Oxidative Phosphorylation – the ATP Synthase

Question 1

What is the function of the ETC?

Answer 1

The ETC converts reducing equivalents from the TCA cycle to a H+ gradient across the inner mitochondrial membrane

Question 1

Explain the proton gradient

Answer 1

Complexes 1, 3 & 4 transport H+ to the intermembrane space, so the [H+] is higher there (i.e. pH lower) than in the matrix.

Energy can be released by the H+ moving back down their gradient.

This is Mitchell’s chemiosmotic theory, and all ionic gradients store energy.

Question 1

The ATP Synthase (Complex V) located in the inner mitochondial membrane

Answer 1

• The ATP synthase is a protein complex that uses the H+ gradient to synthesize ATP.
• It has a “knob-and-stalk” structure
• F₁ (knob) contains the catalytic subunits and sticks into the mitochondrial matrix.
• F₀ (stalk) has a proton channel which spans the membrane.

Hence it is also called the F₁F₀-ATPase

Question 1

Structure of the ATP Synthase

Answer 1

Transmission electron micrograph of mitochondrial membranes show the large complexes on the cristae

Question 2

Crystal structure-derived space-filling models of the ATP Synthase

Answer 2

X-ray crystallography and modelling of the purified protein complex allows generation of images such as those in this slide.

Note that the colours schemes in the 2 figures are different (blue in one ≠blue in the other). I have now turned the crystal structure above upside-down to match the diagram on right.

Question 4

ATP Synthase Subunit Composition

Answer 4

Question 5

Molecular Structure of the F₁F₀-ATPase

Answer 5

Question 6

How does the ATP Synthase use the proton gradient to make ATP?

Answer 6

It is a rotational motor, similar to a watermill but on a nano scale

Question 7

The ATP Synthase is a Rotational Motor - simple overview

Answer 7

• The H+ gradient across the membrane drives H+ passage through F₀
• This causes the c-ring to rotate in the membrane
• This drives rotation of the g-spindle that connects F₀ to F₁
• This causes conformational changes in the ab-hexamer that result in ADP + Pi > ATP + H₂O
• The αβ-hexamer is static – held in place by b and d subunits.

Question 8

How does the proton gradient turn the c-ring?

Answer 8

1. Protons flow from the mitochondrial intermembrane space into the upper half-channel of subunit a.
2. From here, each H+ flows onto a c-subunit (one H+ per c subunit).
3. Protonation takes place at Asp 61:
4. The loss of –ve charge alters the affinity between c and a-subunits.
5. The protonated c subunit is repelled away from subunit a, so causing the cylinder of c subunits to rotate clockwise (as viewed from above).
6. As protons continue to flow, each c subunit becomes protonated at Asp 61.
7. When a protonated subunit c reaches the lower half channel, it is released from Asp 61 and flows down the channel into the mitochondrial matrix.
8. Rotation of the cylinder of c subunits causes rotation of the attached g subunit.

Question 9

A schematic view of the “H+ turbine” and how H+ flow through F₀ drives rotation

Answer 9

As yet there is not too much direct evidence for this appealing model.
It has a ”wheel” of 9-12 hydrophobic “c” subunits, each with one key Asp COOH group. On one side, probably 2 of these exist as COO-, paired with +ve Arg (in “a” subunit). The rest must be neutral in hydrophobic membrane core. There are channels for H+ access to opposite faces of the membrane at each end of the COO- section. Dp will make wheel rotate.

Question 10

3-D view of the F0 Subunit and c-ring turning

Answer 10

The number of c subunits varies between organisms In eukaryotes. One full turn produces 3 ATP (you’ll see why in a moment) but the H+/ATP varies depending on the number of c-subunits in the ring.

Question 11

If 9 c-subunits in the ring, how many protons must go through for 1 full turn of the ring?

Answer 11

If 9 c-subunits in the ring, 9 protons must go through for 1 full turn of the ring

Question 12

How does the c-ring turning converts ADP to ATP?

