5. Enzymology

Question 1

coenzyme

Answer 1

A coenzyme is a substrate for its enzyme, and acts as a transient carrier of specific chemical (functional) groups.

In this role, they are used up during the reaction, but are usually kept at a constant, steady state level in the cell. Coenzymes do not form a permanent part of the enzymes structure.

Most coenzymes are derived from vitamins.

Question 2

cofactors

Answer 2

Cofactors are non-protein groups that bind to the enzyme which are required for activity. Cofactors are not used up during the reaction.

Most cofactors are prosthetic groups or metal ions.

Question 3

What are two important organic molecules that serve as cofactors?

Answer 3

• Flavin (prosthetic group involved in oxidation-reduction reactions)
• Heme (metalloorganic prosthetic group involved in oxygen binding)

Question 4

Even though heme is considered a cofactor, why is hemoglobin not considered an enzyme?

Answer 4

Hemoglobin is not an enzyme, as it does not catalyze any chemical reaction.

Question 5

oxidoreductase

Answer 5

Enzymes that transfer electrons (hydride ions or H atoms).

Question 6

transferases

Answer 6

Enzymes that group transfer reactions.

Question 7

hydrolases

Answer 7

Enzymes that perform hydrolysis reactions (transfer of functional groups to water).

Question 8

lyases

Answer 8

Enzymes that add groups to double bonds, or form double bonds by removal of groups.

Question 9

isomerases

Answer 9

Enzymes that transfer groups within molecules to yield isomeric forms.

Question 10

ligases

Answer 10

Enzymes that form C-C, C-S, C-O, and C-N bonds by condensation reactions coupled to ATP cleavage.

Question 11

enzyme

Answer 11

Enzyme is a catalyst that speeds up chemical reactions so that they occur at a useful rate. As a catalyst, the enzyme is not consumed during the reaction.

Question 12

active site

Answer 12

The distinguishing feature of enzymes as catalysts is that the reaction takes place within the confines of a “pocket” on the enzyme called the active site.

The molecule that binds within the active site is called the substrate.

Question 13

The active site is formed from amino acids that _______________________ and _______________________.

Answer 13

1. Specifically recognize the substrate
2. Accelerate the chemical reaction.

These two features are distinct, i.e. specificity does not guarantee rate acceleration and vice versa.

Question 14

Describe the enzymatic activity of triose phosphate isomerase.

Answer 14

Triose phosphate isomerase in glycolysis catalyzes the conversion of glyceraldehyde-3-phosphate into dihydroxyacetone phosphate via an (unstable) endiol intermediate.

Question 15

Describe what V₀ is and deduce the units of V₀.

Answer 15

The initial velocity, or V₀, is the amount of product produced per unit time at the start of the reaction for a certain concentration. In layman terms, it means the rate when you’ve just combined the enzyme and substrate, and the enzyme is catalyzing the reaction as fast as it can at that particular substrate concentration. During this time, the product concentration is increasingly linearly.

K has units of s-1 whereas [S] has units of M. Thus, V has units of concentration/time, e.g., M/s.

Question 16

What occurs to an enzyme at steady state conditions?

Answer 16

Steady state is when the rate at which the substrate is taken up by the active site is equal to the rate at which product is released by the active site.

In layman terms, the flux in is equal to the flux out.

Question 17

What occurs to the active site (and therefore V) when there are high concentrations of substrate?

Answer 17

At high enough substrate concentrations, all of the active sites are converting S into P. This indicates that the enzyme is saturated, and the rate of reaction will no longer increase when additional substrate is added.

The velocity increases from V₀ to V_max.

Question 18

What occurs to the active site (and therefore V) when there are low concentrations of substrate?

Answer 18

At low substrate concentrations, only a fraction of active sites are converting S into P.

Question 19

What does K_m stand for?

Why is K_m useful?

Answer 19

The substrate concentration that gives you a rate that is halfway to V_max is called the K_m (0.5 x V_max).

