Lecture 10

Question 1

What are two primary ways to increase the rate constant?

Answer 1

– increase temperature and decrease the activation energy.

–This occurs because K is proportional to 1/e. Since increasing temperature decreases e and decreasing activation energy decreases e, a large K will be observed

Question 2

What happens when enzymes decrease activation energy in terms of molecules?

Answer 2

since the barrier is less,

– lowered activation energy increases fraction of molecules that have enough energy to attain transition

Question 3

True or false. Enzymes increase both the forward and reverse direction rates

Answer 3

True

Question 4

How do enzymes work to make the transition state more favorable?

Answer 4

– it forms a more stable transition state by increasing the number of favorable interactions, thus lowering enthalpy of the transition state which is favorable

—> strong binding of transition state
(deltaHcat < deltaHnon)

–At the same time, it could increase the entropy of transition state to make the ordered fashion slightly more favorable

—> proximity and orientation favor formation of transition state. (deltaScat > deltaSnon)

Question 5

How would enzymes increase the entropy difference between reactants and transitions states to make it more favorable?

Answer 5

– They basically force the reactants by binding them together , reducing their entropy. This reduced entropy in the reactants will make the unfavorable difference in entropy between reactants and transition state less, essentially making it more favorable.

Question 6

True or false
Sometimes alternate pathways to the products can be induced by enzymes to lower activation energy, essentially changing chemical structures of catalyzed transition states

Answer 6

True

Question 7

Who discovered the first model of enzyme substrate binding? How about the second?

Answer 7

Emil Fischer. Daniel Koshland

Question 8

What is the current model of enzyme and substrate?

Answer 8

– enzymes and substrates exist in an induced fit mechanism where both substrate and enzyme change slightly upon interaction to maximize the increase in favorability of the reaction and also explaining the binding specificity of catalysis.

–During the induced binding, the substrate is forced into a configuration that better resembles its transition state

Question 9

T or F, a transition state tightly bound to enzyme that is at lower energy level than transition state in free solution

Answer 9

True; enthalpy is decreasing

–> thus energy involved to adpt transition state is lower than without enzyme

Question 10

True or false

catalytic pocket of enzyme is flexible and can change conformation before binding the substrate.

Answer 10

false. it changes conformation once the substrate binds that allows it to become sterically and chemically complementary to the substrate

Question 11

What are characteristics of the active site in enzymes?

Answer 11

– small area in the enzyme that often is a nonpolar environment. Binding occurs via various multiple weak interactions.

Question 12

What are the type of electrostatic catalysis interactions that can occur at the active site? What do they do?

Answer 12

• • hydrogen bonds
• • electrostatic interactions (ion)
• • hydrophobic effects
• • van der waals.
• • they allow strong bindings yet flexibility to promote the formation of the transition state

Question 13

true or false

metal ions can be in the active cite to participate in catalysis

Answer 13

True

Question 14

Why might a nonpolar environment be favored in the active site of an enzyme?

Answer 14

since water may react to some degree with anything in the active site, having a nonpolar environment prevents undesirable hydrolysis reactions

Question 15

What are the different ways enzymes enhance the rate of reactions? (6)

Answer 15

– preferential binding to transition state through noncovalent interactions (electrostatic catalysis, makes change in enthalpy more favorable)

– Distortion of substrate or active site, promoting change to transition state (induced fit)

– Binding of substrates to optimize proximity and orientation (makes change in entropy more favorable by decreasing initial entropy of reactants)

– altering reaction pathway to include alternative intermediate (occurs often in covalent catalysis)

– other mechanisms: General Acid-Base catalysis (GABC) or Metal Ion catalysis

Question 16

What are non covalent bond stabilized catalysis often called?

Answer 16

electrostatic catalysis

Question 17

true or false

covalent catalysis often includes alternative intermediates?

Answer 17

True

Question 18

How are enthalpic stabilization of enzyme transition states generally occur?

Answer 18

– The transition state charges can be stabilized by residues present in the active site, decreasing enthalpy and making it more favorable

Question 19

How does acid vs base catalysis generally work?

Answer 19

– acid catalysis will donate proton from its active site acidic residue to stabilize the negative charge on the substrate

– base catalysis removes an H+ from the transition state of the substrate that will remove any excess positive charge.

Question 20

What are common GABC catalysis?

Answer 20

H, D, E K, and R

Question 21

Where does cleavage occur on the sites on the lysozyme enzyme. What does it typically cleave?

Answer 21

– between the d and e site on the lysozyme cleavage occurs of sugar residues

– it cleaves the peptidoglycan cell wall of bacteria found in tears saliva and mucus

Question 22

What are the specific sugar residues where the cleavage occurs on a lysozyme?

Answer 22

NAG-NAM

Question 23

What is the general mechanism of the lysozyme action?

Answer 23

– a protonated Glu 35 acts as a proton donor in the first part of the cleavage.

