L19: Fundamental Enzyme Kinetics

Question 1

[ES]

Answer 1

Determination of [ES] is difficult or impossible

Fairly constant over much of reaction i.e. reaction is in steady state

Question 2

Vmax

Answer 2

Velocity of reaction at saturating substrate

Question 3

Km

Answer 3

Substrate required to give velocity of vmax/2

True measure of affinity of enzyme for its substrate

Question 4

Reversible enzyme inhibition: competitive inhibition

Answer 4

Inhibitor competes with substrate for same binding site on enzyme

At high substrate conc -> substrate outcompetes inhibitor -> no inhibition

-> Vmax achieved at infinite substrate conc, is not affected by inhibitor

More substrate required to achieve half Vmax -> apprent Km larger than presence of inhibitor

Lineweaver-burk: intersecting lines, intersection on 1/v axis

Question 5

Uncompetitive inhibition

Answer 5

Inhibitor can only bind enzyme-substrate complex and not to the free enzyme

Binding of substrate to enzyme to form ES -> binding site for inhibitor at active site or induces conformational change in enzyme -> creates remote binding site for inhibitor

Enzyme-substrate-inhibitor complex unable to catalyse formation of product from substrate

Even at high substrate conc inhibition occurs -> Vmax reduced

As inhibitor binds to ES-> equilibrium between E & ES in favour of ES -> Km (Kmapp) is smaller in presence of inhibitor

Lineweaver-burk: parallel lines

Question 6

Mixed or noncompetitive

Answer 6

Inhibitor binds to enzyme and ES

Inhibitor may bind more tightly to E than ES -> KI > KIS or the opp. If bound equally well (KI=KIS) -> noncompetitive

Vmax decreased in presence of inhibitor (uncompetitive component)

Competitive dominates (KI Km will increase in presence of inhibitor

Uncompetitive dominates (KIS Kmapp decreased in presence of inhibitor

If neither dominate (KI=KIS) -> Kmapp unaffected by the presence of inhibitor -> =Km

Lineweaver-burk: intersecting lines which intersect to left of 1/v axis

Question 7

Irreversible enzymes inhibition

Answer 7

Involves irreversible formation of covalent bond between inhibitor and enzyme, usually at active site

Usually time dependent and loss of enzyme activity can be monitored

Question 8

Example of irreversible enzyme inhibition

Answer 8

Penicillin: irreversible inhibitor of glycopeptidyl transpeptidase (essential for bacterial cell wall synthesis). A suicide substrate (similar structure to normal substrate -> binds with high affinity to enzyme

Ser-OH in enzyme attacks carbonyl carbon in penicillin as if it were a peptide bond in protein

Once bond between Ser and carbon formed -> cant be broken

Question 9

Allosteric enzyme

Answer 9

Important regulators of metabolism

Regulated by its substrate and activators and inhibitors

Multi-subunit proteins (have quarternary structure). Subunits exist in 2 different structural conformations: T state and R state which are in equilibrium

Question 10

T state

Answer 10

Binds substrate with low affinity

Absence of substrate -> most of subunits in T state -> Km for substrate is high (low affinity)

Allosteric inhibitor -> shifts equilibrium in favour of T -> plot of v vs [subs] even more sigmoidal. At most [subs], v will be lower than in absence of inhibitor

Question 11

R state

Answer 11

Substrate increase -> more substrate binds to enzyme -> equilibrium between T & R states shifted in favour of R state -> Km decreases

At saturating substrate conc all enzyme in R state

Allosteric activator -> shift equilibrium in favour of R

If sufficient activator present -> all of enzyme will be in R state -> plot of v vs [subs] in normal hyperbolic form (Km remains constant and at most [subs], v will be higher than in absence of activator)

Question 12

Michaelis-menten equation

Answer 12

V= Vmax[S]/Km + [S]

Vmax/(1+Km/[S])