BIC Enzymology Lectures
Introduction
Advantages of enzymes
- They are very efficient, you only need a little bit of them
- They are environmentally acceptable (‘natural’)
important in food ingredients and biodegradable
- They operate under mild conditions (water, ambient pH and temperature)
Less waste and less energy requirements.
- Accept unnatural substrates
- Catalyze a broad spectrum of reactions there is almost an enzyme for every reaction
- Enzymes are selective
Chemo-, regio- and stereoselective.
Chemo selective looking at functional groups
- Can be made by fermentation (=cheap)
- Can be modified/optimized (mutations, direct evolution, gene shuffling)
Avoid penicillin resistance: cleave off side chain and attach another.
an enzyme is a biological catalyst (biocatalyst). A catalyst is a compound which enhances the rate
of a chemical reaction without being destroyed or incorporated into the product.
Features of catalyst:
- Makes an alternative reaction path in which less activation
energy (dG‡) is needed
- Equally increases the rate of the backwards and forward
reaction catalysis has no effect on equilibrium position.
There are 2 ways in which it can decrease dG ‡ :
a) By lowering the energy content of the transition state
b) By ground state destabilization
There are organic, inorganic and biological catalysts.
Another way of subdivision is:
- Homogeneous = freely dissolved in solution
- organic catalyst
- organometallic complex
- enzyme in water
- Heterogeneous = solid, in liquid or gaseous environment
- inorganic catalyst
- immobilized enzyme
- enzyme in an organic solvent
The step with the highest Ea is the slowest step.
If R-CH2-Br + OH– R-CH2-OH + Br– (SN2 reaction)
Then the reaction rate is: v = k[R-CH2-Br][OH–]
slow OH-
R C-Br 3 R C+ + Br-3 R C-OH + Br-
N (S 1 reaction)
If 3 fast
Then the reaction rate will be determined by v=k[R 3C-Br]
- Bimolecular : 2 particles collide and react
- Monomolecular: 1 particle dissociates or reacts.
,Determination of the concentration of a compound as a function of time. Ideally you measure both
substrate and product, because sometimes there are side reactions.
Determine reaction rates can be done by using
- Chromatography, measuring color
- GC analysis
- HPLC
- UV spectroscopy (NADH, NAD+), use of lambert beer: A= e*c*d
- Polarimetry (only for chiral compounds)
- NMR spectroscopy
- Titration ml base added vs time; slope = reaction rate.
Chapter 1 : Kinetic data and their interpretation
1st order reaction. For instance : Enzyme inactivation, SN1 reactions, decay
reactions like radioactivity.
The rate of the reaction depends on the concentration of reactant A: the more A, the faster
it goes. But A decreases in time so d[A]/dt is negative so use -d[A]/dt
Rate = -d[A]/dt = k1[A] ln ( [A]t / [A]0 ) = -k1 t
Plotting ln[A]t against t gives a straight line with slope -k1. ln [A]t = -k1 t + C
The half life, t1/2, is defined as the time that is needed to reduce the
concentration of the reactant to 50% of tis original value.
[A]t = 0.5 [A]0 t1/2 = ln(2)/k1.
2nd order reaction. Now the reaction rate depends on both [A] and [B], so
when [A] is not the same as [B] the math is
complicated. After simplification, plotting ln ( [B]/[A] ) against t gives a straight line with slope :
( [B]0 – [A]0 )*k2
Special case of a 2nd order reaction: one reactant is in large excess.
- [A]0 >> [B]0 = pseudo-first order kinetics.
This means that the second order rate equation is simplified to a first order equation.
This is a pseudo first order reaction: [A] is so large compared to [B]
that it is considered to be constant during the reaction.
1 L water = 1 kg. The molar mass of water is 18 g/mol. concentration can be calculated.
Special case of 2nd order reaction: two reactants in equal concentrations
- [A]0 = [B]0
Plotting 1/[A] against t gives k2 as slope.
Reversible reactions
There are 2 reactions running simultaneously
Reaction 1 : and reaction 2:
Together they make :
,Look at slides 15, 16 and 17 to see the full integration. At equilibrium the net reaction rate = 0. This
means that the rate by which A is transformed into B (k1[A]) is equal to the rate by which B is going
back to A (k-1 [B]).
this is the equation eventually, which is the rate equation for a first order
process.