– The Asp52 acts as a stabilizer of positive charge via covalent interactions and the NAG leaves.

– In the secondary step Glu 35 acts as a proton acceptor and Asp52 remains as a charge stabilizer because the second intermediate has a positive charge.
After this final step, the original Nam is regenerated

Question 24

What is different about the E AA in lysozyme compared to a typical E AA?

Answer 24

The pka is elevated to a 6.2 to its ionizable side chain that is typically around 4 (3.9). This allows it to remain protonated in the optimal pH of 5 in the lysozyme.

Question 25

What is the range of optimal activity in the lysozyme? What happens outside of these ranges

Answer 25

– 4-6 with peak at 5.

– pH below 2 the backbone carboxylic acid will be protonated and won’t be able to stabilize the charge?

– at pH above 6, the Glu will be deprotonated and won’t be able to donate a hydrogen that is necessary for the first step

Question 26

What type of bonds do chymotrypsin break?

Answer 26

peptide bonds

Question 27

– What are recognition amino acids for chymotrypsin that allow it to catalyze breakage of peptide bonds?

Answer 27

– bulky groups like phenylalanine, tryptophan, and methionine

Question 28

What is the general mechanism for the cleavage of peptide bonds by chymotrypsin?

Answer 28

– the Serine attacks the desired polypeptide that is non covalently bonded to the active site at its carbonyl on the peptide bond. This is allowed from the H bonding of Serine and Histidine that results in serine becoming more nucleophilic.

– Then the oxyanion hole of the enzyme stabilizes the negative charge that resulted on the peptide as a result of the attack and liberation of electrons onto the carbonyl oxygen. The peptide bond grabs the hydrogen off the histidine and the polypeptide is cleaved. The negative charge on the carbonyl oxygen is removed as the peptide leaves.

– At this step, the N terminal peptide remains bound to the serine and the C terminal peptide is cleaved off

– A water molecule comes in and hydrogen bonds to the histidine, making it more nucleophilic and attacks the carbonyl oxygen on the n terminus peptide.
the serine n terminus peptide attacks the hydrogen on the histidine and this cleaves the final peptide in the original structure

The serine is regenerated in the catalytic site and the process occurs again

Question 29

What is the role of D during the catalytic activity of the chymotrypsin?

Answer 29

– Aspartate group holds His in correct formation for accepting H from Serine

– basically holds His in place so that it can interact w/ Serine

Question 30

True or false

Histidine acts as a base and an acid

Answer 30

True; due to pKa

Question 31

What occurs at the oxyanion hole that makes it favorable for part of the reaction of chymotrypsin cleavage of peptide bonds?

Answer 31

It has the ability to H bond with the charged substrate species through peptide backbones in the enzyme.

Question 32

true or false
oxyanion interactions with the carbonyl oxygen would not be as favorable than with the c-o bond because of shorter bond lengths

Answer 32

false. the c-o has a longer bond length that makes it more favorable H-bond interaction

Question 33

True or false

the transition from the planar to tetrahedral configuration allows for the substrate to interact with the oxyanion hole

Answer 33

True

Question 34

True or false

E35 and H57 act as GABC in serine proteases and lysozyme respectively

Answer 34

false E35 acts in lysozyme and H57 acts in serine proteases.

Question 35

True or false

D52 and S195 act as covalent catalysis in lysozyme and serine proteases respectively

Answer 35

True

Question 36

Which portion of serine proteases acts as an electrostatic stabilizer?

Answer 36

oxyanion hole

Question 37

What does the aspartate do in the catalytic site of chymotrypsin?

Answer 37

– holds His in correct formation for accepting H from Serine

Question 38

How is the oxyanion hole stabilized?

Answer 38

— Hydrogen bonds formed between peptide backbone of enzyme and oxygen atom derive from carbonyl group of activated substrate

Question 39

What are cofactors?

Answer 39

• • small molecules that enzymes need to execute biochemical reactions
• -> apoenzyme + cofactor –> holoenzyme

– either coenzymes or metals

• • can be weakly bounds –> cosubstrates that associate or dissociate during reaction
• • can be tightly bound –> prosthetic groups

Question 40

Example of coenzyme function in catalysis: NAD+

Answer 40

– NAD+ as electron acceptor, oxidizing agent

– accepts hydride ion and convert alcohols to ketones or aldehydes

– NAD+ derived from vitamin Niacin

Question 41

What is the definition of a transition state?

Answer 41

– a particular chemical configuration of the substrate as it evolves toward becoming the product during reaction

– very short lived (transient) chemical state

– usually highest free energy peal of free energy diagram

Question 42

What is electrostatic catalysis?

Answer 42

– enzyme active site stabilizes transition state of reaction by forming electro static interactions w/ substrate

Question 43

What is covalent catalysis ?

Answer 43

– one of 5 ways enzyme will use to catalyze a reaction involves formation of a transient covalent bond between a substrate and a residues in enzyme active site or with a cofactor