Ebook Cambridge international AS and A level Biology coursebook Part 1

(BQ) Part 1 book Cambridge international AS and A level Biology coursebook has contents: Cell structure, biological molecules, enzymes, cell membranes and transport, the mitotic cell cycle, nucleic acids and protein synthesis, transport in plants,... and other contents.

Cambridge International AS and A Level Biology Coursebook Mary Jones, Richard Fosbery, Jennifer Gregory and Dennis Taylor Cambridge International AS and A Level Biology Coursebook Fourth Edition Jones, Fosbery, Gregory and Taylor Mary Jones, Richard Fosbery, Jennifer Gregory and Dennis Taylor Cambridge International AS and A Level Biology Coursebook Fourth Edition University Printing House, Cambridge cb2 8bs, United Kingdom Cambridge University Press is part of the University of Cambridge It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence www.cambridge.org Information on this title: www.cambridge.org © Cambridge University Press 2003, 2014 This publication is in copyright Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press First published 2003 Second edition 2007 Third edition 2013 Fourth edition 2014 Printed in the United Kingdom by Latimer Trend A catalogue record for this publication is available from the British Library isbn 978-1-107-63682-8 Paperback with CD-ROM for Windows® and Mac® Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate Information regarding prices, travel timetables, and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter notice to teachers in the uk It is illegal to reproduce any part of this work in material form (including photocopying and electronic storage) except under the following circumstances: (i) where you are abiding by a licence granted to your school or institution by the Copyright Licensing Agency; (ii) where no such licence exists, or where you wish to exceed the terms of a licence, and you have gained the written permission of Cambridge University Press; (iii) where you are allowed to reproduce without permission under the provisions of Chapter of the Copyright, Designs and Patents Act 1988, which covers, for example, the reproduction of short passages within certain types of educational anthology and reproduction for the purposes of setting examination questions All end-of-chapter questions taken from past papers are reproduced by permission of Cambridge International Examinations Example answers and all other end-of-chapter questions were written by the authors Cambridge International Examinations bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication Contents How to use this book vi Introductionviii Cell structure Why cells? Cell biology and microscopy Animal and plant cells have features in common Differences between animal and plant cells Units of measurement in cell studies Electron microscopy Ultrastructure of an animal cell Ultrastructure of a plant cell Two fundamentally different types of cell End-of-chapter questions Biological molecules 3 5 6 13 19 21 23 27 The building blocks of life 28 Monomers, polymers and macromolecules 29 Carbohydrates29 Lipids36 Proteins39 Water46 End-of-chapter questions 49 Enzymes Mode of action of enzymes Factors that affect enzyme action Enzyme inhibitors Comparing the affinity of different enzymes for their substrates Immobilising enzymes End-of-chapter questions Cell membranes and transport 53 54 57 61 62 64 66 72 Phospholipids73 Structure of membranes 74 Cell signalling 77 Movement of substances into and out of cells 79 End-of-chapter questions 89 The mitotic cell cycle 93 Chromosomes94 Mitosis97 The significance of telomeres 102 Stem cells 103 Cancer103 End-of-chapter questions 106 Nucleic acids and protein synthesis  The structure of DNA and RNA DNA replication Genes and mutations DNA, RNA and protein synthesis End-of-chapter questions Transport in plants 110 111 113 118 118 123 126 The transport needs of plants 127 Two systems: xylem and phloem 128 Structure of stems, roots and leaves 128 The transport of water 134 Transport of mineral ions 146 Translocation146 Differences between sieve tubes and xylem vessels 151 End-of-chapter questions 153 Transport in mammals 157 Transport systems in animals 158 The mammalian cardiovascular system 158 Blood vessels 160 Blood plasma and tissue fluid 164 Lymph164 Blood166 Haemoglobin168 Problems with oxygen transport 171 The heart 173 The cardiac cycle 175 Control of the heart beat 177 End-of-chapter questions 179 iii Cambridge International AS Level and ABiology Level Biology Gas exchange and smoking 185 Gas exchange 186 Lungs186 Trachea, bronchi and bronchioles 187 Alveoli189 Smoking190 Tobacco smoke 190 Lung diseases 190 Short-term effects on the cardiovascular system 193 End-of-chapter questions 195 10 Infectious diseases 198 Worldwide importance of infectious diseases 200 Cholera200 Malaria202 Acquired immune deficiency syndrome (AIDS) 205 Tuberculosis (TB) 209 Measles212 Antibiotics213 End-of-chapter questions 219 11 Immunity iv Defence against disease Cells of the immune system Active and passive immunity Autoimmune diseases – a case of mistaken identity End-of-chapter questions P1 Practical skills for AS 222 223 224 232 237 242 246 Experiments247 Variables and making measurements 247 Estimating uncertainty in measurement 255 Recording quantitative results 255 Constructing a line graph 256 Constructing bar charts and histograms 258 Making conclusions 259 Describing data 259 Making calculations from data 259 Explaining your results 261 Identifying sources of error and suggesting improvements261 Drawings262 End-of-chapter questions 264 12 Energy and respiration 267 The need for energy in living organisms 268 Work269 ATP270 Respiration272 Mitochondrial structure and function 276 Respiration without oxygen 277 Respiratory substrates 278 Adaptations of rice for wet environments 281 End-of-chapter questions 283 13 Photosynthesis 286 An energy transfer process 287 The light dependent reactions of photosynthesis 288 The light independent reactions of photosynthesis290 Chloroplast structure and function 290 Factors necessary for photosynthesis 291 C4 plants 293 Trapping light energy 295 End-of-chapter questions 297 14 Homeostasis 299 Internal environment 300 Control of homeostatic mechanisms 301 The control of body temperature 302 Excretion304 The structure of the kidney 305 Control of water content 312 