Kinetics 4: Complex reactions: pre-equilibrium, steady state, MM

Question 1

Define an elementary step.

Answer 1

An elementary step is a single reactive encounter between species – is the smallest unit of a reaction.

Question 2

Define (kinetic) mechanism.

Answer 2

A proposed set of elementary steps accounting for the features of a reaction.

Question 3

Define an elementary reaction

Answer 3

Consists of a single elementary step

Question 4

Elementary reaction: Cl + O₂ –> ClO + O, with rate constant k

Write the rate of change for each species.

Answer 4

elementary reaction means rate laws can be taken directly from eq since everything is simultaneous

Question 5

The complex reaction below occurs in 2 elementary steps. Write the rate of change of:

• product
• carbocation

Answer 5

• to calculate how [P] changes with time, the diff eqs must be integrated, which is possible here but gets complex as more steps are added, since a given variable appears in multiple eqs.*
• Hard to solve w/o computer, so the following methods remove some of the complex time-dependence by making assumptions about the relative rates of processes.*

Question 6

Two sequential, first-order reactions, with rate constants k₁ and k₂:

A –> B (1) B –> C (2)

Suppose k1 << k2. Draw conc of each species against time on the same axes, and state which step is rate-determining.

Answer 6

1 is the RDS

process 1 has smaller rate constant therefore slower

Question 7

Two sequential, first-order reactions, with rate constants k₁ and k₂:

A –> B (1)

B –> C (2)

Suppose k1 >> k2. Draw conc of each species against time on the same axes, and state which step is rate-determining.

Answer 7

2 is the RDS

process 2 has smaller rate constant therefore slower

Question 8

Study the reaction scheme below. What conditions are required for the pre-equilibrium hypothesis to apply?

Answer 8

• Rate of process 1 >> rate of process 2
• Such that rates of [1] and [-1] are equal (i.e. equilibrium is established between E, H₃O⁺ and EH⁺)
• And therefore process [2] is rate-determining

pre-eq hypothesis is often useful for intermediates involving protonation or deprotonation, since these processes are usually faster than breaking/making bonds to atoms heavier than hydrogen

Question 9

Study the reaction scheme below. Use the pre-equilibrium hypothesis to derive an expression for the rate equation for formation of products.

Answer 9

Rate of process 1 >> rate of process 2, such that rates of [1] and [-1] are equal (in equilibrium), and process [2] is rate-determining:

k₁[E][H₃O]⁺ = k_-1[EH⁺][H₂O]

So [EH]⁺ = (k₁/k_-1)([E][H₃O]⁺/[H₂O])

= k_eq([E][H₃O]⁺/[H₂​O])

r = d[P]/dt = k₂[EH⁺][H₂O]

= k₂k_eq([E][H₃O]⁺)

• pre-eq hypothesis makes final rate eq simpler since it involves only reactants, so can be integrated as in previous sections*
• Could then find experimentally that rate = k_exp([E][H₃O]⁺/[H₂​O]) – comparing with theoretical rate law would show kexp is a composite of the rate constants of 3 steps*

Question 10

What conditions permit using the steady state approximation? Include a graph of concentration against time.

Answer 10

Applicable when, in a complex mechanism, a reactive intermediate reacts as soon as it’s formed, such that its concentration is assumed to be constant.

Only applicable when reaction is in steady state - not in the initial or final phases.

simplifies solving differential eqs since it removes the time dependence

Question 11

Consider the sequential reaction A –> B –> C with rate constants k₁ and k₂.

Use the steady state approximation to derive an expression for the rate of product formation.

Answer 11

as expected, eq indicates B reacts as soon as it’s formed, such that the first step of the reaction is the RDS.

Question 12

In the reaction scheme below, EH⁺ is assumed to be in the steady state. Derive an expression for the rate of formation of products.

