Lecture 6

Question 1

Oxidases:

1) Type of ____ dependent enzyme
2) Directly oxidize substrate ___ incorporating new oxygen atom
3) Do they require NADH or NADPH?
4) How is FAD regenerated?
5) 2 examples

Answer 1

1) Flavin
2) Without
3) No
4) 2 e- reduction O2 to H2O2
5) Glucose oxidase and monoamine oxidase (MAO)

Question 2

Monooxygenases:

1) Type of ____ dependent enzyme
2) Directly oxidize substrate ____ incorporating new oxygen atom
3) Do they require NADH or NADPH?
4) How is FAD regenerated?
5) 2 examples

Answer 2

1) Flavin
2) by
3) Yes
4) 4 e- reduction of O2 to H2O
5) Flavin monooxygenase (FMO) and bayer villiger monooxygenase (BVMO)

Question 3

Why does D-glucose exist primarily as a cyclic hemiacetal?

Answer 3

The intramolecular attack from the nearby oxygen in the alcohol group is kinetically fast so the reaction is highly favorable. Additionally, formation of the 6 membered ring is fast because the 6 membered ring structure is thermodynamically stable.

Question 4

What is an anomer?

Answer 4

Cyclic monosaccharides that differ from one another only at the anomeric carbon (carbon that is either an aldehyde or ketone in the open chain sugar molecule)

Question 5

What is the anomeric effect?

Answer 5

For cyclic glucose molecules, there is either the alpha anomer (OH group on C1 is axial) or the beta anomer (OH group on C1 is equatorial). The alpha anomer occurs in nature at a higher percentage than would be expected due to the anomeric effect. In the cyclic form of glucose, the MO that contains the lone pair on the oxygen in the ring is aligned with the sigma* c-o MO, which produces stereoelectronic stabilization of this orbital which reduces the penalty of having this group in this axial position.

Question 6

Disaccharide vs. glycoside

Answer 6

Sugars are converted to acetals by substitution of the anomeric hydroxyl group. If the substituting group is another sugar, it is a disaccharide. If the substituting group is not a sugar then it is a glycoside.

Question 7

Why is the beta anomer of glucose more stable than the alpha anomer?

Answer 7

Because when OH is placed in the axial position (alpha anomer), it has 1,3 diaxial interactions which cause e-/e- repulsions making the configuration less stable.

Question 8

Compare and contrast inverting and retaining glycosidases.

Answer 8

Inverting glycosidases utilize simultaneous acid and base catalysis in the active site to increase nucleophilicity of water. The two carboxylate groups are 10.5 A apart acccomdating enough space for both the substrate and the water molecule which is the hallmark of the inverting glycosidase.

Retaining glycosidases utilize nucleophilic catalysis in the active site to increase the electrophilicity of the substrate. The two carboxylate groups are only 5.5A apart which accomodates only enouch space for either the substrate or water which is the hallmark of the retaining glycosidase.