The control of blood glucose 315 Urine analysis 319 Homeostasis in plants 320 End-of-chapter questions 325 15 Coordination Nervous communication Muscle contraction Hormonal communication Birth control Control and coordination in plants End-of-chapter questions 329 330 344 349 351 353 358 Contents 16 Inherited change 364 Homologous chromosomes Two types of nuclear division M iosis G n tics Genotype affects phenotype Inheriting genes Multiple alleles Sex inheritance Sex linkage Dihybrid crosses Interactions between loci Autosomal linkage Crossing over The χ2 (chi-squared) test Mutations Gene control in prokaryotes Gene control in eukaryotes End-of-chapter questions 365 367 368 374 374 375 378 378 379 380 382 383 384 386 387 389 391 393 17 Selection and evolution 397 Variation Natural selection Evolution Artificial selection The Darwin–Wallace theory of evolution by natural selection Species and speciation Molecular comparisons between species Extinctions End-of-chapter questions 18 Biodiversity, classification and conservation Ecosyst ms Biodiv rsity Simpson’s Index of Diversity Systematic sampling Corr lation Classification Virus s Threats to biodiversity Why does biodiversity matter? Protecting endangered species Controlling alien species International conservation organisations Restoring degraded habitats End-of-chapter questions 398 402 404 409 19 Genetic technology 462 Genetic engineering Tools for the gene technologist Genetic technology and medicine Gene therapy Genetic technology and agriculture End-of-chapter questions 463 464 475 477 480 487 P2 Planning, analysis and evaluation 490 Planning an investigation Constructing a hypothesis Using the right apparatus Identifying variables Describing the sequence of steps Risk assessment Recording and displaying results Analysis, conclusions and evaluation Pearson’s linear correlation Spearman’s rank correlation Evaluating evidence Conclusions and discussion End-of-chapter questions 491 491 491 492 495 495 495 495 501 503 504 506 507 Appendix 1: Amino acid R groups 512 Appendix 2: DNA and RNA triplet codes 513 Glossary 514 Index 526 Acknowledgements 534 423 CD-ROM CD1 425 426 430 431 433 435 440 441 444 445 451 452 453 455 Advice on how to revise for and approach examinations Introduction to the examination and changes to the syllabus Answers to self-assessment questions Answers to end-of-chapter questions Recommended resources 412 413 416 417 420 CD1 CD16 CD21 CD64 CD128 v Cambridge International AS Level Biology How to use this book Each chapter begins with a short list of the facts and concepts that are explained in it There is a short context at the beginning of each chapter, containing an example of how the material covered in the chapter relates to the ʻreal worldʼ Questions throughout the text give you a chance to check that you have understood the topic you have just read about You can find the answers to these questions on the CD-ROM vi This book does not contain detailed instructions for doing particular experiments, but you will find background information about the practical work you need to in these boxes There are also two chapters, P1 and P2, which provide detailed information about the practical skills you need to develop during your course The text and illustrations describe and explain all of the facts and concepts that you need to know The chapters, and often the content within them as well, are arranged in the same sequence as in your syllabus Important equations and other facts are shown in highlight boxes How to use this book Wherever you need to know how to use a formula to carry out a calculation, there are worked example boxes to show you how to this Definitions that are required by the syllabus are shown in highlight boxes Key words are highlighted in the text when they are first introduced You will also find definitions of these words in the Glossary There is a summary of key points at the end of each chapter You might find this helpful when you are revising Questions at the end of each chapter begin with a few multiple choice questions, then move on to questions that will help you to organise and practise what you have learnt in that chapter Finally, there are several more demanding exam-style questions, some of which may require use of knowledge from previous chapters Answers to these questions can be found on the CD–ROM vii Cambridge International AS Level Biology Introduction viii This fourth edition of Cambridge International AS and A Level Biology provides everything that you need to well in your Cambridge International Examinations AS and A level Biology (9700) courses It provides full coverage of the syllabus for examinations from 2016 onwards The chapters are arranged in the same sequence as the material in your syllabus Chapters to P1 cover the AS material, and Chapters 12 to P2 cover the extra material you need for the full A level examinations The various features that you will find in these chapters are explained on the next two pages In your examinations, you will be asked many questions that test deep understanding of the facts and concepts that you will learn during your course It’s therefore not enough just to learn words and diagrams that you can repeat in the examination; you need to ensure that you really understand each concept fully Trying to answer the questions that you will find within each chapter, and at the end, should help you to this There are answers to all of these questions on the CD-ROM that comes with this book Although you will study your biology as a series of different topics, it’s very important to appreciate that all of these topics link up with each other Some of the questions in your examination will test your ability to make links between different areas of the syllabus For example, in the AS examination you might be asked a question that involves bringing together knowledge about protein synthesis, infectious disease and transport in mammals In particular, you will find that certain key concepts come up again and again These include: ■■ ■■ ■■ ■■ ■■ ■■ cells as units of life biochemical processes DNA, the molecule of heredity natural selection organisms in their environment observation and experiment As you work through your course, make sure that you keep on thinking about the work that you did earlier, and how it relates to the current topic that you are studying On the