Biology for VCE Units 1 and 2 - (Full Colour Interim) Student Edition

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STUDENT EDITION

BIOLOGY FOR VCE UNITS 1&2


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STUDENT WORKBOOK


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Meet the Writing Team

Tracey

Senior Author

Lissa

Tracey Greenwood I have been writing resources for students since 1993. I have a Ph.D in biology, specialising in lake ecology and I have taught both graduate and undergraduate biology. Lissa Bainbridge-Smith I worked in industry in a research and development capacity for 8 years before joining BIOZONE in 2006. I have an M.Sc from Waikato University.

Author

Kent

Kent Pryor I have a BSc from Massey University majoring in zoology and ecology and taught secondary school biology and chemistry for 9 years before joining BIOZONE as an author in 2009.

Author

Richard

Founder & CEO

Richard Allan I have had 11 years experience teaching senior secondary school biology. I have a Masters degree in biology and founded BIOZONE in the 1980s after developing resources for my own students.

Cover Photograph

The southern cassowary (Casuarius casuarius) is one of three cassowary species. It is found in the tropical rainforests of Indonesia, New Guinea, and northeastern Australia. The cassowary is a ratite and related to emus and ostriches. It is the third tallest and second heaviest living ratite and the largest Australian bird, reaching a height of 1.9 m and weighing up to 85 kg. PHOTO: Irawan Subingar

Thanks to:

First edition 2020 Full colour interim edition

ISBN 978-1-98-856646-7

The staff at BIOZONE, including Clare Mansfield for design, Felix Hicks for illustration support, Paolo Curray and James Leggett for IT support, Anu Chauhan for logistics, Allan Young for office handling, and the BIOZONE sales team.

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Purchases of this workbook may be made direct from the publisher:


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Contents

Using This Workbook ............................................. vi Using the Tab System............................................. viii Using BIOZONE's Website.......................................xi Answering Exam Questions......................................x

UNIT 1, AREA OF STUDY 3 (SUPPORT)

45 46 47 48 49

Plant Cells ......................................................... Identifying Structures in a Plant Cell.................. Cell Structures and Organelles.......................... Identifying Organelles........................................ KEY TERMS: Did You Get it?.............................

Crossing the plasma membrane

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Skills for a practical investigation

UNIT 1, AREA OF STUDY 1

Cell size, structure, and function

31 32 33 34 35 36 37 38 39 40 41 42 43 44

Key knowledge................................................... 41 The Cell is the Unit of Life.................................. 42 The Biochemical Nature of Cells........................ 43 Water and Life ................................................... 44 Prokaryotic vs Eukaryotic Cells ........................ 45 Cell Sizes........................................................... 46 Limitations to Cell Size....................................... 47 Cellular Environments ...................................... 48 Types of Cells .................................................... 49 Optical Microscopes ........................................... 50 Preparing a Slide................................................ 52 Calculating Linear Magnification........................ 53 Electron Microscopes......................................... 54 Animal Cells....................................................... 56 Identifying Structures in an Animal Cell............. 57

Activity is marked:

to be done;

when completed

50 51 52 53 54 55 56 57 58 59

Energy transformations 60 61 62 63 64 65

Key knowledge ................................................. Autotrophs and Heterotrophs ............................ Energy Transformations in Cells........................ Plants as Producers .......................................... The Leaf and Photosynthesis ........................... An Introduction to Cellular Respiration ............. KEY TERMS: Did You Get it?.............................

66 67 68 69 70 71 72

Key knowledge ................................................. The Hierarchy of Life: Plants ............................. Specialisation in Plant Cells............................... The Plant Transport System .............................. Uptake at the Root ............................................ Transpiration ..................................................... Investigating Plant Transpiration........................ KEY TERMS: Did You Get It?.............................

77 78 79 70 82 84 86

Functioning systems in plants

87 88 89 90 92 93 95 97

Functioning systems in mammals

73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88

Key knowledge ................................................. 98 The Hierarchy of Life: Mammals ....................... 99 Specialisation in Animal Cells.......................... 100 The Respiratory System ................................. 101 The Lungs ....................................................... 102 Breathing.......................................................... 104 Recording Changes in Breathing and Heart Rate....................................................... 105 Malfunctions of the Respiratory System.......... 106 The Digestive System ..................................... 108 Moving Food Through the Gut......................... 109 The Stomach ................................................... 110 The Small Intestine ......................................... 111 Digestion, Absorption, and Transport .............. 112 The Large Intestine.......................................... 114 Malfunctions of the Digestive System.............. 115 The Circulatory System .................................. 116 Blood................................................................ 117

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Key knowledge.................................................... 1 How Do We Do Science?..................................... 2 Hypotheses and Predictions................................ 3 Types of Data....................................................... 4 Working with Qualitative Data.............................. 5 Collecting Quantitative Data................................ 6 Accuracy and Precision ...................................... 8 Working with Numbers ........................................ 9 Fractions, Percentages, and Ratios .................. 10 Logs and Exponents ......................................... 11 Practising with Data .......................................... 12 Apparatus and Measurement ............................ 13 Biological Drawings............................................ 14 Practising Biological Drawings........................... 16 Using Tables and Graphs .................................. 17 Which Graph to Use? ........................................ 18 Drawing Bar Graphs ......................................... 19 Drawing Histograms .......................................... 20 Drawing Line Graphs ........................................ 21 Correlation or Causation ................................... 23 Drawing Scatter Plots ....................................... 24 Describing Relationships Between Variables..... 25 Mean, Median, and Mode ................................. 26 Spread of Data................................................... 28 Detecting Bias in Samples................................. 29 Investigating Plant Survival in Lab..................... 30 Investigating Factors Affecting Plant Growth .... 32 Test Your Understanding.................................... 34 Structure of a Report ........................................ 37 A Template for Your Investigation ...................... 38 KEY TERMS: Did You Get It? ............................ 40

Key knowledge ................................................. 64 The Plasma Membrane ..................................... 65 Factors Altering Membrane Permeability........... 66 Diffusion ............................................................ 68 Diffusion and Cell Size ...................................... 69 Osmosis ............................................................ 71 Estimating Osmolarity ....................................... 72 Water Relations in Plant Cells .......................... 73 Active Transport................................................. 74 Ion Pumps and Cotransport................................ 75 KEY TERMS: Did You Get it?............................. 76

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

58 59 60 62 63


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Contents

135 Food Webs ...................................................... 136 Constructing a Food Web ............................... 137 Dingo Food Webs............................................ 138 Cave Food Webs.............................................. 139 The Role of Keystone Species......................... 140 Population Density and Distribution ................ 141 Measuring Distribution and Abundance .......... 142 Investigating Survival: Field Based.................. 143 Investigating Distribution and Abundance........ 144 Quadrat Sampling ........................................... 145 Sampling a Rocky Shore Community............... 146 Factors Affecting Population Size.................... 147 Calculating Change in Population Size............ 148 Predation and Population Cycles .................... 149 The Impact of Disease..................................... 150 The Influence of Competition........................... 151 Interspecific Competition ................................. 152 Competition and Species Distribution ............. 153 Field Study of a Rocky Shore........................... 154 Intraspecific Competition.................................. 155 The Effect of Resources on Population Size.... 156 KEY TERMS: Did You Get it? .......................... 157 Review: Unit 1, Area of Study 2 ...................... 158 Synoptic Question: Unit 1, Area of Study 2......

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89 Blood Vessels .................................................. 118 90 The Heart......................................................... 120 91 Dissecting a Mammalian Heart ....................... 122 92 Malfunctions of the Circulatory System........... 124 93 The Excretory System .................................... 126 94 Drawing the Kidney ......................................... 127 95 The Human Urinary System............................ 128 96 Making Urine ................................................... 130 97 Malfunctions of the Excretory System ............ 131 98 Circulation and Respiratory Interaction ........... 132 99 Circulation and Digestive Interaction................ 134 100 Circulation and Excretory Interaction............... 136 101 KEY TERMS: Did You Get it?........................... 138 102 Review: Unit 1, Area of Study 1........................ 139 103 Synoptic Question: Unit 1, Area of Study 1....... 141

Key knowledge ............................................... 143 104 What is Adaptation? ........................................ 144 105 Adaptations of Xerophytes............................... 145 106 Adaptations of Hydrophytes ............................ 146 107 Mangrove Adaptations .................................... 147 108 Plant Adaptations to Fire.................................. 148 109 Adaptations of Insectivorous Plants................. 149 110 Adaptations in Animals.................................... 150 111 Adaptations for Diving ..................................... 152 112 Similar Environments, Similar Adaptations...... 154 113 Biomimicry for Developing New Technologies . 155 114 Homeostasis .................................................. 157 115 Maintaining Homeostasis ................................ 158 116 Negative feedback........................................... 159 117 Positive feedback............................................. 160 118 Negative Feedback and Temperature Control . 161 119 Body Shape and Heat Loss............................. 163 120 Hyperthyroidism and Thermoregulation........... 164 121 Control of Blood Glucose................................. 165 122 Type 1 Diabetes Mellitus.................................. 166 123 Maintaining Water Balance.............................. 167 124 KEY TERMS: Did You Get it? .......................... 168

Organising biodiversity

Key knowledge ............................................... 169 125 Classification Systems: The Old and the New . 170 126 How Do We Assign Species?.......................... 172 127 Classification of Biodiversity ............................ 173 128 Naming an Organism ...................................... 176 129 Why is Biodiversity Important?......................... 177 130 Managing Biodiversity for Conservation........... 179 131 Biodiversity and Bioprospecting....................... 181 132 KEY TERMS: Did You Get it? .......................... 183

Relationships between organisms Key knowledge ............................................... 184 133 Species Interactions ........................................ 185 134 Feeding Relationships..................................... 187

Activity is marked:

to be done;

when completed

UNIT 2, AREA OF STUDY 1 The cell cycle

Key knowledge ............................................... 159 Why Cells Need to Divide ............................... 160 Binary Fission in Prokaryotes.......................... 161 The Cell Cycle in Eukaryotes........................... 162 Mitosis and Cytokinesis .................................. 163 Recognising Stages in Mitosis......................... 164 KEY TERMS: Did You Get it?...........................

222 223 224 225 226 228 229

Asexual reproduction

Key knowledge ............................................... 165 Asexual Reproduction ..................................... 166 Plant Propagation............................................ 167 Grafting............................................................. 168 Micropropagation............................................. 169 Asexual Reproduction by Spores..................... 170 Case Study: Cloning in Horticulture................. 171 KEY TERMS: Did You Get it?...........................

230 231 232 234 235 236 237 238

Sexual reproduction Key knowledge ............................................... 172 Cell Division .................................................... 173 Meiosis............................................................. 174 Meiosis and Variation....................................... 175 Modelling Meiosis............................................ 176 The Advantages of Sexual Reproduction......... 177 KEY TERMS: Did You Get it?...........................

239 240 241 242 243 245 246

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Survival through adaptation and regulation

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UNIT 1, AREA OF STUDY 2

188 189 191 193 194 196 197 198 200 202 203 205 206 207 208 209 210 211 212 215 216 217 218 220


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Contents

Cell growth and cell differentiation 247 248 250 252 253 255 256 258 259 261

216 Sex Linkage..................................................... 217 Inheritance Patterns......................................... 218 Problems Involving Monohybrid Crosses......... 219 Dihybrid Inheritance......................................... 220 Inheritance of Linked Genes............................ 221 Recombination and Dihybrid Inheritance......... 222 Problems Involving Dihybrid Crosses.............. 223 Genetic Screening........................................... 224 KEY TERMS: Did You Get it?........................... 225 Review: Unit 2, Area of Study 2....................... 226 Synoptic Question: Unit 2, Area of Study 2......

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Key knowledge ............................................... 178 What are Stem Cells?...................................... 179 Applications of Stem Cells............................... 180 Stem Cells Give Rise to Other Cells................ 181 Stem Cells and Prenatal Development............ 182 Regulation of the Cell Cycle............................. 183 Disrupting the Cell Cycle................................. 184 KEY TERMS: Did You Get it?........................... 185 Review: Unit 2, Area of Study 1....................... 186 Synoptic Question: Unit 2, Area of Study 1......

UNIT 2, AREA OF STUDY 2

303 304 306 307 308 309 310 311 313 314 316

UNIT 2, AREA OF STUDY 3

Genomes, genes, and alleles

Key knowledge ............................................... 187 Genomes, Genes, and Alleles ......................... 188 Genome Sizes................................................. 189 What is DNA Sequencing?.............................. 190 What was the Human Genome Project?.......... 191 DNA Differences Between Species................. 192 Using DNA to Determine Species Relatedness....................................... 193 Determining Gene Function Using Knockouts. 194 Hunting for a Gene........................................... 195 Screening for Genes......................................... 196 Genomic Analysis and Disease Risk............... 197 KEY TERMS: Did You Get it?...........................

263 264 265 266 267 268 269 270 271 272 273 274

Investigation of an issue

Key knowledge ............................................... 318 227 Recognising Balanced Reporting..................... 319 228 The Misuse of Scientific Data .......................... 320 229 Evaluating the Information .............................. 321 230 Case Study: Genetic Screening in Australia ... 322 231 Communicating Your Findings.......................... 323

Photo Credits and Acknowledgements .................. 324 INDEX .................................................................... 325

Chromosomes

Key knowledge ............................................... 198 Eukaryotic Chromosome Structure.................. 199 Species have Different Chromosome Numbers.................................... 200 Karyotyping...................................................... 201 Abnormal Chromosome Numbers................... 202 Making a Karyogram........................................ 203 KEY TERMS: Did You Get it?...........................

275 276 277 278 280 281 284

295 297

Pedigrees and genetic outcomes Key knowledge ............................................... 213 Pedigree Charts............................................... 214 The Monohybrid Cross..................................... 215 The Test Cross ................................................

Activity is marked:

to be done;

298 299 301 302

when completed

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285 286 287 288 289 290 292 293

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Genotypes and phenotypes

Key knowledge ............................................... 204 Alleles............................................................... 205 Dominant and Recessive Traits........................ 206 Influences on Phenotype................................. 207 Mutations can Alter Phenotype........................ 208 Environment and Phenotype............................ 209 Gene Expression and Phenotype.................... 210 Epigenetic Factors and Phenotype.................. 211 Genes, Environment, and Continuous Variation........................................ 212 KEY TERMS: Did You Get it?...........................


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Using This Workbook

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This first edition of Biology for VCE Units 1 & 2 has been specifically written to meet the content and skills requirements of the Victorian Certificate of Education, Biology, Units 1 and 2. Each chapter introduction provides you with a concise guide to the knowledge and skills requirements for each area of key knowledge. Each key knowledge point is matched to the activity or activities addressing it. Activities complementing practical skills are identified in the chapter introductions by a code (PRAC) and are designed to provide background and familiarity with apparatus, techniques, experimental design, and interpretation of results. A wide range of activities will help you to build on what you already know, explore new topics, work collaboratively, and practise your skills in data handling and interpretation. We hope that you find the workbook valuable and that you make full use of its features. The outline of the chapter structure below will help you to navigate through the material in each chapter.

Introduction

Activities

Literacy

Review

• A check list of key knowledge • A list of key terms

• The KEY IDEA provides your focus for the activity • Annotated diagrams help you understand the content • Questions review the content of the page

• Almost all chapters have a literacy activity based on the introductory key terms list • Several types of activities test your understanding of biological terms

• Create your own summary for review. • Hints help you to focus on what is important. • Your summary will consolidate your understanding of the content in the area of study

Synoptic Question • Synoptic questions conclude the unit and area of study covered in the workbook. • These enable you to practise your written exam skills

Linkages are made between ideas in separate activities

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Structure and organisation of chapters


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Understanding the activity coding system and making use of the online material identified will enable you to get the most out of this resource. The chapter content is structured to build knowledge and skills but this structure does not necessarily represent a strict order of treatment. Be guided by your teacher, who will assign activities as part of a wider programme of independent and group-based work.

Look out for these features and know how to use them: Free response questions allow you to use the information provided to answer questions about the content of the activity, either directly or by applying the same principles to a new situation. In some cases, an activity will assume understanding of prior content.

The activities form most of this workbook. They are numbered sequentially and each has a task code identifying the skill emphasised. Each activity has a short introduction with a key idea identifying the main message of the page. Most of the information is associated with pictures and diagrams, and your understanding of the content is reviewed through the questions. Some of the activities involve modelling and group work.

A TASK CODE on the page tab identifies the type of activity. For example, is it primarily information-based (KNOW), or does it involve modelling (PRAC) or a skill (SKILL)? A full list of codes is given on the following page but the codes themselves are relatively self explanatory.

WEB tabs at the bottom of the activity page alert the reader to the Weblinks resource, which provides external, online support material for the activity, usually in the form of an animation, video clip, photo library, or quiz. Bookmark the Weblinks page (see next page) and visit it frequently as you progress through the workbook.

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The chapter introduction provides you with a summary of the knowledge and skills requirements for the topic, phrased as a set of learning outcomes. Use the check boxes to identify and mark off the points as you complete them. The chapter introduction also provides you with a list of key terms for the chapter, from which you can construct your own glossary as you work through the activities.

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PHOTOCOPYING PROHIBITED

LINK tabs at the bottom of the activity page identify activities that are related in that they build on content or apply the same principles to a new situation.

Š 2020 BIOZONE International


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Using the Tab System

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The tab system is a useful system for quickly identifying related content and online support. Links generally refer to activities that build on the information in the activity in depth or extent. In the example below for the activity "The Cell is the Unit of Life", the weblink 31 describes concept of cells arising from other cells. Activity 34 covers the differences between prokaryotic and eukaryotic cells and 40 describes features of various types of cells. Sometimes, a link will reflect on material that has been covered earlier as a reminder for important terms that have already been defined or for a formula that may be required to answer a question. The weblinks code is always the same as the activity number on which it is cited. On visiting the weblink page (below), find the number and it will correspond to one or more external websites supporting the activity's content. Corrections and clarifications to current editions are also identified on the weblinks page.

Activities are coded

COMP = comprehension of text

SKILL = data handling and interpretation KNOW = content you need to know

PRAC = a paper practical or a practical focus REFER = reference - use this for information REVISE = review the material in the section TEST = test your understanding

Link

Weblinks

Bookmark the weblinks page: www.biozone.com.au/weblink/ VCE1-9407

Access the external URL for the activity by clicking the link

Connections are made between activities in different sections of the syllabus that are related through content or because they build on prior knowledge.

This WEBLINKS page provides links to external websites with supporting information for the activities. Mostly, these are animations and video clips directly relevant to some aspect of the activity on which they are cited. The weblinks page also provides access to annotated 3D models to explore specific aspects of your course. Bookmark this page as it is not accessible directly from BIOZONE's the website.

Annotated 3D models organised under broad topics

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Chapter in the book

Activity in the book

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Hyperlink to the external website page.


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Using BIOZONE's Website Contact us with questions, feedback, ideas, and critical commentary. We welcome your input.

Use Google to search for websites of interest. The more precise your search words are, the better the list of results. Be specific, e.g. "biotechnology medicine DNA uses", rather than "biotechnology".

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Access the BIOLINKS database of web sites directly from the homepage of our website.

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Biolinks is organised into easy-to-use sub-sections relating to general areas of interest. It's a great way to quickly find out more on topics of interest.

Š 2020 BIOZONE International


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Answering Exam Questions Exams require you to demonstrate your understanding of a particular concept by providing a written paragraph or essay. Open answer questions (meaning there is no definitive answer) are designed so that you can demonstrate your level of understanding. The question may give you some guidance as to what you should include in your answer, such as definitions of certain terms or to provide specific examples.

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In order to gain the highest possible mark in these questions, you need to lay out your answer in a clear and logical way so that the examiner can easily see how you have demonstrated your understanding of the topic. The difference between you obtaining a low, mid, or high grade depends on how well you demonstrate your understanding of a concept.

• Defining, drawing, annotating, or giving a description demonstrates a basic understanding of the material. • Explaining how a process works, why it works, and how changes to it may affect an outcome shows a deeper understanding of how the system works in that situation. • Linking biological ideas, comparing and contrasting, analysing, or justifying ideas shows both a deep understanding and an ability to translate that understanding to a new situation.

The following example shows how an answer can be built up from a simple definition, through explanation, to comparisons and linking of ideas. For the animal group mammals, discuss the relationship between the gas exchange system and the internal transport system. You should clearly show how they are linked and the role each plays in supporting a mammal's active lifestyle

How gas exchange is physically linked to transport system.

Description of how oxygen is exchanged and carried through the transport system.

The gas exchange system in mammals is linked to the internal transport system at the gas exchange surface in the alveoli of the lungs. Inhaled air passes through the trachea to the bronchi, to the bronchioles and finally into the alveoli, microscopic saclike structures at the terminal ends of bronchioles. Oxygen from the air diffuses through the gas exchange surface into the blood and binds to haemoglobin in the red blood cells. It is then transported to the body's cells. Haemoglobin greatly increases the amount of oxygen the blood can carry and supports a high oxygen demand (as in mammals).

Carbon dioxide (CO2)is picked up at the respiring tissues and is transported in the blood to the lungs where it diffuses into the alveoli and is exhaled. The diffusion of respiratory gases into and out of the blood proceeds because the diffusion gradients are maintained by the transport of gases to and from the gas exchange surface and continual exchange of air between the lungs and the external environment.

Explanation of how the diffusion of gases is maintained.

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This system supports a mammal's active lifestyle (high metabolic rate) by continually supply oxygen to fuel metabolism and removing CO2 (its waste product). The system can also adapt to increases in oxygen demand by increasing the volume of air breathed (more rapid breathing and deeper breaths) and increasing the speed at which the blood circulates (increasing heart rate to pump more blood per unit time).

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Why the system supports the mammal's lifestyle.

Description of the gas exchange system.

How the system can adjust to changes in demand.


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Skills for a Practical Investigation

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Unit 1 Outcome 3

Key terms

Practical skills assessed for outcome 3

accuracy

Key knowledge and skills

aim

Activity number

c

1

Show an understanding of the biological concepts and key terms as relevant to your investigation.

25 26 27

c

2

Demonstrate an understanding of the characteristics of scientific research methods and techniques for collecting primary qualitative and quantitative data as relevant to your investigation (e.g. microscopy, fieldwork, or observational studies of behaviour). Determine your aim and hypothesis, ask questions, and make predictions that can be tested.

1-5 11 39-41 141-144

c

3

Identify dependent, independent, and controlled variables. Explain your choice of a control, and show awareness of assumptions in your investigation.

5 25 26

c

4

Evaluate your investigative methods and procedures and suggest how any limitations could be overcome through improvements to design, apparatus, or technique. This will include an understanding of precision, accuracy, reliability, and validity of data, and how to minimise bias.

6 11 24 25 26 27 143 144

line graph

c

5

Demonstrate an ability to carry out an investigation safely and ethically.

mean

c

6

Use appropriate means to organise, analyse, and evaluate primary data to identify patterns and relationships. Include reference to sources of error and limitations of data or methods.

14-23

observation

c

7

Make clear, accurate scientific drawings as appropriate to your investigation.

12 13

percentage error

c

8

Evaluate results and draw valid conclusions based on observations and experiments. Resolve any conflicting evidence based on understanding and logical reasoning. Identify and explain anomalies in experimental measurements.

25 26 27 143 144

c

9

Describe options, strategies, or solutions related to survival of an organism or species. Identify and explain key findings as relevant to your investigation. Present your findings in a scientific report.

25 28 29 143

assumption bar graph control

controlled variable correlation

dependent variable fair test

histogram

hypothesis

independent variable

median mode

precision

prediction

primary data

quantitative data

25 143

qualitative data range

reliability sample

scatter plot (graph) secondary data

standard deviation

PASCO

validity (of data) variable

Mathematical and data handling skills Key skills for Units 1 & 2

Activity number

6 7 9 10

1

Demonstrate basic skills in mathematics, including using appropriate units in calculations and an appropriate number of significant figures in reporting.

c

2

Carry out calculations involving rates, percentages, and ratios.

c

3

Find arithmetic means for a range of data. Understand mean, median, and mode.

c

4

Understand and use measures of dispersion, e.g. standard deviation and range. Explain how these measures help you to evaluate the reliability of your data.

23 24

c

5

Select an appropriate format to plot two variables from experimental or other data (including line graphs, scatter plots, histograms, and bar graphs).

15-21

c

6

Show understanding of the principles of sampling as applied to scientific data and analyse random data collected by an appropriate sampling method.

25 142-145

c

7

Demonstrate an understanding of simple probability, e.g. as in genetic inheritance.

214

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c

7 8 22


1 How Do We Do Science?

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Key Idea: The scientific method is a rigorous, dynamic process of observation, investigation, and analysis that helps us to explain phenomena and predict changes in a system. Scientific knowledge is gained through a non-linear, dynamic process called the scientific method. The scientific method

is not a strict set of rules to be followed, but rather a way of approaching problems in a rigorous, but open-minded way. It involves inspiration and creativity, it is dynamic and context dependent, and usually involves collaboration. The model below is one interpretation of the scientific method.

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TASK: Before studying the diagram below, write a paragraph about what you think science is and what it involves. Then study the diagram and points below and state if your views differ from this and in what way. Staple your response to this page. At the end of your course, reexamine what you wrote. Have your ideas changed? Be honest in your first response!

EXPLORE AND DISCOVER

Ask questions

Find inspiration

Share ideas & information

Make observations

Explore the literature

TEST IDEAS Gathering data

Make hypotheses

Expected results

Observed results

Analysing and interpreting data Supporting, contradictory, or surprising data may…

RESULTS AND BENEFITS Develop technologies

…support a hypothesis

…change assumptions

…oppose a hypothesis

Solve problems

Peer review

Repeat investigation

Discussion with peers

Publication

New questions

Theory building

…suggest a new hypothesis

Build knowledge

Satisfy curiosity

ANALYSIS AND FEEDBACK

Inform society

Material adapted from the UC Berkeley's excellent website undsci.berkeley.edu/

Remember what science is and what it is not!

There is a single, linear scientific method

Scientists work without considering applications of their ideas

Scientific ideas are absolute and unchanging

Science is a process through which we can understand what we see

Science is exciting, dynamic, creative, and collaborative

Science has application and relevance in the modern world

Science is ongoing: it moves in a direction of greater understanding

KNOW

1

LINK

2

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WEB

Science is a solitary pursuit

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Science is just collection of facts

Science is a global human endeavour

© 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited


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2 Hypotheses and Predictions

Key Idea: A hypothesis is a tentative, testable explanation for an observed phenomenon. An assumption is something that is accepted as true but is not tested. Scientific hypotheses are tentative testable explanations for observed phenomena. A hypothesis leads to one or more predictions about the way a system will behave so a research hypothesis is usually written to include a testable

prediction, i.e if X is true, then the effect of Y will be Z. For every hypothesis, there is a corresponding null hypothesis: a hypothesis of no difference or no effect. A null hypothesis allows a hypothesis to be tested statistically and can be rejected if the experimental results are statistically significant. Hypotheses are not static, but may be modified as more information becomes available.

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Observations, hypotheses, and predictions

Observation is the basis of formulating hypotheses and predictions in science. An observation may generate a number of plausible hypotheses, and each hypothesis will lead to one or more predictions, which can be tested by further investigation. Observation 1: Some caterpillar species are brightly coloured and appear to be conspicuous to predators such as insectivorous (insect-eating) birds. Predators appear to avoid these species. These caterpillars are often found in groups, rather than as solitary animals.

Observation 2: Some caterpillar species are cryptic in their appearance or behaviour. Their camouflage is so convincing that, when alerted to danger, they are difficult to see against their background. Such caterpillars are often found alone.

Assumptions

Any biological investigation requires you to make assumptions about the system you are working with. Assumptions are features of the system (and investigation) that you assume to be true but do not (or cannot) test. Possible assumptions about the biological system described above include:

insectivorous birds have colour vision; caterpillars that look bright or cryptic to us, also appear that way to insectivorous birds; and insectivorous birds can learn about the palatability of prey by tasting them.

1. Study the example above illustrating the features of cryptic and conspicuous caterpillars, then answer the following:

(a) Generate a hypothesis to explain the observation that some caterpillars are brightly coloured and conspicuous while others are cryptic and blend into their surroundings:

(b) State the null form of this hypothesis:

(c) Describe one assumption being made in your hypothesis:

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Hypothesis:

(d) Based on your hypothesis, generate a prediction about the behaviour of insectivorous birds towards caterpillars:

2. Rewrite your original hypothesis as a research (or working) hypothesis to include your testable prediction:

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3 Types of Data

Key Idea: Data is information collected during an investigation. Data may be quantitative, qualitative, or ranked. Data is information collected during an investigation and it can be quantitative, qualitative, or ranked (below). When

planning a biological investigation, it is important to consider the type of data that will be collected. It is best to collect quantitative data, because it is mathematically versatile and easier to analyse it objectively (without bias).

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Types of data

Quantitative (interval or ratio)

Characteristics for which measurements or counts can be made, e.g. height, weight, number. Summary measures: mean, median, standard deviation.

Qualitative (nominal)

Ranked (ordinal)

Non-numerical and descriptive, e.g. sex, colour, viability (dead/alive), presence or absence of a specific feature. Summary measures: frequencies and proportions

Data are ranked on a scale that represents an order, although the intervals between the orders on the scale may not be equal, e.g. abundance (abundant, common, rare). Summary measures: frequencies and proportions

e.g. Sex of children in a family (male, female)

e.g. Birth order in a family (1, 2, 3)

Discontinuous or discrete data:

Discontinuous (discrete)

Continuous

e.g. Number of children in a family (3, 0, 4)

e.g. Height of children in a family (1.5 m, 0.8 m)

The unit of measurement cannot be split up (e.g. can't have half a child).

Continuous data:

The unit of measurement can be a part number (e.g. 5.25 kg).

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1. For each of the photographic examples A-C below, classify the data as quantitative, ranked, or qualitative:

B: Eggs per nest

A: Skin colour

(a) Skin colour:

(c) Tree trunk diameter:

C: Tree trunk diameter

(b) Number of eggs per nest:

2. Why is it best to collect quantitative data where possible in biological studies?

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3. Give an example of data that could not be collected quantitatively and explain your answer:

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4. Students walked a grid on a football field and ranked plant species present as abundant, common, or rare. How might they have collected and expressed this information more usefully?

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4 Working with Qualitative Data is increased by feeding. Zoo animals, such as primates, are also excellent subjects. By using a record sheet and observing the frequency and duration of behaviours, it is possible to analyse the data more objectively and obtain a semi-quantitative picture of the animal's behaviour. Using a watch, a 'sample' of the animal's behaviour is taken at regular time intervals (e.g. every minute, on the minute). From there, you can make inferences about how these behaviours might relate to survival and how they might change in different situations. You could use the codes listed below to classify the types of behaviour shown by the animal.

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Key Idea: Many studies of animal behaviour involve the collection and systematic recording of qualitative data. Qualitative data is descriptive (non-numerical) data. It is more difficult to analyse and interpret objectively than quantitative data because it is less rigorous (cannot be tested) and is more likely to be biased. However, it can be appropriate to collect qualitative data, as when recording colour changes in a reaction or making observations of behaviour. Your investigation for this area of study may involve gathering data on the behaviour of an individual or species. Common bird species are ideal, especially when resource competition

Species:

Age and sex:

Date:

Weather :

Time:

Season:

Behaviour chart

7. Dr (10 s)

1

21

41

2

22

42

3

23

43

4

24

5

Behaviour codes

R

resting

F

feeding

44

Dr

drinking

25

45

Pr

preening/grooming

6

26

46

Fl

flying

Su

sunning

7

27

47

Wl

walking

8

28

48

Fo

floating

9

29

49

Sw

swimming

Dp

display (e.g. singing)

10

30

50

Ag

aggression (e.g. fighting)

11

31

51

Sb

submission

12

32

52

OS

other social interaction

13

33

53

14

34

54

15

35

55

16

36

56

17

37

57

18

38

58

19

39

59

20

40

60

Location codes airborne

T

in a tree

on a tree trunk

TB

on a tree branch

TL

on leaves

G

on ground

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TT

GG

on ground - grass

GL

on ground - leaf litter

GS

on ground - soil

GM

on ground - marsh

W

in the water

Wu

under water

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TASK: Use your record sheet(s) to produce a report, presenting your results in a table or graph. In your discussion, focus on how the behaviours you observed might relate to the survival of the individual. Š 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

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5 Collecting Quantitative Data

Key Idea: Practical work carried out in a careful and methodical way makes analysis of the results much easier. A major part of any practical investigation is collecting the data. Practical work may be laboratory or field based. Typical laboratory based experiments involve investigating how a biological response is affected by manipulating a particular variable, e.g. temperature. The data collected for a

quantitative practical task should be recorded systematically, with due attention to safe practical techniques, a suitable quantitative method, and accurate measurements to an appropriate degree of precision. If your quantitative practical task is executed well, and you have taken care throughout, your evaluation of the experimental results will be much more straightforward and less problematic.

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Carrying out your practical work

Execution and recording

Preparation

Analysis and reporting

Know how you will take your Familiarise yourself with the equipment measurements and how often. Record and its set it up. Calibrate equipment if your results systematically as you go in necessary to give accurate measurements. a log book. You could record results a hand-written table or in a spreadsheet. If Read through the methods and identify using a data logger, data will be logged. key stages and how long they will take.

Analyse the data. Tables can summarise data. Graphs present the data to show patterns and trends. Statistical tests can determine the significance of results. Present your findings, e.g. as a poster, a digital presentation, or an oral report.

Identifying variables

Experimental controls

A variable is any characteristic or property able to take any one of a range of values. Investigations often look at the effect of changing one variable on another. It is important to identify all variables in an investigation: independent, dependent, and controlled, although there may be nuisance factors of which you are unaware. In all fair tests, only one variable is changed by the investigator.

A control refers to a standard or reference treatment or group in an experiment. It is the same as the experimental (test) group, except that it lacks the one variable being manipulated by the experimenter. Controls are used to demonstrate that the response in the test group is due a specific variable (e.g. temperature). The control undergoes the same preparation, experimental conditions, observations, measurements, and analysis as the test group. This helps to ensure that responses observed in the treatment groups can be reliably interpreted.

Dependent variable

• Measured during the investigation.

Dependent variable

• Recorded on the y axis of the graph.

• Factors that are kept the same or controlled. • List these in the method, as appropriate to your own investigation.

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Independent variable • Set by the experimenter. • Recorded on the graph's x axis.

The experiment above tests the effect of a certain nutrient on microbial growth. All the agar plates are prepared in the same way, but the control plate does not have the test nutrient applied. Each plate is inoculated from the same stock solution, incubated under the same conditions, and examined at the same set periods. The control plate sets the baseline; any growth above that seen on the control plate is attributed to the presence of the nutrient.

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Controlled variables

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Independent variable

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Investigation: effect of light on rate of photosynthesis Background

The aquarium plant, Cabomba aquatica, will produce a stream of oxygen bubbles when illuminated. The oxygen bubbles are a waste product of the process of photosynthesis (overall equation below right), which produces glucose (C6H12O6) for the plant. The rate of oxygen production provides an approximation of photosynthetic rate.

Distance from direct light source (cm) 20

25

30

35

40

No direct light

The method

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 6 x 1.0 g of Cabomba stems were placed into each of 6 test-tubes filled with 10 mL room temperature solution containing 0.2 molL-1 sodium hydrogen carbonate (to supply carbon dioxide).  Test tubes were placed at distances (20, 25, 30, 35, 40 cm) from a 60W light source (light intensity reduces with distance at a predictable rate). One test tube was not exposed to the light source.  Before recording, the Cabomba stems were left to acclimatise to the new light level for 5 minutes. The bubbles emerging from the stem were counted for a period of three minutes at each distance.

1.0 g Cabomba

0.2 molL-1 NaHCO3

6CO2 + 12H2O

Oxygen bubbles

Light

Stems were cut and inverted to ensure a free flow of oxygen bubbles.

C6H12O6 + 6O2 + 6H2O

1. Write a suitable aim for this experiment:

2. Write a possible hypothesis for this experiment:

3. (a) What is the independent variable in this experiment?

(b) What is the range of values for the independent variable?

(c) Name the unit for the independent variable:

(d) How could you better quantify the independent variable? 4. (a) What is the dependent variable in this experiment?

(b) Name the unit for the dependent variable:

(c) What equipment might have made it easier to record the response of the dependent variable accurately? Predict when it would have been most needed:

(d) What is the sample size for each treatment?

(e) What could you change in the design of the experiment to guard against unexpected or erroneous results?

5. Which tube is the control for this experiment?

(a) (b)

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6. Identify two assumptions being made about this system:

8. How might you test the gas being produced is oxygen:

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7. Identify one variable that might have been controlled in this experiment, and how it could have been monitored:


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6 Accuracy and Precision

Key Idea: Accuracy refers to the correctness of a measurement (how true it is to the real value). Precision refers to how close the measurements are to each other. Accuracy refers to how close a measured or derived value is to its true value. Simply put, it is the correctness of the measurement. Precision refers to the closeness of repeated

measurements to each other, i.e. the ability to be exact. A balance with a fault in it could give very precise (repeatable) but inaccurate (untrue) results. Data can only be reported as accurately as the measurement of the apparatus allows and is often expressed as significant figures (the digits in a number that express meaning to a degree of accuracy).

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The accuracy of a measurement refers to how close the measured (or derived) value is to the true value. The precision of a measurement relates to its repeatability. In most laboratory work, we usually have no reason to suspect a piece of equipment is giving inaccurate measurements (is biased), so making precise measures is usually the most important consideration. We can test the precision of our measurements by taking repeated measurements from individual samples. Population studies present us with an additional problem. When a researcher makes measurements of some variable in a study (e.g. fish length), they are usually trying to obtain an estimate of the true value for a parameter of interest, e.g. the mean size (which is correlated with age) of fish. Populations are variable, so we can more accurately estimate a population parameter if we take a large number of random samples from the population.

Accurate but imprecise

Inaccurate and imprecise

The measurements are all close to the true value but quite spread apart.

The measurements are all far apart and not close to the true value.

The measurements are all clustered close together but not close to the true value.

The measurements are all close to the true value and also clustered close together.

Analogy: The arrows are all close to the bullseye.

Analogy: The arrows are spread around the target.

Analogy: The arrows are all clustered close together but not near the bullseye.

Analogy: The arrows are clustered close together near the bullseye.

Significant figures

Significant figures (sf) are the digits of a number that carry meaning contributing to its precision. They communicate how well you could actually measure the data.

For example, you might measure the height of 100 people to the nearest cm. When you calculate their mean height, the answer is 175.0215 cm. If you reported this number, it implies that your measurement technique was accurate to 4 decimal places. You would have to round the result to the number of significant figures you had accurately measured. In this instance the answer is 175 cm.

Precise but inaccurate

Non-zero numbers (1-9) are always significant.

Accurate and precise

All zeros between non-zero numbers are always significant.

0.005704510

Zeros to the left of the first non-zero digit after a decimal point are not significant.

Zeros at the end of number where there is a decimal place are significant (e.g. 4600.0 has five sf). BUT Zeros at the end of a number where there is no decimal point are not significant (e.g. 4600 has two sf).

(a) 3.15985

(d) 1000.0

(b) 0.0012

(e) 42.3006

(c) 1000

(f) 120

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2. State the number of significant figures in the following examples:

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1. Distinguish between accuracy and precision:

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7 Working with Numbers

Key Idea: Using correct mathematical notation and being able to carry out simple calculations and conversions are fundamental skills in biology. Mathematics is used in biology to analyse, interpret, and

compare data. It is important that you are familiar with mathematical notation (the language of mathematics) and can confidently apply some basic mathematical principles and calculations to your data.

Decimal and standard form

In mathematics, universal symbols are used to represent mathematical concepts. They save time and space when writing. Some commonly used symbols are shown below.

Decimal form (also called ordinary form) is the longhand way of writing a number (e.g. 15,000,000). Very large or very small numbers can take up too much space if written in decimal form and are often expressed in a condensed standard form. For example, 15,000,000 is written as 1.5 x 107 in standard form.

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Commonly used mathematical symbols

= Equal to

< The value on the left is less than the value on the right

> The value on the left is greater than the value on the right

In standard form a number is always written as A x 10n, where A is a number between 1 and 10, and n (the exponent) indicates how many places to move the decimal point. n can be positive or negative.

>> The value on the left is much greater than the value on the right

For the example above, A = 1.5 and n = 7 because the decimal point moved seven places (see below).

<< The value on the left is much less than the value on the right

∝ Proportional to. A ∝ B means that A = a constant X B

1 5 0 0 0 0 0 0 = 1.5 x 107

~ Approximately equal to

Conversion factors and expressing units

Small numbers can also be written in standard form. The exponent (n) will be negative. For example, 0.00101 is written as 1.01 x 10-3.

Measurements can be converted from one set of units to another by the use of a conversion factor.

A conversion factor is a numerical factor that multiplies or divides one unit to convert it into another. Conversion factors are commonly used to convert non-SI units to SI units (e.g. converting pounds to kilograms). Note that mL and cm3 are equivalent, as are L and dm3.

0. 0 0 1 0 1 = 1.01 x 10-3

Converting can make calculations easier. Work through the following example to solve 4.5 x 104 + 6.45 x 105.

5. Convert 4.5 x 104 + 6.45 x 105 to decimal form:

In the space below, convert 5.6 cm3 to mm3 (1 cm3 = 1000 mm3):

1.

6. Add the two numbers together:

The value of a variable must be written with its units where possible. SI units or their derivations should be used in recording measurements: volume in cm3 (mL) or dm3 (L), mass in kilograms (kg) or grams (g), length in metres (m), time in seconds (s).

7. Convert to standard form:

For example the rate of oxygen consumption would be expressed: 3 -1 -1

Oxygen consumption ( cm g s )

Note that when using inverse notation, a negative exponent indicates "per", e.g. no. m-2 means no. per metre squared.

Estimates

When carrying out calculations, typing the wrong number into your calculator can put your answer out by several orders of magnitude. An estimate is a way of roughly calculating the answer. It helps you decide if your final calculation is correct.

Numbers are often rounded to help make estimation easier. The rounding rule is, if the next digit is 5 or more, round up. If the next digit is 4 or less, it stays as it is.

Rates

Rates are expressed as a measure per unit of time and show how a variable changes over time. Rates are used to provide meaningful comparisons of data that may have been recorded over different time periods. Often rates are expressed as a mean rate over the duration of the measurement period, but it is also useful to calculate the rate at various times to understand how rate changes over time. The table below shows the reaction rates for a gas produced during a chemical reaction. A worked example for the rate at 4 minutes is provided below the table.

Time (Minute)

Cumulative gas produced (cm3)

Rate of reaction (cm3 min-1)

0

0

0

2

34

17

4

42

4*

Use the following examples to practise estimating:

8

6 10

3. 3.4 x 72 ÷ 15: 4. 658 ÷ 22:

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3

50

1

50

0

* Gas produced between 2- 4 min: 42 cm3 – 34 cm3 = 8 cm3

Rate of reaction between 2- 4 mins:8 ÷ 2 minutes = 4 cm3 min-1

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2. 43.2 x 1044:

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For example, to estimate 6.8 x 704 you would round the numbers to 7 x 700 = 4900. The actual answer is 4787, so the estimate tells us the answer (4787) is probably right.

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8 Fractions, Percentages, and Ratios

Key Idea: Percentages and ratios are alternative ways to express fractions. All forms are commonly used in biology. The data collected in the field or laboratory are called raw data. Data are often expressed in ways that make them easy

to understand, visualise, and work with. Fractions, ratios, and percentages are widely used in biology and are often used to provide a meaningful comparison of sample data where the sample sizes are different.

Ratios

Percentages

• Fractions express how many parts of a whole are present.

• Ratios give the relative amount of two or more quantities, and provide an easy way to identify patterns.

• Percentages are expressed as a fraction of 100 (e.g. 20/100 = 20%).

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Fractions

• Fractions are expressed as two numbers separated by a solidus (/) (e.g. 1/2).

• The top number is the numerator. The bottom number is the denominator. The denominator can not be zero.

• Fractions are often written in their simplest form (the top and bottom numbers cannot be any smaller, while still being whole numbers). Simplifying makes working with fractions easier.

• Ratios do not require units.

• Ratios are usually expressed as a : b.

• Ratios are calculated by dividing all the values by the smallest number. 882 inflated

• Percentages provide a clear expression of what proportion of data fall into any particular category, e.g. for pie graphs.

• Allows meaningful comparison between different samples.

• Useful to monitor change (e.g. % increase from one year to the next). Volume of food colouring (cm3)

Volume of water (cm3)

Concentration of solution (%)

10

0

100

8

2

80

6

4

60

4

6

40

2

8

20

0

10

0

299 constricted

Pea pod shape: Ratio = 2.95 : 1

495 round yellow

In a class of 20 students, five had blue eyes. This fraction is 5/20. To simplify this fraction, divide the numerator and denominator by a common factor (a number which both are divisible by). In this instance the lowest common factor is five (1/4). To add fractions with different denominators, obtain a common denominator, add numerators, then simplify.

158 round green

152 wrinkled yellow

55 wrinkled green

Pea seed shape and colour: Ratio = 9 : 2.8 : 2.9 : 1

Example: Calculating phenotype ratios in Mendelian genetics

Example: Producing standards for a calibration curve.

1. (a) A student prepared a slide of the cells of an onion root tip and counted the cells at various stages in the cell cycle. The results are presented in the table (right). Calculate the ratio of cells in each stage (show your working):

(b) Assuming the same ratio applies in all the slides examined in the class, calculate the number of cells in each phase for a cell total count of 4800.

Cell cycle stage

No. of cells counted

Interphase

140

Prophase

70

Telophase

15

Metaphase

10

Anaphase

5

Total

240

No. of cells calculated

4800

2. Simplify the following fractions:

3. In the fraction example pictured above 5/20 students had blue eyes. In another class, 5/12 students had blue eyes. What fraction of students had blue eyes in both classes combined?

(c) 11/121:

Women

Body mass (kg)

Lean body mass (kg)

Athlete

50

38

56

41

65

46

80

48

95

52

Lean

4. The total body mass and lean body mass for women with different body types is presented in the table (right). Complete the table by calculating the % lean body mass column.

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(b) 84/90:

Normal weight Overweight Obese

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(a) 3/9 :

% lean body mass

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9 Logs and Exponents

Key Idea: A function relates an input to an output. Functions are often defined through a formula that tells us how to compute the output for a given input. Logarithmic, power, and exponential functions are all common in biology. A function is a rule that allows us to calculate an output for any given input. In biology, power functions are often observed in biological scaling, for example, heart beat slows

with increasing size in mammals. Exponential growth is often seen in bacterial populations and also with the spread of viral diseases if intervention does not occur. The 2014 Ebola outbreak is one such example. The numbers associated with exponential growth can be very large and are often log transformed. Log transformations reduce skew in data and make data easier to analyse and interpret.

Exponential function

Log transformations

Power functions are a type of scaling function showing the relationship between two variables, one of which is usually size.

Exponential growth occurs at an increasingly rapid rate in proportion to the growing total number or size.

A log transformation makes very large numbers easier to work with. The log of a number is the exponent to which a fixed value (the base) is raised to get that number. So log10 (1000) = 3 because 103 = 1000.

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Power function

• Power functions are not linear, one variable changes more quickly relative to the other.

Basal metabolic rate

• Examples of power functions include metabolic rate versus body mass (below), or surface area to volume ratio.

BMR = aMb

Body mass

Example: Relationship between body mass and metabolic rate. M = mass and a and b are constants.

• The equation for an exponential function is y = cx. • Exponential growth and decay (reduction) are possible

• Exponential growth is easy to identify because the curve has a J-shape appearance due to its increasing steepness over time. It grows more rapidly than a power function • Examples of exponential growth include the growth of microorganisms in an unlimiting growth environment.

• Both log10 and loge (natural logs or ln) are commonly used. • Log transformations are useful for data where there is an exponential increase or decrease in numbers. In this case, the transformation will produce a straight line plot (below). • To find the log10 of a number, e.g. 32, using a calculator, key in log 32 = . The answer should be 1.51. • Alternatively, the untransformed data can be plotted directly on a log-linear scale. The log axis runs in exponential cycles and the paper makes the log for you.

Log 10 cell numbers

• The equation for an exponential function is y = xc.

• In an exponential function, the base number is fixed (constant) and the exponent is variable.

Cell numbers

• In power functions, the base value is variable and the exponent (power number) is fixed (constant).

Time

Example: Cell growth in a yeast culture in optimal growth conditions.

Time

Example: Yeast cell growth plotted on logarithmic scale.

1. Describe the relationship between body mass and metabolic rate:

(b) What is the purpose of a log transformation?

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3. (a) On what type of data would you carry out a log transformation?

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2. Describe the difference between a power function and exponential growth:

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10 Practising with Data

Key Idea: This activity allows you to practise working with data and applying the skills you have learned in previous activities. 1. Complete the transformations for each of the tables below. The first value is provided in each case.

(a) Photosynthetic rate at different light intensities Average time for leaf disc to float (min)

Reciprocal of time* (min-1)

Time (min)

Pipette arm reading (cm3)

Plant water loss (cm3 min-1)

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Light intensity (%)

(b) Plant water loss using a bubble potometer

0

9.0

25

5

8.0

0.2

25

50

10

7.2

11

93

15

6.2

6

187

20

4.9

100

15

50

0.067

* Reciprocal of time gives a crude measure of rate.

(c) Incidence of cyanogenic clover in different areas

Clover plant type

Frost free area

(d) Frequency of size classes in a sample of eels

Frost prone area

Number

%

Number

Cyanogenic

124

78

26

Acyanogenic

35

Total

159

%

Totals

Size class (mm)

Frequency

Relative frequency (%)

0-50

7

2.6

50-99

23

100-149

59

150-199

98

200-249

50

250-299

30

300-349

3

Total

270

115

Clover

2. Convert the following decimal form numbers to standard form:

(a) 8970

(b) 0.046

(c) 1 467 851

3. Convert the following standard form numbers to decimal form: (a) 4.3 x 10-1

(b) 0.0031 x 10-2 (c) 6.2 x 104

4. (a) The table on the right shows the nutritional label found on a can of chilli beans. Use the information provided to complete the table by calculating the percentage composition for each of the nutritional groups listed:

(b) How much of the total carbohydrates is made up of:

Dietary fibre?

Chilli Beans Nutrition Facts Serving size 1 cup ( 253 g)

Amount per serving Total Fat – Saturated Fat

% Composition

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8g 3g

Sugars?

Total Carbohydrate

22 g

– Dietary Fibre

9g

– Sugars Protein

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4g

25 g

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(c) Manufacturers do not have to state the volume of water, which makes up the remainder of the serving size. What percentage of the can of beans is water?

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11 Apparatus and Measurement

Key Idea: The apparatus used in experimental work must be appropriate for the experiment or analysis and it must be used correctly to eliminate experimental errors. Using scientific equipment can generate experimental errors.

These can be reduced by selecting the right equipment for what you want to measure and by using it correctly. Some error is inevitable, but evaluating experimental error helps to interpret and assess the validity of the results.

Selecting the correct equipment

Recognising potential sources of error

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When measuring physical properties it is vital that you choose equipment that is appropriate for the type of measurement you want to take. For example, if you wanted to accurately weigh out 5.65 g of sucrose, you need a balance that accurately weighs to two decimal places. A balance that weighs to only one decimal place would not allow you to make an accurate enough measurement. Study the glassware (right). Which would you use if you wanted to measure 225 mL? The graduated cylinder has graduations every 10 mL whereas the beaker has graduations every 50 mL. It would be more accurate to measure 225 mL in a graduated cylinder.

Percentage errors

Percentage error is a way of mathematically expressing how far out your result is from the ideal result. The equation for measuring percentage error is: experimental value - ideal value ideal value

x 100

For example, you want to know how accurate a 5 mL pipette is. You dispense 5 mL of water from a pipette and weigh the dispensed volume on a balance. The volume is 4.98 mL. experimental value (4.98) - ideal value (5.0) ideal value (5.0)

x 100

The percentage error = –0.4% (the negative sign tells you the pipette is dispensing less than it should).

It is important to know how to use equipment correctly to reduce errors. A spectrophotometer measures the amount of light absorbed by a solution at a certain wavelength. This information can be used to determine the concentration of the absorbing molecule (e.g. density of bacteria in a culture). The more concentrated the solution, the more light is absorbed. Incorrect use of the spectrophotometer can alter the results. Common mistakes include incorrect calibration, errors in sample preparation, and errors in sample measurement.

A cuvette (left) is a small clear tube designed to hold spectrophotometer samples. Inaccurate readings occur when: • The cuvette is dirty or scratched (light is absorbed giving a falsely high reading). • Some cuvettes have a frosted side to aid alignment. If the cuvette is aligned incorrectly, the frosted side absorbs light, giving a false reading. • Not enough sample is in the cuvette and the beam passes over, rather than through the sample, giving a lower absorbance reading.

1. Assume that you have the following measuring devices available: 50 mL beaker, 50 mL graduated cylinder, 25 mL graduated cylinder, 10 mL pipette, 10 mL beaker. What would you use to accurately measure:

(a) 21 mL:

(b) 48 mL:

(c) 9 mL:

2. Calculate the percentage error for the following situations (show your working): (a) A 1 mL pipette delivers a measured volume of 0.98 mL:

(b) A 10 mL pipette delivers a measured volume of 9.98 mL:

(c) The pipettes used in (a) and (b) above both under-delivered 0.02 mL, yet the percentage errors are quite different. Use this data to describe the effect of volume on percentage error:

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12 Biological Drawings

Key Idea: Good biological drawings provide an accurate record of the specimen you are studying and enable you to make a record of its important features. Drawing is a very important skill to have in biology. Drawings record what a specimen looks like and give you an opportunity

Biological drawings require you to pay attention to detail. It is very important that you draw what you actually see, and not what you think you should see.

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to record its important features. Often drawing something will help you remember its features at a later date (e.g. in a test). Annotated drawings provide explanatory notes about the labelled structures, while plan diagrams label the main structures observed, but provide no additional detail.

Biological drawings should include as much detail as you need to distinguish different structures and types of tissue, but avoid unnecessary detail which can make your drawing confusing.

Attention should be given to the symmetry and proportions of your specimen. Accurate labeling, a statement of magnification or scale, the view (section type), and type of stain used (if applicable) should all be noted on your drawing.

Some key points for making good biological drawing are described on the example below. The drawing of Drosophila (right) is well executed but lacks the information required to make it a good biological drawing.

This drawing of Drosophila is a fair representation of the animal, but has no labels, title, or scale.

All drawings must include a title. Underline the title if it is a scientific name.

Copepod

Single eye

Centre your drawing on the page, not in a corner. This will leave room to place labels around the drawing.

Antenna

Trunk

If you need to represent depth, use stippling (dotting). Do not use shading as this can smudge and obscure detail.

Proportions should be accurate. If necessary, measure the lengths of various parts with a ruler.

Use simple, narrow lines to make your drawings.

Egg sac

Thorax

Your drawing must include a scale or magnification to indicate the size of your subject.

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Setae

Scale 0.2 mm

All parts of your drawing must be labelled accurately. Labeling lines should be drawn with a ruler and should not cross over other label lines. Try to use only vertical or horizontal lines.

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Use a sharp pencil to draw with. Make your drawing on plain white paper.

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Caudal rami

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15 Annotated diagrams

An annotated diagram is a diagram that includes a series of explanatory notes. These provide important or useful information about your subject.

Transverse section through collenchyma of Helianthus stem. Magnification x 450

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Cytoplasm A watery solution containing dissolved substances, enzymes, and the cell organelles.

Primary wall with secondary thickening.

Nucleus A large, visible organelle. It contains most of the cell’s DNA.

Chloroplast These are specialised plastids containing the green pigment chlorophyll. Photosynthesis occurs here.

Vacuole containing cell sap.

Plan diagrams

Plan diagrams are drawings made of samples viewed with the naked eye, hand lens, or under a microscope at low or medium power. They are used to show the distribution of the different tissue types in a sample without any cellular detail. The tissues are identified, but no detail about the cells within them is included. The example here shows a plan diagram produced after viewing a light micrograph of a transverse section through a dicot stem.

Light micrograph of a transverse section through a dicot stem.

Epidermis

Parenchyma between vascular bundles

Vascular cambium Xylem Sclerenchyma (fibre cap) Pith (parenchyma cells)

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Vascular bundle

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Phloem


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13 Practising Biological Drawings

Key Idea: Attention to detail is vital when making accurate and useful biological drawings.

In this activity, you will practise the skills required to translate what is viewed into a good biological drawing.

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Root transverse section from Ranunculus

Root hairs

Epidermal cell

Parenchyma cell

Above: Use relaxed viewing when drawing at the microscope. Use one eye (the left for right handers) to view and the right eye to look at your drawing.

Above: Light micrograph Transverse section (TS) through a Ranunculus root.

Right: A biological drawing of the same section.

Xylem

Scale

0.05 mm

Phloem

1. During your course, you will study the features of cells and also make an investigation related to survival or an organism or species. You may need to identify and draw features of plant or animal tissues with a light microscope. Generally, only large organelles such as the nucleus and chloroplasts are easily seen at the magnifications typical of school microscopes (x 400).

In the space right, make a biological drawing of your own specimen or slide, or practise your drawing by making a plan diagram of the image below.

Below: A light micrograph of a leaf from the beach grass Ammophila below. The leaf is rolled inwards to reduce water loss.

Mesophyll with chloroplasts

Leaf veins

Thick walled packing cells

EII

Waxy leaf epidermis

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Air space

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Hairs

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14 Using Tables and Graphs useful to plot your data as soon as possible, even during your experiment, as this will help you to evaluate your results as you proceed and make adjustments as necessary (e.g. to the sampling interval). The choice between graphing or tabulation in the final report depends on the type and complexity of the data and the information that you are wanting to convey. Usually, both are appropriate.

Presenting data in tables

Presenting data in graphs

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Key Idea: Tables and graphs provide a way to organise and visualise data in a way that helps to identify trends. Tables and graphs are ways to present data and they have different purposes. Tables provide an accurate record of numerical values and allow you to organise your data so that relationships and trends are apparent. Graphs provide a visual image of trends in the data in a minimum of space. It is

Table 1: Length and growth of the third internode of bean plants receiving three different hormone treatments Sample size

Mean rate of internode growth (mm day-1)

Mean internode length (mm)

Control

50

0.60

32.3

Hormone 1

46

1.52

41.6

Hormone 2

98

0.82

38.4

Hormone 3

85

2.06

50.2

1.1

Yield (absorbance at 550 nm)

Treatment

Fig. 1: Yield of two bacterial strains at different antibiotic levels (Âą 95% confidence intervals, n= 6)

1.0

0.9 0.8 0.7 0.6

Sensitive strain Resistant strain

0.5

Tables provide a way to systematically record and condense a large amount of information. They provide an accurate record of numerical data and allow you to organise your data in a way that allows you to identify relationships and trends. This can help to decide the best way to graph the data if graphing is required.

0

0

1

2

3

4

5

Antibiotic concentration (g m-3)

Table titles and row and column headings must be clear and accurate so the reader knows exactly what the table is about. Calculations such as rates and summary statistics (such as mean or standard deviation) may be included on a table.

Graphs are a good way of visually showing trends, patterns, and relationships without taking up too much space. Complex data sets tend to be presented as a graph rather than a table.

Summary statistics make it easier to identify trends and compare different treatments. Rates are useful in making multiple data sets comparable, e.g. if recordings were made over different time periods.

Presenting graphs properly requires attention to a few basic details, including correct orientation and labelling of the axes, accurate plotting of points, and a descriptive, accurate title.

1. Describe the advantages of using a table to present information:

2. (a) What is the benefit of including summary information (e.g. means) on a table?

3. What are the main advantages of presenting data in a graph?

4. Why might you include both graphs and tables in a final report:

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(b) In an experiment, a student recorded water lost from a plant shoot over one hour. The next day, he repeated the experiment to test the effect of wind on water loss, but he had only 45 minutes to let the experiment run. How could the student best present the data in a table to compare the two trials?

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15 Which Graph to Use?

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18

Key Idea: The type of graph you choose to display your data depends on the type of data you have collected. Before you graph your data, it is important to identify what type of data you have. Choosing the correct type of graph can `` One variable is a category

highlight trends or reveal relationships between variables. Choosing the wrong type of graph can obscure information and make the data difficult to interpret. Examples of common types of graphs and when to use them are provided below.

Use a pie graph

`` One variable is a count 23%

Water use key

17%

Use to compare proportions in different categories.

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Cooling water

Irrigation

27%

33%

Commercial /washwater

Drinking supply

`` One variable is a category

`` One variable is continuous data (measurements)

Use a bar or column graph

Average household water consumption in Australian cities Sydney

Use to compare different categories (or treatments) for a continuous variable.

Perth

What type of data have you collected?

Melbourne Adelaide

Canberra Hobart

0 200 400 600 Household consumption (L per year x 1000)

`` One variable is continuous data (measurements)

Use a histogram

Use to show a frequency distribution for a continuous variable.

Frequency

`` One variable is a count

`` Both variables are continuous

Weight (g)

`` The response variable is dependent on the independent variable

Use a line graph

`` The points are connected point to point

12

14

Temperature vs metabolic rate in a rat Line connecting points

Use to illustrate the response to a manipulated variable.

Metabolic rate

10 8 6 4 2 0

0

`` Both variables are continuous

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Use to illustrate the relationship between two correlated variables.

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Line of best fit

2 3 Body length (mm)

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Number of eggs in brood

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Body length vs brood size in Daphnia

`` A line of best fit can be drawn through the points

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Use a scatter plot

`` The two variables are inter-dependent but there is no manipulated variable

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10 20 Temperature (°C)

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16 Drawing Bar Graphs

Key Idea: Bar graphs are used to plot data that is non-numerical or discrete for at least one variable.

Guidelines for bar graphs

Average household water consumption in Australian cities

Bar graphs are appropriate for data that are non-numerical and discrete for at least one variable, i.e. they are grouped into categories. There are no dependent or independent variables. Important features of this type of graph include: Data are collected for discontinuous, non-numerical categories (e.g. colour, species), so the bars do not touch.

263

Perth

330

Melbourne

270

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Sydney

Data values may be entered on or above the bars.

Adelaide

Multiple sets of data can be displayed side by side for comparison (e.g. males and females).

Canberra

Axes may be reversed so that the categories are on the x axis, i.e. bars can be vertical or horizontal. When they are vertical, these graphs are called column graphs.

Species

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Mean (no. m-2)

Site 1

Site 2

Hobart

570

0

100 200 300 400 500 600 Household consumption (litres per year x 1000)

Field data notebook

Total counts at site 1 (11 quadrats) and site 2 (10 quadrats). Quadrats 1 sq m. Site 2

Site 1

No m–2

Species

Total

Ornate limpet Radiate limpet Limpet sp. A Cats-eye Top shell Limpet sp. B Limpet sp. C Chiton

No m–2

Mean

Total Mean

232

21

299

68 420

6 38

344 0

30 34 0

68 16

6 2

16 43

2 4

628 0 12

57 0 1

389 22 30

39 2 3

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Mean abundance of 8 molluscan species from two sites along a rocky shore.

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1. Counts of eight mollusc species were made from a series of quadrat samples at two sites on a rocky shore. The summary data are presented here. (a) Tabulate the mean (average) numbers per square metre at each site in Table 1 (below left). (b) Plot a bar graph of the tabulated data on the grid below. For each species, plot the data from both sites side by side using different colours to distinguish the sites.

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17 Drawing Histograms

Key Idea: Histograms graphically show the frequency distribution of continuous data.

Guidelines for histograms

30 25

Frequency

20

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Histograms are plots of continuous data and are often used to represent frequency distributions, where the y-axis shows the number of times a particular measurement or value was obtained. For this reason, they are often called frequency histograms. Important features of this type of graph include:

Frequency of different mass classes of animals in a population.

1. The weight data provided below were recorded from 95 individuals (male and female), older than 17 years.

(a) Create a tally chart (frequency table) in the frame provided, organising the weight data into a form suitable for plotting. An example of the tally for the weight grouping 55-59.9 kg has been completed for you as an example. Note that the raw data values are crossed off the data set in the notebook once they are recorded as counts on the tally chart. It is important to do this in order to prevent data entry errors. (b) Plot a frequency histogram of the tallied data on the grid provided below.

Weight (kg)

45-49.9

40-44.9

35-39.9

30-34.9

0

25-29.9

A neatly constructed tally chart doubles as a simple histogram.

5

20-24.9

10

15-19.9

The x-axis usually records the class interval. The y-axis usually records the number of individuals in each class interval (frequency).

5-9.9

15

10-14.9

The data are numerical and continuous (e.g. height or weight), so the bars touch.

0-4.9

Mass (g)

Tally

Total

IIII II

7

45-49.9

50-54.9 55-59.9

60-64.9 65-69.9

70-74.9 75-79.9

80-84.9

Lab notebook

85-89.9

Weight (in kg) of 95 individuals

90-94.9

63.4 56.5 84 81.5 73.4 56 60.4 83.5 82 61 55.2 48 53.5 63.8 69 82.8 68.5 67.2 82.5 83 78.4 76.5 83.4 77.5 77 87 89 93.4 83 80 76 56

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65 75.6 76.8 67.8 68.3 63.5 58 58.5 50 92 91.5 88.3 81 72 66.5 61.5 66 65.5 67.4 73 67 71 70.5 65.5 68 90 83.5 73 66 57.5 76

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81.2 83.3 95 105.5 82 73.5 75.2 63 70.4 82.2 87.8 86.5 85.5 87 98 71 76 72.5 61 60.5 67 86 85 93.5 62 62.5 63 60 71.5 73.8 77.5 74

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18 Drawing Line Graphs

Key Idea: Line graphs are used to plot continuous data in which one variable (the independent variable) directly affects

the other (dependent) variable. They are appropriate for data in which the independent variable is manipulated. Metabolic rate of rats at different temperatures

Guidelines for line graphs Metabolic rate (arbitrary units)

12 10 Line connecting points

8

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Line graphs are used when one variable (the independent variable) affects another, the dependent variable. Line graphs can be drawn without a measure of spread (top figure, right) or with some calculated measure of data variability (bottom figure, right). Important features of line graphs include:

14

• The data must be continuous for both variables.

• The dependent variable is usually the biological response. • The independent variable is often time or experimental treatment.

6

4

2

0

0

• The relationship between two variables can be represented as a continuum and the data points are plotted accurately and connected directly (point to point).

• Where no error value has been calculated, the scatter can be shown by plotting the individual data points vertically above and below the mean. By convention, bars are not used to indicate the range of raw values in a data set.

10

15

20

25

30

35

40

Temperature (°C)

Growth rate in peas at different temperatures

Mean growth rate (mm day –1)

• Line graphs may be drawn with measure of error. The data are presented as points (which are the calculated means), with bars above and below, indicating a measure of variability or spread in the data (e.g. standard error, standard deviation, or 95% confidence intervals).

5

0.9

Large bars indicate wide scatter of data either side of the mean

0.8

0.7

0.6 0

8

10

12

14

18

16

20

22

24

Temperature (°C)

1. The results (shown right) were collected in a study investigating the effect of temperature on the activity of an enzyme.

(a) Using the results provided (right), plot a line graph on the grid below:

(b) Estimate the rate of reaction at 15°C:

Lab Notebook

An enzyme’s activity at different temperatures

Temperature (°C)

Rate of reaction

(mg of product formed per minute)

10 20 30 35 40 45 50 60

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1.0 2.1 3.2 3.7 4.1 3.7 2.7 0

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Plotting multiple data sets

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If the two data sets use the same measurement units and a similar range of values for the dependent variable, one scale on the y axis is used.

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Transpiration rate

35

Root uptake

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6

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5

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• If the two data sets use different units and/or have a very different range of values for the dependent variable, two scales for the y axis are used (see example right). The scales can be adjusted if necessary to avoid overlapping plots

Transpiration rate (cm3 h–1)

Transpiration and root uptake rates in peas at different relative humidity

A single figure can be used to show two or more data sets, i.e. more than one curve can be plotted per set of axes. This type of presentation is useful when comparing the trends for two or more treatments, or the response of one species against the response of another. Important points regarding this format are:

3

0

20

The two curves must be distinguished with a key.

30

40

50

60

70

80

90

Root uptake of water (cm3 h–1)

22

100

Relative humidity (percentage)

2. The number of perch and trout in a hydro-electric reservoir were monitored over 19 years. A colony of black shag was also present. Shags feed on perch and (to a lesser extent) trout. In 1960-61, 424 shags were removed from the lake during the nesting season and nest counts were made every spring in subsequent years. In 1971, 60 shags were removed from the lake, and all existing nests dismantled. The results of the population survey are tabulated below. (a) Plot a line graph (joining the data points) for the survey results. Use one scale (on the left) for numbers of perch and trout and another scale for the number of shag nests. Use different symbols to distinguish the lines and include a key.

(b) Use a vertical arrow to indicate the point at which shags and their nests were removed. Mean number of fish per haul

1960

16

1961

1962

1.5

1963

Mean number of fish per haul Trout

Perch

Shag nest numbers

1970

1.5

6

1.5

4

1971

0.5

0.7

1.5

11

5

1972

1

0.8

0

0.8

9

10

1973

0.2

4

0

1964

0

5

22

1974

0.5

6.5

0

1965

1

1

25

1975

0.6

7.6

2

1966

1

2.9

35

1976

1

1.2

10

1967

2

5

40

1977

1.2

1.5

32

1968

1.5

4.6

26

1978

0.7

1.2

28

1969

1.5

6

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Perch

Year (continued)

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Trout

Shag nest numbers

Year

Source: Data adapted from 1987 Bursary Examination

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19 Correlation or Causation?

Key Idea: A correlation is a mutual relationship or association between two or more variables. A correlation between two variables does not imply that one causes change in the other. Researchers often want to know if two variables have any correlation (relationship) to each other. This can be achieved by plotting the data as a scatter graph and drawing a line of

best fit through the data, or by testing for correlation using a statistical test. The strength of a correlation is indicated by the correlation coefficient (r), which varies between 1 and -1. A value of 1 indicates a perfect (1:1) relationship between the variables. A value of -1 indicates a 1:1 negative relationship and 0 indicates no relationship between the variables.

Drawing the line of best fit

You may come across the phrase "correlation does not necessarily imply causation". This means that even when there is a strong correlation between variables (they vary together in a predictable way), you cannot assume that change in one variable caused change in the other.

Some simple guidelines need to be followed when drawing a line of best fit on your scatter plot.

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Correlation does not imply causation

Your line should follow the trend of the data points. Roughly half of your data points should be above the line of best fit, and half below.

Example: When data from the organic food association and the office of special education programmes is plotted (below), there is a strong correlation between the increase in organic food and rates of diagnosed autism. However it is unlikely that eating organic food causes autism, so we can not assume a causative effect here.

25 000

300

Autism Organic food sales

20 000

200

15 000 10 000

100

5000 0

1998

2000

2002

2004

2006

2008

0

The line of best fit should pivot around the point which represents the mean of the x and the mean of the y variables.

Number of individuals diagnosed with autism x1000

Organic food sales ($ millions)

Relationship between organic food sales and autism diagnosis rates in the US

The line of best fit does not necessarily pass through any particular point.

Too steep Good fit Too shallow

Year

1. What does the phrase "correlation does not imply causation" mean?

2. A student measured the hand span and foot length measurements of 21 adults and plotted the data as a scatter graph (right).

(a) Draw a line of best fit through the data:

(b) Describe the results:

Hand span vs foot length in adults

350

(c) Using your line of best fit as a guide, comment on the correlation between handspan and foot length:

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Foot length (mm)

300

150

200

250

300

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20 Drawing Scatter Plots

Key Idea: Scatter graphs are used to plot continuous data where there is a relationship between two interdependent variables. Guidelines for scatter graphs

Body length vs brood size in Daphnia

A scatter graph is used to display continuous data where there is a relationship between two interdependent variables. •

There is no independent (manipulated) variable, but the variables are often correlated, i.e. they vary together in some predictable way.

Line of best fit

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The data must be continuous for both variables.

Number of eggs in brood

80

Scatter graphs are useful for determining the relationship (correlation) between two variables. A relationship does not imply that change in one variable causes change in the other variable.

40

Outlier: a data value that lies outside the main spread of data

20

The points on the graph are not connected, but a line of best fit is often drawn through the points to show the relationship between the variables (this may be drawn by eye or computer generated).

0

0

1

2 Body length (mm)

3

4

1. In the example below, metabolic measurements were taken from seven Antarctic fish Pagothenia borchgrevinski. The fish are affected by a gill disease, which increases the thickness of the gas exchange surfaces and affects oxygen uptake. The results of oxygen consumption of fish with varying amounts of affected gill (at rest and swimming) are tabulated below.

(a) Using one scale only for oxygen consumption, plot the data on the grid below to show the relationship between oxygen consumption and the amount of gill affected by disease. Use different symbols or colours for each set of data (at rest and swimming).

Oxygen consumption of fish with affected gills Fish number

Percentage of gill affected

Oxygen consumption (cm3 g-1 h-1) At rest

Swimming

0

0.05

0.29

2

95

0.04

0.11

3

60

0.04

0.14

4

30

0.05

0.22

2. Describe the relationship between the amount of gill affected and oxygen consumption in the fish:

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0.05

0.08

6

65

0.04

0.18

7

45

0.04

0.20

(a) For the at rest data set:

(b) For the swimming data set:

3. How does the gill disease affect oxygen uptake in resting fish?

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(b) Draw a line of best fit through each set of points. NOTE: A line of best fit is drawn so that the points are evenly distributed on either side of the line.

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21 Describing Relationships Between Variables

Key Idea: The relationship between two variables is indicated by the shape and slope of a line drawn between the points. Scatter plots and line graphs show the relationship between two variables (how y is changing relative to x). The relationship may be linear, exponential, or parabolic (U-shaped). Where there is a positive correlation between variables, the line

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between the plotted points will slope upward and to the right. A negative correlation is shown by a line that slopes downwards and to the right. A horizontal line indicates no relationship between variables. Terms indicating the rate of change between variables, such as sharply, gradually, or constantly, help describe the nature of the relationship.

Transpiration rate

Windspeed

(f) Population number vs time

Population number

(e) Enzyme activity vs pH

Enzyme activity

Rate of photosynthesis

Air temperature

Relative humidity

(d) Photosynthetic rate vs light intensity

Light intensity

(c) Body temperature vs air temperature

Mammalian body temperature

(b) Root uptake vs relative humidity

Root water uptake

(a) Transpiration rate vs windspeed

Time

pH

1. For each of the graphs (b-f) above, give a description of the slope and an interpretation of how one variable changes with respect to the other. For the purposes of your description, call the independent variable (horizontal or x-axis) “variable X” and the dependent variable (vertical or y-axis) “variable Y”. (a) Slope: Positive linear relationship, with constantly rising slope Interpretation: Variable Y

(transpiration rate) increases regularly with increase in variable X (windspeed)

(b) Slope:

Interpretation:

(c) Slope:

(d) Slope:

Interpretation:

(e) Slope:

Interpretation: (f) Slope: Interpretation: © 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

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Interpretation:

KNOW


22 Mean, Median, and Mode

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26

Key Idea: Mean, median, and mode are measures of the central tendency of data. The distribution of the data will determine which measurement of central tendency you use. Measures of a biological response are usually made from more than one sampling unit. In lab-based investigations, the sample size (the number of sampling units) may be as small as three or four (e.g. three test-tubes in each of

four treatments). In field studies, each individual may be a sampling unit, and the sample size can be very large (e.g. 100 individuals). It is useful to summarise data using descriptive statistics. Descriptive statistics, such as mean, median, and mode, can identify the central tendency of a data set. Each of these statistics is appropriate to certain types of data or distribution (as indicated by a frequency distribution).

Variation in data

A: Normal distribution

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25 20

Frequency

Whether they are obtained from observation or experiments, most biological data show variability. In a set of data values, it is useful to know the value about which most of the data are grouped, i.e. the centre value. This value can be the mean, median, or mode depending on the type of variable involved (see below). The main purpose of these statistics is to summarise important features of your data and to provide the basis for statistical analyses.

15 10 5

Type of variable sampled

x

0

Mass (g)

Quantitative (continuous or discontinuous)

Ranked

25

Qualitative

B: Skewed distribution

Mode

The shape of the distribution when the data are plotted

Mode

Frequency

20 15 10

Negative skew: the left tail is longer

5 0

Mass (g)

Skewed peak or outliers present

Two peaks (bimodal)

Mean Median

Median

Modes

25

C: Bimodal (two peaks)

20

Frequency

Symmetrical peak

The shape of the distribution will determine which statistic (mean, median, or mode) best describes the central tendency of the sample data.

15 10 5 0

Mass (g)

Statistic

• Add up all the data entries.

• Measure of central tendency for normally distributed data.

• Divide by the total number of data entries.

In some situations, calculation of a simple arithmetic mean is not appropriate.

• The middle value when data entries are placed in rank order.

• Arrange the data in increasing rank order. • Identify the middle value. • For an even number of entries, find the mid point of the two middle values.

• DO NOT calculate a mean from values that are already means (averages) themselves.

• A good measure of central tendency for skewed distributions.

Mode

• The most common data value. • Suitable for bimodal distributions and qualitative data.

Range

• The difference between the smallest and largest data values. • Provides a crude indication of data spread.

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When NOT to calculate a mean:

• The average of all data entries.

22

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• Identify the category with the highest number of data entries using a tally chart or a bar graph. • Identify the smallest and largest values and find the difference between them.

Remember:

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Median

Method of calculation

• DO NOT calculate a mean of ratios (e.g. percentages) for several groups of different sizes. Go back to the raw values and recalculate. • DO NOT calculate a mean when the measurement scale is not linear, e.g. pH units are not measured on a linear scale.

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Mean

Definition and use

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Total of data entries Number of entries

=

5221

=

180

Case study: height of swimmers

cm

Data (below) and descriptive statistics (left) from a survey of the height of 29 members of a male swim squad.

29

Mean

Height of swimmers (in rank order)

Mode 174 175 185 176 185 177 185 178 186 179 186 180 186 181 188 182 188 183 189 184 185 Median 186 187 188 189

177 177 178 178 178 178 180 180 180 181

Tally

Total

I III IIII III IIII

Raw data: Height (cm)

1 3 5 3 4 0 3 1 0 0 0 3 3 0 2 1

178 180 180 178 176

177 188 176 186 175 181 178 178 176 175 185 185 175 189 174 186 176 185 177 176 188 180 186 177

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174 175 175 175 176 176 176 176 176 177

Height (cm)

III I

III III II I

1. Give a reason for the difference between the mean, median, and mode for the swimmers' height data:

Case study: fern reproduction

Raw data (below) and descriptive statistics (right) from a survey of the number of sori found on the fronds of a fern plant.

Total of data entries Number of entries

=

1641 25

=

66

sori

Mean

Raw data: Number of sori per frond

64 69 71 67

60 70 69 64

64 63 59 63

62 70 70 64

68 70 66

66 63 61

63 62 70

Fern spores

2. Give a reason for the difference between the mean, median, and mode for the fern sori data:

Number of sori per frond (in rank order)

Sori per frond

59 66 60 66 61 67 62 68 62 69 63 69 63 70 Median 63 70 63 70 Mode 64 70 64 70 64 71 64

59 60 61 62 63 64 65 66 67 68 69 70 71

Tally

I I I II IIII IIII

II I I II IIII I

Total

1 1 1 2 4 4 0 2 1 1 2 5 1

10.1 8.0 6.7 9.8 6.2

8.2 8.8 7.7 8.8 8.8

7.7 7.8 8.8 8.9 8.4

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Ladybird mass (mg)

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3. Calculate the mean, median, and mode for the data on ladybird masses below. Draw up a tally chart and show all calculations:


23 Spread of Data

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Key Idea: Standard deviation is used to quantify the variability around the central value and evaluate the reliability of estimates of the true mean. While it is important to know the central tendency (e.g. mean) of a data set, it is also important to know how well the mean represents the data set. This is determined by measuring the

spread of data around the central measure. The variance (s2) or its square root, standard deviation (s) are often used to give a simple measure of the spread or dispersion in data. In general, if the spread of values in a data set around the mean is small, the mean will more accurately represent the data than if the spread of data is large.

Standard deviation

25

Two different sets of data can have the same mean and range, yet the distribution of data within the range can be quite different. In both the data sets pictured in the histograms below, 68% of the values lie within the range x̄ ± 1s and 95% of the values lie within x̄ ± 2s. However, in B, the data values are more tightly clustered around the mean.

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Normal distribution

The standard deviation is a frequently used measure of the variability (spread) in a set of data. It is usually presented in the form x̄ ± s. In a normally distributed set of data, 68% of all data values will lie within one standard deviation (s) of the mean (x̄ ) and 95% of all data values will lie within two standard deviations of the mean (left).

Frequency

20 15

68%

10

2.5%

2.5%

5

95%

0

x-2s

x -1s

x

Size class

x +1s

x+2s

Histogram A has a larger standard deviation; the values are spread widely around the mean.

Histogram B has a smaller standard deviation; the values are clustered more tightly around the mean.

Both plots show a normal distribution with a symmetrical spread of values about the mean.

Frequency

Frequency

Calculating s Standard deviation is easily calculated using a spreadsheet.

2.5%

2.5%

68%

68%

2.5%

2.5%

95% x

x +1s

x+2s

NOTE: you may sometimes see the standard deviation equation written as:

Birth weights (kg)

3.740 3.830 3.530 3.095

3.810 2.640

2.980 3.350 3.780 3.260

1.560 3.910 4.180 3.570

3.800 4.170 4.400 3.770

3.150 3.400 3.380 2.660

3.825 3.130 3.400 3.260

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s=

x-2s

∑(x – x̄)2 n–1

x -1s

x

x +1s

x+2s

This equation gives the same answer as the equation above. The denominator n-1 provides a unbiased sample standard deviation for small sample sizes (large samples can use n).

1. Two data sets have the same mean. The first data set has a much larger standard deviation than the second data set. What does this tell you about the spread of data around the mean in each case? Which data set would be most reliable?

3.220 3.135 3.090 3.830 3.840 4.710 4.050 4.560

3.380 3.690 1.495 3.260

2. The data on the left are the birth weights of 40 newborn babies.

(a) Calculate the mean for the data:

(b) Calculate the standard deviation (s) for the data:

(c) State the mean ± 1s:

(d) What percentage of values are within 1s of the mean?

(e) What does this tell you about the spread of the data?

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x -1s

CL

x-2s

95%

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24 Detecting Bias in Samples

Key Idea: Sampling method can affect the results of a study, especially if it has an unknown bias. Bias refers to the selection for or against one particular group in such as way it can influence the findings of an investigation. Bias can occur when sampling is not random, and certain members of a population are under- or over-represented

This exercise illustrates how random sampling, large sample size, and sampling bias affect our statistical assessment of variation in a population. In this exercise, perch were collected and their body lengths (mm) were measured. Data are presented (left) as a frequency histogram and with descriptive statistics (mean, median, mode and standard deviation).

