Lecture 11 - How fast is that enzyme?

Question 1

Two events related to an enzyme

Answer 1

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

Question 2

K1 and K-1

Answer 2

Relative speeds of K1 (E+S —> ES) and K-1 (ES —> E+S) define how tightly substrate binds

Question 3

K2

Answer 3

The rate of catalysis, K2, relates to energy of activation for the transition state

Question 4

ES –> E + P

Answer 4

first order’ i.e. we assume that every ES that is made has an equal chance of forming products

Question 5

Time course of a reaction

Answer 5

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.

Question 6

Progress curve

Answer 6

A ‘progress curve’ measures the appearance of product (or disappearance of substrate) with time at steady state.

Question 7

Initial reaction velocity

Answer 7

initial reaction velocity (rate) i.e. near time zero – symbol is Vo (or Vi or Vinit). (tangent to near zero)

Question 8

The effect of enzyme concentration on reaction rate

Answer 8

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.

Question 9

The effect of substrate concentration on reaction rate

Answer 9

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.

Question 10

Two kinetic parameters can be identified on a V vs [S] curve

Answer 10

Vmax = maximum velocity possible, when [S] = ∞. 
KM = the substrate concentration at which Vobs = Vmax /2.
KM is called the Michaelis constant.

Question 11

Vo/Vi

Answer 11

Initial reaction rate = velocity of reaction measured at time zero at a particular substrate =

Question 12

V decreases from Vo due to

Answer 12

Using up substrate

Curve because initially fast but then the substrate is being used up over time.

Question 13

Vmax

Answer 13

Vmax = maximum velocity possible, when [S] = ∞. // velocity of reaction when E saturated (when all the enzymes have substrates to act on)

Question 14

Km

Answer 14

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

Question 15

The Michaelis-Menten equation

Answer 15

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.

Question 16

Michaelis-Menten model and assumptions

Answer 16

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]).

Question 17

ES complex converts to E + P with first order kinetics

Answer 17

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

Question 18

When the Michaelis-Menten model fits…

Answer 18

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.

Question 19

Allosteric enzymes do not follow what …

Answer 19

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

Question 20

Allosteric enzymes

Answer 20

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

Question 21

Allosteric enzymes - inhibitors and activators

Answer 21

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

Question 22

The concerted model

Answer 22

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

Question 23

The need for enzymes to tune in

Answer 23

At any given time, some need to turn on and others need to drop out.

Question 24

Phosphofructokinase controls glycolysis

Answer 24

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).

Question 25

Phosphofructokinase conformations

Answer 25

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

Question 26

Pancreatic digestive enzymes

Answer 26

Zymogens are secreted from the pancreas in inactive form. 

Cleavage by proteases in the gut produces active enzymes. 

Temporal and spatial control.

Question 27

At low [S] (i.e. enzyme is unsaturated)…

Answer 27

First order reaction, rate depends on [S]; V (velocity) increases linearly with increasing [S]

Question 28

At high [S] (i.e. enzyme is saturated) ….

Answer 28

(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

Question 29

What is the effect of enzyme on reaction rate (V) and thus also Vmax, when S is in excess?

Answer 29

Linear relationship between V and [E] if you have a big excess of substrates (straight, sloped line)

Question 30

Affecting reaction rate

Answer 30

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

Question 31

Curve and their order of kinetics

Answer 31

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

Question 32

Higher Km means

Answer 32

The worse it binds …
Low km = high affinity between enzyme and substrate
High km = low affinity between enzyme and substrate