K_m is a useful measure of:

• how quickly the reaction rate increases with substrate concentration.
• an enzyme’s affinity for its substrate.

Question 20

Acetylcholine binds to acetylcholinesterase (K_m = 9 x 10^-5), whereas CO₂ binds to carbonic anhydrase (K_m = 1.2 x 10^-2).

Which is the more efficiency or “better” enzyme? Explain why.

Answer 20

Acetylcholinerase.

• Smaller K_m needs less substrate to engage half of the active sites, and is therefore considered to be a better enzyme.
• Larger Km needs more substrate to engage one half of the active sites.

Question 21

What does [S] = K_m indicate?

Answer 21

When [S] = K_m, exactly 50% of active sites are converting S into P.

Question 22

This graph is of an enzyme that is not under steady state conditions. Explain why V₀ for the enzymes are considered the tangents of the curve.

Answer 22

When the substrate concentration is not under steady state conditions, V₀ is the maximum possible reaction velocity or the initial velocity at a particular substrate concentration S, before S starts depleting.

A tangent to each curve taken at time 0 defines the initial velocity, V₀, of each reaction.

Question 23

What is K_cat?

What is the equation for K_cat?

Answer 23

K_cat is the number of substrate molecule each enzyme site converts to product per unit time.

A higher K_cat indicates a more efficient enzyme.

K_{cat =} V_max/[E₀]

Question 24

What is the Michaelis-Menten equation?

Answer 24

Enzymes that display a hyperbolic curve can often be described by an equation relating substrate concentration [S], initial velocity K_m, and maximum velocity V_max, known as the Michaelis-Menten equation.

Question 25

What is t_avg?

Answer 25

t_avg is the average amount of time for the substrate to go through the active site.

Question 26

T/F: K_m should be thought of as the affinity of the substrate for the enzyme’s active site in the steady state.

Answer 26

True. K_m should be thought of as the affinity of the substrate for the enzyme’s active site in the steady state, not as the affinity of the substrate for the active site at equilibrium.

Question 27

What is the significance of k_cat/K_m?

Deduce the units of k_cat/K_m.

Answer 27

k_cat/K_m is an apparent 2nd order rate constant at lowest concentrations of S, and therefore sets the lower limit for any 2nd order elementary rate constant in the reaction mechanism.

kcat/Km has units of M^-1s^-1.

Question 28

Draw a enzyme kinetics function indication K_m and V_max.

Answer 28

Question 29

What is the inverse of the Michaelis-Menton function?

Be able to deduce the equation.

Answer 29

The inverse of the Michaelis Menten functionis called the double-reciprocal or Lineweaver-Burk plot.

Question 30

Draw a Lineweaver-Burk plot, and indicate the following using the Lineweaver-Burk equation:

• x-axis
• y-axis
• slope
• y-intercept

Answer 30

Question 31

What is considered to be “catalytic perfection?”

Answer 31

Catalytic perfection is displayed in enzymes that turnover substrate as fast as the enzyme and substrate can diffuse together.

That is, k_cat/K_m is close to the diffusion-controlled limit, which means as soon as the enzyme and substrate come together, the reaction occurs. Therefore, the rate-limiting step is the diffusion for these enzymes, and not the chemistry.

Question 32

T/F: Catalyst do NOT alter the equilibrium, i.e., they affect ΔG⁰.

Answer 32

False. Catalysts do NOT alter the equilibrium of the reaction, i.e., they do NOT affect ΔG^‘0.

Question 33

How do catalysts speed up chemical reactions?

Be able to draw a kinetic graph that is uncatalyzed and enzyme-catalyzed.

Answer 33

Catalysts speed up chemical reactions by stabilizing the transition state (TS).

The activation energy ΔG‡ is between the two end states substrate and product is lowered by the catalyst, which increases the rate constant.

Question 34

How do enzymes specifically recognize their substrates?

