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NCEA LEVEL 2 BIOLOGY INTERNALS Meet the writing team

Tracey

Senior Author

Kent

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. 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

Lissa Bainbridge-Smith I worked in industry in a research and development capacity for 8 years before joining BIOZONE in 2006. I have a M.Sc from Waikato University. Lissa Author

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. Richard

Founder & CEO

Cover photograph North Island brown kiwi (Apteryx mantelli) is widespread in low numbers throughout the northern two thirds of New Zealand's North Island. This species has adapted to live in a variety of habitats including scrub-like farmland, pine forest, and native forest. However, its preference is for dense subtropical and temperate forest. Females lay 2-3 clutches a year and their chicks hatch fully feathered. The chicks initially feed from the nutrient rich yolk of the egg they hatched from, but begin to supplement this by foraging for food around five days after hatching. PHOTO: Eric Isselee (123RF)

Thanks to: The staff at BIOZONE, including Holly Coon for design and graphics support, Paolo Curray and Malaki Toleafoa for IT support, Debbie Antoniadis and Arahi Hippolite for office handling and logistics, and the BIOZONE sales team. First edition 2017

ISBN: 978-1-927309-60-5 Copyright Š 2017 Richard Allan Published by BIOZONE International Ltd Printed by Wickcliffe Solutions www.wickliffe.co.nz

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

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electrical, mechanical, photocopying, recording or otherwise, without the permission of BIOZONE International Ltd. This book may not be re-sold. The conditions of sale specifically prohibit the photocopying of exercises, worksheets, and diagrams from this book for any reason.


Contents Using This Book ...................................................... v

Using the Tab System ............................................ vii

Using Biozone's Website....................................... viii

AS 2.1 Practical investigation in a biology context Achievement criteria and explanatory notes ...... 1 1 2

How Do We Do Science?..................................... 3

3 4

Accuracy and Precision....................................... 5

5 6

Tallies, Percentages, and Rates............................ 7

7 8

Dealing with Large Numbers ............................... 9

Observations, Hypotheses, and Assumptions..... 4 Working With Numbers........................................ 6 Fractions and Ratios............................................ 8 Apparatus and Measurement ............................ 10

9 Types of Data .................................................... 11 10 Variables and Controls....................................... 12 11 A Case Study: Catalase Activity......................... 13 12 Recording Results.............................................. 14 13 Practising Data Manipulations .......................... 15 14 Constructing Tables............................................ 16 15 Drawing Graphs................................................. 17 16 Interpreting Line Graphs.................................... 19 17 Correlation or Causation? ................................. 20 18 Investigating Plant Growth ................................ 21 19 Designing a Practical Investigation.................... 24

AS 2.2 Analysing the validity of biological Information Achievement criteria and explanatory notes ..... 27 20 Recognising Balanced Reporting ..................... 28 21 The Misuse of Scientific Data............................ 29 22 HPV.................................................................... 30 23 HPV Vaccine ..................................................... 31 24 Presenting Your Findings.................................... 33

AS 2.3 Adaptations of animals to their way of life

34 Moving Food Through the Gut .......................... 48 35 The Stomach and Small Intestine...................... 49 36 The Large Intestine ........................................... 52 37 Summary of the Human Digestive Tract............. 54 38 Mammalian Guts ............................................... 55 39 Comparison between Ruminant and Carnivore . 57 40 Comparison of Tube Guts.................................. 58 41 Cellulose Digestion in a Ruminant .................... 60 42 Cellulose Digestion in a Hindgut Fermenter...... 61 43 Cellulose Digestion in an Insect ........................ 62 44 Digesting Fluids................................................ .. 63 45 Adaptations for Digestion in Fish........................ 64 46 Adaptations for Absorption in Insects ............... 65 47 Adaptations for Absorption in Fish .................... 66 48 Adaptations for Absorption in Mammals ........... 67 49 What You Know So Far:

Digestion and Feeding....................................... 69

50 Essay Style Question: Digestion and Feeding... 70 51 KEY TERMS AND IDEAS:

Digestion and Feeding ...................................... 72

52 The Need for Gas Exchange............................. 73 53 Gas exchange in Insects ................................... 74 54 Insect Adaptations to Gas Exchange in Water... 76 55 Gas Exchange in Fish ....................................... 77 56 Countercurrent Flow.......................................... 79 57 The Human Gas Exchange System................... 80 58 The Lungs and Gas Exchange ......................... 81 59 Breathing in Humans......................................... 83 60 Adaptations for Diving in Mammals .................. 84 61 What You Know So Far: Gas Exchange ............ 86 62 Essay Style Question: Gas Exchange .............. 87 63 KEY TERMS AND IDEAS: Gas Exchange ....... 88 64 Transport and Exchange Systems .................... 89 65 Haemolymph ..................................................... 90 66 Blood.................................................................. 91 67 Blood Vessels..................................................... 92 68 Open and Closed Circulatory Systems.............. 94 69 Single and Double Closed Circulatory Systems ............................................................ 95

Achievement criteria and explanatory notes ..... 35

70 Mammalian Transport System .......................... 97 71 Vertebrate Hearts............................................... 99

25 Obtaining Food.................................................. 37 26 Parasitism ......................................................... 39

72 Dissecting a Mammalian Heart........................ 101 73 Internal Transport and Gas Exchange

27 Predation ........................................................... 40 28 Insect Mouthparts ............................................. 41

Interaction ....................................................... 103

29 The Teeth of Fish .............................................. 43 30 The Teeth of Mammals....................................... 44 31 Mammalian Teeth and Diet ............................... 45 32 Comparing Mammalian Teeth............................ 46

74 Internal Transport and Digestion Interaction ... 105 75 What You Know So Far: Internal Transport...... 107 76 Essay Style Question: Internal Transport......... 108 77 KEY TERMS AND IDEAS: Internal Transport... 109

33 Introduction to Digestion ................................... 47

CODES:

Activity is marked:

to be done

when completed


Contents AS 2.3 Adaptations of plants to their way of life

120 Quadrat-Based Estimates................................ 167 121 Sampling a Rocky Shore Community.............. 168

122 Field Study of a Rocky Shore.......................... 170 123 Transect Sampling........................................... 173

Achievement criteria and explanatory notes .......... 110 78 The Plant Transport System ............................ 112 79 Transpiration ................................................... 114

124 Shoreline Zonation........................................... 175 125 Competition and Species Distribution.............. 176

80 Measuring Transpiration................................... 116 81 Uptake at the Root........................................... 119

126 Niche and Community Patterns....................... 178 127 Altitude Zonation and Physical Environment.... 180

82 Adaptations of Leaves .................................... 121 83 Adaptations of Mesophytes ............................. 122

128 Environment Determines Community Patterns. 182 129 Primary Succession......................................... 184

84 Adaptations of Xerophytes .............................. 123 85 Adaptations of Hydrophytes ............................. 124 86 Mangrove Adaptations .................................... 125 87 What You Know So Far: Transpiration ............. 126 88 Essay Style Question: Transpiration ................ 127 89 KEY TERMS AND IDEAS: Transpiration ......... 128

130 Secondary Succession.................................... 186 131 Patterns of Population Growth......................... 187 132 Succession and Population Densities of Perching Birds.................................................. 189 133 What You Know So Far: Communities.............. 191 134 KEY TERMS AND IDEAS: Communities......... 192

90 Alternation of Generations .............................. 129 91 Reproduction in Ferns....................................... 130 92 Reproduction in Angiosperms.......................... 131 93 Reproduction in Gymnosperms....................... 132 94 Plant Adaptations for Insect Pollination............ 133 95 Pollination Relationships.................................. 134

AS 2.8 Practical microscopy Achievement criteria and explanatory notes ... 193 135 Optical Microscopes......................................... 194 136 Making Biological Drawings............................. 196

96 Wind Pollination................................................ 135 97 Comparing Insect and Wind Pollination............ 136

137 Practising Biological Drawings......................... 198 138 Preparing a Slide............................................. 199

98 Fertilisation Follows Pollination........................ 137 99 Structure of Seeds........................................... 139

139 Staining a Slide................................................ 200 140 Calculating Linear Magnification...................... 201

100 Adaptations for Seed Dispersal....................... 140 101 Summary of Plant Reproduction...................... 142 102 What You Know So Far: Plant Reproduction..... 143 103 Essay Question: Plant Reproduction................ 144 104 KEY TERMS AND IDEAS: Plant Reproduction.. 145

AS 2.6 Patterns in ecological communities Achievement criteria and explanatory notes ... 146

141 Cell Sizes......................................................... 202 142 Examining Cells............................................... 203 143 KEY TERMS AND IDEAS: Practical Microscopy........................................ 205 ANSWERS............................................................... 206

QUESTIONING TERMS AND PHOTO CREDITS.... 231

INDEX .................................................................... 232

105 Guidelines for Investigating an Ecological Pattern.............................................................. 148 106 Components of an Ecosystem......................... 149 107 Habitat.............................................................. 150 108 Feeding Relationships..................................... 151 109 Food Webs....................................................... 152 110 Ecological Niche.............................................. 153 111 Adaptation and Niche....................................... 154 112 Species Interactions in a Beech Forest .......... 156 113 Competition and NIche Size............................ 159 114 Sampling Populations...................................... 160 115 Interpreting Samples........................................ 161 116 Stratification in a Forest................................... 162 117 Physical Factors in a Forest............................. 164 118 Physical Factors on a Rocky Shore................. 165 119 Quadrat Sampling............................................ 166

CODES:

Activity is marked:

to be done

when completed


v

Using This Book BIOZONE's NCEA Level 2 Biology Internals contains material to meet the needs of New Zealand students studying NCEA Biology Level 2 Internal Achievement Standards. The NCEA Level 2 Biology Internals is compliant with Level 7 of the NZ Curriculum (Nature of Science – The Living World) and the NCEA Biology Level 2 Internal Achievement Standards. 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 this resource useful 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 the chapters with multiple sections (Adaptations of animals to their way of life and Adaptations of plants to their way of life. Sections within a chapter share the same structure. They correspond to natural topic breaks within the Achievement Standard.

Introduction

Activities

Review

Test

• A check list of achievement criteria and explanatory notes • A check list of what you need to know • 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

• Create your own summary for review • Hints help you to focus on what is important • Your summary will help you with the essay question

• Essay style questions conclude clusters of related activities • These enable you to practise your essay writing skills

140

Key terms and ideas • Includes a question based on key terms • Other questions test your understanding of the section content

100 Adaptations for Seed Dispersal Key Idea: Seeds must be dispersed from the parent plant to reduce competition for light and nutrients. Seeds may be dispersed by wind, water, or animals. f In order for a seed to survive and grow into a new plant, it must be

transported (dispersed) into an environment where it has enough light, nutrients and water to grow and eventually reproduce. Dispersal also helps to reduce competition for light and nutrients.

102 What You Know So Far: Plant Reproduction

f Plants have evolved many ways to ensure that their seeds are dispersed.

Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question at the end of the chapter. Use the points in the introduction and the hints provided to help you:

This has given them opportunities to expand their range. In some cases the seed itself is the agent of dispersal, but often it is the fruit or an associated attached structure.

f The main agents of seed dispersal are wind, water, and animals, but

others can also be dispersed through explosive discharge or shaking from pods or capsules (e.g. flax, peas, gorse) or through fire (e.g. many Australian plants).

Pollination and fertilisation

HINT: Compare mechanisms for pollination. Describe fertilisation in angiosperms and gymnosperms.

Alternation of generations

HINT: Describe the life cycles of plants and identify the gametophyte and sporophyte generations.

Image right: Coconut seeds germinating after dispersal from the parent plant.

143

Mechanisms of seed dispersal

144

103 Essay Question: Plant Reproduction All plants have adaptations to enable them to reproduce successfully. Although ferns, angiosperms, and gymnosperms share similarities in their life cycles, they have different reproductive strategies. 1. Discuss the different reproductive strategies of fern, angiosperms, and gymnosperms. Your discussion should include identification of sporophyte and gametophyte stages, reference to reliance on water, details of pollination mechanisms if applicable, fertilisation of the egg cell by the sperm, and development of the embryo. You may use extra paper if required.

145

104 KEY TERMS AND IDEAS: Plant Reproduction

1. Use the following word list to complete the sentence below: haploid, generations, meiosis, mitosis, sporophyte The life cycle of plants alternates between a

gametophyte generation and a diploid

generation. This process is called the alternation of

Jenny Ladley

Gametophytes produce gametes by

Wind dispersal Wind dispersed seeds have wing-like or feathery structures that catch the air currents and carry the seeds long distances from the parent plant. Some seeds are light and feathery (dandelions and swan plants) while others (pine, kauri, maple) have winged seeds that help them travel. Plants that rely on wind dispersal produce a large number of seeds. This increases the chances of a seed successfully establishing elsewhere.

Animal dispersal Plants that rely on animals to spread their seeds may have hooks or barbs that catch the animal hair, sticky secretions that adhere to the skin or hair, or fruits that are eaten. In New Zealand, birds are common seed carriers. They consume the seed (contained within a fleshy fruit) and disperse the seeds in their droppings. Seeds can be dispersed over long distances, including to offshore islands. Examples include miro, puriri, and tawa.

alternation of generations

A The phase of a plant life cycle in which the cells of the plant are diploid.

double fertilisation

B The immature male gametophytes of seed plants.

endosperm

C Feature of reproduction in angiosperms. Two sperm nuclei fuse with nuclei in the embryo sac. One fuses with the haploid egg cell nucleus to form the zygote, the other fuses with the diploid endosperm nucleus to form the triploid endosperm.

fertilisation gametophyte germination

HINT: Describe adaptations for successful reproduction, including adaptations for pollination and seed dispersal.

pollen

2. Describe the three main mechanisms of seed dispersal:

D The transfer of pollen from one plant to another for the purpose of fertilisation. E Reproductive unit containing the embryonic plant that forms from the ovule after fertilisation. F The union of male and female gametes to form a zygote. G In angiosperms, this provides nutrients to the developing embryo.

pollination

H The phase of a plant life cycle in which the cells of the plant are haploid.

seed

I

The growth of a seed immediately following a period of dormancy.

J

A feature of the life cycles of plants in which there are distinct sexual haploid and asexual diploid stages.

sporophyte

3. Study the three photos (right) and decide if these reproductive structures are pollinated by wind or by animals. Justify your answer: WEB

KNOW

100

LINK

99

. .

2. Match each term to its definition, as identified by its preceding letter code. Water dispersal When water is used as the means of dispersal the seed simply floats away from the parent plant. Seeds can be transported over long distances, especially if they are transported by ocean currents (e.g mangrove and coconut). If a seed falls from the plant at low tide it can begin to take root and begin growing next to the parent plant. This can increase competition for resources.

Adaptations for reproduction 1. What is the purpose of seed dispersal?

and sporophytes produce spores by

Plant A: daffodil

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Plant B: grass

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REVISE

Plant C: conifer © 2017 BIOZONE International ISBN:978-1-927309-60-5 Photocopying Prohibited

TEST

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TEST


vi

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: The activities form most of this book. 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.

66

2.3 Key terms

Internal transport artery (pl. arteries) atrium (pl. atria) blood bulk flow capillary closed circulatory system double circulatory system haemolymph heart open circulatory system respiratory pigment single circulatory system vein ventricle

gut. Bony fish use pyloric caeca to increase the surface area for absorption. Intestinal spiral valve Examples: cartilaginous fish, e.g. sharks, rays The streamlined shape of sharks (and most other fish) makes it difficult to fit a long intestine into the body. Cartilaginous fish, such as sharks, as well as primitive bony fish, solve this problem by having a short intestine with a spiralling fold in the inner wall. This spiral valve increases the length of time digested material remains in the intestine by effectively increasing the intestine's length. The spiral valve also prevents large masses of indigestible material passing into the intestines.

Achievement criteria and explanatory notes c

A

Demonstrate understanding of adaptation of animals to their way of life: Describe the adaptations and identify aspects of the adaptations that enable each organism to carry out its life process(es) in order to survive in its habitat.

c

M

Demonstrate in-depth understanding of adaptation of animals to their way of life: Provide a biological reason that explains how or why the adaptations enable each organism to carry out its life process(es) in order to survive in its habitat.

c

E

Demonstrate comprehensive understanding of adaptation of animals to their way of life: Link biological ideas the adaptations that enable each organism to carry out its life process(es) in order to survive in its habitat. The discussion may involve justifying, relating, evaluating, comparing and contrasting, and analysing and it must include consideration of the two points from below appropriate to your chosen context. Option 1: Adaptation related to one life process over three taxonomic or functional groups:

i

Compare diversity of adaptation in response to the same demand (e.g. gas exchange) across different taxonomic or functional groups.

ii

Explain limitations and advantages involved in each feature within each organism.

i

Make connections between two life processes within each organism that enhance the effectiveness of both processes.

ii

Explain limitations and advantages involved in each feature.

Explanatory notes

Activity number

Understanding of adaptation to demonstrate in relation to... 1a

One life process over three taxonomic or functional groups of animals, or

c

1b

Two related life processes within one taxonomic or functional group

c

2 •

The ways in which an organism carries out all its life processes, including: relationships with other organisms: competition, predation, mutualism, parasitism, adaptations to the physical environment.

3

Life processes (for animals) selected from:

Food

The number of turns varies between 5 and 50 depending on the species.

Example: most bony fish

Gill cover

Gut of bony fish showing the pyloric caeca, which open into the first part of the intestine

Intestinal wall

Spiral folds increase the length of the intestine.

Body wall (cut) Kidney

Liver

Gonad Stomach

Pyloric caeca

1. (a) How does the spiral valve of cartilaginous fish increase the length of time food remains in the intestine?

(b) How does this affect absorption?

nutrition (adaptations for obtaining and processing nutrients)

c

ii

gas exchange (adaptations for exchanging respiratory gases with the environment) 52 - 63 76

c

iii

internal transport (adaptations for moving materials around the body)

2. Explain how bony fish increase the surface area of the intestine:

64 - 77

WEB

KNOW

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)? A full list of codes is given on the following page but the codes themselves are relatively self explanatory.

Spiral valve

25 - 51 77

i

c

Liver

Many bony fish have pouch-like structures called pyloric caeca, which increase the gut surface area. The caeca are tube-like outgrowths at the junction of the stomach and intestine. They increase gut surface area, and function in enzyme secretion and nutrient absorption. Most bony fish have several hundred pyloric caeca and they are more well developed in carnivores, especially those with short guts.

EII

c

Spiral valve from a nurse shark

Stomach

Pyloric caeca

Option 2: Adaptation across two related life processes within one taxonomic or functional group:

c

Photo: Haplochromis CC 3.0

Achievement criteria for achieved, merit, and excellence

c

Gas exchange alveoli breathing bronchi bronchioles cellular respiration countercurrent flow expiration gas exchange system gills haemoglobin inspiration respiratory gas spiracles tracheae ventilation

Key Idea: Cartilaginous fish increase nutrient absorption by increasing the time food spends in the

Animals possess structural, physiological, and behavioural adaptations that enable them to carry out the life processes essential to survival in their habitat and exploitation of their niche.

Common terms adaptation functional group taxonomic group Nutrition absorption assimilation dentition diet digestion enzyme egestion gut ingestion nutrients

47 Adaptations for Absorption in Fish

Adaptations of animals to their way of life

Achievement Standard

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.

Photo: H Dahimo CC3.0

The chapter introduction provides you with a summary of the achievement criteria and explanatory notes as identified in the Achievement Standard. A check list of what you need to know to meet the knowledge requirements of the standard is provided on the second page of the introduction. Use the check boxes to identify and mark off the points as you complete them. A list of key terms for the chapter is included.

47

LINK

40

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 book.

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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.


1. (a) How does the spiral valve of cartilaginous fish increase the len

vii

Using the Tab System 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. A link may also reflect on material that has been covered (b) How does this affect absorption? earlier as a reminder for important terms that have already been defined. In the example below for the activity "Adaptations for Absorption in Fish", weblink 47 is a short presentation about the digestive system in bony fish and the adaptations of their digestive system to increase efficiency. Activity 40 directs back to an activity comparing the tube guts of mammals, fish, and insects so that you can compare the basic structures of each at2.a Explain glance. The is always the same asof the intestine: how weblinks bony fishcode increase the surface area the activity number on which it is cited. On visiting the weblinks page (below), find the number and it will correspond to one or more external websites providing a video or animation of some aspect of the activity's content. Occasionally, the weblink may provide a bank of photographs where images are provided in colour.

WEB

Activities are coded KNOW = content you need to know

KNOW

47

LINK

40

DATA = data handling and interpretation PRAC = a paper practical or a practical focus REVISE = review the material in the section TEST = test your understanding

Weblinks

Bookmark the weblinks page: www.biozone.co.nz/weblink/ NZL2I-9605 Access the external URL for the activity by clicking the link

Link

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

www.biozone.co.nz/weblink/NZL2I-9605 This WEBLINKS page provides links to external websites with supporting information for the activities. These sites are distinct from those provided in the BIOLINKS area of BIOZONE's web site. For the most part, they are narrowly focussed animations and video clips directly relevant to some aspect of the activity on which they are cited. They provide great support to help your understanding of basic concepts.

Chapter in the book

Hyperlink to the external website page. Activity in the book

Bookmark weblinks by typing in the address: it is not accessible directly from BIOZONE's website Corrections and clarifications to current editions are always posted on the weblinks page

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viii

Using the Website

BIOZONE's web site should be the first stop for biologists. As well as providing all our product information (including shipping dates) and updates, www.biozone.co.nz provides quick access to the latest RSS newsfeeds and podcasts from around the world. The Resource hub also provides quick links to access the websites of publishers of references cited in the workbooks. Perhaps of greatest value to students and teachers is the BIOLINKS area of Biozone's website. The BIOLINKS pages are distinct from Weblinks (which are specific to each workbook edition) and provide a database of well organised hyperlinks pertaining to topics of interest in biology. The database is updated regularly, so that outdated, not operational, or no longer relevant sites are removed and new sites are added as they appear.

Click on a topic to see a list of all its links. Each topic has relevant subtopics to make searching easier and each link has a brief description.

RSS Newsfeeds and Podcasts in the right hand column provide the latest news and information from the world of science.

Click on the link to access the named site. The brief description tells you how the site may be of interest, as well as any country specific bias, if this is relevant.

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Practical investigation in a biology context

PR E V ON IEW LY

No Cla t fo ssr r o Us om e

Achievement Standard

2.1

Key terms

Investigating in science will extend your science knowledge and develop your understanding of the relationship between investigations and scientific theories and models. Part of rigorous investigation involves submitting your findings for peer review and debate.

assumptions bias

Achievement criteria and explanatory notes

conclusion

Achievement criteria for achieved, merit, and excellence

controlled variable data

A

Carry out a practical investigation in a biology context, with supervision: Involves:

Developing a statement of the purpose, linked to a scientific concept or idea, and written as a hypothesis.

For a fair test using a method that describes the independent variable and its range, measurement of the dependent variable and the control of some other key variables.

For a pattern seeking or modelling activity, describing the data that will be collected, range of data/samples, and consideration of other key factors.

independent variable

Collecting recording, and processing data relevant to the purpose of the investigation.

logbook

Interpreting and reporting on the findings of the investigation.

mean

Identifying and including relevant findings from another source.

measurement

Reaching a conclusion based on the processed data, which is relevant to the purpose of the investigation.

c

dependent variable

descriptive statistics fair test graph

hypothesis

median mode

observation

pattern seeking prediction

qualitative data

quantitative data

PASCO

random sampling

c

reliability (of data)

M

Carry out an in depth practical investigation in a biology context, with supervision: Involves:

For a fair test, using a method that describes a valid range for the independent variable, the valid measurement of the dependent variable and the control of other key variables with consideration of factors such as sampling bias and sources of error.

For a pattern seeking or modelling activity, using a method that describes a valid collection of data with consideration of factors such as sampling bias and sources of error.

Collecting recording, and processing data to enable a trend or pattern (or absence of a trend or pattern) to be determined.

Reaching a valid conclusion based on the processed data, which is relevant to the purpose of the investigation.

Discussing the biological ideas relating to the investigation. The discussion should be based on both the findings of the investigation and those from other sources.

E

Carry out a comprehensive practical investigation in a biology context, with supervision: Involves:

Justifying the choices made throughout the sound investigation by evaluating the validity of the method or the reliability of the data.

Explaining the conclusion in terms of the biological ideas relevant to the investigation.

report

sample table

trend (of data) validity

variable

c

No Cla t fo ssr r o Us om e

raw data


No Cla t fo ssr r o Us om e

Explanatory notes: A practical investigation

PR E V ON IEW LY

A practical investigation…

c

1

Should cover the complete investigation process: planning and carrying out the investigation, collecting primary data, processing and interpreting the data, and reporting on the investigation.

c

2

Will involve collecting primary data. You will have opportunities to make changes to your initial method as you work through the investigation.

c

3

May involve manipulating variables (fair test), investigating a pattern or relationship, or the use of models.

c

4

Assessment may be based on a stand-alone investigation or an individual investigation that contributes to a group or class investigation.

AKA

With supervision means…

c

5

The teacher will provide guidelines for the investigation and you will develop and complete the investigation from these guidelines. They may discuss ideas with you and manage the process of sharing findings if you are contributing to a group.

What you need to know for this Achievement Standard Hypotheses and design of the investigation Activities 1 - 3, 8 - 11, 18 , 19, 118

By the end of this section you should be able to:

c

Formulate a working hypothesis from which you can generate predictions about the outcome of your investigation.

Describe a method for collecting valid data for an investigation involving: Fair testing: A single variable is changed while keeping other variables the same.  Pattern seeking: Observing and recording natural events or carrying out an investigation in which the variables cannot easily be controlled.  Modelling: Creating a representation of a system or process to investigate how it works. c 

PASCO

c

For a fair test, identify your dependent and independent variables, their range, and how you will measure them. Identify controlled variables and their significance.

c

For a pattern seeking investigation, identify the data that you will collect and the range of the data or samples. Note/record variables and identify patterns that result from these variables.

c

Determine the amount of data you need to reasonably test your hypothesis and its predictions.

c

Describe any controls in your investigation (if appropriate). Identify any assumptions made in your investigation. Evaluate any sources of error in your design.

Recording, analysing, and presenting data Activities 4 - 7, 9, 12 - 19, 118

By the end of this section you should be able to:

Use and maintain a thorough and accurate logbook

c

Present your data appropriately in a table, including any calculated values.

c

Demonstrate an ability to process raw data, e.g. percentages, total, rates (as appropriate).

c

Present your processed data appropriately in a graph. Recognise any patterns or trends evident.

c

Demonstrate an ability to analyse and interpret data from a fair test or pattern seeking investigation (including the points noted below).

Calculate basic descriptive statistics (e.g. sample mean, median, mode and standard deviation) to summarise trends or differences in your data.

If appropriate, calculate measures of dispersion for your data, related to the true population parameters, e.g. standard error, 95% confidence intervals, and 95% confidence limits.

If required, perform an appropriate statistical test to test the significance of any trends or differences you have observed, e.g. linear regression, chi-squared test student's t test).

c

Use information sources to find information relevant to your study, and reference it appropriately.

c

Write up the findings of your investigation in a scientific report, including methods, results, discussion, and conclusion.

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How Do We Do Science?

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Key Idea: Science is a way of understanding the world based on a rigorous, dynamic process

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of observation, investigation, and analysis.

ffScience is a way of understanding the world we live in: how it formed, the rules it obeys and how it changes over

time. Science distinguishes itself from other ways of understanding by using empirical standards, logical arguments, and skeptical review. Science allows what we understand to change over time as the body of knowledge increases.

ffIt is important to realise that science is a human endeavour and requires creativity and imagination. New research and ways of thinking can be based on the well-argued idea of a single person.

ffScience influences and is influenced by society and technology. As society's beliefs and desires change, what is

or can be researched is also affected. As technology advances what is or can be researched changes. Scientific discoveries advance technology and can change society's understanding about the natural world.

ffScience can never answer questions with absolute certainty. It can be confident of certain outcomes, but only within the limits of the data. Science might help us predict with 99.9% certainty a system will behave a certain way, but that still means there's one chance in a thousand it won't. Exploration and discovery

Questioning, observing, and sharing information

Investigating and testing ideas

Carrying out investigations, comparing results to predictions and developing models that explain the patterns seen.

Benefits and outcomes

Analysis and feedback

Using results to develop technology, solve problems, answer questions and educate.

Review and discussion of results. Publication and repeat investigations.

Science has application and relevance in the modern world

Jigsaw activity

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

Science is exciting, dynamic, creative, and collaborative

Science is a global human endeavour

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

1. The buttons above make five statements about science. Divide your class into five groups, with each group addressing one statement. Discuss the statement as a group and present a brief written summary of what the statement means, what evidence there is to support it, and whether you agree with it. Have one person in each group present the group's views to the class a whole.

2. Why do you think it is important that science is an open-minded, rigorous process?

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Observations, Hypotheses, and Assumptions

Key Idea: Observations are the basis for forming hypotheses and making predictions about systems.

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An assumption is something that is accepted as true but is not tested.

Observations and hypotheses

ffAn observation is watching or recording what is happening. Observation is the basis for forming hypotheses and making predictions. An observation may generate a number of hypotheses (tentative explanations for what we see). Each hypothesis will lead to one or more predictions, which can be tested by investigation.

ffA hypothesis is often written as a statement to include the prediction: "If X is true, then if I do Y (the

experiment), I expect Z (the prediction)". Hypotheses are accepted, changed, or rejected on the basis of investigations. A hypothesis should have a sound theoretical basis and should be testable. Observation 1:

ffSome caterpillar species are brightly

coloured and appear to be highly visible to predators such as insectivorous birds. Predators appear to avoid these caterpillars.

ffThese caterpillars are often found in groups, rather than as solitary animals.

Observation 2:

ffSome caterpillar species have excellent

camouflage. When alerted to danger they are difficult to see because they blend into the background.

ffThese caterpillars are usually found alone.

Assumptions

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

• Insect eating birds have colour vision. • Caterpillars that look bright to us, also appear bright to insectivorous birds. • Birds can learn about the taste of prey by eating them.

Read the two observations about the caterpillars above and then answer the following questions:

1. Generate a hypothesis to explain the observation that some caterpillars are brightly coloured and highly visible while others are camouflaged and blend into their surroundings:

2. Describe one of the assumptions being made in your hypothesis:

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3. Generate a prediction about the behaviour of insect eating birds towards caterpillars:

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Accuracy and Precision

Key Idea: Accuracy refers to the correctness of a measurement (how true it is to the real value).

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Precision refers to how close the measurements are to each other.

The terms accuracy and precision are often used when talking about measurements.

ffAccuracy refers to how close a measured value is to its true value, i.e. the correctness of the measurement.

ffPrecision refers to the closeness of repeated measurements to each other, i.e. the

ability to be exact. For example, a digital device, such as a pH meter (right) will give very precise measurements, but its accuracy depends on correct calibration.

Using the analogy of a target, repeated measurements are compared to arrows shot at a target. This analogy is useful when distinguishing between accuracy and precision. Accurate but imprecise

Inaccurate and imprecise

Precise but inaccurate

Accurate and precise

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 (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.

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

All zeros between non-zero numbers are always significant.

0.005704510

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).

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

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Significant figures

1. Why are precise but inaccurate measurements not helpful in a biological investigation?

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

(a) 3.15985

(d) 1000.0

(b) 0.0012

(e) 42.3006

(c) 1000

(f) 120

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Working With Numbers

Key Idea: Using correct mathematical notation and being able to carry out simple calculations and

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conversions are fundamental skills in biology.

Commonly used mathematical symbols

Estimates

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

ffWhen 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 what answer you should get, and helps you decide if your final calculation is correct.

= Equal to

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

ffNumbers 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.

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

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

ffFor 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.

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

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

~ Approximately equal to

Use the following examples to practise estimating:

4. 43.2 x 1044:

Decimal and standard form

5. 3.4 x 72 ÷ 15:

ffDecimal form (also called ordinary form) is the

6. 658 ÷ 22:

Conversion factors and expressing units ffMeasurements 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.

ffIn 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.

ffFor the example above, A = 1.5 and n = 7 because

ffConversion 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.

the decimal point moved seven places (see below).

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

ffSmall 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.

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

ffConverting can make calculations easier. Work

through the following example to solve 4.5 x 104 + 6.45 x 105.

1. 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): 7.

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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.

ffThe 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). To denote 'per' use a negative exponent, e.g. per second is written as s-1 and per metre squared is written as m-2.

ffFor example the rate of oxygen consumption

2. Add the two numbers together: 3. Convert to standard form:

DATA

should be expressed as:

Oxygen consumption (cm3 g-1 s-1)

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Tallies, Percentages, and Rates

Key Idea: Unprocessed data is called raw data. A set of data is often manipulated or transformed to

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make it easier to understand and to identify important features.

The data collected by measuring or counting in the field or laboratory is called raw data. Raw data often needs to be processed into a form that makes it easier to identify its important features (e.g. trends) and make meaningful comparisons between samples or treatments. Basic calculations, such as totals (the sum of all data values for a variable), are commonly used to compare treatments. Some common methods of processing data include creating tally charts, and calculating percentages and rates. These are explained below.

Tally Chart

Percentages

Rates

Records the number of times a value occurs in a data set

Expressed as a fraction of 100

Expressed as a measure per unit time

0-0.99 0-0.99 1-1.99 1-1.99 2-2.99 2-2.99 3-3.99 3-3.99 4-4.99 4-4.99 5-5.99 5-5.99

TOTAL TOTAL

IIIIII IIIIIIII II IIIIIIII IIIIIIII IIIIIIII IIIIIIII IIII IIIIII IIII

33 66 10 10 1212 33 22

• A useful first step in analysis; a neatly constructed tally chart doubles as a simple histogram.

• Cross out each value on the list as you tally it to prevent double entries.

Example: Height of 6 day old seedlings

Men Men

lean Body Bodymass mass Lean Leanbody body %%lean bodymass mass (kg) (kg) mass mass(kg) (kg) body

Athlete Athlete Lean Lean

70 70 68 68

60 60

56 56

Time Time Cumulative Cumulative Rate Rateof ofsweat sweat (minutes) (minutes) sweat sweatloss loss(mL) (mL) loss loss(mL (mLmin min-1-1))

85.7 85.7

00

00

82.3 82.3

10 10

50 50

55

130 130

88

00

83 Normal Normalweight weight 83 96 Overweight Overweight 96

65 65

78.3 78.3

20 20

62 62

64.6 64.6

30 30

220 220

99

125 125

65 65

52.0 52.0

60 60

560 560

11.3 11.3

Obese Obese

• Percentages express what proportion of data fall into any one 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).

Example: Percentage of lean body mass in men

1. What is raw data?

• Rates show how a variable changes over a standard time period (e.g. one second, one minute, or one hour). • Rates allow meaningful comparison of data that may have been recorded over different time periods.

Example: Rate of sweat loss during exercise in cyclists

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HEIGHT(cm) (cm) TALLY HEIGHT TALLY

2. Why is it useful to process raw data and express it differently, e.g. as a rate or a percentage?

3. Identify the best data transformation in each of the following examples:

(a) Comparing harvest (in kg) of different grain crops from a farm:

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Fractions and Ratios

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Key Idea: Fractions and ratios are widely used in biology and are often used to provide a meaningful

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comparison of sample data where the sample sizes are different.

Fractions

Ratios

ffFractions express how many parts of a whole are

ffRatios give the relative amount of two or more

present.

ffFractions are expressed as two numbers separated by a solidus (/). For example 1/2.

ffThe top number is the numerator. The bottom

number is the denominator. The denominator can not be zero.

Simplifying fractions

ffFractions 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.

ffTo simplify a fraction, the numerator and denominator are divided by the highest common number that divides into both numbers equally.

ffFor example, in a class of 20 students, five had blue eyes. This fraction is 5/20. To simplify this fraction 5 and 20 are divided by the highest common factor (5).

quantities (it shows how much of one thing there is relative to another).

ffRatios provide an easy way to identify patterns. ffRatios do not require units.

ffRatios are usually expressed as a : b.

ffIn the example below, there are 3 blue squares and 1 grey square. The ratio would be written as 3:1.

Calculating ratios

ffRatios are calculated by dividing all the values by the smallest number.

ffRatios are often used in Mendelian genetics to

calculate phenotype (appearance) ratios. Some examples for pea plants are given below.

5 ÷ 5 = 1 and 20 ÷ 5 = 4

ffThe simplified fraction is 1/4.

882 inflated pods

299 constricted pods

To obtain the ratio divide both numbers by 299. 299 ÷ 299 =1 882 ÷ 299 = 2.95 The ratio = 2.95 : 1

Adding fractions

ffTo add fractions the denominators must be the same. If

the denominators are the same the numerators are simply added. E.g. 5/12 + 3/12 = 8/12

ffWhen the denominators are different one (or both)

fractions must be multiplied to give a common denominator, e.g. 4/10 + 1/2. By multiplying 1/2 by 5 the fraction becomes 5/10. The fractions can now be added together (4/10 + 5/10 = 9/10).

495 round yellow

(a) 3/9 :

(b) 84/90:

Cell cycle stage

No. of cells counted

Interphase

140

No. of cells calculated

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(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.

2. Simplify the following fractions:

55 158 round green wrinkled green

For the example above of pea seed shape and colour, all of the values were divided by 55. The ratio obtained was: 9 : 2.8 : 2.9 : 1

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):

152 wrinkled yellow

Prophase

70

Telophase

15

Metaphase

10

Anaphase

5

Total

240

4800

(c) 11/121:

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?

DATA

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Dealing with Large Numbers

Key Idea: Large scale changes in numerical data can be made more manageable by transforming the

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data using logarithms or plotting the data on log-log or log-linear (semi-log) paper.

ffIn biology, numerical data indicating scale can often decrease or increase exponentially. Examples include the exponential growth of populations, exponential decay of radioisotopes, and the pH scale.

ffExponential changes in numbers are defined by a function. A function is simply a rule that allows us to calculate an output for any given input. Exponential functions are common in biology and may involve very large numbers.

ffLog transformations of exponential numbers can make them easier to handle. Exponential function ffExponential growth occurs at an increasingly rapid

rate in proportion to the growing total number or size.

Log transformations

ff 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.

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

ffThe equation for an exponential function is y = cx. ffExponential growth and decay (reduction) are possible.

ff Both log10 (common logs) and loge (natural logs or ln) are commonly used.

ff 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.

ffExponential changes in numbers are easy to identify

because the curve has a J-shape appearance due to its increasing steepness over time.

ffAn example of exponential growth is the growth of a microbial population in an unlimiting, optimal growth environment.

ff To find the log10 of a number, e.g. 32, using a calculator, key in log 32 = . The answer should be 1.51.

ff Alternatively, the untransformed data can be plotted

300,000

1,000,000

Population number

Population number

200,000

150,000

100,000

50,000

Smaller numbers can't be read off the graph

800 400 600 Time (minutes) Example: Cell growth in a yeast culture where growth is not limited by lack of nutrients or build up of toxins. 0

200

1. Why is it useful to plot exponential growth using semi-log paper?

Large numbers are easily accommodated

100,000

10,000

1000

100 10 1

Smaller numbers are easily read off the graph

0

200

400 600 800 Time (minutes) Example: The same yeast cell growth plotted on a loglinear scale. The y axis present 6 exponential cycles

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Further increase is not easily accommodated

250,000

0

directly on a log-linear scale (as below). This is not difficult. You just need to remember that the log axis runs in exponential cycles. The paper makes the log for you.

2. What would you do to show yeast exponential growth as a straight line plot on normal graph paper?

3. Log transformations are often used when a value of interest ranges over several orders of magnitude. Can you think of another example of data from the natural world where the data collected might show this behaviour?

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Apparatus and Measurement

Key Idea: The apparatus used in experimental work must be appropriate for the experiment or

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analysis and it must be used correctly to eliminate experimental errors.

Selecting the correct equipment

Recognising potential sources of error

It is important 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.

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.

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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.

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:

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Types of Data

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Key Idea: Data is information collected during an investigation. Data may be quantitative, qualitative,

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or ranked.

Data is information collected during an investigation. Data may be quantitative, qualitative, or ranked. When planning a biological investigation, it is important to consider the type of data that will be collected. It is best to collect quantitative or numerical data, because it is easier to analyse it objectively (without bias).

Types of data

Quantitative

Qualitative

Ranked

Characteristics for which measurements or counts can be made, e.g. height, weight, number.

Non-numerical and descriptive, e.g. sex, colour, presence or absence of a feature, viability (dead/alive).

Data which can be ranked on a scale that represents an order, e.g. abundance (abundant, common, rare); colour (dark, medium, pale).

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

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

Discontinuous

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

Discontinuous or discrete data 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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e.g. Number of children in a family (3, 0, 4)

Continuous

A: Skin colour

B: Eggs per nest

C: Tree trunk diameter

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

(a) Skin colour:

(c) Tree trunk diameter:

(b) Number of eggs per nest:

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

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

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10 Variables and Controls

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Key Idea: Variables may be dependent, independent, or controlled. A control in an experiment

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allows you to determine the effect of the independent variable.

A variable is a factor that can be changed during an experiment (e.g. temperature). Investigations often look at how changing one variable affects another.

Dependent variable

Types of variables

Dependent variable • Measured during the investigation. • Recorded on the y axis of the graph.

Variables can be classed as:

• Independent • Dependent • Controlled Only one variable should be changed at a time. Any changes seen are a result of the manipulated variable.

Independent variable

Remember! The dependent variable is 'dependent' on the independent variable.

Example 1: When heating water, the water temperature (dependent variable) depends on the time it is heated for (independent variable).

Example 2: Reaction rate (dependent variable) depends on temperature (independent variable). In this case, temperature is the independent variable.

Controlled variable

• Factors that are kept the same.

Independent variable

• Set by the experimenter, it is the variable that is changed. • Recorded on the graph's x axis.

Test plant (nutrient added)

Experimental controls

ffA control is the standard or reference treatment in an experiment. Controls make sure that the results of an experiment are due to the variable being tested (e.g. nutrient level) and not due to another factor (e.g. equipment not working correctly).

ffA control is identical to the original experiment

except it lacks the altered variable. The control undergoes the same preparation, experimental conditions, observations, measurements, and analysis as the test group.

ffIf the control works as expected, it means the

experiment has run correctly, and the results are due to the effect of the variable being tested.

Control plant (no nutrient added)

An experiment was designed to test the effect of a nutrient on plant growth. The control plant had no nutrient added to it. Its growth sets the baseline for the experiment. Any growth in the test plant above that seen in the control plant is due to the presence of the nutrient.

2. Why do we control the variables we are not investigating?

3. What is the purpose of the experimental control?

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1. What is the difference between a dependent variable and an independent variable?

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11 A Case Study: Catalase Activity Key Idea: A simple experiment to test a hypothesis involves manipulating one variable (the

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independent variable) and recording the response.

Investigation: catalase activity

Catalase is an enzyme that converts hydrogen peroxide (H2O2) to oxygen and water. An experiment investigated the effect of temperature on the rate of the catalase reaction.

• 10 cm3 test tubes were used for the reactions, each tube contained 0.5 cm3 of catalase enzyme and 4 cm3 of H2O2. • Reaction rates were measured at four temperatures (10°C, 20°C, 30°C, 60°C). • For each temperature, there were two reaction tubes (e.g. tubes 1 and 2 were both kept at 10°C). • The height of oxygen bubbles present after one minute of reaction was used as a measure of the reaction rate. A faster reaction rate produced more bubbles than a slower reaction rate. • The entire experiment, was repeated on two separate days.

H2O2 (l)

Catalase

H2O (l)

+ O2 (g)

10°C

20°C

30°C

60°C

Tubes 1&2

Tubes 3&4

Tubes 5&6

Tubes 7&8

30°C

Height of oxygen bubbles

4 cm3 H2O2 + 0.5 cm3 catalase enzyme

Tubes 9 & 10 No enzyme

1. Write a suitable aim for this experiment:

2. Write an 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) List the equipment needed to set the independent variable, and describe how it was used:

4. (a) What is the dependent variable in this experiment?

(b) Name the unit for the dependent variable:

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(c) List the equipment needed to measure the dependent variable, and describe how it was used:

5. Which tubes are the control for this experiment? © 2017 BIOZONE International ISBN: 978-1-927309-60-5 Photocopying Prohibited

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12 Recording Results Key Idea: Accurately recording results makes it easier to understand and analyse your data later. A

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table is a good way to record data but dataloggers will also record it automatically.

Ways to record data

Recording your results accurately is very important in any type of scientific investigation. If you have recorded your results accurately and in an organised way, it makes analysing and understanding your data easier. Log books and dataloggers are two methods by which data can be recorded. Log books

Dataloggers

A log book records your ideas and results through your scientific investigation. It also provides proof that you have carried out the work.

A datalogger (also called a data recorder) is an electronic device that automatically records data over time.

• An A4 lined exercise book is a good choice for a log book. It gives enough space to write ideas and record results and provides space to paste in photos or extra material (such as printouts).

• Dataloggers have a variety of sensors to measure different physical properties. Common sensors include light, temperature, pH, conductivity, and humidity.

• Each entry must have the date recorded.

• Dataloggers can be used in both field or laboratory experiments, and can be left to collect data without the experimenter being present.

• Make sure that you can read what you write at a later date. A log book entry is meaningless if it is incomplete or cannot be read.

Day 5

• Information collected by the datalogger can be downloaded to a computer (below) so that the data can be accessed and analysed.

Number of leaves at different urea concentrations

Trial

[urea] 3 x 10-2 3 x 10-3 3 x 10-4 3 x 10-5 1

10

20

17

19

2

8

18

13

15

3

10

15

17

17

If you are not using a datalogger, a table (above) is often a good way to record and present your results as you collect them. Tables can also be useful for showing calculated values (such as rates and means). Recording data in a table as your experiment proceeds lets you identify any trends early on and change experimental conditions if necessary.

2. Why must log book entries be well organised?

3. (a) What is a datalogger?

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1. Why is it important to accurately record your results?

(b) What are the advantages of using a datalogger over manually recording results?

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13 Practising Data Manipulations Key Idea: Percentages, rates, and frequencies are common data manipulations. They are useful for

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summarising results and can make it easier to make comparisons across treatments.

1. Complete the transformations for each of

the tables on the right. The first value, and their working, is provided for each example.

Incidence of red and white clover in different areas Frost free area

124 ÷ 159 = 0.78 = 78%

Totals

Number

%

Number

Red

124

78

26

White

35

Total

159

(a) TABLE: Incidence of red clover in different areas: Working:

Frost prone area

Clover plant type

This is the number of red clover out of the total.

%

115

Plant water loss using a bubble potometer

(b) TABLE: Plant water loss using a bubble potometer: Working:

Time (min)

Pipette arm reading (cm3)

Plant water loss (cm3 min-1)

0

9.0

5

8.0

0.20

10

7.2

15

6.2

20

4.9

(9.0 – 8.0) ÷ 5 min = 0.2

This is the distance the bubble moved over the first 5 minutes. Note that there is no data entry possible for the first reading (0 min) because no difference can be calculated.

Frequency of size classes in a sample of eels

(c) TABLE: Frequency of size classes in a sample of eels: Working:

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

(7 ÷ 270) x 100 = 2.6 %

This is the number of individuals out of the total that appear in the size class 0-50 mm. The relative frequency is rounded to one decimal place.

Body mass composition in women

(d) TABLE: Body composition in women: Working:

(38 ÷ 50) x 100 = 76 %

This is lean body mass. The percentage lean body mass is calculated by dividing lean body mass by total body mass. It is multiplied by 100 to convert it into a percentage.

Women

Body mass (kg)

Lean body mass (kg)

% lean body mass

Athlete

50

38

76

56

41

65

46

80

48

95

52

Lean

Normal weight Overweight Obese

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14 Constructing Tables

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Key Idea: Tables are used to record and summarise data. Tables allow relationships and trends in

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data to be more easily recognised.

ffTables are used to record data during an investigation. Your log book should present neatly tabulated data (right).

ffTables allow a large amount of information to be condensed, and can provide a summary of the results.

ffPresenting data in tables allows you to organise your data in a way that allows you to more easily see the relationships and trends.

ffColumns can be provided to display the results of any data

transformations such as rates. Basic descriptive statistics (such as mean or standard deviation) may also be included.

ffComplex data sets tend to be graphed rather than tabulated.

Features of tables

Tables should have an accurate, descriptive title. Number tables consecutively through a report.

Control values should be placed at the beginning of the table.

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

Treatment

Sample size

Mean rate of internode growth (mm day–1)

Mean internode length (mm)

Mean mass of tissue added (g day–1)

Control

50

0.60

32.3

0.36

Hormone 1

46

1.52

41.6

0.51

Hormone 2

98

0.82

38.4

0.56

Hormone 3

85

2.06

50.2

0.68

Each row should show a different experimental treatment, organism, sampling site etc.

Columns for comparison should be placed alongside each other. Show values only to the level of significance allowable by your measuring technique.

1. What are two advantages of using a table format for data presentation? (a)

(b)

2. Why might you tabulate data before you presented it in a graph?

KNOW

Organise the columns so that each category of like numbers or attributes is listed vertically.

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Independent variable in the left column.

Heading and subheadings identify each set of data and show units of measurement.

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15 Drawing Graphs Key Idea: Graphs are useful for visually displaying numerical data, trends, and relationships between

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variables. Different types of graph are appropriate for different types of data.

ffPresenting graphs properly demands attention to a few basic details, including correct orientation and labelling of the axes, and accurate plotting of points.

ffScatter plots and line graphs are appropriate where data are continuous for both variables. Bar charts are appropriate where data are categorical for one variable. For frequency distributions, histograms are appropriate.

ffFor continuous data with calculated means, points can be connected. On scatter plots, a line of best fit is often drawn.

Guidelines for line graphs

Growth rate in peas at different temperatures

Line graphs are used when one variable (the independent variable) affects another, the dependent variable. Important features include:

ffThe data must be continuous for both variables. The

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

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

ffLine graphs may be drawn with measure of error (right). The data are presented as points (which are calculated means), with error bars above and below, indicating the variability in the data (e.g. standard deviation).

Mean growth rate (mm day-1)

1.6

Bars indicate the scatter of data either side of the mean.

1.2

0.8

0.4 0

10

8

12

14

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20

22

24

Temperature (°C)

Guidelines for scatter graphs

A scatter graph is used to display continuous data where there are two interdependent variables.

Body length vs brood size in Daphnia

ffThe data must be continuous for both variables.

40

Number of eggs in brood

ffThere is no independent variable, but the variables are

often correlated, i.e. they vary together in a predictable way.

ffUseful to determine the relationship between two variables. ffThe 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 computer generated with a value assigned to the goodness of the fit).

ffObvious outliers (points that lie well outside most of the scatter) are usually disregarded from analyses.

Line of best fit

30 20

Outlier

10 0

0

Interpolation: For both line and scatter graphs, the fitted line can be used to find an unknown value inside the set of data points. This is called interpolation.

2

1

3

4

Body length (mm)

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1. Determine what type of graph is appropriate for each of the following examples: (a) Arm span vs height in humans:

(b) Daily energy requirement for different species of canids:

(c) Number of fish of each size in a population:

(d) Volume of water used per person per day in different New Zealand cites:

(e) Mean catalase reaction rate at different temperatures:

(f) Number of eggs per brood in different breeds of chickens:

(g) Mean monthly rainfall vs mean monthly temperature:

2. (a) Use interpolation to determine the mean growth rate of pea seedlings at 17°C:

(b) Use interpolation to determine the number of eggs per brood in a 1.5 mm long Daphnia: © 2017 BIOZONE International ISBN: 978-1-927309-60-5 Photocopying Prohibited

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Guidelines for histograms

30

Histograms are suitable where one variable is continuous and the other is a frequency (counts). These plots produce a frequency distribution, because the y-axis shows the number of times a measurement or value was obtained. Important features of histograms include:

Mass of shrimp in a population

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25

ffData for one variable are discontinuous, non-

numerical categories (e.g. place, colour, species), so the bars do not touch.

ffData values may be entered by the bars if you wish. ffMultiple sets of data can be displayed side by side to compare (e.g. males and females in the same age group).

ffAxes may be reversed so that the categories are on

45-49.9

40-44.9

35-39.9

30-34.9

25-29.9

Mass (g)

Guidelines for bar graphs

Bar graphs are appropriate for data that are non-numerical and discrete (categorical) for one variable. There are no dependent or independent variables. Important features include:

20-24.9

0

y-axis usually records the number of individuals in each class interval.

15-19.9

ffThe x-axis usually records the class interval. The

5

10-14.9

weight) so the bars touch.

10

5-9.9

ffThe data are numerical and continuous (e.g. height or

15

0-4.9

Frequency

20

Size of wetlands associated with Waipa lakes

Mangakaware

1.6

Mangahia

9.2

Turnbull Pond

2.7

Ruatuna

3.8

Rotomanuka

10

Cameron

the x axis, i.e. the bars can be vertical or horizontal. When they are vertical, these graphs are called column graphs.

1.54

0

2

4

6

8

10

Area of associated wetland (ha)

12

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

(a) Choose an appropriate graph type and plot the results (right) on the grid below: (b) Estimate the rate of reaction at 15°C:

Lab notebook

Amylase activity at different temperatures

Temperature (°C)

10

Rate of reaction (mg end product per minute)

1.0

20 30

2.1 3.2

35

3.7

40

4.1

45

3.7

50 60

2.7 0

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16 Interpreting Line Graphs Key Idea: The equation for a straight line is y = mx + c. A line may have a positive, negative, or zero

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slope.

The equation for a linear (straight) line on a graph is y = mx + c. The equation can be used to calculate the gradient (slope) of a straight line and tells us about the relationship between x and y (how fast y is changing relative to x). For a straight line, the rate of change of y relative to x is always constant.

Measuring gradients and intercepts y

The equation for a straight line is written as:

y = mx + c

6

Where :

5

y = the y-axis value

4

m = the slope (or gradient)

The intercept (c) on a graph is where the line crosses the y-axis.

5

3

x = the x-axis value

c = the y intercept (where the line cross the y-axis).

2

1

11

Determining "m" and "c"

1

To find "c" just find where the line crosses the y-axis. To find m:

2

3

4

5

6

7

8

9

10

11

x

For the example above:

1. Choose any two points on the line.

c=1

2. Draw a right-angled triangle between the two points on the line.

m = 0.45 (5 á11)

3. Use the scale on each axis to find the triangle's vertical length and horizontal length.

Once c and m have been determined you can choose any value for x and find the corresponding value for y.

4. Calculate the gradient of the line using the following equation: change in y

For example, when x = 9, the equation would be: y = 9 x 0.45 + 1 y = 5.05

change in x

A line may have a positive, negative, or zero slope

a

b

c

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

0

0

1

2

3

4

5

Positive gradients: the line slopes upward to the right (y is increasing as x increases).

1. For the graph (right):

(a) Identify the value of c:

(b) Calculate the value of m:

(c) Determine y if x = 2:

(d) Describe the slope of the line:

0

0

1

2

3

4

Negative gradients: the line slopes downward to the right (y is decreasing as x increases).

5

0

1

2

3

5

Zero gradients: the line is horizontal (y does not change as x increases).

3 2

1

1

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17 Correlation or Causation Key Idea: A correlation is a mutual relationship or association between two or more variables. A

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correlation between two variables does not imply that one causes change in the other.

Correlation does not imply causation

Drawing the line of best fit

ffYou may come across the phrase "correlation does

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

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.

ffYour line should follow the trend of the data points.

ffRoughly half of your data points should be

ffExample: When data from the organic food

above the line of best fit, and half below.

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

25,000

300

Autism Organic food sales

20,000

200

15,000 10,000

100

5000 0

1998 2000 2002 2004 2006 Year

2008

0

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

ffA line of best fit should pivot around the point

representing the mean of the x and y variables.

Number of individuals diagnosed with autism x1000

Organic food sales ($ millions)

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.

Too steep Good fit

Too shallow

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).

(b) Describe the results:

300

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

250

200

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

Foot length (mm)

Hand span vs foot length in adults

350

150

200

250

Hand span (mm)

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18 Investigating Plant Growth Key Idea: Systematic recording and analysis of results can help draw conclusions about a biological

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response in an experiment.

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

The aim

To investigate the effect of a nitrogen fertiliser on plant growth.

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 century. Crop yields increased and now an estimated 50% of crop yield is attributable to the use of fertiliser. Nitrogen is a very important for plant growth, and several types of nitrogen fertiliser are manufactured to supply it, e.g. urea.

Radishes

Experimental method

Radish seeds were planted in separate identical pots (5 x 5 x 10 cm) and grown together in room conditions. The

seeds were planted into a standard soil mixture, and divided into six groups, each with 5 sample plants, a total of 30 plants in six treatments. The radishes were watered every day at 10 am and 3 pm with 500 ml per treatment per watering.

