Ecology I Course Reader Tutorials PDF

lOMoARcPSD|33505872 Ecology I Course Reader Tutorials Ecology I (Wageningen University & Research) Scannen om te openen op Studeersnel Studeersnel wordt niet gesponsord of ondersteund door een hogeschool of universiteit Gedownload door Sophie Holzhaus (sophie.holz...

lOMoARcPSD|33505872 Ecology I Course Reader Tutorials Ecology I (Wageningen University & Research) Scannen om te openen op Studeersnel Studeersnel wordt niet gesponsord of ondersteund door een hogeschool of universiteit Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorials Note: At the back of the Tutorials section is a short chapter on the application of mathematics within the turorials. Consulting this section may help you with the assignments. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 1 1 Tutorial 1 Demography Preparation Read 'Study material' in EOE (see below). The following assignments from this tutorial must be completed before the actual tutorial: Q 1.1 – Q 1.11. Count on a minimum of 2 hours of preparation time Bring the following to every tutorial - the textbook (EOE) and the study guide - the course registration card - a calculator - graph paper Theme This tutorial focuses on the demographic processes within a population and how the population growth can be calculated from these processes. The theory of population dynamics will be linked to variables that can actually be measured in the field. First, the composition of populations in various age and/or development classes will be addressed. We then analyse the processes of mortality (death) and natality (birth), which will ultimately enable us to calculate the growth of a population. Study material 3 3 ‰ EOE 4th Edition: 5.1 , 5.2 (excluding 'annual life cycles'), 5.3 Corresponding course unit 3 Populations Learning outcomes After studying the material and completing the tutorial, you will be able to explain the following and/or apply the associated competences: - Understand which demographic processes affect the growth of a population. - Understand the information presented in a life-table, and how such a table cap- tures the structure of populations. - Explain and calculate the growth rate of a population. - The relationship between the basic reproductive rate, the growth rate in various time frames and the generation time. - Know the terms cohort, mortality and natality (fecundity), survival, semelparous and iteroparous organisms, basal reproductive rate, generation time and growth rate. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 2 Tutorial 1 Questions Population composition The size (N) of a population at a given time is represented by the following formula (the processes of immigration and emigration have been omitted for simplicity): N time 1 =N time 0 + number of births - number of deaths (1.1) In this tutorial we address the processes of natality and mortality and calculate the growth of populations. We start with a description of the composition of a population. A population can be characterised by dividing the individuals into classes. This classification can be based on age, but for organisms such as insects it can also be based on the stage of development. In a species with a long life expectancy, the age classes will be expressed in years or decades; for short-lived species the age classes will be in days or months. An example of a classification in static age classes is presented in Figure 1.1, which shows two Australian populations of Acacia trees divided into groups of individuals that have germinated in a certain period of time. A group of individuals that has germinated (or was born) in the same time period is called a cohort. Figure 1.1 The distribution across age classes of Acacia burkittii trees at two locations in Australia. Between 1865 and 1970, the South Lake Paddock population was grazed by sheep, and from 1885 to 1970 it was also grazed by rabbits. In 1925 a fence was built which excluded the sheep from the Reserve population, but the fence could not keep rabbits out. Considering this information, the effect of grazing from 1865 is clearly visible in the decreasing number of newly established Acacia trees. However, the effect of fencing after 1925 – a strong increase in newly established Acacia trees – is also visible in the Reserve population. Even after fencing, the effect of grazing by rabbits on the Reserve population is still visible. This is partly because the 1925-1940 age class was much smaller than the 1845-1860 class that became established before rabbit grazing, even though this latter class had survived an additional 75 years (Source: Townsend et al., 2008). Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 1 3 Q 1.1 Which age classes of Acacia trees are affected strongly by grazing? Does graz- ing particularly affect the mortality of the younger or the older trees? From Figure 1.1 can you deduce whether it is the rabbits or the sheep that caused the most significant mortality? Survival and mortality What cannot be deduced from Figure 1.1 or more generally speaking, from a table with a static population structure, is whether a population changes in size. To calculate the change in population size, you also need to know the mortality and the reproductive rate of each cohort in addition to the population composition. Mortality and reproduction are expressed as relative variables: they are the fractions of the population which die or are born, respectively. These fractions can be shown in different ways. Mortality (and survival) can be expressed as a percentage relative to the previous period or as a percentage of the number of individuals at t=0. The same applies to survival. Survival can be expressed as a fraction of the number of individuals at t=0, but also as a percentage of the number of individuals that were still alive at the previous time point. Intuitively, it is clear that the chance of survival of a 20-year-old Acacia is different (but not necessarily greater!) than a 200-year-old Acacia. In practice it is difficult to determine these chances of survival exactly, but we provide a number of methods to estimate this. The first method is to take a sample of a random number of trees from one of the populations. You then determine the age composition of this group (x number of individuals is b years old, y number of individuals is c years old, etc.). Based on the age distribution, you can calculate the fraction of the individuals from a certain age class that survive. The second method is to monitor one cohort, for example all trees that have germinated between 1 January and 31 December 2000, and determine the survival of that group year after year. This method results in a cohort life table (Box 5.2 and Section 5.3 in EoE with the heading 'Cohort life tables') in which the surviving fraction is shown for a group. Caution: It is customary to regard to all individuals germinated or born in the year 2000 as individuals from year "0", and to assume that they all originated from the beginning of that year. For long-lived species, such as Acacia or humans, this second method may be less practical (although life insurance companies benefit from it!), but it will be emphasised in this tutorial. Births The growth of the population is, in addition to the number of deaths, determined by the number of births. In the most basic case, an organism only reproduces once at the end of its life, followed by the death of the individual. We call these semelparous organisms (EOE p. 129, 131). Grasshoppers are one example of a semelparous organism. Eggs hatch into immature nymphs that metamorphose to become grasshoppers, which finally produce a number of eggs and then die. Iteroparous organisms go through multiple reproductive periods during their lifetime. Examples are the Acacia, humans and buf- falo. Various age classes contribute to reproduction, but not all classes contribute equally. The differences in reproductive periods make it difficult to compare iteroparous organisms to semelparous organisms. To solve this problem, we use generations and generation time. In the next part, these concepts will be addressed, linked to the basic reproductive rate and the relative reproductive rate. We start by looking at the growth of a semelparous organism, as captured in a life table. Then we increase the complexity by considering iteroparous organisms. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 4 Tuto orial 1 Pop pulation growth of semelparrous organ hlox nisms: thee annual phlox (Ph drum mmondii) If booth fecundiity and moortality are known, th he growth of the poppulation caan be calcuulated. A liffe table usuually contaiins a colum mn stating thhe reproducction per oriiginal individual in the pop pulation (lxmx). The ssum of all lxmx values is thhe basic reeproductivee rate (R0).. In this tuttorial, we define the basicc reproducttive rate aas the aveerage number off offspring that an avverage indivvidual prod duces throughou me (from birth to deeath). Notee: the ut its lifetim individualss who die before b theyy can produuce offsprinng are included! R 0 is called the 'basicc reproductiive rate' (seee for example att the bottomm of Table 1.1 1 and Tabble 5.2 EOE E, and the text onn page 136)). The word 'rate' actuually indicaates a ratio. Tab ble 1.1 A sim mplified cohoort life table for the annu ual plant speccies Phlox drrummondii (Soource: Townssend et al., 2008). Q 1.22 What effect e will a basic repproductive rate r (R0) off 2.41 for thhe annual phlox p (Table 1.1) have ono the size of o the popullation in thee coming yeear, comparred to the currrent year? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 1 5 The annual phlox is semelparous, i.e. it has discrete generations with no overlap. The population consists of a new generation of plants every year. Q 1.3 Explain the terms generation and generation time. generation: generation time: Using the basic reproductive rate (R0) and the generation time (Tc) we can calculate the growth rate of a population. The growth can be expressed in two ways: (1) as the net number of individuals per unit of time and per individual that are added to or removed from the population (relative growth rate r), or (2) as the factor by which the size of the population changes per unit of time (growth factor R). For a constant population size, for example when a population has reached its carrying capacity (K), r=0 and R=1 therefore apply. Using the growth rate, the size of a population after a certain period can be calculated (usually in intervals of one year, except for small animals and plants with short generation time). The relative growth rate, the growth factor, the basic reproduc- tive rate, and the generation time are related to one another as follows: r = ln R = ln(R0) / T c (1.2 ) where Tc is the generation time of a cohort. Q 1.4 What is the generation time of an annual phlox? Calculate the relative growth rate r for the annual phlox population from Table 1.1. How will the population develop? Population growth of iteroparous organisms: wild boar and buffalo Calculating the growth rate of populations of iteroparous organisms is more complicated. This is because iteroparous organisms have overlapping generations. We cannot simply use the basic reproductive rate (R0) of iteroparous organisms to calculate population growth. Another complicating factor is that in many species only one sex (females) actually produces offspring. In Table 1.2 below we illustrate both of the previously mentioned methods of calculating survival and mortality using wild boar as an example. The left column shows the years (2000 is year 0), and the topmost number in each column indicates how many new individuals were added in that year due to birth (=cohort). Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 6 Tutorial 1 Table 1.2 Modified cohort life table for wild boar, with 5 cohorts, to determine the chances of survival in a population using two methods. At the beginning of year 0 (the first year of life), the first cohort contains 93 individuals born in year 0. At the beginning of year 1 (second year of life), 78 individuals have survived the first year of life (from age 0 to age 1); see text for explanation. Year → 2000 2001 2002 2003 2004 ↓(2000=0) 0 93 1 78 84 2 66 66 99 3 51 48 85 98 4 42 40 69 83 78 5 29 31 52 61 61 6 17 19 38 43 45 7 3 12 28 29 27 8 0 4 14 16 15 In the first method, we interpret the age classes from right to left as follows. At the start of year 4 (in boldface), 78 wild boar were born (0 years old), 83 are 1-year old, 69 are 2-years old, 40 are 3-years old, and 42 are 4-years old. This row is a static life table that immediately shows why it fails to give an accurate representation of reality: the number of 1-year-old wild boar is greater than the number of 0-year old wild boar. A slightly better technique is to count the numbers of individuals per age class in two consecutive years. At the start of year 5 (in boldface), 29 wild boar are 5-years old, 31 are 4-years old, 52 are 3-years old, 61 are 2-years old, and 61 are 1-year old. We have no data about births from year 5. The mortality (expressed as qx) for the 4-year-olds in year 4 (in boldface) is now calculated as (42-29)/42, for 3-year olds as (40-31)/40, etc. In this way we use information from multiple, consecutive cohorts. The surviving fraction (expressed as a percentage relative to the previous year) of the oldest animals (4-year-olds) is 29/42. For the 3-year-olds, the surviving fraction is 31/40, and so forth. In this way, mortality and survival add up to 1. In the second method, we follow only the cohort that was born in 2000 (= year 0). Of that cohort (N=93), 78/93 survived the first year (2000), 66/78 survived the second year (2001), 51/66 survived the third year, and so forth. The mortality (qx) is therefore (93- 78)/93 in the first year of life (2000), (78-66)/78 in the second year, (66-51)/66 in the third year, and so forth. In question 1.9 we will calculate the survival values as a fraction of the total number of individuals at t=0. Q 1.5 Why can the results of the two methods described above differ for the exact same population of wild boar? (Note: a population is not the same as a cohort). Q 1.6 How could you adapt these methods to obtain a better prediction of the chances of survival? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 1 7 In order to gain a clear picture of the chances of survival of the various cohorts, you can also make a graph of the numbers of individuals as a function of their age. This is known as a survivorship curve. To ensure comparability of such curves for different organisms, it is better to express the number of surviving individuals as a percentage of the number at the beginning (=100%). We can distinguish 3 types of survivorship curves (see Figure 5.13, page 138 EOE). Q 1.7 Indicate the most characteristic feature of each of these three types of survivor- ship curves. Type 1: Type 2: Type 3: Q 1.8 Figure 1.2 shows the survivorship curve of a certain animal species. The scale of the y-axis (number of survivors) in Figure A is logarithmic. Sketch the curve for the same information in Figure B for a linear scale on the y-axis. Why is a survivorship curve generally displayed on a logarithmic scale? Figure 1.2. Example of a survivorship curve. Note: the curve in graph A is a straight line. What does that mean? Examples of cohort life tables can be found in Table 1.1 of the reader and in Table 5.2 of EoE. These tables follow only one cohort through time. The second column (ax) represents the number of individuals who reach a subsequent age class. In general, this is the number that has actually been counted. The lx (third column) is the fraction of the original number of animals that are still alive at the beginning of a certain time interval. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 8 Tuto orial 1 Q 1.99 In the table beloww, which is based on the t first cohhort from T Table 1.2 above, a compleete the coluumns for lx (survivorsship rate, see EOE) annd qx (morrtality rate, seee explanation in Tablee 1.1) by callculating thee missing vaalues. Y Year class x ax lx qx 0 93 1.000 0 0 0.161 1 78 0.839 9 0 0.154 2 66 ….. ….. 3 51 ….. ….. 4 42 0.452 2 0 0.310 5 29 0.312 2 0 0.414 6 17 0.183 3 0 0.824 7 3 0.032 2 1 1.000 8 0 0.000 0 By calculating c qx we underrstand how lx is determmined. In liffe tables, qx is often miissing (see Table 1.1 inn the readerr and Table 5.2 in EoE)). After all, for some sppecies it is easier e to coount the nummbers of suurvivors thaan the numbbers of deceeased indivviduals. Thee next page shows a liife table off buffalo inn the Serenggeti National Park in Tanzania (T Table 1.3). Q 1.110 Compleete the coluumn for lx inn Table 1.3, for both male m and female buffallo, by calculaating the misssing valuess. Q 1.111 Draw a survivorshhip curve foor the male and femalee buffalo froom the Sereengeti Nationaal Park in Tanzania. Which W variiable from the life tabble do you plot? Which of the threee types of suurvivorship curves applies here? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 1 9 Table 1.3 Life table for male and female buffalo from the Serengeti National Park in Tanzania (Sinclair, 1977). Males Females Age x ax lx ax lx mx lxmx x(lxmx) Σx(lxmx) 0 892 1.000 916 1.000 0.00 0.000 1 607 0.680 614 0.670 0.00 0.000 2 526 0.590 528 0.576 0.00 0.000 3 523 0.586 518 0.566 0.03 0.017 4 514 507 0.07 5 502 487 0.23 6 493 473 0.23 7 477 0.535 446 0.487 0.41 0.200 8 455 0.510 428 0.467 0.41 0.191 9 411 0.461 396 0.432 0.27 0.117 10 374 0.419 348 0.380 0.27 0.103 11 329 0.369 308 0.336 0.27 0.091 12 293 0.328 241 0.263 0.27 0.071 13 223 0.250 180 0.197 0.20 0.039 14 172 0.193 128 0.140 0.20 0.028 15 112 0.126 83 0.091 0.20 0.018 16 55 0.062 44 0.048 0.17 0.008 17 21 0.024 20 0.022 0.17 0.004 18 15 0.017 6 0.007 0.17 0.001 19 8 0.009 0 0.000 20 0 0.000 Q 1.12 Which two important mortality periods can be distinguished in the life of the buffalo? In which column or columns would you look? The birth rate (fecundity or natality) is represented in life table 1.3 by the column mx. Q 1.13 Define mx and state the unit in which mx is expressed. How old are female buf- falo when they produce offspring? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 10 Tutorial 1 Q 1.14 Do the values of mx in life table 1.3 represent the female offspring, male offspring, or offspring of both sexes? Compare your answer with the data from EoE Table 5.2. Q 1.15 Calculate the values of lxmx and the values of x(lxmx) that are missing in the table. Now calculate the basic reproductive rate (R0). Describe the significance of the number you have found. In what unit is the basic reproductive rate ex- pressed? Q 1.16 Estimate – using the completed column lxmx – the generation time of the female buffalo, within a margin of 3 years. How did you make this estimate? Using the basic reproductive rate, the generation time of the iteroparous African buffalo can be calculated. You can use the following formula for this calculation: Tc = ∑ x(l m x. x) 1.3 ∑l m x x Q 1.17 Calculate the generation time of the female buffalo from Table 1.3, using for- mula 1.3 above. Does the outcome match your estimate from question 1.16? Q 1.18 Now calculate the relative growth rate r of the buffalo population, using for- mula 1.2 above. Does r take males and females into account? In which unit is r expressed? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 1 11 You can now 'predict' the development of the population using the relative growth rate. After one year the size of the population will be N1 = N0*R (1.4) where R = er. To calculate the population size after t years you use Nt = N0 * R t (1.5) Q 1.19 Predict the size of a buffalo population of 56,000 animals after 1 year and after 5 years. What would be the predicted population size if the generation time Tc was half as long? In the Serengeti National Park, the population size was indeed determined at 56,000 at one point. After 5 years, the population had grown to 68,000 buffalo. Q 1.20 Does this number match your calculations from question 1.19? Provide several possible explanations for the difference between your calcu- lated value and the 68,000 animals that were actually counted. Why is it use- ful to calculate the growth rate at one point in time if it cannot accurately pre- dict the growth of the population? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 12 Tutorial 1 Appendix: Variables Abbrevia- Variable tion Unit Number of individuals in a population at time t Nt - The number of individuals from a cohort reaching a subsequent ax - age class The fraction of the original number of animals that is still alive at lx - the beginning of a certain time interval (in this case: one year) The mortality probability of an individual in time interval (year) x qx -..... mx - Number of female descendants per original female individual in lxmx - the population in the relevant time interval (year) x Product of time interval x and the number of female descendants per original female individual in the population in the relevant x(lxmx) - time interval x Product of time interval x and the number of female descendants per original female individual in the population in the relevant Σx(lxmx) - time interval x, added up over all the years that female individuals from the relevant cohort lived Average number of female offspring that an average female individual produces throughout her life = basic reproductive Σ(lxmx) - rate Generation time: Tc year Basic reproductive rate R0 - Change in the size of a population per unit of time (growth factor) R year-1 Net number of individuals that are added to or removed from the population per unit of time per individual (relative growth rate = r year-1 intrinsic growth rate) Maximum net number of individuals that are added to or removed from the population per unit of time and per individual in a popu- rmax year-1 lation (maximum intrinsic growth rate) Authors: Hans de Kroon, Plant Ecology and Nature Conservation Group Arend Brunsting, Resource Ecology Group Ignas Heitkönig, Resource Ecology Group Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 2 13 Tutorial 2 Intraspecific competition and harvesting models Preparation Read 'Study material' in EOE (see below). The following questions must be completed before the tutorial: Q 2.1 – Q 2.10. Bring the following to every tutorial: - the textbook (EOE) and the study guide - the course registration card - a calculator Theme This tutorial focuses on the model-based description of competition within species, il- lustrated by an example of Nile perch (Lates niloticus), a large predatory fish that was introduced to Lake Victoria in Africa. The growth of populations is first described using an exponential model assuming that there are no density effects (i.e. no intraspecific competition). In a logistic growth model, the effects of density are accounted for. Next, a harvesting model based on harvesting a constant amount of fish, i.e. the maximum amount that can be harvested while maintaining a sustainable development of the popu- lation (Maximum Sustainable Yield: MSY) will be addressed. This model will be elabo- rated using the same example of the Nile perch in Lake Victoria. Complications in the use of this harvesting model are illustrated by an example of fishing for herring (Clupea harengus), a commercially important North Sea fish. An alternative harvesting model, based on harvests at a constant effort will be elaborated, again using the example of the Nile perch. Course material ‰ EOE 4th Edition: Section 5.5, Patterns of population growth page 143-147; Box 5.4 Quantitative aspects; 14.7 up to page 437 Figure 14.33 Corresponding course unit Populations Learning outcomes After studying the material, you can describe and use the following concepts: - Changes in population size and growth due to intraspecific competition. - Calculation of the intrinsic growth rate and the models (and Formulas 2.3, 2.4, 2.6 and 2.7) of exponential and logistic growth. - The terms intrinsic growth rate, logistic growth, carrying capacity. - Constructing a harvesting model using the intrinsic growth rate and carrying ca- pacity. - Calculating the Maximum Sustainable Yield (MSY) of a population. - Several reasons why MSY harvesting is risky. - Why a fixed effort harvesting strategy is safer than the MSY strategy. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 14 Tutorial 2 Questions EXPONENTIAL GROWTH In a population with an unlimited amount of space and food, and under optimal condi- tions, each individual will produce a maximum number of offspring. This maximum is determined by intrinsic factors, such as the anatomy and physiology of the organism, and not by extrinsic factors such as environmental conditions. The intrinsic growth rate (r) of the population is reached at a maximum number of offspring per individual and per unit of time and at a minimum mortality. The growth rate of the population (the increase in the number of individuals) is given by multiplying the intrinsic growth rate by the population size (N): dN / dt = r · N (2.1) where N is the number of individuals in the population. The change in this number of individuals is given by integrating this formula (see Figure 5.21 and Box 5.4 in EOE): Nt = N0 · er·t = N0 · Rt (2.2) where t = time and R = er, the reproduction rate (this is where Tutorial 1 ended, see for- mula 1.5). This formula describes the exponential growth. This exponential growth model only describes the development in the number of individuals in a population. However, for many applications (such as fisheries) it is more interesting to know the amount of biomass in a population than to know the number of individuals. The growth in biomass of a population is the result of the development in the number of individuals (as in Formula 2.2) and the growth in size of the individuals. The formulas that describe biomass growth have the same form as those which describe the development in num- bers: dB / dt = r · B (2.3) Bt = B0 · er·t (2.4) where B is the biomass of the population. It is difficult to work with an exponential curve. Therefore, Formula 2.4 is transformed by using the natural logarithm for both sides: ln Bt = ln B0 + r · t (2.5) The exponential growth curve can be illustrated with the emergence of the Nile perch (Lates niloticus) in Lake Victoria in East Africa. The Nile perch was not a native spe- cies in this area and was introduced in the 1950s. In the beginning, the Nile perch popu- lation scarcely increased, but from the 1970s it became more and more numerous (see Table 2.1). The Nile perch population could increase almost without limits because it is a large predatory fish (up to 2 meters in length) which encountered a large food source of small fish in Lake Victoria (especially cichlids: small, colourful perch-like fish which are very popular among aquarium enthusiasts) and experienced hardly any competition from other predatory fish. It therefore played a significant role in the disappearance of many species of fish, probably several hundred, that occurred only in Lake Victoria. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorrial 2 15 Table 2.1 Nile peerch biomasss in Lake Vicctoria (x 1,00 00 tonnes) annd its naturall logarithm. Year Biomass mass) Ln (biom (xx 1,000 tonnees) 1975 2.0 0.700 1976 3.7 1.322 1977 2.2 0.777 1978 5.6 1.722 1979 13.7 2.622 Uganda Kenya 1980 12.9 2.566 Africca Lake 1981 70.1 4.255 Victoria 1982 103.0 4.633 1983 219.3 5.399 Tanzaania 1984 273.6 5.611 1985 307.0 5.733 Figure 2.11 Nile perch (Lates nilotiicus) 1986 584.4 6.377 (image froom FAO) andd the location of Lake Victtoria. Q 2.11 Based on o Table 2.1, draw graaphs of the biomass b of Nile perch in Lake Vicctoria for the period 19755 to 1986. Do D this as fo ollows: a Mannually fit ann exponentiaal curve bassed on the points in the diagram beloow. This is a curve desccribed by Foormula 2.4. b Draaw a straighht line throuugh the poin nts of the naatural logariithm of the catch of Nile N perch inn the diagram below. This straighht line is deescribed by y For- mulaa 2.5. Whicch parameteer from this formula is representedd by the slo ope of the line l and whhat is its biollogical signnificance? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 16 Tuto orial 2 c Whaat are the addvantages of a logarithm mic transforrmation? INTR RASPECIFIIC COMPE ETITION The exponentiall growth deescribed aboove can onlly occur if there are nno restriction ns on growwth. In the example e off Lake Victooria, you caan imagine that when Nile perch were first released, thhe fish encoountered ann enormouss amount off resources (especially y food and space) s that were hardlyy used by other o predattory fish. Moreover, M thhere were soo few Nile perch at thhe beginninng that theyy did not compete witth each othher either. Under U thesee conditionss, exponenttial growth could take place: the growth rate could inccrease lineaarly with thee available biomass b (Foormula 2.3). How wever, at a certain c mom ment a growwing populaation reachees such a laarge size thaat not everyy individuall can complletely satisffy its basic needs n (suchh as food): thhe individu uals in the population p start competing for thhis limitingg resource. We call thhis intraspecific comp petition. Thhe populatioon can no loonger grow indefinitelyy, and the innitial exponential growwth levels offf, and evenntually stopss completelly at a biommass that is eequal to B∞. This is thee maximum m biomass that a popullation can reach r under certain circcumstancess; it is called the carryying capacitty. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 2 17 This density-dependent competition is expressed by adding a factor to the exponential growth equation that indicates which part of the carrying capacity is not yet used (For- mula 2.3): dB/dt = r · B · [1 - (B/B∞)] (2.6) This formula is called the logistic growth equation. The formula is similar to that in Box 5.4 and Figure 5.21 in EOE, except that numbers of individuals (N) are used (in- stead of N∞, K is used). If the biomass (almost) equals zero, then the factor [1 - (B/B∞)] (almost) equals 1, and Formula 2.6 is identical to Formula 2.3: there still is exponential growth. If B increases, the factor decreases and eventually approaches zero when the biomass equals the carry- ing capacity (B∞). The above can also be illustrated by the relative growth rate, i.e. the growth rate divided by the biomass. In the case of exponential growth, this is constant (derived from For- mula 2.3), and is equal to r, the intrinsic growth rate: dB/dt · 1/B = r (2.7). For a logistic growth curve, the relative growth rate decreases with the biomass, which results in: dB/dt · 1/B = r · [1 - (B/B∞)] (2.8). The curves corresponding with Formulas 2.7 and 2.8 are shown in Figure 2.2. We will discuss the specifics of intraspecific competition based on the development of the Nile perch biomass in Lake Victoria. At the beginning (1975) there was sufficient suitable prey for the Nile perch, and other factors were optimal for growth as well. Ta- ble 2.2 shows the development of the biomass from 1987 through 1996. Table 2.2. is a continuation of Table 2.1. r exponential Relative growth rate (1/B*dB/dt) logistic Biomass (B) B∞ Figure 2.2 Relative growth rates of the biomass of populations with exponential and logistic growth. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 18 Tuto orial 2 Tablle 2.2 The Nile perch biomass inn Lake Vicctoria (x 1,0000 tonnes)) and its naatural logarrithm (contiinuation of Table 2.1). Year Biomass Ln (biomaass) ( 1,000 ton- (x nes) 1987 680.0 6.52 1988 906.4 6.81 1989 1020.0 6.93 1990 1107.1 7.01 1991 837.7 6.73 1992 894.8 6.80 1993 1047.2 6.95 gure 2.3 A Nile perch 1.89 meterrs in Fig 1994 980.0 6.89 ngth, caught in the Ethiopian Abaya lake len 1995 1046.3 6.95 in 1989 (Photoo: John Cassselman, Demmeke 1996 983.4 6.89 Addmasu). Q 2.22 The bioomass of Nile perch inn Lake Victooria betweeen 1975 andd 1996 (from m Ta- bles 2.11 and 2.2) iss shown in the t diagram m above. a Untiil which yeear (approxiimately) waas there expponential grrowth? How w can this be seen in the t graph? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 2 19 b Estimate the intrinsic growth rate of Nile perch in Lake Victoria. c How can you derive the unit of the intrinsic growth rate? The carrying capacity of Lake Victoria (in particular the amount of prey fish) became limiting for the growth of the Nile perch population over time due to intraspecific com- petition, as shown in Formula 2.6 and 2.8 and Figure 2.2. Formula 2.6 can also be writ- ten somewhat differently: dB/dt = r · B · [1 - (B/B∞)] = r · B – (r/B∞) · B2 (2.9) The logistic growth formula therefore generates a parabola with a maximum value (Fig- ure 2.4a). (a) (b) (c) exponential exponential logistic Ln (Biomass) Biomass (Bt) B∞ Ln(B∞) growth rate (dB/dt) logistic logistic Absolute B∞ Biomass (B) Time (t) Time (t) Figure 2.4 (a) Absolute growth rate of the biomass of a population with logistic growth. (b) The de- velopment of the biomass over time with exponential and logistic growth on a linear scale and (c) on a logarithmic scale. Q 2.3 Figure 2.4a shows the absolute growth rate as a function of biomass. a Explain why the absolute growth rate of the population initially increases with increasing biomass, while the relative growth rate decreases with B from the beginning (Figure 2.2). b Draw the curve of the absolute growth rate for exponential growth as a function of the biomass in the diagram below. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 20 Tutorial 2 Exponential and logistic growth Absolute growth rate (dB/dt) logistic 0 B∞ Biomass (B) When we plot the biomass of a population (on a linear scale) against time, in case of logistic growth, we get an S-shaped (sigmoid) curve (Figure 2.4b, see also Figure 5.21 EOE). Q 2.4 Determine the carrying capacity B∞ for Nile perch in Lake Victoria and de- scribe the significance of B∞ based on the graph from Question 2.2 and Tables 2.1 and 2.2. Q 2.5 Why is B∞ a discrete density value (for example 219,000 tonnes) only in the- ory, while in practice it is expressed as a density range (for example between 200,000 and 250,000 tonnes)? Q 2.6 Suppose there is another lake that is the same size as Lake Victoria, but in which we find a carrying capacity for Nile perch that is only half that of Lake Victoria. a What could be the cause of this difference in carrying capacity? b If carrying capacity is reached in both systems (twice as high in Lake Victoria as in the fictional lake), is the intraspecific competition in Lake Victoria higher, lower, or equal to that in the other lake? