The function is it decides which cells come in and out and it’s structure would be the phospholipid bilayer
Answer:
Approximately 6944 glucose residues are added enzymatically per second
Explanation:
Cellulose is the main structural polysaccharides in plants. It is composed of unbranched glucose monomer units linked to each other by beta 1-4 glycosidic bonds.
The cell wall and stem of plants cells are composed of cellulose fibers. They provide rigidity and support to the plant.
In the given bamboo plant, the enzymatic addition of glucose units to the growing cellulose fiber chains results in the phenomenal growth rate of the bamboo stem.
Since each glucose unit contributes ~0.5 nm to the length of a cellulose molecule, number of glucose units required for daily growth is calculated as follows:
0.5 nm = 10⁻⁹
0.3 m/0.5 x 10⁻⁹ m = 600000000 units of glucose per day
Number of seconds in a day = 24 * 60 * 60 = 86400 seconds
Number of glucose residues added per second = 600000000/86400
Number of glucose residues added per second = 6944.4 glucose molecules per second
Therefore, approximately 6944 glucose residues are added per second
<h3>Explanation:</h3>
Mitosis actually happens in four phases. The first phase of mitosis through which chromatin condenses into chromosomes, the nuclear case breaks down, centrioles divide, and a spindle begins to form. Microtubule fibres that enable the division of sister chromatids during mitosis and meiosis.
This would probably be C.(liquid water), as while it is not as compact as ice, it still has a bit of molecules in one location, as compared to ice, which is very dense or gas which has minimal molecules in a particular area.