Answer 12

r = d[P]/dt = ((k₁k₂)/(k₁+k₂)) x [E][H₃O]⁺

general process

• identify the intermediate(s)
• write the rate law for each, each reaction producing/consuming the intermediate contributes a term to the expression, those consuming it give negative terms
• set this = 0
• rearrange for [intermediate]_ss
• write rate law for products, sub in [intermediate], simplify

Question 13

In the reaction scheme below, EH⁺ is assumed to be in the steady state. The rate of formation of products is given by:

r = d[P]/dt = ((k₁k₂)/(k₁+k₂)) x [E][H₃O]⁺

Suppose the rate of [2] >> that of [-1]. Simplify the expression.

Answer 13

k₂[EH⁺][H₂O] >> k_-1[EH⁺][H₂O]

Therefore k₂ >> k₁

Approximate denominator (k₁+k₂) = k₂

So rate law becomes

r = d[P]/dt = ((k₁k₂)/(k₂)) x [E][H₃O]⁺

= k₁[E][H₃O]⁺

reaction is effectively 2 sequential steps (since -1 is negligable), with the first being slower, so ​rate depends on [1] only since it’s the RDS

Question 14

In the reaction scheme below, EH⁺ is assumed to be in the steady state. The rate of formation of products is given by:

r = d[P]/dt = ((k₁k₂)/(k_-1+k₂)) x [E][H₃O]⁺

Suppose the rate of [-1] >> that of [2]. Simplify the expression.

Answer 14

k₂[EH⁺][H₂O] << k_-1[EH⁺][H₂O]

Therefore k₂ << k_-1

Approximate denominator (k_-1+k₂) = k_-1

So rate law becomes

r = d[P]/dt = ((k₁k₂)/(k_-1)) x [E][H₃O]⁺

= k_eqk₂[E][H₃O]⁺

NB identical to rate law derived assuming [1] and [-1] form a pre-equilibrium. this is since assuming that rate of [-1] >> [2] assumes an equilibrium is set up, and that [2] is a slow RDS, ie bleeding off of EH+

Question 15

try page 36 on notes, decomposition of N205

Question 16

What is saturation kinetics?

Answer 16

Where, for a fixed quantity of enzyme, as [S] increases, the rate of reaction first increases linearly, then levels off to a limiting maximum value.

Question 17

Write out the Michaelis-Menten scheme.

Answer 17

Question 18

Shown is the Michaelis-Menten scheme.

• Give the expression for K_M
• What units does it have?
• define it

Answer 18

k_M = (k_-1 + k_cat)/k₁)

Units of conc

K_M is the substrate conc which gives ha;f-maximum velocity.

Question 19

Shown is the Michaelis-Menten scheme.

Write the Michaelis-Menten equation, i.e. the expression for the velocity, V (ie rate) of reaction.

Answer 19

V = (k_cat[E]₀[S]) / ([S] + (k_-1 + k_cat)/k₁)

= (k_cat[E]₀[S]) / ([S] + k_M)

where K_M is the michaelis constant

Question 20

Shown is the Michaelis-Menten scheme.

Using the steady state approximation, derive the Michaelis-Menten equation, ie the expression for the velocity, V (ie rate) of reaction.

Answer 20

Assume ES in steady state since intermediate

So d[ES]/dt = k₁[E][S] - k_-1[ES] -k_cat[ES] = 0

Derive an expression for [E], then plug into the above (so that [ES] is the only unknown, given [E]₀ is known)

[E]₀ = [ES] + [E] –> [E] = [E]₀ - [ES]

k₁([E]₀ - [ES])[S] - k_-1[ES] -k_cat[ES] = 0

Rearrange for [ES]

[ES]_ss = (k₁[E]₀[S])/(k₁[S]+ k_-1 + k_cat)

Plug [ES] into the expression for velocity (rate)

d[P]/dt = V = k_cat[ES]

= (k_catk₁[E]₀[S])/(k₁[S]+ k_-1 + k_cat)

divide through by k1

V = (k_cat[E]₀[S]) / ([S] + (k_-1 + k_cat)/k₁)

= (k_cat[E]₀[S]) / ([S] + k_M)

where K_M is the michaelis constant

Question 21

Shown is the Michaelis-Menten equation.

