BCHM 3P01 Lecture Notes - Glycogen, Hydrolysis, Phosphorylase

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fuchsiaweasel270

20 May 2014

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Document Summary

Alpha-1,4 makes the straight chain and alpha-1,6 makes the branching. Vertebrates use this because we have a high metabolic rate. Non-reducing end has the hydroxyl group at carbon 1 occupied. The more of these ends the more that glucose can be mobilized. Dynamically changing (minute by minute) to match energetic needs of organism. 4 enzymes required for breakdown of glycogen: glycogen phosphorylase. Catalyzes bond cleavage in the glycogen molecule by substitution of a phosphate group to a single glucose molecule. Although reaction is reversible, the [] of pi is very high compared to g1p, so the reaction is driven to the right . essentially making this an irreversible reaction. Reaction occurs at non-reducing ends of glycogen chains (ie: ends without a c1-oh) Glucose subunits must be no more than 5 residues away from a branch point in glycogen for this reaction to occur via glycogen phosphorylase. Hydrolyzes phosphate bond causing release of glucose across cell membrane into the blood.

Related Questions

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 Î± (1â6) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the Î±(1â6) bond.

B) Every branch point will have an Î² (1â6) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the Î²(1â6) bond.

C) Every branch point will have an Î± (1â4) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the Î± (1â4) bond.

D) Every branch point will have an Î² (1â4) glycosidic bond that will have to be hydrolyzed. This is handled by the presence of an additional enzyme specific for the Î² (1â4) 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.