Water soluble fertiliser was mixed and added to the 10 am watering on the 1st, 11th and 21st days. The fertiliser concentrations 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 from the pots, washed, and the radish root weighed. Results were tabulated (below):

Tables should have an accurate, descriptive title. Number tables consecutively through the report.

Heading and subheadings identify each set of data and show units of measurement.

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

Independent variable in the left column.

Mass of radish root (g)†

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

0.18

128.3

127.3

127.7

126.8

0.24

23.6

140.3

139.6

137.9

0.30

122.3

121.1

122.6

121.3

4

Total mass

408.5

Mean mass

81.7

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Sample (n)

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

Fertiliser concentration (g L-1)

117.2

DNG* 141.1 123.1

* DNG: Did not germinate

Control values (if present) should be placed at the beginning of the table.

Values should be shown only to the level of significance allowable by your measuring technique.

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Organise the columns so that each category of like numbers or attributes is listed vertically.

Each row should show a different experimental treatment, organism, sampling site etc.

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The students decided to collect more data by recording the number of leaves on each radish plant at day 30:

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Table 2: Number of leaves on radish plant under six different fertiliser concentrations. Number of leaves at day 30

Fertiliser concentration (g L-1)

1

2

3

4

5

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

Sample (n)

Mean

Median

Mode

* DNG: Did not germinate

Calculating simple statistics for a data set

Descriptive statistics can be used to summarise data. The mean, median, and mode can highlight trends or patterns in the data. The mean can be used to compare different groups. You can use more complex statistics to determine if the means of different groups are significantly different. Statistic

Definition and use

Method of calculation

• The average of all data entries. • Measure of central tendency for normally distributed data.

• Add up all the data entries. • Divide by the total number of data entries.

Median

• The middle value when data entries are placed in rank order. • A good measure of central tendency for skewed distributions.

• 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.

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

• Identify the category with the highest number of data entries using a tally chart or a bar graph.

• The difference between the smallest and largest data values. • Gives an indication of data spread.

• Identify the smallest and largest values and find the difference between them.

Mode

Range

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

2. Identify the dependent variable for the experiment:

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

In some situations, a simple arithmetic mean is not appropriate. • DO NOT calculate a mean from values that are already means (averages) themselves.

• 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

When NOT to calculate a mean:

4. (a) One of the radishes recorded in Table 1 did not grow as expected. Record the outlying value here:

(b) Why should this value not be included in future calculations?

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6. Use the grid below to draw a line graph of the experimental results. Use your calculated means from Table 1. Remember to include a title and correctly labelled axes.

7. The students decided to collect more data by counting the number of leaves on each radish plant at day 30. The data are presented in Table 2 (opposite). Use the space below to calculate the mean, median and mode for the number of leaves at each fertiliser concentration. Record the values on table 2.

8. Which concentration of fertiliser appeared to produce the best growth results for leaves?

9. Describe some sources of error for the experiment:

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10. Write a conclusion for the experiment:

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

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19 Designing a Practical Investigation

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Key Idea: If designed well and executed carefully, a well researched practical investigation will provide

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reliable, valid data to test a hypothesis.

This activity will help you to design and undertake a practical investigation of your own. You will use the background information provided and your own scientific knowledge to design an experiment to investigate the effect of light colour (wavelength) on the rate of photosynthesis in an aquatic weed, Cabomba aquatica. Formulate your hypothesis and then plan a scientifically sound experiment to test it. Think carefully how you will collect and analyse your data so your results are meaningful and allow you to make valid conclusions about your findings. You can use this as a template to design your own experiment.

Background information

Action spectrum for photosynthesis

►► Cabomba aquatica (above) is an aquatic plant native to South America, but is commonly found in aquaria all over the world.

►► Cabomba is very easy to grow and maintain. Cabomba is a highly invasive species.

►► Cabomba will produce a stream of oxygen bubbles from its cut stem when illuminated. The oxygen is a waste product of photosynthesis.

►► Stems are snipped just before the experiment is performed to allow the oxygen to escape freely.

►► The rate of oxygen production provides an approximation of photosynthetic rate.

Rate of photosynthesis (as % of rate at 670 nm)

100

(A measure of the effectivenes of different wavelengths in powering photosynthesis)

80 60 40 20 0

400

500

600

Wavelength (nm) ►► There are two categories of photosynthetic pigments: Chlorophylls absorb red and blue-violet light. Carotenoids absorb blue-violet light.

700

►► The effectiveness of different wavelengths in powering photosynthesis can be determined by measuring the rate of oxygen production.

►► The equation for photosynthesis is given below.

6CO2 + 12H2O

Light

C6H12O6 + 6O2 + 6H2O

Timer

Sodium bicarbonate (CO2 supply)

Red, green, and blue cellophane

Boiling tubes

Cabomba

Scissors LINK

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Equipment list

Lamp with 60 W bulb

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Based on the aim of this experiment "To investigate the effect of light colour on photosynthesis rate in Cabomba", the background information provided, your own knowledge about photosynthesis, and the equipment list provided, set up your own experiment using the questions below to guide you. Remember you want to design a fair test where only one variable is changed and all other variables are kept constant.

1. Carry out background research before you begin your experimental design so you have good knowledge of the topic. In the space below note down any information sources you have used. You will need to acknowledge these in your poster.

2. State an aim and a hypothesis for your experiment:

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3. In the space below summarise your method as step by step instructions and draw your experimental set up:


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4. (a) Identify any safety issues associated with your experiment and note the steps required to reduce the risk (if any):

(b) Identify any ethical issues associated with your experiment:

5. (a) What data will you collect?

(b) How often will you collect it?

(c) In the space below draw a template to record your data:

6. (a) Identify any sources of error and limitations in the methods and/or results:

(b) What changes would you make to improve the experiment?

7. Write your conclusions here:

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Analysing the biological validity of information

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Achievement Standard

2.2

Key terms

Socio-scientific issues are controversial social issues relating to science and they are not always reported to the public in a way that is accurate or unbiased. Evaluating the biological validity of information involves applying biological knowledge and recognising inaccuracies.

accurate analyse

Achievement criteria and explanatory notes

biased

Achievement criteria for achieved, merit, and excellence

biological validity

A

c

peer review report

validity

c

Analyse the biological validity of information presented to the public: Involves: Recognising and describing biological features in the information and identifying them as accurate, inaccurate or biased using biological knowledge. Recognising inaccuracies may be demonstrated by making corrections to inaccurate biological features.

Identifying the purpose of the information (e.g. who produced it and the intended audience).

M

Analyse in-depth the biological validity of information presented to the public: Involves giving reasons how or why:

vested interest

c

each biological feature is inaccurate, or contains bias,

inaccuracies and/or bias may have consequences or impacts for the public,

vested interest is conveyed in the information.

E

Comprehensively analyse the biological validity of information presented to the public: Involves:

prioritising, with reasons, aspects of the information in relation to their significance in context,

evaluating the overall impact of the article on the public, based on bias and the balance of accurate and inaccurate features.

Explanatory notes

Biological validity refers to scientifically accurate information that is used in an unbiased way to convey a biological idea.

c

What you need to know for this Achievement Standard Activities 20 - 24

By the end of this chapter you should be able to:

c

Analyse information presented to the public, selecting articles from at least three different genres, e.g. advertisements, documentaries, newspaper articles, historical accounts, videos.

c

Analyse the articles for biological validity. Identify whether the biological information is accurate, inaccurate, or biased. Use the case studies of water fluoridation and vaccination.

c

Demonstrate errors in biological facts by applying your own biological knowledge (i.e. correct mistakes within the article).

c

Present a report analysing the biological validity of the information presented in the articles chosen. Your articles must be submitted as evidence of processing. Your report should:

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inaccurate

Explain the purpose of the articles and identify the biological features that are accurate, inaccurate, or biased.

Identify who produced the article, the intended audience, and if it has been peer reviewed. Decide if the authors have a vested interest and how it is presented to the intended audience.

Explain why or how inaccurate information or bias could influence the way the public view the issue and affect the decisions they make (e.g. about vaccination or fluoridation).

Evaluate the overall impact of the information on the public, based on its accuracy and bias.

Correctly reference all sources of information.


20 Recognising Balanced Reporting

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Key Idea: Balanced reporting provides unbiased information where both the positive and negative

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aspects are presented without a particular emphasis on either.

ffBiological information is presented to the public constantly via print and broadcast media. Some is provided by

government organisations (e.g. Ministry of Health), some is compiled and presented by science reporters, and some is provided by individuals or non-governmental organisations with an interest, but not necessarily expertise, in a topic.

ffUnlike 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).

ffIt 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:

ffIs there more than one side or view to this issue? ffAre all the views presented?

ffHave any compromises been made to reach an outcome?

ffWhat information is presented to the public and is it scientifically correct?

ffIs some information more important than other

information? If so, how is importance assigned?

ffWhat are the consequences to the public if:

• The information presented is poor science? • The information presented is good science?

Student analysis

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In order to recognise and validate biological information presented to you, you need to:

ffIdentify 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?

ffRecognise whether the information is balanced or biased and if it is accurate or inaccurate. You can determine this by using your own biological knowledge.

ffDiscuss the significance of the biological information (e.g. the HPV vaccine may reduce rates of cervical cancer). • What is its overall impact on the public? • Will biased, inaccurate, or accurate articles influence the overall public perception?

ffUnderstand that biological validity means the material presented is based on sound biological principles, and the results are logically derived.

ffBe able to reference information correctly (by giving the title, publisher or journal, date of publication, and authors).

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21 The Misuse of Scientific Data Key Idea: Presenting biased data can influence (negatively or positively) the public's perception about

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a specific topic.

ffScientific 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.

ffSometimes investigations are made that deliberately intend to collect

data to support a particular point of view, 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:

Tooth decay

• 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.

ffDeliberate misuse of scientific data often involves people or groups

Dental fluorosis

Dozenist

using selected parts of a report and matching them to an incomplete knowledge of a particular scientific concept.

ffSome information concerning the artificial fluoridation of water 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-fluoridation statement. The right is a more balanced, but pro-fluoridation statement. • Sodium fluoride and hydrofluorosilicic acid are obtained as by-products of the manufacture of superphosphate fertiliser.

• Fluoride in water is adjusted to 0.7 - 1 mg L-1. At this level there are no verified effects of fluoride on health. Above these levels health effects can be mild to severe as the dose increases. There are many studies that show the benefits of fluoride in water at this level, including reduced tooth decay and cavities.

• Fluoridation first began in the New Zealand in 1954. NZ has natural low levels of fluoride in water.

• In parts of the USA, China, and India natural fluoride in the water is very high and it is removed from the water. Studies from Assam in India where natural fluoride levels can reach 23 mg L-1 have reported effects such as anaemia, stiff joints, painful movement, mottled teeth, and kidney failure.

Pro-fluoridation

Fluoride for water fluoridation is made using almost the same process as fertiliser, using highly toxic and dangerous chemicals. The highly toxic gas hydrogen fluoride is released during fertiliser production, captured and reacted to make hydrofluorosilicic acid or sodium fluoride, which is then dumped into the water supply. Fluoride is poisonous and fluoridation of drinking water can have serious consequences. It can actually damage our teeth, rather than help them. In parts of China and the USA people have suffered pitted teeth, hip fractures, kidney failure and many other health effects. Fluoridation is a hangover from the 1950s, when DDT was considered safe and lead was added to petrol without worry about the long term health effects. Adding fluoride to water is not necessary, as many communities do not add fluoride to water and people in those communities have perfectly healthy teeth.

Water is fluoridated using sodium fluoride and hydrofluorosilicic acid. The concentration added to drinking water is carefully monitored to ensure it stays within the recommended 0.7 to 1.0 parts per million (or 0.7 to 1.0 parts fluoride to 999,999 parts water). At this level there is no scientific evidence for any kind of health issue. The only effect is a possible and mild discolouration of the teeth. In New Zealand natural fluoride levels in water are too low to provide protection against cavities. A large amount of evidence shows those communities with fluoridated water experience less tooth decay than those with un-fluoridated water. Claims that fluoridation of water can lead to cancers and other health issues are often based on evidence from parts of the USA, China, and India where naturally high fluoride levels in the water require it to be removed in order to make the water safe to drink.

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Anti-fluoridation

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22 HPV

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Key Idea: The human papillomavirus (HPV) can cause several cancers, including cervical cancer.

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Cervical cancer is the third most common cancer in New Zealand women aged 22-44.

ffAchievement Standard 2.2 in Biology requires you to analyse

the biological validity of information presented to the public. You must analyse articles and material selected from at least three different genres and, using your own biological knowledge, determine if the information contained within them is accurate, inaccurate, or biased.

HPV general information • HPV stands for human papillomavirus. • There are over 100 strains of HPV. • Both men and women can be infected with HPV. • HPV is easily transmitted through sexual intercourse and can also be contracted by skin to skin contact. • Over 80% of sexually active males and females will become infected with HPV. • HPV can cause a number of cancers including cervical and anal cancer. It also can cause genital warts.

HPV and cervical cancer • Twenty HPV strains cause cancers in the genital region. • Two strain of HPV, 16 and 18, cause 70% of cases of cervical cancer. • Over 99% of patients with cervical cancer also have an HPV infection. • HPV can cause pre-cancerous cervical lesions. If undetected, these can develop into cervical cancer over time (10-30 years).

New Zealand statistics • Each year, 160 New Zealand women are diagnosed with cervical cancer, and 50 will die from the disease. • Cervical cancer is the third most common cancer in New Zealand women aged 25-44. • Regular cervical screening can reduce the risk of cervical cancer by 90%. • The HPV vaccine was added to the New Zealand immunisation schedule in 2008 for females only. In 2017 the immunisation schedule was altered so both males and females aged 9-26 years received the vaccine.

Essential vocabulary Antigen: A molecule that, when it enters the body, triggers the immune system to produce specific antibodies.

Antibody: A large Y-shaped protein produced by the immune system to identify and neutralise antigens, such as the foreign proteins of viruses. Pathogen: A disease-causing agent, e.g. bacterium or virus.

Carcinogen: A cancer-causing agent, e.g. ultra-violet light, some viruses. Immune response: The response of the body to a pathogen. It includes the production of antibodies.

Ed Uthman

The HPV virus

TEM of the human papillomavirus

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advertisements, documentaries, and historical accounts. The following pages present biological information about the Human Papillomavirus (HPV), and the HPV vaccine, Gardasil.

The image above shows the lower half of the uterus and the cervix. The cervix itself has been virtually destroyed by cancer (arrow).

1. The HPV vaccine was originally only given to females. Suggest why the immunisation schedule has been updated to include males:

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NIH

ffSuitable genres include newspaper articles, videos,

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23 The HPV Vaccine Key Idea: A vaccine protects against a specific pathogenic disease. Gardasil is a vaccine developed

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to protect against the human papillomavirus.

ffA vaccine is a preparation of biological material that

The HPV vaccine ffIn New Zealand the Gardasil 9 vaccine is

is deliberately introduced into the body to produce an immune response.

used to protect against HPV.

ffA vaccine contains or resembles the components of

a specific pathogen (disease causing organism). The pathogen is foreign to the body and the immune system detects this and produces antibodies to destroy it. The immune system remembers its response and will respond in the same way if presented with the same foreign material in the future. This immunological memory forms the basis of vaccination programmes. The vaccine's similarity to the specific infectious agent is used to provide future protection against disease.

ffGardasil 9 provides protection against HPV for at least five years.*

ffHPV vaccines are most effective when given to people before they become sexually active.

ffThe vaccine protects against new HPV infections, but provides no benefit to people already infected.

ffGardasil has been approved for use as a vaccine

against HPV by the FDA (Food and Drug Administration). Vaccination is seen as a way to reduce HPV infection and reduce the incidence of HPV-related cancers (including cervical and anal cancer) and genital warts.

Types of vaccine

ffGardasil 9 is 95-100% effective at

preventing pre-cancerous lesions caused by HPV infection.

* Length of protection studies are still being carried out

Subunit vaccines Contain some antigenic part or product of microorganisms. This can be purified from the microbe or (more commonly) synthesised using genetic engineering (these vaccines are called recombinant).

Considerations

Safety: Subunit vaccines are safer than whole agent vaccines because they cannot reproduce in the recipient.

Side effects: Subunit vaccines produce fewer adverse effects than whole-agent vaccines because they contain no extra pathogenic material. HPV Risk: HPV types 16 and 18 are high risk carcinogens. They account for 70% of cervical cancer cases.

Whole-agent vaccines Contain whole, non-virulent microorganisms, which may be weakened or dead.

HPV types 6 and 11 are low risk carcinogens, but they account for 90% of cases of genital warts.

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9-valent means effective against nine strains of HPV.

Gardasil 9 protects against nine strains of HPV, including strains 16 and 18 which cause cervical cancer and types 6 and 11 which cause genital warts.

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CDC

Girl receiving the HPV vaccine

Gardasil is a recombinant subunit vaccine. It contains a recombinant viral protein, but no genetic or other material from HPV itself.

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32 ffThe article below is typical of the kind found in newspapers. It contains a large amount of information. Different

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newspapers can have different ways of reporting, with some being for and others being against particular issues. Each article must be read carefully to determine if there is any bias in the reporting. Using the information on the previous pages practice analysing the newspaper article below for accuracy, bias, and relevance of information.

HPV Vaccine - Allergic Reactions Rare The Local Tribune

By Jill Jones: Wednesday 3 September 2008

The human papillomavirus (HPV) immunisation programme was made available free of charge to 17 and 18 year-old women this month. The vaccine, Gardasil, targets four strains of HPV, a common sexually transmitted infection that causes genital warts and most cases of cervical cancer.

The Ministry of Health say the vaccine is safe and that allergic reactions to it are rare. This is despite recent Australian research which found that young women who received the HPV vaccine were five to 20 times more likely to have severe allergic reaction than girls who received other vaccines. In 2007, a team of Australian researchers at a Sydney hospital studied 114,000 young women vaccinated with Gardasil as part of a New South Wales vaccination programme. Eight young women had anaphylactic reactions to the vaccine. Anaphylactic reactions can cause breathing difficulty, nausea, and rashes, and can be potentially life-threatening if untreated.

They estimated the reaction rate to Gardasil to be 2.6 per 100,000 doses administered. In comparison, a 2003 schoolbased meningitis vaccination programme had a reaction rate of 0.1 per 100,000 doses.

The Sydney researchers said the severe allergic reactions to the HPV vaccine were unusual and manageable, and the vaccine was safe.

The newspaper, and author name of the following article are fictitious, but the text is based on real events and information.

The Ministry of Health deputy director of public health said the vaccine had been tested and analysed by Medsafe. "The vaccine was shown to have an excellent safety profile during large clinical trials in which more than 20,000 people from 30 countries took part." Medsafe is New Zealand's medicines and medical devices safety authority.

Medsafe said it was very rare for people to have serious anaphylactic reactions to vaccinations. The most commonly observed reactions to the vaccine were pain, redness and swelling at the injection site. Less common were fever, vomiting and fainting. HPV was licensed in New Zealand in July 2006. To date there have been no cases of anaphylaxis in New Zealand, but 10 adverse events had been reported - these included whole body tiredness, vomiting, diarrhoea and arm pain.

The vaccine was licensed for use in more than 100 countries. Reactions to the vaccine had been carefully reviewed, the deputy director of public health said. About 160 women are diagnosed with cervical cancer in New Zealand each year and 60 die from it. The immunisation programme was projected to save more than 30 lives a year.

Health providers informed parents and women who were receiving the vaccination about the possible reactions. The Australian researchers said while there were severe allergic reactions to the vaccine, it should not discourage its use.

The Internet as a source of information about Gardasil

Many articles, videos, and websites about Gardasil can be found on the internet by simply typing "Gardasil vaccine" into your search engine. Remember the internet is not regulated and different websites may be pushing their own agenda.

Go to these YouTube sites to see videos about how Gardasil has affected two teenage girls in America. Ashley Ryburn's life ruined by the HPV (Gardasil) vaccine:

http://www.youtube.com/watch?v=6UHy-EkK2xo&feature= PlayList&p=D6117E4A6699E3C3&playnext=1&playnext_ from=PL&index=22

Parents Concerned About Gardasil Vaccine

http://www.youtube.com/watch?v=jdyNAuy6HHo&feature= PlayList&p=D38C30E5401733FD&playnext=1&playnext_ from=PL&index=8

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The Ministry of Health provides information on HPV and Gardasil at http://www.health.govt.nz/our-work/ preventative-health-wellness/immunisation/hpvimmunisation-programme

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24 Presenting Your Findings

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Key Idea: A report should present balanced or unbiased information that is accurate and relevant.

Once you have analysed the examples and information in the previous activity, use the space on this page to plan a report on the use of the Gardasil vaccine or fluoridation. Use bullet points to capture the main points and your ideas. These will provide the scaffold for your full report. The list below provides key points that should be included.

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• Use your own biological, knowledge to decide whether the biological features of the article are accurate, inaccurate, or biased. Identify inaccurate biological information by correcting errors in the article. • What is the purpose of the information and who produced it? • Is the article showing vested interest (the promotion of a particular view for some advantage)? If so, explain why, or how vested interest is presented. • How does inaccurate or biased information influence the way the public perceive the topic? • Identify and prioritise how the information presented could influence a person's decision about the issue. Give reasons for your choices. • Evaluate the overall impact of the article on the public. The evaluation should be based on whether the article is biased or balanced, and upon its biological accuracy.

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Achievement Standard 2.3

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Demonstrate understanding of adaptation of plants or animals to their way of life

Students who are attempting Achievement Standard 2.3 must demonstrate their understanding of the adaptations of plants or animals to their way of life over either:

ffOne life process over three functional or taxonomic groups, or

ffTwo related life processes over one functional or taxonomic group.

This cover page describes the organisation of the next two chapters covering this standard. Information is given on selected life processes for selected groups of animals and plants as explained below.

Although this is an internal standard, we have included essay style questions at the end of each section to help consolidate and test your understanding of the systems you are studying.

Adaptations of animals to their way of life Three life processes in animals are covered:

ffNutrition

ffGas exchange

ffInternal transport

The three taxonomic groups covered are: mammals, fish, and insects. The three functional groups are: carnivores, omnivores, and herbivores.

ffThe contexts for covering one life process over three

functional groups could include comparing the adaptations for nutrition of carnivores, omnivores, and herbivores.

ffThe contexts for covering one life process over

three taxonomic groups could include comparing the adaptations for nutrition of insects, fish, and mammals.

ffThe contexts for covering two related life processes over

one taxonomic or functional group could include gas exchange and internal transport in mammals or digestion and internal transport in mammals.

Adaptations of plants to their way of life Three life processes in plants are covered:

ffTranspiration

ffGas exchange

The taxonomic groups are gymnosperms, angiosperms, and ferns.

The functional groups are xerophytes, mesophytes, hydrophytes, and halophytes.

ffThe contexts for covering one life process over

three taxonomic groups could include comparing the adaptations for reproduction in gymnosperms, angiosperms, and ferns.

ffThe contexts for covering one life process over

three functional groups could include comparing the adaptations for transpiration in (three of) xerophytes, hydrophytes, halophytes, and mesophytes.

ffThe contexts for covering two related life processes

over one taxonomic or functional group could be gas exchange and transpiration in angiosperms.

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ffReproduction


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Adaptations of animals to their way of life

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Achievement Standard

2.3

Key terms

Animals possess structural, physiological, and behavioural adaptations that enable them to carry out the life processes essential to survival in their habitat and exploitation of their niche.

Common terms adaptation functional group taxonomic group

Gas exchange alveoli breathing bronchi bronchioles carbon dioxide cellular respiration countercurrent flow expiration gas exchange system gills haemoglobin inspiration lungs oxygen respiratory gas spiracles tracheae ventilation Internal transport artery (pl. arteries) atrium (pl. atria) blood bulk flow capillary closed circulatory system double circulatory system haemolymph heart open circulatory system respiratory pigment single circulatory system vein ventricle

Achievement criteria for achieved, merit, and excellence

c

A

Demonstrate understanding of adaptation of animals to their way of life: Describe the adaptations and identify aspects of the adaptations that enable each organism to carry out its life process(es) in order to survive in its habitat.

c

M

Demonstrate in-depth understanding of adaptation of animals to their way of life: Provide a biological reason that explains how or why the adaptations enable each organism to carry out its life process(es) in order to survive in its habitat.

c

E

Demonstrate comprehensive understanding of adaptation of animals to their way of life: Link biological ideas the adaptations that enable each organism to carry out its life process(es) in order to survive in its habitat. The discussion may involve justifying, relating, evaluating, comparing and contrasting, and analysing and it must include consideration of the two points from below appropriate to your chosen context. Option 1: Adaptation related to one life process over three taxonomic or functional groups:

c

i

Compare diversity of adaptation in response to the same demand (e.g. gas exchange) across different taxonomic or functional groups.

ii

Explain limitations and advantages involved in each feature within each organism.

Option 2: Adaptation across two related life processes within one taxonomic or functional group:

c

i

Make connections between two life processes within each organism that enhance the effectiveness of both processes.

ii

Explain limitations and advantages involved in each feature.

EII

Explanatory notes

Understanding of adaptation to demonstrate in relation to...

c

1a

One life process over three taxonomic or functional groups of animals, or

c

1b

Two related life processes within one taxonomic or functional group

c

2 •

The ways in which an organism carries out all its life processes, including: relationships with other organisms: competition, predation, mutualism, parasitism, adaptations to the physical environment.

3

Life processes (for animals) selected from:

Activity number

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Nutrition absorption bulk feeding dentition deposit feeding digestion egestion enzyme egestion filter feeding gut ingestion nutrients

Achievement criteria and explanatory notes

25 - 51 74

c

i

nutrition (adaptations for obtaining and processing nutrients)

c

ii

gas exchange (adaptations for exchanging respiratory gases with the environment) 52 - 63 73

c

iii

internal transport (adaptations for moving materials around the body)

64 - 74


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What you need to know for this Achievement Standard Adaptations for nutrition Activities 25 - 51, 74

By the end of this section you should be able to:

EII

Luc Viatour www.Lucnix.be

c

Describe different modes of feeding in in insects, fish, and mammals and the adaptations associated with these (e.g. adaptations of the dentition or mouthparts).

c

Describe the four stages involved in processing food in most animals (ingestion, digestion, absorption, and egestion). Identify where these stages occur in a generalised tube gut.

c

Explain the role of enzymes in digesting food in insects, fish, and mammals. Explain the relationship between the digestive enzymes present in an animal's gut and the animal's diet.

c

Describe similarities and differences in the basic structure of the gut in insects, fish, and mammals. Describe and explain specific adaptations of the gut in chosen examples.

c

Describe adaptations of guts in three functional groups of mammals (e.g. omnivores, carnivores, and herbivores). Relate the regional specialisations of the gut to diet.

c

Compare and contrast adaptations for cellulose digestion in an insect and a mammal. In each case, describe the nature of the mutualistic relationship and the benefits to each party.

c

Describe and explain adaptations for increasing the rate of nutrient absorption in insects, fish, and mammals, such as a human (e.g. gastric caeca, spiral valve, intestinal villi).

c

Describe how different nutrients are absorbed. Distinguish between absorption and assimilation. Explain what assimilated nutrients are used for.

c

Explain the relationship between the digestive system and the internal transport system in mammals, e.g. human, with reference to the role of these interacting systems.

Adaptations for gas exchange Activities 52 - 63, 73

EII

By the end of this section you should be able to:

c

Distinguish between cellular respiration and gas exchange and explain why organisms need to exchange respiratory gases with their environment.

c

Describe how gases are exchanged across gas exchange surfaces. Describe the essential features of gas exchange surfaces and their significance in terms of gas exchange rates.

Describe structural and functional diversity in the gas exchange systems of insect, fish, and mammals, relating specific features to suitability in the environment in each case. Insects: tracheae and spiracles. How is the gas exchange surface ventilated?

Fish: gills and the role of countercurrent exchange. How is the gas exchange surface ventilated? Mammals: lungs and associated structures. How is the gas exchange surface ventilated?

c

Compare and contrast gas transport in insects, fish, and mammals, including reference to the role of respiratory pigments if present.

c

Explain the relationship between the gas exchange and circulatory systems in mammals.

Adaptations for internal transport By the end of this section you should be able to:

EII

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Activities 64 - 77

c

Relate the surface area: volume relationship in different animal taxa to the presence or absence of an internal transport system.

c

Describe the components and functions of a transport system in animals, including the role of blood or haemolymph. Explain the role of respiratory pigments in some taxa.

c

Explain diversity in the structure and function of blood vessels in vertebrates (e.g. fish and mammals), including arteries, capillaries, and veins.

c

Compare and contrast open and closed circulatory systems, and single and double closed circulatory systems. Explain how the type of circulatory system and the efficiency of transport is related to metabolic needs and environment.

c

Compare and contrast the structure and function of hearts in fish and mammals: three chambers in series (fish) and four chambers separating systemic and pulmonary circuits (mammal). Describe the adaptations of the heart and how functional efficiency is achieved in your examples.

c

Explain the relationship between the gas exchange and internal transport systems in mammals.


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25 Obtaining Food Key Idea: All animals are heterotrophs (they feed on other organisms). They display a wide range of

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feeding modes and adaptations to exploit different diets.

All animals are heterotrophs, meaning they feed on other organisms (dead or alive). They display a wide range of feeding modes and show structural and behavioural adaptations related to these feeding modes. Animals may feed on solid or fluid food and may suck, bite, lap, or swallow it whole.

Bulk feeding

Bulk feeding is the most common mode of feeding, especially in large animals. It involves feeding on large food masses which are ingested whole or in pieces after chewing. Bulk feeders may be carnivores, herbivores, or omnivores. Adaptations (inherited specialisations) for bulk feeding include specialised mouthparts for cutting, tearing, or chewing (above and right).

Filter feeding

Filter feeders remove food that is suspended in the water using specialised filtering structures. Filter feeding can be an extremely effective way of feeding. The world's largest animals, the baleen whales, are filter feeders, as is the whale shark, the world's largest fish. There are numerous different adaptations for filter feeding including the comb-like plates (baleen) suspended from the upper jaw of baleen whales and gill rakers in fish.

Deposit feeding

Fluid feeding

diveofficer CC 2.0

Vampire bats

Deposit feeders sift the substrate (e.g. mud, silt) and remove the food particles it contains. Deposit feeding works best in areas of fertile sediment such as rich soils or the mud of estuaries or river beds. Many invertebrates, but relatively few insects, are deposit feeders. Fish such as mullet, catfish, and carp suck up mud and ingest the plant and animal material in it, expelling the inorganic material back out their mouths.

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Fluid feeding is an uncommon form of feeding. However, a number of insect species (e.g. flies, moths, butterflies, aphids), and some fish and mammals are fluid feeders. Fluid feeding involves obtaining nutrients by sucking or lapping the fluids of another organism. These fluids include blood, plant sap, or nectar. Adaptations for fluid feeding include piercing mouthparts (e.g. mosquitoes) or small sharp teeth to remove skin (e.g. vampire bats).

Striped catfish feeding

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1. Use lines to match each of the following modes of feeding with its correct description and example: (a) Filter feeding

Obtaining nutrients from particles suspended in water

Mosquito

(b) Deposit feeding

Obtaining nutrients by eating whole organisms

Stoat

Obtaining nutrients by ingesting only the fluids of another organism

Catfish

Obtaining nutrients by sifting through or ingesting sediment or detritus

Baleen whale

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(c) Bulk feeding (d) Fluid feeding

2. Describe one structural adaptation for obtaining food in each of the following:

(a) A blood sucking mosquito:

(b) A filter feeding whale:

(c) A mammalian predatory carnivore:

(d) A leaf chewing caterpillar:

(e) A vampire bat:

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3. Compare and contrast the types of adaptations for feeding in a bulk feeder, a filter feeder, and a fluid feeder, including examples to illustrate your answer from insects, fish, and mammals:

4. Carp are a deposit feeding fish and are an introduced species in New Zealand. They are responsible for the deterioration of water quality in some lakes and rivers. Based on their method of feeding, suggest why carp have such a negative effect on the environment:

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26 Parasitism Key Idea: Parasitism is the most common type of symbiosis and a specialised way to obtain nutrients.

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Parasites harm their host, but do not kill it.

ffA symbiosis is any form of very close

The emerald cockroach wasp: a small parasitoid with a strategy to subdue its host species.

ecological relationship between two or more species.

ffParasitism is a type of symbiosis in

1 The wasp stings and paralyses the front legs of the cockroach to stop it moving too quickly, then stings the brain, disabling the escape reflex.

which one organism (the parasite) benefits at the expense of the other (the host).

4

ffParasites always harm their host,

The larva pupates inside the now dead host, emerging as fully grown wasp to begins its adult life.

Photo: Radiogaga

but do not usually kill it. The parasite benefits by obtaining nutrition, but there are often other advantages, such as protection.

ffAnimal parasites are highly specialised carnivores, feeding off the body fluids or skin of host species.

Cockroach

ffParasitoids are adapted to spend only

part of their life cycle as a parasite, killing and often consuming the host before leaving as a free-living (nonparasitic) adult. There are many parasitoid species amongst the insects.

Male anglerfish

Animals are not always parasites of other species. Male deep sea anglerfish seek out the much larger females and attach to them by biting and holding on to their skin. They then slowly regress as they merge with the female, becoming little more than a sperm producing appendage.

Sea lampreys attached to host fish

USGS

Vampire bats are sometimes called the only mammalian parasite. They have many features of parasites: they are specialised for feeding on blood (haematophagous) and do not kill their host. However, they do not live on the host, but return to a roost when they have finished feeding.

3 Once Inside the burrow, the wasp lays an egg on the cockroach's abdomen. The larva hatches and lives inside the roach, eating its organs in an order that ensures the host remains alive long enough for the larva to complete development.

NOAA

Vampire bat (Desmodus rotundus)

2 The wasp leads the cockroach to its burrow by tugging on the cockroach's antennae.

Lampreys are jawless fishes that feed by attaching to the flanks of a fish with their rubbery sucker mouthparts. They scrape away the skin with their teeth and then suck the body fluids from the host. Lampreys can swim, but do not need to while attached to the host.

1. (a) Explain how a parasitoid differs from a true parasite:

(b) Using the emerald cockroach wasp as an example, explain the dependence a parasitoid has with its host:

2. Why are animals parasites regarded as highly specialised carnivores?

3. Do you regard the vampire bat as a true parasite? Justify you answer:

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27 Predation Key Idea: A predator is any animal that hunts, captures, and consumes prey. Predators have

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adaptations that help they capture prey.

ffThe adaptations of predators for prey capture and consumption include those associated with sensing and locating prey, capturing and subduing it, and then consuming and digesting the flesh or fluids. New Zealand short tailed bat

DoC: J. Kendrick

Larva in silk nest

Echolocation ('sonar') enables bats to hunt their insect prey while flying at night. Sonar allows the flying bat to locate the exact position of its prey even when both prey and predator are in flight. New Zealand short tailed bats (above) also hunt on the ground amongst the leaf litter but only use sonar for location of prey in the air.

Mnolf cc3.0

Lures

Concealment and ambush are common strategies amongst insect predators. Mantids (above) are well camouflaged against the vegetation in their environment. Unsuspecting prey are ambushed and captured with a rapid movement of the highly specialised, elongated forelegs.

New Zealand glowworm larva

NOAA

Traps and lures allow predators to wait in ambush for their prey. New Zealand glowworms or titiwai are the larvae of a tiny fly. The larva spins a silk nest with many mucus-covered threads hanging down, extending the nest as it grows. It produces a bioluminescent glow to attract prey to the threads and when the prey is trapped in the mucus the larva pulls up the line and eats the prey.

Anglerfish also use lures to attract prey. The lure is a fleshy growth adapted from the dorsal fin and extending over the head. It can be moved independently of the fish and a reflex closes the jaw on the prey as soon as the lure is touched.

Praying mantis

Common dolphins

Speed, agility, and cooperation enable dolphins (left) to hunt together to exploit large schools of fish. The dolphin group will encircle a large school of fish, driving them together into a dense mass or bait ball. The dolphins then take turns to charge through the school to feed.

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1. (a) Describe the role of camouflage in the predatory strategy of ambush predators such as mantids:

(b) Describe some other adaptations commonly associated with prey capture in ambush predators:

2. Describe the adaptations of glowworm larvae for successful prey capture:

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28 Insect Mouthparts Key Idea: Insect mouthparts all consist of the labrum and three sets of modified, paired appendages

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that are adapted for chewing, lapping, and piercing and sucking food.

ffInsect mouthparts are variously adapted to capture, manipulate, or chew food. In some cases, this has involved loss or fusion of some of the paired appendages.

ffIn insects with chewing mouthparts (right and below, top left), the labrum forms an upper lip and helps pull food into the mouth. The mandibles form the first pair of mouthparts and move from side to side, rather than up and down. They are used as jaws to chew, cut, and tear food, but may also be used to carry things, fight, or to mould wax. The maxillae form the second pair of mouthparts and are used for food sensing and handling. The labium is a single structure formed from a fused pair of mouthparts. It acts as a lower lip to close the mouth. Both the maxillae and the labium may have finger-like extensions called palps.

This green locust shows how the mouthparts fit neatly together.

Antenna

Compound eye

Labrum upper lip

Mandible (jaws)

Labial palp

Maxillary palp

Maxilla

Rostrum (Mandibles absent)

Labrum

Grasshopper Chewing

Stylets (composed of mandibles and maxillae) lie in the groove of the heavier labium.

Mandible

Maxillary palp

Proboscis

Modified maxilla

Labium (forms tongue)

Housefly Sponging

Labium (lower lip)

Shield bug Piercing/ sucking

Honey bee (worker) Chewing/sucking

Butterfly Sucking

Proboscis is composed of the labrum, mandibles, and maxillae. Together they form tubes for saliva and sucking in fluids.

Labium (provides a protective sheath for the stylet and does not penetrate.

Labial palp

Maxillary proboscis

Proboscis sheath (labium)

Mosquito Piercing/sucking

A

B

C

D

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1. Match the following list of insects with their correct image (A-F) and their correct mode of feeding: butterfly, mosquito, locust, beetle (larva), aphid, housefly.

E

F

(a) Sponging: Letter(s):

Name:

(c) Sucking:

Letter(s):

Name:

(b) Chewing: Letter(s):

Name:

(d) Piercing/sucking: Letter(s):

Name:

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2. Describe the components of an insectâ&#x20AC;&#x2122;s mouthparts:

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3. Discuss the adaptability of insect mouthparts to different modes of feeding:

(d)

(f)

(i)

(mouthparts only)

(m)

(j)

(a)

(q)

(e)

(g)

(b)

(h)

(c)

(k)

(mouthparts only)

Insect A

Insect B

Insect C

(l)

Insect D

(p)

All photos: EII

(o)

(n)

Insect E

4. For each of the photographs of insects above (A - E), identify the type of insect and the structures that are labelled (a)(q). Note that some of the labelled structures are not mouthparts:

Identity of insect A:

(a)

(b)

(c)

(e)

(f)

(h)

(i)

(k)

(k)

(m)

(n)

(o)

(p)

(q)

Identity of insect B:

(d)

Identity of insect C:

(g)

Identity of insect D:

(j)

Identity of insect E:

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5. Many insects undergo metamorphosis at certain stages in their life cycle. Butterflies start their active life as caterpillars, after which they pass through a pupal stage, to finally emerge as butterflies. Comment on the diets and changes to the mouthparts of caterpillars and their adult forms (butterflies):

(a) Caterpillar diet:

Mouthparts:

(b) Butterfly diet:

Mouthparts:

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29 The Teeth of Fish Key Idea: Fish may have teeth or teeth-like structures on the roof of the mouth, the tongue, and the

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gill arches. Most are homodonts.

Most fish are homodonts, meaning all their teeth are identical, although as a group fish show a variety of teeth types depending on their diet.

ffJawless fish, such as lampreys and hagfish, cannot

chew and instead latch onto a host using an oral disc. They can then bore through the skin to suck fluids from the host.

Manta ray

ffFilter feeders, such as manta rays and whale sharks, often lack teeth and use sieve plates in the mouth to sieve plankton and small fish from the water.

ffPredatory fish, e.g. sharks and piranhas, have teeth

that grip and slice. Few are able to chew and instead bite off chunks of flesh that are swallowed whole. In many cases the teeth are curved inwards towards the mouth to help grip prey.

Piranha

Photo: Drow Male

A shark's upper jaw is not directly connected to the skull. This allows it to push forward as the lower jaw swings open, giving a wider bite than would otherwise be possible.

The teeth of predatory sharks are triangular and serrated at the edges to slice through flesh.

Unlike other fish, shark teeth are not attached to the skeleton and are continually lost and replaced.

The premaxillary, maxillary and mandible of many bony fish are connected so that as the fish opens its mouth, a large cavity is formed that sucks in prey.

Maxillary

Premaxillary

Mandible

Butterfly fish use narrow pincer like jaws to pluck polyps from coral.

Parrotfish have a horny bill, which they use to bite off chunks of coral. It then grinds these up with hard, flat teeth.

Lamprey

1. Identify the fish in this activity with the following types of teeth and jaws:

(a) Narrow pincer like jaws:

(b) Sieve plates:

(c) Serrated, triangular teeth:

(d) No jaws, teeth on an oral disk:

2. Why do the teeth of predatory fish often curve backwards?

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30 The Teeth of Mammals Key Idea: The teeth of mammals are variously specialised for biting, holding, tearing, and chewing a

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range of foods.

The oral cavity and teeth

Dental formulae

The mouth (oral cavity) (below), is formed by the cheeks, hard and soft palate, and tongue. The teeth are very hard structures, specialised for chewing food (mastication). The tongue, which moves food around, and the salivary glands, which produce saliva, act with the teeth to begin digestion.

The oral cavity is divided into quadrants and the number and type of teeth is described by a dental formula. In a human the formula is written as: 2123 2123 indicating the number of teeth in the top and bottom jaw (one side of the mouth) (see left). Numbers indicate incisors: canines: premolars: molars. Other mammals have different formulae:

Adult (permanent) teeth per quadrant

Gingivae (gums)

Incisor: 2

Hard palate

Soft palate

Canine: 1

Uvula

Goat 0033 3133

Pig 3133 3133

Dog 3142 3143

Horse 3143 3143

Palatine tonsil

Premolar: 2

Tongue

Molar: 3

In carnivores, the canine teeth are usually the largest teeth and adapted for holding and tearing, and sometimes used as weapons. Most species have four canine teeth.

1. Describe the purpose of the:

(a) Incisors:

(b) Canine teeth:

(c) Molars:

Molars are adapted to chewing and grinding food. In herbivores they can be very big and are adapted to grind coarse vegetation. In carnivores they are modified for shearing flesh.

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Incisors are adapted for biting, clipping, or nipping food. They are generally large in herbivores and small in carnivores. In elephants the upper incisors develop into the tusks.

2. How does the human tooth pattern compare to the other mammals in this activity?

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31 Mammalian Teeth and Diet Key Idea: Mammals are heterodonts and have different kinds of teeth in their jaws. The diversity of

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dentition amongst the mammals reflects their diet and mode of feeding.

Mammals have a wide range of feeding modes and consequently have a wide range of tooth and jaw structures. Some mammals have relatively generalised dentition, while others are highly specialised, even to the extent of losing teeth entirely (e.g. anteater). Mammals have four different tooth types in their jaw. These are the incisors, canines, premolars and molars. Depending on the diet, the teeth may be highly modified, or lost altogether.

ffCarnivores: The teeth of mammalian carnivores are

Lion

Zebra

specialised to grip prey and slice and shear flesh and bone. The canine teeth are adapted to hold and suffocate prey. The incisors are pointed and the molars modified into carnassials, adapted for shearing rather than chewing.

ffHerbivores: Herbivorous mammals, especially

grazers such as horses, cattle, and sheep, have relatively large incisors for clipping grasses. The canine teeth can be reduced or missing and the molars are large, flat, and adapted for chewing and grinding plant material.

ffOmnivores: The teeth of omnivorous mammals are rather generalised and adapted to biting, tearing, and chewing a variety of food. Human teeth can be considered well adapted to a fully omnivorous diet. The canines in humans are reduced but in other primates they can be large and are important in displays of aggression and in fighting.

Grey whale

Chimpanzee

ffSpecialised diets: A number of mammalian

species have lost their teeth altogether. The jaws of anteaters are fused into a tube through which the tongue flicks in and out. Baleen whales (e.g. blue and humpback whales) have large plates of fibrous baleen hanging from their upper jaw. These are used to filter food (e.g. krill) from a mouthful of water.

1. Describe the diet of each of the mammals in the photographs above. Describe a feature of the dentition and explain how it helps the animal process its diet:

(a) Lion diet: Feature of the dentition:

(b) Zebra diet: Feature of the dentition:

(c) Chimpanzee diet: Feature of the dentition:

(d) Grey whale diet:

Feature of the dentition:

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32 Comparing Mammalian Teeth

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Key Idea: The diet of a mammal can be inferred from the shape and arrangement of its teeth.

The chimpanzee skull shown right is a useful example of generalist dentition. Chimpanzees eat a varied diet and so have a dentition that is adapted to process a wide variety of food.

Canine

Molars

Incisors

Diastema. A gap between the teeth.

Premolars

1. Use the information above and in the activity "The Teeth of Mammals" to answer the questions below.

(a) Identify the type of diet associated with each skull (carnivorous, herbivorous, omnivorous):

(b) Colour in the incisors (yellow), canines (green), premolars (blue) and molars (red). Use the dental formulae in "The Teeth of Mammals" to help you:

Pig

Zebra

Type of diet:

Type of diet:

Type of diet:

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Goat

Lion

Type of diet:

2. Compare the shape of the lion's teeth to those of the horse. Explain why these shapes are so different:

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33 Introduction to Digestion Key Idea: The digestive tract is specialised to maximise the physical and chemical breakdown of

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food (digestion), absorption of nutrients, and elimination of undigested material.

The digestive tract

ffFood provides the source of the energy required to maintain metabolism.

ffThe gut is essentially 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.

Mouth, tongue and teeth

INGESTION Food is taken in through the mouth and swallowed

Salivary glands

Throat (pharynx)

ffUsually, food ingested at the mouth and

pharynx passes through an oesophagus to the stomach (or equivalent).

ffFood is moved through the gut tube by waves of muscular contractions (peristalsis).

ffDuring digestion, food is first physically

broken down by chewing and mixed by the muscular activity of the stomach. It is then broken down in stages by enzymes contained within digestive secretions. This process is called chemical digestion.

ffThe products of chemical digestion are then absorbed across the gut wall. Absorbed molecules can then be assimilated (taken up be the cells). The liver is a large organ associated with the gut but not part of it. Like the pancreas, it contributes digestive secretions but has other non-digestive functions as well.

ffUndigested wastes are egested through an anus (or cloaca in most non-mammals).

Oesophagus

DIGESTION Food is broken down in the mouth, stomach, and small intestine

Liver

Stomach

Gall bladder

ABSORPTION The products of digestion are absorbed across the gut wall

Pancreas

Small intestine

Colon

EGESTION Unwanted material is eliminated by defecation

Rectum

Anus

1. Describe the four stages of food processing in the gut:

(b) How is food chemically broken down?

3. (a) Which organ is associated with the gut, but not physically part of it?

(b) Describe its general role in digestion:

4. (a) In which part of the digestive tract is food absorbed?

(b) What happens to indigestible material?

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2. (a) How is food physically broken down?

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34 Moving Food Through the Gut

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48

Key Idea: Solid food in a tube gut is formed into small lumps that are moved through the gut by waves

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of muscular contraction called peristalsis.

Peristalsis moves food material through the digestive tract

ffFood chewed and mixed with saliva is called a bolus. When swallowed, waves of muscular contractions

(peristalsis) move the bolus along the oesophagus to the stomach. In the stomach, digestive activity reduces the bolus to a slurry called chyme. The chyme enters the small intestine for further digestion.

ffChyme is also moved through the gut by peristalsis. The gut wall has two layers of muscle. The inner circular

muscles contract to narrow the tube, and the outer longitudinal muscles contract to widen and shorten the tube. When one set of muscles contracts, the other relaxes (the muscle are antagonistic).

The process of moving food through the gut by waves of muscular contractions is called peristalsis.

Head this end

Inner circular muscle of the oesophagus

Circular muscles contract behind the plug of food (the bolus)

The bolus enters the stomach, where it is reduced to a slurry. When this material leaves the stomach, peristalsis moves it through the small intestine.

Outer longitudinal muscle of the oesophagus

Bolus movement

Circular muscle

Longitudinal muscle

EII

Bolus

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 as long stands.

1. What is a bolus?

2. What is chyme?

4. In the cross section through an intestine right, label the circular and longitudinal muscle:

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3. Describe how peristalsis moves food through the gut:

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35 The Stomach and Small Intestine Key Idea: The stomach produces acid and a protein-digesting enzyme which breaks down food into

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chyme. The chyme passes to the small intestine which breaks it down further and absorbs it.

In the stomach, food is mixed in an acidic environment to produce a soupy mixture called chyme. The acid in the stomach destroys microbes, denatures proteins, and activates a protein-digesting enzyme precursor. There is very little absorption in the stomach, but small molecules (glucose, aspirin, alcohol) are absorbed directly across the stomach wall into the surrounding blood vessels.

The gall bladder stores bile, an alkaline fluid, which is produced by the liver cells. Fat and acid in the duodenum stimulate release of bile from the gall bladder.

The stomach

Oesophagus

Cardiac sphincter (closes junction between oesophagus and stomach)

Three layered muscular wall mixes the stomach contents to produce chyme. Stretching the stomach wall stimulates gastric secretion.

Bile from liver

Folds (rugae) in the stomach wall allow the stomach to expand to 1 L.

Pyloric sphincter (closes junction between stomach and duodenum)

Stomach

Digestion in the stomach

Pancreas secretes an enzyme-rich alkaline fluid into the duodenum via the pancreatic duct.

Gastric juice

Pancreatic duct

Acts in stomach (optimal pH)

Pepsin (1.5-2.0)

Duodenum (part of small intestine)

Protein â&#x2020;&#x2019; peptides

Detail of a gastric gland (stomach wall)

ffSome cells secrete pepsinogen, a precursor of the enzyme pepsin.

ffSome cells produce hydrochloric acid

(HCl), which activates the pepsinogen.

ffGoblet cells at the neck of the gastric gland secrete mucus to protect the stomach lining from the acid.

ffSome cells secrete a hormone which acts on the stomach to increase gastric secretion.

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Stomach surface

Pepsinogen

Gastric pit

Pepsin

HCl

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The stomach's secretions are produced by gastric glands in the lining of the stomach (gastric means stomach):

Goblet cells

Parietal cell

Chief cell

Hormone secreting cell

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The small intestine ffThe small intestine receives the chyme directly from the stomach. It is divided into three regions, which are

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distinguished by the cell types present: the duodenum, where most chemical digestion occurs, and then the jejunum and the ileum. Most absorption occurs in the jejunum and ileum.

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ffThe intestinal lining is folded into many intestinal villi, which project into the gut lumen (the space enclosed by the gut). The villi increase the surface area for nutrient absorption. The epithelial cells that make up the lining of each villus in turn have a brush-border of many microvilli, which are primarily responsible for nutrient absorption. The membrane of the microvilli is packed with enzymes that break down food molecules for absorption.

ffEnzymes bound to the microvilli of the epithelial cells, and in the pancreatic and intestinal juices, break down fats,

peptides, and carbohydrates (see tables below). The small molecules produced by this digestion are then absorbed into the underlying blood and lymph vessels.

ffTubular exocrine glands and goblet cells secrete alkaline fluid and mucus into the lumen, neutralising the acidity of the chyme entering the small intestine from the stomach and protecting the lining of the intestine from damage. Epithelial cells

Capillaries surround a central lymph vessel.

Epithelial cells migrate toward the tip of the villus to replace lost and worn cells.

Secretion of alkaline fluid and mucus into the lumen. The mucus is protective and the alkaline fluid provides the correct pH environment for the digestive enzymes.

Nutrients are transported away

Photographs below: The intestinal villi are shown projecting into the gut lumen in a scanning electron micrograph (left image) and in a light microscope image (centre image). The microvilli forming the brush border of a single intestinal epithelial cell are shown in the transmission electron micrograph (right image).

TEM

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SEM

Enzymes in the small intestine break down food into small molecules that can be absorbed through the gut wall. Enzymes are present in the pancreatic juice added to the duodenum, in intestinal juice, and bound to the surfaces of the intestinal epithelial cells.

Enzymes in pancreatic juice

Enzymes in duodenum (optimal pH)

Enzymes in intestinal juice (IJ) and epithelium (E)

Enzymes in small intestine (location, optimal pH)

1. Pancreatic amylase (6.7-7.0)

1. Starch → maltose

1. Maltase (E, 6.0-6.5)

1. Maltose → glucose

2. Trypsin* (7.8-8.7)

2. Protein → peptides

2. Peptidases (IJ, E, ~ 8.0)

2. Polypeptides → amino acids

3. Chymotrypsin* (7.8)

3. Protein → peptides

3. Sucrase (E, ~6.0)

3. Sucrose → fructose & glucose

4. Pancreatic lipase (8.0)

4. Fats → fatty acids & glycerol

4. Entropeptidease (IJ 8.0)

4. Activates trypsin*

* secreted in an inactive form

Louisa Howard, Katherine Connollly Dartmouth College

Lumen

*Once activated, trypsin activates chymotrypsin

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1. Summarise the structure and role of each of the following regions of the human digestive tract: (a) Stomach:

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(b) Small intestine:

2. (a) What is the purpose of the hydrochloric acid produced by the parietal cells of the stomach?

(b) Explain why protein-digesting enzymes (e.g. pepsin) are secreted in an inactive form and then activated after release:

3. How does the stomach achieve the mixing of acid and enzymes with food?

4. (a) What is the purpose of the intestinal villi?

(b) What is the purpose of the microvilli (brush border) on intestinal epithelial cells?

(a) Site: Enzyme’s role:

Enzyme:

(b) Site: Enzyme’s role:

Enzyme:

(c) In general, do the enzymes act in acidic or alkaline conditions?

(d) How is this pH environment generated?

6. Suggest why the small intestine is so long:

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5. Identify two sites for enzyme secretion in the small intestine, identify an enzyme produced there, and state its role:


36 The Large Intestine

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Key Idea: The large intestine absorbs water and solidifies the indigestible material before passing it

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out the anus as faeces in a process called egestion.

ffAfter most of the nutrients have been

Colon absorbs water, Na+, and some vitamins. It also incubates bacteria which release B vitamins and vitamin K.

Ascending colon

ffThe appendix is a blind ending sac

Transverse colon

Descending colon

absorbed in the small intestine, the remaining semi-fluid contents pass into the large intestine (caecum (and appendix), colon, and rectum). This mixture includes undigested or indigestible food, 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. It passes the consolidated material to the anus.

Ileum of small intestine

off the caecum. It may have a minor immune function although it is not needed for normal gut function.

See diagram below

ffThe rectum is the final part of the large intestine and stores the waste faecal material before it is discharged out the anus. Fullness in the rectum produces the urge to defaecate. 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.

ffDefaecation is controlled by the anal

sphincters, whose usual state is to be contracted (closing the orifice).

Appendix

Caecum provides a space for the mixing of bacteria with partly digested food from the small intestine to form faeces.

Rectum: The enlarged final segment of the large intestine where faeces are stored and consolidated before elimination.

Anus. The sphincters controlling the anal opening are normally contracted, but relax to allow the expulsion of faeces.

Lining of the large intestine

The lining of the large intestine has a simple epithelium containing tubular glands (crypts) with many mucus-secreting goblet cells. The mucus lubricates the colon wall and helps to form and move the faeces. In the photograph, some of the crypts are in cross section and some are in longitudinal section. Mucus producing goblet cells

Lumen

Goblet cells within crypt Crypt

Connective tissue

Lymph nodule

Circular muscle

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Simple columnar epithelial cells

Note the abundance of pale goblet cells.

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Transverse colon

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Peristaltic movement in the colon

Ascending colon

Descending colon

Some of the muscular contractions in the colon mix the faecal matter.

Faecal matter

Rectum

Up to three times a day, extra strong peristaltic contractions, move faeces from the colon into the rectum (a colonic rush).

During a process called segmentation, alternate sections of muscle contract, moving material back and forth. This causes mixing, but does not propel the material along the colon.

1. Name three structures that make up the large intestine and identify their roles: (a)

(b)

(c)

2. What are the effects of absorbing too little and too much water in the large intestine?

3. What is the purpose of the goblet cells in the colon?

5. (a) Draw arrows on the X-ray image of the colon right to show the direction of movement of material through the colon.

(b) Circle the faecal material in the X-ray.

6. What is the purpose of segmentation in the colon?

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4. How is material moved along the colon?

X-ray of the colon

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37 Summary of the Human Digestive Tract Key Idea: The human digestive tract is similar to other mammals and has several discrete organs

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contributing secretions to assist digestion.

1. In the spaces provided on the diagram below, identify the parts labeled A-L (choose from the word list provided). Match each of the functions described (a)-(g) with the letter representing the corresponding structure on the diagram.

