Chemical Catalysis and Binding

Question 1

catalyst

Answer 1

a catalyst alters the rate of reaction by many orders of magnitude, and at the end of the reaction cycle has returned to its original chemical and electronic state.

A catalyst does not alter the equilibrium of a reaction, just its speed.

Question 2

what is catalytic power? what is the common catalytic power of enzymes?

Answer 2

Catalytic power is defined as the rate of the catalyzed reaction relative to the rate of the uncatalyzed reaction.

Cat power = rate of catalyzed rxn/rate normal rxn

Catalytic power of enzymes is usually in the range of 10⁶ to 10¹⁴

Question 3

relate the equilibrium constant to free energy

Answer 3

∆G’° = -RT ln K’_eq

The larger K’_eq, the more negative ∆G’°

Question 4

ligand

Answer 4

a molecule that binds to a protein through many non-covalent interactions

Substrates bind to Enzymes and are converted to Products

Question 5

energetically, what occurs during substrate binding to an enzyme?

Answer 5

Binding of a ligand to a protein involves the formation of many non-covalent interactions. To achieve this there is an energy penalty in removing bound water molecules from the binding site, and the ligand (desolvation).

Question 6

desolvation

Answer 6

Binding of a ligand to a protein involves the formation of many non-covalent interactions. To achieve this there is an energy penalty in removing bound water molecules from the binding site, and the ligand (desolvation).

Question 7

what is the first step in catalysis?

Answer 7

binding of the ligand

Question 8

what is binding of a ligand to a protein dependent upon?

Answer 8

Binding is a dynamic process, and is concentration dependent, as there is a greater likelihood that two molecules will collide in a productive manner enabling maximal binding interactions to form.

Question 9

what types of bonding play a role in binding a ligand?

Answer 9

many different non-covalent interactions

Question 10

“lock-and-key” hypothesis

Answer 10

The complementarity of the binding surfaces leads to enzymes, receptors and antibodies being very specific for certain ligands

Question 11

“induced fit” hypothesis

Answer 11

Upon initial binding enzymes undergo some level of conformational change to optimize the interactions

Question 12

combine the “lock-and-key” and “induced fit” models

Answer 12

Not all experimental evidence can be adequately explained by using the so-called rigid enzyme model assumed by the “lock and key” theory. For this reason, a modification called the induced-fit theory has been proposed.

The induced-fit theory assumes that the substrate plays a role in determining the final shape of the enzyme and that the enzyme is partially flexible. This explains why certain compounds can bind to the enzyme but do not react because the enzyme has been distorted too much. Other molecules may be too small to induce the proper alignment and therefore cannot react. Only the proper substrate is capable of inducing the proper alignment of the active site.

Question 13

explain stereospecificity in enzyme binding

Answer 13

The asymmetric nature of substrate binding sites, due to inherent chirality (L-amino acids) and complexity of protein enzymes, means they are very stereospecific in the chemistry they catalyze.

Question 14

explain an enzyme’s effect on the equilibrium of a reaction it catalyzes

Answer 14

Enzymes speed up both the forward and backward
reactions equally. Therefore, enzymes do not change the position of equilibirum (i.e. the relative concentrations of X or Y)

(shown: same relative concentration, varying reaction rate)

Question 15

describe the dissociation constant ( K_d )

Answer 15

In chemistry, biochemistry, and pharmacology, a dissociation constant ( K_d ) is a specific type of equilibrium constant that measures the propensity of a larger object to separate (dissociate) reversibly into smaller components, as when a complex falls apart into its component molecules, or when a salt splits up into its component ions. The dissociation constant is the inverse of the association constant. (K_d = 1/K_a) In the special case of salts, the dissociation constant can also be called an ionization constant.

(shown: a two component system where P is a protein and L is a ligand)

Question 16

discribe the association constant ( K_a )

Answer 16

Binding is known as the equilibrium constant of association or association constant, K_a.

The units of K_a are M^-1.

These are the concentrations at equilibrium.

(shown: a two component system where P is a protein and L is a ligand)

Question 17

describe K_d and K_a at equilibrium

Answer 17

K_{<span>a</span>} / K_{<span>d</span>} = K_eq

Question 18

what assumptions are made when modelling enzyme kinematics?