CD-ROM, you will also find some suggestions for other sources of particularly interesting or useful information about the material covered in each chapter Do try to track down and read some of these Practical skills are an important part of your biology course You will develop these skills as you experiments and other practical work related to the topic you are studying Chapters P1 (for AS) and P2 (for A level) explain what these skills are, and what you need to be able to to succeed in the examination papers that test these skills Chapter 12: Energy and respiration protein channel for H+ ions thylakoid membrane of chloroplast or inner membrane of mitochondrion 3H+ high concentration of H+ phospholipid bilayer membrane impermeable to H+ low concentration of H+ ATP synthase – protein transferring energy from H+ ions to ATP synthesis ATP ADP + Pi matrix of mitochondrion or stroma of chloroplast Figure 12.5  ATP synthesis (chemiosmosis) ATP synthase has three binding sites (Figure 12.6) and a part of the molecule (γ) that rotates as hydrogen ions (H+) pass. This produces structural changes in the binding sites and allows them to pass sequentially through three phases: ■■ ■■ ■■ step ADP + Pi step ATP γ binding ADP and Pi forming tightly bound ATP releasing ATP QUESTION 12.1 Write the equation for the reaction catalysed by ATP synthase The role of ATP in active transport Active transport is the movement of molecules or ions across a partially permeable membrane against a concentration gradient Energy is needed, from ATP, to counteract the tendency of these particles to move by diffusion down the gradient All cells show differences in the concentration of ions, in particular sodium and potassium ions, inside the cell with respect to the surrounding solution Most cells seem to have sodium pumps in the cell surface membrane ATP step Figure 12.6  Transverse section (TS) of ATP synthase showing its activity that pump sodium ions out of the cell This is usually coupled with the ability to pump potassium ions from the surrounding solution into the cell The sodium–potassium pump is a protein that spans the cell surface membrane (Figure 4.18, page 87) It has binding sites for sodium ions (Na+) and for ATP on the inner side, and for potassium ions (K+) on the outer side The protein acts as an ATPase and catalyses the hydrolysis of ATP to ADP and inorganic phosphate, releasing energy to drive the pump Changes in the shape of the protein 271 Cambridge International A Level Biology move sodium and potassium ions across the membrane in opposite directions For each ATP used, two potassium ions move into the cell and three sodium ions move out of the cell As only two potassium ions are added to the cell contents for every three sodium ions removed, a potential difference is created across the membrane that is negative inside with respect to the outside Both sodium and potassium ions leak back across the membrane, down their diff usion gradients However, cell surface membranes are much less permeable to sodium ions than potassium ions, so this diff usion actually increases the potential difference across the membrane This potential difference is most clearly seen as the resting potential of a nerve cell (page 333) One of the specialisations of a nerve cell is an exaggeration of the potential difference across the cell surface membrane as a result of the activity of the sodium–potassium pump The importance of active transport in ion movement into and out of cells should not be underestimated About 50% of the ATP used by a resting mammal is devoted to maintaining the ionic content of cells Respiration 272 Respiration is a process in which organic molecules act as a fuel The organic molecules are broken down in a series of stages to release chemical potential energy, which is used to synthesise ATP The main fuel for most cells is carbohydrate, usually glucose Many cells can use only glucose as their respiratory substrate, but others break down fatty acids, glycerol and amino acids in respiration Glucose breakdown can be divided into four stages: glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation (Figure 12.7) Glycolysis is the splitting, or lysis, of glucose It is a multi-step process in which a glucose molecule with six carbon atoms is eventually split into two molecules of pyruvate, each with three carbon atoms Energy from ATP is needed in the first steps, but energy is released in later steps, when it can be used to make ATP There is a net gain of two ATP molecules per molecule of glucose broken down Glycolysis takes place in the cytoplasm of a cell A simplified flow diagram of the pathway is shown in Figure 12.8 In the first stage, phosphorylation, glucose is phosphorylated using ATP As we saw on page 269, glucose is energy-rich but does not react easily To tap the bond energy of glucose, energy must first be used to make the reaction easier (Figure 12.3, page 269) Two ATP molecules are used for each molecule of glucose to make first glucose phosphate, then fructose phosphate, then fructose bisphosphate, which breaks down to produce two molecules of triose phosphate Hydrogen is then removed from triose phosphate and transferred to the carrier molecule NAD (nicotinamide adenine dinucleotide) The structure of NAD is shown in Figure 12.12, page 275 Two molecules of reduced NAD are produced for each molecule of glucose entering glycolysis The hydrogens carried by reduced NAD can easily be transferred to other molecules and are used in oxidative phosphorylation to generate ATP (page 273) The end-product of glycolysis, pyruvate, still contains a glucose (hexose) (6C) ATP fructose phosphate (6C) phosphorylation ATP fructose bisphosphate (6C) The glycolytic pathway molecules of triose phosphate (3C) 2ATP glycolysis glycolysis anaerobic pathways to ethanol or lactate Link reaction aerobic pathways in mitochondria Krebs cycle 2NAD 2H reduced NAD oxidative phosphorylation intermediates 2ATP molecules of pyruvate (3C) –2ATP Figure 12.7 The sequence of events in respiration Figure 12.8 The glycolytic pathway +4ATP +2 reduced NAD Chapter 12: Energy and respiration great deal of chemical potential energy When free oxygen is available, some of this energy can be released via the Krebs cycle and oxidative phosphorylation However, the pyruvate first enters the link reaction, which takes place in the mitochondria (page 275) pyruvate (3C) NAD link reaction reduced NAD CO2 acetyl (2C) CoA The link reaction Pyruvate passes by active transport from the cytoplasm, through the outer and inner membranes of a mitochondrion and into the mitochondrial matrix Here it is decarboxylated (this means that carbon dioxide is removed), dehydrogenated (hydrogen is removed) and combined with coenzyme A (CoA) to give acetyl coenzyme A This is known as the link reaction (Figure 12.9) Coenzyme A is a complex molecule composed of a nucleoside (adenine plus ribose) with a vitamin (pantothenic acid), and acts as a carrier of acetyl groups to the Krebs cycle The hydrogen removed from pyruvate is transferred to NAD pyruvate + CoA + NAD acetyl CoA + CO2 + reduced NAD Fatty acids from fat metabolism may also be used to produce acetyl coenzyme A Fatty acids are broken down in the mitochondrion in a cycle of reactions in which each turn of the cycle shortens the fatty acid chain by a two-carbon acetyl unit Each of these can react with coenzyme A to produce acetyl coenzyme A, which, like that produced from pyruvate, now enters the Krebs cycle The Krebs cycle The Krebs cycle (also known as the citric acid cycle or tricarboxylic acid cycle) was discovered in 1937 by Hans Krebs It is shown in Figure 12.9 The Krebs cycle is a closed pathway of enzymecontrolled reactions ■■ ■■ ■■ Acetyl coenzyme A combines with a four-carbon compound (oxaloacetate) to form a six-carbon compound (citrate) The citrate is decarboxylated and dehydrogenated in a series of steps, to yield carbon dioxide, which is given off as a waste gas, and hydrogens which are accepted by the carriers NAD and FAD (page 275) Oxaloacetate is regenerated to combine with another acetyl coenzyme A For each turn of the cycle, two carbon dioxide molecules are produced, one FAD and three NAD molecules are reduced, and one ATP molecule is generated via an intermediate compound Although part of aerobic respiration, the reactions CoA oxaloacetate (4C) citrate (6C) reduced NAD NAD (4C) (6C) NAD Krebs cycle reduced FAD (4C) reduced NAD FAD (4C) reduced NAD CO2 (5C) NAD CO2 ATP ADP (4C) Figure 12.9  The link reaction and the Krebs cycle of the Krebs cycle make no use of molecular oxygen However, oxygen is necessary for the final stage of aerobic respiration, which is called oxidative phosphorylation The most important contribution of the Krebs cycle to the cell’s energetics is the release of hydrogens, which can be used in oxidative phosphorylation to provide energy to make ATP QUESTION 12.2 Explain how the events of the Krebs cycle can be cyclical Oxidative phosphorylation and the electron transport chain In the final stage of aerobic respiration, oxidative phosphorylation, the energy for the phosphorylation of ADP to ATP comes from the activity of the electron transport chain Oxidative phosphorylation takes place in the inner mitochondrial membrane (Figure 12.10) 273 Cambridge International A Level Biology H+ H+ H+ intermembrane space inner mitochondrial membrane respiratory complex H+ e− e− reduced NAD NAD reduced FAD FAD H+ e− e− ATP synthase matrix ADP + Pi ATP H+ e− oxygen water Figure 12.10 Oxidative phosphorylation: the electron transport chain 274 Reduced NAD and reduced FAD are passed to the electron transport chain Here, the hydrogens are removed from the two hydrogen carriers and each is split into its constituent proton (H+) and electron (e−) The energetic electron is transferred to the fi rst of a series of electron carriers Most of the carriers are associated with membrane proteins, of which there are four types A functional unit, called a respiratory complex, consists of one of each of these proteins, arranged in such a way that electrons can be passed from one to another down an energy gradient As an electron moves from one carrier at a higher energy level to another one at a lower level, energy is released Some of this energy is used to move protons from the matrix of the mitochondrion (Figure 12.11) into the space between the inner and outer membranes of the mitochondrial envelope This produces a higher concentration of protons in the intermembrane space than in the matrix, setting up a concentration gradient Now, protons pass back into the mitochondrial matrix through protein channels in the inner membrane, moving down their concentration gradient Associated with each channel is the enzyme ATP synthase As the protons pass through the channel, their electrical potential energy is used to synthesise ATP in the process called chemiosmosis (Figure 12.5, page 271) Finally, oxygen has a role to play as the final electron acceptor In the mitochondrial matrix, an electron and a proton are transferred to oxygen, reducing it to water The process of aerobic respiration is complete The sequence of events in respiration and their sites are shown in Figure 12.11 The balance sheet of ATP used and synthesised for each molecule of glucose entering the respiration pathway is shown in Table 12.1 Theoretically, three molecules of ATP can be produced from each molecule of reduced NAD, and two molecules of ATP from each molecule of reduced FAD However, this yield cannot be achieved unless ADP and Pi are available inside the mitochondrion About 25% of the total energy yield of electron transfer is used to transport ADP into the mitochondrion and ATP into the cytoplasm Hence, each reduced NAD ATP used ATP made Net gain in ATP −2 +2 link reaction 0 Krebs cycle oxidative phosphorylation Total +2 28 +28 −2 34 +32 glycolysis Table 12.1 Balance sheet of ATP use and synthesis for each molecule of glucose entering respiration Chapter 12: Energy and respiration CO2 glucose O2 glycolysis mitochondrion cell surface membrane of cell mitochondrial envelope cytoplasm pyruvate intermembrane space link reaction matrix inner membrane ADP + Pi ACoA Krebs cycle outer membrane ATP H+ H+ ATP reduced NAD electron transport chain reduced FAD oxidative phosphorylation crista H2O Figure 12.11  The sites of the events of respiration in a cell ACoA = acetyl coenzyme A molecule entering the chain produces on average two and a half molecules of ATP, and each reduced FAD produces one and a half molecules of ATP The number of ATP molecules actually produced varies in different tissues and different circumstances, largely dependent on how much energy is used to move substances into and out of the mitochondria C O O– P O