Word list: liver, small intestine, gall bladder, stomach, salivary glands, colon (large intestine), oesophagus, pancreas, mouth and teeth, anus, rectum, appendix.

Feeding and satiety centers in the hypothalamus regulate eating

B

A

Structures of the human gut

B

F F

B

H

C

I

D

J

E

K

F

L

In the boxes provided, write the letter (A-L) that represents the part of the gut responsible for each of the functions summarised below:

E

G

G

The functions of gut structures

C

D

A

(a) Main region for enzymatic digestion & nutrient absorption

H

(b) Consolidation of the faeces before elimination

I

(c) Main function (humans) is water and mineral absorption (d) Secretes acid and pepsin, stores and mixes food

J

K K

(e) A gland that produces an alkaline, enzyme-rich fluid

L

(f) Produces bile and has many homeostatic functions

(g) Produces saliva which contains the enzyme amylase

2. Identify the regions of the gut illustrated (a)-(c) below. For each photograph, give the identifying features: b

Gastric gland

Villi

c

Lumen

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EII

EII

EII

Bile ducts

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38 Mammalian Guts Key Idea: Mammalian guts show a variety of specialisations to cope with their particular diet.

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Herbivores tend to have longer guts with larger chambers than carnivores.

Small intestine is of medium length.

Omnivore

The specialisations of omnivore guts vary according to the balance of animal and plant material eaten. In all mammals, digestion in the stomach always involves secretion of hydrochloric acid and the protein-digesting enzyme pepsin. No animals produce the cellulase enzymes needed to break down cellulose, but in omnivores, this undigestible cellulose provides bulk in the diet.

Colon is relatively long.

Oesophagus

Stomach capacity in humans is 20-30% of total gut volume.

Caecum: usually poorly developed

Carnivore

Herbivore: ruminant

Vegetation is a low energy diet because most of the energy in plant material is trapped in cellulose, which animals cannot directly digest. In ruminants, e.g. cattle, the stomach is large, with several chambers. One chamber, the rumen, acts as a fermentation chamber, where bacteria break down plant cellulose. Volatile fatty acids released by the microbes provide energy, and digestion of the microbes themselves provide protein.

Herbivore: hindgut digestion In hindgut fermenters, the caecum and colon are expanded for microbial digestion of cellulose. Hindgut fermenters can be divided into caecal fermenters e.g. rabbits, and colonic fermenters e.g. horses. Microbial digestion of cellulose in the hindgut produces volatile fatty acids which are absorbed directly across the colon wall.

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Colon is simple, short, and smooth.

Oesophagus

Small intestine is short and relatively wide.

Stomach capacity 6070% total gut volume. pH 1 or less.

Caecum is poorly developed and may be absent.

Stomach ~70% of total gut volume. pH 5-7.

Oesophagus

Small intestine is long.

Short to medium length caecum.

Colon of medium length.

Small intestine is shorter than in foregut fermenters.

Oesophagus

Stomach is relatively small

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The guts of carnivores, e.g. cats and dogs, are adapted for processing animal flesh, which is a high-nutrient diet. The stomach is often large in order to hold large, infrequent meals. Guts, internal organs, and muscle are all eaten. Regions for microbial fermentation are poorly developed or absent.

Appendix

Caecum is very large

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The colon is very long and pouched.

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1. Use the information on the previous page to complete the table detailing the relative lengths and volumes of the regions of the gut for various mammals: Herbivore: foregut fermenter

Herbivore: hindgut fermenter

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Gut region (length and volume)

Omnivore

Carnivore

Stomach

Small intestine

Caecum

Colon

2. (a) Why is a diet of bulky vegetation of low energy value?

(b) How do many herbivores (specifically grazers and browsers) compensate for the low quality of their diet?

3. Describe the structural differences between the guts of a ruminant and a hindgut fermenter:

(a) What is this nutrient?

(b) Where does it come from and why do hindgut fermenters lose it?

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4. Hindgut fermenters generally lose a nutrient from the hindgut that ruminants easily obtain.

5. Why are chambers for bacterial fermentation of food absent from a carnivore's gut?

6. Why is the stomach volume of a carnivore often very large compared to the rest of the gut?

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39 Comparison Between Ruminants and Carnivores

57

Key Idea: The specific adaptations of the guts and teeth of ruminants and carnivores enable them to

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maximise the energy and nutrient intake from their food.

Ruminant herbivore: Cattle (Bos taurus)

Carnivore: Lion (Panthera leo)

Small temporalis muscle

Large temporalis muscle provides most of the biting force.

Horny pad

Large masseter muscle helps chewing action.

Premolars

Molars are large for grinding

Omasum removes water

Flow of food in the stomach

The teeth and guts of carnivores are adapted for eating animal flesh. The canine and incisor teeth are specialised to bite down and cut, while the carnassials (modified molars/premolars) are enlarged, lengthened, and positioned to act as shears to slice through flesh. Meat is easier to digest than plant material and is higher in energy and food value, so the guts of carnivores are comparatively more uniform and shorter than those of herbivores.

Incisors cut vegetation

Canine

Diastema (toothless space)

Regurgitation, rechewing, and reswallowing

Abomasum is the glandular stomach and secretes gastric juices.

Passage of food

Canines

Incisors

In carnivores, the masseter muscle assists in stabilising the jaw. Chewing is less important.

Premolars

Carnassials (lower is hidden behind upper)

Powerful acidic gastric juices containing pepsin enzyme digest protein. This gastric phase of digestion is similar in all vertebrates.

Oesophagus

Passage of food

Reticulum forms the cud which is returned to the mouth for rechewing.

Rumen contains bacteria to break down cellulose and ciliates to digest starch.

To the duodenum

Strong muscular movements provide physical mixing.

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Dental adaptations

Ruminants depend on a mutualistic relationship with their rumen microorganisms, which digest plant cellulose and provide the ruminant with energy and protein. In return, the rumen provides the microbes with a warm, oxygen-free, and nutrient-rich growing environment. To maximise use of the food eaten, ruminants regurgiatate and rechew the plant material, using their large grinding molars. This increases the surface area for microbial activity.

1. Identify and explain the difference between carnivores and ruminant herbivores with respect to:

(a) The teeth:

(b) The passage of food through the gut:

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40 Comparison of Tube Guts Key Idea: Although the tube guts of many animals have a similar overall plan, specific adaptations

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reflect the way they have solved similar food-processing problems.

Most animals have digestive tubes running between two openings, a mouth and an anus. One-way movement of food allows the gut to become regionally specialised for processing food. Some typical tube guts are illustrated.

Omnivorous mammal Caecum

Liver

ffTube guts follow a general plan, with

Rectum

Stomach

regions for storing, digesting, absorbing, and eliminating the food. Particular specialisations are related to the food type and how it is ingested.

Oesophagus

ffUsually, food ingested at the mouth and

Anus

pharynx passes through an oesophagus to a crop, gizzard or stomach where it is stored and digestion may begin.

ffMost digestion occurs in the intestine

(or midgut in insects). The products of digestion by enzymes are absorbed across the gut wall. Most animals have adaptations to increase the efficiency of nutrient absorption.

Colon

Small intestine

ffUndigested wastes are egested (eliminated) through an anus.

Herbivorous insect

Oesophagus

Salivary glands

Mouth, tongue and teeth

Midgut (stomach) lined with permeable membrane

Herbivorous fish

Malpighian tubules (excretion)

Gall bladder

Intestine

Stomach

Hindgut

Mouth

Rectum

Gizzard

Mouth

Anus

Crop stores food

Gizzard grinds food

Gastric caeca (pouches) of midgut absorb nutrients

Anus

Pyloric caecae off intestines

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1. What is one advantage of having a one-way movement of food through a tube gut?

2. For each of the following regions of a vertebrate gut, name the region in an insect gut with the same functional role and state this role:

(a) Intestine: (b) Stomach:

(c) Teeth:

3. Would you expect the intestine of a carnivorous fish to be longer or shorter than that of a herbivorous fish?

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Recognising organs in a dissection of a rat

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The guts of mammals have a similar basic plan. The dissection of a laboratory rat helps to understand the placement of digestive organs within the body cavity.

Xiphoid cartilage

4

1

5

2

1

3

Kidney

Kidney

2

3

4

Seminal vesicles (male reproductive system)

Mesentery

A: Undisturbed organs in the abdominal cavity

B: Abdominal organs partially dissected

Dissections are often required when studying animal systems. The photographs above show two stages of a rat dissection. Use the information provided on the previous pages to help you identify and label the structures indicated. Most, but not all of these, are digestive organs.

4. In the dissection of the rat (above), label each of the structures indicated in the spaces provided (photo A: 1-4, photo B: 1-5). Some structures are the same, but the same numbers do not necessarily indicate the same structure. List of structures: small intestine, liver, large intestine (colon), ileum of small intestine, duodenum, stomach.

A. 1. 2. 3.

B. 1.

4.

4.

2.

5.

5. (a) Which structures pictured have no role in the digestive system? (b) Which organs are the most prominent on opening the body cavity:

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3.

(c) What can you infer about the role of the mesentery from the photograph on the right?

6. Use the images and information in earlier activities to decide if the rat is a herbivore, carnivore, or omnivore:

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41 Cellulose Digestion in a Ruminant Key Idea: Ruminants rely on a mutualistic relationship with microbes in the rumen to break down the

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cellulose in their diet. They regurgitate and rechew material to maximise digestion.

Grass is a lower energy diet than meat, because mammals lack the enzymes to break the bonds joining the glucose molecules in cellulose. However, many herbivores are able to digest cellulose by making use of mutualistic microorganisms in their gut. Ruminants, such as cattle, goats, and sheep, rely on microbes in the rumen (a large chamber of the stomach) to break down the cellulose. The food is partially digested by the microbes, and is then regurgitated, rechewed, and reswallowed so that the microbes can complete their digestion of the material. The digested slurry then passes to the rest of the gut.

Region of stomach

Functional role

% of total volume

Microbial fermentation of cellulose to produce volatile fatty acids

~84%

Omasum

Removal of water

12%

Abomasum

Gastric digestion

4%

Rumen

Reticulum

The rumen and reticulum act as a single chamber and occupy more than 80% of the volume of the stomach.

3 Regurgitation, rumination (chewing

the cud), and reswallowing. Chewing the cud breaks down plant matter further and stimulates digestion.

Food is eaten quickly and barely chewed. The food is mixed with saliva containing sodium, potassium, phosphate, bicarbonate, and urea, to form a bolus.

Rumen pH 7

4 Once the regurgitated matter has been

Fermentation releases volatile fatty acids (VFAs), as well as methane, hydrogen sulfide, and carbon dioxide. The VFAs are absorbed directly from the rumen and provide up to 70% of the cattle's daily energy needs. The gases are belched out and the remaining matter is regurgitated and rechewed. The VFAs are transported to the liver and converted to glucose.

Rumen contracts rhythmically to mix and move contents in an organised way.

1 The ingested plant material is fermented in the

large rumen by bacteria, which break down the cellulose, and ciliate protozoa, which break down the starch. The microbes use the carbohydrates as well as ammonia and amino acids to grow.

1. What relationship do ruminants form with their bacteria?

2. How does this relationship enable the ruminant to meet its energy needs?

3. What is the purpose of chewing the cud?

rechewed and reswallowed, it bypasses the rumen and passes into the true stomach or abomasum (acid pH) where proteins are digested. The abomasum secretes lysozyme, an enzyme that breaks down bacterial cell walls.

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2

Rumen bacteria are all anaerobes.

4. Suggest why the rumen pH is 7, whereas the abomasum (true stomach) pH is 2?

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42 Cellulose Digestion in a Hindgut Fermenter Key Idea: In hindgut fermenters, breakdown of cellulose by mutualistic microbes occurs in the large

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intestine. Rabbits can reprocess some of this digested material by caecotrophy.

ffUnlike ruminants, hindgut fermenters, such as horses and rabbits, house their cellulose-digesting microorganisms in the hindgut, rather than in a foregut rumen. As in ruminants, the microbes in this mutualistic relationship benefit by obtaining food (carbohydrate) and are provided with a suitable environment for growth and reproduction.

ffRabbits and hares are small active mammals. Like all hindgut fermenters, they require a high throughput of

material, so they eat frequently. Rabbits increase the nutritional gain from their browse by eating soft, nutrient-rich faecal pellets. This is called caecotrophy, and enables the rabbit to meet its requirements for vitamins and protein. Rabbits also produce 'normal' hard pellets which are not eaten. 5 The caecotropes are stored in the fundus of the stomach where digestion of the microbes provides vitamins and protein.

2 On the first pass of food through the gut, fermentation by bacteria in the caecum breaks down the cellulose and releases volatile fatty acids. These VFAs are absorbed and provide up to 40% of the animal's energy requirements.

Oesophagus

1 Grass is eaten

Fundus

Caecum (pH 5.9-6.9) provides a favourable anaerobic environment for the microbes.

Stomach

Second pass (hard faeces will be egested)

6 After a second pass, the now

hard faeces are egested from the anus in the evening when the animal is outside the burrow. These are not eaten.

4 The caecotropes are eaten directly from the

3 The caecum forms the fermented

anus. This behaviour is known as caecotrophy and it is essential to the rabbit's health. The caecotropes are produced and eaten in the morning when the rabbits are in the burrow.

material into soft, high nutrient pellets, called caecotropes.

(a) What does the rabbit gain?

(b) What do the microbes gain?

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1. In the mutualistic relationship between a rabbit and its gut microbes:

2. (a) Not all hindgut fermenters practise caecotrophy. Explain its adaptive advantage to the rabbit:

(b) Predict the consequences to the rabbit if caecotrophy was prevented?

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43 Cellulose Digestion in an Insect

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Key Idea: Cellulose digestion in insects is achieved through a complex symbiosis with both protozoa

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and bacteria.

ffMany insects are herbivores. Many of these feed on leaves where nutriment can be gained by from starch and other molecules in the leaf. Few feed directly on woody material, which is made mostly of cellulose.

ffInsects that feed on cellulose usually do so with the help of bacterial or protozoan symbionts harboured in the

gut (e.g. termites). Some insects, including stick insects, are able to produce cellulase enzymes that break down cellulose, but still rely on bacteria for the most part. It's complicated! A symbiont within a symbiont

Insects that eat wood, such as termites (below), rely on microbial populations to break the chemical bonds between the glucose molecules and provide them with molecules they can use. Chewing mouth parts allow bits of wood to be chewed into pulp.

The very large pouch-like hindgut provides room for the population of flagellated protozoans, which break down the cellulose to sugars that the termite can use.

Termites

Protozoa

Bacteria

Chew and ingest woody material

Engulf wood fibres using phagocytosis

Digest cellulose using cellulase enzymes

Absorb sugars released from protozoan symbionts

Use some sugars and release some to the termite symbiont

Use some sugars and release some to the protozoan symbiont

The gut protozoa feed by engulfing wood fibres by phagocytosis. They do not produce cellulase themselves. They rely on their own mutualistic bacteria to do this.

Excretory tubules (not digestion)

Althepal CC 2.5

Crop

The flagellated protozoans in the termite gut are very mobile. They are more easily kept as gut residents than free bacteria, which would be lost.

1. Describe the features of a termite that allow it to feed on woody material:

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2. Describe how termites derive glucose from wood:

3. Use brief notes to compare and contrast the digestion of cellulose in termites and ruminants:

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44 Digesting Fluids Key Idea: Many insects feed on fluids, including blood, sap and nectar. Adaptations allow high

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volumes of liquids to be stored to extract low concentration nutrients.

ffInsects that feed on fluids must be able to store large volumes during feeding and remove the water from the fluid to concentrate it before it is digested.

Blood (high protein fluid)

Blood is a high protein, low bulk fluid. The problems with processing it include:

Piercing mouthparts inject powerful anticoagulants to keep the blood flowing.

Three sequential regions in the midgut (1) absorb water (2) secrete protease enzymes, and (3) absorb nutrients.

ffCoagulation of blood and blockage of mouthparts during ingestion.

ffStorage of a large quantity of fluid.

Enlarged crop stores the blood, releasing it slowly in smaller amounts into the midgut.

ffSlowing passage of bulk food through the gut so that it can be digested.

Female mosquito

Plant sap and nectar (low protein fluid) Plant sap is a high volume sugary fluid. The problems with processing it include:

The midgut is enlarged and divided into three regions. The first and last parts are greatly coiled and actively remove water.

The middle part of the midgut receives the sap after most of the water has been removed. It secretes carbohydrase enzymes and absorbs nutrients.

ffEliminating large volumes of water and

obtaining sufficient protein and vitamins.

ffStorage of a large quantity of fluid.

ffEnzymes are diluted by the large volumes

Unabsorbed sugars can be passed out of the hindgut in copious amounts as 'honeydew'. This allows bees to process enough food to meet protein requirements.

of fluid ingested.

Bumble bee Other: honeybees, aphids, butterflies

2. (a) Study the graph right. How long after a blood meal does enzyme activity in a mosquito reach its maximum potential?

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1. Describe the problems that fluid feeders face when feeding on fluids and explain how they solve these problems:

(b) Use the graph to help explain the adaptations of a mosquito to a fluid diet:

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Enzyme activity (% of max)

Enzyme activity in mosquito gut

100

50

0

0

2 Days after feeding

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45 Adaptations for Digestion in Fish Key Idea: Some large predatory fish are able to regulate the temperature of their core organs in order

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to increase the rate of digestion and maintain an active lifestyle.

Fish are generally ectothermic, meaning that their body temperature depends on external sources of energy, not metabolism.

ffFor most fish, the temperature of the

body and the environment are the same. However, some large predatory fish, such as tuna and large sharks, can maintain at least some of their body at temperatures above those of the seawater. This is called regional endothermy.

Countercurrent heat exchange

Veins carry metabolically-warmed, deoxygenated blood. Most of the heat is transferred to the arteries.

To heart

26°C

Venous blood loses heat

Heat transfer

Arterial blood is warmed

25°C

ffIncreasing body temperature enables

them to maintain high levels of activity and process greater quantities of food more rapidly.

ffThey maintain these higher temperatures

using a countercurrent network of blood capillaries (right).

17°C

16°C The thermal gradient is maintained because the From gills blood flows in opposite directions. This flow is called countercurrent.

Blood returning to the heart in veins passes through a network of capillaries running the length of the fish. Heat generated by the swimming muscles is transferred from the venous blood to the cold arterial blood passing to the core. This enables the fish to maintain higher core temperatures and minimise heat loss to the environment.

Stomach temperature of bluefin tuna (Thunnus thynnus)

Temperature (oC)

30

20

10

19/8

20/8

21/8

22/8

23/8

Date (day/month)

Surface water temperature Feeding times

The graph above shows the stomach temperature of a bluefin tuna over several days. The tuna was fed semi-frozen food which caused a sudden drop in its stomach temperature. However, the temperature rapidly increased as the food was digested. It was determined that some of the temperature increase came from the hydrolysis (break down) of the food itself, while the rest came from an increase in metabolic activity.

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Tuna stomach temperature

1. Study the graph above. What is the evidence for endothermy in bluefin tuna?

2. What is the advantage of keeping internal organs warm?

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46 Adaptations for Absorption in Insects Key Idea: In insect guts, nutrient absorption can only occur in the midgut. Insects increase the surface

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area for absorption of nutrients with midgut pouches called gastric caeca.

Insect digestive tract

ffIn all animals, survival can depend on

Hindgut lined with impermeable chitin

maximising the nutrient uptake from the food ingested.

Malpighian tubules (excretion)

ffThe rate of nutrient transport across the gut wall depends partly on its surface area. Adaptations for increasing the surface area of the gut are widespread because they increase rates of nutrient absorption so that more food can be processed more quickly.

Foregut Midgut Hindgut

ffIn insects, pouches from the midgut increase the gut surface area.

ffNutrients are absorbed by diffusion or

Secretion of enzymes and absorption of nutrients occurs in the midgut

active transport and are then available for assimilation by the body's cells.

Gastric caeca (midgut pouches) increase surface area of midgut (below)

Insect gastric caeca

In grasshoppers the gastric caeca are midgut pouches just behind the gizzard. The caeca improve absorption by transferring nutrients to the haemolymph ('blood'). The membrane lining the midgut is continually lost, surrounding the faecal pellets when they are passed out. The membrane is replaced by the underlying epithelial cells (like our skin).

Muscles

Dorsal vein

Foregut lined with impermeable chitin

Foregut

Gizzard

Gastric caeca

Midgut

Caeca transfer nutrients to the haemolymph (blood)

A semi-permeable membrane lines the midgut, allowing small molecules to pass through.

Transverse section through locust thorax

1. In which part of the insect gut does absorption occur?

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2. What structure in insects increases the surface area for absorption?

3. What is the adaptive value (the advantage) of increasing the surface area available for nutrient absorption?

4. Explain how insect gastric caeca play the same role as a long small intestine in mammals:

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47 Adaptations for Absorption in Fish Key Idea: Cartilaginous fish increase nutrient absorption by increasing the time food spends in the

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gut. Bony fish use pyloric caeca to increase the surface area for absorption.

Intestinal spiral valve

Examples: cartilaginous fish, e.g. sharks, rays

Spiral valve from a nurse shark

Stomach

Liver

Spiral valve

Food

The number of turns varies between 5 and 50 depending on the species.

Pyloric caeca

Example: most bony fish

Gill cover

Many bony fish have pouch-like structures called pyloric caeca, which increase the gut surface area. The caeca are tube-like outgrowths at the junction of the stomach and intestine. They increase gut surface area, and function in enzyme secretion and nutrient absorption. Most bony fish have several hundred pyloric caeca and they are more well developed in carnivores, especially those with short guts. Gut of bony fish showing the pyloric caeca, which open into the first part of the intestine

Intestinal wall

Spiral folds increase the length of the intestine.

Body wall (cut)

Kidney

Liver

Gonad

Stomach

Photo: H Dahimo CC3.0

Photo: Haplochromis CC 3.0

The streamlined shape of sharks (and most other fish) makes it difficult to fit a long intestine into the body. Cartilaginous fish, such as sharks, as well as primitive bony fish, solve this problem by having a short intestine with a spiralling fold in the inner wall. This spiral valve increases the length of time digested material remains in the intestine by effectively increasing the intestine's length. The spiral valve also prevents large masses of indigestible material passing into the intestines.

Pyloric caeca

(b) How does this affect absorption?

2. Explain how bony fish increase the surface area of the intestine:

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1. (a) How does the spiral valve of cartilaginous fish increase the length of time food remains in the intestine?

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48 Adaptations for Absorption in Mammals Key Idea: The small intestine of mammals is folded into finger-like structures called villi which

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increase the surface area of the intestine for nutrient absorption.

ffThe gut of a mammal is divided into the stomach, and the intestines and associated glands. The digestion of food begins in the mouth, continues in the stomach, and is completed in the small intestine.

ffThe pancreas contributes an alkaline, enzyme-rich fluid to the intestine to break down fats, carbohydrates, and proteins. The alkaline fluid also helps to neutralise acid entering the duodenum from the stomach.

ffThe liver contributes an alkaline fluid containing bile salts to emulsify fats. Emulsify means to make a suspension from two fluids that do not normally mix. The simple molecules that result are absorbed across the intestinal wall.

ffThe intestine is notable for two features. Firstly, the intestinal wall is folded into microscopic finger-like structures called villi. Secondly, each of the epithelial cells lining the intestine has a border of microvilli. These features increase the surface area for absorption of nutrients into the blood.

Stomach begins protein digestion.

Villi

The small intestine is where most chemical digestion occurs and where most nutrients are absorbed.

Intestinal epithelial cell

Glucose and galactose

Active transport (Na+ cotransport)

Amino acids

Dipeptides

Tripeptides

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.

Facilitated diffusion

Active transport of di- and tripeptides is coupled to the downhill movement of H+ across the plasma membrane of the intestinal epithelial cells.

Active transport (Na+ cotransport)

Active transport (proton pump)

Short chain fatty acids

Long chain fatty acids

Monoglycerides

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.

ion

us Diff

Fat soluble vitamins

Monoglycerides and fatty acids associate with bile salts to form lipid spheres, which 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 lipid spheres themselves are not absorbed.

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Lacteal

Artery

Vein

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Gut lumen

Fructose

Lumen

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.

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1. Describe the adaptations of mammals to increase the surface area for nutrient absorption:

2. (a) What is the digestive role of the pancreas?

(b) What is the digestive role of the liver?

3. In what way are fats absorbed and processed differently to amino acids and simple sugars?

4. Describe how each of the following nutrients are absorbed by the intestinal villi:

(a) Glucose:

(b) Fructose:

(c) Amino acids: (d) Di- and tripeptides: 5. How are concentration gradients maintained for the absorption of nutrients by diffusion?

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6. Humans are usually classed as omnivores. Some people assert that humans should be vegetarians (herbivores). Use the information in the activities on nutrition to compare human, carnivore, and herbivore digestive tracts. Discuss if classing humans as omnivores or herbivores is justified, giving your reasoned opinion on the subject:

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49 What You Know So Far: Digestion and Feeding Structure of tube guts

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Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question that follows. Use the points in the introduction and the hints provided to help you:

HINT: Compare the structure of tube guts .

Obtaining food

HINT: Compare and contrast how mammals, fish, and insects obtain and initially process food

Adaptations for absorption

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HINT: Compare and contrast how mammals, fish, and insects absorb nutrients from digested food.

REVISE


50 Essay Style Question: Digestion and Feeding

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All animals have adaptations that enable them to survive. Different animals have evolved different adaptations and methods of obtaining food and processing it to extract nutrients. 1. For the biological process of nutrition, use the taxonomic groups mammals, fish, and insects to answer the following (you may use extra paper if required):

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• Compare and contrast adaptations for the absorption of nutrients from ingested food. • Explain how these adaptations enhance survival.

TEST

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2. For the biological process of nutrition, use the functional groups carnivores, omnivores, and herbivores to answer the following (you may use extra paper if required):

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• Describe the structural features of the digestive system for a mammalian carnivore, omnivore, and herbivore. • Explain how each system is adapted to processing the particular diet of the mammal described.


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51 KEY TERMS AND IDEAS: Digestion and Feeding

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1. (a) Describe two structural adaptations for fluid feeding and bulk feeding:

(b) Give a mammalian example of fluid feeder, bulk feeder, and a filter feeder:

2. (a) Give a general dental formula for a carnivorous mammal:

(b) What general shape would you expect the teeth of carnivorous mammals to be?

3. (a) Give a general dental formula for a herbivorous mammal:

(b) What general shape would you expect the teeth of herbivorous mammals to be?

4. Where do the following four processes occur in a mammal?

(a) Ingestion:

(c) Absorption:

(b) Digestion:

(d) Egestion:

5. (a) What structures from the intestine of a mammal are shown in the photograph (right)?

(b) What is their function?

EII

6. Match the words below with their definitions:

absorption

A A feeding method in which large parts of an organism are eaten at once.

bulk feeding

B The process of obtaining food via trapping it in a system of filters, such as those formed by projections on the gill arches.

dentition

C Finger-like projections lining the surface of the mammalian intestine that increase the surface area for absorption.

deposit feeding

D A feeding method in which soil or sediment is eaten to obtain the food particles it contains.

egestion

filter feeding

E The kind, number, and arrangement of the teeth in the tooth rows.

F The first part of the mammalian intestine comprising the duodenum, jejunum and ileum. It is the main site for digestion and absorption of nutrients.

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digestion

G The part of the mammalian intestine comprising the caecum, colon, and rectum.

ingestion

H The breakdown of food into its constituent molecules.

intestinal villi

I The process of eliminating undigested food from the gut.

large intestine

small intestine

TEST

J The process by which the products of digestion move across a gut lining into the blood or lymph.

K The taking in of fluid or food into the body (by drinking or eating).

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52 The Need for Gas Exchange Key Idea: Cellular respiration (the breaking down of glucose to release energy) requires the exchange

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of gases between the respiring cells and the environment.

Flat organisms can use the body surface as the gas exchange surface, but most have specialised gas exchange systems (below).

Cellular respiration creates a demand for oxygen The inputs and outputs for cellular respiration are shown below for a generalised animal cell. Plant cells also respire, but their gas exchange budget is different because they consume CO2 and produce O2 in photosynthesis.

CO2

Gas exchange is the process by which gases enter and leave the body by diffusion across gas exchange surfaces. To achieve effective gas exchange rates, gas exchange surfaces are thin and have a high surface area. They must also be moist so that the gases can dissolve before diffusing across. The concentration gradients for diffusion are maintained by cellular respiration in the cells.

Water (H2O)

Glucose (C6H12O6)

Energy

O2

Carbon dioxide (CO2)

Cellular respiration takes place in the mitochondria of every cell in the body. Glucose is broken down to harness energy as ATP. It creates a constant demand for oxygen (O2) and a need to eliminate carbon dioxide gas (CO2).

Oxygen (O2)

Gas exchange systems and environment

Spiracles

Insects exchange gases with the air through a system of tubes called tracheae, which penetrate deep into their tissues. Air enters through spiracles (openings) in the body wall.

The lungs of air breathers, such as mammals, are internalised to prevent drying out. They are linked to the air outside by airways that carry the air to and from the gas exchange surface.

1. What is the purpose of gas exchange?

2. Name the respiratory gases:

3. How are gases exchanged with the environment?

In water, the gas exchange membranes are in direct contact with the environment. Fish gills are very efficient at extracting oxygen from water, which compensates for water's relatively low oxygen content.

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Tony Wills, CC 2.5

The environment presents different gas exchange challenges to animals. In air, gas exchange surfaces will dry out. In water, the oxygen content is much lower than in air.

4. Contrast air and water in terms of the challenges they present for gas exchange:

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53 Gas Exchange in Insects

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Key Idea: Insects transport air throughout their bodies via a system of tracheal tubes, which end in

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tiny fluid-filled tracheoles. Spiracles allow air to enter and leave the body.

Insect tracheal tubes

ffInsects transport gases via a system of branching tubes called tracheae (tracheal tubes).

ffGases enter the body through paired openings in the body called spiracles. Filtering devices prevent small particles from clogging the system and valves control the degree to which the spiracles are open.

ffThe tracheal tubes end in tiny (< 1 micron) tracheoles, which are permeable to fluid and respiratory gases.

Respiratory gases dissolve in the fluid and move in to and out of the tissues down their concentration gradients.

ffIn small insects, diffusion is the only mechanism needed to exchange gases, because it occurs so rapidly through the air-filled tubules.

ffLarger, more active insects, such as locusts have a tracheal system that includes air sacs, which can be compressed and expanded to assist in moving air through the tubules.

Tracheal tubes

Air sacs, present in some insects, act as bellows during vigorous body movements.

Closing the valves allows the insect to conserve water when oxygen demands are low.

Entry of air into the tracheal system via the spiracles is regulated by flap-like valves.

Insect muscle fibres (cells)

Every cell in the insectâ&#x20AC;&#x2122;s body is next to or very close to the end of a tracheole.

Carbon dioxide

Tracheole

Oxygen

Spiracle openings on the abdomen

Spiracle

Tracheal tubes

Fluid

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Detail of tracheole endings

CO2 diffuses out of tissues

O2 diffuses into tissues

An insect may have up to 20 spiracles: 8 abdominal pairs and 2 thoracic pairs.

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When the muscles are contracting and more oxygen is being used, fluid is drawn back into the body, exposing more of the tracheole surface to the air.

Oxygen and carbon dioxide dissolve in the fluid at the end of the tracheole and diffuse in and out of the tissues down their concentration gradients.

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A dissection of an insect with a generalised body plan, such as a cockroach (below), provides an appreciation of how extensively the tracheal tubes branch throughout the body, as shown below

Tracheoles

Crop

Trachea

Tracheoles

Fat body

Tracheoles can be seen as silvery lines crossing the dorsal surface of the crop. Removal of the crop reveals the larger tracheal tubes where they join the spiracles.

Careful removal of the fat body allows better viewing of the tracheal system. Even under a low power dissecting microscope, the rings around the tracheal tubes can be seen.

1. (a) How does air enter the insect tracheal system?

(b) What is the purpose of the valves in the spiracles?

(c) What is the adaptive advantage of this?

2. How are oxygen and carbon dioxide exchanged between the air and tissues at the end of insect tracheoles?

3. How is ventilation achieved in a terrestrial insect?

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4. Most insects do not require a respiratory pigment to transport oxygen around the body. Suggest why this is the case:

5. 300 million years ago, insects grew far larger than they do today. Meganeura dragonflies had wingspans of up to 65 cm. Evidence shows the oxygen concentration of the atmosphere at this time reached about 30%. Explain why this would have been one reason why insects grew so much larger than they do today:

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54 Insect Adaptations to Gas Exchange in Water Key Idea: Insects living in water also exchange gases with the environment via tracheae. Air can be

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carried with the insect or it can diffuse into the tracheal system from the water.

Insect gas exchange in water

CO2

ffAquatic insects, like terrestrial insects, exchange gases via the system of airfilled tracheae.

ffAquatic insect larvae rely on diffusion

O2

Dytiscus

Tracheal system

across the body surface, with or without tracheal gills.

Spiracle

ffMany adult insects carry air with them

with them when submerged, trapped by the carapace or by unwettable hairs. This trapped air acts as a diffusion gill.

Oxygen diffuses from water into the tracheae of the gills, and from there into the entire tracheal system.

O2

As the submerged insect respires, the oxygen is gradually used up and the insect must resurface.

O2

Cross section through body

Above: In the diving beetle, Dytiscus, the spiracles open under the wings, where air from the surface is trapped. The trapped air also acts as a diffusion gill, and oxygen diffuses into it from the water and then into the tracheae. The concentration gradient for oxygen diffusion is maintained by metabolism in the tissues.

O2

CO2

Tracheae

Air-filled cavity under the wings

Left: Tracheal gills are flat extensions of the cuticle that increase the surface area across which gases can diffuse. Insects with tracheal gills usually live in fast flowing waters, where the movement helps to ventilate the gills and increase oxygen uptake.

Gill

The tracheal gills of this spiny-gilled mayfly are located on the abdomen. The cuticle of the gills is very thin and gases can diffuse easily across it.

Dytiscid beetles carry air from the surface under the wings and replenish it regularly. They do have a visible bubble of air outside the body.

Fringe of hydrofuge hairs

Hydrophilid beetles use unwettable (hydrofuge) hairs to trap a film of air (called a plastron) against the spiracles. The plastron is often visible as a bubble.

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Stephen Moore

Evanherk (Public Domain)

my.opera.com/Ukwildlife/blog/

Tracheal gills on abdomen

1. Describe three adaptations of freshwater insects for gas exchange while submerged:

(a)

(b)

(c)

2. (a) How does trapped air also act as a diffusion gill in some adult insects?

(b) How is the gradient for diffusion maintained?

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55 Gas Exchange in Fish Key Idea: Fish exchange gases between the blood and the environment using gills. Gills are thin

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filamentous, vascular structures located just behind the head.

ffFish obtain the oxygen they need from the water by means of gills. Gills are membranous structures supported by cartilaginous or bony struts.

ffGill surfaces are very large, but the gill filaments are very thin and respiratory gases are exchanged between the blood and the water by diffusion as the water flows over the gill surface.

ffFish facilitate gas exchange by ventilating the gill surfaces, either by using the gill cover (operculum) as a pump, drawing water past the gill filaments (e.g. bony fish), or by swimming continuously with their mouth open (e.g. sharks).

ffThe percentage of dissolved oxygen in a volume of water is much less than in the same volume of air. Air is 21% oxygen but, in water, dissolved oxygen is about 1% by volume. Efficient uptake of oxygen from the water (as achieved by gills) is necessary for active organisms in an aquatic environment.

ffFish gills are very efficient and can achieve an 80% extraction rate of oxygen from water; over three times the rate of human lungs from air.

Inspiration (mouth open)

Oral valve opens

Breathing in bony fish

Mouth cavity expands, taking in water through the open mouth.

Exspiration (mouth closed)

Oral valve shuts

Mouth cavity contracts to force water across the gills.

Gill cover is closed and moved outwards to assist water intake.

Gill cover is open

Dorsal view of a fish head

Detail of gill filament

Bony bar (branchial arch)

Blood vessels

Gill lamella

Deoxygenated blood Oxygenated blood

Direction of blood flow in gill lamellae

Water flows in opposite direction to blood flow in the gill lamellae (countercurrent flow).

Source: C.J. Clegg & D.G. McKean (1994)

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Water flow

Blood flow

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78 Fish gills

ffThe gills of fish have a great many folds, which are supported and kept apart from each other by the water. This gives them a

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high surface area for gas exchange. The outer surface of the gill is in contact with the water, and blood flows in vessels inside the gill. Gas exchange occurs by diffusion between the water and blood across the gill membrane and capillaries. The operculum (gill cover) allows water to exit and also acts as a pump, drawing water past the gill filaments.

ffBony fish (below left) have four pairs of gills, each supported by a bony arch. The gill cover is involved in ventilating the gills. Cartilaginous fish (e.g. sharks (below right) have five or six pairs of gills. Water enters via the mouth and spiracle and exits through gill slits (there is no gill cover).

ffFish can use two methods of moving water over the gills. Buccal pumping uses the muscles in the mouth to pump water over

the gills, and is common in bony fish. Ram ventilation moves water over the gills by swimming into the water. This is seen is the pelagic sharks (e.g. great white sharks). Many bony fish and sharks can switch between these two methods.

Gill cover (operculum)

1. Describe three features of the fish gas exchange system that facilitate gas exchange:

(a)

(b)

(c)

2. Why is ventilation of the gills necessary?

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3. How does the fish gas exchange system support the active life of fish?

4. Some fish use ram ventilation while others use buccal pumping to move water over the gills. Describe situations when these methods would provide an advantage or a disadvantage:

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56 Countercurrent Flow Key Idea: Water flows past a fish gill in the opposite direction to the blood flowing through it

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(countercurrent flow). This increases the efficiency of O2 extraction from the water.

ffThe structure of fish gills and their physical

arrangement in relation to the blood flow maximises gas exchange rates. A constant stream of oxygenrich water flows over the gill filaments in the opposite direction to the blood flowing through the gill filaments.

% oxygen in water

ffThis is called countercurrent flow (below left) and

60

80

40

Oxygen diffusion

0

10

20

60

40

80

it is an adaptation for maximising the amount of O2 removed from the water. Blood flowing through the gill capillaries encounters water of increasing oxygen content. Therefore, the concentration gradient (for oxygen uptake) across the gill is maintained across the entire distance of the gill lamella and oxygen continues to diffuse into the blood (CO2 diffuses out at the same time).

Water flow

ffA parallel current flow would not achieve the same

% oxygen in blood

oxygen extraction rates because the concentrations across the gill would quickly equalise (below, right).

Blood flow

Gill lamella

Countercurrent flow

100

Water

50

0

Blood

Distance along gill lamellae

Percentage oxygen saturation

Percentage oxygen saturation

100

Parallel current flow

Water

50

Blood

0

At this point, blood and water have the same O2 concentration so no more O2 exchange takes place.

Distance along gill lamellae

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1. Describe countercurrent flow:

2. (a) How does the countercurrent system in a fish gill increase the efficiency of oxygen extraction from the water?

(b) Why wouldn't parallel flow achieve adequate rates of gas exchange?

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57 The Human Gas Exchange System

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Key Idea: The human gas exchange system is made up of specialised cells and tissues, which work

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together to enable the exchange of gases between the body's cells and the environment.

ffThe gas exchange system of mammals (e.g. humans) consists of the passages in the mouth and nose, the

tubes and epithelial tissues of the lungs, and the muscles of the diaphragm and ribcage. Each of these regions is specialised to perform a particular role in the organ system's overall function, which is to exchange respiratory gases (oxygen and carbon dioxide) between the body's cells and the environment.

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).

2. What is the purpose of the hyaline cartilage in the gas exchange system?

3. How are bronchioles similar to insect tracheal tubes?

4. Where does gas exchange take place in the lungs?

5. What is the purpose of ciliated epithelial cells in the trachea?

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1. Which cells form the alveoli?

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58 The Lungs and Gas Exchange Key Idea: Lungs are internal sac-like organs connected to the outside by a system of airways. The

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smallest airways end in thin-walled alveoli, where gas exchange occurs.

ffThe gas exchange system includes all

the structures associated with exchanging respiratory gases with the environment.

ffIn mammals, such as humans, 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.

The trachea transfers air to the lungs. It is strengthened with C-shaped bands of cartilage.

Bronchioles branch from the bronchi and divide into progressively smaller branches. The cartilage is gradually lost as the bronchioles decrease in diameter.

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.

Right lung

The "cardiac notch" in the left lung makes space for the heart.

Left lung

Oxygen enters the blood from air in the alveoli. Carbon dioxide leaves the blood and is breathed out. O2 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

2. What is the purpose of the diaphragm?

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1. What is the purpose of the trachea, bronchi, and bronchioles?

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 alveoli and capillaries The alveolar-capillary membrane

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An alveolus

Alveolar wall

Capillary

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Red blood cell in capillary Connective tissue containing elastic fibres

Connective tissue cell

Alveolus

Surfactant is a phospholipid produced by cells in the alveolar walls.

Interstitial space

Surfactant

Alveolar-capillary membrane

Alveolus

O2

0.5 Îźm

CO2

Nucleus of epithelial cell

Alveolar epithelial wall is very thin to allow diffusion of gases.

Red blood cell

Capillary wall

The diagram above is of a section through an alveolus and surrounding tissue. It illustrates the physical arrangement of the alveoli to the capillaries through which the blood moves. Elastic connective tissue gives the alveoli their ability to expand and recoil.

The alveolar-capillary membrane (or gas exchange membrane) is the layered junction between the alveoli and the capillaries. It is very thin and gases can move freely across this membrane.

4. 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?

5. What is the role of the surfactant in the alveoli?

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59 Breathing in Humans Key Idea: Breathing provides a continual supply of air to the lungs. Quiet breathing uses muscles only

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for inspiration. Forced breathing uses muscles for both inspiration and expiration.

ffIn mammals, the mechanism of breathing (ventilation) provides a continual supply of fresh air to the lungs and

helps to maintain a large diffusion gradient for respiratory gases across the gas exchange surface. Oxygen must be delivered regularly to supply the needs of respiring cells. Similarly, carbon dioxide, which is produced as a result of cellular metabolism, must be quickly removed from the body.

ffAdequate lung ventilation is essential to these exchanges. The volume of gases exchanged during breathing varies according to the physiological demands placed on the body (e.g. by exercise).

ffThe lungs do not have any muscles of their own. Inspiration and expiration are achieved by contraction and relaxation of the muscles surrounding the chest cavity.

Intercostal muscles

Inspiration (inhalation or breathing in) During quiet breathing, inspiration is achieved by increasing the space (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.

1a

External intercostal muscles contract causing the ribcage to expand and move up.

1b

Diaphragm contracts and moves down.

2

Volume of chest cavity increases, lungs expand, and the pressure inside the lungs decreases.

3

Air flows into the lungs in response to the pressure gradient.

Diaphragm contracts and moves down

Expiration (exhalation or breathing out)

1

During quiet breathing, expiration is achieved passively by decreasing the space (thus increasing the pressure) inside the lungs. Air then 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.

In quiet breathing, external intercostal muscles and diaphragm relax. Elasticity of the lung tissue causes recoil. In forced breathing (e.g. during exercise) the internal intercostals and abdominal muscles also contract to increase the force of the expiration.

1. What is the purpose of breathing?

2. What is the difference between quiet breathing and forced breathing?

Volume of chest decreases and the pressure inside the lungs increases.

3

Air flows passively out of the lungs in response to the pressure gradient.

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Diaphragm relaxes and moves up

2

3. Under what circumstance would the body use forced breathing instead of quiet breathing?

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60 Adaptations for Diving in Mammals Key Idea: Diving animals have structural and physiological adaptations that allow them to stay

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submerged for extended periods of time.

ffAll air breathing animals that dive must cope with the problem of oxygen supply to the tissues. This is particularly a problem for mammals which have a high metabolic rate and therefore a high oxygen demand. In addition, resurfacing from dives of 20 m or more brings the risk of decompression sickness (also called the bends).

ffPrimates (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.

Diving mammals ffDolphins, whales, and seals are among the most well adapted divers.

ffThey 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.

ffDuring dives, heart rate

slows and blood flow is redistributed to supply only critical organs.

ffDiving mammals have high

levels of myoglobin and their muscles work well without oxygen.

ffSperm 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).

The problems for humans

Humans are poorly adapted for diving. They lack adequate body fat for long periods underwater and they inhale before diving. Deep diving freedivers (above) may hyperventilate (take a series of deep breaths in and out) before diving. This reduces the CO2 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.

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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2. Freedivers who wish to remain underwater for long periods of time hyperventilate before diving.

(a) Why does hyperventilation suppress the urge to breathe?

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85 (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)

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

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

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Location of oxygen

4. (a) Describe the most striking difference between a seal and a human in terms of the oxygen stores during a dive:

(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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61 What You Know So Far: Gas Exchange

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Adaptations for gas exchange in mammals

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Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question that follows. Use the points in the introduction and the hints provided to help you:

HINT: Describe how mammals exchange oxygen and carbon dioxide with the environment.

Adaptations for gas exchange in fish

HINT: Describe the structure of gills and explain how they extract oxygen from water.

Adaptations for gas exchange in insects

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HINT: Describe the tracheal system of insect.

REVISE

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62 Essay Style Question: Gas Exchange

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All animals have adaptations that enable them to obtain the oxygen they need to carry out their metabolic processes and eliminate waste carbon dioxide. 1. For the biological process of gas exchange use the taxonomic groups mammals, fish, and insects to answer the following (you may need to use extra paper if required):

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• Compare and contrast adaptations for gas exchange. • Explain how these adaptations enhance survival.

TEST


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63 KEY TERMS AND IDEAS: Gas Exchange

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

A Any gas that takes part in respiration. Usually refers to oxygen or carbon dioxide.

breathing

B Gas required for aerobic respiration. Levels average 21% in the atmosphere.

bronchi

C Thin walled sac-like regions in a mammalian lung where gas exchange takes place.

bronchioles

D Large air tubes that branch from the trachea to enter the lungs.

cellular respiration

countercurrent flow expiration

gas exchange gills

E Small air tubes that divide from the bronchi and become progressively smaller. F The exterior opening of the tracheae in arthropods, e.g. insects.

G Series of tube-like structures that conduct air into the body in insects.

H The respiratory organs of most aquatic animals (although not aquatic mammals). I The act of breathing out or removing air from the lungs.

haemoglobin

J A term describing the flow of fluids or air in opposite directions so that diffusion gradients are maintained between the two media and exchanges between them are maximised.

inspiration

K A large iron-containing protein that transports oxygen in the blood of vertebrates.

lungs

L The act of breathing in or filling the lungs with air.

oxygen

M The process by which organisms break down glucose using oxygen to produce usable energy. Carbon dioxide, water, and heat are waste products.

respiratory gas

N The act of inhaling air into and exhaling air from the lungs.

spiracles

O The exchange of oxygen and carbon dioxide across a gas exchange membrane.

tracheae

P Internal gas exchange structures found in vertebrates except fish.

2. Use lines to join the three parts of the sentences below together to form complete sentences. The first column is in order, the centre and right columns are not. Choose the most appropriate joining word to create the sentence.

Gas exchange in terrestrial insects is enabled by tracheae,... Fish and many aquatic insects use gills to extract oxygen from water... Enclosing the lungs inside the body...

Blood flowing through the gills of fish flows countercurrent... This maintains diffusion gradients and extracts the maximum amount...

Lungs use a tidal system (inspiration and then expiration)... Oxygen is transported in red blood cells...

whereas

...oxygen from the water.

which

...mammals use lungs which are enclosed inside the body.

to

...the gas exchange surface drying out.

by

...the water flowing past the gills.

of

...are tubes in the body that supply oxygen directly to the tissues by diffusion.

to

and

3. (a) On the photo of the dissection of a fish's gills, label the gills and operculum. The brachial arch supporting the gill is labelled. (b) Draw arrows on the photo to show the direction of water flow when the fish was in the water.

(c) Explain how these gills are ventilated by the fish:

TEST

...carbon dioxide gas to leave.

prevents

...haemoglobin to the tissues of the body. ...move air in and out.

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Gas exchange surfaces provide a way for oxygen gas to enter the body...

Branchial arch

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64 Transport and Exchange Systems Key Idea: The type of transport system an animal has depends on its requirements for nutrients and

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gas exchange.

Systems for transport and exchange

ffLiving cells need a constant supply of nutrients and oxygen, and continuous removal of wastes. Organisms with

low surface area to volume ratios require systems to facilitate exchanges because diffusion alone will not provide enough nutrients to cells deep within the body.

ffIn all vertebrates and most invertebrates, a circulatory system moves nutrients, wastes, and often gases to where

they are needed. This movement of fluids around the body as a single mass is called mass transport or bulk flow.

A vertebrate transport system

The blood transports nutrients, wastes, hormones, and respiratory gases. It moves by mass transport between the exchange surfaces at the tissues and the gills (or lungs).

The blood circulates within a network of blood vessels, which transport the blood to all regions of the body.

The gut (stomach, intestines) and other internal organs.

Specialised exchange surfaces at the gills or lungs, enable the gases to be exchanged with the environment by diffusion.

In unicellular organisms, such as this amoeba, and small flattened organisms, such as planarians, no part of the organism is very far from the environment. Simple diffusion of molecules from the environment into and out of the organism provides for all the organism's needs.

The heart is a pumping device to circulate blood through a network of blood vessels. The heart may be a simple tube or have several chambers.

In multicellular organisms, nutrients need to be transported to cells deep in the body. Different taxa have different types of system. The type of system present relates to the lifestyle and evolutionary history of the organism.

O2

CO2

Distribution

Wastes

Nutrients

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Nutrients

1. (a) What system delivers nutrients and gases to the body's cells in a large, complex animal?

(b) Why do small simple animals manage to meet their needs for nutrients and gases without this system?

2. In a vertebrate such as a shark, identify the region where:

(a) Nutrients are picked up by the system named in question 1 (a):

(b) Gases are exchanged with the system named in question 1 (a): Š 2017 BIOZONE International ISBN: 978-1-927309-60-5 Photocopying Prohibited

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65 Haemolymph

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Key Idea: The haemolymph in insects is involved in clotting, immunity and moulting. Unlike vertebrate

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blood, it is not usually involved in transport of respiratory gases.

ffHaemolymph is a blood-like substance

Stephen Moore

found in all invertebrates with open circulatory systems. The haemolymph fills the haemocoel (body cavity) and surrounds all cells.

ffAbout 90% of insect haemolymph is plasma,

a watery fluid, which is usually clear. Compared to vertebrate blood, it contains relatively high concentrations of amino acids, proteins, sugars, and inorganic ions.

A few insects (but not many), like the midge larva above, possess haemoglobin as an adaptation to living in low-oxygen substrates.

ffThe remaining 10% of haemolymph

volume is made up of various cell types (haemocytes). These are involved in clotting and internal defence. Unlike vertebrate blood, insect haemolymph lacks red blood cells and (with a few exceptions) lacks respiratory pigment, because oxygen is delivered directly to tissues by the tracheal system.

Haemolymph may make up between 11% and 40% of the total body mass of an insect

Image: Jon Mollivan

Image: Psychonaught

Fluid pressure is used to help insects during moulting (above and left). Pressure of the haemolymph enables arthropods to expand the soft cuticle of the body segments before they harden (sclerotise). This enables them to grow.

Some insects, such as New Zealand's alpine weta (above) can tolerate freezing during winter, when the osmotic pressure of the haemolymph almost doubles. The alpine weta's haemolymph also contains the disaccharide sugar trehalose, which acts as a cryoprotectant, stopping frozen cells from bursting.

1. Describe two common functions of insect haemolymph (which are also performed by mammalian blood): (a)

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(b)

2. Describe one function of mammalian blood not commonly performed by insect haemolymph:

3. Describe one function of insect haemolymph not performed by mammalian blood:

4. Contrast the proportions of cellular and non-cellular components in blood and haemolymph:

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66 Blood Key Idea: Blood transports nutrients, wastes, gases, and hormones around the body. Specific

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adaptations optimise blood function for some animals in extreme environments.

ffBlood is a complex liquid tissue made up of cellular components suspended in a strawcoloured solution called plasma.

ffBlood transports nutrients, respiratory gases, hormones, and wastes. It also has a role in regulating body temperature by distributing heat, defends against infection, and its ability to clot protects against blood loss.

ffThe blood of some animals in polar

environments may even contain non-cellular components such as antifreeze sugars and proteins. These enable them to function effectively at sub-zero temperatures.

Mammalian blood

NON-CELLULAR COMPONENTS

The non-cellular part of the blood is the plasma. It is 92% water and also contains proteins, electrolytes, hormones, and dissolved gases. Plasma makes up 50-60% of blood volume. The water in the plasma transports blood cells, distributes heat and helps maintain blood volume.

CELLULAR COMPONENTS

The formed elements of blood float in the plasma. They include white blood cells, platelets, and red blood cells.

Red blood cells (RBCs) 38-48% of total blood volume. RBCs transport oxygen (carried bound to haemoglobin) and a small amount of carbon dioxide. Red blood cells in mammals are different to all other vertebrates in that they are not nucleated.

White blood cells (WBCs) and platelets 2-3% of the total blood volume. WBCs are involved in internal defence. They include lymphocytes and granulocytes. Platelets are small, membrane-bound cell fragments with a role in blood clotting.

7-8 µm

2 µm

Platelets

Granulocyte

Lymphocyte

Kenneth Catania, Vanderbilt University NSF

Photo:Professor Dr. habil. Uwe Kils CC3.0

Crocodile icefish larva

1. What are three functions of the blood?

Adaptations of haemoglobin have enabled some mammals to live in places with low oxygen content. Llamas and their relatives live at high altitudes with low oxygen pressure. Moles live underground where oxygen is low but carbon dioxide is high. The haemoglobin in some moles has lost the ability to bind the molecule DPG, and instead binds carbon dioxide. This enables the mole to remain active when carbon dioxide levels are high.

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The Antarctic icefish are the only known vertebrates without haemoglobin. They survive because the extremely cold water (as low as -2°C) contains high concentrations of oxygen (cold water holds more oxygen than warmer water). Oxygen diffuses directly across their scaleless skin into the blood and is transported throughout the body. Many icefish contain antifreeze glycoproteins in their blood so the blood to remain fluid at freezing temperatures.

2. Name an adaptation for moles and icefish that allows them to better exploit their habitat:

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67 Blood Vessels Key Idea: The blood vessels of the circulatory system connect the body's cells to the organs that

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exchange gases, absorb nutrients, and dispose of wastes.

ffIn 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) of the blood vessel walls. Vasoconstriction increases blood pressure whereas vasodilation has the opposite effect.

ffVeins 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 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 vesselâ&#x20AC;&#x2122;s role.

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).

Relatively 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): A thin inner layer of epithelial cells called the endothelium lines the artery.

A thick central layer of elastic tissue and smooth muscle that can both stretch and contract.

An outer connective tissue layer has a lot of elastic tissue.

Artery

Thick central layer of elastic tissue and smooth muscle.

Thin endothelium

Artery

The lumen is the space inside the vessel.

1. Why do the artery walls need to be thick with a lot of elastic tissue?

3 (a) What is the effect of vasodilation on the diameter of an artery?

(b) What would the effect of this be on blood pressure?

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2. What is the purpose of smooth muscles in the artery walls?

4. Describe the structure of a capillary and explain the purpose of this structure:

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Nucleus of endothelial cell

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Red blood cell

Dept of Biological Sciences, University of Delaware

Basement membrane

Thin endothelium (one cell thick)

Capillaries

Arterioles carry blood to capillaries

Fat cell

Blood flow

CO2 and wastes move from the tissues into the capillary.

O2 and nutrients move from the capillary into the cells of the tissues.

Capillary

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.

Relatively thick outer layer of connective tissue.

Venules drain capillaries to vein

d oo

One-way valves prevent blood flowing in the wrong direction.

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.

Bl

Thin central layer of elastic and muscle tissue.

Veins

flow

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.

Vein

Thin layer of endothelium

Vein

(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?

7. Why does blood ooze from a venous wound, rather than spurting as it does from an arterial wound?

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68 Open and Closed Circulatory Systems Key Idea: In closed circulatory systems, the blood is always contained within blood vessels. In open

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systems, the haemolymph circulates in the body cavity, not enclosed in vessels.

The open circulatory system

ffMany invertebrates, including insects,

have an open circulatory system. In insects, (unlike other arthropods) this does not transport oxygen, which is delivered directly to the tissues via the system of tracheal tubes.

Haemolymph is pumped towards the head.

Dorsal tubular heart Blood re-enters the with one-way valves heart through ostia Abdomen

Head

ffIn addition to its usual transport

functions (e.g. transporting nutrients from the gut), open circulatory systems allow hydraulic movements of the whole body or its component parts (e.g. butterflies use hydraulic pressure to emerge and to expand their wings after emergence). An open system also dissipates heat efficiently, allowing insects to survive in very hot environments.

