Answer:
(D). Disaccharide.
Explanation:
A disaccharide can be defined as a molecule, made up of two monosaccharides. To form a disaccharide, two monosaccharides join together with the help of an O-glycosodic bond between them.
Sucrose is an example of disaccharide, which is made up by joining of fructose and glucose molecules.
Answer:
Heterozygous dominant
Explanation:
because, as you know, heterozygous is different, and Dd has both dominant and recessive, making it heterozygous rather than homozygous.
Because the first d is dominant (capital D), then it makes the genotype dominant and heterozygous.
Answer:
Regular insulin
Explanation:
Regular insulin is a fast insulin. After being applied, its onset of action takes between half an hour and its maximum effect occurs between two to three hours after application. It is usually used to treat type 1 diabetes mellitus, type 2 diabetes mellitus, gestational diabetes, diabetes complications such as diabetic ketoacidosis (DKA) and hyperglycemic hyperosmolar coma.
Answer:
E.
Explanation:
Im sorry if it's wrong but that's what I would go with
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.
