Cooperativity and Transition State Theory

Question 1

Heme

(or Haem)

Answer 1

Present in myoglobin and haemoglobin
One saturated site for O2 binding
Fe2+ favours octahedral complexes

Question 2

Myoglobin

Answer 2

154 residues, 8 α-helices
Single heme group
One site occupied by solvent.
Porphyrin group with O binding site
Binds O2, favoured at higher temps

Question 3

What is p50?

Answer 3

Partial pressure when we have 50% of binding sites
Equal to Kd

Question 4

What is Kd?

Answer 4

Equilibrium constant for the binding
Kd = [Mb][O2]/[Mb.O2]
Equal to p50

Question 5

Haemoglobin

Answer 5

287 residues, 2 unique proteins
4 heme groups
Dimer of heterodimers

Question 6

Binding for Haemoglobin

Answer 6

There are 4 haems and the presence of an already bonded O2 will make binding a second more favourable.
This means that less Haem needs to be made, as it is more likely to be fully laden.

Question 7

Why does O2 binding make O2 binding more favourable?

Answer 7

Preorders bonding sites, so less entropic change when the next is bound

Question 8

What is the Bohr effect?

Answer 8

Increasing CO2 causes binding to O2 to be less favoured, as CO2 binds instead.
Increasing acidity has the same effect

Question 9

Other grouups binding changes the _ of haem

Answer 9

Conformation

Question 10

What does the Hill-Langmuir equation tell us?

Answer 10

The fractional occupation of binding sites

Question 11

Hill-Langmuir Equation

Answer 11

θ =[L]^n/(Kd+[L]^n ) =1/(1+( Ka/[L] )^n )
Langmuir isotherm?
Theta is fractional occupation

Question 12

What does n tell us in the Hill-Langmuir Equation?

Answer 12

n>1 positively cooperative binding
o Affinity of additional ligands increases
o n for Haem is usually 1.7-3.2
n<1 negatively cooperative binding
o Affinity of additional ligand decreases
n=1 non-cooperative binding
o Affinity is independent of if anything else has bound

Question 13

What is the lock and key model?

Answer 13

Active site is prearranged to have a complementary 3D structure and ending to the substrate

Question 14

What is the induced fit model?

Answer 14

The active site undergoes structural change upon the binding of the substrate.
Actually aligns closer to thte transition state.

Question 15

It is catalytically advantageous for the structure of the undistorted enzyme to align with the _ structure of the substrate(s)

Answer 15

Transition state
Therefore induced fit is a better model

Question 16

Transition state theory equation

Answer 16

kTS = κ (kB x T)/h K‡
* kTS rate constant for passage through transition state
* κ transmission coefficient (accounts for not all activated complex passing through TS) (can assume 1 if not given)
* kB Boltzmann constant (R/NA)

Question 17

What is the transition state?

Answer 17

• TS is in equilibrium with the reactant
• TS is a transient species
• TS decays to form the product

Question 18

Limitations of transition state theory

Answer 18

• Only applies to ‘classical’ systems – quantum not messing with it
• Requires TS to be long lived enough to be able to follow a Boltzmann distribution
• Only applies to relatively low energy reactions (not excited state)

Question 19

Assumptions of the transition state theory

Answer 19

ES and ES‡ are in rapid equilibrium
Rate of reaction is governed by the decomposition of ES‡ to EP
The transient TS has a lifetime in the order of a vibration (~1 ps)

Question 20

What is the enthalpy of activation?

Answer 20

The energy required (by breaking and making bonds) to achieve the transitions state structure

Question 21

What is the entropy of activation?

Answer 21

The change in entropy required to achieve the transition state structure