A slightly different form of NAD has a phosphate group instead of the hydrogen on carbon in one of the ribose rings This molecule is called NADP (nicotinamide adenine dinucleotide phosphate) and is used as a hydrogen carrier molecule in photosynthesis FAD (flavin adenine dinucleotide) is similar in function to NAD and is used in respiration in the Krebs cycle FAD is made of one nucleotide containing ribose and adenine and one with an unusual structure involving a linear molecule, ribitol, instead of ribose ribose H H OH O OH NH2 N N H O– P O CH2 O NH2 N O H NAD is made of two linked nucleotides (Figure 12.12) Both nucleotides contain ribose One nucleotide contains the nitrogenous base adenine The other has a nicotinamide ring, which can accept a hydrogen ion and two electrons, thereby becoming reduced reduced NAD NADH+ + H+ CH2 H Hydrogen carrier molecules NAD + 2H NAD+ + 2H O nicotinamide ring H H N N adenine H O ribose H H OH OH Key replaced by a phosphate group in NADP site which accepts electrons Figure 12.12  NAD (nicotinamide adenine dinucleotide) You not need to learn the structure of this molecule, but may like to compare it with the units that make up DNA and RNA 275 Cambridge International A Level Biology QUESTION 12.3 How does the linkage between the nucleotides in NAD differ from that in a polynucleotide? (You may need to refer back to page 114 to answer this question.) Mitochondrial structure and function 276 In eukaryotic organisms, the mitochondrion is the site of the Krebs cycle and the electron transport chain Mitochondria are rod-shaped or filamentous organelles about 0.5–1.0 μm in diameter Time-lapse photography shows that they are not rigid, but can change their shape The number of mitochondria in a cell depends on its activity For example, highly active mammalian liver cells contain between 1000 and 2000 mitochondria, occupying 20% of the cell volume The structure of a mitochondrion is shown in Figures 1.22 (page 16) and 12.13 Like a chloroplast, each mitochondrion is surrounded by an envelope of two phospholipid membranes (page 74) The outer membrane is smooth, but the inner is much folded inwards to form cristae (singular: crista) These cristae give the inner membrane a large total surface area Cristae in mitochondria from different types of cell show considerable variation, but, in general, mitochondria from active cells have longer, more densely packed cristae than mitochondria from less active cells The two membranes have different compositions and properties The outer membrane is relatively permeable to small molecules, whereas the inner membrane is less permeable The inner membrane is studded with tiny spheres, about 9 nm in diameter, which are attached to the inner membrane by stalks (Figure 12.14) The spheres are the enzyme ATP synthase The inner membrane is the site of the electron transport chain and contains the proteins necessary for this The space between the two membranes of the envelope usually has a lower pH than the matrix of the mitochondrion as a result of the protons that are released into the intermembrane space by the activity of the electron transport chain The matrix of the mitochondrion is the site of the link reaction and the Krebs cycle, and contains the enzymes needed for these reactions It also contains small (70 S) ribosomes and several identical copies of looped mitochondrial DNA ATP is formed in the matrix by the activity of ATP synthase on the cristae The energy for the production of ATP comes from the proton gradient between the intermembrane space and the matrix The ATP can be used for all the energy-requiring reactions of the cell, both inside and outside the mitochondrion Figure 12.14  Transmission electron micrograph of ATP synthase particles on the inner membrane of a mitochondrion (× 400 000) Figure 12.13  Transmission electron micrograph of a mitochondrion from a pancreas cell (× 15 000) Chapter 12: Energy and respiration glucose QUESTIONS 12.4 Calculate the number of reduced NAD and reduced FAD molecules produced for each molecule of glucose entering the respiration pathway when oxygen is available 12.5 Using your answer to Question 12.4, calculate the number of ATP molecules produced for each molecule of glucose in oxidative phosphorylation 12.6 Explain why the important contribution of the Krebs cycle to cellular energetics is the release of hydrogens and not the direct production of ATP 12.7 Explain how the structure of a mitochondrion is adapted for its functions in aerobic respiration ADP reduced NAD 2H NAD ATP 2H pyruvate ethanal ethanol CO2 Figure 12.15  Alcoholic fermentation Respiration without oxygen When free oxygen is not present, hydrogen cannot be disposed of by combination with oxygen The electron transfer chain therefore stops working and no further ATP is formed by oxidative phosphorylation If a cell is to gain even the two ATP molecules for each glucose yielded by glycolysis, it is essential to pass on the hydrogens from the molecules of reduced NAD that are made in glycolysis There are two different anaerobic pathways that solve the problem of ‘dumping’ this hydrogen Both pathways take place in the cytoplasm of the cell In various microorganisms such as yeast, and in some plant tissues, the hydrogen from reduced NAD is passed to ethanal (CH3CHO) This releases the NAD and allows glycolysis to continue The pathway is shown in Figure 12.15 First, pyruvate is decarboxylated to ethanal; then the ethanal is reduced to ethanol (C2H5OH) by the enzyme alcohol dehydrogenase The conversion of glucose to ethanol is referred to as alcoholic fermentation In other microorganisms, and in mammalian muscles when deprived of oxygen, pyruvate acts as the hydrogen acceptor and is converted to lactate by the enzyme lactate dehydrogenase (named after the reverse reaction, which it also catalyses) Again, the NAD is released and allows glycolysis to continue in anaerobic conditions This pathway, known as lactic fermentation, is shown in Figure 12.16 glucose ADP 2H reduced NAD NAD 277 ATP 2H pyruvate lactate Figure 12.16  Lactic fermentation These reactions ‘buy time’ They allow the continued production of at least some ATP even though oxygen is not available as the hydrogen acceptor However, as the products of anaerobic reaction, ethanol or lactate, are toxic, the reactions cannot continue indefinitely The pathway leading to ethanol