Plotting ln ( (Xe – X)/ Xe) against time will give –(k 1+k-1)
This equation will give k-1/k1, so the different values of k1 and k-1 can be
calculated.
Pre-equilibria complicated kinetics, unless you assume that [A*B] is constant
during a large part of the reaction (steady state approach). This means that the
rate by which A·B is formed (k1[A][B]) is equal to the rate by which is it going back (k -1[A·B]) plus the
rate by which it is transformed into C (k 2[A·B]).
The formula is :
There are 2 possibilities
1. Rapid breakdown of A*B k2 >> k-1 so v = k1 [A][B]
2. Slow breakdown of A*B k2 << k1 , k-1 so v = k2 [A][B] =
Interpretation of rate constants: the Arrhenius equation.
Every reaction has to overcome an energy barrier Ea to reach the transition state. At
higher temperature, more particles are able to overcome the energy barrier.
Arrhenius experimentally found the relation between Ea and T.
A = pre-exponential factor
Ea = activation energy (J.mol-1)
R = gas constant (8.3 J.K-1.mol-1)
T = absolute temperature (K)
Ea can be determined by measuring kobs at two (not too) different temperatures and dividing
them so A cancels out (only allowed when T 1 and T2 are not too different!)
Powerpoint slide 27 is how this equation is derived:
Ea needs to be positive, check this!! Ea is in kJ/moles
: Collision theory from statistical mechanics
P = probability factor (not every collision is effective)
Z = collision number (number of collisions per second)
The transition state only has a infinitely small life time, but what if we
assume that it has a defined lifetime and is in equilibrium with starting
materials?
For all equilibria we can write: dG0 = - RT ln K, so for our case we get: dG‡ = - RT ln K‡
dG‡ = dH‡ - TdS‡
eventual equation:
plotting ln (k/T) against 1/T gives a slope of -dH/R.
, Eyring equations: 1 = ,2=
Getting all the activation parameters from the Eyring equation:
1. Determine k at different temperatures
2. Plot ln (k/T) against 1/T to get dH
3. Get dS from
4. Get dG from
Interpretation of activation parameters !!!!!
• DG‡, the Gibbs free energy of activation, determines at which rate a certain reaction will run
at a given temperature
• DH‡, the enthalpy of activation, is a measure for the amount of binding energy that is lost in
the transition state relative to the ground state (including solvent effects) (in the rate
determining step!)
• DS‡, the entropy of activation, is a measure for the difference in (dis)order between the
transition state and the ground state (in the rate determining step!)
for monomolecular reactions: DS‡ 0 J.mol-1.K-1
for a bimolecular reaction: DS‡ << 0 J.mol-1.K-1
(two particles have to come together in the transition state to form one particle, demanding a much
greater order)
see lecture notes !!
dH transition state relative to the ground state
dS difference between transition and ground state
dH can be relatively high if:
- no new bonds to be formed;
- no compensation for the partial cleavage of the C-C bond in the transition state;
- acetonitrile is aprotic, compensation of DH‡ by solvation will be less than in water
aprotic solvent cannot form H-bonds.
Also more steric hindrance = higher dH.
Especially DS‡ is extremely useful to determine the rate limiting step of a reaction mechanism:
DS‡ ~ 0: rate limiting step is monomolecular
DS‡ << 0: rate limiting step is bimolecular
Solvation (solvent effects)
In a bimolecular reaction (A + B C), the solvent has a huge influence on the reaction rate:
both A and B are surrounded by solvent molecules, which have to be ‘pushed aside’ before A and B
can react
the stronger the bond between the reactant and the solvent, the more difficult this is slower
reaction!
If there is more solvation (due to strong dipole-dipole interactions for instance in water), then the
interactions stabilize the compounds. If there is less solvation the ground state is destabilized =
higher ground state energy = lower Ea.
Solvation effects on entropy: loss of DS‡ because of orientation of the substrates, gain of DS ‡ because
of the liberation of water (less solvated transition state). The balance is not easy to predict!
In the case of ions, the ground state is more solvated than the transition state. TS ( ‡) is hardly
solvated due to the spreading of charge.