Answer 18

1. Solutions are behaving ideally, no external factors in the reaction

E + S ⇌ ES ⇌ E + P

Rate₁ = K₁ [E][S]

Rate₂ = K₂ [ES]

2. Enzyme constants are actually constant

[E] remains constant and is not subject to protein synthesis or degredation

K is actually constant and does not change due to some environmental factor

3. S ⟶ P without enzymes is negligible

Question 19

V_max

Answer 19

V_max = k₂[E]_t

The rate of reaction when enzyme is saturated with substrate. Because the enzyme is saturated, adding substrate will not increase the rate of reaction and the reaction is said to be proceeding at its maxium rate (velocity)

Question 20

K_m

Answer 20

K_m is the concentration of substrate which permits the enzyme to achieve half V_max. An enzyme with a high K_m has a low affinity for its substrate, and requires a greater concentration of substrate to achieve V_max.

V_o = V_max/2

Units of M

Question 21

Michaelis-Menten equation

Answer 21

V_o = V_max[S]/K_m+[S]

Question 22

K_cat

Answer 22

K_cat = V_max/[E]_t

The enzyme’s turnover number, the number of substrates and enzyme can turn to product at its maximum speed, in units of s^-1

Question 23

catalytic efficiency

Answer 23

K_cat/K_m

Higher efficiency means faster substrate turnover

Question 24

Describe the Lineweaver-Burke plot in terms of y=mx+b

Answer 24

y = 1/V_o

m = K_m/V_max

x = 1/[S]

b = 1/V_max

Question 25

What is the formula for the Lineweaver-Burke plot?

Answer 25

1/V_o= (K_m/V_max[S])+1/V_max

Question 26

What are the four types of enzyme inhibitors?

Answer 26

1. Competitive inhibitor

competes with substrate for enzyme binding and prevents formation of the ES complex

2. Uncompetitive inhibitor

binds with the substrate at an allosteric site and prevents the enzyme from turning over substrate to product

3. Noncompetitive inhibitor

binds to the enzyme and ES complex equally well and decreases V_max due to less enzyme available, does not effect K_m

4. Mixed inhibitor

binds to the enzyme and the ES complex BUT has different affinities for each. If the inhibitor preferentially binds to the enzyme, K_m increases due to less free enzyme, if the inhibitor binds preferentially to the ES complex, K_m and V_max decrease.

Question 27

General Acid Catalysis

Answer 27

Transition state is stabilised by proton transfer from an acid to the substrate

Question 28

General base catalysis

Answer 28

Transition state is lowered by proton transfer from substrate to a base

Question 29

Concerted general acid-base reaction

Answer 29

Catalyzed reaction involves both general acid catalysis and general base catalysis simultaneously

Question 30

What amino acid side chains act as general acid-base catalysts?

Answer 30

Asp, Glu, His, Cys, Arg, Tyr, Lys

Question 31

specific acid-base catalysis

Answer 31

In specific acid catalysis taking place in solvent S, the reaction rate is proportional to the concentration of the protonated solvent molecules SH⁺. The acid catalyst itself (AH) only contributes to the rate acceleration by shifting the chemical equilibrium between solvent S and AH in favor of the SH⁺ species.

S + AH → SH⁺ + A⁻

Water mediated catalysis

Question 32

What is the difference between general and specific acid-base catalysis?

Answer 32

Specific acid-base catalysis is solvent mediated; general acid-base catalysis includes all acids and bases in the reaction itself, and thus all contribute to the reaction rate.

Longer explanation:

In specific acid catalysis taking place in solvent S, the reaction rate is proportional to the concentration of the protonated solvent molecules SH⁺. The acid catalyst itself (AH) only contributes to the rate acceleration by shifting the chemical equilibrium between solvent S and AH in favor of the SH⁺ species.

S + AH → SH⁺ + A⁻

In general acid catalysis all species capable of donating protons contribute to reaction rate acceleration. The strongest acids are most effective. Reactions in which proton transfer is rate-determining exhibit general acid catalysis