cannot be reversed, and the remaining chemical potential energy of ethanol is wasted The lactate pathway can be reversed in mammals Lactate is carried by the blood plasma to the liver and converted back to pyruvate The liver oxidises some (20%) of the incoming lactate to carbon dioxide and water via aerobic respiration when oxygen is available again The remainder of the lactate is converted by the liver to glycogen Cambridge International A Level Biology Energy values of respiratory substrates Figure 12.17 shows what happens to the oxygen uptake of a person before, during and after taking strenuous exercise Standing still, the person absorbs oxygen at the resting rate of 0.2 dm3 min−1 (This is a measure of the person’s metabolic rate.) When exercise begins, more oxygen is needed to support aerobic respiration in the person’s muscles, increasing the overall demand to 2.5 dm3 min−1 However, it takes four minutes for the heart and lungs to meet this demand, and during this time lactic fermentation occurs in the muscles Thus the person builds up an oxygen deficit For the next three minutes, enough oxygen is supplied When exercise stops, the person continues to breathe deeply and absorb oxygen at a higher rate than when at rest This post-exercise uptake of extra oxygen, which is ‘paying back’ the oxygen deficit, is called the oxygen debt The oxygen is needed for: ■■ ■■ ■■ Most of the energy liberated in aerobic respiration comes from the oxidation of hydrogen to water when reduced NAD and reduced FAD are passed to the electron transport chain Hence, the greater the number of hydrogens in the structure of the substrate molecule, the greater the energy value Fatty acids have more hydrogens per molecule than carbohydrates do, and so lipids have a greater energy value per unit mass, or energy density, than carbohydrates or proteins The energy value of a substrate is determined by burning a known mass of the substance in oxygen in a calorimeter (Figure 12.18) thermometer conversion of lactate to glycogen in the liver reoxygenation of haemoglobin in the blood a high metabolic rate, as many organs are operating at above resting levels Respiratory substrates Although glucose is the essential respiratory substrate for some cells such as neurones in the brain, red blood cells and lymphocytes, other cells can oxidise lipids and amino acids When lipids are respired, carbon atoms are removed in pairs, as acetyl coenzyme A, from the fatty acid chains and fed into the Krebs cycle The carbon– hydrogen skeletons of amino acids are converted into pyruvate or into acetyl coenzyme A rest crucible water substrate oxygen Figure 12.18  A simple calorimeter in which the energy value of a respiratory substrate can be measured exercise recovery oxygen deficit 2.5 Oxygen uptake / dm3 min–1 278 2.0 post-exercise oxygen uptake (oxygen debt) 1.5 1.0 0.5 0 Time / Figure 12.17  Oxygen uptake before, during and after strenuous exercise 10 11 12 13 30 Chapter 12: Energy and respiration The energy liberated by oxidising the substrate can be determined from the rise in temperature of a known mass of water in the calorimeter Typical energy values are shown in Table 12.2 Respiratory substrate Energy density / kJ g −1 carbohydrate 15.8 lipid 39.4 protein 17.0 Table 12.2  Typical energy values Respiratory quotient (RQ) The overall equation for the aerobic respiration of glucose shows that the number of molecules, and hence the volumes, of oxygen used and carbon dioxide produced are the same: C6H12O6 + 6O2 → 6CO2 + 6H2O + energy So the ratio of oxygen taken in and carbon dioxide released is 1 : 1 However, when other substrates are respired, the ratio of the volumes of oxygen used and carbon dioxide given off differ It follows that measuring this ratio, called the respiratory quotient (RQ), shows what substrate is being used in respiration It can also show whether or not anaerobic respiration is occurring RQ = volume of carbon dioxide given out in unit time volume of oxygen taken in in unit time Or, from an equation, RQ = moles or molecules of carbon dioxide given out moles or molecules of oxygen taken in For the aerobic respiration of glucose: CO RQ = O2 6 = 1.0 = When the fatty acid oleic acid (from olive oil) is respired aerobically, the equation is: C18H34O2 + 25.5O2 → 18CO2 + 17H2O + energy For the aerobic respiration of oleic acid: CO RQ = O2 = 18 25.5 = 0.7 Typical RQs for the aerobic respiration of different substrates are shown in Table 12.3 Respiratory substrate carbohydrate lipid protein Respiratory quotient (RQ) 1.0 0.7 0.9 Table 12.3  Respiratory quotients of different substrates QUESTION 12.8 Calculate the RQ for the aerobic respiration of the fatty acid stearic acid (C18H36O2) What happens when respiration is not aerobic? The equation for the alcoholic fermentation of glucose in a yeast cell is: C6H12O6 → 2C2H5OH + 2CO2 + energy CO2 RQ = O2 = =∞ In reality, some respiration in the yeast cell will be aerobic, and so a small volume of oxygen will be taken up and the RQ will be less than infinity High values of RQ indicate that alcoholic fermentation is occurring Note that no RQ can be calculated for muscle cells using the lactate pathway, as no carbon dioxide is produced: glucose (C6H12O6) → lactic acid (C3H6O3) + energy 279 Cambridge International A Level Biology BOX 12.1: Oxygen uptake 280 Oxygen uptake during respiration can be measured using a respirometer A respirometer suitable for measuring the rate of oxygen consumption of seeds or small terrestrial invertebrates at different temperatures is shown in Figure 12.19 Carbon dioxide produced in respiration is absorbed by a suitable chemical such as soda-lime or a concentrated solution of potassium hydroxide (KOH) or sodium hydroxide (NaOH) Any decrease in the volume of air surrounding the organisms results from their oxygen consumption Oxygen consumption in unit time can be measured by reading the level of the manometer fluid against the scale Changes in temperature and pressure alter the volume of air in the apparatus, and so the temperature of the surroundings must be kept constant while readings are taken – for example, by using a thermostatically controlled water bath The presence of a control tube containing an equal volume of inert material to the volume of the organisms used helps to compensate for changes in atmospheric pressure Once measurements have been