Solvent effects in (bio)polymers
Organic polymers or biopolymers like enzymes may have apolar pockets, which lead to:
Introduction
Advantages of enzymes
- They are very efficient, you only need a little bit of them
- They are environmentally acceptable (‘natural’)
important in food ingredients and biodegradable
- They operate under mild conditions (water, ambient pH and temperature)
Less waste and less energy requirements.
- Accept unnatural substrates
- Catalyze a broad spectrum of reactions there is almost an enzyme for every reaction
- Enzymes are selective
Chemo-, regio- and stereoselective.
Chemo selective looking at functional groups
- Can be made by fermentation (=cheap)
- Can be modified/optimized (mutations, direct evolution, gene shuffling)
Avoid penicillin resistance: cleave off side chain and attach another.
an enzyme is a biological catalyst (biocatalyst). A catalyst is a compound which enhances the rate
of a chemical reaction without being destroyed or incorporated into the product.
Features of catalyst:
- Makes an alternative reaction path in which less activation
energy (dG‡) is needed
- Equally increases the rate of the backwards and forward
reaction catalysis has no effect on equilibrium position.
There are 2 ways in which it can decrease dG ‡ :
a) By lowering the energy content of the transition state
b) By ground state destabilization
There are organic, inorganic and biological catalysts.
Another way of subdivision is:
- Homogeneous = freely dissolved in solution
- organic catalyst
- organometallic complex
- enzyme in water
- Heterogeneous = solid, in liquid or gaseous environment
- inorganic catalyst
- immobilized enzyme
- enzyme in an organic solvent
The step with the highest Ea is the slowest step.
If R-CH2-Br + OH– R-CH2-OH + Br– (SN2 reaction)
Then the reaction rate is: v = k[R-CH2-Br][OH–]
slow OH-
R C-Br 3 R C+ + Br-3 R C-OH + Br-
N (S 1 reaction)
If 3 fast
Then the reaction rate will be determined by v=k[R 3C-Br]
- Bimolecular : 2 particles collide and react
- Monomolecular: 1 particle dissociates or reacts.
,Determination of the concentration of a compound as a function of time. Ideally you measure both
substrate and product, because sometimes there are side reactions.
Determine reaction rates can be done by using
- Chromatography, measuring color
- GC analysis
- HPLC
- UV spectroscopy (NADH, NAD+), use of lambert beer: A= e*c*d
- Polarimetry (only for chiral compounds)
- NMR spectroscopy
- Titration ml base added vs time; slope = reaction rate.
Chapter 1 : Kinetic data and their interpretation
1st order reaction. For instance : Enzyme inactivation, SN1 reactions, decay
reactions like radioactivity.
The rate of the reaction depends on the concentration of reactant A: the more A, the faster
it goes. But A decreases in time so d[A]/dt is negative so use -d[A]/dt
Rate = -d[A]/dt = k1[A] ln ( [A]t / [A]0 ) = -k1 t
Plotting ln[A]t against t gives a straight line with slope -k1. ln [A]t = -k1 t + C
The half life, t1/2, is defined as the time that is needed to reduce the
concentration of the reactant to 50% of tis original value.
[A]t = 0.5 [A]0 t1/2 = ln(2)/k1.
2nd order reaction. Now the reaction rate depends on both [A] and [B], so
when [A] is not the same as [B] the math is
complicated. After simplification, plotting ln ( [B]/[A] ) against t gives a straight line with slope :
( [B]0 – [A]0 )*k2
Special case of a 2nd order reaction: one reactant is in large excess.
- [A]0 >> [B]0 = pseudo-first order kinetics.
This means that the second order rate equation is simplified to a first order equation.
This is a pseudo first order reaction: [A] is so large compared to [B]
that it is considered to be constant during the reaction.
1 L water = 1 kg. The molar mass of water is 18 g/mol. concentration can be calculated.
Special case of 2nd order reaction: two reactants in equal concentrations
- [A]0 = [B]0
Plotting 1/[A] against t gives k2 as slope.
Reversible reactions
There are 2 reactions running simultaneously
Reaction 1 : and reaction 2:
Together they make :
,Look at slides 15, 16 and 17 to see the full integration. At equilibrium the net reaction rate = 0. This
means that the rate by which A is transformed into B (k1[A]) is equal to the rate by which B is going
back to A (k-1 [B]).
this is the equation eventually, which is the rate equation for a first order
process.
