Enzyme Kinetics

Question 1

How can you tell that enzymes are catalysts?

Answer 1

Small changes in their abundance, efficiency or distribution in tissue can have major implications for disease.

Question 2

How can you tell that enzymes are specific?

Answer 2

Their catalytic behaviours can be used to diagnose disease (see previous material on isoforms).

Question 3

How can you tell that enzymes control well defined chemical reactions?

Answer 3

They can be use in the laboratory to investigate and develop a wide variety of diagnostic kits and therapeutic drugs.

Question 4

How do we describe enzyme behaviour?

Answer 4

To understand how an enzyme works we need to know how to concentration of substrate (S) affects the rate of the catalysed reaction (V).
2 parameters are used to describe this relationship Vmax and KM (M is subscript).
The concentration of the enzyme is set, only the concentration of substrate is being changed.
Vmax is the maximal rate at which moles of substrate are being converted into product.

Question 5

How is the Michaelis-Menten model is used to explain the relationship between Vmax and KM?

Answer 5

The activation energy barrier makes the enzyme-substrate complex (ES) inherently unstable. This could go either forward or backwards (This is what happens when you reach an equilibrium).
E + S = ES - E + P
(= means the reaction is reversible)
K1 is the forward rate constant for enzyme association with the substrate.
K-1 is the backwards rate constant for enzyme dissociation from the substrate.
K2 is the forward rate consatn of enzyme conversion of substrate to product (P)
The relationship between K1, K-1 and K2 is called the Michaelis Constant, KM as:
KM=(K-1 + K2)/K1

Question 6

How are Vmax and KM measured?

Answer 6

Measure initial reaction velocity,V0 at a known substrate concentration.
Repeat at increasing substrate concentrations.
(When you increase the substrate concentration the rate of the reaction increases and so the V0 is greater).

Question 7

How do you plot initial reaction rates against substrate concentration?

Answer 7

Initial reaction rates V0 are plotted as a function of the substrate concentration, (S).
At infinite substrate concentration (not possible in the lab) the initial reaction rate approaches a maximal rate, Vmax.
The michaelis constant KM is equivalent to the substrate concentration where the intial reaction rate is half-maximal.
BIG “BUT” - It is not straightforward to determine Vmax and KM from such a graph because the kinetics are not linear - reaction velocity never quite reaches true Vmax!

Question 8

How does the Lineweaver-Burk plot - enable accurate determination of Vmax and KM?

Answer 8

Instead of V against (S) we plot 1/V against 1/(S) - a double reciprocal plot is produced.
Vmax = intersection of the straight line with the Y axis
KM = intersection with the X axis

Question 9

What is the definition of Vmax?

Answer 9

Vmax is the maximum velocity of a reaction

Question 10

What are some clinical examples of why Vmax and KM matter?

Answer 10

Glucose homeostasis and maturity - onset diabetes of the young (MODY)
Red Blood Cell
Hexokinase KM = 0.05mM Glucose: Low KM maintains energy production in red blood cells by glycolysis even if glucose levels fall dramatically.
Liver and Pancreas
Glucokinase KM = 5mM Glucose
High KM enables glucose sensing & homeostasis. It’s abundance in liver is regulated by insulin. Excess blood glucose is metabolised.
MODY = loss of glucokinase activity
Loss of insulin - mediated glucose homeostasis.
Normal fasting blood glucose = 5mM
Both catalyse: glucose + ATP - glucose-6-phosphate + ADP

Question 11

Another clinical example of why Vmax and KM matter is Oxygen Regulation of Gene Expression can you explain how this works?

Answer 11

Prolyl Hydroxylases (act as oxygen sensors) When they sense the oxygen levels are low it causes a hypoxia Inducible factor ( a transcription factor) to transcript genes for surviving hypoxia

• red blood cell synthesis
• Blood vessel growth
• Anaerobic survival pathways

Question 12

What conditions does proline hydroxylase require and what does proline hydroxylase enable to happen?

Answer 12

Require O2 as a substrate to regulate the HIF transcription Factor.
High KM for O2 - even higher than atmospheric oxygen levels.
Enables oxygen sensitive oxygen sensing over physiological ranges of PO2.
The erythropoietic (red blood cell synthesis) response to high altitude - a paradigm of oxygen sensing.
The higher a person is, e.g. 45oom above sea level compared to at sea leave, the greater the loss of prolyl hydroxylase activity and the higher the number of samples of a higher percentage of haematocrit.
Monge’s Disease (pronounced “monhay”)
- Excessively high haematocrit
- Loss of arterial oxygen saturation
- Affects 10% of Andean High altitude Native population
- Loss of O2 regulation of Epo production.
Loss of Prolyl Hydroxylase activity in Von Hippel Lindau Syndrome
- Excessive blood vessel growth (angiogenesis)
- Hemangiomas in brain
Subcutaneous hemangiomas track spinal cord.

