Question

1) Why might it be advantageous for a storage polysaccharide to have a branched-chain structure instead of a linear structure?

A) Compared to a linear molecule, a branched-chain polymer has less termini for addition or hydrolysis of glucose units per unit volume of polymer, thereby facilitating both the deposition and mobilization of glucose by providing more sites for enzymatic activity.

B) Compared to a linear molecule, a branched-chain polymer has more termini for addition or hydrolysis of glucose units per unit volume of polymer, thereby complicating both the deposition and mobilization of glucose by providing less sites for enzymatic activity.

C) Compared to a linear molecule, a branched-chain polymer has more termini for addition or hydrolysis of glucose units per unit volume of polymer, thereby facilitating both the deposition and mobilization of glucose by providing more sites for enzymatic activity.

D) Compared to a linear molecule, a branched-chain polymer has less termini for addition or hydrolysis of glucose units per unit volume of polymer, thereby complicating both the deposition and mobilization of glucose by providing less sites for enzymatic activity.

2) Can you foresee any metabolic complications in the process of glycogen degradation? How do you think the cell handles this?

A) Every branch point will have an α (1→6) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the α(1→6) bond.

B) Every branch point will have an β (1→6) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the β(1→6) bond.

C) Every branch point will have an α (1→4) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the α (1→4) bond.

D) Every branch point will have an β (1→4) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the β (1→4) bond.

3) Can you see why cells that degrade amylose instead of amylopectin have enzymes capable of endolytic (internal) as well as exolytic cleavage of glycosidic bonds?

A) Exolytic cleavage breaks the molecule internally, creating additional ends for exolytic attack and thereby banning the mobilization of more glucose per unit time.

B) Endolytic cleavage breaks the molecule internally, creating additional ends for endolytic attack and thereby allowing the mobilization of more glucose per unit time.

C) Exolytic cleavage breaks the molecule internally, creating additional ends for endolytic attack and thereby allowing the mobilization of more glucose per unit time.

D) Endolytic cleavage breaks the molecule internally, creating additional ends for exolytic attack and thereby allowing the mobilization of more glucose per unit time.

4) Why do you suppose the structural polysaccharide cellulose does not contain branches?

A) Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into microfibrils, thereby increasing the rigidity and strength of the microfibrils.

B) Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into microfibrils, thereby decreasing the rigidity and strength of the microfibrils.

C) Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into globules, thereby decreasing the flexibility and strength of the globules.

D) Branches in the molecule would generate side chains that would almost certainly make it difficult to pack the cellulose molecules into globules, thereby increasing the flexibility and strength of the globules.