbers across North America could be skewed based upon irregular patterns and intervals of migration, potentially leading to an incorrect conservation status. The same can be true for predation, as the blue jays that do not migrate are more applicable to be attacked by local cats than are birds that fly away during certain seasons. Part 1, Q4


Based upon the data provided on the website, the blue jay’s survey-wide population seems to be decreasing.

Part 1, Q5 The blue jay’s population has diminished sporadically across the Midwest, Central US, and along the East Coast. Overall, these areas supplant the increases in population across Canada, meaning that the population trajectory of the blue jay is decreasing across North America.

Part 1, Q6 In light of the data that has been presented across multiple sources, it is a feasible conclusion that anthropogenic mortality greatly affects bird populations—“Waters 2013” makes a strong case that mortality rates in birds are much higher than they should be based upon environmental studies. A new experiment should be run that catalogues bird population growth across multiple locations. First, the actual population trajectory should be generated reflecting the survivability of the birds in the population. The next trajectory, however, should be calculated by excluding the anthropogenic mortality rates (which would be calculated by estimates and feline predation as outlined in Water’s article) and visualizing how much the population would have grown without the extra mortality brought on by their human and feline-dominated environment. Additional information, such as fecundity, initial population, density dependence, carry capacity, stage structure, and sex ratio should also be collected to synthesize the best possible population model. Part 2, Q1 After consulting the data, there are 36 species of birds that are of conservation concern on the North American continent. Part 2, Q2


Number of Species Affected by Environmental Threats

Threat

Climate change and severe weather Geological Events Pollution Invasive and other problematic species… Natural system modificationsa Human intrusions and disturbance Biological resource use Transportation/ service corridors Energy production/ mining Agriculture/ Aquaculture Residential/ Commercial Development 0

10

20

30

40

Number of species affected Figure 1 Most species in the study are affected by biological resource use, climate change/ extreme weather, pollution, and invasive species/ genes.

Part 2, Q3 Henslow’s sparrow breeds predominantly in the Eastern United States and extends into Southern Ontario, Canada. This sparrow makes its habitat in areas containing tall, dense grass, and heavy leaf litter during the breading season, but otherwise prefers locations of low foliage density. The destruction of wetlands and grasslands across the continent has restricted potential environments, food, and resources for the Henslow’s sparrow. The composite of earlier harvests, conversion of grasslands and forests, and fire suppression have caused a marked decline in population for this species in recent years. Part 3, Q1 While the author makes a compelling argument, the best course of action would be to continue the trap-neuter-release technique, simply because that making the program more widespread than it already is will alleviate the problem. The reason feral cats exist is because a fertile cat was impregnated and gave birth to kittens. If almost all cats (more so than the current proportion) were neutered/ spayed, than the growth rate of the cat population would decline accordingly. The author notes that TNR programs operating at their current efficiency seem to cancel out the extreme fertility of the cat populations, leading to a constant depletion of birds and other prey-animals from the environment. From an ecological perspective, this means the cat population’s growth rate is equal to one; the population of feral cats killing animals is unchanging. By making this program more expansive than in years past, this ecological impasse can be surmounted so that the growth rate of the feral cat population dips below one. Such a change would mean a decreasing This document is the property of Nerdy Notes (www.nerdy-notes.com). Permission is granted to view this document only to authorized users; under no circumstances can you distribute or transmit this document without our written consent.

population size which, over time and continued effort, would concomitantly decrease the number of feline-induced fatalities in the local environment. If implementing stronger regulations was not working in a timely manner, the cats could be placed in an animal shelter and potentially adopted, both of which would diminish the cats’ ability to indiscriminately kill other animals.


BIO 356 - All Labs and Lab Homeworks

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