Question 13

What are the different types of reversible and irreversible inhibition of enzyme activity?

Answer 13

Reverse Inhibiton can be either competitive inhibition or non-competitive inhibition.
Competitive inhibition is where the inhibitor binds to the active (catalytic) site and blocks substrate access. Orthosteris inhibition - at the same site.
Non-competitive inhibition is where the inhibitor binds to a site other than the catalytic centre; inhibits enzyme by changing its conformation. Allosteric inhibition- at different sites.
Competitive inhibition can also be a form of irreversible inhibition. Inhibition cannot be reversed. Usually involves formation or breakage of covalent bonds in the enzyme complex.

Question 14

What happens to the KM and Vmax in competitive inhibition?

Answer 14

Vmax does not change

KM varies

Question 15

What is a use for competitive inhibition in the clinic?

Answer 15

Methanol Poisoning
Methanol is a substrate for alcohol dehydrogenase (ADH).
Causes severe tissue damage and blindness by conversion to formaldehyde. Also drives metabolic acidosis.
ADH KM for ethanol 20X greater than methanol.
Treat patient with 40% ethanol in combination with dialysis and ventilation.

Question 16

What happens to the KM and Vmax in non-competitive inhibition?

Answer 16

Vmax varies

KM does not change.

Question 17

When is Feedback inhibition commonly used and what is the process of feedback inhibition?

Answer 17

Inhibition of rate limiting enzymes by end product is a common mechanism of allosteric control.
Many enzymes act in series in a biochemical pathway to produce an end product. In many of these cases, the end product itself acts as an inhibitor of the slowest enzyme i.e. the rate limiting enzyme in the process which usually occurs near the beginning of the pathway. This effectively shuts down the subsequent enzymes by limiting their substrate availability. This is an extremely efficient control mechanism as it avoids build ups of intermediates in the pathway which in some cases can be highly reactive and therefore dangerous biochemically.
(The series of enzyme-catalysed reactions constitutes a pathway.
Feedback inhibition - final product blocks an early reaction and shuts down whole series).

Question 18

In what ways do Allosteric enzymes do NOT follow Michaelis-Menten kinetics?

Answer 18

Increasing substrate concentration results in a sigmoidal curve, instead of a hyperbola.
Shows cooperative behaviour. Allosteric factors modulate enzyme kinetic behaviour.
Can be controlled by allosteric inhibitors AND allosteric activators.

Question 19

What is an important example for allosteric regulation and why does myoglobin not show cooperativity?

Answer 19

Binding of oxygen to haemoglobin causes positive co-operativity.
Controlled by:
- H+
- CO2
- 2,3 biphosphoglycerate a side product of glycolysis.

Since it’s an allosteric enzyme, addition of one oxygen molecule increases likelihood of enzyme binding to another oxygen molecule; thus cooperativity is positive (Cooperativity is only for molecules/proteins that contain a quarternary structure) myoglobin is a single stranded protein, thus it doesn’t show cooperativity.

Question 20

How is the Michaelis - Menten relationship expressed as an equation?

Answer 20

It describes the rate of catalysts as a function of substrate concentration. This describes a hyperbola.
V = (Vmax(S)/KM = (S))
can be transformed to
1/V = KM/Vmax X 1/(S) + 1/Vmax
y = m X x + b
which is the equation for a straight line. Now we can accurately determine Vmax and KM.

Question 21

What is the definition of KM?

Answer 21

Km is the concentration in moles of S which gives 1/2 Vmax

KM = (S) at 0.5 Vmax

Question 22

How much substrate does an enzyme with s low KM need to work at half-maximal velocity?

Answer 22

A low Km means that an enzyme only needs a little substrate to work at half-maximal velocity

Question 23

How much substrate does an enzyme with a high KM need to work at half- maximal velocity?

Answer 23

A high KM means that an enzyme needs a lot of substrate to work at half-maximal velocity

Question 24

Do enzymes with the same Vmax automatically have the same KM?

Answer 24

Enzymes can display the same Vmax and have different KM.

KM is a concentration, therefore units are molar or moles/l or moles/dm3

Question 25

There are questions at the end of the slideshow to help study this topic

Answer 25

LOOK AT THEM!!!!!!!!!