Energy and respiration#

Carrier of hydrogen#

• NAD - reduced NAD / $NADH$
• FAD - reduced FAD / $FADH_2$
• NADP - reduced NADP / $NADPH$

Glycolysis#

The splitting of glucose

Location: cytoplasm (glycolysis的产物会进入mitochondria)

在glycolysis中，一个glucose分子最终会被拆成两个pyruvate分子，一个pyruvate分子包含三个碳原子（3C）。

在这个步骤中，ATP会被消耗掉（不是产生），但是能量会在之后的步骤中释放出来，并形成ATP。

Glycolysis的第一步是phosphorylation:

这一步中会raise the energy level of the glucose molecules，为了更容易在下一步中发生反应。

Steps of Glycolysis:

1. glucose中加入两个ATP，经过phosphorylation后形成两个triose phosphate (3C)

2. 两个phosphate group从中间产物上被移除下来，去磷酸化（phosphorylase）环境中的两个$ADP$并形成两个$ATP$

3. 两个氢原子被移除下来，这两个氢原子将两个NAD分子还原，形成两个reduced NAD / NADH

   添加$H$ - reduction

   添加$O$ - oxidation

   中间产物脱下来两个氢被称之为：dehydrogenation (de-hydrogen-a-tion)

   帮助这个反应发生反而酶叫作：dehydrogenase (de-hydrogen-ase)

4. 在从中间产物生成pyruvate的过程中，两个intermediates各自贡献两个$P_i$到环境中的ADP，形成两个ATP

ATP添加到中间反应物时：环境中的ATP经过hydrolysis，产生一个$ADP$和一个phosphate group，phosphate group再添加到中间产物上

ATP在图表上离开中间物时：中间产物贡献phosphate group，与环境中的ADP发生反应形成ATP

• 细胞中的ADP直接接收从中间产物上脱下来的$P_i$被称之为：substrate-linked phosphorylation (为了与之后的oxidative phosphorylation做区分)

在glycolysis中，先是在phosphorylation中消耗了两个ATP，再在后面步骤中产生了四个ATP，所以在glycolysis中净产生了两个ATP：

• the net production of ATP in glycolysis is 2

Purposes of phosphorylation:

• raise the energy level
• make the glucose cannot leave the cell
• decrease the stability of glucose (so it can split into 2 triose phosphate molecules)

Lastly, pyruvate moves across the mitochondrial membranes by active transport (require a small amount of ATP), and enters the mitochondrial matrix.

Link Reaction#

Location: mitochondrial matrix

Link reaction是一个相对而言比较简单的过程，link reaction之所以叫这个名字，是因为它连接了glycolysis和Krebs cycle。

1. decarboxylation: the removal of a carbon dioxide
2. dehydrogenation: the removal of a hydrogen atom ($\mathrm{NAD \rightarrow reduced NAD / NADH}$)
3. the remainder of the molecule combines with coenzyme A (CoA) o produce acetyl coenzyme A (acetyl CoA)

Krebs Cycle#

Krebs cycle also known as citric acid cycle

Location: mitochondrial matrix

The Krebs cycle is a circular pathway of enzyme-controlled reactions:

1. Acetyl coenzyme A (2C) combines with oxaloacetate (4C) to form citrate (6C).

   $$\mathrm{acetyl(2C)\space CoA + oxaloacetate(4C) \rightarrow CoA + citrate(6C)}$$

2. The citrate is decarboxylated and dehydrogenated in a series of steps.

   • release carbon dioxide (as a waste gas)
   • release hydrogens, which are accepted by the carriers NAD and FAD

3. Oxaloacetate is regenerated (为了准备进去下一个循环)

Krebs cycle summary:

1. Carbon Release (Decarboxylation)

   • 2 Carbon Dioxide (2$CO_2$) molecules are produced.

2. Reduction Reactions (Electron Carriers)

   • 1 FAD molecule is reduced → $FADH_2$

   • 3 NAD molecules are reduced → $3NADH$

3. Energy Generation

   • 1 ATP molecule is generated, through substrate-linked reaction

Oxidative Phosphorylation#

Location: inner mitochondrial membrane

氧气只在这一步中参与respiration。

It’s the synthesis of ATP from ADP and Pi

Inner mitochondrial membrane上参与oxidative phosphorylation的结构:

• Electron transport chain ETC 电子传递链

  用来传递电子，里面有很多electron carriers（属于cofactor）

  电子在传递的过程中会：down the energy level，在传递的过程中会释放能量，每个electron carrier都经历了redox reaction.