Body fluids flow freely within the body cavity

In insects haemolymph is pumped by a tubular, or sometimes sac-like, heart through short vessels into large spaces in the body cavity. It bathes the cells and re-enters the heart through holes (ostia). Open systems cannot develop high blood pressure, but muscle contractions help to circulate the haemolymph.

The closed circulatory system

Chambered heart

ffClosed circulatory systems, consisting of

a heart and a network of tube-like vessels, are useful for large, active animals where oxygen and nutrients need to be rapidly and efficiently distributed to the body's cells. Exchanges between the blood and the fluids bathing the cells occurs by diffusion across capillaries.

ffClosed circuits also allow the animal more

Body capillaries

Lungs

control over the distribution of blood flow to different organs and parts of the body by contracting or dilating blood vessels. Closed systems are the most developed in vertebrates where blood remains in the vessels, separated from the body tissues.

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1. Describe the main difference between open and closed circulatory systems:

2. Describe a benefit of a closed transport system to distribute oxygen and nutrients:

3. (a) Describe one advantage of open circulatory system:

(b) Describe one disadvantage of an open circulatory system:

4. Explain why an insect does not bleed in a similar way to a mammal:

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69 Single and Double Circulatory Systems Key Idea: Closed circulatory systems can operate as a single or double circuit. The single circuit system

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of fish operates at lower pressure than the double circuit system of mammals.

Closed, single circulatory systems

ffIn the single circulatory systems of fish, the blood is pumped from the heart to the gills, then directly to the body. ffThe blood loses pressure at the gills and flows at low pressure around the body. The low pressure reduces blood flow through the body and thus reduces the rate of oxygen delivery to the body's cells. However most fish have relatively low metabolic rates and this system adequately meets their needs.

ffWater has much lower oxygen content than air (about 12 parts per million maximum compared with 210,000 ppm in air). This limits the amount of oxygen fish can extract even with efficient gills. The low oxygen content in the water would not support a higher metabolic rate, but because most fish do not use metabolism to maintain body temperature (a large energy cost), a lower metabolic rate still allows a relatively active lifestyle.

Oxygenated blood

Gills

Oxygen moves into the blood

Oxygen moves into the tissues

Sharks

CHAMBERED HEART Ventricle

Capillaries

Body

Atrium

Deoxygenated blood

Bony fish

Direction of blood flow

Closed, double circulatory systems

ffBecause oxygen is relatively abundant in the air, metabolic rates in air breathing animals can be relatively high

(although they are not necessarily so). Double circulatory systems develop higher pressure than single circuit systems, delivering oxygenated blood to the body at a rate sufficient to meet higher metabolic demands.

ffDouble circulatory systems occur in all vertebrates other than fish. They are most efficient in mammals and birds

where the heart is fully divided into two halves and the two circuits are completely divided. These animals rely on metabolism to maintain body temperature, so their metabolic demands are necessarily high.

ffDouble circulatory systems have two distinct circuits, the pulmonary circuit, which circulates blood between the

lungs and the heart, and the systemic circuit, which pumps oxygenated blood to the rest of the body. The return of oxygenated blood from the lungs to the heart means that the blood can be pumped to the rest of the body at the higher pressures needed to supply organs and maintain kidney filtration rates while the blood in the lungs (the pulmonary circuit) remains at a low pressure, suitable for facilitating gas exchange. Lungs

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Left side

Arteries

Right side

Oxygenated blood

CHAMBERED HEART

Veins

Deoxygenated blood

Mammals

Double systems are also found in birds, amphibians, and reptiles. Birds, like mammals, use metabolism to maintain body temperature (a high energy cost) and maintain high metabolic rates. LINK

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1. Compare and contrast the basic structure of the heart and circulatory system in a fish and a mammal:

2. Describe where the blood flows to immediately after it has passed through the gills in a fish:

3. Describe where the blood flows immediately after it has passed through the lungs in a mammal:

4. Explain why the blood in the body of a fish is at a lower pressure than the blood in the body of a mammal:

5. Discuss the relative efficiencies of single and double circulatory systems and why these are suitable to meet the metabolic demands of fish and mammals:

6. (a) Label the diagram of the fish circulatory system below with the following labels: heart, capillaries in gills, capillaries in body, low pressure blood, high pressure blood. (b) Draw arrows on the diagram to show the general direction of blood flow:

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70 Mammalian Transport System Key Idea: The circulatory system is responsible for the transport of nutrients, respiratory gases, and

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wastes in the blood to and from the body's cells via a network of vessels.

Vein

Components of the transport system

WBCs

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.

G Beard

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.

Artery

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).

(a)

(b)

(c)

2. (a) Which component of the blood carries oxygen to the body's cells?

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1. Name three components of the circulatory system and state their function:

(b) Which component of the blood carries metabolic wastes to the kidneys? Š 2017 BIOZONE International ISBN: 978-1-927309-60-5 Photocopying Prohibited

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ffThe blood vessels of the circulatory system (the arteries, arterioles, capillaries, venules, and veins) are organised into specific routes to circulate the blood throughout the body.

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ffMammals have a double circulatory system: a pulmonary system (or circuit), which carries blood between the heart and lungs, and a systemic system (circuit), which has many subdivisions, and carries blood between the heart and the rest of the body.

ffDeoxygenated blood (coloured blue) travels to the right side of the heart via the vena cavae. The heart pumps the deoxygenated blood to the lungs where it releases carbon dioxide and receives oxygen. The oxygenated blood (coloured white) travels via the pulmonary vein back to the heart from where it is pumped to the body.

ffThe arterial system carries blood away from the heart. The venous system returns blood to the heart. Portal systems carry blood between two capillary beds.

VENOUS SYSTEM

ARTERIAL SYSTEM

(a)

Pulmonary vein carries oxygenated blood back to the heart.

Pulmonary artery carries deoxygenated blood to the lungs.

Superior vena cava receives deoxygenated blood from the head and body.

Aorta carries oxygenated blood to the body.

(b)

Left atrium receives oxygenated blood from the lungs.

Right atrium receives deoxygenated blood via the superior and inferior vena cavae.

Left ventricle pumps blood from the left atrium to the aorta.

Right ventricle pumps deoxygenated blood to the lungs.

Hepatic vein carries deoxygenated blood from the liver.

Hepatic portal vein carries deoxygenated, nutrient rich blood from the gut for processing.

Abdominal aorta Parallel to the inferior vena cava, branching to supply the organs of the abdominal cavity.

(c)

Hepatic artery carries oxygenated blood to the liver.

(d)

Mesenteric artery carries oxygenated blood to the gut.

(e)

Renal vein carries deoxygenated blood from the kidneys.

(f)

Renal artery carries oxygenated blood to the kidneys.

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Inferior vena cava receives deoxygenated blood from the lower body and organs.

3. Complete the diagram above by labelling the boxes with the correct organs: lungs, liver, head, intestines, genitals/lower body, kidneys.

4. Use the letters (a) - (f) to identify the organs with a portal system between them:

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71 Vertebrate Hearts Key Idea: Fish have a two chambered heart with the chambers aligned in series, whereas mammals

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have a four chambered heart divided into left and right halves.

Mammalian heart

Fish heart

Ventricle

Atrium

To lungs

From the body

Sinus venosus

Atrioventricular valves

Sinoatrial valves

In mammals, the heart is divided by a septum into left and right halves, each with two chambers, the upper atrium and the lower ventricle. Blood circulates through two separate circuits, with no mixing. Oxygenated blood from the lungs is kept separated from the deoxygenated blood returning from the rest of the body.

Top view of a heart in section, showing valves

(sectioned, anterior view)

Semi-lunar valve of pulmonary artery

Aorta carries oxygenated blood to the head and body Vena cava receives deoxygenated blood from the head and body

Bicuspid (left atrio-ventricular valve)

RV

Aorta

Vena cava

Pulmonary arteries

Septum separates the ventricles

RA Right atrium; receives deoxygenated blood

RV Right ventricle; pumps deoxygenated blood to the lungs

LA Left atrium; receives blood returning to the heart from the lungs

Left ventricle; pumps oxygenated blood to the head and body

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Tricuspid (right atrio-ventricular valve)

Posterior (rear) view of heart

LV

Chordae tendinae non-elastic strands supporting the valve flaps

Semi-lunar valve of aorta

Bicuspid valve

LA

RA

Tricuspid valve prevents backflow of blood into right atrium

LV

Left ventricle

Right ventricle

From the body

Human heart structure

Semi-lunar valve prevents the blood flow back into ventricle.

Left atrium

Right atrium

The fish heart is linear, with an atrium and ventricle in series, and a chamber-like entry point (the sinus) and an exit (the conus). Blood enters the heart via the sinus venosus, then passes into the atrium and ventricle. The ventricle pumps the blood to the gills. One-way valves between the chambers prevent reverse blood flow.

Pulmonary artery carries deoxygenated blood to the lungs

From lungs

From rest of body

To gills

Ventral aorta

To rest of body

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Conus arteriosus

Pulmonary veins

Coronary veins

LV

RV

Coronary artery

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1. In the schematic diagram of the mammalian heart (below), label the four chambers and the main vessels entering and leaving them. Arrows indicate the direction of blood flow. Use circles to mark the position of each of the four valves

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Mammalian heart

(a)

(e)

(b)

(f)

(c)

(g)

(d)

(h)

2. Use the diagram on the previous page to draw a similar labelled schematic of a fish heart.

Fish heart

3. What is the function of the aorta?

4. What is the function of the vena cava?

5. What is the purpose of the valves in the heart?

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6. Compare and contrast the structure of the heart in fish and mammals, relating features of the heart's structure and function to the animalâ&#x20AC;&#x2122;s size, metabolic rate, or environment in each case:

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72 Dissecting a Mammalian Heart Key Idea: A dissection of a sheep's heart allow hands-on exploration of a mammalian heart. It is

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important to be able to identify specific structures during a dissection.

ffThe dissection of a sheep's heart is a common practical activity and allows hands-on exploration of the physical appearance and structure of a mammalian heart. This activity provides photographs of some aspects of a heart dissection to help you in identifying specific structures during your own dissection.

1 Gross anatomy of a sheep's pluck to show heart (dorsal view).

Lobe of right lung

Cut flap of pericardium

Right auricle is a muscular pouch connected to the right atrium.

Lobe of left lung

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.

Did you know?

Right ventricle of heart (dorsal)

2 External ventral view of heart

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.

3 External dorsal view of heart

Aorta

Brachiocephalic artery (cut)

Pulmonary trunk (artery)

Pulmonary veins

Left auricle

Left ventricle

Right ventricle

Left ventricle

Right ventricle

Apex

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Right auricle

Note the surface features of an isolated heart. The narrow pointed end forms the apex, while the wider end, where the blood vessels enter, is the base. On the ventral surface (above) a groove, the interventricular sulcus, marks the division between the left and right ventricles.

On the dorsal surface, 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, at the base, 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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4 Dorsal view of heart

5

Shallow section, ventral view of heart Aorta

Semi-lunar valves lie between the ventricles and the large arteries leaving the heart. They can be difficult to see.

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Pulmonary veins

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

Aorta (from left ventricle)

Pulmonary artery (from right ventricle to lungs but cut)

Left atrium

Right ventricle

Chordae tendinae

Part of left AV valve

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 partner and you have two hearts, make your sections in different ways to maximise the structures you can see.

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 oneway 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.

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 tendinae) run from the cusps to papillary muscles on the ventricular wall.

5. Judging by their position and structure, what do you suppose is the function of the chordae tendinae?

6. What feature shown here most clearly distinguishes the left and right ventricles?

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73 Internal Transport and Gas Exchange Interaction

103

Key Idea: The circulatory and gas exchange systems interact closely to provide the body's tissues

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with oxygen and remove carbon dioxide.

Circulatory system

Gas exchange system

Interaction between systems In mammals, the gas exchange system and cardiovascular system interact to supply oxygen and remove carbon dioxide from the body.

Function

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.

Function

Provides surface for gas exchange. Moves fresh air into the body and stale air out of the body.

Components

Airways: • Pharynx • Larynx • Trachea Lungs: • Bronchi • Bronchioles • Alveoli

Components Heart

Blood vessels: • Arteries • Veins • Capillaries

Diaphragm

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

CO2

Red blood cells are replenished with oxygen from the alveolus and carbon dioxide is released from the blood into the alveolus.

From the lungs to the heart

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The airways of the lungs end at the alveoli, the microscopic air sacs that enable gas exchange.

CO2

O2

O2

Capillary

Red blood cell

The carbon dioxide released from the blood exits the body during exhalation. Inhalation brings in fresh air, containing oxygen.

The gas exchange 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. LINK

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104

The double circulatory system is an adaptation to a high oxygen environment and high metabolic rate. Consistently high metabolic rates allow greater activity and mean the organism does not to rely on the environment for its body heat.

Haemoglobin

Oxygen is transported in red blood cells by the protein haemoglobin. In the capillaries of the lungs or gills, haemoglobin binds oxygen tightly, but releases it in the tissues. Carbon dioxide is carried in the blood as bicarbonate (HCO3-). In the lungs it dissociates back into CO2 and water.

Closed double circulatory systems can develop much higher pressure than single circuit (fish) or open systems (insects). Blood, carrying nutrients and oxygen, can be moved to vital organs or muscle quickly and wastes and carbon dioxide can be quickly transported away.

1. (a) At which point in the body do the respiratory and circulatory systems directly interact?

(b) Explain what is happening at this point:

2. Discuss how the circulatory system and gas exchange system work together to provide a mammal with oxygen and remove carbon dioxide:

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3. Mammals maintain a constant body temperature independently of their environment. What are the advantages of this, what is the cost, and how is it related to the efficiency of the gas exchange and circulatory systems?

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74 Internal Transport and Digestion Interaction

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Key Idea: The circulatory and digestive systems interact to provide the body's tissues with nutrients.

Circulatory system

Digestive system

Interaction between systems

Function

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.

Function

Digest food and absorb useful molecules from it, and eliminate undigested material.

Components

Mouth and pharynx

Components

Oesophagus Stomach

Heart

Liver and gall bladder (accessory organs)

Blood vessels: • Arteries • Veins • Capillaries

Pancreas (accessory organ) Small intestine

Blood

Large intestine

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

Villi

Capillaries

Hepatic portal vein

Small intestine

CO2

O2

Capillary

at hep ic por tal vein

Villi project into the lumen of the small intestine and absorb nutrients. Villi contain capillary networks that receive the nutrients and transport them to the hepatic portal system.

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To

Intestinal epithelial cell

Glucose

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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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.

Suseno

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Human liver

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.

The length of the digestive tract is related to the diet of an organism. Digestive tracts tend to be relatively shorter in carnivores and relatively longer in herbivores. Nutrients absorbed across the intestinal wall are transported in the blood to the body's cells or to the liver for storage.

1. How are nutrients transported in the blood?

2. (a) At which two points in the body do the digestive and circulatory systems directly interact?

(b) Explain what is happening at these points:

3. (a) What happens to blood flow to the digestive tract after a meal?

(b) Explain why the blood from the stomach and intestines drains to the liver via the hepatic portal system rather than just returning directly to the heart:

5. Intestinal villi increase the surface area for what two processes?

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4. In your own words, describe how the circulatory and digestive systems work together to provide the body with nutrients:

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75 What You Know So Far: Internal Transport Structure of the heart

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Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question that follows. Use the points in the introduction and the hints provided to help you:

HINT: Describe the structure of the four chambered mammalian heart.

Comparison of open and closed circulatory systems

HINT: Compare the circulatory systems of insects and vertebrates

System interactions

HINT: Compare the structure, occurrence, and efficiencies of single and double circulatory systems.

HINT: Describe the interactions between the internal transport system and the gas exchange system or digestive system in mammals.

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Double and single circulatory systems

REVISE


76 Essay Style Question: Internal Transport

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Large, complex animals have adaptations for transporting respiratory gases and nutrients around their bodies. 1. For the biological process of internal transport, use the taxonomic groups mammals, fish, and insects to answer the following (you may use extra paper if required):

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• Compare and contrast adaptations for internal transport. • You should focus on: • Heart structure • Type of circulatory system • How the system is adapted to the taxon's way of life and metabolic needs:

TEST

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77 KEY TERMS AND IDEAS: Internal Transport 1. In vertebrates (e.g. mammals and fish): (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?

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2. (a) What component of vertebrate blood is involved in carrying oxygen?

(b) What cell types in blood are involved in defence against pathogens?

3. Identify the blood vessels labelled A and B on the photo (right). Give reasons for your answer:

A

A:

B:

B

4. Match the following words with their definitions: atrium

A The circulatory fluid of insects and other arthropods.

blood

B Circulatory system in which the blood is fully contained within vessels.

closed circulatory system

C Circulatory system in which blood travels from the heart to lungs and back (via the pulmonary circuit) before being pumped to the rest of the body.

double circulatory system

D These cells lack a nucleus, but contain large amounts of haemoglobin.

haemolymph heart

red blood cells

E The central organ of the circulatory system, which pumps blood around the body. A protein carried in blood that is able to bind and transport oxygen. F G Chamber of the heart that pumps blood into the arteries.

respiratory pigment

H 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.

ventricle

I Chamber of the heart that receives blood from the veins.

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5. (a) Add the following labels to the cross section of the heart below: right ventricle, wall of left ventricle, left ventricle, wall of right ventricle.

(b) Explain why the heart above has one ventricle with large muscles and one with thinner walls:

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TEST


Adaptations of plants to their way of life

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Achievement Standard

2.3

Key terms

Gas exchange carbon dioxide guard cells leaf lenticel oxygen stomata water vapour

Reproduction alternation of generations angiosperm cone cross-pollination double fertilisation egg embryo fern fertilisation flower gametophyte germination gymnosperm meiosis mitosis pollen pollen tube pollination seed seed dispersal sporophyte zygote

Plants possess structural, physiological, and behavioural adaptations that enable them to carry out the life processes essential to survival in their habitat and exploitation of their niche.

Achievement criteria and explanatory notes Achievement criteria for achieved, merit, and excellence

c

A

Demonstrate understanding of adaptation of plants to their way of life: Describe the adaptations and identify aspects of the adaptations that enable each organism to carry out its life process(es) in order to survive in its habitat.

c

M

Demonstrate in-depth understanding of adaptation of plants to their way of life: Provide a biological reason that explains how or why the adaptations enable each organism to carry out its life process(es) in order to survive in its habitat.

c

E

Demonstrate comprehensive understanding of adaptation of plants to their way of life: Link biological ideas the adaptations that enable each organism to carry out its life process(es) in order to survive in its habitat. The discussion may involve justifying, relating, evaluating, comparing and contrasting, and analysing and it must include consideration of the two points from below appropriate to your chosen context.

i

Compare diversity of adaptation in response to the same demand across different taxonomic or functional groups (e.g. reproduction in ferns, gymnosperms, angiosperms or transpiration in mesophytes, hydrophytes, and xerophytes).

ii

Explain limitations and advantages involved in each feature within each organism.

Option 1: Adaptation related to one life process over three taxonomic or functional groups:

c

Option 2: Adaptation across two related life processes within one taxonomic or functional group:

c

i

Make connections between two life processes within each organism that enhance the effectiveness of both processes (e.g. gas exchange and transpiration in mesophytes).

ii

Explain limitations and advantages involved in each feature.

EII

Explanatory notes

Understanding of adaptation to demonstrate in relation to...

c

1a

One life process over three taxonomic or functional groups of plants, or

c

1b

Two related life processes within one taxonomic or functional group.

c

2 •

The ways in which an organism carries out all its life processes, including: relationships with other organisms: competition, predation, mutualism, parasitism, adaptations to the physical environment.

3

Life processes (for plants) selected from:

Activity number

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Transpiration adhesion cohesion cohesion-tension hypothesis dicot guard cells halophyte hydrophyte leaf mesophyte monocot potometer root stem stomata transpiration transpiration pull xerophyte xylem

78 - 89

i

transpiration (adaptations for maximising water uptake and limiting water loss)

c

ii

gas exchange (adaptations for exchanging respiratory gases with the environment) 78 79 82

c

iii

reproduction (adaptations for successfully passing on genes to offspring)

c

90  - 104


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What you need to know for this Achievement Standard Adaptations for transpiration Activities 78 - 89

By the end of this section you should be able to:

c

Explain transpiration in a flowering plant, with reference to how water and minerals move through the plant from the roots to the leaves. Demonstrate understanding of the following:

The properties of water that are important in the process, particularly adhesion and cohesion. The cohesion-tension hypothesis for water movement in the xylem.

Transpiration pull as a result of evaporative water loss and a gradient in solute concentration.

c Describe and explain the benefits of transpiration to the plant. Recognise

transpiration as an

inevitable consequence of gas exchange in plants.

c

Using second hand data, explain the effect of environment on transpiration rate, including: humidity, light, air movement, temperature, and water availability.

c

Explain the relationship between transpiration and uptake of water and minerals at the root. Describe the mechanism and pathways for water uptake in plant roots.

c

Explain how plants reduce water loss from stomata while maximising light and carbon dioxide capture for the process of photosynthesis.

Adaptations for transpiration in functional groups.

Capkuckokos CC4

c

Describe and explain the adaptations for transpiration in mesophytes. Explain how their adaptations enable them to successfully exploit equable climates.

c

Describe and explain adaptations to minimise water losses by transpiration in xerophytes. Discuss the challenges to survival posed by arid environments and the various ways in which xerophytes have solved these problems.

c

Describe and explain the adaptations for transpiration in hydrophytes. How are their challenges to survival different to those of mesophytes and xerophytes?

c

Describe and explain adaptations for transpiration in mangroves. How does their saline environment affect their ability to take up water to replace losses by transpiration.

Adaptations for gas exchange Activities 78 - 79, 82 - 86

By the end of this section you should be able to:

Dave Powell USDA

c

Describe the relationship between gas exchange and transpiration in plants.

c

Describe and explain adaptations for gas exchange and transpiration in a mesophyte.

Adaptations for sexual reproduction Activities 90  - 104

By the end of this section you should be able to:

c

Describe alternation of generations as a distinctive feature of plants. Explain the significance of the gametophyte (N) and sporophyte (2N) generations and their relative dominance in ferns and in seed plants (angiosperms and gymnosperms).

c Describe and explain the life cycle and reproduction of a typical fern, including the role of meiosis

P. Camill

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and mitosis, the main dispersal stage, and the reliance of ferns on free water for reproduction.

Kent Pryor

c

Describe the life cycles of angiosperms and gymnosperms. Compare and contrast key features of their life cycles with the life cycle of ferns.

c

Where do mitosis, meiosis, and fertilisation occur in the life cycle and what is the purpose of each of these processes?

c

Compare and contrast the adaptations of the reproductive structures in insect pollinated and wind pollinated angiosperms or in angiosperms and gymnosperms.

c

Describe the adaptations of the reproductive structures in gymnosperms.

c

Explain the relationship between pollination method and the amount and type of pollen produced in seed plants. Describe adaptations to ensure cross-pollination in seed plants.

c

Describe fertilisation in angiosperms and gymnosperms. Include reference to the development of the pollen tube and the significance of the double fertilisation in angiosperms.

c

Explain the purpose of the seed. Describe structural and functional diversity in the seeds of angiosperms. Explain how the seed is activated in germination.

c

Describe and explain diversity in methods of seed dispersal in seed plants.


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78 The Plant Transport System Key Idea: The xylem and phloem form the vascular tissue that moves fluids and nutrients about the

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plant.

ffThe 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.

ffTwo types of vascular tissue make up the plant transport system: xylem, which transports water and minerals, and

phloem, 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.

The vascular tissues or plant "veins" are the phloem and xylem. These tissues are continuous throughout the plant, from the roots to the shoots.

H2O

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.

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

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.

Phloem

Xylem

Vascular cylinder

Root cross section

2. Briefly describe why plants need a transport system:

3. (a) What is the function of xylem?

(b) How does water enter the xylem?

(c) Why is water loss a consequence of gas exchange?

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1. Name the two vascular tissues in plants:

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G

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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.

xP

S

P

G

XV

Micropix cc 3.0

XV (torn) SF

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?

Xylem is dead when mature. Note how the cells have lost their cytoplasm.

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

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.

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 ~80 Âľm

Pits and bordered pits allow transfer of water between cells but there are no end wall perforations.

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5. (a) How does water pass between vessels?

McKDandy cc 2.5

Vessels

No cytoplasm or nucleus in mature cell.

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.


79 Transpiration

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114

Key Idea: Water moves through the xylem primarily as a result of evaporation from the leaves and the

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cohesive and adhesive properties of water molecules.

ffIn a vascular plant, water moves from the roots to the leaves via the xylem. Approximately 99% of the water

absorbed from the soil by a plant's roots is lost by evaporation from the plant's leaves and stems. This loss of water is called transpiration and occurs mostly through pores in the leaf called stomata (sing. stoma).

ffPlants rely on an increase in solute concentration from the roots to the leaves to transport water up the plant. Water flows passively from a low solute concentration (high water concentration) to a high solute concentration (low water concentration). This gradient, called transpiration pull, is primarily responsible for water moving up the plant, although cohesion-tension and root pressure also play a part (see below).

ffThe transpiration stream provides a constant supply of water for essential life processes and helps the plant to maintain an adequate uptake of minerals from the soil. Evaporative water loss also cools the plant.

Air

Water

Evaporative loss of water from the leaves as water vapour

The role of stomata

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.

ffStomata open: gas exchange and

Leaves

transpiration rate increase.

Highest solute concentration Lowest water concentration

ffStomata closed: gas exchange and transpiration rates decrease.

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.

G

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

Soil

Highest water concentration Lowest solute concentration

Water

1. What is transpiration?

2. How does the plant regulate the amount of water lost from the leaves?

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Xylem

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Processes involved in moving water through the xylem Transpiration pull

Leaf

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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.

Cell wall

Plasmodesma

Cytoplasm

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.

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.

Air space

Epidermal cell

Stoma

Xylem vessel

Guard cell

Evaporative loss of water vapour

Symplast pathway (cytoplasm) Apoplast pathway (non-living components)

Water molecule

Water is drawn up the plant xylem

3. Describe one benefit of the transpiration stream for a plant:

4. (a) What would happen if too much water was lost from the leaves?

(b) When might this happen?

5. Describe the three processes that assist the transport of water from the roots of the plant upward:

(b)

(c)

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(a)

6. 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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80 Measuring Transpiration in Plants

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Key Idea: Physical factors in the environment such as humidity, temperature, light level, and air

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movement affect transpiration. Transpiration can be measured with a potometer.

Measuring transpiration

ffTranspiration rate (water loss per unit of time) can be measured using a

simple instrument called a potometer. A basic potometer can easily be moved around so that transpiration rate can be measured under different environmental conditions.

ffPotometers are commonly used to investigate the effect of the following environmental conditions on transpiration rate: • • • • •

Humidity Water supply Temperature Light level Air movement

ffMany plants have adaptations to minimise water loss. Potometers can be used to compare transpiration rates in plants with different adaptations. For example comparing transpiration rates in plants with narrow leaves compared to rates in plants with broad leaves.

How a potometer works

Fresh, leafy shoot

The progress of an air bubble along the pipette is measured at regular intervals

1 cm3 (1 mL) pipette

Bung sealed with petroleum jelly

Rubber bung

A bubble potometer, like the one shown on the left, measures the rate of water loss indirectly. Transpiration rate is measured by measuring how much the bubble moves over a period of time. The movement of the bubble is assumed to be caused by the plant taking up water to replace the water lost by transpiration. In transpiration experiments, it is important that the system is sealed (watertight). You can then be confident that the water losses you are recording are the result of transpiration, not leakage from the system.

Clamp stand

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Flask filled with water

Transpiration experiment ffThis activity describes the results of a plant transpiration experiment investigating the effect of specific environmental conditions on transpiration rate.

ffThe environmental conditions investigated were ambient (standard room conditions), wind (fan), humidity (mist), and bright light (lamp).

ffA potometer was used to measure transpiration.

ffOnce set up, the apparatus was equilibrated for 10 minutes, and the position of the air bubble in the pipette was recorded. This is the time 0 reading.

ffThe plant was then exposed to one of the environmental conditions described above. Students recorded the bubble position every two minutes over a 20 minute period. Results are given in Table 1.

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1. The distance the bubble travelled for each environmental condition is given in Table 1 below. Convert the distance the bubble travelled (in mm) into water loss (mL). For every mm the bubble moved, 0.1 mL of water was lost by transpiration (e.g. 1 mm = 0.1 mL water loss). Determine the water loss for each environmental condition (the first has been completed for you).

Table 1. Water loss under different environmental conditions

Time (min)

0

2

4

6

8

10

12

14

16

18

20

Ambient (mm)

0

0

0

1

1

2

3

3

4

5

5

Ambient (mL)

0

0

0

0.1

0.1

0.2

0.3

0.3

0.4

0.5

0.5

Fan (mm)

0

4

7

9

14

17

25

28

31

33

34

0

0

0

0

1

1

2

2

2

3

3

0

1

3

5

6

7

8

9

10

11

13

Treatment

Fan (mL)

High humidity (mm)

High humidity (mL) Bright light (mm)

Bright light (mL)

2. Using an appropriate graph, plot the water loss (in mL) for each environmental condition on the grid below.

(b) Name the environmental conditions that increased water loss:

(c) How do the environmental conditions in (b) cause water loss?

4. Why is it important that the potometer has no leaks in it?

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3. (a) What is the control for this experiment?


118 Pine

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Aloe (agave)

A conifer

Eucalyptus

Sunflower

An Australian gum tree

A perennial dicot with large leaves

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A succulent

Tropical species with thick, fleshy leaves. Physiology allows it to fix CO2 during the night and keep stomata closed during the day.

Temperate species with thin, needle like leaves and a thick waxy leaf cuticle. Stomata are sunken into pits.

Sub-tropical drought tolerant species with a deep root systems and waxy leaves that hang downwards.

Widespread cultivated North American dicot with a showy flower head and very large soft leaves.

5. Read the information below and use it to design an experiment to measure transpiration rate in different plant species:

Do different plant species lose water at different rates?

ffSome of the adaptations of each of the plants pictured above are described below the photographs. Adaptations that help to reduce water loss include reduced or sunken stomata, thick waxy leaves, and succulence. Deep root systems can help plants take up enough water despite losses by transpiration.

ffUse the information about the use of the potometer to design an experiment to compare rates of water loss (as

measured by water uptake) in the four different plants above. Do not include a full method, but think about the duration of your experiment, assumptions you are making, and how you will control the experimental conditions.

6. (a) Write a hypothesis for the experiment and a prediction of its outcome:

(b) Explain the basis of your prediction:

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My experimental design to compare transpiration loss in different plants

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81 Uptake at the Root Key Idea: Water absorption by the roots from the soil is a passive process that replaces the water lost

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from the leaves. Uptake of mineral ions can be passive or active.

ffPlants constantly take up water, minerals, and ions to compensate

for the continuous loss of water from the leaves by transpiration and to provide the materials they need for the manufacture of food. The uptake of water and minerals is mostly restricted to the younger, most recently formed cells of the roots and the root hairs.

ffWater moves into the roots because the solute concentration is higher in the root tissue than in the soil. Water and ions are taken up by diffusion (osmosis in the case of water) and active transport.

ffSome water moves through the plant tissues via cytoplasmic

connections between cells (the plasmodesmata), but most passes through the free spaces outside the plasma membranes of the cells.

ffRoot hairs (right) provide a large surface area for absorption. They lack the waxy cuticle found on leaves so there is no barrier to water uptake.

Water uptake The uptake of water by roots occurs by osmosis. Water takes three paths through the root tissue:

Water and mineral uptake by roots

Root hairs lack a waxy cuticle, so water enters the root easily

Cortex cells of root

ffMost water travels through all the extra-cellular spaces outside the plasma membranes of cells, e.g. through the cell walls.

Epidermal cell

ffA smaller amount moves through the cytoplasm of cells.

Xylem

ffA very small amount travels through the plant vacuoles.

Stele (vascular cylinder). The outer layer of the stele, the pericycle, is next to the endodermis.

Mineral uptake

Root hair

ffMany minerals are dissolved in water and absorbed passively by diffusion.

Schematic cross-section through a dicot root

ffMinerals in very low

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.

1. (a) Why must plants constantly take up water from the soil?

(b) Describe a benefit of a large water uptake:

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Root hairs are extensions of the root epidermal cells and provide a large surface area for absorbing water and nutrients.

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concentration in the soil must be taken up by active transport by the root cells. This requires energy expenditure by the plant.

Water moves by osmosis

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Paths for water movement through the plant Cell wall

Plasma membrane

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Plasmodesmata

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 apoplast, i.e. the spaces within the cellulose cell walls, the water-filled spaces of dead cells, and the hollow tubes of xylem vessels. A smaller amount moves through the living symplast (the cytoplasm of cells). A very small amount travels through the plant vacuoles.

Epidermis

Cortex

Endodermis

Apoplast Symplast

Zone of lower solute concentration [higher free water] May be due to turgid cells, higher wall pressure or lower concentration of dissolved substances

Pericycle

Xylem

Casparian strip

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 symplast, so the flow of materials into the stele can be regulated.

2. (a) What is the purpose of the root hairs?

(b) What adaptations do root hairs show for their role:

3. (a) What two mechanisms do plants use to absorb nutrients?

(b) Describe the two main pathways by which water moves through a plant:

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4. (a) How does the Casparian strip affect the route water takes into the stele?

(b) Why might this feature be an advantage in terms of selective mineral uptake?

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82 Adaptations of Leaves Key Idea: Plant leaves have adaptations to reduce water loss while maximising the capture of light

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and gas exchanges (entry of carbon dioxide) for photosynthesis.

ffTo photosynthesise, plants require

Waxy cuticle (barrier to diffusion and water loss).

light and carbon dioxide. Chlorophyll in the leaf chloroplasts captures the light energy, while carbon dioxide enters the plant by diffusion through the leaf stomata (sing. stoma).

Upper epidermis (no chlorophyll) Palisade mesophyll

ffWhen carbon dioxide enters the leaf

through the stomata, water vapour also diffuses out and the plant loses water (transpiration). Guard cells around each stoma may open and close the pore to regulate these exchanges.

Chloroplast Spongy mesophyll

CO2

ffPlants have developed many ways to

reduce water loss while still maximising light capture and entry of carbon dioxide into the leaf.

Guard cell

Stoma

Lower Leaf epidermis vein

CO2

H2O, O2

Air space

Stoma

Western New Mexico University Department of Natural Sciences

In many plants, e.g. pine (above), the stomata occur along groves set into the leaf. This raises humidity around the stomata and so reduces water loss.

Most higher plants have a waxy cuticle that prevents water evaporating directly from the leaf surface. This helps reduce water losses when stomata are open.

In arid environments, leaves may be small or reduced to spines (e.g. cacti) and the stem may be photosynthetic. Spines or hairs reduce air flow and raise humidity, reducing evaporation.

1. Identify the cells of a dicot leaf where most of the chloroplasts are found:

(b) Explain how this movement is regulated:

3. Describe three ways plants reduce their water loss:

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2. (a) Describe how gases enter and leave the leaf tissue:

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83 Adaptations of Mesophytes

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Key Idea: Mesophytes are plants adapted to temperate environments with no adaptations to deal

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with environmental extremes (e.g. arid or waterlogged environments).

ffMesophytes are by far the most common

type of plant. They occupy the temperate and tropical zones of the Earth where water is readily available, salt levels are not high, and the temperature is relatively stable and moderate. Mesophytes include plants of many sizes, from grasses to large trees. Many have broad leaves that are adapted to maximise the capture of light, with little need to conserve water.

Transpiration rate is regulated by opening and closing of stomata, but the waxy leaf cuticle is relatively thin. Large amounts of water may be lost through stomata but the plant compensates by having a large water uptake.

Leaves are broad and thin with many stomata on the undersides. Stomata can be opened and closed by guard cells to limit water loss. Having stomata restricted to the leaf underside reduces water loss via transpiration without restricting CO2 entry.

Nucleus of epidermal cell

Stoma with guard cells

Many mesophytes withdraw valuable nutrients from the leaves and then lose them prior to winter. Light levels are often so low during the winter, there is no benefit in presenting leaves to the sun. Leaf loss has the added benefit of preventing damage to the soft tissues and makes large plants less vulnerable to storm damage.

Mesophytes, such as this oak, generally have large root masses that both anchor the plant in the soil and provide a large area for obtaining water to replace losses from transpiration.

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1. Define the term mesophyte:

2. Mesophytes are not adapted to extreme environments, but are well adapted to their own niche. Describe three adaptations of mesophytes, explaining how each helps survival in an equable environment:

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84 Adaptations of Xerophytes Key Idea: Xerophytes are adapted to dry conditions and conserving water. Adaptations include small

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leaves, a thick cuticle, sunken stomata, succulence, and absence of leaves.

ffXerophytes are plants with particular adaptations for conserving water (xeromorphic adaptations).

ffMost 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).

ffMany xerophytes have a succulent morphology. Their

stems are often thickened and retain a large amount of water in the tissues, e.g. Aloe.

ffMany xerophytes have a low surface area to volume ratio, reducing the amount of water lost through transpiration from the plant surface.

Desert xerophytes

Adaptations in cacti

Desert plants, such as cacti (below), must cope with low or sporadic rainfall and high transpiration rates.

Hairs

Leaves modified into spines or hairs to reduce water loss. Light coloured spines reflect solar radiation.

Leaf cross section

The surface tissues of many cacti are tolerant of temperatures in excess of 50°C.

1. Define the term xerophyte:

Stoma

RCN

Cacti have a shallow, but extensive fibrous root system. When in the ground the roots are spread out around the plant.

Fine leaf hairs trap air close to the surface and reduce the rate of water loss.

Sunken stomata

Coastal grasses curl their leaves. Stomata are sunken in pits, creating a moist microenvironment around the pore, which reduces water loss, e.g. marram grass.

Trichome (hair)

WBS

Stem becomes the major photosynthetic organ, plus a reservoir for water storage.

Acacia trees have deep root systems, drawing water from sources deep underground.

Oleander has a thick multi-layered epidermis and the stomata are sunken in trichome-filled pits on the underside of the leaf. These pits restrict water loss.

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Rounded shape reduces surface area.

2. Briefly describe three xeromorphic adaptations of plants that reduce water loss: (a)

(b) (c)

3. How does trapping air close to the areas of water loss reduce the transpiration rate?

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85 Adaptations of Hydrophytes Key Idea: Hydrophytes are adapted to living in water. They require little structural support tissue and

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have few adaptations to reduce water loss.

The leaves of submerged plants are thin to increase the surface area of photosynthetic tissue.

Leaves arranged so that they do not shade each other.

Many hydrophytes have large floating leaves with elongated leaf stalks (petioles). The leaf upper surface has stomata and often a thick waxy cuticle to repel water and help to keep the stomata open and clear.

Aerial (above water) flowers.

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

The water lily Nymphaea alba

Cortex

Vascular bundles

Abundant, large air spaces

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. Submerged plant parts lack any cuticle so can freely take up water, gases, and nutrients.

WBS

Stephen Moore

Air spaces

Myriophyllum’s leaves are well spaced and taper towards the surface to assist with gas exchange and capture of light.

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.

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1. Describe some adaptations of hydrophytes that help them survive in their aquatic environment:

2. (a) Do hydrophytes show any particular adaptations to reduce water losses from transpiration?

(b) Explain your answer:

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86 Mangrove Adaptations Key Idea: Mangroves are halophytes (salt tolerant) and specifically adapted to the high salt, water-

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logged environments of estuaries, tidal flats, and salt marshes.

ffMangroves 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.

Salt crystals

ffThe high salt environment, in which there is little fresh water available, would kill most

other kinds of plants as high salt levels cause water to flow out of the cells. Mangroves, e.g. the New Zealand white mangrove or manawa, 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.

Pneumatophores arise from the cable roots and grow 25-30 cm above the mud surface. 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. Thick, waxy leaves and regulation of stomata opening help to reduce water losses. They may also vary the orientation of their leaves to avoid the harsh midday sun and so reduce evaporation from the leaves.

Oxygen

Lenticels

Water level at high tide

Only the top few centimetres of the mud is oxygenated. Beneath, the mud is anaerobic (lacking oxygen), black, and foul-smelling. A deep root

system is of no use here.

RA

A waxy coating of suberin on the root cells excludes 97% of salt from the water taken up by the roots.

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.

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1. Describe a physiological problem associated with living in a highly saline environment:

2. Describe three adaptive features of mangroves that help them survive in a high salt environment: (a)

(b) (c)

3. Use your knowledge of the transpiration process to explain how mangroves maintain the transpiration stream:

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87 What You Know So Far: Transpiration

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126

Uptake of water and minerals

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Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question at the end of the chapter. Use the points in the introduction and the hints provided to help you:

HINT: Describe how water and minerals enter the plant from the environment. Describe the adaptive role of root hairs.

Transpiration

HINT: Describe transpiration and explain its importance.

Adaptations for transpiration

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HINT: Describe adaptations of three functional groups of plants for transpiration in different environments.

REVISE

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88 Essay Question: Transpiration

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Plants living in different environments have evolved specific adaptations for acquiring and maintaining the flow of water they need to carry out photosynthesis and other metabolic functions.

1. For the biological process of transpiration, discuss the adaptations of (three of) hydrophytes, mesophytes, xerophytes, and halophytes that allow them to survive in their respective environments. You should define these functional groups and cover the range of adaptations they exhibit, as well as any advantages and disadvantages associated with them. You may use extra paper if required.

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89 KEY TERMS AND IDEAS: Transpiration

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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?

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.

halophyte

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.

mesophyte

E

Pores in the leaf surface through which gases and water vapour can pass.

potometer

F

A plant with adaptations to live in water.

stomata

G

Partial explanation for the movement of water up the plant in the transpiration stream.

H

A plant adapted to dry conditions and conserving water.

xerophyte

I

A plant adapted to life in highly saline environments.

xylem

J

A plant adapted to moderate environments where water is generally not limiting.

hydrophyte

transpiration

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?

TEST

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90 Alternation of Generations Key Idea: The life cycle of plants includes alternation between a gametophyte generation (haploid, N)

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and a sporophyte generation (diploid, 2N).

ffThe life cycles of all plants include a gametophyte

In seed plants, the pollen is the gametophyte...

generation (haploid or N phase) and a sporophyte generation (diploid or 2N phase). The two generations alternate, each giving rise to the other. This is called alternation of generations.

ffThe sporophyte and gametophyte generations

are named for the type of reproductive cells they produce. Haploid gametophytes (N) produce gametes by mitosis, whereas diploid sporophytes (2N) produce spores by meiosis. Spores develop directly into organisms. Gametes (egg and sperm) unite during fertilisation to form a zygote which gives rise to an organism.

... and the sporophyte is the plant that we are familiar with.

Mitosis

Diploid (2N) generation

Seed coat from parent sporophyte (2N)

Sporophyte (multicellular diploid 2N)

Zygote (2N)

Sporophyte embryo (2N)

Meiosis

Fertilisation

Gametophyte tissue (N)

Spores (N)

Gymnosperm seed

Gametes (N)

Mitosis

Gametophyte (multicellular haploid N)

Alternating generations can result in multiple generations within one structure. In conifers, the seed contains the parent and new sporophyte generations, and the gametophyte generation.

Mitosis

Haploid (N) generation

2. (a) Give an example of a sporophyte:

(b) Give an example of a gametophyte:

3. Complete the table below to show what produces each structure:

Structure

Produced by Process

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Spores

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1. Describe the alternation of generations in plants:

Gametes

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Zygote

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91 Reproduction in Ferns Key Idea: In ferns, the gametophyte is a small free-living plant that produces gametes, which rely on

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water to unite. The sporophyte (the fern plant) produces haploid, unicellular spores.

ffReproduction in ferns is achieved by the alternating

release of spores and gametes (there are no seeds). Spores from the sporophyte develop in sporangia which are typically found on the underside of the fronds in clusters called sori. Spores are single celled and small enough to be spread by the wind.

ffIf a spore settles in a suitable spot, it develops into a

heart-shaped prothallus (the gametophyte), which is typically only one or two cells thick, small, free-living, and photosynthetic. It produces egg cells and motile sperm cells, which require a film of water to reach the egg cells.

Sori

EII

ffThe division of the zygote begins immediately after

fertilisation to produce the embryonic sporophyte. This becomes rooted in the soil and develops fronds. The gametophyte then disintegrates.

The free living prothallus (the gametophyte) develops from the spore.

Life cycle of ferns

Gametophyte

Sporophyte (a) (

The spores are found in clusters underneath the frond called sori.

(b) (

)

Archegonia (female)

)

Mitosis

(c)

Egg cell (N)

Spores (N)

Sporangia form on the underside of special leaves called sporophylls.

Motile sperm

(d)

Antheridia (male)

Sperm cells (N)

Young sporophyte

The young sporophyte remains attached to the gametophyte as it develops. It depends on it until its rhizoids develop.

(e)

Mitosis

The sperm fertilises the egg to produce a zygote.

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1. (a) - (e) Complete the diagram above using the labels N, 2N, fertilisation, meiosis, mitosis: 2. What feature distinguishes ferns from seed plants (angiosperms and gymnosperms)?

3. (a) Explain why ferns require moist environments in order to reproduce:

(b) Explain why this requirement ultimately limits where they can survive in the long term:

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92 Reproduction in Angiosperms Key Idea: Angiosperms are flowering plants. The sporophyte produces flowers, which house the tiny

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gametophytes. Fertilisation of the gametes gives rise to the seed.

ffAngiosperms are flowering seed plants. In seed plants, pollen

(not water) is the vehicle to carry the male gametes to the female gametes. The leafy plant that bears the flowers represents the sporophyte generation of the life cycle. The flowers contain specialised male and female structures (anthers and ovules), in which the male and female gametophytes are formed.

Endosperm

Embryonic sporophyte

ffThe pollen grain is the male gametophyte and the embryo sac is

the female gametophyte. Each mature pollen grain contains two sperm nuclei and each mature embryo sac contains the egg and two polar nuclei.

Corn (Zea mays) seed

ffThe typical life cycle of angiosperms (below) involves the

The gametophyte generation may be important to a plant for reasons other than producing gametes. Endosperm (3N) derived from male and female gametophytes provides nutrients for embryonic sporophyte.

formation of gametes (egg and sperm) from the haploid gametophytes, the fertilisation of the egg by a sperm cell to form the zygote, the production of fruit around the seed, and the germination of the seed and its growth by mitosis.

Life cycle of angiosperms (flowering plants)

Sporophyte (a) (

)

Stigma

Gametophyte

Anther produces pollen (N)

(b) (

)

(c)

Anther

Pollen contains two sperm nuclei (N)

Megaspore mother cell produces the embryo sac, which contains the egg (N)

Ovule produces megaspore mother cell (2N)

Pollination - pollen is transferred to the stigma

Pollen tube transfers sperm to egg

Ovule

Polar nuclei (N+N)

Egg (N)

Seed

(e)

Embryo

Zygote

Mitosis

Endosperm

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Seedling

Pollen tube

One sperm nucleus fuses with the egg cell forming the zygote. The other fuses with the polar nuclei to form the endosperm.

(d)

1. (a) - (e) Complete the diagram above using the labels N, 2N, fertilisation, meiosis, mitosis:

2. (a) In which part of a flowering plant do the male gametophytes develop? (b) In which part of a flowering plant do the female gametophytes develop? 3. Why don't angiosperms need a moist environment for fertilisation to occur?

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93 Reproduction in Gymnosperms

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Key Idea: Conifers (cone-bearing plants) are typical gymnosperms. The sporophytes produce cones

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which produce either pollen or egg cells (not both). Conifers are wind pollinated.

ffThe seeds of gymnosperms are naked and develop on the surface of scales or leaves, which are often modified into cones, as in cycads and conifers (cone bearing seed plants). Conifers usually bear the smaller male and the much larger female cones on the same plant.

ffThe vast majority of gymnosperms are wind pollinated. In conifers,

the wind-carried pollen sticks to a drop of fluid within a tiny hole (the micropyle) at the end of the ovule. When the fluid evaporates, it draws the pollen into the ovule. The pollen develops a pollen tube, which grows towards the egg cell (this can take up to 15 months).

ffAfter fertilisation, a zygote forms which develops into an embryo then

into a seed, which is often 'winged'. Some conifers require the intense heat from a fire to open the cones and release the seeds.

Male (left) and female (right) cones

Life cycle of conifers (cone-bearing plants)

Sporophyte (a) (

Gametophyte (b) (

)

)

Pollination - pollen is carried by the wind.

(c)

Male cone

Microspore mother cells (2N) give rise to pollen.

Micropyle

Pollen (N)

Pollen tube

Fertilisation

(d)

Egg cell (N)

Ovule

Megaspore mother cell (2N)

Female cone

Megagametophyte (N) within the ovule (dotted line)

Seed

Cone scale

Embryo (2N)

Wing

Mitosis

Mitosis

(e)

Seed

Megagametophyte (N)

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Seedling

1. (a) - (e) Complete the diagram above using the labels N, 2N, meiosis, mitosis: 2. Briefly describe how the sperm cell reaches the egg cell in conifers:

3. Which structure in a conifer is equivalent to the embryo sac of angiosperms? 4. What is the role of wind in the complete life cycle of a conifer?

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94 Plant Adaptations for Insect Pollination Key Idea: The flowers of many angiosperms are adapted to attract animal pollinators, which carry

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pollen from flower to flower, increasing the probability of fertilisation.

ffFlowering plants (angiosperms) are highly successful

Petals guide insects towards the pollen or nectar at the centre of the flower. This ensures pollen is transferred in the most efficient way.

organisms. The egg cell is retained within the flower of the parent plant and the male gametes (contained in the pollen) must be transferred to it by pollination before fertilisation can occur. In most angiosperms, the male and female parts occur on the same plant. Some of these plants can self-pollinate, but most have mechanisms that make this difficult or impossible. This means the pollen must be carried to another flower or plant for fertilisation to occur. This helps to ensure cross pollination and therefore reduces inbreeding.

ffTo make this happen, many angiosperms have developed

flowers to attract animal couriers, including insects, birds, and mammals. Insects are the most effective pollinators. Flowers attract insects with brightly coloured petals, scent, and offers of food such as nectar and pollen.

Many flowers produce pigments that reflect ultraviolet light, which can be seen by many insects, thus attracting specific pollinators.

Cross section of an insect pollinated flower

Anther: Top portion of stamen. the stamen, the male pollen. organ of reproduction.

Carpel (female)

part of the carpel. Pollen grains will germinate only

Style: Supports if they land here. the stigma.

Filament: The slender Filament: stalk of the stamen that supports the anther.

Ovary: The the carpel where Ovary:the The base of the carpel where the ovules develop.

Petals: Collectively, these form the corolla. Petals: Together Often brightly coloured.

form the corolla. Often brightly

ovules develop.

Sepals: Together form the calyx. Usually green, but coloured. sometimes the same colour as the petals.

Ovules: Egg Ovules: cells. These are eggs andthey once fertilised, When fertilised become the seeds. The become the seeds. ovule skin becomes the

Sepals: Collectively form the in colour.

Nectary: Plants produce a calyx. sugary liquid called Often nectar to green attract insects to the flower.

seed coat or testa.

Receptacle: The swollen base of the flower. Nectary: Produces Sometimes it forms the succulent tissue of the fruit.

An entire female part is the carpel. There may be one Receptacle: Base or more carpels per flower.

of the flower.

ii. The female part of the flower is the

(b) Name two adaptations that angiosperms use to attract pollinators to their flowers:

(c) How do angiosperms guide insects towards the pollen or nectar?

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nectar.

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1. (a) What is the function of the anther? (b) What is the function of the stigma? (c) i. The male part of the flower is the 2. (a) Why do flowering plants need to attract pollinators?

Supports the

anther.

Style: The structure that supports base ofthe stigma.

Stamen (male)

Anther: Top part of the Produces the

Stigma: Receives the pollen. Stigma: The receptive

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95 Pollination Relationships Key Idea: Many angiosperms have a mutualistic relationship with their pollinators in which the plants

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achieve pollination by rewarding insects with food such as nectar or pollen.

ffPollination of flowers by insects is usually mutualistic. Mutualistic

relationships involve exchanges between two species so that each species benefits. The benefits are not always equal for each party, because each species acts in its own interests. In the case of insect pollination, the insect benefits from the energy in the plant nectar or pollen it consumes. The plant benefits by having its gametes transferred to another plant. About 87.5% of all flowering plants are pollinated by animals, with the vast majority being insect pollinated. Positive and negative effects of non-specific pollinators

Orchid flowers (above) are highly variable in their structure and often highly specialised. They often have only one specific insect pollinator. This relationship is the result of the plant and its pollinator evolving together.

Plant species A

+ve effect

+

â&#x20AC;&#x201C;ve effect

Pollinator

+ve effect

Plant species B

Magnolias are an ancient plant group with generalised flowers (above). They evolved before bees, and beetles are their main pollinators. Magnolias produce large amounts of pollen, some of which is food for the beetles.

Most insect pollinators are generalists, meaning they do not form pollination relationships with specific plants. Honey bees, for example, pollinate many different kinds of plants. This can be of negative value to a particular plant, as the energy expended in producing pollen and nectar is wasted if the bee does not next fly to a plant of the same species. Plants of the same species often synchronous their flowering, this helps to ensure the insect will visit (and pollinate) flowers of the same species consecutively.

1. Define mutualism:

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2. Describe the benefits of mutualism to both the flower and the pollinator:

3. (a) Identify one advantage and one disadvantage of a having specific pollinator:

(b) Identify one advantage and one disadvantage of having a generalist pollinator:

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96 Wind Pollination Key Idea: The flowers of wind pollinated plants are usually small and pale coloured. Vast quantities of

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pollen are needed to achieve pollination.

The structure of wind pollinated flowers

ffWind pollinated plants do not need to attract animal pollinators,

Oat spikelets

so they do not invest energy in producing large, brightly coloured flowers or nectar. Wind pollinated flowers typically have many small, drab flowers, and may lack petals altogether, while the anthers and stigma are large and hang clear of the surrounding structures.

ffWind pollination is very inefficient and most pollen falls to the ground

Grass flowers are arranged in spikelets, each with one or more florets, e.g. oats, above.

within 100 m of the flower. The anthers must therefore produce large amounts of pollen to ensure cross-pollination with other plants. A spikelet

A floret

Stigma

The large, feathery stigma hangs clear of the flower.

Anther hangs well clear of the leaf scale and produces lightweight pollen which is easily carried away

Anther

Filament

Stamen

Awn is a bristle-like extension of the floret. It is typical of grasses.

Exposed reproductive parts of one floret.

Large quantities of smooth, dry pollen are produced but there is no scent or nectar.

Modified leaves enclose both the spikelet and the individual florets.

Bulbous structures at the base help open the floret

Roger Griffith

Wind pollinated angiosperms range from relatively small, such as this grass plant, to very large, such as oak trees. They are often found in temperate regions.

1. Describe two features of the flowers of wind pollinated plants: (a) (b)

2. Identify one advantage and one disadvantage of wind pollination:

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The flowers of wind pollinated trees are small and numerous, and often have green petals. They often open before the leaves so that pollen can be dispersed without obstruction by leaves.

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Only around 10% of flowering plants are wind pollinated, but this 10% includes the grasses, which are one of the most successful of all angiosperm groups.

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97 Comparing Insect and Wind Pollination

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Key Idea: Insect and wind pollinated plants differ in where they invest their energy for pollen dispersal.

Insect pollinated flowers

Wind pollinated flowers

Large brightly coloured petals to attract insect pollinators.

Small petals, usually pale or dull green.

Often sweetly scented.

No scent.

Contain nectaries to produce nectar as a reward for pollinators.

No nectaries.

Pollen is often sticky, to ensure it in securely attached to pollinators.

Pollen is small and light to be carried to by the wind.

Moderate to small amounts of pollen produced.

Pollen produced in vast amounts to ensure at least some grains reach another plant of the same species.

Anthers and stigma inside the petals and place to ensure pollinators brushes against them.

Anthers and stigma extend beyond the petals.

Stigma has a sticky coat to ensure pollen sticks to it.

Stigma is feathery or net-like.

Hardyplants

(a) Appearance of the flowers:

(b) Production of scent and nectar:

(c) Position of the reproductive parts (stigma, stamens):

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1. Using the photographs and the table above to help you, contrast wind and insect pollinated flowers with respect to each of the following characteristics. For each, give reasons for the differences observed:

2. Where do insect pollinated plants put their energy for pollen dispersal compared to wind pollinated plants?

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98 Fertilisation Follows Pollination Key Idea: After pollination has been achieved, both angiosperm and gymnosperm pollen produce a

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pollen tube that transfers sperm nuclei to the egg cell.

ffFertilisation is the union of male and female gametes to produce a

Male pine cone

fertilised egg (zygote).