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Figure 1. Frequency histogram for the complete perch data set (N= 689)

relative to others in the population. Small sample sizes can also bias the results, which may then not accurately reflect the population as a whole. Bias can be reduced by random sampling (sampling in which all members of the population have the same chance of being selected). Using appropriate collection methods and apparatus can also reduce bias.

x

x +1s

x+2s

40

Figure 1. Frequency histogram for the complete perch data set (N= 689)

50 30

x -1s x histogram x +1s x+2s Figurex-2s 1. Frequency for the complete perch data set (N= 689) x -1s

x-2s

40 50 20

x

x +1s

x+2s

30 20 0

25

29 33 37 41

20 10

45 49 53 57 61 65 69

Length (mm)

Mean: 48 mm

Median: 47 mm

Length (mm)

25

29 33 37 41

45 49 53 57 61 65 69

2

Lengthhistogram (mm) Figure 2. Frequency for the N=30 perch data set (random sampling)

1 3

Figure 2. Frequency histogram for the N=30 perch data set (random sampling)

0 2 3

25

30

35

1 2

0 1

2. (a) Compare the results for the two small data sets (Figures 2 and 3). How close are the mean and median to each other in each sample set?

40 45 50 Length (mm)

55

60

65

(b) Compare the standard deviation for each sample set:

(c) Describe how each of the smaller sample sets compares to the large sample set (Figure 1):

25 Figure 30 3. 35Frequency 40 45 histogram 50 55for the 60 N=50 65 Length (mm) perch data set (biased sampling)

0 14 25 45 5049.555mm 60 65 Mean:30 49.2335 mm 40 Median: 12 Mode: 38 mm Standard Length (mm) deviation: 11.37 10 Figure 3. Frequency histogram for the N=50 8 perch data set (biased sampling) 6 14 Figure 3. Frequency histogram for the N=50 4 12 perch data set (biased sampling) 2 10 14 0 8 1246 48 50 52 54 56 58 60 62 64 66 68 6 10 Length (mm) 4 8 2 6 0 4 46 48 50 52 54 56 58 60 62 64 66 68 2 Length (mm) 0 46 48 50 52 54 56 58 60 62 64 66 68 Length (mm)

Mean: 61.44 mm Mode: 64 mm

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Median: 63 mm Standard deviation: 3.82

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Frequency Frequency

Frequency

03

Frequency

Figures 2 and 3 show results for two smaller sample sets drawn from the same population. The data collected in Figure 2 were obtained by random sampling but the sample was relatively small (N = 30). The person gathering the data displayed in Figure 3 used a net with a large mesh size to collect the perch.

1. What do the histogram and summary statistics of figure 1 tell you about the distribution of perch from this sample:

40 30 10

10 Figure 2. Frequency histogram for the7.81 N=30 0 Mode: 45 mm Standard deviation: 41 set 49 53 sampling) 57 61 65 69 45(random 33 37data 25 29 perch

Frequency Frequency

Figure 1 shows the results for the complete data set. The sample set was large (N= 689) and the perch were randomly sampled.

(d) Why do you think the two smaller sample sets look so different to each other?

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Frequency

Frequency Frequency

x -1s

x-2s

50

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25 Investigating Plant Survival in the Lab

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Key Idea: When designing an experiment, consider the variables (dependent, independent, and controlled), any assumptions, and what analysis may be required. The figure below provides a basic experimental design to investigate the effect of pH on the growth of a plant species adapted to living in a bog. It is not intended as a

full methodology, but offers some points to consider. The analysis for this experiment would include a calculation of the mean for each treatment group and a plot of the data with a calculated measure of spread (e.g. 95% confidence intervals). Depending on trends indicated by the plotted data, a regression or an ANOVA might also be appropriate.

Observation

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Two students noticed an abundance of a common plant (species A) in a boggy area of land near their school. They tested the soil pH in the area and found it to be quite low (around pH 5). Garden soil was about pH 6.5-7.0.

Research hypothesis

If species A is well adapted to grow in soil pH of ~5, then it will grow more vigorously (as measured by wet weight after 20 days) at pH 5 than at higher or lower pH.

Fluorescent strip lighting

Experiment

The students designed an experiment to test the prediction that the plants would grow best at pH 5. The basic design is described but it is not intended to be a full methodology.

Control of variables

Fixed (controlled) variables: These are controlled and the same across all treatments.

 Lighting regime (quantity and quality)  Age and history of plants  Type and volume of soil  Pot size and type (dimensions, material)  Watering regime (volume per day, frequency)

Watering regime

Watering regime

150 mL per day water at pH 3

150 mL per day water at pH 5

Dependent variable: This is the biological response.

 Plant growth rate (g day-1) calculated from wet weight of entire plants (washed and blotted) after 20 days

Independent variable: This is the factor that is being manipulated in the experiment

 pH of the water provided to the plants

Control: In this experiment, one treatment with the assumed ideal pH for plant growth (pH 7) serves as the control. For other experimental designs, the control is the treatment that lacks the variable of interest. Other variables: Factors that cannot be controlled.

 Genetic variation between plants (uncontrollable but

Notes on preparation and data collection  60 seeds were germinated on damp blotting paper. Of these, 24 in a similar stage of germination (10 mm shoot) were chosen for the experiment.  Each seedling was weighed to the nearest 0.1 g before planting into each of the 24 test pots..  All treatments were arranged on a lab bench in the centre of an internal lab (no windows).  Growth rate per day was estimated from the total wet weight of each plant at the end of 20 days.

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assessed by having six plants per treatment)

 Temperature (all plants received the same room temperature regime but this was not controlled)

Assumptions Features of the experiment assumed to be true but were not (or could not be) tested.  A pH of 7 is a good indicator of the ideal growth pH for most non-acid adapted plants.  All plants are essentially no different to each other in their growth response at different pH levels.  The soil mix, light quality and quantity, and temperature are adequate for healthy continued growth.  Watering volume is adequate. This could be tested with a trial experiment beforehand.

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Watering regime 150 mL per day water at pH 9

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Watering regime (control) 150 mL per day water at pH 7

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The importance of sample size Choosing an appropriate sample size (the number of samples you will take) is very important in the design of any investigation. ď ˝ The sample size must be large enough to provide enough unbiased, reliable data to test your hypothesis and its predictions. ď ˝ The sample size will be limited by the time and resources available to collect and analyse the data (the sampling effort).

1

3 5

Replication in experiments

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Replication refers to the number of times you repeat your entire experimental design (including controls), at the same time. It is not the same as increasing the sample size (n).

2

Replication is an important feature of rigorous scientific studies. Variation between the replicates are analysed and accounted for statistically. Replication accounts for unforeseen effects operating in the set-up. It is important when the response to treatments is likely to vary because of uncontrollable nuisance factors (e.g. in plant field trials).

4

In this experiment of the effects of nutrient level on plant growth (above), the sample size is 5 (n = 5), there are three treatments (three nutrient levels) and four replicates (12 pots in all; not all are visible). The plot is randomised.

1. With reference to the experiment described opposite, explain the importance of each of the following:

(a) Only one plant in each pot:

(b) Obtaining sample plants from germination, not directly from the field:

2. (a) The 12 pots in the experiment opposite were arranged on a bench in the lab. In the space right, show how you might organise the pots to maximise the collection of unbiased data:

(b) What is the sample size in this experiment:

n =

(c) How many treatments are there?

(d) What was the rationale for using treatment of pH 7 as the control?

3. Explain the best way to take account of the natural genetic variability between individuals when designing an experiment:

4. (a) Explain the purpose of replication in experiments:

(b) Describe factors that would limit your ability to replicate your experimental design:

(c) Suggest how you could compensate for the lack of true replication:

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26 Investigating Factors Affecting Plant Growth

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Key Idea: A fair test is when only one variable is changed and all other variables are kept constant. Conclusions based on results are more likely to be valid when the test is fair.

The experiment below describes a fair test for analysing the effect of urea on duckweed growth. Use this activity to test your knowledge about how to analyse and interpret results.

The Aim

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To investigate the effect of urea concentration on the growth of duckweed (Lemna minor).

Hypothesis

If plants need nitrogen to grow, growth rate of Lemna will increase with increasing nitrogen concentration.

Leaves

Image: Kjetil Lenes

Roots

Background

Lemna minor (duckweed) is is a small plant

commonly found floating in the water of drains and pond edges. It is a small (1-3 mm), free-floating plant with 2-4 leaves held flat against the water's surface and a single root. Lemna minor grows very rapidly. In ideal conditions doubling time is as little as 3 days.

Lemna minor (common duckweed)

Experimental method

Duckweed plants

Solutions of urea were made up to concentrations of 3 x 10-2, 3 x 10-3, 3 x 10-4, and 3 x 10-5 mol L-1. For each urea concentration, 80 mL of the appropriate solution was pipetted into three separate beakers. Ten duckweed plants, each with one leaf, were placed in each beaker. The beakers were arranged randomly and placed together in direct sunlight. The number of plants in each beaker was counted and recorded eight times over the next three weeks.

No. of leaves

3 x10-2

3 x10-3

3 x10-4

3 x10-5

Ten plants, each with one leaf, in each beaker

Day 9

No. of leaves

3 x10-3

3 x10-4

3 x10-5

1

10

21

25

21

2

10

24

18

3

12

31

25

Day 8

No. of leaves

3 x10-2

3 x10-3

3 x10-4

3 x10-5

1

10

20

17

19

2

8

18

13

3

10

15

17

Day 12

3 x10-2

Day 18

Day 5

Dilute urea solution

No. of leaves

3 x10-2

3 x10-3

3 x10-4

3 x10-5

1

8

28

24

24

15

2

8

23

17

17

17

3

10

20

22

20

No. of leaves

Day 15

No. of leaves

3 x10-2

3 x10-3

3 x10-4

3 x10-5

1

10

20

37

22

20

2

8

28

19

23

2

7

28

3

10

34

29

31

3

7

No. of leaves 3 x10-2

3 x10-3

3 x10-4

3 x10-5

Day 21

3 x10-2

3 x10-3

3 x10-4

3 x10-5

10

23

28

22

23

21

23

32

28

31

1

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Day 1

100 mL

No. of leaves 3 x10-2

3 x10-3

3 x10-4

3 x10-5

30

34

21

29

31

25

1

7

36

30

34

1

6

38

2

7

27

22

26

2

7

26

3

8

25

29

22

3

6

25

WEB

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Data provided by F. Hicks

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1. Use the raw data on the previous page to complete the table below:

Concentration of urea

Mean number of leaves Day 1

Day 5

Day 8

Day 9

Day 12

Day 15

Day 18

Day 21

3 x 10-2 mol L-1 3 x 10-3 mol L-1 3 x 10-4 mol L-1

PR E O V N IE LY W

3 x 10-5 mol L-1

2. Plot the average number of leaves per day for each concentration on the grid below:

3. (a) What is the sample size for each treatment in this experiment? (b) Describe the results:

(c) Do the results support the hypothesis?

(d) What relevance might the result have to survival of Lemna in different environments?

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27 Test Your Understanding

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Key Idea: Systematic recording and analysis of results can help identify trends and draw conclusions about a biological response in an experiment.

Using the information below, analyse results and draw conclusions about the effect of a nitrogen fertiliser on the growth of radish plants.

The Aim

PR E O V N IE LY W

To investigate the effect of a nitrogen fertiliser on the growth of radish plants.

Hypothesis

If plants need nitrogen to grow, radish growth will increase with increasing nitrogen concentration.

Background

Inorganic fertilisers revolutionised crop farming when they were introduced during the late 19th and early 20th century. Crop yields soared and today it is estimated around 50% of crop yield is attributable to the use of fertiliser. Nitrogen is a very important element for plant growth and several types of purely nitrogen fertiliser are manufactured to supply it, e.g. urea.

Radishes

Experimental method

This experiment was designed to test the effect of nitrogen fertiliser on plant growth. Radish seeds were planted in separate identical pots (5 cm x 5 cm wide x 10 cm deep) and grown together in normal room temperature (22°C) conditions.

To investigate the effect of nitrogen on plant growth, a group of students set up an experiment using different concentrations of nitrogen fertiliser. Radish seeds were planted into a standard soil mixture and divided into six groups, each with five sample plants (30 plants in total).

The radishes were watered every day at 10 am and 3 pm with 1.25 L per treatment. Water soluble fertiliser was mixed and added with the first watering on the 1st, 11th and 21st days. The fertiliser concentrations used were: 0.00, 0.06, 0.12, 0.18, 0.24, and 0.30 g L-1 with each treatment receiving a different concentration. The plants were grown for 30 days before being removed, washed, and the root (radish) weighed. Results were tabulated below:

Table 1: Mass (g) of radish plant roots under six different fertiliser concentrations (data given to 1 dp). Mass of radish root (g)†

Fertiliser concentration (g L-1)

1

2

3

4

5

0

80.1

83.2

82.0

79.1

84.1

0.06

109.2

110.3

108.2

107.9

110.7

0.12

117.9

118.9

118.3

119.1

117.2

0.18

128.3

127.3

127.7

126.8

DNG*

0.24

23.6

140.3

139.6

137.9

141.1

0.30

122.3

121.1

122.6

121.3

123.1

Sample (n)

Total mass

Mean mass

408.5

81.7

1. Identify the independent variable for the experiment and its range:

2. What is the sample size for each concentration of fertiliser? LINK

TEST

22

* DNG: Did not germinate

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† Based on data from M S Jilani, et al Journal Agricultural Research

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3. One of the radishes recorded in Table 1 did not grow as expected and produced an extreme value. Record the outlying value here and decide whether or not you should include it in future calculations:

4. Complete the table on the previous page by calculating the total mass and mean mass of the radish roots:

PR E O V N IE LY W

5. Use the grid below to draw a line graph of the experimental results. Use your calculated means and remember to include a title and correctly labelled axes.

6. The students recorded the wet mass of the root (the root still containing water) in their table. What mass should they have actually recorded to get a better representation of the effect of the fertiliser on root mass?

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7. Why would measuring just root mass not be a totally accurate way of measuring the effect of fertiliser on radish growth?

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8. Describe some other measurements the students could have taken to make their experiment more complete:


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9. Complete Table 2 by calculating the mean, median and mode for each concentration of fertiliser: The students decided to further their experiment by recording the number of leaves on each radish plant:

Table 2: Number of leaves on radish plant under six different fertiliser concentrations.

1

Sample (n) 2

3

4

5

Mean

Median

Mode

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Fertiliser concentration (g L-1)

Number of leaves

0

9

9

10

8

7

0.06

15

16

15

16

16

0.12

16

17

17

17

16

0.18

18

18

19

18

DNG*

0.24

6

19

19

18

18

0.30

18

17

18

19

19

* DNG: Did not germinate

10. (a) Identify the outlier in the table above:

(b) Recalculate the mean if the outlier was included:

(c) Calculate the standard deviation for the fertiliser concentration affected by an outlier: With the outlier included:

Without the outlier included:

(d) Compare the results in (c). What can you conclude about how accurately the mean reflects the data set when the outlier is included?

11. Which concentration of fertiliser appeared to produce the best growth results? 12. Describe some sources of error for the experiment:

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13. Write a conclusion for the experiment with reference to the aim, hypothesis, and results:

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14. The students decided to repeat the experiment (carry it out again). How might this improve the experiment's results?

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28 Structure of a Report

Key Idea: A scientific report is an account of research presented in a standardised way. Scientific reports allow researchers to present their work to peers for review. Scientific reports present the findings from a scientific investigation. Reports follow a standardised structure, clearly outlining how the investigation was carried out and presenting

1. Title (and author) Provides a clear and concise description of the project.

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For further advice on poster presentations, see the excellent NC State University web site linked via Weblinks

results and conclusions. Reports are formal documents, so should be written in past tense, using active voice and clear language, and avoiding jargon. A poster (below) is a concise way to present a scientific investigation. An effective poster is focussed on its message, uses graphics with minimal text, and presents material in a clear, easily followed sequence.

2. Introduction

Includes the aim, hypothesis, and background to the study

4. Results

3. Materials and Methods A description of the materials and procedures used.

An account of results including tables and graphs. This section should not discuss the result, just present them.

5. Discussion

Image courtesy: Adam Luckenbach, NC State University

An discussion of the findings in light of the biological concepts involved. It should include comments on any limitations of the study.

6. Conclusion

A clear statement of whether tor not the findings support the hypothesis. In abbreviated poster presentations, these sections may be combined.

7. References & acknowledgements An organised list of all sources of information. Entries should be consistent within your report. Your teacher will advise you as to the preferred format.

1. Explain the purpose of each of the following sections of a report. The first has one been completed for you:

(a) Introduction: Provides the reader with the background to the topic and the rationale for the study

(b) Methods:

(c) Results:

(d) Discussion:

(e) References and acknowledgements:

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2. What are the key features of a poster presentation?

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29 A Template for Your Investigation

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38

Key Idea: Careful planning and execution are essential to carrying out a valid scientific investigation. Use the following template as a guide to help you carry out your own investigation relating to the survival of an organism.

What do you want to investigate? Talk to your teacher about your project. Do you have the equipment and time to do this topic?

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Aim:

It may be a lab based fair test, a field study, or a behavioural study. For each of the stages of your investigation, create notes to help provide a structure to your final report. They will act as a checklist to ensure you have not forgotten anything.

Background research:

Prepare well so that you have a good knowledge of the topic before you begin. Consult the web, journals, textbooks and people with expertise in your topic.

Research hypothesis:

This is a proposed explanation for your observations that is usually stated along with a testable prediction, e.g. if plants need light to photosynthesise, they will grow more vigorously in the light than in the dark.

Design of the investigation and method:

Outline the basic design of your investigation. If your investigation is an experiment, make sure it a fair test, in which you change only one independent variable to see the effect on the dependent variable (response) you are interested in.

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Write the method out as a procedure that someone else could follow to get the same results. If your investigation is an experiment, think about sample size and controls. If it is field based, consider sample size and size of your sampling unit (e.g. quadrat size). What type of data will you collect?

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39 Record your data in a table, logbook, or spreadsheet as you collect it. Does the data make sense (does it help answer the question you have asked)?

Note any problems with the method, or any changes you made as you worked through the investigation.

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Data collection:

Results (data analysis and presentation):

Decide how to present and summarise your results (e.g. table or graph). Are there any outliers? If so, should you include these in the analysis? Why or why not? Include measures of variability to help you evaluate the reliability of your data. Will you use a statistical test?

Discussion:

Discuss your findings. Use key facts from the background information to explain your results (including unexpected ones). How could your method have been improved?

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References & acknowledgements:

Your conclusions summarise how your results support or contradict your hypothesis.

Include all of your sources of information as a reference list or bibliography. Your teacher will tell you what format to use.

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Conclusion:


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30 KEY TERMS: Did You Get It?

1.

Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

accuracy

A

How close repeated measures are to each other.

biological drawing

B

A variable whose values are determined by another variable.

C

A standard (reference) treatment that helps to ensure that the responses to the other treatments can be reliably interpreted.

D

The variable manipulated by the experimenter in an experiment.

control

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dependent variable

independent variable mean

precision

E

The sum of the data divided by the number of data entries (n).

F

A diagram drawn to accurately show what has been seen by the observer.

G

The correctness of a measurement, i.e. how close it is to the real value.

2. A balance has a calibration error of +0.04 g. A student weighs out 11.71 g of sodium hydroxide. Calculate the percentage error (show your working):

3. The table (below right) shows the rate of sweat production in an athlete on a stationary cycle. Complete the table to:

(a) Convert the cumulative sweat loss to cm3:

(b) Determine the rate of sweat loss (cm3min-1):

Cumulative sweat loss (mm3)

0

0

(c) Plot a double axis graph on the grid below to show cumulative sweat loss in cm3 and rate of sweat loss against time.

10

50,000

20

130,000

(d) Describe how the rate of sweat loss changes over time:

30

220,000

60

560,000

Cumulative sweat loss (cm3)

Rate of sweat loss (cm3 min-1)

TEST

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Time (minutes)

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Cell size, structure and function

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Unit 1 Outcome 1

Key terms

Cells are the unit of life

cell wall

Key knowledge

centrioles

chloroplast cilia

1

Recognise cells as the basic unit of life on Earth and understand the basic principles of the cell theory.

c

2

List the basic biochemical components of cells. Appreciate the role of water in life on Earth and summarise its biologically important properties.

c

3

Describe the characteristics of living organisms and explain why viruses do not fulfil the criteria for being living cells.

c

4

Describe the main differences between eukaryotic and prokaryotic cells.

30 33

c

5

Recognise the cells of fungi, plants, protists, animals, and bacteria by their characteristic features.

30 37

c

6

Use drawings and electron micrographs to compare and contrast the structure of prokaryotic cells and eukaryotic cells.

30 33

electron microscope

endoplasmic reticulum (ER) eukaryotic cell flagella

30

c

cytoplasm

electron micrograph

Activity number

Golgi apparatus

31 32 30

light (=optical) microscope lipid

lysosome

magnification

mitochondrion nucleic acid nucleolus

EII

WMU

nucleus

organelles

Limitations to cell size

plasma membrane

Key knowledge

c

7

Describe the range of cell sizes. Express cell sizes in different units of measurement (mm, Âľm, nm).

c

8

Explain the importance of surface area to volume ratio in limiting cell size and describe the role of cellular organelles in creating cellular compartments with specific functions.

c

9

Describe the requirements of cells in terms of their immediate environment. Explain how unicellular and multicellular organisms meet the challenges for surviving in different environments.

protein

ribosome

rough ER (rER)

smooth ER (sER) stain

vacuole

The ultrastructure of plant and animal cells Key knowledge

c

10

Understand the structure and basic principles of light (optical) microscopes. Contrast light and electron microscopy in terms of magnification and resolution.

c

11

PRAC

c

12

Compare and contrast the ultrastructure of plant cells and animal cells in terms of their organelles. Identify these organelles in drawings and in light and electron micrographs. Include reference to cell wall, cytoplasm, vacuole, nucleus, rough and smooth endoplasmic reticulum, Golgi apparatus, lysosomes, ribosomes, centrioles, plasma membrane, chloroplasts, mitochondria, and cilia and flagella.

34 35 36

37 38

Activity number 38 41

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resolution

Use dissecting and compound light microscopes to locate prepared material and focus images. Calculate the linear magnification of images viewed with a microscope.

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prokaryotic cell

Activity number

38 39 40

42-46


31 The Cell is the Unit of Life

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Key Idea: All living organisms are composed of cells. Cells are broadly classified as prokaryotic or eukaryotic. The cell theory is a fundamental idea of biology. This idea,

The cell theory

that all living things are composed of cells, developed over many years and is strongly linked to the invention and refinement of the microscope in the 1600s.

All cells show the functions of life

The idea that cells are fundamental units of life is part of the cell theory. The basic principles of the theory are: ►► All living things are composed of cells and cell products.

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►► New cells are formed only by the division of pre-existing cells.

Cells use food (e.g. glucose) to maintain a stable internal environment, grow, reproduce, and produce wastes. The sum total of all the chemical reactions that sustain life is called metabolism.

►► The cell contains inherited information (genes) that are used as instructions for growth, functioning, and development.

►► The cell is the functioning unit of life; all chemical reactions of life take place within cells.

Living things

Prokaryotic (bacterial) cells

• Autotrophic or heterotrophic • Single celled • Lack a membrane-bound nucleus and membrane-bound organelles • Cells 0.5-10 µm • DNA a single, circular chromosome. There may be small accessory chromosomes called plasmids. • Cell walls containing peptidoglycan.

Cells

Movement Respiration Sensitivity Growth Reproduction Excretion Nutrition

Viruses are non-cellular

• Non-cellular. • Typical size range: 20-300 nm. • Contain no cytoplasm or organelles. • No chromosome, just RNA or DNA strands. • Enclosed in a protein coat. • Depend on cells for metabolism and reproduction (replication). Influenzavirus

Eukaryotic cells

• Cells 30-150 µm • Membrane-bound nucleus and membrane-bound organelles • Linear chromosomes

Plant cells

Animal cells

Protist cells

Fungal cells

• Exist as part of multicellular organism with specialisation of cells into many types.

• Exist as part of multicellular organism with specialisation of cells into many types.

• Mainly single-celled or exist as cell colonies.

• Rarely exist as discrete cells, except for some unicellular forms (e.g. yeasts)

• Autotrophic (make their own food): photosynthetic cells with chloroplasts.

• Lack cell walls.

• Some are heterotrophic.

1. What are the characteristic features of a prokaryotic cell? 2. What are the characteristic features of a eukaryotic cell?

3. Why are viruses considered to be non-cellular (non-living)?

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Amoeba cell

Yeast cell

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White blood cell

• Rigid cell walls containing chitin. • Heterotrophic.

• Cell walls of cellulose.

Generalised plant cell

• Plant-like, but lack chlorophyll.

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• Heterotrophic (rely on other organisms for food).

• Some are autotrophic and carry out photosynthesis.

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32 The Biochemical Nature of Cells

Key Idea: The main components of a cell are water and compounds of carbon, hydrogen, nitrogen, and oxygen. Water is the main component of cells and organisms, providing an aqueous environment in which metabolic reactions can occur. occur. Apart from water, most other substances in cells are compounds of carbon, hydrogen, oxygen, and nitrogen.

Chloroplasts in plant cells

Plant epidermis

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Centrioles

Carbon can combine with many other elements to form a large number of carbon-based (or organic) molecules. The organic molecules that make up living things can be grouped into four broad classes: carbohydrates, lipids, proteins, and nucleic acids. In addition, a small number of inorganic ions are also components of larger molecules.

Proteins have an enormous number of structural and functional roles in plants and animals, e.g. as enzymes, structural materials (such as collagen), in transport, and movement (e.g. cytoskeleton and centrioles).

Inorganic ions: Dissolved ions participate in metabolic reactions and are components of larger organic molecules, e.g. Mg2+ is a component of the green chlorophyll pigment in the chloroplasts of green plants.

Water is a major component of cells: many substances dissolve in it and metabolic reactions occur in it. In plant cells, fluid pressure against the cell wall provides turgor, which supports the cell.

Animal cell

Chromosome

Nucleotides and nucleic acids Nucleic acids encode information for the construction and functioning of an organism. ATP, a nucleotide derivative, is the energy carrier of the cell.

Plant cell

Plant cell wall

Carbohydrates form the structural components of cells, e.g. cellulose cell walls (arrowed). They are important in energy storage and they are involved in cellular recognition.

Chloroplast membranes

Lipids provide a concentrated source of energy. Phospholipids are a major component of cellular membranes, including the membranes of organelles such as chloroplasts and mitochondria.

(a) Carbohydrates:

(b) Lipids:

(c) Proteins:

(e) Inorganic ions:

(d) Nucleic acids:

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1. Summarise the role of each of the following cell components:

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33 Water and Life

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Key Idea: Water forms bonds between other water molecules and also with ions allowing water to act as a medium for transporting molecules and the biological reactions of life. Water (H2O) is the main component of living things, and typically makes up about 70% of any organism. Water δ-

Small –ve charge

Water molecules stick together Hydrogen bonds form between water molecules so they stick together. This property of cohesion allows water to form drops and is behind surface tension.

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O

is important in cell chemistry as it takes part in, and is a common product of, many reactions. Water can form bonds with other water molecules, and also with other ions (charged molecules). Because of this chemical ability, water is regarded as the universal solvent.

H

H

H

Hydrogen bonds

O

O

H

O

H

δ+

H

H

δ+

Small +ve charges

A water molecule is polar, meaning it has a positively and a negatively charged region. Water molecules have a weak attraction for each other, forming large numbers of weak hydrogen bonds with other water molecules.

Water also forms bonds with other polar molecules or ions, so acts as a solvent for many chemicals.

Water molecules stick to other molecules Water forms hydrogen bonds with other polar molecules. This property of adhesion is behind capillary action (wicking). Water is a universal solvent Other substances dissolve in water because water surrounds other charged molecules and prevents them from clumping together. Water has special thermal properties It takes a lot of energy before water will change temperature, so water heats up and cools down slowly. Water has a high boiling point and is liquid at room temperature. The liquid environment supports metabolism.

It takes a lot of energy to transform water from the liquid to the gas phase. In sweating, the energy is provided by the body, so sweat has a cooling effect.

Euglena

Water provides the medium for metabolic reactions. Water can also act as an acid (donating H+) or a base (receiving H+) in chemical reactions.

Water provides an aquatic environment for organisms. Ice is less dense than water and floats, insulating the water underneath and maintaining the habitat.

Water is colourless, with a high transmission of visible light. Light penetrates aquatic environments, allowing photosynthesis to continue at depth.

The high surface tension of water supports small objects and is essential for transport processes such as blood flow in animals and water transport in plants.

1. Summarise seven important biological roles of water related to its distinctive properties: (a)

(b) (c)

(e) (f) (g)

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(d)

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2. What aspect of water's molecular structure is responsible for its important biological properties:

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34 Prokaryotic vs Eukaryotic Cells

Key Idea: Cells are classified as either prokaryotic or eukaryotic and are distinguished on the basis of their size, internal organisation, and complexity. Cells are divided into two broad groups based on their

Prokaryotic cells

size and organisation. Prokaryotic cells (all bacteria and archaea) are small, single cells with a simple internal structure. Eukaryotic cells are larger, more complex cells. All multicellular and some unicellular organisms are eukaryotic.

Eukaryotic cells Eukaryotic cells have a membrane-bound nucleus, and other membrane-bound organelles.

Prokaryotic cells are often also called bacterial cells. Examples of bacterial cells include E. coli and Staphylococcus aureus.

Plant cells, animal cells, fungal cells, and protists are all eukaryotic cells.

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Prokaryotic cells lack a membrane-bound nucleus or any membrane-bound organelles.

They are small (generally 0.5-10 µm) single cells (unicellular).

They are relatively unstructured and have little cellular organisation (their DNA, ribosomes, and enzymes are free floating within the cell cytoplasm).

Eukaryotic cells are large (30-150 µm). They may exist as single cells or as part of a multicellular organism. Multiple linear chromosomes consisting of DNA and associated proteins. They are more complex than prokaryotic cells, with more structure and internal organisation.

Single, circular chromosome of naked DNA.

Prokaryotes have cell walls, but it is different to the cell walls that some eukaryotes have.

Nuclear membrane absent. Single, naked chromosome is free in cytoplasm within a nucleoid region.

Simple cell structure (limited organisation)

Membrane-bound organelles are absent

Nuclear membrane present

Chromosomes contained in nucleus

Complex cell structure (high degree of organisation)

Nucleoid region (pale)

Presence of membranebound organelles

Peptidoglycan cell wall

The image above shows a bacterium called Escherichia coli

A human white blood cell (above) is an example of an animal cell

1. List three features of a prokaryotic cell: (a)

(b) (c)

2. List three features of a eukaryotic cell: a) (b) (c)

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(d) Name an example of a prokaryote:

(d) Name examples of eukaryotic cells:

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35 Cell Sizes

Key Idea: Cells vary in size (2-100 µm), with prokaryotic cells being approximately 10 times smaller than eukaryotic cells. Cells can only be seen properly when viewed through the magnifying lenses of a microscope. The images below show

a variety of cell types, including a multicellular microscopic animal and a virus (non-cellular) for comparison. For each of these images, note the scale and relate this to the type of microscopy used.

Unit of length (international system) Unit

Metres

Equivalent

1 metre (m)

1m

= 1000 millimetres

1 millimetre (mm)

10

-3

m

= 1000 micrometres

10

-6

m

= 1000 nanometres

10

-9

m

= 1000 picometres

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Human white blood cell

Parenchyma cell of flowering plant

Eukaryotic cells

(e.g. plant and animal cells) Size: 10-100 µm diameter. Cellular organelles may be up to 10 µm.

1 micrometre (µm) 1 nanometre (nm)

Prokaryotic cells

Size: Typically 2-10 µm length, 0.2-2 µm diameter. Upper limit 30 µm long.

Viruses

Micrometres are sometime referred to as microns. Smaller structures are usually measured in nanometres (nm) e.g. molecules (1 nm) and plasma membrane thickness (10 nm).

Size: 0.02-0.25 µm (20-250 nm)

50 µm

1.0 mm

Paramecium is a protozoan commonly found in ponds.

CDC

3 µm

SEM of Giardia, a protozoan that infects the small intestines of many vertebrate groups.

Coronavirus is the virus responsible for SARS. 10 nm

Daphnia is a small crustacean found as part of the zooplankton of lakes and ponds.

n

10 µm

RCN

c

Salmonella is a bacterium found in many environments and causes food poisoning in humans.

Elodea is an aquatic plant. In these leaf cells, the chloroplasts (c) can be seen around the inner edge of the cells.

100 µm

Onion epidermal cells: the nucleus (n) is just visible.

50 µm

1. Using the measurement scales provided on each of the photographs above, determine the longest dimension (length or diameter) of the cell/animal/organelle indicated in µm and mm. Attach your working: (a) Daphnia:

µm

mm

(e) Chloroplast:

(b) Giardia:

µm

mm

(f) Paramecium: µm

mm

µm

mm

(g) Salmonella: µm

mm

µm

mm

(h) Coronavirus: µm

mm

2. Mark and label the examples above on the scale below according to their size:

0.1 nm

1 nm WEB

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10 nm LINK

41

100 nm

mm

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(d) Elodea leaf cell:

1 mm

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(c) Nucleus

µm

10 mm

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36 Limitations to Cell Size

Key Idea: As a cell's surface area to volume ratio declines, the cell becomes less efficient at acquiring the nutrients it needs. In order to function, a cell must obtain the raw materials it needs and dispose of the waste products of metabolism. These exchanges must occur across the plasma membrane.

In a spherical cell, the cell volume increases faster than the corresponding surface area. As the cell becomes larger, it becomes more and more difficult for it to obtain all the materials it needs to sustain its metabolism. This constraint ultimately limits the size of the cell.

Solving the size problem

Cells have a wide range of sizes. Large eukaryotic cells may reach 100 mm in diameter, whereas bacteria typically only reach a tenth of that. Eukaryotic cells can remain efficient at larger sizes in part because they contain organelles, which concentrate associated materials (such as the reactants and enzymes in a metabolic pathway) into specific regions for specific purposes. These cellular compartments enable efficiency of function.

One way of increasing a cell's surface area while retaining the same volume is to elongate the cell. An elongated sphere (an ellipsoid, e.g. a rod shaped cell) has a greater surface area than a sphere of the same volume. In this way a cell can grow larger while still gaining the materials it needs. The cells of multicellular organisms are often highly specialised to maximise SA: V. The three images below are all to scale.

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Cell size and functional efficiency

Oxygen

Carbon dioxide

r

Nitrogenous wastes

Ellipsoid V = 2 cm3 SA= 8.8 cm2

Sphere V = 2 cm3 SA= 7.65 cm2

Disc shaped ellipsoid V = 2 cm3 SA= 14.98 cm2

Food

The plasma membrane, which surrounds every cell, regulates movements of substances into and out of the cell. For each square micrometre of membrane, only so much of a particular substance can cross per second.

Aerobic cellular respiration occurs within the mitochondria. The mitochondrion itself has regions in which different reactions occur.

White blood cell

Skeletal muscle cells

Red blood cell

By flattening the ellipsoid along one axis and stretching it along the other two to form a disc, the surface area can be further increased while keeping the volume the same.

The membrane-bound compartments of the Golgi are responsible for modifying and packaging proteins for secretion.

The nucleus contains the information for regulating the cell's activities. The reactions of photosynthesis are isolated within chloroplasts.

(a) 2 µm:

(c) 10 µm:

(b) 5 µm:

(d) 30 µm:

2. (a) What happens to the SA:V ratio of a spherical cell as its volume increases?

N AS OT SR F OO OR M US E

1. Use the formula 4πr2 (where π = 3.14) to calculate the surface area of a spherical cell with a radius (r) of:

(b) What are the consequences of this to the cell's exchange rates with the environment?

3. Describe two ways in which eukaryotic cells can efficiently obtain the raw materials they need for metabolism, even as they become larger:

(b)

© 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

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(a)

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37 Cellular Environments

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Key Idea: Cell specialisation and organisation has allowed organisms to survive in a wide variety of environments. Organisms are found in a wide variety of physically different environments. Each environment has its own challenges (e.g. ion concentration, temperature, pH, oxygen availability), and to survive, organisms must maintain a relatively stable internal environment so their vital life processes can be

carried out. The evolution of multicellularity (consisting of many cells) provided some independence from the external environment, because specialised cells could be organised into protective tissues, such as skin. Cell specialisation and organisation has enabled multicellular organisms to grow larger and provide the cells within their bodies with the optimum environment for cellular function. Typical ion concentrations in cytosol and blood of a mammal (millimolar)

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The challenge of different environments

Alpine and aerial Low temperatures and low oxygen

Freshwater Environments with low concentrations of solutes (influx of water into cells)

Volcanic vents

Ion

Concentration in cytosol (mM)

Concentration in blood (mM)

Potassium

139

4

Sodium

12

145

Chloride

4

116

Bicarbonate

12

29

Amino acids

138

9

Magnesium

0.8

1.5

Calcium

<0.0002

1.8

Terrestrial High evaporation rates on land (potential risk of dehydration)

Marine Seas, oceans, estuaries

Cave

Subterranean Dark, low temperature environments

Lake

Geothermal Hot, acidic environments in and around hot springs and mud pools

Single celled organisms occur throughout the biosphere. Some species are adapted to extreme conditions, such as high temperature or low pH.

Deep ocean trench High pressure environments at the bottom of trenches. Lack of sunlight

Soil Underground in soil

Collections of specialised cells are organised into tissues, organs, and organ systems. Multicellularity enables the organism to create a relatively stable cellular environment.

1. Describe one challenge to cell survival presented by each of the following environments:

(a) Freshwater:

(b) Ocean trench:

(c) Terrestrial:

N AS OT SR F OO OR M US E

2. What are the biological consequences of multicellularity?

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3. What do the differences between the ion concentrations in the cytosol (within the cell) and the blood (in the table above) tell you about the functioning of a multicellular organism such as a mammal?

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38 Types of Cells

Key Idea: Cells come in a wide range of shapes and sizes. In multicellular organisms, cells are adapted for a specific role. Cells come in a wide range of types and forms. The images

below show a selection of cell types from the five kingdoms. Multicellular organisms typically have many specialised cell types, each of which performs a specific function.

Plant cells

Cell wall made of cellulose

Root hair cell

Nucleus

Cell wall

Cell wall

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Cell wall

Guard cells

Kristian Peters

Chloroplasts

Palisade mesophyll cells

Nucleus

Epidermal cell

Epidermal cell

Guard cells and epidermal cells

Root hair cell

Animal cells

Sperm cell (male)

Neutrophil

Egg cell (female)

Erythrocyte

Osteocyte

Nucleus

Openstax College

Eosinophil

Lymphocyte

Monocyte

Muscle fibres (cells)

Blood cells

Muscle cells

Nuclei

Bone cells (osteocytes)

Reproductive cells

Protists (single cells or colonies) Individual cells

Spine

Nucleus

Nucleus

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Amoeba cell

Paramecium cell

Volvox colony

Fungal cells

Nucleus

Cell wall of chitin

Nucleus

Daughter colony

NOAA

Cilia

Scenedesmus colony

Bacterial cells

Microsporangium

Flagellum

New cell wall

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Yeast cells

Dartmouth college

Hyphae

Staphylococcus cell (dividing)

Fungi cells

Campylobacter cell

1. Identify one distinguishing feature of each of the following cell types, based on what is (or is not) labelled above:

(a) Plant cells:

(c) Fungal cells:

(b) Bacterial cells:

(d) Animal cells:

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2. Both plants and animals have a large number of specialised cell types. Why do you think this is?

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39 Optical Microscopes

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Key Idea: Optical microscopes use light focussed through a series of lenses to magnify objects up to several 100 times. The light (or optical) microscope is an important tool in biology and using it correctly is an essential skill. High power compound light microscopes use visible light and a

combination of lenses to magnify objects up to several 100 times. The resolution of light microscopes is limited by the wavelength of light and specimens must be thin and mostly transparent so that light can pass through. No detail will be seen in specimens that are thick or opaque.

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Typical compound light microscope

RCN

(a)

Word list: In-built light source, arm, coarse focus knob, fine focus knob, condenser, mechanical stage, eyepiece lens, objective lens

Stoma in leaf epidermis

(e) (f)

(b)

(g) (h)

(c)

What is Magnification?

(d)

Magnification refers to the number of times larger an object appears compared to its actual size.

A specimen viewed with a compound light microscope must be thin and mostly transparent so that light can pass through it. No detail will be seen if specimens are thick or opaque. Modern microscopes are binocular, i.e. they have two adjustable eyepieces.

Magnification is calculated as follows: Objective lens power

(i)

Attached light source (not always present)

X

Eyepiece lens power

What is Resolution?

Resolution is the ability to distinguish between close together but separate objects. Examples of high and low resolution for separating two objects viewed under the same magnification are given below.

(j)

(k) (l)

High resolution

AntWeb.org CC 3.0

Low resolution

(m)

Dissecting microscope

Sarah Greenwood CC 4.0

Word list: Focus knob, stage, eyepiece lens, objective lens, eyepiece focus

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A dissecting microscope has two separate lens systems, one for each eye. Such microscopes produce a 3-D view of the specimen and are sometimes called stereo microscopes for this reason.

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Strongylognathus

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Dissecting microscopes are a special type of binocular microscope used for observations at low total magnification (X4 to X50), where a large working distance between the objectives and stage is required.

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Photos: EII

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Dissecting microscopes are used for identifying and sorting organisms, observing microbial cultures, and dissections.

These onion epidermal cells are viewed with standard bright field lighting. Very little detail can be seen (only cell walls) and the cell nuclei are barely visible.

Dark field illumination is excellent for viewing specimens that are almost transparent. The nuclei of these onion epidermal cells are clearly visible.

1. Label the two photographs on the previous page, the compound light microscope (a) to (h) and the dissecting microscope (i) to (m). Use words from the lists supplied for each image. 2. Determine the magnification of a microscope using:

(a) 15 X eyepiece and 40 X objective lens:

(b) 10 X eyepiece and 60 X objective lens:

3. Describe the main difference between a compound light microscope and a dissecting microscope:

4. What type of microscope would you use to:

(a) Count stream invertebrates in a sample:

(b) Observe cells in mitosis:

5. (a) Distinguish between magnification and resolution:

(b) Explain the benefits of a higher resolution:

6. Below is a list of ten key steps taken to set up a microscope and optimally view a sample. The steps have been mixed up. Put them in their correct order by numbering each step: Focus and centre the specimen using the high objective lens. Adjust focus using the fine focus knob only.

Adjust the illumination to an appropriate level by adjusting the iris diaphragm and the condenser. The light should appear on the slide directly below the objective lens, and give an even amount of illumination. Rotate the objective lenses until the shortest lens is in place (pointing down towards the stage). This is the lowest / highest power objective lens (delete one).

Place the slide on the microscope stage. Secure with the sample clips.

Fine tune the illumination so you can view maximum detail on your sample.

Turn on the light source.

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Focus and centre the specimen using the medium objective lens. Focus firstly with the coarse focus knob, then with the fine focus knob (if needed).

Focus and centre the specimen using the low objective lens. Focus firstly with the coarse focus knob, then with the fine focus knob. Focus the eyepieces to adjust your view.

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Adjust the distance between the eyepieces so that they are comfortable for your eyes.


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40 Preparing a Slide

Key Idea: Correctly preparing and mounting a specimen on a slide is important if structures are to be seen clearly under a microscope. A wet mount is suitable for many slides. Specimens are often prepared in some way before viewing in order to highlight features and reveal details. A wet mount is a temporary preparation in which a specimen and a drop of fluid are trapped under a thin coverslip. Wet mounts are

used to view thin tissue sections, live microscopic organisms, and suspensions such as blood. A wet mount improves a sample's appearance and enhances visible detail. Sections must be made very thin for two main reasons. A thick section stops light shining through making it appear dark when viewed. It also ends up with too many layers of cells, making it difficult to make out detail.

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Preparing a specimen Upper epidermis peeled away

Onions provide good tissue samples when preparing a simple wet mount for viewing with a light microscope. A segment is cut from a thick leaf from the bulb. The segment is then bent towards the upper epidermis and snapped so that just the epidermis is left attached. The epidermis can then be peeled off to provide a thin layer of cells for viewing.

KP

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Upper epidermis

Sections through stems or other soft objects need to be made with a razor blade or scalpel, and must be very thin. Cutting at a slight angle to produce a wedge shape creates a thin edge. Ideally specimens should be set in wax first, to prevent crushing and make it easier to cut the specimen accurately.

Applying a simple stain

Mounting a specimen

If a specimen is already mounted, a drop of stain can be placed at one end of the coverslip and drawn through using filter paper (below). Water can be drawn through in the same way to remove excess stain. Stains help to highlight structures. Simple temporary stains include iodine, which stains starchy structures blue-black, and methylene blue, which stains nuclei blue.

Mounted needle

Mounting fluid

Coverslip

Stain or water

Specimen

Coverslip

Filter paper

Specimen

Microscope slide

The thin section of tissue is placed in the centre of a clean glass microscope slide and covered with a drop of mounting fluid (e.g. water, glycerol, or stain). A coverslip is placed on top using a mounted needle to support and lower it over the specimen. This excludes air from the mount.

Viewing

Locate the specimen or region of interest at the lowest magnification. Focus using the lowest magnification first, before switching to the higher magnifications.

1. Why must sections viewed under a microscope be very thin?

3. Why would no chloroplasts be visible in an onion epidermis cell slide?

4. What is the purpose of a stain?

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2. What is the purpose of the coverslip?

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5. Why is it necessary to focus on the lowest magnification first, before switching to higher magnifications?

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41 Calculating Linear Magnification

Key Idea: Magnification is how much larger an object appears compared to its actual size. It can be calculated from the ratio of image height to object height. Microscopes produce an enlarged (magnified) image of an object allowing it to be observed in greater detail than is possible with the naked eye. Magnification refers to the number of times larger an object appears compared to its

53

actual size. Linear magnification is calculated by taking a ratio of the image height to the object's actual height. If this ratio is greater than one, the image is enlarged. If it is less than one, it is reduced. To calculate magnification, all measurements are converted to the same units. Often, you will be asked to calculate an object's actual size, in which case you will be told the size of the object and the magnification.

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Calculating linear magnification: A worked example

1

2

Measure the body length of the bed bug image (right). Your measurement should be 40 mm (not including the body hairs and antennae).

1.0 mm

Measure the length of the scale line marked 1.0 mm. You will find it is 10 mm long. The magnification of the scale line can be calculated using equation 1 (below right).

The magnification of the scale line is 10 (10 mm / 1 mm) *NB: The magnification of the bed bug image will also be 10x because the scale line and image are magnified to the same degree.

3

Calculate the actual (real) size of the bed bug using equation 2 (right):

The actual size of the bed bug is 4 mm (40 mm / 10 x magnification)

Microscopy equations

1. Magnification =

measured size of the object actual size of the object

size of the image 2. Actual object size = magnification

1. The bright field microscopy image on the left is of onion epidermal cells. The measured length of the onion cell in the centre of the photograph is 52,000 µm (52 mm). The image has been magnified 140 x. Calculate the actual size of the cell:

x 140

2. The image of the flea (left) has been captured using light microscopy. (a) Calculate the magnification using the scale line on the image:

(b) The body length of the flea is indicated by a line. Measure along the line and calculate the actual length of the flea:

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0.5 mm

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3. The image size of the E.coli cell (left) is 43 mm, and its actual size is 2 µm. Using this information, calculate the magnification of the image:

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42 Electron Microscopes

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Key Idea: Electron microscopes use the short wavelengths of electrons to produce high resolution images of extremely small objects. Electron microscopes (EMs) use a beam of electrons, instead of light, to produce an image. The higher resolution of EMs is due to the shorter wavelengths of electrons. There

are two basic types of electron microscope: scanning electron microscopes (SEM) and transmission electron microscopes (TEM). In SEMs, the electrons are bounced off the surface of an object to produce detailed images of the external appearance. TEMs produce very clear images of specially prepared thin sections.

Scanning electron microscope (SEM)

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Transmission electron microscope (TEM)

The transmission electron microscope is used to view extremely thin sections of material. Electrons pass through the specimen and are scattered. Magnetic lenses focus the image onto a fluorescent screen or photographic plate. The sections are so thin that they have to be prepared with a special machine, called an ultramicrotome, which can cut wafers to just 30 thousandths of a millimetre thick. It can magnify several hundred thousand times.

Olaboy Olaboy

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The scanning electron microscope scans a sample with a beam of primary electrons, which knocks electrons from the sample's surface. These secondary electrons are picked up by a collector, amplified, and transmitted onto a viewing screen or photographic plate, producing a 3-D image. A microscope of this power easily obtains clear images of very small organisms such as bacteria, and small particles such as viruses. The image produced is of the outside surface only. Electron gun

Electron gun

Primary electron beam

Electron beam

Electromagnetic condenser lens

Electromagnetic lenses

Specimen

Vacuum pump

Electromagnetic objective lens

Vacuum pump

Electron collector

Electromagnetic projector lens

TEM

SEM

Eyepiece

Amplifier

Viewing screen

Fluorescent screen or photographic plate

TEM photo showing the Golgi (G) and a mitochondrion (M).

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Three HIV viruses budding out of a human lymphocyte (TEM).

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G

Secondary electrons

SEM photo of stoma and epidermal cells on the upper surface of a leaf.

Image of hair louse clinging to two hairs on a Hooker’s sealion (SEM).

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M

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Specimen

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Light microscope

Transmission electron microscope (TEM)

Scanning electron microscope (SEM)

light

electrons

electrons

Wavelength:

400-700 nm

0.005 nm

0.005 nm

Lenses:

glass

electromagnetic

electromagnetic

Specimen:

living or non-living supported on glass slide

non-living supported on a small copper grid in a vacuum

non-living supported on a metal disc in a vacuum

Maximum resolution:

200 nm

1 nm

10 nm

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Radiation source:

Maximum magnification:

1500 x

250 000 x

100 000 x

Stains:

coloured dyes

impregnated with heavy metals

coated with carbon or gold

Type of image:

coloured, surface or section

monochrome, section

monochrome, surface only

1. Explain why electron microscopes are able to resolve much greater detail than a light microscope:

2. Which type of microscope [TEM, SEM, compound light microscope, or dissecting microscope] would you use for each of the following scenarios. Explain your choice in each case:

(a) Distinguishing extinct plant species on the basis of pollen surface features:

(b) Resolving the ultrastructure of a chloroplast:

(c) Performing a count of white blood cells from the blood of a person with an infection:

(d) Counting the heart rate and rate of limb beating in a water flea (Daphnia):

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3. Identify which type of electron microscope (SEM or TEM) or optical microscope (compound light microscope or dissecting) was used to produce each of the images in the photos below (A-H):

Plant vascular tissue

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Body lice

G

EII

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Kidney cells

F

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Head louse

Plant epidermal cells

C EII

B EII

A

Mitochondrion

Tongue papilla

H

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Cardiac muscle


43 Animal Cells

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Key Idea: Animal cells are eukaryotic cells. They have many features in common with plant cells, but also have a number of unique features. Animal cells, unlike plant cells, do not have a regular shape. In fact, some animal cells (such as phagocytes) are able to alter their shape for various purposes (e.g. engulfing

foreign material). The diagram below shows the structure and organelles of a liver cell. It contains organelles common to most relatively unspecialised human cells. Note the differences between this cell and the generalised plant cell. The plant cells activity provides further information on the organelles listed here but not described.

Vacuoles: Fluid filled compartments.Smaller than those found in plant cells.

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Lysosome: A sac bounded by a single membrane. They are pinched off from the Golgi apparatus and contain enzymes that break down food and foreign matter. Specialised lysosomes are generally absent from plant cells.

Smooth endoplasmic reticulum (sER): Endoplasmic reticulum (ER) without ribosomes. It is a site for lipid and carbohydrate metabolism. Nucleolus: A dense, solid structure composed of crystalline protein and nucleic acid. They are involved in ribosome synthesis.

Tight junctions: Join cells together in the formation of tissues. Nuclear pore: A hole in the nuclear membrane allowing the nucleus to communicate with the rest of the cell.

Ribosomes: These small structures may be free in the cytoplasm or associated with the ER.

Nucleus

Nuclear membrane: Double layered Cytoplasm

Rough endoplasmic reticulum (rER): ER with attached ribosomes. It is a site of protein synthesis.

Plasma membrane: Lipid bilayer with embedded proteins. It regulates the passage of substances into and out of the cell. Centrioles: Structures associated with nuclear division, about 0.25 µm in diameter. Centrioles are absent in higher plant cells and some protists.

Golgi apparatus: A series of flattened, disc-shaped sacs, stacked one on top of the other and connected with the ER. The Golgi stores, modifies, and packages proteins. It ‘tags’ proteins so that they go to their correct destination.

Mitochondrion (pl. mitochondria): A specialised organelle bounded by a double membrane system. It is the site of cellular respiration.The number in a cell depends on its metabolic activity.

Generalised animal cell

1. The two light micrographs (left) show several types of animal cells. Identify the features indicated by the letters A-C:

EII

A

Neurones (nerve cells) in the spinal cord

C

A:

B:

C: 2. White blood cells are mobile, phagocytic cells, whereas red blood cells are smaller than white blood cells and, in humans, lack a nucleus.

(a) In the photomicrograph (lower, left), circle a white blood cell and a red blood cell:

(b) With respect to the features that you can see, explain how you made your decision.

Blood smear showing different cell types

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3. Name one structure or organelle present in generalised animal cells but absent from plant cells and describe its function:

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44 Identifying Structures in an Animal Cell

Key Idea: The position of the organelles in an electron micrograph can result in variations in their appearance. Transmission electron microscopy (TEM) is the most frequently used technique for viewing cellular organelles.

When viewing TEMs, the cellular organelles may have quite different appearances depending on whether they are in transverse or longitudinal section.

1. Identify and label the structures in the animal cell below using the following list of terms: cytoplasm, plasma membrane, rough endoplasmic reticulum, mitochondrion, nucleus, centriole, Golgi apparatus, lysosome

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(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

2. Which of the organelles in the EM above are obviously shown in both transverse and longitudinal section?

3. Why do plants lack any of the mobile phagocytic cells typical of animal cells?

(a) It has a role in producing and secreting proteins:

(b) It is metabolically very active:

5. What features of the lymphocyte cell above identify it as eukaryotic?

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4. The animal cell pictured above is a lymphocyte. Describe the features that suggest to you that:

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45 Plant Cells

Key Idea: Plant cells are eukaryotic cells. They have features in common with animal cells, but also several unique features. Eukaryotic cells have a similar basic structure, although they may vary tremendously in size, shape, and function. Certain features are common to almost all eukaryotic cells, including their three main regions: a nucleus, surrounded by a watery

cytoplasm, which is itself enclosed by the plasma membrane. Plant cells are enclosed in a cellulose cell wall, which gives them a regular, uniform appearance. The cell wall protects the cell, maintains its shape, and prevents excessive water uptake. It provides rigidity to plant structures but permits the free passage of materials into and out of the cell. Mitochondrion: 1.5 µm X 2–8 µm. They are the cell's energy transformers, converting chemical energy into ATP.

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Generalised plant cell

Starch granule: Carbohydrate stored in amyloplasts (specialised storage organelles).

Plasma membrane: Located inside the cell wall in plants, 3 to 10 nm thick.

Endoplasmic reticulum (ER): A network of tubes and flattened sacs. ER is continuous with the nuclear membrane and may be smooth or have attached ribosomes (rough ER).

Large central vacuole: usually filled with an aqueous solution of ions. Vacuoles are prominent in plants and function in storage, waste disposal, and growth.

Chloroplast

Chloroplast: Specialised organelles, 2 µm x 5 µm, containing the green pigment chlorophyll. Chloroplasts contain dense stacks of membranes within a fluid which is much like cytosol. They are the sites for photosynthesis and are found mainly in leaves.

Nuclear pore: 100 nm diameter Nuclear membrane: a double layered structure. Nucleus: A conspicuous organelle 5 µm diameter. Nucleolus

The vacuole is surrounded by a special membrane called the tonoplast.

Cell wall: A semi-rigid structure outside the plasma membrane, 0.1 µm to several µm thick. It is composed mainly of cellulose. It supports the cell and limits its volume.

Ribosomes: These small (20 nm) structures manufacture proteins. They may be free in the cytoplasm or associated with the surface of the endoplasmic reticulum.

ob er ts

Golgi apparatus

Middle lamella (seen here between adjacent cells left): The first layer of the cell wall formed during cell division. It contains pectin and protein, and provides stability. It allows the cells to form plasmodesmata (P), special channels that allow communication and transport to occur between cells.

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P

Alison Roberts

Cytoplasm: A watery solution containing dissolved substances, enzymes, and the cell organelles and structures.

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1. (a) What are the functions of the cell wall in plants?

2. What distinguishes the tonoplast and the plasma membrane?

3. (a) What structure takes up the majority of space in the plant cell?

(b) What are its roles?

4. Identify two structures in the diagram that are not found in animal cells: WEB

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(b) Why is the middle lamella of the cell wall important?

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46 Identifying Structures in a Plant Cell

Key Idea: The position and appearance of the organelles in an electron micrograph can be used to identify them. 1. Study the diagrams on the other pages in this chapter to familiarise yourself with the structures found in eukaryotic cells. Identify the 11 structures in the cell below using the following word list: cytoplasm, smooth endoplasmic reticulum, mitochondrion, starch granule, chromosome, nucleus, vacuole, plasma membrane, cell wall, chloroplast, nuclear membrane

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(a)

(b)

(c)

(d) (e) (f)

(g) (h) (i)

(j)

(k)

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TEM

2. State how many cells, or parts of cells, are visible in the electron micrograph above: 3. Describe the features that identify this cell as a plant cell:

4. (a) Explain where cytoplasm is found in the cell:

(b) Describe what cytoplasm is made up of:

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(a)

(b)

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5. Describe two structures, pictured in the cell above, that are associated with storage:

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47 Cell Structures and Organelles

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Key Idea: Each type of organelle in a cell has a specific role. Not all cell types contain every type of organelle. The diagram below provides spaces for you to summarise

information about the organelles found in eukaryotic cells. The log scale of measurements (top of next page) illustrates the relative sizes of some cellular structures.

(a) Name: Plasma

membrane the cell Function: Encloses cell contents and regulates movement of substances into and out of cell. Location: Surrounds

Lipid bilayer: a double layer of phospholipids

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Protein

Small subunit

(b) Name:

Large subunit

Location:

Function:

(c) Name:

Location:

Function:

(d) Name: Smooth

& rough endoplasmic reticulum

Location: Penetrates

the whole cytoplasm

Function of smooth ER:

Plant cell

Transport pathway

Ribosomes

Function of rough ER:

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Rough

Smooth

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Flattened membrane sacs

Budding vesicles

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Plasma membrane

Golgi

Ribosome

Chloroplast

Nucleus

Animal cell

Plant cell

Leaf section

Leaf

DNA

0.1 nm

1 nm

10 nm

Mitochondrion

100 nm

1 mm

10 mm

Secretory vesicle budding off the trans face

apparatus

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(e) Name: Golgi

Location: Cytoplasm Function:

associated with smooth ER

Cisternae

Transfer vesicles enter from the smooth ER

Middle lamella

(f) Name:

Location:

Function:

Pectins

Hemicelluloses

Pores

Cellulose fibres

Double membrane

(g) Name:

Location:

Function:

Genetic material

Stacks of membranes

Nucleolus

Outer membrane Inner membrane

(h) Name:

Location: Within

the cytoplasm

Function:

Inner membrane

(i) Name: Matrix

Location: Function:

Folded inner membranes

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Outer membrane

Lamellae

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Stroma (fluid)


48 Identifying Organelles

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Key Idea: Cellular organelles can be identified in electron micrographs by their specific features. Electron microscopes produce a magnified image at high

resolution (distinguish between close together but separate objects). The transmission electron microscope (TEM) images below show the ultrastructure of some organelles.

1. (a) Name the circled organelle: (b) Which kind of cell(s) would this organelle be found in?

(c) Describe the function of this organelle:

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2. (a) Name this organelle (arrowed):

(b) State which kind of cell(s) this organelle would be found in:

(c) Describe the function of this organelle:

3. (a) Name the large, circular organelle:

(b) State which kind of cell(s) this organelle would be found in:

(c) Describe the function of this organelle:

(d) Label two regions that can be seen inside this organelle.

(b) State which kind of cell(s) this organelle is found in:

(c) Describe the function of this organelle:

(d) Name the dark ‘blobs’ attached to the organelle you have labelled:

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4. (a) Name and label the ribbon-like organelle in this photograph (arrowed):

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(b) State which kind of cell(s) this organelle would be found in:

(c) Describe the function of this organelle:

(d) Label three features relating to this organelle in the photograph.

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BF

5. (a) Name this large circular organelle (arrowed):

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49 KEY TERMS: Did You Get It?

1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code. cell wall

A Organelle responsible for producing the cell’s ATP. It appears oval in shape with an outer double membrane and a convoluted interior membrane. Contains its own circular DNA.

chloroplast

B

cytoplasm

With reference to cells, lacking a distinct nucleus and with no membrane-bound organelles DNA is present as a single, circular, naked chromosome.

C Cell types with a distinct membrane-bound nucleus and membrane-bound organelles. D A chemical that binds to parts of the cell and allows those parts to be seen more easily under a microscope.

magnification

E

How many times larger an image is than the original object.

F

The watery contents of the cell within the plasma membrane, but excluding the contents of the nucleus.

PR E O V N IE LY W eukaryotic

mitochondrion nucleus

G A structural and functional part of the cell usually bound within its own membrane. Examples include the mitochondria and chloroplasts.

optical microscope

H Lipid bilayer membrane surrounding the cell. Proteins are embedded in it and are responsible for the passage of material into and out of the cell.

organelle

I

Membrane-bound region within a eukaryotic cell where the chromosomes are found.

J

Microscope that uses lenses to focus visible light waves passing through an object into an image.

K

A structure, present in plants and bacteria, which is found outside the plasma membrane and gives rigidity to the cell.

L

The ability to distinguish between close together but separate objects.

M

An organelle found in photosynthetic organisms such as plants, which contains chlorophyll and in which the reactions of photosynthesis take place.

plasma membrane prokaryotic resolution stain

2. (a) Identify organelle 1:

(b) The organelle in (a) is found in a plant cell / animal cell / both plant and animal cells (circle the correct answer).

(c) Identify organelle 2:

(d) The organelle in (c) is found in a plant / animal cell / plant and animal cell (circle the correct answer).

1

2

3. Using the formulae: cuboid SA = 2(lh + lw + hw), cuboid volume = lwh, calculate the surface area to volume ratio of the following cell shapes:

(a) A cubic cell 6 mm x 6 mm x 6 mm:

(b) A cuboid cell 1 mm x 12 mm x 5 mm:

(c) Which of these cells would exchange substances with its environment most efficiently and why:

4. Match the start of the sentences on the left below to the end of the sentences on the right: ...such as photosynthesis or respiration.

A cell is enclosed by a plasma membrane...

…a cell wall of cellulose.

Animal cells do... Eukaryotic cells contain many different types of organelle... Each organelle carries out a specific function in the cell... Prokaryotic cells... © 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

...do not contain membrane-bound organelles. ...units of life.

....not have a cell wall.

...some of which are composed of membranes. ...made of a phospholipid bilayer

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Plant cells have…

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Cells are the basic...

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Crossing the plasma membrane

PR E O V N IE LY W

Unit 1 Outcome 1

Key terms

The characteristics of the plasma membrane

active transport

Key knowledge

carrier protein

channel protein

concentration gradient diffusion

c

1

Describe the role of the plasma membrane as a partially permeable boundary between the internal and external environments of the cell.

50

c

2

Describe the basic structure of the plasma membrane, including the phospholipid bilayer and the presence and general role of membrane proteins. Recognise that internal membranes, e.g. the membranes of membranous organelles, have the same basic structure.

50

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3

To show your understanding of the properties of the plasma membrane, describe and explain the effect of temperature and solvents on membrane permeability.

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4

PRAC

facilitated diffusion hypertonic hypotonic ion pump

Activity number

isotonic

Investigate factors affecting membrane structure and permeability. How might your findings be relevant to survival in different environments?

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osmolarity osmosis

partially permeable (=semi-permeable) passive transport phospholipid

plasma membrane plasmolysis

Mnolf

surface area: volume ratio

Modes of transport across membranes

turgor

Key knowledge

Activity number

c

5

Describe the movement of molecules across membranes by diffusion and facilitated diffusion, identifying them as passive transport processes. Explain factors affecting diffusion rates across membranes: membrane thickness, surface area, and concentration gradient.

52

c

6

Recall the importance of surface area to volume ratio in limiting cell size. Describe the effect of decreasing SA:V ratio on how efficiently diffusion can deliver materials to the interior of cells.

53

c

7

Investigate factors affecting diffusion rates in model cells.

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8

Explain the movement of water across membranes by osmosis. Understand the terms hypotonic, isotonic, and hypertonic with respect to solutions of differing solute concentration. Explain the effects that solutions of different solute concentration can have on plant and animal cells.

54

c

9

Explain turgor and plasmolysis in plant cells. Understand the role of the cell wall in turgor pressure and the role of turgor pressure in plant support.

54

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10

PRAC

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11

Distinguish between passive transport and active transport, identifying the involvement of membrane proteins and energy in active transport processes.

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c

12

Describe active transport across membrane using ion pumps, e.g. the sodiumpotassium cotransporter (symport).

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Investigate the effects of solutions of different solute concentration on plant or animal cells.

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50 The Plasma Membrane

PR E O V N IE LY W

Key Idea: The plasma membrane is composed of a lipid bilayer with proteins moving freely within it. It is the partially permeable (also called semi-permeable or selectively permeable) boundary between the internal and external cell environments. All cells have a plasma membrane, which forms the outer limit of the cell. A cell wall, if present, lies outside this, and it is quite distinct from it. Cellular membranes are also found inside eukaryotic cells as part of membranous organelles. The currently accepted model of the plasma membrane describes a lipid bilayer with proteins embedded within it, called the fluid-mosaic model (below). The plasma membrane is a partially permeable barrier. It allows the passage of some molecules but not others. Many of the proteins embedded in the membrane are involved in the movement of molecules by transporting specific molecules (often large molecules or ions) across the membrane, often against their concentration gradients.

Molecular model showing how phospholipid molecules naturally orientate to form a bilayer.

Simple membrane structure

External cellular environment

CO2

Cholesterol molecule regulates membrane consistency.

Fatty acid tail is hydrophobic

Phosphate head is hydrophilic

Lipid soluble molecules, e.g. gases and steroids, can move through the membrane by diffusion, down their concentration gradient.

Water molecules pass between the phospholipid molecules by osmosis.

Internal cellular environment

Based on a diagram in Biol. Sci. Review, Nov. 2009, pp. 20-21

Phospholipid

Phospholipids naturally form a bilayer.

Carrier proteins permit the passage of specific molecules by facilitated diffusion or active transport.

Channel proteins form a pore through the hydrophobic interior of the membrane to enable water soluble molecules to pass by facilitated diffusion.

1. List the important components of the plasma membrane:

2. Identify the kind of molecule on the diagram above that:

(a) Can move through the plasma membrane by diffusion:

(b) Forms a channel through the membrane:

3. (a) On the diagram (right) label the hydrophobic and hydrophilic ends of the phospholipid and indicate which end is attracted to water:

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(b) How do the properties of phospholipids contribute to their role in forming the structural framework of membranes?

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51 Factors Altering Membrane Permeability

Key Idea: Temperature and solvents can disrupt the structure of cellular membranes and alter their permeability. Membrane permeability can be disrupted if membranes are subjected to high temperatures or solvents. At temperatures above the optimum, the membrane proteins become

denatured. Alcohols, e.g. ethanol, can also denature proteins. In both instances, the denatured proteins no longer function properly and the membrane loses its selective permeability and becomes leaky. In addition, the combination of alcohol and high temperature can also dissolve lipids.

The aim and hypothesis To investigate the effect of temperature on membrane permeability. The students hypothesised that the amount of pigment leaking from the beetroot cubes would increase with increasing temperature.

PR E O V N IE LY W

Beetroot cubes

Experimental method

Raw beetroot was cut into uniform cubes using a cork borer with a 4 mm internal diameter. The cubes were trimmed to 20 mm lengths and placed in a beaker of distilled water for 30 minutes. Five cm3 of distilled water was added to 15 clean test tubes. Three were placed into a beaker containing ice. These were the 0°C samples. Three test tubes were placed into water baths at 20, 40, 60, or 90°C and equilibrated for 30 minutes. Once the tubes were at temperature, the beetroot cubes were removed from the distilled water and blotted dry on a paper towel. One beetroot cube was added to each of the test tubes. After 30 minutes, they were removed. The colour of the solution in each test tube was observed by eye and then the absorbance of each sample was measured at 530 nm. Results are given in the table below.

Background

Plant cells often contain a large central vacuole surrounded by a membrane called a tonoplast. In beetroot plants, the vacuole contains a water-soluble red pigment called betacyanin, which gives beetroot its colour. If the tonoplast is damaged, the red pigment leaks out into the surrounding environment. The amount of leaked pigment relates to the amount of damage to the tonoplast.

Absorbance of beetroot samples at varying temperatures

Temperature (°C)

Absorbance at 530 nm

Observation

Sample 1

Sample 2

Sample 3

0

No colour

0

0.007

0.004

20

Very pale pink

0.027

0.022

0.018

40

Very pale pink

0.096

0.114

0.114

60

Pink

0.580

0.524

0.509

90

Red

3

3

3

Mean

1. Why is it important to wash the beetroot cubes in distilled water prior to carrying out the experiment?

2. (a) Complete the table above by calculating the mean absorbance for each temperature:

(b) Based on the results in the table above, describe the effect of temperature on membrane permeability:

(c) Explain why this effect occurs:

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Method for determining effect of ethanol concentration on membrane permeability

Ethanol concentration (%)

Sample 1

Sample 2

Sample 3

0

0.014

0.038

0.038

6.25

0.009

0.015

0.023

Absorbance at 477 nm

Mean

PR E O V N IE LY W

Beetroot cubes were prepared the same way as described on the previous page. The following ethanol concentrations were prepared using serial dilution: 0, 6.25, 12.5, 25, 50, and 100%. Eighteen clean test tubes were divided into six groups of three and labelled with one of the six ethanol concentrations. Three cm3 of the appropriate ethanol solution was placed into each test tube. A dried beetroot cube was added to each test tube. The test tubes were covered with parafilm (plastic paraffin film with a paper backing) and left at room temperature. After one hour the beetroot cubes were removed and the absorbance measured at 477 nm. Results are given in the table, right.

Absorbance of beetroot samples at varying ethanol concentrations

12.5

0.010

0.041

0.018

25

0.067

0.064

0.116

50

0.945

1.100

0.731

100

1.269

1.376

0.907

3. What was the purpose of the 0% ethanol solution in the experiment described above?

4. (a) Why do you think the tubes were covered in parafilm?

(b) How could the results have been affected if the test tubes were not covered with parafilm?

5. (a) Complete the table above by calculating the mean absorbance for each ethanol concentration: (b) Plot a line graph of ethanol concentration against mean absorbance on the grid (above):

(c) Describe the effect of ethanol concentration on the membrane permeability of beetroot:

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6. How does ethanol affect membrane permeability?

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52 Diffusion

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Key Idea: Diffusion is the movement of molecules from higher concentration to a lower concentration (i.e. down a concentration gradient). The molecules that make up substances are constantly moving about in a random way. This random motion causes

molecules to disperse from areas of high to low concentration. This dispersal is called diffusion and it requires no energy. Each type of molecule moves down its own concentration gradient. Diffusion is important in allowing exchanges with the environment and in the regulation of cell water content.

What is diffusion?

Factors affecting the rate of diffusion Concentration gradient

The rate of diffusion is higher when there is a greater difference between the concentrations of two regions.

The distance moved

Diffusion over shorter distance occurs at a greater rate than over a larger distance.

The surface area involved

The larger the area across which diffusion occurs, the greater the rate of diffusion.

Barriers to diffusion

Thick barriers have a slower rate of diffusion than thin barriers.

Temperature

Particles at a high temperature diffuse at a greater rate than at a low temperature.

PR E O V N IE LY W

Diffusion is the movement of particles from regions of high concentration to regions of low concentration (down a concentration gradient). Diffusion is a passive process, meaning it needs no input of energy to occur. During diffusion, molecules move randomly about, becoming evenly dispersed.

High concentration

Low concentration

Concentration gradient

If molecules can move freely, they move from high to low concentration (down a concentration gradient) until evenly dispersed.

Glucose

Lipid soluble solutes

Inorganic ion

Carrier protein

Channel protein

Carrier-mediated facilitated diffusion

Channel-mediated facilitated diffusion

Molecules move directly through the membrane without assistance. Example: O2 diffuses into the blood and CO2 diffuses out.

Carrier proteins allow large lipid-insoluble molecules that cannot cross the membrane by simple diffusion to be transported into the cell. Example: the transport of glucose into red blood cells.

Channels (hydrophilic pores) in the membrane allow inorganic ions to pass through the membrane. Example: K+ ions exiting nerve cells to restore resting potential.

2. What do the three types of diffusion described above all have in common?

3. How does facilitated diffusion differ from simple diffusion?

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1. What is diffusion?

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Simple diffusion

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53 Diffusion and Cell Size becomes larger, its surface area compared to its volume is smaller. Diffusion is no longer an effective way to transport materials to the inside. This places a physical limit on the size a cell can grow, with the effectiveness of diffusion being the controlling factor. Larger organisms overcome this constraint by becoming multicellular.

Single-celled organisms

Multicellular organisms

Single-celled organisms (e.g. Amoeba), are small and have a large surface area relative to the cell’s volume. The cell's requirements can be met by the diffusion or active transport of materials into and out of the cell (below).

Multicellular organisms (e.g. plants and animals) are often quite large and large organisms have a small surface area compared to their volume. They require specialised systems to transport the materials they need to and from the cells and tissues in their body.

Oxygen

In a multicellular organism, such as an elephant, the body's need for respiratory gases cannot be met by diffusion through the skin.

PR E O V N IE LY W

Key Idea: Diffusion is less efficient in cells with a small surface area relative to their volume than in cells with a large surface area relative to their volume. When an object (e.g. a cell) is small it has a large surface area in comparison to its volume. Diffusion is an effective way to transport materials (e.g. gases) into the cell. As an object

Carbon dioxide

Wastes

Food

A specialised gas exchange surface (lungs) and circulatory (blood) system are required to transport substances to the body's cells.

The plasma membrane, which surrounds every cell, regulates movements of substances into and out of the cell. For each square micrometre of membrane, only so much of a particular substance can cross per second. The diagram below shows four hypothetical cells of different sizes. They range from a small 2 cm cube to a 5 cm cube. This exercise investigates the effect of cell size on the efficiency of diffusion.

2 cm cube

3 cm cube

4 cm cube

5 cm cube

1. Calculate the volume, surface area and the ratio of surface area to volume for each of the four cubes above (the first has been done for you). When completing the table below, show your calculations.

3 cm cube

4 cm cube

5 cm cube

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Volume

Surface area to volume ratio

2 x 2 x 6 = 24 cm2

2 x 2 x 2 = 8 cm3

(2 cm x 2 cm x 6 sides)

(height x width x depth)

24 to 8 = 3:1

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2 cm cube

Surface area

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Cube size

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2. Create a graph, plotting the surface area against the volume of each cube, on the grid on the right. Draw a line connecting the points and label axes and units. 3. Which increases the fastest with increasing size: the volume or the surface area?

PR E O V N IE LY W

4. Explain what happens to the ratio of surface area to volume with increasing size.

5. The diffusion of molecules into a cell can be modelled by using agar cubes infused with phenolphthalein indicator and soaked in sodium hydroxide (NaOH). Phenolphthalein turns a pink colour when in the presence of a base. As the NaOH diffuses into the agar, the phenolphthalein changes to pink and thus indicates how far the NaOH has diffused into the agar. By cutting an agar block into cubes of various sizes, it is possible to show the effect of cell size on diffusion.

(a) Use the information below to fill in the table on the right:

Cube

1

2

3

Cube 1

2 cm

Cube 2

NaOH solution

1 cm

2. Volume not pink (cm3)

4 cm

Cube 3

Region of no colour change

Agar cubes infused with phenolphthalein

1. Total volume (cm3)

Region of colour change

3. Diffused volume (cm3) (subtract value 2 from value 1) 4. Percentage diffusion

Cubes shown to same scale

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(b) Diffusion of substances into and out of a cell occurs across the plasma membrane. For a cuboid cell, explain how increasing cell size affects the effective ability of diffusion to provide the materials required by the cell:

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6. Explain why a single large cell of 2 cm x 2 cm x 2 cm is less efficient in terms of passively acquiring nutrients than eight cells of 1 cm x 1 cm x 1 cm:

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54 Osmosis

Key Idea: Osmosis is the diffusion of water molecules from a lower solute concentration to a higher solute concentration across a partially permeable membrane. Osmosis is the diffusion of water molecules from regions of lower solute concentration (higher free water concentration) to regions of higher solute concentration (lower free water concentration) across a partially permeable membrane. A

partially permeable membrane allows some molecules, but not others, to pass through. Water molecules will diffuse across a partially permeable membrane until an equilibrium is reached and net movement is zero. The plasma membrane of a cell is an example of a partially permeable membrane. Osmosis is a passive process and does not require any energy input.

Osmotic potential

PR E O V N IE LY W

Demonstrating osmosis

The presence of solutes (dissolved substances) in a solution increases the tendency of water to move into that solution. This tendency is called the osmotic potential or osmotic pressure. The more total dissolved solutes a solution contains, the greater its osmotic potential.

Osmosis can be demonstrated using dialysis tubing in a simple experiment (described below). Dialysis tubing, like all cellular membranes, is a partially permeable membrane. A sucrose solution (high solute concentration) is placed into dialysis tubing, and the tubing is placed into a beaker of water (low solute concentration). The difference in concentration of sucrose (solute) between the two solutions creates an osmotic gradient. Water moves by osmosis into the sucrose solution and the volume of the sucrose solution inside the dialysis tubing increases.

Describing solutions

Osmosis is important when handling body tissues for medical transport or preparation. The tissue must be bathed a solution with an osmolarity (a measure of solute concentration) equal to the tissue's to avoid a loss or gain of fluid in the tissue. Solutions separated by a partially permeable membrane are often described in terms of their solute concentration concentrations relative to one another.

The dialysis tubing acts as a partially permeable membrane, allowing water to pass freely, while keeping the sucrose inside the dialysis tubing. Glass capillary tube

Dialysis tubing containing sucrose solution

Dialysis tubing (partially permeable membrane)

Isotonic solution: Having the same solute concentration relative to another solution (e.g. the cell's contents).

Water molecule

Sucrose molecule

Hypotonic solution: Having a lower solute concentration relative to another solution.

Zephyris

Hypertonic solution: Having a higher solute concentration relative to another solution.

Water

Net water movement

The red blood cells above were placed into a hypertonic solution. As a result, the cells have lost water and have begun to shrink, losing their usual discoid shape.

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1. What is osmosis?

2. (a) In the blue box on the diagram above, draw an arrow to show the direction of net water movement.

(b) Why did water move in this direction?

3. What would happen to the height of the water in the capillary tube if the sucrose concentration was increased?

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55 Estimating Osmolarity of Cells

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Key Idea: A cell placed in a hypotonic solution will gain water while a cell placed in a hypertonic solution will lose water. The osmolarity (a measure of solute concentration) of a cell or tissue can be estimated by placing part of the cell or

tissue into a series of solutions of known concentration and observing if the tissue loses (hypertonic solution) or gains (hypotonic solution) water. The solution in which the tissue remains unchanged indicates the osmolarity of the tissue.

The aim

To investigate the osmolarity of potatoes by placing cubes of potato in varying solutions of sucrose, C12 H22O11 (table sugar).

PR E O V N IE LY W

The method

Fifteen identical 1.5 cm3 cubes of potato where cut and weighed in grams to two decimal places. Five solutions of sucrose were prepared in the following range (in mol L-1): 0.00, 0.25, 0.50, 0.75, 1.00. Three potato cubes were placed in each solution for two hours, stirring every 15 minutes. The cubes were then retrieved, patted dry on blotting paper and weighed again.

Potato cubes

Kent Pryor

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The results

Sucrose concentration (mol L-1)

Potato sample

Initial mass (I) (g)

Final mass (F) (g)

1

5.11

6.00

2

5.15

6.07

3

5.20

5.15

1

6.01

4.98

2

6.07

5.95

3

7.10

7.00

1

6.12

5.10

2

7.03

6.01

3

5.11

5.03

1

5.03

3.96

2

7.10

4.90

3

7.03

5.13

0.00

1. Complete the table (left) by calculating the total mass of the potato cubes, the total change in mass, and the total % change in mass for all the sucrose concentrations: 2. Use the grid below to draw a line graph of the sucrose concentration vs total % change in mass:

Total

Change (C) (F-I) (g)

% Change (C/I x 100)

0.25

Total

Change (C) (F-I) (g)

% Change (C/I x 100)

0.50

Total

Change (C) (F-I) (g)

% Change (C/I x 100)

0.75

% Change (C/I x 100)

1.00 Total Change (C) (F-I) (g) % Change (C/I x 100) LINK

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1

5.00

2

5.04

3.95

3

6.10

5.02

4.03

4. Identify which of the solutions are hypotonic and which are hypertonic.

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Change (C) (F-I) (g)

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3. Use the graph to estimate the osmolarity of the potato (the point where there is no change in mass):

Total

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56 Water Relations in Plant Cells

Key Idea: Plant cells in a hypertonic solution lose water and undergo plasmolysis. In a hypotonic solution, they gain water creating turgor pressure. Osmosis across the partially permeable cell membrane is the main way by which water enters and leaves the cell. When

the external concentration of free water molecules is the same as that of the cytoplasm there is no net movement of water. Changing the tonicity of the external environment will cause a net movement of water into or out of the cell as water moves down its concentration gradient.

Osmosis and tonicity

PR E O V N IE LY W

When the watery contents of a plant cell push against the cell wall they create turgor (tightness) which helps to provide support for the plant body. When cells lose water, there is a loss of cell turgor and the plant will wilt. Complete loss of turgor from a cell is called plasmolysis and is irreversible. Two systems (cell and environment) with the same effective osmotic pressure are termed isotonic and there is no net movement of water molecules. However, when there is an osmotic gradient between the cell and environment there will be a net movement of water molecules down their concentration gradient. The diagram below shows two different situations: when a plant cell is in a hypertonic solution and when it is in a hypotonic solution.

Cell in hypertonic salt solution

Wilted plant (cells have lost turgor)

Plant cells are turgid

Cell in pure water (hypotonic)

Cell wall bulges outward but prevents cell rupture

Cell wall is freely permeable to water.

Water

Cell contents more dilute than the external environment

The cytoplasm has a lower solute concentration than outside. Water leaves the cell.