  电子释放的一部分能量会把氢离子$H^+$从matrix运输到intermembrane space，然后建立matrix和intermembrane space之间的concentration gradient

  氢离子会在ATP synthase中通过facilitated diffusion回到matrix，在回来的过程中（ATP synthase中）会推动ATP的合成

• ATP synthase (also known as stalked particle)

1. Reduced NAD and reduced FAD transfer and release hydrogen atoms to ETC

2. $\mathrm{H \rightarrow H^+ + e^-}$

3. As the electron moves along the ETC, energy is released. 电子在通过ETC时会释放能量

   Some of the energy is used to actively move protons from the mitochondrial matrix into the intermembrane spase. 一部分能量用来运输氢离子

   This produces a higher concentration of proton in the intermembrane space and creates a concentration gradient of protons between the intermembrane space and the mitochondrial matrix 建立浓度梯度差

4. Protons diffuse back into the mitochondrial matrix by facilitate diffusion down their concentration gradient through ATP synthase (it serves as transport protein and enzyme) 氢离子通过协助运输回到matrix时，会推动ATP的生成

5. As the protons pass through the ATP synthase, ATP is produced from ADP and Pi by ATP synthase in chemiosmosis

   chemiosmosis是ATP生成的过程

6. Oxygen is the final electron acceptor

   $$\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O}$$

Summary#

答题的时候需要提到：

• glycolysis产生了两个ATP
• oxidative phosphorylation produces most ATP

Yeast and Plant#

在yeast and plant中发生的无氧呼吸又被叫作：Ethanol Fermentation

• dehydrogenation / glycolysis
• decarboxylation
• substrate-linked phosphorylation

这整个过程，除了glycolysis，只有两步：

1. $\mathrm{pyruvate} \rightarrow \mathrm{CO}_2 + \mathrm{ethanal\text{乙醛}}$

   pyruvate分离出一个二氧化碳，生成ethanal

2. $\mathrm{ethanal} + \mathrm{NADH} \xrightarrow{\mathrm{alcohol\ dehydrogenase}} \mathrm{ethanol} + \mathrm{NAD}$

   ethanal被还原（ethanal是hydrogen acceptor），在alcohol dehydrogenase的催化下，生成ethanol

Respiration的目的是产生ATP，但是只有glycolysis会产生ATP，之后步骤的发生目的是：

• 为了使reduced NAD能被氧化成NAD
• glycolysis需要NAD被还原（reduce），所以需要有后续步骤回收NAD

Animal#

Muscle cells can respire anaerobically for a short time. 剧烈运动时（肺和心脏无法提供足够的氧气时），肌肉细胞可以在无氧的环境下进行呼吸作用，释放能量。

无氧呼吸可以释放能量，但是释放的很少：

• Only glycolysis occurs

• The net production of ATP is 2 (每个glucose只产生两个ATP)

• Fewer substrate-linked phosphorylation occurs

• Lactate is energy-rich

• Oxygen is not available as final electron acceptor

• Oxidative phosphorylation dose not occur

• … but most ATP is produced in oxidative phosphorylation

一个glucose分子被拆成两个一样的lactate分子（并最终形成lactic acid）。

1. dehydrogenation
2. substrate-linked phosphorylation

Oxygen Bebt / Excess Post-exercise Oxygen Consumption#

Happens in mammals

After strenuous exercise, lactate is produced due to anaerobic respiration:

• oxygen debt is built up: Lactic acid builds up in muscles and blood
• or EPOC (Excess Post-exercise Oxygen Consumption) is required

Lactate can be oxidised, and converted back to pyruvate in liver, so it can then be fed into the Krebs cycle and generate ATP.

Removing oxygen debt:

1. continue to have a faster heart rate
2. transport lactate in the blood from the muscles to the liver
3. continue to have deeper and faster breathing for oxygen supply
4. aerobic respiration in liver

Alternatively, the lactate can be converted to the glycogen and stored in liver.

As a contrast, ethanol is only a waste product, it cannot be further metabolised

Ethanol Fermentation in Rice#

Rice is a staple crop (主食作物), 为了是产量最大化，水稻通常种在稻田（paddies）中。

Paddies - are fields where the ground is intentionally flooded.

大多数杂草不能长在稻田中，所以：This reduction in competition for light and mineral salts increases yields.