Female pine cone

ffBefore the egg and sperm can fuse in fertilisation, the pollen (containing the male gametes) must be transferred from the male reproductive structures to the female reproductive structures in pollination. Plants rarely self-pollinate, although they can be made to do so.

ffMechanisms to ensure cross-pollination include:

• Having separate male and female plants (e.g. willow).

• Separate male and female flowers or cones (e.g. pine).

• Male and female parts of the flower develop at different times. For example, avocados exhibit an unusual flowering behaviour that promotes cross-pollination. In type A cultivars, the flowers open as females during the morning of the first day, and on the second day they open as males in the afternoon. In type B cultivars, female flowers open on the afternoon of the first day, but open as male flowers on the morning of the next day.

Avocado flowers

Open flower

Closed flower

• Being self-incompatible (pollen will not fertilise the same plant).

ffCross pollination increases genetic diversity and may provide plants with an adaptive advantage if environmental conditions change.

ffOnce pollination has occurred, the pollen grain can develop to produce the pollen tube allowing the sperm nuclei to enter the ovule and fertilisation to take place.

Fertilisation in angiosperms

Pollen wall composed of extremely hard material called sporopollenin.

ffIn angiosperms, pollen lands on the sticky stigma and

germinates, producing a pollen tube that extends to the ovary.

ffThe pollen tube is directed by chemicals (usually calcium) to the ovule. It enters the ovule through a small gap called the micropyle.

ffA double fertilisation takes place. One sperm nucleus fuses

Sperm cells

Tube nucleus

Pollen tube

with the egg to form the zygote. A second sperm nucleus fuses with the two polar nuclei within the embryo sac to produce a triploid endosperm (3N).

Germinating pollen grain

ffThe endosperm plays an important role by supplying nutrients to the developing embryo and helping to regulate seed development and germination.

Anther

ffThere are usually many ovules in an ovary, therefore many pollen grains (and fertilisations) are needed for the entire ovary to develop.

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Pollen tube grows down to ovary Stamen

Ovary wall Ovule

Embryo sac Polar nuclei Egg Micropyle

SEM image of pollen tubes growing from pollen grains.

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Two sperm nuclei

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Fertilisation in gymnosperms ffIn conifers, the pollen lands on a drop of fluid covering the micropyle and is drawn down into the ovule as the

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fluid evaporates.

ffThe pollen tube germinates shortly afterwards, but may take 15 months to reach the egg cell.

ffThe pollen grain produces the sperm-producing cell, which produces two sperm nuclei. One nucleus fuses with the egg cell while the other degenerates.

ffAbout 4% of seeds contain multiple embryos due to more than one egg cell in the ovule developing.

ffIn conifers (and all gymnosperms), the seed is made up of the seed coat, the embryo, and the remains of the female gametophyte (megagametophyte), which provides nourishment to the developing embryo.

Micropyle liquid

Pollen tube

Sperm nuclei

Egg nuclei

Egg cell

Female gametophyte

Micropyle

Scale from female cone

Pollen grain

Ovule

Female cone

1. Distinguish between pollination and fertilisation:

2. (a) Why is cross pollination important in plants?

(b) Describe some mechanisms plants use to ensure cross pollination:

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3. (a) What event distinguishes fertilisation in angiosperms and gymnosperms?

(b) Describe what occurs during this event and name the structure formed in addition to the embryo:

(c) What is the purpose of the structure named in 2(b)?

(d) what is the functionally equivalent structure in a gymnosperm?

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99 Structure of Seeds Key Idea: A seed contains the dormant embryonic plant and its food store. The adaptations of seeds

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ensure their survival through dormancy and germination at an appropriate time.

Seed structure

ffAfter fertilisation has occurred in angiosperms, the ovary

Seed structure in a monocot and dicot

develops into the fruit and the ovules within the ovary become the seeds.

Testa or seed coat

ffA seed is an entire reproductive unit, containing the

Endosperm

embryonic plant in a state of dormancy. During the last stages of maturing, the seed dehydrates until its water content is only 5-15% of its weight. The embryo stops growing and remains dormant until the seed germinates (begins to grow and develop).

Plumule

Cotyledon

ffEvery seed contains an embryo comprising a rudimentary

Radicle

shoot (plumule), root (radicle), and one or two seed leaves (cotyledons). The embryo and its food supply are encased in a tough, protective seed coat or testa.

Root cap

ffIn monocots, the endosperm provides the food supply,

Dicot seed (garden bean: Phaseolus vulgaris)

Monocot seed (maize: Zea mays)

whereas in most dicot seeds, the nutrients from the endosperm are transferred to the large, fleshy cotyledons.

Seeds only germinate in favourable conditions

ffSeeds only germinate when the environmental conditions are favourable

to growth. If plants germinated during unfavourable conditions (e.g. to dry or too cold) they are unlikely to survive. As a result, seeds have several adaptations to provide them with an adaptive advantage. These include:

Germinating seed

• Layers of hardened tissue to prevent the seed drying out. • Tissue to provide nutrients for embryonic growth. • Ability to remain dormant until environmental conditions become favourable for germination and growth.

ffGermination requires rehydration of the seed and reactivation of the

metabolism. The seed absorbs water through the seed coat (testa) and micropyle. As the dry substances in the seed take up water, the cells expand, metabolism is reactivated, and embryonic growth begins. Activation stimulates the synthesis of various hormones and enzymes. Starch in the cotyledons or endosperm is hydrolysed to produce sugars. The mobilised food stores are then delivered to the developing roots and shoots.

2. (a) What is the adaptive advantage of germination being triggered by favourable environmental conditions?

(a)

(b)

(c)

(b) What adaptation ensures that a seed does not germinate until there is enough water in the environment?

(d) (e)

(f)

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1. Identify the parts of the monocot seed (a) - (f) right:

Seed coat

Word list: Cotyledon, plumule, radicle, testa, root cap, endosperm

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100 Adaptations for Seed Dispersal Key Idea: Seeds must be dispersed from the parent plant to reduce competition for light and nutrients.

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Seeds may be dispersed by wind, water, or animals.

ffIn order for a seed to survive and grow into a new plant, it must be

transported (dispersed) into an environment where it has enough light, water, and nutrients to germinate, survive, and eventually reproduce. Efficient dispersal helps to reduce competition with the parent plant for light, water, and nutrients.

ffPlants have evolved many ways to ensure that their seeds are

dispersed. This has given them opportunities to expand their range. In some cases the seed itself is the agent of dispersal, but often it is the fruit or an associated attached structure.

ffThe main agents of seed dispersal are wind, water, and animals, but

others can also be dispersed through explosive discharge or shaking from pods or capsules (e.g. flax, peas, gorse) or through fire (e.g. many Australian plants).

Image right: Coconut seeds germinating after dispersal from the parent plant.

Jenny Ladley

Mechanisms of seed dispersal

Wind dispersal Wind dispersed seeds have wing-like or feathery structures that catch the air currents and carry the seeds long distances from the parent plant. Some seeds are light and feathery (dandelions and swan plants) while others (pine, kauri, maple) have winged seeds that help them travel. Plants that rely on wind dispersal produce a large number of seeds. This increases the chances of a seed successfully establishing elsewhere.

Animal dispersal Plants that rely on animals to spread their seeds may have hooks or barbs that catch the animal hair, sticky secretions that adhere to the skin or hair, or fruits that are eaten. In New Zealand, birds are common seed carriers. They consume the seed (contained within a fleshy fruit) and disperse the seeds in their droppings. Seeds can be dispersed over long distances, including to offshore islands. Examples include miro, puriri, and tawa.

Water dispersal When water is used as the means of dispersal the seed simply floats away from the parent plant. Many seeds dispersed by fresh water are very small, so that they float. For seeds that are dispersed by ocean currents, the seed must be very buoyant and capable of surviving in the salt water. Some (coconut) have woody, waterproof coverings and can remain at sea for a considerable time. Others (mangrove) are partly developed propagules and can take root as soon as they settle in a suitable substrate.

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1. What is the purpose of seed dispersal?

2. Describe at least one typical adaptation of seeds dispersed by each of the following mechanisms:

(a) Wind:

(b) Animal:

(c) Water:

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For the examples below, describe the method of dispersal and the adaptive features associated with the method:

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3. Pinus seed is surround by a scale shaped like a long wing: (a) Dispersal mechanism:

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(b) Adaptive features:

KP

4. Whau (Entelea aborescens) thorny capsules each hold about 100 seeds:

(a) Dispersal mechanism:

(b) Adaptive features:

5. Karamu (Coprosma robusta) produces succulent, bright fruits:

(a) Dispersal mechanism:

(b) Adaptive features:

6. Totara (Podocarpus totara) produces a fleshy modified cone below the seed: (a) Dispersal mechanism:

(b) Adaptive features:

Kahuroa

(a) Dispersal mechanism:

(b) Adaptive features:

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7. New Zealand flax (Phormium spp.) produces seeds in pods:

8. New Zealand mangrove (Avicennia marina subsp. australasica) produces buoyant seeds that float on water:

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(a) Dispersal mechanism:

(b) Adaptive features:

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101 A Summary of Plant Reproduction Key Idea: Ferns, angiosperms, and gymnosperms have a number of differences and similarities in

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how they reproduce.

1. Use the table below to summarise the similarities and differences in sexual reproduction in ferns, angiosperms, and gymnosperms:

Fern

Angiosperm

Gymnosperm (conifer)

Dominant generation

Alternate generation

Method of transferring male gametes

Reproductive structure

Sporangia

Single or double fertilisation?

Production of seeds?

Method of seed/spore dispersal

Reliance on water with respect to reproduction

2. One of the principal trends evidence in plant life cycles is the increasing independence on water for reproduction. Explain what feature of the male gamete (sperm or pollen) illustrates this.

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3. (a) Complete the diagram below adding the following labels: spores, gametes, zygote. (b) Add the labels N, 2N, haploid, and diploid to the appropriate places in the diagram:

Sporophyte

Fertilisation

Meiosis

Gametophyte

TEST

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102 What You Know So Far: Plant Reproduction Pollination and fertilisation

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Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question at the end of the chapter. Use the points in the introduction and the hints provided to help you:

HINT: Compare mechanisms for pollination. Describe fertilisation in angiosperms and gymnosperms.

Alternation of generations

HINT: Describe the life cycles of plants and identify the gametophyte and sporophyte generations.

Adaptations for reproduction

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HINT: Describe adaptations for successful reproduction, including adaptations for pollination and seed dispersal.

REVISE


103 Essay Question: Plant Reproduction

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All plants have adaptations to enable them to reproduce successfully. Although ferns, angiosperms, and gymnosperms share similarities in their life cycles, they have different reproductive strategies.

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1. Discuss the different reproductive strategies of ferns, angiosperms, and gymnosperms. Your discussion should include identification of sporophyte and gametophyte stages, reference to reliance on water, details of pollination mechanisms if applicable, fertilisation of the egg cell by the sperm, and development of the embryo. You may use extra paper if required.

TEST

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104 KEY TERMS AND IDEAS: Plant Reproduction

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1. Use the following word list to complete the sentence below: haploid, generations, meiosis, mitosis, sporophyte The life cycle of plants alternates between a

Gametophytes produce gametes by

gametophyte generation and a diploid

generation. This process is called the alternation of

.

and sporophytes produce spores by

.

2. Match each term to its definition, as identified by its preceding letter code. alternation of generations

A The phase of a plant life cycle in which the cells of the plant are diploid.

double fertilisation

B The immature male gametophytes of seed plants.

endosperm

C Feature of reproduction in angiosperms. Two sperm nuclei fuse with nuclei in the embryo sac. One fuses with the haploid egg cell nucleus to form the zygote, the other fuses with the diploid endosperm nucleus to form the triploid endosperm.

fertilisation

gametophyte germination pollen

D The transfer of pollen from one plant to another for the purpose of fertilisation.

E Reproductive unit containing the embryonic plant that forms from the ovule after fertilisation. F The union of male and female gametes to form a zygote.

G In angiosperms, this provides nutrients to the developing embryo.

pollination

H The phase of a plant life cycle in which the cells of the plant are haploid.

seed

I The growth of a seed immediately following a period of dormancy.

sporophyte

J A feature of the life cycles of plants in which there are distinct sexual haploid and asexual diploid stages.

3. Study the three photos (right) and decide if these reproductive structures are pollinated by wind or by animals. Justify your answer:

Plant B

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Plant A

Plant C

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TEST


Patterns in ecological communities

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Achievement Standard

2.6

Key terms

Ecological communities exhibit observable patterns as result of physical and/or biotic factors. Patterns include changes over time (succession), along an environmental gradient (zonation, stratification), or differences in distribution related to different biotic or abiotic environments.

abiotic (physical) factor biotic factor

Achievement criteria and explanatory notes

community

Achievement criteria for achieved, merit, and excellence

competition

c

A

Investigate a pattern in an ecological community, with supervision: Describe observations or findings and use those findings to identify the pattern (or absence of a pattern) in an ecological community, relating this pattern to an environmental factor, and describing how the environmental factor might affect chosen species within the community.

c

M

Investigate in-depth a pattern in an ecological community, with supervision: Provide a reason for how or why the biology of one of the chosen species relates to the pattern (or absence of a pattern). The biology involves structural, behavioural, or physiological adaptations of the organism that are related to the environmental factor and to an interrelationship with an organism of another species (e.g. competition, predation, or mutualism).

c

E

Comprehensively investigate a pattern in an ecological community, with supervision: Use an environmental factor and the biology of interrelated organisms of different species to explain the pattern (or absence of a pattern). The explanation may involve elaborating, applying, justifying, relating, evaluating, comparing and contrasting, and analysing.

distribution

ecological succession environment

environmental gradient food chain food web

mutualism parasitism predation

primary succession producer quadrat sample

Habitat News

secondary succession

BH

Explanatory notes: Patterns in ecological communities

stratification

An investigation of a pattern in an ecological community involves ...

c

1

Analysing and interpreting information about the ecosystem.

transect

c

2

This information may come from direct observations, collection of field data tables, graphs, resource sheets, photographs, videos, websites, or texts.

c

3

A community pattern may include succession, zonation, or stratification, or some other pattern such as differences in distribution or abundance of organisms in relation to an environmental factor.

c

4

Environmental factors may include abiotic (physical) factors or biotic factors.

c

5

Abiotic factors are measurable factors such light, temperature, exposure time, or wind speed, but may also include aspects of the physical environment that are by quantified by multiple factors, such as altitude or depth.

c

6

Biotic factors. These are factors generated by living organisms, including predation, competition, mutualism and other species relationships.

c

7

Biology of the organisms refers to adaptations of organisms that relate to the pattern being investigated and may include (but is not restricted to) interrelationships such as competition, predation, or mutualism.

c

8

Assessment may be based on a stand-alone investigation or an individual investigation that contributes to the a group or class investigation.

trophic position zonation

With supervision meansâ&#x20AC;Ś

c

9

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symbiosis (pl. symbioses)

The teacher will provide guidelines for the investigation and you will develop and complete the investigation from these guidelines. They may discuss ideas with you and manage the process of sharing findings if you are contributing to a group.


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What you need to know for this Achievement Standard Collecting information about communities Activities 105, 106, 114, 115, 117 - 121, 123

By the end of this section you should be able to:

Conrad Pilditch

c

Describe and justify how you will investigate a named community pattern. Information could be gathered in the field, or analysing prepared information from various media.

c

Describe common methods and their appropriate use for sampling communities: including:

• Quadrats: Know how to determine the size of quadrat to use and how many samples to take. • Transects: Distinguish between belt and point transects and know when to use a transect.

c

Know how information about physical factors in the environment (e.g. light, temperatures) can be obtained, e.g. using sensors, meters, or data loggers.

Recognising patterns in communities Activities 105, 116 - 118, 122, 124 - 128, 129, 130

Conrad Pilditch

By the end of this section you should be able to:

c

Explain how environmental gradients result in species distribution that reflects adaptation and tolerance to physical factors.

c

Know that changes in composition of the community can occur along an environmental gradient and over time.

c

Identify factors that may account for differences in the abundance or distribution of organisms in similar communities, e.g. mowed or unmowed pasture.

c

Describe and explain causative factors in the common distributional patterns in communities: Stratification: the vertical layering in plant communities.

PASCO

Zonation: the spatial distribution of species into specific zones.

c

Recognise ecological succession as a community pattern in time. Distinguish between primary and secondary succession and identify factors important in the process.

Explaining community patterns Activities 106 - 109, 116 - 118, 122, 124 - 134

PASCO

By the end of this section you should be able to:

c

Describe and explain the community pattern you are investigating in terms of the relevant abiotic (physical) and biotic factors: • Abiotic factors include light, temperature, altitude, precipitation, salinity, dissolved oxygen. • Biotic factors include predation, competition, and symbioses (parasitism, mutualism).

Understand terms describing the interactions of species in communities. Use them appropriately to explain factors determining community patterns (see biotic factors above).

c

Describe and explain the adaptations of organisms that relate directly or indirectly to the community pattern you are investigating. For example the adaptations may relate to:

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c

• tolerance of an abiotic factor (e.g. exposure time on the shore)

• the organism's relationship with another in the community (e.g. beech regeneration in relation to availability of mutualistic mycorrhizal fungi).


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105 Guidelines for Investigating an Ecological Pattern Key Idea: Achievement standard 2.6 requires you to analyse data and explain any pattern in relation

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to the biology of the organisms involved and the environment in which they live.

ffInvestigating a pattern in an ecological community requires you to identify a pattern such as zonation or

stratification (patterns in space), or succession (a pattern in time). Other patterns could include observed differences in abundance and distribution between populations of organisms in different environments.

ffThe reason for the pattern must be discussed. This involves relating the biology of the organisms involved and any abiotic factors involved to the pattern observed.

Simple example of investigating zonation

1

2

Observe and describe a pattern: The pattern appears to be a zonation pattern, with different organisms living in different parts of the rocky shore.

Obtain data on the organisms present: The organisms present and their adaptations to their particular environment. For example, barnacles have watertight shells. Seaweed lacks a waterproof epidermis and dries out when out of the water.

3 Obtain data on the

environment: e.g limit of high and low tides, maximum temperature at high and low tides.

Bare rock

Barnacles

4 Discuss reasons for

Seaweed

the zonation pattern: Seaweed needs to remain in water as it dries out quickly, thus remains at the low tide level. Barnacles can live higher up the tide line because they can seal their shells and avoid desiccation.

Simple example of investigating succession

Pine and poplar forest

2

Obtain data on the organisms present: List the organisms present and their adaptations to their particular environment. E.g. grassy and weedy species quickly colonise open ground. Trees take longer to grow.

environment: e.g the temperature, soil type, reason for original disturbance.

4 Discuss reasons for

Poplar

Pine tree

Grassy plain

Linking with Achievement Standard Biology 2.1 ffThe collection of data can be used as part of Biology 2.1: Carry out a practical investigation in a biology context. In this case, collecting data to show a pattern.

ffFor example using a transect and quadrat sampling along a rocky shore to show the distribution of organisms. Data gathered could include direct counts of the organisms along the transect to show where the edges of the zones occur. Justification of the methods used and evaluation of the results is required. The biological reasons for the pattern could then be explained as part of AS2.6.

REFER

3 Obtain data on the

the succession pattern: Grassy/weedy species germinate quickly and grow quickly, covering disturbed ground. Larger plants take longer to produce seeds and grow. They will eventually over top the grasses, producing more seed and dominating the plain.

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1

Observe and describe a pattern: succession. The grassy plain in front of the forest is being populated by new forest growth.

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106 Components of an Ecosystem Key Idea: An ecosystem is made up of all the living (biotic) and non-living (abiotic) factors within a

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particular environment.

What is the biosphere?

ffThe biosphere is a narrow belt around the Earth containing all the Earth’s living organisms. It extends from

the bottom of the oceans to the upper atmosphere. Broad scale life-zones or biomes within the biosphere are characterised according to the main vegetation type. Within these biomes, ecosystems form natural units made up of the non-living, physical environment (the atmosphere, water, and soil) and the community.

The ecosystem

The concept of the ecosystem was developed to describe the way groups of organisms are predictably found together in their physical environment. A community comprises all the organisms within an ecosystem. The structure and function of a community is determined by the physical (abiotic) and biotic factors, which determine species distribution and survival.

Physical environment abiotic factors:

Atmosphere

• • • •

Wind speed and direction Light intensity and quality Precipitation and humidity Air temperature

Community: biotic factors

Soil

Nutrient availability Soil moisture and pH Composition Temperature

• • • •

Dissolved nutrients pH and salinity Dissolved oxygen Temperature

1. Distinguish between biotic and abiotic factors:

Interact in the community as: Competitors, parasites, pathogens, symbionts, predators, herbivores

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123Rf Tomas Sobek

• • • •

• Producers • Consumers • Detritivores • Decomposers

Water

2. Use one or more of the following terms to describe each of the features of a beech ecosystem listed below: Terms: population, community, ecosystem, physical factor.

(a) All the beech trees present:

(c) All the organisms present:

(b) The entire forest:

(d) The humidity:

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107 106

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107 Habitat Key Idea: The environment in which an organism lives is called its habitat. The tolerance range will

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determine the optimum position of organisms in their habitat.

The habitat is where an organism lives

The natural environment in which an organism lives is its habitat. It includes all the physical and biotic factors in that occupied area. Habitats vary widely in scale as do the physical factors that influence them.

A habitat may be vast and relatively homogeneous, as is the open ocean. Predatory barracuda (above) occur around reefs and in the open ocean.

For sessile organisms, such as this fungus, a suitable habitat may be defined by the environment in a small area, such as on this decaying log.

For microbial organisms, such as those in the ruminant gut, the habitat is defined by the chemical environment within the rumen (R) of the host, in this case, a cow.

Tolerance range determines distribution in the habitat

Each species has a tolerance range for factors in its environment (below). However, the members of a population are individually different and so vary in their tolerance. Organisms are usually most abundant where the conditions for their survival and reproduction are optimal. Outside this optimal ecological space, the environment is less favourable for survival and there are fewer individuals. For any species, there will also be environments that are unavailable to them. Examples of abiotic factors influencing niche size:

Too little

Rainfall

Too cold

Temperature

Too hot

Tolerance range

Zone of death or avoidance

Number of organisms

Too much

Zone of physiological stress

Most individuals

Optimum range

Zone of physiological stress

Zone of death or avoidance

Few individuals

No individuals

Unavailable

Marginal

Preferred

Marginal

Unavailable

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1. In which part of an organism's range will competition for resources be most intense and why?

Species tolerant of large environmental variations tend to be more widespread than organisms with a narrow tolerance range. The Atlantic blue crab (left) is widespread along the Atlantic coast from Nova Scotia to Argentina. Adults tolerate a wide range of water salinity ranging from almost fresh to highly saline. This species is an omnivore and eats anything from shellfish to carrion and animal waste.

Wendy Kaveney

2. Suggest an advantage to being able to tolerate variations in a wide range of environmental factors?

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107 106 112

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108 Feeding Relationships Key Idea: An ecosystem has a trophic structure based on the hierarchy of feeding relationships

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between organisms. Organisms occupy a trophic level based on what they eat.

ffThe hierarchy of feeding relationships in an ecosystem (its trophic structure), determines the pathways for energy flow and nutrient cycling. Organisms are assigned to trophic (feeding) levels based on the way they obtain their energy.

ffThe sequence of organisms, each of which is a food source for the next, is a food chain. Some consumers occupy

Callum O'Hagan

several trophic levels, and some may occupy different trophic levels at different stages in their life.

Millipede

Producers (autotrophs) e.g. plants, make their own food from simple inorganic substances, commonly by photosynthesis. The producers ultimately support all other trophic levels.

Consumers (heterotrophs) e.g. animals, get their energy from other organisms. Consumers are ranked according to the trophic level they occupy, i.e. first order, second order.

Detritivores and decomposers process dead organic matter (DOM) from all trophic levels. This helps recycle the material back into the soils. Detritivores ingest DOM whereas decomposers break it down with extracellular enzymes.

1. The diagram below represents the basic elements of a food chain. In the questions below add to the diagram the features that indicate the flow of energy through the community of organisms.

(a) State the original energy source for this food chain:

(b) Draw arrows on the diagram below to show how energy flows through the organisms in the food chain. One is done for you.

(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 by way of respiration.

Respiration

Herbivores Trophic level: 2

Carnivores Trophic level: 3

Detritivores and decomposers

Carnivores Trophic level: 4

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Producers Trophic level: 1

2. What A could 2 you infer about the trophic level(s) of the kingfisher if it was found to eat both katydids and frogs?

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109 Food Webs

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152

Key Idea: Food chains can be put together to form food webs. The complexity of a food web depends

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on the number of foods chains and trophic levels in it.

ffIn any community, no species exists independently of others. All

Key to food web (below)

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 organisms in an ecosystem.

Flow of nutrients from the living components to detritus or the nutrient pool. Consumer–resource interactions.

ffThe 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. 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.

Losses of each food web component from the system and external input of limiting nutrients.

A simple food web (a)

Koura

Smelt

(e)

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.

(b)

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)

Nutrient pool

Energy flows through ecosystems in the high energy chemical bonds in organic matter.

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Nutrients cycle between the atmosphere, the Earth’s crust, water, and living organisms.

1. (a) - (e) Complete the food web above by adding in the following labels: carnivore, herbivore, autotroph, detritus, detritivore: 2. Why do most food chains have fewer than six links?

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109 108

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110 Ecological Niche

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Key Idea: An organism's ecological niche describes its functional position in its environment.

The niche is the functional role of an organism

The ecological niche (or niche) of an organism describes its functional position in its environment. The full range of environmental conditions under which an organism can exist describes its fundamental niche.

ffInteractions with other species, e.g. competition, usually force organisms to occupy a space that is narrower than this. This is called the realised niche.

ffCentral to the niche concept is the idea that two species with exactly the same niche cannot coexist, because they would compete for the same resources and one would exclude the other. More often, species compete for only some of the same resources. These competitive interactions limit population sizes and influence distributions.

The physical conditions influence the habitat. A factor may be well suited to the organism, or present it with problems to be overcome.

Adaptations enable the organism to exploit the resources of the habitat. The adaptations take the form of structural, physiological and behavioral characteristics of the organism.

Physical conditions • • • • • • • • •

Substrate Humidity Sunlight Temperature Salinity pH Exposure Altitude Depth

Adaptations for:

• Locomotion • Activity pattern • Tolerance to physical conditions • Predator avoidance • Defence • Reproduction • Feeding • Competition

Resources offered by the habitat

• Food sources • Shelter • Mating sites • Nesting sites • Predator avoidance

Resource availability is affected by the presence of other organisms and interactions with them: competition, predation, parasitism, and disease.

The habitat provides opportunities and resources for the organism. The organism may or may not have the adaptations to exploit them fully.

(a) Ecological niche:

(b) Fundamental niche:

(c) Realised niche:

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1. Define the following:

2. What factors contribute to making the realised niche narrower than the fundamental niche?

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111 Adaptation and Niche Key Idea: The adaptations of a species help it to function in its niche. Different adaptations in different

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species enhance niche differentiation and allow different species to coexist.

ffThe adaptations (adaptive features) of species are the result of selection pressures on them throughout their evolution. They may involve an animalâ&#x20AC;&#x2122;s structure (morphology), its physiology, or its behaviour.

ffAdaptations enable an organism to function most effectively in its niche, enhancing its exploitation of its environment, and contributing to its survival and successful reproduction.

ffThe following examples describe some of the adaptations of New Zealand's two native bat species (pekapeka). The New Zealand long tailed bat (Chalinolobus tuberculatus)

A long tail and short ears distinguish this species from the short tailed bat (next page).

Echolocation identifies surroundings and insect targets, allowing the bat to home in on prey.

Fine silky hair provides insulation, reducing the need to generate internal heat.

Strong fliers, long-tailed bats have a large home range (100 km2).

Photo: DoC: J Kendrick

The long tailed bat has a large wingspan (25-28 cm) for its size, enabling it to fly up to 60 km hr-1 and cover its large home range (up to 100 km2)

A long tail is used to scoop insects in flight.

The hind feet and legs are small and delicate. A small thumb projects free of the wrist and has a long curved claw. They are well suited to roosting and clinging to trees to rest.

Adult weight: 8-11 g

Habitat, behaviour, and reproduction of the NZ long tailed bat ffChalinolobus tuberculatus is widespread in native and exotic forests

throughout the North and South Islands. It roosts in hollow trees, under bridges, and in caves. It is an aerial insectivore, feeding solely on flying insects (moths, midges, mosquitoes, and beetles). It flies at dusk in a characteristic zigzag pattern to catch flying insects.

ffC. tuberculatus roosts in colonies of up to several hundred and hibernates in

DoC

autumn and winter when food is scarce. During hibernation, metabolism is depressed and the animal is inactive, expending very little energy. The period of gestation is 6-8 weeks long and females give birth in summer to a single pup. The pup is large relative to the size of the mother and flying becomes energy expensive and difficult for her as the pregnancy progresses. The juveniles grow rapidly, flying at 4-6 weeks and reaching adult size soon after.

(a) The presence of hair:

(b) The zigzag flight pattern observed in Chalinolobus:

(c) The long tail observed in Chalinolobus:

(d) The omnivorous diet of Mystacina:

(e) Hibernation in Chalinolobus during autumn and winter:

(f) The presence of a long curved claw on the hind feet:

(g) An ability to navigate by echolocation: WEB

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1. Examples of adaptations in New Zealand bat species are listed below. Identify them as predominantly structural, physiological, and/or behavioural:

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155 km2)

is smaller than Home range (60 the long tailed bat. Most individuals are active within a 5 km2 range.

Thick fur provides insulation.

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The New Zealand lesser short tailed bat (Mystacina tuberculata)

Tail is not enclosed in the skin between the legs.

Claws uniquely possess talons which aid movement on the ground.

The wings of the short tailed bat fold up into a protective sheath when foraging on the forest floor. Wingspan is similar to the long tailed bat.

Tongue is long and bristly to aid nectar feeding.

Adult weight 11-16 g

Habitat, behaviour, and reproduction of the NZ lesser short tailed bat Island, although populations are scattered. It is nocturnal (active at night) and omnivorous, feeding off forest fruits, nectar, pollen, and insects. Its stout feet are adapted to moving around on the forest floor and it has well developed teeth. Mystacina roosts singly or communally in hollow trees and does not hibernate, although it may become torpid (inactive) in winter.

ffOf its total time spent foraging, only 30% is spent catching low flying

insects. About 40% is spent feeding from plants, and the remaining 30% is spent hunting on the forest floor.

ffM. tuberculata pollinates a number of native plants, including wood rose

Foraging on the forest floor

(Dactylanthus), perching lilies, kiekie, and pohutukawa. The bats disperse seeds in their droppings. The wood rose relies primarily on the NZ shorttailed bat for pollination. Extinction of the short tailed bat would likely result in extinction of the wood rose.

Bat photos this page DoC: CR Veitch

ffMystacina tuberculata occupies native forests, mostly in the central North

mating season (February-May), attracting females with ultrasonic calls. The embryo enters a state of delayed implantation so the single young is born in summer. The young weigh only 5 g but, like long tailed bats, develop quickly, flying by 4 weeks of age and fully grown by 3 months.

Wood rose

DoC: J Barkle

ffShort tailed bats are lek breeders and males occupy roosts during the

2. What adaptations of Mystacina tuberculata aid its ground foraging?

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3. Chalinolobus and Mystacina are similar sized bats occupying similar habitats. Describe the structural, physiological and behavioural adaptations that help reduce the competition between them:


112 Species Interactions in a Beech Forest

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156

Key Idea: Within any ecological community, there are numerous and complex interactions between

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different species, with many species being directly or indirectly dependent on many others.

ffThe particular characteristics of ecosystems and their communities arise as a result of the physical environment, the species that live there, and the complex interactions occurring between those species.

ffIn this activity, we examine species interactions within a New Zealand beech forest community in the northern

South Island. Black beech (Nothofagus solandri) is a key species in the structure and function of this community because a large number of species depend directly or indirectly on it for survival. The interdependence of native species is finely balanced. The introduction of non-native competitors and predators, such as German wasps, mice, and stoats, disturbs established relationships and threatens community diversity.

Black beech community

Consumer Beech scale

l D. Ke

ay Pe t e r B r

ly,

UC

Consumers Sooty moulds

Parasite Red mistletoe

J.L a

dley, U C

Consumer Bellbird

Eri

c H u e ti n g

Consumer Tui

ca

re

C

u c h a n a n la nd

y, U

P. B

Parasite Tinder fungus

Ke

rr

Landca

dl

re

Consumer Wasp

pe

dia

cc 3 .0 )

Parasite (probably) Black orchid

S. Ke rr

In s k

Parasite/pathogen Bootlace mushroom

i (w

i ki

Mycorrhiza Amanita fungus

Black beech, Nothofagus solandri, is a producer, a host plant for parasites, and a partner in mutualistic relationships.

Photo of black beech: David Kelly, ŠUniversity of Canterbury

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Red mistletoe

Photo: Jenny Ladley, ©University of Canterbury

Photo: Stu Phillips CC 3.0

Photo: Shirley Kerr, Kaimai Bush

Red mistletoe is one of three species of beech mistletoes. Mistletoes are hemiparasites, they can photosynthesise (they are green), but they take water and nutrients from the host tree root system by penetrating the host xylem tissues. Bellbird and tui are native honeyeaters and are able to open the flowers of red mistletoe to feed on the nectar. In doing so, they pollinate the flowers.

Beech sooty scale is a sap sucking insect. Its mouthparts are embedded in the beech tree phloem and excess sap passes out a waxy tube attached to its anus as energy-rich honeydew. Honeydew is an important food source for birds, insects, and sooty mould fungi. Research indicates that the sap-sucking activities of the scale insect stimulate the tree to produce more sap, benefitting the whole community.

Huperei, the black orchid

Amanita nothofagi (right) is a fungus endemic to New Zealand and associated only with southern beech. It forms mutualistic mycorrhizae (fungal-plant associations) with black beech roots. The mycorrhizae provide the fungus with direct access to the carbohydrates made by the plant. The plant benefits from the increased capacity to absorb water and minerals via the fungal mycelium, which connects a large volume of soil to the roots.

Harore, the bootlace mushroom

Not all fungi are beneficial to the beech. Harore, (bottom left) the bootlace mushroom, is a fungal pathogen of beech. Its hyphae invade new roots, gaining access to the plant's tissues. This fungus is itself apparently parasitised by with huperei (top left), or black orchid, which lacks leaves and chlorophyll and apparently receives nutrients via the bootlace fungus. The relationship is complex and needs further study.

Eric Heuting

Tui

Sooty mould fungi grow on waste honeydew forming a dark covering over the tree. Sooty mould is actually a complex of many species and provides cover for the scale insects. Together, the sooty mould fungi are probably the largest consumer of honeydew. The photo above shows black beech, but other Nothofagus species can be blackened with sooty mould also.

Photo: Michael (inski) CC 3.0

Anal tube of beech sooty scale with drop of honeydew

Photo: Peter Bray, Airborne Honey Ltd

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ffIntroduced common and German wasps have

flourished in New Zealand and can reach extraordinarily high numbers in some beech forests (e.g. in the Nelson area). In some regions wasp biomass exceeds that of birds.

ffThe wasps prey on native insects and compete

aggressively for the honeydew food source, reducing the amount available to other species.

ffDepletion of the honeydew and a reduction in

the abundance of small insects and spiders has detrimental flow on effects on the birds that depend on the honeydew and invertebrates for food. © 2017 BIOZONE International ISBN: 978-1-927309-60-5 Photocopying Prohibited

German wasp

Wasps can carry around 15 µL of honeydew at a time. The amount of honeydew wasps take is so much than they can reduce the standing crop of honeydew by more than 90% for five months of the year.

Landcare Research

How do wasps affect the beech community?

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Amanita nothofagi and beech: an essential partnership


1. Use the information in this activity to provide an example from the beech community of the four interactions listed below:

Interaction

Sp. 1

Sp. 2

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Description

+

+

A close ecological relationship in which both parties benefit by gaining a resource or service.

Exploitation: predation

+

A relationship in which one species (the predator) kills and eats another (the prey).

One species (the parasite) benefits by exploiting the other the (host). The host is harmed but not usually killed.

Two or more species contest a resource such as food. All parties are harmed (receive less of the resource).

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Example

Exploitation: parasitism

Competition

+

2. Explain why the common and German wasps have such a large impact on a beech forest community:

3. Identify a useful function performed by tuis as they feed on mistletoe nectar:

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4. Using the information in this activity, write a short (200 word) essay discussing the role of species interactions in the structure, function, and diversity of a beech forest community. Your discussion should emphasise the interdependence of the species present and describe and explain threats to community function. You may use more paper if you wish:

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113 Competition and Niche Size Key Idea: Niche breadth is influenced by competition. Slight differences in the niches of competing

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species allow them to occupy the same general area of the environment.

Competition occurs both between species (interspecific) and within species (intraspecific). Such interactions affect the breadth (size) of an organism's niche. If there is intense competition between two species, e.g. because of the introduction of a new species into an area, one may exclude the other.

1

The fundamental niche

Realised niche of species A

The tolerance range represents the fundamental niche a species could exploit. The actual or realised niche of a species is usually narrower than this because of competition with other species.

Possible tolerance range

2

Intraspecific competition

Broader niche

Competition is strongest between individuals of the same species, because their resource needs exactly overlap. When intraspecific competition is intense (as when population density increases), individuals are forced to exploit resources at the extremes of their tolerance range. This leads to expansion of the realised niche.

If two (or more) species compete for some of the same resource, their resource use curves will overlap. Within the zone of overlap, competition will be intense and selection will favour niche specialisation so that one or both species occupy a narrower niche. Niche differences between species have evolved to reduce competition. In the absence of competition, niches tend to be broader.

Narrower niche

Amount eaten

3

Possible tolerance range

Interspecific competition

Narrower niche

Zone of overlap

Species A

Species B

Resource use as measured by food item size

1. (a) Define interspecific competition:

(b) Define intraspecific competition:

3. Identify the type of competition occurring in:

(a) Photograph A:

(b) Photograph B:

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A

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2. Contrast the effect of intraspecific and interspecific competition on niche breadth and explain reasons for the difference:

B

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114 Sampling Populations Key Idea: The method used to sample a population must be matched to the type of organism being

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sampled, its distribution, and the environment.

ffIn most field studies, it is not possible to measure

or count every member of a population. Instead, the population is sampled in a way that provides a fair (unbiased) representation of the organisms present and their distribution. This is usually achieved through random sampling, in which every possible sample of a given size has the same chance of selection.

ffMost practical exercises in community ecology involve

BH

collecting or recording data from living organisms. The technique you use to sample must be appropriate to the community being studied and the information you want to obtain. You must also think about the time and equipment available, the organisms involved, and the impact your study might have on the environment.

Point sampling is time efficient and good for determining species abundance and community composition. However, organisms in low abundance may be missed.

Point sampling: systematic grid

Transects are well suited to determining changes in community composition along an environmental gradient but can be time consuming to do well.

Line transects

Photo: www.coastal planning.net

Which sampling method?

Random point sampling

Belt transects

Marine ecologists use quadrat sampling to estimate biodiversity before activities such as dredging. Quadrats are suited to determining community composition and features of population abundance.

Quadrats are also good for assessments of community diversity and composition but are largely restricted to plants and immobile animals. Quadrat size must also be appropriate for the organisms being sampled.

Random quadrats

First sample marked

Second sample: proportion recaptured

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RA

Mark and recapture is useful for highly mobile species which are otherwise difficult to record. However, it is time consuming to do well. Radio-tracking is a commonly used alternative.

Line transects are well suited to determining changes in community composition along an environmental gradient. When placed randomly, they provide a quick measure of species occurrence.

1. Name a sampling technique that would be appropriate for determining each of the following:

(a) A highly mobile species:

(b) The community composition of a rocky shore:

(c) A change in community composition from low altitude to high altitude: WEB

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115 Interpreting Samples Key Idea: If sample data are collected without bias and in sufficient quantity, even a simple analysis

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can provide useful information about community composition.

1. A group of students surveyed the vegetation on Mount Taranaki. Starting at an altitude of 500 m, they ran out a 25 m transect line and simply identified and counted the number of plants touching it. They did this at five locations up the mountain from 500 m to 1500 m. They labelled the plants as tree (T), shrub (S) or herb (H).

Altitude (m)

Species

(a) Name the dominant plant at each altitude:

No.

500

Kamahi (T) Rata (T) Tawa (T) Rimu (T)

7 10 2 3

750

Rata (T) Kamahi (T) Rimu (T)

5 13 1

1000

Mountain totara (T) Kaikawaka (T) Kamahi (T)

7 2 15

(c) Which plant has the greatest range in altitude?

1250

Leatherwood (S) Coprosma (S)

20 5

(d) State the relationship between altitude and plant height:

1500

Everlasting daisy (H) Red tussock (H)

3 20

(b) At what altitude is the "treeline"?

2. The figure (below) shows changes in vegetation cover along a 2 m vertical transect up the trunk of an oak tree. Changes in the physical factors light, humidity, and temperature along the same transect were also recorded. Quadrat 5

Red stem moss Fern moss

Quadrat 4

Snake moss

Quadrat 3

Star moss

Quadrat 2

Broad leaved star moss

Eye brow moss Tree moss

Quadrat 1

Lichens (various species)

QUADRAT

50 Percentage cover

1

2

3

4

5

Height (m)

0.4

0.8

1.2

1.6

2.0

Light (arbitrary units)

40

56

68

72

72

Humidity (percent)

99

88

80

76

78

Temperature (°C)

12.1 12.2

13

100

14.3 14.2

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(a) At which height were mosses most diverse and abundant?

(b) What plant type predominates at 2.0 m height?

(c) What can you deduce about the habitat preferences of most mosses and lichens from this study?

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116 Stratification in a Forest

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Key Idea: Forest stratification is a vertical layering pattern arising as a result of tree species growing

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to particular heights.

ffForest communities throughout the world show stratification. This is a pattern of vertical layering that is created by the different heights to which the species present grow.

ffStratification patterns vary for different forest types and for different regions. The final forest composition depends on altitude, light levels, soil type, drainage, and the past history of the area.

ffNew Zealandâ&#x20AC;&#x2122;s lowland podocarp-broadleaf forests (below) show five distinctive layers (plus epiphytes and lianes). Canopy trees that grow taller than the canopy layer are called emergents.

Epiphyte

Canopy

Tree fern layer

The topmost layer of podocarps (native gymnosperms) is usually uneven and 20-60 m in height. The canopy intercepts most of the direct sunlight. Canopy trees that grow taller than the canopy layer are called emergents. Species in this layer include kauri, matai, miro, rimu, totara, and kahikatea.

A layer of large tree ferns (Cyathea), including mamaku (black tree fern) and silver fern.

This lower level of smaller trees is not always present. In forests where the large podocarps are absent, it may make up the canopy. It forms a semi-continuous layer of angiosperms (called broadleafs) under the podocarps. Species in this layer include tawa, kamahi, taraire, and puriri.

Epiphytes and lianes

Epiphytes are plants that have no contact with the soil but grow in crevices in the branches and trunks of larger trees. Lianes are rooted in the ground, but clamber into the canopy. This layer includes ferns and orchids.

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A layer of plants 1-3 m tall. During their evolution, these plants were grazed by moa and many show adaptations (such as a divaricating or branching growth habit) to deter moa browsing. Species in this layer include kawakawa, horopito, tutu, rangiora.

Ground layer

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Subcanopy

Shrub layer

A covering of leaf litter and rotting material on the ground provides good growth conditions for plants in this layer. Seedlings less than 1 m tall and shade adapted, low growing plants such as ferns and mosses are found here as well as fungi and lichens. This layer may incorporate seedlings of the understorey and canopy species. Light is limited.

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Disturbance and forest stratification

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ffSmall-scale disturbances, such as a large canopy

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tree falling down through disease or bad weather, can alter the physical environment around it. For example, its absence allows more light to filter down to the lower levels of the forest allowing smaller plants to grow more quickly in the improved physical conditions.

ffLikewise logging of large trees also changes the

stratification landscape, although the effects are generally larger as more trees are removed than by small-scale natural disturbances.

Photos right: Tree falls leave gaps in the canopy and allow sunlight to filter through to lower levels of the forest.

1. Use the descriptions of the forest layers on the previous page to identify the boundary of each layer (do not include epiphytes and lianes). Draw lines on the forest diagram to show these and label each layer. 2. (a) What physical factor is limited at the ground layer?

(b) How might this determine the type of plants found in the ground layer?

3. Explain why a forest with a strong pattern of stratification might provide a greater diversity of habitats and support a greater species diversity than a forest without such a vertical structure:

4. (a) Explain the effects on a forest community when a single large tree is removed from a forest (e.g. through tree fall):

(b) Predict the impact of deliberate removal (logging) of emergents and large canopy trees on the community composition and how existing and colonising species might respond to this:

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117 Physical Factors in a Forest Key Idea: The physical conditions on a forest floor are usually very different to those in the canopy.

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This is largely a consequence of stratification in the forest community.

ffIn a vertically stratified community such as a mature forest, the community structure produces a gradient in physical factors.

ffIn a forest, the light quantity and quality, humidity, wind speed, and temperature change gradually from canopy to

forest floor. This is a consequence of the modification of the physical environment by the species present at each layer.

Light: 70% Wind: 15 km h–1 Humid: 67%

Canopy

Light: 50% Wind: 12 km h–1 Humid: 75%

Light: 12% Wind: 9 km h–1 Humid: 80%

A datalogger (above) fitted with suitable probes was used to gather data on wind speed (Wind), humidity (Humid), and light intensity (Light) from canopy to floor. Light intensity is given as a percentage of full sunlight.

Light: 6% Wind: 5 km h–1 Humid: 85% Light: 1% Wind: 3 km h–1 Humid: 90% Light: 0% Wind: 0 km h–1 Humid: 98%

Leaf litter

Mature podocarp forests are complex communities with a vertical structure or stratification, which divides the vegetation into layers.

(a) Light intensity:

(b) Wind speed:

(c) Humidity:

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1. Describe the trend from the canopy to the leaf litter for the following factors and explain why they change:

2. (a) How does the vertical gradient in physical factors in a mature forest arise?

(b) How is this different to the situation on a rocky shore (opposite)?

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118 Physical Factors on a Rocky Shore

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Key Idea: Within any particular habitat, physical factors may vary from one place to the next.

ffGradients in abiotic (physical) factors are found in almost every environment. They influence habitats and create

microclimates, and can be very important in determining community patterns. Accurate measurements of abiotic factors (e.g. temperature and light intensity) help us to understand the patterns we observe in communities.

ffThis activity examines the physical gradients present on a rocky shore, where the community often shows zones

of species distribution. On the rocky shore, the gradient in abiotic factors is created by distance from the low water mark (exposure time). In this way it differs from the vertical gradient in the physical environment seen in a stratified forest, which is created largely by the community of organisms itself. Salin: 42 g L–1 Temp: 28°C Oxy: 20% Exp: 12 h

ark (m) water m w lo e e th on abov

Cliff

3.0

Salin: 39 g L–1 Temp: 28°C Oxy: 30% Exp: 10 h

Salin: 38.5 g L–1 Temp: 26°C Oxy: 42% Exp: 8 h

Salin: 37 g L–1 Temp: 22°C Oxy: 57% Exp: 6 h

Salin: 36 g L–1 Temp: 19°C Oxy: 74% Exp: 4 h

Salin: Temp: Oxy: Exp:

35 g L–1 17°C 100% 0h

2.0

1.0

0.0

Enlarged below

Sea

Elevati

Pool

HWM

13

12

11

10

LWM

9

8

7

Distanc

e from

Boulders

A

B

C

6

5

low wate

r mark

4

(m)

3

2

1

0

The diagram above shows a profile of a rock platform at low tide. The high water mark (HWM) shown is the average height of the spring tide. The low water mark (LWM) is the average low point of the spring tide. The rock pools on the platform vary in size, depth, and position. Rock pools at different elevations trap seawater for 10-12 hours. Pools near the HWM are exposed for longer periods of time than those near the LWM. The difference in exposure time causes a gradient in physical conditions over a horizontal and/or vertical distance. In the diagram above, physical factors in the pools include salinity, or the amount of dissolved salts (g) per litre (Salin), temperature (Temp), dissolved oxygen compared to that of open ocean water (Oxy), and exposure, or the amount of time isolated from the seawater (Exp).

(a) Salinity:

(b) Temperature:

(c) Dissolved oxygen:

(d) Exposure:

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1. Describe the change in physical environment (environmental gradient) from the LWM to the HWM for:

2. How might the mechanical force of wave action and surface temperature when exposed differ at each of the labelled points A, B, and C in the inset diagram of the boulders?

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119 Quadrat Sampling

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Key Idea: Quadrat sampling involves a series of random placements of a frame of known size over an

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area of habitat to assess the abundance or diversity of organisms in the area.

ffQuadrat sampling is a method by which organisms

Quadrat

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.

Guidelines for quadrat use:

1. The area of each quadrat must be known exactly and quadrats should be the same shape. The quadrat does not have to be square (it may be rectangular, hexagonal etc.).

Area being sampled

1

2 3 4

5

6 7

8 9 10

1 2 3 4

2. Enough quadrat samples must be taken to provide results that represent the total population.

5

3. The population of each quadrat must be known exactly. Species must be distinguished from each other. It must be decided beforehand what the count procedure will be and how organisms on the quadrat boundary will be counted.

7

6 8 9

10

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 can be achieved by random sampling (right).

A random number table gives the coordinates for a quadrat's position.

The area to be sampled is divided up into a grid pattern with indexed coordinates.

0

Sampling a centipede population

1. Determine the average number of centipedes per quadrat:

2. Calculate the estimated average density of centipedes per square metre. Use the formula given below: Total number of individuals counted

Estimated = average density Number of quadrats X area of each quadrat

3. Looking at the data for individual quadrats, describe in general terms the distribution of the centipedes in the sample area:

1

0

0

0

0

2

1

2

0

0

1

0

2

0

0

0

1

0

1

0

0

0

1

3

1

0

1

3

2

1

0

2

1

2

0

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In 1967 a researcher named M. Lloyd sampled centipedes in Wytham Woods, near Oxford, England. A total of 37 hexagonâ&#x20AC;&#x201C;shaped quadrats were used, each with a diameter of 30 cm and area of 0.08 m2 and arranged as shown in the diagram (right). Use the data in the diagram to answer the following questions.

2

The number in each hexagon indicates how many centipedes were caught in that quadrat.

Centipede

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120 Quadrat-Based Estimates Key Idea: The size and number of quadrats used during a survey must be sufficient to represent the

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community studied but not take too much time to use or analyse.

What size quadrat to use?

ffThe simplest description of a plant community in a habitat is a list of the species that are present. This qualitative

assessment of the community has the limitation of not providing any information about the relative abundance of the species present. Quadrats can be used to provide this information. The quadrat must be large enough to be representative of the community, but not so large as to take a very long time to use.

A quadrat covering an area of 0.25 m2 is suitable for most low growing plant communities, such as this alpine meadow, fields, and grasslands.

Larger quadrats (e.g. 1 m2) are needed for communities with shrubs and trees. Quadrats as large as 4 m x 4 m may be needed in woodlands.

Small quadrats (0.01 m2 or 100 mm x 100 mm) are appropriate for lichens and mosses on rock faces and tree trunks.

How many quadrats?

Describing vegetation

The number of quadrats used is also important. In species-poor or very uniform habitats, a small number of quadrats will be sufficient. In species-rich or diverse habitats, more quadrats will be needed to ensure that all species are represented adequately.

Density (number of individuals per unit area) is a useful measure of abundance for animal populations, but for plants, where it can be difficult to determine where one plant ends and another begins, abundance is often assessed using percentage cover. Here, the percentage of each quadrat covered by each species is recorded, either as a numerical value or using an abundance scale such as the ACFOR scale.

Determining the number of quadrats

â&#x20AC;˘ Plot the cumulative number of species recorded against the number of quadrats already taken. â&#x20AC;˘ The point at which the curve levels off indicates the suitable number of quadrats required.

The ACFOR abundance scale

A = Abundant (30% +)

C = Common (20-29%)

Number of species

F = Frequent (10-19%)

O = Occasional (5-9%) R = Rare (1-4%)

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Number of quadrats

Abundance scales such as ACFOR are subjective, but it is not difficult to determine which category each species falls into.

1. What is the main consideration when determining appropriate quadrat size?

2. What is the main consideration when determining number of quadrats?

3. Why is percentage cover more appropriate than density for estimating the abundance of plant species?

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121 Sampling a Rocky Shore Community Key Idea: The estimates of a population gained from using a quadrat may vary depending on where

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the quadrats are placed. Larger samples can account for variation.

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. 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.

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. The quadrats should be to selected in a random manner. It is not sufficient to simply guess or choose your own. 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. 4. Choose quadrats randomly 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 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 (e.g. 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.

Discuss the merits and problems of the suggestions above with members of the class (or group). You may have counting criteria of your own. Consider other factors that could cause problems with your count.

6. Carry out the sampling 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 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 quadrat area

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:

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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):

Barnacle: Oyster borer: Chiton: Limpet: Compare your estimated population density to the actual population density for each species:

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119 122

Algae:

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Coordinates for each quadrat

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169

Table of random numbers

Oyster borer

1: 2:

Snakeskin chiton

Limpet

Coralline algae

B

C

D

22 32 31 46 43 56

31 15 56 36 42 14

62 63 36 13 45 31

22 43 64 45 35 14

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Plicate barnacle

A

Use the random number table above to select quadrats randomly from the grid above. Chose 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 to the column number.

3: 4: 5:

6:

TOTAL

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Example: 5 2

refers to the 5th row and the 2nd column.


122 Field Study of a Rocky Shore

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170

Key Idea: Field studies collect physical and biological data about aspects of community structure or

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function and can reveal reasons for differences in patterns of distribution or abundance.

The aim

To investigate the differences in abundance of intertidal animals on an exposed rocky shore and a sheltered rocky shore.

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

Prevailing direction of wind and swell

1km

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 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.

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 students recorded the number of animals in each quadrat.

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 New Zealand's rocky shores, although the ornate limpet has a slight preference to exposed shores.

The black nerite (snail) is found throughout northern rocky shores and extends across most tidal zones. It is more common on exposed rocky shores.

ou G ra h a m B

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ld

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Photo: K Pryor

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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.


171 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. In the space below, tabulate the data to show the total number of each species and the mean number of animals per quadrat:

Site A

Brown barnacle Oyster borer Columnar barnacle Plicate barnacle Ornate limpet Radiate limpet Black nerite

Field data notebook Count per quadrat. Quadrats 1 m2

1

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

Brown barnacle Oyster borer Columnar barnacle Plicate barnacle Rock oyster Ornate limpet Radiate limpet Black nerite

Brown barnacle

7 2 56 11 7 7 13 6

Oyster borer

6 3 57 11 8 8 14 5

Columnar barnacle

7 1 58 13 8 5 11 3

Plicate barnacle

5 3 55 10 6 6 10 1

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

7 1 36 8 6 3 13 3

Radiate limpet

Black nerite

Mean number of 2 Site animals per m A Median value Modal value

Total number of animals

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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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:

(b) Explain why more oyster borers 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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6. (a) Explain why more brown barnacles and plicate barnacles were found at site A:

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123 Transect Sampling Key Idea: Transect sampling is useful for providing information on species distribution along an

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environmental gradient.

A transect is a line placed across a community of organisms.

Point sampling

Sample points

The usual practice for small transects is to stretch a string between two markers. The string is marked off in measured distance intervals, and the species at each marked point are noted.

Some transects provide information on the vertical, as well as horizontal, distribution of species (e.g. tree canopies in a forest).

Continuous belt transect

Continuous sampling

The sampling points along the transect may also be used for the siting of quadrats, so that changes in density and community composition can be recorded.

Interrupted belt transect

Belt transects are essentially a form of continuous quadrat sampling. They provide more information on community composition but can be difficult to carry out.

4 quadrats across each sample point

Transects provide information on the distribution of species in the community. This is of particular value where environmental factors change over the sampled distance. This change is called an environmental gradient (e.g. up a mountain or across a seashore).

2. What kind of information does transect sampling provide?

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1. What is transect sampling?

3. What are some similarities and differences between a belt transect and sampling using quadrats?

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LINK

LINK

WEB

127 126 123

KNOW


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4. Kite graphs are an ideal way in which to present distributional data from a belt transect (e.g. abundance or percentage cover) along an environmental gradient. Usually, they involve plots for more than one species. This makes them good for highlighting probable differences in habitat preference between species. Kite graphs may also be used to show changes in distribution with time (e.g. with daily or seasonal cycles).

