overview of lecture topics by rc 1) substrate diffusion and catalysis 2) biological reaction...
TRANSCRIPT
Overview of Lecture Topics by RC
1) Substrate diffusion and catalysis
2) Biological reaction kinetics
3) Biological reactions that change the environment
Lec 1 Substrate diffusion and catalyis
• What are bacteria? The advantage of being small.• What do bacteria do? Catalyse exergonic redox
reactions.• Substrate diffusion to the cell: Can we predict the
random diffusion of a molecule? Not of one, but of many molecules.
• What is the kinetics of substrate diffusion (1st order kinetics, Fick’s law).
• What is diffusion driven by? Order, Concentration gradient,
• How come that many molecules seem to have a behaviour but an individual molecule does not?
• What is the principle of catalysing redox reaction? The enzyme does not bind to the substrate.
What do bacteria do?
• Catalyse exergonic redox reactions
• Exergonic, (downhill) reactions loose Gibbs Free Energy. (DG=negative)
• Bacteria utilise a portion of the free energy released for growth processes and multiplication
G
S
P
ΔG
What does ΔG (Gibbs Free Energy Change) mean?
• Spontaneous reactions are downhill reactions
• Energy of substrates is higher than of products
• Products are more stable than Substrate (stable = low energy)
• The driving force = hill height = difference in G = Delta (Δ) G ΔG = G (prod.) – G (substr.)
• The reaction is driven by the loss in G Change in G is a negative value (e.g. Δ G = -256 kJ/mol)
G
S
P
ΔG
Does the Δ G allow to predict the reaction rate ?
• No• But for ΔG = zero, reaction
rate will be zero• For ΔG = positive, reaction
would go backwards (PS)• The rate is determined by
the activation energy (AE)• In biological reactions the
AE is largely determined by the presence of enzymes (lower AE)
• Now how do they do that?
G
S
P
ΔG
AE
The ΔGo of redox reaction is related to the ΔEo of e- donor and acceptor
• ΔEo is the difference in redox potential of the half reactions
• Couple with lower Eo will become e- donor
• If not reaction would reverse
• ΔGo = n F Δ Eo • Microbes that use electron donors of a very low
Eo (e.g. H2 and an electron acceptor of a very high level e.g. O2) have a lot of free energy available for growth
-Eo
e- don.
ΔE
e- acc.
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What is enzyme catalysis? The rate of a spontaneous reaction is
increased.
If the reaction is already spontaneous, why
do we need an enzyme?
The activation energy needs to be
overcome.
Spontaneous = downhill = exergonic
Enzymes can not catalyse uphill =
endergonic reactions
If catalysed uphill reactions proceed in the
reverse direction (Products Substrates)
G
Reaction path
G
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How do enzymes catalyze ?A. By lowering the activation energy barrier!
B. By binding to the substrate like a lock to a key???
This is the simplified textbook explanation, but how can we
visualise it?
How can an enzyme convert a substrate to product by
binding to it?
Binding means stabilising (lower Energy level) and hence
slowing down reaction rates.
Example: Antibody binding to antigen.
Antibody is a protein designed by the immune system to bind and “neutralise” a foreign substances
(the antigen).
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Why don’t antigens catalyze?
Will an antibody against the substrate be able
to catalyse the reaction?
No
What is the difference between an antigen
and an enzyme.
Both bind, one catalyses the other does not.
Does the enzyme have perhaps a special
mechanism (lever) that can break the
substrate into product(s) ?
No
P
S
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How do enzymes catalyse ?
P
Example: Substrate = stick, Product = broken stick
For the stick to be broken it must go through a
transition state (T)
How does the enzyme (stickase, of course) catalyse by binding to the stick?
Binding to the stick stabilises rather than activates the substrate not making reaction easier
The simplified concept of the enzyme binding to the substrate does not make sense
Let’s have a closer look at the energy diagram for clues.
S
T
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High Energy=
Unlikely, reactive, activated
Low Energy=
Likely to exist, stable
G
Reaction path
P
S
P
S
T
needed to lower the activation energy level:a way to make the transition (T) state more likely
Note T not an intermediate just a “deformed molecule” that makes
forward and backward reaction equally likely
How do enzymes catalyse ?
