Lecture 2 - Enzyme Kinetics II

Question 1

simple kinetics I:

Answer 1

one simple reaction (A → B) which is referred to be “first order” - where A disappears as B is produced

Question 2

simple kinetics II:

Answer 2

•Rate constant (k) of reaction does not change
•But with time, the velocity of the reaction reduces as the reactant A is used up and B becomes a substrate for the reverse reaction

the reaction reaches equilibrium when rates for forward and reverse reactions are EQUAL and the overall rate for the reaction is zero

Question 3

simple kinetics II:

Answer 3

two reactants - we assume that the rate of forward reaction is linearly proportional to the concentrations of A and B, and the reverse reaction is linearly proportional to the concentration of C (A + B <—> C)

this is a second order reaction as its based on the concentration of two reactants

Question 4

equilibrium:

Answer 4

equilibrium is reached when the net rate of reaction is zero. equilibrium constant tells us the extent of the reaction, NOT its speed.

Question 5

basic problem of enzyme kinetics:

Answer 5

the reaction rates saturate

Question 6

enzyme catalysis:

Answer 6

• overall this is an irreversible reaction (could be reversible)
• enzyme binds to substrate to form intermediates
• intermediates enzyme substrate then decomposes irreversibly to yield the product and the enzyme is not changed

Question 7

assumption I:

Answer 7

if we assume [substrate] is greater than the [enzyme], then the enzyme will mostly be present as the intermediate enzyme substrate and [ES] will remain constant

Question 8

rates of enzyme reactions:

Answer 8

•The initial rate increases with enzyme concentration [E]
•The initial rate increase with substrate concentration [S] but levels off (saturates) to a maximum value (Vmax)

Question 9

assumption II:

Answer 9

steady state assumption - the rate of ES formation = rate of ES dissociation

(steady state is reached when the concentration of intermediate (ES) is constant)

Question 10

the rate ‘v’ is given by the rate at which ES breaks down:

Answer 10

v = k2 [ES]

Question 11

initial rate depends on:

Answer 11

enzyme and substrate concentrations

Question 12

KM:

Answer 12

is the michael is constant (units of concentration)

Question 13

maximum rate, V-max is:

Answer 13

V-max = k2 [Eo]

Question 14

V-max:

Answer 14

maximum catalytic rate at full saturation

Question 15

how can values for KM & V-max be determined?

Answer 15

values of V-max and KM can be determined by least-squares fitting of initial rate data to this equation

Question 16

Michaelis-Menten equation:

Answer 16

V-max [S] / KM + [S]

Question 17

Lineweaver-Burke Plot:

Answer 17

•More accurate estimation of Vmax than the saturation plot

•Vmax from the y intercept (1/Y)

•km from the negative x axis (-1/X) or from slope

Question 18

disadvantage of Lineweaver-Burke Plots:

Answer 18

Disadvantage - the plot is dominated by points at low
[S] i.e. high 1/[S]

At low [S], rate measurements may be less accurate

Question 19

Ki:

Answer 19

dissociation constant for inhibitor

Question 20

how does KM & V-max change when in the presence of competitive inhibitors?

Answer 20

V-max remains the same but KM increases

Question 21

how can inhibition be overcame with competitive-inhibitors?

Answer 21

through increasing [S]

Question 22

example of non-competitive inhibitor:

Answer 22

Pyruvate Kinase catalyses final step of Glycolysis

Question 23

kinetics of non-competitive inhibitor:

Answer 23

Increasing [S] cannot overcome inhibition

Less E available, V-max is lower, KM remains the same for available E

Question 24

how does V-max & KM change with non-competitive plots:

Answer 24

V-max is decreased and KM is unchanged

Question 25

other plots:

Answer 25

Eadie Hofstee Plot v against v/[S] - linear plot
GRADIENT is KM, Intercept on y-axis is Vmax

Hanes or Dixon Plot: •Plot [S]/v against [S]. Linear plot
•Gradient = 1/Vmax •Intercept on y-axis = KM/Vmax

Question 26

example competitive inhibitors:

Answer 26

substrate “mimic”

A competitive inhibitor of dihydrofolate reductase - Used to treat cancer

Substrate for dihydrofolate reductase role in purine & pyrimidine biosynthesis.

Question 27

The Active Site of an Enzyme:

Answer 27

1.The active site is the region that binds the substrates (&
cofactors if any)

2.It contains the residues that directly participate in the making & breaking of bonds (these residues are called catalytic groups)

3.The interaction of the enzyme and substrate at the active site promotes the formation of the transition state

4. The active site is the region that most directly lowers G‡ of the reaction - resulting in rate enhancement of the reaction

Question 28

common features of active sites:

Answer 28

1.The active site is a 3-dimensional cleft formed by groups that come from different parts of the amino acid sequence

2. The active site takes up a relatively small part of the total volume of an enzyme. Why are enzymes so big? Scaffolding, regulatory sites, interaction sites for other proteins, & channels

3.Active sites are clefts or crevices - exclude H2O

4.Substrates are bound to enzymes by multiple weak attractions (electrostatic interactions, hydrogen bonds, Van der Waals forces, & hydrophobic interactions can be Involved)

5.The specificity of binding depends on the precisely defined arrangement of atoms at the active site

Question 29

specificity:

Answer 29

• the specificity of an enzyme is due to the precise interaction of substrate with the enzyme
• this is a result of the intricate three-dimensional structure of the enzyme protein

Question 30

enzyme inhibition by DIPF:

Answer 30

DIPF (nerve gas) reacts with Ser in acetylcholinesterase, affecting nerve function