taken at a series of temperatures, a graph can be plotted of oxygen consumption against temperature The same apparatus can be used to measure the RQ of an organism First, oxygen consumption at a particular temperature is found (x cm3 min−1) Then the respirometer is set up with the same organism at the same temperature, but with no chemical to absorb carbon dioxide The manometer scale will show whether the volumes of oxygen absorbed and carbon dioxide produced are the same When the volumes are the same, the level of the manometer fluid will not change and the RQ = When more carbon dioxide is produced than oxygen absorbed, the scale will show an increase in the volume of air in the respirometer (by y cm3 min−1) The RQ can then be calculated: CO2 x + y RQ = − x O2 Conversely, when less carbon dioxide is produced than oxygen absorbed, the volume of air in the respirometer will decrease (by z cm3 min−1) and the calculation will be: CO2 x − z RQ = − O2 x Another way of investigating the rate of respiration of yeast is to use a redox dye such as a solution of dichlorophenolindophenol (DCPIP) (Box 13.1 and page 289) or of methylene blue These dyes not damage cells and so can be added to a suspension of yeast cells When reduced, these blue dyes become colourless The rate of change from blue to colourless is a measure of the rate of respiration of the yeast This technique can be used to investigate the effect of various factors on yeast respiration, such as temperature, substrate concentration or different substrates cm3 syringe screw-clip three-way tap non-vertebrates to be studied glass beads gauze platform soda-lime or KOH or NaOH soda-lime or KOH or NaOH experimental tube Figure 12.19  A respirometer control tube capillary U-tube containing manometer fluid Chapter 12: Energy and respiration QUESTION 12.9 Outline the steps you would take to investigate the effect of temperature on respiration rate Adaptations of rice for wet environments Although rice can grow in dry conditions, it is often grown in ‘paddies’ – fields where the ground is intentionally flooded Rice can tolerate growing in water, whereas most of the weeds that might compete with it are not able to so (Figure 12.20) Most plants cannot grow in deep water because their roots not get enough oxygen Oxygen is required for aerobic respiration, which provides ATP as an energy source for active transport and other energy-consuming processes such as cell division Nor, if the leaves are submerged, can photosynthesis take place, because there is not enough carbon dioxide available This happens because gases diffuse much more slowly in water than they in air Moreover, the concentrations of dissolved oxygen and dissolved carbon dioxide in water are much less than they are in air This is especially true in rice paddies, where the rich mud in which the rice roots are planted contains large populations of microorganisms, many of which are aerobic and take oxygen from the water Some varieties of rice respond to flooding by growing taller As the water rises around them, they keep growing upwards so that the top parts of their leaves and flower Figure 12.20  Rice growing in Madagascar The blocks of rice were planted at different times and are at different stages of growth spikes are always held above the water This allows oxygen and carbon dioxide to be exchanged through the stomata on the leaves The stems of the rice plants contain loosely packed cells forming a tissue known as aerenchyma (Figure 12.21) Gases are able to diffuse through the aerenchyma to other parts of the plant, including those under the water This is supplemented by air that is trapped in between the ridges of the underwater leaves These leaves have a hydrophobic, corrugated surface that holds a thin layer of air in contact with the leaf surface Nevertheless, the cells in the submerged roots still have to use alcoholic fermentation at least some of the time Ethanol can therefore build up in the tissues Ethanol is toxic, but the cells in rice roots can tolerate much higher levels than most plants They also produce more alcohol dehydrogenase, which breaks down ethanol This allows the plants to grow actively even when oxygen is scarce, using ATP produced by alcoholic fermentation QUESTION 12.10 List the features that make rice adapted to grow when partly submerged in water aerenchyma vascular tissues air space cells of the cortex Figure 12.21  Photomicrograph of a cross-section of a rice stem near its tip, with a leaf base around it Lower down, the stem is completely hollow (× 140) 281 Cambridge International A Level Biology Summary ■■ 282 Organisms must work to stay alive The energy input necessary for this work is either light, for photosynthesis, or the chemical potential energy of organic molecules Work includes anabolic reactions, active transport and movement ■■ Some organisms, such as mammals and birds, use thermal energy released from metabolic reactions to maintain their body temperature Reactions that release energy must be harnessed to energy-requiring reactions This involves an intermediary molecule, ATP ATP can be synthesised from ADP and phosphate using energy, and hydrolysed to ADP and phosphate to release energy ATP therefore acts as an energy currency in all living organisms ■■ Respiration is the sequence of enzyme-controlled steps by which an organic molecule, usually glucose, is broken down so that its chemical potential energy can be used to make the energy currency, ATP In aerobic respiration, the sequence involves four main stages: glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation ■■ ■■ In glycolysis, glucose is first phosphorylated and then split into two triose phosphate molecules These are further oxidised to pyruvate, giving a small yield of ATP and reduced NAD Glycolysis occurs in the cell cytoplasm When oxygen is available (aerobic respiration), the pyruvate passes to the matrix of a mitochondrion In a mitochondrion, in the link reaction, pyruvate is decarboxylated and dehydrogenated and the remaining 2C acetyl unit combined with coenzyme A to give acetyl coenzyme A The acetyl coenzyme A enters the Krebs cycle in the mitochondrial matrix and donates the acetyl unit to oxaloacetate (4C) to make citrate (6C) The Krebs cycle decarboxylates and dehydrogenates citrate to oxaloacetate in a series of small steps The oxaloacetate can then react with