Question

Explain the difference between conversion efficiency and capture efficiency in predator-prey

interactions.

Answer 1

Now consider the case where there is more than one predator – that is, there are P predators and there are V prey. Recall that each predator has no influence on the consumption rates of other predators. The predators are just like molecules randomly bumping into prey molecules.

The total rate of predation now is:

f(V)⋅P=αVPf(V)⋅P=αVP

The term α is also known as the capture efficiency. Consider a case where every predator “bumps into” all prey in the landscape each time step, but only consumes 3% of those prey. Now consider a predator that consumes 50% of those prey per time step? Which predator is more efficient.

The increase in total predators due to prey consumption now is:

g(V)⋅P=βV⋅Pg(V)⋅P=βV⋅P

Does this make sense?

The term β is related to the conversion efficiency of the predator (how effieciently the predator converts its food into the production of new predators). Consider a case where every predator consumes 100 prey per time step (α⋅V=100α⋅V=100), but the predator population growth increases by only 0.01 relative to a scenario with no prey. Now consider a predator that consumes 1 prey per time step, and growth increases by 0.1… Which predator is more efficient at converting prey biomass into population growth