Answer 12

• Clockwise rotation of the γ subunit (driven by the c subunit cylinder) causes a conformational change of the 3 β subunits:
• In the first form (Open) the binding of products (ATP) or substrates (ADP and Pi) to the b subunit is unfavourable (the eqm is > off).
• In the second form (Loose) binding of ADP and Pi from the mitochondrial matrix to the β subunit is favourable (eqm is > on).
• In the third form (Tight) the b subunit binds the ADP and Pi very tightly and converts them (reversibly) into ATP. Note that the Tight form cannot release its bound nucleotides.
• Further rotation of the spindle returns the b subunit to the Open form so product is released (molecules are only released from the b subunit when it is in the Open conformation).

Question 13

When seen form the IMS side, which direction do the C-ring and spindle turn?

Answer 13

The c-ring and spindle turn clockwise when seen from the IMS side

Question 14

The b-subunit cycles through conformations:

Answer 14

Loose (L) > Tight (T) > Open (O)

Question 15

Describe

Answer 15

• The conformational changes are driven by the g-spindle and therefore by energy from the H+ gradient.
• 3 protons are required for each conformational change (~9 for a full turn = 3 conformational changes).
• Remember: it’s the spindle that is rotating, within the alpha-beta complex, not the beta subunits rotating.

Question 16

The asymmetrical g-spindle forces the b-subunits into different shapes that squeeze ADP and Pi together

Answer 16

3 sides of the gamma-spindle all have different shapes and force the beta-subunit they are in contact with into a different conformation.

Question 17

ATP Synthase or F1F0ATPase? ATP can drive the rotation in reverse if the H+ gradient is removed

Answer 17

All enzymes are theoretically reversible, i.e. the reaction can go either way depending on the concentrations of substrates, products and the free energy.

If the part of the enzyme which drives the clockwise rotation is removed (the c cylinder, subunit a and subunit b₂) and ATP is supplied, the g subunit will rotate backwards (anticlockwise), driven by the hydrolysis of ATP!

Question 18

Describe the control of respiration

Answer 18

· Oxidation of NADH & phosphorylation of ADP are tightly controlled (usually tightly coupled).

· The delivery of ATP must be flexible to accommodate changes in activity.

· The rate of respiration is controlled by the ATP mass action ratio:

[ATP]/[ADP].[Pi]

· ADP level is an important contributor to this.

· The build up of the proton motive force (PMF) prevents electron flow through the chain.

· Important sites of control are thought to be Cyt c Oxidase and the Adenine Nucleotide Translocase.

Question 19

Actually, it’s not easy to work this out as other processes use energy from the H+ Gradient

Answer 19

The H⁺ gradient is used to power exchange reactions for ATP, ADP, Pi and TCA cycle metabolites across the inner mitochondrial membrane.

Question 20

Other uses of the Mitochondrial proton gradient: Brown Bears and Mitochondrial Function

Answer 20

• Brown Bears are hibernating animals.
• They have a lot of brown adipose tissue that is rich in mitochondria.
• The mitochondria contain a special protein in the inner membrane that can be activated as an uncoupler.

The mitochondrial uncoupling proteins (UCP) dissipate the proton gradient (they are a type of anion transporter).

UCP-1 is also called thermogenin.

Question 21

STOICHIOMETRIES IN RESPIRATION

Answer 21

Question 22

Explain the Energy Yield from Ox Phos

Answer 22

• Because these processes dissipate some of H+ gradient, the energy yield is less than expected.
• This affects the P/O ratio (ATP formed per ½ O₂)
• 2 e- are required for each ½ O2 reduced to H₂O
• 3 H+ translocated for each ATP synthesized
• For each NADH (2e-) 10 H+ exported = 2.5 ATP
• For each FADH2 (2e-) 6 H+ exported = 1.5 ATP

Question 23

Aerobic Glucose Utilization

Answer 23

• 2 ATP from glycolysis / glucose
• 2 GTP from TCA cycle /glucose
• 2 NADH from glycolysis (but energy lost on > mito)
• 2 NADH from Pyruvate DH / glucose
• 6 NADH from TCA cycle / glucose
• 2 FADH2 from TCA cycle / glucose

2 + 2 + (2 x 1.5) + (8 x 2.5) + (2 x 1.5) = 30 ATP / glucose

i.e. much more ATP than from anaerobic respiration

(2 NADH are formed but cannot be used for energy in the absence of oxygen, so they are recycled by Lactate Dehydrogenase)