another acetyl coenzyme A from the link reaction ■■ Dehydrogenation provides hydrogen atoms, which are accepted by the carriers NAD and FAD These pass to the inner membrane of the mitochondrial envelope, where they are split into protons and electrons ■■ In the process of oxidative phosphorylation, the electrons are passed along a series of carriers Some of the energy released in oxidative phosphorylation is used to move protons from the mitochondrial matrix to the intermembrane space The movement of electrons sets up a gradient of protons across the inner membrane of the mitochondrial envelope Protons pass back into the matrix, moving down their concentration gradient through protein channels in the inner membrane An enzyme, ATP synthase, is associated with each of the proton channels ATP synthase uses the electrical potential energy of the proton gradient to phosphorylate ADP to ATP At the end of the carrier chain, electrons and protons are recombined and reduce oxygen to water ■■ In the absence of oxygen as a hydrogen acceptor (in alcoholic and lactic fermentations), a small yield of ATP is made through glycolysis, then dumping hydrogen into other pathways in the cytoplasm which produce ethanol or lactate The lactate pathway can be reversed in mammals when oxygen becomes available The oxygen needed to remove the lactate produced during lactic fermentation is called the oxygen debt ■■ The energy values of respiratory substrates depend on the number of hydrogen atoms per molecule Lipids have a higher energy density than carbohydrates or proteins The respiratory quotient (RQ) is the ratio of the volume of oxygen absorbed and the volume of carbon dioxide given off in respiration The RQ reveals the nature of the substrate being respired Carbohydrate has an RQ of 1.0, lipid 0.7 and protein 0.9 Oxygen uptake, and hence RQ, can be measured using a respirometer Chapter 12: Energy and respiration End-of-chapter questions What does not occur in the conversion of glucose to two molecules of pyruvate? A hydrolysis of ATP B phosphorylation of ATP C phosphorylation of triose (3C) sugar D reduction of NAD [1] Where does each stage of aerobic respiration occur in a eukaryotic cell? A B C D Link reaction Krebs cycle Oxidative phosphorylation cytoplasm mitochondrial cristae cytoplasm mitochondrial matrix mitochondrial matrix cytoplasm mitochondrial cristae mitochondrial matrix mitochondrial cristae mitochondrial matrix mitochondrial matrix mitochondrial cristae [1] The diagram summarises how glucose can be used to produce ATP, without the use of oxygen glucose X Y in mammals Z in yeast Which compounds are represented by the letters X, Y and Z ? A B C D X Y Z ethanol lactate pyruvate pyruvate pyruvate ethanol ethanol lactate lactate pyruvate lactate ethanol 283 [1] Distinguish between: a an energy currency molecule and an energy storage molecule b decarboxylation and dehydrogenation [2] [2] [Total: 4] State the roles in respiration of: a NAD b coenzyme A c oxygen [1] [1] [1] [Total: 3] Copy and complete the table to show how much ATP is used and produced for each molecule of glucose respired in the various stages of respiration ATP used ATP produced Net gain in ATP glycolysis link reaction Krebs cycle oxidative phosphorylation Total [5] Cambridge International A Level Biology a Explain why the energy value of lipid is more than twice that of carbohydrate b Explain what is meant by respiratory quotient (RQ) c Copy and complete the table to show the respiratory substrates with each of the given RQs Respiratory substrate [2] [2] RQ 1.0 0.7 0.9 [3] d Measurements of oxygen uptake and carbon dioxide production by germinating seeds in a respirometer showed that 25 cm3 of oxygen was used and 17.5 cm3 of carbon dioxide was produced over the same time period i Calculate the RQ for these seeds ii Identify the respiratory substrate used by the seeds e Dahlia plants store a compound called inulin, which is a polymer of fructose The structure of fructose is shown in the diagram CH2OH CH2OH O C H 284 [2] [1] C H HO C C OH H OH Calculate the RQ when inulin is hydrolysed and then respired aerobically [2] [Total: 12] Copy and complete the following passage describing the adaptations of rice for growing with its roots submerged in water The stems and leaves of rice plants have very large in tissue called ., which allow oxygen to pass from the air to the The roots are very shallow, giving them access to the higher concentration of in surface water When oxygen concentrations fall, the roots can oxidise glucose through This produces ., which is toxic However, the root cells are tolerant of higher concentrations of this than are most cells and they also contain high concentrations of the enzyme to break it down [7] Chapter 12: Energy and respiration In aerobic respiration, the Krebs cycle is regarded as a series of small steps One of these steps is the conversion of succinate to fumarate by an enzyme, succinate dehydrogenase a State the role played by dehydrogenase enzymes in the Krebs cycle and explain briefly the importance of this role in the production of ATP b An investigation was carried out into the effect of different concentrations of aluminium ions on the activity of succinate dehydrogenase The enzyme concentration and all other conditions were kept constant The graph below shows the results of this investigation [3] Rate of fumarate production / mmol min−1 65 60 55 50 45 40 35 285 20 40 60 80 100 Concentration of aluminium ions / μmol 120 With reference to the graph: i describe the effect of the concentration of aluminium ions on the rate of production of fumarate ii suggest an explanation for this effect Cambridge International AS and A Level Biology 9700/04, Question 7, October/November 2007 [2] [2] [Total: 7] ... 17 3 The cardiac cycle 17 5 Control of the heart beat 17 7 End-of-chapter questions 17 9 iii Cambridge International AS Level and ABiology Level Biology Gas exchange and smoking 18 5 Gas exchange... End-of-chapter questions Transport in plants 11 0 11 1 11 3 11 8 11 8 12 3 12 6 The transport needs of plants 12 7 Two systems: xylem and phloem 12 8 Structure of stems, roots and leaves 12 8 The transport... graticule scale and the stage micrometer scale 0 .1 stage micrometer scale (marked in 0.0 1mm and 0 .1 mm divisions) 0.2 Cambridge International AS Level Biology WORKED EXAMPLE Calculating the magnification