大多数植物不能长在水中是因为它们的根无法获得足够的氧气来进行有氧呼吸。水稻为了在水中适应，它做出了以下进化：

• internodes of stem grow fast

• root cells can carry out anaerobic respiration due to low oxygen concentration in water

• gas diffuses down to root cells through aerenchyma

  aerenchyma [ae-ren-chy-ma] 通气组织 - plant tissue containing air spaces, both root and stem have this structure

• some leaves trap air underwater due to ridges on leaves

  ridges可以捕捉一层空气膜来增加氧气的获取量

• root cells can tolerate (high concentration) ethanol (produced from anaerobic respiration)

  • … produce ethanol dehydrogenase to break down ethanol

Comparison Table#

Respiratory Quotient#

respiratory quotient (RQ) 呼吸商 - the ratio of the volume of carbon dioxide produced to the volume of oxygen used

For the aerobic respiration of glucose:

$$\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O}$$

使用了6个氧分子，产生了6个二氧化碳分子

$$\begin{aligned} \mathrm{RQ} &= \frac{\text{amount of }\mathrm{CO}_2}{\text{amount of }\mathrm{O}_2} \\ &= \frac{6}{6} \\ &= 1.0 \end{aligned}$$

如果知道了其他呼吸作用的反应式后，对应的RQ也可以计算出来：

For example, when the fatty acid oleic acid (from olive oil) is respired aerobically, the equation is:

$$\mathrm{C}_{18}\mathrm{H}_{34}\mathrm{O}_2 + 25.5\mathrm{O}_2 \longrightarrow 18\mathrm{CO}_2 + 17\mathrm{H}_2\mathrm{O}$$

So, for the aerobic respiration of oleic acid:

$$\begin{aligned} \mathrm{RQ} &= \frac{\mathrm{CO}_2}{\mathrm{O}_2} \\ &= \frac{18}{25.5} \\ &= 0.7 \end{aligned}$$

如果有一个respiratory substrate是$\mathrm{C_{18}H_{36}O_{2}}$，且是aerobic respiration，那么它的反应方程式是：

1. 根据respiratory substrate中碳原子的数量补齐$CO_2$的数量：

   $$\mathrm{C}_{18}\mathrm{H}_{36}\mathrm{O}_2 + ?\mathrm{O}_2 \longrightarrow 18\mathrm{CO}_2 + ?\mathrm{H}_2\mathrm{O}$$

2. 按照同样的方法，可以根据respiratory substrate中的$H$的数量推断出$H_2O$的数量：

   $$\mathrm{C}_{18}\mathrm{H}_{36}\mathrm{O}_2 + ?\mathrm{O}_2 \longrightarrow 18\mathrm{CO}_2 + 18\mathrm{H}_2\mathrm{O}$$

3. 反应式左边的氧原子数量必须等于右边的氧原子的数量：

   $$\mathrm{2 + n \times 2 = 18 \times 2 + 18 \times 1}$$

   $O_2$的数量最终为26

4. 最终：

   $$\mathrm{C}_{18}\mathrm{H}_{36}\mathrm{O}_2 + 26\mathrm{O}_2 \longrightarrow 18\mathrm{CO}_2 + 18\mathrm{H}_2\mathrm{O}$$

计算RQ：

Measurement of RQ by Respirometer#

Respirometer [respir - o - meter]

Respirometer由一个u形管连接两个试管组成：

• experimental tube里面主要放进行呼吸作用的对象，比如小型虫子（non-vertebrates）
• control tube里放与相同呼吸对象体积的，不会参与呼吸作用的物体，比如glass beads

一个计算RQ的测试分为两轮，第一轮中测试氧气被消耗掉的量，第二轮中测试消耗的氧气和产生的二氧化碳之间的体积差。

第一轮测试

1. 在两试管内添加用于吸收二氧化碳的物质，例如：soda lime（固体）或者KOH / NaOH（液体）
2. 等待一定时间
3. 观察并记录体积变化（标记为$X$）

第二轮测试

1. 设置好系统后，直接等待一定时间（不需要去除二氧化碳）
2. 观察并记录结果（标记为$Y$）

计算RQ

NOTE

Example:

第一轮测试：

• 液体上升 1 mL

• 体积减少 1 cm3

  记 X为 −1

第二轮测试：

• 液体上升 0.5 mL

• 体积减少 0.5 cm3

  记 Y为 −0.5

计算：

$$\begin{aligned} RQ &= \frac{X - Y}{X} \\ &= \frac{-1 - (-0.5)}{-1} \\ &= 0.5 \end{aligned}$$