Water

Cell contents less dilute than the external environment

Cytoplasm

Plasma membrane

Water

Water

Water

Plasmolysis in a plant cell

Tonicity is a measure of the osmotic pressure of a solution. In a hypertonic solution, the external free water concentration is lower than the free water concentration of the cell. Water leaves the cell and, because the cell wall is rigid, the cell membrane shrinks away from the cell wall. This is called plasmolysis and the cell becomes flaccid.

The cytoplasm has a higher solute concentration than outside. Water enters the cell, putting pressure on the cell wall.

Water

Turgor in a plant cell

In a hypotonic solution, the external free water concentration is higher than the cell cytoplasm. Water enters the cell, causing it to swell tight. A wall (turgor) pressure is generated when the cell contents press against the cell wall. Turgor pressure increases until no more water enters the cell (the cell is turgid).

1. Identify the outcome of the following situations:

(a) A plant cell is placed in a hypertonic solution:

(b) A plant cell is placed in a hypotonic solution:

(c) A plant cell in an isotonic solution:

(b) Discuss the role of cell turgor in plants:

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2. (a) Explain the role of cell wall pressure in generating cell turgor in plants:

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57 Active Transport

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Key Idea: Active transport uses energy to transport molecules against their concentration gradient across a partially permeable membrane. Active transport is the movement of molecules (or ions) from

regions of low concentration to regions of high concentration across a cellular membrane by a transport protein. Active transport needs energy to proceed because molecules are being moved against their concentration gradient.

The energy for active transport comes from ATP (adenosine triphosphate). Energy is released when ATP is hydrolysed (water is added) forming ADP (adenosine diphosphate) and inorganic phosphate (Pi).

Active

Active

PR E O V N IE LY W

Transport (carrier) proteins in the membrane are used to actively transport molecules from one side of the membrane to the other (below).

Passive

It requires energy to actively move an object across a physical barrier.

A ball falling is a passive process (it requires no energy input). Replacing the ball requires active energy input.

Active transport can be used to move molecules into and out of a cell.

Active transport can be either primary or secondary. Primary active transport directly uses ATP for the energy to transport molecules. In secondary active transport, energy is stored in a concentration gradient. The transport of one molecule is coupled to the movement of another down its concentration gradient, ATP is not directly involved in the transport process.

Sometimes the energy of a passively moving object can be used to actively move another. For example, a falling ball can be used to catapult another (left).

Active transport

1

ATP binds to a transport protein.

2

A molecule or ion to be transported binds to the transport protein.

3

ATP is hydrolysed and the energy released is used to transport the molecule or ion across the membrane.

4

The molecule or ion is released and the transport protein reverts to its previous state.

Transport protein

High molecule concentration

ATP

ATP

H 2O

ADP

H

P

OH

Molecule to be transported

Low molecule concentration

2. Where does the energy for active transport come from?

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1. What is active transport?

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3. What is the difference between primary active transport and secondary active transport?

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58 Ion Pumps and Cotransport

Key Idea: Ion pumps are transmembrane proteins that use energy to move ions and molecules across a membrane against their concentration gradient. Sometimes molecules or ions are needed in concentrations that diffusion alone cannot supply to the cell, or they cannot diffuse across the plasma membrane. In this case ion

Sodium-potassium pump

Cotransport (the sodium-glucose symport)

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Extracellular fluid or lumen of gut

pumps move ions (and some molecules) across the plasma membrane. The sodium-potassium pump (below, left) is found in almost all animal cells and is also common in plant cells. The concentration gradient created by ion pumps is often coupled to the transport of other molecules such as glucose across the membrane (below right).

Na+

Na+

Na+

Na+

Na+

K+ K+ binding site

Na+

Diffusion of sodium ions

+ + + + + + + + +

Glucose

Na+

Plasma membrane

– – – – – – – – –

Na binding site

Carrier protein

+

ATP

Na

+

Cell cytoplasm

K+

K+

3 Na+ are pumped out of the cell for every 2 K+ pumped in

Na+

Sodium-potassium (Na+/K+) Pump

Cotransport (coupled transport)

The Na+/K+ pump is a protein in the membrane that uses energy in the form of ATP to exchange sodium ions (Na+) for potassium ions (K+) across the membrane. The unequal balance of Na+ and K+ across the membrane creates large concentration gradients that can be used to drive transport of other substances (e.g. cotransport of glucose). The Na+/K+ pump also helps to maintain the right balance of ions and so helps regulate the cell's water balance.

A gradient in sodium ions drives the active transport of glucose into intestinal epithelial cells. The specific transport protein couples the return of Na+ down its concentration gradient to the transport of glucose into the intestinal epithelial cell. Glucose diffuses from the epithelial cells and is transported away in the blood. A low intracellular concentration of Na+ (and therefore the concentration gradient) is maintained by a sodium-potassium pump.

1. Why is ATP required for membrane pump systems to operate?

2. (a) Explain what is meant by cotransport:

(b) How is cotransport used to move glucose into the intestinal epithelial cells?

(c) What happens to the glucose that is transported into the intestinal epithelial cells?

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3. (a) The sodium-potassium pump uses primary/secondary (delete one) active transport. (b) The sodium-glucose symport uses primary/secondary (delete one) active transport.

(c) Describe one consequences of the extracellular accumulation of sodium ions:

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59 KEY TERMS: Did You Get It?

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1. Match each term to its definition, as identified by its preceding letter code.

active transport carrier protein

concentration gradient

A A partially permeable phospholipid bilayer forming the boundary of all cells. B Movement of substances across a biological membrane without energy expenditure. C The passive movement of molecules from high to low concentration. D A type of passive transport facilitated by transport proteins.

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diffusion

facilitated diffusion hypertonic ion pump

E A membrane-bound protein involved in the transport of a specific molecule across the membrane either by active transport or facilitated diffusion. F The energy-requiring movement of substances across a biological membrane against a concentration gradient. G The tendency of water molecules to move into a solution.

osmolarity

H A transmembrane protein that moves ions across a plasma membrane against their concentration gradient.

osmosis

I Passive movement of water molecules across a partially permeable membrane down a concentration gradient.

osmotic potential

J The gradual difference in the concentration of solutes in a solution between two regions. In biology, this usually results from unequal distribution of ions across a membrane.

passive transport

K Having a greater concentration of solutes relative to another solution.

plasma membrane

L A measure of the solute concentration of a solution.

2. The diagrams below depict what happens when a red blood cell is placed into three solutions with differing concentrations of solutes. Describe the tonicity of the solution (in relation to the cell) and describe what is happening: A

B

C

3. Consider the two diagrams below. For each, draw in the appropriate box what you would expect to see after one hour. Particle with diameter of 5 nm

Particle with diameter of 20 nm

After one hour:

Soluble particles placed in at high concentration

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Partially permeable membrane with pores of 10 nm.

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Container of water at 20° C

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Energy transformations

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Unit 1 Outcome 1

Key terms

Obtaining energy for life

alcoholic fermentation

Key knowledge

anaerobic metabolism ATP

autotrophic

Calvin cycle

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1

Distinguish between autotrophs and heterotrophs in terms of their nutrition.

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2

Giving examples, explain the difference between photosynthetic autotrophs and chemosynthetic autotrophs in terms of their source of energy and carbon.

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3

Describe the role of ATP (adenosine triphosphate) as an energy carrier in cells. Compare and contrast cellular respiration and photosynthesis as energy transformation processes. Include reference to the relationship between the raw materials and products of the two processes.

61

carbohydrate

cellular respiration chemotrophic chlorophyll

Photosynthesis

chloroplast

electron transport chain

Activity number

Activity number

Key knowledge

c

4

Describe the ecological role and importance of plants as producers. Summarise photosynthesis in a simple word equation identifying raw materials and end products. Identify the form in which the food is produced and what it is used for.

60 61 62

glycolysis

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5

Describe features of chloroplasts that are related to their role in photosynthesis.

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grana

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6

Describe the features of leaves that facilitate photosynthesis and describe adaptations to maximise photosynthetic rates in different environments.

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ethanol

fermentation

heterotrophic Krebs cycle

lactic acid fermentation

light dependent reactions

light independent reactions mitochondrion

EII

photosynthesis stroma

Cellular respiration

thylakoids

Activity number

Key knowledge

7

Explain why organisms need to respire, recalling the universal role of ATP in metabolism. Recognise that organisms can respire aerobically (using oxygen) and anaerobically (in the absence of oxygen).

64

c

8

Describe aerobic cellular respiration and summarise it in a word equation. Identify the raw materials and end products and relate these to an organism's nutrition and its need to exchange respiratory gases with its environment.

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c

9

Recall the structure of a mitochondrion and its role in cellular respiration. Identify the main steps in the complete oxidation of glucose: glycolysis, Krebs cycle, and electron transport chain.

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10

Describe anaerobic (without oxygen) pathways for ATP generation: lactic acid fermentation in mammalian muscle, alcoholic fermentation in yeast and plant roots, and anaerobic respiration by prokaryotes (e.g. sulfur bacteria). Compare the energy yield of aerobic and anaerobic pathways for ATP generation.

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Teacher's note: Fermentation is an anaerobic pathway but it is not anaerobic respiration, which, by definition, involves an electron transport chain with an electron acceptor other than oxygen, e.g. nitrate, sulfate, or elemental sulfur. Fermentation utilises substrate-level phosphorylation and not an electrochemical gradient to generate ATP.


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60 Autotrophs and Heterotrophs simple inorganic substances using the free energy in sunlight or chemical energy. Heterotrophs feed on other organisms to obtain their energy. They depend either directly on other organisms (dead or alive) or their by-products (e.g. faeces, cell walls, or food stores).

1. Distinguish between photoautotrophs, chemoautotrophs, and chemoheterotrophs in terms of their sources of energy and carbon:

Autotrophic nutrition (autotrophs)

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BH

Key Idea: Heterotrophs (feeders on others) feed on other things to gain energy and carbon. Autotrophs (self feeders) use light or chemical energy to make their own food. The nutritional mode of an organism describes how it obtains its energy and carbon. Autotrophs make their food from

(a) Photoautotroph

Source of energy:

Source of carbon:

Cyanobacterium: Anabaena

Green plant

Source of energy:

Source of carbon:

Photoautotrophs (photosynthetic organisms) use light as their energy source, and carbon dioxide as a source of carbon to make their own food. They include bacteria, cyanobacteria, algae (photosynthetic protists), and green plants.

(b) Chemoautotrophs

UC Berkley

(c) Chemoheterotroph

Source of energy:

Methanococcus

Chemoautotrophs (chemosynthetic organisms) use inorganic compounds (e.g. elemental hydrogen) as a source of energy, and CO2 as a source of carbon. Most are bacteria or archaea that live in hostile environments, such as geothermal and deep sea vents, e.g. Methanococcus (above) uses hydrogen to reduce CO2 to methane.

Source of carbon:

Heterotrophic nutrition (chemoheterotrophs)

Brett Taylor cc 2.0

Barfooz and Josh Grosse cc 3.0

2. Describe the three main nutritional modes of chemoheterotrophs:

NJR ZA cc 3.0

CDC PD

3. What is the main difference between parasites and saprotrophs in the way in which obtain their nutrition?

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Heterotrophs need an organic source of carbon and energy (this is usually glucose). Many bacteria and many protists, e.g. Paramecium (above left) are heterotrophic. All fungi (above right) are chemoheterotrophs and most are decomposers (saprotrophs), obtaining nutrition from the extracellular digestion of dead organic material.

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All animals are heterotrophs, relying on glucose (from plants, dead material, or other animals) for energy and carbon. Parasites, e.g. tapeworms (above left), live on or within their living host organism for part or all of their life. Bacteria, fungi, protists, and animals all have parasitic representatives. The ingestion of solid or liquid organic material from other organisms (holozoic nutrition) is the main feeding mode of animals. Š 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited


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61 Energy Transformations in Cells

Key Idea: The energy from sunlight is captured and stored as glucose, which powers the production of ATP in the process of cellular respiration. Hydrolysis of ATP provides the energy for the chemical reactions in living systems.

Light energy

Oxygen

Note: Heterotrophs depend on organic molecules (food) to provide the glucose for cellular respiration.

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Photosynthesis is a chemical process that captures light energy and transforms it to the chemical energy in carbohydrate (glucose).

Energy flow in the cell of an autotroph is shown below. Note that ATP has a central role in acting as an energy carrier to power metabolic reactions. Some of the energy is lost as heat during these reactions.

Photosynthesis

Glucose

Other uses of glucose

*

Oxygen

Fuel

ADP + Pi

Pi

30.7 kJ per mole

Carbon dioxide + water

Respiration

ATP

The hydrolysis of ATP provides the energy for metabolic reactions. Each mole of ATP hydrolysed releases 30.7 kJ of energy. Some energy is stored in chemical bonds, while some is lost as heat.

Cellular respiration is a chemical process in which the step-wise breakdown of glucose provides the energy to for high energy ATP from ADP and inorganic phosphate (Pi).

Water

A photosynthetic plant cell

Heat energy

It takes energy to break bonds, so how does the hydrolysis of ATP provide energy for metabolic reactions? The hydrolysis of ATP is linked to the formation of a reactive intermediate, which can be used to do work. The reactions that make the energy in ATP available occur virtually simultaneously, so the reaction is simplified to omit the intermediates:

Carbon dioxide

A+B

AB + heat energy ADP + Pi

ATP

1. How does ATP acts as a supplier of energy to power metabolic reactions?

2. (a) Identify the ultimate source of energy for most autotrophs:

(b) Identify a group of autotrophic organisms that do not use this source of energy:

3. Identify the ultimate source of energy for most heterotrophs:

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4. Outline the differences between photosynthesis and cellular respiration, including reference to the raw materials used and the waste products produced:

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62 Plants as Producers

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80

Key Idea: Photosynthesis by producers (photoautotrophs) directly and indirectly sustains heterotrophic life on Earth. Most life on Earth is solar-powered. Plants, algae, and some bacteria are photoautotrophs, using special pigments, called chlorophylls, to absorb light of specific wavelengths and capture the light energy. The light energy is used in a process called photosynthesis. Photosynthesis is of fundamental

importance to living things because it releases free oxygen gas, absorbs carbon dioxide (a waste product of cellular metabolism), and importantly transforms sunlight energy into chemical energy initially stored in simple carbohydrates called sugars. The chemical energy stored in sugars fuels the reactions that sustain life. Heterotrophs rely on producers directly or indirectly for their energy.

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Photosynthesis

Photosynthesis uses sunlight, water and carbon dioxide. It produces oxygen and glucose.

The final product is a carbohydrate called glucose which is used to produce other carbohydrates such as sucrose, starch, and cellulose.

Sunlight: Light is absorbed by cells in the plant's leaves.

Oxygen

Carbon dioxide

The high energy electrons are added to carbon dioxide to make carbohydrate. This is called fixing carbon.

Organelles called chloroplasts inside the plant's cells contain chlorophyll pigments, which absorb sunlight energy and use it to drive the fixation of carbon in photosynthesis.

Chlorophyll molecules are bound to the inner membranes of the chloroplast.

Water

The energy absorbed by the chlorophyll is used to split water into hydrogen and oxygen atoms. This process is called photolysis.

Carbon dioxide molecule

Hydrogen atom

Water molecule

The hydrogen atoms carry electrons that provide the energy for the next step of photosynthesis.

Oxygen atom

Oxygen is a by-product of splitting water.

Light

6CO2 + 12H2O

1. Write the overall word equation for photosynthesis:

Chlorophyll

C6H12O6 + 6O2 + 6H2O

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Photosynthesis equation

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2. Describe three things of fundamental biological importance provided by photosynthesis:

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81 Photosynthesis is not a single process but two complex processes (the light dependent and light independent reactions) each with multiple steps.

Requirements for photosynthesis

Production of carbohydrate (light independent reactions of Calvin cycle) occur in the fluid stroma of chloroplast. This is commonly called carbon 'fixation'.

LI

Energy capture (light dependent reactions) occurs in the inner membranes (thylakoids) of the chloroplast.

LD

RA

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Plants need only a few raw materials to make their own food:  Light energy from the sun  Chlorophyll absorbs light energy  CO2 gas is reduced to carbohydrate  Water is split to provide the electrons for the fixation of carbon as carbohydrate

The photosynthesis of marine algae supplies a substantial portion of the world’s oxygen. The oceans also act as sinks for absorbing large amounts of carbon dioxide.

Macroalgae, like this giant kelp, are important marine producers. Algae living near the ocean surface get access to light used in photosynthesis (the red wavelength).

On land, vascular plants (such as trees with transport vessels) are the main producers of food. Plants at different levels in a forest receive different intensity and quality of light.

3. (a) What form of energy is used to drive photosynthesis?

(b) What is the name of the molecule that captures this energy?

(c) Where in the plant cell does photosynthesis take place?

(d) What form of energy is your answer to (a) converted into?

(e) What happens in each of the two phases of photosynthesis?

4. (a) Primary production is the production of carbon compounds from carbon dioxide, generally by photosynthesis. Study the graph (right) showing primary production in the oceans. Describe what the graph is showing: Ocean primary production Primary production (mgC ⁄ m3 ⁄ day)

0

2

4

6

8

0

(b) Explain the shape of the curves described in (a):

(c) About 90% of all marine life lives in the photic zone (the depth to which light penetrates). Suggest why this is so:

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Depth (m)

50

100

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63 The Leaf and Photosynthesis

Key Idea: Leaves are the main site for photosynthesis in plants. Leaf structure aids the photosynthetic process. The main function of leaves is as photosynthetic organs in which light energy is captured for use in photosynthesis. Leaves are green because they reflect the green wavelengths of light not involved in photosynthesis. The structure of leaves

Dicot leaf structure Cuticle forms a barrier to the diffusion of gases

PR E O V N IE LY W

Respiring plant cells use oxygen (O2) and produce carbon dioxide (CO2). These gases move in and out of the plant and through the air spaces by diffusion. Flowering plants have many air spaces between the cells of the stems, leaves, and roots. These air spaces are continuous and gases are able to move freely through them and into the plant’s cells via the stomata (sing. stoma).

maximises the capture of sunlight energy and facilitates the diffusion of gases used and produced by photosynthesis into and out of the leaf tissue. Gases enter and exit the leaf by way of stomata (small pores) on the leaf. Inside the leaf, large air spaces and the loose arrangement of the spongy mesophyll provides a large surface area for gas exchange.

When the plant is photosynthesising, the situation is more complex. Overall there is a net use of CO2 and a net production of oxygen. CO2 uptake (in photosynthesis) maintains a gradient in CO2 concentration between the inside of the leaf and the atmosphere. Oxygen is produced in excess of respiratory needs and diffuses out of the leaf. These exchanges are indicated by the arrows on the diagram.

Upper epidermis (lacks chloroplasts) Palisade mesophyll cell with chloroplasts Spongy mesophyll cell with chloroplasts Leaf vein (xylem and phloem)

O2

Substomatal air space

CO2

Lower epidermis

CO2

O2

Entry and exit of gases CO2 through the stomata

Guard cell

O2

Stoma (pore)

Net gas exchanges in a photosynthesising dicot leaf

Stoma

Nucleus of epidermal cell

Guard cell

Stoma

The surface of the leaf epidermis of a dicot illustrating the density and scattered arrangement of the pores or stomata. In dicots, stomata are usually present only on the lower leaf surface.

Alternating the placement of leaves is an adaptation to gather the most amount of light without shading leaves on the same plant. Leaves are most often found only on the outer edges of a plant or tree.

Guard cells on each side of a stoma (pl. stomata) regulate the entry and exit of gases and water vapour. Stomata permit gas exchange but are also the major routes for water loss.

1. Describe two adaptive features of leaves to reduce water loss: (a)

(b)

2. Identify the region of a dicot leaf where most of the chloroplasts are found:

(b) How is this movement regulated?

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4. (a) How do gases enter and leave the leaf tissue?

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3. What is the purpose of the air spaces in the leaf tissue?

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The amount of light falling on a plant affects its ability to photosynthesise. The greater the intensity of the light, the greater the rate of photosynthesis, but also the greater the risk of the leaf drying out or burning. Plants that grow in the shade have different leaf structures to help deal with this while optimising photosynthesis.

Sun plant

Shade plant

A sun leaf, when exposed to high light intensities, can absorb much of the light available to the cells. Intense light

A shade leaf can absorb the light available at lower light intensities. If exposed to high light, most would pass through.

Palisade mesophyll layer often 2-3 cells thick

Palisade mesophyll layer only 1 cell thick

Low light intensity

Thick leaves

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Thin leaves

Chloroplasts are mostly restricted to palisade mesophyll cells (few in spongy mesophyll).

Chloroplasts occur throughout the mesophyll (as many in the spongy as in the palisade mesophyll).

Sun leaves

Plants adapted for full sunlight have high levels of respiration. Sun plants include many weed species found on open ground. They expend much more energy on the construction and maintenance of thicker leaves than do shade plants. The benefit of this investment is that they can absorb the higher light intensities available and grow more quickly.

Shade plants

Shade plants typically grow in forested areas, partly shaded by the canopy of larger trees. They have lower rates of respiration than sun plants, mainly because they build thinner leaves. The fewer number of cells need less energy for their production and maintenance.

5. Explain why plants have most of their leaves on the outer edges of the plant (rather than hidden within the branches):

6. (a) Compare the respiration rates of sun and shade plants:

(b) Compare the rate of photosynthesis between sun and shade plants and explain the consequences of this:

(c) Why is it unwise to try to grow a shade plant in a sunny place in the garden?

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7. The diagram on the right shows the effects of the light environment on leaf size:

Sunny

(a) What happens to leaf size as we move from wet and shady to sunny and dry?

(b) Why would plants produce small leaves in sunny, dry areas?

Shady

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Wet

Dry


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64 An Introduction to Cellular Respiration

Key Idea: Cellular respiration is the process in which energy is released from molecules via a series of chemical reactions. Energy is released in cells by the breakdown of sugars and other substances in cellular respiration. During aerobic respiration oxygen is consumed and carbon dioxide is

To carry out aerobic respiration, the body needs oxygen and must remove carbon dioxide. Gas exchange surfaces provide a way for these respiratory gases to enter and leave the body by diffusion. Some organisms use the body surface as the gas exchange surface, but many have specialised gas exchange structures (e.g. lungs, gills, or stomata).

Carbon dioxide gas

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Glucose is an important form of energy for most life forms. It contains a large amount of energy (16 kJ g-1 or 2870 kJ mol-1) which is used drive the production of ATP.

released. These gases need to be exchanged with the environment by diffusion. Diffusion gradients are maintained by transport of gases away from the gas exchange surface (e.g. the cell surface). In anaerobic pathways, ATP is generated but oxygen is not used.

Oxygen gas

Every cell of an organism's body respires. Aerobic cellular respiration creates a constant demand for oxygen and a need to eliminate carbon dioxide gas.

Water (H2O)

Carbon dioxide (CO2)

Glucose

(C6H12O6)

Energy

Oxygen (O2)

Living cells require energy for the activities of life. In eukaryotes, mitochondria are the main site where glucose is broken down to release energy. The mitochondrion provides a membranous compartment in which the reactions of cellular respiration can take place. In the process, oxygen is used to make water and carbon dioxide is released as a waste product.

1. (a) Write the word equation for aerobic cellular respiration:

(b) Write the chemical equation for aerobic cellular respiration:

2. Where does cellular respiration occur in a eukaryotic cell?

3. Distinguish between cellular respiration and gas exchange:

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4. (a) What gases are involved in cellular respiration?

(b) Why must these gases be continuously supplied to or transported away from the cell?

(c) By which transport process do these gases move through the gas exchange surface?

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5. What is the main difference between aerobic and anaerobic pathways for ATP generation?

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Aerobic and anaerobic pathways for ATP production

A sulfate reducing bacterium

B Lactic acid fermentation

C Alcoholic fermentation

D Anaerobic respiration

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A Aerobic respiration

Brewer's yeast

Aerobic respiration produces the energy (as ATP) needed for metabolism. The rate of aerobic respiration is limited by the amount of oxygen available. In animals and plants, most of the time the oxygen supply is sufficient to maintain aerobic metabolism. Aerobic respiration produces a high yield of ATP per molecule of glucose (path A).

During maximum physical activity, when oxygen is limited, anaerobic metabolism provides ATP for working muscle. In mammalian muscle, metabolism of a respiratory intermediate produces lactate, which provides fuel for working muscle and produces a low yield of ATP. This process is called lactic acid fermentation (path B).

In most energy-yielding pathways the initial source of chemical energy is glucose. The first step, glycolysis, is an almost universal pathway. The paths differ in what happens after glucose has been converted to the molecule pyruvate.

The process of brewing utilises the anaerobic metabolism of yeasts. Brewers yeasts preferentially use anaerobic metabolism in the presence of excess sugars. This process, called alcoholic fermentation, produces ethanol and CO2 from the respiratory intermediate pyruvate. It is carried out in vats that prevent entry of O2 (path C)

Many bacteria and archaea are anaerobic, using molecules other than oxygen (e.g. nitrate or sulfate) as a terminal electron acceptor of their electron transport chain. These electron acceptors are not as efficient as oxygen (less energy is released per oxidised molecule) so the energy (ATP) yield from anaerobic respiration is generally quite low (path D).

Lactic acid

B

CO2

Oxygen

Pyruvate

Glycolysis

Glucose

Krebs cycle

Electron transport chain

A

Water

Anaerobic Aerobic

C

2 ATP

2 ATP

34 ATP

CO2

Sulfate

Ethanol

The theoretical maximum yield of 38 ATP per mole of glucose has recently been revised down to 32 ATP (28 from the ETC). All figures will be updated in subsequent revisions.

Electron transport chain

<<34 ATP

D

Sulfides

6. Distinguish between anaerobic pathways in eukaryotes (e.g. yeasts) and anaerobic respiration in anaerobic microbes:

7. When brewing alcohol, why is it important to prevent entry of oxygen to the fermentation vats?

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8. Explain why aerobic respiration is energetically more efficient than fermentation and anaerobic respiration:

9. (a) How many ATP molecules are produced from one glucose molecule during aerobic respiration?

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(b) Extension: If one mole of glucose contains 2870 kJ of energy, and one mole of ATP releases 30.7 kJ of energy during a reaction, what is the percentage of energy in glucose that is available for the body to use?


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65 KEY TERMS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code

ATP

A Obtaining energy from other living organisms or their dead remains. B The organelle found in the cells of all green plants and where photosynthesis takes place. It contains the green pigment chlorophyll and other pigments involved with photosynthesis.

autotrophs

cellular respiration

PR E O V N IE LY W

C The biochemical process that uses light energy to convert carbon dioxide and water into glucose molecules and oxygen

chlorophyll

D The organelle responsible for producing a cell's energy.

chloroplast

E The metabolic process in which the chemical energy in complex organic molecules is coupled to ATP production.

fermentation

F An anaerobic process in which pyruvate is converted to lactic acid or to ethanol and carbon dioxide.

heterotrophic

G A photosynthetic pigment that strongly absorbs red and blue-violet light and appears green in colour.

mitochondrion

H

photosynthesis

I

Organisms that manufacture their own food from simple inorganic substances.

A nucleotide derivative, which acts as the cell’s energy carrier.

2. Complete the schematic diagram of photosynthesis below:

Raw material

(a)

(b)

Process

Main product

(c)

Solar energy

By-product

(e)

(d)

3. (a) Name the process described in the equation (right):

C6H12O6 + 6O2

(b) Where does this process occur?

(c) Name the type of energy molecule that is produced in this process:

6CO2 + 6H2O + Energy

(a) What is the net production of sugars at the compensation point?

(b) What happens to the rate of sugar production as light intensity increases:

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Rate of formation and usage of sugars

4. The graph on the right shows the rate of sugar production against light intensity during photosynthesis.

Production of sugars by photosynthesis

Compensation point

Net gain in sugars

Loss of sugars in respiration

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Light intensity

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Functioning systems in plants

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Unit 1 Outcome 1

Key terms

Transport tissues in plants

apoplast

Key knowledge

cohesion-tension hypothesis

1

Describe the hierarchy of organisation in a vascular plant, with reference to how its cells are specialised into tissues and tissues into organs for the purposes of water uptake, water transport, and regulation of water loss.

c

2

Name the cells contributing to xylem in angiosperms and their roles in the tissue. Describe the features of mature xylem and identify xylem tissue in light micrographs of stem and root sections.

68

c

3

PRAC

Dissect stems longitudinally and transversely to show the position and structure of the xylem vessels.

51

solute stem

symplast

66 67

c

potometer root

Activity number

transpiration

transpiration rate

transpiration stream vascular tissue xylem

PASCO

Transport of water through the plant

Mckdanday

Activity number

Key knowledge

69 70

c

4

Describe the movement of water through the plant from the roots to the air. Include reference to the pathways for water movement (i.e. the non-living apoplast and the living symplast) and their relative importance.

c

5

Explain why water loss in plants is an inevitable consequence of the gas exchanges necessary for photosynthesis. Describe how water is lost from the plant and explain how plants regulate the rate of this water loss.

70

c

6

Explain the movement of water through the plant (the transpiration stream) in terms of osmosis, gradients in solute concentration, and the cohesion-tension hypothesis. Identify the benefits and disadvantages of transpiration to the plant.

70

c

7

Identify the environmental factors affecting the rate of transpiration and describe and explain their effects.

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8

PRAC

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Use a potometer to estimate transpiration rates in different plants or different conditions. Interpret data from investigations of transpiration.


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66 The Hierarchy of Life: Plants

Key Idea: The structural organisation of plants, like all multicellular organisms, is hierarchical. Components at each level of organisation are part of the next level. Plants, like all multicellular organisms, are organised in a hierarchy of structural levels, where each level builds on the one below it. Higher levels of organisation are more complex

than lower levels and often exhibit new (emergent) properties. Hierarchical organisation enables specialisation so that individual components perform a specific function or set of related functions. This specialisation enables the organism to function more efficiently. The diagram below explains this hierarchical organisation for a plant.

The chemical level

cellular level 3 The Cells are the basic structural and

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1

All the chemicals essential for maintaining life, e.g. water, ions, fats, carbohydrates, amino acids, proteins, and nucleic acids.

functional units of an organism. Cells are specialised to carry out specific functions, e.g. mesophyll cell (below).

DNA

2

Atoms and molecules

7

The organelle level

Molecules associate together to form the organelles and structural components of cells, e.g. chloroplast (above).

The organism

The cooperating organ systems make up the organism, e.g. a tomato plant.

6 The organ system level

Groups of organs with a common function form an organ system, e.g. the shoot system in plants (the aboveground parts of the plant, left).

Mesophyll tissue

4

The tissue level

Groups of cells with related functions form tissues, e.g. mesophyll tissue (above). The cells of tissue often have a similar origin.

organ level 5 The An organ is made up of two or more

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types of tissues to carry out a particular function. Organs have a definite form and structure, e.g. leaf (left).

1. Briefly describe how hierarchical organisation and cell specialisation allow an organism to function more efficiently:

2. Assign each of the following new (emergent) properties to the level at which you think they first appear: (a) Metabolism: LINK

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(b) Response:

(c) Self replication:

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67 Specialisation in Plant Cells

Key Idea: The specialised cells in a plant have specific features associated with their particular roles. The cell is the functioning unit structure from which living organisms are made. In multicellular organisms (organisms

made up of more than one cell), cell differentiation produces specialised cells with specific functions. The differentiation of cells gives rise to specialised cell types that fulfil specific roles in the plant, e.g. support, transport, or photosynthesis.

A specialised cell is a cell with the specific features needed to perform a particular function in the organism.

Cells in the leaves of plants are often green because they contain the pigment chlorophyll which is needed for photosynthesis.

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Cell specialisation occurs during development when specific genes (a specific section of DNA that codes for protein) are switched on or off.

Vascular tissue transports water and sugar around the plant. Vascular tissue (the plant veins) is made up of collections of cells specialised to transport different substances.

Multicellular organisms have many types of specialised cells. These work together to carry out the essential functions of life.

The size and shape of a cell allows it to perform its function. The number and type of organelles in a cell is also related to the cell's role in the organism.

Some cells are strengthened to provide support for the plant, allowing it to keep its form and structure.

Specialised cells come together to form tissues with a specific functional role, e.g. water transport. In plants, simple tissues contain only one cell type, whereas complex tissues contain more than one cell type.

Plants have root-hair cells so they can obtain water and nutrients (mineral ions) from the soil.

Leaf surface

Stoma

EII

Guard cell

Specialised guard cells surround the pores (stomata) on plant leaves. The guard cells flanking the pore control the opening and closing of stomata and prevent too much water being lost from the plant.

The semi-rigid cell wall gives many plant cells a regular shape. This Tradescantia epidermis has polyhedral epidermal cells with stomata flanked by guard cells and four more irregular epidermal cells.

Phloem

RCN

Aiofthestorm PD

Xylem

The vascular tissues of plants (xylem and phloem) are complex tissues that transport materials through the plants. The main components of the tissues are supported by packing and strengthening cells.

1. (a) What is a specialised cell?

(b) How does cell specialisation occur?

2. (a) Name the specialised cell that helps to prevent water loss in plants:

(b) How does this cell prevent water from being lost in plants?

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3. How do specialised root hairs help plants to absorb more water and minerals from the soil?

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68 The Plant Transport System which transports sugars. Xylem is highly specialised for its role and its transporting tissues are dead when mature. The transport of water is passive process and does not require energy, so the organelles usually found in cells to support metabolic activity are absent in xylem. Most of the xylem tissue consists of specialised transporting cells, although small numbers of other cell types are present as well.

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Key Idea: The xylem and phloem form the vascular tissue that moves fluids and nutrients about the plant. The transport system of plants moves water and nutrients around the plant in order to meet the plant's needs for metabolic processes such as photosynthesis and growth. Two types of vascular tissue make up the plant transport system: xylem, which transports water and minerals, and phloem,

The vascular tissues or plant "veins" are the phloem and xylem. These tissues are continuous throughout the plant, from the roots to the shoots.

Loss of water vapour (H2O) through the stomata is a consequence of gas exchange. The plant can reduce water loss by closing the stomata but this also stops photosynthesis because the carbon dioxide (CO2) cannot enter the leaf. If the plant cannot replace the water it loses it will wilt and die.

H2O

Leaf cross section

Vascular bundles

Water (and minerals) are transported around the plant in the xylem. Water enters the xylem at the roots by osmosis, which requires no energy.

Stem cross section

Phloem Xylem

Vascular cylinder

Root cross section

Water and minerals are absorbed from the soil by the root system. A large water uptake enables the plant to take up the minerals it needs, as these are often in low concentration in the soil. Some minerals enter the root passively by diffusion and some are taken up by active transport.

1. Name the two vascular tissues in plants:

(b) How does water enter the xylem?

(c) Why is water loss a consequence of gas exchange?

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3. (a) What is the function of xylem?

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2. Briefly describe why plants need a transport system:

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XV

G

xP

S

P

SF

Micropix cc 3.0

XV

G

XV (torn)

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Root hair

A plant root-hair is a tube-like outgrowth of a plant root cell. Their long, thin shape greatly increases their surface area. This allows the plant to absorb water and minerals efficiently.

This image shows xylem and phloem (P) in a sedge stem. Five cell types make up xylem in flowering plants: tracheids, vessels (XV), parenchyma (xP), sclereids and fibres (SF).

Water loss occurs via pores (stomata) on the leaf surface. Specialised cells called guard cells (G) around the stomata (S) open and close the stomata to regulate water loss.

4. (a) What cells conduct the water in xylem?

Water moves through the continuous tubes made by the vessel elements of the xylem.

(b) What other cells are present in xylem tissue and what is their role?

Smaller tracheids are connected by pits in the walls but do not have end wall perforations

SEM

(b) How does water pass between tracheids:

(c) Which cell type do you think provides the most rapid transport of water and why?

(d) Why do you think the tracheids and vessel elements have/need secondary thickening?

LM

As shown in these SEM and light micrographs of xylem, the tracheids and vessel elements form the bulk of the xylem tissue. They are heavily strengthened and are involved in moving water through the plant. The transporting elements are supported by parenchyma (packing and storage cells) and sclerenchyma cells (fibres and sclereids), which provide mechanical support to the xylem. Vessel element

Vessels connect end to end. The end walls of the vessels are perforated to allow rapid water transport.

6. How can xylem vessels and tracheids be dead when mature and functional?

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Tip of tracheid

Diameter up to 500 Âľm Secondary walls of cellulose are laid down after the cell has elongated or enlarged and lignin is deposited to add strength. This thickening is a feature of tracheids and vessels.

Diameter ~80 Âľm

Pits and bordered pits allow transfer of water between cells but there are no end wall perforations. No cytoplasm or nucleus in mature cell.

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Xylem is dead when mature. Note how the cells have lost their cytoplasm.

Tracheids are longer and thinner than vessels.

Vessel elements and tracheids are the two water conducting cell types in the xylem of flowering plants. Tracheids are long, tapering hollow cells. Water passes from one tracheid to another through thin regions in the wall called pits. Vessel elements are much larger cells with secondary thickening in different patterns (e.g. spirals). Vessel end walls are perforated to allow efficient conduction of water.

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5. (a) How does water pass between vessels?

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Vessels


69 Uptake at the Root

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Key Idea: Water uptake by the root is a passive process. Mineral uptake can be passive or active. Plants need to take up water and minerals constantly. They must compensate for the continuous loss of water from the leaves and provide the materials the plant needs to make

food. The uptake of water and minerals is mostly restricted to the younger, most recently formed cells of the roots and the root hairs. Water uptake occurs by osmosis, whereas mineral ions enter the root by diffusion and active transport. Pathways for water movements through the plant are outlined below. 1. (a) What two mechanisms do plants use to absorb nutrients?

Water and mineral uptake by roots Cortex cells of root

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Root hairs have a thin cuticle, so water enters the root easily

(b) Describe the two main pathways by which water moves through a plant:

Epidermal cell

Xylem Xylem

Stele (vascular cylinder). The outer layer of the stele, the pericycle, is next to the endodermis. Root hair

Schematic cross-section through a dicot root

The endodermis is the central, innermost layer of the cortex. It is a single layer of cells with a waterproof band of suberin, called the Casparian strip, which encircles each cell.

2. Plants take up water constantly to compensate for losses due to transpiration. Describe a benefit of a large water uptake:

Water moves by osmosis

Root hairs are extensions of the root epidermal cells and provide a large surface area for absorbing water and nutrients.

3. (a) How does the Casparian strip affect the route water takes into the stele?

Paths for water movement through the plant

Cortex

Endodermis

Pericycle

Xylem

Non-living spaces Cytoplasm

Zone of lower solute concentration [higher free water] May be due to turgid cells, higher wall pressure or lower concentration of dissolved substances

Zone of higher solute concentration [lower free water] May be due to less turgid cells, lower wall pressure or higher concentration of dissolved substances

Some dissolved mineral ions enter the root passively with water. Minerals that are in very low concentration in the soil are taken up by active transport. At the waterproof Casparian strip, water and dissolved minerals must pass into the cytoplasm, so the flow of materials into the stele can be regulated.

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(b) Why might this feature be an advantage in terms of selective mineral uptake?

Casparian strip

The uptake of water through the roots occurs by osmosis, i.e. the diffusion of water from a lower solute concentration to a higher solute concentration. Most water travels through the non-living spaces within the cellulose cell walls, the water-filled spaces of dead cells, and the hollow tubes of xylem. Smaller amounts move through the cytoplasm of cells and the plant vacuoles.

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Epidermis

Plasma membrane

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Plasmodesmata

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70 Transpiration

Key Idea: Water moves through the xylem primarily as a result of evaporation from the leaves and the cohesive and adhesive properties of water molecules. Plants lose water all the time. Approximately 99% of the water a plant absorbs from the soil is lost by evaporation from the leaves and stem. This loss, mostly through stomata, is called transpiration and the flow of water through the plant is called the transpiration stream. Plants rely on a

gradient in solute concentration that increases from the roots to the air to move water through their cells. Water flows passively from soil to air along this gradient of increasing solute concentration. The gradient is the driving force for the movement of water up a plant. Transpiration has benefits to the plant because evaporative water loss cools the plant and the transpiration stream helps the plant to take up minerals. Factors contributing to water movement are described below. Water

The role of stomata

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Air

Evaporative loss of water from the leaves as water vapour

Leaves

Water loss occurs mainly through stomata (pores in the leaf). The rate of water loss can be regulated by specialised guard cells each side of the stoma, which open or close the pore.

Highest solute concentration Lowest water concentration

Stomata open: gas exchange and transpiration rate increase. Stomata closed: gas exchange and transpiration rates decrease.

G

Water flows passively from a low solute concentration (high water concentration) to a high solute concentration (lower water concentration). This gradient is the driving force in the transport of water up a plant.

S

G

EII

Water

The continuous flow of water is called the transpiration stream. It is primarily responsible for water moving up the plant.

Solute particle

Xylem

Soil

Highest water concentration Lowest solute concentration

Water

(b) Describe one benefit of the transpiration stream for a plant:

2. How does the plant regulate the amount of water lost from the leaves?

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1. (a) What is transpiration?

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Processes involved in moving water through the xylem Transpiration pull

Water is lost from the air spaces by evaporation through stomata and is replaced by water from the mesophyll cells. The constant loss of water to the air (and production of sugars) creates a solute concentration in the leaves that is higher than elsewhere in the plant. Water is pulled through the plant along a gradient of increasing solute concentration.

Leaf

Cell wall

Cytoplasm

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Plasmodesma Vacuole

Cohesion-tension

The transpiration pull is assisted by the special cohesive properties of water. Water molecules cling together as they are pulled through the plant. They also adhere to the walls of the xylem (adhesion). This creates one unbroken column of water through the plant. The upward pull on the cohesive sap creates a tension (a negative pressure). This helps water uptake and movement up the plant.

Xylem vessel

Air space

Guard cell

Epidermal cell

Stoma

Root pressure

Water entering the stele from the soil creates a root pressure; a weak 'push' effect for the water's upward movement through the plant. Root pressure can force water droplets from some small plants under certain conditions (guttation), but generally it plays a minor part in the ascent of water.

Evaporative loss of water vapor

Symplast pathway (cytoplasm)

Apoplast pathway (non-living components)

Water molecule

Water is drawn up the plant xylem

3. (a) What would happen if too much water was lost from the leaves?

(b) When might this happen?

4. Describe the three processes that assist the transport of water from the roots of the plant upward: (a)

(b)

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(c)

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5. The maximum height water can move up the xylem by cohesion-tension alone is about 10 m. How then does water move up the height of a 40 m tall tree?

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71 Investigating Plant Transpiration

Key Idea: The relationship between the rate of transpiration and the environment can be investigated using a potometer. This activity describes a typical experiment to investigate the

effect of different environmental conditions on transpiration rate using a potometer. You will present and analyse the results provided.

The potometer

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A potometer is a simple instrument for investigating transpiration rate (water loss per unit time). The equipment is simple to use and easy to obtain. A basic potometer, such as the one shown right, can easily be moved around so that transpiration rate can be measured under different environmental conditions. Some physical conditions investigated are:

• Humidity or vapour pressure (high or low) • Temperature (high or low)

• Air movement (still or windy) • Light level (high or low) • Water supply

It is also possible to compare the transpiration rates of plants with different adaptations e.g. comparing transpiration rates in plants with rolled leaves vs rates in plants with broad leaves. If possible, experiments like these should be conducted simultaneously using replicate equipment. If conducted sequentially, care should be taken to keep the environmental conditions the same for all plants used.

The aim

Fresh, leafy shoot

The progress of an air bubble along the pipette is measured at 3 minute intervals.

1 cm3 pipette

Sealed with petroleum jelly Rubber bung

Flask filled with water

To investigate the effect of environmental conditions on the transpiration rate of plants.

Background

Plants lose water all the time by evaporation from the leaves and stem. This loss, mostly through pores in the leaf surfaces, is called transpiration. Despite the adaptations plants have to help prevent water loss (e.g. waxy leaf cuticle), 99% of the water a plant absorbs from the soil is lost by evaporation. Environmental conditions can affect transpiration rate by increasing or decreasing the gradient for diffusion of water molecules between the plant and its external environment.

Clamp stand

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This experiment investigated the influence of environmental conditions on plant transpiration rate. The experiment examined four conditions: room conditions (ambient), wind, bright light, and high humidity. After setting up the potometer, the apparatus was equilibrated for 10 minutes, and then the position of the air bubble in the pipette was recorded. This is the time 0 reading. The plant was then exposed to one of the environmental conditions. Students recorded the location of the air bubble every three minutes over a 30 minute period. The potometer readings for each environmental condition are presented in Table 1 (next page).

A class was divided into four groups to study how four different environmental conditions (ambient, wind, bright light, and high humidity) affected transpiration rate. A potometer was used to measure transpiration rate (water loss per unit time). A basic potometer, such as the one shown left, can easily be moved around so that transpiration rate can be measured under different environmental conditions.

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The apparatus

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Table 1. Potometer readings (in mL water loss) Time (min)

0

3

6

9

12

15

18

21

24

27

30

Ambient

0

0.002

0.005

0.008

0.012

0.017

0.022

0.028

0.032

0.036

0.042

Wind

0

0.025

0.054

0.088

0.112

0.142

0.175

0.208

0.246

0.283

0.325

High humidity

0

0.002

0.004

0.006

0.008

0.011

0.014

0.018

0.019

0.021

0.024

Bright light

0

0.021

0.042

0.070

0.091

0.112

0.141

0.158

0.183

0.218

0.239

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Treatment

1. (a) Plot the potometer data from Table 1 on the grid provided: (b) Identify the independent variable: 2. (a) Identify the control:

(b) Explain the purpose of including an experimental control in an experiment:

(c) Which factors increased water loss?

(e) Explain why the plant lost less water in humid conditions:

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(d) How does each environmental factor influence water loss?

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72 KEY TERMS: Did You Get It?

1. (a) What is the name given to the loss of water vapour from plant leaves and stems?

(b) What plant tissue is involved in this process?

(c) Is this tissue alive or dead?

(d) Does this process require energy?

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2. Match each term to its definition, as identified by its preceding letter code.

cohesion-tension

A

Device used for investigating the rate of transpiration.

guard cells

B

The loss of water vapour by plants, mainly from leaves via the stomata.

C

Specialised cells each side of the stoma, which open or close the pore.

D

Vascular tissue that conducts water and mineral salts from the roots to the rest of the plant.

E

Pores in the leaf surface through which gases and water vapour can pass.

F

Groups of cells with related functions form these.

G

Partial explanation for the movement of water up the plant in the transpiration stream.

H

Tissue that conducts dissolved sugars in vascular plants. Largely made up of sieve tubes and companion cells.

I

A process by which a cell becomes adapted to carry out a particular role.

J

A structure made up of two or more types of tissues to carry out a particular function.

organ

phloem

potometer

specialisation stomata tissue

transpiration xylem

3. An experiment was performed to investigate transpiration from a hydrangea shoot in a potometer. The experiment was set up and the plant left to stabilise (environmental conditions: still air, light shade, 20°C). The plant was then placed in different environmental conditions and the water loss was measured each hour. Finally, the plant was returned to original conditions, allowed to stabilise and transpiration rate measured again. The data are presented below: Experimental conditions

Temperature (°C)

Humidity (%)

Transpiration rate (g h-1)

(a) Still air, light shade, room temperature

20°C

70

1.20

(b) Moving air, light shade

20°C

70

1.60

(c) Still air, bright sunlight

23°C

70

3.75

19.5°C

100

0.05

(d) Still air and dark, moist chamber

(a) What conditions acted as the control in this experiment?

(b) Which factors increased transpiration rate and why?

(c) Why did the plant have such a low transpiration rate in humid, dark conditions?

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Functioning systems in mammals

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Unit 1 Outcome 1

Key terms

Cells, tissues, and organs

alveoli

Key knowledge

arteries

assimilation

c

1

Describe the hierarchy of organisation in a mammal. Explain how the cells are specialised into tissues and tissues into organs that carry out a specific role, e.g. in gas exchange, internal transport, digestion, or excretion.

c

2

Choosing from the respiratory, digestive, circulatory, or excretory system: Identify components of the system, including specialised cells and their role in tissues (e.g. epithelial tissue, blood) and organs (e.g. lungs, stomach, heart, kidneys).

atrium blood

Activity number

blood vessel breathing

73

74

75 80 87 93

bronchi

capillaries

circulatory system digestion

digestive system disease

egestion

WMU

excretion

excretory system

Mammalian body systems

gas exchange

Key knowledge (select points corresponding to your system of choice)

glomerulus

c

4

PRAC

Investigate factors affecting breathing rate. How are breathing rate and heart rate interconnected and what is the biological significance of this?

78

c

5

Describe malfunctions of the respiratory system, e.g. asthma, emphysema, lung cancer, or bronchitis, and their biological consequences.

79

c

6

The digestive system Describe the structure and function of the digestive system, including reference to how food is moved through the gut, how it is broken down by secretions of specialised regions, and how the products of digestion are absorbed.

c

7

Describe malfunctions of the digestive system, e.g. gastric reflux and infection, and their biological consequences.

c

8

The circulatory system Describe the structure and function of the circulatory system, including how blood is moved around the body and the role of the heart and different blood vessels.

87-90

c

9

PRAC

Dissect a mammalian heart to investigate its structure and function. Account for differences in thickness between the left and right ventricles.

91

c

10

Describe malfunctions of the circulatory system, e.g. coronary artery disease, heart "attack", and diseases of the veins, and their biological consequences.

c

11

The excretory system Describe the structure and function of the excretory system, including how blood is filtered by the kidney and how excretory products are disposed of by the body.

c

12

Describe malfunctions of the excretory system, e.g. kidney failure and kidney stones, and their biological consequences.

c

13

Explain how the body system you chose is interconnected to other body systems and how this related to survival of the organism.

organ

organ system

red blood cells

respiratory system small intestine

specialised cell stomach tissue ultrafiltration urine veins ventricle

80-85

86

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lungs

nephron

75 76 77

The respiratory system Describe the structure and function of the respiratory system, including reference to how gases are exchanged between the external environment and the blood.

kidney

large intestine

Activity number

3

internal transport intestinal villi

EII

c

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heart

istock

93-96

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73 The Hierarchy of Life: Mammals

Key Idea: Structural organisation in multicellular organisms is hierarchical, with new properties arising at each level. Organisation and the emergence of novel properties in complex systems are two of the defining features of living organisms. As we saw for plants, organisms are organised

according to a hierarchy of structural levels. At each level, new properties arise that were absent at the simpler level. Hierarchical organisation allows specialised cells to group together into tissues and organs to perform a specific function. This improves efficiency in the organism.

The diagram below explains this hierarchical organisation for a mammalian example - a human.

The cellular level Cells are the basic structural and functional units of an organism. Cells are specialised to carry out specific functions, e.g. cardiac (heart) muscle cells (below).

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3 2

DNA

1

The organelle level

Molecules associate together to form the organelles and structural components of cells, e.g. the nucleus (above).

Atoms and molecules

The chemical level

All the chemicals essential for maintaining life, e.g. water, ions, fats, carbohydrates, amino acids, proteins, and nucleic acids.

organ system level 6 The Groups of organs with a common function form an organ system, e.g. cardiovascular system (left).

The tissue level

4

Groups of cells with related functions form tissues, e.g. cardiac (heart) muscle (above). The cells of a tissue often have a similar origin.

7 The organism

The cooperating organ systems make up the organism, e.g. a human.

5 The organ level

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An organ is made up of two or more types of tissues to carry out a particular function. Organs have a definite form and structure, e.g. heart (left).

1. Assign each of the following emergent properties to the level at which it first appears: (a) Metabolism:

(b) Behaviour:

(c) Replication:

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74 Specialisation in Animal Cells

Key Idea: There are many different types of animal cells, each with a specific role in the body. Animal cells are often highly modified for their specific role. There are over 200 different types of cells in the human

body. Animal cells lack a cell wall, so they can take on many different shapes. Some, e.g. white blood cells, are even mobile. The shape, size, and even the internal structure of a specialised cell reflects its functional role in the body.

►► Specialised cells often have modifications or exaggerations to a normal cell feature to help them do their job. For example, nerve cells have long, thin extensions to carry nerve impulses over long distances in the body.

Thin, flat epithelial cells line the walls of blood vessels (arrow). Large fat cells store lipid.

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Fat cell

►► Specialisation improves efficiency because each cell type is highly specialised to perform a particular task. They may have more (or fewer) of a particular organelle in order to perform their role most efficiently.

Some nerve cells are over 1 m long.

There are many types of blood cell, each with a specific task.

National Cancer Institute

TEM: Nerve cells

TEM: Cellular projections of intestinal cell

Cross section of muscle fibres

RBC

Louisa Howard, Katherine Connollly Dartmouth College

Some animal cells can move or change shape. Muscle cells, called fibres, are able to contract (shorten) as protein fibres within the cell ratchet past each other. This action causes the movement of limbs, and of organs, such as the heart and intestine.

Cells that line the intestine have extended cell membranes. This increases their surface area so that food (nutrients) can be absorbed quickly and efficiently. Red blood cells (RBCs) have no nucleus so they have more room inside to carry oxygen around the body.

Nerve cells conduct impulses in the form of changes in membrane potential. Impulses are carried from receptors (e.g eye) to effectors (e.g. muscle) allowing responses to the environment.

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1. What is the advantage of cell specialisation in a multicellular organism?

(a) Muscle fibre:

(b) Intestinal cell:

(c) Nerve cell:

(d) Red blood cell: LINK

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2. For each of the following specialised animal cells, name a feature that helps it carry out its function:

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75 The Respiratory System

Key Idea: The respiratory system is made up of specialised cells and tissues, which work together to enable the exchange of gases between the body's cells and the environment. The gas exchange system consists of the passages of the mouth and nose, the trachea, and the tubes and air sacs of

the lungs. Cooperation with the muscles of the diaphragm and ribcage contribute to its function. Each region is specialised to perform a particular role in the organ system's overall function, which is to exchange respiratory gases (O2 and CO2) between the body's cells and the environment.

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Goblet cells in the nasal cavity produce mucus, which traps dust particles. Ciliated epithelial cells sweep the mucus towards the throat where it is swallowed. The trachea is also lined with goblet cells and ciliated epithelium.

Cartilage

Rings of hyaline cartilage provide support for the trachea, bronchi, and the larger bronchioles.

Bronchiole

Trachea

Rib

Alveolar duct

iole

nch

Bro

The lungs contain air spaces surrounded by alveolar epithelial cells, forming alveoli (air sacs), where gas exchange takes place. The alveoli receive air from tubes, called bronchioles.

Alveolus

Bronchioles

Diaphragm

KP

The lungs have a soft, spongy texture made up of the epithelium of the alveoli. Bronchioles form a network of small tubes to transport gases to and from the alveoli. The larger bronchioles are supported by connective tissue (e.g. cartilage).

1. Name three types of cells in the respiratory system and their function: (a)

2. Which cells form the alveoli? 3. What is the purpose of the hyaline cartilage in the respiratory system?

4. Where does gas exchange take place in the lungs? Š 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

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76 The Lungs

Key Idea: Lungs are internal sac-like organs connected to the outside by a system of airways. The smallest airways end in thin-walled alveoli, where gas exchange occurs. The respiratory system includes all the structures associated

with exchanging respiratory gases with the environment. In mammals, the gas exchange organs are paired lungs connected to the outside air by way of a system of tubular passageways: the trachea, bronchi, and bronchioles.

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The trachea transfers air to the lungs. It is strengthened with C-shaped bands of cartilage.

The trachea divides into two bronchi. These are also supported by cartilage bands.

The right lung is slightly larger than the left. It takes up 55-60% of the total lung volume.

Bronchioles branch from the bronchi and divide into progressively smaller branches. The cartilage is gradually lost as the bronchioles decrease in diameter.

Right lung

Left lung

The "cardiac notch" in the left lung makes space for the heart.

O2

Oxygen enters the blood from air in the alveoli. Carbon dioxide leaves the blood and is breathed out. CO2

The diaphragm is a dome shaped muscle that is the body's main breathing muscle. When it contracts, it moves down, and air is drawn into the lungs.

The walls of the smallest bronchioles lack cartilage but have a large amount of smooth muscle.

The smallest respiratory bronchioles subdivide into the alveolar ducts from which arise the alveoli.

The alveoli are the site of gas exchange. They provide a large surface area (70 m2) for the exchange of respiratory gases by diffusion between the air in the lungs and the blood in the capillaries. The alveoli deflate after each breath out. A phospholipid surfactant helps to prevent collapse of the alveoli by decreasing surface tension.

Alveolar cross section at top of next page

1. What is the purpose of the trachea, bronchi, and bronchioles?

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2. What is the purpose of the diaphragm?

3. (a) Explain how the basic structure of the human gas exchange system provides such a large area for gas exchange:

(b) What is the significance of the close arrangement of alveoli and lung capillaries?

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The gas exchange membrane The gas exchange membrane

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Cross section through an alveolus Cross section through an alveolus Alveolar macrophage (defensive role)

Alveolar wall

Connective tissue cell

Alveolus

Monocyte (defensive role)

Surfactant secreted by type II pneumocytes

Capillary

Interstitial space

Alveolar epithelial wall 0.5 Âľm

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Nucleus of type I pneumocyte

Surfactant is a phospholipid produced by type II pneumocytes in the alveolar walls.

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Gas exchange membrane

Alveolus

O2

Red blood cell in capillary

Connective tissue containing elastic fibres

Capillary

CO2

Epithelial basement membrane

Red blood cell

Capillary basement membrane

The diagram above illustrates the physical arrangement of the alveoli to the capillaries through which the blood moves. The alveolus is lined with alveolar epithelial cells called pneumocytes. Phagocytes (monocytes and macrophages) are also present to protect the lung tissue. Elastic connective tissue gives the alveoli their ability to expand and recoil.

Capillary endothelium

The gas exchange membrane is the layered junction between the alveolar epithelial cells, the endothelial cells of the capillary, and their associated basement membranes (thin connective tissue layers under the epithelia). Gases move freely across this membrane.

4. Describe the structure and purpose of the gas exchange membrane:

5. The diagram below shows the different types of cells and their positions and occurrence in the lungs. Use it to answer the following questions: Ciliated cells

Serous and mucous glands

Hyaline cartilage

Smooth muscle

Elastic fibres

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Respiratory

Conducting

Height Goblet of the cells epithelium

(a) Why does the epithelium become very thin in the respiratory zone?

(b) Why would elastic fibres be present in the respiratory zone, while hyaline cartilage is not?

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77 Breathing

Key Idea: Breathing provides a continual supply of air to the lungs to maintain the concentration gradients for gas exchange. Different muscles are used in inspiration and expiration to force air in and out of the lungs. Breathing (ventilation) provides a continual supply of oxygen-

1. Explain the purpose of breathing:

rich air to the lungs and expels air high in carbon dioxide. Together with the cardiovascular system, which transports respiratory gases between the alveolar and the cells of the body, breathing maintains concentration gradients for gas exchange. Breathing is achieved by the action of muscles.

Breathing and muscle action

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Muscles can only do work by contracting, so they can only perform movement in one direction. To achieve motion in two directions, muscles work as antagonistic pairs. Antagonistic pairs of muscles have opposing actions and create movement when one contracts and the other relaxes. Breathing in humans involves two sets of antagonistic muscles. The external and internal intercostal muscles of the ribcage, and the diaphragm and abdominal muscles.

2. In general terms, how is breathing achieved?

Inspiration (inhalation or breathing in)

3. (a) Describe the sequence of events involved in quiet breathing:

During quiet breathing, inspiration is achieved by increasing the thoracic volume (therefore decreasing the pressure inside the lungs). Air then flows into the lungs in response to the decreased pressure inside the lung. Inspiration is always an active process involving muscle contraction. Intercostal muscles

External intercostal muscles contract causing the ribcage to expand and move up. Diaphragm contracts and moves down.

(b) What is the essential difference between this and the situation during forced breathing:

Thoracic volume increases, lungs expand, and the pressure inside the lungs decreases.

Air flows into the lungs in response to the pressure gradient.

Diaphragm contracts and moves down

Expiration (exhalation or breathing out)

4. During inspiration, which muscles are:

(b) Relaxed:

In quiet breathing, expiration is a passive process, achieved when the external intercostals and diaphragm relax and thoracic volume decreases. Air flows passively out of the lungs to equalise with the air pressure. In active breathing, muscle contraction is involved in bringing about both inspiration and expiration.

Intercostal muscles

5. During forced expiration, which muscles are:

(a) Contracting:

(b) Relaxed:

6. Explain the role of antagonistic muscles in breathing: Diaphragm contracts and moves down

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In quiet breathing, external intercostals and diaphragm relax. The elasticity of the lung tissue causes recoil.

In forced breathing, the internal intercostals and abdominal muscles contract to compress the thoracic cavity and increase the force of the expiration. Thoracic volume decreases and the pressure inside the lungs increases.

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(a) Contracting:

Air flows passively out of the lungs in response to the pressure gradient.

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Diaphragm relaxes and moves up

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78 Recording Changes in Breathing and Heart Rate

Key Idea: Breathing rate and heart rate both increase during exercise to meet the body's increased metabolic demands. During exercise, the body's metabolic rate increases and the demand for oxygen increases. Oxygen is required for cellular respiration and ATP production. Increasing the rate of breathing delivers more oxygen to working tissues and

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enables them to make the ATP they need to keep working. An increased breathing rate also increases the rate at which carbon dioxide is expelled from the body. Heart rate also increases so blood can be moved around the body more quickly. This allows for faster delivery of oxygen and removal of carbon dioxide.

In this practical, you will work in groups of three to see how exercise affects breathing and heart rate. Choose one person to carry out the exercise and one person each to record heart rate and breathing rate. Heart rate (beats per minute) is obtained by measuring the pulse (right) for 15 seconds and multiplying by four. Breathing rate (breaths per minute) is measured by counting the number of breaths taken in 15 seconds and multiplying it by four.

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Gently press your index and middle fingers, not your thumb, against the carotid artery in the neck (just under the jaw) or the radial artery (on the wrist just under the thumb) until you feel a pulse.

Measuring the radial pulse

CAUTION: The person exercising should have no known pre-existing heart or respiratory conditions.

Procedure

Resting measurements

Measuring the carotid pulse

Have the person carrying out the exercise sit down on a chair for 5 minutes. They should try not to move. After 5 minutes of sitting, measure their heart rate and breathing rate. Record the resting data on the table (right).

Heart rate (beats minute-1)

Breathing rate (breaths minute-1)

Resting

Exercising measurements

Choose an exercise to perform. Some examples include step ups onto a chair, skipping rope, jumping jacks, and running in place. Begin the exercise, and take measurements after 1, 2, 3, and 4 minutes of exercise. The person exercising should stop just long enough for the measurements to be taken. Record the results in the table.

1 minute

2 minutes 3 minutes 4 minutes

Post exercise measurements

After the exercise period has finished, have the exerciser sit down in a chair. Take their measurements 1 and 5 minutes after finishing the exercise. Record the results on the table.

1 minute after

5 minutes after

1. (a) Graph your results on separate piece of paper. You will need to make left and right vertical axes, one for heart rate and another for breathing rate. When you have finished answering the questions below, attach it to this page.

2. (a) Describe what happened to heart rate and breathing rate after exercise:

(b) Why did this change occur?

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(b) Analyse your graph and describe what happened to heart rate and breathing rate during exercise:

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79 Malfunctions of the Respiratory System

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whether they prevent air reaching the alveoli (obstructive) or whether they affect the gas exchange tissue itself (restrictive). Such diseases ultimately have similar effects in that gas exchange rates are too low to meet metabolic requirements. Chronic obstructive diseases, such as emphysema and chronic bronchitis, are commonly known as COPD (chronic obstructive pulmonary disease) and are associated with an inflammatory response of the lung to noxious particles or gases, most commonly tobacco smoke.

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Key Idea: Respiratory diseases severely limit gas exchange function by preventing air getting to the gas exchange surface or damaging the gas exchange surface itself. Respiratory diseases are diseases of the gas exchange system, including diseases of the lung, bronchial tubes, trachea, and upper respiratory tract. Respiratory diseases include mild and self-limiting diseases such as the common cold, to life-threatening infections such as tuberculosis. Noninfectious respiratory diseases are categorised according to

Obstructive lung disease

In obstructive lung diseases, the air cannot get to the gas exchange surface.

Mucus

Capillary

Chronic bronchitis Excess mucus blocks airway, leading to inflammation and infection. There is often an associated persistent cough. Air pollution and cigarette smoking are common causes.

Contracting muscle narrows airway

Airway

Asthma Thickening of bronchiole wall and increase in muscle. Bronchioles narrow, restricing airflow. Asthma is a hypersensitive reaction to environmental triggers such as dust or cold. Emphysema Destruction of capillaries and structures supporting the small airways and lung tissue. Air pollution and cigarette smoking are common causes.

Cross sections through a bronchiole with various types of obstructive lung disease

Normal lung tissue

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In Australia lung disease causes around 20,000 deaths a year, with 40% of those being caused by lung cancer. Most lung cancer cases are linked to lifestyle choices, with the vast majority of these (80-90%) being associated with long term exposure to tobacco smoke, particularly cigarette smoke.

Asbestosis

Restrictive lung diseases are caused by scarring of the lung tissue. The scarring may make it difficult for the lung to inflate and limit the exchange of respiratory gases. Restrictive lung diseases are usually the result of exposure to inhaled substances (especially dusts) in the environment. Asbestosis is a restrictive lung disease and is often associated with lung cancer.

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Serious trauma may cause air to enter the pleural space between the lungs and the chest wall (arrowed). This increases the pressure within the pleural space and reduces the lung's potential volume, making it unable to inflate properly (a so-called collapsed lung). Treatment may require the use of a syringe to remove the air from the chest cavity.

Emmanueim

James Heilman

Cancerous tissue

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Asthma (x 1000)

100

300

80 200

60 40

100

20 0

0

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Numbers of cases of lung disease Australia 2012 400

0-14 15-24

25-34 35-44

45-54

55-64

65-74

Chronic obstructive pulmonary disease (x 1000)

In Australia nearly 300,000 people are hospitalised with lung disease every year. 47% of these are due to respiratory infections. The graph right shows the agerelated incidence of asthma and chronic obstructive pulmonary disease or COPD (emphysema and chronic bronchitis) in Australia in 2012.

75+

Age

1. Explain how obstructive lung diseases prevent efficient gas exchange:

2. Describe the cause and the effect on the airways of:

(a) Asthma:

(b) Chronic bronchitis:

(c) Emphysema:

3. Describe the general trend in asthma and COPD rates in Australia by age:

4. Tobacco smoking is a major risk factor in the occurrence of emphysema, chronic bronchitis, and cancers of the respiratory tract. Smoking causes the lung tissue to lose its elasticity and tar from the tobacco smoke clogs the airways and damages the alveoli (right). Use the diagram to help you to explain why smoking reduces the gas exchange capacity of the lung tissue:

Non-smoker

Cilia

Normal alveoli

Thin layer of mucus

Cells lining airways

Smoker

Smoke particles

Coalesced alveoli

Extra mucus produced

5. Based on the graph, below right, what evidence is there to link COPD and smoking:

Smoke particles indirectly destroy the walls of the lung’s alveoli.

Cancerous cell

Death rates due to COPD (Australia) in people 55+, 1964–2011 and smoking rates, 1945–2011, by sex

80 COPD deaths males

70 60

300

50

Smoking males

200

40 30

100

Smoking females

20 10

COPD deaths females

0 1945

1955

1965

1975

1985

1995

2005

0

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Data: Mortality from asthma and COPD in Australia. Australian Institute of Health and Welfare 2014

Percentage smoking

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6. What causes restrictive lung disease and how does it affect the lungs?

Deaths per 100,000

400


80 The Digestive System

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Key Idea: The digestive tract is specialised to maximise the digestion of food, absorption of nutrients, and elimination of undigested material. The human digestive system (gut) is a tubular tract, which is regionally specialised into a complex series of organs and glands that work in sequence to maximise the efficiency with which food is processed. Collectively, the organs of the

digestive tract carry out the physical and chemical breakdown (digestion) of food, absorption of nutrients, and elimination of undigested material. The gut is a hollow, open-ended, muscular tube, and the food within it is essentially outside the body, having contact only with the cells lining the tract. External to the digestive tract are several accessory organs and glands, which add enzymes to the food to aid digestion.

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Gastric gland

In the stomach, gastric glands contain parietal cells, which produce hydrochloric acid, and chief cells, which produce enzymes to break down protein.

Salivary glands

Cells lining the walls on the small intestine (the intestinal epithelium) have microscopic extensions of the plasma membrane called microvilli. These form a brush border that increases the surface area for absorption of food molecules. Under lower power microscopy, it appears as a fuzzy edge.

Microvilli

Oesophagus

Louisa Howard, Katherine Connollly Dartmouth College

Villi

Epithelial cells

Liver

ach

Goblet cells

m

Gall bladder

In the small intestine, the intestinal epithelial cells and mucus-producing goblet cells make up the epithelium lining the gut wall. The wall is folded into finger like projections called villi (sing. villus). These further increase the surface area of the intestine.

Sto

Pancreas

La

rge intestine

Nephron

Small intestine

Smooth muscle

Villus

Intestinal lumen

The intestinal epithelium is supported by underlying connective tissue. Two layers of smooth muscle, one running lengthwise and one running around the gut, encircle the tube, contracting in waves to move food through the intestine.

3. What is the purpose of the smooth muscle surrounding the intestine?

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2. How are villi formed?

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1. Name three cell types of the digestive system and state their function:

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81 Moving Food Through the Gut

Key Idea: Solid food is chewed into a small mass called a bolus and swallowed. Further digestion produces chyme. Food is moved through the gut by waves of muscular contraction called peristalsis.

Ingested food is chewed and mixed with saliva to form a small mass called a bolus. Wave-like muscular contractions called peristalsis moves the food, first as a bolus and then as semifluid chyme, through the digestive tract as described below.

Peristalsis

Head this end

The process of moving food through the oesophagus by waves of muscular contractions is called peristalsis.

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Inner circular muscle

Circular muscles contract behind the bolus

Circular muscle

Bolus

Outer longitudinal muscle

The bolus enters the stomach, where the digestive activity there reduces it to a slurry called chyme. The chyme enters the small intestine for further digestion.

Longitudinal muscle

EII

Longitudinal muscles contract ahead of the bolus, causing the tube to shorten and widen to receive the food mass.

Cross section through the small intestine

A cross section through the small intestine shows the outer longitudinal and inner circular muscles involved in peristalsis. In a cross sectional view, the longitudinal muscles appear circular because they are viewed end on, whereas the circular muscle appears in longitudinal section.

Bolus movement

Peristaltic movement in the colon

Ascending colon

Transverse colon

Some of the muscular contractions in the colon mix the faecal matter.

Descending colon

Waste material will form faeces

Rectum

Up to three times a day, extra strong peristaltic contractions, move faeces from the colon into the rectum.

1. Describe how peristalsis moves food through the gut:

2. What are the two main functions of peristalsis?

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X-ray of the colon

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3. Draw arrows on the X-ray of the colon (above right) to show the direction of movement of the waste matter through the colon. Circle the areas of waste matter. LINK

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82 The Stomach

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Key Idea: The stomach produces acid and a protein-digesting enzyme, which break food down into a slurry, called chyme. The stomach is a hollow, muscular organ between the esophagus and small intestine. In the stomach, food is mixed in an acidic environment to produce a semi-fluid mixture

called chyme. The low pH of the stomach destroys microbes, denatures proteins, and activates a protein-digesting enzyme precursor. There is very little absorption in the stomach, although small molecules (glucose, alcohol) are absorbed across the stomach wall into the surrounding blood vessels. Oesophagus

Cardiac sphincter (closes the junction between oesophagus and stomach). Prevents food moving back up oesophagus.

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Three layered muscular wall mixes the stomach contents to produce chyme. Stretching the stomach wall stimulates gastric secretion.

The gall bladder stores bile, which is produced by the liver cells. Fat and acid in the duodenum stimulate release of bile from the gall bladder.

Bile from liver

See detail below

Folds (rugae) in the stomach wall allow the stomach to expand to 1 L.

Stomach

Pyloric sphincter (closes junction between stomach and duodenum).

Pancreas secretes an enzyme-rich alkaline fluid into the duodenum via the pancreatic duct.

Pancreatic duct

Duodenum (part of small intestine)

Detail of a gastric gland (stomach wall) Stomach surface

Gastric pit

Pepsinogen (activated by HCl)

Pepsin

HCl

Stomach secretions Gastric juice

Goblet cells

Pepsin (optimal pH 1.5-2.0) Acts on proteins and breaks them down into peptides (short chains of amino acids).

secrete mucus to protect the stomach lining from the acid.

1. What is the role of the stomach?

Parietal cell secretes HCl

Chief cell - secretes pepsinogen

Gastrin secreting cell. Gastrin is a hormone that increases HCl production.

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Acid (HCl) secretion

3. How does the stomach achieve the mixing of acid and enzymes with food?

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2. What is the purpose of the hydrochloric acid produced by the parietal cells of the stomach?

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83 The Small Intestine

Key Idea: The small intestine is the site of further enzymic break down of food and absorption of nutrients. Intestinal villi and microvilli increase surface area for nutrient absorption. The small intestine is divided into three sections: the duodenum, where most chemical digestion occurs, and the

jejunum and the ileum, where most absorption occurs. The small intestine's role is to complete the chemical digestion of food and absorb nutrient molecules into the blood. The presence of intestinal villi in the small intestine greatly increase the surface area for absorption.

The small intestine is a tubular structure between the stomach and the large intestine. It receives the chyme (food and enzyme mixture) directly from the stomach.

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Epithelial cells

Capillaries surround a central lymph vessel

The intestinal lining is folded into many intestinal villi (shown right), which project into the gut lumen (the space enclosed by the gut). They increase the surface area for nutrient absorption. The epithelial cells of each villus in turn have a brush-border of many microvilli, which increase the surface area further.

Secretion of alkaline fluid and mucus into the lumen.

Enzymes bound to the surfaces of the epithelial cells, and in the pancreatic and intestinal juices, break down peptides and carbohydrate molecules (tables below). The breakdown products are then absorbed into the underlying blood and lymph vessels. Tubular exocrine glands and goblet cells secrete alkaline fluid and mucus into the lumen.

Enzymes bound to

the surfaces of the epithelial cells break down peptides.

Epithelial cells migrate toward the tip of the villus to replace lost and worn cells.

Nutrients are transported away

Lumen

Enzymes in pancreatic juice

1. Pancreatic amylase (6.7-7.0) 2. Trypsin (7.8-8.7) 3. Chymotrypsin (7.8) 4. Pancreatic lipase (8.0)

1. Starch → maltose 2. Protein → peptides 3. Protein → peptides 4. Fats → fatty acids & glycerol

Enzymes in intestinal juice

Enzymes in small intestine (optimal pH)

Action

1. Maltase (6.0-6.5) 2. Peptidases (~ 8.0)

1. Maltose → glucose 2. Polypeptides → amino acids

Louisa Howard, Katherine Connollly Dartmouth College

Action

Pennsylvania State University College of Medicine

Enzymes in duodenum (optimal pH)

Villi

Mucosa

Muscle layers

TEM

The intestinal villi are shown projecting into the gut lumen in the light microscope image (left). The microvilli forming the brush border of a single intestinal cell are shown in the transmission electron micrograph (right). The mucosa consists of three layers: the epithelium, underlying connective tissue, and thin muscle layer.

1. (a) Name the three regions of the small intestine:

2. What is the purpose of the intestinal villi?

3. Where are enzymes found in the small intestine?

4. In general do the pancreatic enzymes act in acidic or alkaline conditions? © 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

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(b) Identify a functional difference between these regions:

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84 Digestion, Absorption, and Transport

Key Idea: Food must be digested into components small enough to be absorbed by the body's cells and assimilated. Nutrient absorption involves both active and passive transport. Digestion breaks down food molecules into small molecules that can pass through the intestinal lining into the underlying blood and lymph vessels. For example, starch is broken down first into maltose and short chain carbohydrates such as dextrose, before being hydrolysed to the simple sugar

glucose (below). Breakdown products of other foodstuffs include amino acids (from proteins), and fatty acids, glycerol, and acylglycerols (from fats). The passage of these molecules from the gut into the blood or lymph is called absorption. Nutrients are then transported directly or indirectly to the liver for storage or processing. After they have been absorbed nutrients can be assimilated, i.e incorporated into the substance of the body itself.

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Digestion of starch

Starch digestion begins in the mouth. The teeth grind the solid mass, which increases its surface area and mixes in the amylase, produced by the salivary glands.

Food such as bread contains carbohydrates in the form of starch.

Starch

Starch is hydrolysed into smaller components. Amylase acts on the a-1,4 glycosidic bonds to produce short chain carbohydrates and the disaccharide maltose.

Amylase (enzyme)

Maltose

Amylase is inactivated in the acid environment of the stomach.

Glucose

Amylase is also produced by the pancreas. The hydrolysis of carbohydrate continues as the stomach contents (called chyme) passes into the small intestine.

Maltose is hydrolysed into glucose by the enzyme maltase, produced by the intestinal epithelial cells. Glucose can then be absorbed into the bloodstream.

1. The experiment on the right demonstrates why food must be digested before it can be absorbed.

Modelling nutrient absorption

(a) Why was there no starch in the distilled water?

Dialysis tubing (partially-permeable membrane)

Distilled water

(b) Use this model to explain why food must be digested before it can be absorbed.

Solution containing starch and glucose

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To model absorption of nutrients in the small intestine, a dialysis tubing was filled with a solution of starch and glucose. The outside of the tubing was washed with distilled water to remove any starch or glucose that spilled on to the outer surface during filling. The tubing was placed into a beaker of distilled water. After one hour, the distilled water was tested for the presence of starch or glucose. Only glucose was present in the distilled water. NOTE: Dialysis tubing comes in many different pore sizes. It only allows molecules smaller than the size of the pore to pass through.

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Nutrient absorption by intestinal villi

Gut lumen

Fructose

Intestinal epithelial cell

Facilitated diffusion

Glucose and galactose

Glucose and amino acids are actively transported by cotransport proteins along with sodium (sodium symport). This maintains a sodium gradient which helps with the absorption of water.

Active transport (Na cotransport)

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+

Amino acids

Active transport (Na+ cotransport)

Active transport of di- and tripeptides is coupled to the downhill movement of H+ across the plasma membrane of the intestinal epithelial cells.

Dipeptides

Active transport (proton pump)

Tripeptides

Short chain fatty acids

Diffusion

Once the monoglycerides and fatty acids are absorbed, triglycerides are re-formed and transported to the liver as protein-coated aggregations in the lacteals of the lymphatic system.

Long chain fatty acids

Monoglycerides

n

usio

Diff

Fat soluble vitamins

Cross section through a villus, showing how the products of digestion are absorbed across the intestinal epithelium into the capillaries or into the lacteals of the lymphatic system. The nutrients are delivered to the liver.

Monoglycerides and fatty acids associate with bile salts to form lipid spheres called micelles. Micelles hold the poorly soluble fatty acids and monoglycerides in suspension and transport them to the surface of the epithelial cells where they can be absorbed. The micelles themselves are not absorbed.

Lacteal

Artery

Vein

2. Explain the roles of amylase and maltase in starch digestion:

3. Salivary and pancreatic secretions contain amylase. Why do two digestive organs produce the same enzyme?

(a) Glucose:

(b) Fructose:

(c) Amino acids: (d) Di- and tripeptides: 5. What is the role of micelles in the absorption of lipids?

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4. Describe how each of the following nutrients are absorbed by the intestinal villi:

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6. How are concentration gradients maintained for the absorption of nutrients by diffusion:


85 The Large Intestine

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Key Idea: The large intestine absorbs water and solidifies the indigestible material before passing it to the rectum. Undigested waste are egested as faeces from the anus. After most of the nutrients have been absorbed in the small intestine, the remaining semi-fluid contents pass into the

Transverse colon

The rectum stores the waste faecal material before it is discharged out the anus. Fullness in the rectum produces the urge to defecate. If too little water is absorbed, the faeces will be watery as in diarrhoea. If too much water is absorbed the faeces will become compacted and difficult to pass.

Ascending colon

Descending colon

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After most of the nutrients have been absorbed in the small intestine, the remaining semi-fluid contents pass into the large intestine (appendix, cecum, and colon). This mixture includes undigested or indigestible food, (such as cellulose), bacteria, dead cells, mucus, bile, ions, and water. In humans and other omnivores, the large intestine's main role is to reabsorb water and electrolytes and consolidate the undigested material for egestion (elimination) from the anus.

large intestine (consisting of the appendix, caecum, colon, and rectum). The large intestine's main role is to reabsorb water and electrolytes and to consolidate the waste material into faeces, which are eliminated from the anus in a process called egestion.

Colon absorbs water, Na+, and some vitamins. It also incubates bacteria which produce vitamin K.

Ileum of small intestine

See diagram below

Rectum: The enlarged final segment of the large intestine where faeces are stored and consolidated before elimination.

Caecum

Defaecation is controlled by the anal sphincters, whose usual state is to be contracted (closing the orifice). Defaecation is under nervous control.

Appendix is a blindending sac that may have a minor immune system function.

Anus. The sphincters controlling the anal opening are normally contracted, but relax to allow faeces to be expelled.

Lining of the large intestine

The lining of the large intestine has a simple epithelium containing tubular glands (crypts) with many mucussecreting cells. The mucus lubricates the colon wall and helps to form and move the faeces. In the photograph, some of the crypts are in XS and some are in LS. Mucus producing goblet cells

Lumen

Simple columnar epithelial cells Crypt

Goblet cells within crypt

Connective tissue

Lymph nodule

EII

Circular muscle

1. What is the main purpose of the large intestine?

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Note the abundance of pale goblet cells.

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2. What are the effects of absorbing too little and too much water in the large intestine?

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86 Malfunctions of the Digestive System

Key Idea: Malfunctions of the digestive system may be caused by infections, blockages, or parts of the system failing to work correctly. The digestive system is a collection of organs and any one of these failing to carry out its task correctly can cause digestive

problems (e.g. reflux caused by a faulty cardiac sphincter). Failure to eat a healthy diet can lead to some digestive problems (e.g. constipation caused by a lack of fibre). Infection of the intestinal track by pathogens, e.g. Salmonella or cholera, can cause severe and acute diarrhoea.

GERD

Cardiac sphincter

Cardiac sphincter fully closed

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Gastroesophageal reflux disease (GERD) occurs when the cardiac sphincter of the stomach does not close properly. This allows food and digestive juices to reenter the oesophagus. This can cause a burning sensation in the throat and chest, which is sometimes called heartburn.

Normal

Food and digestive juices

Steven Fruitsmaak

Cardiac sphincter does not fully close

X-ray showing stomach juices moving up the oesophagus (arrowed).

GERD

Cholera is treated with oral rehydration salts (ORS) containing glucose and electrolytes, which are taken up by active transport by the intestinal cells. This reverses the osmotic gradient and allows blood volume to be restored.