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorrial 2 21 HAR RVESTING G An im mportant quuestion, esppecially for fisheries, is i how to caalculate thee largest posssible harveest that will not cause the populaation to colllapse. The answer a to thhis questionn will be illlustrated byy the exampple of Nile perch p in Lakke Victoria. The maxim mum amoun nt that can beb harvestedd depends ono the popuulation grow wth, the popuulation stabbility and th he sta- bilityy of the envvironment. HAR RVESTING G A CONSTA ANT AMOU UNT In thhe commonlly used harvvesting moddel that we explain here, the Maxximum Susstain- able Yield (MS SY) is calcuulated. Thiss assumes that t the maaximum quaantity to bee har- vesteed is equal to t the grow wth of the poopulation ovver the periood betweenn harvestingg peri- ods: harvesting the surpluss productioon. The annu ual net grow wth of the ppopulation can c be harveested yearlyy. The growwth of a poppulation throoughout a year y can be calculated using the growth g rate (dB/dt). Thhe result of harvesting h the t exact am mount by w which the po opula- tion has h grown results r in a population growth ratee of zero: noo increase oor decrease in i the size of o the popuulation. The growth (or net recruitm ment) in turrn depends oon the denssity of the population p a shown inn Formula 2.6 as 2 and Fig gure 2.4. Thhat is why w we first exaamine the reelationship between grrowth and density, and then analysse the effectts of harvestting. Q 2.77 In the diagram d bellow draw a graph of thhe growth rate r (dB/dt) of Nile perrch in Lake Victoria V as a function of o the densiity (B, as inn Figure 2.44). Use the equa- tion foor logistic growth (Foormula 2.6 6) and use r=0.5668 year-1 and B∞= U the folloowing valuees for B: 1000, 200, 400,, 600, 800, 1000, 990,0000 tonnes. Use and 12000 x 1,000 tonnes. t What is the sig gnificance of negative ooutcomes? Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 22 Tutorial 2 Q 2.8 What is the MSY of this population and at what value of B will it be reached? Indicate this maximum harvestable biomass by means of a line on the graph from Question 2.7. Is the MSY a quantity or a rate? We can now calculate the development of the biomass of the population as follows. Starting from a certain biomass, growth is calculated using the logistic equation (see the figure in Question 2.7) After this, an amount of biomass is harvested. The biomass in the following year is given by calculating biomass + growth - harvest. For the biomass in the following year, you again calculate the growth and harvest, and so on. Q 2.9 Calculate the biomass of Nile perch for three consecutive years for three har- vesting pressures. Start with a biomass of 495,000 tonnes and use the following harvesting pressures: the MSY (from Question 2.8) and harvests that are 20% lower and 20% higher than the MSY. Use the data from the table below. harvest = MSY harvest = 0.8 · MSY harvest = 1.2 · MSY Time biomass growth harvest biomass growth harvest biomass growth harvest (years) (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 tonnes) tonnes / tonnes / tonnes) tonnes / tonnes / tonnes) tonnes / tonnes / year) year) year) year) year) year) 1 495 495 495 2 3 Q 2.10 The following table shows the values for biomass, growth and harvest for years 4 through 10. This is the continuation of the table you filled out in Question 2.9. In the diagram below, plot the biomass for each of the three harvesting pressures against time. Describe the development of the biomass of the popula- tion. What biomass will be reached after approximately 15 years in each sce- nario? harvest = MSY harvest = 0.8 · MSY harvest = 1.2 · MSY Time biomass growth harvest biomass growth harvest biomass growth harvest (years) (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 (x 1,000 tonnes) tonnes / tonnes / tonnes) tonnes / tonnes / tonnes) tonnes / tonnes / year) year) year) year) year) year) 4 495 140 140 577 136 112 409 136 168 5 495 140 140 601 134 112 376 132 168 6 495 140 140 623 131 112 340 127 168 7 495 140 140 641 128 112 298 118 168 8 495 140 140 657 125 112 248 105 168 9 495 140 140 670 123 112 185 85 168 10 495 140 140 681 121 112 102 52 168 Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorrial 2 23 Q 2.111 We now w focus again on the graph g with the growthh rate of thee Nile perch h you createdd in Question 2.7. Draw w a line indiicating a harrvest that iss 20% lowerr than the MSSY. a Usinng this grapph, derive the t equilibrrium biomass of the population at a this harvvesting level. Assume ana initial bio omass of 1)) 200,000 toonnes and an a ini- tial biomass b of 2) 1,000,0000 tonnes. b Indicate the equuilibrium points p betweeen harvest and growthh and divid de the X-axxis into threee biomass domains leeft and righht of the equuilibrium points. Indicate whetheer dB/dt is positive orr negative for f those doomains and draw vecttors along thhe X-axis that t indicatee the directtion of the sshift in biom mass. Howw do the equuilibrium pooints differ?? Q 2.112 Based ono the previous questioon and the graph g from Question 2.7, considerr why it is so risky to haarvest the MSY, M or even to harveest 20% lesss than the MSY. M Explainn why this risk resultss from the characteristtics of the lleftmost eq quilib- rium pooint. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 24 Tutorial 2 HARVESTING WITH CONSTANT EFFORT The policy on fisheries focuses on sustainability. However, the preceding has shown that sustainability is difficult to achieve by striving for MSY. Even with a safety margin, for example by catching only 80% of the MSY, this strategy is still risky. A system in which the fishery is managed according the total amount of fish that can be caught (quota system) is always affected by the above-mentioned disadvantages. This can be seen, for example, in the catches of herring in the Northeast Atlantic (Figure 2.5). Until the end of the 1960s, the catches were large (usually between 2,500 and 3,000 million tonnes). Over time however, more and more boats fishing for longer peri- ods of time were needed to achieve this catch. The herring catch collapsed in the late 1960s. It was only after a sharp reduction in fishing effort, during which the herring fishery was stopped completely in some countries between 1977 and 1982, that the her- ring population began to recover. There is also another way to regulate fisheries: by changing the fishing effort (see Sec- tion 14.7 and Fig. 14.32 in EOE). If we assume that a certain fraction of the biomass can always be harvested with a certain fishing effort (fixed effort), this means that a popula- tion decrease will automatically result in a smaller harvest. Herring catches Northeast Atlantic Ocean 4000 3500 3000 Catch (in 106 kg) 2500 2000 1500 1000 500 0 1950 1960 1970 1980 1990 2000 2010 Year Figure 2.5 Herring catches in the Northeast Atlantic from 1950 to 2013 (data and figure: FAO). Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorrial 2 25 Q 2.113 The efffects of fishhing accordding to the fixed f effort principle oon the Nile perch p populattion is show wn in the diagram belo ow: Nile peerch harvestts corresponnding with thhree fishingg efforts (exxpressed ass 0.1, 0.3 annd 0.8 x thhe availablee bio- mass) are a indicatedd. a Coppy the logisstic growth curve from m Question 2.7 onto thhis diagram m and indiccate the eqquilibrium points. Use vectors tto show hoow the bioomass deveelops near thhese points. b Use the graph tot determinee the chang ge in the bioomass over time at the three o 100,000 ttonnes and from harvvesting presssures, startting from a biomass of 700,,000 tonnes. c Explain why thhe fixed effoort strategy is i safer thann the MSY sstrategy (connsider whenn both strategies can faiil). Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 26 Tutorial 2 Appendix: Variables Variable Abbreviation Unit Number of individuals in a population N - Absolute growth rate dB/dt kg/year Biomass of the population B kg Carrying capacity (numbers of individuals) K ( = N∞) - Carrying capacity (biomass) B∞ kg Intrinsic growth rate = maximum relative growth rate r year-1 (fractional increase or decrease) Maximum Sustainable Yield (MSY) dB/dt kg/year Harvesting rate dB/dt kg/year Relative growth rate dB/dt · 1/B year-1 Reproduction rate (factor of increase or decrease) R year-1 Time t year Authors: Leo Nagelkerke (Aquaculture and Fisheries Group) Wim van Densen (Aquaculture and Fisheries Group) in collaboration with Hans de Kroon (Plant Ecology and Nature Conservation) Arend Brunsting (Resource Ecology) Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 3 27 Tutorial 3 Interspecific competition Preparation Read study material in EOE (see below). The following questions from this tutorial must be completed before the tutorial: Q 3.1 – Q 3.11. Bring the following to the tutorial: - The textbook (EOE), the study guide, a calculator - The course registration card Theme Besides intraspecific competition, there often is competition between species, referred to as interspecific competition. This kind of competition is also relevant in agricultural systems. For example, interspecific competition is very important in intercropping sys- tems, in which two or more crops are simultaneously grown on the same field. To study competition and productivity in intercropping systems, the Wageningen agronomist C.T. de Wit developed an experimental approach: the replacement experiment. In this tutorial, both the design of a replacement experiment and the analysis of the results with the help of a replacement diagram will be addressed. The change in size of a population is also influenced by the presence of competing spe- cies. In the second part of this tutorial it will be explained how the influence of inter- specific competition can be incorporated in the logistic growth equation. The Lotka- Volterra model for interspecific competition, composed of one growth equation for each of the two competing species, will be explained. Based on this model it will be deter- mined whether species in a certain habitat are able to survive alongside each other (co- existence) or that ultimately only one of the two species will remain, while the other goes extinct (competitive exclusion). The role of niche differentiation regarding the out- come of competition will be discussed. Course material ‰ EOE 4th Edition: 6.1 page 155 - 168, Box 6.2 Corresponding course unit 4 Interspecific competition and species richness Learning outcomes After studying the material, you can describe and use the following concepts: - The design of a ‘De Wit replacement experiment’. - Development and interpretation of a replacement diagram. - Competition coefficient. - The Lotka-Volterra model for interspecific competition. - The significance of isoclines and the interpretation of the combined isoclines for the outcome of interspecific competition. - Niche differentiation. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 28 Tutorial 3 Questions The De Wit replacement experiment In some instances, individuals from different species cause less competitive pressure than individuals from the same species. This principle is made use of in intercropping systems: the simultaneous cultivation of two or more crops on the same plot. This method of cultivation is frequently used in 'low-input' systems in tropical and sub- tropical areas, but is hardly used on a large scale in temperate areas with a high degree of mechanisation. In the 1950s, C.T. de Wit developed an experimental approach, known as 'the replace- ment design', to measure the competition between two plant species and to determine whether an intercropping system has a higher yield potential than the individual crops in pure stands. The basic principle of this method is that species are sown or planted at a constant total plant density. Pure stands of both species serve as reference, while one or more intercropping systems with the same total plant density, but differing in mixing ratio, are also established. Figure 3.1 shows a schematic representation of such a re- placement design. Pure stand A Intercropping A-B Pure stand B (50/50) Figure 3.1 Schematic representation of a replacement design (planted in rows). Besides the two pure stands, Figure 3.2 also shows three of the many possible outcomes of a 50:50 combination. The size of each circle indicates the size of an individual plant. Each of the three outcomes is also shown in a replacement diagram (Figure 3.3). The share of crop A (zA), ranging from 0 to 1, is shown on the x-axis. If zA = 0, this indicates a pure stand of crop B (on the left in the diagram). In a pure stand of crop A, zA = 1 (on the right in the diagram). For the 50:50 intercropping, zA = 0.5. The relative yield (RY) of both crops is shown on the y-axis. This is calculated by dividing the yield of a crop grown in an intercropping system by the yield of the same crop in pure stand. Q 3.1 In the first diagram, determine which of the intersecting lines represents the relative yield of crop A and which line represents the relative yield of crop B. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 3 29 Pure stand A Pure stand B Intercropping A-B (50/50): several possible outcomes. Figure 3.2 Schematic representation of a replacement experiment with crop A and crop B. The bottom row shows several possible results of the 50/50 combination. Q 3.2 In the first diagram, a straight line is found for both crops. What can you de- duce from these lines regarding the ratio between interspecific and intraspecific competition? (Perhaps Figure 3.2 will give you an idea). By adding the relative yields of both crops in the intercropping system, the relative yield total (RYT) is obtained. RYT provides information about the productivity of the inter- cropping system and is shown with the topmost line in the replacement diagram. Figure 3.3 Replacement diagrams to display the results of a replacement experiment. Diagrams corresponding to the results in Figure 3.2 Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 30 Tutorial 3 Q 3.3 What does the straight line in the first diagram indicate about the value of RYT and consequently about the productivity of the intercropping system? In the second replacement diagram, regardless of the ratio between crop A and crop B, the RYT is also equal to 1. However, we find curved lines for both the RY of crop A and the RY of crop B. Q 3.4 Explain the meaning of both curved lines, both from the perspective of crop A and from the perspective of crop B, in terms of the ratio between intraspecific and interspecific competition. Q 3.5 Analyse the competitive relationships in the third replacement diagram and de- termine whether the intercropping system is over-yielding. Provide a possible explanation for your observation. An expansion of the logistic growth equation In the previous tutorial we discussed the logistic growth model and saw how it takes intraspecific competition into account. For each species the logistic growth model is characterised by the intrinsic growth rate (r) and the carrying capacity (K). To study in- terspecific competition, we will analyse an experiment with Paramecium. Three species of Paramecium were cultivated separately in test tubes with bacteria as a source of food; the following values were found: r [1/day] K [# per ml] species 1 1.045 450 species 2 1.105 130 species 3 1.099 127 To determine the extent to which the presence of one species influences the develop- ment of the other species, the organisms were also reared together: species 1 with spe- cies 2, and species 1 with species 3. The initial density of both species and the amount of food were identical to that in the pure cultures. The number of Paramecium were counted every day; the results are shown in Figure 3.4. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 3 31 Figure 3.4 Change in the number of Paramecium in two competition experiments (left: species 1 and species 2; right: species 1 and species 3). Q 3.6 For both figures: determine whether there is coexistence or competitive exclu- sion. To understand these results, we will initially focus on the competition experiment with the combination of species 1 and species 2. The two species compete for the same food source (bacteria) in the test tube. The growth rate of species 1 now not only depends on the number of individuals of species 1, but also on the number of individuals of species 2. Any difference in competitiveness between both species is expressed in the competi- tion coefficient. This coefficient reflects the relationship between interspecific and in- traspecific competitiveness. For example, the competition coefficient α12 indicates how, for individuals of species 1, the interspecific competitiveness of one individual of spe- cies 2 relates to the intraspecific competitiveness of one individual of species 1. In other words, the first index (in this case species 1) indicates on which species an effect is ex- erted, while the second index indicates by which species (in this case species 2) the in- terspecific competition is exerted. Q 3.7 In the mixed culture of Paramecium species 1 and 2, the competition coeffi- cient α12 is 3.3. Describe what an α12 of 3.3 means from the perspective of spe- cies 1. Also explain why the competition coefficient is greater than 1. When we compare the value of α12 = 3.3 with the outcome of competition in Figure 3.4, it is obvious that a competition coefficient on its own does not say anything about the final outcome of competition; despite the much larger individual competitiveness of species 2, it is species 1 that ultimately prevails. To determine how this is possible, we have expanded the logistic growth equation from tutorial 2 in such a way that, in addition to intraspecific competition, interspecific com- petition is also taken into account. Multiplying the density of species 2 (N2) with com- Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 32 Tutorial 3 petition coefficient α12 indicates the constraint that species 2 exerts on species 1, ex- pressed in 'species 1 equivalents'. This variable is included in the last term of the logistic growth equation to calculate the residual fraction of the carrying capacity: ( K1 − N1 − α12 N 2 ) K1 (3.1) For species 1, this results in the following growth equation: dN1 ( K − N1 − α 12 N 2 ) = r1 N1 1 dt K1 (3.2) Figure 3.5 shows how the growth rate of species 1, in the absence of species 2, depends on population size. This relationship was discussed extensively in tutorial 2. Figure 3.5 Relationship between population size and absolute growth rate of Parame- cium species 1 in the absence of competing species. Q 3.8 For densities of N1 = 50, 250 and 450, calculate the growth rate of species 1 if species 2 is present at a density of N2 = 30. Indicate the results in the figure. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 3 33 Q 3.9 Explain why the presence of species 2 at a density of N1 = 350 results in a growth rate that is essentially zero. Q 3.10 At which density will species 2 have to be present to prevent an increase of species 1, even for a minimal density of species 1? Is it plausible that species 2 will ever reach this density? What conclusion – precisely formulated – can you draw from this? Q 3.11 Suppose that species 2 could exploit the exact same niche as species 1 and util- ise this food with the same efficiency. What values would you expect for carry- ing capacity K2 and competition coefficient α21? Apparently the conditions mentioned in the question above do not apply, because (as mentioned before) K2 = 130. Moreover, a value of 0.4 was found for α21. This indicates that both species differ more from each other than is assumed above. This could be the result of aspects such as niches that are not fully overlapping, a different nutritional need and/or a difference in utilisation efficiency. The Lotka-Volterra model To understand why only species 1 ultimately prevails, we need to expand our analysis. After all, species 2 also has its own dynamics. Besides intraspecific competition, this will also be determined by the interspecific competition exerted by species 1. To study this interaction, we add a similar equation for species 2 to equation 3.2 (equation 3.3): dN1 ( K − N1 − α 12 N 2 ) = r1 N1 1 dt K1 (3.2) dN 2 ( K − N 2 − α 21 N1 ) = r2 N 2 2 dt K2 (3.3) These two equations together are referred to as the Lotka-Volterra model for inter- specific competition. Using this model you can predict the course of population growth of both species, starting from certain initial population sizes (N0) and with known values of α, r and K. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 34 Tutorial 3 Q 3.12 Which of the four factors mentioned above ultimately determine whether both species can coexist or whether there will be competitive exclusion? Support your answer with reasoning, and at the end of this tutorial determine whether you were correct. To gain insight into the outcome of competition, we do not focus on the change of both populations over time, but rather focus on the equilibrium that will establish over time. For this purpose we construct a graph with the density of species 1 on the x-axis and the density of species 2 on the y-axis. Every possible combination of densities of species 1 and species 2 can be represented with a point in this figure. Q 3.13 Indicate where the following combinations of densities are on the graph in Fig- ure 3.6: Species 1 [per ml] Species 2 [per ml] Point F: 40 60 Point G: 300 60 Point H: 150 80 Now it is important to determine whether the size of a species population will increase or decrease for a given density combination. This can simply be determined by using the growth equations for species 1 and species 2 (equations 3.2 and 3.3). Species 1 Species 2 r [1/d] 1.045 1.105 K [#/ml] 450 130 α12 resp. α21 3.3 0.4 Q 3.14 In the table above, the main characteristics of species 1 and species 2 are sum- marised. Use these values to calculate the growth rate at points G and H for both species. Also calculate the new population sizes that are reached after one day and add them to the table below. Numbers on day 0 Growth rate Numbers on day 1 Point N1-day 0 N2-day 0 dN1/dt dN2/dt N1-day 1 N2-day 1 F 40 60 +19.7 +27.5 59.7 87.5 G 300 60 H 150 80 Q 3.15 Show the growth rate of both species with a vector (arrow) in Figure 3.6, where the direction of the arrow indicates whether the population is increasing or de- creasing and the length of the arrow indicates the rate at which this occurs. Why is a change in species 1 shown with a horizontal arrow? Explain why a population increases at one point and decreases at the other. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 3 35 160 140 120 N2 [#/ml] 100 80 60 40 20 0 0 100 200 300 400 500 N1 [#/ml] Figure 3.6 Density combinations of two Paramecium types. To gain more insight, it is useful to determine those density combinations of species 1 and species 2 at which the population of a species remains exactly the same. The zero isocline of species 1 is the line with combinations of population sizes of species 1 and species 2 at which the population growth of species 1 is exactly zero: dN1 ( K − N1 − α 12 N 2 ) = r1 N1 1 dt K1 = 0. (3.4) This happens when the possibility for further expansion of species 1 is exactly zero. This point is reached when the total population size of species 1 and species 2, ex- pressed in 'species 1 equivalents' (=N1+α12N2), is exactly equal to the carrying capacity of species 1: ( K1 − N1 − α12 N 2 ) = 0. ⇒ N1 + α12 N 2 = K1 (3.5) Since there is a linear relationship between N1 and N2 in equation 3.5 (check this!), only two points are needed to draw the zero isocline. For this purpose we determine the in- tersections with both axes: N1 = 0 Æ N2 = K1/α12 (EOE Figure 6.7a) and N2 = 0 Æ N1 = K1 (EOE Figure 6.7a) Q 3.16 On Figure 3.6, draw the zero isocline for Paramecium species 1 in mixed cul- ture with species 2. For this: calculate the points at which the line intersects the y- and x-axes (where N1 = 0 and N2 = 0, respectively) and connect both points with a straight line. Intersection with y-axis: N1 = 0. N2 = Intersection with x-axis: N2 = 0. N1 = Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 36 Tutorial 3 Q 3.17 Why do the previously drawn horizontal arrows all run in the same direction as the zero isocline? Q 3.18 On the same figure draw the zero isocline for species 2. The method is analo- gous to the calculations for species 1: First calculate the intersections of the zero isocline with the x- and y-axis based on the second equation from the Lotka-Volterra model (equation 3.3), and connect the calculated intersections with a straight line. Intersection with y-axis: N1 = 0. N2 = Intersection with x-axis: N2 = 0. N1 = Vector analysis Using the figure we just constructed, we can predict what will eventually happen to both populations: only one of the two populations prevails (competitive exclusion), or a sta- ble equilibrium develops in which both species coexist. Q 3.19 In Figure 3.6, determine the composite vector for points F, G and H. Based on these vectors, you can deduce the final equilibrium of the mixed population. If the final outcome is not yet clear to you, determine the composite vector for a number of additional combined densities of species 1 and species 2. Which equilibrium is ultimately reached? Does this correspond with the progression of the populations in Figure 3.1? Is this an unstable or a stable equilibrium? Analysis of intersections Another way to derive the outcome of competition is based on the intersections of both zero isoclines with the x- and y-axis. If we apply this to the above example, we can start with the intersections with the y-axis (N1 = 0). For species 1: N1 = 0 Æ N2 = K1/α12 = 450./3.3 = 136.4 For species 2: N1 = 0 Æ N2 = K2 = 130. In this case, comparison of both intersections shows that: K1/α12 > K2 Æ K1 > α12K2 The above notation is informative, since the terms on the left and right side of the > sign have clear significance in terms of competition. Q 3.20 Use the first equation of the Lotka-Volterra model (equation 3.2) to determine the significance of both terms (possibly look back at question 3.10). Now de- scribe the significance of K1 > α12K2 in words. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 3 37 The intersection with the x-axis can then be analysed in an identical manner. In this case, for the zero isocline of species 1 and species 2: For species 1: N2 = 0. Æ N1 = K1 = 450. For species 2: N2 = 0. Æ N1 = K2/α21 = 130./0.4 = 325. In this case, therefore: K2/α21 < K1 Æ K2 < α21K1 Q 3.21 Based on this situation and using the second equation of the Lotka-Volterra model (equation 3.3), what do you conclude? Q 3.22 Combine the results of questions 3.20 and 3.21 and use this to explain why species 1 will be the only one that survives. Explain the possible relevance of niche differentiation to this survival. Q 3.23 Why do, in this case, both the intrinsic growth rate (r) and the initial densities of both species (N0) only affect the time course of population development and not the outcome of competition? In the example above, only one species ultimately remains. The competitiveness of this species is greater than that of the other species in such a way that the second species completely disappears and the first species expands to its carrying capacity. In the event of competition between two species, however, several outcomes are possible. For ex- ample, both species can continue to exist side-by-side in a steady state. This is called coexistence (e.g. Paramecium species 1 and 3 in Fig 3.4). This occurs when – for both species –intraspecific competition exerted by an individual of the same species is greater than the interspecific competition exerted by an individual of the other species. In other words, this happens when individuals from the same species compete more strongly with each other than with individuals from other species. One of the possible causes of this is niche differentiation, where both species partially use different re- sources, or retrieve the resources from a different location (e.g. rooting depth for plant species). A special situation occurs when the interspecific competition between two species is stronger than the intraspecific competition within both species. In such a case, either species can completely replace the other. In this case, starting density (N0) and intrinsic growth rate (r) co-determine which species will prevail. In Figure 3.7 the described cases are schematically shown. The zero isocline of species 1 is shown with a solid line, while the zero isocline of species 2 is shown with a dotted line. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 38 Tutorial 3 N2 N1 N2 N1 Figure 3.7 Zero isoclines of species 1 (solid line) and species 2 (dotted line) in competition. Q 3.24 Analyse both cases with the vector method. Which of the cases described above corresponds with each of the two figures? Check your answer by analys- ing the intersections of the zero isoclines with the x- and y-axes. For both cases also determine whether the intersection of the zero isoclines represents a stable or unstable equilibrium. Author: Lammert Bastiaans (Centre for Crop Systems Analysis) in collaboration with Arend Brunsting (Resource Ecology Group) Hans de Kroon (Plant Ecology and Nature Conservation) Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 4 39 Tutorial 4 Predation Preparation Read the study material in EOE (see below). The following questions from this tutorial must be completed before the tutorial: Q 4.1 - 4.8 Bring the following to every tutorial: - the textbook (EOE) and the study guide - the course registration card - a calculator Course unit Theme In tutorial 3 we examined the Lotka-Volterra equations for interspecific competition. An- other example of interspecific interaction is the relationship between predators and their prey. This is addressed in more detail in this tutorial. First we discuss the Lotka-Volterra equations for predator-prey interactions. These describe the change in numbers of preda- tors and prey in the simplest model. Next we discuss three more complex links between predation rate and the density of the prey population (the functional response) and their significance for the balance between predator and prey. Then we look at the behaviour of a predator who can choose between two prey species. We will see how functional responses can be used to determine the prey preference of the predator. Finally, we will see how changing the prey preference can influence fluctuations in the predator and the prey popu- lations. ‰ Course material EOE: 7.13, 7.43, 7.53 (Underlying dynamics and Predator-prey cycles), Box 7.1 (p. 200-201), Summary (p. 214) Note: Functional response is not described in Essentials of Ecology but is ad- dressed in this tutorial and is mandatory exam material. Corresponding course unit: 3 Populations Learning outcomes After studying the material, you can: - Derive variations in the number of predators and their prey from the Lotka-Volterra model for predator-prey interactions (you need to understand the symbols, but you do not have to memorise the formulas). - Draw three types of functional response and indicate the consequences of the func- tional response for changes in the size of the prey population. - Explain the effect of functional response type III on fluctuations in numbers of predators and their prey. - Indicate how functional responses can be used in the determination of prey prefer- ence. - Explain what 'switching' means for fluctuations in the prey and predator populations. Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 40 Tutorial 4 Questions INTRODUCTION: PREDATORS AND PREY This tutorial focusses on the interactions between predators and their prey. When we think about predators, the large predators of the African plains often come to mind. But preda- tors at a micro level exist in much larger numbers all around us. Experimental studies on predation are often performed on insects and mites (about 80% of all species are insects and mites) because these small animals can easily be grown in large numbers and are therefore ideal for studies under controlled conditions. After laboratory experiments, field experiments can be performed to test the laboratory data under natural conditions. Preda- tory mites have often been used in experimental work (see an example in Figure 7.21 in EOE). These preda- tors are about 0.5-0.7 mm in size and live from prey such as spider mites, rust mites or small insects. In ad- dition, predatory mites are used on a large scale in the biological control of pests. Special forms of predation are grazing and parasitizing prey organisms. Figure 4.1 - Predatory mite (right) consuming a spider mite (left). Q 4.1 Identify a type of prey from the examples below and describe to which class the predators belong. Use ‘true predators’, ‘grazers’ and ‘parasites’ as classes. (see EOE Section 7.1 (p. 186) for the definitions of these classes of predators). Predator ‘Prey’ Predator class Cow Dolphin Parasitic wasp Tapeworm Wolf Parasitic plant (mistletoe) Tick Tawny owl LOTKA-VOLTERRA EQUATIONS Not only is the number of prey influenced by predators, but the number of predators is also determined by the size of the prey population. The dynamics of predator-prey systems are described in EOE (Box 7.1 page 200) by two differential equations developed by Lotka and Volterra. The prey equation reads: dN/dt = rN - aPN. (4.1) r is the intrinsic growth rate of the prey population; N is the prey density and P is the predator density; a is the captured fraction of the prey population per predator per unit of time (‘attack coefficient’). Gedownload door Sophie Holzhaus ([email protected])  lOMoARcPSD|33505872 Tutorial 4 41 The factors dN/dt, r and N were previously encountered in the exponential and logistic growth equations. The difference with the logistic growth equations is that the term (K – N)/K for the intraspecific competition has been omitted and a term (- aPN) has been added. The predator equation reads: dP/dt = faPN - qP. (4.2) Besides the term (aPN), there are two new factors: f is the number of predators that can be produced by consumption of one unit of prey (conversion efficiency) q is the mortality of the predator (time –1). Q 4.2 a Explain in words what the term (aPN) means. b Derive the unit of the first term of the predator equation from the dimen- sions of f, a, P and N. As with the Lotka-Volterra model for interspecific competition (Tutorial 3), we can de- rive the dynamics of the predator-prey system from a vector diagram on which the zero- isoclines of both populations are drawn. Q 4.3 For a predator-prey system, the following parameter values have been deter- mined: r = 1.012 a = 0.10 q = 0.66 f = 0.30. Calculate the intersections of the zero isoclines of prey and predator with the x- axis and the y-axis and draw the zero isoclines on the figure below. Which combination of densities results in a constant predator-prey system? N* = P* = 16 14 Predator Density (P) 12 10 8 6