Answer 34

• The enzyme active site is formed from amino acid side chains that form many weak, non-covalent interactions with the substrate (hydrogen bonding, salt bridge, hydrophobic, etc.).
• Further, the shape of the active site accommodates the substrate molecule.

Thus, a potential substrate molecule must have the correct shape to fit into the active site, and importantly, all the H bonding donors/acceptors and charged groups must have precisely positioned partners within the active site.

Question 35

Suppose that an enzyme binds to two substrates. During an enzyme-catalyzed reaction, why does the enzyme re-position the two substrates?

Answer 35

In the enzyme catalyzed reaction, binding of A and B to the enzyme’s active site positions the molecules precisely so that the functional groups of A and B can react. Therefore, the active site greatly increases the probability of a reaction.

In fact, by bringing two molecules together in an enzyme’s active site, we can expect rate enhancements on the order of 10⁵ to 10⁸.

Question 36

What is the “effective concentration” of an enzyme?

Answer 36

The rate enhancement of an enzyme is considered to be the “effective concentration” of an enzyme.

The “effective concentration” is the hypothetical concentration the reactant would have to be, free in solution, to experiences the same collisional frequency.

Question 37

What are the two models describing the interaction between a substrate and an enzyme?

Answer 37

1. Lock and key theory
2. Induced fit theory

Question 38

What is the lock and key hypothesis?

Answer 38

In 1894, Emil Fisher hypothesized that enzymes are structurally complementary t their substrates, and fit together like a lock and key.

This theory was disproved because it could not explain how enzymes accelerate chemical reactions. In fact, if the enzyme was to bind strongly to the substrate, the enzyme-substrate would be lower in energy, and the activation barrier would be much higher than in an uncatalyzed reaction.

Question 39

How does a catalyst decrease the activation barrier?

That is, should an enzyme bind strongly to a substrate or its transition state?

Answer 39

The main function of a catalyst is to decrease the activation barrier. Therefore, enzymes should bind strongly to the transition state, thus offsetting the higher activation energy with the binding energy.

Question 40

Enzymes recognize substrates through multiple weak bonds and interactions. How are these bonds strengthened upon binding of the enzyme and substrate?

Answer 40

Weak interactions between the enzyme and substrate are optimized and strengthened in the transition state (TS), thus compensating the large activation barrier with the binding energy.

Question 41

Draw enzyme kinetic graphs of the following reactions:

1. No enzyme
2. Lock and key theory
3. Induced fit theory

Answer 41

Question 42

What are two experiments that provide evidence for the induced fit theory?

Answer 42

Experiment 1: Some TS analogs bind the enzyme with 10² to 10⁶ fold higher affinity than the substrate.

Experiment 2: Monoclonal antibodies were produced to recognize TS analogs. These mAbs werethen able to catalyze the cleavage of esters with 10² to 10³ fold rate enhancements!

Question 43

What are the three types of enzymatic catalysis?

Answer 43

1. General acid/base catalysis (vs. specific)
2. Metal ion catalysis
3. Covalent catalysis

Question 44

Describe a general base catalysis.

Answer 44

In specific base catalysis, a water hydroxide anion serves as the base.

Question 45

Describe a general acid catalysis.

Answer 45

In specific acid catalysis, a water hydronium ion serves as the acid.

Question 46

What kind of catalysis is taking place in this picture?

Describe what is occuring.

Answer 46

General base catalysis

• Uncatalyzed hydrolysis of an ester involves an unstable charge distribution in the TS.
• This reaction can be catalyzed by the acetate ion, by stabilizing the unfavorable charge distribution that develops on the attacking water molecule in the TS. This is called general-base catalysis. In specific base catalysis, a water hydroxide anion serves as the base.

Question 47

What kind of catalysis is taking place in this picture?

Describe what is occuring.

Answer 47

General acid catalysis

• Uncatalyzed hydrolysis of an acetal (a carbon with two single bonded oxygen atoms) involves an unstable charge distribution in the TS.
• This reaction can be catalyzed by the acetic acid, by stabilizing the unfavorable charge distribution that develops on the oxygen atom in the TS.

