The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. Given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth, calculate the number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule.

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

b is the correct answer

d. cellulose and starch