Predator population vs prey population2/22/2024 We monitored the communities immediately before and after imposing the disturbance and four days after resuming the pre-disturbance dilution regime to infer resistance and recovery properties. increased dilution simulating density-independent mortality as press or pulse disturbances coupled with resource deprivation) on bacterial abundance, diversity and community structure in the absence or presence of a protist predator. Using model microbial communities, we experimentally tested the effects of imposed disturbances (i.e. Manipulating multiple factors (biotic and abiotic) in controlled settings and measuring multiple descriptors of multi-trophic communities could enlighten our understanding of the context dependency of ecological disturbances. Many biodiversity studies rely on endpoint measurements instead of following the dynamics that lead to those outcomes and testing ecological drivers individually, often considering only a single trophic level. ?100? lions supports in balance ?5,000? zebras and vice versa.Ecological disturbances are important drivers of biodiversity patterns. These are the three equilibrium solutions of the system, and if we analyze them we can say:Īt ?(L,Z)=(0,0)? both populations are at ?(0,0)?.Īt ?(L,Z)=(0,10,000)? the zebra population is stable at ?10,000?, but the lion population is at ?0?.Īt ?(L,Z)=(100,5,000)? the system is stable. To turn the remaining two pairs into equilibrium solutions, we’ll solve each pair as a system of linear equations. The first pair, ?L=0? and ?Z=0?, is the solution ?(0,0)?, so the pairs become The third pair, ?Z=5,000? and ?Z=0?, doesn’t include a value for ?L?, so we can eliminate that pair completely and focus on just the other three: We’re looking for each pair to generate an equilibrium solution ?(L,Z)?. So if we pair ?L=0? with both ?Z=0? and ?Z+50L=10,000?, and then in addition we pair ?Z=5,000? with both ?Z=0? and ?Z+50L=10,000?, we get the following combinations: We need to test both solutions from the first equation with both solutions from the second equation. Setting both equations equal to ?0? gives To find equilibrium solutions, we’ll factor both equations. The equation for lions ?dL/dt? has a positive ?LZ?-term, but the equation for zebras ?dZ/dt? has a negative ?LZ?-term, which means this is a predator-prey system in which the lions are the predators and the zebras are the prey. How to determine whether the system is cooperative, competitive, or predator-preyĭoes the system of lions and zebras represent a cooperative, competitive, or predator-prey system? What are the equilibrium solutions of the system and what do they mean? The size ?a? of population ?x? supports in balance the size ?b? of population ?y?, and vice versa.Īn equilibrium solution ?(a,0)? means that population ?x? is stable at size ?a?, but population ?y? is at ?0?.Īn equilibrium solution ?(0,b)? means that population ?y? is stable at size ?b?, but population ?x? is at ?0?. Solving the system for an equilibrium solution ?(x,y)? will help you draw conclusions about both populations.Īn equilibrium solution ?(0,0)? means that both populations are at ?0?.Īn equilibrium solution ?(a,b)? means that the system is stable. How to interpret population stability from the equilibrium solutions On the other hand, because the equation for ?dx/dt? contains the higher-degree term ?bx^2?, it means that the predator population is effected by the prey population, as well as another factor, like carrying capacity. Because there are no higher-degree terms in the equation for ?dy/dt?, the prey population is only effected by the predators. ?x? is the predator population because ?cxy? is positive, and ?y? is the prey population because ?-gxy? is negative. Sometimes one or both equations will contain higher-degree terms. What it means when the system contains higher-degree terms In a predator-prey system, the equation whose ?xy?-term is positive represents the predator population the equation whose ?xy?-term is negative represents the prey population. If one sign is positive and the other is negative, the system is predator-prey. If both signs are negative, the system is competitive. If both signs are positive, the system is cooperative.
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