TEM

SEM

CDC

Cholera is an acute intestinal infection caused by the bacterium Vibrio cholerae (right). Infection results in severe diarrhoea, which may be fatal. The cholera bacterium produces a toxin, which increases the permeability of the intestinal epithelium to chloride ions and results in copious, painless, watery diarrhoea. This can lead to severe dehydration, electrolyte imbalance, kidney failure, and death within hours if untreated.

Dartmouth Electron Microscope Facility

Cholera and diarrhoea

1. (a) What is GERD and what causes it?

(b) Why would a sufferer of GERD experience chest pains?

(b) How does cholera affect the intestinal lining?

(c) How is the disease treated?

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2. (a) Explain why cholera can be fatal:

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87 The Circulatory System

Key Idea: The circulatory system is responsible for the transport of nutrients, respiratory gases, and wastes in the blood to and from the body's cells via a network of vessels. The circulatory system comprises the heart, arteries, veins,

capillaries, and blood. The blood transports oxygen and nutrients to the cells, carbon dioxide to the lungs, and metabolic wastes to the kidneys. It is also moves cells of the immune system (the leucocytes) about the body.

WBCs

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RBCs

Red blood cells (RBCs) carry oxygen to the body's cells. The oxygen binds to the protein haemoglobin, which gives RBCs their colour. The larger white blood cells (WBCs) or leucocytes are part of the immune system and also circulate in the blood.

Vein

G Beard

Artery

Blood is a liquid tissue. Blood cells are suspended in a watery material called plasma, which carries dissolved materials, e.g. blood proteins, electrolytes (salts) and nitrogenous waste.

Heart

Blood moves through blood vessels, the smallest of which are the capillaries. These are only one cell thick, allowing oxygen and other molecules to easily move out of or into the blood from the cells of the body's tissues.

Vein

The heart is the central organ of the circulatory system. It it composed primarily of cardiac muscle which contracts rhythmically to pump blood around the body.

Artery

Blood is transported away from the heart in arteries, blood vessels with thick walls of elastic connective tissue and smooth muscle. Blood returns to the heart in veins, which have thinner walls but a larger lumen (inside space).

1. Name three components of the circulatory system and state their function:

(c)

2. (a) Which component of the blood carries oxygen to the body's cells?

(b) Which component of the blood carries metabolic wastes to the kidneys? WEB

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(b)

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(a)

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88 Blood

Key Idea: Blood transports nutrients, wastes, hormones, and respiratory gases around the body. Blood is a complex connective tissue made up of cellular components suspended in matrix of liquid plasma. It makes up about 8% of body weight. If a blood sample is taken, the cells can be separated from the plasma by centrifugation. The cells (formed elements) settle as a dense red pellet below the transparent, straw-coloured

plasma. Blood performs many functions. It transports nutrients, respiratory gases, hormones, and wastes and has a role in thermoregulation through the distribution of heat. Blood also defends against infection and its ability to clot protects against blood loss. The examination of blood is also useful in diagnosing disease because the cellular components of blood are normally present in specified ratios. Deviations from these ratios may indicate disease.

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Mammalian blood

CELLULAR COMPONENTS The cellular (or formed) elements of blood float in the plasma and make up 40-50% of the total blood volume.

White blood cells and platelets make up 2-3% of the total blood volume.

NON-CELLULAR COMPONENTS The non-cellular part of the blood is the plasma, a watery matrix making up 50-60% of blood volume. Most of the blood is water. It transports dissolved substances, provides cells with water, distributes heat, and maintains blood volume. Most of the plasma is water, but it also contains dissolved proteins, glucose, amino acids, vitamins, minerals, urea, uric acid, CO2, hormones, and antibodies.

Lymphocyte

White blood cells are involved in internal defence. Lymphocytes are important in immunity and make up 24% of the white cell count.

Platelets are small, membranebound cell fragments with a role in blood clotting.

Platelet

All EM iamges: Dartmouth College

Red blood cells (RBCs) account for 38-48% of total blood volume. RBCs transport oxygen (carried bound to haemoglobin) and a small amount of carbon dioxide. Unlike other blood cells, RBCs have no nucleus and lack most organelles. They are packed full of haemoglobin protein.

RBC

Blood plasma

1. Describe one feature distinguishing red and white blood cells in mammalian blood:

2. Describe two functions of the blood plasma:

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4. Explain why the blood can be called a "liquid tissue":

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3. What is the function of platelets in the blood?

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89 Blood Vessels of the blood vessel walls. Vasoconstriction increases blood pressure whereas vasodilation has the opposite effect. Veins are the blood vessels that return blood to the heart from the tissues. The smallest veins (venules) return blood from the capillaries to the veins. Veins and their branches contain about 59% of the blood in the body. The structural differences between veins and arteries are mainly associated with differences in the relative thickness of the vessel layers and the diameter of the lumen (space within the vessel). These, in turn, are related to the vessel’s functional role.

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Key Idea: The blood vessels of the circulatory system connect the body's cells to the organs that exchange gases, absorb nutrients, and dispose of wastes. In vertebrates, arteries are the blood vessels that carry blood away from the heart to the capillaries within the tissues. The large arteries that leave the heart divide into medium-sized (distributing) arteries. Within the tissues and organs, these distributing arteries branch to form arterioles, which deliver blood to capillaries. Blood flow to the tissues is altered by contraction (vasoconstriction) or relaxation (vasodilation)

Arteries

Arteries, regardless of size, can be recognised by their well-defined rounded lumen (internal space) and the muscularity of the vessel wall. Arteries have an elastic, stretchy structure that gives them the ability to withstand the high pressure of blood being pumped from the heart. At the same time, they help to maintain pressure by having some contractile ability themselves (a feature of the central muscle layer).

Thin outer layer connective tissue.

Bloo

d flo w

Arteries nearer the heart have more elastic tissue to resist the higher pressures of the blood leaving the left ventricle. Arteries further from the heart have more muscle to help them maintain blood pressure. Between heartbeats, the arteries undergo elastic recoil and contract. This tends to smooth out the flow of blood through the vessel.

Arteries have three regions (left): 1.

A thin inner layer of epithelial cells called the endothelium lines the artery.

2.

A thick central layer of elastic tissue and smooth muscle that can both stretch and contract.

3.

An outer connective tissue layer has a lot of elastic tissue.

Artery

Thin endothelium

Artery

Thick central layer of elastic tissue and smooth muscle.

1. Why do the artery walls need to be thick with a lot of elastic tissue?

2. What is the purpose of smooth muscles in the artery walls?

(b) What would the effect of this be on blood pressure?

4. Describe the structure of a capillary and explain the purpose of this structure:

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3 (a) What is the effect of vasodilation on the diameter of an artery?

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119 Red blood cell

Nucleus of endothelial cell

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Thin endothelium (one cell thick)

Fat cell Blood flow Capillary

CO2 and wastes move from the tissues into the capillary.

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O2 and nutrients move from the capillary into the cells of the tissues.

Dept of Biological Sciences, University of Delaware

Basement membrane

Capillaries are small blood vessels with a diameter of 4-10 µm. Red blood cells (7-8 µm) can only just squeeze through. Blood flow is very slow through the capillaries (less than 1 mm per second) allowing the exchange of nutrients and wastes between the blood and tissues. Capillaries form large networks, especially in tissues and organs with high metabolic rates.

Capillaries

Arterioles carry blood to capillaries

Venules drain capillaries to vein

d oo

One-way valves prevent blood flowing in the wrong direction.

flow

Bl

Thin central layer of elastic and muscle tissue.

Thicker outer layer of connective tissue.

Veins

Veins are made up of the same three layers as arteries but they have less elastic and muscle tissue, a relatively thicker external layer, and a larger, less defined lumen.

Vein

Thin layer of endothelium

Vein

Although veins are less elastic than arteries, they can still expand enough to adapt to changes in the pressure and volume of the blood passing through them. Blood flowing in the veins has lost a lot of pressure because it has passed through the narrow capillaries. The lower pressure flow means that many veins, especially those in the limbs, have valves to prevent backflow of the blood as it returns to the heart.

(a) Thickness of muscle and elastic tissue:

(b) Size of the lumen (inside of the vessel):

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5. Contrast the structure of veins and arteries for each of the following properties:

6. What is the role of the valves in assisting the veins to return blood back to the heart?

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7. Why does blood ooze from a venous wound, rather than spurting as it does from an arterial wound?


90 The Heart

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Key Idea: Humans have a four chambered heart divided into left and right halves. It acts as a double pump. The heart is the centre of the human cardiovascular system. It is a hollow, muscular organ made up of four chambers (two atria and two ventricles) that alternately fill and empty of blood, acting as a double pump. The left side (systemic

circuit) pumps blood to the body tissues and the right side (pulmonary circuit) pumps blood to the lungs. The heart lies between the lungs, to the left of the midline, and is surrounded by a double layered pericardium of connective tissue, which prevents over distension of the heart and anchors it within the central compartment of the thoracic cavity. Top view of a heart in section, showing valves

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Human heart structure (sectioned, anterior view)

Semi-lunar valve of pulmonary artery

Semi-lunar valve of aorta

Aorta carries oxygenated blood to the head and body Vena cava receives deoxygenated blood from the head and body

Bicuspid valve

Pulmonary artery carries deoxygenated blood to the lungs

LA

RA

Tricuspid (right atrioventricular valve)

Tricuspid valve prevents backflow of blood into right atrium

RV

Bicuspid (left atrioventricular valve)

Anterior view of heart to show coronary arteries

Superior vena cava

Chordae tendinae nonelastic strands supporting the valve flaps

Aorta

LV

Pulmonary artery

Right coronary artery

Semi-lunar valve prevents the blood flow back into ventricle.

Septum separates the ventricles

The heart is not a symmetrical organ. Although the quantity of blood pumped by each side is the same, the walls of the left ventricle are thicker and more muscular than those of the right ventricle. The difference affects the shape of the ventricular cavities, so the right ventricle is twisted over the left.

Key to abbreviations

RA

Right atrium: receives deoxygenated blood via the anterior and posterior vena cava

RV

Right ventricle: pumps deoxygenated blood to the lungs via the pulmonary artery

LA

Left atrium: receives blood returning to the heart from the lungs via the pulmonary veins

LV

Left ventricle: pumps oxygenated blood to the head and body via the aorta

Pulmonary veins

Pulmonary veins

Left coronary artery

Right cardiac vein

RV

Inferior vena cava

Descending aorta

LV

Left cardiac vein

Coronary arteries: The high oxygen demands of the heart muscle are met by a dense capillary network. Coronary arteries arise from the aorta and spread over the surface of the heart supplying the cardiac muscle with oxygenated blood. Deoxygenated blood is collected by cardiac veins and returned to the right atrium via a large coronary sinus.

1. In the schematic diagram of the heart, below, label the four chambers and the main vessels entering and leaving them. The arrows indicate the direction of blood flow. Use large coloured circles to mark the position of each of the four valves. (a)

(b)

(f)

(c)

(g)

(d)

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(e)

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121 aorta, 100 mg Hg

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Pressure changes and the asymmetry of the heart The heart is not a symmetrical organ. The left ventricle and its associated arteries are thicker and more muscular than the corresponding structures on the right side. This asymmetry is related to the necessary pressure differences between the pulmonary (lung) and systemic (body) circulations.

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The graph below shows changes in blood pressure in each of the major blood vessel types in the systemic and pulmonary circuits (the horizontal distance not to scale). The pulmonary circuit must operate at a much lower pressure than the systemic circuit to prevent fluid from accumulating in the alveoli of the lungs. The left side of the heart must develop enough “spare” pressure to enable increased blood flow to the muscles of the body and maintain kidney filtration rates without decreasing the blood supply to the brain. 120

Blood pressure during contraction (systole)

100

Pressure (mm Hg)

80

60

40

The greatest fall in pressure occurs when the blood moves into the capillaries, even though the distance through the capillaries represents only a tiny proportion of the total distance travelled.

Blood pressure during relaxation (diastole)

radial artery, 98 mg Hg

20

arterial end of capillary, 30 mg Hg

0

aorta

arteries

A

capillaries

B

veins

vena pulmonary cava arteries

Systemic circulation horizontal distance not to scale

C

D

venules pulmonary veins

Pulmonary circulation horizontal distance not to scale

2. What is the purpose of the valves in the heart?

3. The heart is full of blood, yet it requires its own blood supply. Suggest two reasons why this is the case: (a) (b)

4. Predict the effect on the heart if blood flow through a coronary artery is restricted or blocked:

5. Identify the vessels corresponding to the letters A-D on the graph above: A:

B:

C:

D:

(b) Relate this to differences in the thickness of the wall of the left and right ventricles of the heart:

7. What are you recording when you take a pulse?

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6. (a) Why must the pulmonary circuit operate at a lower pressure than the systemic system?


91 Dissecting a Mammalian Heart

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Key Idea: Dissecting a sheep's heart allows hands-on exploration of a mammalian heart. The dissection of a sheep's heart is a common practical activity and allows hands-on exploration of the appearance

and structure of a mammalian heart. A diagram of a heart is an idealised representation of an organ that may look quite different in reality. You must learn to transfer what you know from a diagram to the interpretation of the real organ.

1 Gross anatomy of a sheep's pluck (the thoracic organs) to show a dorsal view of the heart.

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Lobe of right lung

Cut flap of pericardium

Lobe of left lung

Right auricle is a muscular pouch connected to the right atrium. It is also called the right atrial appendage.

Thymus (large in young animals)

The heart and the roots of the great vessels (vena cavae, pulmonary artery and aorta) are contained within a double-walled sac called the pericardium. It is filled with fluid and protects the heart in its central position in the body cavity.

Right ventricle of heart (dorsal)

2 External ventral view of heart

Did you know? The term auricle is Latin for ear and it describes the ear-like look of the small muscular pouches (one left and one right) that lead to the atria.

Probe

3 External dorsal view of heart

Aorta

Pulmonary trunk (artery)

Brachiocephalic artery (cut)

Left auricle

Pulmonary veins

Right auricle

Left ventricle

Right ventricle

Apex

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Left ventricle

Right ventricle

On the dorsal surface of the heart, above, locate the large thin-walled vena cavae and pulmonary veins. You may be able to distinguish between the anterior and posterior vessels. On the right side of the dorsal surface (as you look at the heart) at the base of the heart is the right atrium, with the right ventricle below it.

1. Use coloured lines to indicate the interventricular sulcus and the base of the heart. Label the coronary arteries.

2. On this photograph, label the vessel indicated by the probe.

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Note the main surface features of an isolated heart. The narrow pointed end forms the apex of the heart, while the wider end, where the blood vessels enter is the base. The ventral surface of the heart (above) is identified by a groove, the interventricular sulcus, which marks the division between the left and right ventricles.

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4 Dorsal view of heart

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Shallow section, ventral view of heart

5 Semi-lunar valves lie between the ventricles and the large arteries leaving the heart. They can be difficult to see.

Aorta

Pulmonary veins

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Left auricle

Chordae tendinae, right ventricle

Left ventricle

Right ventricle

Thick wall of left ventricle

3. On this dorsal view, label the vessel indicated. Palpate

the heart and feel the difference in the thickness of the left and right ventricle walls.

4. This photograph shows a shallow section to expose the

right ventricle. Label the vessel in the box indicated.

6 Frontal sections of heart to show chambers

Part of left AV valve

Pulmonary artery (from right ventricle to lungs but cut) Left atrium

Aorta (from left ventricle)

Part of left AV valve

Right atrium

Right ventricle

Chordae tendinae Papillary muscles

The white and blue dotted arrows indicated blood flow from the RV and LV respectively.

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Group work If you are working with a lab partner and you have two hearts to study, make your sections in different ways to maximise the structures you can see.

The atrioventricular (AV) valves of the two sides of the heart are similar in structure except that the right AV valve has three cusps (tricuspid) while the left atrioventricular valve has two cusps (bicuspid or mitral valve). Connective tissue (chordae tendineae) run from the cusps to papillary muscles on the ventricular wall.

5. Judging by their position and structure, what do you

6. What feature shown here most clearly distinguishes the left and right ventricles?.

suppose is the function of the chordae tendinae?

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If the heart is sectioned and the two halves opened, the valves of the heart can be seen. Each side of the heart has a one-way valve between the atrium and the ventricle known as the atrioventricular valve. They close during ventricular contraction to prevent back flow of the blood into the lower pressure atria.


92 Malfunctions of the Circulatory System

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Key Idea: Malfunctions of the circulatory system may be caused by abnormalities at birth, blockages, or disease. Failure of the heart or circulatory system can cause oxygen starvation in cells of the body.

Malfunctions of the circulatory system include blockages of the veins or arteries (often by blood clots), failure of valves to work properly (which affects blood flow), and heart attacks, which can be serious, and sometimes fatal.

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Myocardial infarction A myocardial infarction (heart attack) occurs when an area of the heart is deprived of its blood supply, resulting in tissue damage or death. Symptoms of infarction include a sudden onset of chest pain, breathlessness, nausea, and cold clammy skin. Damage to the heart may be so severe that it leads to heart failure and even death (myocardial infarction is fatal within 20 days in 40 to 50% of all cases). Formation of a blood clot, which blocks the congested artery.

Atheroma

Myocardial infarct. Area of heart tissue supplied by the coronary artery dies.

Atherosclerosis Atherosclerosis (hardening of the arteries) is caused by hardened deposits of fats and cholesterol (atheromas) in the inner walls of the arteries. Blood flow becomes restricted and there is an increased risk of blood clots (thrombosis). Aneurysm An aneurysm is a weakening of the aorta wall, causing it to balloon out. Aneurysms are usually the result of heart disease and chronic high blood pressure.

Hypertension Hypertension is abnormal and persistent high blood pressure. Chronic hypertension causes serious problems in the heart and increases the risks that atheroma will dislodge.

Deep vein thrombosis

Restrictions in veins can cause blood clots (thrombosis). If a clot forms, it can be dislodged and travel to the heart or lungs where it can cause serious complications, such as a pulmonary embolism. Clots can form in the veins during times of immobility (e.g. sitting during long haul flights).

Hellerhoff

Thrombosis in a vein

Varicose veins

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Varicose veins occur when the valves in the veins fail to work correctly, letting blood pool. They are most commonly associated with the veins in the legs. Although varicose veins are not usually life threatening, they can be painful.

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Varicose veins

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1. Describe each of the following circulatory malfunctions, and explain their effects on the body: (a) Aneurysm:

(b) Myocardial infarction:

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(c) Hypertension:

(d) Atherosclerosis:

(e) Varicose veins:

(f) Deep vein thrombosis:

2. Study the information on myocardial infarction, below right, and answer the following questions:

(a) What is the effect of smoking on the relative risk of myocardial infarction?

Relative risk of myocardial infarction (heart attack) by current tobacco exposure

Smoker: >24 gday-1

Smoker: 15-24 gday-1

(b) Compare the effect of stopping smoking with the effect of never having smoked on the risk of myocardial infarction:

Smoker: 1-14 gday-1

Ex-smoker

Men Women

Never smoked

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1

2

Relative risk

(c) Compare the effect of smoking by sex on the relative risk of myocardial infarction:

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93 The Excretory System

Key Idea: The excretory system is responsible for removing metabolic wastes from the body by filtering the blood. The central organs of the excretory system in humans and other mammals are the kidneys. They act as a selective filter of the blood, removing metabolic wastes while retaining

useful substances, such as valuable ions and glucose. The kidneys receive blood under high pressure via the arterioles from the renal artery. This high pressure forces blood plasma out of the capillaries, forming a fluid called filtrate, which is then modified as it passes through the kidney.

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Blood is filtered in the kidneys by the glomerulus, a dense knot of capillaries. Blood is forced through them at high pressure, a process known as ultrafiltration. The filtrate is collected in the Bowman's capsule which surrounds the glomerulus.

Vena cava

Dorsal aorta

The filtrate moves from Bowman's capsule to the convoluted tubules. These are lined with cuboidal epithelial cells, which have a brush border of microvilli to enhance absorption of substances from the filtrate. The glomerulus, capsule, and tubules form the functional unit structure of the kidney, the nephron.

Renal artery

Renal vein

Kidney

The thousands of filtering elements of the kidney (the nephrons) are aligned and organised in a particular way in the kidney. The glomeruli and convoluted tubules are found in the outer region or cortex, while the "loop of Henle" is found in the inner region of medulla.

Medulla

Cor tex

Ureter

Bladder

The filtrate passes to the renal ducts and then to the ureter and finally to the bladder. The kidney itself is bean shaped and is around 10 cm long in humans.

1. What is the purpose of the microvilli in the epithelial cells of the convoluted tubules?

2. (a) How is filtrate formed?

(b) How is the filtrate modified?

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3. The circulation rate of blood through the renal artery is about 1.2 L min-1, about one quarter of the heart's total output. Why does so much blood need to pass through the kidneys every minute?

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94 Drawing the Kidney visualise the three dimensional shape of the structure under view. Observational drawing from a microscope is a skill that must be developed. It requires relaxed viewing in which the image is viewed with one eye, while the other eye attends to the drawing. Attention should be given to the symmetry and proportions of the structure, accurate labelling, statement of magnification and sectioning, and stain used, if this is appropriate. In this activity, you will practise the skills required to translate what is viewed into a good biological drawing.

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Key Idea: Drawing from a dissection or histological preparation requires practise and an understanding of structure. Many observational studies made using microscopes will require you to make accurate representations of what you see. Observations need to be made at both low (X40) and high magnifications to identify the finer structure of the tissue. Tissue sections will usually be provided as longitudinal (LS) or traverse sections (TS). When you have access to both TS and LS images from the same specimen, it is also possible to

Biological drawings

TASK 1

►► Biological drawings should include as much detail as you need to distinguish different structures and types of tissue, but avoid unnecessary detail.

Complete a biological drawing of the kidney shown below left including the labels on the photo. At this level, the layers of the kidney can be seen while the detail of each layer cannot.

►► Tissue preparations are rarely neat and tidy and there may be areas where it is difficult to distinguish detail. In these cases you will need to infer detail where possible from adjacent cells.

TASK 2

►► Avoid shading as this can obscure detail.

►► Labelling involves interpretation based on your knowledge. Labels should be away from the drawing with label lines pointing to the structures identified. ►► Add a title and details of the image such as magnification.

Renal pelvis

Complete a plan diagram of a TS through the cortical region (cortex) of a kidney nephron (below). The ovoid glomerular cluster of capillaries can be seen in the upper centre of the photograph, separated from the Bowman's capsule by a space. The capsule is lined with squamous epithelium and surrounded by tubules formed by cuboidal epithelial cells. You can infer this from the position and spacing of the cell nuclei.

Ureter

Medulla

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Light micrograph of a transverse section through a kidney nephron to show glomerulus and convoluted tubules.

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Renal capsule

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Cortex

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95 The Human Urinary System

Key Idea: Mammalian kidneys contain thousands of repeating functional units, or nephrons, for filtering the blood. Each kidney is functionally independent of the other. The mammalian urinary system consists of the kidneys and bladder, and their associated blood vessels and ducts. The kidneys have a plentiful blood supply from the renal artery. The blood plasma is filtered by the kidneys to form urine. Urine is produced continuously, passing along the ureters

Kidneys in-situ (rat)

Sagittal section of kidney (pig)

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Urinary system

to the bladder, where it is stored. Each day the kidneys filter about 180 L of plasma. Most of this is reabsorbed, leaving a daily urine output of about 1 L. By adjusting the composition of the fluid excreted, the kidneys help to maintain the body’s internal chemical balance. Mammalian kidneys are very efficient, producing a urine that is more or less concentrated to varying degrees depending on requirements and individual adaptations of the species.

Tough connective tissue calyces (drains) collect the urine and pass it to the ureter

Lung

Heart

Rib

Vena cava

Dorsal aorta

Kidney

Left kidney

Urethra

EII

Bladder

The kidneys of humans (above), rats (dissection, above centre), and many other mammals (e.g. pig above right) are distinctive, bean shaped organs that lie at the back of the abdominal cavity to either side of the spine. They are partly protected by the lower ribs (see kidneys in-situ above centre).

Human kidneys are ~100-120 mm long and 25 mm thick. A cut through in a sagittal plane (see photo above right), reveals numerous tough connective tissue calyces. These collect the urine from the papillae where it is discharged and drain it into the ureter.

A smooth fibrous membrane (renal capsule) covers the kidney and protects it against trauma and infection.

The kidneys and their blood supply

Vena cava returns blood to the heart

Dorsal aorta supplies oxygenated blood to the body.

Kidney produces urine and regulates blood volume.

Adrenal glands are associated with, but not part of, the urinary system.

Renal vein returns blood from the kidney to the venous circulation.

EII

Right kidney

Ureter

Renal artery carries blood from the aorta to the kidney.

Ureters carry urine to the bladder.

(a) Kidney:

(b) Ureters:

(c) Bladder:

(d) Urethra:

(e) Renal artery:

(f) Renal vein:

(g) Renal capsule:

2. Calculate the percentage of the plasma reabsorbed by the kidneys: WEB

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1. State the function of each of the following components of the urinary system:

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129 Internal structure of the human kidney

Nephrons are arranged with all the collecting ducts pointing towards the renal pelvis.

Ureter

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Cortex

Inner medulla is organised into pyramids.

Urine flows from the pyramids towards the ureter. Urine collects in a space near the ureter called the renal pelvis, before leaving the kidney via the ureter.

Renal pelvis

Urine flow

Ureter

EII

Outer cortex contains the renal corpuscles and convoluted tubules.

Medulla

Nephron

The inside of the kidney appears striped because the nephrons are all aligned so that urine is concentrated as it flows towards the ureter. The outer cortex and inner medulla can be seen in the low power LM of the kidney, far right.

3. Under normal circumstances, how many kidneys does a person have?

4. A kidney contains 1.5 million nephrons (filtering units). A person only needs 300,000 working nephrons to survive.

(a) What percentage of nephrons actually need to be working for a person to survive?

(b) Why is this important to someone with a damaged kidney?

5. Explain the importance of the kidney's location near the lower part of the ribcage:

6. The graph below shows the volume of urine collected from a subject after drinking 1000 cm3 of distilled water. The subject’s urine was collected at 25 minute intervals over a number of hours.

3

Volume of urine (cm )

400 350 300 250 200 150 100

50 0

0

25

50

Drink 1000 cm3 distilled water

75

100

125

150

175

200

225

Time (minutes)

(a) Describe the changes in urine output during the experiment:

(b) Explain the difference in the volume of urine collected at 25 minutes and 50 minutes:

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96 Making Urine

Key Idea: The nephron is the functional unit of the kidney creating and using a salt gradient in the kidney to modify the composition of the urine it produces. The kidney nephron, is a selective filter element, made up of a renal corpuscle (glomerulus and capsule) and its associated tubules and ducts. Ultrafiltration, i.e. forcing fluid and dissolved substances through a membrane by pressure,

occurs in the first part of the nephron, across the membranes of the capillaries and the glomerular capsule. The formation of the glomerular filtrate depends on the pressure of the blood entering the renal corpusule and is precisely regulated so that filtration rate per day stays constant. After formation of the initial filtrate, the urine is modified through secretion and reabsorption according to physiological needs at the time.

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Nephron structure and function

Renal corpuscle: Blood is filtered and the filtrate enters the convoluted tubule (enlargement below). The filtrate contains water, glucose, urea, and ions, but lacks cells and large proteins.

1

Distal convoluted tubule: Further modification of the filtrate by active reabsorption and secretion of ions.

4

2

cap sule

Glomerulus

n's

Blood Filtrate (urine) Blood vessels around nephron

a Bowm

Renal corpuscle = Glomerulus + Bowman's capsule

Proximal convoluted tubule: Reabsorption of ~ 90% of filtrate, including glucose and valuable ions.

5

Loop of Henle: Transport of salt and passive movement of water creates a salt gradient through the kidney. The water is transported away by blood vessels around the nephron.

3

Collecting duct: Water leaves the filtrate (urine) by osmosis, making it more concentrated. The salt gradient established by the loop of Henle allows water to be removed along the entire length of the collecting duct.

1. Summarise the main activities in each of the five regions of the nephron: (a) Renal corpuscle:

(b) Proximal (near) convoluted tubule:

(c) Loop of Henle:

(d) Distal (far) convoluted tubule:

(e) Collecting duct:

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3. Explain the importance of the salt gradient in the kidney:

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2. Why are the nephrons all aligned in the same way in the kidney?

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97 Malfunctions of the Excretory System urine). The symptoms of urinary infection include pain while urinating, soreness in the lower back, or an abnormal volume or composition or urine. In some cases, kidney malfunction may be the end result of a more widespread problem (e.g. kidney failure can result from chronic high blood pressure). Most malfunctions can be overcome, although treatments may be difficult and painful. Some, like kidney dialysis, are costly, continual, and very limiting. Others, like transplants, may offer a better long term solution.

NIH

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Key Idea: Malfunctions of the excretory system range from life threatening problems, such as kidney failure, to more minor problems, such as kidney stones and incontinence. Malfunctions of the excretory system can range from relatively harmless and inconvenient conditions, e.g. leakage of urine, to life threatening conditions, e.g. kidney disease. Changes to urine composition or volume may indicate a problem with the kidneys or bladder, or a metabolic disorder (e.g. diabetes mellitus involves excessive production of high-glucose

A kidney stone (renal calculus) forms as crystalline deposits of calcium and other salts. Kidney stones can be left in place with few ill effects, but they can be also be disrupted and broken up using high frequency sound waves.

Incontinence is a tendency for the bladder to leak urine. Stress incontinence can occur when the muscles of the pelvic floor are weakened, e.g. after childbirth, allowing the urethra to open. Incontinence can also occur with loss of mental capability (e.g. dementia).

Urinalysis is the analysis of urine to diagnose disease. Normal, fresh urine is pale to dark yellow or amber in colour and clear. Cloudiness or a red or red-brown colour may indicate signs of kidney disease or failure.

Kidney failure

Kidney failure

Kidney failure may be acute or chronic. Acute kidney failure occurs when the kidneys suddenly stop functioning properly. Causes include dehydration, sudden illness or infection, or trauma. Chronic kidney failure is the progressive loss of kidney function over a period of months or years. Causes often include untreated diabetes or high blood pressure, or kidney disease.

The consequences of kidney failure

When the kidneys fail, toxins accumulate in the blood, and electrolyte and fluid levels become unbalanced. Kidney failure results in anaemia, low bone density (calcium is leached from the bones to restore blood calcium), cardiovascular failure, and nerve damage.

Patient undergoing kidney dialysis. The blood is run through a dialysis machine, which takes over the function of the kidney, removing wastes and excess water.

End-stage kidney failure, when 90% of kidney function is lost, is the result of untreated kidney failure. Without appropriate treatment, e.g. dialysis or a kidney transplant, end-stage kidney failure is fatal.

(b) Identify three consequences of kidney failure:

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2. (a) Distinguish between acute and chronic kidney failure:

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1. Identify some symptoms of a urinary infection:

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98 Circulation and Respiratory Interaction

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Key Idea: The circulatory and respiratory systems interact to provide the body's tissues with oxygen and remove carbon dioxide.

Circulatory system

Respiratory system

Interaction between systems In mammals, the respiratory system and cardiovascular system interact to supply oxygen and remove carbon dioxide from the body.

Function

Function Provides surface for gas exchange. Moves fresh air into and stale air out of the body.

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Delivers oxygen (O2) and nutrients to all cells and tissues. Removes carbon dioxide (CO2) and other waste products of metabolism. CO2 is transported to the lungs.

Components

Airways: • Pharynx • Larynx • Trachea Lungs: • Bronchi • Bronchioles • Alveoli Diaphragm

Components

Heart Blood vessels: • Arteries • Veins • Capillaries Blood

Head and upper body

Bronchiole

Oxygen (O2) from inhaled air moves from the lungs into the circulatory system and is transported within red blood cells to the heart. The heart pumps the blood to the body where O2 is released and carbon dioxide (CO2) is picked up. The blood returns to the heart and is pumped to the lungs where CO2 is released into the lungs to be breathed out.

Lung

Heart

Lung

Capillaries

Lower body

From the heart to the lungs

The airways of the lungs end at the alveoli, the microscopic air sacs that enable gas exchange.

CO2

Red blood cells are replenished with oxygen from the alveolus and carbon dioxide is released from the blood into the alveolus.

Red blood cell

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87

The respiratory system and the circulatory system come together at the gas exchange membrane formed by the junction of the alveolus and the capillary wall. Oxygen and carbon dioxide diffuse easily across this thin barrier.

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O2

The carbon dioxide released from the blood exits the body during exhalation. Inhalation brings in fresh air, containing oxygen.

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O2

Capillary

From the lungs to the heart

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CO2

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Haemoglobin

Lung

In emphysemic lungs, normal tissue has been replaced by large spaces (circled)

The response to exercise illustrates the close link between the circulatory and respiratory systems. During exercise, breathing rate increases to provide more oxygen, which is carried by the blood to supply respiration and ATP generation in working muscles. Heart rate increases to increase the rate at which oxygen is delivered to the tissues and carbon dioxide is returned to the lungs.

Oxygen is transported in red blood cells by the protein haemoglobin. In the capillaries of the lungs (high oxygen), haemoglobin binds oxygen tightly. In the tissues, higher carbon dioxide levels cause haemoglobin to release its oxygen. Carbon dioxide is carried in the blood as bicarbonate (CO2 + H2O → H2CO3 → H+ + HCO3-). In the lungs, this dissociates back into carbon dioxide and water.

James Heilman, MD cc 3.0

Lung

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133

Diseases of the respiratory system affect the ability of the circulatory system to supply oxygen to the body's cells and remove CO2. For example, emphysema (shown in X-ray of chest cavity above) reduces the ability of the lungs to provide enough oxygen to meet needs. This leaves a person short of breath and with very little energy because oxygen delivery to the tissues is inadequate.

1. (a) What happens to the rate of blood flow during exercise?

(b) What happens to the breathing rate during exercise?

(c) How do the circulatory and respiratory systems interact to accommodate the extra oxygen requirements of an exercising person?

2. Explain how a lung disease affects the body, including the circulatory system:

3. (a) At which point in the body do the respiratory and circulatory systems directly interact?

(b) Explain what is happening at this point:

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4. In your own words, describe how the circulatory system and respiratory system work together to provide the body with oxygen and remove carbon dioxide:


99 Circulation and Digestive Interaction

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134

Key Idea: The circulatory and digestive systems interact to provide the body's tissues with nutrients.

Circulatory system

Digestive system

Interaction between systems

Function

Function Digest food and absorb useful molecules from it, and eliminate undigested material.

Components

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Delivers oxygen (O2) and nutrients to all cells and tissues. Removes carbon dioxide (CO2) and other waste products of metabolism.

In mammals, the digestive system and cardiovascular system interact to supply nutrients to the body.

Mouth and pharynx Oesophagus Stomach Liver and gall bladder (accessory organs) Pancreas (accessory organ) Small intestine Large intestine

Components

Heart Blood vessels: • Arteries • Veins • Capillaries Blood

Hepatic vein

Food digested in the stomach and small intestine and then absorbed is passed to the circulatory system. The capillaries around the stomach and intestines collect nutrients and then drain to the hepatic portal vein, which carries the blood directly to the liver. The liver then processes this nutrient rich blood, e.g. glucose is stored as glycogen. The hepatic vein then transports nutrients from the liver to supply the other tissues of the body.

Liver

Stomach

Villus

Capillaries

Hepatic portal vein

Small intestine

To

Intestinal epithelial cell

O2

Glucose

Capillary

Villi project into the lumen of the small intestine and absorb nutrients. Villi contain capillary networks which receive the nutrients and transport them to the hepatic portal system.

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CO2

at hep ic por tal vein

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Glucose and other nutrient molecules are passed to the blood and transported to other parts of the body. Oxygen passes to the intestinal cells, while carbon dioxide passes into the blood.

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Cirrhosis of the liver

Blood flow to the digestive tract increases steadily after a meal and remain elevated for about 2.5 hours, reaching a maximum after about 30 minutes. During exercise, blood flow in the digestive tract is reduced as it is redirected to the muscles.

Nutrients, e.g. minerals, sugars, and amino acids, are transported in the blood plasma to the liver. The liver receives nutrient-rich deoxygenated blood from the digestive system via the hepatic portal vein and oxygen rich blood from the hepatic artery.

CDC

Human liver

Suseno

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135

Scarring of the liver tissue, or cirrhosis, can result in portal hypertension (high blood pressure). The scarred tissue obstructs blood flow in the liver. This causes pressure to build up in upstream blood vessels, resulting in swelling and possible haemorrhage.

1. How are nutrients transported in the blood?

2. Explain how a liver cirrhosis affects the circulatory system:

3. (a) At which two points in the body do the digestive and circulatory systems directly interact?

(b) Explain what is happening at these points:

4. (a) What happens to blood flow to the digestive tract after a meal?

(b) Explain why it is often recommended that a person should exercise within 2.5 hours of eating, or eat within half an hour of exercising to gain a most benefit from the exercise (in terms of muscle development):

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5. In your own words, describe how the circulatory and digestive systems work together to provide the body with nutrients:


100 Circulation and Excretory Interaction

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136

Key Idea: The circulatory and urinary systems interact to remove wastes from the body's tissues and helps maintain blood volume and pressure.

Circulatory system

Function

In mammals, the urinary system and cardiovascular system interact to remove metabolic wastes from the body.

Function Filters blood, retaining useful molecules and removing harmful ones. Regulates blood volume and ion content.

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Delivers oxygen (O2) and nutrients to all cells and tissues. Removes carbon dioxide (CO2) and other waste products of metabolism.

Urinary system

Interaction between systems

Components

Components

Heart Blood vessels: • Arteries • Veins • Capillaries Blood

Kidneys Ureters Bladder Urethra

Metabolic wastes generated in the cells move into the blood plasma and are transported to the kidneys. Blood is forced at high pressure through the capillaries of the glomerulus, producing a filtrate, which is collected in the surrounding Bowman's capsule. Both useful and harmful molecules are contained in the filtrate although large proteins and blood cells are excluded.

Glucose

Artery

Vein

Capillaries

H2O

H+

Toxin

Convoluted tubule lumen

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Epithelial cell

LINK

93

Capillary

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Na+

CL

Cl-

The convoluted tubules through which the filtrate travels are surrounded by capillaries (above). As the filtrate moves along the tubules, epithelial cells reabsorb useful molecules or ions, e.g. glucose and sodium ions, back into the blood in the capillaries (left). Unwanted ions (e.g. H+) and toxins are also secreted into the filtrate by active transport and eliminated in the urine.

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Sodium chloride crystals

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Polycystic kidney disease

The kidneys regulate blood volume by regulating the ion content of the blood. Retention of ions help retain water (via osmosis). Various hormones carried in the blood stimulate the kidneys to increase water retention.

The regulation of blood volume is important for the body. Optimal blood volume enables the blood to flow at the correct rate through the capillaries, helps maintain the correct concentrations of electrolytes, and maintains the pressure needed for glomerular filtration.

Chronic kidney disease makes it difficult for the body to regulate blood volume and this can promote hypertension. Hypertension can lead to atherosclerosis of the arteries and in turn heart disease, heart attack, or peripheral arterial disease.

1. (a) Explain why regulating salt content of the blood in turn regulates blood volume:

(b) How would the active secretion of hydrogen ions into the filtrate (urine) help to regulate blood pH?

2. (a) Explain why kidney disease can make regulating blood volume and pressure difficult:

(b) Explain why kidney disease can also lead to heart disease and heart attacks:

3. Use the graph right to answer the following questions:

(a) What is the effect of reducing glomerular filtration rate on the risk of cardiovascular disease?

(b) How many times greater is the risk of cardiovascular disease at a GFR of <15 compared to a GFR of >60?

(c) What might cause reduced GFR?

40

Rate of cardiovascular event per 100 person years

Risk of cardiovascular disease vs glomerular filtration rate

30

21.8

20

11.29

10

2.11

0

3.65

>60 45-59 30-44 15-29 <15 Estimated glomerular filtration rate (GFR) (mL min-1 1.73 m2)

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4. In your own words, describe how the circulatory system and urinary system work together to remove wastes and regulate the volume, pressure, and composition of the blood:

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101 KEY TERMS: Did You Get It?

1. (a) What process moves food through the gut?

(b) In what region of the digestive system does most nutrient absorption occur?

(c) What structures from the small intestine of a mammal are shown in the photograph (right)?

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(d) What is their function?

EII

2. (a) What type of blood vessel transports blood away from the heart?

(b) What type of blood vessel transports blood to the heart?

(c) What type of blood vessel enables exchanges between the blood and tissues?

3. (a) What component of blood is involved in carrying oxygen?

(b) What cell types in blood are involved in defence against pathogens? 4. (a) Name the excretory organ of mammals:

(b) Name its selective filtering element:

5. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code. alveoli

A A process involving forcing fluid under pressure through capillaries to form a filtrate.

atrium

B A collection of one of more specialised cells and extracellular material that perform a specific role in the body, e.g. support or lining.

blood

C The exchange of oxygen and carbon dioxide across the gas exchange membrane.

breathing

D The act of inhaling air into and exhaling air from the lungs.

bronchi

E Internal gas exchange structures found in all vertebrates except fish.

excretion

F Any gas that takes part in the respiratory process.

gas exchange glomerulus heart

G The process by which metabolic wastes are removed from the body.

H A functional structure, often made up of several tissues, which forms a specific role in the body. I Microscopic structures in vertebrate lungs, which arise at the terminus of the smallest bronchioles. The site of gas exchange.

J A liquid connective tissue made up of numerous cell types within a fluid matrix, which acts as a circulatory fluid transporting respiratory gases, nutrients, and wastes.

organ

K Chamber of the heart that pumps blood into the arteries.

respiratory gas

L The knot of capillaries forming the first part of a nephron and where the initial filtrate is produced.

tissue

M Chamber of the heart that receives blood from the veins.

ultrafiltration

N Large air tubes that branch from the trachea to enter the lungs.

ventricle

O The central organ of the circulatory system. It acts as a double pump to transport blood around the body.

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lungs

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102 Review: Unit 1 Area of Study 1

Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered:

Crossing the plasma membrane

HINT: Diffusion, osmosis, and active transport

Cell size, structure, and function

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HINT: Prokaryote vs eukaryote cells, cell size, organelles, and microscopy

Energy transformations

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HINT: Photosynthesis and cellular respiration, and the role of ATP

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140 Functioning systems in plants:

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HINT: How is water transported in plants

Functioning systems in mammals:

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HINT: Relationships between the respiratory, digestive, circulatory, and urinary systems.

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103 Synoptic Question: Unit 1 Area of Study 1

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1. Plant and animal cells are eukaryotic cells, and share several common features. They also have features unique to each cell type. Annotate the following diagrams to compare and contrast the features of plant and animal cells and their functions. You may use extra paper if required. In your answer you should: • Identify each cell type • Identify and describe features common and unique to each • Briefly state the functions of the structures and organelles labelled

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2. Describe the characteristics of the plasma membrane and discuss how substances are transported across it. Include simple diffusion, facilitated diffusion, osmosis, and active transport. You may use extra paper if required.

TEST


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3. The many organ systems of the body are constantly interacting. For a named organ system describe the structure of the organ system with reference to its cells, tissues, and organs. Explain how the organ system interacts with another named organ system, and the biological consequences of a system malfunction.

4. In each of the regions shown below use the spaces to describe how the cells of a plant are specialised for the uptake, transport and loss of water. Drawings may help your description. Leaves

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Roots

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Stems

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Survival via adaptation and regulation

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Unit 1 Outcome 2

Key terms

Adaptations and survival

adaptation

Key knowledge

antidiuretic hormone (ADH)

1

Explain what is meant by an adaptation and give some examples. Recognise adaptations as the result of evolution in species, not changes made to individuals during their lifetimes.

c

2

Distinguish between structural (morphological), physiological, and behavioural adaptations and explain how they enhance survival and enable life to exist in a range of different environments. Describe specific examples of adaptations in some named species.

105-111

c

3

Describe examples of how unrelated organisms can appear similar because of their similar adaptations to environment.

112

diabetes mellitus effector

glucagon

homeostasis

104

c

behavioural adaptation biomimicry

Activity number

hyperthermia

hyperthyroidism hypothalamus hypothermia insulin

Noel Elhardt

negative feedback

physiological adaptation positive feedback receptor

Adaptations as models for biomimicry

response

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stimulus

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thermoregulation

Explain how the adaptations of organisms have provided models for biomimicry. Describe examples of the use of biomimicry to solve human challenges, e.g. in engineering, electronics, architecture, and materials science.

Maintaining homeostasis

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Explain what is meant by homeostasis. Explain why organisms need to regulate their exchanges with the changing environment in order to maintain homeostasis and describe examples.

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Use the stimulus-response model to explain how organisms monitor and respond to their environment. Using an example, explain the role of feedback loops (particularly negative feedback) in homeostatic regulation.

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Homeostatic mechanisms in humans Key knowledge

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Describe examples of homeostatic regulation in humans, including the control of blood glucose, control of body temperature (thermoregulation), and control of water balance. For each, explain the role of receptors, effectors, and negative feedback in the homeostatic regulation.

Homeostatic failures in humans Key knowledge c

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Describe malfunctions to homeostatic mechanisms and their consequences to include type 1 diabetes and hyperthyroidism. Explain the malfunction in each case and how the normal negative feedback regulation is disrupted.

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structural adaptation

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104 What is Adaptation?

Key Idea: An adaptation is any heritable trait that equips an organism for its functional role in the environment (its niche). An adaptation is any heritable characteristic (trait) that equips an organism for its niche, enhancing its exploitation of the environment and contributing to its survival and

Microbial digestion in the expanded fore-stomach improves energy gain from poor forage. Kangaroos can survive without water if food is green.

Robust, high crowned molar teeth. The molars are replaced as they wear down as an adaptation to a diet of abrasive grasses.

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The red kangaroo is the largest living marsupial. It is powerfully built and adapted for high speed, hopping locomotion, and survival in dry habitats. They are active mainly at night, roaming in small groups over a home range of 8 km2 (or larger when resources are scarce). Dominant males establish exclusive mating rights by boxing.

successful reproduction (fitness). Adaptations may be structural (morphological), physiological, or behavioural. The adaptations of species are the result of evolution in particular environments. Traits that are not helpful to survival and reproduction will not be favoured and will be lost.

Dense, fine fur provides insulation against excessive heat loss or gain. Fur is reflective, especially on the flanks. Thin skin well supplied with blood vessels, especially on the forelimbs, to assist heat loss by evaporation.

Females may breed all year. They may have a joey at heel, one in the pouch, and a dormant embryo ready to replace the pouch offspring as soon as it leaves. In unfavourable conditions, embryos can be reabsorbed by the mother if resources are scarce.

Licking the pads of the front paws assists in cooling by evaporation.

Hind limbs much more heavily muscled and longer than the forelimbs. High speed hopping achieves speeds of up to 50 km h–1.

Stout, tapering tail acts as a fifth limb in slow five-point movement. In bipedal hopping the tail acts as a counterweight.

Long foot bones help balance. The second and third digits are fused and the fourth digit is much larger and longer than the others. Strongly developed, elongated, load-bearing heel.

Conserving energy

When a kangaroo hops at high speed (left), it stores energy in its elastic tendons and uses that energy to help power its next jump. This allows for energy savings of about 30%. When the animal's foot hits the ground, the tendon is stretched. When the foot leaves the ground, elastic recoil provides additional energy for a more forceful jump. This adaptation is the inspiration for an energy efficient robot called the bionic kangaroo. The technology can be used to build energy efficient machines capable of storing and reusing energy.

1. (a) What is an adaptation?

(b) How do adaptations contribute to an organism's fitness?

(a) Structural adaptation:

(b) Physiological adaptation:

(c) Behavioural adaptation:

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2. For the red kangaroo, describe a structural, physiological, and behavioural adaptation. For each example, explain how the adaptation assists the kangaroo to survive in its environment:

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105 Adaptations of Xerophytes

Key Idea: Xerophytes are plants with adaptations that allow them to conserve water and survive in dry environments. Plants adapted to dry conditions are called xerophytes. Xerophytes are found in a number of environments, but all

show adaptations to conserve water. These adaptations include small, hard leaves, an epidermis with a thick cuticle, sunken stomata, succulence, and permanent or temporary absence of leaves.

Most xerophytes are found in deserts, but they may be found in humid environments, provided that their roots are in dry micro-environments (e.g. the roots of epiphytic plants that grow on tree trunks or branches).

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Many xerophytes have a succulent morphology. Their stems are often thickened and retain a large amount of water in the tissues, e.g. Aloe. Many xerophytes have a low surface area to volume ratio, reducing the amount of water lost through transpiration.

Salt tolerant plants and many alpine species may show xeromorphic features in response to the lack of free water and high transpirational losses in these often windy, exposed environments.

Desert xerophytes in Arizona have a succulent photosynthetic stem

Adaptations in cacti

Hairs

Desert plants, such as cacti (below), must cope with low or sporadic rainfall and high transpiration rates. Leaves modified into spines or hairs to reduce water loss. Light coloured spines reflect solar radiation.

Rounded shape reduces surface area.

Acacia trees have deep root systems, allowing them to draw water from sources deep underground.

An outer surface coated in fine hairs traps air close to the surface and reduces the transpiration rate.

Stem becomes the major photosynthetic organ, plus a reservoir for water storage.

The surface tissues of many cacti are tolerant of temperatures in excess of 50°C.

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Trichome (hair)

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Cacti have a shallow, but extensive fibrous root system. When in the ground the roots are spread out around the plant.

Stoma

Stoma

Grasses on coastal sand dunes (e.g. marram grass, above) curl their leaves. Stomata are sunken in pits, creating a moist microclimate around the pore, which reduces transpiration rate.

Oleander has a thick multi-layered epidermis and the stomata are sunken in trichome-filled pits on the leaf underside which restrict water loss.

1. What is a xeromorphic adaptation?

2. Describe three xeromorphic adaptations of plants that reduce water loss:

(b) (c)

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3. How does creating a moist microclimate around the areas of water loss reduce the transpiration rate?

4. How does a low surface area to volume ratio in a plant such as a cactus reduce water loss?

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106 Adaptations of Hydrophytes

Key Idea: Hydrophytes are adapted to living in water. They require little structural support tissue because they are

The leaves of submerged plants are thin to increase the surface area of photosynthetic tissue.

supported by the denser medium of water and have few adaptations to reduce water loss. Leaves arranged so that they do not shade each other.

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Flowers are above water so still accessible to pollinators.

Many hydrophytes have large floating leaves with elongated leaf stalks (petioles).

Cross section through the petiole

Hydrophytes are supported by the water, so they need very little structural support tissue. There are large air spaces in the stems. There is no lignin (strengthening material) in the vascular tissue.

Water milfoil Myriophyllum spicatum

Hydrophytes have a reduced root system. This is related to the relatively high concentration of nutrients in the sediment and the plant’s ability to remove nitrogen and phosphorus directly from the water.

Vascular bundles

The water lily Nymphaea alba

Cortex

Abundant, large air spaces

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Air spaces

Myriophyllum’s submerged leaves are well spaced and taper to the surface to help with gas exchange and distribution of sunlight.

The floating leaves of water lilies (Nymphaea) have a high density of stomata on the upper leaf surface so they are not blocked by water.

Cross section through Potamogeton, showing large air spaces which assist with flotation and gas exchange.

(b) Thin cuticle:

(c) High stomatal densities on the upper leaf surface:

2. Why do hydrophytic plants retain an aerial (above water) flowering system?

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1. Explain how the following adaptations assist hydrophytes to survive in an aquatic environment: (a) Large air spaces within the plant's tissues:

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107 Mangrove Adaptations

Key Idea: Mangroves are salt tolerant (halophytes) and specifically adapted to the high salt, water-logged environments of estuaries, tidal flats, and salt marshes. Mangroves are halophytes, a group of plants with adaptations

for growth in seawater or salty, water-logged soil. They grow in the upper part of the intertidal zone, but also extend further inland to form salt marshes and other coastal wetland communities. Australia has 36 species of mangroves.

Mangroves grow from the upper part of the intertidal zone to the high water mark, forming some of the most complex and productive ecosystems on Earth. The high salt environment would kill most other kinds of plants as high salt levels cause water to flow out of the cells. Mangroves overcome this by storing salt in their cell vacuoles and maintaining a high concentration of solutes in the cytoplasm of their cells. This reverses the osmotic gradient and maintains the transpiration stream.

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Salt crystals

Pneumatophores are specialised "breathing" roots that grow 25-30 cm above the mud surface. They allow the mangrove to obtain oxygen. They are composed of spongy tissue with numerous air spaces. Oxygen enters the pneumatophores through lenticels (pits) in the waterproof bark. It diffuses through the spongy tissue to the rest of the plant.

Salt may be secreted through salt glands in the surface layer of the leaves or stored in older leaves before they fall.

Oxygen

Lenticels

Water level at high tide

A waxy coating of suberin on the root cells excludes 97% of salt from the water taken up by the roots.

Only the top few centimetres of the mud contains oxygen. Beneath, the mud is anaerobic (lacking oxygen), black, and foul-smelling. A deep root system is of no use here.

Cable roots radiate from the trunk, about 20-30 cm below the surface. Growing off these radial roots are fine feedingroots (not shown), which create a stable platform.

Prop roots that descend from the trunk act like buttresses, providing additional support for the tree in the soft mud and supplement the oxygen uptake from the pneumatophores.

The mangrove propagule is a partially developed seedling adapted for dispersal in water. It is able to quickly take root once it reaches a suitable site.

1. What two physical adaptations of mangroves provide support for the plant in the soft mud? (a) (b)

3. Describe a physiological problem associated with living in a high-salt substrate:

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2. Which adaptation enables mangroves to tolerate anaerobic conditions in the substrate?

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4. Describe three methods by which various mangrove species solve the problem of a high salt environment: (a)

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108 Plant Adaptations to Fire fire, or leave offspring that will germinate after the fire. In general, plants cope with fire in two ways. Some re-sprout from protected buds after above-ground parts have been burned away. Other plants may die completely, relying on the germination of fire resistant seeds to recolonise.

Capsule

Epicormic regrowth from eucalypt

Some Australian eucalypt trees have epicormic buds. These are dormant buds protected from fire because they lie deep beneath the thick bark. They sprout after the fire, allowing the vegetative regeneration of branches from their trunks.

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Key Idea: Some Australian plants have adaptations that allow them to survive and even benefit from episodes of fire. Some Australian plants have adaptations associated with surviving and even benefiting from fires. Species living in ecosystems prone to fire must be able to survive the

A lignotuber

A lignotuber is a woody swelling of the root crown. It protects the plant stem from being destroyed by fire. It is safe from fire and can store water and food to support regrowth. The same lignotuber can survive many fires.

The seeds of eucalypts are often held in an insulated capsule, which opens only in response to heat of the fire. The smoke from the fire stimulates the seeds of annual plants, which may have been dormant for many years, to germinate.

New growth: After a fire, new growth emerges from the top of the stem.

Fire tolerant grasses, (such as the one below) are burnt in bush fires, but quickly regenerate so long as they are not heavily grazed. After the fire, new growth emerges from the stubble. The growing tips are protected during a fire by soil or the clumped nature of the grass.

Dead leaves: Dead leaves form a skirt at the base of the crown. These burn rapidly and fiercely but do not damage the growing tip in the crown.

New growth

Fire-blackened stem: Although it looks damaged, the fibrous stem provides special protection from the fire and the important vascular tissue survives well. The plant sheds its leaves each year, but the bases remain attached to the stem and produce a thick gum that glues the whole lot together into a very effective fire guard.

Burnt stubble: Old, dry leaves are burnt off easily during the fire, but the intensity of the flame is not great and does not damage the growing tissue.

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1. How do lignotubers and epicormic buds aid survival of fire-tolerant species?

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2. Bush fires are natural events, but humans have increased the frequency of fire occurrence when clearing land. This has meant that species that would usually mature and set seed before the next fire do not have the chance to do so. What would happen to these seed-producing species if fire frequency in an area increased markedly?

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109 Adaptations of Insectivorous Plants

Key Idea: Insectivorous plants capture and digest small invertebrates to meet their nitrogen requirements. Plants that live in acid bogs or in nutrient-poor soils often obtain extra nutrients (particularly nitrogen) by capturing and digesting small invertebrates. These plants are called insectivorous (or sometimes carnivorous). They The leaf curls up to form a temporary stomach in which digestion occurs

Sundew (left): Sundews (Drosera) capture their prey in the sticky secretions on their leaf hairs. These make a trap like flypaper. The hairs bend over the prey and restrain it and the entire leaf cups around to enclose the prey. Sundews are found throughout most of Australia, apart from the arid centre.

Insects are attracted to the pitcher's colourful and prominent lip region by sweet secretions just over the rim.

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Long hairs tipped with sticky droplets

photosynthesise to make their own sugars, but obtain some of their nitrogen and minerals from animal tissues. The traps are leaf modifications and usually contain special glands that secrete digestive enzymes. Insectivorous plants are usually small because their nutrient poor environment does not support the growth of large plants.

Insects climb over the lip to find themselves on a nearly vertical surface made slippery by waxy secretions. This causes them to fall into the digestive fluid below.

Sundew

Each leaf has a spring-like hinge of thin-walled cells down its midrib. When triggered, these cells rapidly lose water causing the two halves of the leaf to close together.

Spines line the edge of the leaf, creating a cage when the leaf folds together.

Gland cells line the lower part of the inside of the pitcher. They secrete digestive enzymes and are sometimes involved in the absorption of food.

Venus flytrap

Insects touch these trigger hairs on the leaf surface

Venus fly trap: The Venus fly trap consists of two, lobed modified leaves that can rapidly close together to trap prey (usually small insects). The trigger for closing is a touch on the sensory hairs of the leaves.

Pitcher plant

The digestive fluid that fills the lower part of the pitcher contains at least two potent, protein splitting enzymes. One is similar to pepsin, the enzyme found in vertebrate stomachs.

Pitcher plant: This plant is a passive trap. Prey fall into the water collected at the base of the pitcher and then drown. The digestive enzymes produced by the plant slowly digest the prey’s tissues. Pitcher plants grow in the tropical rainforests of Cape York peninsula.

1. (a) What does an insectivorous plant gain from digesting insects?

(b) How this is an advantage to the plant in its habitat?

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2. (a) Describe one cost (disadvantage) to the plant of producing insect-trapping modified leaves:

(b) Why are insectivorous plants not usually found growing in nutrient rich soils where ordinary plants are present?

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3. For one of the named plants above, describe the modifications to the leaf structure that enable capture of insects:

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110 Adaptations in Animals

Key Idea: Structural, physiological, and behavioural adaptations contribute to increased fitness. The adaptations of animals (like all organisms) are the product of evolution and contribute to their fitness (survival and successful reproduction). Animals must find food, avoid being eaten, and survive environmental extremes (e.g. cold,

wet, dry). The adaptations of two very different taxa, butterflies and lace monitors (goannas), illustrate the range and diversity of animal adaptations, both to physical environment and to niche. Butterflies are a globally successful group of insects adapted to a wide range of conditions. Goannas are highly successful top predators, often in harsh environments.

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Butterfly adaptations

Basking: Butterflies must bask in the sun to raise their body temperature so that they can fly. Heat capture is maximised by holding their wings outstretched and angled towards the sun to increase energy capture. Their wings are often dark coloured because dark colours absorb more of the sun's energy than light colours.

Damaged wing Eye spot

Fright and distraction: Eye spots on the wings of some butterflies can trick predators into thinking it is a much larger animal, and deter an attack. Eye spots are often located at the posterior of the animal, so if the butterfly is attacked, the head is not targeted. Damage to the wings is unlikely to be fatal, whereas an attack to the head is likely to be fatal.

Surviving winter: Butterfly species have a number of strategies to survive low winter temperatures. Strategies include dormancy in adult butterflies to save energy, or laying eggs in late summer or autumn so that eggs hatch in spring when food is plentiful. Some species overwinter as caterpillars or a chrysalis insulated within vegetation or underground.

Viceroy

Camouflage: Some butterfly species use camouflage to blend into their surroundings. Colours and patterns on the wings make it difficult for their predators to see them, allowing the butterfly to avoid being eaten while at rest or feeding.

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Monarch

Mimicry: Some species do not try to hide and are very brightly coloured. This colouration is a warning to predators that they are poisonous (e.g. monarch butterfly) and should not be eaten. Other species take advantage of this and mimic the warning colours. For example, the viceroy butterfly is not poisonous, but survives by mimicking the Monarch's colouration. Predators treat it as poisonous and leave it alone. Some species mimic other animals such as wasps and snakes.

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1. Some butterflies use camouflage as a passive defence against being eaten, while others, such as the monarch butterfly are very brightly coloured. Briefly describe how these two very different adaptations aid survival.

(a) What is a disadvantage of this?

(b) What adaptations do butterflies have to overcome this disadvantage?

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2. Butterflies must use the Sun's energy to raise their body temperature so that they can fly.

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Adaptations of goannas or monitor lizards (Varanus spp.) Upper jaw can move independently of the rest of the skull to help swallow prey whole. Sharp, recurved teeth help hold and handle prey.

The skin of species in arid regions is highly reflective to reflect heat and help thermoregulate.

Strong muscular tail used in defence. The base of the tail may become thickened as a fat store to provide energy.

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Strong neck and jaw muscles aid holding, shaking, and subduing prey. They are strict carnivores eating a range of available animal species, including carrion.

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Forked tongue collects scent particles which are delivered to the Jacobson’s chemosensory organ on the roof of the mouth.

Asian water monitor

Long, sharp claws are used for tearing at prey

Monitors are active during the day in all seasons, maintaining high body temperatures (up to 38°C) through basking. Gular pumping (using throat movements to pump air in and out of the lungs) helps ventilate the lungs during activity so high activity can be maintained.

Water monitor basking

Goannas are found in a wide variety of habitats, from aquatic to arid semi-desert. In hot environments, gaping and panting, accompanied by movements of the throat region (gular movements) when the mouth is open, aid evaporative cooling by increasing convective heat loss.

Gould's monitor

Raised twolegged stance, accompanied by hissing deters attackers.

Nile monitor gaping with gular ventilation

3. Describe a structural, physiological, and behavioural adaptation of goannas. For each, explain how the adaptation assists survival in the environment and contributes to fitness:

(a) Structural adaptation:

(b) Physiological adaptation:

(c) Behavioural adaptation:

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4. The Australian marsupial mole (Notoryctes genus, right) is a burrowing mammal found only in Australia. They have evolved adaptations to aid their underground lifestyle. Research some of the adaptations of the marsupial mole and list them here. Make sure you categorise the adaptations as structural, physiological, or behavioural.


111 Adaptations for Diving

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from dives of 20 m or more brings the risk of decompression sickness (or the bends). Primates (including humans) are one of the few orders of mammals without diving representatives. The problems that humans encounter when diving while breathing compressed air (e.g. the bends and nitrogen narcosis) are the result of continuing to breathe during the dive. Animals adapted for diving do not do this.

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Key Idea: Diving animals have structural and physiological adaptations that allow them to stay submerged for extended periods of time. All air breathing animals that dive must cope with the problem of oxygen supply to the tissues. This is particularly a problem for mammals and birds, which have a high metabolic rate and therefore a high oxygen demand. In addition, resurfacing

Diving birds

Diving reptiles

Penguins show many of the adaptations typical of diving birds. During dives, their heart rate slows, and blood is diverted to the head, heart, and eyes.

Sea turtles such as the green sea turtle have low metabolic rates, and tolerate low oxygen. They use the lining of the mouth and the cloaca for gas exchange and can overwinter at 10-15 m depth for several months.

The problems for humans

Humans are poorly adapted for diving. They lack adequate body fat for long periods underwater and they inhale before diving. Snorkellers (above) and Asian pearl divers may hyperventilate (take a series of deep breaths in and out) before diving. This reduces the carbon dioxide content of the blood and suppresses the urge to breathe. Blood oxygen levels can fall dangerously low before the urge to breathe is felt. If the diver faints, voluntary control is lost and normal, involuntary breathing resumes, resulting in drowning. Divers using compressed air (SCUBA) may stay submerged for much longer. However, they must take care when ascending in order to equalise the pressure of their blood gases with those on the surface.

Diving mammals

Dolphins, whales, and seals are among the most well adapted divers. The exhale before diving and, at depth, the lungs are compressed so that only the trachea contains air. This stops nitrogen entering the blood and prevents the bends when surfacing. During dives, heart rate slows and blood flow is redistributed to supply only critical organs. Diving mammals have high levels of myoglobin and their muscles work well without oxygen. Sperm whales are the deepest divers (recorded at 3000 m). Weddell seals dive to 1000 m for 40 minutes or more. During these dives, heart rate drops to 4 or 5 beats per minute (4% of the rate at the surface).

1. (a) Describe an advantage gained from breathing out before diving:

(b) Explain how this behaviour is different from a human diving (unaided by equipment):

(c) Describe the adaptive advantage in reducing heart rate during a dive:

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(a) Why does hyperventilation suppress the urge to breathe?

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2. Snorkel divers who wish to remain underwater for long periods of time hyperventilate before diving.

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(b) Why is hyperventilation a dangerous practice when diving?

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3. The ability to remain submerged for long periods of time depends on the ability to maintain the oxygen supply to the tissues. This depends on oxygen stores. The table below compares the amount of oxygen in different regions of the body during a dive in a small seal and a human (not on scuba). In the spaces provided in the table, calculate the amount of oxygen (in mL) per kilogram of bodyweight for both the seal and the human. Graph the result in the space provided.

Location of oxygen in the body

Seal (30 kg)

Amount of oxygen (mL)

Human (70 kg)

Oxygen (mL kg-1)

Amount of oxygen (mL)

Alveolar air

55

720

Blood

1125

1000

Muscle

270

240

Tissue water

100

200

Total

1550

51.67

Oxygen (mL kg-1)

30.86

2160

Relative amounts of oxygen (mL kg-1 bodyweight) in a seal and a human

Relative amount of oxygen (mL kg -1)

60

Seal

50

Human

40 30 20 10

0

Alveolar air

Blood

Muscle

Tissue water

Total

Location of oxygen

(b) With respect to diving adaptations, suggest why this is the case:

5. (a) Where does the seal store a large proportion of its oxygen during a dive?

(b) How does this compare to a human?

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4. (a) Describe the most striking difference between a seal and a human in terms of the oxygen stores during a dive:


112 Similar Environments, Similar Adaptations

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Key Idea: Unrelated species often evolve similar adaptations to overcome the same environmental challenges. Sometimes, genetically unrelated organisms evolve similar adaptations in response to the particular environmental

challenges they face. The adaptations may result in different species with a very similar appearance. Although the organisms are not closely related, evolution has produced similar solutions in order to solve similar ecological problems.

Similar adaptations in unrelated animals

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Similar adaptations in unrelated plants

Cactus

Euphorbia

The North American cactus and African Euphorbia species shown above are both xerophytes. They have evolved similar structural adaptations to conserve water and survive in a hot, dry, desert environment. Although they have a similar appearance, they are not related. They provide an excellent illustration of how unrelated organisms living in the same environment have independently evolved the same adaptations to survive. Their appearance is so similar at first glance that the Euphorbia is often mistaken for a cactus. Both have thick stems to store water and both have lost the presence of obvious leaves. Instead, they have spines or thorns to conserve water (a leafy plant would quickly exhaust its water reserves because of losses via transpiration). In cacti, spines are highly modified leaves. In Euphorbia, the thorns are modified stalks. It is not until the two flower that their differences are obvious.

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Flying squirrel (rodent)

Colugo (related to lemurs)

The ability for mammals to glide between trees has evolved independently in unrelated animals. The characteristics listed below for the sugar glider are typically found in the gliding mammals shown here.

Sugar glider (marsupial)

Tail acts as a stabiliser and an air brake during flight.

The animals are nocturnal and large eyes help them see at night.

Skin is stretched between the front and back legs to form a wing like flap. This allows them to glide up to 50 m between trees.

By moving their limbs the animal has some control about the direction it flies in.

1. Explain why the North American cactus and African Euphorbia species have evolved such similar adaptations:

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2. Suggest why gliding between trees (rather than walking) is an advantage to the gliding mammals described above:

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3. The echidnas and hedgehogs are an example of unrelated organisms that have both evolved spines. Explain the advantage of this adaptation to both these animals and why might it have evolved?

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113 Biomimicry for Developing New Technologies

Key Idea: Biomimicry uses the structures and processes of the natural world as inspiration for new technologies. The inspiration for many new technologies comes from studying organisms, systems, and structures in nature. This is

called biomimicry. Studying organisms and natural process provides ideas on how to develop and improve technologies for human use. The applications are varied, ranging from energy production through to medicine.

Sharkskin biomimicry

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Sharkskin is rough and tough because it is covered in overlapping scales called dermal denticles (background image shows an EM of sharkskin). The structure of sharkskin has been the inspiration for several technological and medical innovations.

Performance swimwear Grooves running the length of each denticle allow water to slip past the fish, reducing hydrodynamic drag when swimming. This inspired the development of the controversial sharkskin swimsuit, which was used in the 2008 Summer Olympics, but is now banned from toplevel competition because of the competitive advantage provided by reduced drag.

Sharklet™ catheter tubing The structure of sharkskin prevents attachment and growth of microbes. It has inspired improvements in surgical catheters, which are a common source of infections in hospitals. Sharklet™ tubing has a textured surface, which resists bacterial colonisation within the tube, reducing infection rates and improving patient outcomes.

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Antifouling coatings In the marine industry, an elastic silicone product mimicking sharkskin is painted onto the hull of boats to improve hydrodynamics, allowing ships to move through the water more efficiently. The product also prevents the growth of marine organisms without the damaging environmental effects of traditional chemical antifouling agents.

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Peacock butterfly

Many butterflies have wings that appear to change colour in flight, a property known as iridescence. Butterfly wings are covered in thousands of tiny scales, each with two or three layers of their own. When light hits the different layers, it is reflected back many times producing a bright, intense colour. This arrangement has been mimicked in electronic displays to improve the quality. Thousands of microscopic mirrors beneath the surface of a monitor's glass take the white light and reflect back a colour image. The intensity changes with the level of light.

The initial bite of a mosquito is quite painless. This is because their mouthparts are made up of several moving parts that are able to pierce the skin with minimal pain. Japanese researchers have developed a medical needle based on the mosquito mouthpart. The needle uses pressure to stabilise and pierce the skin painlessly. In the UK, a neuroprobe (used in brain surgery) has been developed using a similar design. It requires less force to manipulate it, so reduces the chances of accidently causing damage during surgery.

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Geckos have rows of tiny hairs on their feet allowing them to walk on smooth surfaces. Each hair has 1000 microscopic pads at the tip, each generating a van der Waals force that adheres the gecko to the surface. Although the individual forces are weak, collectively they are very strong, and a gecko can hold its entire weight vertically using only a single toe. This ability has inspired the design of powerful adhesives which have the potential to be used under water, in space, or as a replacement for sutures and staples in hospitals.

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New technology The Japanese bullet train can reach speeds of 300 km h-1. However, a downside of the speed is that the train produces a loud sonic boom when it exits a tunnel. Engineers had been trying to reduce the noise as it disturbs people living near the tracks. They looked to nature for a solution and noticed that the pointed beaks of kingfisher birds created very little splash when they dived into water after prey. When this shape was applied to the front of the newer bullet trains, they no longer produced a boom. Air flowed past the train instead of being trapped (and pushed) in front of it.

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Inspiration

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Termite mounds maintain their temperature within 1°C even though external temperatures vary greatly. The physical mass of the mound insulates against temperature changes. A shaft and vent system, which can be opened and closed, allows warm air to escape and draws cooler air in. Termite mounds were the inspiration for the energy efficient Eastgate Centre in Zimbabwe. Outside air drawn into the building is either warmed or cooled by the building mass depending on which is hotter. An extensive duct system moves air through the building. The building uses only 35% of the energy required by a conventional building. Velcro is a famous example of biomimicry. Swiss engineer George de Mestral noticed burdock burrs were stuck very tightly to his dog's coat. Looking at them under a microscope he saw that the burrs had many tiny hooks allowing them to stick tightly to the fur. He wondered if this could be replicated to make a "zipperless" fastener. His research resulted in velcro, two strips of fabric, one covered in thousands of tiny hooks and the other with thousands of tiny loops. The two strips can be pressed together to hold firmly, but are also easy to separate, and so are reusable.

Velcro viewed under a microscope

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The surface of a lotus leaf has a nanoscopic, textured, waxy construction. The structure makes the leaves highly water repellent and, as water bounces and rolls off the leaf, it collects and removes solid particles, such as dust, cleaning the leaf. This "superhydrophobicity", or "lotus effect", has the potential for use in self-cleaning products, such as paints, roof tiles, coatings, and fabrics. A major advantage of such a technology would be a substantial reduction in the use of chemical cleaners and water used for cleaning.

Lotus leaf

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(a) Aeroplane wing:

(b) Sonar navigation:

(c) Computer printer:

(d) Solar cells:

(e) Glow sticks:

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1. Look at the following design and engineering ideas. Are they based on biomimicry? If your answer is yes, research and record what their natural inspiration was:

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114 Homeostasis

Key Idea: Homeostasis refers to the (relatively) constant physiological state of the body despite fluctuations in the external environment.

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Organisms maintain a relatively constant physiological state, called homeostasis, despite changes in their environment. Any change in the environment to which an organism responds is called a stimulus and, because environmental stimuli are not static, organisms must also adjust their behaviour and physiology constantly to maintain homeostasis. This requires the coordinated activity of the body's organ systems. Homeostatic mechanisms prevent deviations from the steady state and keep the body's internal conditions within strict limits. Deviations from these limits can be harmful. An example of homeostasis occurs when you exercise (right). Your body must keep your body temperature constant at about 37.0°C despite the increased heat generated by activity. Similarly, you must regulate blood sugar levels and blood pH, water and electrolyte balance, and blood pressure. Your body's organ systems carry out these tasks. To maintain homeostasis, the body must detect stimuli through receptors, process this sensory information, and respond to it appropriately via effectors. The responses provide new feedback to the receptor. These three components are illustrated below.

How homeostasis is maintained Muscles and glands

Sense organ (e.g. eye)

Receptor

Effector

Detects change and sends a message to the control centre.

Responds to the output from the control centre.

Brain and spinal cord

Control centre

Receives the message and coordinates a response. Sends an output message to an effector.

The analogy of a thermostat on a heater is a good way to understand how homeostasis is maintained. A heater has sensors (a receptor) to monitor room temperature. It also has a control centre to receive and process the data from the sensors. Depending on the data it receives, the control centre activates the effector (heating unit), switching it on or off. When the room is too cold, the heater switches on. When it is too hot, the heater switches off. This maintains a constant temperature.

1. What is homeostasis?

(a) Receptor:

(b) Control centre:

(c) Effector:

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2. What is the role of the following components in maintaining homeostasis:

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Key Idea: An organism's internal environment must be kept in balance in order maintain the essential processes of life. Many organ systems work together to maintain homeostasis. An organism must constantly regulate its internal environment in order to carry out essential life processes such as growth, tissue repair, and reproduction. Changes outside normal levels for too long can cause the body systems to malfunction,

The digestive system is responsible for the digestion (break down) and absorption of food. Ultimately, it provides the nutrients required by all body systems.

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The circulatory system distributes gases, nutrients, and other substances (e.g. hormones) around the body in the blood.

and can result in illness or death. Homeostasis relies on monitoring all the information received from the internal and external environment and coordinating appropriate responses. This often involves different interdependent organ systems working together to provide for the needs of all the body’s cells. This makes it possible for organisms to function effectively in often highly variable external environments.

Once food has been digested in the digestive system, the nutrients absorbed (taken up) into the blood of the circulatory system.

Unabsorbed digestive matter expelled from the digestive system as faeces.

Food provides the energy and nutrients needed to provide energy to all of the body's systems.

Heart

Urine is produced by the kidneys of the urinary system. It contains the waste products of metabolism.

The urinary system has multiple roles including disposal of nitrogencontaining and other waste products from the body, regulating ion balance, and controlling the volume and pressure of the blood.

The respiratory system exchanges respiratory gases with the environment, providing a supply of oxygen to the body's cells and removing carbon dioxide (a waste product of cellular respiration).

Cells

After absorption, nutrients are carried in the blood and delivered to cells all around the body (solid arrow). The incorporation of these nutrients into the tissues is called assimilation. Wastes (dashed arrow) move from the cells back into the blood and are transported and removed.

During physical activity, the body's demand for oxygen increases to meet the increased metabolic needs of the muscles. The rate of carbon dioxide production also increases. Breathing rate increases to supply sufficient oxygen to meet the higher demands. Physical activity generates heat. Without homeostatic mechanisms to dissipate the extra heat generated by muscular activity, the enzymes involved regulating metabolic pathways would become damaged and metabolism would slow or stop.

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Pyruvate dehydrogenase (above) is an enzyme involved in cellular respiration.

2. What can happen to the body's enzymes if temperature is not closely regulated?

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1. Why is maintaining a steady state important?

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3. Why do we say that the body systems work interdependently to maintain homeostasis?

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116 Negative Feedback

Key Idea: Negative feedback mechanisms detect departures from a desired set point and act to restore the steady state. Negative feedback is a regulatory mechanism that maintains the body's homeostasis by detecting deviations from a certain set point and acting to restore those set point conditions.

Negative feedback mechanisms act to dampen variations and so have a stabilising effect on biological systems. Most body systems achieve homeostasis through negative feedback. Body temperature, blood glucose levels, and blood pressure are all controlled by negative feedback mechanisms.

Negative feedback and control systems

Negative feedback in temperature regulation The diagram (left) shows how temperature is regulated by negative feedback mechanisms.

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Corrective mechanisms activated, e.g. sweating

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A stressor, e.g. exercise, takes the internal environment away from optimum.

Return to optimum

Stress, e.g. exercise generates excessive body heat

Stress is detected by receptors and corrective mechanisms (e.g. sweating or shivering) are activated.

Normal body temperature

Stress, e.g. cold weather causes excessive heat loss

Corrective mechanisms act to restore optimum conditions.

Corrective mechanisms activated, e.g. shivering

Negative feedback in calcium homeostasis

Negative feedback in stomach emptying

Blood calcium is regulated by several hormones, including parathyroid hormone (PTH). Low blood Ca2+ stimulates release of PTH. When blood Ca2+ is restored, PTH secretion stops.

Empty stomach. Stomach wall is relaxed.

ch receptors are de Stret act iva te

Low Ca2+

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PTH stimulates kidneys to reabsorb more calcium into the blood from the urine.

B

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Smooth muscle in the stomach wall contracts. Food is mixed and emptied from the stomach.

Normal Ca2+

Food is eaten

Release of PTH from the parathyroid glands

Food enters the stomach, stretching the stomach wall.

tre tch

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PTH stimulates release of calcium from bone

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(a) State the stimulus at A:

(b) Name the effector in this system: (c) What is the steady state for this example?

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State the response at B:

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2. On the diagram of stomach emptying:

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1. How do negative feedback mechanisms maintain homeostasis in a variable environment?

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117 Positive Feedback

Key Idea: Positive feedback mechanisms in biological systems amplify a response to achieve a particular outcome. Positive feedback mechanisms amplify (increase) a response in order to achieve a particular result. Examples of positive feedback include fruit ripening, fever, blood clotting, childbirth (labour) and lactation (production of milk). A positive feedback

Positive feedback and childbirth

mechanism stops when the end result is achieved (e.g. the baby is born, a pathogen is destroyed by a fever, or ripe fruit falls off a tree). Positive feedback is less common than negative feedback in biological systems because the escalation in response is unstable. Unresolved positive feedback responses (e.g. high fevers) can be fatal.

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The cycle is repeated until the baby is born (right) and pressure is released from the cervix.

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Oxytocin stimulates more contractions, forcing the baby's head harder onto the cervix.

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Oxytocin in carried in the blood to the uterus.

Baby moves into the birth canal.

Cervix

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A hormone called oxytocin is released from the pituitary gland into the blood.

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The pressure results in a nerve impulse being sent to the brain.

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Baby's head pushes against the cervix. Pressure receptors in the cervix are stimulated.

Positive feedback also occurs in blood clotting. A wound releases signal chemicals that activate platelets in the blood. Activated platelets release chemicals that activate more platelets, so a blood clot is rapidly formed.

Ethylene is a gaseous plant hormone involved in fruit ripening. It accelerates the ripening of fruit in its vicinity so nearby fruit also ripens, releasing more ethylene. Over-exposure to ethylene causes fruit to over-ripen (rot).

Infection can reset the set-point of the hypothalamus to a higher temperature. The body temperature then increases above the normal range, resulting in a fever. Fever is an important defence against infection.

1. (a) Why is positive feedback much less common than negative feedback in body systems?

(b) Why can positive feedback be dangerous if it continues on for too long?

(c) How is a positive feedback loop normally stopped?

(d) Predict what could happen if a person's temperature increased uncontrollably during a fever?

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118 Negative Feedback and Temperature Control

Key Idea: The temperature regulation centre in humans is in the hypothalamus. Thermoregulation involves negative feedback mechanisms and involves several body systems. The temperature regulation centre of the human body is in the hypothalamus of the brain which has a ‘set-point’ temperature of 36.7°C. The hypothalamus acts like a thermostat, regulating

body temperature through negative feedback. Changes in core body temperature or in skin temperature are registered by the hypothalamus, which then coordinates the nervous and hormonal responses to counteract the changes and restore normal body temperature. When normal temperature is restored, the corrective mechanisms are switched off.

Counteracting heat gain

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Counteracting heat loss

The hypothalamus detects a fall in skin or core temperature below 35.8°C and coordinates responses that generate and conserve heat.

The hypothalamus monitors any rise in skin or core temperature above 37.5°C and coordinates responses that increase heat loss.

Sweating occurs. This cools the body by evaporation.

Increased metabolic rate produces heat.

Body hairs become raised (goosebumps) and increase the insulating air layer around the body.

Decreased metabolic rate. This reduces the amount of heat generated by the body.

In extreme cold, two hormones (adrenaline and thyroxine) increase the energy-releasing activity of the liver.

Body hairs become flattened against the skin. This reduces the insulating air layer around the body and helps heat loss.

The flow of blood to the skin decreases, keeping warm blood near the core (where the vital organs are).

The flow of blood to the skin increases. Warm blood from the body core is transported to the skin and the heat is lost from the skin surface.

Shivering (fast contraction and relaxation of muscles) produces internal heat.

Factors causing heat loss 

Cold external temperature

Warm external temperature

High humidity

 Insufficient insulation (e.g. not wearing enough clothing)

Excessive fat deposits

Wearing too much clothing

Intense physical activity

Being wet or in cold water

Dehydration or being in "shock"

(b) What is its role in thermoregulation:

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 Wind

1. (a) Where is the temperature regulation centre in humans located?

Factors causing heat gain

2. State two mechanisms by which body temperature could be reduced after intensive activity:

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Regulating blood flow to the skin

Thermoreceptors Thermoreceptors are simple sensory receptors that are located in the skin and respond to changes in temperature. When they detect a temperature change they send that information (as nerve impulses) to the hypothalamus. Hot thermoreceptors detect an increase in skin temperature above 37.5°C and cold thermoreceptors detect a fall below 35.8°C.

