Question

1. lakes in hoosier national forest

location # baby fish # bald eagles (fish eaters) # alligator snapping turtle (fish eaters) # insect larva (fish food)
indian lake 8000 2 29 20,000
celina lake 8000 1 0 50,000
tipsaw lake 8000 1 19 15,000

1. which lake in hoosier national forest has the most fish - eating predators?
2. which lake is likely to have a stable population of fish?
3. which lake is likely to experience exponential growth followed by many fish dying? how can you tell?

Explanation:

Step1: Identify fish - eating predators

For Indian lake, number of fish - eating predators = 2 (bald eagles)+29 (alligator snapping turtles)=31.
For Celina Lake, number of fish - eating predators = 1 (bald eagles)+0 (alligator snapping turtles)=1.
For Tipsaw Lake, number of fish - eating predators = 1 (bald eagles)+19 (alligator snapping turtles)=20.

Step2: Determine lake with most fish - eating predators

Since 31>20>1, Indian lake has the most fish - eating predators.

Step3: Analyze stable fish population

A stable fish population occurs when the number of predators is balanced with the amount of food. Celina Lake has the least number of predators (1) and a relatively high amount of fish food (50,000 insect larva), so it is likely to have a stable fish population.

Step4: Analyze exponential growth and die - off

Tipsaw Lake has a moderate number of predators (20 in total) and a relatively low amount of fish food (15,000 insect larva). Initially, with a large number of baby fish (8000), there could be exponential growth. But as the fish grow and the food source is limited while predators are present, many fish are likely to die.

Answer:

1. Indian lake
2. Celina Lake
3. Tipsaw Lake. There are a large number of baby fish initially for potential exponential growth, but a relatively low amount of fish food and a moderate number of predators which could lead to fish dying.