Lecture 11 - How fast is that enzyme? Flashcards
Two events related to an enzyme
Binding and catalysis
To model enzyme catalysis we use a simple system in which an enzyme, E, converts a single substrate, S, to a single product, that is instantly released. Furthermore we assume that the conversion is irreversible
K1 and K-1
Relative speeds of K1 (E+S —> ES) and K-1 (ES —> E+S) define how tightly substrate binds
K2
The rate of catalysis, K2, relates to energy of activation for the transition state
ES –> E + P
first order’ i.e. we assume that every ES that is made has an equal chance of forming products
Time course of a reaction
The complex ES is necessary for reaction, so [ES] at any time will govern the rate.
‘Steady state’ refers to time during which [ES] does not change, and plateus/goes flat.
Progress curve
A ‘progress curve’ measures the appearance of product (or disappearance of substrate) with time at steady state.
Initial reaction velocity
initial reaction velocity (rate) i.e. near time zero – symbol is Vo (or Vi or Vinit). (tangent to near zero)
The effect of enzyme concentration on reaction rate
If there is sufficient excess of substrate, then as the amount of enzyme is increased, the rate of reaction increases.
Differently put: Vo, initial velocity, is proportional to [E], enzyme concentration, when substrate is in excess.
The effect of substrate concentration on reaction rate
Usually data are collected with a fixed amount of enzyme and variable amount of substrate.
As [S], concentration of substrate, is increased, the initial rate Vo (also called Vobs)
Increases but in a linear way first BUT as all the enzyme active sites become occupied, the rate of reaction stops increasing.
Two kinetic parameters can be identified on a V vs [S] curve
Vmax = maximum velocity possible, when [S] = ∞. KM = the substrate concentration at which Vobs = Vmax /2. KM is called the Michaelis constant.
Vo/Vi
Initial reaction rate = velocity of reaction measured at time zero at a particular substrate =
V decreases from Vo due to
Using up substrate
Curve because initially fast but then the substrate is being used up over time.
Vmax
Vmax = maximum velocity possible, when [S] = ∞. // velocity of reaction when E saturated (when all the enzymes have substrates to act on)
Km
KM = the substrate concentration at which Vobs = Vmax /2. KM is called the Michaelis constant.
A(n inverse) measure of the affinity of the enzyme for the substrate. Defined as the substrate concentration required for the reaction to proceed at half Vmax i.e. Km=[S] when V=1/2 Vmax.
Km indicates the enzyme’s ability to bind substrate
Always positive value
The Michaelis-Menten equation
The Vobs vs. [S] curve is described by the Michaelis- Menten equation:
Vmax [S]
Vobs = ––––––––– Know this equation.
KM + [S]
Many enzymes obey Michaelis-Menten behaviour, and this is how we determine their kinetic parameters.
Michaelis-Menten model and assumptions
Simplifying assumptions:
Product is not converted back to substrate.
Haldane’s steady state assumption: the rate of ES formation equals the rate of its breakdown; that is
d[ES]
–––––– = 0
dt
3. Measuring initial rate insures [S] does not change significantly (and [S] is much greater than [E]).
ES complex converts to E + P with first order kinetics
Single molecule events, like radioactive decay, occur with a set probability, giving first order kinetics.
If each ES complex has the same chance of going through the transition state, the ES —> E + P step will follow first order kinetics.
Every ES complex has the same chance to getting to E+P, chance gets lower and lower the higher the activation energy is
When the Michaelis-Menten model fits…
Some assumptions:
All ES complexes have same rate of reaction.
[S] is in vast excess to [E].
Haldane’s steady state assumption: the rate of ES
formation equals the rate of its breakdown.
Initial rate is measured. That is, early enough that [S] does
not change significantly.
The reverse reaction does not occur.
Allosteric enzymes do not follow what …
Michaelis-Menten equation
Vobs vs.[S] plot is a sigmoidal, not hyperbolic.
Respond more steeply to intermediate changes in [S].
Evolve at regulatory points in metabolic pathways.
Recall haemoglobin. Aspartate transcarbamylase (A TCase) and phosphofructokinase are allosteric enzymes.Respond to binding effectors away from the active site just as haemoglobin responds to BPG. Allosteric and cooperative therefore have an R and T state which is why it is difficult to model it with this equation. Because one of the assumptions was that every enzyme acts the same, R ones running fast and T ones running slow so the Michaelis Menten model is not complex enough to capture this.
