Proteins and enzymes, Section 4, relating structure to function (Dr. Taylorson)

Question 1

What is induced fit ?

Answer 1

The induced-fit theory states that the binding of a substrate or some other molecule to an enzyme causes a change in the shape of the enzyme so as to enhance or inhibit its activity.

Question 2

What does molecular recognition depend on ?

Answer 2

complementarity of shape and charge distribution between the ligand and the protein.

Question 3

What shapes can the binding sites have ?

Answer 3

For macromolecules, binding sites can be concave, convex, flat of maybe grooves.
For smaller molecules, they can be clefts, pockets or cavities.

Question 4

Where are catalytic sites often situated in quaternary structure ?

Answer 4

At the interface between subunits or domains in the protein.

Question 5

What types of residues usually make up binding sites ?

Answer 5

A higher than average amount of hydrophobic AAs.

Question 6

How is the NRG for driving binding events usually provided ?

Answer 6

By the displacement of water molecules from the ligand binding site.

Question 7

What is Kd ?

What does it express ?

Answer 7

Kd is the dissociation constant
Kd = [L]*[P] / [P.L]
It represents the affinity of the ligand for the protein.
A low Kd means a tight binding and a high Kd means a weak binding.

Question 8

What is the Kd of the Biotin ligand for avidin protein ?

Of Ca2+ for calmodulin ?

Answer 8

1. 10E-15 M

1. 10E-5 M

Question 9

How is O2 transported in the bloodstream ?

Answer 9

It is bound to proteins, haemoglobin and myoglobin, because it is v poorly soluble in the bloodstream.

Question 10

What is the structure of myoglobin ?

Answer 10

Predominantly helical (80%), compact single polypeptide of 153 AAs. Helices A to F.
It contains a ham group comprising protoporphyrin IX w/ a bound iron in the ferrous state. The iron has 6 coordination bonds, 4 to the flat porphyrin ring system and two perpendicular to it. One is associated w/ the proximal histidine on the F-helix and the other is free to bind oxygen. 
In deoxymyoglobin, the interaction of the iron with the proximal histidine induces a puckering of the porphyrin ring system.

Question 11

Describe the binding of O2 to hemoglobin.

Answer 11

O2 binds to the iron in the haem. This pulls the iron into the plane of the ring of the porphyrin and moves the F-helix.
The binding results in a transfer of an e- from the ferrous ion to O2 creating a superoxide ion. This ion is stabilized by a H-bond donated by the distal histidine on the E-helix.
The H-bond pulls the O2 into the plane of the porphyrin ring and the whole structure changes shape slightly via mvnt of the H-helix to tighten up around the bound O2 molecule.

Question 12

What are the relative affinities of myoglobin and haemoglobin for O2 ?

Answer 12

Myoglobin has a v high affinity for oxygen and is described by a hyperbolic curve.
Haemoglobin has a lower affinity for oxygen and is described by a sigmoidal curve.

Question 13

What are the subunits of haemoglobin ?

Answer 13

Haemoglobin has two alpha and to beta subunits. Each subunit has a bound harm containing an iron in the ferrous state.
Whilst sequentially dissimilar, each subunit is structurally v similar to myoglobin.
Key residues, likes the proximal and distal histidine chains, as well as the haem binding pocket interactions, are conserved.

Question 14

What did the studies of Max Perutz in the 1960s-70s show concerning haemoglobin ?

Answer 14

That O2 can bind haemoglobin in 2 possible state: T and R. O2 has a much higher affinity for the R state and stabilizes the R state.
Haemoglobin is predominantly in the T state.
Upon binding O2 in the T state, it switches to the R state. This causes the subunits to change relative to one another.

Question 15

What is the overall effect of O2 binding to one of the haemoglobin subunits ?

Answer 15

It causes the bound subunit to rotate relative to the other subunits in the structure and the overall shape of the molecule changes from T to R.

Question 16

Why is this switching from T to R state in myoglobin important ?