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P
Example: Substrate = stick, Product = broken stick
For the stick to be broken it must go through a
transition state (T)
The enzyme (stickase, of course) binds to T
stabilises Tmakes T more likelyhigher quantities of T availablehigher likelyhood for P to form(P cannot form when T is in ultra low concentration)
Enzymes catalyse by binding to the transition state and making it more likely.
S
T
How do enzymes catalyse ?
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Significant product formation
depends on availability of T.
Non catalysed reactions are slow
because [T] is low
By binding to T enzymes increase
the chances of T to exist, hence
speed up the reaction.
Binding does not mean, holding on
to T but releasing T rapidly, either
as S or as P.
G
Reaction path
P
S
P
S
T
How do enzymes catalyse ?
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The enzyme substrate complex
(ES) is not a transitional state but a
true, existing, defined intermediate
Intermediates are in a “valley”
Transition states are on a “peak”G
Reaction path
P
S
P
S
T
ES
How do enzymes catalyse ?
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How far will the catalysed reacton go?
With decreasing substrate concentration
the energy content of the substrates sinks.
Increasing product concentrations lift the
energy content of the products.
The reaction continues until the difference
in G (Delta G) is zero.
Then the energy level of substrate equals
that of the product
Reaction is at equilibrium
Rate of backwards reaction equals that of
forward reaction.
The ratio [P]/[S] now represents the
equilibrium constant keq.
`
G
Reaction path
P
S
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The ratio [P]/[S] now represents the
equilibrium constant keq.
An original very exergonic reaction needs
lots of P to accumulate until equilibrium is
reached
at equilibrium [P]/[S] is very high (e.g.
10,000)
endergonic reactions have low keq (e.g.
0.00001)
The enzyme does not affect the
spontaneity or reversibility of reaction but
the energetics does.
Not surprisingly the reaction driving force is
related to keq ΔG =RT ln keq
`
G
Reaction path
PS
The dynamic equilibrium
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Where does the energy come from to
overcome activation energy?
An enzyme/substrate to be more precise
enzyme/T complex forms hydrogen bonds
and hydrophobic interaction bonds.
The binding energy released is the energy
source of lowering the activation energy
(analogy of magnets in stickase)
H bonds of T with H2O are replaced by
bonds with E (dry bonding)
T
Binding Energy
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What is the effect of activation energy on reaction rate?
The overal rate constant (k) of the reaction
depends directly on the activation energy:
k= (B/P*T)*e -DG(activ.)/RT
B/P=Bolzman/Planck constant
E.g. lowering the activation energy by
5.7kJ will increase rate 10 fold
T
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What is the effect of activation energy on reaction rate?
After we have concluded:
that the enzyme (E) catalyses the substrate (S) conversion by
forming an Enzyme-substrate complex
Let us see how we can derive the kinetic behaviour of the enzyme
reaction from first principles.
The widely accepted enzyme kinetics model is the Michaelis
Menten model.
T
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For the overall enzyme reaction a number of rate constants need to be
considered:
The rate of conversion of E and S to ES is k1
k-1 is the rate constant for ES going back to E and S
k2 is the rate for conversion of ES to E and P
In enzyme assays P is negligible (startup velocity of reaction) k-2 is
not included.
(Rate constants mean first order kinetics rate constants as explained below.)
Foundation of Michaelis-Menten Kinetics
k1E + S ES E + P
k-1
k2
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Foundation of Michaelis Menten kinetics
Rate constant of first order reaction predicts that the rate is proportional
to the substrate concentration
The rate for ES to go to E +P is given by:
ES (mM) * k2 (h-1) (mM/h)
k1
k-1
k2E + S ES E + P
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Foundation of Michaelis Menten kinetics
At substrate saturation no free enzyme is available ([E]=0)
overall rate is determined by k2 ( kcat = k2)
(btw: kcat= vmax/(total enzyme concentration (Et, E+ES))
The ratio of ES formation over ES disintegration is km (formula?)