V = (k_cat[E]₀[S]) / ([S] + k_M)

Dependence of V on [S] is complex, so V is measured for the initial rate.

When substrate conc is low:

• Simplify the equation
• State the order of reaction in [S] and [E]₀
• State the rate constant and its order
• Give the biological interpretation

Answer 21

[S] << K_M so simplify denominator:

V = (k_cat[E]₀[S]) / k_M

1st order in [S] and [E]₀

2nd order rate constant k_cat/K_M

There is more free enzyme than substrate-bound enzyme, so rate is limited by substrate availability, hence first order in [S].

Question 22

Shown is the Michaelis-Menten scheme.

Write the expression for V_max and define it.

Answer 22

V_max = k_cat[E]₀

Maximum velocity which occurs when all enzyme is saturated, i.e. 0 order in [S].

Question 23

Shown is the Michaelis-Menten equation.

V = (k_cat[E]₀[S]) / ([S] + k_M)

Dependence of V on [S] is complex, so V is measured for the initial rate.

When substrate conc is high:

• Simplify the equation
• State the order of reaction in [S] and [E]₀
• State the rate constant and its order
• Give the biological interpretation
• State the RDS

Answer 23

[S] >> K_M so simplify denominator:

V = (k_cat[E]₀[S]) / [S] = k_cat[E]₀ = V_max

0 order in [S], 1st order in [E]₀

1st order rate constant k_cat

Velocity is independent of S, due to saturation of V: enzymes all substate-bound so adding S has no effect.

RDS = ES –> E+ P

Question 24

Shown is the Michaelis-Menten equation.

V = (k_cat[E]₀[S]) / ([S] + k_M)

Substitute in the expression for V_max and use this to interpret the meaning of K_m.

Answer 24

V_max = k_cat[E]₀

V = (V_max[S]) / ([S] + k_M)

Substate conc [S]_1/2 results in half the max velocity

V_max/2 = (V_max[S]_1/2) / ([S]_1/2 + K_M)

So 1/2 = ([S]_1/2) / ([S]_1/2 + K_M)

So ([S]_1/2 + K_M) = 2[S]_1/2

Thus [S]_1/2 = K_M

So K_M is the substrate conc which gives half-maximum velocity.

Question 25

What is the interpretation of large and small K_M?

Answer 25

• Large Km: large [S] is needed to achieve half-maximum rate, since enzyme has weak affinity for substrate
• Small Km: little [S] is needed to achieve half-maximum rate, since enzyme has strong affinity for substrate

Question 26

Shown is the Michaelis-Menten equation.

V = (k_cat[E]₀[S]) / ([S] + k_M)

Show how it can give a straight-line plot, and state the intercept and slope (give the version not using, and the one using, Vmax).

Answer 26

V = (k_cat[E]₀[S]) / ([S] + k_M)

1/V = ([S] + k_M) / (k_cat[E]₀[S])

= K_M/(k_cat[E]₀) x (1/[S]) + 1/(k_cat[E]₀)

Plot 1/V against 1/[S], giving straight line with slope K_M/(k_cat[E]₀) and intercept 1/(k_cat[E]₀)

V_max version

recall Vmax = [S][E]₀

V = (k_cat[E]₀[S]) / ([S] + k_M)

1/V = ([S] + k_M) / (k_catV_max)

= (K_M/V_max)(1/[S]) + 1/V_max

Plotting 1/V against 1[S] gives a straight line with slope (K_M/V_max) and intercept 1/V_max

Question 27

Shown is the Michaelis-Menten equation.

V = (k_cat[E]₀[S]) / ([S] + k_M)

Plotting 1/V against 1/[S] gives a straight line with slope K_M/(k_cat[E]₀) and intercept 1/(k_cat[E]₀), or a straight line with slope (K_M/V_max) and intercept 1/V_max.

Why are these double-reciprocal plots prone to error?

Answer 27

Error in low [S] values are magnified, since in this region 1/[S] is high. So the position of the line, and thus the slope and intercept used to deduce k_cat and K_M, is overly influenced by these low [S] values.