Constriction of a small blood vessel. A red blood cell (R) is in the centre of the vessel (TEM).

To regulate heat loss or gain from the skin, the blood vessels beneath the surface constrict (become narrower) to reduce blood flow or dilate (expand) to increase blood flow. When blood vessels are fully constricted there may be as much as a 10°C temperature gradient from the outer to inner layers of the skin. Extremities such the hands and feet have additional vascular controls, which can reduce blood flow to them in times of severe cooling.

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The hair erector muscles, sweat glands, and blood vessels are the effectors for mediating a response to information from thermoreceptors. Temperature regulation by the skin involves negative feedback because the output is fed back to the skin receptors and becomes part of a new stimulus-response cycle. Photo above left shows vasodilation and sweating in response high temperature or exertion. Photo above right shows vasoconstriction and goosebumps in response low temperature or inactivity.

3. Describe the role of each of the following in regulating body temperature:

(a) Shivering:

(b) The skin:

(c) Nervous input to effectors:

(d) Hormones:

4. What is the purpose of sweating and how does it achieve its effect?

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5. Explain how negative feedback is involved in the regulation of body temperature:

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6. How do the blood vessels help to regulate the amount of heat lost from the skin and body?

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119 Body Shape and Heat Loss

Key Idea: Body shape influences how quickly heat is lost or retained. Death can occur if body temperature deviates too far either side of the set point. The human body works best at around 37°C and prolonged fluctuations either side of this set point can be fatal. Body

shape influences surface area to volume ratio and therefore also how easily heat is retained or lost. Tall, thin bodies lose heat faster than shorter, stockier bodies. This is related to where different human populations have evolved and has implications for survival in certain conditions.

Body shape influences heat loss

Warmer climate

Body shape influences how heat is retained or lost. Animals with a lower surface area to volume ratio will lose less body heat per unit of mass than animals with a high surface area to volume ratio. In humans there is a negative relationship between surface area and latitude. Indigenous people living near the equator tend to have a higher surface area to volume ratio so they can lose heat quickly. In contrast, people living in higher latitudes near the poles have a lower surface area to volume ratio so that they can conserve heat.

Cooler climate

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Taller, thinner body shape.

Shorter, stocky (wider) body shape.

Longer limbs to increase the surface area for heat loss.

Relationship between ratio of surface area to body mass and latitude

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Shorter limbs reduce the surface area for heat loss.

Data source: C.B Ruff (1994) Morphological Adaptation to Climate in Modern and Fossil Hominids.

Body surface area/mass (cm2 ⁄ kg)

320

Equator

Pole

Increasing latitude

Failure to thermoregulate

A failure of the body's thermoregulatory mechanisms may result in either hyperthermia or hypothermia. Hyperthermia (overheating) occurs when excess heat is produced and cannot be dissipated quickly enough. It is often the result of drugs that interfere with normal thermoregulation.

Proportionately longer legs for their height.

Proportionately shorter legs for their height.

Like all organisms, human populations have evolved in response to environment. Human populations in tropical regions tend to have higher surface area to volume ratios than population in colder regions, where a squatter body shape assists in heat retention. Exposure to cold water results in hypothermia far more quickly than exposure to the same temperature of air. This is because water is much more effective than air at conducting heat away from the body. In the graph (below), hypothermia resulting in death is highly likely in region 1 and highly unlikely in region 2.

Hypothermia occurs when the body cannot generate enough heat and the core body temperature drops below 35°C. The body reduces blood supply to the extremities to help keep the core warm. Over time, vasoconstriction fails, and warm blood rushes to the extremities. Victims feel too hot so they remove their clothes. Death follows because heat loss is increased. This strange behaviour is called paradoxical undressing.

1. Why would having a high surface area to volume ratio be an advantage in a hot climate?

Water exposure and survival times

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Body shape: short and stocky

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Body shape: tall and thin

10 20 15 Water temperature (°C)

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2. (a) With reference to the graph (right), which body shape has best survival at 15°C?

Survival time (hours)

10

25

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120 Hyperthyroidism and Thermoregulation

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Key Idea: Hyperthyroidism refers to over-production of the hormone thyroxine by the thyroid gland. This in turn can affect the body's ability to thermoregulate properly. The thyroid gland is a large endocrine gland found in the neck. The thyroid gland produces a number of hormones

The thyroid gland

Voice box (larynx)

Thyroxine levels are regulated by negative feedback

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The thyroid gland is a butterfly shaped endocrine gland located just below the Adam's apple at the front of the wind pipe (trachea). The thyroid secretes several hormones, collectively called thyroid hormones, but the main hormone produced is thyroxine. Thyroxine is also called T4. Thyroid hormones have many functions including regulating metabolism, growth and development, and body temperature.

involved in regulating many aspects of metabolism, including temperature regulation. Hyperthyroidism (overactive thyroid) is a medical condition in which the thyroid gland produces too much of the hormone thyroxine. One of the effects of too much thyroxine is that it increases body temperature.

Thyroid gland

Wind pipe (trachea)

Hyperthyroidism and temperature regulation

Thyroxine (T4) production is controlled by negative feedback. This mechanism involves two parts of the brain, the hypothalamus and the pituitary gland. Low body temperature stimulates the hypothalamus to secrete thyroid releasing hormone (TRH), which in turn stimulates cells in the anterior pituitary to secrete thyroid stimulating hormone (TSH). TSH acts on the thyroid gland, causing it to produce thyroid hormones, including thyroxine. Thyroxine binds to target cells, increasing their metabolic activity, resulting in heat production. High levels of circulating thyroid hormones inhibit production of TRH and TSH. As a result, thyroid secretion is reduced. When the level of thyroid hormones drops below a certain threshold, TRH and TSH production begins again.  Temperature

One of the effects of thyroxine is that it speeds up metabolic activity in cells. The increase in metabolic activity also results in the production of heat and, under normal conditions, this is one of the mechanisms by which the body raises body temperature. The negative feedback mechanism for thyroxine production can be disrupted by hyperthyroidism, a condition where too much thyroxine is produced by the thyroid gland. As a result this can disrupt temperature regulation.

The most common cause of hyperthyroidism is Graves' disease. In Graves' disease, the negative feedback loop is bypassed because a protein called thyroid stimulating immunoglobulin (TSI) binds directly to the thyroid and stimulates thyroxine production. In this instance, thyroxine production is independent of thyroid stimulating hormone (TSH) production, so the negative feedback regulation of its production is ineffective. Photo above: People with hyperthyroidism often have goitre, an enlarged thyroid gland.

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Hypothalamus

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TRH

Anterior pituitary

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TSH

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T4

Target cells

 Metabolic activity

 Temperature

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1. Briefly outline how negative feedback of thyroxine production is involved in temperature regulation:

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2. Why do high levels of thyroxine not inhibit its production from the thyroid gland in a person with Graves' disease?

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121 Control of Blood Glucose

Key Idea: The endocrine portion of the pancreas produces two hormones, insulin and glucagon, which maintain blood glucose at a steady state through negative feedback. Blood glucose levels are controlled by negative feedback involving two hormones, insulin and glucagon. These hormones are produced by the islet cells of the pancreas, and act in opposition to control blood glucose levels. Insulin lowers blood glucose by promoting the uptake of glucose

by the body's cells and the conversion of glucose into the storage molecule glycogen in the liver. Glucagon increases blood glucose by stimulating the breakdown of stored glycogen and the synthesis of glucose from amino acids. Negative feedback stops hormone secretion when normal blood glucose levels are restored. Blood glucose homeostasis allows energy to be available to cells as required. The liver has a central role in these carbohydrate conversions.

Blood sugar (mmol L-1)

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Blood insulin levels Blood glucose levels

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Negative feedback in blood glucose regulation Negative feedback in blood glucose regulation

Blood glucose can be tested using a finger prick test. The glucose in the blood reacts with an enzyme electrode, generating an electric charge proportional to the glucose concentration. This is displayed as a digital readout.

beta cells

alpha cells

Stimulates α cells to secrete glucagon

Stimulates β cells to secrete insulin

Uptake of glucose by cells. Conversion of glucose to stored glycogen or fat in the liver.

Rise in BG

Normal blood glucose (BG) level 3.9-5.6 mmol

Fall in BG

Breakdown of glycogen to glucose in the liver.

L -1

Decreases blood glucose

Release of glucose into the blood

(b) Identify the stimulus for the release of glucagon:

(c) How does glucagon increase blood glucose level?

(d) How does insulin decrease blood glucose level?

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1. (a) Identify the stimulus for the release of insulin:

2. Explain the pattern of fluctuations in blood glucose and blood insulin levels in the graph above:

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122 Type 1 Diabetes Mellitus

Key Idea: Diabetes mellitus is a condition in which blood glucose levels are too high and glucose appears in the urine. In type 1 diabetes, the insulin-producing cells of the pancreas are destroyed and insulin is not produced. Diabetes mellitus (often called diabetes) is a condition in which blood glucose is too high because the body's cells

The effects of diabetes mellitus Insulin promotes the uptake of glucose from the blood by the cells of the body, where it is used as an energy source. However, in type 1 diabetes no insulin is produced, so blood glucose levels remain high. High blood glucose irritates and damages blood vessels and nerves. This can cause a number of problems including numbness, loss of vision as cells in the eye are damaged, gangrene, and failure of wounds to heal.

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Type 1 diabetes mellitus

cannot take up glucose in the normal way. Diabetes mellitus is characterised by large volumes (diabetes) of sweet (mellitus) urine and extreme thirst. In type 1 diabetes, the insulin producing cells of the pancreas are destroyed and no insulin is produced. Patients must have regular insulin injections to stabilise blood glucose levels.

The body does not produce insulin. Type 1 diabetes is also called insulin dependent diabetes. Age at onset: Early in life (often in childhood).

Symptoms: The cells cannot take up glucose so blood glucose is high. The glucose in the blood exceeds the reabsorption capacity of the kidney and glucose spills over in the urine resulting in large volumes of sweet (high glucose) urine, thirst, hunger, weight loss, fatigue, and infections. Cause: The beta cells are destroyed by the body's own immune system. A genetic predisposition and environmental factors (e.g. a viral infection) may be the trigger. Treatment: There is no cure. Blood glucose is monitored regularly. Insulin injections combined with dietary management keep blood sugar levels stable.

When glucose is not available, the body's cells metabolise fat as an alternative energy source. This can also cause problems such as a fall in blood pH (which can be fatal), and damage to blood vessels and the development of cardiovascular disease.

Without insulin, cells cannot take up glucose and so lack an energy source for metabolism. Production of urine from the kidneys increases to clear the body of excess blood glucose. Glucose is present in the urine. There is constant thirst. Weight is lost despite hunger and overeating. Inability to utilise glucose leads to muscle weakness and fatigue.

Type 1 diabetics must regularly check their blood glucose levels and administer insulin injections if it is too high.

Fats are metabolised for energy leading to a fall in blood pH (ketosis). This is potentially fatal.

Ulcer on the leg of a diabetic

High sugar levels in blood and urine promote bacterial and fungal infections of the bladder and urinogenital tract.

Foot

1. (a) What is type 1 diabetes?

(b) Explain how the usual negative feedback mechanisms for blood glucose homeostasis are disrupted in a person with type 1 diabetes:

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2. How do regular insulin injections help a person with type 1 diabetes to maintain their blood glucose homeostasis?

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123 Maintaining Water Balance

Key Idea: Antidiuretic hormone (ADH) helps maintain water balance by regulating water absorption by the kidneys. The body regulates water balance in response to fluctuations in fluid gains and losses. One mechanism by which fluid balance is maintained is by varying the volume of water absorbed by the kidneys and thereby the concentration of the

urine produced. This involves a hormone called antidiuretic hormone (ADH). Osmoreceptors in the hypothalamus monitor blood osmolarity (water content) and send messages to the pituitary gland which regulates the amount of ADH released. ADH promotes the reabsorption of water from the kidney collecting ducts, producing a less or more concentrated urine. Too much water in blood

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Too little water in blood

Receptors in the hypothalamus detect there is not enough water in the blood (blood volume is too low and blood is too concentrated).

ADH is secreted from the pituitary gland. The amount released depends on how concentrated the blood is (the more concentrated the blood, the more ADH is released).

Receptors in the hypothalamus detect there is too much water in the blood (blood volume is too high and blood is too dilute). ADH secretion from the pituitary reduces or stops.

ADH travels via the circulatory system to the kidneys.

 ADH

 ADH

 Kidney permeability to water

 Kidney permeability to water

 Water absorption from urine

 Water absorption from urine

 Urine volume

 Urine volume

 Urine concentration

 Urine concentration

Blood volume increases until the levels return to normal.

Blood volume decreases until the levels return to normal.

1. What effect does ADH have on the kidneys?

2. How do negative feedback mechanisms operate to regulate blood volume and urine output?

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3. Predict whether a high fluid intake would increase or decrease ADH production:

(b) How would diabetes insipidus be treated?

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4. (a) Diabetes insipidus is a type of diabetes caused by the a lack of ADH. Based on what you know about the role of ADH in kidney function, describe the symptoms of this disease:

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124 KEY TERMS: Did You Get It?

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168

1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code. adaptation ADH

biomimicry

A A destabilising mechanism in which the output of the system causes an escalation in the initial response. B Any change in the environment that is capable of generating a response in an organism. C A medical condition in which levels of blood glucose are elevated.

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diabetes mellitus homeostasis

D A hormone involved in regulating the amount water absorbed by the kidneys and the concentration of the urine produced (acronym).

E The regulation of body temperature.

insulin

F An adaptation involving the morphology of an organism.

negative feedback

G A heritable characteristic that contributes to an organism's survival and fitness.

H The design of technologies that are modelled on biological entities and processes.

positive feedback receptor stimulus

I

A sensory structure that responds to a stimulus in the internal or external environment of an organism.

J The hormone that lowers blood glucose by promoting the uptake of glucose by the body's cells. K Regulation of the internal environment to maintain a stable, constant condition.

structural adaptation thermoregulation

L A mechanism in which the output of a system acts to oppose changes to the input of the system. The net effect is to stabilise the system and dampen fluctuations.

2. Test your knowledge about feedback mechanisms by studying the two graphs below, and answering the questions about them. In your answers, use biological terms appropriately to show your understanding.

A

+

B

+

Type of feedback mechanism:

Mode of action:

Mode of action:

Biological examples of this mechanism:

Biological examples of this mechanism:

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TEST

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Type of feedback mechanism:

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Organising biodiversity

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Unit 1 Outcome 2

Key terms

Classification of biodiversity

biodiversity

Key knowledge

binomial nomenclature biopiracy

c

1

Explain what is meant by biodiversity. Describe how biodiversity, both past and present, is classified into taxonomic groups. Explain the basis for this classification with reference to shared morphological and molecular characteristics.

125 126 127

c

2

Explain how molecular techniques have provided new approaches for the classification of biodiversity. Understand the difference it has made to the classification of extinct species from fossil evidence and to how we determine species relationships in past and present taxa.

125 127

c

3

Describe the biological classification of species with reference to the taxonomic ranks of domain, kingdom, phylum, class, order, family, genus, and species.

127

c

4

Understand and explain why it is often difficult to assign organisms to any one species, or to identify new species.

126

c

5

Describe the scientific naming of species using binomial nomenclature. Understand the benefits of a scientific name and the problems associated with the use of common names to identify species.

bioprospecting class

common name conservation domain family

genus

kingdom

molecular features

Activity number

morphological features

127 128

order

phylum

scientific name shared derived characteristics species taxon

taxonomic rank

Managing the Earth's biodiversity

Activity number

Key knowledge

6

Explain how the biodiversity of life on Earth represents a resource and discuss the economic, aesthetic, and ecological reasons for its conservation, including the role of biodiversity in providing the ecosystem services on which humans depend.

129

c

7

Identify threats to Australia's biodiversity and recognise the control of these threats is part of any conservation strategy. Describe methods for the conservation of vulnerable and endangered species and their habitats. Compare and contrast in-situ and ex-situ methods and comment on their relative costs and benefits.

130

c

8

Explain what is meant by bioprospecting and describe its dependence on a reservoir of biodiversity. Describe examples to show your understanding of how the Earth's biodiversity can provide new food sources and medicines.

131

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c


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125 Classification Systems: The Old and The New

Key Idea: The classification of biodiversity into groups, or taxa, is constantly being updated in light of new information. The variety of life on Earth is called biodiversity and its classification into formal groups is called taxonomy. Taxonomy, as with all science, is constantly changing as new information is discovered. With the advent of DNA sequencing technology, scientists began to analyse the genetic make-up

Whittaker 1969

Woese et al. 1977

Woese et al. 1990

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A changing view of classification

of many bacteria. In 1996, these analyses confirmed that life comprises three major lineages (domains), not two as was the convention. The recognised lineages are the Bacteria, the Eukarya, and the Archaea. The new classification better reflects the evolutionary history of life on Earth. Molecular evidence has since led to the reclassification of many other taxa, including birds, reptiles, many plants, and primates.

Before DNA sequencing, taxonomists divided life into five kingdoms based mainly on visible characteristics (morphology). The five kingdom system places all prokaryotes in one kingdom, with protists, fungi, plants, and animals being the other four. This system is dated and seriously at odds with molecular evidence. In particular, it does not fairly represent the diversity or evolutionary history of the prokaryotic organisms or unicellular eukaryotes.

A new view of the world

In 1996, scientists deciphered the full DNA sequence of the thermophilic bacterium Methanococcus jannaschii. The data supported the hypothesis of three major evolutionary lineages and gave rise to a modified six kingdom classification. This was further revised to the current three domain system (below), which more properly represents the phylogeny of life on Earth.

Domain Bacteria

Five kingdoms

Three domains

Eubacteria

Bacteria

Archaebacteria

Archaea

Monera

Protista

Protista

Fungi

Fungi

Plantae

Plantae

Animalia

Animalia

Eukarya

Domain Archaea

Bacteria and cyanobacteria

Six kingdoms

Archaeans

Domain Eukarya

Animals

Amoebae

Slime moulds

Fungi

Plants

Ciliates

Flagellates

Microsporidia Diplomonads

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NIAID

UC Berkeley

Last universal common ancestor

Domain Archaea

Lack a distinct nucleus and cell organelles. Present in most of Earth's habitats and vital to its ecology. Includes well-known pathogens, many harmless and beneficial species, and the cyanobacteria (photosynthetic bacteria containing the pigments chlorophyll a and phycocyanin).

Methanococcus jannaschii was the first archaean genome to be sequenced. The sequencing identified many genes unique to Archaea and provided strong evidence for three evolutionary lineages. Although archaeans may resemble bacteria, they posses several metabolic pathways that are more similar to eukaryotes. Other aspects of their structure and metabolism, such their membrane lipids and respiratory pathways, are unique. Although once regarded as organisms of extreme environments, such as volcanic springs, archaeans are now known to be widespread, including in the ocean and soil.

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Domain Eukarya

Complex cell structure with organelles and nucleus. The three domain classification recognises the diversity and different evolutionary paths of the unicellular eukaryotes (formerly Protista), which have little in common with each other. The fungi, animals, and plants form the remaining lineages.

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Domain Bacteria

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Scientists can use the tools at their disposal, including anatomical and molecular evidence, to construct a branching diagram to illustrate how the species in a taxonomic group are related. These 'phylogenetic trees' represent a hypothesis for the evolutionary history of the taxon and will be supported to greater or lesser extents by the evidence. Increasingly, DNA and protein analyses are indicating that many traditional phylogenies do not accurately portray evolutionary relationships. This is true at all levels of classification including class (e.g. reptiles and birds) and order e.g. primates. To illustrate this, the traditional classification of primates is compared with the revised classification based on cladistic analysis (a newer method of determining evolutionary relationships based on molecular evidence).

A classical taxonomic view

Hominidae

A view based on molecular evidence

Pongidae

Hominidae

The ‘great apes’

Ponginae

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Homininae

Humans

Chimpanzees

Gorillas

Orangutans

Humans

Chimpanzees

1.4%

Gorillas

Orangutans

1.8%

A small genetic difference indicates a recent common ancestor

3.6%

A greater genetic differences indicates that two taxa are more distantly related

On the basis of overall anatomical similarity (e.g. bones and limb length, teeth, musculature), apes were grouped into a family (Pongidae) that is separate from humans and their immediate ancestors (Hominidae). The family Pongidae (the great apes) is not monophyletic (of one phylogeny with one common ancestor and all its descendents), because it stems from an ancestor that also gave rise to a species in another family (i.e. humans). This traditional classification scheme is now at odds with schemes derived after considering genetic evidence.

Based on the evidence of genetic differences (% values above), chimpanzees and gorillas are more closely related to humans than to orangutans, and chimpanzees are more closely related to humans than they are to gorillas. Under this scheme there is no true family of great apes. The family Hominidae includes two subfamilies: Ponginae and Homininae (humans, chimpanzees, and gorillas). This classification is monophyletic: the Hominidae includes all the species that arise from a common ancestor.

1. (a) Outline the evidence supporting a three domain classification system:

(b) In what respects is the five kingdom classification system inadequate?

(c) How has molecular evidence contributed to the taxonomic revisions described in this activity:

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3. What evidence has led to the reclassification of the primates?

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2. Suggest why classifications based on molecular evidence might provide a more likely phylogeny than one based on appearance alone:


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126 How Do We Assign Species?

Key Idea: Assigning species on the basis of appearance alone can be inaccurate. Molecular analysis allows distinct species with a similar appearance to be distinguished. Humans have always tried to identify and classify the diversity of organisms around them. Traditional classifications were based on similarities in morphology, but modern taxonomy relies heavily on biochemical data as well in an

attempt to classify species on the basis of their evolutionary relationships. Assigning an organism to a particular species is not a trivial matter. Organisms within a single species may have very different appearances and, conversely, different species may be morphologically indistinguishable. How we identify and define species has important implications because we cannot conserve what we cannot recognise.

Different species may look the same

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Assigning species based on morphology alone can lead to mis-classification. Some organisms may be morphologically indistinguishable but DNA analysis or a close examination of their biology may prove them to be different species. DNA analysis is becoming increasingly important in distinguishing species.

Meyer A, PLoS Biology CC2.5

Top row: H. erato

Example: The butterfly genus Heliconius comprises 39 species, many of which mimic each others' patterns and colours. Heliconius is a species complex, a group of closely related or sibling species with similar appearances and unclear boundaries between them. Their appearance varies with geographical location, but different species often adopt a similar pattern in the same location (left) making species determination based on appearance alone extremely difficult.

Bottom row: H. melpomene

Molecular analysis can distinguish cryptic species DNA variation occurs within species as well as between species, so scientists must determine what level of variation is acceptable within a species before a new species classification is made. If such boundaries were not set, every molecular variation observed would result in the classification of a new species.

African forest elephant African bush elephant Loxodonta africana

Yathin S Krishnappa CC3.0

African forest elephant Loxodonta cyclotis

Thomas Breuer cc2.5

Example: Molecular studies have been important in identifying cryptic species, i.e. two or more distinct species disguised under one species name. The African bush elephant and the African forest elephant were once considered subspecies, but recent genetic analysis has confirmed they are separate species, which diverged from each other 2-7 million years ago. Analysis of morphological differences, including skull anatomy, supports this.

Conservation and species assignment

The recognition of species complexes and cryptic species has important implications to our estimates of biodiversity, and affects decisions made in species conservation and in the management of economically important species (including pests and medically important organisms). DNA analyses indicate that cryptic species are far more common than previously thought, providing a strong argument for whole ecosystem (in-situ) management of biodiversity.

Ron Pastorino CC3.0

Tony Wills CC3.0

A. muscaria muscaria Amanita m. var. formosa

The High Fin Sperm Whale CC3.0

Example: The fly agaric mushroom (Amanita muscaria, left) comprises several cryptic species. Genetic analyses in 2006 and 2008 showed at least four groups or clades that are genetically distinct enough to be considered separate species. The varieties identified left include the type specimen (far left) and two of the subspecies now considered to be separate species.

Amanita m. var. guessowii

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1. In what way have traditional methods of classification based on morphology been inadequate?

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2. What are the implications of species complexes and cryptic species in conservation?

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127 Classification of Biodiversity group reptiles excludes the birds, which are more closely related to the crocodiles than the crocodiles are to lizards and snakes. In modern classification, distinction of taxa is not so much based on morphological similarities but on evolutionary relationships (phyletics). Classification is also now more commonly based on molecular studies, including genetic and protein analyses. Both morphological and molecular studies should provide evidence that is not contradictory.

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Key Idea: Organisms can be placed into groups based on their shared characteristics. Taxonomy is the science of grouping organisms based on shared characteristics. Smaller, more precise groups can be grouped together into larger less precise groups, forming a hierarchy of taxonomic ranks. There are various ways in which organisms can be classified, but not all fairly represent their evolutionary relationships. For example, the traditional

Taxonomic ranks

Traditionally, living organisms are classified into a hierarchy of seven main taxonomic ranks (although commonly today the rank of "domain" is often included above kingdom as the most encompassing group, making eight major taxonomic ranks). The taxonomic ranks in descending order are: domain, kingdom, phylum, class, order, family, genus, species. Other ranks such as subclass, may also be included. There are also some differences between naming conventions depending on the organism, e.g. animal phyla are equivalent to plant divisions.

Domain

Phylum (or division)

Kingdom

Order

Genus

Family

Class

Species

The example below shows how as we move through the taxonomic ranks, the organisms we are grouping become more exclusive based on the characteristics of the group. In this case, we are looking at the classification of the grey wolf Canis lupus. Kingdom: Animalia

1

2

3

4

4

5

6

2

3

5

6

7

8

9

Domain: Eukarya

4

2

Class: Mammalia

3

6

4

2

4

9

9

2

7

Phylum: Chordata

6 9

9

2

6

6

6

9

9

9

Order: Carnivora

Family: Canidae

Species: Canis lupus

Genus: Canis

1. The table below shows part of the classification for humans using the eight major levels of classification. For this question, use the example of the classification of the grey wolf, above, and the red kangaroo, on page 175, as a guide.

(a) Complete the list of the taxonomic ranks on the left hand side of the table below:

(b) Complete the classification for humans (Homo sapiens) on the table below.

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Family

Human classification

Hominidae

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1. 2. 3. 4. 5. 6. 7.

Taxonomic rank

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174 Why are birds dinosaurs?

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Defining groups of organisms and evaluating their ancestry using morphological features alone can be problematic because similarities in structure may not necessarily be the result of shared ancestry. This problem can be overcome by only considering the shared derived characteristics, i.e. the characteristics of two of more taxa that are present in their most recent common ancestor but not in older ancestors. Tracing the evolution of derived character states can more accurately identify the evolutionary history of a taxon. The ancestry of birds below illustrates this. Although birds are commonly regarded as a single taxon (and in modern terms they are) birds are simply the last in the lineage of the dinosaurs. Recent analysis of the protein structure of fossil collagen from Tyrannosaur fossils puts birds and dinosaurs in the same taxon.

Features shared by birds and dinosaurs

Birds

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3

Tyrannosauridae

Features of each group

Coelurosauria

6

Pubic foot (end of bone), stiffened tail

5

Reversed first toe, pubis longer than ischium, long arms

1

2

6

Allosauroidea

Tetanurae

Ceratosauroidea

Theropoda

4

Sauropoda

Saurischia

3

Hollow bones, 3 main fingers and toes, bowed femur

5

2

4

Archaeopteryx (transitional between birds and dinosaurs)

3

1

Lizard hipped pelvis

2

Ornithischia

6

2

Dinosauria

2

2+ sacral vertebrae, offset thumb, inturned femoral head (top of thigh bone)

Archosauria

1

Socketed teeth, Antorbital fenestra (opening in the skull)

Crocodiles

5

4

Velociraptor (a late Cretaceous dinosaur)

2. Construct an acronym or mnemonic to help you remember the principal taxonomic ranks (DKPCOFGS):

3. Classification has traditionally been based on similarities in morphology, but new biochemical methods are now widely used to determine species relatedness. What contribution are these techniques making to the science of classification?

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4. Explain why defining the boundaries between specific taxa can be problematic:

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5. Explain how grouping organisms based on shared derived characteristics can help explain their evolutionary history:

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Classification of the red kangaroo Below is the classification scheme for the red kangaroo. Only one of each group is subdivided in this chart showing the levels that can be used in classifying an organism. Not all possible subdivisions (e.g. sub-order) have been shown here. The only natural category is the species, often separated into races, or sub-species, which generally differ in appearance. The

Animalia Animals; one of 5 kingdoms

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Kingdom:

classification below follows that in “The Australian Museum Complete Book of Australian Mammals” R. Strahan (Ed.) 1983. Cornstalk Publishing. ISBN 0 207 14454 0. It differs from many classifications in raising the marsupials to a sub-class (rather than an order) and dividing the sub-class into the orders Polyprotodonta and Diprotodonta.

Phylum:

Sub-phylum:

Chordata

Animals with a notochord (supporting rod of cells along the upper surface) Sea squirts, salps, lancelets, and vertebrates

23 other phyla

Vertebrata

Animals with backbones Fish, amphibians, reptiles, birds, mammals

Class:

Mammalia

Sub-class:

Metatheria or Marsupialia

Order:

Diprotodonta

Superfamily:

Macropodoidea

Family:

Macropodidae

Genus:

Macropus

Species:

rufus

Animals that suckle their young on milk from mammary glands Placentals, marsupials, monotremes

Mammals whose young develop for some time in the female's reproductive tract and then migrate to a pouch. Marsupials

Herbivorous marsupials Possums, rat kangaroos, honey possums, cuscuses and bushtails, koala, pygmy possums, wombats, gliders, kangaroos, wallabies

Possess extremely long and powerful hindlegs Potoroos, bettongs, rat-kangaroos, kangaroos, wallabies

Kangaroo and wallaby family with 50 species of kangaroos, wallabies, rock-wallabies, nailtail wallabies, forest wallabies, tree kangaroos, pademelons, quokka, swamp wallabies

13 other species

Subspecies: Several subspecies have been described but their validity is not well established.

Red kangaroo Macropus rufus

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Synonyms: plains kangaroo, marloo, blue-flier (female).

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Red kangaroo

Description: One of the largest living marsupials and powerfully built. Colouration varies, often regionally; red to blue-grey above, and distinctly white below. Females are usually blue-grey and are often called blue-fliers. In some regions of Australia, both sexes are reddish-brown.

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10 other genera

Ground-dwelling kangaroos and wallabies consisting of 14 species in this genus: red kangaroo, eastern grey, western grey kangaroo, wallaroo, several species of wallaby


128 Naming an Organism

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176

Key Idea: Every organism is given a two part scientific name that is unique to that organism. Many organisms have common names that people use in everyday language. However common names change from place to place and language to language. Common names may also apply to more than one organism, which causes

confusion when people from different parts of the world are referring to particular organisms. For example, the mountain lion, cougar, puma, catamount, and panther are all names given to the the same animal, Puma concolor. To solve this problem each species is give a two part (binomial) name, called the scientific name, that is unique to that species.

Names and meanings

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Most species have a common name as well a scientific name. Common names may change from place to place as people from different areas name species differently based on both language and custom. Scientifically, every species is given a classification that reflects its known lineage (i.e. its evolutionary history). The last two (and most specific) parts of that lineage are the genus and species names. Together these are called the scientific name and every species has its own that is specific to that species. The term given to this two-part naming system is binomial nomenclature. When typed the name is always italicised. If handwritten, it should be underlined. The genus name is always written with a capital letter, but the species name is not.

The animal Rangifer tarandus is known as the caribou in North America, but as the reindeer in Europe. The scientific name is unambiguous.

Scientific names are normally based on Latin and sometimes Greek. The name often, but not always, describes the organism. For example, Hippopotamus amphibius is derived from Greek - hippos meaning horse and potamos meaning river, giving "river horse". Amphibius meaning two lives Hippopotamus amphibius lives both in the water and out of it.

Spongiforma squarepantsii

Tom Bruns

In 2011 researchers from San Francisco State University discovered a sponge-like mushroom growing in the Malaysian rain forest. They named it Spongiforma squarepantsii after the cartoon character SpongeBob Squarepants.

The Bengal tiger

Panthera tigris tigris (India)

The Siberian tiger

Dave Pape PD

A monotypic species has no distinct populations but, for some species, scientists recognise morphologically and genetically distinct subspecies (in animals) or varieties (in plants, algae, and fungi). Subspecies could interbreed but do not generally do so because they are isolated by geography or habitat (e.g. subspecies of tiger). Subspecies are identified by a third name after the species name (but never with the subspecies name alone). The type specimen carries the same specific and subspecific name. Species with defined subspecies are called polytypic species.

Sbj1976 CC4.0

What are subspecies?

Panthera tigris altaica (Russia)

1. (a) What is the two part naming system for classifying organisms called?

(b) What are the two parts of the name?

(b)

3. Which is the type specimen in the tiger examples above?

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2. Give two reasons why the classification of organisms is important: (a)

(a) Ceratotherium simum:

(b) Canis Lupus: WEB

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4. Describe what is wrong with the way each of the following scientific names is written:

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129 Why is Biodiversity Important?

Key Idea: Maintaining biodiversity enhances ecosystem stability and functioning and also provides economic and aesthetic benefits to humans. Ecosystems provide both material and non-material benefits to humans. These benefits, or ecosystem services, are best provided by healthy, diverse systems. As a general rule, high diversity ecosystems are ecologically more stable (constant

The aesthetic benefits of biodiversity

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Ecological reasons for maintaining biodiversity

in character over time) and resistant to disturbance (resilient) than systems with low diversity. Maintaining diversity therefore has economic benefits to humans through their use of ecosystem resources and through the generation of income from tourism to ecologically significant areas. Although often discussed as individual benefits, the ecological, aesthetic, and economic reasons for maintaining biodiversity overlap.

Evidence from both experimental and natural systems indicates that the most diverse ecosystems are generally the most stable, most probably because the complex network of species interactions has a buffering effect against change. Maintaining biodiversity is therefore critical to maintaining key ecological functions such as nutrient cycling and water purification. Ecosystems include many interdependent species (e.g. flowering plants and their pollinators, hosts and parasites). The loss of even one species can detrimentally alter ecosystem dynamics, especially if the species is a keystone species (a species with a pivotal role in ecosystem functioning). Genetic diversity is an important part of ecosystem stability and resilience. Losing genetic diversity increases the risk of extinction and the ecosystem degradation. Genetic diversity effectively represents genetic resources, i.e. those genes in living organisms that may have benefits to humans (e.g. medicinal plants). Once an organism is extinct, those genetic resources are also lost.

Many people enjoy looking at, or spending time in areas of natural beauty. Aesthetically pleasing landscapes provide satisfaction and enjoyment to individuals. Nature can also provide inspiration for artists, photographers, and writers, as well as economic benefits from tourism. As countries become more populated and development increases, it becomes more important to maintain and protect natural landscapes as areas of natural beauty. The images below show two coral reefs. Reef A has high biodiversity, many different species are represented and it is full of colour and life. Reef B is an area of coral bleaching and supports far fewer species. Imagine you were a tourist paying to visit the reef. Which one would you rather visit?

A

B

Rainforest

Monoculture of soy beans

Rainforests represent the highest diversity systems on Earth. Whilst they are generally resistant to disturbance (resilient), once degraded (e.g. by deforestation), they have little ability to recover. Monocultures, which provide the majority of the world's food supply, represent very low diversity systems and are particularly susceptible to diseases, pests, and disturbance.

1. What is the most likely reason for high diversity ecosystems being more stable than ecosystems with low biodiversity?

3. (a) What is a genetic resource?

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2. Describe two aesthetic benefits of maintaining biodiversity:

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(b) Many medicines are derived from plants. How does tropical rainforest destruction reduce genetic resources?

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Economic reasons for maintaining biodiversity A variety of economic benefits (goods and services) are generated by biodiversity. These benefits are commonly called ecosystem services and are split into four categories: provisioning, regulating, and cultural services, which directly affect people, and supporting services, which maintain the other three services. The provisioning services are sometimes referred to as goods because they can be sold and their economic value can easily be calculated as they have a monetary value. Estimating the total economic value of total ecosystem services is difficult and contentious, but some estimates amount to many trillions of dollars per year.

Regulating services

(Products obtained from ecosystems) • Food • Water • Fuel wood • Fibres • Biochemicals • Genetic resources

(Benefits obtained from the regulation of ecosystem processes) • Climate regulation • Disease regulation • Water regulation • Water purification • Pollination

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Provisioning services

Food production Ecotourism

Pollination

Soil formation

Cultural services

Supporting services

(Nonmaterial benefits people obtain from ecosystems) • Spiritual and religious • Recreation and ecotourism • Aesthetic • Inspirational • Educational • Cultural heritage

(Services necessary for the production of all other ecosystem services) • Soil formation • Nutrient cycling • Primary production

The economic cost of soil depletion

Soil depletion refers to the decline in soil fertility due to the removal of nutrients. Some definitions of soil depletion also include the physical loss (erosion) of soil.

The increase of continuous monoculture farming practices has contributed to a rapid loss of nutrients from the soil over the last few decades. Farmers must make an economic choice: spend money on fertilisers to add nutrients back to the soil or do nothing and suffer the economic consequences of low crop yields.

4. The 17th century Irish potato famine is an example of how low biodiversity can threaten our food supply. Farmers planted only one potato variety with limited genetic diversity. Most potato crops were destroyed by the fungal disease late blight and, exacerbated by the political environment at the time, there was widespread famine. How could have this situation have been prevented?

Blight affected potato

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5. Summarise the economic benefits of maintaining biodiversity:

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130 Managing Biodiversity for Conservation

Key Idea: The conservation of biodiversity involves management using in-situ or ex-situ strategies. In 1992, the Convention on Biological Diversity was adopted in Rio de Janeiro. Its aims are to conserve biodiversity, use biodiversity in a sustainable way, and ensure that the benefits of genetic resources are shared equitably. Conservationists

have a number of strategies to protect at-risk species and help threatened species to return to sustainable population sizes. Ex-situ methods operate away from the natural environment and are useful where species are critically endangered. In-situ conservation methods manage ecosystems to protect diversity within the natural environment.

Martin Pot

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In-situ conservation: strategies operating as part of the natural environment

Numbat, a vulnerable species.

Habitat protection and restoration: Parks and reserves focus on ecosystem conservation. These areas aim to preserve habitats with special importance and they may be intensively managed through pest and weed control programmes, revegetation, and reintroduction of threatened species. A “research by management” approach is associated with careful population monitoring and management to return threatened species to viable levels. In this photograph, volunteers restore a habitat corridor which makes it easier for species to move between separated regions of habitat.

Protection through legislation: In Australia, the Environment Protection and Biodiversity Conservation Act (1999) provides a legal framework to protect and manage important flora, fauna, ecological communities and heritage places. Internationally, the Convention on International Trade in Endangered Species (CITES) is an agreement between governments which aims to ensure that international trade in species of wild animals and plants does not threaten their survival. Unfortunately, even CITES does not guarantee species are safe from illegal trade.

Ian Smith

Ex-situ conservation: strategies operating outside the natural environment

Captive breeding and relocation: Captured individuals are bred under protection. If breeding is successful and suitable habitat is available, captive individuals may be relocated to the wild where they can establish natural populations. Taronga Zoo runs a captive breeding programme for the Regent honeyeater, above.

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Salvinia weed invades waterways

At least 18 exotic mammals, including cats, foxes, pigs, and goats, have established feral populations in Australia. Cats and foxes are predators of native animals and have been implicated in many species declines. Introduced rodents compete with native species with similar niches. The activities of introduced invertebrates and plants have also contributed to the decline in Australia’s native species. German wasps arrived in timber shipments in 1978 and have since spread widely, preying on native insects and attacking soft fruits. About 25% of introduced plants are considered actual or potential serious weeds. Salvinia (left) is an aggressive invader despite control measures.

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DoC-RM

Threats to Australia's flora and fauna

Botanic gardens play a critical role in plant conservation. They maintain seed banks, nuture rare species, maintain a collection of plants, and help to conserve indigenous plant knowledge. They also have an important role in research and education. Seed banks are especially important as a store of genetic diversity, especially of ancient varieties

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The role of zoos: Many zoos specialise in captive breeding programmes for endangered species and participate in global programmes to retain the genetic diversity of captive bred species. Several zoos and sanctuaries, including Taronga and Healesville are involved in recovery programmes for the Tasmanian devil.

Blue Mountains Botanic Gardens

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IS

Luis A de la Parra

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St Helens wax flower

A species is presumed extinct if it has not been located during the preceding 50 years (or 10 years after thorough searching). Endangered species, such the St Helens wax flower, are in danger of extinction. Their survival in the wild is unlikely if the factors causing their decline continue. Vulnerable species are those that will become endangered in the next 25 years if the factors causing their decline continue.

Dainty green tree frog

Palm cockatoo (Probosciger aterrimus): This spectacular bird has a very restricted natural distribution, found at the very tip of Cape York in Queensland. Their large size, magnificent head crest, and red face patch make them much sought after by illegal bird traffickers and collectors. Palm cockatoos are currently not listed as threatened but could move into that category if habitat loss in the area continues.

Amphibians: Globally, amphibians are in serious decline as a result of habitat losses and disease (e.g. amphibian chytrid disease). Australia's amphibian fauna includes a large number of endemic species. More than 30 species are endangered or critically endangered and many more are threatened or at risk. In general, too little is known about their true diversity and their biology to identify risks and halt their decline.

1. Explain the impact of each of the following on Australian native species:

(a) Habitat destruction:

(b) Pests, predators and weeds:

(c) Hunting and collection:

2. There are a number of strategies for the management of native species. For each of those listed below, briefly describe its role in enhancing species survival in the wild:

(a) Protection of habitat:

(b) Habitat restoration:

(c) Pest control:

(d) Captive breeding and relocation:

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3. Suggest why species should be preserved from extinction:

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4. Species can be ranked by their conservation status (e.g. critically endangered, endangered, vulnerable).Although they are not endangered, the plight of vulnerable species also concerns conservationists. Suggest why:

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131 Biodiversity and Bioprospecting

Key Idea: Bioprospecting searches nature for useful organisms or compounds that can be commercialised. Bioprospecting is the systematic search for and development of new sources of useful products from nature, e.g. chemical compounds or microorganisms. Bioprospecting looks for ways to commercialise biodiversity. Bioprospecting is not new; most of our medical drugs are based on chemicals

found in nature. In the late 1700s, English doctor and scientist, William Withering noticed that patients suffering from a serious heart condition became better when given a traditional herbal remedy. He went on to isolate the active compound from the foxglove plant, which he called digitalis (or digoxin). It is now listed by WHO as one of the most important heart medications in the health system.

Biodiversity and new compounds

Number of natural products isolated from marine organisms

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2000

1000

0

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An

Ec

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a zo yo

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M

Br

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About 60% of drugs used in cancer treatments and 75% used against infectious diseases are derived from natural sources. Traditionally, terrestrial sources have been the most exploited for compounds, due largely to the ease of access. Increasingly though, marine organisms are being investigated for useful chemistry. About half of the compounds now under study come from marine sources.

4000

Number of natural products isolated between 1985 and 2008

Recall that humans conserve biodiversity for ecological, aesthetic, and economic reasons. One of the economic arguments for biodiversity conservation is the important role of biodiversity as a source of valuable commodities, such as medicines, new foods, and materials. William Fenical of the Scripps Institution of Oceanography states that "biodiversity translates to genetic uniqueness, which in turn results in the expression of diverse biochemical processes producing metabolic products ...". With much of the world's biodiversity under threat, it is likely that there are many useful compounds, unique to certain organisms, that could soon be lost.

Phylum

Sources of new compounds

Kristian Peters CC 3.0

CDC

The plant sweet wormwood (Artemisia annua) has been used in traditional Chinese medicine for over two thousand years. It produces the chemical artemisinin, which is a potent antimalarial agent.

Clostridium botulinum is a bacterium that produces the most powerful toxin known (both synthetic and natural). Just 1.5 nanograms per kilogram in humans is lethal. The toxin stops muscles contracting, causing paralysis. This makes it useful as a medical treatment for various muscular conditions.

Mammals are unable to manufacture essential polyunsaturated fatty acids. They must be taken in with the diet. Species of microalgae, including species of the genus Chlorophyta, are used to produce fatty acids which are added to food or developed as dietary supplements.

(b) Identify three cases of bioprospecting and describe the product and its uses in each case:

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1. (a) What is bioprospecting?

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Bioprospecting and biopiracy Patent laws are designed to protect the investment a person or company puts into the creation of something new. They stop other people or companies taking someone else's idea and making money from it for a period of time so that the patent holder has time to develop and sell the idea (or product) to recoup the investment they put into it. Biopiracy is a term used to describe the taking of a traditional resource (e.g. a traditional medicine or plant type), applying a patent to it, and then stopping any traditional use of the resource, or selling it back to the people who traditionally used it in the first place. This can happen relatively easily, especially when resources are acquired from countries that have traditionally thought of those resources as communal, with no stated ownership.

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The case of the Enola bean

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Neil Palmer

In 1996 Larry Proctor, the president of POD-NERS a USA based seed company, bought a bag of commercially available beans (Phaseolus vulgaris) from Mexico. The beans, a variety known in Mexico as mayacoba, have seeds with various shades of yellow. Proctor took them back to the USA, and planted only the beans with a specific shade of yellow. After several generations of self pollination, the beans produced a variety with seeds that were a stable shade of yellow. He then took out a patent on the "new" Enola variety, just two years after buying the original beans. With the patent, POD-NERS was able to block the importation of any yellow beans from Mexico to the USA. This was despite the fact that the mayacoba variety had been exported from Mexico to the USA for generations. An outraged Mexican government appealed the patent and it was eventually revoked in 2008.

Hoodia gordonii

Winfried Bruenken

Hoodia gordonii is a succulent remarkably similar to cacti, although it is unrelated. It is native to South Africa and Namibia. Hoodia gordonii has been used by the San people of Namibia as an appetite suppressant on long hunting trips. The San have occupied parts of Namibia and the surrounding lands for around 20,000 years. In 1977, the South African council for science and industrial research isolated the active compound for suppressing appetite (called P57) and realised its potential as a weight loss supplement. It patented P57 in 1996. The production rights were sold to Phytopharm and Pfizer with no acknowledgement of the San people's traditional knowledge. Eventually, after work by the group WIMSA, an agreement was reached for a benefit-sharing scheme with the San people. However trials of P57 have found it is difficult to extract and metabolised too quickly in the body to be commercially useful.

2. Explain why bioprospecting is more efficient than trying to a develop drug treatment for a disease from first principles:

3. Estimate how many natural products were isolated from marine life between 1985 and 2008:

4. (a) What is biopiracy?

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(b) Explain how the patent system can encourage biopiracy:

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132 KEY TERMS: Did You Get It?

1. Test your vocabulary by matching each term to its correct definition, as identified by its preceding letter code

biodiversity

A The process of discovery and commercialisation of products based on biological resources.

binomial nomenclature

B The situation in which traditional or indigenous knowledge or resources are used for profit without acknowledgement, permission from, or compensation to, the indigenous or traditional owners of that knowledge or resource.

bioprospecting

C Any defined unit in the classification of organisms.

common name

D Structures of organisms that are the result of common ancestry are this.

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biopiracy

E A division of the living world above phylum but lower than domain.

conservation

F The largest division of living things, including the Eukarya, Archaea, and Bacteria.

domain

G A general, non-definitive name used for an organism.

kingdom

H A term describing the variation of life at all levels of biological organisation.

scientific name

I A definitive name for an organism, which is unchanging from country to country. J A taxonomic group denoting organisms that are so genetically alike they can interbreed to produce viable offspring.

shared derived characteristics

K The active management of natural populations in order to rebuild numbers and ensure species survival.

species

L The two part scientific system that is used to identify organisms to genus and species.

taxonomic rank

2. Give some reasons for maintaining areas of both high biodiversity (e.g rainforest) and low biodiversity (e.g. farmland).

(a) Rainforest:

(b) Farmland:

3. List the following in the correct order (least specific to most specific): kingdom, order, class, species, domain, family, genus, phylum:

(b) Compare your groupings and evolutionary history with others in your class. Have they come up with similar groupings to you? Discuss why and why not.

A

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B

C

D

E

F

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4. (a) Study the shapes below. On a separate sheet divide them into groups and explain the basis for your groups. Use your groupings to draw a family tree or evolutionary history of the shapes (i.e explain how they are all related to each other). Staple your sheet to this page.

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Relationships between organisms

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Unit 1 Outcome 2

Key terms

Relationships between species

abundance

Key knowledge

amensalism autotroph

c

1

birth rate (natality) carnivore

Activity number

Describe how species within ecosystems can interact in ways that are harmful, neutral, or beneficial to one or both parties. Define and describe examples of amensalism, commensalism, mutualism, predation, parasitism, and competition. Identify which of these relationships is classified as an exploitation.

133

carrying capacity commensalism competition consumer

death rate (mortality) density

distribution ecosystem

Sagt

exploitation

Food webs and keystone species

herbivore

Key knowledge

heterotroph

c

2

Explain how the interdependent relationships between species are represented by food chains and food webs. Distinguish between producers, consumers, detritivores, and decomposers. Describe the role of each of these in energy and biomass transfers in ecosystems.

134 135

c

3

Construct food chains and food webs for named communities.

136-138

c

4

Explain what is meant by a keystone species and describe examples. Explain how keystone species are pivotal to ecosystem function because of their role in some critical ecosystem function, e.g. as a top predator or seed dispersal agent.

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Factors affecting population density and distribution

Activity number

intraspecific competition keystone species mutualism omnivore

parasitism

population

population size

Key knowledge

predation producer sample

140 141

c

5

Define abundance. Distinguish between the density, distribution, and size of a population of a species within an ecosystem.

c

6

Understand that ecosystems have a certain carrying capacity, which may not be static, but may vary with the availability of resources offered by the environment.

146

c

7

Describe and explain the impact of factors affecting the density, distribution, and size of populations within ecosystems. Include reference to the availability of resources, including food and space, predation, competition, disease, chance environmental events, births, deaths, and migration.

146 148 149-152 154 155

c

8

Use a simple equation to calculate change in a population's size based on changes in births, deaths, and migration.

c

9

With reference to your chosen practical investigation, describe how the distribution and abundance of organisms in an ecosystem can be measured, including reference to quadrats and transects.

c

10

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transect

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migration

Use appropriate sampling methods to investigate factors affecting the distribution and abundance of one or more species in an ecosystem.

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interspecific competition

Activity number

141 144 145 153

141-145


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133 Species Interactions

Key Idea: Every species interacts with others. The interactions usually, but not always, benefit at least one of the species. Species interact with other species. The nature and outcome of these interactions structures communities and leads to interdependence. Interactions range from those that are beneficial to all parties, to those where only one species

benefits. In a relationship involving exploitation, one party benefits at the expense of another. Sometimes, neither party benefits from the interaction. Such interactions include amensalism (below) or competition, a relationship in which two parties (which may be the same or different species) directly or indirectly contest the same resource (e.g. food).

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Type of interaction between species

Mutualism

A

B

Benefits

Benefits

Both species benefit from the association.

Examples: Tick bird on zebra removes parasites and alerts zebra to danger, while tick bird gains access to food. Flowering plants and their insect pollinators have a mutualistic relationship. Flowers are pollinated and the insect gains food (below).

Commensalism

A

B

Benefits

Unharmed

One species benefits and the other is unaffected by the association.

Examples: Remora are fish with special sucker organs to attach to other marine animals such as sharks or turtles. The remora saves energy by hitching a ride on the other animal, and the shark or turtle is unharmed.

Exploitation

Amensalism

A

Unaffected

Predation

Parasitism

B

A

B

A

B

Harmed

Benefits

Harmed

Benefits

Harmed

One species incidentally harms the second species but does not obtain any benefit from the interaction.

Examples: Algal blooms can lead to the death of fish and other aquatic organisms by depleting the water of oxygen or producing toxins. However, the algae do not benefit from the deaths of the fish.

Predator kills the prey outright and eats it.

The parasite lives in or on the host, taking (usually all) its nutrition from it. The host is harmed but usually not killed.

Examples: Lion preying on wildebeest or praying mantis (below) consuming insect prey. The adaptations of predators and prey are the result of their close ecological relationship throughout their evolution: predators have adaptations to capture prey and prey have adaptations to avoid capture.

Examples: Pork tapeworm in a pig's gut. Some plants (e.g. mistletoes) are semi-parasitic (hemiparasites). They photosynthesise but rob the host plant of nutrients and water.

Luc Viatour www.Lucnix.be

Honeybee and flower

Remora attached to turtle Algal bloom

Mantid eats cricket

Pork tapeworm

1. Summarise your knowledge of species interactions by completing the following, entering a (+), (–), or (0) for each species, and writing a brief description of each relationship. Codes: (+): species benefits, (–): species is harmed, (0): species is unaffected.

Interaction

Species

A

B

Description of relationship

(c) Amensalism (d) Parasitism (e) Predation (f) Competition

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(b) Commensalism

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(a) Mutualism

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2. The honeyeaters are a diverse family of small to medium-sized nectar-feeding birds common in Australia. Many Australian plant species, including proteas and myrtles, are pollinated by honeyeaters.

(a) Identify this type of interaction:

(b) Describe how each species is affected (benefits/harmed/no effect):

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Examples of interactions between different species are illustrated below. For each example, identify the type of interaction, and explain how each species in the relationship is affected.

3. The squat anemone shrimp, also known as the sexy shrimp, lives among the tentacles of sea anemones, where it gains protection and scavenges scraps of food from the anemone. The anemone is apparently neither harmed nor benefitted by the shrimp's presence. (a) Identify this type of interaction:

(b) Describe how each species is affected (benefits/harmed/no effect):

4. Dingoes will kill and scavenge a range of species. In groups of two or more, they can attack and kill large animals, such as kangaroos, but will also scavenge carrion, such as this dingo with a fish on Fraser Island. (a) Identify this type of interaction:

(b) Describe how each species is affected (benefits/harmed/no effect):

Marc Tarlock cc 2.0

5. The Australian paralysis tick, Ixodes holocyclus, lives attached to the skin of mammalian hosts, commonly bandicoots, koalas, possums, and kangaroos, where it sucks body fluids and causes irritation. Most native species are immune to the tick's toxins but it can cause paralysis in susceptible species.

(a) Identify this type of interaction:

(b) Describe how each species is affected (benefits/harmed/no effect):

6. Large herbivores expose insects in the vegetation as they graze. The cattle egret, which is widespread in tropical and subtropical regions, follows the herbivores as they graze, feeding on the disturbed insects when the herbivore moves away. (a) Identify this type of interaction:

(b) Describe how each species is affected (benefits/harmed/no effect):

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7. Explain the similarities and differences between a predator and a parasite:

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134 Feeding Relationships this reason, food chains usually have fewer than six links. An organism is assigned to a trophic level based on its position in the food chain, but they may occupy different trophic levels in different food chains or during different stages of their life. Arrows link the organisms in a food chain. The direction of the arrow shows the flow of energy through the trophic levels. Most food chains begin with a producer, which is eaten by a primary consumer (herbivore). Higher level consumers (carnivores and omnivores) eat other consumers.

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Key Idea: A food chain is a model to illustrate the feeding relationships between organisms. Organisms in ecosystems interact by way of their feeding (trophic) relationships. These interactions can be shown in a food chain, which illustrates how energy, in the form of food, passes from one organism to the next. The levels of a food chain are called trophic levels. Energy flows through trophic levels rather inefficiently, with only 5-20% of the energy at one level being transferred to the next (the rest is lost as heat). For

Producers Trophic level: 1

Respiration

Herbivores Trophic level: 2

Carnivores Trophic level: 3

Carnivores Trophic level: 4

Detritivores and decomposers

The diagram above represents the basic elements of a food chain. In the questions below, you are asked to add to the diagram the features that indicate the flow of energy through the community of organisms. 1. (a) State the original energy source for this food chain:

(b) Draw arrows on the diagram above to show how the energy flows through the organisms in the food chain. (c) Label each of the arrows with the process that carries out this transfer of energy. (d) Draw arrows on the diagram to show how the energy is lost as heat by way of respiration. 2. (a) What happens to the amount of energy available to each successive trophic level in a food chain?

(b) How does this limit the number of links in a food chain?

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3. Discuss the trophic structure of ecosystems, including reference to food chains and trophic levels:

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4. What could you infer about the trophic level(s) of the kingfisher if it was found to eat both katydids and frogs?

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135 Food Webs

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Key Idea: A food web shows how the food chains of a community are interconnected. The complexity of a food web depends on the number of organisms, food chains, and trophic levels present. In any community, no species exists independently of others. All organisms, dead or alive, are potential sources of food for other organisms. Within a community, there are hundreds of

feeding relationships, and most species participate in several food chains. The different food chains in an ecosystem tend to form food webs, a complex series of interactions showing the feeding relationships between the organisms in an ecosystem. A food web model (below) can be used to show the trophic linkages between different organisms in a community and can be applied to any ecosystem.

The complexity of food webs varies

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Key to food web (below)

The complexity of feeding relationships in a community contributes to its structure and specific features. A simple community, like those that establish on bare soil after a landslide, will have a simpler web of feeding relationships than a mature forest.

Flow of nutrients from the living components to detritus or the nutrient pool.

Consumer–resource interactions. Losses of each food web component from the system and external input of limiting nutrients.

A simple food web

(a)

Organisms whose food is obtained through the same number of links belong to the same trophic level.

Kingfisher

Species interact in complex ways as competitors, predators, and symbionts.

Yabby

Tadpole

(b)

(e)

Only 5-20% of usable energy is transferred to the next trophic level. For this reason, food chains rarely have more than six links.

Autotrophic protist

(c)

(d)

Nutrients cycle between the atmosphere, the Earth’s crust, water, and living organisms.

Nutrient pool

Energy flows through ecosystems in the high energy chemical bonds in organic matter.

1. (a) - (e) Complete the food web above by adding the labels: carnivore, herbivore, autotroph, detritus, detritivore.

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3. In what way are different communities and different ecosystems linked?

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2. Why would a newly established community have a much simpler food web than a more established mature community?

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136 Constructing a Food Web

Key Idea: The many food chains in a community can be organised into food webs to show all the feeding interactions. A food web depicts all the interconnected food chains in an ecosystem. For the lake community below, assemble the organisms into a food web to illustrate their trophic interactions. Remember that species are assigned to trophic levels on the basis of what they eat, with the first trophic

level (the producers), ultimately supporting all other levels. Consumers are ranked according to the trophic level they occupy, although some may feed at several different trophic levels. For this example, the detritus (accumulated dead organic matter) has been omitted, but it provides a rich source of nutrients for many organisms, including yabbies and ducks.

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Feeding requirements of lake organisms

Daphnia (Daphnia spp.)

Autotrophic protists

Mallard duck (Anas platyrhynchos)

Small freshwater crustacean that forms part of the zooplankton. It feeds on planktonic algae by filtering them from the water with its limbs.

e.g. Chlamydomonas (left), Euglena (right) Microscopic, autotrophic protists. Two of many species that form the phytoplankton.

Feed mostly on water plants. Sifts fallen seed, aquatic plants and small animals from the muddy margins of the lake.

Yabby (Cherax destructor)

These freshwater crayfish filter water to extract zooplankton (Daphnia), feed on macrophytes, and stir up detritus to scavenge from it.

Macrophytes (various species)

Kingfisher (Alcedo)

A variety of flowering aquatic plants are adapted for being submerged, freefloating, or growing at the lake margin.

Although not restricted to the lake community, kingfishers feed on small fish, yabbies, and tadpoles.

Brown trout (Salmo trutta)

Diving beetle (Dytiscus)

Diving beetles (adults and larvae) feed on aquatic insect larvae and adult insects blown into the lake community.

Feed on zooplankton, freshwater crayfish, aquatic insect larvae, and insects blown into the water from the surrounding area.

Tadpole (immature frog, Litoria spp.) Feed on algae and very small zooplankton. Adult frogs feed on terrestrial (land dwelling) invertebrates.

Mosquito larva (Culex spp.)

Pelicans feed on fish of varying sizes, including smaller trout and silver perch.

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Platypus (Ornithorhynchus anatinus) A unique nocturnal mammal that feeds on insects, molluscs, worms, other invertebrates and small vertebrates.

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Pelican (Pelecanus conspicillatus)

Humans (Homo sapiens)

Humans most commonly exploit the lake community food sources by taking trout and yabbies. Traditionally, aboriginal populations have fed on yabbies and fish.

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Silver perch (Bidyanus bidyanus)

Omnivorous, feeding on a range of invertebrates, including crustaceans, aquatic insects, and molluscs, but also some vegetation.

The larvae of most mosquito species, e.g. Culex, feed on planktonic algae and small protozoans before passing through a pupal stage and undergoing metamorphosis into adult mosquitoes.

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1. From the information provided for the lake food web components on the previous page, construct ten different food chains (using their names only) to show the feeding relationships between the organisms. Some food chains may be shorter than others and most species will appear in more than one food chain. An example has been completed for you. Example 1:

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Macrophyte

Yabby

Kingfisher

Brown trout

(a) (b) (c)

(d) (e) (f)

(g) (h) (i) (j)

2. (a) Use the food chains that you have created above to help you to draw up a complete food web for this community. Use only the supplied information to draw arrows showing the flow of energy between species. (NOTE: Only energy from (not to) the detritus is required) (b) Label each species with the following codes to indicate its trophic level and status: Indicate: • Diet type: P = Producer, H = Herbivore, C = Carnivore, O = Omnivore (Note: based on the information given). • Position in the food chain as a consumer (1st, 2nd, 3rd, 4th order consumer): 1–4 (does not include producers). Example: Mosquito larva is H1

Trophic level 4

Trout

Silver perch

Trophic level 3

Tadpole

Diving beetle

Trophic level 2

Daphnia

Mosquito larva

Trophic level 1

Autotrophic protists

Humans

Platypus

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Kingfisher

Detritus

Trophic level 5

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137 Dingo Food Webs

Key Idea: The habitats and trophic interactions of dingoes vary regionally, depending on the community composition. Dingoes are widespread in Australia as top predators. Dingoes are members of different communities in different

parts of Australia, and the food webs they are part of also vary regionally. Samples taken from six different locations (below) show how the prey taken by dingoes varies from one location to another. Location of sampling sites Kakadu National Park Barkley Tableland Fortesque River Harts Ranges Eldunda Simpson Desert Nullarbor Plain George's Creek Nature Reserve Kosciusko National Park Victorian Highlands Nadgee Nature Reserve

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1 2 3 4 5 6 7 8 9 10 11

Wet-dry tropics North Australia

6722 faeces

Dusky rat 33.8 Magpie goose 32.5 Agile wallaby 15.1 Northern ringtailed possum 9.7 Grass species 7.1 Feral water buffalo 5.8 Feral pig 3.5 Unidentified matter 2.6 Antilopine wallaroo 1.8 Northern brown bandicoot 1.4 Feral cattle 1.3 Bird (undetermined species) 0.7 Insect (undetermined species) 0.7 Beetle 0.4 Semi-arid north-west Australia

413 faeces/ stomachs

Red kangaroo/euro 80.6 Cattle 11.4 Sheep 8.0 Bird (undetermined species) 5.6 Reptile (undetermined species) 3.4 Insects (undetermined species) 2.9 Echidna 2.2 Dingo 1.7 Feral cat 0.5 Bat (undetermined species) 0.2 Fish (undetermined species) 0.2 Red fox 0.2 Rothschild's rock wallaby 0.2

Arid & semi-arid central Australia

1480 stomachs

Rabbit 37.9 Cattle 23.3 Long-haired rat 17.6 Red kangaroo 10.2 Central netted dragon 7.8 Small mammal (undetermined sp.) 3.8 House mouse 3.6 Grasshopper 2.7 Bearded dragon 2.2 Zebra finch 2.1 Bird (undetermined species) 2.0 Feral cat 1.8 Galah 1.8 Budgerigar 1.5

Cool coastal mountains 2063 faeces/ stomachs SE Australia

Swamp wallaby 17.9 Wallaby (undetermined spp.) 15.8 Wombat 15.0 Animal remains (unidentified) 11.5 Rabbit 10.5 Common ringtail possum 8.0 Waterbird (undetermined sp.) 7.7 Red-neck wallaby 5.3 Possum (undetermined species) 5.1 Rat (undetermined species) 5.0 Little penguin 4.4 Fish (undetermined species) 3.7 Mutton bird 3.6 Echidna 3.3

Adapted from: Corbett, L. 1995. The dingo in Australia and Asia, Appendix C: pp. 183-186. University of NSW Press

Arid south-west Australia

131 faeces/ stomachs

Rabbit 63.4 Red kangaroo 32.1 Cattle 7.6 Red fox 3.8 Little crow 2.3 Bobtail skink 1.5 Feral cat 1.5 Centipede/millipede 0.8 Dingo 0.8 Grasshopper 0.8

Humid coastal mountains 1993 faeces eastern Australia

Swamp wallaby Bush rat Red-necked wallaby Brushtail possums Bandicoots

30.5 12.2 11.1 6.9

(long-nosed, Southern brown) 6.8 Rabbit 6.4 Antichinuses (brown, dusky) 5.8 Parma wallaby 4.6 Common ringtail possum 4.4 Ring-necked pademelon 3.8 Echidna 3.5 Long-nosed potoroo 1.7 Greater glider 1.5

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The table below shows the diet of dingoes in six major Australian habitats. The numbers represent the proportion (% occurrence) of stomachs or faeces that contained each prey species. A total of 12,802 stomachs and faeces were sampled over a 20 year period (1966-1986).

Despite the immense range of potential prey species across Australia, only ten species formed almost 80% of a dingo's diet. Dingoes are specialists, rather than generalists, with respect to dietary intake.

1. Sites 3 and 7 (above) yielded a small number of prey species in the samples taken. Suggest the likely reason for this:

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2. Name three prey species that are taken by dingoes in the ‘Cool coastal mountains, SE Australia’ (sites 9, 10, and 11), that are restricted to that type of environment (i.e. not represented in the prey taken at other sites):

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3. What evidence is there from the data on the previous page, that dingoes engage in cannibalism?

4. Which general kind of prey makes up most of the dingoes’ diet? 5. At some sample sites, the dingoes’ prey included domesticated animals. (a) Which prey items represent domesticated animals?

(b) What kind of environmental conditions have encouraged dingoes to make these animals part of their diet?

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6. (a) Which of the sample sites has the least reliable data for indicating the diet of dingoes in its area?

(b) Explain your choice:

7. In this study, the diet of dingoes was determined by the sampling methods of examining large numbers of stomach contents and faeces.

(a) Explain which of these two methods should prove the most reliable for positive identification of prey species:

(b) Suggest two reasons why the researchers did not simply follow the dingoes and watch what they ate as a way of gathering dietary information on the dingoes:

Reason 1: Reason 2:

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8. Using the data on the previous page, choose one of the ‘regional ecotypes’ (e.g. wet-dry tropics, north Australia) and produce a food web in the space below. Use only the first five positively identified prey species (in most cases, do not include unidentified species). This activity will require you to carry out some research into what the prey species eat.

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138 Cave Food Webs (6) is a predator of the dung beetle, the millipede, and the cricket. Inside the cave, the horseshoe bat (7) roosts and breeds, leaving the cave to feed on flying insects. The bats produce vast quantities of guano (droppings). The guano is eaten by the blind cave beetle (8), millipede (4) and springtail (9). These invertebrates are hunted by the predatory cave spider (10). In underground pools, the bat guano supports the growth of bacteria (11). Flatworms (12) and isopods (13) feed on the bacteria and are eaten by the blind cave shrimp (14). The blind cave fish (15) is the top predator in this system, feeding on isopods and the blind cave shrimps.

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Key Idea: The food webs of cave ecosystems are fragile and based on only a few resources. Cave environments lack the light that sustains most ecosystems. Despite this, a wide range of animals are adapted to live there. Some animals, such as bats, are not permanent cave-dwellers, but rest and breed there. Around the entrance of the cave, the owl (1) preys on the mouse (2) which itself feeds on the vegetation outside the cave. The owl and the mouse leave droppings that support the cave dung beetle (3) and millipede (4). The cave cricket (5) scavenges dead birds and mammals near the entrance. The harvestman

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Guano and detritus layer

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1. Using the lake food web activity as a guide, construct a food web for the cave ecosystem on a separate sheet of paper. For animals that feed outside the cave, do not include this outside source of food. As in the lake food web, label each species with the following codes to indicate its diet type (e.g. producer, detritivore, herbivore, carnivore, omnivore) and its position in the food chain if it is a consumer (1st, 2nd, 3rd, 4th order consumer). Staple your finished web to this page. 2. Which major trophic level is missing from the cave food web? 3. How is energy imported into the cave’s food web?

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4. How might energy be removed from the cave ecosystem?

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5. In many parts of the world, cave-dwelling bat species are endangered, often taken as food by humans or killed as pests. Explain how the cave food web would be affected if bat numbers were to fall substantially:

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139 The Role of Keystone Species

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Key Idea: All organisms within an ecosystem contribute to its structure and functioning, but keystone species have a disproportionate effect on ecosystem processes. Although every species has a role in ecosystem function, some have a disproportionate effect on ecosystem processes and stability (how unchanging the ecosystem is over time). These species are called keystone species and they are important

because they play a pivotal role in the way the ecosystem works, e.g. as top predators or by recycling nutrients. The loss of a keystone species can have a large and rapid impact on the structure and function of an ecosystem, changing the balance of relationships and leading to instability. This has important implications for ecosystem management because many keystone species are endangered.

Why are keystone species important?

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A keystone species is one that plays a unique and crucial role in the way an ecosystem functions. Often, but not always, keystone species are top predators. The role of the keystone species varies from ecosystem to ecosystem, but the loss of a keystone species from any ecosystem has a domino effect, and a large number of species can be affected. This can lead to can rapid ecosystem change or the collapse of the ecosystem completely.

Ochre starfish: Paine removed these in his study to see what the effect would have on the rocky shore community.

Keystone species in action

The idea of the keystone species was first hypothesised in 1969 by Robert Paine. He studied an area of rocky seashore, noting that diversity seemed to be correlated with the number of predators (ochre starfish) present (i.e. diversity declined as the number of predators declined).

Top predators, such as Australia's dingo, are often keystone species. Many conservationists regard dingoes as a functional replacement for native predators that are now extinct, such as the Tasmanian tiger. Dingoes have a varied diet and are a major constraint on introduced species, such as a foxes and pigs, thereby helping to maintain native mammal diversity.

The endangered southern cassowary is a key species in Australia's wet tropics. They are obligate frugivores, and their gut passes seeds, unharmed, into a pile of manure. More than 200 plant species depend on the cassowary to disperse their seeds, yet their populations are all declining. Their loss would also mean the loss of an ecological role.

1. Why are keystone species so important to ecosystem function?

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Keystone: critical food source

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Keystone: seed dispersal

All species of banksias produce large amounts of nectar, and are a vital component of food chains in the Australian bush. In the Avon Wheatbelt region of Western Australia, the acorn banksia is the sole source of nectar for honeyeaters at certain times of the year. The loss of this plant species would also mean the loss of honeyeaters from the region.

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Keystone: predator

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The term keystone species comes from the analogy of the keystone in a true arch (above). An archway is supported by a series of stones, the central one being the keystone. If the keystone is removed the arch collapses.

To test this he removed the starfish from an 8 m by 2 m area of seashore. Initially, the barnacle population increased rapidly before collapsing and being replaced by mussels and gooseneck barnacles. Eventually the mussels crowded out the gooseneck barnacles and the algae that covered the rocks. Limpets that fed on the algae were lost and the number of species present in the study area dropped from 15 to 8.

istockphotos.com

Keystone

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Marjorie Lundgren cc3.0

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Australian keystone species

Tiger shark

Many sharks are top predators and are keystone species in the waters around Australia. One shark species inhabiting Shark Bay (WA) is the tiger shark. It doesn't even have to kill its prey to exert an effect on ecosystem structure. The presence of the tiger shark causes marine herbivores such as green turtles and dugongs to avoid the area or to spend less time grazing because they are looking out for the sharks. As a result, the seagrass meadows thrive and support many more species than would be possible if they were grazed intensively by herbivores. As a result, biodiversity in Shark Bay is high. Fishing is the main threat to tiger sharks as they hunted for their flesh, fins, and skin. Finning, although largely banned in Australian waters, still continues illegally.

Cockatoo grass

Grey-headed flying fox

The grey-headed flying fox (Pteropus poliocephalus) is found in a variety of habitats along the east coast of Australia, including Victoria. The greyheaded flying fox feeds on the fruit and nectar of over 180 species of trees, including Australian natives Eucalyptus, Banksia, palms, and myrtles. It will fly up to 50 km each night looking for food and this allows it to fulfill an important ecological role by dispersing the pollen and seeds of a wide range of plants. Its role is especially important in the subtropical rainforests as it is the only mammalian species to consume nectar and fruit in these regions. The species is under threat from the loss of foraging and roosting habitat and control measures by horticulturists to prevent crop losses.

Cockatoo grass (Alloteropsis semialata) is found through tropical savannas in northern and north eastern Australia. Cockatoo grass is an early developer in the wet season, providing a food source to many animal species before other plant species are available. Cockatoo grass is considered to be a keystone species because at certain times of the year it is the only food source available for two endangered species, the golden-shouldered parrot and the Northern bettong, a small marsupial. Young cockatoo grass is a preferred food source cattle and pigs, so it is easily overgrazed, leaving little for the wild species that rely on it. Conservation efforts are made to protect stands of cockatoo grass in some areas.

(a) Acorn banksia:

(b) Southern cassowary:

(c) Dingo:

(d) Tiger shark:

(e) Cockatoo grass:

(f) Grey-headed flying fox:

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3. Elephants are keystone species on the African savanna, maintaining the grassland by pulling down the scattered trees that grow on the savanna for food (right). Predict what would happen if the elephants did not carry out this role:

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2. For each species below, summarise the features of its ecology that contribute to its position as a keystone species:


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140 Population Density and Distribution

Key Idea: Population density is the number of organisms of one species in a specified area. Distribution describes how the organisms are distributed relative to each other. A population is defined as all the members of the same species in a particular geographical area (and therefore

Random distribution

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Low density

capable of interbreeding). Populations have characteristics, such as density, distribution, and age structure, not shown by individuals. Population density is the number of individuals per unit area or volume. The way the individuals are spaced in the physical environment is called the population distribution.

In low density populations, individuals are spaced well apart. There are only a few individuals per unit area or volume (e.g. highly territorial, solitary mammal species).

High density

In high density populations, individuals are crowded together. There are many individuals per unit area or volume (e.g. colonial organisms, such as many corals).

A random distribution describes an irregular spacing between individuals. The presence of one individual does not directly affect the location of any other. Random distributions are uncommon in animals but are often seen in plants.

Clumped distribution

Clumped distributions occur when individuals are grouped in patches (often around a resource). The presence of one individual increases the probability of finding another close by. Such distributions occur in herding and social species.

Uniform distribution

Tigers are solitary animals, found at low densities.

Termites form well organised, high density colonies.

Regular distribution patterns occur when individuals are evenly spaced within the area. The presence of one individual decreases the probability of finding another individual very close by. The gannets above are also at a high density.

1. (a) How could the distribution of resources lead to organisms having a clumped distribution pattern?

(b) How could a group social behaviour lead to organisms having a clumped distribution pattern?

4. Describe an example of each of the following types of distribution pattern:

(a) Clumped:

(b) Random (more or less):

(c) Uniform (more or less): WEB

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3. What type of environment would encourage uniform distribution?

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2. What type of behaviour might encourage a uniform distribution of an animal species?

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141 Measuring Distribution and Abundance

Key Idea: Random sampling using an appropriate technique provides unbiased information about the distribution and abundance of one or more species in a community. Most practical exercises in ecology involve collecting data about the distribution and abundance of one or more species in a community. Most studies also measure the physical factors in the environment as these may help to explain the

patterns of distribution and abundance observed. The use of random sampling methods, in which every possible sample of a given size the same chance of selection, provides unbiased data. As long as the sample size is large enough and the sampling technique is appropriate to the community being studied, sample data enables us to make inferences about aspects of the whole population.

Distribution and abundance

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Ecological sampling collects data about where organisms are found and how they are distributed in the environment. This information can be used to determine the health and viability of a population and its ecosystem. When investigating populations it is useful to monitor: Species distribution (where the species are located)

Species abundance (how many of a species there are)

Systematic (grid)

Area sampling using quadrats A quadrat is a sampling tool that provides a known unit area of sample (e.g. 0.5 m2). Quadrats are placed randomly or in a grid pattern on the sample area. The presence and abundance of organisms in these squares is noted. Quadrat sampling is appropriate for plants and slow moving animals and can be used to evaluate community composition.

Environmental gradient

Line transects A tape or rope marks the line. The species occurring on the line are recorded (all along the line or at regular points). Lines can be chosen randomly (left) or may follow an environmental gradient. Line transects have little impact on the environment and are good for assessing the presence/absence of plant species. However, rare species may be missed.

0.5 m

Belt transects A measured strip is located across the study area and quadrats are used to sample the plants or animals at regular intervals along the belt. Belt transects provide information on abundance and distribution as well as presence/absence. Depending on the width of the belt and length of the transect, they can be time consuming.

Mark and recapture sampling Animals are captured, marked, and released. After a suitable time, the population is resampled. The number of marked animals recaptured in a second sample is recorded as a proportion of the total. Mark and First sample: Second sample: recapture is useful for highly mobile species marked proportion which are otherwise difficult to record. recapture However, it is time consuming to do well.

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2. Name a sampling technique that would be appropriate for determining: (a) Percentage cover of a plant species in pasture:

(b) Change in community composition from low to high altitude on a mountain:

(c) Association of plant species with particular soil types in a nature reserve:

3. Why is it common practice to also collect information about the physical environment when sampling populations?

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Point sampling Individual points are chosen (using a grid reference or random numbers applied to a map grid) and the organisms are sampled at those points. Point sampling is most often used to collect data about vegetation distribution. It is time efficient and good for determining species abundance and community composition, however, organisms in low abundance may be missed.

1. Distinguish between distribution and abundance:

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Sampling designs and techniques

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The methods used to sample communities and their constituent populations must be appropriate to the ecosystem being investigated. Communities in which the populations are at low density and have a random or clumped distribution will require a different sampling strategy to those where the populations are uniformly distributed and at higher density. There are many sampling options (below), each with advantages and drawbacks for particular communities.

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142 Investigating Survival: Field Based

Key Idea: A field study should be based on random sampling and collection of sufficient data to test the hypothesis. The figure below provides an example and some ideas for

designing a field study. It provides a framework, which can be modified for most simple comparative field investigations. For reasons of space, the full methodology is not included.

Observation A student read that a particular species of native pill millipede is extremely abundant in forest leaf litter, but a search in the litter at the edge of a eucalypt woodland revealed only very low number of millipedes.

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Hypothesis and aim

Hypothesis: If native millipedes are adapted to a niche in eucalypt leaf litter, more should be found in the middle of a eucalypt woodland than at the edges.

Pill millipede Cyliosoma sp.

Eucalypt woodland

Sampling sites 2

8

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8m

6

3

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20 m

6

A sampling programme was designed to test the prediction that native millipedes would be more abundant in the leaf litter near the middle of a eucalypt woodland than in the leaf litter at the edge of the woodland.

3

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Sampling programme

5

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Sampling sites numbered 1-8 at evenly spaced intervals on a 2 x 2 m grid within an area of 20 m x 8 m. In m pattern or transect is situations like this, where20 a grid used, the choice of grid or transect position must be made randomly.

Sampling equipment: leaf litter light trap

Light from a battery operated lamp drives the invertebrates down through the leaf litter.

Large (diameter 300 mm) funnel containing leaf litter resting on a gauze platform.

Gauze allows invertebrates of a certain size to move down the funnel.

Equipment and procedure

Sites: For each of the two sampling areas (middle and edge of the woodland), an area 20 m x 8 m was randomly chosen and marked out in a 2 x 2 m grid. Eight sampling sites were selected, evenly spaced along the grid. • The general area for the study chosen was selected on the basis of the large amounts of leaf litter present. • Eight sites were chosen as the largest number feasible to collect and analyse in the time available. • The two areas of woodland were sampled on sequential days.

Capture of millipedes: At each site, a 0.4 x 0.4 m quadrat was placed on the forest floor and the leaf litter within the quadrat was collected. Millipedes and other leaf litter invertebrates were captured using a simple gauze lined funnel containing the leaf litter from within the quadrat. A lamp was positioned over each funnel for 2 hours and the invertebrates in the litter moved down and were trapped in the collecting jar. • After 2 hours each jar was labelled with the site number and returned to the lab for analysis.

• The litter in each funnel was bagged, labelled with the site number and returned to the lab for weighing. • The number of millipedes at each site was recorded. • The numbers of other invertebrates (classified into major taxa) were also noted for reference.

• Average (mean) millipede abundance was calculated from the counts from the eight sites.

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Aim: To determine if data collected from invertebrate populations in the middle and at the edge of a eucalypt forest support this hypothesis, and to attempt to explain any patterns evident in the data.

Ethical considerations

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Collecting jar placed in the litter on the forest floor traps the invertebrates that fall through the gauze and prevents their escape.

• Minimise damage to the habitat when sampling. • Minimise handling to reduce stress to the organism. • Where possible, return all the collected organisms to the sites they were collected from. • Do not sample from the same site continually as this may affect population size.

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Sampling strategies

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In most ecological studies, it is not possible to measure or count all the members of a population. Instead, information is obtained through sampling in a manner that provides a fair (unbiased) representation of the organisms present and their distribution. This is usually achieved through random sampling. Sometimes researchers collect information by non-random sampling, a process that does not give all the individuals in the population an equal chance of being selected. While faster and cheaper to carry out than random sampling, non-random sampling may not give a true representation of the population.

Group 1

Group 2

Systematic sampling

Stratified sampling

Opportunistic sampling

Samples from a larger population are selected according to a random starting point and a fixed, periodic sampling interval. For the example above, the sampling period is every fourth individual. Systematic sampling is a random sampling method, provided the periodic interval is determined beforehand and the starting point is random.

Stratified sampling divides the population into subgroups before sampling. The strata should be mutually exclusive, and individuals must be assigned to only one stratum. Stratified sampling is used to highlight a specific subgroup within the population. Individuals are then randomly sampled from the strata to study.

A non-random sampling technique in which subjects are selected because of they are easily accessible to the researcher. Opportunistic sampling excludes a large proportion of the population and is usually not representative of the population. It is sometimes used in pilot studies to gather data quickly and with little cost.

Example: Selecting individuals from a patient list.

Example: Dividing the population into males and females.

Example: Selecting 13 people at a cafe where you are having lunch.

