Chapter 11 (Part 2)

Question 1

What is the role of a lysozyme?

Answer 1

To destroy bacterial cell walls.

Question 2

How do lysozymes complete their intended function (generally speaking; this is not a mechanistic question)?

Answer 2

By degrading peptidoglycans in the bacterial cell wall.

Question 3

Lysozymes hydrolyze the ____ linkage between ______ and ______ residues in the polysaccharide portion of the peptidoglycan.

Answer 3

Hydrolyze the beta-1,4 linkage between n-acetylglucoasamine (NAG) and n-acetyluric acid (NAM) residues

Question 4

Lysozyme has a striking groove along 1 whole face. What is this groove used for?

Answer 4

Binding the substrate.

Question 5

How many saccharide residues in how many regions can the binding cleft accommodate?

Answer 5

6 saccharides in 6 regions (that are referred to as subsites and given letters A-F)

Question 6

NAG residues bind in subsites __, __, & __ of lysozyme, but NAM residues bind in subsides __, __, & __.

Answer 6

A, C, and E.

B, D, and F.

Question 7

What occurs when the substrate is in the groove of lysozyme?

Answer 7

The enzyme hydrolyzes the beta-1,4 linkage between the D (NAM) and E (NAG) residues

Question 8

When the 6 residues lay in the groove of lysozyme, the NAM sugar in subsite D is distorted into a _______ conformation.

Answer 8

Half-chair

Question 9

What provides the energy necessary to distort the NAM residue in subsite D into a half-chair conformation?

Answer 9

The other 5 residues bind in the normal chair conformation, and the binding energy from those other hexoses in the other subsites provides the energy necessary.

Question 10

Modified hexoses normally exist in the _____ conformation.

Answer 10

Chair

Question 11

What is special about the location of Glu35 in lysozyme? Describe what this location allows Glu35 to do.

Answer 11

It is in a nonpolar pocket, an environment that raises the pKa of the carboxyl group on its side chain. This allows Glu35 to stay protonated at unusually high pH values, like around 7.

Question 12

The unusual pKa of Glu35 allows it to act as an ______ catalyst in the lysozyme mechanism by _________.

Answer 12

Acid catalyst, by protonating the O1 atom (which is part of the 1,4 linkage connecting the D & E rings)

Question 13

The protonation of O1 in the lysozyme mechanism is a good way to promote the cleavage of the __-__ bond and convert the O into a _____ group on the __ residue, which is now released as __ product(s).

Answer 13

Cleavage of the C1-O1 bond
Convert the O into a hydroxyl group on the E residue
Released as 1 product

Question 14

In the lysozyme mechanism, oxonium ion has a large (+,-) charge. This charge electrostatically interacts with the side chain of ____, which is (protonated, deprotonated) and (+,-) charged, stabilizing the oxonium ion’s transition state.

Answer 14

Oxonium has a large positive charge

Interacts with the side chain of Asp52, which is deprotonated and negatively charged

Question 15

Why does a nucleophilic attack between Asp52 and the oxonium ion occur in the lysozyme mechanism?

Answer 15

Because Asp52 is a good nucleophile while carbon 1 of the transition state oxonium is electron deficient (an electrophile)

Question 16

Once a nucleophilic attack occurs between Asp52 and the oxonium ion in the lysozyme mechanism, what binds to the active site in place of the E ring (which has departed)?

Answer 16

A water molecule

Question 17

Once water binds to the active site in the lysozyme mechanism, how is it converted into a good nucleophile?

Answer 17

By Glu35, which now acts as a base catalyst by pulling a hydrogen off H2O

Question 18

After water is converted into a good nucleophile (hydroxide) by Glu35 in the lysozyme mechanism, which bond does it attack?

Answer 18

The hydroxide ion attacks the bond between C1 of the D ring and the carboxyl group of Asp52

Question 19

When the lysozyme mechanism is complete, Glu35 and Asp52 are reset. In what state is each found? Protonated or deprotonated?

Answer 19

Glu35 is reset to the protonated state.

Asp35 is reset to the deprotonated state

Question 20

What is the function of a serine protease?

Answer 20

To hydrolyze the peptide bond of polypeptide chains

Question 21

Name 2 examples of a serine protease.

Answer 21

Trypsin, chymotrypsin, elastase, some proteases involved in activation of the blood clotting cascade

Question 22

The serine protease enzymes share 3 important active site residues. What are they called collectively? What are the 3 residues?

