Lecture Notes for

Chapter 15 Oxidative Phosphorylation

Essential Biochemistry Third Edition

Charlotte W. Pratt | Kathleen Cornely

Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.

15-1 The Thermodynamics of Oxidation-Reduction Reactions

oxidant reductant

還原 = 獲得電子 = 與氫鍵結增加

Reduction potential indicates a substance’s

tendency to accept electrons

Half reaction

將還原電位量化 還原 = 分子接受電子 電位 = 伏特的改變 還原電位 = 分子接受電子後伏特的改變

The free energy change can be

calculated from the change in

reduction potential

Rule

• 不同分子有不同還原電位

• 還原電位高 = 分子拉電子力強

• Δ ε °’ = ε °’ acceptor – ε °’

donor

• 為方便記憶，均以獲得電子方向

之還原電位進行計算

• Donor : NADH，ε °’ = -0.315

• Acceptor : Q，ε °’ = +0.045

• 故Δ ε °’ =0.045 – (-0.315) =

0.36

• 淨反應如下：

The actual reduction potential depends on

the actual concentrations of oxidized and

reduced species.

The Nernst Equation

R = Gas Constant = 8.3145 J‧mol-1‧K-1 T = temperature in Kelvin

n = # of electrons F = Faraday’s constant = 96,485 J‧V-1‧K-1

實際還原電位受氧化態與還原態濃度影響，以Nernst equation計算之，其中：

換算為自由能：

亦即還原電位為正值時，Δ G°’為負值，可自發之放熱反應

Δ ε °’= +0.360

Δ ε °’= +0.190

Δ ε °’= +0.580

15-2 Mitochondrial Electron Transport

• Respiration: 還原能最終傳遞至氧氣，使其還原為H2O，因此消耗氧氣而獲得能量，稱之為呼吸作用

Electron transport takes place in the

mitochondrion.

© 2014 John Wiley & Sons, Inc. All rights reserved.

粒線體結構 可通透

不可通透

生產ATP所必需原料： • NADH • ADP • Pi

The malate-aspartate shuttle

transports reducing agents across

the inner mitochondrial membrane.

何種運送機制可使NADH由粒線體外轉移至粒線體內?

NADH除了由TCA生產，亦可由粒線體外轉換而補充:

A different transport system is used to

move ATP from the matrix to the cytosol.

ATP translocase

:負責ATP換出，ADP換入

Symport protein

:利用H+的濃度差同向運送Pi進入

Oxidative phosphorylation

© 2014 John Wiley & Sons, Inc. All rights reserved.

Complex I 接受來自NADH之電子，排出H+

Complex II 接受來自FADH2之電子

Complex III 匯整電子流，排出H+

Complex IV 將O2還原為水，排出H+

Complex I binds ubiquinone.

• Complex I 結構解析

– 使電子由NADH轉移到輔酶Q

– 主要利用FMN(B2)與Fe-S 為輔因子

– 嵌在粒線體內膜中，故結構上有許多α -helix二級結構

© 2014 John Wiley & Sons, Inc. All rights reserved.

FMN can pick up two electrons from

NADH.

© 2014 John Wiley & Sons, Inc. All rights reserved.

第一步：FMN由NADH+H+獲得2電子形成FMNH2

Iron-sulfur clusters

undergo

one-electron

transfer reactions.

© 2014 John Wiley & Sons, Inc. All rights reserved.

第二步：一系列的Fe-S輔因子協助電子傳送

第三步：輔酶Q接過2電子，形成QH2

Proton wire (質子線路)

a series of hydrogen-bonded protein groups plus water molecules that form a

chain through which a proton can be rapidly relayed.

Complex I Function

移出2H+

移出2H+

共計4H+

Structure of Complex II

(succinate dehydrogenase)

Succinate dehydrogenase，即為TCA循環中的一個產生FADH2之enzyme FAD接收 Succinate之電子，產生FADH2

FADH2之電子經由Fe-S傳至CoQ使其還原為CoQH2

但FADH2還原電位幾乎為0，故淨反應放熱不足以產生ATP，亦無法排出質子

Complex I電子來自____，而Complex II電子來自____，主要都傳給______

TCA循環中哪一個酶與氧化磷酸化有關?

Complex III transfers electrons from

ubiquinol to cytochrome c

Heme prosthetic group One-electron reduction

Cytochrome 細胞色素

• 利用含鐵血紅素(Heme b)輔基傳送電子之蛋白

• 鐵離子可於還原態Fe2+與氧化態Fe3+切換，傳送一個電子

• QH2有兩個電子，故需要 2 Cytochrome c

cytochrome b

Iron–Sulfur Protein

(2Fe–2S cluster)

cytochrome c1

Structure of Mammalian Complex III

The Q cycle

• 由QH2傳進2個電子，但cytochrome c一次只能收1個電子，故由cytochrome b保留，再傳回Q形成‧Q-

• 另一個QH2再傳進2個電子

，由 cytochrome b再傳至‧Q-，與2H+形成新的QH2

移出2H+ (屬於I)

移出2H+

移出2H+ 共計4H+

Summary for Complex III function

• 每Round釋出2H+，兩Round釋出4H+，運送2電子

Complex IV oxidizes cytochrome c

and reduces O2.