1. (a) Explain the importance of recognising assumptions when designing a field study:

(b) List assumptions that may have been made during this study regarding:

The location of the areas sampled:

Size of the areas sampled:

Accuracy of the results:

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(a) Systematic sampling:

(b) Stratified sampling:

(c) Opportunistic sampling:

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2. A student wants to investigate the incidence of asthma in their school. Describe how they might select samples from the school population using:


143 Investigating Distribution and Abundance

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Key Idea: Sampling populations in-situ can reveal patterns of distribution, which can be attributed to habitat preference. These investigations are common in ecological studies. The aim

Use this activity to practise analysing data from a field study in which the aim was to identify and describe an existing pattern of species distribution.

Pill millipede Cyliosoma

To investigate the effect of fallen tree logs on the distribution of pill millipedes in a forest.

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Background

Millipedes consume decaying vegetation and live in the moist conditions beneath logs and in the leaf litter of forest floors. The moist environment protects them from drying out as their cuticle is not a barrier to water loss.

Experimental method

The distribution of millipede populations in relation to fallen tree logs was investigated in a small forest reserve. Six logs of similar size were chosen from similar but separate regions of the forest. Logs with the same or similar surrounding environment (e.g. leaf litter depth, moisture levels) were selected. For each log, eight samples of leaf litter at varying distances from the fallen tree log were taken using 30 cm2 quadrats. Samples were taken from two transects, one each side of the log. The sample distances were: directly below the log (0 m), 1.5 m, 2.5 m, and 3.5 m from the log. It was assumed that the conditions on each side of the log would be essentially the same. The leaf litter was placed in Tullgren funnels and the invertebrates extracted. The number of millipedes in each sample was counted. The raw data are shown below.

Experimental setup

TRANSECT 1

TRANSECT 2

Distance from log (m)

3.5

2.5

1.5

Distance from log (m)

0

0

1.5

2.5

3.5

Tree log

Centre-line of log (where possible)

cm2

30 quadrats

Environmental conditions for each transect position either side of the log were assumed to be equal.

Raw data for tree log and millipede investigation

Distance from log (m)

0

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Distance from log (m)

Tree log Transect 4

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Tree log

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1. Plot column graphs on the grids below to show the distribution of millipedes at each log. Plot transect 1 and 2 data separately for each graph: Log 2

Log 3

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Log 1

Log 4

Log 5

Log 6

2. Study the graphs and answer the following questions:

(a) Is there a relationship between distance from the tree log and the number of millipedes found?

(b) What physical factors might account for this?

(a) Do you think this assumption is valid?

(b) Explain the reason you have given in (a):

(c) How could you test the assumption?

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3. During this investigation an assumption was made that the environmental factors were the same on each side of the log.


144 Quadrat Sampling

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Key Idea: Quadrat sampling involves a series of random placements of a frame of known size over an area of habitat to assess the abundance or diversity of organisms. Quadrat sampling is a method by which organisms in a certain proportion (sample) of the habitat are counted directly. It is used when the organisms are too numerous to count in total. It can be used to estimate population abundance (number), density, frequency of occurrence, and distribution. Quadrats may be used without a transect when studying a relatively uniform habitat. In this case, the quadrat positions are chosen randomly using a random number table. The general procedure is to count all the individuals (or estimate their percentage cover) in a number of quadrats of known size and to use this information to work out the abundance or percentage cover value for the whole area.

Quadrat

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Area being sampled

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Estimated = average density

Total number of individuals counted

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Number of quadrats X area of each quadrat

4 5

Guidelines for quadrat use:

6

1. The area of each quadrat must be known. Quadrats should be the same shape, but not necessarily square.

7

2. Enough quadrat samples must be taken to provide results that are representative of the total population.

9

3. The population of each quadrat must be known. Species must be distinguishable from each other, even if they have to be identified at a later date. It has to be decided beforehand what the count procedure will be and how organisms over the quadrat boundary will be counted.

8

10

Quadrats are applied to the predetermined grid on a random basis. This can be achieved by using a random number table.

The area to be sampled is divided up into a grid pattern with indexed coordinates

4. The size of the quadrat should be appropriate to the organisms and habitat, e.g. a large size quadrat for trees.

5. The quadrats must be representative of the whole area. This is usually achieved by random sampling (right).

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1

Sampling a centipede population

A researcher by the name of Lloyd (1967) sampled centipedes in Wytham Woods, near Oxford in England. A total of 37 hexagon–shaped quadrats were used, each with a diameter of 30 cm (see diagram on right). These were arranged in a pattern so that they were all touching each other. Use the data in the diagram to answer the following questions. 1. Determine the average number of centipedes captured per quadrat:

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The number in each hexagon indicates how many centipedes were caught in that quadrat.

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4. Describe one factor that might account for the distribution pattern:

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3. Looking at the data for individual quadrats, describe in general terms the distribution of the centipedes in the sample area:

Each quadrat was a hexagon with a diameter of 30 cm and an area of 0.08 square meters.

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1

0

2

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2. Calculate the estimated average density of centipedes per square metre (remember that each quadrat is 0.08 square metres in area):

2

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145 Sampling a Rocky Shore Community

Key Idea: The estimates of a population gained from using quadrat sampling may vary depending on where the quadrats are placed. Larger samples can account for variation. The diagram (next page) represents an area of seashore with

its resident organisms. The distribution of coralline algae and four animal species are shown. This exercise is designed to prepare you for planning and carrying out a similar procedure to practically investigate a natural community.

1. Decide on the sampling method

For the purpose of this exercise, it has been decided that the populations to be investigated are too large to be counted directly and a quadrat sampling method is to be used to estimate the average density of the four animal species as well as that of the algae.

for counting need to be established. You must decide before sampling begins as to what to do about individuals that are only partly inside the quadrat. Possible answers include:

(a) Only counting individuals that are completely inside the quadrat. (b) Only counting individuals with a clearly defined part of their body inside the quadrat (such as the head). (c) Allowing for ‘half individuals’ (e.g. 3.5 barnacles). (d) Counting an individual that is inside the quadrat by half or more as one complete individual.

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2. Mark out a grid pattern

Use a ruler to mark out 3 cm intervals along each side of the sampling area (area of quadrat = 0.03 x 0.03 m). Draw lines between these marks to create a 6 x 6 grid pattern (total area = 0.18 x 0.18 m). This will provide a total of 36 quadrats that can be investigated.

3. Number the axes of the grid

Only a small proportion of the possible quadrat positions are going to be sampled. It is necessary to select the quadrats in a random manner. It is not sufficient to simply guess or choose your own on a ‘gut feeling’. The best way to choose the quadrats randomly is to create a numbering system for the grid pattern and then select the quadrats from a random number table. Starting at the top left hand corner, number the columns and rows from 1 to 6 on each axis.

6. Carry out the sampling

To select the required number of quadrats randomly, use random numbers from a random number table. The random numbers are used as an index to the grid coordinates. Choose 6 quadrats from the total of 36 using table of random numbers provided for you at the bottom of the next page. Make a note of which column of random numbers you choose. Each member of your group should choose a different set of random numbers (i.e. different column: A–D) so that you can compare the effectiveness of the sampling method.

Column of random numbers chosen: ______

NOTE: Highlight the boundary of each selected quadrat with coloured pen/highlighter.

5. Decide on the counting criteria

Before counting the individuals of each species, the criteria

Carefully examine each selected quadrat and count the number of individuals of each species present. Record your data in the spaces provided on the next page.

7. Calculate the population density

4. Choose quadrats randomly

Discuss the merits and problems of the suggestions above with other members of the class (or group). You may even have counting criteria of your own. Think about other factors that could cause problems with your counting.

Use the combined data TOTALS for the sampled quadrats to estimate the average density for each species by using the formula: Density =

Total number in all quadrats sampled

Number of quadrats sampled X area of a quadrat

Remember that a total of 6 quadrats are sampled and each has an area of 0.0009 m2. The density should be expressed as the number of individuals per square metre (no. m –2).

Plicate barnacle:

Snakeskin chiton:

Oyster: borer

Coralline algae:

Limpet:

8. (a) In this example the animals are not moving. Describe the problems associated with sampling moving organisms. Explain how you would cope with sampling these same animals if they were really alive and very active:

(b) Carry out a direct count of all 4 animal species and the algae for the whole sample area (all 36 quadrats). Apply the data from your direct count to the equation given in (7) above to calculate the actual population density (remember that the number of quadrats in this case = 36):

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Barnacle: Oyster borer: Chiton: Limpet: Compare your estimated population density to the actual population density for each species:

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Plicate barnacle

Oyster borer

Snakeskin chiton

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62 63 36 13 45 31

22 43 64 45 35 14

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Table of random numbers

A

The table above has been adapted from a table of random numbers from a statistics book. Use this table to select quadrats randomly from the grid above. Choose one of the columns (A to D) and use the numbers in that column as an index to the grid. The first digit refers to the row number and the second digit refers to the column number. To locate each of the 6 quadrats, find where the row and column intersect, as shown below: Example:

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Coordinates for each quadrat

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refers to the 5th row and the 2nd column

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146 Factors Affecting Population Size

Key Idea: Carrying capacity is the maximum number of organisms a particular environment can support. An ecosystem's carrying capacity, i.e. the maximum number of individuals of a given species that the resources can sustain indefinitely, is limited by the ecosystem's resources. Factors affecting carrying capacity of an ecosystem can be biotic (e.g. food supply) or abiotic (e.g. water, climate, and

Limiting factors

available space).The carrying capacity of an ecosystem is determined by the most limiting factor and can change over time (e.g. as a result of seasonal changes). Below carrying capacity, population size increases because resources are not limiting. As the population approaches carrying capacity (or exceeds it) resources become limiting and environmental resistance increases, decreasing population growth.

Environmental resistance

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The effect of limiting factors and the type of factor that is limiting may change over time. The graph, right, shows how the carrying capacity of a forest environment varies based on changes to the limiting factors:

2

The population overshoots the carrying capacity.

3

The environment is damaged due to large numbers and food becomes more limited, lowering the original carrying capacity.

4

The population becomes stable at the new carrying capacity.

5

The forest experiences a drought and the carrying capacity is reduced as a result.

6

The drought breaks and the carrying capacity rises but is less than before because of habitat damage during the drought.

Water, space, food

Population size

1

A population moves into the forest and rapidly increases in numbers due to abundant resources.

Limiting factors:

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4

3

Overshoot

Carrying capacity

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5

1

Population

Time

Factors affecting population size

Density dependent factors The effect of these on population size is influenced by population density.

Density independent factors The effect of these on population size does not depend on population density.

They include:

They include catastrophic events such as:

►► Competition

►► Volcanic eruptions, fire

►► Predation

►► Drought, flood, tsunamis

►► Disease

►► Earthquakes

Density dependent factors tend to be biotic and are less important when population density is low.

Density independent factors tend to be abiotic. They regulate population size by increasing death rates.

They regulate population size by decreasing birth rates and increasing death rates.

1. What is carrying capacity?

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2. How does carrying capacity limit population numbers?

3. What limiting factors have changed at points 3, 5, and 6 in the graph above, and how have they changed? (a) 3:

(b) 5:

(c) 6:

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147 Calculating Change in Population Size

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Key Idea: Population size increases through births or immigration and decreases through deaths and emigration. Populations are dynamic and the number of individuals in a population may fluctuate considerably over time. Populations gain individuals through births or immigration, and lose

individuals through deaths and emigration. For a population in equilibrium, these factors balance out and there is no net change in the population abundance. When losses exceed gains, the population declines. When gains exceed losses, the population increases.

Births, deaths, immigration (movements into the population) and emigration (movements out of the population) are events that determine the population size. Population growth depends on the number of individuals added to the population from births and immigration, minus the number lost through deaths and emigration. This is expressed as:

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Calculating change in population numbers Bir

on

ths

Im

Em

D)

igr

s(

th ea

• Numbers per unit time, e.g. 20,150 live births per year. The birth rate is termed the natality, whereas the death rate is the mortality.

(I)

mi

(B)

The difference between immigration and emigration gives net migration. Ecologists usually measure the rate of these events. These rates are influenced by environmental factors (see below) and by the characteristics of the organisms themselves. Rates in population studies are commonly expressed in one of two ways:

ti gra

D

ati

on

(E)

• Per capita rate (number per head of population), e.g. 122 live births per 1000 individuals per year (12.2%).

Limiting factors

Limiting factors, i.e. factors or resources that control organism or population growth, affect population size. Availability of food, predation pressure, or available habitat may all limit population growth.

The human population is estimated to peak at around 10 billion by 2050 as a result of multiple factors, including falling birth rates. Humans have the technology and production efficiency to solve many resource problems and so might appear exempt from limiting factors, but declining availability of water and land for food production is likely to constrain population growth, at least regionally.

1. Define the following terms used to describe changes in population numbers:

(a) Death rate (mortality):

(b) Birth rate (natality):

(c) Net migration rate:

2. Explain how the concept of limiting factors applies to population biology:

3. Using the terms, B, D, I, and E (above), construct equations to express the following (the first is completed for you):

(a) A population in equilibrium:

(b) A declining population:

(c) An increasing population:

B + I = D + E

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4. A population started with a total number of 100 individuals. Over the following year, population data were collected. Calculate birth rates, death rates, net migration rate, and rate of population change for the data below (as percentages):

(a) Births = 14: Birth rate =

(b) Net migration = +2: Net migration rate =

(c) Deaths = 20: Death rate =

(d) Rate of population change =

(e) State whether the population is increasing or declining:

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5. The human population is around 7.3 billion. Describe and explain two limiting factors for population growth in humans:

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148 Predation and Population Cycles

Key Idea: Predator and prey populations frequently show regular population cycles. The predator cycle is often based on the intrinsic population cycle of the prey species. It was once thought that predators regulated the population numbers of their prey. However, we now know that this is

A case study in predator-prey numbers

Bamboo plants are home to many insect species, including ladybirds and aphids. Aphids feed off the bamboo sap, and the ladybirds are predators of the aphids (below).

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In some areas of Northeast India, a number of woolly aphid species colonise and feed off bamboo plants. The aphids can damage the bamboo so much that it is no longer able to be used by the local people for construction and the production of textiles.

not usually the case. Prey species are more likely to be regulated by other factors such as the availability of food. However, predator population cycles are often regulated by the availability of prey, especially when there is little opportunity for switching to alternative prey species.

Giant ladybird beetles (Anisolemnia dilatata) feed exclusively on the woolly aphids of bamboo plants. There is some interest in using them as biological control agents to reduce woolly aphid numbers, and limit the damage woolly aphids do to bamboo plants.

The graph below shows the relationship between the giant lady bird beetle and the woolly aphid when grown in controlled laboratory conditions.

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Aphid population

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Number of giant ladybirds

Number of aphids

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1. (a) On the graph above, mark (using different coloured pens) where the peak numbers of woolly aphids and giant ladybirds occurs:

(b) Do the peak numbers for both species occur at the same time?

(c) Why do you think this is?

(b) Explain your answer:

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2. (a) Is the trend between the giant ladybirds woolly aphids positive or negative (circle one).

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Source: Majumder & Agarwala (2013) World Journal of Zoology 8 (1): 55-61

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149 The Impact of Disease

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Key Idea: Disease regulates population numbers by increasing the death rate within a population. Disease is a limiting factor on population size. Disease regulates population size by increasing death rates within a

population. Infectious disease is a density dependent factor and has more impact when population density is high and the disease can be spread more easily between individuals. In small populations, disease can be the cause of extinction.

Case study: the lions of the Ngorongoro Crater

Devil facial tumour disease in the Tasmanian devil

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The Ngorongoro Conservation Area in Tanzania is home to a large number of savanna species, including the Crater lion prides. The area is rich in resources, including prey species for the lions, and has the capacity to sustain about 120 lions. All members of the Crater lion population are descended from 15 founders, and the population has a low genetic diversity relative to lions in the nearby Serengeti. In recent times, the population size has been around 60, well below the carrying capacity of the area.

Facial tumour

120

Number of lions

100

Carrying capacity

Tasmanian devil with DFTD

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Menna Jones, cc 2.5

Population numbers of the Crater lions in the Ngorongoro Conservation Area, Tanzania

Tasmanian devils are highly susceptible to catching a type of infectious cancer called devil facial tumour disease (DFTD). Unlike most cancers, DFTD is contagious and it is easily spread between devils when they bite each other. The disease causes larger tumours to grow on the face and mouth. These prevent the animal from eating and it usually starves to death.

60 40 20

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1972 1977

1982

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1997

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Food and water have always been in plentiful supply to the crater lions, and their cubs have high survival rates. Researchers believe disease is the main factor regulating the size of the crater lion population. In 1962, an outbreak of blood-sucking stable flies resulted in the crater lion population crashing from near carrying capacity to 12. The flies fed off the lions and caused extensive skin infections that prevented the lions from hunting, and many died of starvation. Although lion numbers recovered well in the 1970s, a series of canine distemper virus outbreaks (transferred from hyaenas) have kept the lion population low.

Researchers estimate that 60% of animals in the wild have been killed by the disease since 1996, although in high density populations there may be up to 100% mortality in 12-18 months. This has reduced the genetic diversity (the variety of genes) of the devil population markedly. Low diversity in the immune genes in particular means that the devils are unable to fight the disease. Currently, at least 80% of the devil population is infected although the disease is concentrated in the eastern part of Tasmania. The Tasmanian devil may become extinct within 25 years if the disease cannot be halted. Research includes genomic analysis to help find a genetic pathway for control and captive breeding involving partially immune individuals.

1. (a) What is the main factor regulating population size of the crater lions?

(b) What evidence is there to support this?

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2. (a) What effect is disease having on the population numbers of Tasmanian devils?

(b) How does population density contribute to the impact of the disease in local populations?

(c) Why are the Tasmanian devils at risk of extinction?

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150 The Influence of Competition for the same limited resource (e.g. food or space). Competition harms both competitors. It can occur between members of the same species (intraspecific competition) or between members of different species (interspecific competition).

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Key Idea: Species interact with other living organisms in their environment. Competition occurs within and between species when they utilise the same limited resources. Competition occurs when two or more organisms compete

Competition occurring between members of the same species is called intraspecific competition. Male kangaroos (above) fight each other to mate with a female or for access to limited resources such as water.

Competition harms both competitors. The negative effects of competition limit population numbers because resources are limited and growth, reproduction, and survival are affected.

A complex system of interactions occurs between the different species living on the Great Barrier Reef. Population numbers will be limited by competition for resources, such as food and space on the reef.

Examples of limited resources

Hyaena

Brocken Inaglory cc3.0

Jackal

Vulture

Space can be a limited resource These sea anemones are competing for space in a tidal pool. Some species defend areas, called territories, which contain the resources they need.

Suitable mates can be hard to find Within a species, individuals may compete for a mate. These male whitetail deer are fighting to determine which one will mate with the females.

Food is usually a limited resource In most natural systems, there is competition for food between individuals of the same species, and between different species with similar diets.

(b) Why does competition occur?

(c) Why does competition have a negative effect on both competitors?

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1. (a) What is competition?

2. Explain the difference between intraspecific and interspecific competition and give an example of each:

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151 Interspecific Competition

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Key Idea: Interspecific competition is competition between individuals of different species. It can affect the distribution of species sharing the same general environment. Organisms occupy a functional role in the ecosystem called the niche. This is defined by their habitat, feeding relationships, and interactions with other species. In natural systems, competition with other species may force organisms to occupy a narrower niche than they would otherwise. This

is called the realised niche. Naturally coexisting species have evolved slightly different niche requirements even if many of their resource needs are the same. However, when two species with very similar ecological requirements are brought into direct competition through the introduction of a foreign species, one usually benefits at the expense of the other (the competitive exclusion principle). The introduction of alien species is implicated in the decline of many native species.

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Competitive exclusion in barnacles

High tide mark

Chthamalus Realised niche

Settling Balanus larvae die from desiccation at low tide

Inset enlarged right

Chthamalus Potential niche

Chthamalus adults

Low tide mark

Balanus Potential niche = realised niche

Settling Chthamalus larvae are crowded out by Balanus

Two barnacle species, Balanus balanoides and Chthalamus stellatus, coexist in the same region of the Scottish coast. The barnacles naturally show a stratified distribution, with Balanus concentrated on the lower shore, and Chthalamus on the upper shore. When Balanus were experimentally removed from the lower strata, Chthalamus spread into that area. However, when Chthalamus were removed from the upper strata, Balanus did not extend any further up the shore than usual.

Balanus adults

Squirrel glider

Mahogany glider

Pfinge cc 2.0

Sugar glider

Figaro Public Domain

Niche overlap and coexistence in gliders

In Australia, gliders of the genus Petaurus occupy very similar niches. All are nocturnal, require tree hollows for nesting, and feed on insects, nectar, pollen, honeydew, and plant sap. The sugar glider is found in a wide range of habitats including rainforest, wet and dry eucalypt forest, and woodland. Where its range overlaps with the larger squirrel glider, sugar gliders are restricted to wetter, denser forests, whereas the squirrel glider occupies drier, open eucalypt forest. The mahogany glider has a very restricted distribution in Queensland. It is the largest of the gliders and, like the squirrel glider, requires open woodland. Although it can coexist with sugar gliders, it never coexists with the squirrel glider.

1. Describe the evidence for the barnacle distribution being the result of competition:

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2. In what aspects of their niches do the sugar, squirrel, and mahogany gliders overlap?

3. How do sugar gliders and squirrel gliders avoid directly competing for the same resources when their ranges overlap?

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4. Mahogany gliders can coexist with sugar gliders but not with squirrel gliders. Can you suggest a reason for this?

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152 Competition and Species Distribution

Key Idea: Interspecific competition is reduced when different species exploit slightly different resources. Competition is most intense between members of the same species because their habitat and resource requirements are identical. Interspecific competition is usually less

intense, although many species exploit at least some of the same resources. Different species with similar ecological requirements may reduce direct competition by exploiting the resources within different microhabitats or by exploiting the same resources at different times of the day or year.

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Canopy

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Catch insects in flight

Lf

St

Wt

Catch insects in flight

Gleaning from foliage

Extracts insects from bark

Secondary tree layer

Bt

Rf

Gleaning from foliage

Catch insects in flight

Understorey

Omnivorous

Ground

Ys

Gt

Insectivorous

Adapted from: Recher et al., 1986. A Natural Legacy: Ecology in Australia. Maxwell Macmillan Publishing Australia.

Reducing competition in a eucalypt forest

The diagram above illustrates how a layered forest structure provides the opportunities and resources for species with similar foraging niches to coexist. Different layers of the forest allow insectivorous birds to specialise in foraging at different heights and in different ways. The similar sized striated and brown thornbills feed at different heights, as do the leaden flycatcher and the rufous fantail. Adaptations reflect their feeding specialisations. The ground-dwelling yellow-throated scrubwren and the larger ground thrush have robust legs and feet, while the white-throated treecreeper has long toes and large curved claws, specialising in removing insects from the bark. The swifts are extremely agile fliers capable of catching insects on the wing.

Key to bird species

Rf Rufous fantail

Lf Leaden flycatcher

Bt Brown thornbill

Gt Ground thrush

PHOTO: Greg Miles cc 2.0

PHOTO: JJ Harrison cc 3.0

PHOTO: Jim Bendon cc 2.0

PHOTO: JJ Harrison cc 3.0

swift Wt White-throated treecreeper Sw Spine-tailed PHOTO: Ron Knight cc 2.0 PHOTO: Lip Kee cc 2.0 thornbill Ys Yellow-throated scrubwren St Striated PHOTO: Bernard Dupont cc 2.0 PHOTO: JJ Harrison cc 3.0

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1. How do the insectivorous birds in a layered forest avoid direct competition for the same resources?

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2. In forests where shrubs are absent or sparse, only the striated thornbill is present and in shrub habitats with few trees there are few striated thornbills and the brown thornbills are common. Suggest why this is the case:

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153 Field Study of a Rocky Shore

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Key Idea: Field studies collect physical and biological data that measure aspects of community structure or function. Many biological investigations require the collection of data from natural communities. Biotic data may include the density

or distribution of organisms at a site. Recording physical (abiotic) data of the site allows the site to be compared with others. The investigation below looks at the populations of animals found on an exposed and a sheltered rocky shore. The aim To investigate the differences in the abundance of intertidal animals on an exposed rocky shore and a sheltered rocky shore.

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Sample site A: Exposed rocky shore. Frequent heavy waves and high winds. Smooth rock face with few boulders and relatively steep slope towards the sea.

Coastline

Background

The composition of rocky shore communities is strongly influenced by the shore's physical environment. Animals that cling to rocks must keep their hold on the substrate while being subjected to intense wave action and currents. However, the constant wave action brings high levels of nutrients and oxygen. Communities on sheltered rocky shores, although encountering less physical stress, may face lower nutrient and oxygen levels.

Prevailing direction of wind and swell

1km

To investigate differences in the abundance of intertidal animals, students laid out 1 m2 quadrats at regular intervals along one tidal zone at two separate but nearby sites: a rocky shore exposed to wind and heavy wave action and a rocky shore with very little heavy wave action. The animals were counted and their numbers in each quadrat recorded.

Sample site B: Sheltered rocky shore. Small, gentle waves and little wind. Jagged rock face with large boulders and shallower slope leading to the sea.

All photos: C. Pilditch except where indicated

Rocky shore animals

The columnar barnacle is found around the high to mid tide level but can extend lower in suitable areas. It is uncommon on soft substrates and prefers moderately exposed shore lines.

The plicate and brown barnacles can be found together on exposed rocky shores. On more sheltered shores, the columnar barnacle is more prevalent.

The rock oyster often grows on steeply sloped or vertical surfaces and tends to flourish in harbours, as settlement on rocks is inhibited by even moderate wave action.

Limpets are found throughout rocky shores, although the ornate limpet has a slight preference to exposed shores.

The black nerite (snail) is found throughout rocky shores and extends across most tidal zones. It is more common on exposed rocky shores.

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The oyster borer is carnivorous and preys on barnacles such as the brown barnacle and the plicate barnacle. Numbers of oyster borers may be lower when there are fewer barnacles as prey.


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1. Underline an appropriate hypothesis for this field study from the four possible hypotheses below: (a) Rocky shore communities differ because of differences in wave action.

(b) Rocky shore communities differ because of the topography of the coastline.

(c) The physical conditions of exposed rocky shores and sheltered rocky shores are very different and so the intertidal communities will also be different.

(d) Rocky shore communities differ because of differences in water temperature.

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2. During the field study, students counted the number of animals in each quadrat and recorded them in a note book. Complete the table with the total number of each species at each site, the mean number of animals per quadrat, and the median and mode for each set of samples per species. Remember, in this case, there can be no 'part animals' so you will need to round your values to the nearest whole number:

Site A

Field data notebook Count per quadrat. Quadrats 1 m2

1

Brown barnacle Oyster borer Columnar barnacle Plicate barnacle Ornate limpet Radiate limpet Black nerite

39 6 6 50 9 5 7

2

3

4

5

6

7

8

38 7 8 52 7 6 7

37 4 14 46 8 4 6

21 3 10 45 10 8 8

40 7 9 56 6 6 4

56 8 12 15 7 7 6

36 9 8 68 6 5 8

41 2 11 54 10 6 9

Site B

7 2 56 11 7 7 13 6

Brown barnacle Oyster borer Columnar barnacle Plicate barnacle Rock oyster Ornate limpet Radiate limpet Black nerite

Brown barnacle

Oyster borer

6 3 57 11 8 8 14 5

7 1 58 13 8 5 11 3

Columnar barnacle

5 3 55 10 6 6 10 1

Plicate barnacle

8 2 60 14 2 5 14 4

Rock oyster

5 2 47 9 4 7 12 5

Ornate limpet

7 1 58 9 8 9 9 2

Radiate limpet

7 1 36 8 6 3 13 3

Black nerite

Total number of animals

Mean number of 2 Site animals per m A Median value

Mean number of 2 Site animals per m B Median value Modal value

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Total number of animals

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Modal value


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3. Use the grid below to draw a column graph of the mean number of species per 1 m2 at each sample site. Remember to include a title, correctly labelled axes, and a key.

4. (a) Compare the mean, median, and modal values obtained for the samples at each site:

(b) What does this tell you about the distribution of the data:

5. (a) Which species was entirely absent from site A?

(b) Suggest why this might be the case:

6. (a) Explain why more brown barnacles and plicate barnacles were found at site A:

7. (a) Comment on the numbers of limpets at each site:

(b) What does this suggest to you about their biology:

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(b) Explain why more oyster borers were found at site A:

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154 Intraspecific Competition

Key Idea: Intraspecific competition is usually intense as individuals are competing for exactly the same resources. As a population grows, the resources available to each individual become fewer and intraspecific competition (competition between members of the same species) increases. When the demand for a resource (e.g. light) exceeds supply, that resource becomes a limiting factor to the number of individuals the environment can support

(the carrying capacity). Populations respond to resource limitation by reducing growth rate (e.g. lower birth rates or higher mortality). The response of individuals to limited resources varies. In many invertebrates and some vertebrates, individuals reduce their growth rate and mature at a smaller size. In many vertebrates, territories space individuals apart according to resource availability and only those individuals able to secure a territory will be able to breed.

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Intraspecific competition

Scramble competition in pigeons

Contest competition in wolves

Direct competition for food between members of the same species is called scramble competition. Where scramble competition is intense, none of the competitors gets enough food to survive.

Display Display of of aa sulfur sulphur crested crested cockatoo cockatoo

In some cases, competition is limited by hierarchies within a social group. Dominant individuals receive enough food, but individuals low in the hierarchy must contest what is left and may miss out.

Individuals may compete for breeding sites, mates, or food. In sulfur crested cockatoos (above), a raised crest display can indicate that the cockatoo is looking for a mate or that it is defending its territory.

Competition between tadpoles of Rana tigrina

Both figures: Smith & Smith: Ecology and field biology, 2001

Minimum mass for metamorphosis (0.75 g)

1.0

Mean body mass (g)

0.8

5

0.6

40

60

160

0.4 0.2

0

0

1

2

3

4 5 6 Time (weeks)

7

8

9

10

Food shortage reduces both individual growth rate and survival, and population growth. In some organisms, where there is a metamorphosis or a series of moults before adulthood (e.g. frogs, crustacean zooplankton, and butterflies), individuals may die before they mature. The graph (left) shows how the growth rate of tadpoles (Rana tigrina) declines as the density increases from 5 to 160 individuals (in the same sized space).

 At high densities, tadpoles grow more slowly, take longer to reach the minimum size for metamorphosis (0.75 g), and have less chance of metamorphosing into frogs. Tadpoles held at lower densities grow faster to a larger  size, metamorphosing at an average size of 0.889 g. In some species, such as frogs and butterflies, the  adults and juveniles reduce the intensity of intraspecific competition by exploiting different food resources.

(a) Individual growth rate:

(b) Population growth rate:

(c) Final population size:

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1. Using an example, predict the likely effects of intraspecific competition on each of the following:

2. In the tank experiment with Rana (above), the tadpoles were contained in a fixed volume with a set amount of food:

(a) How did Rana tadpoles respond to resource limitation?

(b) Categorise the effect on the tadpoles as density-dependent / density-independent (delete one).

(c) Do you think the results of this experiment are representative of what happens in a natural population? Explain:

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150 146 154

KNOW


155 The Effect of Resources on Population Size

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216

Key Idea: Organisms may need to exploit several different habitats to get the resources they need to survive. Some animals range over a variety of habitats, partly in order to obtain different resources from different habitats, and sometimes simply because they are forced into marginal

habitats by competition. Dingoes are found throughout Australia, in ecosystems as diverse as the tropical rainforests of the north to the arid deserts of the central Australia. Within each of these ecosystems, they may frequent several habitats or microhabitats.

Habitats provide resources

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Species may tolerate wide variations in a range of physical and biotic factors. As a result of this tolerance range, the habitat that is occupied by members of a species may be quite variable.

The important thing is that the habitat provides resources for the organisms that live there. These resources include water, food, shelter, and places to raise offspring.

Some habitats can be richer in resources than others and are usually described with reference to their main features. For example, riverine habitats (rivers and creeks containing water and thick vegetated cover) provide water, food, and cover (right).

Dingo habitats

Dingoes (right) are wild dogs found throughout Australia. The table on the far right gives information about five dingo packs at one location, including how much of their territory is made up of riverine areas. Kangaroos are the main prey for these dingoes.

% of total territory made up of riverine areas

Dingo pack name

Territory area (km2)

Pack size

Dingo density

Pack A

113

12

10.6

Pack B

94

12

14

Pack C

86

3

2

Pack D

63

6

12

Pack E

45

10

14

10

1. Calculate the density of each of the dingo packs per 100 km2 using the equation below, and record it in the table above. The first one has been done for you.

Density = pack size á territory area x 100

(b) Describe the relationship between dingo density and amount of riverine area:

(c) Can you explain why this relationship might occur?

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2. (a) Plot a scatter graph (right) of dingo density per 100 km2 versus how much of their territory is made up of riverine areas for each pack.

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156 KEY TERMS: Did You Get It?

1. A simple food chain for a cropland ecosystem is pictured below. Label the organisms with their trophic status (e.g. primary consumer). Mouse

Corn snake

Hawk

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Corn

Trophic status:

2. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

A

A sequence of steps describing how an organism derives energy from the ones before it.

B

A mutually beneficial interaction between individuals of different species.

C

The maximum number of organisms of a species an environment can support indefinitely.

D

Any of the feeding levels that energy passes through in an ecosystem.

E

An organism that obtains its carbon and energy from other organisms.

F

Interaction in which a resource is contested.

G

A complex series of interactions showing the feeding relationships between organisms in an ecosystem.

mutualism

H

Exploitation involving an organism and its host. The host is detrimentally affected by the relationship but is not usually killed.

parasitism

I

A technique in which every possible sample of a given size has the same chance of selection.

J

A species that has a disproportionate effect on ecosystem stability and function because of their pivotal role in some aspect of ecosystem functioning such as nutrient cycling.

carrying capacity competition consumer

food chain food web

keystone species

random sampling trophic level

3. Study the graph of population growth for a hypothetical population below and answer the following questions: (a) Estimate the carrying capacity of the environment:

7,500

5,000

2,500

(c) What happened at point B? A 0

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(b) What happened at point A?

Population number

10,000

B

(d) What factors may have caused this?

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Time

TEST


157 Review: Unit 1 Area of Study 2

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218

Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered:

Regulating body systems

HINT: Give examples of how homeostasis maintains the body's systems.

Mechanisms of homeostasis

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HINT: What is homeostasis and how is it maintained?

Classification of organisms

REVISE

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HINT: Morphological and genetic characteristics are used to classify organisms.

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219

Population sizes

HINT: What is biodiversity and why is it important?

HINT: What factors regulate population size?

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Biodiversity

Competition

HINT: Describe the different types of species interactions and the effects on each party.

HINT: Describe the different types of competition. What do species compete for?

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Species interactions


158 Synoptic Question: Unit 1 Area of Study 2

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220

1. Homeostasis refers to the (relatively) constant physiological state of the body despite fluctuations in the external environment. Using the control of blood glucose as an example, discuss why it is important that animals maintain a constant internal environment. You may use extra paper if needed. Your answer should include: • The components of the control system involved The role of hormones and negative feedback in maintaining blood glucose levels How disease can disrupt blood glucose homeostasis

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• •

TEST

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2. Describe the environments of hydrophytes and xerophytes and discuss the adaptations that allow them to survive in their respective environments.

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221

(b) Discuss how high biodiversity systems, such seen in the reef system (right), are important in the search for new resources:

(c) The image on the right shows a low biodiversity reef system. Discuss the consequences of impoverished, low diversity systems on the likely success of a bioprospecting initiative:

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Bruno de Giusti

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3. (a) Bioprospecting in the oceans has resulted in the discovery and development of many new medicines. What is bioprospecting?

4. The graphs below show the growth curves for two species of microbes (A and B) grown separately, and then together. Treatment 2. Microbe B grown alone

Time (weeks)

No. of microbes

Treatment 3. Microbes A & B grown together

Time (weeks)

Time (weeks)

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No. of microbes

No. of microbes

Treatment 1. Microbe A grown alone

(a) What type of interspecific relationship is observed when the two species of microbes are grown together?

(b) Give reasons for the answer you have given in (a):

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The cell cycle

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Unit 2 Outcome 1

Key terms

Derivation of cells through the cell cycle

anaphase

Key knowledge

asexual reproduction binary fission cell cycle

cell division

c

1

Explain how all cells arise from preexisting cells through completion of the cell cycle.

c

2

Describe the various roles of mitotic cell division, in the life cycles of organisms: including asexual reproduction, growth, and repair and replacement of tissues.

Activity number 31 159

159 160

cytokinesis

DNA replication eukaryote

first gap phase (G1) interphase M phase

metaphase mitosis

EII

EII

prokaryote prophase

Reproduction of prokaryotic cells by binary fission

S phase

Key knowledge

c

3

Describe asexual reproduction by binary fission in prokaryotes. Identify distinct stages including elongation of the cell, duplication of the genetic material, cross wall formation, and division of the cell into two.

160

c

4

Understand that the length of the cell cycle in prokaryotes varies and that reproduction, and therefore population increase, can be very rapid.

160

telophase

Key events in the eukaryotic cell cycle Key knowledge

Activity number

5

Describe the cell cycle in eukaryotes, including reference to: DNA replication (S), growth (G1 and G2), and mitosis (M phase).

161

c

6

Describe the events in each of the main stages of mitosis: prophase, metaphase, anaphase, and telophase.

162

c

7

Describe and explain cytokinesis in both plant cells and animal cells.

162

c

8

Recognise stages of mitosis in light and electron micrographs. Calculate mitotic index (ratio of cells in mitosis vs interphase) in an actively growing tissue, e.g. root tip tissue. How would this compare to a tissue that is not growing?

163

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c

CL

second gap phase (G2)

Activity number


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159 Why Cells Need To Divide

Key Idea: Mitotic cell division has three primary functions: growth of the organism, replacement of damaged or old cells, and asexual reproduction (in some organisms). Mitotic cell division produces daughter cells that are genetically identical to the parent cell. It has three purposes: growth, repair, and reproduction. Multicellular organisms grow

from a single fertilised cell into a mature organism that may consist of several thousand to several trillion cells. Repair occurs by replacing damaged and old cells with new cells. Some unicellular eukaryotes (such as yeasts) and some multicellular organisms (e.g. Hydra) reproduce asexually by mitotic division.

Embryo

Adult

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Zygote

Asexual reproduction

Growth

Some simple eukaryotic organisms reproduce asexually by cell division. Yeasts (such as baker's yeast) can reproduce by budding. The parent cell buds to form a daughter cell (right) which eventually separates from the parent cell. Prokaryotes divide by binary fission, a different but superficially similar process.

Multicellular organisms develop from a single fertilised egg cell (zygote) and grow by increasing cell numbers. Cells complete a cell cycle, in which the cell copies its DNA and then divides to produce two identical cells. During the period of growth, the production of new cells is faster than the death of old ones. Organisms, such as the 12 day old mouse embryo (above, middle), grow by increasing their total cell number and the cell become specialised as part of development. Cell growth is highly regulated and once the mouse reaches its adult size (above, right), physical growth stops and the number of cell deaths equals the number of new cells produced.

Parent cell

Daughter cell

Jpbarrass

Brocken Inaglory

Damaged limbs

Broken bone

Repair

Mitotic cell division is responsible for the repair and replacement of damaged cells in multicellular organisms. When you break a bone, or graze your skin, new cells are generated to repair the damage. Some organisms, like the sea star (above right) are able to generate new limbs if they are broken off.

(a) Growth of an organism:

(b) Replacement of damaged cells:

(c) Asexual reproduction:

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1. Use examples to explain the role of cell division in:

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161 160

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31

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160 Binary Fission in Prokaryotes

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224

Key Idea: Binary fission involves division of the parent body into two, fairly equal, parts to produce two identical cells. New prokaryotic cells arise through the division of existing ones in a process called binary fission. Binary fission is a form of asexual reproduction. It is carried out by most prokaryotes, some eukaryotic organelles, such as

chloroplasts, and some unicellular eukaryotes (although the process is different in eukaryotic cells). The time required for a bacterial cell to divide, or for a population of bacterial cells to double, is called the generation time. Generation times may be quite short (20 minutes) in some species and as long as several days in others. Most bacteria reproduce asexually by binary fission (left). The cell’s DNA is replicated and each copy attaches to a different part of the plasma membrane. When the cell begins to pull apart, the replicated and original chromosomes are separated. Binary fission in bacteria does not involve mitosis or cytokinesis.

Cell wall

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DNA

Plasma membrane

Cell elongation occurs and the chromosome is duplicated.

Cross wall forming

This gram positive coccus (right) is in the process of binary fission. A cross wall (arrow) has formed.

The cell wall and cell membrane begin to grow inwards forming a cross wall.

Generation time (minutes)

Population size

0

1

20

2

40

4

60

8

80

CDC

The ingrowing cell walls meet and two identical cells are formed.

This Salmonella typhimurium bacterium (left) has completed cell division. The separation between the two cells can be clearly seen (arrow).

1. What is binary fission?

2. Explain why the formation of the cross wall is important in binary fission:

100 120 140 160

3. Explain the term generation time:

180 200 240 280

5. State how many bacteria were present after: (a) 1 hour: (b) 3 hours: (c) 6 hours:

300 320 340 360 WEB

160

CL

4. A species of bacteria reproduces every 20 minutes. Complete the table (left) by calculating the number of bacteria present at 20 minute intervals.

260

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220

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161 The Cell Cycle in Eukaryotes

Key Idea: The cell cycle of eukaryotes has four main phases. There are also three checkpoints that regulate the progression to the next phase of the cell cycle. The cell cycle can be divided into interphase and M phase. The cell spends 90% of its time in interphase. Interphase is the time when the cell is not in M-phase. It can be further

divided into three phases: G1, S-phase, and G2. M-phase includes mitosis (the separation of replicated DNA) and cytokinesis (the division of the cell into two new cells). During the cell cycle, three checkpoints must be passed. These ensure the cell has the correct size, nutrients, or proteins to enter and complete the next phase.

The cell cycle

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Interphase

Cells spend most of their time in interphase. Interphase is divided into three stages (right): The first gap phase. The S-phase. The second gap phase.

During interphase the cell grows, carries out its normal activities, and replicates its DNA in preparation for cell division. Interphase is not a stage in mitosis.

Second gap phase: Rapid cell growth and protein synthesis. Cell prepares for mitosis.

S phase: Chromosome replication (DNA synthesis).

G2

S

M

Mitosis: Nuclear division

Mitosis and cytokinesis (M-phase)

Mitosis and cytokinesis occur during M-phase. During mitosis, the cell nucleus (containing the replicated DNA) divides in two equal parts. Cytokinesis occurs at the end of M-phase. During cytokinesis the cell cytoplasm divides, and two new daughter cells are produced.

During interphase, the cell grows and acquires the materials needed to undergo mitosis. It also prepares the nuclear material for separation by replicating it.

G1

First gap phase: Cell increases in size and makes the mRNA and proteins needed for DNA synthesis.

During interphase the nuclear material is unwound. As mitosis approaches, the nuclear material begins to reorganise in readiness for nuclear division.

Cytokinesis: The cytoplasm divides and the two cells separate. Cytokinesis is part of M phase but distinct from nuclear division.

During mitosis the chromosomes are separated. Mitosis is a highly organised process and the cell must pass "checkpoints" before it proceeds to the next phase.

(a) Interphase:

(b) Mitosis:

(c) Cytokinesis:

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1. Briefly outline what occurs during the following phases of the cell cycle:

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162 Mitosis and Cytokinesis

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226

Key Idea: Mitosis is part of the cell cycle in which an existing cell (the parent cell) divides into two (the daughter cells). Mitosis results in the separation of the nuclear material and division of the cell. It does not result in a change of chromosome number and the daughter cells are identical to

Interphase

the parent cell. Although mitosis is part of a continuous cell cycle, it is often divided into stages to help differentiate the processes occurring. Mitosis is one of the shortest stages of the cell cycle. Cytokinesis (the division of the newly formed cells) is part of M-phase but distinct from nuclear division.

The animal cell cycle and stages of mitosis Early prophase

DNA condenses into distinct chromosomes. The nuclear membrane breaks down.

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Interphase refers to events between mitosis. The cell replicates the nuclear material ready for mitosis.

Nucleus

1

Centrosomes move to opposite poles

Centrosome (forms spindle)

Nuclear membrane

Cytokinesis

Division of the cytoplasm. When cytokinesis is complete, there are two separate daughter cells.

2

Late prophase

Chromosomes appear as two chromatids held together at the centromere. Spindle fibres begin to form.

Homologous pair of replicated chromosomes

6

Telophase

Two new nuclei form. A furrow forms across the midline of the parent cell, pinching it in two.

Some spindle fibres attach to and organise the chromosomes on the equator of the cell. Some spindle fibres span the cell.

3

Metaphase

Spindle

5

Late anaphase

4

Anaphase

Other spindle fibres lengthen, pushing the poles apart and causing the cell to elongate.

Spindle fibres attached to chromatids shorten, pulling the chromatids apart.

3. (a) What is the purpose of the spindle fibres?

(b) Where do the spindle fibres originate?

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2. What must occur before mitosis takes place?

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1. What is the purpose of mitosis?

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Cytokinesis

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dsworth Center- New York State Department of Health

In plant cells (below right), cytokinesis (division of the cytoplasm) involves construction of a cell plate (a precursor of the new cell wall) in the middle of the cell. The cell wall materials are delivered by vesicles derived from the Golgi. The vesicles join together to become the plasma membranes of the new cell surfaces. Animal cell cytokinesis (below left) begins shortly after the sister chromatids have separated in anaphase of mitosis. A ring of microtubules assembles in the middle of the cell, next to the plasma membrane, constricting it to form a cleavage furrow. In an energy-using process, the cleavage furrow moves inwards, forming a region of separation where the two cells will separate.

Cleavage furrow

Constriction by microtubules

Animal cell

Plant (onion) cells

Cleavage furrow

Cytokinesis in an animal cell

Cell plate forming

Cytokinesis in a plant cell

4. Summarise what happens in each of the following phases:

(a) Prophase:

(b) Metaphase:

(c) Anaphase:

(d) Telophase:

(b) Describe the differences between cytokinesis in an animal cell and a plant cell:

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5. (a) What is the purpose of cytokinesis?


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163 Recognising Stages in Mitosis

Key Idea: The stages of mitosis can be recognised by the organisation of the cell and chromosomes. Although mitosis is a continuous process it is divided into four

stages (prophase, metaphase, anaphase, and telophase) to more easily describe the processes occurring during its progression. 14

The mitotic index measures the ratio of cells in mitosis to the number of cells counted. It is a measure of cell proliferation and can be used to diagnose cancer. In areas of high cell growth the mitotic index is high such as in plant apical meristems or the growing tips of plant roots. The mitotic index can be calculated using the formula:

12

The growing tip of a root has a high mitotic index. Further from the root, cell division is reduced and the mitotic index is lower.

10 8

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Mitotic index (%)

The mitotic index

Mitotic index =

6 4

Number of cells in mitosis

2

Total number of cells

0

0.0

0.5

1.0

1.5

2.0

Distance from root cap (mm)

(a)

(b)

(c)

2. (a) The light micrograph (right) shows a section of cells in an onion root tip. These cells have a cell cycle of approximately 24 hours. The cells can be seen to be in various stages of the cell cycle. By counting the number of cells in the various stages it is possible to calculate how long the cell spends in each stage of the cycle. Count and record the number of cells in the image that are in mitosis and those that are in interphase. Cells in cytokinesis can be recorded as in interphase. Estimate the amount of time a cell spends in each phase. Stage

No. of cells

% of total cells

Photos: RCN

1. Use the information in the previous activity to identify which stage of mitosis is shown in each of the photographs below:

(d)

Onion root tip cells

Estimated time in stage

Interphase Mitosis Total

3. What would you expect to happen to the mitotic index of a population of cells that loses the ability to divide as they mature?

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(b) Use your counts from 2(a) to calculate the mitotic index for this section of cells.

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164 KEY TERMS: Did You Get it?

1. (a) Label the cell cycle right with the following labels: G1, G2, M, S, cytokinesis.

(b) Briefly describe what happens in each of the following phases:

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G1:

C

G2:

M phase:

S phase:

2. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

binary fission

A The stage in the cell cycle between divisions. B

cell cycle

cell division cytokinesis interphase mitosis

The phase of a cell cycle involving nuclear division in which the replicated chromosomes in a cell nucleus are separated into two identical sets.

C Process by which a parent cell divides into two or more daughter cells.

D A type of asexual reproduction in prokaryotes where the genetic material is replicated and the parent cell splits into two identical cells. E

The changes that take place in a cell in the period between its formation as a product of cell division and its own subsequent division.

F

The division of the cytoplasm of parent eukaryotic cell into two daughter cells during the late stages of cell division.

3. For what three reasons do cells divide?

4. How many chromosomes are in the daughter cells after mitosis in cells with the following number of chromosomes:

(a) 22:

(b) 48:

(a) 3 hours:

(b) 10 hours:

(c) 48 hours:

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5. If a bacterial cell undergoes binary fission every 30 minutes how many cells will be present in:

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6. A region of plant meristem (growing region) was examined on a slide under a microscope. It was found that 65 cells were in the process of mitosis and 216 were in interphase. What is the mitotic index of the plant meristem?

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Asexual reproduction

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Unit 2 Outcome 1

Types of asexual reproduction

Key terms

Activity number

Key knowledge

asexual reproduction budding

c

1

Describe the types of asexual reproduction by fission in unicellular eukaryotes.

165

bulb

c

2

Describe asexual reproduction in animals including budding in sponges and cnidarians, and fragmentation in flatworms.

165

c

3

Describe the diversity of mechanisms for asexual reproduction by vegetative propagation in plants. Include reference to natural vegetative structures such as bulbs, rhizomes, and tubers, and methods used by plant growers to produce plant clones, such as cuttings, grafting, and micropropagation.

c

4

Describe asexual production by spores in prokaryotes and fungi (e.g. the bread mould Rhizopus). Appreciate the complexity of fungal life cycles with many species producing both sexual and asexual spores. Recognise that the reproduction in plants involves the production of haploid spores by meiosis but that this is part of their sexual life cycle.

callus clone

cutting diploid

explant fission

fragmentation graft

166-168

169

haploid

micropropagation

plant tissue culture rhizome

root stock scion

spore tuber

vegetative propagation

Benefits and drawbacks of sexual reproduction Key knowledge

Activity number

c

5

Describe the biological advantages of asexual reproduction in eukaryotes, e.g fungi and plants. Include reference to survival, dispersal, rapid growth in favourable environments, and propagation of rare varieties.

165-169

c

6

Describe the biological disadvantages of asexual reproduction in eukaryotes, including reference to the loss of genetic diversity.

166 168

c

7

Describe the biological advantages of spore formation in prokaryotes. Why would bacteria produce spores when they already produce so rapidly by binary fission?

Applications of cloning

Activity number

Key knowledge

Define the term clone. Using examples, describe and explain the issues associated with plant cloning in agriculture and horticulture. Recognise that cloning provides a way to rapidly propagate successful varieties but this carries risks associated with loss of genetic diversity.

168 170

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165 Asexual Reproduction

Key Idea: Asexual reproduction produces offspring that are identical to the parent (the offspring are clones). In most forms of asexual reproduction, the parent splits, fragments, or buds to produce offspring identical to itself. Parthenogenesis is a special type of asexual reproduction in

which unfertilised eggs give rise to clones. Asexual organisms do not need to find a mate, and this saves them energy. Although asexual reproduction is rapid and highly efficient, all the offspring are genetically identical. If environmental conditions change there is little ability to adapt.

Fragmentation

Budding

Unicellular eukaryotes can reproduce asexually by splitting in two. This process involves mitosis and cytokinesis, and is not to be confused with the binary fission of bacteria which does not involve mitosis and cytokinesis. Fission is shown below for an Amoeba. Note that some multi-nucleated amoebas and some life cycle stages of parasitic protozoa, such as the malarial parasite Plasmodium, undergo multiple fission. The nucleus divides repeatedly before the final division of the cytoplasm to produce many new cells. Repeated cycles of multiple fission rapidly produce large numbers of offspring.

Some cnidarians, sponges and flatworms can reproduce by fragmentation. In this process, the organism spontaneously divides into fragments. Each fragment develops into a mature, fully grown individual identical to the original organism. In flatworms, a physiological gradient exists resulting in polarisation of the body. One pole represents the anterior (head) region, and the other pole the posterior (tail) region. This ensures that each new organism develops normally (e.g. they don't have two heads).

Sponges and most cnidarians (e.g. Hydra) can reproduce by budding. A small part of the parent body separates from the rest and develops into a new individual, which is smaller than the parent. This new individual may remain attached as part of the colony, or the bud may constrict at its point of attachment and be released as an independent organism.

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Fission in unicellular eukaryotes

The photo below shows Hydra budding. The new individuals are budding from the main body of the parent animal. The photograph shows the bulge and constriction where each new offspring will separate to form an independent individual.

Single cell

Feeding tentacles

Nucleus dividing

Bud forming

Cytoplasm dividing

Division into two cells

Constriction

Budding offspring

Hydra vulgaris

Parent

1. How does binary fission in unicellular eukaryotes differ from that of prokaryotes?

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2. What are the genetic consequences of asexual methods of reproduction?

3. (a) Why does multiple fission produce offspring more rapidly than simple binary fission?

(b) Explain the advantage of multiple fission to an intracellular parasite such as Plasmodium:

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166 Plant Propagation

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Key Idea: Vegetative propagation in plants produces new, genetically identical individuals (clones) without the need for seeds or spores. Many flowering plants reproduce asexually by vegetative propagation, the process by which new individuals arise without the production of seeds or spores. Plants can spread Remains of the parent plant

quickly by vegetative means when conditions are favourable. By doing this they do not need to produce flowers, pollen, or seeds (processes with large energy costs). Humans exploit the vegetative abilities of plants and many crop strains are hardly ever grown from seed. Vegetative propagation enables successful varieties to be propagated indefinitely.

Natural vegetative structures in plants

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Bulbs A bulb is really just a typical shoot compressed into a shortened form. Fleshy storage leaves are attached to a stem plate and form concentric circles around the growing tip. New roots form from the lower part of the stem.

Tuber

Tubers are the swollen part of an underground stem or root, usually modified for storing food. The potato (below) is a stem tuber, as indicated by the presence of terminal and lateral buds.

Food is stored in fleshy “scale” leaves

Potato stem tuber

Stem plate to which the leaves are attached

Dahlia root tuber

Growth in crocus, a typical corm

Root tubers, e.g. dahlias (above), lack terminal and lateral buds. Both stem and root tubers can give rise to new individuals.

Shoot Corm

'Eye' (lateral bud)

Underground stem containing stored food

Adventitious roots

In a corm, food is stored in stem tissue. Corms look like bulbs, but if you cut a corm in half you see a mass of homogenous tissue rather than concentric rings of fleshy leaves as in a bulb. Cyclamen, gladiolus, and crocus (above) are corms.

University of Florida

Iris rhizome

In rhizomes, as in corms, food is stored in the horizontal, underground stem. Rhizomes tend to be thick, fleshy or woody, and bear nodes with scale or foliage leaves and buds. Growth occurs at the buds on the ends of the rhizome or nearby nodes. Ginger, irises and lilyof-the-valley are rhizomes.

Strawberry plants send out runners. These are above-ground, trailing stems that form roots at their nodes.

1. What is meant by vegetative propagation?

Tussock grasses propagate from horizontal stems that grow from buds at the tip of the rhizomes.

The roots growing out of these potatoes will form new tubers and eventually independent plants.

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Garlic propagates via a bulb. Axillary buds (called cloves) form beside the existing bulb.

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2. Describe one major advantage (to the plant) of having a stem, tuber, or bulb filled with stored food:

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233 Propagating plants from cuttings

1

A leaf and accompanying axial bud are cut from the parent stock.

2

The cutting is placed in a growth medium containing rooting hormones and a new plant grows.

Melburnian

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Cuttings are sections of a parent plant, which are removed and grown as new individuals. These individuals will be clones (genetically identical copies) of the parent plant. The plant hormone auxin accumulates at the base of the stem triggering the formation of roots. Adding synthetic auxins to the end of the cutting promotes greater root development.

The spiny daisy (Acanthocladium dockeri) is a critically endangered daisy found in South Australia. It was first collected in 1860 in New South Wales, but by 1990, it was thought to be extinct. In 1999, five plants were found on a farm in South Australia. Cuttings were taken and the resulting plants have been successfully established around Banrock Station in South Australia.

3. Describe how humans have benefited from the vegetative propagation of plants:

4. Describe how plants benefit by reproducing vegetatively:

5. (a) Explain how vegetative propagation of plant material can be used as a strategy in species conservation:

(b) Suggest why stocks of rare plants could be reproduced this way, rather than using seed:

(c) Describe a potential disadvantage of vegetative propagation of plants that are endangered by disease:

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6. The cavendish banana is the most common banana sold in supermarkets today. It is sterile and is propagated vegetatively. Occasionally a plant produces a "sport", a branch arising from a mutation that may produce a variation in leaves or fruit. Explain how this sport may be propagated to produce a new variety of banana:


167 Grafting

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Key Idea: Grafting is the joining of the tissues of one plant to the tissues of another (usually a variety of the same species). Grafting is an important horticultural technique. It allows two or more different plants to be joined so that the combined plant has characteristics not fully present in either parent plant. Grafting is a way to produce a plant with a desirable

suite of characteristics quickly, without needing to breed those features into a new variety. Plant breeders can focus on producing a plant with good root stock without worrying about the effect on the fruit or leaves and vice versa. Then the superior fruiting plant can be grafted on to the superior root stock, producing a plant with features of both.

Multiple graftings

Grafting involves joining structures from two or more plants, usually varieties of the same species. Typically a twig section (scion) from one plant is joined to the shoot of another (root stock). Grafting is a very common technique in the production of fruit and landscape trees.

Multiple grafts can be made on to a single root stock. A large tree can have virtually every branch grafted. Many people like to grow fruit trees but do not have the room to grow more than one or two. It is now possible to buy fruit trees with two, three, four, or more fruit types grafted onto the one plant. Grafting more than one plant variety together can also help pollination because many plants are not self fertile and need a pollinating plant nearby.

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Grafting

Scion

This apple tree has been grafted to produce two varieties. On the left is a red skinned variety, whereas the right branches produce a green skinned variety.

1

A scion is prepared by taking a cutting. The scion is then grafted to another plant (root stock).

2

The graft is covered in wax to prevent infection and held together with twine or raffia.

Bob Embleton

Root stock

Grafting process 1

3

2

4

Root stock

Scion

Scion

Incision into parent plant

A scion is removed from the parent plant prior to grafting.

Scion being grafted onto the stem of the root stock.

The graft is sealed and covered to prevent water loss and infection.

The graft is then labelled for future reference and monitoring.

1. What is grafting?

2. Explain the difference between the scion and the root stock:

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4. Explain why the grafting of multiple plant types can be useful:

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3. Explain how grafting can be used to produce plants with dual qualities:

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168 Micropropagation

Key Idea: Micropropagation is the propagation of multiple clones from one piece of plant tissue. Micropropagation (plant tissue culture) is a method used for cloning plants. It is widely used for the rapid multiplication of commercially important plant varieties with superior genotypes, as well as in the recovery programmes for endangered plant species. Micropropagation is possible

because differentiated plant cells have the potential to give rise to all the cells of an adult plant (a property called totipotency). Micropopagation has many advantages over traditional methods of plant propagation but it is very labour intensive. Its success is affected by factors such as the composition of the culture media, selection of the original parent material, hormone levels, lighting, and temperature.

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The process of micropropagation

Callus

(undifferentiated cell mass)

Growth media

1

An explant (in this case it is an axial bud) is removed from a disease-free stock plant. Explants are commonly taken from cotyledons, axial buds, or roots.

2

Explant tissue is cultured in a sterile nutrient medium until a callus forms. The callus is transferred to another test tube containing growth hormones, and shoots are encouraged to grow.

3

Shoots develop from the callus and begin to photosynthesise. The new shoots are removed from the callus and placed in individual culture media. The process is repeated every few weeks so that one explant will give rise to many plants.

Joydeep

Callus

Radiata pine clones

Micropropagation is extensively used in the forestry industry in many countries to produce uniform trees for timber.

Micropropagation is useful for quickly propagating genetically engineered plants using material taken from a callus containing genetically engineered material.

Banana plants, which are sterile, are produced by micropropagation, as they cannot be grown from seed.

1. What is the general purpose of micropropagation?

2. (a) Explain what a callus is:

3. Describe some advantages and disadvantages of micropropagation:

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(b) How can a callus be stimulated to initiate root and shoot formation?

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169 Asexual Reproduction by Spores

Key Idea: A spore is an asexual reproductive unit produced for dispersal. Spores are usually unicellular and haploid. Many organisms, although not animals, produce spores as part of their normal reproductive life cycle. A spore is a unicellular, usually haploid, reproductive unit. It is a means of

dispersal, reproduction, and in some cases survival. Asexual reproduction by spores is common in fungi and some, including bread mould, rarely have a sexual phase in the life cycle at all. Huge numbers of spores may be produced enabling rapid growth when food sources are available.

Asexual reproduction in bread mould, Rhizopus

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Aerial hypha

Haploid means having a single set of unpaired chromosomes.

Feeding hyphae

Aerial hypha produces a sporangium containing asexual haploid (N) spores by mitosis

Curtis Clark cc 3.0

Sporangium

Clonal sporangia in Rhizopus

The asexual life cycle of bread mould involves the release of vegetative spores from sporangia.

Hyphae grow and spread through the food source

When conditions are favourable, the sporangium will burst, releasing the spores

Conidium

Spores germinate producing hyphae

Conidia in Penicillium mould

Conidia are asexual spores produced by mitosis. They germinate to produce clones of the parent.

In Rhizopus, asexual reproduction by haploid spores is the main method of reproduction. In asexual reproduction, haploid hyphae produce upright sporangia, which release numerous haploid spores. These spores are dispersed widely and each germinates to form a new fungal body.

Sori

Endospores are small resistant spores produced by some bacteria during unfavourable conditions. A small daughter cell is drawn into the original cell and forms a very thick wall to become an endospore. The endospore, which can survive environmental extremes for many years, is released when the original cell breaks down.

Kalyan Varma

Martin Wu

Endospore

All fungi disperse by spores, which are produced during both sexual and asexual reproduction. Commonly a fruiting body (e.g. puffball mushroom above) is produced in which the spores either develop or are shed. Spores are dispersive structures and are easily caught by the wind and spread over large distances.

Plants also produce spores, but they are always produced by meiosis. In ferns, haploid spores are produced on the underside of the fronds in clusters of sporangia called sori. They germinate to produce a small, free living structure called a prothallus. The haploid prothallus produces gametes for the sexual phase of the life cycle.

3. How are asexual spores produced in Rhizopus? 4. What is the purpose of an endospore in bacteria?

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2. What is the role of spores in the asexual reproduction of fungi?

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1. What is a spore?

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170 Case Study: Cloning in Horticulture production and distribution of plants and fruits. However it has also created some problems. Many ancient varieties of fruit have been lost in the move to mass produce a small number of cultivars. Also, as many millions of plants are clones, they all have the same vulnerabilities, e.g. to disease. This makes it extremely important to preserve older plant varieties and maintain the genetic diversity necessary to produce vigorous, disease resistant varieties in the future.

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Key Idea: The use of cloning in horticulture has produced limited genetic diversity in some common fruit varieties. Cloning has been used in horticulture for many hundreds of years. Many of the most commonly eaten fruits including apples, oranges, grapes, and bananas are clones, produced by the propagation of one or two varieties for hundreds (or thousands in the case of the banana) of years. The cloning of fruit varieties has been extraordinarily successful in the mass

Cloning and horticulture:

Imagine you were walking through an orchard and found a fruit tree seedling that had been missed by the orchardist's mowing and spraying machines. You dig it up and plant it at home. When it fruits, you find it produces the best tasting, longest lasting fruit of its type ever.

Wanting to capitalise on your discovery, you plant the seeds from one of the fruits, but they do not germinate. Closer inspection of seeds from other fruit on the tree finds the seeds never fully form. The tree cannot be regrown from seed; it is a one-off chance event and sterile. However, there is another way to produce more of this unique fruit tree - cloning it using cuttings and grafting. Some cuttings are grown directly into trees, others are grafted on to the root stock of another variety of the fruit. Soon, thousands of new trees are produced and distributed to orchards. Within a few years, your new fruit variety is the dominant one grown worldwide. Within decades it is virtually the only fruit of that type grown. Then disaster strikes. The fruit is susceptible to a soil fungus found in a country different to the variety's origin. Unknowingly, the fungus has been spread worldwide in orchard equipment. Within a year of discovering the fungus, half of the trees are failing. Within two years, three quarters of the fruit trees worldwide are dead. Another year and it is no longer economical to even plant the variety as the plants never reach maturity. Meanwhile, as a result of to the fruit's dominance, nothing is left to replace it. For years there is a massive shortage of the fruit, until older varieties can be re-established.

Sound fictional? It has happened at least once and may not be far from happening again.

1. Discuss the advantages and disadvantages of cloning in horticulture:

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In 1820, a mutation was found in an orange tree in Bahia, Brazil. The mutant tree produced a sweet, seedless fruit with an undeveloped fruit near the base, giving the characteristic "navel". Cuttings were grafted on to citrus root stock and new plants were grown. The new plants were taken to the United States and grown in California. The fruit was received well by orchardists. It is now the single most popular type of orange variety in the world. However, because it is seedless, all the plants are clones grown by grafting onto root stock. Since its discovery, there have been a few new mutations that have produced new varieties of navel orange.

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The Granny Smith apple originated in Eastwood, New South Wales, around 1870. It is named after its discoverer Maria Ann Smith, who also originally propagated it. How exactly she found the original tree is unclear but it may have come from a variety of crab apple. The Granny Smith became extremely popular during WWII due to its long shelf life, making it easy to export. It is not grown commercially from seed as the fruit produced varies and is usually very tart in taste. As a result, all the trees are clones directly related to the original tree in Eastwood. It is still in the top twenty most popular apples in the USA.

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There are many varieties of banana. All commercial bananas are descendents of the two banana species Musa acuminata and M. balbisiana.The most common banana, the Cavendish, is a triploid (three sets of chromosomes) variant of the banana species Musa acuminata. This makes it sterile. All the Cavendish plants worldwide are clones. The Cavendish replaced the Gros Michel, which was devastated by Panama disease, around the 1950s. The Cavendish is now being badly affected by a new strain of the disease and scientists are trying to find or develop a variety to replace the Cavendish should the need arise.

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171 KEY TERMS: Did You Get It?

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1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

asexual reproduction clone

B

An asexual means of reproduction by which plants reproduce themselves using vegetative structures such as bulbs and tubers.

C The production and large numbers of clones from a small piece or parent plant material.

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cutting

A A section of plant that is removed and grown as an individual.

grafting

micropropagation scion

spore

vegetative propagation

D Reproduction that does not involve meiosis or fertilisation. The offspring are genetically identical to the parent.

E

A organism that is genetically identical to the one from which it originated.

F

The process of removing a part of a plant and attaching to a different plant to form a fused individual.

G A unicellular haploid reproductive unit capable of growing into a new organism. H The piece of plant material (e.g. stem) that is grafted to a root stock.

2. Describe the following methods of asexual reproduction in animals:

(a) Fission:

(b) Fragmentation:

(c) Budding:

3. Explain the difference between producing cuttings and grafting in plants:

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4. What is the difference between a spore in plants and fungi and an endospore in bacteria?

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5. Explain how micropropagation is used in the production of new plants and its advantages and disadvantages:

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Sexual reproduction

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Unit 2 Outcome 1

The consequences of sexual reproduction

Key terms

Key knowledge

anaphase I

anaphase II

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Describe the role of sexual reproduction in producing offspring with a unique genetic makeup. Summarise the role of mitosis and meiosis in the life cycle of a sexually reproducing organism, recognising the role of mitosis in growth and repair and meiosis and fertilisation in producing genetically variable offspring.

c

2

In a general way, explain how sexual reproduction (meiosis and fertilisation) results in the genetically unique offspring of two parents.

crossing over diploid

fertilisation gamete

Activity number 172

172 175

haploid

homologous pair independent assortment interphase meiosis

metaphase I

metaphase II prophase I

prophase II

Meiosis

recombination

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telophase II

Describe the key events in the two divisions of meiosis, including: Meiosis I: prophase I, metaphase I, anaphase I, and telophase I Meiosis II: prophase II, metaphase II, anaphase II, and telophase II

173

Explain how the events in meiosis lead to production of haploid gametes (or haploid spores in plants) from diploid cells.

c

4

Explain the significance of the following events in meiosis: • Crossing over between homologous chromosomes in prophase I of meiosis. • Recombination of alleles as a result of crossing over. • Independent assortment of homologous pairs in metaphase I. • The non-dividing centromere in metaphase I.

173-175

The advantages of sexual reproduction

Activity number

Key knowledge

5

With reference to examples, describe and explain the advantages of sexual reproduction in which all the offspring are genetically unique. How does this compare with the benefits afforded to organisms of rapid population growth by asexual reproduction?

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telophase I

Activity number

Key knowledge

sexual reproduction

EII


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Mitosis produces two identical daughter cells from a parent cell. Meiosis is a special type of cell division, it produces sex cells (gametes or spores) for sexual reproduction. In sexual reproduction, sex cells from two parents combine to form a new individual that is genetically different to its parents.

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Key Idea: There are two types of cell division in eukaryotes, mitosis and meiosis, but only meiosis produces cells that are genetically different to the parent cell. New cells are formed when existing cells divide. There are two forms of cell division in eukaryotes, mitosis and meiosis.

The sex cells in humans, called eggs and sperm, are produced by meiosis. Events occurring during meiosis creates gametes with unique combinations of gene variants and so creates genetic variability.

Sexual reproduction rearranges and reshuffles the genetic material into new combinations. This is why family members may look similar, but they'll never be identical (except for identical twins).

The 2N (diploid) number refers to the cells each having two whole sets of chromosomes. For a normal human embryo, all cells will have a 2N number of 46.

Female embryo 2N

Mitosis produces genetically identical cells. This characteristic allows the body to produce cells to heal itself when it is damaged, and is also responsible for the production of the cells required for growth.

Gametes are produced by meiosis; a special division which reduces the chromosome number to half that of a somatic cell. The 1N (haploid) number indicates a single set of chromosomes.

Female adult 2N

Many mitotic divisions

Jpbarrass

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172 Cell Division

Many mitotic divisions give rise to the adult. Mitosis continues throughout life for cell replacement and repair of tissues. e.g. blood cells are replaced at a rate of two million per second, and a layer of skin cells is constantly lost and replaced about every 28 days.

Meiosis

Egg 1N

Mitotic division is responsible for growth of body cells (somatic growth) to the adult size.

Male embryo 2N

Fusion of the sperm and the egg in fertilisation produces a diploid zygote. This cell will give rise to a new individual through growth and differentiation.

Male adult 2N

Many mitotic divisions

Embryo 2N

Zygote 2N

Meiosis

Sperm 1N

Adult 2N

Mitosis

Mitosis

Cell division and the life cycle of an organism

1. (a) Where does mitosis take place in animals?

(b) Describe the roles of mitosis in the human body:

(c) In mitosis, the daughter cells are genetically different to the parent cell. True or False (delete one)

2. (a) Where does meiosis take place in animals?

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(b) What is the purpose of meiosis?

(c) In meiosis, the sex cells are genetically different to the parent cell. True or False (delete one) LINK

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173 Meiosis

Key Idea: Meiosis is a special type of cell division. It produces sex cells (gametes) for the purpose of sexual reproduction. Meiosis involves a single chromosomal duplication followed by two successive nuclear divisions, and results in a halving

What are homologous chromosomes?

Paternal chromosome

Interphase

When a cell is not dividing (interphase) the chromosomes are not visible, but the DNA is being replicated. The cell shown in the diagram (left) is 2N, where N is the number of copies of chromosomes in the nucleus. N = one copy of each chromosome (haploid). 2N = two copies of each chromosome (diploid).

2N

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A pair of chromosomes with the same gene sequence. One comes from the mother (maternal) and one comes from the father (paternal).

of the diploid chromosome number. Meiosis occurs in the sex organs of animals and the sporangia of plants. If genetic mistakes (gene and chromosome mutations) occur here, they will be passed on to the offspring (they will be inherited).

Maternal chromosome

Meiosis starts here

Homologous chromosomes pair up: Prior to cell division, the chromosomes condense into visible structures. Replicated chromosomes appear as two sister chromatids held together at the centromere. Homologous chromosomes pair up (synapsis). Crossing over may occur at this time making sister chromatids differ from one another.

Prophase I

Meiosis I (reduction division) The first division separates the homologous chromosomes into two intermediate cells.

Spindle apparatus forms.

Independent assortment: Homologous pairs line up in the middle of the cell independently of each other. This results in paternal and maternal chromosomes assorting independently into the gametes. The centromere does not dissociate so the sister chromatids remain together.

Metaphase I

Anaphase I

Homologous pairs separate, pulled apart by the spindle fibres

Telophase I

Intermediate daughter cells form.

Intermediate cells

Prophase II

Spindle apparatus forms. Chromosomes migrate towards the metaphase plate.

Meiosis II (mitotic division)

The second division is merely a mitotic one in nature, where the chromatids are pulled apart, but the number of chromosomes remains the same. This allows large numbers of gametes to be produced.

Metaphase II

Chromosomes line up on the metaphase plate.

Centromere divides and sister chromatids (now individual chromosomes) are separated.

Anaphase II

Separate gametes are produced

N

N

Telophase II

N

N

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2. Describe the behaviour of the chromosomes in the second division of meiosis:

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1. Describe the behaviour of the chromosomes in the first division of meiosis:

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174 Meiosis and Variation

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Key Idea: Meiosis produces genetic variation via the processes of crossing over and independent assortment. Meiosis creates genetic variation in the sex cells through crossing over and independent assortment. Crossing over

refers to the mutual exchange of pieces of chromosome (and their genes) between homologous chromosomes. In independent assortment, homologous chromosomes are randomly distributed to the gametes.

Crossing over and recombination

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Chromosomes replicate during interphase, before meiosis, to produce replicated chromosomes with sister chromatids held together at the centromere (see below). When the replicated chromosomes are paired during the first stage of meiosis, non-sister chromatids may become entangled and segments may be exchanged in a process called crossing over. Crossing over results in the recombination of alleles (variations of the same gene) producing greater variation in the offspring than would otherwise occur. No crossing over

Sister chromatids

Crossing over

Recombined chromosomes

Recombined chromatids

Non-sister chromatids

Centromere

Non-sister chromatids

Possible gametes with no crossing over

Possible gametes with crossing over

Independent assortment

Independent assortment is the random alignment and distribution of chromosomes during meiosis. Independent assortment is an important mechanism for producing variation in gametes. During the first stage of meiosis, replicated homologous chromosomes pair up along the middle of the cell. The way the chromosomes pair up is random. For the homologous chromosomes right, there are two possible ways in which they can line up resulting in four different combinations in the gametes. The intermediate steps of meiosis have been left out for simplicity.

or

1

2

3

4

1. How does independent assortment increase the variation in gametes?

(b) How does crossing over increase the variation in the gametes (and hence the offspring)?

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2. (a) What is crossing over?

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175 Modelling Meiosis

Key Idea: We can simulate crossing over, gamete production, and the inheritance of alleles during meiosis using ice-block sticks to represent chromosomes. This practical activity simulates the production of gametes (sperm and eggs) by meiosis and shows you how crossing

Background

over increases genetic variability. This is demonstrated by studying how two of your own alleles are inherited by the child produced at the completion of the activity. Completing this activity will help you to visualise and understand meiosis. It will take 25-45 minutes. Dominant: Tongue roller

Dominant: Right hand

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Each of your somatic cells contain 46 chromosomes. You received 23 chromosomes from your mother (maternal chromosomes), and 23 chromosomes from your father (paternal chromosomes). Therefore, you have 23 homologous (same) pairs. For simplicity, the number of chromosomes studied in this exercise has been reduced to four (two homologous pairs). To study the effect of crossing over on genetic variability, you will look at the inheritance of two of your own traits: the ability to tongue roll and handedness. Chromosome #

Phenotype

Genotype

10 10 2 2

Tongue roller Non-tongue roller Right handed Left handed

TT, Tt tt RR, Rr rr

Recessive: Non-roller

Trait

Record your phenotype and genotype for each trait in the table (right). NOTE: If you have a dominant trait, you will not know if you are heterozygous or homozygous for that trait, so you can choose either genotype for this activity.

Label four sticky dots with the alleles for each of your phenotypic traits, and stick it onto the appropriate chromosome. For example, if you are heterozygous for tongue rolling, the sticky dots with have the alleles T and t, and they will be placed on chromosome 10. If you are left handed, the alleles will be r and r and be placed on chromosome 2 (right).

2. Randomly drop the chromosomes onto a table. This represents a cell in either the testes or ovaries. Duplicate your chromosomes (to simulate DNA replication) by adding four more identical ice-block sticks to the table (below). This represents interphase.

Phenotype

Genotype

Handedness

Tongue rolling

BEFORE YOU START THE SIMULATION: Partner up with a classmate. Your gametes will combine with theirs (fertilisation) at the end of the activity to produce a child. Decide who will be the female, and who will be the male. You will need to work with this person again at step 6. 1. Collect four ice-blocks sticks. These represent four chromosomes. Colour two sticks blue or mark them with a P. These are the paternal chromosomes. The plain sticks are the maternal chromosomes. Write your initial on each of the four sticks. Label each chromosome with their chromosome number (right).

Recessive: Left hand

Your genotype

t

T

Paternal chromosome

Your initials

r

r

Maternal chromosome

Paternal chromosome

LB

LB

LB

LB

10

10

2

2

Homologous pair

Chromosome number

Homologous pair

3. Simulate prophase I by lining the duplicated chromosome pair with their homologous pair (below). For each chromosome number, you will have four sticks touching side-by-side (A). At this stage crossing over occurs. Simulate this by swapping sticky dots from adjoining homologs (B). (A)

T

r

T

t

t

r

r

r

r

LB 2

T

LB

LB LB LB LB

r

2

r

LB 2

LB 2

r

10 10 10 10

LB LB LB LB 2

2

2

2

(B)

T

LB

10

10

LB L B 10 1 0

t t LB

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T

t

T

t

LB LB LB LB

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10 10 10 10

r

r

r

r

LB LB LB LB 2

2

2

2

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4. Randomly align the homologous chromosome pairs to simulate alignment on the metaphase plate (as occurs in metaphase I). Simulate anaphase I by separating chromosome pairs. For each group of four sticks, two are pulled to each pole.

T

t

T

t

t

T

t

T

LB LB

LB LB

LB LB

10 10

10 10

10 10

10 10

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LB LB

r

r

r

LB LB 2

r

r

LB LB

LB LB

2

2

r

r

2

2

r

LB LB

2

2

2

5. Telophase I: Two intermediate cells are formed. If you have been random in the previous step, each intermediate cell will contain a mixture of maternal and paternal chromosomes. This is the end of meiosis 1.

Now that meiosis 1 is completed, your cells need to undergo meiosis 2. Carry out prophase II, metaphase II, anaphase II, and telophase II. Remember, there is no crossing over in meiosis II. At the end of the process each intermediate cell will have produced two haploid gametes (below). Intermediate cell 2

t

Intermediate cell 1

t

T

T

r

r

r

Paternal chromosome

LB

r

LB 2

Haploid (N) gamete

10

Maternal chromosome

10

LB

Maternal chromosome

LB 2

LB 2

t

t

r

r

r

T

10

LB 2

LB 2

LB 10

LB 10

LB 2

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LB 2

10

LB

LB

T

r

LB 2

LB LB 10 10

6. Pair up with the partner you chose at the beginning of the exercise to carry out fertilisation. Randomly select one sperm and one egg cell. The unsuccessful gametes can be removed from the table. Combine the chromosomes of the successful gametes. You have created a child! Fill in the following chart to describe your child's genotype and phenotype for tongue rolling and handedness.

Handedness Tongue rolling

Phenotype

Genotype

CL

Trait

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176 The Advantages of Sexual Reproduction

Key Idea: Sexual reproduction produces genetic variation in the offspring. Variation allows a population to adapt to a changing environment. Sexual reproduction produces genetic variation (diversity of genotypes) and therefore contributes to variation in the phenotype (appearance) of organisms. In a genetically

variable population, some individuals have better survival and produce more offspring than others. If the environment changes, different phenotypes may be favoured. In contrast, populations of asexually reproducing organisms are clones. The lack of variation in the population makes every individual equally susceptible to unfavourable environmental change.

Jose Luis Cernadas Iglesias cc 2.0

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Variation and disease resistance

Sexual reproduction produces variability in the offspring and so provides a greater chance that any one of the offspring will survive an environmental challenge, such as a disease outbreak. Individuals with better survival and reproduction (higher fitness) will predominate. Those with less favourable phenotypes will become less common.

Earth100

HIV test: test 2 is positive

Resistance to the pathogen HIV varies within the human population. A genetic mutation (change in the DNA) confers resistance to HIV. People who inherit two copies of the mutation (one from each parent) are often immune to the HIV virus and survive infection. In this way, favourable mutations can spread quickly through a population.

Wild tomatoes

Wild tomatoes (Lycopersicon) show a large amount of natural variation in the Pto gene for resistance to the bacterial pathogen Pseudomonas. There are many forms of the gene (alleles) in the tomato population, so the resistance to the pathogen is maintained even when the pathogen gains favourable mutations.

Variation from recombining alleles

Variation is greater when organisms reproduce by sexual reproduction

During meiosis, alleles are recombined in new combinations. Some combinations of alleles may be better suited to a particular environment than others. This variability is produced without the need for mutation, however beneficial mutations in separate lineages can be quickly combined through sexual reproduction. The environment is always 'testing' new combinations of alleles. Sexual reproduction is always producing new variations for testing in a changing environment.

A

B

Mutation A

Mutation B

AB

AB

AB

C

AB

AB

AB

ABC

s

Mutation C

ABC

ABC

1. How does sexual reproduction contribute to variation?

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2. Why is variation important in populations?

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3. How can sexual reproduction contribute to disease resistance in a population as a whole?

KNOW


177 KEY TERMS: Did You Get It?

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246

1. Identify the process shown on the right: 2. What is the purpose of this process?

3. (a) Describe the two events that have occurred at point A on the diagram:

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A

(b) How do these two mechanisms increase genetic variation?

4. Complete the diagram below (a) - (d) by drawing the gametes formed:

(a)

(b)

Gametes

(c)

(d)

6. Circle the correct answer for the following question: Which of the following cells undergo mitosis: A. Egg cells

B. 4 haploid daughter cells

B. Kidney cells

C. 2 diploid daughter cells

C. Bacterial cells

D. 4 diploid daughter cells

D. All of the above

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5. Circle the correct answer for the following question: Meiosis results in: A. 2 haploid daughter cells

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Cell growth and cell differentiation

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Unit 2 Outcome 1

Types and functions of stem cells

Key terms

Key knowledge

adult stem cell cancer

c

1

Explain what is meant by a stem cell. Describe the properties of stem cells, including self-renewal and potency.

