Oxidative Phosphorylation 2

Question 1

How can the transfer of 2e- from NADH to O2 be written as

Answer 1

1. NADH + H+ + ½ O2 NAD+ + H2O

Question 2

How would you calculate gibbs energy for the transfer of 2e-

Answer 2

1. DGo’ = –nFDEo’ = –nF[Eo’(acceptor) – Eo’(donor)]

Question 3

How do we determine DG available from H+ gradient across mitochondrial membrane

Answer 3

1. DG = RT ln (c2/c1) + ZFDy
2. (c2/c1) = concentration ratio for the ion that moves
3. Z = absolute value of its charge
4. F = Faraday (96.5 kJ mol-1 V-1)
5. Dy = electrical potential difference across membrane (V)

Question 4

In actively respiring mitochondria what is DG from H+ gradient per NADH

Answer 4

1. DpH is about 0.75 pH units
2. Dy is about –0.15 V
3. DG = –0.74 kJ mol-1 + –14.48 kJ mol-1 = –15.21 kJ mol-1
4. 10 H+ are available from each NADH, so DG = –152.1 kJ mol-1 per NADH

Question 5

Where does most of the free energy available from the H+ gradient come from

Answer 5

1. Most of the free energy available from the H+ gradient (the proton-motive force) in mitochondria is derived from Dy (–14.48 kJ mol-1) rather than DpH (–0.74 kJ mol-1)

Question 6

What is the efficiency of energy conversion by electron transport system based on energy difference

Answer 6

1. Efficiency of energy conversion by electron transport system is about 70%:
2. 152 kJ mol-1 (generated from proton gradient) / 220 kJ mol-1 (energy required to transfer electrons?)≃ 0.7
3. Energy difference of about 70 kJ mol-1 is lost as heat and contributes to thermogenesis
4. DG (as opposed to DGo’) of NADH is likely to be more negative than –220 kJ mol-1

Question 7

What is the efficiency of energy conversion by electron transport system based on energy difference based on number of ATP molecules made

Answer 7

1. estimated DG for ATP synthesis in mitochondrial matrix is –40 kJ mol-1
2. Hence theoretical yield per NADH oxidized is 152 kJ mol-1 / 40 kJ mol-1 ≃ 3.8 ATP
3. Actual yield per NADH oxidised is about 2.5 ATP, so efficiency is 2.5 / 3.8 ≃ 66%

Question 8

Describe the main components of ATP synthase

Answer 8

1. F1
2. F0
3. Stalk
4. Associated polypeptides

Question 9

Describe the subunit composition of F1 and its function

Answer 9

1. a3b3yδe
2. Beta- contains the ATP synthase site
3. δ- forms the gate coupling the F0 proton channel with F1

Question 10

Describe the subunit composition of F0 and its function

Answer 10

1. 4-5 types of subunit including 6-10 copies of DCCD-binding proteolipid
2. DCCD-binding proteolipid oligomer forms the proton channel

Question 11

Describe the subunit composition of the stalk and its function

Answer 11

1. One copy each of OSCP and F6

2. Required to bind F0 to F1

Question 12

Describe the subunit composition of the associated polypeptides and its function

Answer 12

1. IF1 and F(B)

2. IF1 inhibits ATP- hydrolyses and binds to the F1 B subunit

Question 13

Describe the difference between the bacterial and yeast F1 component

Answer 13

1. The bacterial ATP synthase complex contains a subunit called ε that associates with the F1 component. – Sits in cytoplasm as does not have mitochondria
2. The δ subunit of the yeast ATP synthase complex is homologous to the bacterial ε subunit.
3. Equivalent subunits with different names e.g. yeast δ = bacterial ε

Question 14

What does the OSCP subunit stand for

Answer 14

1. The OSCP subunit in the yeast ATP synthase complex is the oligomycin-sensitivity-conferring protein.
2. C sub unit is part that rotates
3. Rotary motor- One direction – synthesises ATP, Other direction- hydrolyses ATP

Question 15

What is the main role of F1 and F0

Answer 15

1. F1= catalytic activity

2. F0= H+ channel

Question 16

What are the general structures of the F0 and F1 subunit

Answer 16

1. F0 is water insoluble transmembrane protein composed of 10-12 subunits
2. F1 is water soluble peripheral membrane protein composed of 5 types of subunits

Question 17

What can oligomycin do

Answer 17

1. Inhibits ATP synthase by binding to a subunit of F0 (not OSCP)
2. This means it interferes with H+ transport through F0

Question 18

What are the 3 functional units of ATP synthase

Answer 18

1. Stator:
   2) Rotor:
   3) Headpiece:

Question 19

Describe the stator

Answer 19

1. a subunit with half-channels for H+ to enter and exit FO, plus stabilizing arm (b, d, h and OSCP
2. The stator holds the headpiece in place so that it does not turn with the rotor.

