Answer:
c = Enzymatic activity
e = Attachment to ECM and cytoskeleton
f = Signal reception and relay
g = Transport
h = Intercellular junctions
i = Cell-cell recognition
a = Phospholipid bilayer
b = Fibers of extracellular matrix (ECM)
d = Microfilaments of cytoskeletons
Explanation:
c) Enzymatic activity is an indication of the amount of active enzymes present to increase a reaction rate
e) Attachment to ECM and cytoskeleton is enabled by integrins that take signals from the ECM and control signaling pathways of the cell
f) Signal reception and relay is the transduction of signal
g) Transport is the movement of matter in and out of the cell through the cell membrane
h) Intercellular junctions are the contact regions between adjacent cells and plasma membrane
i) Cell-cell recognition is power of the cell to decipher the different neighboring cells in the cell's environment
a) Phospholipid bilayer consists of a hydrophobic interior and a hydrophilic exterior
b) Fibers of extracellular matrix (ECM) are ECM associated the cell
d) Microfilaments of cytoskeletons consist of actin and aid in cellular movement.
Animals obtain nitrogen by eating plants which has absorbed nitrogen from the soil.
Answer:
vultures, pretty much anything that is a decomposer like worms and stuff
Explanation:
Answer:
After DNA replication, each chromosome now consists of two physically attached sister chromatids. After chromosome condensation, the chromosomes condense to form compact structures (still made up of two chromatids). ... The two copies of a chromosome are called sister chromatids
Explanation:
Answer:
a. 8.1 milligrams
b. 40.07 hours
c. 8.859 milligrams
Explanation:
If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time can be modeled by

Where A(t) = amount of Valium remaining in the blood after t hours
t = time or duration after the drug is taken
a. we have to calculate the amount of drug remaining in the bloodstream after 12 hours


= 10×0.81253
= 8.1 milligrams
b. In this part we have to calculate the time when A(t) = 5 milligrams


0.5 = 
Now we take natural log on both the sides of the equation.
ln(0.5) = ln(
-0.69314 = -0.0173t
t = 
t = 40.0658
≈ 40.07 hours
c. In this part we have to calculate the rate, by which amount of drug will decay in the bloodstream after 7 hours.


= 10×0.8859
= 8.859 milligrams