178

c

2

Describe the types and functions of stem cells in human development, distinguishing between embryonic stem cells (ESC) and adult stem cells (ASC).

178

c

3

Distinguish between the potency of different stem cell types, including totipotent, pluripotent, and multipotent stem cells. Understand that as cells become more specialised during division of the zygote (fertilised egg) they lose their ability to produce different cell types.

178

c

4

Discuss the potential use of stem cells in the development of medical therapies. Explain how ESC and ASC could be used, what technical difficulties must be overcome, and what ethical issues are associated with each.

cell cycle

differentiation ectoderm

embryonic stem cell endoderm

germ layer

mesoderm

Activity number

multipotent

178 179

pluripotent potency

self renewal stem cell

totipotent tumour zygote

Matthias Zepper

Stem cell differentiation

EII

Activity number

Key knowledge

c

5

Describe the consequences of stem cell differentiation during human prenatal development. Identify the three germ layers of the embryo and explain how they arise. Identify the types of tissues formed from each germ layer.

c

6

Distinguish between an embryo and a fetus.

Disruptions to the cell cycle

180 181

181

Activity number

Key knowledge

7

Describe how the cell cycle is regulated according to requirements, including the role of specific checkpoints.

c

8

Explain how disruptions to the cell cycle can occur when cell cycle checkpoints are ignored or bypassed. Identify the causes of cell cycle disruption, including genetic predisposition and mutagens, and describe the consequences, e.g. uncontrolled cell division, cancer, and abnormal embryonic development

182

181 183

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178 What are Stem Cells?

Key Idea: Stem cells are undifferentiated cells found in multicellular organisms. They are characterised by the properties of self renewal and potency. A zygote can differentiate into all the cell types of the body because its early divisions produce stem cells. Stem cells are unspecialised cells that can divide repeatedly while remaining

unspecialised. They give rise to the many cell types that make up the tissues of a multicellular organism. For example, the stem cells in bone marrow specialise to produce all the cell types that make up blood. These multipotent (or adult) stem cells are found in most organs, where they replace old or damaged cells and replenish cells throughout life.

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Properties of stem cells

Stem cell

Differentiated cells

Potency: The ability to differentiate (transform) into specialised cells. There are different levels of potency that depend on the type of stem cell.

Types of stem cells

Self renewal: Stem cells have the ability to divide many times while maintaining an unspecialised state.

Totipotent

Pluripotent

4 cell divisions

Multipotent

3 cell divisions

The zygote and its first few divisions into the morula (~16 cell stage)

The inner cell mass of the blastocyst (~128 cells)

Bone marrow or umbilical cord blood

Totipotent stem cells

Pluripotent stem cells

Multipotent stem cells

These stem cells can differentiate into all the cells in an organism. Example: In humans, the zygote and its first few divisions. The tissue at the root and shoot tips of plants is also totipotent.

These stem cells can give rise to any cells of the body, except extra-embryonic cells (e.g. placenta and chorion). Example: Embryonic stem cells.

These adult stem cells can give rise to a limited number of cell types, related to their tissue of origin. Example: Bone marrow stem cells, epithelial stem cells, bone stem cells (osteoblasts).

1. Describe the two defining features of stem cells: (a) (b)

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2. Describe the potency of stem cells and where they are found: (a) Totipotency:

(b) Pluripotency:

(c) Multipotency:

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249 Embryonic stem cells Trophoblast becomes the placenta

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Blastocyst cavity (blastocoele)

Adult stem cells

Inner cell mass or embryoblast becomes the embryo

Harvested bone marrow

►► Embryonic stem cells (ESC) are derived from the inner cell mass of blastocysts (above). Blastocysts are embryos that are about five days old and consist of a hollow ball of 50-150 cells.

►► Adult stem cells (ASC) are undifferentiated cells found in several types of tissues (e.g. brain, bone marrow, fat, and liver) in adults, children, and umbilical cord blood.

►► Cells derived from the inner cell mass are pluripotent. They can become any cells of the body, with the exception of placental cells.

►► Unlike ESCs, they are multipotent and can only differentiate into a limited number of cell types, usually related to the tissue of origin.

►► When cultured without any stimulation to differentiate, ESC retain their potency through multiple cell divisions. This means they have great potential for therapeutic use in regenerative medicine and tissue replacement.

►► There are fewer ethical issues associated with using ASC for therapeutic purposes, because no embryos are destroyed.

►► However, the use of ESC involves the deliberate creation and destruction of embryos and is therefore is ethically unacceptable to many people.

►► For this reason, ASC are already widely used to treat a number of diseases including leukaemia and other blood disorders.

3. Distinguish between embryonic stem cells and adult stem cells with respect to their potency:

4. Suggest how stem cells could be potentially useful for treating diseased or damaged organs:

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5. Why are there ethical issues with the use of embryonic stem cells in research and medicine?

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6. New research has produced pluripotent stem cells from differentiated adult cells. These are called induced pluripotent stem cells. Why would using these cells in medicine generate fewer ethical issues than using other stem cell types?


179 Applications of Stem Cells

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250

Key Idea: Stem cells have many potential medical applications, but technical difficulties must be overcome first. Stem cell research is at an early stage and there is still much to be learned about the environments that cells require in order to differentiate into specific cell types. The ability of stem cells to differentiate into any cell type means that they have potential applications in cell therapy and in tissue engineering

to replace diseased or damaged cells. One of the problems with using stem cells to treat diseased tissues is the response of the recipient's immune system. The immune system has evolved to destroy foreign objects. When stem cells cultured from another person are introduced to a recipient, they are attacked by that person's immune system. There are a few possible ways around this, but none are simple.

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How to use stem cells

1. Use donor stem cells to repair tissues or organs

2. Reprogramme patient's cells before implantation

Problem: immune system will attack the donor's cells.

Problem: Some diseases are the result of defective genes. Stem cells from the patient will carry these defective genes.

Solution: Firstly a donor with a tissue match is selected (the types of cell surface proteins on donor and recipient cells are the same or very similar). This reduces the risk of immune rejection. Secondly the recipient will need to take immunosuppressant drugs to stop their immune system attacking the donated cells.

Another way of getting around the problem is to encase the donated cells in a protective shell, isolating them from the body. This is being investigated with respect to pancreatic cells and diabetes.

Solution: If the disease is due to a genetic fault (e.g. cystic fibrosis), then the stem cells will need to be genetically corrected before use (otherwise the disease may reoccur). Stem cells are isolated and cultured in the laboratory in the presence of the corrected gene. The culture is screened for cells that have taken up the gene. These are then transplanted back into the patient, without any immune rejection.

Stem cell

Donor stem cells removed and type matched.

Corrected gene mixed with cell culture.

Patient

Donor

Cells surrounded in protective shell to prevent immune rejection.

Stem cells cultured in lab

Immunosuppressant drugs taken to stop rejection of cells.

Stem cells with corrected genes are transplanted back into patient.

Recipient

Patient

1. Identify a problem with using stem cells from a donor to treat a recipient patient:

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2. Explain why stem cells with a defective gene must be corrected before reimplanting them into the patient:

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3. Umbilical cord blood is promoted as a rich source of multipotent stem cells for autologous (self) transplants. Can you see a problem with the use of a baby's cord blood to treat a disease in that child at a later date?

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251

Stem cells for Stargardt's disease

Stem cells for type 1 diabetes

Stargardt's disease is an inherited form of juvenile macular degeneration (a loss of the central visual field of the eye). The disease is associated with a number of mutations and results in dysfunction of the retinal pigment epithelium (RPE) cells, which nourish the retinal photoreceptor cells and protect the retina from excess light. Dysfunction of the RPE causes deterioration of the photoreceptor cells in the central portion of the retina and progressive loss of central vision. This often begins between ages 6 and 12 and continues until a person is legally blind. Trials using stem cells have obtained promising results in treating the disease.

Type 1 diabetes results from the body's own immune system attacking and destroying the insulin producing beta cells of the pancreas. In theory, new beta cells could be produced using stem cells. Research is focused on how to obtain the stem cells and deliver them effectively to the patient. Many different techniques are currently being investigated. Most techniques use stem cells from non-diabetics, requiring recipients to use immunosuppressant drugs so the cells are not rejected.

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A study published in 2014 described a method for treating type 1 diabetes in mice using fibroblast cells taken from the skin of mice.

2

1 Human embryonic

stem cells (hESC) are cultured in the lab to develop into retinal pigment epithelium (RPE) cells.

Cells treated with chemicals to reprogramme them into endoderm-like cells. These are a type of embryonic cell from which the organs develop.

1

RPE cells are injected just 2 The below the retina of the eye and above the choroid (the layer containing the blood vessels).

3

The RPE cells develop and replace the patient's damaged RPE cells, restoring vision.

Fibroblast cells collected from the skin of mice.

3 Using different

chemicals, the endoderm-like cells were induced to produce insulin.

Retina

Choroid

RPE layer

4 The cells were then injected into

model diabetic mice and behaved as fully functioning beta cells.

4. (a) Explain the basis for correcting Stargardt's disease using stem cell technology:

(b) Suggest why researchers derived the RPE cells from embryos rather than by reprogramming a patient's own cells:

(c) What advantage is there in reprogramming a patient's own cells and when would this be a preferable option?

(b) How can this be treated with stem cells?

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5. (a) What causes type 1 diabetes?

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180 Stem Cells Give Rise to Other Cells

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252

Key Idea: All cells arise from preexisting cells. All cells can be traced back to a particular type of stem cell. The zygote is the ultimate stem cell and is able to give rise to all other cell types (it is totipotent). As the zygote divides, and

the cells produced become more specialised, they lose their ability to differentiate into specific cell types. Adult stem cells are only able to produce a limited range of differentiated cells, related to the tissue of origin.

Stem cells and blood cell production

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New blood cells are produced in the red bone marrow, which becomes the main site of blood production after birth, taking over from the fetal liver. All types of blood cells develop from a single cell type: called a multipotent stem cell. These cells are capable of mitosis and of differentiation into 'committed' precursors of each of the main types of blood cell. Each of the different cell lines is controlled by a specific growth factor. When a stem cell divides, one of its daughters remains a stem cell, while the other becomes a precursor cell, either a lymphoid cell or myeloid cell. These cells continue to mature into the various specialised cell types.

Lymphoid precursor cell

Function

T lymphocyte

Immunity

Natural killer lymphocyte

White blood cells (leukocytes) are part of the immune system. They defend the body against infectious disease and foreign materials.

Matures in thymus

B lymphocyte

Lymphocytes

Specialised white blood cells involved in the specific immune response.

Multipotent stem cell

Neutrophil

Monocytes and macrophages

Red bone marrow

Basophil

Granulocytes

Specialised white blood cells which destroy foreign material (e.g. bacterial cells) by phagocytosis.

Eosinophil

Myeloid precursor cell

Blood cells are produced in the red bone marrow.

Red blood cells

Gas exchange

Transports oxygen around the body.

Platelets

Blood clotting

Form clots to prevent excessive bleeding.

(a) A neutrophil cell:

(b) A T lymphocyte:

2. How many cell types can a myeloid precursor differentiate into? 3. What controls the differentiation of stem cells into specialised cells? 4. What happens to the daughter cells when a stem cell divides?

5. Why is the zygote "the ultimate stem cell"?

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1. Describe the pathway for the production of the following cell types:

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181 Stem Cells and Prenatal Development

Key Idea: Three distinct cell layers are produced early in embryonic development. These give rise to specific cell types. There are some 230 different types of cells in humans. How they formed and how they got to where they are can seem very confusing when looking at an adult human. Even within one organ, there are many different kinds on cells, e.g. in a

small piece of muscle tissue there are muscle cells (called fibres), fat cells, nerve cells, cells of the blood vessels, and the blood cells themselves. Studying the development of very young embryos enables us to understand how all these cell types are related, and to see how cellular differentiation produces all the specialised cells of the body.

Development of the germ layers

Eight cell stage (after 3 days)

Zygote

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►► Cell specialisation begins four days after fertilisation at the blastocyst stage. The cells are pluripotent.

►► Gastrulation is the next stage of embryonic development during which cells in the singlelayered blastula migrate and are reorganised into a three layered embryo or gastrula. The three germ layers, the endoderm, mesoderm, and ectoderm, are organised in their proper locations. ►► The three germ layers (or cell lines) are the forerunners for all adult cells and tissues. At this stage, the cells are now multipotent and considered adult stem cells.

►► Because three germ layers form, all vertebrate embryos are said to be triploblastic.

Ectoderm

Mesoderm

Blastula (hollow ball stage)

Endoderm

Gastrulation

Gastrula (after 16 days)

The endoderm

The mesoderm

►► The endoderm is the innermost germ layer.

►► The mesoderm is the middle germ layer, between the ectoderm and endoderm.

►► In the early embryo, the endoderm forms the embryonic gut.

►► The mesoderm differentiates to give rise to the muscles, circulatory system (heart and blood vessels), urinogenital system, dermis (inner skin layer), skeleton, and other supportive and connective tissue.

►► It differentiates to form the digestive system, glands, and part of the respiratory system.

Muscle fibres

Thyroid

Bone

Alveoli

Blood vessel

Gastrula

The ectoderm

►► The ectoderm is the outermost germ layer.

Nerve cell

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Skin

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►► In the fully developed embryo and adult human the ectoderm forms the brain and the nervous system, epidermis of skin (including hair, sweat glands and nails), tooth enamel, some endocrine glands, cornea and lens of the eye.

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Pancreas

Cornea and lens

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Embryo and fetus

Embryo The embryo is the earliest stage of development from fertilisation to the eighth week of development. During this stage, development of the organ systems (brain, heart, internal organs) occurs. Limb buds appear by the sixth week.

Fetus

Embryo

Fetus

Ed Uthman

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By the end of the eighth week the major systems are present. The fetus is more or less dedicated to the growth of these systems in preparation for birth. The fetus will begin to move and exercise limbs. During the first 3 months of human development, the embryo is particularly sensitive to agents (teratogens), such as alcohol, that can cause developmental abnormalities. Errors in the cell cycle resulting in abnormal cell division are fatal very early in development, usually at the blastocyst stage.

1. How long after fertilisation does cell specialisation occur? 2. How does the gastrula form?

3. What does triploblastic mean?

4. Identify the three layers of the gastrula and give examples of what each of them give rise to: (a)

(b)

(c)

5. (a) What are the main events occurring in the embryo?

(b) What are the main events occurring in the fetus?

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6. Explain why the cells of the gastrula are considered to be adult stem cells (multipotent):

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7. Suggest why the early stages of development are the most vulnerable to teratogens:

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182 Regulation of the Cell Cycle

Key Idea: The cell cycle is regulated by checkpoints, which ensure the cell has met certain conditions before it continues to the next phase of the cell cycle. The cell cycle is an orderly sequence of events, but its duration varies enormously between cells of different species and between cell types in one organism. For example, human

intestinal cells normally divide around twice a day, whereas cells in the liver typically divide once a year. However, if these tissues are damaged, cell division increases rapidly to repair the damage. Progression through the cell cycle is controlled by regulatory checkpoints, which ensure the cell has met the conditions required to successfully complete the next phase.

Checkpoints during the cell cycle

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There are three checkpoints during the cell cycle. A checkpoint is a critical regulatory point in the cell cycle. At each checkpoint, a set of conditions determines whether or not the cell will continue into the next phase. For example, cell size is important in regulating whether or not the cell can pass through the G1 checkpoint.

G2 checkpoint: Pass this checkpoint if:  Cell size is large enough.  Replication of chromosomes has been successfully completed.  Proteins required for mitosis have been synthesised.

G2

S

The cell cycle

G1 checkpoint Pass this checkpoint if:  Cell size is large enough.  Sufficient nutrients are available.  Signals from other cells have been received.

M

Metaphase checkpoint Pass this checkpoint if:  All chromosomes are attached to the mitotic spindle.

G1

The G1 checkpoint is the critical regulatory point in cells. At this checkpoint, the cell decides whether to commit to the cell cycle or to enter an arrested phase called G0. Once sufficient nutrients or cell size is reached and the checkpoint is passed the cell is committed to replication of the nuclear material. Most cells that pass G1 complete the cell cycle.

EII

These yeast cells must reach a certain size before committing to replication of their DNA

The G2 checkpoint determines if synthesis was completed correctly, that the necessary proteins for mitosis have been synthesised, and that the cell has reached a size suitable for cell division. Damage to the DNA prevents entry to M phase. The entry into M phase is controlled by a protein called cyclin B, which reaches a concentration peak at the G2-M phase boundary.

EII

Chromosome actively transcribing genes to make proteins

The metaphase checkpoint, or spindle checkpoint, checks that all the chromatids are attached to the spindle fibres and under the correct tension. At this point, cyclin B is degraded, ultimately resulting in the sister chromatids separating and the cell entering anaphase, pulling the chromatids apart. The cell then begins cytokinesis and produces two new daughter cells.

(b) Why is this checkpoint important?

3. What would happen if the cell cycle was not regulated?

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2. (a) What is the purpose of the metaphase checkpoint?

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1. What is the general purpose of cell cycle checkpoints?

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183 Disrupting the Cell Cycle

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Key Idea: The cell cycle is strictly controlled. Failure of the cell checkpoints can lead to unregulated cell division. Cell cycle checkpoints have evolved to ensure that a cell divides in a controlled way. A cell that ignores its checkpoints may not have enough material to divide correctly or it may divide more often than it should. This can result in abnormal development of the embryo or the formation of a tumour

(cancer). Cancer cells are cells in which the checkpoints have failed so that the cell divides continually without regulation. Carcinogens are agents capable of causing cancer. Roughly 90% of carcinogens are also mutagens, i.e. they damage DNA. Any one of a number of cancer-causing factors (including defective genes) may interact to disrupt the cell cycle and result in cancer.

Genetic predisposition

The cell cycle is controlled by genes in a cell's DNA. A change to those genes (a mutation) can cause the normal regulatory mechanisms of the cell cycle to fail. Any substance that causes a mutation is called a mutagen. Mutagens include ionising radiation, viruses, solvents such as benzene, and tobacco smoke.

Genetic predisposition occurs when the DNA carries a gene variant (an allele) that is likely (but not guaranteed) to behave in a certain way, especially under certain circumstances. Genetic predisposition is important in the occurrence of many diseases, but especially cancers. Lifestyle choices are important in the development of disease in people with a genetic predisposition because they can offset risk. A lifestyle (e.g. diet) that reduces disease risk may keep an individual free of a disease, even if they have a predisposition.

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DNA, mutations, and the cell cycle

Carcinogens are mutagens that cause cancer (malignant tumours). Cancerous cells form when the genes controlling cell growth and multiplication are changed by carcinogens into oncogenes (genes that can cause cancer). Normally, damaged cells will fail to meet the checkpoints required for cell division to continue and are destroyed. However, cancerous cells evade the cell cycle checkpoints and divide rapidly, forming a tumour. Some tumours (or cell masses) are benign, meaning they do not invade surrounding tissues or spread. However, cancers are termed malignant tumours, because they invade surrounding tissues and can spread throughout the body.

The effect of mutagens on DNA UV Light

The BRCA1 and BRCA2 genes

BRCA1 and BRCA2 are genes that produce regulatory proteins involved in controlling the cell cycle. Specific inherited mutations of these genes increase the risk of both breast and ovarian cancer in females. Generally, about 12% of women will experience breast cancer in their lifetime. However, about 60% of women carrying a mutated BRCA1 gene develop breast cancer and 39% develop ovarian cancer. Similarly 45% of women carrying the mutated BRCA2 gene will develop breast cancer and 17% ovarian cancer.

Thymine dimer

DNA of tumour suppressor gene

Tumour

An example of how mutagens cause damage to the genes controlling the normal cell cycle is shown above. After exposure to UV light (a potent mutagen), adjacent thymine bases in DNA become cross-linked to form a 'thymine dimer'. This disrupts the normal base pairing and throws the controlling gene's instructions into chaos.

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Breast tissue

The photo above shows a large tumour in the breast of a patient with breast cancer. The breast was surgically removed as part of treatment. Note how the cancerous tissue has grown rapidly compared to the normal breast tissue surrounding it.

(b) Identify some common mutagens:

(c) What is the role of carcinogens in the development of cancer?

2. In a general way, explain how a tumour typically forms?

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1. (a) What is a mutagen?

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Cancer: cells out of control

Normal cell

Damaged DNA

Tumor-suppressor genes When damage occurs, the tumor-suppressor gene p53 commands other genes to bring cell division to a halt.

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Cancerous transformation results from changes in the genes controlling normal cell growth and division. The resulting cells become immortal and no longer carry out their functional role. Two types of gene are normally involved in controlling the cell cycle: proto-oncogenes, which start the cell division process and are essential for normal cell development, and tumour-suppressor genes, which switch off cell division. In their normal form, both kinds of genes work as a team, enabling the body to perform vital tasks such as repairing defective cells and replacing dead ones. But mutations in these genes can disrupt these finely tuned checks and balances. Proto-oncogenes, through mutation, can give rise to oncogenes (genes that lead to uncontrollable cell division). Mutations to tumour-suppressor genes initiate most human cancers. The best studied tumoursuppressor gene is p53, which encodes a protein that halts the cell cycle so that DNA can be repaired before division.

A mutation in one or two of the controlling genes causes a benign (nonmalignant) tumor. As the number of controlling genes with mutations increases, so too does the loss of control until the cell becomes cancerous.

If the damage is too serious to repair, p53 activates other genes that cause the cell to self-destruct. If repairs are made, then p53 allows the cell cycle to continue.

Benzo(a)pyrene from tobacco smoke changes G to T

The panel, right, shows the mutagenic action of some selected carcinogens on four of five codons (3-base sequences) of the p53 gene.

The diagram right shows a single lung cell that has become cancerous. It no longer carries out the role of a lung cell, and instead takes on a parasitic lifestyle, taking nutrients from the body and contributing nothing in return. The rate of cell division is greater than in normal cells in the same tissue because there is no resting phase between divisions.

Proto-oncogenes Genes that turn on cell division. The mutated form or oncogene leads to unregulated cell division.

Aflatoxin from moldy grain changes G to T

T

T

- - G G C - - - - - - ATG - - - - - - A AG - - - - - - C G G - - - - - - AGG 245

246

247

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- - C C G - - - - - - TAC - - - - - - T T C - - - - - - G C C - - - - - - T C C

TT

Features of cancer cells

DNA molecule

UV exposure changes CC to TT

Given a continual supply of nutrients, cancer cells can go on dividing indefinitely and are said to be immortal.

Deamination changes C to T

T

The bloated, lumpy shape is readily distinguishable from a healthy cell, which has a flat, scaly appearance. Metabolism is disrupted and the cell ceases to function constructively.

Cancer cells may have unusual numbers of chromosomes.

Cancerous cells lose their attachments to neighbouring cells.

3. (a) What is the impact of genetic predisposition to disease on disease risk?

(b) How could this risk be offset?

4. How do cancerous cells differ from normal cells?

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5. Explain how the cell cycle is normally controlled, including reference to the role of tumour-suppressor genes:

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6. With reference to the role of oncogenes, explain how the normal controls over the cell cycle can be lost:


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184 KEY TERMS: Did You Get It?

1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code. cancer

cell cycle

A A type of cell that possesses the qualities of self renewal and potency (ability to give rise to other types of cell). B

C Able to give rise to any cells of the body, except extra-embryonic cells. D A general term for any cell mass in which the cells grow in an uncontrolled way and invade nearby tissue.

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multipotent

Able to differentiate into all the cell types in an organism.

pluripotent stem cell

totipotent zygote

E

The changes that take place in a cell in the period between its formation as a product of cell division and its own subsequent division.

F

Able to give rise a limited number of cell types, related to their tissue of origin.

G

The initial cell formed from the union of two gametes. A new organism is produced by means of sexual reproduction.

2. Give an example of each of the following stem cell types:

(a) Multipotent:

(b) Pluripotent:

(c) Totipotent:

3. Match the beginning of the sentences with their correct ending: Stem cells can give rise to many...

...causes a mutation in the DNA.

Self renewal is the ability to divide many...

...cell types that make tissues and organs.

Embryonic stem cells are derived...

...sequence, e.g. changing a G to a C.

Gastrulation occurs when the blastula...

...times while maintaining an unspecialised state.

The cell cycle is controlled by three...

...from the inner cell mass of the blastocyst.

A mutation is any change in the DNA...

....develops an infolding after about 16 days.

A mutagen is any substance that...

...checkpoints, the G1, G2, and metaphase checkpoints.

4. (a) Label the cell cycle right with the following labels: G1 checkpoint, G2 checkpoint, metaphase checkpoint.

(b) Briefly describe what happens in each of the following checkpoints:

G1 checkpoint:

C

M phase checkpoint:

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G2 checkpoint:

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185 Review: Unit 2 Area of Study 1

Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered: The cell cycle

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HINT: Cell cycle and mitosis

Asexual reproduction

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HINT: Vegetative propagation

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260 Sexual reproduction

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HINT: Meiosis and variation

Cell Growth and cell differentiation

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HINT: Stem cells and their applications

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186 Synoptic Question: Unit 2 Area of Study 1

1. Briefly describe the process of binary fission in prokaryotes:

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2. Eukaryotic cells undergo mitosis before cell division,

(a) What is the purpose of mitosis?

(b) Use the space below to draw a labelled sequence of mitosis, including the important stages:

3. Some animals and most plants are able to reproduce asexually. (a) What is asexual reproduction?

(b) Describe a method of asexual reproduction for a named animal and explain how asexual reproduction is advantageous for the animal:

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4. In 1994, an off-duty field officer from the National Parks and Wildlife Service of New South Wales, Australia discovered a new type of conifer in the Wollemi National Park that became know as the Wollemi pine. Only about 100 are known in the wild, making the Wollemi Pine very rare and at risk of extinction. However, over the last two decades the pine has been propagated vegetatively by cuttings, and these have been distributed to many botanical gardens around the world. (a) What is vegetative propagation?

(b) Explain how using cuttings can help prevent the extinction of a plant:

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5. Meiosis is a type cell division in sexually reproducing organisms:

(a) What is the purpose of meiosis?

(b) How is meiosis different to mitosis?

(c) How does meiosis increase the variation in the gametes?

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6. Stem cells have the potential to be an important medical tool for the replacement of damaged or diseased tissue or organs. Explain why stem cells are potentially useful and describe some of the ethical issues associated with their use:

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Genomes, genes, and alleles

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Unit 2 Outcome 2

Genomes, genes, and alleles

Key terms

Activity number

Key knowledge

allele

DNA sequencing

c

1

Distinguish between a genome, gene, and allele.

gene

c

2

Understand that the genome is the sum total of an organism's DNA contained in a haploid set of chromosomes. State how the genome is measured and describe the variation in genome size between different organisms.

187 188

c

3

Recognise that there is no clear relationship between genome size, number of genes, and organism complexity.

188

gene knockout

genetic screening Human Genome Project (HGP)

Maggie Bartlett, NHGRI

Genomic research

Activity number

Key knowledge

c

4

Explain what is meant by DNA sequencing, Explain the role of automated sequencing in the feasibility of large scale genome projects.

189

c

5

Recognise the Human Genome Project (HGP) as an international collaboration to sequence the human genome. Describe the goal of the HGP and its achievements. Recognise that the HGP and subsequent genome projects were made possible by advancements in DNA sequencing technology.

190

c

6

Outline the genomic research since the HGP, with reference to the following:

The sequencing of the genes of many organisms, including many model organisms such as Drosophila and mice, and commercially important species, such as rice and yeast (Saccharomyces).

Comparing species relatedness and evolutionary relationships.

Determining gene function, e.g by using gene knockouts in model organisms.

193

Using genomic information, including patterns in sequence variations, for the early detection and diagnosis of disease.

194-196

188

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genome

187


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187 Genomes, Genes, and Alleles

Key Idea: A genome is the entire haploid amount of genetic material, including all the genes, of a cell or organism. Eukaryotes can have different versions of a gene (alleles) because they have two copies of each gene. The genome refers to all the genetic material in one haploid set of chromosomes. The genome contains all of the information the organism needs to function and reproduce.

Every cell in an individual has a complete copy of the genome. Within the genome are sections of DNA, called genes, which code for proteins. Collectively, genes determine what an organism looks like (its traits). Eukaryotes have two copies of each gene (one inherited from each parent), so it is possible for one individual to have two different versions of a gene. These different versions are called alleles.

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The location and size of the genome varies between organisms Protein coat

NIH

Agrobacterium

MSU

Koala: a eukaryote

Genome size

Small

Large

Number of genes

Few

Many

The viral genome is contained within the virus's outer protein coat. Viral genomes are typically small and highly variable. They can consist of single stranded or double stranded DNA or RNA and contain only a small number of genes.

In bacteria (prokaryotes) most of the DNA is located within a single circular chromosome, which makes them haploid (i.e. one allele) for most genes. Many bacteria also have small accessory chromosomes called plasmids, which carry genes for special functions such as antibiotic resistance and substrate metabolism.

In eukaryotes, most of the DNA is located inside the cell's nucleus. A small amount resides in the mitochondria and chloroplasts (in plants). The DNA is arranged into linear chromosomes and most eukaryotes are diploid, with two sets of chromosomes, one from each parent.

The HPV genome consists of a double stranded circular DNA molecule ~8000 bp long.

The Agrobacterium genome is 5.7 Mb long and consists of a linear chromosome and two plasmids, one of which enables it to infect plants.

The koala genome is ~3.37 Mb long in 8 chromosomes (the diploid number is 16). This genome size is similar to humans.

Measuring genomes

Genome size is often expressed as the number of base pairs. The unit most often used to show the size of a genome is the megabase (Mb). Note: 1 megabase = 1 million base pairs. The image right is of the φX174 bacteriophage, a virus that infects bacterial cells. Its entire genome is only 5375 base pairs long (0.005375 Mb) and it contains only nine genes, coding for nine different proteins. At least 2000 times this amount of DNA would be found in a single bacterial cell. Half a million times the quantity of DNA would be found in the genome of a single human cell.

Spikes on protein coat

Model of φX174 bacteriophage

1. Define the following terms:

(a) Genome:

(b) Gene:

(c) Allele:

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2. Describe the general trend for genome size and gene number for viruses, bacteria, and eukaryotic organisms:

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3. Explain why an individual eukaryote can have different versions of a gene (allele) but viruses and bacteria do not:

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Human papillomavirus (HPV)

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188 Genome Sizes

Key Idea: The sequenced genomes of many organisms indicate that genome size varies greatly between organisms. Genome projects determine the DNA sequence of an organism’s entire genome. Once completed, genome sequences are analysed by computer to identify genes. This information can be used to investigate gene function, compare species relatedness, and better understand disease.

Gene sequences are often entered into online databases that can be searched by anyone wishing to find out information about a particular gene. Genome sizes and the number of genes per genome vary greatly. While eukaryotes have larger genomes than viruses or bacteria, within the eukaryotes, there is no predictable relationship between genome size, number of genes, and complexity of the organism.

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*Mb = megabase pairs or 1,000,000 bp

Artist's impression

Bacterium

(Escherichia coli)

Yeast

(Saccharomyces cerevisiae)

Human

Rice

(Homo sapiens)

(Oryza sativa)

Genome size: 13 Mb Number of genes: 6000

Genome size: 3000 Mb Number of genes: < 22,500

Genome size: 466 Mb (indica) and 420 Mb (japonica) Number of genes: 46,000

E. coli has been used as a laboratory organism for over 70 years. Various strains of E. coli are responsible for several human diseases.

Yeast was the first eukaryotic genome to be completely sequenced. Yeast is used as a model organism to study human cancer.

The completion of the human genome has allowed advances in medical research, especially in cancer research.

A food staple for much of the world’s population. The importance of rice as a world food crop made sequencing it a high priority.

Mouse

Fruit fly

Japanese canopy plant

T2 phage

Genome size: 2500 Mb Number of genes: 30,000

Genome size: 150 Mb Number of genes: 14,000

Genome size: 149,000 Mb

Genome size: 160,000 bp Number of genes: Approx. 300

New drugs destined for human use are often tested on mice because more than 90% of their proteins show similarities to human proteins.

Drosophila has been used extensively for genetic studies for many years. About 50% of all fly proteins show similarities to mammalian proteins.

alpsdake PD

Dr Graham Beards

Genome size: 4.6 Mb* Number of genes: 4403

(Mus musculus)

(Drosophila melanogaster)

(Paris japonica)

This rare native Japanese plant has the largest genome sequenced so far (15% larger than any previous estimate for a eukaryote). Plants with very large genomes reproduce and grow slowly.

T2 phage is one of a group of related T-even phages that infect bacteria. Analysis of these phages indicates a small core genome with variations being the result of genetic transfers during evolution.

1. For each organism below, calculate how much smaller or larger the genome is than the human genome:

(a) Japanese canopy plant:

(b) E. coli:

(c) T2 phage:

4. Why is the sequencing of major food crops, such as rice, important?

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3. Why was the sequencing of the mouse genome important to humans?

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2. Plants with very large genome sizes are at higher risk of extinction. Can you suggest why?

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189 What is DNA Sequencing?

Key Idea: DNA sequencing techniques are used to determine the nucleotide (base) sequence of DNA. DNA is composed of a linear sequence of building blocks called nucleotides, each of which contains one of four types of bases. DNA sequencing determines the sequence (order) of these bases within a DNA molecule and therefore provides information about the entire genome or particular genes. The Sanger method is a manual DNA sequencing method based

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1% modified T, C, G, or A nucleotides are added to each reaction vessel. Only one kind of modified nucleotide is added per reaction vessel to cause replication to stop at random T, C, G, or A sites.

on the use of modified nucleotides, which prematurely stop DNA replication (copying) when they are incorporated into the nucleotide sequence. Sanger sequencing, or some variation of it, was widely used in early sequencing projects, including the Human Genome Project, and is still used for smaller projects. Most larger sequencing projects now use computer driven automated sequencing methods. These are called next-generation (Next-Gen) sequencing methods.

TC

TT

TG

TA

Each test tube shows the variety of fragments produced by each reaction. A radioactive primer (starter sequence) is attached to each fragment. The sequencing reaction for thymine also includes normal nucleotides.

T 1

T

T

T

T

T

T

T

T

T

G

G

G

G

G 2

G

G

G

G

A

A 3

A

A

C

C

A

A

A

A

C

C 4

C

C

C

C 5

C

C

C

C

A

A

G

G

G

A 6

G

A T 9

A

7

C A

Modified thymine is added at random to each forming fragment, which stops the DNA growing any longer.

T

A

A 8

C 10

T

C

G

A

DNA fragments move in this direction

-ve

Gel is read in this direction

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+ve

The fragments are placed on an electrophoresis gel to separate them so they can be read.

In Sanger (chain termination) sequencing, four separate reactions are run, each containing a modified nucleotide mixed with its normal counterpart, as well as the three unmodified nucleotides. When a modified nucleotide is added, synthesis stops. Different lengths of DNA are produced and these are separated using gel electrophoresis to reveal the original sequence. The DNA fragments will move when a current is applied to the gel. The smallest fragments move the fastest and the largest are the slowest.

1. What is the purpose of DNA sequencing?

2. (a) On the diagram circle the shortest fragment:

(b) Analyse the gel diagram, above right, and number the bands 1-10 from smallest to largest:

(c) Write the sequence of the copied DNA:

3. (a) What is the purpose of the modified nucleotides?

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4. Why must this method of DNA sequencing use four separate reaction vessels?

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(b) Why is only 1% of the reaction mix modified nucleotides?

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190 What was the Human Genome Project?

Key Idea: The human genome has been sequenced. The sequencing information is being used to determine gene function, to identify disease-causing genes, and to help develop treatments to some diseases. The Human Genome Project (HGP) was an international project carried out to determine the sequence of the human

genome and identify the genes that it contains. It was completed in 2003. Now that the genome sequence has been determined, researchers are trying to determine the functions of the genes and proteins it codes for. This information will be used in a number of fields including medicine, biotechnology, and studies of evolutionary relationships.

Genes can be mapped

Key

Rhesus blood type

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Information provided from sequencing the human genome can be used to find out information about the role and location of specific genes and how they are inherited. The variability of genes between individuals can also be studied. Researchers are particularly interested in locating (mapping) genes that cause disease. Gene mapping has been used to find the genes responsible for range of disorders including cystic fibrosis and colour blindness. Gene maps may also help scientists locate genes that may play a role in common disorders such as asthma, heart disease, diabetes, and cancer. Some examples of mapped genes are given on the right.

The Cancer Genome Atlas Project aims to understand the molecular basis of cancer through genome analysis. Fifty different types of cancer are being studied.

Variable regions

Regions reflecting the unique patterns of light and dark bands seen on stained chromosomes

ABO blood type

Down syndrome, critical region

Chromosome:

MN blood type

Structure of nails and kneecaps

21

Shape of red blood cells

Production of amylase enzyme Duffy blood type

Skin structure

9

4

1

Using the HGP to treat cancer

High throughput assay

Gene mapping allows researchers to identify mutations that can cause cells to become cancerous.

Understanding the role of specific genes in cancer can help researchers develop therapeutic drugs. High throughput screening of potential new drugs means many drugs can be developed and tested for effectiveness. Researchers aim to develop drugs to target specific types of cancer.

The new drugs may be more effective at treating cancer than current treatments and may have fewer side effects.

1. What was the purpose of the Human Genome Project?

(b) Why is gene mapping important in understanding inherited disease?

(c) How can gene mapping help researchers treat cancer?

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2. (a) What is gene mapping?

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191 DNA Differences Between Species

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268

with more differences. Often one gene is used to make the comparison. It is good practice to choose a gene that is large enough to show difference between species, but is still small enough so that any differences can be easily compared. The sequences below are for a gene called cytochrome oxidase I (COXI). It is involved in cell respiration.

Read in this direction

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Key Idea: The sequence variations between species can give information about their relationships. These sequence variations can be visualised on an electrophoresis gel. Comparing the DNA sequences of several species can provide information about their relatedness. Species with fewer differences are more closely related than species

T A G T

T

A

G

C

T

Cow

A

G

C

Sheep

T

A

G

C

Goat

T

A

G

C

Horse

1. For each of the species above:

(a) Determine the sequence of synthesised DNA in the gel in the photographs above. The synthesised DNA is what is visible on the gel. Cow: synthesised DNA: Sheep: synthesised DNA: Goat: synthesised DNA:

Horse: synthesised DNA: Based on the number of differences in the DNA sequences:

(b) Identify the two species that are most closely related:

(c) Identify the two species that are the least closely related:

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A

B

C

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E

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Calibration

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2. Determine the relatedness of each individual (A-E) using each banding pattern on DNA profiles (below). When you have done this, complete the phylogenetic tree by adding the letter of each individual.

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192 Using DNA to Determine Species Relatedness

Key Idea: Bioinformatics allows researchers to quickly compare sequences between species and determine their evolutionary relationships. The development of high throughput technologies has enabled researchers to process large amounts of DNA

sequence information. Bioinformatics uses computer science and mathematics to collect, analyse, and store the biochemical information. DNA or protein sequences can be retrieved from databases and compared to investigate the evolutionary relationships between species.

An overview of the bioinformatics process

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A gene of interest is selected for analysis.

High throughput 'Next-Gen' sequencing technologies allow the DNA sequence of the gene to be quickly determined.

... G A G A A C T G T T T A G A T G C A A A A...

Organism 1 ... G A G A A C T G T T T A G A T G C A A A A... Organism 2 ... G A G A T C T G T G T A G A T G C A G A A... Organism 3 ... G A G T T C T G T G T C G A T G C A G A A...

Organism 4 ... G A G T T C T G T T T C G A T G C A G A G...

Powerful computer software can quickly compare the DNA sequences of many organisms. Similarities and differences in the DNA sequence can help to determine the evolutionary relationships of organisms. The blue boxes indicate differences in the DNA sequences.

Once sequence comparisons have been made, the evolutionary relationships can be displayed as a phylogenetic tree. The example (right) shows the evolutionary relationships of the whales to some land mammals.

Camels Pigs

Peccaries

Bioinformatics has played an important role in determining the origin of whales and their transition from a terrestrial (land) form to a fully aquatic form. This phylogenetic tree was determined by comparing repetitive DNA fragments (retroposons) in whales and some of their closest relatives. Retroposons and their locations are predictable and stable, so they make reliable markers for determining species relationships. If two species have the same retroposons in the same location, they probably share a common ancestor.

Chevrotains (mouse deer)

Pecorans (horned ruminants)

Toothed whales Baleen whales

Data source: Nikaido et.al, PNAS 1996

Hippopotamuses

1. How has bioinformatics helped scientists determine the evolutionary relationship of organisms?

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2. The diagram above shows the relatedness of several mammals as determined by DNA sequencing of 10 genes: (a) Which land mammal are whales most related to?

(b) Mark with an arrow on the diagram above where whales and the organism in (a) last shared a common ancestor.

(c) Pigs were once considered to be the most closely related land ancestor to the whales. Use the phylogenetic tree above to describe the currently accepted relationship.

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193 Determining Gene Function Using Knockouts

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Key Idea: Gene function can be determined by making a gene non-functional and identifying the effect on the organism. This is called gene knockout. One way to find out what a gene does is to stop it from working. This is achieved by producing an individual with a non-functional version of the gene. One technique for doing

this, called gene knockout, involves altering a target gene so that it is non-functional. The gene's function is determined by comparing an individual with a functioning gene to a gene knockout individual with the non-functioning gene. Gene knockout is used to study diseases caused by the loss of gene function and for drug development.

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Creating gene knockout mice

The embryos formed are chimaeras (they have cells from different origins).

Implant chimaeric blastocysts into surrogate female.

Several thousand strains of knockout mice have now been bred. The knockout mouse (above left) was created as a model for obesity. Knockout mice have also been developed to study cystic fibrosis and cancer.

Grow a colony of ESCs

Select ESCs with the non-functional gene

NIH

Gene knockout mice are commonly used to determine the effect of genes that humans and mice both have in common. Mice are used because they are the most closely related laboratory animal species to humans to which the gene knockout technique can easily be applied.

Lexicon Genetics/HGRI

Construct a non-functional version of the target gene and introduce it to ESCs.

Inject ESCs into 3 day old blastocysts.

Test first generation offspring for the non-functional gene. Breed those with the non-functional gene.

Second generation offspring are heterozygous for the non-functional gene. Selective breeding produces third generation offspring that are homozygous for the non-functional gene.

Sickle cell

Normal cell Openstax College CC 3.0

Grow embryonic stem cells (ESCs) in culture.

Collect fertilized eggs from female mouse.

Knockout mice are used to develop and screen drug treatments for human disease. Examples include a treatment for sickle cell disease and a treatment for diabetes.

3. Why are mice good models for studying human disease?

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2. What are some applications of gene knockout?

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1. How is gene knockout used to determine gene function?

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194 Hunting for a Gene Key Idea: Huntington's disease is caused by a repeating section of DNA. The longer the repeating pattern, the earlier the disease tends to appear and the worse its symptoms. Huntington's disease (HD) is a genetic neuro-degenerative disease that normally does not affect people until about the age of 40. Its symptoms usually appear first as a shaking of the hands and an awkward gait. Later manifestations of the disease include serious loss of muscle control and mental function, often ending in dementia and ultimately death.

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All humans have the huntingtin (HTT) gene, which in its normal state produces a protein with roles in gene transcription, synaptic transmission, and brain cell survival. The mutant gene (mHTT) causes changes to and death of the cells of the cerebrum, the hippocampus, and cerebellum, resulting in the atrophy (reduction) of brain matter. The gene was discovered by Nancy Wexler in 1983 after ten years of research working with cell samples and family histories of more than 10,000 people from the town of San Luis in Venezuela, where around 1% of the population have the disease (compared to about 0.01% in the rest of the world). Ten years later the exact location of the gene on the chromosome 4 was discovered. The identification of the HD gene began by looking for a gene probe that would bind to the DNA of people who had HD, and not to those who didn't. Eventually a marker for HD, called G8, was found. The next step was to find which chromosome carried the marker and where on the chromosome it was. The researchers hybridised human cells with those of mice so that each cell contained only one human chromosome, a different chromosome in each cell. The hybrid cell with chromosome 4 was the one with the G8 marker. They then found a marker that overlapped G8 and then another marker that overlapped that marker. By repeating this many times, they produced a map of the genes on chromosome 4. The researchers then sequenced the genes and found people who had HD had one gene that was considerably longer than people who did not have HD, and the increase in length was caused by the repetition of the base sequence CAG.

New research has shown that the mHTT gene activates an enzyme called JNK3, which is expressed only in the neurones and causes a drop in nerve cell activity. While a person is young and still growing, the neurones can compensate for the accumulation of JNK3. However, when people get older and neurone growth stops, the effects of JNK3 become greater and the physical signs of HD become apparent. Because of mHTT's dominance, an affected person has a 50% chance of having offspring who are also affected. Genetic testing for the disease is relatively easy now that the genetic cause of the disease is known. While locating and counting the CAG repeats does not give a date for the occurrence of HD, it does provide some understanding of the chances of passing on the disease.

NYWTS

The HD mutation (mHTT) is called a trinucleotide repeat expansion. In the case of mHTT, the base sequence CAG is repeated multiple times on the short arm of chromosome 4. The normal number of CAG repeats is between 6 and 30. The mHTT gene causes the repeat number to be 35 or more and the size of the repeat often increases from generation to generation, with the severity of the disease increasing with the number of repeats. Individuals who have 27 to 35 CAG repeats in the HTT gene do not develop Huntington disease, but they are at risk of having children who will develop the disorder. The mutant allele, mHTT, is also dominant, so those who are homozygous or heterozygous for the allele are both at risk of developing HD.

American singer-songwriter and folk musician Woody Guthrie died from complications of HD

1. Describe the physical effects of Huntington's disease:

2. Describe how the mHTT gene was discovered:

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3. Discuss the cause of Huntington's disease and its pattern of increasing severity with each generation:

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195 Screening for Genes

Key Idea: Gene screening uses DNA probes to identify the presence and location of specific genes. It can be used to detect the presence of mutations associated with disease. Once a gene sequence is known, DNA probes can be used to determine if an individual has that specific gene (or a specific mutation). A DNA probe is a fragment of DNA or RNA, which

is complementary to a known DNA sequence (the target sequence) and so will bind to it. The probe has a radioactive or fluorescent label so it can be visualised (e.g. under UV light) when it is bound to its target sequence. This technique is called genetic screening and is used to screen individuals for gene mutations associated with specific diseases.

Why screen for disease?

Many diseases have a genetic origin and they often run in families. If one person has a heritable genetic mutation associated with a certain disease, this Information can be useful to other family members. It can raise awareness of the condition and allow others who may not be showing symptoms of the disease to seek testing and treatment if required.

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Many diseases are caused by specific gene mutations, and there are many reasons people choose to undergo gene screening tests.  Identify

people who may be at an increased risk of developing a certain disease.

 Determine

the severity of a known disease.

 Identify

carriers of certain diseases and determine the likelihood of the disease developing in their future children.

Genetic screening can also be useful for couples making a decision about whether to have a child or not. Genetic testing may be offered when someone in the family is a carrier for a particular condition. The couple can then analyse and discuss the risks of their children developing the disease, and make an informed decision about whether or not to have children.

 Help

doctors decide the best medicines or treatment to use on an individual.

Screening can be carried out at different times, for example on gametes, embryos, children, or adults.

Healthy brain

Alzheimer’s brain

Normal breast tissue

NIH

CF patients need therapy to clear the lungs of mucus

People with the BRCA gene mutation have an increased risk of developing breast cancer. Some families with a history of breast cancer screen for the BRCA gene. If the test is positive, a number of steps can be taken to reduce the risk of cancer developing. These include increased screening for early detection, use of drug therapies, and surgical removal of the breasts.

Cystic fibrosis (CF) is caused by a mutation to the CFTR gene. It causes excess mucus to be produced, which builds up in the lungs making it difficult to breathe. People with one copy of the mutation are carriers (but have no symptoms), but individuals with two copies are affected. When two carriers have children they may choose to screen the fetus for CF. Early diagnosis and treatment can greatly improve life expectancy.

NIH

Cancerous tissue

Alzheimer’s is an irreversible brain disorder in which the brain cells become damaged. Memory, thinking, and behaviour are affected. Several genes are associated with Alzheimer’s, but the most common is the ApoE gene for which there are several alleles. People with two copies of the type 2 allele are at lower risk of developing Alzheimer’s than people with two copies of the type 4 allele.

1. (a) How can gene screening help in the early detection of disease?

(b) What are the advantages of early disease detection?

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2. A family has a history of a certain genetic disease. Why might members of this family want to be screened for the disease before they start a family of their own?

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196 Genomic Analysis and Disease Risk

Key Idea: DNA sequence information can be used to identify disease risk and produce targeted, personalised treatments. All individuals carry many small (single nucleotide) variations in their DNA sequences. These variations include specific mutations as well as differences in the DNA outside genes. This variation can be used to identify areas of DNA related to

genetic diseases and match a person to a drug treatment. The key is to identify variation in the DNA that is related to specific disease types or drug susceptibility. Modern computing technology has provided the tools for this, analysing vast amounts of DNA sequence information and evaluating disease risk based on statistical probabilities.

Identifying the link between DNA and disease Scientists analysed a 3000 base pair piece of DNA near the gene and identified 13 nucleotide variations.

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To study the genetic component of a disease, researchers require two main study groups: people with the disease and people without the disease. A genome wide association study (GWAS) is carried out on each individual in each group. The study looks for single nucleotide variations to see if there are differences between the control group and the afflicted group. There are over 100 million such variations spread throughout the human genome and there are many ways to identify them, including DNA sequencing. Once the variations are identified, their frequencies are analysed to see if any are significantly different between the groups. Variations found to be associated with a disease do not always just appear in people with the disease. Often they can be found in people without the disease. Linking a disease to a specific DNA profile is therefore a matter of probability and risk analysis.

3

Single nucleotide variations

4

% haplotypes in population

5

Genomic analysis and targeted therapy

Using genomic information to evaluate disease risk is a relatively new field. Screening for a known gene mutation (such as BRCA) is relatively straightforward, but many diseases are very complex and involve large numbers of interacting genes. However, genetic information is being increasingly used with success to select appropriate therapies for people with certain diseases, as outlined below for the treatment of asthma.

Albuterol binds to receptors in a lung cell's plasma membrane and causes the smooth muscle in the airways to relax.

Other

Every person has a genetic profile relating to the receptor protein gene. Five haplotypes are relatively common.

A

E

D

6

1

Albuterol

Albuterol is used to relieve the symptoms of asthma. It works well on some people but not on others.

The variations occur in specific combinations called haplotypes. 12 different haplotypes have been identified.

B

C

A

Poor

B

Good

C

Fair

D

None

E

Very good

Some of the haplotypes seem to be related to the effect of albuterol on people. Some haplotypes are found mainly in people who have a poor response to albuterol (D), while others are found in people who respond well to the drug (B, E).

This data can be used by doctors when prescribing albuterol.

7

2

The receptor protein is encoded by a specific gene. Scientists wondered if genetic differences in or near this gene affected how well albuterol worked.

In the future, analysis of sequence variations across the entire genome could be used to identify disease risk and target specific diseases with the most appropriate drug therapy.

Patient responsive to albuterol.

(b) How are these sequence variations analysed?

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1. (a) How do sequence variations help to evaluate an individual's risk of disease?

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2. How can genomic analysis be used to improve the treatment outcomes of patients?

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197 KEY TERMS: Did You Get It?

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1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

allele

A The analysis of a person's DNA to detect the presence of a gene or genes associated with a specific disease.

DNA sequencing

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B An international research project to determine the DNA sequence of the entire human genome.

gene

C One of a number of alternative versions of a gene.

gene knockout

D A segment of DNA that codes for a particular protein.

gene mapping

E The process of establishing the location of genes on a chromosome.

genetic screening

F The complete haploid set of DNA in an organism.

G A genetic technique in which one of an organism's genes is made non-functional.

genome

Human Genome Project

H The process of determining the precise order of nucleotide bases within a DNA molecule.

2. Three partial DNA sequences are shown below for the turkey, emu, and ostrich.

Turkey

Emu

Ostrich

(a) Identify the sequence differences between the three species (the first one has been done for you):

Ostrich A T G G C C C C C A A C A T T C G A A A A T C G C A C C C C C T G C T C A A A A T T A T C A A C Emu

ATGGCCCCTAACATCCGAAAATCCCACCCT CTACTCAAAATCATCAAC

Turkey A T G G C A C C C A A T A T C C G A A A A T C A C A C C C C C T A T T A A A A A C A A T C A A C

(b) How many differences between the ostrich and emu?

(c) How many differences between the ostrich and turkey?

(d) Which two species are most closely related?

(e) Explain your answer:

(f) Looking at the photographs, does the sequence information support similarities based on appearance? Explain:

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Chromosomes

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Unit 2 Outcome 2

The role of chromosomes

Key terms

Key knowledge

aneuploidy chomatid

c

1

Describe the role of chromosomes as structures that package DNA. Describe how DNA is organised in a eukaryotic chromosome, including the role of histone proteins. Appreciate that the extent of packaging affects the accessibility of the DNA to the enzymes that read and translate the DNA's information into proteins.

198

c

2

Describe how the number and size of chromosomes and the number of genes they carry varies between different organisms. Explain how this variability might arise.

199

c

3

Distinguish between autosomes and sex chromosomes. Explain what is meant by a homologous pair of chromosomes and appreciate that the sex chromosomes are not homologous.

chromosome

Down syndrome exon

gene

histone

homologous chromosome

Activity number

intron

199 200

karyogram karyotype

Klinefelter syndrome locus (pl. loci) maternal

non-disjunction paternal

EII

syndrome

Turner syndrome

Karyotypes

EII

Activity number

Key knowledge

c

4

Define the term karyotype. Describe how an individual's complete set of chromosomes can be prepared and displayed in an organised way to produce a karyogram.

200

c

5

Explain how chromosomal abnormalities can arise when chromosomes fail to separate during meiosis (non-disjunction). Show in a diagram how nondisjunction in meiosis can produce abnormal gametes and lead to an individual having an incorrect number of chromosomes in their cells.

201

c

6

Explain how a karyogram can be used to identify chromosomal abnormalities, including human aneuploidies, e.g. Down syndrome (autosomal trisomy), and Turner and Klinefelter syndromes (sex chromosome aneuploidies).

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Create a karyogram by matching the size and banding pattern of individual chromosomes. What is the phenotype of the person?


198 Eukaryotic Chromosome Structure

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Key Idea: Eukaryotic DNA is located in the cell's nucleus. DNA molecules are very long and must be tightly packaged in an organised fashion in order to fit into the cell's nucleus. Eukaryotes package their DNA as discrete linear

chromosomes. The number of chromosomes varies from species to species. The extent of DNA packaging changes during the life cycle of the cell, but classic chromosome structures (below) appear during metaphase of mitosis.

Eukaryotic chromosomes are formed from the coiling of chromatin into organised structures. They appear during cell division.

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Chromosome

Chromatids (2)

In eukaryotes, chromosomes are located in the nucleus.

Nucleosome

Histone tail

DNA is complexed with protein to form chromatin. The DNA is packaged in an organised way, wrapped around groups of 8 histone proteins to form nucleosomes. This loosely packed 'beads on a string' arrangement is how most of the DNA exists for much of the cell cycle.

DNA

Histones may be modified by a number of processes, including addition of methyl, acetyl, or phosphate groups to the histone tail. Depending on the type of modification, the chromatin may pack together more tightly or more loosely, affecting the cell's ability to express genes.

Nucleosome = 8 histones and 2 turns of DNA

Gene (protein coding region). Genes on a chromosome can only be expressed (read and translated into proteins) when the DNA is unwound. DNA has a double helix structure. It is made up of many building blocks called nucleotides joined together.

Transcription (reading) start sequence Exon (coding region)

Intron (non-coding region)

Transcription stop sequence

1. Where is the DNA located in eukaryotes?

2. Why does DNA need to be packaged up to fit inside a cell nucleus?

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4. Suggest why a cell coils up its chromosomes into tight structures during mitosis:

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3. How do histone proteins help in the coiling of DNA?

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199 Species Have Different Chromosome Numbers

Key Idea: The number and size of chromosomes is highly variable between species. The number of chromosomes does not reflect the "genetic complexity" of an organism. Chromosomes carry the genetic material (DNA) and every eukaryotic species has at least two. The number of chromosomes is generally fixed for a given species (a few

have variable numbers) but does not reflect how genetically "complex" a species is because chromosomes may have been joined or split during the course of a species' evolution. So far, the largest number of chromosomes known in a eukaryote is found in the primitive adder's tongue fern, Ophioglossum reticulatum, which has 631 pairs.

Sex chromosomes

The number of chromosomes between species is extremely variable. Even closely related species can have large differences in the number of chromosomes. For example horses, donkeys, and zebras (genus Equus) have 64, 62, and 44 chromosomes respectively.

The sex of an organism is controlled in most cases by the sex chromosomes provided by each parent. In humans, males are the heterogametic sex because each somatic cell has one X and one Y chromosome. The determination of sex is based on the presence or absence of the Y chromosome; without it, an individual will develop into a female. In mammals, the male is always the heterogametic sex, but this is not necessarily the case in other taxa. In birds and butterflies, the female is the heterogametic sex, and in some insects the male is simply X whereas the female is XX.

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Chromosome numbers for different species

Organism

Chromosome number (2N)

Vertebrates cat rat rabbit human chimpanzee gorilla cattle dog turkey goldfish

38 42 44 46 48 48 60 78 82 94

Invertebrates horse roundworm fruit fly Drosophila housefly honey bee Hydra Plants broad bean cabbage garden pea rice Ponderosa pine orange potato

2 8 12 32 or 16 32

12 18 14 24 24 18, 27, or 36 48 Y

Zebra (2N = 44)

Zebra N = 22 (below) 2N = 44

Y

Zebra (2N = 44)

Donkey (2N = 62)

Donkey N = 31 (below) 2N = 62

The human X X chromosome is much larger than the Y and contains many more genes. As these genes are not present on the Y chromosome, they will always be expressed in males.

Horse (2N = 64)

The Y chromosome has lost approximately 96% of its original estimated genetic material. It currently possesses 200 genes, 72 of which produce proteins.

X

Donkey (2N = 62)

X

Horse (2N = 64)

1. (a) What is the difference in the diploid chromosome number between humans and chimpanzees?

(b) Which animal has the minimum number of chromosomes possible?

(c) Which plant has the same number of chromosomes as a gorilla? 2. Why is the number of chromosomes in an organism not a reflection of their genetic complexity?

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3. Explain what determines the sex of the offspring at the moment of conception in humans:

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4. Explain why, in male humans, many genes on the X chromosome will always be expressed:

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200 Karyotyping

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Key Idea: The karyotype is the number and appearance of chromosomes in the nucleus of a eukaryotic cell. The karyotype can be pictured in a standard format, called a karyogram, in which the chromosomes are ordered by size. The diagram below shows the preparation of a karyogram. Karyotyping begins with 'freezing' the nuclei of cultured white blood cells in metaphase of mitosis. A photograph

of the chromosomes is then cut up and the chromosomes are organised on a grid to produce the karyogram. The homologous pairs are placed together, identified by their shape, length, and banding pattern after staining. In humans, the male karyotype has 44 autosomes (non-sex chromosomes), and an X and a Y chromosome (44 + XY). The female karyotype has two X chromosomes (44 + XX).

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Preparing a karyotype Lymphocytes (a type of white blood cell) are treated with a drug to cause them to go into cell division (mitosis) and are grown for several days in culture. They are treated with another drug to halt the cycle at the metaphase stage.

Red blood cells

RCN

White blood cell

A sample of cells is taken from the person under investigation. This may be from the amniotic fluid surrounding a fetus or from a blood sample from an adult or child.

The cells are centrifuged and treated with a solution to make the cells swell up and their chromosomes spread out.

Microscope slide with blood smear

A drop of the cell suspension in preservative is spread on a microscope slide, dried and stained with a dye that causes a banding pattern to appear on each chromosome.

The stained white blood cells are viewed under a microscope and a clearly arranged spread of chromosomes is photographed.

Female karyotype: 44 + XX Chromosomes are arranged in groups according to size, shape and banding pattern.

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200 201 202

Chromosome 22 was originally identified as the smallest chromosome, but later research showed that 21 was smaller. The numbering wasn't changed because chromosome 21 was already identified as the chromosome that can lead to Down syndrome.

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Male karyotype: 44 + XY

The photograph is cut up so that each chromosome is separate from the others. This may be done manually or electronically with the aid of a computer. The chromosomes are arranged into homologous pairs according to size and shape.

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Karyotypes: Cytogenetics Dept, Waikato Hospital

Homologous pair

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Typical layout of a human karyogram

2

3

4

X

5

6

7

8

EII

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1

Y

This SEM shows the human X and Y chromosomes. Although these two are the sex chromosomes, they are not homologous.

9

10

11

12

17

18

13

14

19

20

15

16

21

22

Y

X

EII

Variable region

A scanning electron micrograph (SEM) of human chromosomes clearly showing their double chromatids.

1. (a) What is a karyogram?

(b) What information can it provide?

2. On the male and female karyograms on the previous page, number each homologous pair of chromosomes using the diagram above as a guide. 3. Circle the sex chromosomes (X and Y) in the karyogram of the female and the male.

4. Write down the number of autosomes and the arrangement of sex chromosomes for each sex: (a) Female:

No. of autosomes:

Sex chromosomes:

(b) Male:

No. of autosomes:

Sex chromosomes:

5. State how many chromosomes are found in a:

(a) Normal human (somatic) body cell:

(b) Normal human sperm or egg cell:

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6. Distinguish between autosomes and sex chromosomes:

8. Why are the X and Y chromosomes not homologous?

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7. What features of the chromosomes allow them to be paired up for a karyogram?


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201 Abnormal Chromosome Numbers

Key Idea: Abnormal chromosome numbers are associated with specific disorders, called syndromes. Occasionally sister chromatids fail to separate correctly during meiosis (an error called non-disjunction). This produces gametes with incorrect numbers of chromosomes, so that when gametes unite, the zygote will have more or less than

the correct number of chromosomes. Aneuploidy is the condition where the total chromosome number is not an exact multiple of the normal haploid set for the species (the number may be more, e.g. 2N+2, or less, e.g. 2N–1). Polysomy is aneuploidy involving specific chromosome(s) e.g. trisomy 17 in which there are 3 copies (2N+1) of chromosome 17. A

Non-disjunction is the failure of homologous chromosomes or sister chromatids to separate correctly during meiosis. It can occur in either meiosis I (failure of homologues to separate) or meiosis II (failure of sister chromatids to separate). Non-disjunction in meiosis I is the most damaging as it disrupts the chromosome allocation in all the possible gametes. Nondisjunction during meiosis II (right) results in half the gametes being abnormal.

n

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Non-disjunction

C n-1

n+1

Non-disjunction

Meiosis I

Turner syndrome (XO)

D

Gametes

Klinefelter syndrome (XXY)

Cytogenetics Department, Waikato Hospital

Down syndrome (trisomy 21)

Meiosis II

Down syndrome is the most common of the human aneuploidies. The incidence rate in humans is about 1 in 800 births for women aged 30 to 31 years, (the rate increases rapidly with maternal age, a phenomenon known as the maternal age effect). The most common form of this condition arises when meiosis fails to separate the pair of chromosome number 21s in the eggs that are forming in the woman's ovaries.

Turner syndrome results from the nondisjunction of the sex chromosomes during meiosis. The individual has only one sex chromosome (X) and lacks either another X or a Y. The karyotype (above) has a total of 45 chromosomes (one less than the normal 46). The incidence rate is 1 in 5000 live female births. In most cases, the X chromosome comes from the mother, and faulty sperm production causes the lack of a Y or second X chromosome.

PHOTOS: Cytogenetics Department, Waikato Hospital

When an abnormal gamete fuses with a normal gamete in fertilisation, the zygote will have an abnormal number of chromosomes.

n

B

Klinefelter syndrome also results from the non-disjunction of the sex chromosomes during meiosis. The individual has an extra sex chromosome (X), to produce a total complement of XXY. The karyotype (pictured here) of a Klinefelter syndrome individual shows a total complement of 47 including XXY sex chromosomes. The incidence rate is an average of 1 in 1000 live male births, with a maternal age effect.

1. What is non-disjunction?

2. (a) Place a circle around the sex chromosomes of the Klinefelter and Turner syndrome karyograms above.

(b) For Turner syndrome:

State the chromosome configuration:

Sex:

(c) For Klinefelter syndrome:

State the chromosome configuration:

Sex:

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3. Down syndrome (trisomy 21) is a viable phenotype, but most trisomies are lethal. Suggest why an extra chromosome 21 might be tolerated when other trisomies are not (hint: chromosome 21 is actually the smallest autosome):

A:

B:

C:

D:

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4. For each of the gametes shown above (A-D) write down the total number of chromosomes present in the zygote and if the zygote is normal or abnormal. Assume fertilisation occurs with a normal gamete:

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202 Making a Karyogram

Key Idea: A karyogram can be created by matching the size and banding pattern of individual chromosomes. Each chromosome has specific distinguishing features. Chromosomes are stained in a special technique that gives them a banded appearance in which the banding pattern represents regions containing up to many hundreds of genes.

Cut out the chromosomes below and arrange them on the record sheet in order to determine the sex and chromosome condition of the individual whose karyotype is shown. The karyograms presented on the previous pages and the hints on how to recognise chromosome pairs can be used to help you complete this activity.

Distinguishing Distinguishing characteristics characteristics of of chromosomes chromosomes Offset from centre

Central

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Near the end

Banding pattern

Satellite endings

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Centromere position

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Chromosome length

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1. Cut out the chromosomes on page 281 and arrange them on the record sheet below in their homologous pairs. 2. (a) Determine the sex of this individual: male or female (circle one) (b) State whether the individual's chromosome arrangement is: normal or abnormal (circle one)

(c) If the arrangement is abnormal, state in what way and name the syndrome displayed:

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13

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Sex chromosomes

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203 KEY TERMS: Did You Get It?

1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code. aneuploidy autosome

A Single piece of DNA that contains many genes and associated regulatory elements and proteins. B

chromatid

One of two identical DNA strands forming a replicated chromosome and held together by the centromere.

C The number and appearance of chromosomes in the nucleus of a eukaryotic cell.

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chromosome karyogram

D The condition of having a chromosome number that is not an exact multiple of the normal diploid (2N) condition. E

An error during meiosis in which homologous chromosomes or sister chromatids do not separate correctly.

F

The chromosome that carries the gene for determination of sex in individual organisms.

G

The rearrangement of a micrograph of chromosomes into a standard image and format.

polysomy

H

Condition in which some of the chromosomes are present in more than the diploid number.

sex chromosome

I

A non-sex chromosome.

karyotype

non-disjunction

2. For each of the karyograms shown below:

(a) Determine the sex of the individual:

(b) Determine if the karyotype shown is normal/abnormal (circle any abnormalities)

i) Sex:

ii) Sex:

Normal / abnormal

Normal / abnormal

3. The diagram on the right the shows a genetic cross for sex chromosomes in two sets of gametes. Some of the gametes and the possible sex chromosome combinations in the zygote have been left out.

(b) How many of the zygotes have the normal sex chromosome complement?

XY Gametes

XY

XY

(c) Write the ploidy of the zygotes labelled (i) and (ii)

(i)

(ii)

TEST

XX

XY

Zygote (i)

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(a) Complete the diagram by filling in the circles:

Female

X

X

(ii)

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Male

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Genotypes and phenotypes

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Unit 2 Outcome 2

Genotypes and phenotypes

Key terms

Key knowledge

allele

continuous variation

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1

Recall the difference between genes and alleles. Demonstrate understanding of the terms used in studying inheritance: allele, locus, trait, heterozygous, homozygous, genotype, phenotype, cross. Use symbols to represent genotypes and distinguish between alleles in genetic crosses.

c

2

Giving examples, distinguish between recessive and dominant traits.

204 205

Factors determining phenotype

Activity number

DNA methylation

dominant allele/trait epigenetics

Activity number

genotype

heterozygous

histone modification homozygous locus

mutation

phenotype

polygenes (=multiple genes) recessive allele/trait trait

Key knowledge

c

3

Explain how genetic make-up (genotype), environmental factors, and epigenetic factors contribute to produce the phenotype of an organism.

206

c

4

Recall that sexual reproduction contributes to variation in the genotype (and phenotype) of offspring in sexually reproducing organisms. Explain how changes to the DNA sequence through mutation can produce new alleles and result in an altered phenotype.

176 207

c

5

Describing examples in both plants and animals, explain how the environment of an organism during or after development can alter the expression of the genotype and produce variable phenotypes.

c

6

Recognise epigenetic factors as modifications to DNA that do not affect the DNA sequence itself. Explain how these modifications can make genes more or less likely to be expressed and how these changes can affect phenotype.

Polygenic inheritance and continuous variation Key knowledge

Using examples, describe polygenic inheritance and explain how it contributes to continuous variation in a population. Suitable examples include variation in skin colour or height in humans. Recognise the contribution of environment to the continuous variation observed for phenotypes determined by the inheritance of multiple genes (polygenes).

209 210

Activity number 211

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204 Alleles

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Key Idea: Eukaryotes generally have paired chromosomes. Each chromosome contains many genes and each gene may have a number of versions called alleles. Sexually reproducing organisms usually have paired sets

of chromosomes, one set from each parent. The equivalent chromosomes that form a pair are termed homologues. They carry equivalent sets of genes, but there is the potential for different versions of a gene (alleles) to exist in a population.

Homologous chromosomes

3

In sexually reproducing organisms, most cells have a homologous pair of chromosomes (one coming from each parent). This diagram shows the position of three different genes on the same chromosome that control three different traits (A, B and C).

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4 4

5

2

r

Chromosomes are formed from DNA and proteins. DNA tightly winds around special proteins to form the chromosome.

Having two different versions (alleles) of gene A is a heterozygous condition. Only the dominant allele (A) will be expressed. Alleles differ by only a few bases.

R

3

4

1

S

When both chromosomes have identical copies of the dominant allele for gene B the organism is homozygous dominant for that gene.

When both chromosomes have identical copies of the recessive allele for gene C the organism is said to be homozygous recessive for that gene.

S

c

c

c

c

Maternal chromosome originating from the egg of the female parent.

1

5

2

This diagram shows the complete chromosome complement for a hypothetical organism. It has a total of ten chromosomes, as five, nearly identical pairs (each pair is numbered). Each parent contributes one chromosome to the pair. The pairs are called homologues or homologous pairs. Each homologue carries an identical assortment of genes, but the version of the gene (the allele) from each parent may differ.

A gene is the unit of heredity. Genes occupying the same locus or position on a chromosome code for the same phenotypic character (e.g. eye colour). Paternal chromosome originating from the sperm of the male parent.

1. Define the following terms used to describe the allele combinations in the genotype for a given gene:

(a) Heterozygous:

(b) Homozygous dominant:

(c) Homozygous recessive:

2. For a gene given the symbol ‘A’, name the alleles present in an organism that is identified as:

(a) Heterozygous:

(b) Homozygous dominant:

(c) Homozygous recessive:

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4. Discuss the significance of genes existing as alleles:

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3. What is a homologous pair of chromosomes?

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205 Dominant and Recessive Traits

Key Idea: A phenotype refers to the observable characteristics of an organism. A variant of a phenotypic characteristic is a trait. Traits may result from dominant or recessive alleles. Traits are particular variants of phenotypic (observed physical) characters. For example, a phenotypic character is eye colour, a trait is blue eye colour. Traits may be controlled by

one gene or many genes and can show continuous variation, e.g. height in humans, or discontinuous variation, e.g. flower colour in pea plants. What trait appears depends on the alleles present. Dominant alleles will produce a dominant trait. Recessive alleles will only produce a recessive trait if both alleles present are recessive.

Some of the best know experiments in phenotypes are the experiments carried out by Gregor Mendel (right) on pea plants. During one of the experiments (shown below) he noticed how traits expressed in one generation disappeared in the second generation, but reappeared in the third generation. In his experiments Mendel used true breeding plants. When self-crossed, true breeding organisms produce offspring with the same phenotypes as the parents.

Mendel's experiments

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Mendel studied seven phenotypic characters of the pea plant: • Flower colour (violet or white) • Pod colour (green or yellow) • Height (tall or short) • Position of the flowers on the stem (axial or terminal) • Pod shape (inflated or constricted) • Seed shape (round of wrinkled) • Seed colour (yellow or green)

P1

In this diagram R (round) is dominant to r (wrinkled).

rr

RR

Wrinkled

F1

Rr

Rr

Rr

One of the experiments crossed true breeding round seed plants with true breeding wrinkled seed plants.

Round

Rr

Rr

Rr

F2

rr

RR

Rr

Rr

Rr

rr

RR

Rr

Rr

Out of the thousands of seeds produced, all were round, none were wrinkled.

Rr

Mendel then crossed the F1 offspring together. The wrinkled seed reappeared in the second generation. He saw similar results with all the other phenotypic characters he studied.

How can this be explained?

Mendel was able to explain his observations in the following way:

►► Traits are determined by a unit, which passes unchanged from parent to offspring (we now know these units are genes). ►► Each individual inherits one unit (gene) for each trait from each parent (each individual has two units).

►► Traits may not physically appear in an individual, but the units (genes) for them can still be passed to its offspring.

1. (a) Define a trait:

(b) Define true breeding:

2. (a) What was the ratio of smooth seeds to wrinkled seeds in the F2 generation?

(b) Why did the wrinkled seed trait not appear in the F1 generation?

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206 Influences on Phenotype

Key Idea: An organism's phenotype is influenced by the effects of the environment during and after development, even though the genotype remains unaffected. The phenotype encoded by genes is a product not only of the genes themselves, but of their internal and external environment and the variations in the way those genes

are controlled (epigenetics). Even identical twins have minor differences in their appearance due to epigenetic and environmental factors such as diet and intrauterine environment. Genes, together with epigenetic and environmental factors determine the unique phenotype that is produced. Genotype

Epigenetics

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Sources of variation in organisms

Alleles

Sexual reproduction

Histone modification

Phenotype

Single nucleotide variations

DNA methylation

Mutations

The phenotype is the product of the many complex interactions between the genotype, the environment, and the chemical tags and markers that regulate the expression of the genes (epigenetic factors).

Competition

Nutrition

Pathogens

Predators

Drugs

Toxins

Physical environment

Environment

Plasticity and polyphenism

Polyphenism is the expression of different phenotypes in a species due to environmental influences. Examples include sex determination in reptiles and changes in pigmentation in the wings of some butterfly species as the seasons change. The amount of change in a phenotype due environmental influences is called its phenotypic plasticity. Plants show a high amount of phenotypic plasticity because they are unable to move and must therefore adjust to a wide range of environmental changes throughout their lives. Animals generally show a much lower level of phenotypic plasticity, although plasticity in behaviour is often very high.

All worker bees and the queen bee (circled left) in a hive have the same genome, yet the queen looks and behaves very differently from the workers. Bee larvae fed a substance called royal jelly develop into queens. Without the royal jelly, the larvae develop into workers.

1. (a) What are some sources of genetically induced variation?

(b) What are some sources of environmentally induced variation?

3. Explain why behaviour is often highly plastic:

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2. Explain why genetically identical twins are not always phenotypically identical:

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4. Give an example in which an organism's environment produces a marked phenotypic change:

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207 Mutations can Alter Phenotype

Key Idea: Changes to the DNA sequence are called mutations. Mutations are the ultimate source of new genetic information, i.e. new alleles. Mutations are changes to the DNA sequence. They arise through errors in DNA copying and involve alterations in the DNA from a single base pair to large parts of chromosomes. Bases may be inserted into, substituted, or deleted from

the DNA. Mutations are the source of all new alleles. Most often mutations are harmful, but occasionally they can be beneficial. Sometimes they produce no phenotypic change and are silent. An example of a mutation producing a new allele is described below. This mutation causes a form of genetic hearing loss (called NSRD). It occurs in the gene coding for the protein connexin 26.

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Peptide chain formed (amino acid sequence)

1

Leu

2

Gly

Gly

Val

Asn

42

28

The gene that codes for the peptide chain connexin 26.

The code is read in 3-base sequences called codons. One codon codes for one amino acid.

C T G G G G G G T G T G A A C

Normal DNA sequence

Chromosome 13

The most common mutation in this gene is the deletion of the 35th base (guanine (G)). Deleting the guanine changes the code and results in a short chain, which cannot function.

The mutation causes a shortened peptide chain.

Peptide chain formed

3

Leu

Gly

28

Val

Stop

42

G

C T G G G G G T G T G A A C A

Connexin 26 peptide chains

Six connexin 26 peptide chains join to form a connection between cells. If these proteins are not made correctly, they cannot form the final structure.

Mutated sequence

4

TGA codes for stop

The deletion creates a recessive allele. A person is deaf when they carry two recessive alleles.

Sickle cell

In humans, there are three major blood groups, A, B, and O, controlled by the alleles A, B, and O. The alleles produce enzymes that modify carbohydrates (sugars) on the surface of blood cells. Different blood groups have different carbohydrates on the cell surface. It is believed that The A allele group evolved first, followed by O, and then B.

1. What is a mutation?

Red blood cells are packed with the oxygencarrying protein haemoglobin, encoded by the HBB gene. A mutation to the HBB gene produces an allele that causes the haemoglobin to distort the red blood cells into a sickle shape. The homozygous condition is lethal but the mutation persists because heterozygotes are more resistant to malaria.

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In mammals, production of the enzyme lactase, which digests the sugar lactose in milk, stops after childhood. In humans of European, East African, or Indian descent, a mutation about 10,000 years ago produced a dominant allele for the lactase gene. This kept the lactase gene active, allowing adults to continue to benefit from eating or drinking dairy products.

2. The NSRD mutation is a harmful mutation. Why might someone with this mutation not actually be affected?

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3. Why would the appearance of the persistent lactase allele be an advantage?

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Key Idea: The environment can play a big part in an organism's eventual phenotype. Environmental factors, including physical factors such as temperature and biotic factors such as presence of predators,

The effect of other organisms For some animals, the presence of other individuals of the same species may control sex determination.

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The effect of temperature

can influence how genes are expressed. Factors such as heat or chemicals can turn genes on (genes are expressed) or off. When and for how long the genes are expressed can have large effects on an organism's eventual phenotype.

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208 Environment and Phenotype

The sex of some animals is determined by incubation temperature during embryonic development. Examples include turtles, crocodiles, and the American alligator. In some species, high incubation temperatures produce males and low temperatures produce females. In other species, the opposite is true. Temperature regulated sex determination may be advantageous by preventing inbreeding (since all siblings will tend to be of the same sex).

Some fish species, including some in the wrasse family (e.g. Coris sandageri, right), show this phenomenon. The fish live in groups consisting of a single male with attendant females and juveniles. In the presence of a male, all juveniles become females. When the male dies, the dominant female will undergo changes in physiology and appearance to become a male.

Female

Male

Chemical signal

Non-helmeted form

Colour-pointing in breeds of cats and rabbits (e.g. Siamese, Himalayan) is a result of a temperature sensitive mutation in one of the enzymes in the metabolic pathway from tyrosine to melanin. The dark pigment is only produced in the cooler areas of the body (face, ears, feet, and tail), while the rest of the body is a paler version of the same colour, or white.

Some organisms respond to the presence of harmful organisms by changing their morphology or body shape. When the water flea Daphnia is exposed to predatory phantom midge larvae it develops a helmet and/or tail spine and also produces young with the same defensive structures. These responses are mediated through chemicals produced by the predator.

Helmeted form

The helmet makes Daphnia more difficult to attack and handle by the predatory midge.

1. Describe an example to illustrate how genotype and environment contribute to phenotype:

2. (a) How is helmet and spine development in Daphnia a response to environment?

(b) How does the phenotypic response help the animal survive?

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3. Why are the darker patches of fur in colour-pointed cats and rabbits found only on the face, paws and tail:

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The effect of altitude Severe stunting (krummholz)

The effect of chemical environment Growth to genetic potential

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Cline

Increasing altitude can stunt the phenotype of plants with the same genotype. In some conifers, e.g. Engelmann spruce, plants at low altitude grow to their full genetic potential, but become progressively more stunted as elevation increases, forming krummholz (gnarled bushy forms) at the highest sites. A continuous gradation in a phenotypic character within a species, associated with a change in an environmental variable, is called a cline.

The chemical environment can influence the expressed phenotype in plants and animals. In hydrangeas, flower colour varies according to soil pH. Flowers are blue in more acidic soils (pH 5.0-5.5), but pink in more alkaline soils (pH 6.0-6.5). The blue colour is due to the presence of aluminium compounds in the flowers and aluminium is more readily available when the soil pH is low.

4. Describe an example of how the chemical environment of a plant can influence phenotype:

5. Vegetable growers can produce enormous vegetables for competition. How could you improve the chance that a vegetable would reach its maximum genetic potential?

6. (a) What is a cline?

(b) What are the physical factors associated with altitude that could affect plant phenotype?

(c) On a windswept portion of a coast, two different species of plant (species A and species B) were found growing together. Both had a low growing (prostrate) phenotype. One of each plant type was transferred to a greenhouse where "ideal" conditions were provided to allow maximum growth. In this controlled environment, species B continued to grow in its original prostrate form, but species A changed its growing pattern and became erect in form. Identify the cause of the prostrate phenotype in each of the coastal grown plant species and explain your answer:

Plant species A:

Plant species B:

(d) Which of these species (A or B) would be most likely to exhibit clinal variation?

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209 Gene Expression and Phenotype

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Key Idea: The environment alters how the DNA is coiled and how the DNA is read and expressed by the cell. The nucleosomes in the chromatin can package the DNA in the nucleus tightly (as heterochromatin) or more loosely (as euchromatin). The extent of packaging affects whether or not

the DNA can be expressed (read and translated into proteins). DNA packaging is affected by modification of the organising histones and by DNA methylation. Methylation of the DNA is usually reset during fertilisation, although sometimes the methylation is carried over to the next generation. Methylation and epigenetics

DNA methylation

Methylation and histone modifications can be passed on from a cell's DNA to its daughter cells during DNA replication. In this way, any environmental or other chemical effects encountered by a cell can be passed on to the next generation of cells.

Cytosine

NH2

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NH2

N H

CH3

Methylation

N

N

N H

O

DNA sequence

O

CH3

5’ ---CGCGCAC---3’ 3’ ---GCGCGTG---5’

Cytosine methylation is an important process in DNA packaging and gene expression. Cytosine methylation can alter gene expression by enzymes binding to the DNA or it may cause the chromatin to bind tightly together so that genes cannot be copied by the cellular machinery.

Methylation is very important during embryonic development when a lot of methylation takes place and a lot of cells are being produced. Once a cell has differentiated into a specific cell type, all the daughter cells inherit the modifications and remain the same type of cell. The study of these modifications and how the environment influences them is called epigenetics. Epimeans 'on top of' or 'extra to'. Methylation (together with histone modification) are part of epigenetics because they don't directly change the DNA sequence.

