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Transcript
3/27/2017
Oxidative Phosphorylation
Pratt and Cornely, Chapter 15
Goal: ATP Synthesis 1
3/27/2017
Overview
• Redox reactions
• Electron transport chain
• Proton gradient
• ATP synthesis
• Shuttles
Standard Reduction Potential
½ O2 + 2 e + 2 H+ H2O
• “potential to be reduced”
• Listed according to oxidized compound being reduced
• High RP means that the compound is a strong oxidizer (ie O2)
• Its conjugate is a strong reducing agent (ie NADH)
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Half Reactions
• Reduction potential written in terms of a reduction half reaction
• Aox  Ared
• Example: Calculate Eo for reduction of FMN by NADH, given the Eo of FMN to be ‐0.30V
• Then calculate Go’
NADH + FMN  NAD+ + FMNH2
Redox reactions: electricity
•
•
•
•
NAD+: Eo’ = ‐.32
FMN: Eo’=‐.30
Eo’ = +0.02V
Calculate Go’:2 e‐
transfer
• Go’ = ‐nFEo’ = ‐2(96485 J/mol
V)(0.02V)
= ‐3.9 kJ/mol
3
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Numerous Redox Substrates
• Chain of increasing reduction potential (potential to accept e‐)
• Mobile carriers vs. within Complex
• Types of redox groups
– Organic cofactors
– Metals (iron/sulfur clusters)
– Cytochromes
– O2
Overall chain from NADH
•
•
•
•
•
•
•
NADH
Complex I
Q
Complex III
Cytochrome c
Complex IV
O2
4
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Within Complexes
• FAD/FMN
– Prosthetic group
– Bridge from 2 e‐ donors to 1 e‐ donors
• Iron‐sulfur clusters
Coenzyme Q: Mobile Carrier
• 1 or 2 e‐ carrier, 1 e‐
acceptor/donor
• Ubiquinone is a mobile carrier
• Can diffuse through nonpolar regions easily
• “Q pool” made by Complex I and Complex II (and others)
5
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Cytochrome c
• Mobile carrier
• Protein/heme
• 1 electron carrier
Oxygen: the final electron acceptor
• Water is produced—has very low reactivity, very stable
• Superoxide, peroxide as toxic intermediates
• Overall reaction
NADH + H+ + ½ O2  NAD+ + H2O
6
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Flow Through Complexes
7
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Compartmentalization
Protonmotive Force
• NADH + H+ + ½ O2  NAD+ + H2O + 10 H+ pumped
• succinate + ½ O2  fumarate + H2O + 6 H+ pumped
8
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Complex I
• NADH  Q through
– FMN
– Iron‐sulfur clusters
• “Q pool”
• 4 protons pumped
– Proton wire
Complex III
•
•
•
•
QH2  cytochromes
4 protons pumped
Through Q cycle
Problem 10: An iron‐
sulfur protein in Complex III donates an electron to cytochrome c. Use the half reactions below to calculate the standard free energy change. How can you account for the fact that this process is spontaneous in the cell?
FeS (ox) + e‐  FeS (red) Eo’ = 0.280 V
Cyt c (Fe3+) + e‐  cyt c (Fe2+) Eo’ = 0.215 V
9
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Complex IV
• Cytochromes  O2
• Stoichiometry of half of an oxygen atom
10
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Complex II (and others)
• Non‐NADH sources
– Complex II (part of the citric acid cycle)
– Fatty acid oxidation
– Glycolysis NADH shuttle (Glycerol‐3‐phosphate) • Bypasses Complex I – Loss of 4 protons pumped
Two paths of NADH into Matrix
• NADH of glycolysis must get “into” matrix
• Not direct
• Needs either
– malate‐aspartate shuttle (liver)
– Glycerol‐3‐phosphate shuttle (muscle)
• Costs 1 ATP worth of proton gradients, but allows for transport against NADH gradient
11
3/27/2017
Glycerol‐3‐phosphate Shuttle
• Glycerol phosphate shuttle (1.5 ATP/NADH)
• Produces QH2
• Operational in some tissues/circumstances
Net ATP Harvest from Glucose
• Glycolysis = 2 ATP
– Plus 3 or 5 ATP from NADH
– What leads to difference in this case?
• Pyruvate DH = 5 ATP
• Citric Acid Cycle = 20 ATP
• Total: 30‐32 ATP/glucose
12
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Overall
• Chemiosmosis
• 10 protons shuttled from matrix to intermembrane space
• Makes pH gradient and ion gradient Protonmotive Force and Oxidative Phosphorylation
• Flow of electrons is useless if not coupled to a useful process
– Battery connected to wire
• Proton gradient across mitochondrial membrane
13
3/27/2017
Proton Gradient
• Gradient driven by concentration difference + charge difference
– Assume pH 0.5 and 170mV membrane potential
Using the Gradient
• Coupled to ATP synthesis
• Uncouplers used to show link of oxygen uptake and ATP synthesis
14
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Uncouplers
• “Uncouple” protonmotive force from ATP synthase
– DNP pKa / solubility perfectly suitable
Respiratory Poisons
• Other respiration poisons
– Cyanide—binds Complex IV in place of oxygen
15
3/27/2017
Complex V: ATP Synthase
• Molecular motor
• Rotor: c, , 
– Proton channel
Proton Channel
• Protons enters channel between rotor and stator • Rotor rotates to release strain by allowing proton to enter matrix
• 8‐ 10 protons = full rotation
– Species dependent
16
3/27/2017
• “Stalk” () moves inside the“knob”—
hexameric ATP synthase
• Knob held stationary by “b”
17
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Hexameric Knob
Binding‐Change Mechanism
• Stalk causes ATP synthase to have three different conformations: open, loose, tight
• In “tight” conformation, energy has been used to cause an energy conformation that favors ATP formation
18
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Problem 39
• How did these key experiments support the chemiosmotic theory of Peter Mitchell?
– The pH of the intermembrane space is lower than the pH of the mitochondrial matrix.
– Oxidative phosphorylation does not occur in mitochondrial preparations to which detergents have been added.
– Lipid‐soluble compounds inhibit oxidative phosphorylation while allowing electron transport to continue.
Energy Accounting
• ATP costs 2.7 protons
– 8 protons produces 3 ATP
• NADH pumps 10 protons when 2 e‐ reduce ½ O2
– 4 protons in Complex I, 4 protons in Complex III, and 2 protons in Complex IV
• P/O ratio‐‐# of phosphorylation per oxygen atom
– 10H+/NADH (1 ATP/2.7 H+) = 3.7 ATP/NADH
– 6H+/QH2 (1 ATP/2.7 H+) = 2.3 ATP/QH2
• In vivo, P/O ratio closer to 2.5 and 1.5 due to other proton “leaking”
– i.e. importing phosphate
19
3/27/2017
Active Transport of ATP
• ATP must go out, ADP and Pi must go in
• Together, use about 1 proton of protonmotive
force
Regulation of Oxidative Phosphorylation
• Electron transport is tightly coupled to ATP production
– Oxygen is not used unless ATP is being made
– Avoid waste of fuels
– Adding ADP causes oxygen utilization
– Respiratory control
20
3/27/2017
Problem 47
• A culture of yeast grown under anaerobic conditions is exposed to oxygen, resulting in dramatic decrease in glucose consumption. This is called the Pasteur effect. Explain.
• The [NADH]/[NAD+] and [ATP]/[ADP] ratios also change when an anaerobic culture is exposed to oxygen. Explain how the ratios change and what effect this has on glycolysis and the citric acid cycle in yeast.
21