Question 20

Describe the rotor

Answer 20

1. c + g + d + e rotate as H+ enter and exit c-ring

2. the rotor is responsible for translating proton-motive force into protein conformational changes in the headpiece.

Question 21

Describe the headpiece

Answer 21

1. hexameric a3b3 unit responsible for ATP synthesis
2. a3b3 hexamer contains the 3 catalytic sites, on outer edge of each b subunit
3. g subunit extends inside a3b3 hexamer and is attached to the c ring

Question 22

What ae the 3 phases of the binding change mechanism

Answer 22

1. Translocation of protons carried out by F0
2. Catalysis of formation of the phosphoanhydride bond of ATP carried out by F1
3. Coupling of the dissipation of the proton gradient with ATP synthesis which requires interaction of F1 and F0

Question 23

What are the different ATP binding affinities of the 3 beta subunits

Answer 23

1. Tight: ATP bound
2. Loose: ADP and Pi bound
3. Open: ATP is released and accept ADP and phosphate

Question 24

Why are there 3 different beta conformations

Answer 24

1. g directly contacts all three b subunits, but each interaction is distinct giving rise to 3 different b conformations

Question 25

What drives the conformational change that transitions the subunits to different sites

Answer 25

5. The number of H+ required for each 120° turn of the γ subunit depends on the number of subunits in the c ring; the yeast mitochondrial ATP synthase complex appears to require ∼3 H+ for each ATP synthesized

Question 26

What drives the conformational change that transitions the subunits to different sites

Answer 26

1. Driven by free energy supplied by proton flow
2. The impermeability of the inner mitochondrial membrane allows an electrochemical gradient to be established across the membrane during the H+ translocation associated with electron transport
3. The only way for the H+ to reenter the matrix is through the F0 portion of the proton-translocating ATP synthase
4. The electrochemical gradient builds up until the free energy required to transport H+ balances the free energy of electron transport
5. Electron transport then must cease
6. ATP synthesis by dissipating the electrochemical gradient allows the electron transport to continue.

Question 27

How many H+ are required for each 120 degrees turn of Y subunit

Answer 27

1. The number of H+ required depends on the number of subunits in the c ring;
2. the yeast mitochondrial ATP synthase complex appears to require ∼3 H+ for each ATP synthesized

Question 28

Describe the first experiment that shows that ATP synthase rotates

Answer 28

1. Took catalytic part of ATP -gamma sub-unit and attached long fluorescent actin filament to the gamma sub-unit
2. Attached to glass slide
3. Fed it ATP
4. Rotary motion can be seen from fluorescence
5. The g subunit rotates in 120º steps.
6. The fluorescence micrographs show the position of the fluorescent actin polymer at 133 ms intervals.

Question 29

Describe the second experiment that shows that ATP synthase rotates

Answer 29

1. F1 component as a nanomotor driving ATP synthesis in absence of electrochemical proton gradient
2. Attach a magnetic bead to the gamma subunit
3. Made a chamber with electromagnets
4. Fed atp
5. Use electric field to drive synthesis or hydrolysis – can reverse the field to change direction of rotation
6. ATP synthesis or ATP hydrolysis can be catalysed by a3b3 headpiece simply as function of rotational direction of g subunit imposed by electromagnets
7. Proof that rotational motion of g subunit drives ATP synthesis, independent of electrochemical H+ gradient

Question 30

How does H+ movement cause rotation of g subunit?

Answer 30

1. Two-channel model
2. H+ neutralizes D59 allowing c subunit to rotate 36° into hydrophobic membrane
3. Proton binds to aspartate side chain – neutralises positive charge- can move into hydrophobic membrane
4. This rotation allows D59 in a different c subunit to access the second half-channel in the a subunit and exit the channel because of the low H+ concentration on matrix side
5. Carousel analogy: each H+ must ride once around the c ring carousel to exit into matrix