1. Chapter 4 Biotics 1) What is a community? 2) What factors influence community structure? 3) How do we determine which species wins in competition?

2. What is a community? Defined in various way by various people.... Might be just a list of all the species in an area Which area: Benthic, Pelagic, Littoral Are these really separate communities? The concept of community generally implies interaction among species

3. What is community structure? Again, depends on who you ask... In general, provides information both about which species are there, and their relative abundance M.R. Allen Unpublished data

4. Some terms to know: Competition – a mutually negative (-/-) interaction between two or more individuals over acquisition of a limiting resource. Predation – the consumption of all or part of one living organism by another (+ / -). Parasitism – an interaction in which one organism (the parasite) gains resources at the expense of the other (the host) (+ / -).

5. Exploitative competition — competition through removal of resources Interference competition — competition through direct interference Allelopathy – the direct effect of competitors through the release of chemicals that inhibit other species May discuss competition, predation and parasitism separately, but what we really care about is community structure. More terms to know:

6. interspecific competition — competition involving more than one species intraspecific competition — competition between members of a single species More terms to know: Trophic level - a subset of a community. Includes species that acquire energy in the same way.

7. Trade-offs – exist when benefit gained by change in one trait comes at a cost paid out through change in another trait. Rarely is any one species good at all things Some species are good at taking up P or N Some species are good at avoiding being eaten Algae Compete – mostly by exploitative competition Competition can occur at any trophic level as long as resources are limiting

8. How do you know who wins in competition? Early experiments (prior to 1970s) used circular logic. Put two species together on a limiting resource, let it run for many generations. The superior competitor is the one that will displace all other species The species that displaced all other species was declared to be the superior competitor

9. Tilman ’ s Mechanistic Theory of Resource Competition Tilman 1982 (book)—several paper prior to that resulted in hundreds of studies to date Assessed competitive ability of algae based on their physiology. Is there a way to tell who will win before you put two species together?

10. You as the experimenter decide how much new medium (= nutrients) are going in. Used chemostats—established equilibrium conditions http://www.unifr.ch/biol/ecology/ebert/lab/chemostat2.jpg http://www.news.cornell.edu/Chronicle/00/12.7.00/predator-prey.JPG

11. R* -- the resource availability at which the reproductive rate equals the mortality rate Species with the lowest R* should win at competition because it reduces the resource to a level at which other species have a mortality rate that is higher than their growth rate When loss > growth, that species is eliminated from the culture

12. Under these conditions, the growth rate of the population can be approximated by Monod ’ s (1950) equation:   max S K s + S =   max * S K s + S =

13.  = growth rate (increase in biomass per unit biomass per unit time)  MAX = maximum growth rate at unlimited supply of substrate S = substrate (nutrient) concentration K s = ½ saturation constant, concentration of S that produces ½  MAX (i.e. S when  =  MAX /2)  max * S K s + S = Where:

14. Look at it in a graph: Can use these physiological uptake rates to predict winner in competition for 1 limiting resource.

15. Assume two species of algae both have same  MAX (4/day) One species has K s = 2 the other has K s = 10 Table of  values SKs = 2Ks = 10 000000 11.330.36 22.000.67 42.671.14 83.201.78 163.562.46 323.763.05 643.883.46 1283.943.71 2563.973.85

16. Both approach  MAX, but the one with the lower K s gets there faster.  max

17. Set mortality rate (since this is a chemostat, experimenter controls loss rate) mortality rate

18. The point at with the growth and loss curves cross is R* mortality rate R*

19. This is a very simple technique. Can make more complicated examples Which species wins under a high mortality rate? Under a low mortality rate? What if they have different death rates?

20. Grew each alone in the chemostat under P- limited conditions Tilman tested his theory using diatoms from Lake Michigan Asterionella and Cyclotella

21. No matter where you draw the mortality curve, Asterionella should always out compete Cyclotella under conditions of P-limitation He ran the experiment 9 times with Asterionella and Cyclotella together under P-limited conditions and Asterionella always displaced Cyclotella PO 4 (  mol/L) 

22. Ran this experiment 6 times—Cyclotella won 5 out of 6 times. Diatoms can also be limited by Si. Asterionella is a good competitor for P, but it is relatively poor for Si No matter where you draw the mortality curve, Cyclotella should always outcompete Asterionella under conditions of Si limitation. SiO 2 (  mol/L) 

23. Who wins depends on ratio of two nutrients. Asterionella is a better competitor for P Cyclotella is a better competitor for Si But, the lake water contains many nutrients

24. To determine point of switch in nutrient limitation, find K s when  Si =  P 1 species, 2 nutrients At what concentration is Asterionella limited by P and at what concentration is it limited by Si? =  maxSi * [SiO 2 ] K SiO2 + [SiO 2 ]  maxP * [PO 4 ] K PO4 + [PO 4 ]

25. =  maxSi * [SiO 2 ] K SiO2 + [SiO 2 ]  maxP * [PO 4 ] K PO4 + [PO 4 ] First, assume  maxSi =  maxP and cancel = [SiO 2 ] K SiO2 + [SiO 2 ] [PO 4 ] K PO4 + [PO 4 ]

26. = [SiO 2 ] K SiO2 + [SiO 2 ] [PO 4 ] K PO4 + [PO 4 ] = (K SiO2 +[SiO 2 ])[PO 4 ]* [SiO 2 ]* (K PO4 + [PO 4 ]) = *K SiO2 ) +([PO 4 ]*[SiO 2 ])([PO 4 ] ([SiO 2 ]* K PO4 ) + ([SiO 2 ]*[PO 4 ]) Cancel ([SiO 2 ]*[PO 4 ])

27. *K SiO2 )([PO 4 ]= ([SiO 2 ]* K PO4 ) K SiO2 [PO 4 ] = [SiO 2 ] K PO4

28. These are their “optimal” Si:P ratio. Supply ratio at which they are not P limited or Si Limited For Asterionella—3.9/0.04 = 97.5 For Cyclotella—1.4/0.25 = 5.6 Si:P LowHigh Asterionella P-limited Asterionella Si-Limited 97.5 6 Cyclotella P-limited Cyclotella Si-Limited

29. Intra-specific Competition—competition within one species Either in the lab or in the field, set up replicate enclosures of several treatments: Species 100121 Species 2 1 2100 Total12221 Treatment12345 Inter-specific Competition—competition between more than one species How do you measure competition among animals? Some species can be grown in a chemostat, can use similar technique

30. “ Winner ” of competition is measured as: Species that displaces the other Species that most influences the fitness of the other Hu and Tessier 1995

31. Concepts to Know Know the different interspecific interactions stressed in class Understand the Monod growth rate equation and how it is used to predict algal growth rates. Know how to interpret the results from Tilman ’ s Mechanistic Theory of Resource Competition

32. Practice question Food concentration (mg Carbon/L) 102345678910 Growth rate (per day) 0.25 0.50 1.0 0.75 0.0 Brachionus rubens Brachionus calyciflorus Plotted below are the Monod growth curves for two species of rotifers, Brachionus rubens, and Brachionus calyciflorus grown individually in a chemostat on a diet of Chlamydomonas. If the death rate is set at 0.25/day, which species is the superior competitor?______________ What if death rate is set at 0.75/day?________________________ Explain your choices.