• Cytochrome c 一次轉移1個電子至Complex IV

• Complex IV 將O2還原為水，使Cytochrome c氧化，故又稱為Cytochrome c oxidase

© 2014 John Wiley & Sons, Inc. All rights reserved.

氧化磷酸化反應中，最終的電子接受者為何?

More on Complex IV Function

• 1NADH轉移2個電子，最終使Cyto c轉移2個質子出去

• 另外2個質子與氧結合，

½ O2 H2O

© 2014 John Wiley & Sons, Inc. All rights reserved.

質子梯度(Proton gradient)的形成

© 2014 John Wiley & Sons, Inc. All rights reserved.

Complex I排出4質子 Complex III排出4質子

1分子NADH+H+ 提供2電子

Complex IV排出2質子

[H+]↑，pH↓

[H+]↓，pH↑

1分子NADH參與呼吸鏈，可由粒線體基質排出多少質子?

15-3 Chemiosmosis

ADP + Pi ATP

Δ G°’= +30.5kJ/mol

如何轉移能量形成ATP?

Chemiosmosis links electron transport

and oxidative phosphorylation

• Chemiosmotic theory 化學滲透假說:

粒線體內膜內外的質子梯度 proton gradient 推動反應

• Proton-motive force :

質子梯度差趨使質子由高濃度往低濃度移動，產生之動能進而合成ATP

Computing the free energy change

for the imbalance of protons.

化學能推動 電位差推動

Z = Ion’s Charge

Δy = Membrane Potential

© 2014 John Wiley & Sons, Inc. All rights reserved.

整合

15-4 ATP Synthase (Complex V)

ATP synthase

F1球體：伸進粒線體基質

F0 跨膜結構

ATP synthase rotates as it translocates

protons

a subunit

b subunit

c subunit X 12

F1：α *3, β *3, γ *1, δ *1, ε *1

ATP synthase rotates as it

translocates protons.

• H+ 由a subunit通道進入c subunit.

• 結構誘導H+單方向轉動，帶動整個c subunit單方向旋轉

• 每個c subunit有1個H+

結合位，1個H+進入即帶動旋轉1個c subunit而釋出1個H+

© 2014 John Wiley & Sons, Inc. All rights reserved.

ATP synthase rotates as it translocates

protons

1個質子轉動30°，4個質子轉動1/3圈，帶動γ subunit轉動120°

γ rotates in steps of 120° 4 proton = 1 ATP

僅β subunit具催化活性

The binding change mechanism explains

how ATP is made

• Binding change mechanism: 隨著γ subunit的轉動，αβ subunit有三種構型，要ADP與Pi結合才使結構得以旋轉，稱之Binding change mechanism

• Loose-binding (L)： 疏鬆結構，此時ADP與Pi可就定位

• Tight-binding (T)： 緊密結構，此時將ADP與Pi壓縮為ATP

• Open (O)： 開放結構，此時ATP釋出

ADP與Pi於α β subunit處於何種狀態才可接上ATP synthase?

O: Open T: Tight L: Loose

Step 1 Loose state ADP and Pi bind to β

Step 2 Tight state ADP+Pi → ATP

Step 3 Open state ATP release

© 2014 John Wiley & Sons, Inc. All rights reserved.

Quantifying Oxidative Phosphorylation

• 氧化磷酸化反應，以P:O敘述氧消耗與磷酸化的比例

• P:O ratio = # phosphorylation of ADP per # of oxygen

atoms reduced (每個氧原子還原使ADP磷酸化比例)

– 1原子氧還原等於產生10個H+

– 1ADP磷酸化需要4個H+

– 同樣以H+計算，則使氧消耗與磷酸化得到共同的換算依據

© 2014 John Wiley & Sons, Inc. All rights reserved.

• ADP phosphorylation ∝ Oxygen atoms reduced

1 NADH (2e-)

→Complex I(4H+ out)

→Complex III(4H+ out)

→Complex IV(2H+ out) and ½ O2 reduced

4H+ = 1ATP, 10H+ = 2.5 ATP, Oxygen atom = 1

• So the P:O ratio of 1NADH = 2.5

The P:O ratio describes the stoichiometry of

oxidative phosphorylation

The P:O ratio describes the stoichiometry of

oxidative phosphorylation

• 1 QH2 (2e-)

→Complex II

→Complex III(4H+ out)

→Complex IV(2H+ out) and ½ O2 reduced

4H+ = 1ATP, 6H+ = 1.5 ATP, Oxygen atom = 1

So the P:O ratio of 1QH2 = 1.5

一分子NADH/QH2分別可產生多少ATP? NADH/NADPH/QH2/FADH2何者不使用於呼吸鏈反應?

The rate of oxidative phosphorylation

depends on the rate of fuel catabolism

•Reduced cofactors - (NADH and QH2) 不足

•Binding change mechanism – ADP與Pi不足

•pH調控之ATP synthase inhibitor

•基質內pH值偏高→失去與ATP synthase結合之能力→質子濃度差促進ATP合成