An example of a kite graph

Scale

0

5 organisms

Distance above water line (m)

A line equals 0 organisms

10

Field data notebook

Numbers of barnacles (4 common species) showing distribution on a rocky shore

Height above low water (m)

Plicate barnacle

Barnacle species Brown Columnar barnacle barnacle

Sheet barnacle

0

0

0

65

10

0

0

12

3

32 55

0 0

0 0

0 0

4

100

18

0

0

5

50

124

0

0

6

30

69

2

0

7

0

40

11

0

8

0

0

47

0

9 10

0 0

0 0

59 65

0 0

0 1

2

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The data on the right were collected from a rocky shore field trip. Four common species of barnacle were sampled in a continuous belt transect from the low water mark, to a height of 10 m above that level. The number of each of the four species in a 1 m2 quadrat was recorded. Plot a kite graph of the data for all four species on the grid below. Choose a scale that takes account of the maximum number found at any one point and allows you to include all the species on the one plot. Include the scale on the diagram.

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124 Shoreline Zonation Key Idea: Zonation is a community pattern in which there are visible strata or zones in an

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ecosystem caused by changes in abiotic factors along an environmental gradient.

ffZonation refers to the division of an ecosystem into distinct zones that

experience similar abiotic conditions. In a more global sense, zonation may also refer to the broad distribution of vegetation according to latitude and altitude. Zonation is particularly clear on rocky seashores, where assemblages of different species form a banding pattern approximately parallel to the waterline.

Zonation patterns generally reflect the vertical movement of seawater up or down the shore. Sheer rocks can show marked zonation as a result of tidal changes with little or no horizontal change in species distribution. The profiles below show zonation patterns on an exposed rocky shore (left profile) with an exposed sandy shore for comparison (right profile). They are not intended to represent specific locations.

ELT = Extreme spring low tide mark, MLT = Mean spring low tide mark, MHT = Mean spring high tide mark, EHT = Extreme spring high tide mark.

Rocky shore, Raglan, west coast NZ

Where several species are indicated within a band, they occupy the entire zone, not just the position where their number appears.

2

21

3

4

EHT

22

Upper shore

1

Exposed sandy shore

23

10

MHT

5

7

6

24

12

25

8

Middle shore

11

13

9

14

MLT

26

Lower shore

17

15

18

16

ELT

19

20

27 28

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21 Sandhopper 22 Ocean sand crab 23 Tuatua 24 Ghost shrimp 25 Cockle 26 Acorn worm 27 Sand urchin 28 Eel-grass

Eulittoral zone

Sandy shore

1 Banded periwinkle 2 Lichen 3 Black snail 4 Brown barnacle 5 Black mussel 6 Plicate barnacle 7 Sea star 8 Rock crab 9 Snakeskin chiton 10 Oyster borer 11 Limpet 12 Topshell or catseye 13 Tubeworms 14 Sea squirt 15 Barnacle Balanus 16 Sea urchin (kina) 17 Coralline algae 18 Red algae 19 Neptuneâ&#x20AC;&#x2122;s necklace 20 Kelp

Sublittoral zone

Rocky shore

Littoral fringe

Exposed rock shore

Key to New Zealand species

Denise Fort

ffIn New Zealand, exposed rocky shores occur along much of the coastline.

1. (a) Identify two abiotic factors that might influence species distribution on a rocky shore:

(b) Identify two biotic factors that might influence species distribution on a rocky shore:

2. Describe the zonation pattern on a rocky shore:

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126 125 124

KNOW


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125 Competition and Species Distribution Key Idea: Interspecific competition can affect the distribution of species sharing the same general

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environment causing them to occupy a narrower range of habitat than they could tolerate.

ffDirect competition between different species (interspecific

competition) in naturally occurring populations is usually less intense than competition between the same species (intraspecific competition) because coexisting species have evolved slight differences in their realised niches, even though their fundamental niches may overlap. This has been well documented in barnacle species (below).

Photo: Conrad Pilditch

Epopella

Chamaesipho

ffHowever, when two species with very similar niche

requirements are brought into direct competition through the introduction of a foreign species, one usually benefits at the expense of the other, which is excluded.

In New Zealand regions where the columnar barnacle (Chamaesipho columna) and the plicate barnacle (Epopella plicata) overlap, they compete for space and position.

High tide mark

Chthamalus Fundamental niche

A

Inset enlarged right

Settling Balanus larvae dry out and die at low tide

Chthamalus adults

Low tide mark

Balanus Fundamental = realised niche

Settling Chthamalus larvae are crowded out by Balanus

On the Scottish coast, two barnacle species, Balanus balanoides and Chthalamus stellatus, coexist in the same general environment. The barnacles show a zonal distribution. Balanus is 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 failed to establish any further up the shore than usual.

Balanus adults

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1. Explain why interspecific competition is generally less intense than intraspecific competition:

2. (a) In the example of the barnacles (above), describe what is represented by the zone labelled with the arrow A:

(b) Explain the evidence for the barnacle distribution being the result of competitive exclusion:

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113

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177

Interspecific competition may account for changes in the distribution of species

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ffIn New Zealand, the introduction of foreign species is a

Left: Gambusia affinis are sometimes called mosquito fish in the belief that they controlled mosquitoes. Below: New Zealand black mudfish

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likely factor in the competitive displacement and decline of many native species. For example, the large number of common and German wasps in honeydew beech forests is implicated in the decline of kaka and other native birds, which depend on honeydew for a food source.

ffThese introductions alter community structure and change patterns of species distribution and abundance.

ffIt can be difficult to definitively show that competition

from one species has caused the decline of another, but it is can be inferred if the range of the native species decreases and that of the introduced competitor shows a corresponding increase.

Grey duck

Mallard duck

In New Zealand, the mallard duck, Anas platyrhynchos is at least partly responsible for the decline in native Pacific grey duck (Anas superciliosa). Mallards are bigger than greys and can physically bully them in competing for food. They also breed more profusely, with an average clutch size of 11 eggs (compared with 8 in greys). Mallard males are also sexually aggressive and will interbreed with greys to form fertile hybrids so that the pool of “pure bred” greys is diminished.

Gambusia were first introduced into the ponds of the Auckland Botanical Gardens in the 1930s to reduce mosquito populations. Their range has continued to increase due to further introductions and natural population spread and they now compete against native fish species (e.g. black mudfish and galaxiids) for food and habitat. Gambusia are aggressive and will attack native fish and consume their eggs. They breed quickly (50 offspring per brood) and reach sexual maturity in three weeks (compared with ~1 year in mudfish) so may displace native fish in some habitats. The black mudfish is particularly vulnerable because its North Island distribution overlaps with Gambusia's, although it is long lived, nocturnal (Gambusia are diurnal), and is well able to survive in a seasonally dry habitat, which Gambusia cannot.

3. In general terms, how can we tell that the introduction of a new species is affecting another species?

4. (a) Explain how the biology of the introduced Gambusia may give it a competitive advantage over black mudfish:

(b) Describe two aspects of the black mudfish's biology that may enable it to persist despite competition from Gambusia:

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5. What aspect of the grey duck's biology makes them particularly vulnerable as a species to the introduction of mallards?

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126 Niche and Community Patterns Key Idea: An organism's realised niche is determined by its adaptations to and tolerance for both

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biotic and abiotic factors.

ffUnderstanding the niche of organisms in a particular community is helpful when trying to explain community patterns such as zonation.

ffOn a rocky shore, species are distributed according to their tolerance to various abiotic factors such as temperature, salinity, time exposed to air, and intensity of wave action.

ffWithin their tolerance zone, they may be further constrained by their interactions with other organisms including

Zonation of some intertidal species at Taylor's Mistake, Banks Peninsula. The curve represents the percentage exposure to air at different levels.

A

EHT MHT

MLT

ELT

Snakeskin chiton

m

100

igor_NZ: www.flickr.com/photos/igor_nz

predators and competitors. Some aspects of the niches of several intertidal organisms are described below.

Chitons have eight hinged shell parts, surrounded by a flexible girdle. They lie flat on contoured rock, minimising wave exposure, and grip tightly with their muscular foot to prevent drying out. Chitons have a very abrasive radula, scraping algae from the surface of rocks. Like limpets, they make a home scar in the rock and return to it after feeding.

80

60

F

D

C

E

50

G

40

D

B

70

Acorn barnacles are found throughout the intertidal zone. They are surrounded by a shell of hard plates, extending feathery appendages when submerged to feed off material suspended in the water. The smaller species here is the columnar barnacle, surrounding the large plicate barnacle.

3

90

C

2

A

Percentage of time exposed to air

Columnar barnacle

1

H

B

Plicate barnacle

0

20

KNOW

LINK

LINK

126 105 124

KEY

ELT = Extreme spring low tide mark MLT = Mean spring low tide mark

0

WEB

10

The common rock crab (rerere) is an agile intertidal crab, usually found sheltering under rocks, between the high and low water of neap tides (these are tides with the least difference between high and low tidal marks). They emerge from crevices and from under rocks at night to forage, mainly for plant food although they scavenge for scraps. They are most active when the tide is in, as they are not particularly tolerant of long dry periods.

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30

Common rock crab Hemigrapsus edwardsii

MHT = Mean spring high tide mark

EHT = Extreme spring high tide mark

All photos on this page except Sypharochiton, thanks to Conrad PilditchUniversity of Waikato. Zonation pattern based on a study by Knox (1953).

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Black nerite Nerita melanotragus

Oyster borer Haustrum scobina

E

Grooved topshell Diloma aethiops

H

G

Radiate limpet Cellana radians

F

The black nerite (matangarahu) is a herbivorous snail, grazing on lichens and algae in the intertidal zone. It is extremely hardy, surviving out of water for prolonged periods. In this photograph, there is also a grooved topshell (maihi). This species is also herbivorous and is most common lower on the shore, under rocks and in crevices, often in groups.

The oyster borer (kaikai tio) is a carnivore, drilling into the shells of mussels and barnacles using chemicals and the mechanical action of its radula. A layer of mucus around their shell helps to prevent desiccation. They are often very numerous and will form large groups, which also conserves moisture.

The radiate limpet (ngakihi) is a grazing univalve, moving about when covered by the tide to feed on young seaweeds. They make a home groove in the rock to exactly fit their shell and they return to this spot before the tide goes out. A limpet's shell has a flat profile that can withstand very turbulent water. When not feeding, they clamp strongly to the rock using mucus and their muscular foot, sealing the shell edge to prevent predation and desiccation.

1. Study the diagram opposite and list the species in order from least to most amount of time exposed to air:

2. (a) Name two abiotic factors that rocky shore organisms must cope with:

(b) Name two biotic factors that rocky shore organisms must cope with:

3. Describe adaptations of the following organisms to the abiotic environment of the rocky shore:

(a) Radiate limpet:

(b) Common rock crab:

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4. Using the information provided, explain how the niches of the rocky shore species contribute to the pattern of zonation:


127 Altitude Zonation and Physical Environment

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180

Key Idea: Changes in physical factors associated with increasing altitude produce marked bands of

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vegetation (zonation).

ffMount Taranaki lies on the west coast of New Zealand's North Island.

The area surrounding the summit within a radius of 9.6 km is designated as a National Park and it is home to a number of species found nowhere else in New Zealand. The vegetation on the mountain shows marked zonation in response to physical and biotic factors. With increasing altitude, the vegetation changes in composition, growth form, and height. This zonation pattern gives a basis for describing the vegetation types in the region.

ffThe two diagrams below illustrate profiles taken along a transect line

Mt Taranaki zones: montane to alpine

3000

Profile of vegetation on Mount Taranaki

Summit (2518 m )

2500

Altitude above sea level (m)

K Pryor

beginning at the Egmont National Park boundary and extending to the summit. One shows changes in vegetation zones with increasing altitude. These zones are characterised by particular species or species assemblages. Illustrated below the vegetation profile are the changes in temperature and rainfall with altitude. The photograph right illustrates some of the vegetation changes occurring with altitude on Mount Taranaki. A simplified plan view is shown on the next page. The zones are clearly visible from a distance.

Key to vegetation zones

Alpine barren Herbfields-blue tussock Tussock zone Sub-alpine scrub zone Montane zone Lower-montane zone Lowland zone Semi-coastal zone

2000 1500 1000

Fanthams Peak (1970 m )

The Beehives

846 m

952 m

500

0

1

15

4

5

6

10

9000

6000 4500

3

LINK

9

7500

6

Precipitation

Photograph is not necessary for interpretation of the graph

0 1 Low altitude LINK

2

3

4

5

6

7

Horizontal distance from National Park boundary (km) LINK

LINK

105 115 123 128

8

9

S. Fairweather

3000

SF

-3

8

Temperature

9

0

7

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Average temperature (oC)

3

Profile of temperature and rainfall on Mount Taranaki

12

KNOW

2

Annual precipitation (mm)

0

10 Summit

1500 0

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Simplified plan view of the vegetation distribution on Mount Taranaki

KEY

N

Egmont National Park

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Rimu-rata forest Scrub and tree fernland and logged forest Kamahi-mountain totara forest

a

Mountain totara-kaikawaka forest Kamahi-leatherwood scrub

Rata-kamahi forest Leatherwood scrub Red tussock-herbfield-mossfield Gravel-ice-snowfield Swamps and bogs

c

Sto

ny

Riv

er

a

c

b

Manganui

Rivers

River

Transect line for vegetation profiles Vegetation boundaries within the major altitudinal zones

a

0 1

2

3

4

5 km

am tre

(All data and diagrams for this exercise were adapted from: Clarkson & Irwin Vegetation of Egmont National Park New Zealand National Parks S cientific Series No. 5, DSIR, 1986)

Transect line

b

Fanthams Peak (1970 m)

S ni pu Ka

Summit (2518 m)

1. Identify three physical factors that change with altitude on the slopes of Mt Taranaki:

2. Study the graph on the previous page and describe how the following factors change with increasing altitude:

(a) Temperature:

(b) Precipitation:

3. Name another ecosystem showing a zonation pattern over an environmental gradient:

4. From the graph on the previous page, identify the temperature and precipitation ranges for the montane zone:

5. The tussock zone is characterised by the presence of red tussock. Study the profile (previous page) and: (a) Determine the altitude range for the tussock zone (in metres):

(b) Suggest a possible reason why the tussock zone is not found at lower altitude:

(c) Describe the probable physical factor that prevents the tussock zone extending to a higher altitude:

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128 Environment Determines Community Patterns

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182

Key Idea: A community pattern is influenced by the local environment including climate and geological

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features as well as biotic factors such as the interactions between the species present.

ffAbiotic (physical) and biotic (living) factors both influence community patterns.

ffThe community structure on Mount Taranaki changes with altitude, and is heavily influenced by abiotic factors.

The final composition of the community is influenced not only by climate, but also by rock type, soil composition, drainage, and aspect.

ffBiotic interactions also play a role in community structure, influencing the distribution and abundance of species within their tolerance range.

Changes in community structure with altitude on Mount Taranaki

24 22 20

Ra

Height (m)

18

Ri

16

16

14

14

Ka

12

10

Hi

8

Bm Black maire

Rimu

Kw Kaikawaka

Hi

Hinau

bl

Tf

Treeferns

Lw Leatherwood

Ka

10 8

4

4

2

2

2

0

0

550 m

4

Lw

2

In

Lw

In

4

Lw In

2

0

Inaka

Bm

Ka

To

4

Tf

Broadleaf (Griselinia)

In

6

Tf

Mountain totara

Ka Kamahi

6

6

Height (m)

Mi

8

Ka

To

Ri

Mi Miro

12

Ka

10

Ra Rata

Kw

4

Red tussock

2

0

To

Ka

Ka

bl

0

825 m

Kw

To

1065 m

Blue tussock

0

1310 m

1525 m

1830 m

Changes in species abundance with altitude on Mount Taranaki

A

70

Leatherwood

60

Everlasting daisy

Rata

50 40 30

Mountain totara

Rimu

20 10

500 m

750 m

Emergent trees Rimu Rata Tawa

LINK

LINK

Coprosma sp.

Kaikawaka

Tawa

0

KNOW

B

Kamahi

Canopy trees

Kamahi Mountain totara Kaikawaka

LINK

105 123 127

1000 m 1250 m Altitude (metres) Shrubs & scrub Leatherwood Coprosma sp.

Blue tussock Moss

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Relative cover (%)

80

Red tussock

1500 m

Tussocks

Red tussock

1750 m

Herbfield

Everlasting daisy Moss Blue tussock

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Forstera

The alpine herb fields of Mt Taranaki are dominated by herbaceous plants. Mountain and everlasting daisies, and the leatheryleaved hebe Forstera are common.

Celmisia

Introduced browsers, primarily hares in the tussocklands, preferentially eat palatable native plant species. This allows less palatable species, such as mountain daisies (Celmisia) to flourish.

Everlasting daisy

New Zealand montane and alpine areas have few insect pollinators. Flowering plants in these regions have generalised pollination systems, so that many insect species can pollinate them.

All photos on this page: C. Gemmill

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183

Egmont red tussock

Mt Taranaki's biota has evolved in isolation from other New Zealand alpine areas. As a result, many species, such as the Egmont red tussock (left), are endemic to the region. These communities are fragile and susceptible to the activities of introduced species, which alter community composition. Kamahi (right) is a preferred food for possums. Significant possum damage to kamahi can cause it to die back. Other plant species then exploit the gaps left by kamahi, which may not regenerate, particularly if pollination rates are low because of loss of flowers.

Kamahi

Phil Bendle

C.Gemmill

Biotic factors influencing community patterns

1. The upper diagram opposite illustrates the changes in the structure of the vegetation communities on Mount Taranaki at six different altitudes. Only major species present are labelled. Study the profile and describe a general trend in:

(a) Height of the vegetation with increasing altitude: (b) Vegetation diversity (number of species) with increasing altitude: 2. Describe the change in community structure (species presence/absence) between 550 m and 825 m:

3. The graph opposite shows changes in species abundance with increasing altitude. The various species fall into categories (e.g. emergents) according to their size and stature.

(a) Mark on the diagram (with vertical lines) the upper limits of: the emergents, canopy trees, shrub/scrub, and tussock.

(b) State the lower altitude limit of the herbfield species:

(c) Identify the single most abundant species in each of the following catergories:

Emergent trees:

Canopy trees:

Shrub/scrubs:

(e) Identify the species present at sample points A and B on the graph:

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(d) Identify the species with the greatest tolerance to high altitude and suggest why it becomes more abundant at the upper end of its range:

Point A: Point B: 4. In a general way, explain how introduced species might alter the natural community composition of Mount Taranaki:

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129 Primary Succession

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184

Key Idea: Primary succession is a type of ecological succession occurring in a region where there is

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no pre-existing vegetation or soil, such as rock faces and volcanic flows.

ffEcological succession is the process by which communities

change over time. Succession takes place as a result of complex interactions of biotic and abiotic factors. The activities of early colonising (pioneer) species modify the physical environment. This in turn alters the biotic community, which further alters the physical environment and so on. Each successive community makes the environment more favourable for the establishment of the next species assemblage in the succession.

ffPrimary succession is the biological colonisation and

development of a region with no pre-existing vegetation or soil. A primary succession may take hundreds of years, but the time scale depends strongly on factors such as distance from vegetated areas and availability of pollinating species such as birds and insects.

ffExamples of primary succession include the emergence of new

volcanic islands, new coral atolls, or regions where the previous community has been destroyed by a volcanic eruption.

ffMany texts show a sequence of colonisation beginning with

K Pryor

lichens, mosses, and liverworts, then progressing to ferns, grasses, shrubs, and finally a climax community of mature vegetation forms which remains relatively stable until a new disturbance occurs. In reality, this sequence is very rare.

Community composition changes over time. Species diversity and community complexity increase.

Present community

Some species in the past community were outcompeted and/or did not tolerate altered abiotic conditions. Other species persist. The early community of colonising species is simple in structure with a low species diversity.

TIME

The activities of the present community modify the abiotic environment altering light intensity and quality, air temperature, soil composition, and available water and nutrients.

Lichens and annual herbs

Bare rock

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129

LINK

LINK

130 131

Grasses and small shrubs including nitrogen fixers

Future community

The activities of the present community will allow species of the future community to become established. The mature cor climax community is relatively stable. It has a higher species diversity and a more complex structure than earlier communities. Community processes such as nutrient cycling are more efficient.

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Past community

Fast growing trees

Slower growing broadleaf and evergreen species

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185

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Succession sequence for Mount Tarawera Bare ground

1. (a) Define primary succession:

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Mt. Tawawera erupted in 1886. Existing vegetation was buried under ash and there was no vegetation present for 10 years.

(b) Describe some situations in which primary succession would occur:

Early colonisers

Organic acids secreted by lichens (right) broke down rock to produce humus and eventually soil. Toetoe, bracken fern and tutu began to appear next on the lower slopes. Lichens, grasses, annual herbs, and hardy shrubs are typical early colonising species.

2. What is a climax community?

3. Describe the general trend in community composition as a successional sequence progresses:

Nitrogen fixers

4. (a) Identify some early colonisers during the early phase of establishing a community on bare rock:

Š Maurice Grout

(b) Describe two important roles of the species that are early colonisers of bare slopes:

(c) Explain why these species are replaced in later stages of the succession:

Trees established

By the 1990s, tutu had spread and a kamahi forest covered the upper slopes. Kamahi (right) is common in New Zealand forests, where it often exists with other broadleaf trees.

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30 years after the eruption, the nitrogen-fixing shrub tutu (right) was spreading up the mountain's slopes. Over time, the soil became enriched and other plant species became established on the lower slopes. Nitrogen fixation is an important process in succession, enriching the soil and allowing subsequent species to establish.


130 Secondary Succession

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186

Key Idea: Secondary succession occurs when a previously vegetated area is cleared by a fire or

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landslide, leaving behind soil, roots, and seeds.

ffA secondary succession occurs when land is cleared

of vegetation after a fire or landslide. These events do not involve the loss of soil or seeds, and root stocks are often undamaged. As a result, secondary succession tends to be more rapid than primary succession, although the time scale depends on the species involved, the soil composition, and climate.

ffSecondary succession events may occur over a wide area (such as after a forest fire), or in smaller areas where single trees have fallen.

Large storms can destroy large areas of beech forest (right). Many trees may fall, and their broken logs can remain on the forest floor for decades. Catastrophic events, such as storms or landslides, destroy areas of beech forest about every 10 years.

Background

• Beech (Nothofagus species) often act as pioneer species. However, once established, mature beech forest regenerates itself. • Beech seeds and seedlings grow rapidly in open sites where there is plenty of light.

• Beech is a mast species: in some years, very little seed is produced, whereas in other years, vast quantities of seed are produced.

1

2

Fallen trees play an important role in the ecosystem diversity of a forest. They support a variety of invertebrates, bacteria, and fungi.

Photo: Shirley Kerr

Lancewood

Rimu

Kamahi

Fern

Coprosma

1. How is secondary succession different to primary succession?

Manuka and bracken ferns (above) grow quickly in the absence of beech. Their growth produces large areas of shade, and prevents beech seeds (which require a high light environment) from growing. This sequence of events allows other forest types, such as kamahi, to develop.

If a catastrophic event occurs in a year when very little beech seed has been produced, the beech forest may fail to regenerate. Other forest types will develop in its place (left).

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3

BH

The mycorrhizal fungus, Cortinarius bellus, is found associated with beech trees. Its hyphae (fine filaments) spread out from the beech roots into the soil and greatly increase the surface area over which the plant can absorb nutrients. The fungus benefits by having direct access to the plant's carbohydrate.

2. Why does secondary succession usually take place more rapidly than primary succession?

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130 129 131 132

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131 Strategies for Survival in Different Environments

187

Key Idea: Species able to rapidly increase their numbers dominate early in a successional sequence,

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whereas species with longer lifespans tend to dominate in later successional stages.

ffAn ecological succession is characterised by a change

Important definitions

in species composition over time. Different types of species predominate at different stages in a succession.

ffThe life history strategy of a species describes how it

allocates energy to survival, growth, and reproduction. Species that produce many offspring rapidly and reach maturity quickly are called r-selected species because their biotic potential predominates in their life history (right). Species that reproduce less frequently, have fewer offspring, and take longer to reach maturity are called K-selected species, because they exist near the carrying capacity (K) of the environment and competitive ability and efficiency are important aspects of their life history.

Biotic potential (r)

The capacity of a species to increase its numbers.

Carrying capacity (K)

The maximum population size of the species that the environment can sustain indefinitely.

Survivorship

The proportion of a population surviving to a given age

ffThe differing characteristics of r- and K-selected

species allow them to exploit different types of environments most effectively. Disturbed environments favour fast growing r-selected species, whereas established communities favour K-selected species.

Dandelion (r-selected)

Beech (K-selected)

Features of r-selected species

Features of K-selected species

Climate

Variable and/or unpredictable

Climate

Fairly constant and/or predictable

Mortality

Density independent

Mortality

Density dependent

Survivorship

Large, early losses of offspring

Survivorship

Constant or late loss of older individuals

Population size

Typically shows large fluctuations

Population size

Fairly constant, near carrying capacity.

Competition

Variable, often lax. Generalist niche.

Competition

Usually keen. Specialist niche.

Selection favours

Rapid development, high biotic potential, early reproduction, small body size, single reproduction.

Selection favours

Slower development, larger body size, greater competitive ability, delayed reproduction, repeated reproduction.

Length of life

Short (usually less than one year)

Length of life

Longer (greater than one year)

Leads to:

Productivity

Leads to:

Efficiency

In general, r-selected species dominate in disturbed habitats (early successional communities). Their generalist nature allows them to quickly colonise new environments. Examples include algae, bacteria, rodents, many insects, and most annual plants.

K-selected species begin to replace r-selected species, and dominate later stages of an ecological succession and the climax community. This is because they are better adapted to the many specialist niches available. Examples include large mammals, birds of prey, and large, long-lived plants.

Carrying capacity, K

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r-selected species These species rarely reach carrying capacity (K). Their populations are in nearly exponential growth phases for much of the year. Early growth, rapid development, and fast population growth are important.

Population numbers (N)

Population numbers (N)

Carrying capacity, K

K-selected species

These species exist near carrying capacity (K) for most of the time. Competition and the efficient use of resources are important.

Time

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188

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1. Why do r-selected species predominate in early successional communities?

2. Why do K-selected species tend to predominate in climax communities?

3. Describe factors that might cause a change in the predominance of K-selected species in a climax community:

4. Based on the information given below, decide whether the following organisms are r-selected or K-selected species and use a bar to mark their likely approximate position and duration on the succession timeline:

Fauna

Manuka (Leptospermum scoparium) A scrub-like tree, typically 2â&#x20AC;&#x201C;5 m tall, with dense branching and small prickly leaves. Hardy and grows well in exposed, wet or dry sites as long as there is high light. Tolerant of very poor soil and flowers and seeds prolifically. Seeds will not generate if shaded. Life span <60 years.

http://en.wikipedia.org/wiki/Image:Manukaflowers.jpg

r-selected or K-selected (delete one)

Long-tailed blue butterfly (Lampides) Self-introduced, highly adaptable butterfly. Prefers open habitat, but will live wherever there are suitable larval food plants (legumes such as clover or gorse). Larvae are opportunists and cannibalistic and will pupate early if short of food. Generally live <1 year.

Davidvraju cc 4.0

Tawa (Beilschmiedia tawa) A dominant canopy or sub-canopy tree in many lowland forests reaching ~30 m. Slow growing and long lived (mature trees 200 years+).Produces large succulent fruits which are food for kereru and kokako, which disperse the seeds.

Tree weta (Hemideina) Large, heavy insects up to 40 mm long. They live in tree holes and are nocturnal, feeding on lichens, leaves, flowers, seed-heads, and fruit. Long lived, maturing at ~2 years and notable for their cold tolerance.

r-selected or K-selected (delete one)

Early

r-selected or K-selected (delete one)

Melodi2 cc 4.0

Rudolph89 cc 3.0

Early-mid

Mid

r-selected or K-selected (delete one)

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Flora

Mid-late

Late

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189

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132 Succession and Population Growth in Birds Key Idea: The type of plant community can affect the animals that live there. Succession of one plant

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community into another can cause changes in the animal community.

ffMuch of New Zealand's native forest has been cleared to produce grassland for grazing. Over time, in more remote areas, some of this grassland has returned to shrubland by secondary succession.

ffMany of the original native birds have been lost from these formerly forested areas and have been replaced by introduced species or self-introduced species.

Central Otago

ffA study was carried out in Central Otago to examine the

effect of succession from grassland to woodland on New Zealand birds. A map of the study sites is shown right.

ffIt was hypothesised that the density of ground dwelling

grainivores would decrease and the density of tree living insectivores would increase as the grassland returned to shrubland.

Lake Wanaka

ffThree sites were used at Bendigo, Blackstone Hill and Cambrian.

Bendigo Cambrian

ffIn each site, transects were carried out along grasslands,

an s st tain n Du oun m

intermediate zones, and woody shrubland.

ffThe population densities of various birds were estimated by

walking the transect and recording bird calls and sightings. The density was estimated by recording the distance from the transect for each observation (a distance transect).

Blackstone Hill

ffThe birds observed included greenfinch, goldfinch, redpoll, yellowhammer, grey warbler, and silver eye.

ct

je

ob

to

ce

Redpoll: ground dwelling grainivore

Francis C. Franklin / CC-BY-SA-3.0

Chaffinch: Grainivore

st

an

Silvereye: tree dwelling Insectivore

Goldfinch: Grainivore

Di

fir0002 https://commons.wikimedia.org/ wiki/Commons:GNU_Free_Documentation_License,_version_1.2

Object

se

Yellowhammer: Grainivore

Woodiness level

Blackstone Hill

Cambrian

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Greenfinch: Grainivore

lin

e

Mean height (m)

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Bendigo

Smalljim CC 3.0

ct

Distance sampling method. The number of objects and distance of the object perpendicular to the transect is recorded. A simple formula can then be used estimate the density of the population.

Site

Blackbird: Insectivore

Tr an

Grassland

0.17

Intermediate

0.42

Shrubland

1.26

Grassland

0.30

Intermediate

0.37

Shrubland

0.52

Grassland

0.51

Intermediate

0.39

Shrubland

0.89

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190

Goldfinch

Greenfinch

Redpoll

Density (ha-1)

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2

From: How does woody succession affect population densities of passerine birds in New Zealand drylands? D Wilson, G Norbury, S Walker. www.newzealandecology.org. 9 April 2014

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ffThe population density of the birds compared to the frequency of woody species is shown in the graphs below:

1

0

0

200

400

0

600

Density (ha-1)

Yellowhammer

400

600

0

Chaffinch

1.0

1.0

0.5

0.5

0.5

0

200

400

0.0

Dunnock

1.0

1.0

0.5

0.5

0.0

0.0

0

200

400

600

Woody species frequency (%)

400

600

0.0

0

600

200

Blackbird

1.0

0.0

Density (ha-1)

200

200

400

600

0

200

400

600

Silvereye

Grey warbler (Insectivore)

2 1 0

0

200

400

0

600

200

400

600

Woody species frequency (%)

Woody species frequency (%)

* Because of the sampling method the fequency of a woody species can be greater than 100%

Bendigo

Blackstone Hill

Cambrian

1. In general how does the height of vegetation change from grassland to shrubland?

2. Why would it be hypothesised that ground dwelling grainivores would decrease and the density of tree living insectivores would increase as the grassland returned to shrubland?

4. What might this mean for the restoration of shrubland?

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3. Study the graphs above. Is there any evidence to suggest the hypothesis is supported? Explain your answer:

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191

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133 What You Know So Far: Communities Community patterns

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Summarise what you know about this topic so far under the headings provided. You can draw diagrams or mind maps, or write short notes to organise your thoughts in preparation for the essay question at the end of the chapter. Use the points in the introduction and the hints provided to help you:

HINT: Describe and explain changes in community composition along an environmental gradient.

Collecting data from ecological communities

HINT: Describe common methods for sampling communities.

Ecological succession

HINT: Describe the common types of species interactions and provide examples.

HINT: Describe and explain both primary and secondary succession.

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Species interactions

REVISE


192

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134 KEY TERMS AND IDEAS: Communities

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1. Match the following words with their definitions:

abiotic factor biotic factor

A The description of how a species is spread throughout an area and is interrelated with the density of a population.

B An interaction between organisms which is beneficial to all parties involved.

community

competition distribution

ecological succession

C A trend in physical and biological features of an environment from one region to another. D Exploitation involving an organism and its host. The host is detrimentally affected by the relationship but is not usually killed.

A sequence of organisms in which each organism is a source of food for the next. E

F A vertical pattern of layering in a forest that changes from the ground to the canopy.

environmental gradient

G The term used in ecology for any contribution to the environment by a living organism.

food chain

H An interaction between organisms exploiting the same resource.

mutualism

I The progression from colonisation of a newly formed or cleared area to a climax community.

parasitism

J A sub-set of a whole used to estimate population parameters (the values that might have been obtained if every individual or response was measured).

predation

K A term for any non-living part of the environment, e.g. rainfall, temperature.

sample

L Exploitation in which one organism kills and eats another, usually of a different species.

stratification

M A naturally occurring group of different species living within the same environment and interacting together.

2. Use lines to join the three parts of the sentences below together to form complete sentences. The first column is in order, the centre and right columns are not. Choose the most appropriate joining word(s) to create the sentence. Methods commonly used to sample communities...

by

A transect along an environmental gradient...

include

Zonation is common...

with

The occurrence and distribution of organisms in the community…

is often

… abiotic and biotic factors determines their position along the environmental gradient.

The adaptations of species...

to

… more restricted than they might potentially occupy.

Their region of occupation…

creates

This reflects their realised niche…

and the

… community position to which they are best adapted.

They are often forced to this position…

may reveal

… altitude and on rocky shores.

… a community pattern of zonation. … quadrats and transects.

… a banding pattern.

(b) What abiotic factors are likely to be important in this ecological pattern:

(c) What other factors might be important in determining the distribution and abundance of organisms present?

TEST

Bcasterline, Public Domain

3. (a) Identify the ecological pattern shown in the photograph right:

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… competition and predation.

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Practical microscopy

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Achievement Standard

2.8

Key terms

Investigating biological material at the microscopic level requires specific skills associated with selecting and preparing biological material for viewing, using a light microscope correctly to view specimens, and producing a biologically accurate record of what you observe.

biological drawing biological stain

Achievement criteria and explanatory notes

compound microscope

dissecting (stereo) microscope

Achievement criteria for achieved, merit, and excellence

c

light (=optical) microscope

linear magnification magnification

A

Investigate biological material at the microscopic level: Involves:

preparing biological material for viewing under a light microscope,

viewing biological material using a light microscope to enable detail of cell structures and components to be determined,

recording observations of biological material in biological drawings,

identifying observed specialised features and relating them to the function of the cells/tissues.

resolution

Investigate in-depth biological material at the microscopic level: Involves giving reasons for how or why observed features enable the cells to carry out their function(s) effectively.

wet mount

Explanatory notes

c

M

c

1

Biological material for viewing includes two different plant tissues and one unicellular organism.

c

2

To produce an accurate drawing, preparation of material may include: staining, use of cavity slides, use of methyl cellulose for live specimens, epidermal tear, and cutting sections.

c

3

A biological drawing follows the accepted conventions to record observations consistent with the biological material being viewed.

c

4

Specialised features may include: arrangement of cells or cell types with a tissue, shape of cell, presence or absence of organelles,number of distribution of organelles within a cells. Drawings may be annotated to include notes about specialised features.

c

5

Relating observed specialised features to the function of the cell or tissue must include: identifying the feature or organelle, stating its function, and giving reasons for how or why it contributes to the function of the cell or tissue.

Activities 135 - 143

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What you need to know for this Achievement Standard By the end of this chapter you should be able to:

c

Understand the basic principles of optical microscopy. Use dissecting and compound light microscopes to locate material and focus images.

c

Distinguish between magnification and resolution and understand their significance.

c

Explain the role of sample preparation (including stains and wet mounts) in light microscopy. Demonstrate an ability to use simple staining techniques to show features of cells.

c

Demonstrate an ability to prepare a temporary mount for viewing with a light microscope.

c

Calculate the linear magnification of images viewed with a microscope.

c

Use microscopy to identify specialised features of cells or tissues and relate the features to the function of the cell or tissue under observation.

c

Demonstrate an ability to record observations of biological material as accurate biological drawings. Use accepted biological conventions to produce the drawing.


135 Optical Microscopes

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194

Key Idea: Optical microscopes use a series of lenses to magnify objects up to several 100 times. Total

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magnification depends on the magnifications of the objective and eyepiece lenses.

1. Label the two photographs below, the compound light microscope (a) to (h) and the dissecting microscope (i) to (m). Use words from the lists supplied for each image.

RCN

(a)

Stoma in leaf epidermis

Typical compound light microscope

In-built light source, arm, coarse focus knob, fine focus knob, condenser lens, mechanical stage, eyepiece lens, objective lens (e) (f)

(b)

(g)

(c)

(h)

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 in specimens that are thick or opaque. Modern microscopes are binocular, i.e. they have two adjustable eyepieces.

(d)

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. 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.

(i)

(j)

(k) (l)

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RCN

Knob for the adjustment of the microscope on the arm.

Attached light source (not always present).

Drosophila

Dissecting microscope

(m)

Focus knob, stage, eyepiece lens, objective lens, eyepiece focus

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135 140 142

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John Green

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Types of microscopy

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.

What is magnification?

What is resolution?

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

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.

Objective lens power

X

Eyepiece lens power

Low resolution

High resolution

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. What type of light microscope, dissecting microscope (DM) or compound microscope (CM), would you use to:

(a) Count stream invertebrates in a sample? (b) Observe cells in mitosis?

4. Below is a list of 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).

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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.

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).

Viewing from the side, lower the shortest objective lens to its lowest point or until it is almost touching the specimen. Turn on the light source.

Focus and centre the specimen using the low objective lens, moving the lens away from the specimen. Focus firstly with the coarse focus knob, then with the fine focus knob.


136 Biological Drawings

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196

Key Idea: Biological drawings follow standard conventions or rules that should be followed to produce

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an accurate record of the specimen you are observing

ffDrawing is a very important skill to have in biology. Drawings record what a specimen looks like and give you an opportunity to record its important features. Often drawing something will help you remember its features at a later date (e.g. in a test).

ffBiological 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.

ffBiological 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.

ffAttention should be given to the symmetry and proportions of your

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

ffSome key points for making good biological drawings 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

Place your drawing on the left of the page. This will leave room to place all the labels to the right of 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.

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Egg sac

Thorax

Caudal rami

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

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

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LINK

136 137

Setae

Scale

0.2 mm

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

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Annotated diagrams

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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)

Cytoplasm - Solution of dissolved substances, enzymes, and organelles. Vacuole containing cell sap.

Chloroplast - Organelles containing chlorophyll where photosynthesis occurs.

Nucleus - A large organelle containing most of the cell's DNA. Primary wall with secondary thickening.

Plan diagrams

Plan diagrams are drawings made of samples viewed 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.

Parenchyma between vascular bundles

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Epidermis

0.5 mm

Phloem

Vascular cambium Xylem

Sclerenchyma (fiber cap)

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Pith (parenchyma cells)

Vascular bundle


137 Practising Biological Drawings

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198

Key Idea: Good biological drawings provide an accurate record of the specimen you are studying.

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Attention to detail and recording what you see are vital

Root transverse section Root transverse section through Ranunculus 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

Phloem

0.05 mm

1. The image below is a labelled photomicrograph (x50) showing a partial section through a dicot root. Use this image to construct a plan diagram: Epidermis

Phloem

Xylem

Cortex

PRAC

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Pericycle

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138 Preparing a Slide Key Idea: Correct preparation of a slide is important if structures are to be seen clearly under a

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microscope. A wet mount is suitable for most slides.

ffSpecimens 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.

ffSections 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. Upper epidermis peeled away

KP

Onions make good subjects for preparing a simple wet mount. A square 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 for viewing.

KP

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.

Mounted needle

Mounting fluid

Coverslip

Specimen

Microscope slide

Mounting: The thin layer is placed in the centre of a clean glass microscope slide and covered with a drop of mounting liquid (e.g. water, glycerol, or stain). A coverslip is placed on top using a mounted needle to support and lower it gently over the specimen. This avoids including air in the mount.

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

2. What is the purpose of the coverslip?

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

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

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KNOW


200

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139 Staining a Slide Key Idea: Staining the material being viewed under a microscope can make cell structures easier to

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see. Different stains affect the visibility of different organelles.

ffStains and dyes can be used to highlight specific

Some commonly used stains

Stain

Final colour

Used for

Iodine solution

blue-black

Starch

Crystal violet

purple

Gram staining

Aniline sulfate

yellow

lignin

Methylene blue

blue

Nuclei

Mnolf

CDC: Dr Lucille K. Georg

components or structures. Most stains are non-viable, and are used on dead specimens, but harmless viable stains can be applied to living material. Stains contain chemicals that interact with molecules in the cell. Some stains bind to a particular molecule making it easier to see where those molecules are. Others cause a change in a target molecule, which changes their colour, making them more visible.

Iodine solution stains starch-containing organelles, such as potato amyloplasts, blue-black.

Methylene blue is a common temporary stain for animal cells, such as these cheek cells. It stains DNA and so makes the nuclei more visible.

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.

Irrigation

Specimen

Coverslip

Viable (or vital) stains do not immediately harm living cells. Trypan blue distinguishes living and dead cells, and is used to study fungal hyphae.

1

Filter paper

2

1. What is the main purpose of using a stain?

2. What is the difference between a viable and non-viable stain?

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The micrographs 1 and 2 (right) taken with a light microscope show how adding a stain can enhance viewing or certain structures. The top image has no stain applied. Only the cell wall is visible. Adding iodine makes the cell wall and nucleus stand out.

3. Identify a stain that would be appropriate for improving identification of the following:

(a) Fungal hyphae:

(c) Lignin in a plant root section:

(b) Starch in potato cells:

(d) Nuclei in cheek cells:

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140 Calculating Linear Magnification Key Idea: Magnification is how much larger an object appears compared to its actual size. It can be

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calculated from the ratio of image height to object height.

Linear magnification

ffMicroscopes produce an enlarged (magnified) image of

an object allowing it to be observed in greater detail than is possible with the naked eye. 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.

1.0 mm

A worked example 1

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

2

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 (right).

1. Magnification = measured size of the image

actual size of object

The magnification of the scale line is 10 (10 mm / 1 mm) *The bed bug image is also magnified 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):

2. Actual object size = size of the image magnification

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

1. In the bright field microscope image (left), the measured length of the onion epidermal 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:

RCN

X 140

0.5 mm

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EII

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:

EII

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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141 140

DATA


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141 Cell Sizes Key Idea: Cells vary in size (2-100 µm), with prokaryotic cells being approximately 10 times smaller

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than eukaryotic cells.

Unit of length (International System)

Cells are extremely small and can only be seen properly when viewed through the magnifying lenses of a microscope. The diagrams below show a variety of cell types alongside viruses (non cellular particles).

Parenchyma cell of flowering plant

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

Unit

Metres

Equivalent

1 metre (m)

1m

= 1000 millimetres

10

-3

m

= 1000 micrometres

1 micrometre (µm)

10

-6

m

= 1000 nanometres

1 nanometre (nm)

10 -9 m

1 millimetre (mm)

Human white blood cell

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

= 1000 picometres

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

50 µm

Viruses Size: 0.02-0.25 μm (20-250 nm)

Giardia are protozoa that infect the small intestines of many vertebrate groups.

1.0 mm

Paramecium is a protozoan commonly found in ponds and stagnant water.

10 µm

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

n

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

RCN

CDC

SEM

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

c

100 µm

50 µm

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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: µm

mm

(d) Elodea leaf cell:

µm

mm

(b) Giardia:

µm

mm

(e) Chloroplast:

µm

mm

µm

mm

(f) Paramecium: µm

(a) Daphnia:

(c) Nucleus

mm

2. List the five specimens shown above in order of cell size (largest to smallest):

3. Study the scale of your ruler and state which of the five specimens you would be able to be seen with your unaided eye?

DATA

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142 Examining Cells Key Idea: Identifying structures and organelles in specimens under the microscope is an important

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part of microscopy.

ffExamining cells through a light microscope can be tricky. High magnification and inconsistent slides (e.g. sections that are too thick or poorly stained) can produce a lot of distortion and make it hard to distinguish details.

Thale cress, leaf epidermal cells Size: 30 x 20 µm Habitat: Terrestrial

ffStaining may enhance the visibility of some structures (e.g. nucleus) in a cell.

ffSome specimens have structures that are relatively easy

Nucleus

to see. For example Elodea (below) has very thin leaves with large cells, making it easy to see the chloroplasts even under low magnification.

ffMobile specimens such as Paramecium (below right) can

be viewed using methyl cellulose to slow their movement. Elodea leaf cells Size: 50 x 25 µm Habitat: Freshwater

Chloroplasts: Site of photosyntheis

Guard cells

Magnification (X 250)

Stoma: Pore allowing gas exchange

Food vacuoles

Contractile vacuole

Nucleus: Contains the cell’s DNA (not always visible)

Magnification (X 400)

Macronucleus: Controls metabolic activity.

Cilia: Hair like structures assist cell movment.

Oral groove: Food is moved to base of oral groove by cilia, forming a food vacuole.

Food: Bacteria and small protists.

Micronucleus: meiosis and sexual reproduction.

Anal pore

Magnification (X 300)

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Cell wall: Composed of cellulose. Gives the cell shape and limits its volume.

Paramecium Size: 240 x 80 µm Habitat: Freshwater, sea water

1. List the three specimens shown above in order of cell size (largest to smallest):

2. Why can it be difficult to view slides under a microscope?

3. Why would methyl cellulose be used when viewing live Paramecium?

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204

B

(b) What is the function of this structure?

(c) Identify the structure labelled B:

(d) What is the function of this structure?

(e) What is the length of structure B?

(f) Calculate the magnification of the image.

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4. (a) Identify the structure labelled A:

A

A

Kristian Peters

100 µm

B

Differential phase contrast image of two amoebae Thyme-moss leaf cells (Amoeba protetus)

5. (a) Identify this organism:

A

(b) What feature(s) helped you make your identification?

(c) Identify the organelle labelled A:

(d) Circle a second example of this type of organelle:

(e) Identify the structures indicated by the blue arrows and describe their purpose:

Barfooz CC 3.0

25 µm

6. (a) Identify the organelle labelled A:

B

(b) Describe the function of this organelle:

(c) Identify the cell labelled B.

(d) What is the function of this cell?

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A

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205

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143 KEY TERMS AND IDEAS: Microscopy

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1. Match the following words with their definitions: biological drawing

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

biological stain

B The ability to distinguish between close together but separate objects, e.g. the greater the resolution of a microscope the smaller the objects viewed can be.

compound microscope

C A microscope that uses multiple lenses to produce a magnified image of the object being observed.

dissecting (stereo) microscope

D An optical microscope designed for low magnification for observing the surface of samples. It uses two separate optical paths to produce a stereo image.

light (=optical) microscope

E A chemical that binds to parts of the cell and allows those parts to be seen more easily under a microscope.

magnification

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

resolution

G A drawing to accurately represent the biological material being observed.

2. Identify and label the following parts of the microscope in the photo below: stage, objective lens, condenser lens, slide, in-built light source, eyepiece lens.

3. Calculate the magnification of the organisms shown below:

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(b) Scenedesmus, a colonial alga (this one has 4 cells)

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(a) Amoeba, a unicellular protist

TEST


206

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Answers

CHAPTER

PAGE

Practical investigation in a biology context: 206 Analysing the biological validity of information 208 Adaptations of animals to their way of life 209 Adaptations of plants to their way of life 217 Patterns in ecological communities 222 Practical microscopy 227

Practical investigation in a biology context 1. How Do We Do Science? (page 3) 1. Students' own responses.

2. Science needs to be open to new ideas because the techniques and information available are continually changing as knowledge and expertise grows. It must also be rigorous so that new information can be tested and assessed for how well it fits both the observations and the existing theories and models. If new data explain phenomena better than existing theories and models, these may require modification so it is important that the methods used to test new information stand up to scrutiny.

5. Tallies, Percentages, and Rates (page 7)

1. Raw data is data that has been collected but has not been processed in any way.

2. Processing data allows important trends to be seen and make meaningful comparisons.

3. (a) Kilograms per hectare (kg ha-1) (b) mL per hour per square metre leaf surface (mL hr-1 m-2)

6. Fractions and Ratios (page 8)

1. (a) 140/5 = 28, 70/5 = 14, 15/5= 3, 10/5 = 2, 5/5 = 1 Ratio: 28:14:3:2:1 (b) Cell cycle stage

No. of cells counted

No. of cells calculated

Interphase

140

2800

Prophase

70

1400

Telophase

15

300

Metaphase

10

200

Anaphase

5

100

Total

240

4800

2. (a) 1/3

(b) 14/15

(c) 1/11

3. (5/20 x 3 = 15/60) + (5/12 x 5 = 25/60) = 20/60 = 2/3. 2/3 ÷ 2 = 1/3

2. Observations, Hypotheses and Assumptions (page 4)

7. Dealing With Large Numbers (page 9)

2. The yeast populations numbers would need to be log transformed before plotting.

– Bright colour patterns might signal to potential predators that the caterpillars are distasteful. – Inconspicuous caterpillars are good to eat and their cryptic coloration reduces the chance that they will be discovered and eaten.

2. Possible assumptions: – Birds and other predators have colour vision. – Birds and other predators can learn about the palatability of their prey by tasting them.

3. Prediction 1: That birds will avoid preying on brightly coloured, conspicuous caterpillars. Prediction 2: Naive (inexperienced) birds will learn from a distasteful experience with an unpalatable caterpillar species and will avoid them thereafter. Prediction 3: Birds will prey readily on cryptically coloured caterpillars if these are provided as food.

3. Accuracy and Precision (page 5)

1. Precise but inaccurate data will give measurements that are either higher or lower than the true value of the variable you are wanting to measure.

2. (a) 6 (b) 2 (c) 1

(d) 5 (e) 6 (f) 2

60 mm ÷ 0.48 mm = 125 x magnification.

(page 6)

1. Semi-log paper reduces the space needed when graphing large numbers as it compresses the y (or x) axis into a logarithmic scale.

3. Exponential growth of bacterial populations, pH, energy release at different earthquake magnitudes, the number of radioactive atoms in a sample over time.

8. Apparatus and Measurement (page 10) 1. (a) 25 mL graduated cylinder (b) 50 mL graduated cylinder (c) 10 mL pipette

2. (a) ((0.98-1)/1) x 100 = -2% (b) ((9.98-10)/10) x 100 = -0.2%

9. Types of Data (page 11)

1. (a) Skin colour: ranked data. (b) Number of eggs: quantitative data, discontinuous. (c) Tree trunk diameter: quantitative data, continuous.

2. Quantitative data are more easily analysed in a rigorous and meaningful way, e.g. by using descriptive statistics.

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1. There are several hypotheses that could be generated to explain these observations:

3. Examples include: biological sex, viability (dead or alive), species, presence or absence of a feature, flower colour. These data are categorical; no numerical value can be assigned to them.

10. Variables and Controls (page 12)

1. An independent variable is set by the person running the experiment. It is the variable that is manipulated. The dependent variable is the response that is measured during the experiment (the response variable). The value of the dependent variable is dependent on the independent variable.

1. 45,000 + 645,000

2. 690,000

3. 6.9 x 105

4. 43,000 i.e. 43 x 1000

6. 33, i.e. 66 ÷ 2

2. Variables not being investigated must be controlled to make the investigation or test fair. If more than one variable is changed at a time, the effect to the investigation cannot be accurately and fairly attributed. Comparison between investigations is also not possible.

7. 5600 mm3

3. Controls determine if an experiment has run correctly or

5. 15 i.e. 3 x 70 = 210 ÷ 10 = 21. 210 ÷ 20 = 10.5 and 15 is halfway between 21 and 10.5

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207 not. If the control works as expected, you can assume the experimental results are valid.

14. Constructing Tables (page 16)

2. If the rate of catalase activity is dependent on temperature then more bubbles will be produced at higher temperatures.

2. To identify trends in the data. To help decide the best method of graphing the data.

3.

15. Drawing Graphs (page 17)

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1. To investigate the effect of temperature on the rate of catalase activity.

1.

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11. A Case Study: Catalase Activity (page 13)

(a)-(b) any two of the following: – Tables provide a systematic record of information. – Tables provide a way of condensing data. – Tables provide a summary of results. – Tables show trends and relationships in the data.

(a) Independent variable: Temperature. (b) 10-60°C in uneven steps: 10°C, 20°C, 30°C, 60°C. (c) °C (d) A way to maintain the test-tubes at the set temperatures, e.g. water baths. Equilibrate all reactants to the required temperatures before adding enzyme to the reaction tubes.

4. (a) Height of oxygen bubbles. (b) mm (c) Ruler. Place vertically alongside the tube and read off the height (directly facing).

1.

(a) Scatter graph (b) Bar graph (c) Histogram (d) Bar graph (e) Line graph (f) Bar graph (g) Scatter graph

5. Tubes 9 and 10 were the controls.

2. (a) 0.8 mm d-1 (b) 20 eggs

12. Recording Results (page 14)

3. (a) Line graph

1. Accurately recording results makes it easier to analyse and interpret the data later on.

Rate of reaction of amylase at different temperatures

Reaction rate (mg product formed per minute)

2. Logbook should be well organised to minimise entry errors and ensure that data can be found easily at a later date. It also enables historic methods and data to be reviewed.

3. (a) A datalogger is an electronic device that automatically records data over time. (b) Dataloggers can be set to record data automatically. They can also be more precise than many manual recording devices and, if calibrated correctly, more accurate.

13. Practicing Data Manipulations (page 15)

1. (a) Incidence of cyanogenic clover in different regions: Clover type

Frost prone

No

No.

%

Totals

%

4 3 2

1

0

10 20 30 40 50 60

Temperature (°C)

(b) Rate of reaction at 15°C = 1.6 mg product min-1

Cyanogenic

124

78

26

18

150

Acyanogenic

35

22

115

82

150

159

100

141

100

300

16. Interpreting Line Graphs (page 19)

(b) Plant water loss (missing values only) Time

Pipette reading (cm3)

Water loss (cm-3 min-1)

10

7.2

0.16

15

6.2

0.20

20

4.9

0.26

1. (a) 1 (c) 5 (b) 2 (d) Positive, straight

17. Correlation or Causation (page 20)

1. A correlation between variables does not mean that one variable causes or is even related to the other.

2. (a)

Hand span vs foot length in adults

(c) Frequency of size classes of eels:

350

Size class (mm)

Frequency

Relative freq. (%)

0-50

7

2.6

23

8.5

59

21.9

150-199

98

36.3

200-249

50

18.5

250-299

30

11.1

300-349

3

1.1

Total

270

100.0

Foot length (mm)

50-99

100-149

300

200

(d)

Women

250

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Total

Frost free

5

Body mass (kg)

Lean mass (kg)

% lean body mass

Athlete

50

38

76

Lean

56

41

73.2

Normal weight

65

46

70.8

Overweight

80

48

60

Obese

95

52

54.7

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150

200

250

300

Hand span (mm)

(b) There is a general trend in that a larger hand span will have a longer foot length. (c) There is a positive relationship between hand span and foot length but it is not very strong. (r2 = 0.5 calc. in Excel)

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208 18. Investigating Plant Growth (page 21)

3. Suggested method: • Add 0.5 grams of sodium bicarbonate to the boiling tube and fill with water.

2. Mass of the radish root.

Place the Cabomba in water in the boiling tube. Carefully trim the stem under the water with the scissors.

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1. Fertiliser concentration. Range: 0.0-0.30 g L-1 in steps of 0.06 g L-1.

3. 5

4. (a) Outlying value: 23.6. (b) This value should not be used in calculations as it is likely to be anomalous.

Place red cellophane over the lamp. Switch off other lights (red light should be the only light source). Place the lamp 30 cm from the boiling tubes and start the timer. Do not measure the volume of gas produced for the first minute while the plant acclimatises.

After one minute, place a syringe and rubber bung or funnel attached to a tube over the boiling tube to collect any gas produced. Measure the amount of gas produced for the next ten minutes.

Repeat the procedure for the green and blue cellophane.

Repeat the entire experiment two more times.

5. Missing values recorded below: Fertiliser concn

Total mass

Mean mass

0.0

408.5

81.7

0.06

546.3

109.3

0.12

591.4

118.28

0.18

510.1

127.5

0.24

558.9

139.7

0.30

610.4

122.1

6. Completed graph below.

150 140

Mass (g) of radish roots under six different fertiliser concentrations

Time (min)

Volume of gas produced (ml) (trial 1) Red

130 120

1

110

2

100

3

90 80

4

0.0 0.06 0.12 0.18 0.24 0.32 Concentration of nitrogen fertiliser (gL-1)

7.

Fertiliser concn

Mean

Median

Mode

0.0

8.6

9

9

0.06

15.6

16

16

0.12

16.6

17

17

0.18

18.2

18

18

0.24

18.5

18

No mode

0.30

18.2

18

No mode

8. 0.24 g L-1 fertiliser (ignoring outlier).

9. Not all plants in sample may have received the same amount of fertiliser/water. Plants in centre of group may be more shaded/protected.

