06. Enzyme Kinetics

Question 1

What is meant by the velocity of a reaction? Give the 2 definitions

Answer 1

It is the rate of reaction
- Rate of product formation over time
- Rate of reactant depletion over time

Question 2

For a simple reaction :

A –> B,
what is the rate equation?

Answer 2

Rate = k[A], where k is a rate constant

Question 3

Sketch the curve of velocity against substrate concentration for an enzyme- catalysed reaction

Answer 3

See notes

Question 4

Why does the velocity-[substrate] curve plateau at high [S] ?

Answer 4

At high [S], enzymes are fully saturated and maximum activity is reached.

Question 5

what 2 steps is an enzyme-catalysed reaction split into?

Answer 5

1. Binding step (binding of S to E to form ES complex)
2. Catalytic step (conversion to products)

Question 6

Write an equation(w the reversible arrows etc) to display the the kinetics of an enzyme-catalysed reaction.

Answer 6

E + P ⇌ ES –> P

*refer to notes for rate constant

Question 7

The Michaelis Menten equation describes the shape of the graph of velocity against [S] for enzyme-catalysed reaction. What is the Michaelis-Menten equation?

Answer 7

See notes

V0 = (V_max[S]) / (Km+[S])

Question 8

What are the 5 assumptions made in formation of Michaelis Menten equation?

Note : (not impt, assumptions made when deriving eqn and we dont need to know how to derve)

Answer 8

1. [S]&raquo_space;> [E], binding step is fast and thus catalytic step is rate limiting
2. Since initial velocity is measured at early time points, [P] ≈ 0 and so E + P –> ES is negligible (rate constant K_-2 is negligible)
3. Reaction is in equilibrium and in steady state, [ES] is constant.
   - note: reaction can be in unsteady state though in equilibrium
4. Enzyme only exists as free enzyme and ES complex
5. When [ES] = [E_T], it means all total available enzymes has binded to substrate to form ES complex, thus V0 = Vmax.

Question 9

What is the simplest rate equation for velocity of enzyme-catalysed reaction?

Answer 9

V0 = k2[ES]

Question 10

In what way is the plot of V0 against [S] ( Michaelis Menten ) useful?

Answer 10

Plots of V0 versus [S] reveals the relation between Vmax, Km and
reaction orders.

Question 11

Using Michaelis Menten equation, what is V0 of the reaction and order of reaction (if applicable) if :

1. [S] «< Km?
2. [S] = Km?
3. [S]&raquo_space;> Km?

Answer 11

1. V0 = (Vmax/Km) x [S], order of reaction = 1
   (V0 increases linearly with [S])
2. V0 = 1/2 Vmax
3. V0 = Vmax, order of reaction = 0
   (V0 is not dependent on [S])

• can refer to notes

Question 12

What is Km?

Answer 12

1. the substrate concentration where 1/2(V_max) is achieved

Question 13

What does Km tell us?
What does high/low Km mean?

What is the units of Km?

Answer 13

It tells us about the binding step, where Km signifies the affinity of the substrate to enzyme to form ES complex.

High Km : low affinity to form ES (cuz need more substrate to reach 1/2 V_max)

Low Km: high affinity of E to S to form ES

Units of Km: depends on qn but M/µM etc

Question 14

What is Kcat?

Answer 14

Kcat is the turnover number, which describes the amount of substrate converted into product per enzyme molecule per unit time.

Question 15

What is the equation for Kcat, and what is it units?

Base units:
concentration : mM
time : s

Answer 15

Kcat = Vmax / [E_T]

V_max units : mM/s
[E_T] units: mM

Kcat units: S^-1

Question 16

What is the formula for catalytic efficiency and what does it tell us about the rate of reaction?

Answer 16

catalytic efficiency = Kcat/Km
- Need to consider both how fast 1 enzyme convert substrate to product AND affinity.

Higher catalytic efficiency = higher v
lower catalytic efficiency = lower v

Question 17

What are the units for catalytic efficiency?

base units:
- concentration : Molar, mM
- time : s

Answer 17

Kcat : s^-1
Km : mM

units: (mM)^-1 . s^-1

Question 18

What does the inhibitor do in competitive inhibition?
How does Km and V_max change?

Answer 18

The inhibitor binds to active site it of free enzyme and blocks it, disrupting the binding of substrate to enzyme.

• V_max unchanged as at high [S] ([S]»>[E]) , maximal velocity can still be reached since substrate outcompete inhibitor
• Km increases, lower affinity of S binding to E (usually inhibitor have a higher affinity to bind to E than S)

Question 19

What is the Michaelis Menten equation for competitive inhibition?

Answer 19

See notes for btr view

V0 = (V_max[S]) / (αKm + [S])

• multiply by α as competitive inhibition increases Km

Question 20

for competitive inhibition, the Km is given by αKm. What is the formula for α?

Answer 20

α = 1+ [I]/Ki

Ki is the dissociation constant of the EI complex. Ki = [E][I]/[EI]

Question 21

What does the inhibitor do in uncompetitive inhibition?

How does Km and V_max change?

Answer 21

The inhibitor binds to an allosteric site on the ES complex, and alters the conformation/shape of the enzyme. This disrupts catalytic activity and substrate cannot be turned into products.

• V_max decreases as ES-I complex cannot release product while the inhibitor is bound (affecting catalysis)
• Km decreases as the substrate is stuck into enzyme and cannot be converted into products –>increased affinity for substrate to bind to enzyme

Question 22

What is the Michaelis Menten equation for uncompetitive inhibition?

Answer 22

See notes for btr view

V0 = ((V_max/α’)[S]) / ((Km/α’) + [S])

• divide Vmax and Km by α’ as uncompetitive inhibition decreases both Km and v_max

Question 23

In is the Michaelis Menten equation for uncompetitive inhibition, there is introduction of the constant α’. What is the equation for α’?

Answer 23

α’ = 1+ ([I]/[Ki’])

Where Ki’ is the dissociation constant for ESI complex.

Ki’ = ([ES][I])/[ESI]

Question 24

What does the inhibitor do in non-competitive inhibition?

In the 2 kinds of non-competitive inhibition, how does Vmax and Km change?

Answer 24

the inhibitor binds to the active site of free enzyme AND to the ES complex.

1. Pure non-competitive inhibition
   - Vmax decreases, Km unchanged as rate of binding to free enzyme = rate to binding of ES complex.
2. Mixed non-competitive inhibition
   V_max decreases (always).

• When rate of binding to free enzyme > rate of binding to ES: Km increases.
• When rate of binding to ES > rate of binding to free enzyme : Km decreases.

Question 25

What is the Michaelis Menten equation for (all types of) non-competitive inhibition?

Answer 25

See notes for btr view

V0 = ((V_max/α’)[S]) / (Km(α/α’)) + [S])

• divide Vmax and Km by α’ as uncompetitive inhibition decreases both Km and v_max

Question 26

The Line Weaver Burk plot is the double reciprocal graph of Michalis Menten plot of V0 against [S]. Why is the Line Weaver Burk plot preferred when trying to find out the Km of a substrate/Vmax experimentally?

Answer 26

Line Weaver Burk:
- Easier to tell the value of V_max : for MM plot, hard to tell where the graph plateaus off

• lower [S] is needed and easier to prepare for Line Weaver Burk : if use MM method, concentration of [S] may not be sufficient –> MM method need to find V_max –> 1/2 V_max –> Km