Answer:
When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (P) is 0
Explanation:
According to Levin's model (1969):
<em>dP/dt = c - e</em>
where P represents the proportion of occupied patches.
<em>c</em><em> </em>and <em>e </em>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:
- if dP/dt >0, then P is increasing with time
- if dP/dt <0, then P is decreasing with time
- if dP/dt = 0, then P is remaining the same with time.
The rate dP/dt is calculated by the difference between colonization or occupancy rate (<em>c</em>) and extinction rate (<em>e</em>).
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.
Answer:
The dipeptide would digest faster. The dipeptide would digest slower or not at all.
They talk about it because they have to you know and it keeps them from getting people
Mike is able to maintain this sleep schedule because He has a hereditary condition that confers unnaturally precise CLOCK gene expression (Option A).
<h3>What is an adaptation?</h3>
Adaptations are phenotypic features that naturally appear in a population as a consequence of genetic variation which are expressed to determine the level of the trait.
In this case, it is expected that Mike exhibit certain genetic features associated with his circadian rhythm.
In conclusion, Mike is able to maintain this sleep schedule because He has a hereditary condition that confers unnaturally precise CLOCK gene expression (Option A).
Learn more about adaptation and genetics here:
brainly.com/question/1381142
#SPJ1