10. Nitrogen fertiliser increases the growth of the root mass of radish plants, but only up to a limit, with peak performance reached at 0.24 g L-1 of fertiliser. The fertiliser also increases the number of leaves per plant (up to a limit) which is likely related to the overall increase in growth of the plant. 11. Replication decreases the likelihood of chance events affecting results or may identify true results that may have been attributed to chance. It helps to remove uncontrollable factors and adds weight to the findings

19. Designing a Practical Investigation (page 24) 1. Student's own research.

2. Aim: To investigate the effect of colour on photosynthetic rate in Cabomba.

5. (a) The volume of gas (or number of bubbles) produced in a set time. (b) Volume or bubble every minute, or volume or bubble over the entire length of time (e.g ten minutes). (c)

Hypothesis: If different wavelengths of light are variably effective at driving photosynthesis then photosynthetic rate in Cabomba will vary depending on the wavelength of light to which it is exposed.

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Green

Blue

5 6 7 8 9

10

Total volume

6. (a) Errors and limitations depend on the student's work. Some errors might involve the counting of bubbles or measuring of the gas produced, the amount of light from the lamp, the effect of the cellophane (red light bulbs vs red cellophane?). (b) Using coloured light bulbs will produce more accurate wavelengths. A heat sink or water bath for the light may reduce the effect of heat on the investigation, the distance of the light bulb to the boiling tube will be important (brightness of light), etc. Another repeat could also be useful (4 trials in total). 7. Student's own conclusions based on their investigation.

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Mean root mass (g)

4. (a) Cabomba should not be released into the natural environment after the experiment, dispose of it carefully. Be careful that the heat from the lamp does not affect the cellophane (use a max 60W incandescent light bulb. (b) As in (a) Cabomba should not be released into the natural environment. It must be disposed of at the end of the investigation.

Analysing the biological validity of information 22. HPV (page 30)

1. Both men and women can be infected with HPV and the new vaccine covers the HPV strains that cause genital warts as well as those that cause cervical cancer. It is in therefore in the public health interests to vaccinate males as well as females so that spread of these strains is limited.

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209

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Adaptations of animals to their way of life 25. Obtaining Food (page 37)

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1. (a) Filter feeding: Obtaining nutrients from particles suspended in the water. Baleen whale. (b) Deposit feeding: Obtaining nutrients by sifting and/or ingesting sediment or detritus. Catfish. (c) Bulk feeding: Obtaining nutrients by eating whole organisms. Stoat. (d) Fluid feeding: Obtaining nutrients by ingesting only the fluids of another organism. Mosquito.

2. Glowworm larvae use a combination of structural, physiological, and behavioural adaptations to capture flying insects (prey) using sticky mucus lures. The larvae spin a silk nest from which they hang sticky, mucus-covered threads or fishing lines (behavioural and structural adaptation). The larvae then produce a bioluminescent glow (a physiological adaptation) attracting insect prey, which then become trapped in the fishing lines. The glowworm larvae then pull the prey rapidly upwards to be eaten (a behavioural adaptation). The larvae can extend the nest as they grow and can reconstruct and repair their threads as required (behavioural and structural). Glowworms are restricted to habitats without much air movement because wind would tangle the threads and prey capture would be unsuccessful. They are adapted to a niche as predators of flying insects in sheltered environments where lures and traps work most effectively.

3. Bulk feeding, filter feeding, and fluid feeding are quite different feeding modes and require specific adaptations to enable food to be caught and ingested. Bulk feeders feed on large food masses, which are ingested whole or in pieces after cutting or chewing. Adaptations are therefore associated with creating pieces that are small enough to be taken into the mouth. In insects, this involves holding and sometimes cutting the food with mandibles and using other mouthparts to manipulate the food and push it into the mouth. In mammals, specialised teeth cut the food into manageable pieces which can then be chewed and swallowed. Bulk feeding fish often swallow food whole, or cut off large pieces with razor-sharp teeth (sharks). Filter feeding involves straining water to retain the food particles. Mammalian filter feeders include whales, which have baleen (specialised keratin) plates. When a mouthful of water is taken in and squeezed out through the plates, krill or other planktonic organisms are trapped and swallowed. Some sharks (e.g. whale sharks) are also filter feeders using a similar system of tiny teeth and filter pads. Filter feeding has a low energy cost, so filter feeders can reach very large sizes. Fluid feeders need to pierce the epidermis of a plant or animal to extract fluids, which they must be able to suck or lap up. They therefore have sharp piercing tubular mouthparts (as in mosquitoes and sap sucking insects) or sharp teeth, which break the epidermis so that the fluid can be lapped from the surface (vampire bats).

4. Carp are deposit feeders and loosen sediment and plant material from the beds of lakes and rivers. This damages fragile benthic communities and increases the amount of suspended material in the water (reduces water quality).

26. Parasitism (page 39)

1. (a) Parasitoids spend only part of their life cycle as parasites and often consume the host. Parasites live their entire life in or on the host and do not usually kill it. (b) The parasitoid is dependent on a specific host for a crucial part of its life cycle (usually the larval phase). The emerald cockroach wasp relies on a cockroach to provide food for its larva. The cockroach dies as a result of the interaction. Without the cockroach, the emerald cockroach wasp could not complete its life cycle. 2. Parasites feed on other animals and adapted to feeding on only a part of the animal they parasitise, e.g. blood or skin, without killing it.

3. Vampire bats act like parasites in that they feed off a host, harming but not killing it. However, unlike true parasites, they are free-living. They do not live on they host but return to a roost after feeding.

27. Predation (page 40)

1. (a) Camouflage enables the predator to remain undetected by the prey until it has the opportunity to strike successfully. (b) Other adaptations of ambush predators include structural modifications to allow long strike reach (long legs and tongues) and behaviour to allow precise, very slow movements to approach prey without being detected (e.g. chameleons, mantids).

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28. Insect Mouthparts (page 41) 1.

(a) Sponging: D housefly (b) Chewing: A locust, E beetle (larva) (c) Sucking: C butterfly (d) Piercing/sucking: B mosquito, F aphid

2. Labrum (anterior lip), mandibles (paired mouthpart, usually for biting), maxillae (one or two pairs and modified in different groups), and labium (lower lip).

3. Insect mouthparts are highly versatile, all consisting of the labrum and three sets of paired appendages which have been variously modified (through evolution) to pierce, suck, chew, or sponge.

4.

Insect A: Housefly type (a) Maxillary palp (b) Rostrum (c) Labium

Insect B: Butterfly (moth might be acceptable, although the antennae are butterfly-like). (d) Antenna (e) Maxillary proboscis (f) Compound eye

Insect C: Mosquito (g) Antennae (h) Proboscis sheath (i) Leg

Insect D: Bee (actually a honeybee) (j) Galea (k) Glossa (tongue) (l) Labial palp

Insect E: Cockroach (any generalised orthopteran is acceptable, including grasshopper). (m) Antenna (n) Maxilla (o) Mandible (p) Maxillary palp (q) Labrum

5. (a) Caterpillar diet: Foliage, sometimes flowers. Mouthparts: Relatively generalised chewing type, with strong jaws (mandibles). (b) Butterfly diet: Nectar. Mouthparts: Mouthparts greatly rearranged to a specialised maxillary proboscis, with loss or fusion of the other mouthparts.

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2. (a) Mosquito: Mouthparts specialised to form a piercing tube to puncture skin and suck up fluids. (b) Filter-feeding whale: Comb-like plates (baleen) to collect krill from large volumes of water. (c) Mammalian predatory carnivore: Specialised types of teeth for biting, holding, shearing and chewing food. (d) Leaf chewing caterpillar: Cutting and chewing mouthparts (mandibles). (e) Vampire bat: Small sharp teeth for breaking skin and long tongue to lap blood.

29. The Teeth of Fish (page 43)

1.

(a) Butterfly fish (b) Manta ray and other filter feeders (c) Sharks and other predatory fish e.g. piranha (d) Lampreys

2. Predatory fish often have backwards pointing teeth to help them securely grip prey.

30. The Teeth of Mammals (page 44)

1. (a) To bite and tear food (b) Hold and tear prey/food. Sometimes used as weapons. (c) Chewing/grinding food. Modified in carnivores to shear flesh.

2. The human tooth pattern is different in that humans have only four incisors in total and a reduced number of premolars.

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210 31. Mammalian Teeth and Diet (page 45)

4.

Circular muscle

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1. (a) Lion diet: Meat Feature of dentition: pointed incisors for gripping and tearing, carnassials to shear flesh, canines hold and throttle prey. (b) Zebra diet: Grass Feature of dentition: large incisors for clipping grass and large molars for grinding. (c) Chimpanzee diet: Omnivorous. Feature of dentition: Generalised dentition. Teeth are relatively unspecialised, but are good for biting and chewing most foods. (d) Grey whale diet: Krill. Feature of dentition: Teeth are absent. Large plates of baleen hang from upper jaw and act as a sieve. Food is retained and water is squeezed out of the mouth by the tongue.

32. Comparing Mammalian Teeth (page 29) 1. (a) Zebra: Grass Pig: Omnivorous Lion: Meat Goat: Grass and other plant material. (b)

Incisors

Canines

Premolars

Molars

Longitudinal muscle

35. The Stomach and Small Intestine (page 49)

1. (a) Stomach: A three layered muscular wall around a lumen (cavity) that holds food. The muscles produce the movements to mix the food into chyme. Rugae (folds) allow expansion of volume. Gastric glands specialised to produce acid, mucus, and protein digesting enzyme precursor. (b) Small intestine: Divided into the duodenum, jejunum, ileum. Most of the chemical digestion occurs in the duodenum. Most of the absorption occurs in the jejunum and ileum where fingerlike villi project into the lumen and provide large surface area for absorption of nutrients. Intestinal glands produce mucus to protect gut mucosa from damage and alkaline fluid to provide an appropriate pH for intestinal and pancreatic enzymes. A double wall of circular and longitudinal muscle helps move the food through the intestine by peristalsis.

2. (a) HCl activates pepsinogen to form active pepsin. It also helps to kill any bacteria in food and denatures proteins. (b) The enzymes are secreted as inactive precursors in order to prevent their activity in the site of production and release (where they would damage the tissue). Once in the gut lumen, they can be activated to digest the food (the gut lining itself is protected by mucus). 3. Muscles in the stomach wall mix acid, enzymes, and food.

2. The teeth of a lion (a carnivore) are pointed, helping to grip and tear flesh. The canines are long to help hold and grip prey and the carnassials overlap to help shear flesh and bone. The teeth of horses are adapted to eating grass and so the incisors are flat at the top and angle outwards to the front of the mouth to assist in clipping grass. The molars line up and meet together to grind grass.

33. Introduction to Digestion (page 47)

1. Ingestion (eating), Digestion (food is broken down into parts that can be absorbed), Absorption (food molecules are absorbed across the gut wall), Egestion (waste material is defected).

2. (a) Food is physically broken by chewing and the muscular activity of the stomach wall. (b) Food is chemically broken down by enzymes in the digestive secretions. 3. (a) The liver (b) Produces digestive secretions (bile). 4 (a) The small intestine (b) It is egested via the anus.

34. Moving Food Through the Gut (page 48)

1. A bolus is a small mass of chewed food material mixed with saliva.

2. Chyme is the semi-fluid slurry that enters the small intestine from the stomach.

3. Contraction of circular muscle behind the food squeezes it through the gut. Contraction of longitudinal muscles in front of the food shortens and widens the gut to receive the food.

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4. (a) Intestinal villi increase the surface area of the gut for absorption. (b) The brush border forms a high surface area membrane for nutrient absorption and contains its own enzymes that help in the digestive process. 5. (a) and (b): any two of the following in any order: Site: Stomach Enzyme: pepsin Role: Digestion of proteins to polypeptides. Site: Pancreas Enzyme: pancreatic amylase Role: Digestion of starch to maltose.

Site: Pancreas Enzymes: trypsin/chymotrypsin Role: Digestion of proteins to polypeptides.

Site: Pancreas Enzyme: pancreatic lipase Role: Digestion of fats to fatty acids and glycerol. Site: Pancreas Enzymes: peptidases Role: Digestion of polypeptides to amino acids.

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Site: Intestinal mucosa Enzymes: peptidases. Role: Digestion of polypeptides to amino acids.

Site: Intestinal mucosa Enzymes: maltase, lactase, sucrase Role: Digestion of carbohydrates (maltose, lactose, sucrose respectively) into their constituent parts. (c) Alkaline (d) Through the secretion of alkaline fluids from the intestinal glands, the pancreas, and liver (bile). 6. The small intestine is long to accommodate the regional specialisation that is needed to digest different types of food molecules and then absorb them across the intestinal wall. If it was too short, food might pass through without being digested and absorbed.

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211 36. The Large Intestine (page 52)

3. Ruminants have a large stomach divided into separate chambers for fermenting. Fermented food is passed to the small intestine for absorption. The caecum is relatively small. Hindgut fermenters have a small stomach and short intestine in comparison. Food is passed to the caecum or colon for fermentation In rabbits the fermented material is egested and re-eaten to gain the nutrients from it.

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1. Given in sequential order: (a) Caecum: Provide a space to mix bacteria and partly digested food to form faeces. (b) Colon: Absorbs water, electrolytes (e.g. sodium ions), and some vitamins. Incubates bacteria, which produce B vitamins and vitamin K. (c) Rectum: Enlarged region at the end of the large intestine where faeces are consolidated before elimination.

2. Absorbing too little water (material moving too quickly through the gut) causes diarrhoea (watery faeces). Absorbing too much water (material moving too slowly through the gut) results in constipation (hard faeces).

3. Goblet cells produce mucus to lubricate the colon wall and help to move and form the faeces.

4. Material moves by peristalsis (waves of muscular contraction).

5. (a) and (b)

4. (a) Protein. (b) Protein comes from the digestion of microbes. Hindgut fermenters lose protein because their microbes are located in the caecum and colon and no digestion occurs here. 5. Carnivores digest their food using enzymes produced in digestive secretions. They do not use bacteria for digestion, so bacterial fermentation chambers are not needed.

6. Carnivores may eat infrequently so they have a large stomach to hold as much food as possible when they do eat.

39. Comparison Between Ruminants and Carnivores (page 57)

1. (a) The teeth (any of): – Carnivores molars/premolars modified into carnassials for shearing. Herbivore teeth are flatter with large grinding surfaces, specialised for grinding and chewing. – Diastema in herbivores provides space to manipulate bulky food within the oral cavity. – In herbivores, there may be a horny pad against which the incisors act to clip vegetation. – The canines in carnivores are very large, they are modified to hold prey and tear flesh. (b) In a carnivore the food passes into the stomach and then into the intestine and rest of the gut. The food travels through the gut only once.

6. Segmentation mixes the material in the colon. (Segmentation also occurs in the small intestine for the same purpose).

37. Summary of the Human Digestive Tract (page 54)

1.

Structures as follows: A Mouth and teeth B Salivary glands C Oesophagus D Liver E Stomach F Pancreas

Region responsible for each stated function as follows: (a) I - small intestine (e) F - pancreas (or B) (b) J - rectum (f) D - liver (c) H - colon (g) B - salivary gland (d) E - stomach

G Gall bladder H Colon (or large intestine) I Small intestine J Rectum K Appendix L Anus

2 (a) Lining (mucosa) of the stomach. Feature: gastric gland. (b) Villi lining lumen of the small intestine (duodenum). Feature: Fingerlike villi project into the lumen. Layer of muscle visible below connective tissue of villi. (c) Liver. Feature: Bile ducts.

38. Mammalian Guts (page 55) 1. See table top below.

2. (a) Much of the energy value in plant material is trapped in cellulose which cannot be digested by mammals. (b) Herbivores compensate for low energy values by having large and long guts that carry a large amount of food and give a long time for digestion.

In a ruminant herbivore the food passes into the first and second chambers of stomach and then is regurgitated back into the mouth for rechewing. The food is reswallowed and passes to the third and fourth chambers of the stomach before passing to the rest of the gut.

40. Comparison of Tube Guts (page 58)

1. Different parts of the gut can become regionally specialised to carry out specific tasks.

2. (a) Midgut. Function: digestion and absorption. (b) Gizzard. Function: storage and initial processing. (c) Mandibles. Function: Chewing food. 3. Shorter.

4. A. 1 Liver 2 Small intestine 3 Stomach 4 Large intestine (colon) B. 1 Stomach 2 Duodenum 3 Large intestine (colon) 4 Ileum (small intestine is acceptable) 5 Liver

2. (a) Kidney and seminal vesicles. (b) Liver and small and large intestines. (c) The mesentery is a membrane that holds the gut in place (suspends it from the abdominal wall).

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X-ray of the colon

3. Omnivore (long gut, simple stomach)

Gut region (length and volume)

Herbivore: foregut fermenter

Herbivore: hindgut fermenter

Omnivore

Carnivore

Stomach

Large - 70% gut volume.

Relatively small compared to gut

Medium size - 20-30% of gut volume

Large - 60-70% gut volume

Small intestine

Long

Long (but shorter than a foregut fermenter)

Medium length (shorter than a herbivore but longer than a carnivore)

Short and relatively wide

Caecum

Short to medium length

Large

Usually poorly developed

Poorly developed. May be absent

Colon

Medium to long length

Very long and pouched

Relatively long compared to rest of gut

Short and smooth.

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212 41. Cellulose Digestion in a Ruminant (page 60)

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1. A mutualistic relationship.

1. (a) The spiral valve increases the time food is in the gut by producing a lengthened internal surface (food must twist through the spiral rather than pass straight through). (b) Increasing the time food is in the gut increases the amount of nutrients that can be absorbed from the food.

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2. The ruminant receives a source of energy (volatile fatty acids) and vitamins and protein from digesting the microbes.

47. Adaptations for Absorption in Fish (page 66)

4. The rumen contains the microbes which would be killed in a low pH environment. The abomasum has a low pH which activates enzymes for digestion of proteins and kills and digests any bacteria.

42. Cellulose Digestion in a Hindgut Fermenter (page 61)

1. (a) Energy via volatile fatty acids along with vitamins and nutrient. (b) A continual supply of sugars from cellulose. A constant and equable environment for growth and reproduction.

2. (a) Caecotrophy increases the rabbit's nutritional gain from its food by obtaining extra nutrients in the caecotropes. (b) The rabbit would miss out on essential nutrients and vitamins. Its general health would decline.

43. Cellulose Digestion in an Insect (page 62)

1. Termites chew the wood to make a pulp for ingestion. Then mutualistic protozoans in their pouch-like hindgut digest the cellulose in the wood for them.

2. The symbiotic protozoans in the hindgut phagocytose the wood fibres and bacteria within the protozoans use cellulase (enzyme) to break down the cellulose, releasing glucose. Some of the glucose used by the protozoan but some is provided to the termite.

3. Both termites and ruminants rely on a mutualistic relationship with microbes to digest the cellulose in their diet. In ruminants, the products of microbial digestion are directly available to the ruminant from the bacteria. In termites, the symbiosis is more complex, because the termite receives sugars from gut protozoa, which themselves rely on bacteria to digest the cellulose. In both cases, the gut microbes receive a source of energy (as cellulose) and a place to live.

44. Digesting Fluids (page 63)

1. Fluid feeding requires the storage of large volumes and they must remove the water to concentrate the nutrients and ensure that digestive enzymes are not diluted. These problems are solved by having a gut with regions specialised to absorb water, then secrete enzymes, and then absorb the nutrients. If the fluid is blood, coagulation and blockage of mouthparts is overcome by injecting anticoagulants (mosquitoes).

2. (a) 2 days after feeding. (b) The graph shows that a lot of processing of the fluid occurs before maximal enzyme activity. This would associated with removing the water in the blood in the first part of the midgut. The enzymes can then act without being diluted.

45. Adaptations for Digestion in Fish (page 64)

1. The temperature of the stomach drops every time the fish are given frozen food. However it quickly recovers to a temperature of around 28 oC, showing there must be some internal mechanism for heating the internal organs.

2. Keeping the internal organs warm speeds up digestion, allows larger quantities of food to be processed, and enables them to maintain a for a higher level of activity.

46. Adaptations for Absorption in Insects (page 65) 1. The midgut.

2. Gastric caeca

3. Increasing the surface area of the gut increases the area available for nutrient absorption and increases the rate of nutrient absorption. Efficient nutrient absorption enables the maximum gain from the food.

4. Both gastric caeca and the small intestine of mammals increase the surface area for the absorption of nutrients across the gut lining.

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2. In bony fish, pyloric caeca are outpockets of the small intestine which create a greater surface area for nutrient absorption by projecting into the gut lumen.

48. Adaptations for Absorption in Mammals (page 67) 1. The wall of the intestine is folded into structures called villi. The cells lining the villi have a brush border of microvilli.

2. (a) Secretes alkaline fluid containing digestive enzymes into the duodenum to break down fats, carbohydrates, and proteins. The alkaline fluid also helps to neutralise acid entering the duodenum from the stomach. (b) Secretes an alkaline fluid containing bile salts to emulsify fats. The alkaline fluid also helps to neutralise acid entering the duodenum from the stomach.

3. Fats are absorbed into the lymph lacteals and transported to the thoracic duct of the lymph. Amino acids and simple sugars are absorbed from the intestinal epithelia cells into the capillaries

4.

(a) Active transport (Na+ cotransport) (b) Facilitated diffusion (c) Active transport (Na+ cotransport) (d) Active transport (proton pump)

5. Nutrients are constantly being transported away in the blood and lymph so the diffusion gradients are maintained.

6. This question is a student's own response. However, based on the structure of the human gut, humans are typical omnivores. The stomach is simple and of moderate volume. The small intestine is relatively shorter than that of a typical herbivore, the caecum is poorly developed, and the colon is relatively long (large amount of water to absorb). The dentition, with rather generalised different types of teeth, is typical of an omnivore and humans produce a wide range of digestive enzymes (to break down all major food groups). Students will have their own justifications.

49. What You Know So Far: Digestion and Feeding (page 69)

There are no answers. Summary is entirely student based.

50. Essay Style Question: Digestion and Feeding (page 70)

NOTE: This is a suggested answer only. What should be noted are the elements (the parts of the answer) it contains and how the answer is constructed.

1. Absorption of nutrients occurs mainly in the small intestine in mammals and fish and the midgut in insects. All three taxa have adaptations that increase the surface area for absorption of nutrients. Insects increase the surface area of the gut with gastric caeca. These are tube like extensions of the midgut. Cartilaginous fish increase the time it takes food to pass through the gut (and therefore the amount of time for absorption to occur) with a spiral valve. Spiral valves are folds the intestinal lining. The more folds, the greater the surface area. The small intestine of mammals is folded into microscopic fingerlike villi, which greatly increase surface area. The intestinal cells lining the villi also have a brush border of microvilli (small extensions of the plasma membrane) to further increase surface area and are packed with enzymes to break down food molecules which can then be easily absorbed. The small intestine of mammals can be extremely long, especially in herbivores. These adaptations increase the surface area for absorption of nutrients. Increasing surface area enhances survival by maximising the amount of energy and nutrients absorbed from food, which can be used for growth, development, and reproduction. It also reduces the time required for foraging and thus reduces energy use and time spent in possibly hazardous environments obtaining food.

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3. Chewing the cud breaks down the plant matter and stimulates digestion.

2. Carnivores, omnivores, and herbivores have very different diets so have very different digestive structures to cope with each diet. Carnivores eat meat, herbivores eat plant material, and omnivores eat both plants and meat. Eating meat requires the capture of prey. To assist this, carnivores

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213 able to supply oxygen requirements. Insects could therefore be larger and still meet their oxygen demands.

54. Insect Adaptations to Gas Exchange in Water (page 76)

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have teeth specialised for biting, hold, tearing, and shearing. The canine teeth are large to hold and throttle prey. The molars (carnassials) are specialised to shear flesh and bone. Meals may be infrequent but large. To accommodate this the stomach of carnivore is often large in comparison to the rest of the digestive system. Meat is very nutritious and the small intestine is short as digestion and nutrient absorption is relatively quick. Herbivores have teeth that are usually flat and adapted to crushing and grinding plant food. Animals cannot digest the cellulose in plants directly and form mutualistic relationships with bacteria to aid digestion. Foregut fermenters (e.g. ruminants) have a modified stomach to accommodate microorganisms which digest the plant material. Hindgut fermenters (e.g. rabbits) accommodate microbes in the caecum to digest cellulose. Omnivores have teeth that are adapted to bite, tear, and chew plant and meat material. The small intestine is often still quite long but does not have adaptations for housing digestive microbes. The caecum (appendix) is normally small and undeveloped.

51. KEY TERMS AND IDEAS: Digestion and Feeding (page 72)

1. (a) Fluid feeding: Piercing mouthparts or teeth designed for shaving through skin. Mouthparts for sucking or lapping fluids. Stomach to hold large quantities of fluid and eliminate water. Bulk feeding: Mouthparts to bite and chew food. Stomach to hold large quantities of food. (b) Fluid feeder: vampire bat Bulk feeder: lion (most mammals). Filter feeder: baleen whales (e.g. humpback). 2.

3.

(a) 3,1,4,2/3,1,4,3 (b) Carnivore teeth are generally pointed or triangular. (c) 3,1,4,3/3,1,4,3 (top incisors and canines may be missing). (d) Molars would be flat or possibly ridged on top. Incisors would be flat on top but possibly chisel shaped to assist in clipping grass and vegetation. (a) Ingestion: mouth (b) Digestion: stomach (c) Absorption: small intestine (water large intestine) (d) Egestion: anus.

4. (a) Intestinal villi (b) Nutrient absorption mainly although intestinal epithelial cells themselves do have membrane-bound digestive enzymes also.

5. Absorption (J), bulk feeding (A), dentition (E), deposit feeding (D), digestion (H), egestion (I), filter feeding (B), ingestion (K), intestinal villi (C), large intestine (G), small intestine (F).

52. The Need for Gas Exchange (page 73)

1. Gas exchange provides oxygen for cellular respiration and removes the carbon dioxide released.

2. Oxygen and carbon dioxide.

3. By diffusion across a gas exchange surface.

1. (a)-(c) any of: – Tracheal gills e.g. aquatic insect larvae such as mayfly larvae. – Trapped air beneath the wings e.g. Dytiscus. – A plastron e.g. adult hydrophilid beetles. – Siphons to the water surface e.g. mosquito larvae.

2. (a) Oxygen from the trapped air enters the spiracles. Oxygen from the water diffuses into the air bubble, replacing the used oxygen. (b) The gradient is maintained by the oxygen use in the tissues (which lowers the oxygen level in the bubble).

55. Gas Exchange in Fish (page 77)

1. (a)-(c) any of: – Greatly folded surface of gills (high surface area). – Gills supported and kept apart from each other by the gas exchange medium (water). – Water flow across the gill surface is opposite to that of the blood flow in the gill capillaries (countercurrent), facilitating oxygen uptake. – Pumping mechanism of operculum aids movement of the water across the gas exchange surface. 2. Ventilation of the gills keeps fresh water moving over the gill surface so the gradient for diffusion of gases is maintained.

3. The gas exchange system makes use of countercurrent flow to enhance diffusion of respiratory gases. As blood flows through the gill capillaries (gaining oxygen) it encounters blood of increasing oxygen content, so a diffusion gradient is maintained across the entire gill surface.

4. Ram ventilation is a good strategy when swimming because water moves over the gills as a consequence of locomotion. When a fish is resting (or hiding) buccal pumping keeps water moving over the gills without the fish having to swim around.

56. Countercurrent Flow (page 79)

1. Blood passing through the gills flows in the opposite direction to the oxygen rich water. This maintains an exchange gradient allowing the exchange of gases. 2. (a) It maintains the diffusion gradient. As blood flows through the gill capillaries (gaining oxygen) it encounters blood of increasing oxygen content, so a diffusion gradient is maintained across the entire gill surface. (b) Diffusion of gases into the blood would stop too soon. Parallel flow would result in rapid equilibration of oxygen saturation between the blood and the water and diffusion across the surface would stop.

57. The Human Gas Exchange System (page 80) 1. The alveolar epithelial cells.

4. Air tends to dry out gas exchange membranes so they must be internal to remain moist. In water, the oxygen content is much lower that air. Gas exchange therefore must be very efficient to get the oxygen required.

2. The hyaline cartilage provides support for the trachea, bronchi and larger bronchioles so that they do not collapse.

53. Gas Exchange in Insects (page 74)

4. Gas exchange in the lungs takes place across the alveoli.

2. In insect tracheae, gases move by diffusion directly into the tissues. Gases diffuse into and out of the fluid at the end of the tracheole, and the fluid acts as the medium for gas exchange into the tissues.

3. Insects use rhythmic body movements to ventilate the gas exchange surface. This helps to move the air in and out of the tracheae.

4. The tracheae reach deep inside the animal, delivering oxygen to all the tissues. Transport of oxygen is direct so insects don't need to carry oxygen in their haemolymph.

5. The tracheal system occupies a large volume of the insect mass. With a higher level of atmospheric oxygen, tracheal tubes could be narrower and occupy less space and still be ©2017 BIOZONE International

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1. (a) Air enters through the spiracles. (b) The valves allow the spiracles to be opened and closed. (c) Closing the valves allows the insect to conserve water when oxygen demands are low.

3. The hyaline cartilage provides support for the trachea, bronchi and larger bronchioles so that they do not collapse.

5. Ciliated epithelial cells sweep mucus towards the throat so that it can be swallowed.

58. The Lungs and Gas Exchange (page 81)

1. The trachea, bronchi, and bronchioles provide a system of airways that connect the lungs to the outside environment.

2. The diaphragm is a muscle. When it contracts it moves down, (thoracic volume increases) and air is drawn into the lungs.

3. (a) The structural arrangement (lobes, each with its own bronchus, dividing many times to end in numerous alveoli) provides an immense surface area for gas exchange. (b) Gas exchange occurs in the alveoli, across the alveolar walls where they form a gas exchange membrane with the adjacent lung capillaries.

4. (a) The epithelium becomes very thin in the respiratory zone PHOTOCOPYING PROHIBITED


214

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so that gas exchange can readily occur between the capillaries and the air in the alveoli (no diffusion barrier). (b) Elastic fibres allow the alveoli to expand and recoil with breathing. If hyaline cartilage was present they would not be able to do this (cartilage is too stiff).

There are no answers. Summary is entirely student based.

62. Essay Style Question: Gas Exchange (page 87)

NOTE: The answer below is a suggested answer. What should be noted are the elements (the parts of the answer) it contains and how the answer is constructed.

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61. What You Know So Far: Gas Exchange (page 86)

5. Surfactant reduces surface tension in the alveoli so that they don't collapse and stick together after each breath out.

59. Breathing in Humans (page 83)

1. Breathing ventilates the lungs, renewing the supply of fresh (high oxygen) air while expelling air high in CO2 (gained as a result of gas exchanges in the tissues).

2. Quiet breathing uses muscle activity only for inspiration (expiration is passive). Forced breathing uses muscle activity for inspiration and expiration.

3. Forced breathing would be used during exercise when the body must increase the rate of gas exchange.

60. Adaptations for Diving in Mammals (page 84)

1. (a) One of: Breathing out before diving prevents nitrogen from the lungs entering the blood (avoiding the narcotic effects of nitrogen and the formation of nitrogen bubbles in the blood on ascent from the dive). Breathing out before diving also helps to reduce buoyancy, assisting the descent. (b) A human breathes in before diving. (c) Reducing heart rate during a dive has the effect of slowing blood flow and metabolic rate which conserves oxygen during the dive.

1. Mammals, fish, and insects have very different gas exchange systems but all rely on respiratory gases (oxygen and carbon dioxide) moving into and out of the body by diffusion. Mammals extract oxygen from the air using lungs, whereas fish extract oxygen from the water using gills. Insects pipe oxygen from the air directly to their tissues via a system of tracheal tubes. The gradient for diffusion of gases is maintained by ventilation of the gas exchange surface. Mammals ventilate the lungs by breathing, and air is moved into and out of the lungs along a pressure gradient caused expansion and recoil of the chest cavity. In fish, the gills are ventilated by movements of the gill cover or by swimming. Insects pump air into their tracheal system using muscular movements of the abdomen.

In mammals, oxygen diffuses from the air in the lungs into the blood across thin walled structures in the lungs called alveoli. Carbon dioxide diffuses in the opposite direction and is breathed out. The respiratory gases are transported in the blood to the working tissues where oxygen is delivered and CO2 removed. In fish gills, blood flows countercurrent to the flow of water which maintains a gradient for diffusion of respiratory gases along the entire length of the gill. O2 diffuses from the water into the blood and CO2 diffuses in the opposite direction. In the insect tracheal system, air diffuses into the tracheal tubes through spiracles or openings on the side of the abdomen. Respiratory gases diffuse into and out of the fluid at the ends of the tube. Oxygen diffuses into the body cells and CO2 diffuses out. The tracheal system is very efficient for small organisms so insects do not need to transport respiratory gases in their blood. Fish and mammals both transport gases in the blood and oxygen is carried bound to haemoglobin, an oxygen-carrying pigment. This increases the capacity of the blood to carry oxygen and enables higher metabolic rates to be maintained.

The gas exchange systems of insects, fish, and mammals are adapted to the environment in which these animals live. The internalised lungs of mammals remain moist and functional in the dry environment of air. Fish gills are external but supported and kept moist by their aquatic environment. The countercurrent flow of water through the gill lamella extracts oxygen very efficiently, which is necessary because oxygen is much less available in water than in air. Insect tracheae are adapted to provide air directly to the tissues without reliance on a transport system and without losing too much water to the environment. Aquatic insects have adapted this system for use in water by carrying air with them to act as a diffusion gill while they swim.

2. (a) Hyperventilation causes large amounts of CO2 to be removed from the blood. CO2 levels in the blood (rather than low oxygen) are responsible for inducing the urge to breathe. When CO2 levels are very low, the urge to breathe is suppressed. (b) Oxygen levels can fall low enough to cause unconsciousness before there is an urge to breathe. Once unconscious, normal breathing mechanisms take over and the diver will inhale water and drown. 3. Missing answers in the table (oxygen in mL kg-1) given:

Location Seal Air in lungs 1.83 Blood 37.5 Muscle 9.0 3.4 Tissue water 3.33 Total 51.67

Human 10.3 14.3 2.9 30.86

Amount of oxygen (mL) per kg of bodyweight in a seal and a human

60

Human

50

63. KEY TERMS AND IDEAS: Gas Exchange (page 88)

1. Alveoli (C) breathing (N), bronchi (D), bronchioles (E), cellular respiration (M), countercurrent flow (J), expiration (I), gas exchange (O), gills (H), haemoglobin (K), inspiration (L), lungs (P), oxygen (B), respiratory gas (A), spiracles (F), tracheae (G).

40

30

20

10

0

Air in lungs

Blood

Muscle

Tissue water

Total

Location of oxygen

4. (a) The seal has very little of its oxygen store in the lungs whereas humans keep a large proportion of their oxygen store in the lungs during a dive. (b) The seal exhales before diving. Seals also have very high oxygen stores (as myoglobin). 5. (a) In the blood and muscle during a dive. (b) A human stores a large proportion of its oxygen in the lungs during a dive.

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2. Gas exchange surfaces provide a way for oxygen gas to enter the body and carbon dioxide gas to leave. Gas exchange in terrestrial insects is enabled by tracheae, which are tubes in the body that supply oxygen directly to the tissues by diffusion. Fish and many aquatic insects use gills to extract oxygen from water whereas mammals use lungs, which are enclosed inside the body. Enclosing the lungs inside the body prevents the gas exchange surface drying out. Blood flowing through the gills of fish flows countercurrent to the water flowing past the gills. This maintains diffusion gradients and extracts the maximum amount of oxygen from the water. Lungs use a tidal system (inspiration and then expiration) to move air in and out. Oxygen is transported in red blood cells by haemoglobin to the tissues of the body.

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Amount of oxygen (mL)

Seal

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215 1. (a) and (b)

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narrow capillary vessels (with their high resistance to flow). Arterial blood spurts out rapidly because it is being pumped directly from the heart and has not yet entered the capillary networks. Branchial arch

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Gills

Operculum

2. A closed system is able to direct blood towards area where it is needed or bypass other areas (e.g. when the body is hot blood is directed to the skin to help cool the body down).

Water flow

(c) The fish opens its mouth and water enters. The operculum bulges letting the water move to the opercular cavity. The mouth closes and the operculum opens, moving water out past the gills.

64. Transport and Exchange Systems (page 89)

1. (a) The internal transport system (circulatory system). (b) In small, simple animals the cells are close enough to the surface that respiratory gases and nutrients can diffuse directly across the body wall and into and out of the cells. 2. (a) The gut (intestines) (b) Tissues and the gills or lungs

4. Insect haemolymph will leak from the body cavity in a diffuse way under low pressure because it is not contained in vessels.

69. Single and Double Circulatory Systems (page 95)

1. A fish has a simple two chambered heart and the blood is pumped first to the gills and then flows on to the rest of the body under low pressure. In mammals, the heart is four chambered and divided and blood is pumped to the lungs and returns, oxygenated, before being pumped out to the rest of the body.

2. To the body

65. Haemolymph (page 90)

1. (a) and (b), any two in any order: Clotting wounds, internal defence, transport of nutrients.

2. Vertebrate blood carries oxygen to the body tissues. Insect hemolymph generally does not.

3. Either of: Maintain body pressure to assist with moulting and antifreeze for overwintering.

4. 10% of the hemolymph is cells whereas 40-50% of blood is cells.

66. Blood (page 91)

1.

3. (a) Allows hydraulic movements of parts of the body. Dissipates heat efficiently. (b) Haemolymph remains at low pressure, limiting the circulation of haemolyph throughout the body.

Any three of: – Transports nutrients, wastes, and hormones. – Helps regulate body temperature. – Contains clotting agents to prevent blood loss. – Defends against infection.

2. Icefish blood contains antifreeze glycoproteins to keep blood fluid at low temperatures. Moles live in a high CO2 environment. Their haemoglobin binds CO2, effectively removing it from the blood allowing them to tolerate high CO2 levels.

67. Blood Vessels (page 92)

1. Blood in arteries travels at high pressure, especially close to the heart. The thick wall and elastic structure of arteries enables them to withstand (and maintain) the high pressure.

2. The smooth muscle in the artery wall can contract to maintain blood pressure within the artery and also to smooth the flow of blood through the artery (preventing surges associated with the contraction and relaxation of the heart).

3. (a) Vasodilation increases the diameter of the artery. (b) Blood pressure would be reduced.

4. Capillaries are very small blood vessels and have a very small diameter (4-10 µm). Their walls are only one cell thick. They form large networks throughout the body and are especially numerous in tissues and organs with high metabolic rates. The structure of capillaries allows gases, nutrients and metabolic wastes to be efficiently exchanged between the blood in the capillaries and the cells of the body's tissues.

5. (a) Veins have less elastic and muscle tissue than arteries. (b) Veins have a larger lumen than arteries. 6. One way valves help to return venous blood to the heart by preventing backflow of blood away from the heart.

7. Venous blood oozes out in an even flow from a wound because it has lost a lot of pressure after passing through the ©2017 BIOZONE International

3. To the heart

4. Blood flows at low pressure around the body of a fish because it loses pressure as it flows through the gills. In a mammal, blood is pumped from the heart to the lungs and then back to the heart so it receives a pressure boost before being pumped to the body.

5. A single circuit system as in a fish could be regarded as less efficient than a double circuit system because the blood is pumped to the gills, loses pressure, and then flows around the rest of the body at low pressure, supplying oxygen and removing carbon dioxide at a relatively low rate. However, this system is adequate to meet the metabolic demands of fish, which have lower metabolic rates than mammals primarily because they do not need to expend energy in maintaining body temperature. The double circuit system of mammals is more efficient in that blood is pumped first to the lungs and then returns to the heart before being pumped to the rest of the body. The boost in pressure by this system supplies oxygen and removes carbon dioxide at a rate that is sufficient to supply the higher metabolic requirements of mammals, which expend large amounts of energy in maintaining body temperature.

6. (a) and (b) below. Capillaries in gills

Capillaries in body

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68. Open and Closed Circulatory Systems (page 94)

1. Unlike closed systems, open circulatory systems do not have a complete circuit of vessels for blood to pass through from the heart to the body and back again. In open systems, there is no distinction between the blood in the vessels and that bathing the tissues.

High pressure blood

Heart

Low pressure blood

70. The Circulatory System (page 97)

1. (a) The blood transports gases, nutrients, and metabolic wastes to different areas of the body. (b) Blood vessels (arteries, veins, and capillaries). These form the conduits for the transport of blood around the body. (c) The heart is a large muscular pump. It pumps blood around the body through the system of blood vessels.

2. (a) Red blood cells (RBCs) carry oxygen to the body's cells. (b) Metabolic wastes are transported in blood plasma.

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216 (d) Gut (intestines) (e) Kidneys (f) Genitals/lower body

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3. (a) Head (b) Lungs (c) Liver

5. Chordae tendineae hold the valves between the atrium and ventricle closed during contraction of the ventricle.

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4. D ad C

3. Aorta 4. Pulmonary artery

71. Vertebrate Hearts (page 99)

1.

Schematic of human heart. Labels (a)-(h) (a) Pulmonary artery (e) Aorta (b) Vena cava (f) Pulmonary vein (c) Right atrium (g) Left atrium (d) Right ventricle (h) Left ventricle

Valves marked on diagram

Right atrium

Semi-lunar valve

1. (a) The respiratory and circulatory systems interact at the junction of the alveolus and the capillary wall (the gas exchange membrane). (b) Gas exchange is occurring at this point. Oxygen from the alveoli diffuses into the blood in the capillary and is transported around the body. Carbon dioxide in the blood in the capillary diffuses into the alveoli and is expelled from the lungs by expiration.

Aorta

Left atrium

Tricuspid

Bicuspid

Right ventricle

Left ventricle

Semi-lunar valve

Septum separates two halves of the heart

2. Schematic diagram of fish heart.

Atrium

Ventricle

73. Internal Transport and Gas Exchange Interaction (page 103)

Sinus venosus

3. Aorta carries oxygenated blood to head and body.

4. Vena cava receives blood from the head and body.

5. Valves prevent the back flow of blood into the atria and ensure blood flows in the correct direction.

6. Mammals have a double circuit system with a four chambered heart. The pulmonary circuit pumps blood through the right side of the heart to the lungs where the blood is oxygenated. The systemic circuit receives blood from the lungs and pumps it through the left side of the heart to the body, where it supplies the tissues. Mammals are endothermic (maintain a high constant body temperature) and this has a high energy demand. The double pump is an efficient system for enabling exchanges (oxygen, nutrients, wastes) when there is a high metabolic rate and high oxygen demand. In fish, the heart is essentially two chambered, with a single atrium and ventricle. The sinus venosus is a third 'entry' point and sometimes designated a chamber. The chambers are in series. There are no distinct pulmonary and systemic circuits. The blood flows from the heart at low pressure to the gills for oxygenation and then supplies the body before returning to the heart. Fish gills achieve high rates of oxygen extraction from the water, allowing these animals to achieve large sizes and rapid swimming speeds in spite of the lack of a separate pulmonary circuit.

72. Dissecting a Mammalian Heart (page 101) 1.

Base

Interventricular sulcus Coronary arteries

2. The circulatory system and gas exchange system interact to provide the body's tissues with oxygen and remove carbon dioxide. The gas exchange system brings oxygen into the body during inspiration and removes carbon dioxide during expiration. The circulatory system transports oxygen, bound to haemoglobin within red blood cells, to the body's tissues where it is released to supply the raw material for cellular respiration. The blood picks up the waste carbon dioxide from this process and transports it back to the lungs where it can be expelled from the body. The gas exchange and circulatory systems interact directly at the gas exchange membrane, where exchange of gases occurs by diffusion.

3. Maintaining a constant body temperature independently of the environment allows mammals to continue to remain active despite high or low environmental temperatures. This means they can exploit a wide range of habitats in a wide range of climatic zones. This strategy is costly in terms of energy expenditure however. Most mammals must constantly forage to fuel their high metabolic rate, which can be difficult through periods when food supplies are short. The high efficiency of the gas exchange and circulatory systems in delivering oxygen to fuel metabolism and removing carbon dioxide (its waste product) enables mammals to maintain their high metabolic rate and adjust their rates of gas exchange to accommodate changes in physical activity.

74. Internal Transport and Digestion Interaction (page 105) 1. Nutrients (e.g. minerals, sugars, and amino acids) are transported in the blood plasma.

2. (a) The digestive and circulatory systems interact where blood vessels come into contact with the gastrointestinal tract, and at the liver, where the nutrient-rich blood is processed. (b) Nutrients from digestion are transported across the intestinal epithelium of the small intestine to the blood. The nutrients are then transported to the liver via the hepatic portal vein for processing (storage and conversion).

3. (a) Blood flow to the digestive tract increases after a meal. (b) Blood leaving the gut is very high in nutrients, only some of which are immediately required. Transporting blood directly to the liver allows a large proportion of these nutrients to be processed and stored if necessary for later release as required.

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Pulmonary artery

6. The thickness of the ventricle walls (the right ventricle wall is relatively thin while the left ventricle wall is much thicker and very muscular).

4. The circulatory system and digestive system interact to provide the body's tissues with nutrients. The digestive system digests and processes food into smaller molecules that can be absorbed by the body's cells. The nutrients then pass into the blood and are distributed by the circulatory system to the body's cells and tissues, providing them with the components they need to metabolise.

5. Digestion and absorption of nutrients.

75. What You Know So Far: Internal Transport (page 107)

There are no answers. Summary is entirely student based.

2. Vena cava Š2017 BIOZONE International

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217 76. Essay Style Question: Internal Transport (page 108) NOTE: The answer below is only a suggested answer. What should be noted are the elements (the parts of the answer) it contains and how the answer is constructed.

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78. The Plant Transport System (page 112)

1. The two vascular tissues in plants are xylem and phloem.

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1. Transport systems may be open (blood contained within vessels) or closed (circulatory fluid not completely enclosed within vessels). Closed systems may be classed as double or single circuit, depending on whether the blood returns to the heart after being oxygenated.

Adaptations of plants to their way of life

Insects have an open circulatory system. A simple dorsal heart pumps haemolymph (a mix of blood and interstitial fluid) through short vessels into the body cavity, where it bathes the insect's organs before reentering the heart through ostia (holes). Unlike the blood of vertebrates, the insect open circulatory system does not transport oxygen around the body. It is a low pressure system and movement of the insect's body helps the fluid to circulate through the body.

Mammals and fish have closed circulatory systems, in which the blood flows through a series of blood vessels and does not mix with interstitial fluids. Fish have a single circulatory system in which the blood flows from the heart to the gills and then flows at low pressure around the body. This system is relatively slow at providing oxygen to systems, but is adequate for many fish which often have relatively lower metabolic requirements. The fish heart has a linear structure with the sinus (entry point), atrium, ventricle, and conus arteriosus (exit) in series.

Mammals have a double circulatory system in which blood flows from the heart to the lungs and back to the heart (the pulmonary circuit), before being pumped at high pressure to the rest of the body (the systemic circuit). The heart is divided into separate halves each with an atrium and a ventricle (unlike a fish heart). The right ventricle is smaller and less muscular than the left and pumps blood to the lungs, while the more muscular left side of the heart pumps blood to the body at high pressure. The high pressure circuit delivers oxygenated blood to the muscles quickly to meet the high metabolic needs of most mammals.

77. KEY TERMS AND IDEAS: Internal Transport (page 109)

2. Plants require a transport system to move water and nutrients around the plant.

3. (a) The function of the xylem is to transport water and minerals around the plant. (b) Water enters the xylem at the roots by osmosis. (c) Plants must open their stomata for photosynthesis. As a result, water is lost (as water vapour) through stomata.

4. (a) The water conducting cells of the xylem are the vessel elements and the tracheids. (b) Other cells present in xylem tissue include parenchyma (packing and storage cells) and sclerenchyma cells (fibres and sclereids) which provide mechanical support.

5. (a) The end walls of the vessel elements have perforations (holes) that allow water to pass through from one vessel to the next. (b) Water passes horizontally between tracheids via thin regions in the wall called pits. (c) Water transport is faster in vessel elements because the end-on-end arrangement of the pits in the vessel elements allows water to move quickly and unimpeded. In tracheids, the water moves horizontally, so transfer is slower. (d) The thickening provides support and stops collapse of the water conducting vessels so that water conduction is not impeded. 6. Xylem forms a continuous tube for the passive process of water transport. Because the transport is passive the cell does not need to be alive.

79. Transpiration (page 114)

1. Transpiration is the evaporative loss of water from the leaves and stems of a plant.

2. Water loss is regulated by the opening and closing of stomata.

3. (a) Vein. Thin wall and large lumen. (b) Artery. Thick wall with a lot of muscle and smaller lumen.

4. (a) The plant would lose water from the cells and wilt. (b) During a prolonged period without water (e.g. a drought).

4. Atrium (I), blood (H), closed circulatory system (B), double circulatory system (C), haemolymph (A), heart (E), red blood cells (D), respiratory pigment (F), ventricle (G)

Left ventricle

4. Water moves by osmosis in all cases. In any order: (a) Transpiration pull: Photosynthesis and evaporative loss of water from leaf surfaces create higher solute concentrations (lower water concentration) in the leaf cells than elsewhere, so water moves down its concentration gradient towards the site of evaporation (the stomata). (b) Cohesion-tension: Water molecules cling together and adhere to the xylem, creating an unbroken water column through the plant. The upward pull on the water creates a tension that facilitates movement of water up the plant. (c) Root pressure provides a weak push effect for upward water movement.

Right ventricle

5. Water is moved up the tree by a combination of cohesiontension, transpiration pull, and root pressure. Together these processes can move water up to heights of more than 40 m.

1. (a) Artery

(b) Vein (c) Capillary

5. (a)

Wall of left ventricle

Wall of right ventricle

(b) One ventricle (left) is thicker because it must generate higher pressure to supply the body's organs and maintain kidney filtration rates. The other ventricle (right) does not need to be as thick because the pulmonary circuit operates at lower pressure, facilitating gas exchanges at the lungs.

©2017 BIOZONE International

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2. (a) Haemoglobin (b) Leucocytes or white blood cell (lymphocyte is acceptable)

3. Any one of: – Transpiration stream enables plants to absorb the minerals they need (minerals are absorbed with the water and are often in low concentration in the soil). – Transpiration helps cool the plant.

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218 Table 1. Water loss under different environmental conditions

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Time (min)

0

2

4

6

8

10

12

14

16

18

20

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Treatment

Ambient (mm)

0

0

0

1

1

2

3

3

4

5

5

Ambient (mL)

0

0

0

0.1

0.1

0.2

0.3

0.3

0.4

0.5

0.5

Fan (mm)

0

4

7

9

14

17

25

28

31

33

34

0

0.4

0.7

0.9

1.4

1.7

2.5

2.8

3.1

3.3

3.4

0

0

0

0

1

1

2

2

2

3

3

0

0

0

0

0.1

0.1

0.2

0.2

0.2

0.3

0.3

Bright light (mm)

0

1

3

5

6

7

8

9

10

11

13

Bright light (mL)

0

0.1

0.3

0.5

0.6

0.8

0.9

1.0

1.1

1.3

Fan (mL)

High humidity (mm)

High humidity (mL)

0.7

80. Measuring Transpiration in Plants (page 116)

2. See completed graph below.

3. (a) Passive absorption of minerals along with the water they are dissolved in and active transport. (b) Main pathways for water absorption are through the spaces within the cell walls, the xylem vessels, and through the cytoplasm of the cells. Only a very small amount travels through the plant vacuoles.

1. See completed table above.

Ambient Fan High humidity Bright light

4.0

4. (a) The Casparian strip represents a waterproof barrier to water flow through the endodermis into the stele. It forces the water to move into the cells. (b) This feature enables the plant to better regulate its uptake of ions, i.e. take up ions selectively. The movement of ions through the cell walls cannot be regulated because the flow does not occur across any partially permeable membranes.

3.0 2.0

1.0 0

0

4

8 12 Time (minutes)

16

20

3. (a) The control is the potometer set-up in ambient conditions. (b) The fan and bright light both increased water loss above the ambient conditions. (c) Moving air and bright light increase transpiration rate because they increase the rate of evaporation from the leaves. In bright light, stomata are also wide open to maximise CO2 entry for photosynthesis. 4. If the potometer had leaks, water could be lost from the system unrelated to transpiration. The observer would record a transpiration rate that was higher than that actually occurring.

5. Experimental design should include: – As large a sample size as practicable. Each species should have at least 3-5 samples being tested. – Leaf area. The water loss per square cm should be compared per plant species not just the overall loss. – The experiment should be run for an adequate period of time e.g. 24 hours. – Each plant should receive the same experimental conditions (amount of sunlight/wind etc.).

6. (a) If transpiration rate is related to leaf area, the plant with the greatest leaf area (the sunflower) will lose water at the greatest rate. Prediction: Sunflowers will have the greatest rate of transpiration per square centimetre. (b) Sunflowers have large leaves and they lack a waxy layer on the leaves to reduce water loss.

81. Uptake at the Root (page 119)

1. (a) Plants need to constantly replace the water always being lost by transpiration. (b) Large water uptake allows plants to take up sufficient quantities of minerals from the soil. These are often in very low concentration in the soil and a low water uptake would not provide adequate quantities. 2. (a) The root hairs provide a large surface area for water and nutrient uptake. ©2017 BIOZONE International

82. Adaptations of Leaves (page 121) 1. Mesophyll cells

2. (a) Gas exchange occurs through small pores on the leaf surface called stomata. Carbon dioxide enters and oxygen and water exit. (b) Gas exchange is regulated by guard cells which surround each stoma. (When the guard cell is turgid they swell, opening the stomatal pore. When water moves out of the guard cells, they lose turgor and become flaccid, causing the pore to close). 3.

– – – –

Regulation of stomata by guard cells stomata set in grooves waxy cuticle spines or hairs to reduce air movement

83. Adaptations of Mesophytes (page 122)

1. Mesophytes are plants adapted to equable climates with no adaptations for dealing with environmental extremes.

2. – Leaves are broad and thin, with many stomata: ensures a large amount of light is gathered and plenty of stomata for the entry of carbon dioxide for photosynthesis. Also ensures a constant uptake and flow of water via the transpiration stream. – Stomata on the lower surface help to reduce water loss via transpiration without restricting entry of CO2. – Large root masses: Ensures a large amount of water is absorbed from the soil as well as anchoring the plant as it increases in size. – Many lose their leaves in winter: Saves energy (no energetic benefit from presenting them to the sun) and prevents storm damage.

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Water loss (mL)

Water loss (in mL) under different environmental

(b) They lack a waterproof waxy cuticle so there is no barrier to water entry. They are long and thin.

84. Adaptations of Xerophytes (page 123)

1. A xerophyte is a plant adapted to conserving water and living in arid conditions.

2. (a)-(c), any three in any order: – Modification of leaves to reduce water loss (e.g. spines, curling, leaf hairs). – Rounded, squat shape of plant body to reduce surface area for water loss.

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219

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– Stomata in sunken pits lined with hairs to raise local humidity. – Waxy leaves to prevent water loss.

temperature is not extreme and rainfall is relatively consistent throughout the year. Adaptations include large leaves with many stomata to capture sunlight and maintain high photosynthetic rates, and large root systems to anchor the plant and take up water to replace losses from transpiration. The leaves of mesophytes have a waxy cuticle which reduces water loss directly from the leaf surface but is relatively thin. Having most stomata on the leaf undersides (shaded) reduces water loss via transpiration somewhat without restricting gas exchanges. In generally, mesophytes compensate for relatively high transpirational losses by having an extensive root system and a large water uptake, so they are susceptible to drought.

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3. Trapping air close to the surface helps produce a moist microenvironment (raises the humidity around the stomata). There is therefore less tendency for water to leave the plant (reduced diffusion gradient).

85. Adaptations of Hydrophytes (page 124)

2. (a) No (b) Any water is lost by transpiration will be replaced by the water the plant is growing in. Submerged parts of the plant without a waterproof cuticle will be able to absorb water directly from the environment.

86. Mangrove Adaptations (page 125)

1. Salt enters the roots from the environment (since the plant tissues are higher in water than the saline environment). Excess salt results in dehydration, prevents water uptake, and is toxic at high levels.

2. (a)-(c) Any order: • Storing the salt in older leaves which are then shed from the plant. • Preventing salt from entering the roots via special tissue which allows water entry, but excludes salt. • Secreting salt through tiny glands in the surface layer of their leaves.

3. Mangroves maintain the transpiration stream by storing salt in the cell vacuoles and maintaining a high concentration of solutes in the cytoplasm. This maintains a very low free water concentration within the plant so that water will move into the plant tissues by diffusion.

87. What You Know So Far: Transpiration (page 126) There are no answers. Summary is entirely student based.

88. Essay Style Question: Transpiration (page 127)

NOTE: The answer below is only a suggested answer. What should be noted are the elements (the parts of the answer) it contains and how the answer is constructed.