Allosteric enzymes instead produce a sigmoidal curve
Allosteric enzymes
By definition, ‘allosteric’ enzymes respond to effectors binding away from the active site.
Binding accompanies a change of shape, which in turn changes enzymatic activity.
They often have multiple subunits and display cooperative behaviour.
Both cooperativity and allostery depend on the enzyme switching between active and inactive forms.
Allosteric enzymes control metabolic pathways
Allosteric enzymes - inhibitors and activators
Allosteric inhibitors and activators bind to an allosteric enzyme to a site that is not the active site (allosteric site)
V on y axis and S on the X axis
Activator - shifts the curve to the left and speeds things up
Inhibitor - shifts the curve to the right and slows things down
The concerted model
Monod, Wyman and Changeux (MWC), 1965.
Subunits can be in an inactive tense (T) or active relaxed (R) state.
All subunits must be in the same state (either T or R).
Binding each successive substrate shifts the equilibrium in favour of R.
Inhibitors stabilise the T state.
Activators stabilise the Rstate.
The more substrate bound the more likely it is to be in the R state, the emptier it gets the more likely it is to be in the T state
The need for enzymes to tune in
At any given time, some need to turn on and others need to drop out.
Phosphofructokinase controls glycolysis
Phosphorylates fructose-6- phosphate (F6P) to fructose bisphosphate.
Inhibited if cell has plenty of ATP, i.e. when glycolysis is not needed for energy.
Homotetramer is cooperative when inhibited by ATP or phosphoenolpyruvate (PEP).
Phosphofructokinase conformations
T-state is more compact, stabilised by PEP, an abundant intermediate of glycolysis.
R-state is stabilized by substrate F6P and ADP.
Arginine 162 and Glutamic acid 161 swap positions in active site.
In the R state with the positive charges at its Ns in arginine, goes into the active site and makes favourable electrostatic interactions with the phosphate of F6P, so substrate binds well because of the interaction between the phosphate and the positive charge on the arginine meanwhile the glutamic acid is sitting and pointing towards the allosteric effector. As the enzyme switches to the T state by binding of PEP, as an allosteric effector of by loss of substrate from other active sites (standard cooperatively - one subunit is responding to what is going on in another subunit, stabilising T state pulls the arginine and Glu161 flips its positivon and the carboxylate group within Glu161 carries a negative charge which won’t bind well to the phosphate of F6P as we get electrostatic replusion and the substrate just doesn’t bind to the active site (turns the enzyme off until it switched back to the R state). Turning the enzyme on has to do with the arrangement of the polypeptide chain, swapping two residues wither sticking into the active site or reaching over towards the allosteric effector site with big effects on the binding affinity and the Km
Pancreatic digestive enzymes
Zymogens are secreted from the pancreas in inactive form.
Cleavage by proteases in the gut produces active enzymes.
Temporal and spatial control.
At low [S] (i.e. enzyme is unsaturated)…
First order reaction, rate depends on [S]; V (velocity) increases linearly with increasing [S]
At high [S] (i.e. enzyme is saturated) ….
(no more enzymes that can do work = saturated)
Zero order reaction, rate independent of [S]; No (or only a tiny) increase in V (velocity) with increasing [S] —> Vmax reached
What is the effect of enzyme on reaction rate (V) and thus also Vmax, when S is in excess?
Linear relationship between V and [E] if you have a big excess of substrates (straight, sloped line)
Affecting reaction rate
Enzyme concentration (V vs [E] curve):
- excess of substrate
- Vo proportional to [E] i.e. as the amount of enzyme increases, the rate increases - faster reaction if more active sites are available
Substrate concentration (V vs [S] curve):
- fixed amount of enzyme i.e. [E] is limiting
- as the amount of substrate increases the rate increases linearly but as the active sites become occupied the rate of reaction stops increasing
- asymptotes to Vmax - forms because active sites that aren’t there are trying to be filled
Curve and their order of kinetics
Initially first order kinetics as the rate is dependent on the concentration of the substrate
Later it is zero order kinetics because the rate does not depend on concentration of substrate
Higher Km means
The worse it binds …
Low km = high affinity between enzyme and substrate
High km = low affinity between enzyme and substrate