Gene expression changes as an organism develops. During the development of the embryo there are many genes that are switched on and off as development of tissues and organs proceeds. Some of this control is achieved by DNA methylation.

Enzymes can add or remove methyl groups from cytosine bases and so activate or silence genes. Often methylation is affected by changes in the environment and this provides the developing organism with a rapid response mechanism.

Degree of methylation

Methylation and gene expression

Fertilisation

Methylation in germ cells

Germ cells develop

Sperm

Somatic cells in adult

Zygote

Demethylation in early embryo

Germ cells develop

Eggs

Gametogenesis

Pre-implantation

Post-implantation

Development

(b) How does DNA methylation affect gene expression?

(c) What is the role of methylation during development?

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1. (a) What is DNA methylation?

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2. Why is anything the pregnant mother does particularly likely to affect gene expression in the embryo?

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210 Epigenetic Factors and Phenotype

Key Idea: The environment or experiences of an individual can affect the development of following generations. Studies of heredity have found that the environment, experiences, or lifestyle of an ancestor can have an effect on future generations. Certain environments or diets can affect the methylation and packaging of the DNA (rather

than the DNA itself) determining which genes are switched on or off and so affecting the development of the individual. These effects can be passed on to offspring, and even on to future generations. It is thought that these inherited effects may provide a rapid way to adapt to particular environmental situations, such as famine or chronic stress.

Epigenetics and inheritance

How can epigenetic effects be proved to be inherited? Anything the mother is exposed to will affect her and will also expose the fetus. In a female fetus, egg cells develop in the ovaries, so a third generation will be exposed also. If a fourth generation proves to be affected, then there is evidence that the epigenetic effect is inherited.

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The destruction of New York's Twin Towers on September 11, 2001, traumatised thousands of people. In those thousands were 1700 pregnant women. Some of them suffered (often severe) posttraumatic stress disorder (PTSD), others did not. Studies on the mothers who developed PTSD found very low levels of the cortisol in their saliva. Cortisol helps the body adapt to stress. The children of these mothers also had much lower levels of cortisol than those whose mothers had not suffered PTSD. The environment of the mother had affected the offspring.

Mother (F0) Fetus (F1)

Egg cells (F2)

Mice and environmental effects

The effect of the environment and diet of mothers on later generations exposed to a breast cancer trigger (a carcinogenic chemical) was investigated in rats fed a high fat diet (HFD) or a diet high in the hormone oestrogen (HOD). The length of time taken for breast cancer to develop in later generations after the trigger for breast cancer was given was recorded and compared. The data are presented below. F1 = daughters, F2 = granddaughters, F3 = great granddaughters.

Cumulative percentage rats with breast cancer (high fat diet, HFD) F1%

F 2%

F 3%

Weeks since trigger

HFD

Control

HFD

Control

HFD

Control

6

0

0

5

0

3

0

8

15

0

20

5

3

20

10

22

8

30

5

10

25

12

22

18

50

20

20

30

14

22

18

50

30

25

40

16

29

18

60

30

25

40

18

29

18

60

40

40

42

20

40

18

65

40

50

60

22

80

60

79

50

50

60

Data source: Science News April 6, 2013

Cumulative percentage rats with breast cancer (high oestrogen diet, HOD) F1%

F 2%

F 3%

HOD

Control

HOD

Control

HOD

Control

6

5

0

10

0

0

0

15

10

30

20

40

20

50

20

50

30

75

40

80

45

80

60

10

0

10

0

30

15

15

20

12

38

19

30

30

14

50

22

30

40

16

50

22

30

40

18

60

35

40

40

20

60

42

50

50

22

80

55

50

50

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Weeks since trigger

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1. Use the data on the previous page to complete the graphs below. The first graph is done for you: % F1breast with breast cancer % F1 with cancer (HFD)(HFD)

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60 40

20

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60

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% rats with breast cancer

High fat diet diet 80 High fat Control Control

100 100

% rats with breast cancer

80

% rats with breast cancer

% rats with breast cancer

100 100

80 80 % rats with breast cancer

High fat diet Control

40 40 20 20

6

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100100 80 80 % rats with breast cancer

High fat diet Control

60 60 40 40 20 20

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68 8 10 10 12 12 14 14 16 16 18 18 22 22 24 Weeks after exposure to cancer Weeks after exposure to cancer triggertrigger

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8 10 12 14 16 18 22 Weeks after exposure to cancer trigger

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60 40 20

0 6 8 8 10 1012 1214 1416 1618 18 22 2224 24 6 8 10 12 14 16 18 22 Weeks after exposure to cancer trigger Weeks after exposure to cancer trigger Weeks after exposure to cancer trigger 2. (a) Which generations are affected by the original mother eating a high fat diet? 0

24

100

6 8 8 10 1012 1214 1416 1618 18 22 2224 24 Weeks exposure to cancer trigger Weeks afterafter exposure to cancer trigger % F1 with breast cancer (HFD)

6

0

(b) Which generations are affected by the mother eating a high oestrogen diet?

(c) Which diet had the longest lasting effect?

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% rats with breast cancer

% F1 with breast cancer (HFD)

60 60

0

0

24

3. What do these experiments show with respect to epigenetic changes and inheritance?

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% rats with breast cancer

100100

68 8 10 10 12 12 14 14 16 16 18 18 22 22 24 Weeks after exposure to cancer Weeks after exposure to cancer triggertrigger

% rats with breast cancer

0

6

% rats with breast cancer

0

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211 Genes, Environment, and Continuous Variation

Key Idea: Many phenotypes are affected by multiple genes. Many phenotypes are controlled by more than one gene, a situation known as polygeny or polygenic inheritance. As there are many genes and therefore many alleles

controlling the phenotype, there are a large range of possible phenotypes. Combined with environmental effects, this produces continuous variation within the population. Two examples in humans are skin colour and height.

Polygenic traits

Polygenic traits are usually identified by ►► Traits are usually quantified by measuring rather than counting.

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►► Two or more genes contribute to the phenotype.

►► Phenotypic expression is over a wide range (often in a bell shaped curve).

►► Polygenic phenotypes include skin colour, height, eye colour, and weight.

It is estimated that skin colour is controlled by at least eight genes (probably more). There are various ways to compare skin colour. One is shown right, in which there are seven shades ranging from very pale to very dark. Most individuals are somewhat intermediate in skin colour.

Very pale

Light

Medium light

Medium

Medium dark

Dark

Black

1

2

3

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6

0

The table below shows a cross between three genes involved in skin colour, A, B, and C, each with two alleles (AaBbCc x AaBbCc). This is sufficient to give the seven shades of skin colour shown above. The shaded boxes indicate their effect on skin colour when combined. No dominant allele results in a lack of dark pigment (aabbcc). Full pigmentation (black) requires six dominant alleles (AABBCC). Note that for three genes with two alleles each there are 23 x 23 = 8 X 8 = 64 possible genotypes. How many would there be if eight genes were included in the table? ABC

ABc

AbC

Abc

aBC

aBc

abC

abc

ABC

AABB CC

AABB Cc

AABb CC

AABb Cc

AaBB CC

AaBB Cc

AaBb CC

AaBb Cc

ABc

AABB Cc

AABB cc

AABb Cc

AABb cc

AaBB Cc

AaBB cc

AaBb Cc

AaBb cc

AbC

AABb CC

AABb Cc

AAbb CC

AAbb Cc

AaBb CC

AaBb Cc

Aabb Cc

Aabb Cc

Abc

AABb Cc

AABb cc

AAbb Cc

AAbb cc

AaBb Cc

AaBb cc

Aabb Cc

Aabb cc

aBC

AaBB CC

AaBB Cc

AaBb CC

AaBb Cc

aaBB CC

aaBB Cc

aaBb CC

aaBb Cc

aBc

AaBB Cc

AaBB cc

AaBb Cc

AaBb cc

aaBB Cc

aaBB cc

aaBb Cc

aaBb cc

abC

AaBb CC

AaBb Cc

Aabb CC

Aabb Cc

aaBb CC

aaBb Cc

aabb CC

aabb Cc

abc

AaBb Cc

AaBb cc

Aabb Cc

Aabb cc

aaBb Cc

aaBb cc

aabb Cc

aabb cc

Gametes

A dark knight to rescue the fair maiden? Read a fairy tale and inevitably the princess is faircomplexioned and light of hair. The knight sent to save her is darker or tanned, with wild black or dark hair. Really? Its just a fairy tale, isn't it?... Well, not entirely. Research shows that, in any human population, women on average have a lighter skin colour than men. In European populations, women have skin 15.2% lighter than men while, in African populations, women have skin 11.1% lighter than men. The same research shows men are generally attracted to women with lighter skin, whereas women are attracted to men with darker skin. Why? Although preference is undoubtedly linked to cultural values, there is an evolutionary advantage for women to have lighter skin. It enables them to absorb more UV light, which helps manufacture the vitamin D needed to absorb calcium from the diet. High calcium levels are needed for strong pelvic development, development of the fetal skeleton, and to produce milk to feed the growing infant. So, is there something in those old fairy tales after all?

1. (a) What is polygeny?

(b) How does polygeny contribute to continuous variation?

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2. Identify two continuous phenotypes in humans and explain why the traits are continuous:

3. Study the cross between the A, B, and C genes above. Write down the frequencies of the seven phenotypes (0-6):

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4. Explain the differences between continuous and discontinuous variation, giving examples to illustrate your answer:

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5. From a sample of no less than 30 adults, collect data (by request or measurement) for one continuous variable (e.g. height, weight, shoe size, or hand span). Record and tabulate your results in the space below, and then plot a frequency histogram of the data on the grid below:

Raw data

Tally chart (frequency table)

Frequency

Variable:

(a) Calculate the following for your data and attach your working:

Mean:

Mode:

Median:

Standard deviation:

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(b) Describe the pattern of distribution shown by the graph, giving a reason for your answer: (c) What is the genetic basis of this distribution?

(d) What is the importance of a large sample size when gathering data relating to a continuous variable?

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212 KEY TERMS: Did You Get It?

1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code.

allele

A

Possessing two different alleles for a gene, one inherited from each parent.

continuous variation

B

Chromosome pairs, one paternal and one maternal, of the same length, centromere position, and staining pattern with genes for the same characteristics at corresponding loci.

C

Possessing two identical alleles of a gene, one inherited from each parent.

D

One of a number of alternative versions of a gene.

E

Specific phenotypic characteristic, e.g. red hair.

homozygous

F

Observable characteristics in an organism.

homologous chromosomes

G

An allele that will only express its trait in the absence of the dominant allele.

mutation

H

Allele that expresses its trait irrespective of the other allele.

I

Variation between individuals of a population in which differences are the result of polygenic interactions and the phenotypic variation exhibits a normal distribution.

J

A change to the DNA sequence of an organism.

K

The study of chemical changes to the DNA that alter gene expression without altering the DNA sequence.

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dominant

epigenetics

heterozygous

phenotype recessive trait

2. A breeder has two guinea pigs, one with black fur and the other with white. The two are bred together and all the offspring are black. Two of the offspring are then crossed. Four offspring are born one is white the rest are black.

(a) What phenotype is being investigated here?

(b) Which phenotypic trait is dominant?

3. A student decides to investigate the width of a flower for a school project. She measures the diameter of forty of her chosen flower type from the local botanical gardens. The raw data are shown below: Flower diameter (mm)

65, 40, 40, 66, 64, 42, 39, 67, 68, 38, 43, 60, 67, 38, 37, 37, 43, 60, 63, 67, 68, 67, 30, 34, 43, 44, 38, 71, 72, 70, 67, 69, 69, 31, 37, 39, 40, 44, 43, 73

Tally chart of flower diameter (mm)

(b) Graph the data on the grid provided:

(c) Described the shape of the graph and suggest a possible reason for the shape:

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(a) Tally the data into a frequency table:

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Pedigrees and genetic outcomes

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Unit 2 Outcome 2

Pedigree charts and patterns of inheritance

Key terms

autosomal dominant trait

autosomal recessive trait dihybrid cross

Key knowledge

Activity number

c

1

Analyse pedigree charts to trace the inheritance patterns of particular traits, including autosomal dominant, autosomal recessive, X-linked, and Y-linked traits.

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2

Develop your own set of guidelines for interpreting pedigree charts based on a range of examples.

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embryo selection genetic cross

genetic screening genotype

linked genes

monohybrid cross pedigree

phenotype

recombination

Genetic crosses

sex linkage

X-linked trait

Activity number

Key knowledge

Y-linked trait

c

3

c

4

c

5

Use Punnett squares to determine genotypes and predict the outcomes of genetic crosses involving autosomal and sex linked monohybrid inheritance.

Understand use of the monohybrid test cross to determine unknown genotypes.

Use Punnett squares to solve problems involving dihybrid inheritance of unlinked (independent), autosomal genes for two independent characteristics.

Inheritance of linked genes

214 216 217 218 215

219 222

Activity number

Key knowledge

c

6

Explain what is meant by genetic linkage (linked genes) and explain its biological consequences. Describe the dihybrid inheritance of linked genes.

220

c

7

Explain how recombination of alleles can occur as a result of crossing over between linked genes. Explain how recombination can be identified in the offspring of genetic crosses involving the inheritance of two characteristics.

221

Genetic testing

Activity number

Key knowledge

8

Recall how the knowledge gained by the Human Genome Project is being used for the purposes of identifying specific alleles or their markers (genetic screening). Describe the nature and uses of genetic testing for screening of embryos and adults, e.g. for a specific disease. Recall that genetic screening increasingly involves looking for patterns in genomic variations that are associated with increased disease risk.

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9

Discuss the social and ethical implications of genetic screening, including issues of embryo selection, saviour siblings, privacy, and discrimination.

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213 Pedigree Charts

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III 2 the inheritance 3 6 7 make used1to study of4 genetic5 disorders and it possible to follow the genetic history of an individual.

Key Idea: Pedigree charts are a way to graphically illustrate inheritance patterns over a number of generations. They are

Pedigree charts

Key to Symbols

A pedigree chart is a diagram that shows the occurrence and appearance of a particular gene or trait from one generation to the next. In humans, pedigree charts are often used to analyse the inheritance of heritable conditions. In domestic animals, pedigree charts are often used to trace the inheritance of characteristics in selective breeding programmes for horses and dogs.

Normal female

Sex unknown

Normal male

Died in infancy

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Affected female

Identical twins

Affected male

Pedigree charts use symbols to indicate an individual's particular traits. The key (right) explains the meaning of the symbols. Particular individuals are identified by their generation number and their order number in that generation. For example, II-6 is the sixth person in the second row. The arrow indicates the person through whom the pedigree was discovered (i.e. who reported the condition).

Non-identical twins

Carrier (heterozygote)

I, II, III

1, 2, 3 Children (in order of birth)

I

If the chart on the right were illustrating a human family tree, it would represent three generations: grandparents (I-1 and I-2) with three sons and one daughter. Two of the sons (II-3 and II-4) are identical twins, but did not marry or have any children. The other son (II-1) married and had a daughter and another child (sex unknown). The daughter (II-5) married and had two sons and two daughters (plus a child that died in infancy).

1

Generations

2

II

1

2

1

2

3

4

5

6

III

For the particular trait being studied, the grandfather was expressing the phenotype (showing the trait) and the grandmother was a carrier. One of their sons and one of their daughters also show the trait, together with one of their granddaughters (arrow).

3

4

5

6

7

Key to Symbols

Normal female

The pedigree of lactose intolerance

Sex unknown

Lactose intolerance is the inability to digest the milk sugar lactose. It occurs because some people do not produce lactase, the Normal male in infancy enzyme needed to break down lactose. The pedigree chart below was one of the original studies to determine Died the inheritance pattern of lactose intolerance. Researchers concluded that because two lactose tolerant parents can produce a lactose intolerant child, Affected female lactose intolerance must be a recessively inherited condition (it needs two copies of the gene for lactose intolerance to show up). Identical twins

Affected male

?

?

I

Non-identical twins

Carrier (heterozygote)

1

2

I, II, III

1, 2, 3 Children (in order of birth)

II

?

?

1

2

Generations

?

4

3

5

6

iii

1

2

3

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7

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V

KEY

2

1 Lactose tolerant male

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Lactose intolerant male

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Lactose tolerant female

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Lactose intolerant female

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IV

? Lactose tolerance unknown

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1. Autosomal recessive traits Albinos lack pigment in the hair, skin and eyes. This is an autosomal recessive trait.

I

1

2

3

4

II

(b) Why must the parents (II-3) and (II-4) be carriers of a recessive allele:

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(a) Write the genotype for each of the individuals on the chart using the following letter codes: PP normal skin colour; Pnormal, but unknown if homozygous; Pp carrier; pp albino.

Albinism in humans

2. Sex linked recessive traits Haemophilia is a disease where blood clotting is affected. A person can die from a simple bruise (which is internal bleeding). The clotting factor gene is carried on the X chromosome.

(a) Write the genotype for each of the individuals on the chart using the codes: XY normal male; XhY affected male; XX normal female; XhX female carrier; XhXh affected female: (b) Why can males never be carriers?

Haemophilia in humans

I

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1

3. Autosomal dominant traits An unusual trait found in some humans is woolly hair (not to be confused with curly hair). Each affected individual will have at least one affected parent.

(a) Write the genotype for each of the individuals on the chart using the following letter codes: WW woolly hair; Ww woolly hair (heterozygous); W- woolly hair, but unknown if homozygous; ww normal hair

2

3

4

Woolly hair in humans

I

1

2

II

(b) Describe a feature of this inheritance pattern that suggests the trait is the result of a dominant allele:

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4. Sex linked dominant traits A rare form of rickets is inherited on the X chromosome. All daughters of affected males will be affected. More females than males will show the trait.

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A rare form of rickets in humans

(b) Why will more females than males be affected?

1

II 1

III

5. Using the examples on this these two pages, make up your own set of guidelines for interpreting pedigree charts. How do you distinguish an autosomal inheritance pattern from an X-linked one? What are the features of autosomal recessive inheritance? Of autosomal dominant? Of X-linked dominant traits and X-linked recessive traits. Attach your summary to this page.

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(a) Write the genotype for each of the individuals on the chart using the following letter codes: XY normal male; XRY affected male; XX normal female; XR– female (unknown if homozygous); XRXR affected female.

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214 The Monohybrid Cross

Key Idea: The outcome of a cross depends on the parental genotypes. A true breeding parent is homozygous for the gene involved. Examine the diagrams depicting monohybrid (single gene) inheritance. The F1 generation by definition describes

the offspring of a cross between distinctly different, truebreeding (homozygous) parents. A back cross refers to any cross between an offspring and one of its parents. If the back cross is to a homozygous recessive, it is diagnostic, and is therefore called a test cross.

Monohybrid cross F1 Homozygous white

Heterozygous purple

Heterozygous purple

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Homozygous purple

Monohybrid cross F2

Parents:

PP

X

Parents:

pp

X

Pp

Gametes:

P

p

P

Pp

p

P

P

Male gametes

Female gametes

p

p

F1

Genotypes: All

Pp

Phenotypes: All purple

75% purple PP, Pp 25% white pp

A true-breeding organism is homozygous for the gene involved. The F1 offspring of a cross between two true breeding parent plants are all purple (Pp).

A cross between the F1 offspring (Pp x Pp) would yield a 3:1 ratio in the F2 of purple (PP, Pp, Pp) to white (pp).

1. Study the diagrams above and explain why white flower colour does not appear in the F1 generation but reappears in the F2 generation: Dihybrid cross Homozygous yellow-round

Parents:

Gametes:

X

YR

YyRr X Monohybrid cross

1. Fill in the Punnett square (below right) to purple Homozygous white show theHeterozygous genotypes of the F2 generation. Offspring (F2)

pp

Possible fertilizations

YR Parents:

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Yr

PP

yr

yR

X

Pp

Yr yR yr

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3. Express these numbers as a ratio:

Male gametes

Green-wrinkled

for the F2

Homozygous Female purple gametes Heterozygous purple

Gametes:

Yellow-wrinkled

YyRr

YR

Yellow-round Gametes:

Green-round

yr

F1 all yellow-round

Monohybrid cross

2. In the boxes below, use fractions to Parents: indicate the numbers of each phenotype Pp X produced from this cross.

Homozygous green-wrinkled

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A dihybrid cross studies the inheritance patterns of two genes. In pea seeds, yellow color (Y) is dominant to green (y) and round shape (R) is dominant to wrinkled (r). Each true breeding parental plant has matching alleles for each of these characters (YYRR or yyrr). F1 offspring will all have the same genotype and phenotype (yellow-round: YyRr). 2. Complete the crosses below:

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215 The Test Cross

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Key Idea: If an individual's genotype is unknown it can be determined using a test cross. It is not always possible to determine an organism's genotype by its appearance because gene expression is complicated by patterns of dominance and by gene interactions. The test cross was developed by Mendel as a wayThe to establish basis ofthea genotype of an organism with the dominant phenotype for

a particular trait. The principle is simple. The individual with the unknown genotype is bred with a homozygous recessive individual for the trait(s) of interest. The homozygous recessive can produce only one type of allele (recessive), so the phenotypes of the offspring will reveal the genotype of the unknown parent (below). The test cross can be used test cross to determine the genotype of single genes or multiple genes.

Parent 1 Unknown genotype (but with dominant traits)

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Parent 2 Homozygous recessive genotype (no dominant traits)

?

?

?

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X

a

b

a

b

The common fruit fly (Drosophila melanogaster) is often used to illustrate basic principles of inheritance because it has several genetic markers whose phenotypes are easily identified. Once such phenotype is body colour. Wild type (normal) Drosophila have yellowbrown bodies. The allele for yellow-brown body colour (E) is dominant. The allele for an ebony coloured body (e) is recessive. The test crosses below show the possible outcomes for an individual with homozygous and heterozygous alleles for ebony body colour. A. A homozygous recessive female (ee) with an ebony body is crossed with a homozyogous dominant male (EE).

B. A homozygous recessive female (ee) with an ebony body is crossed with a heterozygous male (Ee).

Female gametes

Male gametes

Female gametes

Male gametes

Cross A:

(a) Genotype frequency:

(b) Phenotype frequency:

100% Ee

100% yellow-brown

Cross B:

(a) Genotype frequency:

(b) Phenotype frequency:

50% Ee, 50% ee

50% yellow-brown, 50% ebony

1. In Drosophila, the allele for brown eyes (b) is recessive, while the red eye allele (B) is dominant. How would you set up a two gene test cross to determine the genotype of a male who has a normal body colour and red eyes?

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2. List all of the possible genotypes for the male Drosophila:

(a) What is the genotype of the male Drosophila?

(b) Explain your answer:

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3. 50% of the resulting progeny are yellow-brown bodies with red eyes, and 50% have ebony bodies with red eyes.

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216 Sex Linkage

Key Idea: Many genes on the X chromosome do not have a match on the Y chromosome. In males, which are XY, a recessive allele on the X chromosome will be expressed. Sex linkage refers to the way genes on the sex chromosomes are inherited and expressed. In humans, the sex chromosomes are X and Y, but sex linkage usually involves genes on the X chromosome, which has many more genes than the Y

chromosome. X-linked recessive traits are usually seen only in males (XY) and occur rarely in the females (XX) because females may be heterozygous (carriers). X-linked dominant traits do not necessarily affect males more than females. In humans, recessive sex linked genes are responsible for a number of heritable disorders in males. Y-linked disorders are rare and usually associated with infertility.

Parents

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Haemophilia is a recessive disorder linked to the X-chromosome that results in ineffective blood clotting when a blood vessel is damaged. The most common type, haemophilia A, occurs in 1 in 5000 male births. Any male who carries the gene will express the phenotype. Haemophilia is extremely rare in women. 1. A couple wish to have children. The woman knows she is a carrier for haemophilia. The man is not a haemophiliac. Use the notation Xh for haemophilia and XH for the dominant allele to complete the diagram on the right including the parent genotypes, gametes and possible fertilisations. Write the genotypes and phenotypes in the table below.

Genotypes

Carrier female

Normal male

X

Gametes

Phenotypes

Male children

Possible fertilisations

Paternal gametes

Paternal gametes

Maternal gametes

3. The gene for red-green colour vision is carried on the X chromosome. If the gene is faulty, colour blindness (Xb) will occur in males. Red-green colour blindness occurs in about 8% of males but in fewer than 1% of females.

A colour blind man has children with a woman who is not colour blind. The couple have four children: 1 non colour blind son, 1 colour blind son, 2 non colour blind daughters. Describe the mother's:

(a) Genotype: (b) Phenotype:

(c) Identify the genotype not possessed by any of the children:

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Maternal gametes

Parents

Non colour blind female

Gametes Possible fertilisations

X

Colour blind male

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(b) The man is not a haemophiliac. Determine the probability that their first male child will have haemophilia. Use the Punnett squares to help you:

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Paternal gametes

Paternal gametes

Maternal gametes

2. (a) A second couple also wish to have children. The woman knows her maternal grandfather was a haemophiliac, but neither her mother or father were. Determine the probability she is a carrier (XHXh) Use the Punnett squares, right, to help you:

Maternal gametes

Female children

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Dominant allele in humans A rare form of rickets in humans is determined by a dominant allele of a gene on the X chromosome (it is not found on the Y chromosome). This condition is not successfully treated with vitamin D therapy. The allele types, genotypes, and phenotypes are as follows:

Allele types

Genotypes

XR = affected by rickets

XRXR, XRX = Affected female

X = normal

XRY

= Affected male

XX, XY

= Normal female, male

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Phenotypes

As a genetic counsellor you are presented with a married couple where one of them has a family history of this disease. The husband is affected by this disease and the wife is normal. The couple, who are thinking of starting a family, would like to know what their chances are of having a child born with this condition. They would also like to know what the probabilities are of having an affected boy or affected girl. Use the symbols above to complete the diagram right and determine the probabilities stated below (expressed as a proportion or percentage). 4. Determine the probability of having:

(a) Affected children:

(b) An affected girl:

(c) An affected boy:

Affected husband

Normal wife

X

Parents

Gametes

Possible fertilisations

Children

Another couple with a family history of the same disease also come in to see you to obtain genetic counselling. In this case, the husband is normal and the wife is affected. The wife's father was not affected by this disease. Determine what their chances are of having a child born with this condition. They would also like to know what the probabilities are of having an affected boy or affected girl. Use the symbols above to complete the diagram right and determine the probabilities stated below (expressed as a proportion or percentage).

Affected wife (whose father was normal)

Normal husband

Parents

X

Gametes

5. Determine the probability of having:

(a) Affected children:

(b) An affected girl:

(c) An affected boy:

Possible fertilisations

Children

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6. Why are males much more likely to inherit X-linked recessive disorders than females?

(a) (b)

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7. From what you know about sex linkage, what two features could you use to detect a Y-linked disorder in a pedigree?

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217 Inheritance Patterns

Key Idea: Sex-linked traits and autosomal traits have different inheritance patterns. Complete the following monohybrid crosses for different types

of inheritance patterns in humans: autosomal recessive, autosomal dominant, sex linked recessive, and sex linked dominant inheritance. Female parent phenotype:

Male parent phenotype:

P

eggs

p

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1. Inheritance of autosomal recessive traits Example: Albinism Albinism (lack of pigment in hair, eyes and skin) is inherited as an autosomal recessive allele (not sex-linked). Using the codes: PP (normal) Pp (carrier) pp (albino) (a) Enter the parent phenotypes and complete the Punnett square for a cross between two carrier genotypes. (b) Give the ratios for the phenotypes from this cross. Phenotype ratios:

2. Inheritance of autosomal dominant traits Example: Woolly hair Woolly hair is inherited as an autosomal dominant allele. Each affected individual will have at least one affected parent. Using the codes: WW (woolly hair) Ww (woolly hair, heterozygous) ww (normal hair) (a) Enter the parent phenotypes and complete the Punnett square for a cross between two heterozygous individuals. (b) Give the ratios for the phenotypes from this cross. Phenotype ratios:

P

sperm

p

Female parent phenotype:

Male parent phenotype:

W

sperm

w

Female parent phenotype:

Inheritance of haemophilia is sex linked. Males with the recessive (haemophilia) allele, are affected. Females can be carriers.

Using the codes: XX (normal female) XXh (carrier female) XhXh (haemophiliac female) XY (normal male) XhY (haemophiliac male) (a) Enter the parent phenotypes and complete the Punnett square for a cross between a normal male and a carrier female.

w

W

3. Inheritance of sex linked recessive traits Example: Haemophilia

eggs

(b) Give the ratios for the phenotypes from this cross.

Male parent phenotype:

X

eggs

Xh

X

sperm

Y

Phenotype ratios:

4. Inheritance of sex linked dominant traits Example: Sex linked form of rickets A rare form of rickets is inherited on the X chromosome.

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Phenotype ratios:

Male parent phenotype:

X

X

R

eggs

X

R

sperm

Y

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Using the codes: XX (normal female); XY (normal male) XRX (affected heterozygote female) XRXR (affected female) XRY (affected male) (a) Enter the parent phenotypes and complete the Punnett square for a cross between an affected male and heterozygous female. (b) Give the ratios for the phenotypes from this cross.

Female parent phenotype:

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218 Problems Involving Monohybrid Crosses

Key Idea: For monohybrid crosses involving autosomal unlinked genes, the offspring appear in predictable ratios.

Test your understanding of monohybrid crosses by solving these problems involving the inheritance of a single gene.

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1. A dominant gene (W) produces wire-haired texture in dogs; its recessive allele (w) produces smooth hair. A group of heterozygous wire-haired individuals are crossed and their progeny are then test-crossed. Determine the expected genotypic and phenotypic ratios among the test cross progeny:

2. In sheep, black wool is due to a recessive allele (b) and white wool to its dominant allele (B). A white ram is crossed to a white ewe. Both animals carry the black allele (b). They produce a white ram lamb, which is then back crossed to the female parent. Determine the probability of the back cross offspring being black:

3. A recessive allele, a, is responsible for albinism, an inability to produce or deposit melanin in tissues. Humans and a variety of other animals can exhibit this phenotype. In each of the following cases, determine the possible genotypes of the mother and father, and of their children:

(a) Both parents have normal phenotypes; some of their children are albino and others are unaffected:

(b) Both parents are albino and have only albino children:

(c) The woman is unaffected, the man is albino, and they have one albino child and three unaffected children:

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4. Two mothers give birth to sons at a busy hospital. The son of the first couple has haemophilia, a recessive, X-linked disease. Neither parent from couple #1 has the disease. The second couple has an unaffected son, despite the fact that the father has haemophilia. The two couples challenge the hospital in court, claiming their babies must have been swapped at birth. You must advise as to whether or not the sons could have been swapped. What would you say?

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219 Dihybrid Inheritance

Key Idea: A dihybrid cross studies the inheritance pattern of two genes. In crosses involving unlinked autosomal genes, the offspring occur in predictable ratios. There are four types of gamete produced in a cross involving two genes, where the genes are carried on separate

Homozygous white, long hair Parents: The notation P is only used for a cross between true breeding (homozygous) parents.

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Homozygous black, short hair

chromosomes and are sorted independently of each other during meiosis. The two genes in the example below are on separate chromosomes and control two unrelated characteristics, hair colour and coat length. Black (B) and short (L) are dominant to white and long.

BBLL

Parents (P)

BL

BL

Gametes

BL

bbll

X

BL

bl

bl

bl

bl

Gametes: Only one type of gamete is produced from each parent (although they will produce four gametes from each oocyte or spermatocyte). This is because each parent is homozygous for both traits.

Possible fertilisations

Offspring (F1)

BbLl

Female gametes

Offspring (F2) Possible fertilisations

X

BbLl

F1 offspring: There is only one kind of gamete from each parent, therefore only one kind of offspring produced in the first generation. The notation F1 is only used to denote the heterozygous offspring of a cross between two true breeding parents.

BL

Bl

bL

bl

Male gametes

BL

F2 offspring: The F1 were mated with each other (selfed). Each individual from the F1 is able to produce four different kinds of gamete. Using a grid called a Punnett square (left), it is possible to determine the expected genotype and phenotype ratios in the F2 offspring. The notation F2 is only used to denote the offspring produced by crossing F1 heterozygotes.

Bl

Each of the 16 animals shown here represents the possible zygotes formed by different combinations of gametes coming together at fertilisation.

bL

bl

Genotype BBLL BbLL BBLl BbLl

The offspring can be arranged in groups with similar phenotypes:

Phenotype

9 black, short hair

A total of 9 offspring with one of 4 different genotypes can produce black, short hair

BBll A total of 3 offspring with one of 2 different genotypes can produce black, long hair

3 white, short hair

bbLL bbLl

bbll

3 black, long hair

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Bbll

A total of 3 offspring with one of 2 different genotypes can produce white, short hair

Only 1 offspring of a given genotype can produce white, long hair

1 white, long hair

1. Complete the Punnett square above and use it to fill in the number of each genotype in the boxes (above left). LINK

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220 Inheritance of Linked Genes

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Key Idea: Linked genes are genes found on the same chromosome and tend to be inherited together. Linkage reduces the genetic variation in the offspring. Genes are said to be linked when they are on the same chromosome. Linked genes tend to be inherited together. The likelihood of crossing over between linked genes depends on Parent 1 (2N)

how close together they are on the chromosome (if the are close, crossing over is less likely). In genetic crosses, linkage is indicated when a greater proportion of the offspring are of the parental type (than would be expected if the alleles were on separate chromosomes and assorting independently). Linkage reduces the genetic variation in the offspring. Parent 2 (2N)

Overview of linkage

White body, white eyes aabb

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Dark body, blue eyes AaBb

Chromosomes before replication

A

a

B

b

Genes are linked when they are found on the same chromosomes. In this case A (body colour) and B (eye colour) are linked.

a

a

b

b

a

a

a

a

b

b

b

b

Chromosomes after replication

A

A

a

a

B

B

b

b

X

Each replicated chromosome produces two identical gametes. One of each is shown here.

Meiosis

A

a

B

b

Gametes

A

a

A

a

B

b

B

b

Offspring

AaBb AaBb Dark body, blue eyes

a

a

b

b

a

a

a

a

b

b

b

b

aabb aabb White body, white eyes

Possible offspring

Only two kinds of genotype combinations are possible. They are they same as the parent genotype.

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2. Explain how linkage decreases the amount of genetic variation in the offspring:

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1. What is the effect of linkage on the inheritance of genes?

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221 Recombination and Dihybrid Inheritance

Key Idea: Recombination is the exchange of alleles between homologous chromosomes as a result of crossing over. Recombination increases the genetic variation in the offspring. The alleles of parental linkage groups can separate in crossing over so that new associations of alleles are formed in the gametes. Offspring formed from these gametes are called recombinants and show combinations of characteristics not

seen in the parents. In contrast to linkage, recombination increases genetic variation in the offspring. Recombination between the alleles of parental linkage groups is indicated by the appearance of non-parental types in the offspring, although not in the numbers that would be expected had the alleles been on separate chromosomes (and therefore assorting independently).

Overview of recombination

Parent 2 (2N)

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Parent 1 (2N)

Dark body, blue eyes AaBb

White body, white eyes aabb

A

a

B

b

Chromosomes before replication

a

a

b

b

Chromosomes after replication

Crossing over has occurred between these chromosomes

A

A

a

a

a

a

a

b

b

b

Because this individual is homozygous, if a these genes cross over there is no change to the allele b combination.

X

B

b

B

b

Meiosis

A

A

b

B

a

a

Gametes

b

B

A

a

a

a

B

b

b

b

a

a

a

a

b

b

b

b

A

a

a

a

b

b

B

b

Offspring

Non-recombinant offspring These two offspring show allele combinations that are expected as a result of independent assortment during meiosis. Also called parental types.

1. Describe the effect of recombination on the inheritance of genes:

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Aabb

aaBb

Dark body, white eyes

White body, blue eyes

Recombinant offspring

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aabb

White body, white eyes

These two offspring show unexpected allele combinations. They can only arise if one of the parent's chromosomes has undergone crossing over.

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AaBb

Dark body, blue eyes

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222 Problems Involving Dihybrid Crosses

Key Idea: For dihybrid crosses involving autosomal unlinked genes, the offspring appear in predictable ratios.

Test your understanding of dihybrid inheritance by solving problems involving the inheritance of two genes.

1. In rabbits, spotted coat S is dominant to solid colour s, while for coat colour, black B is dominant to brown b. A brown spotted rabbit is mated with a solid black one and all the offspring are black spotted (the genes are not linked).

(a) State the genotypes:

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Parent 1: Parent 2: Offspring:

(b) Use the Punnett square to show the outcome of a cross between the F1 (the F2):

(c) Using ratios, state the phenotypes of the F2 generation:

2. The Himalayan colour-pointed, long-haired cat is a breed developed by crossing a pedigree (truebreeding), uniform-coloured, long-haired Persian with a pedigree colour-pointed (darker face, ears, paws, and tail) short-haired Siamese.

The genes controlling hair colouring and length are on separate chromosomes: uniform colour U, colour pointed u, short hair S, long hair s.

Persian

(a) Using the symbols above, indicate the genotype of each breed below its photograph (above, right).

(b) State the genotype of the F1 (Siamese X Persian):

(c) State the phenotype of the F1:

(e) State the ratio of the F2 that would be Himalayan:

(f) State whether the Himalayan would be true breeding:

(g) State the ratio of the F2 that would be colour-point, short-haired cats:

Siamese

Himalayan

(d) Use the Punnett square to show the outcome of a cross between the F1 (the F2):

3. A Drosophila male with genotype CucuEbeb (straight wing, grey body) is crossed with a female with the genotype cucuebeb (curled wing, ebony body). The phenotypes of the F1 were recorded and the percentage of each type calculated. The percentages were: Straight wings, grey body 45%, curled wings, ebony body 43%, straight wings, ebony body 6%, and curled wings grey body 6%. (a) Is there evidence of crossing over in the offspring?

(b) Explain your answer:

(c) Determine the genotypes of the offspring:

Straight wing Cucu Grey body, Ebeb

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Curled wing cucu Ebony body, ebeb

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223 Genetic Screening

Key Idea: Genetic screening is used to identify specific (often harmful) genes in embryos, children, and adults. It raises many ethical issues, especially for embryo testing. The genetic screening of gametes, embryos, children, and adults for some diseases is now possible. Genetic screening has many applications including in the detection

and treatment of diseases. Whilst genetic screening has many positive applications, it raises a number of ethical issues. This is particularly the case for the screening of embryos and fetuses because it may result in the destruction of embryos and fetuses if they have genetic defects, or even an undesirable genotype (e.g. the wrong sex).

Embryo screening

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Why carry out genetic screening?

Carrier testing: A person with a family history for a disease may want to be tested to see if they carry the gene for that disease. The result may influence whether or not they choose to have children.

Egg

Sperm

The sperm fertilises the egg outside of the body (in vitro). When the embryo reaches the eight cell stage, one cell is removed and its genome is screened.

The genetic material is tested for the gene of interest.

Diagnostic testing: A person may have symptoms typical of a particular genetic disorder. Genetic screening is used to determine if the person has the gene associated with a particular disease or not.

Embryos that test negative for the genetic disorder are implanted into the mother.

Embryos that test positive for the genetic disorder are not implanted into the mother.

Embryos produced by in vitro fertilisation (IVF) are tested for genetic abnormalities to ensure only healthy embryos are transplanted into the mother. This is called pre-implantation genetic diagnosis (PGD).

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Pharmacogenetics: Genetic screening can be used to help decide what type, or dose, of medicine will be best for an individual. Targeted treatment can increase the chances of the medicine working.

Newborn screening: Newborns are screened for a range of metabolic disorders (e.g. phenylketonuria). If a disease is detected, treatment can begin immediately and the child's prognosis is improved. Š 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

Embryos can also be screened in vivo (inside the mother) for genetic disorders. For example, Down syndrome is often first detected by ultrasound, and then confirmed by genetic testing.

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USAF

Sometimes PGD is used to select embryos with a genetic disorder. This most often occurs when parents have a particular disorder (e.g. genetic deafness) and they want their child to have it too.

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Arguments for genetic screening

 Testing allows potential carriers to be screened for a disease so they can decide whether they have children or not. This is important for diseases that do not show any symptoms until later in life (e.g. Huntington's disease).

 Genetic tests can only tell you if you carry a gene for an associated disorder. They cannot predict when and if you will develop the disease, or to what extent. Testing therefore carries the risk of causing unnecessary anxiety.

 An individual's privacy may be compromised by testing. The knowledge that you may develop a genetic disorder in the future could be used against you (e.g. medical insurance could be declined or an employer may no longer want to employ you).

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 Researchers can study individuals with the gene(s) associated with a disease and this may help them to develop a treatment or cure for that disease.

Arguments against genetic screening

 Knowing a person's genetic make-up can be used to optimise drug therapies and improve treatment outcomes.

 Knowing the risk of developing a disease allows informed decisions to be made about medical options. For example, breast cancer can be treated, so an individual may decide to increase screening to increase the chance of early detection. They may choose to reduce risk factors (e.g. breast removal if they are at high risk of developing breast cancer).

 The discovery of a genetic defect in an unborn child provides an opportunity to come to terms with the situation and prepare for the delivery and ongoing care of a special needs child.

 Designer babies could be produced where parents pick certain characteristics they want their child to have. This is already seen in countries where more value is placed on the birth of a boy child than a girl, and unwanted female fetuses are terminated.

 The discovery of a genetic defect in an unborn child may lead to the decision to terminate the pregnancy, an action some people believe is morally wrong because they feel it devalues human life.

1. Describe some of the benefits of genetic screening:

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2. PGD is controversially used produce saviour siblings. Saviour siblings are created using IVF to provide an organ or stem cells to a child who has a certain disease. During the process, the embryos are tested to make sure that they are healthy and that they are compatible with the older sibling. Any undesirable embryos are destroyed. Using saviour siblings as a an example, discuss the ethics of genetic screening:

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224 KEY TERMS: Did You Get It?

1. Study the two pedigree charts below.

Pedigree chart A

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Pedigree chart B

(a) What type of inheritance pattern is shown in A?

(b) Give a reason for your answer:

(c) What type of inheritance pattern is shown in B?

(d) Give a reason for your answer:

2. The following dihybrid cross shows the inheritance of colour and shape in pea seeds. Yellow (Y) is dominant over green (y) and a round shape (R) is dominant over the wrinkled (r) form.

(a) Describe the appearance (phenotype) of pea seeds with the genotype YyRr:

(b) Complete the Punnett square below when two seeds with the YyRr genotype are crossed. Indicate the number of each phenotype in the boxes on the right.

Female gametes

Yellow-round Green-round

Yellow-wrinkled

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Male gametes

Green-wrinkled

TEST


225 Review Unit 2: Area of Study 2

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Summarise what you know about this topic under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts. Use the images and hints to help you and refer back to the introduction to check the points covered: Genomes, genes and alleles

HINT: Describe the structure and functions of homologous chromosomes. How can karyograms be used to detect abnormal chromosome numbers?

REVISE

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HINT: How has the information from the Human Genome Project been used to understand gene structure and function?

Chromosomes

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315 Genotypes and phenotypes

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HINT: How can genotype and environmental factors alter phenotype?

Pedigrees and genetic outcomes

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HINT: How can pedigree charts and genetic testing be used to determine the likelihood of inheriting a genetic disease? What ethical considerations are involved?


226 Synoptic Question: Unit 2 Area of Study 2

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316

1. Discuss the importance of genome sequencing with respect to:

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• Comparing relatedness between species • Determining gene function • Detecting and diagnosing human disease

2. The preparation of a karyogram involves arranging the chromosomes of an individual into homologous pairs in order.

(b) Study the karyogram on the right. Circle the sex chromosomes:

(c) State the sex of this individual:

(d) Determine if the karyotype shown is normal/abnormal:

(e) Explain the reason for the answer you have given in (d):

(b) What type of disorder is Down syndrome?

(c) Explain the cause of Down syndrome:

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3. (a) What would you expect to see in the karyogram of an individual with Down syndrome?

Cytogenetics Dept, Waikato Hospital

(a) Name some applications of this process:

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4. Define phenotype and genotype:

5. Using examples, discuss how phenotype can be affected by: (a) Genotype:

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(b) Environment:

(c) Epigenetic factors:

6. Explain how the number of genes associated with a phenotypic trait affects its appearance:

7. In guinea pigs, black hair (B) is dominant over white hair (b) and short hair (L) is dominant over long hair (l). In order to produce consistent and reliable coat colour in guinea pigs, breeders use true breeding parents. A breeder has a male guinea she knows to be BBLL and a female she knows to be bbll. She wants to use them to begin a breeding programme to produce a pure breeding white, short haired guinea pig.

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Explain how the desired pure breeding animal can be produced and how its ability to breed true can be proved. Support your answer with Punnett squares and the correct use of the allele letters given (you may use extra paper if required):

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Investigation of an issue

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Unit 2 Outcome 3

Choosing an issue for investigation

Key terms

Activity number

anecdote

For this assessment task, you will be required to investigate a question of interest related to an issue in genetics or reproductive science. This could be related to human cloning, genetic modification of organisms, the use of forensic DNA databanks, assisted reproductive technologies, or genetic testing.

bias

data

evidence opinion

223 230

You will be required to communicate your findings in an appropriate format, such as a formal report, oral communication, or digital presentation.

reliability

The purpose of this chapter is to provide you with some scenarios in which you can test your own evaluation of presented information. This will help you critically evaluate the information you come across when conducting your own research.

validity

Effective science communication

Activity number

Key knowledge

c

1

Understand the characteristics of effective science communication. The biological information you present must be accurate, the biological concepts and models must be clearly explained, it must be clear why the findings are important, and the information presented must be appropriate for purpose and audience.

231

c

2

When explaining biological concepts specific to the investigation be sure to define key terms and use appropriate biological terminology.

231

c

3

Make appropriate use of data, models, and theories when organising and explaining biological phenomena and concepts. Understand and outline the limitations of the data and models used.

229

The nature of evidence

Activity number

Key knowledge

4

Describe and understand the nature of evidence, including the distinction between opinion, anecdote, and evidence, weak and strong evidence, and scientific and non-scientific ideas.

228 229

c

5

In your research evaluate the validity, reliability, and authority of data, including acknowledging any possible errors or bias.

228 229

Influencing factors Key knowledge 6

Activity number

In your report, include an evaluation of the social, economic, ethical, and legal factors relevant to your issue.

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c

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227 Recognising Balanced Reporting

Key Idea: Balanced reporting provides unbiased information where both the positive and negative aspects are presented without a particular emphasis on either. The reporting of scientific information should always be

unbiased and a statement of fact backed by evidence. Unfortunately, a lot of information reported today is highly biased, lacks scientific rigor, or is interpreted incorrectly, resulting in the public being misled about many issues.

►► Biological information is presented to the public constantly via print and broadcast media. Some is provided by government organisations, some is compiled and presented by science reporters, and some is provided by individuals or nongovernmental organisations with an interest, but not necessarily expertise, in a topic.

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►► Unlike peer reviewed publications (e.g. journal articles), these sources are not reviewed by experts before being made available to the public. The information presented may be inaccurate (containing scientific errors) or biased (presenting only one view). ►► It is important to use your own biological knowledge to critically review and analyse media for biological validity. The decisions made by individuals in a democratic society about biological issues can be heavily dependent on the quality of the information provided. Inaccurate or biased information can lead to poor decision making, whereas accurate information promotes informed debate.

Points to consider when analysing biological information

The following points will help you to critically analyse articles about biological issues and determine whether or not they present a fair, unbiased view, and contain biologically valid information:

►► Is there more than one side or view to this issue? ►► Are all the views presented?

►► Have any compromises been made to reach an outcome?

►► What information is presented to the public and is it scientifically correct? ►► Is some information more important than other

information? If so, how is importance assigned?

►►

What are the consequences to the public if: • the information presented is poor science? • the information presented is good science? • the information presented is anecdotal (unreliable and based on hearsay?

Student analysis

In order to recognise and validate biological information presented to you, you need to:

►► Identify and explain the purpose of the biological information that is presented to the public:

• Does the person/group presenting the information have a particular agenda (issue they want to push) or bias?

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►► Recognise whether the information is balanced or biased and if it is accurate or inaccurate. You can determine this by using your own biological knowledge. ►► Discuss the significance of the biological information (e.g. can GM plants fertilise other plant and pass the modifications on?). • What is its overall impact on the public? • Will biased, inaccurate, or accurate articles influence the overall public perception?

►► Understand that biological validity means the material presented is based on sound biological principles, and the results are logically derived.

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►► Be able to reference information correctly (by giving the title, publisher or journal, date of publication, and authors).

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228 The Misuse of Scientific Data

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Key Idea: Presenting biased data can influence (negatively or positively) the public's perception about a specific topic. The way information is presented can be very influential. Reporters with a particular agenda may use particularly emotive words in their reports in order to sway opinion. You

should remember that all the information you gather should be checked at the source of the information if at all possible (i.e use first hand information and data, not second, third, or fourth hand). This allows you to see and interpret the information to gain your own understanding and opinion.

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►► Scientific investigations and experiments are carried out to enhance our understanding of how different systems (biological, physical, geological, or chemical) work. Many of these investigations involve equipment, data, and results that are highly technical and can be easily misunderstood by people outside the field of study. ►► Sometimes investigations are made that deliberately intend to collect data to support a particular point of view, i.e. they are biased. This can be done by asking questions or manipulating an investigation in such a way that only biased or skewed results will be obtained. For example, consider the following questions:

Do you support the use of dihydrogen monoxide as a solvent for sucrose in carbonated beverages?

Do you support water being used to dissolve sugar in soft drink?

Both these questions ask exactly the same thing, but could get different results if asked in a survey.

►► Deliberate misuse of scientific data often involves people or groups using selected parts of a report and matching them to an incomplete knowledge of a particular scientific concept.

Case study: Genetic modification of plants

Some information concerning the genetic modification of food crops is presented below. Beneath are two statements, both using the same information but written in different styles. The left is emotive and written as an anti-GM statement. The right is a more balanced, but pro-GM statement. All DNA uses the same four bases A, T, C, G. This means DNA from one organism can be spliced into the DNA of another. Therefore genes from a bacterium can be spliced into a plant, because the cellular machinery than reads the DNA is the same for both organisms. • The splicing of genes from one organism to another occurs regularly in nature. Bacteria pick up DNA from the environment. The bacterium Agrobacterium tumefaciens transfers DNA to plants forming crown galls. • Crop plants can be modified to carry genes that will protect them from insect pests or increase yields.

• There are concerns over the effects of eating crop plants modified to carry toxins that protect them from insect pests. There are also concerns over increased resistance in insect pests. • Between 1996 and 2013 the land area devoted to growing GM crops increased from 17,000 km2 to 1,750,000 km2. 96% of all soybeans planted in the USA are genetically modified.

Pro-GMO crops

Genetic modification allows scientists to create organisms in a way that does not occur in nature. These scientists swap genes around in plants and animals, playing God, and creating untested products for human consumption. These modified plants can spread easily through the environment and contaminate non GM areas so that soon people will be forced to eat GM crops, even without knowing it. Outrageously, the crop seed for these GM plants is owned by giant, faceless international conglomerates, which have forced farmers to grow their GM crops, increasing the amount of GM covered land 100 fold in the last twenty years. There is almost no chance for small farmers to grow non-GM crops, especially for those growing soybeans because almost 100% of soybeans are genetically modified. This means there is no choice for farmers or consumers. Even worse, these GM crops carry poisons that are eaten by humans, poisoning them with toxic DNA.

Genetic modification allows scientists to improve the growth and production of crop plants. Because DNA is the same in all organisms, genes can be transferred between organisms. This occurs naturally in bacteria. Scientists use the bacteria's ability to pick up and transfer genes to insert genes to target organisms, e.g. crop plants. In this way, a gene that may produce a useful product in one organism that is not directly useful to humans can be induced to produce the same product in another organism that is directly useful to humans. For example, the bacterium Bacillus thuringiensis has a gene that produces a natural insecticide. This gene can be transferred to crop plants so that they produce the insecticide. This saves farmers millions of dollars in lost crops and spraying costs. It improves crop yield and means less spray is put into the environment. The insecticide is not toxic to humans. This system is so successful that in twenty years farmers have increased the amount of land they devote to growing GM crops from 17,000 km2 to over 17 million km2.

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Anti-GMO crops

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229 Evaluating the Information involved, and it can sometimes be difficult to determine which information is based on scientific data and which is not. The information in this activity outlines some of the main arguments about genetic screening and provides links to other resources. While evaluating the information you must use your biological knowledge to evaluate the quality of the information. Some points to consider are presented below.

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Key Idea: The evaluation of biological information requires critical analysis of its validity and bias. Your investigation of an issue requires you to investigate biological concepts, identify opinions, ethical implications and justify your conclusions. In order to do this, you must review a range of information related to your chosen topic. Socio-scientific issues can be very difficult to research because there are two or more, often very different, opinions

Points to consider when evaluating biological information

►► In order to form an opinion about a socio-scientific issue, you must use your biological knowledge to critically evaluate information. Some points to consider include: ►► Validity of the information.

• The currency of the information. Is it up to date?

• Is the information peer reviewed?

• Has the information been accepted by the scientific community?

►► Does the information present an unbiased view?

• Is information presented in a fair, unbiased way?

• Is the information presented clouded by the attitudes, beliefs, or values of the person or group providing the information?

• The information presented must be based on fact and not emotions.

Keep a log book or portfolio of the information you have reviewed. This will be used by your teacher to verify you have sufficiently researched the topic, and that the work you submit is your own. It is important to remember that not all biological information presented to the public is peer reviewed (has been reviewed by experts). Sometimes the information presented may be inaccurate (containing scientific errors) or biased (only one view is presented). It is important to use your own biological knowledge to critically review and analyse information for biological validity.

Finding information

There are many resources that can be used to obtain information about your issue, e.g. journal articles, blogs, news articles, and videos. Most are available on the internet if your use the appropriate wording in your chosen search engine. For example, typing "genetic screening" into your search engine will provide some general information on genetic screening. Typing "genetic screening in Australia" will provide information about genetic screening in Australia. Be aware of the site from which you obtain information. Is it reputable or just someone's own website with their own unverifiable ideas? Be cautious with video clips (e.g. YouTube). Again, these often present a personal view with little backing. Check the comments to see what others have said. Often the comments may point out errors (if any) in the video. Journals are peer-reviewed. That is, the information is checked by experts in the topic area of the reported article. This makes the information highly reliable. However, journals are often highly technical and can be difficult to understand, especially for people outside the area of expertise.

Periodicals or technical magazines, e.g. National Geographic or Popular Mechanics, are useful sources of reliable information. As they are written for the general public they make understanding the technical information much easier. © 2020 BIOZONE International ISBN: 978-1-98-856646-7 Photocopying Prohibited

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Online sites that are specific for a topic need to be carefully scrutinised for validity. Stay away from conspiracy sites as these often dramatise stories and mix them with incorrect science. Government sites often have the most current and reliable data based on information from skilled advisers.

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Newspaper articles are a good starting point as a source of generally reliable information, but beware of the newspaper's particular leaning. Tabloids often sensationalise stories, while some newspapers may have left or right political leanings, which can skew the focus of a story.

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230 Case Study: Genetic Screening in Australia

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Key Idea: Genetic screening can be useful, but it can also lead to a life in limbo. As genetic screening becomes more commonplace, news relating to it is often reported in mainstream media such as television, internet, and print. The top article presented below gives information about uses of genetic screening. The newspaper article (bottom) is fictitious, but is based on

information that has appeared recently in the mainstream media. It will give you an idea of the type of information you are likely to find during your research. Read the newspaper article and decide whether you think the view it presents is balanced or biased. Use your biological knowledge to determine if the information is based on scientific evidence, or statements that are unsupported by fact.

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Genetic screening in Australia

In Australia genetic screening tests are carried out by a number of genetic screening facilities and medical centres.

When a chromosomal abnormality (such as Down syndrome) is suspected from the results of a prenatal ultrasound, the parent(s) may elect to have a chromosome test (karyotype) to confirm the diagnosis. In the case of Down syndrome, there would be an extra chromosome 21 present (far right).

The karyotype (right) shows this individual has Down syndrome (Trisomy 21).

Some cancers (such as breast cancer) have a strong familial (inherited) pattern of occurrence. Surveys and genetic tests may be conducted in a patient's family to see if there is such a familial pattern. If women test positive for a breast cancer gene, they may undergo more frequent screening to increase the chance of early cancer detection. In some cases, they may opt to have their breasts surgically removed to reduce their chance of developing cancer.

Living with uncertainty The Newspaper

Tumour

This mammogram (left) shows the presence of a tumour in the breast. Women with increased genetic risk of getting breast cancer may be screened more often than the general population.

The newspaper, author name, and names used in this article are fictitious, but the text is based on real events and information.

By Sarah McKinney: 09 February 2013

In Australia, newborn babies are routinely screened for a number of genetic disorders, including cystic fibrosis (CF), a disease that affects one in every 2800 Australian babies. In 2009, newborn Amy Jones' blood sample tested positive for genetic changes associated with CF. She was intensely monitored, and attended regular checkups at the hospital to assess her condition. For over two years no symptoms developed, so further tests were ordered which came back negative for the CF gene. However it hasn't stopped her parents worrying that CF could still develop. Amy's mother did not return to paid work after Amy's diagnosis. She and her husband worried that if Amy was

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Amy's tests were carried out by an approved laboratory using accredited tests, but many direct-to-consumer tests being sold over the internet are not clinically validated. These tests may not accurately predict disease, and the result may mean many people are making decisions based on the occurrence of a disease that may never eventuate.

Ryan Anderson, a bioethics expert in the field of genetic screening, urges consumers to consider the confidentiality and privacy issues associated with genetic testing. Mr Anderson has concerns that health and life insurance providers will soon be able to ask their clients to routinely undergo genetic testing, and use the information to grant or deny cover and set the cost of that cover. This approach will disadvantage many people who may have a gene for a specific disease, but who will never develop that disease.

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People who are told they have genes for potentially fatal diseases, but don't show any symptoms, are called patientsin-waiting. This situation is causing many unforeseen consequences.

placed in daycare she would pick up normal childhood infections, which could develop into life-threatening respiratory infections if a person had CF. Amy is now four years old and still has no symptoms, but her parents feel as if they are sitting on a time bomb, constantly wondering if she will develop CF.

Until researchers find a way to accurately predict the occurrence and onset of genetic diseases, many people like Amy are waiting in limbo looking for signs of a disease that may not emerge.

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Genetic testing, analysis of a person's DNA to detect the presence of a gene associated with a particular disease, was once limited to medical professionals. Today, access to gene testing is widespread and relatively inexpensive. Many companies sell gene testing kits directly to the public. A customer sends a cheek or saliva swab to the laboratory for analysis, and they receive a report about their genetic risk of a certain disease. However, as more people find that they carry potentially dangerous genes, many are left in limbo, waiting for a disease that may not occur.

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231 Communicating Your Findings

Key Idea: A report's findings must be communicated in a clear and ordered way. Communicating your findings is as important as the research

Things to think about: • Who is your report targeting?

• Does the report contain

information that is not needed or not obviously related to the main idea? Be concise but precise.

• What is the report's main focus (what idea or concept are your trying to explain)?.

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Once you have gathered your information and data, use the space on this page to plan your report. Use bullet points to capture the main points and your ideas. These will provide the scaffold to construct and write your full report. The list (right) provides key points to think about when planning your report.

you have carried out. Poorly presented information cannot be easily understood. Each idea you present must be clearly identified and related to the other information in your report.

• For written reports, check your

• Does your report contain

conflicting information (have you said something in one part and said the opposite elsewhere)?

• How are you going to format the

spelling and grammar, are key words spelt correctly. Do your sentences make sense? Is your argument presented logically?

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report? What tables or graphs are you going to include and what do they show?

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Appendix

Questioning terms in biology

The writing team would like to thank the following people and organisations for their contributions to this edition: • Adam Luckenbach and the North Carolina State University for use of the poster presentation image • UC Berkley • Dartmouth College • Pennsylvania State University College of Medicine • Ian Smith for the photo of the green and golden bell frog Kristian Peters • Luis A de la Parra, for his photograph of the palm cockatoo • The late Stephen Moore for his photos of aquatic invertebrates PASCO for their photographs of probeware • University of Florida for the image of the strawberry runners • Dr Roger Wagner, Dept of Biological Sciences, Uni of Delaware, for the LS of a capillary Martin Pot for the numbat photo • Dr Conrad Pilditch, Uni. of Waikato, for images of rocky shore invertebrates. • openstax college • G. Beard • Aiofthestorm • National Cancer Institute • Ed Uthman • Cytogenetics Dept. Waikato Hospital for the karyograms • Dr Graham Beards • Lexicon Genetics/HGRI • Maggie Bartlett, NHGRI • alpsdake • Georgetown University Hospital, Washington, D.C. for the photo of the bone marrow transplant • Greenpeace for the GMO images

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The following terms are often used when asking questions in examinations and assessments.

Image Credits

Annotate:

Add brief notes to a diagram, drawing or graph.

Apply:

Use an idea, equation, principle, theory, or law in a new situation.

Calculate: Find an answer using mathematical methods. Show the working unless instructed not to.

Compare: Give an account of similarities between two or more items, referring to both (or all) of them throughout. Construct: Represent or develop in graphical form. Contrast:

Show differences. Set in opposition.

Define:

Give the precise meaning of a word or phrase as concisely as possible.

Derive:

Manipulate a mathematical equation to give a new equation or result.

Describe: Define, name, draw annotated diagrams, give characteristics of, or an account of. Design:

Produce a plan, object, simulation or model.

Determine: Find the only possible answer. Discuss:

Show understanding by linking ideas. Where necessary, justify, relate, evaluate, compare and contrast, or analyse.

Distinguish: Give the difference(s) between two or more items. Draw:

Represent by means of pencil lines. Add labels unless told not to do so.

Estimate:

Find an approximate value for an unknown quantity, based on the information provided and application of scientific knowledge.

Evaluate:

Assess the implications and limitations.

Explain:

Provide a reason as to how or why something occurs.

Identify:

Find an answer from a number of possibilities.

Illustrate:

Give concrete examples. Explain clearly by using comparisons or examples.

Interpret:

Comment upon, give examples, describe relationships. Describe, then evaluate.

List:

Give a sequence of answers with no elaboration.

Measure:

Find a value for a quantity.

Outline:

Give a brief account or summary. Include essential information only.

Predict:

Give an expected result.

Solve:

Obtain an answer using numerical methods.

State:

Give a specific name, value, or other answer. No supporting argument or calculation is necessary.

Suggest:

Propose a hypothesis or other possible explanation.

Summarise: Give a brief, condensed account. Include conclusions and avoid unnecessary details.

We also acknowledge the photographers who have made images available through Wikimedia Commons under Creative Commons Licences 2.0, 2.5, 3.0, or 4.0: Zephyris • JPbarrass • J Brett Taylor • Josh Grosse • NJRZA • Barfooz and John Grosse • Kjetil Lenes • Olaboy • Micropix • McKDandy • James Heilman • Emmanuiem • LWalsh84 • Nephron • Steven Fruitsmaak • Hellerhoff • Suseno • Robert Kerton (CSIRO) • Genet • Chiswick Chap • Quartl • D. Gordon E. Robertson • Lip Kee Yap • Ben133uk • Ralf Pfeifer • Natural Philo • David Brazier • FontanaCG • Ragesoss • RM Hunt • Ildar Sagdejev • Meyer A • Thomas Breuer • Jeffmock • Fdardel • Yathin S. Krishnappa • Tony Wills • Ron Pastorino • The High Fin Sperm Whale • Tom Bruns • Sbj1976 • Kristian Peters • Neil Palmer • Winfried Bruenken • Sagt • JJ Harrison • Marc Tarlock • Albert kok • Marjorie Lundgren • Andrew Mercer • Marshal Hedin • Menna Jones • Brocken Inaglory • Bruno de Giusti • Jpbarrass • dsworth New York State Department of Health • Melburnian • Bob Embleton • Curtis Clark • Joydeep • Martin Wu • Kalyan Varma • Earth100 • Matthias Zepper • Emmanuelm • Lip Kee • Bernard Dupont • Greg Miles • JJ Harrison Figaro (Public Domain) • Bfinge • Jim Bendon • Ron Knight • Gnangarra Contributors identified by coded credits are: BF: Brian Finerran (Uni. of Canterbury), BH: Brendan Hicks (Uni. of Waikato), BOB: Barry O’Brien (Uni. of Waikato), CDC: Centers for Disease Control and Prevention, Atlanta, USA, DH: Don Horne, DoC: Dept. of Conservation (NZ), DoC-RM: Rod Morris, (DOC), EII: Education Interactive Imaging, GW: Graham Walker, IS: Ian Smith, KP: Kent Pryor LC: Landcare (NZ), MSU: Michigan State University NIAID: National Institute of Allergy and Infectious Diseases, NIH National Institute of Health, NOAA: National Oceanic & Atmospheric Administration, RA: Richard Allan, RCN: Ralph Cocklin, TG: Tracey Greenwood, WBS: Warwick Silvester (Uni. of Waikato), WMU: Waikato Microscope Unit. USAF United States Air Force

We acknowledge our use of royalty-free images, purchased by BIOZONE International Ltd from the following sources: iStock images, Dollar Photo Club, Corel Corporation from various titles in their Professional Photos CD-ROM collection; IMSI (International Microcomputer Software Inc.) images from IMSI’s MasterClips® and MasterPhotosTM Collection, 1895 Francisco Blvd. East, San Rafael, CA 94901-5506, USA; ©1996 Digital Stock, Medicine and Health Care collection; ©Hemera Technologies Inc, 1997-2001; © 2005 JupiterImages Corporation www.clipart.com; ©1994., ©Digital Vision; Gazelle Technologies Inc.; ©1994-1996 Education Interactive Imaging (UK), PhotoDisc®, Inc. USA, www.photodisc.com. We also acknowledge the following clipart providers: TechPool Studios, for their clipart collection of human anatomy: Copyright ©1994, TechPool Studios Corp. USA (some of these images have been modified); Totem Graphics, for clipart; Corel Corporation, for vector art from the Corel MEGAGALLERY collection.

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Interpret data to reach stated conclusions.

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Analyse:

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Index Bowman's capsule 126, 130 Breathing rate 105 Breathing, forced 104 Breathing, quiet 104 Bronchi 102 Bronchiole 101-103 Butterfly adaptations 150

Cohesion-tension 94 Collecting duct 129 Colon 114 Column graphs, guidelines 18-19 Commensalism 186 Competition 209 - and species distribution 211 - barnacles 210 - gliders 210 - niche 210 - tadpoles 215 - hierarchies 215 - interspecific 209 - intraspecific 209 - scramble 215 Concentration gradient 74 Conservation 179 - and biodiversity 179, 180 - ex-situ 179 - in-situ 179 - of species 172 Constructing graphs 17 Continuous data 4 Continuous variation 295 Control, experimental 30 Controlled variable 6, 30 Convoluted tubule 130 Correlation 23 Cotransport 75 Crossing over 242 Cryptic species 172 Cuttings 233 Cystic fibrosis 272 Cytokinesis 225-227 Cytoplasm 56, 58

E Ecosystem services 178 Electron micrograph 54, 57, 59, 62 Electron microsope 54 Electron transport chain 85 Electrophoresis gel 266-268 Embryo 254 Embryo screening 311 Embryonic stem cell 249, 270 Emphysema 106 Endoplasmic reticulum 56, 58 Energy in cells 79 Epigenetics 292-293 Error bars 17, 21 Ethanol 85 Ethics, genetic screening 312 Eukarya 170 Eukaryote 45, 56 Eukaryotic chromosome structure 276 Excretory system 116 - and circulatory system 136 - malfunctions of 131 Exercise, effect on heart rate 103 Expiration, of the lungs 102 Experimental control 6, 30 Exponential functions 11

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A Active transport 74-75 Adaptation 144 - for diving 152 - behavioural 144, 150-152 - in animals 150 - insectivorous plants 149 - butterflies 150 - hydrophytes 146 - kangaroos 144 - mangroves 147 - monitor lizards 151 - physiological 144, 150-152 - plant adaptations to fire 148 - structural 144, 150-152 - to similar environments 154 - xerophytes 145 Adult stem cell 249 Alcoholic fermentation 85 Alleles 264, 286-287, 302 Alveoli 101-103 - structure 103 Alzheimer's disease 272 Amensalism 186 Anaerobic fermentation 85 Analysing biological information 319 Aneuploidy 280 Aneurysm 124 Animal adaptations 144, 150-152, 154 Animal cell 42, 56 Antagonistic muscles 104 Apoplast 92 Archaea 170 Arteries 118 Arterioles 118 Asbestosis 104 Asexual reproduction 231, 232 - animals 231 - bacteria 236 - cuttings 233 - fungi 236 - plants 232, 236-237 - spores 236 - unicellular eukaryotes 231 Assumptions 3 Asthma 104, 273 Atherosclerosis 122 ATP 79, 85 Autosome 278 Autotrophic 78

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G Gametes 240, 280 Gas exchange system 101-103 Gastroesophageal reflux disease 115 Gel electrophoresis 266 Gene 264, 276 - sex linked 304-305 Gene expression 292 Gene function 270-271 Gene knockout 270 Gene linkage 308-309 Gene mapping 267-271 Gene probe 271 Genetic diversity, disease 208 Genetic marker 271 Genetic predisposition 257 Genetic screening 272, 311-312, 322 Genetically modified organism 322 Genome 264-265 Genomic analysis 273 Genotype 287-288, 301- 302 Genotype frequency 303 GERD 115 Germ layers 254 Glomerulus 126, 130 Glucagon 165 Glucose 84-85, 112-113, 165 Glycolysis 85 Goblet cell 101, 108 Golgi apparatus 56, 58 Grafting 234, 237 Granny Smith apple 237 Graphs, types 18 Graves disease 164 Grey wolf classification 173 Guard cell 89 Guinea pig coat colour 307

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D Data 4 - manipulation 12 - plotting 17-24 Deep vein thrombosis 124 Density dependent factors 205 Density independent factors 205 Density, of populations 196 Dependent variable 6, 30 Descriptive statistics 26 Diabetes mellitus 166 - and stem cells 250-251 Diaphragm 101-102 Diarrhoea 114-115 Diffusion 68-69 Digestion 110-113 Digestive system 108-115, 158 - and circulatory system 134 - malfunctions 115 Dihybrid cross 307 Dingo 191, 194, 216 - habitat 216 - population size 216 Discontinuous data 4 Disease - Alzheimer's 272 - cancer 256-257, 272 - cystic fibrosis 272 - genetic diversity 208 - of the circulatory system 122 - of the digestive system 115 - of the excretory system 131 - of the respiratory system 106 - of the heart 123 - populations 208 Distribution, of populations 196 Diving, adaptations 152 DNA 276 - disease risk 273 - haplotype 273 - methylation 292-293 - packaging 277 - probe 272 - screening 273 - sequencing 266, 268 - Sanger method 266 DNA and chromosomes 276 DNA, distinguishing species 268 Dominant alleles 286-287, 302 Down syndrome 280

F Facilitated diffusion 68 Fair test 32, 34 Fermentation 85 Fetus 255 Fever, positive feedback 160 Field study 198-199 - of a rocky shore 212 Filtrate 130 Fire, plant adaptations 148 Five kingdom classification 170 Fluid mosaic model 65 Food chain 187 Food web 188 - cave 193 - dingo 191 - lake 189 Forced breathing 104 Fractions 10 Frequency distribution, perch 29 Fruit ripening, positive feedback 160 Fungal cell 42 Fungi, asexual reproduction 236

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B Back cross 301 Bacteria 170 Bacterial cell 45 Bacterial cell wall 45 Banana 235, 237 Bar graphs, guidelines 18, 19 Bias 29 - in reporting 319-321 Binary fission 224 Binomial nomenclature 176 Biodiversity 177, 181 - benefits 177-178 - conservation 179 - economics 178 - resilience 177 Bioinformatics 269 Biological drawing 14-16, 127 Biological information 319, 321, 323 Biomimicry 155, 156 Biopiracy 182 Bioprospecting 181 Blood 115 - components 117 - granulocyte 252 - lymphocyte 252 - platelets 117 - plasma 116-117 - red blood cells 117, 252 - white blood cells 117 Blood cell differentiation 252 Blood clotting, positive feedback 160 Blood glucose regulation 165 Blood groups 289 Blood plasma 116-117 Blood vessels 118

C Cacti adaptations 145 Calcium homeostasis 159 Calvin cycle 81 Camouflage, butterflies 150 Cancer 256-257, 272 - of the lung 106 Cancer Genome Atlas Project 267 Capillaries 119 Carbohydrate 43, 80 Carcinogen 256 Carrier proteins, membranes 65 Carrier-mediated diffusion 68 Carrying capacity 205, 215 Cartilage, hyaline 101 Casparian strip 92 Cell - animal 42, 56 - bacterial 42, 45 - eukaryotic 45 - plant 42, 58 - protist 42 - size 46, 47 Cell cycle 225-226, 255 - cancer 257 - checkpoints 225, 255 - disruption 257 Cell division 223, 225-227, 240 - binary fission 224 - growth and repair 223 - asexual 223 - cytokinesis 225-227 - interphase 225 - meiosis 240-241 - mitosis 225-226, 240 - self renewal 248 Cell sizes, compared 46 Cell specialisation - goblet cell 101, 108 - in animals 100 - guard cell 89 - intestinal epithelial cell 108 - muscle 100 - nerve cell 100 - parietal cells 108 - red blood cell 100, 116 - root hair 91 - tracheid 91 - vessel element 91 - white blood cell 117 Cell theory 42 Cell types 42 Cell wall, bacterial 45 Cell wall, plant 58 Cellular environments 48 Cellular respiration 84, 85 Centrioles 56 Channel proteins, membranes 65 Channel-mediated diffusion 68 Chemoautotroph 78 Childbirth, positive feedback 160 Chlorophyll 80 Chloroplast 58, 80 Cholera 114 Chromosome number 277 Chromosomes 267, 276, 278, 286 - homologous 286 - non disjunction 280 - sex 277, 279 Chronic bronchitis 106 Chronic obstructive pulmonary disease 106 Circulatory system 116, 158 - interactions, digestive system 134 - interactions, excretory system 136 - interactions, respiratory system 132 - malfunction of 124 Classification 170-171, 173 Clone 231, 235, 237 Cloning, and horticulture 237


326 Meiosis 240-241 - crossing over 242 - independent assortment 242 - modelling 242 - non-disjunction 280 - recombination 242 Membrane proteins 65 Membrane permeability 66, 67 Membrane structure 65 Membranes, role of 65 Mendel, Gregor 287 Micropropagation 235 Microscopy techniques 51, 52 Microscopy - electron 54 - mounting samples 52 - optical 50 Middle lamella 58 Millipedes, sampling 198, 200 Mimicry, butterflies 150 Mitochondrion 56, 58, 84 Mitosis 225-226, 228, 240 Mitotic index 228 Mode 26 Monitor lizards, adaptations of 151 Monohybrid cross 301 Mounting samples, microscopy 52 Multicellularity 48 Muscle cell, function 100 Muscles, antagonistic 104 Mutagen 257 Mutation 241, 257, 271, 289 Mutualism 186 Myocardial infarction 124

- vascular tissue 90, 91 - and cuttings 233 - asexual reproduction 232, 236-237 - banana 237 - grafting 234 - Granny smith apple 237 - micropropagation 235 - navel orange 237 - vegetative propagation 232 Plasma membrane 56, 58, 65 Plasmolysis 73 Platelets 117 Polygeny 295 Polyphenism 288 Polysomy 280 Population cycles 207 Population distribution 196 Population growth 206 Population size, dingo 216 Populations 205 - density dependent factors 205 - density independent factors 205 - disease 208 - limiting factors 206 Positive feedback 160 Potometer 95 Power function 11 Predation 186, 207 Predator-prey cycle 207 Predictions 3 Prenatal development 254 Prokaryote 45 Prokaryotes, binary fission 224 Propagation 232 -234 Proteins 43 Protist cell 42 Punnett square 307

- and competition 211 Species interactions 186, 209, 211 Species relationships 269 Species, chromosome number 277 Species, cryptic 172 Stains, in microscopy 52 Standard deviation 28 Starch granule 58 Starch, digestion 112 Statistics, descriptive 26 Stem 90 Stem cells 249, 252-254 - adult 249, 252 - and germ layers 253 - applications 250-251 - diabetes 250-251 - embryonic 249 - potency 248-249, 252 - self renewal 248 Stem cell differentiation 252 Stomach 110 - enzymes 111 Stomach emptying 159 Stomata 82, 89, 91, 93 Stroma 81 Sun plant 83 Surface area:volume ratio 47, 69 Symplast 92

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H Habitat, dingo 216 Haplotype 273 Heart 116 - blood flow 120 - blood pressure 121 - diseases of 124 - dissection 122 - structure 120 Heart attack 124 Heart rate 105 Heterotroph 79 Heterozygous condition 286 Hierarchy of life, animals 99 Hierarchy of life, plants 88 Histograms, guidelines 18, 20 Histone modification 276 Holozoic nutrition 78 Homeostasis 157-158 Homologous chromosomes 286 Homozygous condition 286 Human classification 173 Human Genome Project 267 Human karyogram 279 Huntington's disease 271 Hyaline cartilage 101 Hydrolysis 79 Hydrophytes, adaptations of 146 Hypertension 124 Hyperthermia 163 Hyperthyroidism 164 Hypertonic 71 Hypothermia 163 Hypothesis 3 Hypotonic 71

L Lactic acid fermentation 85 Lactose 289, 299 Large intestine 109, 114 Leaf structure 82, 83 Leucocyte 116 Life, functions 42 Light dependent reactions 81 Light independent reactions 81 Light microscopy 50 Limiting factors 205 Line graphs, guidelines 18, 21-22 Linear magnification, calculating 53 Lipid 43, 65 Log transformation 11 Loop of Henle 130 Lung disease 106 Lungs 101-102 Lysosome 56 M Magnification, defined 53 Mangroves, adaptations of 147 Mark and recapture sampling 197 Mean 26 Median 26

O Observation 3 Obstructive lung disease 106 Optical microscopes 50 Organ 97 Organ system 97 Organ system interactions 132-136 Organelles 56, 58, 60 Organisation, levels of 88, 99 Osmosis 71-72 Osmotic potential 71

P Pancreas, enzymes 111 Parasite 79 Parasitism, ecological interaction 186 Parietal cell 108 Passive transport 68 Pedigree chart 299 Percentage error 13 Percentages, calculation 10 Peristalsis 108-109 Phenotype 287-289, 292, 295, 301 - and environment 288, 290, 293 - height in humans 295 - skin colour in humans 295 Phenotypic frequency 303 Phenotypic plasticity 288 Phloem 89, 90, 91 Phospholipid 65 Photoautotroph 78 Photosynthesis 7, 79, 80, 81 Pitcher plant, adaptations of 149 Plant cell 42, 58 - specialisation 89 Plant - adaptations 145-149 - cell wall 58 - leaves 83 - root structure 91, 92, - transpiration 93 - transport system 90

Q Quadrat sampling 197-200, 202-203 Qualitative data 4, 5 Qualitative variables 5 Quantitative investigation 6, 30 Quantitative variables 6 Quiet breathing 104 R Random sampling 29, 197 Ranked variables 4 Rates, calculation 4, 9, 105 Ratios 10 Recessive alleles 286-287, 302 Recombination 242, 309 Red blood cell 98, 116 -117 Red kangaroo classification 175 Renal corpuscle 128 Replication of experiments 31 Report, structure 37 Reproduction, asexual 231 Reproduction, sexual 240 Resolution, defined 50 Respiration 79, 84 Respiratory system 101, 158 - and circulatory system 132 Restrictive lung disease 106 Ribosome 56, 58 Rocky shore community 203, 212 Rocky shore, field study 212 Root 92 Root hair cell 89, 91, 92 Root pressure 94 Root system 90

S Sampling 199-200 - bias 29 - techniques 197-199 - rocky shore 203 Sanger sequencing 266 Saprotroph 79 Scatter plot, guidelines 18, 24 Scientific method 2 Sex chromosomes 277 Sex determination 277, 290 Sex linkage 304 Sexual reproduction 240, 245 - advantages 245 Shade plant 83 Shared derived characteristics 174 Significant figures 8, 9 Small intestine 109, 111 - absorption 113 - enzymes 111 Sodium-potassium pump 75 Species abundance 197 Species complex 172 Species conservation 172 Species distribution 197

U Ultrafiltration 130 Unicellular eukaryote 42 Urinalysis 131 Urinary system 158 - structure 128 Urine 130

V Vacuole 58 Variables, experimental 6, 30 Variables, types 6 Variation, continuous 295 Variation, sources of 288 Varicose veins 124 Vascular tissue 89, 90, 91 Vasoconstriction 118 Vasodilation 118 Vegetative propagation 232 Veins 118-119 - varicose 124 Ventilation 104 Venus fly trap, adaptations of 149 Vessel element 91 Virus 42

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K Kangaroo adaptations 144 Karyogram 278-279, 281 Karyotype 278, 280 Keystone species 194, 195 Kidney disease 131 Kidney stone 131 Kidney 126 - Bowman's capsule 126, 130 - collecting duct 130 - function 126 - glomerulus 126, 130 - nephron 130 - structure 126, 129 - urine 130 - water balance 167 Klinefleter syndrome 280 Krebs cycle 85

N Navel orange 237 Negative feedback 158, 162 - blood flow 162 - blood glucose regulation 165 - calcium homeostasis 159 - insulin 165 - stomach emptying 159 - thermoregulation 159, 161 - thyroxine 164 - water balance 167 Nephron 130 Niche 210 Non-disjunction 280 Nuclear membrane 56, 58 Nuclear pore 43, 56, 58 Nucleic acid 43 Nucleolus 56, 58 Nucleotide 43 Null hypothesis 3

W Water 43, 44 Water balance 167 White blood cell 116-117

XYZ Xerophytes, adaptations of 145 Xylem 89, 90, 91, 92 Zygote 240

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I Independent assortment 242 Independent variable 6, 30 Inheritance patterns 306 Insectivorous plants 149 Inspiration 102 Insulin 165 Interactions of organ systems 132-136 Interphase 227-228 Interspecific competition 211-212 Intestinal epithelial cell 108 Intraspecific competition 210, 217 Investigation, qualitative 5 Investigation, quantitative 6 Ion pumps 75 Isotonic 71

T Tables, constructing 17 Tally chart 20 Tasmanian devil 208 Taxonomic rank 173 Taxonomy 170, 173 Test cross 302 Thermoreceptors 162 Thermoregulation 159, 161 - body shape 163 - hyperthyroidism 164 Three domain classification 170 Thylakoid discs 81 Thyroid gland 164 Thyroxine 164 Tight junction 56 Trachea 101-102 Tracheid 91 Trait 287 Transect sampling 197 Transformation, of data 11 Transpiration 90, 93-94 - factors affecting 95-96 Transpiration pull 94 Transpiration rate 95 Transpiration stream 93 Transport, passive 68 Tropic levels 187, 190 True breeding 287, 301 Turgor pressure 73 Turner syndrome 280 Type 1 diabetes 166

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