1. All four functional groups are covered in this answer. Students only need to cover three. Only adaptations relating to the biological process of transpiration are discussed.

In all higher plants, gas exchange and loss of water (transpiration) occur largely through the stomata. Plants must be able to obtain carbon dioxide for photosynthesis and, for this, the stomata must be open. Water loss is therefore a necessary consequence of gas exchange. The number and location of stomata and the adaptations of the leaves in particular are related to the particular adaptations of the plant to its environment. Hydrophytes are plants that are adapted to living in water (e.g. water lilies). Mesophytes are plants that live in relatively equable conditions and have few adaptations for conserving water. The majority of plants are mesophytes. Xerophytes live in dry, arid regions and are adapted to conserve water (e.g. cacti and succulents). Halophytes are plants adapted to grow in highly saline environments. Hydrophytes have no adaptations to reduce transpiration because water loss is not a problem for them. The stomata occur on the upper surface of floating leaves where gas exchange will not be blocked by the water. This upper surface may have a thick waxy cuticle to repel water and keep the stomata clear. The undersides of the leaves do not have stomata. The submerged parts of the plant lack any cuticle so water, gases, and nutrients can diffuse into and out of the plant freely. Because hydrophytes have no adaptations to reduce transpirational losses, they are very susceptible to drying out, e.g. when water bodies shrink during summer. Mesophytes live in a wide range of areas where the ©2017 BIOZONE International

Xerophytes, e.g. cacti, live in arid areas or where there is little free water (e.g. coastal). Their adaptations are related to reducing water losses in an environment where water is very scarce. In general, this involves reducing surface area and creating a high humidity environment for stomata. In cacti, the leaves are much reduced (to spines or hairs) and the stems are succulent. Spines are light coloured, reflecting sunlight, and reducing evaporative losses. The spines also reduce air movement around the plant, increasing the relative humidity and reducing the concentration gradient for water loss without restricting entry of carbon dioxide. Some xerophytes have leaves that curl up, keeping the stomata away from air currents and creating a moist microenvironment that reduces the gradient for diffusion of water vapour from the leaf. Some xerophytes (oleander) have a thick, multi-layered epidermis with the stomata sunken in pits on the leaf underside. Again, these adaptations increase humidity around the stoma and reduce water loss. Extensive fibrous root systems help the xerophyte take up any available water but, overall, they rely on reducing their water losses to the environment. Halophytes (e.g. mangroves) have to cope with a highly saline substrate and lack of fresh water so their adaptations are related to excluding or excreting salt and reducing water losses. To enable them to take up the water they require, they maintain a very high concentration of solutes in their vacuoles. The low free water concentration there then maintains the transpiration stream. The roots have a waxy coating of suberin, which also helps to exclude most of the salt from the water the plant takes up. The thick waxy leaves help to reduce water loss and the plant may also orientate its leaves to avoid direct midday sun. These adaptations help to limit water losses to the environment.

89. KEY TERMS AND IDEAS: Transpiration (page 128)

1.

(a) Transpiration. (b) Xylem tissue. (c) The tissue is dead. (d) No energy is required, it is a passive process.

2. Cohesion-tension (G), guard cells (C), halophyte (I), hydrophyte (F), mesophyte (J), potometer (A), stomata (E), transpiration (B), xerophyte (H). xylem (D).

3. (a) Control conditions: Still air, light shade, room temperature. (b) Factors increasing transpiration rate: High wind, high light, high temperature, low humidity. All increase evaporation from the leaves. (c) Humid conditions reduce evaporative loss, dark conditions stop photosynthetic production of sugars, reducing leaf solute concentration. Both reduce transpiration by reducing the concentration gradient for water movement.

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1. – Air spaces aid flotation and provide large surface area for diffusion of gases from the aerial parts of the plant to the roots. – Absence of cuticle on submerged parts means the plant can take up water, nutrients, and gases by diffusion. – No stomata on lower leaf surfaces. – High stomatal densities on the upper leaf surface allow diffusion of CO2 into the inside of the leaf.

90. Alternation of Generations (page 129)

1. Plants cycle through a diploid sporophyte generation (which produces haploid spores by meiosis) and a haploid gametophyte generation (which produces the haploid gametes by mitosis.

2. (a) Sporophyte: A tree, shrub or flower, e.g. kauri, tulip, fern, tawa, manuka. (b) Gametophyte: pollen, prothallus in fern plants. 3. Spores Produced by: Sporophyte Process: Meiosis

Gametes Produced by: Gametophyte Process: Mitosis

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220 96. Wind Pollination (page 135)

Zygote: Produced by: Gametes Process: Fertilisation

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Any two of the following: – Large amount of pollen. – Small pollen grains (easily dispersed by the wind). – Anthers point away from the stalk and stand free of the plants (pollen is easily caught and dispersed by wind).

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91. Reproduction in Ferns (page 130)

1.

(a) 2N (b) N (c) Meiosis (d) Fertilisation (e) Mitosis

2. Seed plants produce seeds from which the new plant grows. In ferns the new plant grows from the prothallus, which was produced by spores from a fern plant.

3. (a) Fern gametophytes produce motile sperm cells, which must swim to the egg cell. A film of water is required for this stage of reproduction to occur. (b) Ferns are limited to areas where water is relatively abundant because the prothallus is thin and susceptible to drying out, and the sperm need water to reach the egg. Ferns are not therefore well adapted to very dry environments.

92. Reproduction in Angiosperms (page 131)

1.

(a) 2N (b) N (c) Meiosis (d) Fertilisation (e) Mitosis

2. (a) The anther.

(b) The ovule.

3. The gametes are carried by pollen directly from the male to female parts of the flower. The pollen is either transported by wind or animals.

93. Reproduction in Gymnosperms (page 132)

1.

(a) 2N (b) N (c) Meiosis (d) Meiosis (e) Mitosis

2. Advantage: Pollinators are not required, so energy is not expended in producing large flowers or nectar (the energy is put into producing large mounts of pollen).

Disadvantage (any of): Pollen is not carried directly to other flowers. Much of it will fall to the ground and be wasted.

97. Comparing Insect and Wind Pollination (page 136)

1. (a) Insect pollinated flowers tend to be large and/or brightly coloured, often with landing platforms and patterns of light or dark which act as nectar guides for foraging insects. In contrast, wind pollinated flowers are small, numerous, and pale, often with a wispy appearance caused by the feathery stigmas which project out of the flower to catch wind blown pollen. (b) Insect pollinated flowers produce scents and foods to attract their pollinating insects. In contrast, wind pollinated flowers produce no scent or nectar (it would be a waste of energy to do so). (c) The reproductive parts of insect pollinated flowers are often well within the flower petals but above the nectary so that insects must push past the reproductive parts to reach the energy rich food reward. In wind pollinated flowers, the reproductive parts stand free of the flower so pollen can be carried on wind currents. 2. Insect pollinated flowers tend to put their energy into making nectar, producing scent, and producing large attractive flowers. Wind pollinated plants put their energy into making large quantities of pollen.

98. Fertilisation Follows Pollination (page 137)

1. Pollination is the transfer of pollen from the male anthers to the female stigma. Fertilisation is the formation of the embryo by the fusion of a sperm nucleus with the egg.

2. Pollen alights on a drop of fluid in the micropyle. The fluid evaporates bringing the pollen close to the egg cell. The pollen tube develops and the sperm cell moves through it to the egg cell.

3. The megagametophyte.

4. Wind carries the pollen from the male cones to the female cones.

94. Plant Adaptations for Insect Pollination (page 133) 1. (a) The anther produces the pollen. (b) The stigma receives the pollen. (c) Male: stamen, Female: carpel.

2. (a) The pollinators act as couriers of pollen, transferring it to the stigma of other flowers. This helps ensure successful fertilisation and cross pollination. (b) Any two of: Colourful flowers, nectar, pollen, scent. (c) The shape of the petals and markings on them direct pollinators to the centre of the flower where the pollen or nectar is.

95. Pollination Relationships (page 134)

1. A close ecological relationship in which both parties benefit.

2. The pollinator gains energy from the nectar or pollen eaten, while the flower benefits from having its pollen transferred directly to the stigma of another flower.

3. (a) Advantage: Specific pollinators will only transfer pollen from one plant to another of the same species so pollen isn't wasted. Disadvantage: There may only be one or two pollinators for that plant. (b) Advantage: There are many different pollinators available to transfer pollen. Disadvantage: A generalist pollinator may not transfer pollen from one plant to another of the same species, so pollen may be wasted.

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2. (a) Limiting self pollination and ensuring cross pollination prevents the negative effects of inbreeding and increases variability in the offspring and therefore adaptability in a changing environment. (b) Any of: – Male and female flowers on different plants. – Physically separating the male and female parts in the same flower. – Different maturation times for male and female gametes. – Gametes from the same plant incompatible. 3. (a) The double fertilisation (b) The double fertilisation results in the formation of the triploid endosperm as well as the embryo. (c) The endosperm supplies nutrients to the developing embryo and helps regulate embryonic development and seed germination. (d) Megagametophyte (female gametophyte)

99. Structure of Seeds (page 139) 1. (a) Testa (b) Endosperm (c) Cotyledon

(d) Plumule (e) Radicle (f) Root cap

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1.

2. (a) It ensures that the seed germinates only when environmental conditions are most suitable for the seedlings growth and survival (primarily available water and warm temperatures). (b) Germination does not occur until the seed has soaked up enough water through the testa (seed coat) to reactivate metabolism.

100. Adaptations for Seed Dispersal (page 140)

1. Seed dispersal ensures the seeds are transported away from the parent plant where they can germinate in a favourable area and not compete with the parent plant for light, water, and nutrients.

2. (a) Wind: Light, feathery or winged seeds that can catch the air currents. (b) Animal: Barbs, hooks, or sticky secretions that attach the seed to an animal's coat. Fleshy fruits that are eaten by animals and then passed out in the droppings.

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221 (c) Water: Very small seeds that can float on water. For seeds dispersed by ocean currents, buoyant with a woody, waterproof coverings which enable them to float in the salty water for long periods.

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There are no answers. Summary is entirely student based.

103. Essay Style Question: Transpiration (page 144)

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3. Pinus (a) Mechanism: Wind. (b) Adaptive features: Large seeds attached to “wings” which catch the air currents, carrying the seed away from the parent plant.

102. What You Know So Far: Plant Reproduction (page 143)

5. Karamu (Coprosma robusta) (a) Mechanism: Animal. (b) Adaptive features: The female trees have tight clusters of small succulent fruits. The brightly coloured fruit are highly prized by birds such as the kereru, bellbird, and tui which help disperse the seed.

1. Ferns, angiosperms,, and gymnosperms all show alternation of generations in which a haploid gametophyte generation alternates with a diploid sporophyte generation. The sporophyte generation produces haploid spores by meiosis and the gametophyte generation produces haploid gametes by mitosis. In all these three taxa the sporophyte generation is dominant (the large plant we see) and the gametophyte is smaller. In ferns, the gametophyte is a free-living (independent) organism (the prothallus) but in angiosperms and gymnosperms the male and female gametophytes are tiny. The male gametophyte is reduced to a pollen grain and the female gametophytes are retained within and dependent on the sporophyte.

Fern reproduction depends on the availability of free water because the motile sperm must swim across the surface of the prothallus (gametophyte) to fertilise the egg. In angiosperms and gymnosperms, the male gametophyte is the pollen grain and is carried by wind or animals to the egg cell and is not dependent on free water. The fern sporophyte produces spores in sori by meiosis and these germinate to form the haploid gametophyte (prothallus) which produces both egg and sperm. The union of these in fertilisation produces the young fern sporophyte, which depends on the prothallus for a short time until it develops its own roots.

Angiosperms are flowering plants. Flowers may be female, male or both. The male parts include the anther which produces the pollen (male gametophyte). The female parts include the stigma, which receives the pollen, and the ovaries in which the seeds will develop. The ovary sits below the stigma. Angiosperms may be wind or animal pollinated. Wind pollinated angiosperms have small flowers and produce vast amounts of pollen to ensure some lands on the stigma. Animal pollinated flowers have large colourful petals and produce nectar and scent to attract pollinators. The pollinators transfer the pollen from the anther to the stigma, preferably of another plant of the same species.

Gymnosperms produce both male and female cones. The male cone produces pollen (male gametophyte) and the female cone produces the egg cell (female gametophyte). The majority of gymnosperms are wind pollinated.

In an angiosperm, pollen on the stigma grows a pollen tube that follows a chemical gradient to the ovary. The pollen produces two sperm nuclei. One fuses with the egg cell nucleus to form the zygote which will develop into the seed embryo (diploid). The other fuses with the two polar nuclei in the embryo sac to form the (triploid) endosperm. The double fertilisation is a feature of angiosperms. The ovary swells to form a fruit, which may be large and fleshy to attract animals, which eat the fruit and disperse the seeds.

In gymnosperms, such as conifers, the pollen lands on a drop of water in the micropyle, and is drawn down to the ovule. It forms a pollen tube. A sperm-producing cell is formed which in turn produces two sperm nuclei. One fuses with the egg cell nuclei and forms the zygote, the other degenerates. The seed develops in the cone and, when mature, is carried from the parent plant, usually by wind. The seeds of conifers often

6. Totara (Podocarpus totara) (a) Mechanism: Animal (b) Adaptive features: The fleshy "foot" attracts animals which eat the cone and seed and eliminate the seed (with faeces) elsewhere.

7. New Zealand flax (Phormium spp.) (a) Mechanism: Wind. (b) Adaptive features: Pods split and shed copious numbers of light, flaky seeds, which are easily dispersed by the wind.

8. New Zealand mangrove (Avicennia marina. (a) Mechanism: Water. (b) Adaptive features: Seed is a propagule that is a partly developed seedling. It is buoyant and floats until encountering suitable substrate when it can quickly take root.

101. A Summary of Plant Reproduction (page 142) 1. Table at the bottom of the page

2. The sperm of ferns, mosses and algae are flagellated, indicating an ability to swim through water. The pollen of gymnosperms and angiosperms is not flagellated, but is small and light suitable for transport by wind or insects, indicating an independence of water.

3. (a) and (b) diagram below: Zygote

Fertilisation

Gametes

Sporophyte 2N

Diploid

Meiosis

Haploid

Gametophyte N

NOTE: The answer below is only a suggested answer. What should be noted are the elements (the parts of the answer) it contains and how the answer is constructed.

Spores

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4. Whau (Entelea aborescens) (a) Mechanism: Wind. (b) Adaptive features: Spiny seed capsules stop the seeds being eaten. The capsules split open at the top exposing the seed compartments and the wind buffets the capsules spilling and scattering the seeds. The seeds stored low down in the capsules are retained for months and may only escape as the capsules disintegrate.

Fern

Angiosperm

Gymnosperm

Dominant generation

Sporophyte

Sporophyte

Sporophyte

Alternate generation

Gametophyte

Gametophyte

Gametophyte

Transfer of male gametes

Water

Wind or animal

Mostly wind (except cycads which are beetle pollinated)

Reproductive structure

Sporangia

Flowers

Cones

Single or double fertilisation?

Single

Double

Single

Production of seeds?

No

Yes

Yes

Method of seed/spore dispersal

Wind

Wind or animal

Majority wind, some vertebrate dispersed

Reliance on water for reproduction

Yes

No

No

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222

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have winged structures to aid in their dispersal from the parent plant.

1. (a) Carnivore (c) Detritus (b) Detritivore (d) Autotroph (e) Herbivore (when young) 2. Most energy is lost from the system and very little is transferred to the next level. After five or six links there is very little energy left in the system (not enough energy available to support the organisms in another level).

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104. KEY TERMS AND IDEAS: Plant Reproduction (page 145)

109. Food Webs (page 152)

1. The life cycle of plants alternates between a haploid gametophyte generation and a diploid sporophyte generation. This process is called the alternation of generations. Gametophytes produce gametes by mitosis and sporophytes produce spores by meiosis.

2. Alternation of generations (J), double fertilisation (C), endosperm (G), fertilisation (F), gametophyte (H), germination (I), pollen (B), pollination (D), seed (E), sporophyte (A).

3. Plant A: Wind pollinated (conifer). Cones open and release the pollen, which is carried away by wind.

Plant B: Animal pollinated (dicot). The daffodil has a large showy flower, which would attract insects to the flower to obtain nectar. In doing so, the insects would collect and disperse pollen to other plants.

Plant C: Wind pollinated (monocot grass). Small, drab flowers which hang away from the flower stems to catch the wind, which will carry the pollen away.

Patterns in ecological communities 106. Components of an Ecosystem (page 149)

110. Ecological Niche (page 153)

1. (a) The functional position of an organisms in the community. Its description includes how an organism uses available resources and alters those resources for other species. (b) The full range of environmental conditions under which a species can exist, i.e. the full extent of its niche. (c) The range of the niche actually occupied by the organism due to interactions with the environment and other organisms. 2. Competition with other species for resources.

111. Adaptation and Niche (page 226) 1.

(a) Structural (b) Behavioural (c) Structural (d) Physiological

(e) Physiological (f) Structural (g) Physiological/behavioural

2. The wings fold up into protective sheaths when on the ground, talons aid movement on the ground.

3. There are many adaptive features that reduce competition between the two species of bats. The following points can be used as a scaffold to construct an excellence answer:

1. The biotic factors are all the living factors (organisms) in an environment and the abiotic factors are all the non-living (physical) factors, e.g. temperature.

Diet: Physiological. Long tailed bat feeds exclusively on aerial insects (carnivore), the lesser short tailed bat is an omnivore (most of its food is from plants).

2. (a) Population (b) Ecosystem

Activity pattern: Behavioural. Long tailed bat is active at dusk. Lesser short tailed bat is active at night (nocturnal).

Roosting-feeding behaviour: Behavioural. Long tailed bat roosts in trees, caves, and under bridges. It feeds off aerial insects. The lesser short tailed bat roots in hollow trees and does much of its foraging on the forest floor.

Hibernation/torpor: Behavioural/physiological. The long tailed species hibernates through late autumn and winter to conserve energy for the birth of young in the spring. The lesser short tailed bat does not hibernate although it is much less active during the winter months.

(c) The community (d) Physical factor

107. Habitat (page 150)

1. Competition will be most intense in the optimum range. This is the preferred niche, so many individuals will compete for resources in this area.

2. A tolerance for variations in a range of environmental factors allows an organism to occupy a wider range of habitats than would be possible if they had more restricted tolerance. They can then be more widespread (occur over a wider geographic range) and the species will be more secure (its long term persistence is more assured).

1. Mutualism: Nothofagus fungus forms a mutualistic mycorrhizal relationship with the roots of black beech trees. Red mistletoe depends on tui and bellbirds to pollinate its flowers and the birds receive nectar in return.

1. (a) The sun. (b)-(d) Refer to the diagram below.

lost

st

Exploitation: parasitism: Bootlace mushroom is a parasite of black beech roots. Black orchid is a parasite of bootlace mushroom. Red mistletoe (hemiparasite) takes water and nutrients from the beech trees although they make their own food.

Competition: Wasps compete with native birds for invertebrates. Sooty mould, wasps, and native birds compete for honeydew.

st

t

Ene rgy

lo

gy

lo

er

Exploitation: predation: Wasps prey on native insects and spiders.

los

En

gy er En

rgy

Ene

Energy lost

Respiration

Producers

Herbivores

Carnivores

Carnivores

Trophic level: 1

Trophic level: 2

Trophic level: 3

Trophic level: 4

Dead organisms from every level are decomposed

Detritivores and decomposers

Some secondary consumers feed directly off decomposer organisms

2. The kingfisher moves between trophic levels 3 or 4 at different times (depending on the prey it is after).

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108. Feeding Relationships (page 151)

SUN Photosynthesis

112. Species Interactions in a Beech Forest (page 234)

2. The wasp species compete directly with native animals for food resources, e.g. nectar, honeydew, and invertebrates. The honeydew is an important food source for many organisms in the black beech community. Wasps can reach such large numbers that their biomass can exceed that of all other consumers so they severely reduce the volume of honeydew (and other foods) available to others. Wasps also consume many invertebrate prey (so much so that some insect and spiders are not able to successfully leave any young). This progressively depletes the invertebrate fauna and further reduces the food available to native species.

3. They act as pollinators, transferring the mistletoe pollen between plants.

4. The beech forest community is characterised by a large number of complex interactions between different species PHOTOCOPYING PROHIBITED


223 115. Interpreting Samples (page 161)

1. (a) 500 m Rata, kamahi 750 m Kamahi 1000 m Kamahi 1250 m Leatherwood 1500 m Red tussock (b) 1000 m (or between 1000m and 1250 m) (c) Kamahi (500 m to 1000 m) (d) Plant height decreases as altitude increases.

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with a mutual interdependence on the black beech and the food and habitat resources it provides. A large number of species in the beech forest community, including the beech scale insect, the sooty moulds, native birds, and insects (native and introduced) depend on the beech sap directly or indirectly as a food source. The beech scale insect feeds on the sugary sap and excretes the excess as honeydew, which forms an important food source for insects, birds, and the sooty mould fungus. The activities of the scale insect are also thought to stimulate the beech to produce more sap, so benefitting the entire community.

The beech trees also support a number of fungal species including the Nothofagus fungus with which it has a mutualistic mycorrhizal relationship and the bootlace mushroom which is a parasite of the tree's tissues. Beech mistletoes are also dependent on the beech, taking water and mineral nutrients from the beech and using it as support in the canopy. Native bellbirds and tui depend on the mistletoe for food and the mistletoe relies on the birds to pollinate its flowers.

The complexity of the interactions in the beech forest and the interdependence of so many species make the community vulnerable to threats. Competition from the large number of German and common wasps is threatening the availability of the honeydew resource to native consumers. Wasps deplete honeydew and insect resources so much that the survival of native species is threatened. A decline in native birds through lack of food (and other threats such as predation by rats) then threatens the survival of species like the mistletoes that depend on particular bird species to pollinate their flowers. Over time, the community biodiversity will decline and species may even be lost entirely.

2. (a) 0.4 m (b) Lichens (various species). (c) Most mosses need higher moisture and lower light and temperatures than lichens.

116. Stratification in a Forest (page 162) 1. See diagram below.

2. (a) Light. (b) The plants found in the ground layer will be shade tolerant plants that are unable to withstand high light levels, high wind, or low humidity.

3. Vertical structure in a community provides complexity and heterogeneity, with different layers providing different physical and biotic conditions. These differences create different microhabitats and provide opportunity for different species to avoid competition by exploiting resources in different layers. Because of this, the forest will be able to support a wider range of different species, so species diversity will be higher than in a forest with a simple structure.

2 In intraspecific competition, individuals all require the same resources, so resources become scarce, forcing individuals to exploit resources outside their preferred resource range (broadening their niche). In interspecific competition, species become (through evolution) more specialised in their exploitation of resources in order to avoid or minimise competition (e.g. hunting at a different times). This results in narrower, more differentiated niches.

4. (a) When a large tree is removed, a gap in the canopy is created allowing more light to penetrate to the forest floor. This will accelerate the growth of sub-canopy and canopy species, which will compete to fill that gap. The fallen log will remain on the forest floor providing habitat and food for other species of plants and for invertebrates. (b) Deliberate removal of emergent trees has some of the same effects as tree falls, allowing more light to penetrate and encouraging the growth of seedlings to fill the gap. If too many emergents are removed however, too much light may penetrate and encourage the growth of fast growing species that are tolerant of high light conditions (weed species). Loss of too many emergents also causes a loss of some of the forest's vertical structure, reducing habitat and species diversity and leading to a detrimental simplification of the community structure (and ultimately loss of efficiency in some ecosystem processes such as nutrient recycling).

3. (a) Intraspecific competition (b) Interspecific competition.

117. Physical Factors in a Forest (page 164)

113. Competition and Niche Size (page 159)

1. (a) Interspecific competition: Competition between members of different species. (b) Intraspecific competition: Competition between members of the same species.

114. Sampling Populations (page 160) 1. (a) Mark and recapture. (b) Belt transect, random quadrats. (c) Line transect, belt transect.

1. Environmental gradients from canopy to the leaf litter: (a) Light intensity: decreases. Foliage becomes more dense and light is absorbed progressively by the plants in different layers. (b) Wind speed: decreases. Foliage acts as a wind barrier. (c) Humidity: increases. Foliage traps water and reduces drying winds.

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2. (a) The gradient in physical factors arises primarily as a result of the vegetation growing to different heights (biotic) which alters the physical environment (e.g. by

Canopy

Sub-canopy

Tree fern layer

Shrub layer Ground layer Š2017 BIOZONE International

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224 at low densities. Extension: Groups could combine data to see if they get a more representative sample (i.e. closer to the direct count).

122. Field Study of a Rocky Shore (page 170)

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creating shade, reducing wind speed, and altering light regimes). (b) The gradient on a rocky shore is primarily caused by physical factors (time of air and water exposure).

118. Physical Factors on a Rocky Shore (page 165) 1.

(a) Salinity: Increases from LWM to HWM (b) Temperature: Increases from LWM to HWM (c) Dissolved oxygen: Decreases from LWM to HWM (d) Exposure: Increases from LWM to HWM

2. Mechanical force of wave action: Point B will receive the full force of waves moving inshore, Point A will receive only milder backwash, Point C will experience some surge but no direct wave impacts.

Surface temperature: Points A and B will experience greater variations in rock temperature depending on whether the tide is in or out, day or night, water temperature, wind chill. Point C is more protected from some of these factors and will not experience the warming effect of direct sunlight.

119. Quadrat Sampling (page 166)

1. Mean number of centipedes captured per quadrat: Total number centipedes ÷ total number quadrats 30 individuals ÷ 37 quadrats = 0.811 per quadrat.

2. Number per quadrat ÷ area of each quadrat 0.811 ÷ 0.08 = 10.1 centipedes per m2 3. Clumped or random distribution.

120. Quadrat-Based Estimates (page 167)

1. Quadrat must be large enough to be representative of the community.

2. Habitat heterogeneity (diverse habitats require more samples than a less diverse habitat to be representative).

3. Plants may overlap each other and it can be difficult to determine where one plant starts and another ends, so a percentage cover is more suitable for assessing plant communities.

1. Hypothesis (c): The communities of intertidal animals differ between exposed rocky shores and sheltered rocky shores.

2. See table top of next page.

3. See graph top of next page.

4. (a) The mean, medians, and modes are all similar. (b) The data are normally distributed.

5. (a) Rock oyster (b) Site A is open to the swell, which dislodges the oysters. Site B is more sheltered.

6. (a) Brown and plicate barnacles have a preference for exposed rocky shores. (b) Oyster borers are predators of brown and plicate barnacles so are more abundant when brown and plicate barnacles are also abundant.

7. (a) There are relatively high numbers of limpets at each site. (b) Limpets have a wide range of tolerance and are therefore able to live in many areas.

123. Transect Sampling (page 173)

1. Transect sampling involves placing a line across a community and sampling at specific points along the transect line.

2. Transect sampling provides information about changes in or distribution of a community over an environmental gradient.

3. Similarities: Both provide information about a community with a specific area.

Differences: Continuous belt transects provide information along a gradient (sampling points are next to each other), quadrats do not necessarily have to be next to each other. Belt transects may also measure vertical distribution.

4.

Brown barnacle

121. Sampling a Rocky Shore Community (page 168)

The results per se are not particularly important, but it is important to understand the method and its limitations. The results will vary depending on a group’s agreed criteria for inclusion of organisms in a given quadrat (e.g. when and how an organism is counted when it is partly inside a quadrat). Note: Some algae are almost obscured by other species or have other algae on top of them.

Represents 10 individuals

Columnar barnacle

6. Typical results (total for each category) are:

Plicate barnacle

Direct A B C D count Barnacle : 9 6 6 16 80 Oyster borer: 0 0 1 1 3 Chiton: 1 0 1 0 3 Limpet: 0 3 0 0 6 Algae: 27 18 15 13 101

7. Typical results for calculated population density based on A-D and on a direct count (question 8b):

NB: Area of 6 quadrats = (0.03 x 0.03) x 6 = 0.0054 m2 Area of total sample area = 0.18 x 0.18 = 0.0324 m2

8. (a) Once the quadrats have been laid, the animals moving from one quadrat to another might be counted twice. The quadrat could involve the placement of physical barriers between each quadrat. There is a possibility of exposing the entire area and photographing it for later analysis. (b) Densities calculated on direct count in the last column of the table above. Students should be aware of the dangers of extrapolating data from a small sample. Including or excluding single individuals can have a large effect on the density calculated, particularly where species are present ©2017 BIOZONE International

Sheet barnacle

0

1

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Direct A B C D count Barnacle: 1667 1111 1111 2963 2469 Oyster borer: 0 0 185 185 93 Chiton: 185 0 185 0 93 Limpet: 0 556 0 0 185 Algae: 5000 3333 2778 2407 3117

2

3

4

5

6

7

8

Height above low water mark (m)

9

10

124. Shoreline Zonation (page 175)

1. (a) Two of: Intensity of wave action, temperature, salinity, or oxygen level (in pools), exposure time. (b) Presence or absence of competing species, presence or absence of predators. 2. Zonation bands are broad and are approximately parallel to the waters edge.

125. Competition and Species Distribution (page 176)

1. Interspecific competition occurs between different species, which naturally will have slightly different niches, even if there is considerable overlap in resource use. When in competition they will be able to exploit slightly different ranges of resources or use resources in different places or at

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Table: Total and mean numbers of intertidal animals at two rocky shore sites. Oyster borer

Columnar barnacle

Plicate barnacle

Rock oyster

Ornate limpet

Radiate limpet

Black nerite

Total number of animals

308

46

78

386

0

63

47

55

Mean animals per m2

39

6

10

48

0

8

6

7

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Brown barnacle

Site A

Site B

Median value

38.5

6

9.5

51

0

7.5

6

7

Modal value

-

7

8

-

-

-

6

-

Total number of animals

52

15

427

85

49

50

96

29

Mean animals per m2

7

2

53

11

6

6

12

4

Median value

7

2

56.5

10.5

7.5

6.5

12.5

3.5

Modal value

7

-

58

-

8

5

14

-

Number of rock based animals per m2 of rocky shore

60

Number of animals (m2)

50

Site A

Site B

40 30 20 10 0

Oyster borer

Columnar barnacle

Plicate barnacle

different times. When members of the same species compete for resources, it is intense because their resource needs are exactly the same.

2. (a) A represents the realised niche of Chthamalus. (b) Cthalamus stellatus exists in a tidal zone above Balanus balanoides. When B. balanoides is removed C. stellatus moves into that area. This strongly suggests that B. balanoides is competitively excluding C. stellatus from the lower tidal area. 3. We can infer that the introduction of a new species is affecting an existing species if the distribution and/ or abundance of the existing species declines and the abundance and/or distribution of the introduced species shows a corresponding increase.

4. (a) Gambusia are aggressive and will attack native fish and eat their eggs (studies have shown them fin-nipping natives). The can breed much more quickly than black mudfish, producing many young, which mature quickly. (b) Black mudfish are long lived, they are nocturnal (so can avoid the dirurnal Gambusia), and they are adpated to survive periods of seasonal drying, which Gambusia cannot (mudfish aestivate in the mud). Aestivation (dormancy) is an adaptation to seasonally dry habitats, such as drains. 5. They interbreed to produce fertile hybrids.

126. Niche and Community Patterns (page 178)

1. Grooved topshell, common rock crab, plicate barnacle, radiate limpet, oyster borer, columnar barnacle, snakeskin chiton, black nerite.

2. (a) Any two of: Exposure time, temperature (higher when exposed), intensity of wave action, salinity. (b) Predation (including grazing), competition (for food and space). 3. (a) Radiate limpet: Shell has a flat profile that enables the animal to withstand turbulent water (wave action). Muscular foot and production of mucus enables it to

©2017 BIOZONE International

Rock oyster

Ornate limpet

Radiate limpet

Black nerite

clamp to the rock and avoid water loss (desiccation). (b) Common rock crab: Nocturnal, forage at night and hide in crevices and under rocks during the day avoiding desiccation. Intolerant of exposure so behaviour keeps them in regions where they will not lose water.

4. Organisms will occupy zones appropriate to their tolerance for air exposure, their feeding requirements, and the presence of other intertidal organisms. Organisms that can tolerate a longer period of exposure are found in the high to mid water zone whereas those less tolerant of exposure are found in the mid to low water zone. The differences in tolerance and niche lead to species being restricted specific regions and creates the zones we see on the shoreline.

127. Altitude Zonation and Physical Environment (page 180) 1. Temperature, rainfall, and soil type.

2. (a) Temperature declines with increasing altitude. (b) Precipitation increases with increasing altitude.

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Brown barnacle

3. Rocky shore ecosystem.

4. Annual temperature range: 6.4–8.5°C. Annual precipitation range: 3250–6750 mm.

5. (a) Altitude range for tussock zone: 1250-1600 m. (b) Tussock zone does not extend lower (probably) because of competition with sub-alpine scrub. (c) Any one of: Cold (low temperature), unstable slopes (scree), poor nutrient availability.

128. Environment Determines Community Patterns (page 182) 1. (a) Vegetation height declines with increasing altitude. (b) Number of species present generally declines with increasing altitude.

2. Between 550 and 825 m: Loss of large emergents (rimu and rata), tree ferns, and hinau at the higher altitude. Black maire and totara appear at the higher altitude. The height of the PHOTOCOPYING PROHIBITED


226 100

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Large tree emergent line

Tree line

Scrub line

Tussock line

Herbfield line

90

A

B

Kamahi

70 60 50

Rata

20

10

Tawa

0

500 m

Blue tussock

Everlasting daisy

Mountain totara

40

Rimu

30

Leatherwood

■ ●

Kaikawaka

1000 m

1250 m

▲ ●

1500 m

1750 m

Altitude (metres)

largest trees decreases by ~8 m at 825 m.

2. (a) See diagram above. (b) Lower limit of herbfield spp: approx. 1450-1475 m. (c) Emergent trees: Rata Canopy trees: Kamahi Shrub/scrub: Leatherwood (d) Blue tussock has greatest tolerance to high altitude. Other species drop out as altitude increases. Species competing with the blue tussock are probably increasingly intolerant of the harsher conditions and the blue tussock can therefore become more abundant. (e) Species at A: Rata, kamahi, rimu, tawa Species at B: Moss, blue tussock, everlasting daisy. 3. Introduced browser species such as hares and possums will preferentially eat palatable native plants (tussocks and kamahi) reducing their numbers, and ignore less palatable species such as mountain daisies and pepperwood. The less grazed plants will increase in abundance.

129. Primary Succession (page 184)

1. (a) The biological colonisation and development of a region where there is no pre-existing vegetation or soil. (b) Glacial retreat, exposed slip, new volcanic island.

2. The climax community is the mature, relatively stable community that results from a succession. Climax vegetation depends on local climate and region.

3. A general trend for increasing species diversity, greater community complexity, and greater community efficiency (as more species interactions are involved).

4. (a) Lichens, mosses, liverworts, and hardy annual herb species are often the first to colonise bare ground. (b) Early colonisers chemically and physically erode the rock (producing the beginnings of soil) and add nutrients by decay and nitrogen fixation. (c) As later successional species establish, they begin to shade out and outcompete many of the earlier species (which are adapted to high light levels of open ground).

130. Secondary Succession (page 186)

1. Primary succession refers to the colonisation of regions where there is no preexisting community (e.g. rocky slope, exposed slip, new volcanic island). Changes in the community occur in stages until a climax community is reached. Secondary succession follows the interruption of an established climax community (e.g. logging, pasture reverting to bush). 2. Secondary succession proceeds more rapidly because there is no loss of soil or seed stores. Many plants may still be able to grow despite the disturbance and the climax community will reestablish faster because soil and nutrients are already available.

©2017 BIOZONE International

Red tussock

Moss

750 m

Coprosma sp.

131. Strategies for Survival in Different Environments (page 187) 1. The characteristics of r-selected species (e.g. highly productive, rapid population growth rate, short life spans, opportunistic) are well suited to rapid saturation of a new environment where there are no established competitors. Lichens and annual plants are typical early colonisers and are frequently adapted to high light environments with poor/ no soil. These conditions are met in early successional environments.

2. Climax communities have a complex structure with high species diversity and narrow niches. K-selected species have longer life spans, lower productivity, and occupy specialist niches. Their characteristics make them able to complete within a more complex community structure where productivity is less important than efficiency.

3. Generally land clearance, e.g. logging or burning of established climax vegetation. this increases light levels and produces scoured or bare areas where colonising species can quickly establish.

4.

Long-tailed blue

Weta

Manuka

E

Tawa

E-M

M

M-L

L

132. Succession and Population Growth in Birds (page 189)

1. Height of vegetation increases from grassland to shrubland.

2. This would be a reasonable hypothesis because, as shrubs increased in number and density, the habitat and food for ground dwelling grainvores could be expected to decrease and the habitat and food for tree living insectivores could be expected to increase.

3. There is some evidence to support this hypothesis although it is not strong and, for some species, the results contradict the hypothesis. Most species show no significant change in density with the increase in density of woody species and the blackbird and silvereye even show a decrease in density as shrubland increases (contrary to what you would expect). However, goldfinch, greenfinch, redpoll, and yellowhammer (all grainvores) do show a decrease in density as the percentage of shrubland increases, supporting the hypothesis. Generally, the hypothesis seems to be supported for grainvores but not for insectivores.

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Relative cover (%)

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80

4. Retaining some open grassland areas when shrubland is being restored might be a good precaution to ensure food and habitat for populations of grain-eating perching birds.

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There are no answers. Summary is entirely student based.

134. KEY TERMS AND IDEAS: Communities (page 193)

1. Student's own drawing. Points to note: – Drawing should state what the structure is, its sectional orientation, and the magnification. – Regions of cells should be indicated, but not cells themselves. – Label lines are straight and accurate. – Labels are accurate.

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1. Abiotic factor (K), biotic factor (G), community (M), competition (H), distribution (A), ecological succession (I), environmental gradient (C), food chain (E), mutualism (B), parasitism (D), predation (L), sample (J), stratification (F).

137. Practising Biological Drawings (page 198)

Plan diagram through a dicot root (x 50). Section completed from visible parts.

2. Methods commonly used to sample communities include quadrats and transects. A transect along an environmental gradient may reveal a community pattern of zonation. Zonation is common with altitude and on rocky shores. The occurrence and distribution of organisms in the community creates a banding pattern. The adaptations of species to abiotic and biotic factors determines their physical position along the environmental gradient. Their region of occupation is often more restricted than they might potentially occupy. This reflects their realised niche and the community position to which they are best adapted. They are often forced to this position by competition and predation.

Epidermis

Phloem Xylem

Pericycle

Cortex

3. (a) Zonation (b) Time (duration) of exposure above the low water mark, temperature. (c) Predation and interspecific competition.

138. Preparing a Slide (page 199)

1. Thin sections allow light to pass through so features can be more easily seen. Thin sections also reduce the layers of cells (making it easier to see details).

135. Optical Microscopes (page 194) 1.

(a) Eyepiece lens (b) Arm (c) Coarse focus knob (d) Fine focus knob (e) Objective lens (f) Mechanical stage (g) Condenser lens

(h) In-built light source (i) Eyepiece lens (j) Eyepiece focus (k) Focus knob (l) Objective lens (m)Stage

2. (a) 600X magnification

(b) 600X magnification

3. (a) DM (dissecting microscope). (b) CM (compound microscope).

4. Steps for setting up a microscope 1. Turn on the light source. 2. Rotate the objective lenses until the shortest lens is in place (pointing down towards the stage). This is the lowest power objective lens. 3. Place the slide on the microscope stage. Secure with the sample clips. 4. 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. 5. Viewing from the side, lower the shortest objective lens to its lowest point or until it is almost touching the specimen. 6. Focus and centre the specimen using the low objective lens. Focus firstly with the coarse focus knob, then with the fine focus knob. 7. Fine tune the illumination so you can view maximum detail on your sample. 8. 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). 9. Focus and centre the specimen using the high objective lens. Adjust focus using the fine focus knob only.

2. The coverslip helps to smooth out the specimen and exclude air bubbles that may obscured features.

3. The onion epidermal cells do not take part in photosynthesis, so they do not contain chloroplasts.

139. Staining a Slide (page 200)

1. Stains are mostly used to enhance specific features of a sample (e.g. specific organelles). 2. Viable stains are harmless and can be used on living samples. Non-viable staining is used on cell or tissue preparations which are dead.

3.

(a) Trypan blue (b) Iodine solution (c) Aniline sulfate (d) Methylene blue

140. Calculating Linear Magnification (page 201)

1. Actual size = image size ÷ magnification = 52 000 µm ÷ 140 = 371 µm (0.37 mm) 2. (a) Actual length of scale line = 10 mm Given length of scale line = 0.5 mm 10 ÷ 0.5 = 20 x magnification (b) Measured length = 42 mm Magnification = 20 x Actual length = 42 ÷ 20 = 2.1 mm 3.

43 mm = 43 000 µm Magnification = size of the image ÷ actual size of object = 43 000 µm ÷ 2 µm = 21 500 x magnification

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Practical microscopy

141. Cell Sizes (page 202) 1. (a) Daphnia: (b) Giardia: (c) Nucleus (d) Elodea: (e) Chloroplast: (f) Paramecium:

1800 15 10 50 7 250

µm µm µm µm µm µm

1.8 mm 0.015 mm 0.01 mm 0.050 mm 0.007 mm 0.25 mm

2. Largest to smallest: Daphnia, Paramecium, onion epidermal cell (~200 mm long), Elodea leaf cell, Giardia

3. Visible to the unaided eye: Daphnia

©2017 BIOZONE International

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1. Paramecium, Elodea leaf cell, thale cress leaf epidermis cell

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2. High magnification and inconsistent slides (e.g. sections that are too thick or poorly stained) can causes distortion and make details difficult to distinguish.

3. Methyl cellulose slows the movement of simple motile organisms so that they can viewed more easily. Otherwise they quickly move in and out of the field of view.

4.

(a) Cell wall (b) Gives the cell shape and rigidity and limits its volume. (c) Chloroplast (d) The site of photosynthesis. (e) 10 mm (f) 5 mm ÷ 0.01 mm = 500 x magnification

5. (a) Paramecium (b) Cilia on the outside of the cell, contractile vacuoles (c) Contractile vacuole (d)

(e) Cilia. Movement - they propel the organism through the medium.

6. (a) Nucleus (b) Carries the genetic instructions for everything that is made or done in the cell. Controls the activity of the cell. (c) Guard cell (d) Regulate the opening and closing of the leaf stomata.

143. KEY TERMS AND IDEAS: Practical Microscopy (page 205)

1. Biological drawing (G), biological stain (E), compound microscope (C), dissecting (stereo) microscope (D), light (optical) microscope (F), magnification (A), resolution (B).

2.

Eyepiece lens

Objective lens

Condenser lens In-built light source

3. (a) Amoeba (longest dimension taken): 60 mm ÷ 0.48 mm = 125 x magnification. (b) Scenedesmus 24 mm ÷ 0.03 mm = 800 x magnification.

©2017 BIOZONE International

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Stage Slide

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Appendix A

Questioning terms in biology

Photo credits

The following terms are often used when asking questions in examinations and assessments. Interpret data to reach stated conclusions.

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.

We also acknowledge the photographers that have made their images available through Wikimedia Commons under Creative Commons Licences 2.0, 2.5, 3.0 or 4.0: • AKA • Dozenist • Ed Uthman • Luc Viatour • diveofficer • radiogaga • mnoff • althepal • haplochromis • H dahimo • Tony Willis • Evenherk • my.opera.com/Ukwildlife/blog • Professor Dr. habil • Kenneth Catania • Suseno • Roger Griffith • Hardyplants • Habitat News • Wendy Kaveney • E Heuting (Ericthereddish) • inski • Stu Philips • Igor_NZ • Rudolph89 • Davidvraju • Melodi2 • Francis C. Franklin • Smalljim • CDC: Dr Lucille K. Georg • Kristian Peters • Barfooz

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.

© 2017 BIOZONE International

• PASCO • Dartmouth College Electronic Microscope Facility • Stephen Moore for his photos of aquatic invertebrates • Western New Mexico University Department of Natural Sciences for their photo of the pine stomata • Jenny Ladley (University of Canterbury) for the photos of tui • C. Pilditch for photos of rocky shore animals • Callum O'Hagan for the photo of the NZ native forest • DoC J Kendrick for NZ long tailed bat photo • DoC CR Veitch for the photo of the short tailed bat • DoC J Barkle for photo of the wood rose • David Kelly (University of Canterbury) for the photos of the black beech • Peter Bray - Airborne Honey Ltd • Shirley Kerr (www.kaimaibush.co.nz) for fungi photos • Landcare Research • www.coastalplanning.net for the image of a marine quadrat • Phil Bendle for Mt Taranaki flora photos • Chrissen Gemmill for use of plant photos (Mount Taranaki) • John Green

Contributors identified by coded credits are: BH: Brendan Hicks (Uni. of Waikato) CDC: Centers for Disease Control and Prevention, Atlanta, USA, EII: Education Interactive Imaging, KP: Kent Pryor, NIH: National Institutes of Health, NOAA: National Oceanic and Atmospheric Administration www.photolib. noaa.gov, RA: Richard Allan, RCN: Ralph Cocklin, USGS:United States Geological Survey, WBS: Warwick Silvester (Uni. of Waikato),

Royalty free images, purchased by Biozone International Ltd, are used throughout this workbook and have been obtained from the following sources: iStock, Corel Corporation from their Professional Photos CD-ROM collection; IMSI (Intl Microcomputer Software Inc.) images from IMSI’s MasterClips® and MasterPhotos™ Collection, 1895 Francisco Blvd. East, San Rafael, CA 94901-5506, USA; ©1996 Digital Stock, Medicine and Health Care collection; © 2005 JupiterImages Corporation www.clipart.com; ©Hemera Technologies Inc, 1997-2001; ©Click Art, ©T/Maker Company; ©1994., ©Digital Vision; Gazelle Technologies Inc.; PhotoDisc®, Inc. USA, www.photodisc.com. • TechPool Studios, for their clipart collection of human anatomy: Copyright ©1994, TechPool Studios Corp. USA (some of these images were modified by Biozone) • Totem Graphics, for their clipart collection • Corel Corporation, for use of their clipart from the Corel MEGAGALLERY collection • 3D images created using Bryce, Vue 6, Poser, and Pymol

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Analyse:

We acknowledge the generosity of those who have provided photographs for this edition:

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Index

B Balanced reporting 28 Bar graph, rules for 18 Bat, adaptations to niche 154-155 Beech tree community 156-158 Bias, in reporting 28 Biological drawings 196-197 Biological information, - analysis 28 - reporting 33 Biome 149 Biosphere 149 Biotic factor - and community patterns 183 - defined 149 Blood - mammalian 91, 97 - role in gas transport 91 Blood vessels 92-93, 97 Breathing, in humans 83 Bronchi 80-81 Bronchioles 80-81 Bulk feeding 37 Bulk flow 89 C Caecotrophy 61 Cancer, role of HPV 30 Capillary 93 Carnivore gut 57

Catalase activity, experiment 13 Cell size, calculating 202 Cell types 203-204 Cellulose digestion 60-62 Circulatory system, - mammalian 103-106 - types 94-95 Climax community 184 Closed circulatory system 94 Cohesion-tension 114-115 Colon 52-53 Competition - and niche size 159 - and species distribution 176-177 - defined 158 - interspecific 176-177 Competitive exclusion 176 Consumer 151 Continuous data 11 Control, experimental 12 Controlled variable 12 Conversion factors 6 Correlation 20 Countercurrent flow - in gas exchange 79 - in body temperature 64 D Data loggers 14 Data transformation 15 Data, types 11 Datalogger 164 Decimal form 6 Decomposer 151 Decompression sickness 84 Dental formulae 44 Dentition, mammalian 4-46 Dependent variable 12 Deposit feeding 37 Descriptive statistics 22 Detritivore 151 Diffusion 76 Digestion, - adaptations 60-62, 64 - in insects 62-63 - in fish 64 - in mammals 55-61 - of cellulose 60-62 - ruminant 60 - stages 47 Digestive organs, rat dissection 59 Digestive system - insect 58 - fish 58 - mammalian 47-58, 106-107 Discontinuous data 11 Discrete data 11 Dispersal of seeds 140 Dissection - mammalian heart 101-102 - rat digestive organs 59 Disturbance, and stratification 163 Diversity, tube guts 58 Diving mammals, adaptations of 84 Double circulatory system 95, 98 Double fertilisation - in angiosperms 137 Drawings, biological 196-198 E Ecological niche 153 Ecological patterns, investigating 148 Ecological succession 184-186 Ecosystem components 149 Endosperm 137 Enzyme - catalase activity 13 - digestive 50 Error, sources of 10 Estimation 6 Š 2017 BIOZONE International

Exchange systems 89 Experiment, plant growth 21-23 Experimental control 12 Experimental design 24-26 Expiration, of lungs 83 Exponential functions 9

F Fair test, definition 25 Feeding relationships 151 Feeding 37 Fern life cyc 130 Fertilisation, of plants 137-138 Filter feeding 37 Fish, - adaptation for absorption 66 - adaptations for digestion 64 - gas exchange in 72-73 - heart structure 99 - teeth 43 Flower structure 133 Flowering plants - fertilisation 137 - reproduction 131 Fluid feeder 63 Fluid feeding 37 Fluoridation of water 28-29 Food chain 151 Food web 152 Food, methods of obtaining 37 62-163 Forest stratification Fractions 8 Fundamental niche 153, 159

G Gametophyte, defined 130 Gardasil vaccine 31-32 Gas exchange in plants 114, 121 Gas exchange membrane 82 Gas exchange surfaces - properties of 73 Gas exchange 73 - adaptations in animals 74-82 - in fish 72-73 - in mammals 80-82, 103-104 Gas transport system 89 Gastric caeca 65 Germination of seeds 139 Gills 73, 76-78 Gradients on line graphs 19 Graph, types 17-20 Gut adaptations 57 Guts - diversity of 58 - mammalian 55, 57 Gymnosperm - fertilisation 138 - life cycle 132 H Habitat 150 Haemoglobin 91 Haemolymph 90 Halophytes, adaptations of 125 Heart, mammalian 97, 99 Heart, fish 99 Heterotroph 37, 151 Hindgut fermenter 61 Histogram, rules for 18 Homodont 43 HPV 30 Human heart, structure of 99 Human papillomavirus (HPV) 30 Humans, gas exchange 80-82 Hydrophytes, adaptations of 124 Hypothesis 4

Insect mouthparts 41 Insect - adaptations for absorption 65 - adaptations for digestion 62-63 - adaptations for gas exchange 76 - gas exchange in 74-76 Inspiration, of lungs 83 Interactions, between species 156-158 Intercepts on line graphs 19 Internal transport systems 89 Internal transport, mammals 103-106 Interpreting samples 161 Interspecific competition 159, 176-177 Intestinal villi 50, 67 Intraspecific competition 159 KL K-selected species 187

Large intestine 52-53 Leaf structure 121 Leaves, adaptations 119, 121-125 Life cycle, of plants 129-132 Light microscope 194-195 Line graphs 17, 19 Line of best fit 17, 20 Linear magnification 201 Log book 14 Log transformations 9 Lungs 73, 80-82

M Magnification 195, 201 Mammal, heart structure 99, 101-102 Mammalian blood 91 Mammalian gut, structure of 55 Mammals - adaptation for absorption 67 - adaptation for digestion 60-61 - adaptations for diving 84 - cellulose digestion 60-61 - digestive system 47-57 - gas exchange in 73, 80-82 - teeth 44-46 Mangroves, adaptations of 125 Mark and recapture sampling 160 Mass transport 89 Mathematical symbols 6 Mean 22 Median 22 Meiosis, plant life cycle 129-132 Mesophyte 122 Microscope, optical 194-195 Microscopy, stains 200 Mineral uptake, in plants 119 Mitosis, plant life cycle 129-132 Mode 22 Mount Taranaki, zonation 180-181 Mounting, biological samples 199 Mouthparts, adaptations 41-42 Mutualism - and pollination 134 - cellulose digestion 60-61 - defined 158

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A Abiotic factor 149 Absorption - adaptations for 65-68 - adaptations in fish 66 - adaptations in insects 65 - adaptations in mammals 67 Abundance scale 167 Accuracy, of data 5 ACFOR abundance scale 167 Adaptations - for absorption 50, 65-68 - for digestion 60-62, 64 - for diving 84 - for feeding 37, 45-46, 63 - for gas exchange in animals 74-82 - for gas exchange in plants 114 - for plant reproduction 129-132 - for pollination 133-136 - in New Zealand bats 154-155 - mammalian gut 57-59 - of halophytes 125 - of hydrophytes 124 - of mangroves 125 - of mesophytes 122 - of plants 121-125, 140 - of predators 40 - of seeds 140 - of xerophytes 123 Alternation of generations 129 Altitude, - and community patterns 180-183 - and zonation 180-183 Alveolar-capillary membrane 82 Alveoli 81-82 Analysing biological information 28 Angiosperms, - life cycle 131 - fertilisation 137 Animal pollination 133-134, 136 Animals, gas exchange in 73-82 Annotated diagram 197 Apparatus, use of 10 Aquatic invertebrates - gas exchange in 76 Arterial system 98 Artery 92, 97 Assumptions 4 Autotroph 151

IJ Independent variable 12 Insect haemolymph 90

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N Niche size, effect of competition 159 Niche - adaptations to 154-155 - and community patterns 178-179 - ecological 153 O Observations 4 Open circulatory system 94 Optical microscope 194-195


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231

P Parallel current flow 79 Parasitism 39, 158 Percentage error, calculating 10 Percentages, calculating 7 Peristalsis 48, 53 Phloem 112-113 Physical factor 149 Physical factors - and community patterns 180-182 - measurement of 164-165 - rocky shore 165 - role in zonation 180-181 Plan diagram 197 Plant adaptations - for water uptake 119 - for pollination 133-136 - for seed dispersal 140 - leaves 121-125 Plant fertilisation 137-138 Plant growth, experiment 21-23 Plant reproduction 129-132 - summary 142 Plant root structure 119 Plant transport system 112-113 Plant, life cycles 129-132 Plastron 76 Pneumatophore 125 Point sampling 160 Pollination methods 133-136 Pollination relationships 134 Potometer 116 Precision, of data 5 Predation 40 - defined 158 Primary succession 184-185 Producer 151 Prothallus 130 Pulmonary system 98 Pyloric caeca, in fish 66 Q Quadrat sampling 160, 166 - exercise in 168-169 Quadrat-based estimates 167 Qualitative data 11 Quantitative data 11

R r-selected species 187 Radio tracking 160 Random sampling 160 Range, of data 22 Ranked data 11 Rates, calculating 7 Ratios 8 Raw data 7 Realised niche 153 Red blood cells 97 - and oxygen transport 91 Relative abundance 167 Reporting biological information 33 Reproduction, in plants 129-132 Resolution, defined 195 Respiratory system - mammalian 103-104 Results, recording of 14 Rocky shore - field study 170-172 - physical factors 165 - zonation 175, 178 Root hairs 112-113 Root pressure 114-115 Root, uptake at 119-120 Ruminant digestion 60 Ruminant gut 57

S Sample preparation, for microscopy 199-200 Sampling methods 160 Sampling - physical factors 164 - populations 160 - quadrat 166 - transect 173-174 Scatter graph, rules for 17 Scientific method 3 Secondary succession 186 Seed dispersal 140 Seed plants, reproduction in 131-132 Seed structure 139 Significant figures 5 Single circulatory system 95 Slide preparation 199-200 Small intestine 49-50 Sori 130 Species distribution, effect of competition 176-177 Species interactions 156-158 Spiracles 74 Spiral valve, in fish 66 Sporophyte, defined 129 Stains, microscopy 200 Standard form 6 Stomach 49 Stomata, and gas exchange 112-114, 121 Stratification, forest 162-163 Succession 148, 184-186 - population growth 187, 189-190 - species composition 187 Symbiosis, in digesting wood 62 System interactions, - mammalian 103-106 Systemic system 98 T Tables, constructing 16 Tally 7 Teeth - fish 43 - mammals 44-46 The bends 84 Tolerance range 150, 80-81 Tracheal gills 76 Tracheid 113 Transect sampling 160, 173-174 Transpiration 114-115 - measuring 116-118 Transpiration pull 114-115 Transpiration stream 114-115 Transport system - mammalian 97-98 - open vs closed 94 - plants 112-113 - single vs double 95-96 Trophic structure 151 Tube gut, comparison 58-59 V Vaccinations, types of 31-32 Vaccine, HPV 31-32 Variables, types 12 Vasoconstriction 92 Vasodilation 92 Vein 93 Venous system 98 Vertebrate hearts 99 Vessel element 113

W Water uptake, in plants 119-120 Wet mount 199 Wind pollination 135-136

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XYZ Xerophytes, adaptations of 123 Xylem 112-113

Zonation 148 Zonation - altitude 180-181 - shoreline 175, 178-179

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Optimum range 150

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