Question 48

Describe how the following amino acid side chains can function in general acid/base catalysis:

1. Glu/Asp
2. Lys/Arg
3. Cys
4. His
5. Ser
6. Tyr

Answer 48

Question 49

How do metal ions facilitate catalysis?

Answer 49

Metal ion cofactors can be positioned to stabilize charged TS and/or intermediates.

Question 50

What are the two requirements for covalent catalysis?

Answer 50

1. The catalyst must be a better nucleophile than water
2. The covalent intermediate must be less stable to hydrolysis than the substrate

Nucleophiles are chemical species that donates an electron pair to form a chemical bond in relation to a reaction. In order to be a better nucleophile, the chemical species must have more electrons.

Question 51

How does covalent catalysis affect the enzyme kinetics graph?

Answer 51

In covalent catalysis, the reaction has been “broken up” into pieces, each with a lower ∆G≠ than the uncatalyzed reaction.

Again, the ∆G⁰ has not been affected.

Question 52

Describe why chymotrypsin, an enzyme found in the digestion system, is considered a serine protease.

Answer 52

Chymotrypsin is considered a serine protease, because serine acts as the catalyst at its active site.

The hydroxyl group of the serine amino acid serves as the nucleophile (covalent catalysis), and specifically cleaves peptide bonds that are adjacent to hydrophobic amino acids.

Question 53

Chymotrypsin displays which of the following enzymatic catalysis methods?

1. General acid/base catalysis (vs. specific)
2. Metal ion catalysis
3. Covalent catalysis

Question 54

What are the seven steps of chymotrypsin enzymatic catalysis?

Answer 54

1. E meets S
2. Formation of the ES complex
3. Short-lived intermediate and backbone cleavage
4. Acyl-enzyme intermediate
5. Fate of the intermediate
6. Short-lived intermediate and second product
7. Enzyme-product 2 complex

Question 55

What is the first step of chymotrypsin?

Answer 55

Question 56

What is the second step of chymotrypsin?

Answer 56

Question 57

What is the third step of chymotrypsin?

Answer 57

Question 58

What is the fourth step of chymotrypsin?

Answer 58

Question 59

What is the fifth step of chymotrypsin?

Answer 59

Question 60

What is the sixth step of chymotrypsin?

Answer 60

Question 61

What is the seventh step of chymotrypsin?

Answer 61

Question 62

What are allosteric enzymes?

Answer 62

Certain enzymes function as regulatory checkpoints along various pathways.

Their function is considerably more complex because they contain multiple active sites.

Allosteric enzymes fine-tune a pathway.

Question 63

What is the difference between homoallosteric and heteroallosteric enzymes?

Answer 63

Homoallosteric enzymes are enzymes act as both a regulator and a substrate of a pathway.

Heteroallosteric enzymes are enzymes that regulate a pathway, but are not a substrate of the pathway.

Question 64

What is the typical structure of allosteric enzymes?

How are they activated?

Answer 64

Allosteric enzymes are typically made up of both catalytic domains (C) and regulatory domains (R).

Positive effector molecules activate catalysis at the catalytic domain by binding at the regulatory domain.

Question 65

An allosteric enzyme may also be activated by its own substrate-homotropic cooperativity.

Describe the kinetic curve of such an enzyme.

Answer 65

A sigmoidal velocity curve is the result.

Note the nonlinear velocity curve at low S values.

Question 66

A heterotropic positive or negative modulator can significantly influence an allosteric enzyme.

Describe the kinetic curve of such an enzyme. Include positive and negative modulation.

Answer 66

Note how much the Km can change by simply adding a positive or negative modulator.

Question 67

Why is allosteric modulation so important?

Answer 67

Small changes in effector or substrate concentrations can result in large changes in enzyme activity. That is, conversion of S to P can be regulated like a switch.