km = (k2 + k-1)/k1
enzyme with low km: ES formation faster than disintegration
k1
k-1
k2E + S ES E + P
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Derivation of MM kinetics from first principles
At steady state: rate of ES disintegration =rate of ES formation
k-1*ES + K2*ES = k1*ES (Et=E+ES)
k-1ES + K2ES = k1(Et-ES)S multiply out right
k-1ES + K2ES = k1EtS - k1ESS k1ESS
k-1ES + K2ES + k1ESS = k1EtS bracket out ES
ES (k-1+ K2 + k1S ) = k1EtS solve for ES
ES = k1EtS / (k-1+ K2 + k1S )
ES =
k1
k-1
k2E + S ES E + P = km
k-1+ K2k1
k1EtS
k-1+ K2 + k1S
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ES =
ES =
k1
k-1
k2E + S ES E + P
k1EtS
k-1+ K2 + k1S
cancel k1
= EtS
k-1+ K2k1
+S
= km k-1+ K2
k1
Et S
km+S (as vo = k2 *ES ES = vo/k2 )
Et S
km+S
vo k2
=
k2Et S
km+Svo =
vmax S
km+S
(as vmax =k2*Et)
(k2 is also called kcat)vo =
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S
km + Svo = vmax
(students don’t need to be able to derive it but to know the final equation)
Vo = the initial velocity (no products present)
S = [substrate]
km = half saturation constant, also called kS
vmax = maximum velocity under substrate saturation
when overall reaction only depends on k2
The Michaelis Menten Model has been derived:
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S
km + Svo = vmax
The Michaelis Menten (MM) model is not only useful for
enzymatic reactions but overall microbial reactions
such as algal blooms or microbial growth in bioreactors.
After we have seen where the most widely used
biological kinetic mode comes from…
let us compare 2 types of microbial kinetics (MM and
exponential kinetics) with traditional chemical kinetics
(zero and first order).
Microbial Reaction Kinetics:
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S
km + Svo = vmax
The substrate (reactant) of bioprocesses can disappear
according to different time courses, such as linear
disappearance, first order, exponential, mixed kinetics
(such as Michaelis Menten) or incompletely such as for
product inhibition or approaching of the thermodynamic
equilibrium.
Students should be able to sketch the different types
and explain which kinetic types occur in which
processes.
Microbial Reaction Kinetics:
Zero Order kinetics
• constant velocity• velocity independent of S• v = K (M/s)• Ex: limited access to S (O2
diffusion to food)• water loss• enzyme reactions at high S
PS
Time
S
V
V
First Order Kinetics
• velocity decreases over time (parallel to [S])
• velocity is determined by and hence proportional to S
• v = S * K (s-1)• K is the rate constant
• Ex: Most chemical reactions • Radioactive decay (half
time)
P
S
Time
S
V
V
Exponential Kinetics
• velocity exponentially increasing over time
• independent of S but related to P• Can give appearance of sudden
increase in P• v ~ P * K (s-1)• a product must enhance reaction
velocity (e.g. heat, chain reaction)
Biological examples: • bacterial spoilage (e.g. milk)
multiplication of catalyst• Auto-oxidation of fats (radicals
propagation)
P
S
Time
P
V
V
Michaelis Menten Kinetics
• Two reaction phases:• 1: zero order, [Enzyme] limiting• 2: first order, [S] limiting• Why do we get first and zero order if
S or E is limiting?• [S] changes, [E] does not change!• v = vmax * S / (s + kS)
Examples:• Most biochemical reactions • Microbial reaction in the
environment when S is low (pollution, groundwater, soil, ocean)
PS
Time
S
V
V
Phase 1Phase 2
Phase 1 Phase 2
Kinetics Summary• At least 4 different types of
reaction can occur in biological environments
• Development of reaction rate can increase, decrease or stay the same.
• There are clear mechanistic reasons for reaction behaviour
• Significance: When knowing the reaction kinetics the behaviour biological material (e.g. food, bioreactors, body) can be predicted
• 2nd order reaction (dependence on two substrates) is similar to 1st order and neglected here
Time
S
V
Time
0
1Enz Exp
Decide Order or Kinetics
Time
S
V
Time
0
1Enz Exp
1. Decide which order the reaction kinetics is:
• v constant, S decrease linear Zero order
• v increasing exponential• v continuously slowing (1st, 2nd,
3rd order)• v constant , then slowing
MM kinetics