Answer 16

Heamoglobin binds O2 at high pp (partial pressures) of O2 and releases it at low pp of O2.
Myoglobin does not release would not release O2 effectively at the tissues and haemoglobin would not pick up much O2 in the lungs.
Haemoglobin solves this problem by undergoing the transition for low affinity T state to a higher affinity R state.

Question 17

What is cooperative binding ?

Answer 17

It is the kind of binding exhibit in haemoglobin: the first O2 molecule binds weakly to a subunit in the T state. This leads to a conformational change that is communicated to adjacent subunits making it easier for additional O2 molecules to bind.

Question 18

What is an allosteric protein and why is haemoglobin such a protein ?

Answer 18

An allosteric protein changes shape as a result of the binding of ligands referred to as modulators.

Question 19

Give an example of structural proteins.

Answer 19

F-actin and tubulin are both polymeric structural proteins.

Question 20

What are enzymes ?

Answer 20

Enzymes are molecules that bind substrate and catalyst heir chemical reactions.

Question 21

What is the transition state ?

Answer 21

The highest point in free NRG of reaction pathway from substrate to product.
This is where the bonds are being made and broken and represents a reactions that has reached completion.
It exists for about 10E-15 sec (time it takes for a bond vibration).

Question 22

How do enzymes enhance chemical reactions ?

Answer 22

By lowering the activation NRG and thereby speeding up the rate of the reaction, raising the free NRG of the ground state (by binding the substrate in a way that destabilizes it) or providing an alternative lower NRG path for the reaction.

Question 23

Why do enzymes have pH optima ?

Answer 23

Because the catalytic gourds must be correctly ionized for the binding event to happen.
E.g. superoxide dismutase: +vely charged active site attracts the -vely charged substrate, which is directed to the active site by repulsive forces generated by the large -vely charged field presented y the rest of the protein.

Question 24

Why is positioning and orientation in the active site so important ?

Answer 24

Optimal positioning of the reactive groups effectively increases the concentration species in the confines of the active site.
Orientation can also add a massive factor of catalytic efficiency e.g. holding two molecules next to each other in the AS in the perfect orientation w/o any special chemistry.

Question 25

Why is removal of water from the AS not just thermodynamically important for the binding of substrate ?

Answer 25

Binding substrate often displaces the water from the AS. This leads to local mvnts of AA side chains that often result in correct alignment and orientation of the substrate relative to the catalytic residues.

Question 26

75% of metabolic reactions driven by enzymes fall into 4 general types of chemical transformations.
What are these ?

Answer 26

Redox reactions;
Addition/elimination reactions (e.g. to form double bonds);
The cleavage of esters, amides or esters by reactions w/ water. The formation of esters, amides and acetals requires the removal of water;
Removal of a single C atom from a molecule by loss of a CO2.

Question 27

Why do enzymes recruit cofactors ?

Answer 27

Because reactivity of the AS is limited.

Question 28

What are coenzymes ?

Answer 28

Cofactors that are organic compounds and assist in catalytic chemistry.

Question 29

What are the two different types of cofactors ?

Answer 29

Essential metal ions and coenzymes.
Essentially metal ions can be activator ions (weakly bound) and AS ions (tightly bound).
Coenzymes can be cosubstrates (weakly bound) or prosthetic groups (tightly bound).

Question 30

Give an example of a metalloenzyme.

Answer 30

Cytochrome oxidase, which is bound to an iron.

Question 31

Give an example of a prothetic group.

Answer 31

Rhodopsin has retinal as a prothetic group.

Question 32

What are the 6 types of enzymes ?

Answer 32

Hydrolases
Oxidoreductases
Transferases
Isomerases
Lyases
Ligases

Question 33

What is the different between transition states and intermediate states ?

Answer 33

We cannot determine the structure of transition states because they cannot be isolated !

Question 34

Are all residues in the AS involved in catalysis and binding ?

Answer 34

Absolutely not. Some residues can be involved in catalysis and others in binding but these are two separate events measured by distinct constants.