8 - Enzymes III: Enzyme kinetics

Question 1

Use of enzyme kinetics

Answer 1

• expressing the catalytic activity of an enzyme in solution
• purity (specific activity)
• efficiency in relation to different substrates……
• ……active site structure
• characterising the efficiencies of isoenzymes
• effectiveness of inhibitors and mode of action
• …….active site structure

Question 2

Ludwig Wilhelmy (1812-1864)

Answer 2

worked on basic chemical kinetics (no enzymes).

1850 – rate of acid hydrolysis of sucrose is proportional to sucrose concentration at a constant acid

Question 3

Adrian Brown(1852-1919)

Answer 3

Hydrolysis of sucrose by yeast.
1902 – at low substrate (sucrose) concs (and constant [E]) the rate is proportional to substrate concentration.
However, at high substrate the rate becomes independent – rate is being influenced by the nature of invertase (enzyme)-sucrose (substrate) complex formation

Question 4

Leonor Michaelis and Maud Menten

Answer 4

• Provided a mathematical solution describing the rectangular hyperbolic nature of the rate vs. [substrate] plot
• Published in 1913 and still used today
• Forms the basis for most of our understanding of enzyme mechanisms and kinetics i.e. how enzymes interact with their substrates and inhibitors

Question 5

The Michaelis-Menten equation, Vmax and KM

Answer 5

V =Vmax [S] / KM+ [S]
• The model is not perfect - makes several other assumptions.
• Rate (V) vs. [S] plot is a rectangular hyperbolic curve

Question 6

What is Vmax?

Answer 6

• Velocity (V) is a rate = e.g. nmol min-1
• Vmax is the maximum possible rate of the reaction at that particular enzyme concentration. It’s a velocity so units are still e.g. nmol min-1
• ONLY AN ESTIMATE – have to extrapolate due to infinite [S] required

Question 7

What is KM?

Answer 7

• Substrate concentration ([S]) is a concentration = e.g. nmol L-1
• KM is the substrate concentration at which the velocity of the reaction is half Vmax. It’s a concentration so units are e.g. nmol L-1
• Also a descriptor of the affinity of the active site for the substrate (for most enzymes)

Question 8

Vmax and KM

Answer 8

• KM is a constant which defines the interaction between enzyme and substrate molecules
• KM is independent of the amount of enzyme and substrate present
• Vmax is constant only for a given enzyme concentration; it is independent of the amount of substrate present but dependent on the amount of enzyme

Question 9

Michaelis-Menten; the assumptions and limitations

Answer 9

1 – the reaction is a single substrate reaction – things get more complicated with multi-substrate reactions e.g. one substrate can alter the binding of the other to the enzyme. Also different binding sites possible.
E +S -> (K1 ) ES -> (K2) E + P
2 – k2 in the simplified equation above assumes that there is no reverse component to this step of the mechanism but:
1. REVERSE CATALYSIS – a number of enzymes can catalyse the conversion of the product back into substrate.
2. PRODUCT INHIBITION

Question 10

The importance of using initial reaction rates; mitigating reverse catalysis and product inhibition

Answer 10

• Use initial velocity to plot on MM graph – overcomes issues of reverse
catalysis and product inhibition (product negligible in early stages of reaction

Question 11

Revisiting the rectangular hyperbolism; low substrate concentrations

Answer 11

At low [S]At low [S]

Question 12

Revisiting the rectangular hyperbolism;high substrate concentrations

Answer 12

At high [S]&raquo_space;KM,
Enzyme active sites occupied
Rate (V) becomes independent of [S]

Question 13

What do KM and Vmax tell us about an enzyme?

Substrate/inhibitor affinity

Answer 13

• KM = measure of affinity of the enzyme active site for a substrate; or the [S] at which half of the enzyme active sites are occupied
• Low KM for substrate A = low substrate concentration needed to occupy half of enzyme active sites – enzyme has HIGH AFFINITY for substrate
• High KM for substrate B = high substrate concentration needed to occupy half of enzyme active sites – enzyme has LOWER AFFINITY for substrate
• Drug design – drug/inhibitor needs lower KM than substrates so it binds more effectively to enzyme active site

Question 14

Isoenzymes

Answer 14

different (related) enzymes that catalyse the same reaction e.g. alcohol dehydrogenase (ADH).

Question 15

What do KM and Vmax tell us about an enzyme?

Studying the catalytic efficiencies of isoenzymes

Answer 15

• Humans – 6 classes of ADH – each class split into several more forms depending on, among other things, subunit composition in the dimer (8 different subunits)
• Altered properties such as isoelectric point or surface charge isoelectric focussing or native electrophoresis

Question 16

Turnover number (Kcat)

Answer 16

the number of substrate molecules broken down per molecule of enzyme per second

Question 17

What do KM and Vmax tell us about an enzyme?
Turnover number (Kcat)

Answer 17

• At none limiting (i.e. very high) substrate concentrations, Kcat = Vmax/[ET] where [ET] = [E] + [ES] is the total number of enzyme molecules at the start of the reaction (no [EP] at the start).
The turnover number can be thought of as a measure of the efficiency of the enzyme that can be compared between totally unrelated enzymes with different substrates
• All you are doing here is standardizing things by dividing by the total
initial number of enzyme molecules to express in terms of molecules substrate broken down per molecule of enzyme per second.

Question 18

Reciprocal plots

Answer 18

Reciprocal plot - The Eadie-Hoftsee plot

Double reciprocal plot - The Lineweaver-Burk plot :Hans Lineweaver & Dean Burk (1934)