Enzyme Kinetics Pt. 1

Question 1

Functions of Enzymes

Answer 1

• higher reaction rates
• milder reaction conditions
• greater specificity
• capacity for control

Question 2

Active Site

Answer 2

• region of an enzyme binding the substrate
• usually clefts/crevices in protein
• amino acids and cofactors are held in precise arrangement with respect to substrate structure
• amino acids in active site define specificity

Question 3

Lock and Key Model

Answer 3

• ligand binding site is rigid and complementary to ligand shape

Question 4

Induced Fit Model

Answer 4

• flexible interaction between ligand and active site induced conformational change (adaptation leads to perfect fit)
• enhances reaction mechanism and enables more specific fit
• stabilises transition state

Question 5

Substrate Specificity

Answer 5

• geometric specificity = active site complementary to structure of substrate
• electronic specificity = amino acids in active site interact with substrate so that only the substrate can favorably bind to form the enzyme substrate complex

Question 6

Geometric Specificity

Answer 6

• selective about chemical groups of the substrate
• more stringent requirements
• varying degrees of geometric specificity
• few enzymes are absolutely specific for one substrate
• some work on a group of related molecules

Question 7

Electronic Specificity

Answer 7

• opposite charges attract

- hydrophobic and hydrophilic attractions

Question 8

Stereospecificity

Answer 8

• enzymes are highly specific in binding chiral substrates and in catalysing their reactions
• discriminates between enantiomeric substrates
• stereospecificity is due to enzymes active site
• absolute stereospecificity

Question 9

CIP Rules for Chirality

Answer 9

1. orient lowest priority group (lowest atomic number) facing away
2. number 3 groups by increasing priority
3. determine rotation of groups in decreasing priority
   - clockwise: R
   - counterclockwise: S

Question 10

Oxidoreductases

Answer 10

redox reactions, catalyse H+ and O atom transfer

Question 11

Transferase

Answer 11

transfer of functions groups to other compounds

Question 12

Hydrolyses

Answer 12

catalyse the hydrolytic cleavage of C-O, C-N, C-C and some other bonds like anhydride and phosphoric groups

Question 13

Lyases

Answer 13

cleave C-C, C-O, C-N bonds by elimination, cleaving double bonds or rings or adding groups to double bonds

Question 14

Isomerases

Answer 14

catalyses geometric/structural changes in a molecule (isomerisation)

Question 15

Ligases

Answer 15

catalyses joining of 2 molecules couple with hydrolysis of diphosphate bond

Question 16

Cofactors

Answer 16

• some enzymes require small molecules during catalysis
• cofactors are metal ions/organic molecules
• organic molecules: coenzymes
• cosubstrates: behaves like substrate and leaves after the reaction
• prosthetic group: tightly bound and remains after reaction

Question 17

Vitamins as Coenzyme Precursors

Answer 17

• many organisms are unable to synthesise parts of essential cofactors
• substrates are present in organisms diet
• vitamins that are precursors are water soluble

Question 18

Chemical Kinetics

Answer 18

• study of reaction rates
• binding affinities
• enzyme mechanism
• influence of conditions on rate

Question 19

Rate of Reaction

Answer 19

Proportional to frequency of reaction molecule collision

rate = k [A]^a[B]^b

Question 20

Reaction Order

Answer 20

Power dependance of rate on concentration of species involved
Sum of powers in rate equation
1st order: one molecular changes to each other
2nd order: two molecules reacting

Question 21

1st order reactions

Answer 21

• direct proportionality between rate and concentration

- natural log of concentration is directly proportional to time (straight line down)

Question 22

2nd order reactions

Answer 22

• rate proportional to the square of concentration of one reactant

Question 23

Half-Life

Answer 23

time for substrates concentration to fall to 1/2 their initial value

• 1st order half life = ln2/k
• 2nd order half life = 1/k[A]o
• independent of inital substrate concentration

Question 24

Rate Determining Step

Answer 24

• slowest intermediate step in the reaction that determines the rate of the total reaction
• RDS determines the overall rate equation

Question 25

Collision Theory

Answer 25

1. properly oriented

2. sufficiently energetic collision

Question 26

Boltzmann Maxwell Graph

Answer 26

Average kinetic energy of molecules is proportional to absolute temperature
Higher temperature means more molecules have KE greater than or equal to Ea

Question 27

Arrhenius Equation

Answer 27

k = Ae^-Ea/RT
A = pre exponential/frequency factor is fraction of sufficiently energetic collisions that lead to reaction
Ea (kj/mol) = minimum amount of energy needed for a reaction to occur upon collision

Question 28

Arrhenius Graph

Answer 28

ln k = ln A - Ea/RT
slope = -Ea/R
y intercept = ln A
high activation energy = steeper slope at a rate sensitive to temperature

Question 29

Transition State

Answer 29

Enzymes enhance the rate of reaction by stabilising the TS of the reaction. Active site is more favorable to the transition state so the substrate is forced to adopt this.
TS is a metastable compound sharing features of both the reactants and products
Some reactant bonds are being broken while product bonds are being formed
NOT the same as the intermediate state and not able to be isolated or purified

Question 30

Erying Equation

Answer 30

Transition state is in rapid equilibrium with reactants and eq constant K (dagger)

Question 31

Enzymatic Rate of Reaction

Answer 31

Rate lowered by lowering the activation barrier between reactants/transition state, increasing fraction of reactants able to achieve the TS
Kinetic barrier lowered to the same extent for forward and reverse reactions as ground state of free substrates/products remain the same
Catalyst doesn’t affect position of equilibrium

Question 32

Reducing the Activation Barrier

Answer 32

By forming H bonds or hydrophobic contacts, enzymes are increase rate by 10

• stabilising transition state
• destabilising ground state of enzyme bound substrates and products
• introduces how reaction pathway with a different TS with lower free energy
• initial interaction is noncovalent and uses H bonds, hydrophobic interactions, VDW forces etc
• favorable interaction between E and S results in a favorable intrinsic binding energy
• there is a loss of entropy as two species become one, and the substrate is more ordered

Question 33

Transition State Analogues

Answer 33

• have high affinity for the active site, higher than the substrate as they are an exact fit
• act as inhibitors
• used to elucidate mechanism of reaction

Question 34

Catalytic Antibodies

Answer 34

• antibodies tightly bind their antigen but without altering its chemical nature
• theoretically, if an antibody binds a TS molecule it may be expected to catalyse a corresponding chemical reaction by forcing substrates into the TS geometry

Question 35

Hammond Postulate

Answer 35

If 2 states (eg. TS and unstable intermediate) occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganisation of the molecular structures
Species that are sequential on the reaction coorinate + similar in energy are similar in structure
- Exothermic reaction: TS resembles reactants
- Endothermic reaction: TS resembles products

Question 36

Desolvation

Answer 36

• substrate binding to enzyme causes surrounding water to be replaced by enzyme
• make substrate more reactive by destabilizing substrate charge
• lowers substrate entropy

Question 37

Strain & Distortion

Answer 37

• active site conformational change to fit TS

- this substrate distortion increases reactivity

Question 38

Proximity

Answer 38

• rate increase due to 2 reactants being brought together
• enzymes do this by providing a docking site + micro environment allowing proper substrate orientation for reaction
• contributes to the loss of substrate freedom of movement and loss of entropy
• increase chances of reaction by increasing effective substrate concentration

Question 39

Km

Answer 39

• Michaelis constant
• higher value means there is lower affinity of the enzyme for the substrate
• need more substrate to achieve 1/2 the Vmax value