lecture 44 - enzyme kinetics

Question 1

what do enzyme kinetics help us understand? (2)

Answer 1

• the mechanisms of catalysis (e.g. metabolic processes)
• e.g. can compare two different substrates with one enzyme or two different enzymes with one substrate
• can assess and compare inhibitors (important for drug design)

Question 2

what is the equilibrium constant for the conversion of substrate into product via enzymes? is it directly affected by enzymes?

Answer 2

• Keq = [P] / [S]

- not affected by enzyme

Question 3

if we were to plot product formation over time for a susbtrate, what is the initial slope?

Answer 3

• the initial slope (given by d[P]/dt at t=0) is known as the initial velocity
• here we assume that [P] = 0, meaning there is no reverse reaction to interfere with product formation

Question 4

if we were to plot the initial velocity vs substrate concentration, what graph would this resemble?

Answer 4

theta vs [L] binding curve

Question 5

what is the Vmax?

Answer 5

the max initial velocity for a substrate (umol/s)

Question 6

what is Michaelis constant (Km) ?

Answer 6

• related to substrate affinity (similar to Kd for ligand binding)
• it is the [S] at Vmax/2

Question 7

is initial velocity for a substrate sensitive to the concentraion of substrate?
why or why not?
what does this mean for the initial velocity at different substrate concentrations?

Answer 7

• yes Vo is sensitive to [S]
• because the enzyme is highly saturated with substrate
• at low [S], Vo is proportional to [S]
• at high [S], Vo = Vmax

Question 8

what are the 2 assumptions for michaelis constant?

Answer 8

(1) assume fast equilibrium for E + S = ES

(2) assume equilibrium is slower than 1 for ES = E+ P

Question 9

what is the rate limiting step for the enzymatic conversion of substrate to product? why?

Answer 9

• ES = E + P

- Vo is proportional to [ES]

Question 10

what is Michaelis Menten equation?

Answer 10

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

Question 11

for Michaelis Menten equation, what is the relationship btwn Vo and [S]?

Answer 11

rectangular hyperbolic realtionship

Question 12

what are the assumptions for Michaelis Menten equation? (3)

Answer 12

(1) initial rate, [P] = 0, means theres no reverse reaction
- i.e. E + S = ES = E +P, with only k1, k-1 and k2

(2) fast equilibrium for E + S = ES
-simple binding process ([S]&raquo_space;> [E])
-d[ES]/dt = 0 (steady-state assumption)
-Vo = Vmax[S] / Km + [S] = kcat [E]T [S] / Km + [S]
where kcat = rate limiting constant = k2

(3) Km = k2 + k-1 / k1
- k2 + k-1 = rate constant for the breakdown of ES
- k1 = rate constant for the formation of ES
- if k2

Question 13

for enzyme kinetic measurements, is a simple Vo vs [S] a good source? why or why not

Answer 13

• no
• at low [S] our instrament may not be sensitive enough
• at high [S] solubility may interfere with results

Question 14

what do we do to increase the accuracy of enzyme kinetic measurements?

Answer 14

-replot as a linearized from: (Lineweaver-Bark Plot)
1/Vo = Km/Vmax * 1/[S] + [S]/Vmax[S], cancel out [S] to get:
1/Vo = Km/Vmax * 1/[S] + 1/Vmax
-plot 1/Vo vs 1/[S]
-slope = Km/Vmax
-y-intercept = 1/Vmax
-NOTE: Vmax and Km come from extrapolation of the data

Question 15

what can knowledge of the Km and Vmax of an enzyme with a variety of substrates provide?

Answer 15

information about how the substrate binds to the enzyme and how the enzyme catalyzes the reaction

Question 16

what does Kms1 < Kms2 and Vmaxs1 = Vmaxs2 suggest?

Answer 16

suggests substrate 1 has a higher affinity than substrate 2

Question 17

what does Kms1 ~ Kms2 and Vmaxs1 > Vmaxs2 suggest?

Answer 17

suggests that substrate 1 has a lower activation energy (lower ΔG transition state) than substrate 2
-i.e. the enzyme is better able to stabilize transition state 1 than transition state 2