When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (p) is
a. 1.5
b. 1
c. 2.0
d. 0

When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (P) is 0

Explanation:

According to Levin's model (1969):

dP/dt = c - e

where P represents the proportion of occupied patches.

c and e are the local immigration and extinction probabilities per patch.

Thus, the rate of change of P, written as dP/dt, tells you whether P will increase, decrease or stay the same:

The rate dP/dt is calculated by the difference between colonization or occupancy rate (c) and extinction rate (e).

c is then calculated as the number of successful colonizations of unoccupied patches as a proportion of all available patches, while e is the proportion of patches becoming empty. Notice that P can range between 0  and  1.

As a result, if the patch occupancy rate (c) equals the patch extinction rate (e), then patch occupancy P equals to 0.

you are certainly talking about this picture:

a: is the label of the white square, it is showing the  replication fork, formed by helicase (d) ==> so it is the site of dna separation and replication 

b: is a new dna brand that is following directly the replication fork, so it is continuous ==> built continuously from 5' to 3

c: dna polymerase cannot replicate dna on its own, it needs the of their proteins, and one of them plays the role of a clamp which fixes  the dna with the dna template and increases  the processivity of the enzyme. ==>   stops polymerase from falling off the  template.

d: is the enzyme responsible for forming the replication fork (helicase) ==> unwinds and separates dna into two strands

e: is the delayed strand, it is built  in the opposite sense of the helicase, so dna will be built in fragments ==> built  fragments from 5' to 3 joins

and finally f: is the enzyme that builds  the new dna strand (dna polymerase) ==>   adds nucleotides to make a new dna helix.

credit: gradman555p98x9p

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