Answer 22

The catalytic triad. Made up of a serine (Ser195), histidine (His57), and aspartate (Asp102)

Question 23

Catalytic Triad Mechanistic Question.

How is Ser195 originally made into a strong nucleophile?

Answer 23

The imidazole ring of His57 abstracts a proton from the Ser195 side chain

Question 24

Catalytic Triad Mechanistic Question.

How is His57 originally made into a stronger base to facilitate the removal of the Ser195 proton?

Answer 24

The buried carboxylate ion of Asp102 forms a low-barrier hydrogen bond with N1 of His57, making His57 a stronger base

Question 25

Catalytic Triad Mechanistic Question.

Following deprotonation, what does Ser195 do?

Answer 25

Deprotonated Ser195 attacks the alpha-carbon carbonyl of the substrate peptide bond, forming the preliminary tetrahedral intermediate

Question 26

Catalytic Triad Mechanistic Question.

(T/F) All 3 residues are working together only through base catalysis.

Answer 26

False. All 3 residues work together in a combination of covalent catalysis and base catalysis.

Question 27

Catalytic Triad Mechanistic Question.
Following formation of the first tetrahedral intermediate, the leaving group amine is (protonated, deprotonated) by N3 of His57

Answer 27

Protonated

Question 28

Catalytic Triad Mechanistic Question.

After the leaving group amine is protonated by N3 of His57, what happens to the tetrahedral intermediate?

Answer 28

It decomposes to the acyl-enzyme intermediate

Question 29

Catalytic Triad Mechanistic Question.

The first product is formed after a new __- terminus is formed at the site of _____.

Answer 29

N-terminus is formed at the site of peptide bond cleavage

Question 30

Catalytic Triad Mechanistic Question.

(T/F) Once the new N-terminal is released from the enzyme, a water molecule takes its place.

Answer 30

True. The water molecule is then deprotonated into a hydroxide ion by a His residue

Question 31

Catalytic Triad Mechanistic Question.

How is the second tetrahedral intermediate formed?

Answer 31

Following a nucleophilic attack by the hydroxide ion on the electron-deficient carbonyl carbon

Question 32

Catalytic Triad Mechanistic Question.

In the second tetrahedral intermediate, water takes the place of the original _____.

Answer 32

Amide group

Question 33

Catalytic Triad Mechanistic Question.

How is the second tetrahedral intermediate resolved?

Answer 33

His57 facilitates the reaction by acting as an acid catalyst, donating a proton to the Ser195 side chain, releasing Ser195 from the substrate.

Question 34

Catalytic Triad Mechanistic Question.
Following the release of Ser195 from the substrate, what is formed on the original carbonyl carbon? What functional group is it known as?

Answer 34

A new carbonyl group is generated on the original carbonyl carbon to create a carboxylic acid.

Question 35

Catalytic Triad Mechanistic Question.

What is the second product of the mechanism?

Answer 35

A new C-terminus in the substrate where the peptide bond used to be.

Question 36

Catalytic Triad Mechanistic Question.
After formation of the new C-terminus in the substrate following resolution of the second tetrahedral intermediate, what occurs to the new C-terminus?

Answer 36

The new C-terminal is released, and the enzyme is reset to go through another round.

Question 37

Although acid-base catalysis and covalent catalysis are important aspects of the serine protease mechanism, the enzymes take greatest advantage of ______.

Answer 37

Preferential binding of the transition site.

Question 38

In the serine protease mechanism, what is the “oxyanion hole?”

Answer 38

The oxyanion hole region contains good hydrogen bond donors from backbone amides of Gly193 and Ser195. It is potentially a good place for an oxyanion to bind. When the substrate initially binds to the active site, it is held in place and constrained from migrating into the oxyanion hole.

Question 39

Catalytic Triad Mechanistic Question.
After formation of the first tetrahedral intermediate, why does the carbonyl O of the peptide bond to be cut need to be made more anionic?

Answer 39

A conformational distortion in the enzyme active site causes that O to move back into the oxyanion hole. There it forms 2 new hydrogen bonds with backbone amides. The backbone amide group of the substrate can also hydrogen bond with the carbonyl group of Gly193. The formation of these 3 additional H-bonds between the enzyme and the transition state is critical to provide driving force for the reaction