Plotting ln ( (Xe – X)/ Xe) against time will give –(k 1+k-1)
This equation will give k-1/k1, so the different values of k1 and k-1 can be
calculated.
Pre-equilibria complicated kinetics, unless you assume that [A*B] is constant
during a large part of the reaction (steady state approach). This means that the
rate by which A·B is formed (k1[A][B]) is equal to the rate by which is it going back (k -1[A·B]) plus the
rate by which it is transformed into C (k 2[A·B]).
The formula is :
There are 2 possibilities
1. Rapid breakdown of A*B k2 >> k-1 so v = k1 [A][B]
2. Slow breakdown of A*B k2 << k1 , k-1 so v = k2 [A][B] =
Interpretation of rate constants: the Arrhenius equation.
Every reaction has to overcome an energy barrier Ea to reach the transition state. At
higher temperature, more particles are able to overcome the energy barrier.
Arrhenius experimentally found the relation between Ea and T.
A = pre-exponential factor
Ea = activation energy (J.mol-1)
R = gas constant (8.3 J.K-1.mol-1)
T = absolute temperature (K)
Ea can be determined by measuring kobs at two (not too) different temperatures and dividing
them so A cancels out (only allowed when T 1 and T2 are not too different!)
Powerpoint slide 27 is how this equation is derived:
Ea needs to be positive, check this!! Ea is in kJ/moles
: Collision theory from statistical mechanics
P = probability factor (not every collision is effective)
Z = collision number (number of collisions per second)
The transition state only has a infinitely small life time, but what if we
assume that it has a defined lifetime and is in equilibrium with starting
materials?
For all equilibria we can write: dG0 = - RT ln K, so for our case we get: dG‡ = - RT ln K‡
dG‡ = dH‡ - TdS‡
eventual equation:
plotting ln (k/T) against 1/T gives a slope of -dH/R.
, Eyring equations: 1 = ,2=
Getting all the activation parameters from the Eyring equation:
1. Determine k at different temperatures
2. Plot ln (k/T) against 1/T to get dH
3. Get dS from
4. Get dG from
Interpretation of activation parameters !!!!!
• DG‡, the Gibbs free energy of activation, determines at which rate a certain reaction will run
at a given temperature
• DH‡, the enthalpy of activation, is a measure for the amount of binding energy that is lost in
the transition state relative to the ground state (including solvent effects) (in the rate
determining step!)
• DS‡, the entropy of activation, is a measure for the difference in (dis)order between the
transition state and the ground state (in the rate determining step!)
for monomolecular reactions: DS‡ 0 J.mol-1.K-1
for a bimolecular reaction: DS‡ << 0 J.mol-1.K-1
(two particles have to come together in the transition state to form one particle, demanding a much
greater order)
see lecture notes !!
dH transition state relative to the ground state
dS difference between transition and ground state
dH can be relatively high if:
- no new bonds to be formed;
- no compensation for the partial cleavage of the C-C bond in the transition state;
- acetonitrile is aprotic, compensation of DH‡ by solvation will be less than in water
aprotic solvent cannot form H-bonds.
Also more steric hindrance = higher dH.
Especially DS‡ is extremely useful to determine the rate limiting step of a reaction mechanism:
DS‡ ~ 0: rate limiting step is monomolecular
DS‡ << 0: rate limiting step is bimolecular
Solvation (solvent effects)
In a bimolecular reaction (A + B C), the solvent has a huge influence on the reaction rate:
both A and B are surrounded by solvent molecules, which have to be ‘pushed aside’ before A and B
can react
the stronger the bond between the reactant and the solvent, the more difficult this is slower
reaction!
If there is more solvation (due to strong dipole-dipole interactions for instance in water), then the
interactions stabilize the compounds. If there is less solvation the ground state is destabilized =
higher ground state energy = lower Ea.
Solvation effects on entropy: loss of DS‡ because of orientation of the substrates, gain of DS ‡ because
of the liberation of water (less solvated transition state). The balance is not easy to predict!
In the case of ions, the ground state is more solvated than the transition state. TS ( ‡) is hardly
solvated due to the spreading of charge.
Solvent effects in (bio)polymers
Organic polymers or biopolymers like enzymes may have apolar pockets, which lead to:
