NCEA Level 2 Biology Externals by BIOZONE International

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The kakapo, Strigops habroptilus Kakapo are endemic to New Zealand. They are the world's heaviest and only flightless parrot and, like many parrots, are long-lived. Kakapo are nocturnal and the only parrot to breed using a polygynous lek system. During the breeding season males establish a court on a hilltop or ridge and produce a deep booming call to attract a mate. Although once common, kakapo are now critically endangered.

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.

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NCEA LEVEL 2 BIOLOGY EXTERNALS

Cover photograph

PHOTO: James T. Reardon

and explanatory notes 81 59 Alleles 83 60 Sources of Variation 84 61 Examples

Genetic Variation 86 62 The Role of Mutations in Populations ............. 88 63 Mutations Can Produce New Alleles ............... 89 64 Somatic and Gametic Mutations ..................... 90 65 Meiosis ............................................................ 91 66 Meiosis and Variation 92 67 Modelling Meiosis 94 68 Mendel's Laws of Inheritance 96 69 Dominant and Recessive Traits 97 70 The Monohybrid Cross 99 71 The Test Cross 100 72 Practising Monohybrid Crosses 101 73 Codominance in Alleles 102 74 Codominance in Multiple Allele Systems 103 75 Incomplete Dominance 105 76 Lethal Alleles 106 77 Problems Involving Monohybrid Inheritance 107 78 Dihybrid Cross 108 79 Inheritance of Linked Genes 110 80 Recombination and Dihybrid Inheritance 112 81 Problems Involving Dihybrid Inheritance 114 82 What You Know So Far: Sources of Variation 115 83 NCEA Style Question: Sources of Variation 116 CODES: Activity is marked: to be done when completed

AS 2.5 Genetic Variation and Change criteria of

Achievement

Contents Using This Resource v Using the Tab System vii How To Scaffold an NCEA Style Answer viii AS 2.4 Life Processes at the Cellular Level Achievement criteria and explanatory notes 1 1 The Structure of Plant Cells 3 2 Identifying Structures in Plant Cells 5 3 The Structure of Animal Cells 6 4 Identifying Structures in Animal Cells 8 5 Structure Relates to Function 9 6 Cell Processes 11 7 Identifying Organelles in Micrographs ............. 13 8 Cell Structures and Organelles ....................... 15 9 The Structure of Membranes .......................... 16 10 Diffusion in Cells .............................................. 18 11 Osmosis in Cells 20 12 Water Relations in Plants 21 13 Diffusion and Cell Size 22 14 Temperature and Membrane Permeability 24 15 Active Transport 25 16 Ion Pumps 26 17 Cytosis 28 18 Active and Passive Transport Summary 30 19 What You Know So Far: Cell Structure and Transport 31 20 NCEA Style Question: Cell Structure and Transport 32 21 KEY TERMS AND IDEAS: Cell Structure and Transport 34 22 Reactions in Cells 35 23 Enzymes 36 24 Enzymes and Activation Energy 37 25 How Enzymes Work 38 26 Cofactors and Inhibitors Affect Enzyme Activity 39 27 Enzyme Reaction Rates 40 28 Investigating Catalase Activity 41 29 ATP 42 30 Energy Transformations in Cells 43 31 Cellular Respiration: Inputs and Outputs 44 32 Steps in Cellular Respiration .......................... 46 33 Measuring Respiration .................................... 48 34 Anaerobic Metabolism in Plants and Fungi .... 49 35 Anaerobic Metabolism in Animals .................... 50 36 Leaf Structure and Photosynthesis 51 37 The Structure of Chloroplasts 53 38 Photosynthesis: Inputs and Outputs 54 39 Steps in Photosynthesis 56 40 Factors Affecting Photosynthesis 57 41 Investigating Photosynthetic Rate ................... 59 42 Modelling Photosynthesis and Cellular Respiration ...................................................... 60 43 What You Know So Far: Energy Transformations 63 44 NCEA Style Question: Photosynthesis 64 45 NCEA Style Question: Cellular Respiration 65 46 KEY TERMS AND IDEAS: Energy Transformations 66 47 DNA Replication 67 48 Details of DNA Replication 68 49 The Eukaryotic Cell Cycle 69 50 Checkpoints in the Cell Cycle 70 51 Functions of Mitosis 71 52 Mitosis 72 53 Mitosis and Cytokinesis 73 54 Modelling Mitosis 75 55 Putting it all Together: Metabolism! 76 56 What You Know So Far: Cell Division 78 57 NCEA Style Question: Cell Division 79 58 KEY TERMS AND IDEAS: Cell Division 80

APPENDIX 1: Hints For Drawing Graphs 190 APPENDIX 2: Questioning Terms 191 PHOTO CREDITS...........................................191 INDEX 192

102 Constructing a DNA Model 148

.............................

121 Effect of Substitution Mutation on Phenotype 177

111 Protein Shape is related to Function 162

123 Skin Cancer: A Non-Inherited Mutation 180

92 The Founder Effect 131

109 Amino Acids Make Up Proteins 160

125 Genes and Environment Interact 184

117 Interrupting Metabolic Pathways 170

...................................................

105 What is the Genetic Code? 154

106 Cracking the Genetic Code 156

124 Environment and Phenotype 181

AS 2.7 Gene Expression

................................

114 NCEA Style Question: Nucleic Acids and Proteins 165

91 Natural Selection in Pocket Mice 129

100 Nucleotides 145

84 NCEA Style Question: Inheritance of Alleles 118

104 Transcription 153

93 Population Bottlenecks 133

108 Protein Synthesis Summary 159

Sources of Variation 120

116 Metabolic Pathways 168

97 KEY TERMS AND IDEAS: Changes in Gene Pools 140

CODES: Activity is marked: to be done when completed

94 Genetic Drift 135

128 KEY TERMS AND IDEAS: Genotype and Phenotype 189

118 Influences on Phenotype 172

112 Globular and Fibrous Proteins 163

.........................

Achievement criteria and explanatory notes 141

120 Mutation, Genotype, and Phenotype 174

127 NCEA Style Question: Genotype and Phenotype 187

95 What You Know So Far: Changes in Gene Pools 137

99 The Structure of Chromosomes 144

115 KEY TERMS AND IDEAS: Nucleic Acids and Proteins 167

101 DNA and RNA 146

90 Gene Pool Model 127

110 The Structure of Proteins 161

87 How Natural Selection Works 123

85 KEY TERMS AND IDEAS:

96 NCEA Style Question: Changes in Gene Pools 138

Contents

119 Mutagens 173

122 A Case Study: Cystic Fibrosis 178

88 Types of Natural Selection 125

.............................

103 Genes Code for Proteins 152

113 What You Know So Far: Nucleic Acids and Proteins 164

89 Modelling Natural Selection 126

86 Processes in Gene Pools 121

98 The Role of DNA in Cells 143

107 Translation 157

126 What You Know So Far: Genotype and Phenotype 186

• Other questions test your understanding of the section content

• A check list of what you need to know

9 Phospholipids naturally form a bilayer in aqueous solutions.(watery) Cholesterol molecule maintains membrane integrity, preventing it becoming too fluid or too firm. CO H2O Glucose Na Phosphate head is (waterisFatty(waterhydrophilicloving)acidtailhydrophobichating) Water molecules pass freely between the phospholipid molecules by osmosis. Lipid soluble molecules, e.g. gases and steroids, can move through the membrane by diffusion, down their concentration gradient. Carrier proteins permit the passage of specific molecules by facilitated diffusion or active transport. Ion pumps are a type of carrier protein. Channel proteins (including ion channels) form pores through the hydrophobic interior of the membrane so that water soluble molecules can pass by facilitated diffusion. Key Idea The plasma membrane encloses the cell and regulates the entry and exit of substances into the cell. It is a phospholipid bilayer with embedded proteins moving freely within it. f The cell surface (or plasma) membrane encloses the cell's contents. The fluid-mosaic model of membrane structure describes a phospholipid bilayer within which various proteins can move about freely. f The plasma membrane is a dynamic structure and is actively involved in cellular activities. It is selectively permeable, allowing some molecules, but not others, to pass. Proteins in the membrane enable the cell to regulate the movement of materials into and out of the cell. This enables the cell to obtain what it needs for its metabolism. 1. Describe the fluid mosaic model of membrane structure: 2. (a) What are the two main roles of the plasma membrane? (b) How is its structure related to these roles? 3. On the diagram (right) label the hydrophobic and hydrophilic ends of the phospholipid and indicate which end is attracted to water: 69 14 REVISE 78 © 1994-2016 BIOZONE International Photocopying Prohibited What You Know So Far: Cell Division56 DNA replication HINT: Steps in DNA replication including base matching. Include a diagram of DNA replication. The cell cycle and mitosis HINT: Describe the stages in the cell cycle and the events of mitosis. Note the differences in cytokinesis between animal cells and plant cells. 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 NCEA style essay question that follows. Use the points in the introduction and the hints provided to help you: TEST 79 © 1994-2016 BIOZONE International Photocopying Prohibited DNA replication occurs prior to mitosis and cell division. It produces new DNA by semi-conservative replication. 1. What is the purpose of DNA replication? 2. Explain how the parent DNA is copied by semi-conservative replication (you may use extra paper if required). Your answer should include: • the units that make up DNA the process of replication • a labelled diagram to support your answer NCEA Style Question: Cell Division57 TEST 80 ISBN: 978-1-927309-51-3 Photocopying Prohibited 2. Match the statements in the table below to form complete sentences, then put the sentences in order to make a coherent paragraph about DNA replication and its role: The enzymes also proofread the DNA during replication......is required before mitosis can occur. DNA replication is the process by which the DNA molecule... ...by enzymes. Replication is tightly controlled... ...to correct any mistakes. After replication, the chromosome... ...and half new (daughter) DNA. DNA replication... ...during mitosis The chromatids separate... ...is copied to produce two identical DNA strands. A chromatid contains half original (parent) ... ...is made up of two chromatids. Write the complete paragraph here: cell semi-conservativemitosisDNAcytokinesischromosomecyclereplication A The process of producing two identical copies of DNA from one original DNA molecule. B The sequence of events that a cell completes prior to and including its division. C The physical process of cell division, which divides the cytoplasm of a parental cell into two daughter cells. D An organised structure of protein and DNA found in the nucleus of eukaryotic cells. It contains most of the cell's genetic material containing most of the DNA. E A term to describe the DNA replication process in which each double stranded DNA molecule contains one original strand and one new strand. F The process of nuclear division in cells. 3. DNA replication occurs during the S (synthesis) phase of the cell cycle The light micrograph (right) shows a section of cells in an onion root tip. These cells have a cell cycle of approximately 24 hours. The cells can be seen to be in various stages of the cell cycle. By counting the number of cells in the various stages it is possible to calculate how long the cell spends in each stage of the cycle. Count and record the number of cells in the image which are undergoing mitosis and those that are in interphase. Estimate the amount of time a cell spends in each phase. StageNo. of cells% of cellstotal timeEstimatedinstage InterphaseMitosisTotal 100 Onion root tip cells 1. Match each term to its definition, as identified by its preceding letter code. KEY TERMS AND IDEAS: Cell Division58

Activities

• Annotated diagrams help you understand the content

Sections within a chapter share the same structure. They correspond to natural topic breaks within the Achievement Standard.

Using This Resource

Test

Introduction

• These enable you to practise your NCEA exam skills

• Hints help you to focus on what is important

• Includes a question based on key terms

• A check list of explanatorycriteriaachievementandnotes

• Your summar y will help you with the NCEA style question

Review

• NCEA activitiesclustersquestionsstyleconcludeofrelated

The outline of the chapter structure below will help you to navigate through the material in each chapter.

The Structure of Membranes

Key terms and ideas

• Questions review the content of the page

• Create your own summar y for review

BIOZONE's NCEA Level 2 Biology Externals contains material to meet the needs of New Zealand students studying NCEA Biology Level 2 External Achievement Standards. The NCEA Level 2 Biology Externals is compliant with Level 7 of the NZ Curriculum (Nature of Science – The Living World) and the NCEA Biology Level 2 External Achievement Standards. A wide range of activities will help you to build on what you already know, explore new topics, work collaboratively, and apply your understanding. We hope that you find this resource useful and that you make full use of its features.

• The KEY IDEA provides your focus for the activity

• A list of key terms

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.

interior

The Structure of Membranes9 hydrophobic of the membrane so that water soluble molecules can pass by facilitated diffusion.

Biological ideas relating to life processes Activity number Select biological ideas relating to each of life process from... c Reasons for similarities and differences between plant and animal cells such as size and shape, and type and number of organelles present. 1 - 9 c ii Movement of materials, including diffusion, osmosis, and active transport. 9 - 18 c iii Enzyme activity (although specific names of enzymes is not required). 22 - 28 c iv Factors affecting the process (e.g. photosynthesis, cellular respiration, cell 14 transport processes, enzyme activity, cell division). 26 - 28 33 40 41 53 c Details of the processes as they relate to the overall functioning of the 30 - 32 cell (although the names of specific stages are not required). 38 -39 47 48 51 54 Explanatory notes: Life processes at the cellular level Activity number Life processes for plant and animal cells to include the following... c 1 Photosynthesis: The process that converts light energy into chemical energy. 36 - 42 c 2 Cellular respiration: The oxidation of complex organic substances to produce usable energy as ATP. 29 - 35 c 3 Cell division: DNA replication and mitosis as part of the cell cycle. 47

E Demonstrate comprehensive understanding of life processes at the cellular level: Link biological ideas about life processes at the cellular level. The discussion may involve justifying, relating, evaluating, comparing and contrasting, or analysing.

1. Describe the fluid mosaic model of membrane structure: 2. (a) What are the two main roles of the plasma membrane? (b) How is its structure related to these roles? 3. On the diagram (right) label the hydrophobic and hydrophilic ends of the phospholipid and indicate which end is attracted to water: LINK 6 WEB 9 LINK 14 Achievement Standard 2.4 Key terms Cell semi-conservativemitosisDNAcytokinesischromosomecellCellphotosynthesismitochondrionmetabolismmetabolicphaselightphaselightKrebsglycolysisfermentationenzymechainelectrondenaturationchloroplastcellularcatalystCalvinATPanaerobicaerobicactiveEnzymesratiosurfaceplasmapassivepartiallyosmosisionexocytosisendocytosisdiffusionactivematerialsMovementorganelleeukaryoticstructureoftransportpumppermeabletransportmembranearea:volumeandenergysitecyclerespirationtransportcycledependentindependentpathwaydivisioncyclereplication Life

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.

Look out for these features and know how to use them:

Phospholipids naturally form a bilayer in aqueous solutions.(watery) Cholesterol molecule maintains membrane integrity, preventing it becoming too fluid or too firm. CO2 H2O Glucose Na Phosphate head is (waterisFatty(waterhydrophilicloving)acidtailhydrophobichating) Water molecules pass freely between the phospholipid molecules by osmosis. Lipid soluble molecules, e.g. gases and steroids, can move through the membrane by diffusion, down their concentration gradient. Carrier proteins permit the passage of specific molecules by facilitated diffusion or active transport. Ion pumps are a type of carrier protein. Channel proteins (including ion channels) form pores through the

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.

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.

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.

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.

f The plasma membrane is a dynamic structure and is actively involved in cellular activities. It is selectively permeable, allowing some molecules, but not others, to pass. Proteins in the membrane enable the cell to regulate the movement of materials into and out of the cell. This enables the cell to obtain what it needs for its metabolism. processes at the cellular level -

Key Idea: The plasma membrane encloses the cell and regulates the entry and exit of substances into the cell. It is a phospholipid bilayer with embedded proteins moving freely within it.

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.

f The cell surface (or plasma) membrane encloses the cell's contents. The fluid-mosaic model of membrane structure describes a phospholipid bilayer within which various proteins can move about freely.

M Demonstrate in-depth understanding of life processes at the cellular level: Use biological ideas to give reasons how or why life processes occur the cellular level.

55 Plant and animal cells are eukaryotic cells. They have a number of features in common but also several distinguishing features. Cells exchange substances with their environment to maintain the reactions of life (metabolism). These reactions are catalysed by enzymes. Achievement criteria and explanatory notes Achievement criteria for achieved, merit, and excellence c A Demonstrate understanding of life processes at the cellular level: Define and use annotated diagrams or models to describe life processes at the cellular level. Describe characteristics of, or provide an account of, life processes at the cellular level.

2. (a) What are the two main roles of the plasma membrane? (b) How is its structure related to these roles?

Bookmark weblinks by typing in the address: it is not accessible directly from BIOZONE's website

Link

Weblinks

Using the Tab System

Corrections and clarifications to current editions are always posted on the weblinks page

Chapter in the in the book

KNOW = content you need to know

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 earlier as a reminder for important terms that have already been defined. In the example below for the activity "The Structure of Membranes", the weblink 9 provides information about plasma membranes. Activity 6 directs back to an overview of cell processes. The weblinks code is always the same as 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.

Activities are coded

PRAC = a paper practical or a practical focus

DATA = data handling and interpretation

Activitybook

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.

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

Bookmark the weblinks activityAccessNZL2E-9513www.biozone.co.nz/weblink/page:theexternalURLforthebyclickingthelink

www.biozone.co.nz/weblink/NZL2E-9513

REVISE = review the material in the section

Hyperlink to the external website page.

LINKWEB LINK

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

TEST = test your understanding KNOW

Generally the question is designed to require an open answer (meaning there is no definitive answer) in which you can demonstrate your level of understanding. The question may give you guidance as to what you should include in your answer, such as definitions of certain terms or to provide examples of particular phenomena.

In order to gain the highest possible mar k in these questions you need to lay out your answer in a clear and logical way, so that the examiner can easily see how you have demonstrated your understanding of the topic.

How mutationspoint affect the DNA explained.sequence

ATT GGC GCG CTT ATT TGT G Insert G between CC cause aa1 aa2 aa10 aa11 aa 12 aa13 frame shift to the right

A mutation is a change to the base pairs in a DNA sequence. Point mutations occur when just one base pair is affected.

• Linking biological ideas, compar ing and contrasting, analysing, or justifying ideas is excellence level.

How mutationspoint cause reading frame shifts.

The effect of reading frame shifts on the amino acid sequence

The external NCEA exams require you to demonstrate your understanding of a particular concept by providing a wr itten paragraph or essay.

How to Scaffold an NCEA Style Answer

The following example question shows how an answer can be built up from a simple definition, through explanation, to comparisons and linking of ideas.

Point mutations to the DNA sequence may be insertions, substitutions, or deletions of a base pair. Insertions and deletions cause reading frame shifts in the mRNA which result in changes to the amino acid sequence downstream of where the mutation occurred. Substitutions, the replacement of one base pair with another, can result in a change to just one amino acid.

Mutations in DNA affect proteins. Discuss the effect of point mutations on proteins, demonstrating how they change the DNA sequence and comparing the effect of changes:

• Explaining how a process works, why it works, and how it changes to it may affect an outcome is merit level.

ATT GGC AGC TTA TTT GTG Substitution of second C for A aa 1 aa2 aa14 aa4 aa5 aa6

ATT GGC GCT TAT TTG TG Second C deleted causes frame shift aa1 aa2 aa7 aa8 aa9 to the left

ATT GGC CGC TTA TTT GTG original strand aa 1 aa2 aa3 aa4 aa5 aa6

Link amino acid sequence to DNA (and sequence.mRNA)

The effect of these changes to the protein depends on where the mutation occurred and whether it caused a reading frame shift. Insertions and deletions add or remove a base causing all other base pairs to shift up or down one place e.g:

Because a base substitution affects only one codon (one amino acid) and many codons may code for the same amino acid, a base substitution may not affect the protein depending on the nature of the substitution (i.e the change may be silent).

• Defining, drawing, annotating, or giving a description of a process is achievement level only.

The amino acid sequence determines the way the protein folds up. Changes to the sequence affect the final structure and functioning of the protein.

Link structureproteinto amino acid sequence.

The difference between you obtaining an achievement, merit, or excellence grade depends on how well you demonstrated your understanding of a concept.

Definition of mutation provided. Point mutation defined.

Select biological ideas relating to each of life process from...

PREVIEWONLYNotforClassroomUseNotforClassroomUse

c iii Enzyme activity (although specific names of enzymes is not required). 22 28

Life processes for plant and animal cells to include the following...

Life processes at the cellular level

Biological ideas relating to life processes numberActivity

c 2 Cellular respiration: The oxidation of complex organic substances to produce usable energy as ATP. 29 35

Plant and animal cells are eukaryotic cells. They have a number of features in common but also several distinguishing features. Cells exchange substances with their environment to maintain the reactions of life (metabolism). These reactions are catalysed by enzymes.

Achievement criteria for achieved, merit, and excellence

c v Details of the processes as they relate to the overall functioning of the 30 32 cell (although the names of specific stages are not required). 38 -39 47 48 52 53

c M Demonstrate in-depth understanding of life processes at the cellular level: Use biological ideas to give reasons how or why life processes occur at the cellular level.

c i Reasons for similarities and differences between plant and animal cells such as size and shape, and type and number of organelles present. 1 - 9

c 3 Cell division: DNA replication and mitosis as part of the cell cycle. 47 55

c E Demonstrate comprehensive understanding of life processes at the cellular level: Link biological ideas about life processes at the cellular level. The discussion may involve justifying, relating, evaluating, comparing and contrasting, or analysing.

c iv Factors affecting the process (e.g. photosynthesis, cellular respiration, cell 14 transport processes, enzyme activity, cell division). 28 33 41 50

c ii Movement of materials, including diffusion, osmosis, and active transport. 9 18

Explanatory notes: Life processes at the cellular level numberActivity

Achievement criteria and explanatory notes

c A Demonstrate understanding of life processes at the cellular level: Define and use annotated diagrams or models to describe life processes at the cellular level. Describe characteristics of, or provide an account of, life processes at the cellular level.

Achievement Standard 2.4 Key terms Cell semi-conservativemitosisDNAcytokinesischromosomecellCellphotosynthesismitochondrionmetabolismmetabolicphaselightphaselightKrebsglycolysisfermentationenzymechainelectrondenaturationchloroplastcellularcatalystCalvinATPanaerobicaerobicactiveEnzymesratiosurfaceplasmapassivepartiallyosmosisionexocytosisendocytosisdiffusionactivematerialsMovementorganelleeukaryoticstructureoftransportpumppermeabletransportmembranearea:volumeandenergysitecyclerespirationtransportcycledependentindependentpathwaydivisioncyclereplication

c 1 Photosynthesis: The process that converts light energy into chemical energy. 36 42

c Define active transport and explain why energy is needed to move molecules and ions against their concentration gradient.

Describe the structure of cellular membranes in relation to the movement of substances within and between cells. Define selectively permeable.

Structure of plant and animal cells

c Identify phases in the cell cycle and explain the importance of each stage.

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

Movement of materials

c Describe DNA replication and explain its role in preparing the cell for division.

EIIJeffMnolfPodos What you need to know for this Achievement Standard PREVIEWONLYNotforClassroomUseNotforClassroomUse

c Describe and explain enzyme denaturation.

c Describe and explain the fixation of carbon in green plants by photosynthesis, including the inputs and outputs of the light dependent and light independent phases.

c Use examples to explain how enzymes regulate sequential steps in metabolic pathways.

Define the term concentration gradient and explain its significance to the movement of materials within and between cells.

c Explain the movement of materials by active transport mechanisms including ion pumps, endocytosis, and exocytosis.

Enzymes and energy transformations

c Describe and explain the breakdown of glucose by cellular respiration including glycolysis, the Krebs cycle, and the electron transport chain.

Recognise different types of specialised plant and animal cells and explain how the cell's features relate to its functional role.

c Describe mitosis and cytokinesis, including the cellular outcome and the behaviour of the chromosomes. What factors determine whether or not a cell enters mitosis?

Activities 47 - 58

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

c Describe ATP production by fermentation when oxygen is absent.

c Explain the role of enzymes in DNA replication and the significance of DNA's anti-parallel nature.

Describe the structure of a plant cell and an animal cell.

Activities 22 46

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

c Describe how enzymes work, including reference to the active site and activation energy.

Identify, describe, and give reasons for similarities and differences between plant and animal cells. Include reference to cell size and shape and number and type of organelles present.

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

c Describe factors that affect the rate of photosynthesis.

Cell division

Activities 10 - 21

c Describe factors affecting the activity of enzymes (and therefore reactions they catalyse). Include reference to substrate concentration, enzyme concentration, pH, and temperature.

Activities 1 9, 19 21

c Compare and contrast cellular respiration and photosynthesis as energy transformation processes. Explain the universal role of ATP in cells.

c Describe the role of mitosis in organisms.

c Define the term passive transport. Explain the movement of materials by diffusion, including facilitated diffusion and the role of membrane proteins in moving material across membranes.

c Explain the movement of water by osmosis and explain its significance in terms of the tonicity of the cell.

Cell cytoskeleton

Amyloplast

Features of a plant cell

Cytoplasm

Convert chemical energy into ATP.

(located within cytoplasm) provides structure and shape to a cell, is responsible for cell movement and provides intracellular transport of organelles and other structures.

f Cellulose cell wall

Plasma membrane

Ribosomes

Key Idea: Plant cells are eukaryotic cells. They share many features in common with animal cells, but they also have several unique features.

Specialised plastids containing the green pigment chlorophyll. They are the sites for photosynthesis.

Tonoplast: A special surroundingmembranethevacuole.

The Structure of Plant Cells1 LINK 2 LINK 3 WEB 1

Nuclear membrane

Large central vacuole: Plant vacuoles contain cell sap; an aqueous solution of dissolved food material, ions, waste products, and pigments. Functions include storage of water and ions, waste disposal, and growth.

Plant cells share many structures and organelles in common with animal cells, but they also have several features not seen in animal cells. Features which can be used to identify a plant cell include the presence of:

f Chloroplast

GolgiNucleolusapparatus

Cellulose cell wall

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

Chloroplast

Plant cells are eukaryotic cells. Features which identify plant cells as eukaryotic cells include:

Nucleus contains most of a cell's DNA.

Stores, modifies, and packages proteins.

PREVIEWONLYNotforClassroomUseNotforClassroomUse

A semi-rigid structure outside the plasma membrane, made mainly from cellulose. It protects the cell, maintains its shape, and prevents excessive water uptake. It provides rigidity to plant structures but still allows materials to pass into and out of the cell.

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

f Amyloplast

Located inside the cell wall in plants. Controls the movement of substances in and out of the cell.

f Presence of membrane-bound organelles (e.g. mitochondrion, Golgi apparatus, endoplasmic reticulum).

Manufacture proteins.

f Presence of a membrane-bound nucleus.

Mitochondrion

Nuclear pore

reticulumEndoplasmic(ER)

f Large vacuole (often centrally located)

Specialised plastids which synthesise and store starch.

CC4Capkuckokos

(b) What are its roles?

(a) Chloroplasts:

Chloroplasts are found in cells in the green parts of the plant, such as the leaves and sometimes the stem. These parts are exposed to light and are photosynthetic. In leaves, they are found in palisade and spongy mesophyll cells.

Plant cells are surrounded by cellulose cell walls. The cellulose supports the cell (and the plant). Cellulose is a polysaccharide, made up of repeating glucose units. The cell wall also contains the polymer lignin, especially in woody parts of the plant.

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

5. What is the function of cellulose and lignin?

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Mnolf

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

4. (a) In which parts of the plant are chloroplasts found?

(b) Plasma membrane:

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

Plants have different types of plastids. They have roles in storing fats, protein, starch (amyloplasts), pigments, and tannins, as well as carrying out photosynthesis (chloroplasts).

(b) Why are they found there?

Amyloplasts (above) and chloroplasts are types of organelles called plastids

6. Briefly describe the functions of the following:

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5. Identify two structures in the cell above that are associated with storage. Describe what they store and any other roles:

(b) How many of these organelles are present in the labelled cell above?

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1. Study the diagram of a plant cell in the previous activity. Identify and label the ten structures in the transmission electron micrograph (TEM) of the cell above. Use the following list of terms to help you: nuclear membrane, cytoplasm, endoplasmic reticulum, mitochondrion, starch granule, nucleus, vacuole, plasma membrane, cell wall, chloroplast.

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4. (a) Name the organelle where photosynthesis occurs:

(b)(a)

3. Describe the features that identify this cell as a plant cell:

Identifying Structures in Plant Cells

Key Idea: The position and appearance of the organelles in an electron micrograph can be used to identify them.

2. State how many cells, or parts of cells, are visible in the electron micrograph above:

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

The structure of a relatively unspecialised animal cell (a liver cell)

Cytoplasm

The Structure of Animal Cells

Mitochondria

KNOW 6

Microtubular structures associated with nuclear division.

A hole in the nuclear membrane. It ofthecommunicationallowsbetweennucleusandtherestthecell.

f Irregular shape

is ER an interconnected network of membrane with ribosomes attached to its surface. Proteins destined for transport outside of the cell are made here.

Ribosomes

Nuclear pore

Rough reticulumendoplasmic(roughER)

A large containingorganellemostof the cell’s DNA. Within the nucleus, is the nucleolus (n). It is involved in ribosome synthesis.

f Presence of a membrane-bound nucleus.

Lysosome

An interconnected network of membranes. A site for lipid and hormonemetabolism,carbohydrateincludingsynthesis.

f Lack of chloroplasts

Animal cells are eukaryotic cells. Features that identify animal cells as eukaryotic cells include:

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

Smooth reticulumendoplasmic(smoothER)

These contain enzymes that break down foreign material.

f Presence of membrane-bound organelles (e.g. mitochondria, Golgi apparatus, endoplasmic reticulum, lysosomes).

organellesOval-shapedbounded by a double membrane.

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Small projections which increase the surface area for absorption (not all animal cells have these).

Nucleus

f Lack of a cell wall

Key Idea: Animal cells are eukaryotic cells. They have features in common with plant cells, but they also have several unique features.

Golgi apparatus

f Centrioles

Features of an animal cell

Animal cells share many of the same structures and organelles that plant cells have, but they also have several features not seen in plant cells. Features which can be used to identify an animal cell include:

Microvilli

Centrioles

These manufacture proteins. They may be free in the cytoplasm or associated with the surface of the endoplasmic reticulum.

Phospholipid bilayer with associated proteins and lipids. Controls the movement of substances in and out of the cell.

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Plasma membrane

C:B:A:

SEM: Skin cells

SEM: Blood cells

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

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

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

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

Many animal cells are specialised to carry out specific functions within the body. As a result, the morphology and physiology of animal cells are highly varied. Some examples are presented here.

3 Name and describe a structure or organelle present in generalised animal cells but absent from plant cells: 4. Briefly describe the function of the following: (a) Microvilli: (b) Ribosomes: (c) Golgi apparatus: (d) Rough endoplasmic reticulum (rER): EIIPhotos: White blood cells and red blood cells (blood smear) Neurones (ner ve cells) in the spinal cord A B C PREVIEWONLYNotforClassroomUseNotforClassroomUse

Nerve cell

SEM: Egg cell

2. What features on the cell above identify it as an animal cell?

3. (a) Where is the plasma membrane located on an animal cell?

(b) What is the function of the plasma membrane?

4. The cell above contains a large amount of rough ER. What does this tell you about the cell? (h)(g)(f)(e)(d)(c)(b)(a)

Key Idea: The number and types of organelles present in a cell can reflect the cell's function. Organelles seen in an electron micrograph can vary in appearance depending on position.

Identifying Structures in Animal Cells

1. Study the diagram of an animal cell in the previous activity to become familiar with the features of an animal cell. Use your knowledge to identify and label the structures in the transmission electron micrograph (TEM) of the cell above. Use the following list of terms to help you: cytoplasm, plasma membrane, rough endoplasmic reticulum, mitochondrion, nucleus, centriole, Golgi apparatus, lysosome.

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KNOW 8

When the cells are turgid (tight), the uneven thickening makes the cells bend, opening the stoma (leaf pore).

Tube produced by the pollen. Sperm cell travel down the tube to the egg cell.

Cell type: Xylem vessel Function: Transports water and minerals

Open pore

1. For each of the cells (b) to (f) pictured above, describe how their structure relates to their function:

LigninPerforations

Relates to Function

Cell type: Sieve tube member of phloem Function: Transports sap

Cell type: Pair of guard cells Function: Open and close stoma, regulating entry and exit of gases from the leaf.

Tube nucleuscell

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(f)(e)(d)(c)(b)(a)

Guard cells: Curved, unevenly thickened cells. When the cells are turgid, the thickening makes them bend to open the stoma. When flaccid, the pore is closed. This way, the cells regulate the entry and exit of gases from the leaf.

StructureProhibited

Cell type: Pollen tube cell Function: Enables the sperm cell to reach the egg cell in flowers

Perforated cell end movementallows of sap.

Elongated cellnearChloroplastscell.locatedouteredgeoftogatherlight.VacuoleChloroplast

Cell wall waterprooflackscuticleHighsurface areaVacuole Nucleus

Examples of specialised plant cells

Cell type: Root hair cell Function: Absorbs water and nutrients

Cell type: Palisade mesophyll cell Function: Photosynthesis

Pollen grain Sperm cellsSievecellCompanion

Key Idea: Eukaryotic cells come in a wide range of shapes and sizes. Each cell type has specific features that enable it to fulfil its particular functional role in the organism.

cellparenchymaPhloemmembertube

Elongated, End(strengthened)lignifiedcells.ofcellperforated to allow movement of water.

thickeningUneven

Produce microscopic hairs that increase surface area for absorbing water and mineral ions.

membranesReceptor with light pigmentssensitive Midpiece Tail

Cell type: Muscle cell (fibre)

Cell type: White blood cell (neutrophil)

Neutrophil. It is spheroid but can change shape (amoeba-like) which enables it to engulf bacteria and other foreign material and perform its role in defence.

Function: Detects and responds to light

Head

Cilia wave in a way that moves mucus and debris up the trachea to the throat.

Cell type: Red blood cell (erythrocyte)

Function: Internal defence

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ISBN: 978-1-927309-51-3

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Cell type: Sper m cell

Function: Contracts to create movement

Myofibrils of muscle fibre

Consists of a head carrying the genetic material, a midpiece with many mitochondria, and a tail for movement.

Cell type: Ciliated epithelial cells

Cell type: Retinal cell

NucleusBacteria

Nerve fibre

Large spheroid cell. Able to change shape and exhibit amoeba-like behaviour. Engulfs and destroys foreign material.

Function: Transfers genetic material to egg cell

endingsNerve

Function: Moves mucus and trapped debris out of airways

Cylindrical shape with banded myofibrils. Capable of contraction (shortening) to move bones at joints.

2. For each of the cells (b) to (f) pictured above, describe how their structure relates to their function: (f)(e)(d)(c)(b)(a)

Examples of specialised animal cells

Function: Carries oxygen in the blood and offloads it at the tissues

Long cell that receives light energy and converts it into electrical nerve signals.

Small, biconcave shape can squeeze through capillaries. No nucleus so the cell is packed with haemoglobin.

Cilia Contains fewNocarrymoleculeshaemoglobinmanytooxygennucleusandorganelles

Processes in an animal cell

Photosynthesis (plant cell)

Transport in and out of the cell Organelle involved: plasma membrane

f To function, cells must exchange materials with their environment. They must maintain the right balance of fluid and ions to maintain cell volume and carry out the reactions of life. They need to obtain the raw materials to build molecules, and they need to get rid of materials they don't want. Many processes in cells therefore involve the transport of materials into and out of the cell. The diagram below describes some of the processes in cells.

OrganelleCytosis involved: plasma membrane

Protein Organellessynthesisinvolved: nucleus, rough endoplasmic reticulum, free ribosomes. Genetic information in the nucleus is translated into proteins by ribosomes.

Can engulf solids or fluid to bring them into the cell (endocytosis) or fuse with the Golgi secretory vesicles to expel material from the cell (exocytosis).

Simple diffusion and active transport move substances across the plasma membrane.

978-1-927309-51-3 Photocopying Prohibited

f The cell can be compared to a factory with an assembly line. Organelles in the cell provide the equivalent of the power supply, assembly line, packaging department, repair and maintenance, transport system, and the control centre. The sum total of all the processes occurring in a cell is known as metabolism

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Organelle involved: chloroplast Chloroplasts capture light energy and convert it into useful chemical energy (as sugars).

KNOW 11

Plant cells carry out photosynthesis

ISBN:

Cell Processes

OrganelleAutolysisinvolved: lysosome

Destroys unwanted cell organelles and foreign material.

OrganellesSecretion involved: Golgi apparatus, plasma membrane

Cellular Organellesrespirationinvolved : cytoplasm, mitochondria

Centrioles are microtubular structures that are involved in key stages of cell division. They are part of a larger organelle called the centrosome. The centrosomes of higher plant cells lack centrioles.

Glucose is broken down, supplying the cell with energy to carry out the many other reactions involved in metabolism.

Key Idea: Cells perform the processes essential to life. Each cellular organelle carries out one or more processes that contribute to the functioning of the cell as a whole.

Chloroplast

Cell Organellesdivisioninvolved: nucleus, centrioles

The Golgi produces secretory vesicles (small membrane-bound sacs) that are used to modify and move substances around and export them from the cell (e.g. hormones, digestive enzymes).

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Membranes in cells

is a cell like an assembly line in a factory?

The nucleus is surrounded by a doublemembrane structure called the nuclear envelope, which forms a separate compartment containing the cell's genetic material (DNA).

Explain what is meant by metabolism and describe an example of a metabolic process:

1. For each of the processes listed below, identify the organelles or structures associated with that process (there may be more than one associated with a process):

(h) Transpor t in/out of cell:

(a) Secretion:

2. In what way

The inner membrane of a mitochondrion provides attachments for enzymes involved in cellular respiration. It allows ion gradients to be produced that can be used in the production of ATP.

3.0CCItayba

Identify two examples of intracellular membranes and describe their functions: (b)(a) PREVIEWONLYNotforClassroomUseNotforClassroomUse

The Golgi apparatus is a specialised membrane-bound organelle which compartmentalises the modification, packing, and secretion of substances such as proteins and hormones.

(b) Respiration: (c) Endocytosis: (d) Protein synthesis: (e) Photosynthesis: (f) Cell division: (g) Autolysis:

Identifying Organelles in Micrographs Idea: Cellular organelles be identified in electron micrographs by their specific features.

can

The photographs on this page were taken using a transmission electron microscope (TEM). They show the ultrastructure of some organelles. Use the information on the previous pages to identify the organelles and help answer the following questions.

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(b) Which kind of cell(s) would this organelle be found in?

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

(b) Which kind of cell(s) would this organelle be found in?

1. (a) Identify this organelle (arrowed):

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

(b) Describe the function of this organelle:

7 Key

2. (a) Name the circled organelle:

(b) Which kind of cell(s) would this organelle be found in?

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

KNOW 14 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited Key Idea: Cells contain a variety of organelles that carr y out specialised functions within the cell. Not all cell types contain every type of organelle. Cell Structures and Organelles8 Transport pathwayRibosomes Flattened membrane sacs Smooth vesiclesBuddingRough (c) Function:Location:Name: Plant cell Lipid bilayer: a double layer of phospholipids Protein (b) Function:Location:Name: subunitLargesubunitSmall (a) Function:Location:Name: Plasma EnclosesSurroundsmembranethecellcellcontents and regulates movement of substances into and out of cell. (d) FunctionLocation:Name: of smooth ER: Function of rough ER: Smooth & rough endoplasmic reticulum Penetrates the whole cytoplasm LINK 1 LINK 3 PREVIEWONLYNotforClassroomUseNotforClassroomUse

15 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited (f) Function:Location:Name: PoresGenetic material Nucleolus Double membrane PectinsHemicellulosesCellulose fibres Middle lamella Lamellae OuterInnermembranemembraneStacks membranesofStroma(fluid) membraneInnermembranesFoldedmembraneOuterinner (fluid)Matrix (g) Function:Location:Name: Cisternae Secretory vesicle budding off the trans face Transfer vesicles enter from the endoplasmicsmoothreticulum (e) Function:Location:Name: Golgi Cytoplasmapparatusassociated with smooth ER (h) Function:Location:Name: Within the cytoplasm (i) Function:Location:Name: Cellulose cell wall PREVIEWONLYNotforClassroomUseNotforClassroomUse

Phospholipids naturally form a bilayer in aqueous solutions.(watery)

Channel proteins (including ion channels) form pores through the hydrophobic interior of the membrane so that water soluble molecules can pass by facilitated diffusion.

Glucose

Key Idea: The plasma membrane encloses the cell and regulates the entry and exit of substances into the cell. It is a phospholipid bilayer with embedded proteins moving freely within it.

roles of

The Structure of Membranes9

2. are the two main the plasma membrane?

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

(waterisFatty(waterhydrophilicloving)acidtailhydrophobichating)

1. Describe the fluid mosaic model of membrane structure:

How is

Na+

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

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Cholesterol molecule maintains membrane integrity, preventing it becoming too fluid or too firm. CO2 2O

(a) What

Water molecules pass freely between the phospholipid molecules by osmosis.

Phosphate head is

Carrier proteins permit the passage of specific molecules by facilitated diffusion or active transport. Ion pumps are a type of carrier protein.

(b) its structure

related to these roles?

Small membrane.throughdiffusemoleculesunchargedcaneasilythe

Large polar molecules cannot directly cross the membrane. Transport (by facilitated diffusion or active transport) involves carrier proteins.

Lipid unimpeded.theintomoleculessolublediffuseandoutofmembrane

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of water movement.

17 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited CO2 Benzene EthanolH2O Glucose Cl Ca2+ Na+ O2 What can cross a lipid bilayer? Gases Hydrophobicmolecules Small moleculespolar Largemoleculespolar moleculesCharged

7. Explain why ions have to pass through a protein channel in the lipid bilayer:

6. In the diagram above, identify the molecule(s) that:

4. What is the purpose of carrier proteins in the membrane?

Ions can transportedbe across the membrane via channel proteins, e.g. ion channels (passive) or ion pumps (active).

(a) Can diffuse through the plasma membrane on their own:

Small polar molecules are small enough to diffuse channels)Aquaporinsthrough.(waterincrease rate

5. What is the purpose of channel proteins in the membrane?

(b) Can diffuse through the membrane via channel proteins:

The surface area involved

Photocopying Prohibited

What is diffusion?

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

Glucose

f Most diffusion in biological systems occurs across membranes. Simple diffusion occurs directly across a membrane, whereas facilitated diffusion involves helper proteins. Neither requires the cell to expend energy.

KNOW 18

Sometimes the rate of diffusion across the membrane is too low to meet the cell's needs for a particular molecule.

Channel-mediated facilitated diffusion provides channels that allow inorganic ions to pass through the membrane by creating hydrophilic pores in the membrane, e.g. sodium ions entering nerve cells.

Barriersdiffusionto

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

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

Concentrationgradient

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

Carrier-mediated facilitated diffusion allows the transport of large lipid-insoluble molecules through the plasma membrane. Each carrier protein is specific to the molecule being transported. Carrier proteins allow molecules that cannot cross the membrane by simple diffusion to be transported into the cell. An example is the transport of glucose into red blood cells.

solutessolubleLipid

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

High concentration Low concentration gradient

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Facilitated diffusion by channels

Simple diffusion across membranes can occur by molecules moving directly through the membrane without any assistance e.g. at the alveolar surface of the lung, O2 diffuses into the blood and CO2 diffuses out.

ISBN: 978-1-927309-51-3

Factors affecting the rate of diffusion

Carrier protein

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

ionInorganicChannelprotein

Facilitated diffusion by carriers

Simple diffusion

Key Idea: Diffusion is the movement of molecules from high to low concentration. Diffusion through plasma membranes can be facilitated by various transport proteins.

Themoveddistance

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

Concentration

Diffusion in Cells

20°ofContainerwateratC

Soluble particles placed in at diameterParticleconcentrationhighwithof5nm

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1. What is diffusion?

5. Cells create and maintain concentration gradients to keep particles (ions etc.) moving in one direction. They do this by using molecules up or transporting them away. Use diagrams to help explain how a concentration gradient maintains movement in one direction and why net movement stops if the concentration gradient is lost:

3. Describe two proper ties of a membrane that would facilitate rapid diffusion:

Particle diameterwithof20 nm

After one hour:

2. Compare and contrast facilitated diffusion and simple diffusion across a membrane:

membranepermeablePartially with pores of 10 nm.

4. Consider the two diagrams below. For each, draw in the appropriate box what you would expect to see after one hour.

The presence of solutes (substances that dissolve) in a solution increases the tendency of water to move into that solution. This tendency is sometimes called the osmotic potential or osmotic pressure. The greater the solution's concentration the greater the osmotic potential. The osmotic potential is the pressure required to prevent the flow of water into the solution.

f When two solutions have the same solute concentration they are called isotonic Idea lower solute solute partially

Net water movement

f Hypotonic solutions have a lower solute concentration compared to another solution. A cell placed into a hypotonic solution will gain water and swell (possibly bursting).

f Any solution of water and a solute (e.g. glucose) has a concentration of free water molecules lower than that of pure water. Water always diffuses from regions of lower solute concentration (higher free water concentration) to regions of higher solute concentration (lower free water concentration).

: Osmosis is the diffusion of water molecules from a

permeable membrane. Osmosis in Cells11 WEB 11 LINK 9 LINK 10 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Tonicity is a measure of the difference between two solutions separated by a partially permeable membrane. A solution is described relative to another:

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concentration across a

Osmosis using dialysis tubing

moleculeGlucose

Osmotic potential

tubecapillaryGlass moleculeWater Water

Cells and tonicity

f A solution of glucose inside dialysis tubing and placed in a beaker of water (below left) will form an osmotic gradient. The glucose solution will gain water from the contents of the beaker. The dialysis tubing acts as a partially permeable membrane, allowing water to freely pass through while keeping the glucose inside the dialysis tubing.

Dialysis tubing (partially permeable membrane)

Dialysis glucosecontainingtubingsolution

f Hypertonic solutions have a higher solute concentration compared to another solution. A cell placed into a hypertonic solution will lose water via osmosis and shrink.

1. In the blue box below the diagram above, draw an arrow to show the direction of the net water movement. 2. (a) Study the diagram above. What will happen to the water level in the glass tube over time? (b) What would happen if a more concentrated solution of glucose was used? 3. Explain why you would see the results you predicted in question 2 (b): 4. What would happen to a cell if it was placed into an isotonic solution? Key

concentration to a higher

The cytoplasm has a higher concentsoluteration than outside. Water enters the cell, putting pressure on the cell wall.

Turgor in a plant cell

Cell contents less dilute than the environmentexternal

Osmosis and tonicity

Plasmolysis in a plant cell

1. Identify the outcome of the following situations:

(a) A plant cell is placed in a hypertonic solution:

2. What is the turgor pressure in a fully plasmolysed cell?

3. Why does a plant with flaccid cells wilt?

Water Relations in Plants

Cell wall bulges outward but prevents cell rupture Water Water Water

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Cell wall is freely permeable to water.

membranePlasmaCytoplasm Cell environmentmorecontentsdilutethantheexternal

The cytoplasm has a lower concentsoluteration than outside. Water leaves the cell.

f A plant cell provides support for a plant when water causes the cell contents to create pressure against the cell wall (turgor). Losing water reduces turgor and the plant wilts. Complete loss of turgor is irreversible. Water crosses the partially permeable plasma membrane of the cell by osmosis. When the external water concentration is the same as that of the cell there is no net movement of water. Two systems (cell and environment) with the same water concentration are termed isotonic. The diagram below illustrates two different situations: when a plant cell is in a hypertonic solution and when it is in a hypotonic solution.

Cell in hypertonic salt solution

In a hypotonic solution, the external water concentration is higher than the cell cytoplasm. Water enters the cell, causing it to swell tight. A wall (turgor) pressure is generated when the cell contents press against the cell wall. Turgor pressure increases until no more water enters the cell (the cell is turgid).

Plant cells are turgid

(b) A plant cell is placed in a hypotonic solution:

Key Idea: Plant cells in a hypertonic solution lose water and undergo plasmolysis. In a hypotonic solution they gain water, creating the turgor pressure that helps support the plant.

Wilted plant (cells have lost turgor)

In a hypertonic solution, the external water concentration is lower than the water concentration of the cell. Water leaves the cell and, because the cell wall is rigid, the cell membrane shrinks away from the cell wall. This is called plasmolysis and the cell becomes flaccid.

WaterWater Water Water Water

Cell in pure water (hypotonic)

Single-celled organisms

A specialised gas exchange surface (lungs) and circulatory (blood) system are required to transport substances to the body's cells.

PRAC 22 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited Diffusion in organisms of different sizes 1. Calculate the volume, surface area and the ratio of surface area to volume for each of the four cubes above (the first has been done for you). When completing the table below, show your calculations. 2 cm cube3 cm cube 4 cm cube 5 cm cube Cube size Surface area Volume Surface area to volume ratio 2 cm cube 2 x 2 x 6 = 24 cm2 (2 cm x 2 cm x 6 sides) 2 x 2 x 2 = 8 cm3 (height x width x depth) 24 to 8 = 3:1 3 cm cube 4 cm cube 5 cm cube Key Idea: Diffusion is less efficient in cells with a small surface area relative to their volume than in ones with a large surface area relative to their volume. Diffusion and Cell Size13

In a multicellular organism, such as an elephant, the body's need for respiratory gases cannot be met by diffusion through the skin.

Multicellular organisms (e.g. plants and animals) are often very large, and larger organisms have a smaller surface area compared to their volume. They require specialised systems to transport the materials they need to and from the cells and tissues in their body.

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

Single-celled organisms (e.g. Amoeba), are small and have a large surface area relative to the cell’s volume. The cell's requirements can be met by the diffusion or active transport of materials into and out of the cell (below).

Multicellular organisms

OxygenFood

WastesdioxideCarbon

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cell of 2 cm x 2 cm x 2 cm is less efficient at passively acquiring nutrients than 8 cells of 1 cm x 1 cm x 1 cm: Cube 123 1. Total volume 2.(cm3)Volume not pink (cm3) 3. (1.volumeDiffused–2.)(cm3)4.Percentagediffusion solutionNaOH Region of no colour change 4 cm 2 cm Cube 2 Cube 1 Cube 3 Cubes shown to same scale 1 cm Region of colour change Agar phenolphthaleininfusedcubeswith PREVIEWONLYNotforClassroomUseNotforClassroomUse

5. The diffusion of molecules into a cell can be modelled by using agar cubes infused with phenolphthalein indicator and soaked in sodium hydroxide (NaOH). Phenolphthalein turns pink in the presence of a base. As the NaOH diffuses into the agar, the phenolphthalein changes to pink and indicates how far the NaOH has diffused into the agar. It is possible to show the effect of cell size on diffusion by cutting an agar block into cubes of various sizes.

4. Explain what happens to the ratio of surface area to volume with increasing size.

(b) Diffusion of substances into and out of a cell occurs across the plasma membrane. For a cuboid cell, explain how increasing cell size affects the ability of diffusion to provide the materials required by the cell:

(a) Use the information below to fill in the table on the right:

3. Which increases the fastest with increasing size: the volume or the surface area?

6. Explain why a

2. Use the information you calculated in question 1 to create a graph of the surface area against the volume of each cube, on the grid on the right. Draw a line connecting the points and label axes and units.

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

The aim and hypothesis

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

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

Key Idea: High temperatures can disrupt the structure of cellular membranes and alter their permeability, making them leaky. 14 Absorbance of beetroot samples at varying temperatures Temperature(°C) Absorbance at 530 nm Mean ObservationSample 1Sample 2Sample 3 0 No colour00.0070.004 20 Very pale pink0.0270.0220.018 40 Very pale pink0.0960.1140.114 60 Pink 0.5800.5240.509 90 Red 3 3 3 cubesBeetroot

Temperature and Membrane Permeability

Experimental method

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

Membrane permeability can be disrupted if membranes are subjected to high temperatures. At temperatures above the optimum, the membrane proteins become denatured (they lose their structure). The denatured proteins no longer function properly and the membrane loses its selective permeability and becomes leaky.

Background

Raw beetroot was cut into uniform cubes using a cork borer with a 4 mm internal diameter. The cubes were trimmed to 20 mm lengths and placed in a beaker of distilled water for 30 minutes. 5 cm3 of distilled water was added to 15 clean test tubes. Three were placed into a beaker containing ice. These were the 0°C samples. Three test tubes were placed into water baths at 20, 40, 60, or 90°C and equilibrated for 30 minutes. Once the tubes were at temperature, the beetroot cubes were removed from the distilled water and blotted dry on a paper towel. One beetroot cube was added to each of the test tubes. After 30 minutes, they were removed. The colour of the solution in each test tube was observed by eye and then the absorbance of each sample was measured at 530 nm. Results are given in the table below.

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

(b) Contrast the two ways in which the energy can be supplied to do work:

f The energy for active transport comes from the molecule ATP (adenosine triphosphate). Energy is released when ATP is hydrolysed (water is added) forming ADP (adenosine diphosphate) and inorganic phosphate (Pi). The energy is used by a transport protein to move a target molecule across a membrane against its concentration gradient.

The molecule or ion is released and the transport protein reverts to its previous state.

f The work of active transport is performed by specific transport proteins. These proteins move molecules across a membrane from a lower to a higher concentration. ATP may be used directly by a transport protein to move a molecule or ion across a membrane (below and top right), or it may be used indirectly to create a concentration gradient, which can be then used to couple the passive movement of one molecule to the movement of another molecule against its concentration gradient (right).

H2O proteinTransport Molecule to be transported H OH 1 2 3 4 ATP ATP ATP and active transportADP P LINK 6 LINK 16 LINK 29 WEB 15 PREVIEWONLYNotforClassroomUseNotforClassroomUse

f Active transport is the energy-using process of moving molecules (or ions) against their concentration gradient.

ATP is hydrolysed and the energy released is coupled to the transport of the molecule or ion across the membrane.

2. (a) Where does the energy for active transport come from?

A ball falling is a passive process. Replacing the ball requires active energy input.

Active Transport15

Energy is needed to move an object over a physical barrier.

Active Active Passive

A molecule or ion to be transported binds to the transport protein.

Sometimes the energy of a passively moving object can be used to actively move another, e.g. a falling ball can be used to catapult another.

1. Define active transpor t:

ATP binds to a transport protein.

Key Idea: Active transport is a process that uses energy to transport molecules across the plasma membrane against their concentration gradient. The energy comes from ATP.

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Sodium-potassium (Na+/K+) Pump

Ion Pumps

f The sodium-potassium pump (below, left) is found in almost all animal cells and is also common in plant cells. The concentration gradient created by ion pumps is often coupled to the transport of other molecules, such as glucose or sucrose, across the membrane (below right).

f Membrane proteins have a role in moving molecules into and out of cells, either when they move down their concentration gradient by facilitated diffusion or when they are moved against their concentration gradient by active transport.

A specific carrier protein controls the entry of glucose into the intestinal epithelial cells from the gut where digestion is taking place. The energy for this is provided indirectly by a gradient in sodium ions. The carrier 'couples' the return of Na+ down its concentration gradient to the transport of glucose into the cell. The process is therefore called cotransport. A low intracellular concentration of Na+ (and therefore the concentration gradient for transport) is maintained by a sodium-potassium pump.

Key Idea: Ion pumps are transmembrane proteins that directly or indirectly use energy to move ions and molecules across a plasma membrane against their concentration gradient.

Plasma membrane + ++++++++ Cell ExtracellularcytoplasmfluidorlumenofgutExtracellularfluidorlumenofgut Sodium-potassium pump Cotransport (the sodium-glucose symport) K+ K+ Na+ Na+ Na+ Na binding+siteK+bindingsite Na+ K+ ATP Glucose Diffusion of sodium ions Na+ Na+ Na+ Na+ Na+ proteinCarrier 3 Na+ pumpedareout of the cell for every 2 K+ pumped in 1. What is an ion pump? 2. Explain why ion pumps require energy to carry out their functional role: LINK 6 WEB 16 LINK 29 LINK 15 PREVIEWONLYNotforClassroomUseNotforClassroomUse

f Special membrane proteins, called ion pumps, directly or indirectly use energy to transport ions across the membrane against a concentration gradient.

The Na+/K+ pump is a protein in the membrane that uses energy in the form of ATP to exchange sodium ions (Na+) for potassium ions (K+) across the membrane. The unequal balance of Na+ and K+ across the membrane creates a large concentration gradient that can be used to drive transport of other substances (e.g. cotransport of glucose). The Na+/K+ pump also helps to maintain the right balance of ions and so helps regulate the cell's water balance.

f The transport of molecules against their concentration gradient requires an input of energy. This is called active transport.

Cotransport (coupled transport)

space Proton

3. Identify the energy sources for the three carrier proteins described in this activity:

6. In nerve cells, there is a difference in charge across the plasma membrane (the inside of the cell is less positive than the outside of the cell). Explain how the sodium-potassium pump achieves this difference in charge:

5. Describe the function of the proton pump in photosynthesis:

electronsarechloroplastsmembranesthylakoidofpoweredbypassed

(b) How is cotranspor t used to move glucose into the intestinal epithelial cells?

ThylakoidStroma pumps in the to them by chlorophyll.

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4 (a) Explain what is meant by cotranspor t:

+ + + + + + Cell Extracellularcytoplasmfluid H+ H+H+ H+ H+ ATP electronsor Hydrogen ion proteinCarrierPlasmamembrane Proton pump The proton (hydrogen) pump and its role in photosynthesis and respiration membraneThylakoid ADP ATP HH++ H+ChlorophyllChlorophyll Hydrogen pump ATP synthase

(a) Proton pump:

(b) Sodium-potassium pump:

Proton pumps create a potential difference across a membrane by using energy (ATP or electrons) to move H+ from one side of the membrane to the other. This difference can be coupled to the transport of other molecules. In cell respiration and the light reactions of photosynthesis (below), the energy for moving the H+ comes from electrons, and the flow of H+ back across the membrane drives the synthesis of ATP by a membrane-bound enzyme called ATP synthase.

Vesicle from the Golgi carrying molecules for export moves to the perimeter of the cell. 1 2 From Golgi apparatus

Exocytosis is important in the transport of neurotransmitters (NT) into the junction (synapse) between nerve cells to transmit nervous signals.

f Most cells carry out cytosis, which is a form of active transport involving the in- or outfolding of the plasma membrane. Cytosis is possible because the plasma membrane is flexible.

Cytosis

Plasma membrane

Fungi and bacteria use exocytosis to secrete digestive enzymes, which break down substances extracellularly so that nutrients can be absorbed (by endocytosis).

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Golgi apparatus forming vesicles Nerve cell

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f Cytosis results in the bulk transport of materials into or out of the cell. It is achieved through the localised activity of protein filaments (e.g. microtubules) in the cell cytoskeleton. It is this cytoskeletal movement that requires energy.

f Engulfment of material is termed endocytosis. Endocytosis typically occurs in protozoans and certain white blood cells of the mammalian defence system (e.g. neutrophils, macrophages).

f Exocytosis is the reverse of endocytosis and involves the release of material from vesicles or vacuoles that have fused with the plasma membrane. Exocytosis is typical of cells that export material (secretory cells and neurones).

1. What is the purpose of exocytosis?

Exocytosis

3 The contents of the vesicle are expelled into the extracellular space.

The transport of Golgi vesicles to the edge of the cell and their expulsion from the cell occurs through the activity of the cytoskeleton. This requires energy (ATP).

Key Idea: Endocytosis and exocytosis are active transport processes. Endocytosis involves the cell engulfing material. Exocytosis involves the cell expelling material.

Vesicle fuses with the membrane.plasma

Exocytosis (below) is an active transport process in which a secretory vesicle fuses with the plasma membrane and expels its contents into the extracellular space. In multicellular organisms, various types of cells (e.g. endocrine cells and neurones) are specialised to manufacture products, such as proteins, and then export them from the cell to elsewhere in the body or outside it.

The vesicle carries molecules into the cell. The contents may then be digested by enzymes delivered to the vacuole by lysosomes.

Pinocytosis (or ‘cell-drinking’) involves the non-specific uptake of liquids or fine suspensions into the cell to form small pinocytic vesicles. Pinocytosis is used primarily for absorbing extracellular fluid. Examples: Uptake in many protozoa, some liver cells, and some plant cells.

Plasma membrane

Endocytosis is a type of active transport in which the plasma membrane folds around a substance to transport it across the plasma membrane into the cell. As with exocytosis, the localised activity of the cytoskeleton is involved and it is possible because of the flexibility of the plasma membrane.

Describe two examples of exocytosis in cells: 3. Describe the following types of endocytosis: (a) Phagocytosis: (b) Receptor mediated endocytosis: (c) Pinocytosis:Exocytosisandendocytosisrequireenergybecausetheyinvolvemovementofcytoskeletalproteins. PREVIEWONLYNotforClassroomUseNotforClassroomUse

Phagocytosis (or ‘cell-eating’) involves the cell engulfing solid material to form large phagosomes or vacuoles (e.g. food vacuoles). It may be non-specific (above) or receptor-mediated (centre photo). Examples: Feeding in Amoeba, phagocytosis of foreign material and cell debris by neutrophils and macrophages.

Vesicle buds inwards from the plasma membrane

Material (solids or fluids) that are to be brought into the cell are engulfed by an infolding of the plasma membrane.

Endocytosis

Receptor mediated endocytosis is triggered when certain metabolites, hormones, or viral particles bind to specific receptor proteins on the membrane so that the material can be engulfed. Examples: The uptake of lipoproteins by mammalian cells and endocytosis of viruses (above).

CDC CollegeDartmouth CollegeDartmouthtobeginningandReceptorsparticleHIVpitform

EDGF

(d) Secretion of digestive enzymes from cells of the pancreas:

(e) Synthesis of ATP via membrane-bound ATP synthase:

Key

1. Identify each of the processes (A-G) described in the diagram above in the spaces provided. Indicate whether the transport process is active or passive by using A for active and P for passive.

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A B C

use

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Active and Passive Transport Summary Idea out of the cell either passive transport, which does not energy, or by active transport which requires energy, usually as ATP.

Diffusion involving a carrier system (channel proteins or carrier proteins) but without any energy expenditure.

A protein in the plasma membrane that uses energy (ATP) to exchange sodium for potassium ions (3 Na+ out for every 2 K+ in). The concentration gradient can be used to drive other active transport processes.

2. Identify the transport mechanism involved in each of the following processes in cells:

: Cells move materials into and

Fluid or a suspension is taken into the cell. The plasma membrane encloses some of the fluid to form a small vesicle, which then fuses with a lysosome and is broken down.

(a) Uptake of extracellular fluid by liver cells:

Vesicles bud off the Golgi or ER and fuse with the plasma membrane to expel their contents into the extracellular fluid.

A type of endocytosis in which solids are taken into the cell. The plasma membrane encloses one or more particles and buds off to form a vacuole. Lysosomes fuse with it to digest the contents.

3. In general terms describe the energy requirements of passive and active transport:

Molecules of liquids, dissolved solids, and gases move into or out of the cell without any expenditure of energy. These molecules move down their own concentration gradients.

Golgi

(b) Capture and destruction of a bacterial cell by a white blood cell:

Diffusion of water across a partially permeable membrane. It causes cells in fresh water to take up water.

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Cell structure and function

Active and passive transport processes

HINT: Include differences between plant and animal cells and features of specialised cells that help them carry out their function.

Organelle structure and function

HINT: Include the structure and function of the plasma membrane and important organelles such as mitochondria and chloroplasts,

HINT: Include reference to membrane proteins and their roles.

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 NCEA style essay question that follows. Use the points in the introduction and the hints provided to help you:

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NCEA Question: Cell

• discuss features unique to plant and animal cells

In your answer you should:

Style

• briefly describe and contrast the role of the plasma membrane and cell wall

• discuss features common to plant and animal cells

1. Plant and animal cells are eukaryotic cells, and share several common features. They also have features unique to each cell type. Compare and contrast the features of plant and animal cells. You may use extra paper if required:

(b) Active transpor t:

2. Organisms use a variety of transport processes to move materials in and out of the cell. These include osmosis and active Describetransport.theprocesses of osmosis and active transport.

(a) Osmosis:

3. Explain why osmosis and active transport are needed in a cell and describe examples of how they might be used in an organism. You may use extra paper if required:

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F The energy-requiring movement of substances across a biological membrane against a concentration gradient.

Cells are the basic...

Activemembrane...transport involves membrane proteins, which couple the energy provided by ATP...

H Passive movement of water molecules across a partially permeable membrane down a concentration gradient.

A phospholipid is made up of a...

A cell is enclosed by a plasma membrane...

Activemembrane...transport requires the input of energy...

1. Match each term to its definition, as identified by its preceding letter code.

2. (a) Identify organelle 1: (b) The organelle is found in a plant/animal cell/plant and animal cell (delete two).

(d) The organelle is found in a plant/animal cell/plant and animal cell (delete two).

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C The passive movement of molecules from high to low concentration.

Facilitated diffusion involves proteins in the plasma

Proteins are embedded...

E A nucleotide comprising a purine base, a pentose sugar, and three phosphate groups, which acts as the cell's energy carrier.

A A partially-permeable phospholipid bilayer with embedded proteins, which forms the boundary of all cells

Passive transport involves the movement of molecules

(b)(a)

B Movement of substances down a concentration gradient without energy expenditure.

D A transmembrane protein that moves ions across a plasma membrane against their concentration gradient.

1 2

...made of a phospholipid bilayer with embedded proteins ...some of which are composed of membranes.

3. Match the statements in the table below to form a complete paragraph. The left hand column is in the correct order, the right hand column is not.

....to the movement of molecules or ions against their concentration gradient.

...can be active or passive. ...directly across the lipid bilayer of the membrane ...which help molecules or ions to move through. ...whereas passive transport does not.

...such as photosynthesis or respiration. ...hydrophilic head and a hydrophobic tail. ...units of life. ...in the plasma membrane.

KEY TERMS AND IDEAS: Cell Structure & Transport active plasmapassiveosmosisorganelleionendocytosisdiffusion(ATP)adenosinetransporttriphosphatepumptransportmembrane

...high concentration to low concentration (down a concentration gradient)

Transport of molecules though the plasma

Simplefrom... diffusion can occur...

Photocopying Prohibited

Eukaryotic cells contain different types of organelle... Each organelle has a specific function in the cell...

G Active transpor t in which molecules are engulfed by the plasma membrane, forming a phagosome or food vacuole within the cell.

I A structural and functional par t of the cell, usually bound within its own membrane. Examples include the mitochondria and chloroplasts.

Energy

One large molecule

f The energy released from catabolic reactions can be used to drive other metabolic processes.

(a) Protein synthesis:

(d) DNA synthesis:

ReactionsProhibited Cells

Key Idea: Anabolic reactions build complex molecules and str uctures from simpler ones. Catabolic reactions break down larger molecules into smaller molecules.

2. (a) What is a catabolic reaction?

Energy

Many moleculessmall

(b) ATP conversion to ADP:

1. What is an anabolic reaction?

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3. Identify the following reactions as either catabolic or anabolic:

f There are two categories of metabolic reactions: anabolic and catabolic.

f Anabolic reactions are reactions that result in the production (synthesis) of a more complex molecule from smaller components or smaller molecules. During anabolic reactions, simple molecules are joined to form larger, more complex molecules.

f Metabolism refers to all of the chemical reactions carried out within a living organism to maintain life. All metabolic reactions are controlled by enzymes.

Plants carry Photosynthesischloroplastsorganellesphotosynthesisoutincalled(left).is an anabolic process because it converts carbon dioxide and water into glucose. Energy from the sun is required to drive photosynthesis.

Catabolic reactions

f Catabolic reactions are reactions that break down large molecules into smaller components.

f Catabolic reactions are the opposite of anabolic reactions.

f Anabolic reactions need a net input of energy to proceed. They are called endergonic reactions

Cellular respiration is a catabolic reaction. Glucose is broken down in a series of reactions and used to form to form ATP (energy) releasing carbon dioxide and water. The ATP is used to fuel other activities in the cell. Most of the reactions of cellular respiration take place in the mitochondria (left).

f Catabolic reactions involve a net release of energy. They are called exergonic reactions

Anabolic reactions

(b) Why are catabolic reactions considered to be opposite to anabolic reactions?

One large molecule

For a reaction to occur, the reactants must collide with sufficient speed and with the correct orientation. Enzymes enhance reaction rates by providing a site for reactants to come together in such a way that a reaction will occur. They do this by orientating the reactants so that the reactive regions are brought together. They may also destabilise the bonds within the reactants making it easier for a reaction to occur.

Substrates collide with an enzyme's active site

2H2O2 2H2O + O2

Amylase is a digestive enzyme produced in the salivary glands and pancreas in humans. However, it acts in the mouth and small intestine respectively to hydrolyse starch into sugars.

Trypsin is a protein-digesting enzyme. It is produced in an inactive form (called trypsinogen) and secreted by the pancreas into the lumen of the small intestine (where it will work). It is activated in the intestine by the enzyme enteropeptidase to form trypsin. Active trypsin can convert more trypsinogen to trypsin.

An extracellular enzyme is an enzyme that functions outside the cell from which it originates (i.e. it is produced in one location but active in another).

Examples: Amylase and trypsin.

The substrate is the chemical that an enzyme acts on. A specific enzyme acts on a specific substrate.

f Most enzymes are globular proteins. Enzymes are biological catalysts because they speed up biochemical reactions, but the enzyme itself remains unchanged. The substrate in a reaction binds to a region of the enzyme called the active site, which is formed by the folding of the enzyme's amino acid chain (its tertiary structure).

Many metabolic processes produce hydrogen peroxide, which is harmful to cells. Catalase converts hydrogen peroxide into water and oxygen (below) to prevent damage to cells and tissues.

Enzymes can be defined based on where they are produced relative to where they are active.

Example: Catalase.

1. What is an enzyme's active site and how it is it formed?

Incorrect orientationreactant=noreaction orientatesEnzyme the reactants making the morereactionlikely. X

Enzymes

Extremes of temperature or pH can alter the enzyme's active site and lead to loss of function. This is called denaturation

f Enzymes control metabolic pathways. One enzyme will act on a substance to produce the next reactant in a pathway, which will be acted on by a different enzyme.

f Enzymes are often named after their substrate and/or the type of reaction they catalyse together with the suffix -ase. For example, lipase breaks downs lipid molecules, glucose oxidase catalyses the oxidation of glucose.

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An intracellular enzyme is an enzyme that performs its function within the cell that produces it. Most enzymes are intracellular enzymes, e.g. respiratory enzymes.

Catalase

2. How do substrate molecules come into contact with an enzyme's active site?

Enzymes have an active site to which specific substrates bind. The shape and chemistry of the active site is specific to an enzyme, and is a function of the polypeptide's complex tertiary structure. The active site

Enzymes can be intracellular or extracellular

Key Idea: Enzymes are biological catalysts. The active site is critical to this functional role.

3. Why would a protein digesting enzyme such as tr ypsin need to be activated extracellularly (outside the cell)?

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3. Enzymes are not used up in the reactions that they catalyse. Why is this important?

Start Finish HighLow energyHigh

Transition(unstable)state

Energy barrier

Direction of reaction

storedenergyofAmount

1. (a) What is the transition state?

Product energyLowenergyLow

f An enzyme increases the rate of a reaction but is not used up by the reaction. Nor does the enzyme change the energies of the original reactants or the products.

(b) What is the activation energy of a chemical reaction?

Key Idea: Enzymes lower the activation energy required for a reaction to proceed. This helps speed up the reaction rate.

With enzyme: Ea is reduced by the presence of the enzyme and the reactants form products more readily.

Ea is the activation energy required for the reaction to begin. chemicalsthein

2. Why is it important that the rate of biological reactions is increased by enzymes?

Ea Ea

f Chemical reactions in cells are accompanied by energy changes. The amount of energy released or taken up is directly related to the tendency of a reaction to run to completion (for all the reactants to form products).

Reactants

Without enzyme: The energy required for the reaction to proceed (Ea) is high without the enzyme present.

f Any reaction needs to raise the energy of the reactants to an unstable transition state before the reaction will proceed (below). The amount of energy needed to do this is the activation energy (Ea). Enzymes lower the Ea by destabilising bonds in the reactants so that they are more reactive. Without enzymes, biological reactions would occur at rates that would be too slow to sustain life.

2. Why are a number of enzymes required for a metabolic pathway?

Photocopying Prohibited

Products are released Product Substrates Enzyme Products Substrate Enzyme Substrates drawn into the active site. Enzyme catalyses the formation of bonds.

Product releasedis

How Enzymes Work

End releasedproductEnzyme

The binding of the substrate causes the enzyme to change shape and the catalytic parts of the enzyme contact the substrate. The reaction can then occur.

Induced fit model of enzyme activity

Active

Anabolic reaction: The enzyme catalyses the formation of bonds and a more complex product is released. The product may be the reactant for the next step in making a more complex molecule. Example: In photosynthesis, a series of linked, enzyme-catalysed steps uses ATP to build glucose from CO2 and water.

Catabolic reaction: The enzyme catalyses the breaking of bonds, releasing simpler products. The products may be further broken down in subsequent steps. Example: In cellular respiration, a series of linked, enzyme-catalysed steps breaks down glucose to CO2 and water, forming ATP.

The end product is released its previous shape.

Substrate molecules 1 2b

shapechangesEnzymeEnzyme 2 site of the enzyme. site

Enzyme

f Complex molecules are usually built up and broken down in a series of steps, (called metabolic pathways). Each step in the process is catalysed by a different enzyme. The product of one step forms the reactant for the next.

Product Substrates Products Substrate Enzyme Substrate drawn into the active site. Enzyme catalyses the breaking of bonds.

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Enzymes in anabolic and catabolic reactions

Enzyme

Key Idea: Enzymes catalyse reactions by providing a reaction site for a substrate. The model that describes the behaviour of enzymes the best is the induced fit model.

1. (a) Describe the induced fit model of enzyme action: (b) In the induced fit model, the enzyme returns to its original shape after the reaction. Why is this important?:

f An early model to explain enzyme activity described the enzyme and its substrate as a lock and key, where the substrate fitted neatly into the active site of the enzyme. Evidence showed this model to be flawed and it has since been modified to recognise the flexible nature of enzymes (the induced fit model).

f The current induced fit model for enzyme action is shown below. The shape of the enzyme changes when the substrate interacts with the active site. The reactants become bound to the enzyme by weak chemical bonds, which can weaken bonds within the reactants themselves, allowing the reaction to proceed more readily.

Key Idea: Some enzymes require cofactors to function. Enzyme activity can be reduced or stopped by inhibitors.

a-amylase

Enzyme Hg

Some enzymes need cofactorsInhibitors

Enzyme

The enzyme a-amylase is present in saliva where it starts the hydrolysis of starch into the simple sugars maltose and glucose. To work correctly, it needs the ions Ca2+ and Cl Cl increases the binding of Ca2+ by 100 times. It also shifts the optimum pH for amylase from 6 to 6.8.

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cofactors.Substratemolecules

Many drugs work by irreversible inhibition of a pathogen's enzymes. Penicillin (below) and related antibiotics inhibit transpeptidase, the enzyme that forms some of the linkages in the bacterial cell wall. Susceptible bacteria cannot complete cell wall synthesis and so cannot divide. Human cells are unaffected by the

slow or stop enzyme activity

Enzyme inhibitors slow or stop enzyme activity. They are important in controlling metabolic pathways, but they can also act as poisons if they bind irreversibly. Competitive inhibitors compete with the substrate for the active site. Non-competitive inhibitors bind to the enzyme outside the active site and change the enzyme's shape so that the substrate cannot bind.

1. What are enzyme cofactors?

binds outside the active site and substrate cannot bind.

Competitive inhibitor binds to the active site and the substrate cannot bind.

Cl ion

Ca2+ion

and Inhibitors Affect Enzyme Activity26

Carbonic anhydrase is an important molecule in the transport of CO2 into and out of the cell. It contains a central Zn2+ ion as a prosthetic group (a prosthetic group is tightly bound to the enzyme).

Inhibitor, e.g. mercury

Zinc ion

Carbonic anhydrase Drugs

CofactorsProhibited

While some enzymes are complete and active as a protein only molecule, others require cofactors to function. Cofactors are non-protein chemicals that bind to an enzyme, completing the active site and making the enzyme functional. Many vitamins and metal ions act as

Enzyme changesEnzyme shape

2. Distinguish between competitive and non-competitive inhibition:

Temperature (°C) activityEnzyme denaturesEnzymeToo cold to 1002030405060operatepH Pepsin Trypsin activityEnzyme Acid Alkaline Urease 1 2 345 6 789 10

(b) How does this allow a cell to change the rate of substrate use?

2. (a) Describe the change in the reaction rate when substrate concentration is increased (assuming a fixed amount of enzyme):

Substrate concentration

Substrate concentration reactionofRate

Enzyme concentration

3. Study the graph comparing the effect of competitive and non- competitive enzyme inhibitors on reaction rate

Enzyme Reaction Rates27

reactionofRate

inhibitorCompetitive

Pepsin: Trypsin: Urease:

5. Like all proteins, enzymes are denatured by extremes of pH. Within these extremes, most enzymes have a pH range for optimum activity.

With ample substrate

No inhibitor

(a) What is meant by optimum temperature for enzyme activity?

4 Although an increase in temperature increases reaction rate, few enzymes remain functional at temperatures above 50 60°C. The rate at which enzymes are denatured increases with temperature.

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(b) Which one would you find in the stomach (pH 2)?

In the graphs below, the rate of reaction (enzyme activity) is plotted against factors that affect enzyme performance.

Photocopying Prohibited

(a) What is the general effect of an enzyme inhibitor : (b) How can the effect of a competitive inhibitor be overcome?

Key Idea: Enzymes have a narrow range of conditions in which they work optimally. Outside this range, activity decreases and the enzyme may lose its structure.

1. (a) Describe the change in the reaction rate when enzyme concentration is increased (assuming substrate is not limiting):

inhibitorNon-competitive

(a) State the optimum pH for each of the enzymes:

reactionofRate

(b) Mark the optimum temperature on the graph left with a vertical line.

(b) Explain why the rate changes the way it does:

With fixed amount of enzyme

5. State one extra reaction that should have been carried out by the students:

Hypothesis and prediction

Water in the 50 thedisplacedcylindercm3isbyoxygen.

Potato cubes + excess H2O2 oxygenreleasedtransfersTube

Catalase Activity

Aim

Reaction rate increases with amount of enzyme present, so a greater mass of potato (therefore more enzyme) will produce a greater reaction rate.

The students cut raw potato into cubes with a mass of one gram. These were placed a conical flask with excess H2O2 (right). The reaction was left for five minutes and the volume of oxygen produced recorded. The students recorded the results for three replicates each of 1, 2, 3, 4, and 5 cubes of potato below:

To investigate the effect of potato mass (and therefore enzyme concentration) on the rate of H2O2 (hydrogen peroxide) decomposition.

2. Plot the mass of the potato vs the rate of production on the grid (right):

Key Idea: The effect of increasing the amount of enzyme in a reaction can be measured indirectly by measuring the volume of reaction products.

1. Complete the table by filling in the mean volume of oxygen produced and the rate of oxygen production.

releasedOxygen by the reaction

A 50 cm3 cylinder is upturned in a small dish of water, excluding the air.

Timed for 5 minutes.

4. Why did the students add excess H2O2 to the reaction?

Masspotatoof(g) Volume oxygen (cm3) (5 minutes) Mean Mean rate of O2 (cmproduction 3 min-1) Test 1Test 2 Test 3 524232542120203141515210991656 LINK 27 WEB 28 PREVIEWONLYNotforClassroomUseNotforClassroomUse

6. (a) The students decide to cook some potato and carry out the test again with two grams of potato. Predict the result:

Method

3. Relate the rate of the reaction to the amount of enzyme present.

(b) Explain this result:

Ribose

• A purine base (adenine)

Note! The phosphate bonds in ATP are often referred to as being high energy bonds. This can be misleading. The bonds contain electrons in a high energy state (making the bonds themselves relatively weak). A small amount of energy is required to break the bonds, but when the intermediaries recombine and form new chemical bonds a large amount of energy is released. The final product is less reactive than the original reactants.

(b) Where is the energy stored in ATP?

3. Why does the conversion of ATP to ADP help keep us warm?

• Three phosphate groups.

• A pentose sugar (ribose)

f The ATP molecule (right) is a nucleotide derivative. It consists of three components;

f The bonds between the phosphate groups of ATP are unstable, very little energy is needed to break them. The energy in the ATP molecule is transferred to a target molecule (e.g. a protein) by a hydrolysis reaction. Water is split during the reaction and added to the terminal phosphate on ATP, forming ADP and an inorganic phosphate molecule (Pi).

Key Idea: ATP transfers energy to where it is needed in the cell. Hydrolysis of the phosphate group releases energy, which can be used to do work.

Adenosine triphosphate (ATP)

Adenine

f ATP acts as a store of energy within the cell. The bonds between the phosphate groups contain electrons in a high energy state, which store a large amount of energy that is released during a chemical reaction. The removal of one phosphate group from ATP results in the formation of adenosine diphosphate (ADP).

2. (a) What is the biological role of ATP?

ATP29

Phosphate groups

1. What are the three components of ATP?

f When the Pi molecule combines with a target molecule, energy is released. Most of the energy (about 60%) is lost as heat (this helps keep you warm). The rest of the energy is transferred to the target molecule, allowing it to do work, e.g. joining with another molecule (right).

Pi Pi AA PiAA AA B B B A Reactants A and B have low energy B Reactant A combines with Pi raising its energy C The gainedenergybyAlets it react with B Structure of an ATP molecule PPPAdenosine O H H + Adenosine triphosphate (ATP) Adenosine diphosphate (ADP) Water PPAdenosine HPOH + + Pi Adenine + ribose = adenosine LINK 30 WEB 29 LINK 32 PREVIEWONLYNotforClassroomUseNotforClassroomUse

How does ATP provide energy?

Note: Heterotrophs depend on organic molecules (food) to provide the glucose for cellular respiration.

Carbondioxide+water Water Heat

3. (a) What is the ultimate source of energy for plants?

(b) What process do animals (and plants) use to extract the energy from this source?

Key Idea: Photosynthesis uses energy from the sun to produce glucose. Glucose breakdown produces ATP, which is used by all cells to provide the energy for metabolism.

4. (a) What is the ultimate source of energy for animals?

The hydrolysis of ATP provides the energy for metabolic reactions. Each mole of ATP hydrolysed releases 30.7 kJ of energy. Some energy is stored in chemical bonds, while some is lost as heat.

Carbon

f Heterotrophs (organisms that cannot make their own food) obtain their glucose by eating plants or other organisms.

(b) In what way is the ADP/ATP system like a rechargeable battery?

Photosynthesis is a chemical process that captures light energy and transforms it into the chemical energy in carbohydrate (glucose).

* Other

Light

(b) What process do plants use to store or fix this energy?

f During cellular respiration ATP is formed through a series of chemical reactions. The ATP provides the energy to drive life's essential processes.

2. What is the immediate source of energy for refor ming ATP from ADP?

Energy Transformations in Cells

A plantphotosyntheticcell

1. (a) How does ATP supply energy to power metabolism?

Photosynthesis

Cellular respiration Glucose uses of glucose Oxygenenergy dioxideenergy Oxygen

Cellular respiration is a chemical process in which the phosphatefromformprovidesbreakdownstep-wiseofglucosetheenergytohighenergyATPADPandinorganic(Pi).

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f During photosynthesis light energy is converted into chemical energy in the form of glucose. Glucose is used by plants and animals to provide the energy for cellular respiration.

+ADPPi ATP Fuel

The general equation for cellular respiration 6CO2 + 6H2O + energyC6H12O6 + 6O2

(b) The link reaction:

31 LINK 32 WEB 31 LINK 33 34 ATP2 ATP2 ATP Krebscycle Glycolysis reactionLink transportElectronchain Matrix: enzymes for the Krebs cycle reside here. NADH +FADH2 Pyruvate coenzymeAcetylA Glucose Cristae:

5. Describe two functions of the Krebs cycle in cellular respiration: (ii)(i)

f Glycolysis and the Krebs cycle supply electrons to the electron transport chain (ETC). The conversion of pyruvate (the end product of glycolysis) to acetyl CoA links glycolysis to the Krebs cycle. Most of the ATP generated in cellular respiration is produced in the electron transport chain. folded inner

4. Describe three functions of glycolysis in cellular respiration: (iii)(ii)(i)

2. Write the general equation for cellular respiration in words:

(a) Glycolysis:

Key Idea: During cellular respiration, the energy in glucose is transferred to ATP in a series of enzyme controlled steps.

3. What is the total number of ATP produced from one glucose molecule?

1. Describe precisely in which part of the cell the following take place:

f Cellular respiration involves three metabolic stages (plus a link reaction) summarised below. The first two stages are the catabolic pathways that decompose glucose and other organic fuels. In the third stage, the electron transport chain accepts electrons from the first two stages and passes these from one electron acceptor to another. The energy released at each stepwise transfer is used to make ATP.

chainelectronEnzymesmembrane.forthetransportresidehereMitochondrionoutermembraneNADH MitochondrionCytoplasm PREVIEWONLYNotforClassroomUseNotforClassroomUse

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Cellular Respiration: Inputs and Outputs

KNOW 44

(d) Electron transport chain:

International

muscles CarbohydratesinmusclesGlucose ATP ADP Pi contractionMuscle transferEnergy PREVIEWONLYNotforClassroomUseNotforClassroomUse

6. Explain how the energy in glucose is converted to useful energy in the body. Use the example of muscle contraction to help illustrate your ideas:

f A molecule's energy is contained in the electrons within the molecule's chemical bonds. During a chemical reaction, energy (e.g. heat) can break the bonds of the reactants.

GlucoseGlycolysis 2 ATP 2 ATP34 ATP Link reactionKrebschainElectroncycletransportOxygen Water energypotentialChemical

f When the reactants form products, the new bonds within the product will contain electrons with less energy, making the bonds more stable. The difference in energy is usually lost as heat. However, some of the energy can be captured to do work.

A model for ATP production and energy transfer from glucoseA model for ATP use in the

f Glucose contains 16 kJ of energy per gram (2870 kJ mol-1). The step-wise breakdown of glucose through a series of chemical reactions yields ATP. In total, 38 ATP molecules can be produced from 1 glucose molecule.

8. ATP is used at a rate of about 109 molecules of ATP per cell every 2 minutes. Calculate how many ATP molecules are being generated every second per cell to replace this number:

How does cellular respiration provide energy?

7. (a) One mole of glucose contains 2870 kJ of energy. The hydrolysis of one mole of ATP releases 30.7 kJ of energy. Calculate the percentage of energy that is transformed to useful energy in the body. Show your working.

(b) Use your calculations above to explain why shivering keeps you warm and muscular exertion causes you to get hot:

f ATP is produced during cellular respiration by both substrate level phosophorylation and oxidative phosphorylation.

f The energy from the electrons is used to pump H+ ions across the inner membrane from the matrix into the intermembrane space. These flow back to the matrix via the membrane-bound enzyme ATP synthase, which uses their energy to produce 34 ATP per glucose molecule

Link reaction

f Substrate level phosphorylation of ADP to ATP occurs during glycolysis and the Krebs cycle. An enzyme transfers a phosphate group directly from a substrate to ADP to form ATP.

f Acetyl coenzyme A is attached to a 4C intermediate and coenzyme A is released.

f The electrons are transferred to NAD+ to form the coenzyme NADH. NADH carries hydrogens to the electron transport chain.

Key Idea: There are four major steps to cellular respiration. The majority of ATP is produced in the electron transport chain. Enzymes catalyse each reaction of cellular respiration.

Steps in Cellular Respiration

f 1 ATP is made by substrate level phosphorylation and 2 CO2 are produced per turn.

f The Krebs cycle turns twice per glucose molecule (once per pyruvate molecule).

f The intermediate is remade in a cyclic series of enzyme-controlled reactions that remove carbons and produce more NADH and FADH2 for the electron transport chain.

Photocopying Prohibited

Krebs cycle

2 ATP 4CO2 2 ATP 2CO2 2NADH2NADHcoenzymeAcetylA Electron transport2chainGlucosePyruvate 34 (viaATPETC) H2O 2H+ +1/2O2 H+ NADH FAD+FADH2 NAD+ H+ H+ H+ H+ e Glycolysis 2FADH6NADH2 Krebs cycle

Glycolysis

Electron transport chain (ETC)

f The electrons are coupled to H+ and oxygen at the end of the ETC to form water.

f The process is called oxidative phosphorylation.

f Electrons carried by NADH and FADH2 are passed to a series of electron carrier enzymes embedded in the inner membrane of the mitochondria.

f Glycolysis uses 2 ATP but produces 4 ATP molecules. This net 2 ATP is made by transferring a phosphate directly from a substrate to ADP (substrate level phosphorylation).

f Oxidative phosphorylation occurs at the end of the electron transport chain. Electrons are transferred (by NADH and FADH2) from electron donors (e.g. glucose) to the electron transport chain and finally to electron acceptors (e.g. oxygen) in redox reactions. The energy released is used to produce ATP from ADP.

32 73

f The link reaction removes CO2 from pyruvate and adds coenzyme A, producing the 2C molecule acetyl coenzyme A, which enters the Krebs cycle. NADH is also produced for use in the electron transport chain.

f An enzyme strips two electrons from glucose to produce two pyruvate molecules, each of which can enter the Krebs cycle.

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4. What is the purpose of NADH and FADH2 in cellular respiration?

5. How are glycolysis and the Krebs cycle linked to the electron transport chain?

(b) How many ATP are produced this way during cellular respiration (per molecule of glucose)?

1. (a) What is substrate level phosphorylation?

(b) How many ATP are produced this way during cellular respiration (per molecule of glucose)?

7. Use the space below to draw a summary of cellular respiration, including the location, inputs, and outputs of each stage (glycolysis, the link reaction, Krebs cycle, and the electron transport chain):

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6. Describe how ATP is produced in the electron transport chain and explain the importance of the proton gradient:

2. (a) What is oxidative phosphor ylation?

3. Which parts of cellular respiration produce CO2?

Measuring Respiration

(b) Plot the rate of respiration on the grid, below right.

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(a) Calculate the rate of respiration and record this in the table. The first two calculations have been done for you.

f Differential respirometers (not shown) use two connected chambers (a control chamber with no organisms and a test chamber). Changes in temperature or atmospheric pressure act equally on both chambers. Observed changes are therefore only due to respiration.

2. A student used a simple respirometer (like the one above) to measure respiration in maggots. Their results are presented in the table (right). The maggots were left to acclimatise for 10 minutes before the experiment was started.

bubbleColoured tubeCapillary

Key Idea: Oxygen consumption and carbon dioxide production in respiring organisms can be measured with a respirometer.

f Soda lime or potassium hydroxide is placed into the chamber to absorb any CO2 produced during respiration. The respirometer measures O2 consumption.

1. Why does the bubble in the capillary tube move?

Caution is required when handling KOH as it is caustic. Wear protective eyewear and gloves.

SodaseedsGerminatinglime(or KOH) pellets (CO2 absorbant) metal cage

f The coloured bubble in the capillary tube moves in response to the change in O2 consumption. Measuring the movement of the bubble (e.g. with a ruler or taped graph paper) allows the change in volume of gas to be estimated.

130

3. Why would it have been better to use a differential respirometer? 0 5 25 5 65 20 25 160

f Changes in temperature or atmospheric pressure may change the readings and give a false measure of respiration when using a simple respirometer.

Perforated

Measuring respiration with a simple respirometer

f Respiring organisms (e.g. germinating seeds) are placed into the chamber and the screw clip is closed. The position of the coloured bubble is measured (time zero reading).

(d) Why was there an acclimatisation period before the experiment began?

A respirometer measures the amount of oxygen consumed and the amount of carbon dioxide produced during cellular respiration. The diagram below shows a simple respirometer Appendix

15 95

Screw clipClamp stand Scale

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(minutes)Time bubbleDistancemoved(mm) (mmRatemin-1) 0

f In alcoholic fermentation, the electron acceptor is ethanal (acetaldehyde) which is reduced to ethanol in two steps with the release of carbon dioxide (CO2). Enzymes (E) catalyse both steps.

f Yeasts will respire aerobically when oxygen is available and the availability of sugar is low. When oxygen is absent, especially when sugar is plentiful, they will use alcoholic fermentation to generate ATP. At ethanol levels above 12-15%, the ethanol is toxic and this limits their ability to use this pathway indefinitely.

1. Contrast and explain the efficiencies of alcoholic fermentation and aerobic respiration in terms of ATP generated:

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The alcohol and CO2 produced from alcoholic fermentation form the basis of the brewing and baking industries. In baking, the dough is left to ferment and the yeast metabolises sugars to produce ethanol and CO2. The CO2 causes the dough to rise.

f The root cells of plants also use fermentation as a pathway when oxygen is unavailable but the ethanol must eventually be converted back to respiratory intermediates and respired

(b) Explain why fer mentation in useful for making alcoholic beverages:

are used to produce almost all alcoholic beverages (e.g. wine and beers). The yeast used in the process breaks down the sugars into ethanol (alcohol) and CO2. The alcohol produced is a metabolic by-product of fermentation by the yeast.

Key Idea: Alcoholic fermentation is an anaerobic process in plants and fungi, which passes electrons directly to the final electron acceptor ethanal.

Oxygen is generally scarce in waterlogged soils, so aerobic respiration is inhibited in the roots of plants in oxygen-poor soils. Carbohydrates are broken down by fermentation to produce ATP. Some plants (e.g. rice) are well adapted to low oxygen, flooded soils.

Glucose C6H12O6 2 x pyruvate CH3COCOO Alcoholic fermentation (higher plants, yeast) 2 ADP 2 ATP yield 2 NAD+ 2 NADH + 2H+ Ethanal CH3CHO CO2 Ethanol CH3CH2OH

Alcoholic fermentation

3. When do root cells generate ATP anaerobically and why?

aerobically.Yeasts

f Organisms can generate ATP when oxygen is absent by using a molecule other than oxygen as the terminal electron acceptor for the pathway.

Anaerobic Metabolism in Plants and Fungi

2. (a) Explain why fer mentation in useful for rising dough:

Wintec LINK 33 WEB 35 E PREVIEWONLYNotforClassroomUseNotforClassroomUse

Photocopying Prohibited

f Importantly, this pathway operates alongside the aerobic system (even when oxygen is present) to enable greater intensity and duration of muscle activity. It is an important mechanism for balancing the distribution of substrates and waste products, especially when pyruvate is building up faster than it can be metabolised.

1. What is the key difference between alcoholic fermentation and lactic acid fermentation?

2 ADP2 ATP yield Glucose C6H12O6

f The conversion of pyruvate to lactate is reversible. The pyruvate-lactate interconversion is catalysed by an enzyme (E) called lactate dehydrogenase. Hence it is called the lactate 'shuttle'.

f Lactate can be metabolised in the muscle itself or it can enter the circulation and be taken up by the liver to replenish carbohydrate stores. It moves from its site of production to regions within and outside the muscle where it can be respired aerobically.

Lactic acid CH3CHOHCOO +H+ + NAD+

f Mammalian skeletal muscle can produce ATP anaerobically using lactic acid fermentation. In this pathway, the electron acceptor is pyruvate, the end product of glycolysis. The pyruvate is reduced to lactic acid, which dissociates to form lactate and H+.

Anaerobic Metabolism in Animals

(b) Study the pathway of lactic acid fermentation above and explain how the lactate produced in working muscle could provide fuel for aerobic respiration in other tissues:

2 NAD+ 2 NADH + H+

Key Idea: Lactic acid fermentation is an anaerobic process that passes electrons directly to pyruvate as the final electron acceptor. It occurs in the muscles of mammals.

2 x pyruvate CH3COCOOH pyruvate CH3COCOOH + NADH +H+

Lactic acid (animalfermentationtissues)

3. Compare the efficiency of lactic acid fermentation to aerobic respiration in terms of ATP produced:

f During moderate exercise, lactate released from other tissues is the main fuel source for the heart. Possibly up to 75% of the lactate produced in muscle during exercise is used in this way. There is also evidence to support the hypothesis that lactate is an important fuel source in the brain.

2. (a) Explain the importance of the lactate shuttle during moderate exercise:

Lactic acid fermentation

2. Identify the region of a dicot leaf where most of the chloroplasts are found:

CO2

1 (a) How do gases enter and leave the leaf tissue?

3. What is the purpose of the air spaces in the leaf tissue?

(b) How is this movement regulated?

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Entry and exit of gases through the stomata

SEM of black walnut leaf showing epidermis (E), palisade mesophyll (Mp), and spongy mesophyll (Ms) with its numerous air spaces.

CO2

f The main function of leaves is as photosynthetic organs in which light energy is captured for use in photosynthesis. Leaves are green because they reflect the green wavelengths of light not involved in photosynthesis.

f When the plant is photosynthesising, the situation is more complex. Overall there is a net use of CO2 and a net production of oxygen. CO2 uptake (in photosynthesis) maintains a gradient in CO2 concentration between the inside of the leaf and the atmosphere. Oxygen is produced in excess of respiratory needs and diffuses out of the leaf. These exchanges are indicated by the arrows on the diagram.

O2 O2CO2 O2

f The structure of leaves maximises the capture of sunlight energy and facilitates the diffusion of gases used and produced by photosynthesis into and out of the leaf tissue. Gases enter and exit the leaf through stomata (pores) in the leaf. Guard cells each side a stoma lose or gain water to control the size of the pore and regulate the movement of gases. Inside the leaf, large air spaces and the loose arrangement of the spongy mesophyll provides a large surface area for gas exchange.

Cuticle forms a barrier to the diffusion of gases

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LowerairSubstomatal(xylemLeafcellSpongycellPalisadechloroplasts)mesophyllwithchloroplastsmesophyllwithchloroplastsveinandphloem)spaceepidermisGuardcellStoma(pore)

f Respiring plant cells use oxygen (O2) and produce carbon dioxide (CO2). These gases move in and out of the plant and through the air spaces by diffusion. Flowering plants have many air spaces between the cells of the stems, leaves, and roots. These air spaces are continuous and gases are able to move freely through them and into the plant’s cells via the stomata (sing. stoma).

Key Idea: Leaves are the main site for photosynthesis in plants. Leaf structure aids the photosynthetic process.

LeafProhibitedStructure and Photosynthesis

4. The theoretical output of O2 from photosynthesis is the same as the theoretical input of O2 to respiration. Explain why there is a net output of O2 from plants.

Upper epidermis (lacks

Dicot leaf structure

Net gas exchanges in a photosynthesising dicot leaf

Palisade mesophyll layer only 1 cell thickLow intensitylight

Thin leaves

Photocopying Prohibited

Shade leaves

A shade leaf can absorb the light available at lower light intensities. If exposed to high light, most would pass through.

Shade leaves

A shade leaf can absorb the light available at lower light intensities. If exposed to high light, most would pass through.

A sun leaf, when exposed to high light intensities, can absorbmuch of the light available to the cells.

(b) Compare the rate of photosynthesis between sun and shade plants and explain the consequences of this:

Palisade mesophyll layer only 1 cell thickLow intensitylight

Thin leaves

Shade plant

Chloroplasts are mostly restricted to palisade mesophyll cells (few in spongy mesophyll).

Wet SunnyShady Dry

A sun leaf, when exposed to high light intensities, can absorbmuch of the light available to the cells.

Thick leaves

Palisade mesophyll layer often 2-3 cells thick

Chloroplasts occur throughout the mesophyll (as many in the spongy as in the palisade mesophyll).

Intense light

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The amount of light falling on a plant affects its ability to photosynthesise. A greater intensity of light enables a higher rate of photosynthesis, but increases the risk of the leaf drying out or burning. The leaves of plants are adapted to their environment (e.g. sun or shade) to maximise photosynthesis and minimise water loss and sun damage.

Plants adapted for full sunlight have high levels of respiration. Sun plants include many weed species found on open ground. They expend much more energy on the construction and maintenance of thicker leaves than do shade plants. The benefit of this investment is that they can absorb the higher light intensities available and grow more quickly.

(b) Why would plants produce small leaves in sunny, dry areas?

(a) What happens to leaf size as we move from wet and shady to sunny and dry?

Sun plant

Chloroplasts are mostly restricted to palisade mesophyll cells (few in spongy mesophyll).

Palisade mesophyll layer often 2-3 cells thick

7. The diagram on the right shows the effects of the light environment on leaf size:

Sun plant

ISBN: 978-1-927309-51-3

Thick leaves

Chloroplasts occur throughout the mesophyll (as many in the spongy as in the palisade mesophyll).

Shade plant

Intense light

Sun leaves

5. Explain why plants have most of their leaves on the outer edges of the plant (rather than hidden within the branches):

6. (a) Compare the respiration rates of sun and shade plants:

Shade plants typically grow in forested areas, partly shaded by the canopy of larger trees. They have lower rates of respiration than sun plants, mainly because they build thinner leaves. Less energy is needed to produce and maintain a smaller number of cells.

A mesophyll contains 50-100 chloroplasts!

Chloroplasts are aligned with their broad surfaces parallel to the cell wall to maximise the surface area for light absorption.

Chloroplast is enclosed by an inner and outer membrane

1. image a chloroplast

Label the transmission electron microscope (TEM)

f This structure creates distinct regions in which the separate stages of photosynthesis occur. One stage depends on light and the other doesn't.

How do plants maximise the amount of light that can be absorbed? Image: Dartmouth College (b)(a)(c) (f)(e)(d)(c)

leaf cell

Thylakoidmembranes.membranes provide a large surface area for light absorption. They are organised so as not to shade each other.

Starch granule

Lipid droplet

The Structure of Chloroplasts

f Pigments on the thylakoid membranes called chlorophylls capture light energy by absorbing light of specific wavelengths. Chlorophylls reflect green light, giving leaves their green colour.

ChloroplastsCellwall

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Key Idea: The complex membranous str ucture of chloroplasts creates compartments in which the separate stages of photosynthesis can occur.

f Photosynthesis takes place in disk-shaped organelles called chloroplasts (4-6 µm in diameter). The inner structure of a chloroplast is characterised by a system of membrane-bound compartments called thylakoids arranged into stacks called grana. These are linked by membranous structures called stroma lamellae. These membranes are located within a fluid stroma.

PetersKristian

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(Grana sing. Liquidgranum)stroma

Stroma lamellae connect grana. They account for 20% of the thylakoid

below: 2. What does chlorophyll do?

(C3H7O6P)

The light dependent phase occurs in the thylakoid membranes of the grana. outer membrane

Triose phosphate

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f Only red and blue wavelengths of light are absorbed for photosynthesis, the rest are reflected, producing the green colour of most plant leaves.

(b) Where does the light independent phase of photosynthesis occur?

Light independent phase (LIP):

Rubisco is the central enzyme in the LIP of photosynthesis (carbon fixation) catalysing the first step in the Calvin cycle. However it is remarkably inefficient, processing just three reactions a second. To compensate, rubisco makes up almost half the protein content of chloroplasts.

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(b) Rubisco is the most abundant protein on Earth. Suggest a reason for this:

CO2 from the air provides raw materials for glucose production.

Photosynthesis can be summarised in the equation:

Key Idea: Photosynthesis transfor ms the energy in light into chemical energy. It uses carbon dioxide and water and produces glucose and oxygen.

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LDP LDP LIP

The apparently extra 6 H2Os are to show that O2 comes from the H2O, not the CO2.

Monosaccharides (e.g. glucose) and other carbohydrates, lipids, and amino acids.

The second phase of photosynthesis occurs in the stroma and uses the NADPH and the ATP to drive a series of enzyme controlled reactions (the Calvin cycle) that fix carbon dioxide to produce triose phosphate. This phase does not need light to proceed.

Sunlight

Photosynthesis: Inputs and Outputs

f Photosynthesis is of fundamental importance to life. It uses the pigment chlorophyll to capture light energy, which is used to drive the chemical reactions that transform carbon dioxide and water into glucose and oxygen.

3. Explain why most photosynthetic organisms are green:

6CO2 + 12H2O C6H12O6 + 6O2 + 6H2O

2. (a) What is the role of the enzyme Rubisco?

NADPH CO2

ATP toConverted (waste)H2O

In the first phase of photosynthesis chlorophyll captures light energy which is used to split water, producing O2 gas (waste) and H+ ions that are transferred to the molecule NADPH. ATP is also produced.

1. (a) Where does the light dependent phase of photosynthesis occur?

ChlorophyllLight

Light dependent phase (LDP):

Experiment A: 6CO2 + 12H2O + sunlight energy C6H12O6 + 6O2 + 6H2O

6. Name the products that triose phosphate is converted into:

8. What is the function of each of the following in photosynthesis: (a) ATP:

(b) NADPH:

4. Use the information on the opposite page to fill in the diagram below, including the raw material (inputs), products (outputs), and processes.

7. Describe what happens during: (a) The light dependent phase of photosynthesis: (b) The light independent phase of photosynthesis:

NADPNADPH+ADP By-productsmaterialsRaw (e) Main product (d)(b)(a) (c) energySolar Light dependent phase (f) (g)Process:Location: Light independent phase (h) Process: (i) Location: ATP PREVIEWONLYNotforClassroomUseNotforClassroomUse

Experiment B: 6CO2 + 12H2O + sunlight energy C6H12O6 + 6O2 + 6H2O

5. In two experiments, radioactively-labelled oxygen (shown in blue) was used to follow oxygen through the photosynthetic process. The results of the experiment are shown below:

From these results, what would you conclude about the source of the oxygen in: (a) The carbohydrate produced?

(b) The oxygen released?

(e) Water :

Stages of the Calvin cycle

Photocopying Prohibited

Reduction: ATP driven reactions form intermediates then NADPH driven reactions form triose phosphate (TP).

Carbon fixation: The enzyme (E) RuBisCo joins CO2 with RuBP to form glycerate 3-phosphate (GP).

Steps in Photosynthesis

Regeneration: Some triose phosphate leaves the chloroplast and forms sugars while the rest continues through the cycle to eventually reform RuBP.

2. How is the H+ gradient produced and used in the light dependent phase?

f The light dependent phase (LDP) of photosynthesis (shown right) uses the pigment chlorophyll to capture photons of light and use their energy to excite electrons. The chlorophyll molecules are embedded within two multi-protein enzyme complexes, called photosystems, which are embedded in the thylakoid membranes.

Light dependent phase

f ATP and NADPH are passed to the stroma where they power the reactions of the Calvin cycle.

4. What would happen to the reactions of the light independent phase if the plant was suddenly put in the dark and why?

3. What is the purpose of Rubisco in the light independent phase?

1. What is the role of photosystem II in the light dependent phase?

f Photosystem II splits water (forming H+ and O2) to obtain electrons. These are first used to pump H+ across the thylakoid membrane into the thylakoid space creating a H+ gradient. The electrons are then passed to photosystem I which boosts their energy and uses them to produce the hydrogen carrier NADPH (similar to NADH in respiration).

Light independent phase (Calvin cycle)

Key Idea: Photosynthesis involves two linked phases. The light dependent phase captures light energy and stores it as ATP. The light independent phase uses ATP to fix carbon.

f The Calvin cycle (right) is a series of reactions driven by ATP and NADPH. It generates triose phosphate (TP), the 3C precursor for hexose sugars such as glucose, and regenerates ribulose 1,5 bisphosphate (RuBP), the 5C molecule needed for the first step of the cycle.

ReductionfixationCarbonRegeneration E Thylakoid space Stroma ATPNADPHADPNADP+ 2H2O 4H+ + O2 H+ H+H+ eePhotosystem II Photosystem I synthaseATP complexCytochrome energyLight PREVIEWONLYNotforClassroomUseNotforClassroomUse

f The H+ gradient is used to drive the production of ATP via the enzyme ATP synthase which is also embedded in the thylakoid membrane.

f Three CO2 molecules are needed for the production of one TP molecule. Two TP molecules are used to form one glucose molecule.

39 LINK 38 WEB 39 Glucose and other products6 ATP 3 ATP 6 ADP 3 ADP 12 12NADP+NADPH 3 x CO2 6 x 3C 6 x 3C 3 x 5C 1 x 3C 5 x 3C 3 x 5C 3 x 6C GP TP RuBP

The effect of different limiting factors on the rate of photosynthesis in cucumber plants

Figure B: The effect of different light intensities at two temperatures and two CO2 concentrations. In each of these experiments, either CO2 level or temperature was changed at each light intensity in turn.

rate

(b) An increase in light intensity:

7123456HighCO2at 30oC

f Manipulating abiotic (non living) factors can maximise crop yields by maximising photosynthetic rate and reducing losses to pests, disease, and competition. Control of the abiotic conditions can be achieved by using a controlled environment such as a greenhouse.

B: Light intensity, CO2, and temperature photosynthetic

Key Idea: The two main factors that affect the rate of photosynthesis are the availability of light and carbon dioxide.

f The photosynthetic rate is the rate at which plants make carbohydrate. It is dependent on environmental factors, the most important being the availability of light and carbon dioxide (CO2). Temperature is important, but its influence is less clear because it depends on the availability of the other two limiting factors (CO2 and light) and the temperature tolerance of the plant.

Figure A: The effect of different light intensities on photosynthetic rate. The temperature and carbon dioxide (CO2) level are kept constant.

High CO2 at 20oC Low CO2 at 30oC Low CO2 at 20oC

f Temperature, carbon dioxide (CO2) concentration, and light intensity are optimised to maximise the rate of photosynthesis and therefore production. Greenhouses also allow specific abiotic factors to be manipulated to trigger certain life cycle events such as flowering. CO2 enrichment dramatically increases the growth of greenhouse crops providing that other important abiotic factors (such as nutrients) are not limiting.

2402001601208040280 7

(a) An increase in CO2

A: Light intensity vs photosynthetic

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f The effect of CO2 and light can be tested by altering one of the factors while holding the others constant (a controlled experiment). In reality, a plant in its natural environment is subjected to variations in many different environmental factors, all of which will directly or indirectly influence the rate at which photosynthesis can occur.

40 Glasshouse environments can artificially boost CO2 levels

Units of light intensity (arbitrary scale)

concentration:

Factors Affecting Photosynthesis

rate-23(mmphotosynthesisofRateCO2cmh-1) -23(mmphotosynthesisofRateCO2cmh-1)908070605040

of each

Units of light intensity (arbitrary scale) 123456

1. What is the effect of the following photosynthetic rate?

High wind increases water loss from stomata

(b) How can you tell this from the graph?

5. For plants that live in arid (hot and dry) climates, explain the adaptive advantage of having stomata positioned in pits or grooves instead on the leaf surface:

Plants must balance the need for gas exchange against the loss of water from open stomata. One way to reduce water loss is to reduce leaf size or place the stomata at the bottom of pits in the leaves. Cacti (above left) have reduced their leaves to spines, which reduces air flow around the stem. The stomata are located in the bottom of grooves that run vertically down the plant. Marram grass (above right) rolls its leaves and has stomata in pits. These adaptations reduce water losses more than they reduce CO2 uptake so the plant can continue photosynthesising during the day.

Plants acquire CO2 from the atmosphere through their stomata. They also lose water through their stomata when they are open, so conditions that cause them to close their stomata to reduce water loss also reduce CO2 uptake and lower the photosynthetic rate. Such conditions include increased wind speed and water stress. Glasshouses are used to manipulate the physical environment to maximise photosynthetic rates.

2. Why would photosynthetic rate decline when the CO2 level is reduced?

Photocopying Prohibited

Plants have adaptations to reduce water loss from stomata

3. (a) Use figure B (previous page) to determine if CO2 or temperature had the greatest effect on photosynthetic rate:

4. Why would hot and windy conditions reduce photosynthetic rates?

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Method

Photosynthetic rate can be investigated by measuring the uptake of carbon dioxide (CO2) and production of oxygen (O2) over time. Measuring the rate of oxygen production provides an approximation of photosynthetic rate.

f 0.8-1.0 g of Cabomba stem were weighed. The stem was cut and inverted to ensure a free flow of oxygen.

2. Use the data to draw a graph on the grid above of the bubbles produced per minute vs light intensity:

: Measuring the production of oxygen in provides a simple means of measuring the rate of photosynthesis. Investigating Photosynthetic Rate41

f The stem was placed into a beaker filled with a 20°C solution of 0.2 mol L-1 sodium bicarbonate (NaHCO3) to supply CO2. An inverted funnel and a test tube filled with the NaHCO3 solution collected the gas produced.

3. Although the light source was placed set distances from the Cabomba stem, light intensity in lux was recorded at each distance rather than distance per se. Explain why this would be more accurate:

To investigate the effect of light intensity on the rate of photosynthesis in an aquatic plant, Cabomba aquatica

Hypothesis

Experimental set up

f The beaker was placed at distances (20, 25, 30, 35, 40, 45 cm) from a 60W light source and the light intensity measured with a lux (lx) meter at each interval. One beaker was not exposed to the light source (5 lx).

f Before recording data, the stem was left to acclimatise to the new light level for 5 minutes. Bubbles were counted for a period of three minutes at each distance.

See Appendix

Background

Aim

TheinLightresultsintensitylx(distance) Bubbles counted in three minutes Bubblesminuteper 5 0 13 (45 cm) 6 30 (40 cm) 9 60 (35 cm) 12 95 (30 cm) 18 150 (25 cm) 33 190 (20 cm) 35 Key

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4. The sample of gas collected during the experiment was tested with a glowing splint. The splint reignited when placed in the gas. What does this confirm about the gas produced? Idea

If photosynthetic rate is dependent on light intensity, more oxygen bubbles will be produced by Cabomba per unit time at higher light intensities.

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1. Complete the table (left) by calculating the rate of oxygen production (bubbles of oxygen gas per minute):

Carbon: Hydrogen: Oxygen:

Note: You can either work by yourself or team up with a partner. If you have beads or molecular models you could use these instead of the shapes on the next page.

(d) State the end products of photosynthesis:

f During photosynthesis and cellular respiration, molecules are broken down and recombined to form new molecules.

Carbon Hydrogen Oxygen Water

(a) State the starting reactants in photosynthesis:

(f) Use the atoms you have cut out to make the end products of cellular respiration.

f In this activity you will model the inputs and outputs of each of these processes using the atoms (carbon, hydrogen, and oxygen) on the next page. We have placed the atoms in boxes to make it easier to cut them out.

(b) State the total number of atoms of each type needed to make the star ting reactants:

(e) State the total number of atoms of each type needed to make the end products of cellular respiration: Carbon: Hydrogen: Oxygen:

Carbon: Hydrogen: Oxygen:

dioxideCarbonGlucose

(g) Name the end products of cellular respiration that are utilised in photosynthesis:

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(g) What do you notice about the number of C, H, and O atoms on each side of the photosynthesis equation?

(f) Use the atoms you have cut out to make the end products of photosynthesis.

(d) State the end products of cellular respiration:

f At the end of this activity you will be able to see how the reactants (starting molecules) are recombined to form the final products.

(a) State the starting reactants in cellular respiration:

(b) State the total number of atoms of each type needed to make the star ting reactants: Carbon: Hydrogen: Oxygen:

3. Write the equation for cellular respiration here:

(h) Name the energy source for this process and add it to the model you have made:

2. Write the equation for photosynthesis here:

(e) State the total number atoms of each type needed to make the end products of photosynthesis:

Modelling Photosynthesis and Cellular Respiration

Key Idea: Modelling photosynthesis and cellular respiration using paper cut outs will help you better understand the net chemical changes occurring.

1. Cut out the atoms and shapes on the following page. They are colour coded as follows:

This page is left blank deliberately

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HINT: Include reference to the activation energy and the current model of enzyme activity.

HINT: Name the factors affecting enzyme activity. Include reference to enzyme denaturation if it is relevant.

How enzymes work

HINT: Include reference to where the different phases of photosynthesis occur.

Factors affecting enzyme activity

HINT: Include reference to where the reactions of cellular respiration take place.

Photosynthesis

Cellular respiration

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NCEA Style Question: Photosynthesis

• an outline of the light dependent and light independent reactions.

1. Photosynthesis is a process in which plants use light to produce carbohydrate molecules. Discuss the process of photosynthesis and how its rate is affected by specific environmental factors (you may use extra paper if required):

Your discussion should include:

TEST 64

• the raw material(s) for photosynthesis and the final product(s)

• the location of photosynthesis in plant cells

• the role of enzymes in photosynthesis

• factors that affect the rate of photosynthesis

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• the role of enzymes in these processes

• the purpose of aerobic cellular respiration

Your discussion should include:

1. Aerobic cellular respiration is essential for providing energy (as ATP) for plant and animal cells. ATP can also be generated by anaerobic pathways in some circumstances. Discuss aerobic cellular respiration in plants and animals and contrast it with the anaerobic pathways they use for ATP generation. Describe situations in which each process is used:

• the raw material(s) for aerobic cellular respiration and anaerobic ATP generation and the final product(s)

• the location of these processes in plant and animal cells

• situations in which aerobic or anaerobic pathways are used. Cellular Respiration

NCEA Style Question:

• the anaerobic generation of ATP

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A word equation:

3. Label the following features of a chloroplast on the diagram below: inner membrane, outer membrane, granum, stroma, thylakoid disc, stroma lamellae

Foil

KEY TERMS AND IDEAS: Energy Transformations

A chemical equation: (b) Where does photosynthesis occur?

2. (a) Write the process of photosynthesis as:

C The phase in photosynthesis in which chemical energy is used for the synthesis of carbohydrate. Also called the Calvin cycle.

(f)

1. Complete the diagram of cellular respiration below by filling in the boxes:

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Glycolysis reactionLinkCO2 No.InputATP Waste Waste Waste No. ATP No. ATP(a) (b) (c) (d) (e) (g)

Krebs cycle light dependent phase light independent phase White

4. In the image below, par t of a variegated leaf is covered with aluminium foil. On the outline of the leaf below, draw or shade the parts of the leaf where you would expect photosynthesis to occur:

D Chain of enzyme-based redox reactions, which passes electrons from high to low redox potentials. The energy released is used to pump protons across a membrane and produce ATP. transport chain

electron

A Part of a metabolic pathway involved in the chemical conversion of carbohydrates to CO2 and water to generate usable energy as ATP. It begins when acetyl coenzyme A is attached to oxaloacetate.

(h)

5. Match each term to its definition, as identified by its preceding letter code

B The phase in photosynthesis when light energy is conver ted to chemical energy

Chromatid

1. What is the purpose of DNA replication?

Centromere links sister joinedconsistsReplicatedchromatidschromosomeoftwochromatidsatthecentromere.

Enzymes unzip the DNA and create a swivel point to unwind the DNA helix. 5'3'

Key Idea: Before a cell can divide, the DNA must be copied. DNA replication produces two identical copies of DNA. A copy goes to each new daughter cell.

DNA that will become the other chromatid

The centromere keeps sister chromatids together in an organised way until they are separated prior to nuclear division.

The 3' and 5' tags show the direction of the antiparallel strands. The directionality is important for replication.

3. (a) What does a replicated chromosome look like?

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f In eukaryotes, DNA in the nucleus is packaged with proteins into organised structures called chromosomes.

DNA that will become one chromatid

4. Explain what semi-conservative replication means:

3'5'3'5'

Enzymes catalyse the addition of bases during replication of the DNA strand. Enzymes proofread and correct any copying errors made during DNA replication.

DNANew

Each resulting DNA molecule is made up of one parent strand and one daughter strand of DNA. This is why DNA replication is called semi-conservative

DNA chromatidswithchromosomecreatesreplicationatwoidentical

chromosomeParent

DNA replication is called semi-conservative. This is because each resulting DNA molecule is made up of one parent strand and one daughter strand of DNA.

DNA replication duplicates chromosomes

Parent chromosome before replication. It is a double stranded molecule.

2. What would happen if DNA was not replicated before cell division?

f After the DNA has replicated, each chromosome is made up of two chromatids, which are joined at the centromere. The two chromatids will become separated during cell division to form two separate chromosomes.

Original ‘parent’ ‘daughter’ DNA

f Before a cell can divide, its DNA must be copied (replicated). DNA replication ensures that the two daughter cells that are produced receive identical genetic information.

(b) What is the purpose of the centromere?

DNAProhibitedReplication

DNA replication is semi-conservative

f DNA replication takes place between cell divisions. It is a semi-conservative process, meaning that each chromatid contains half original (parent) DNA and half new (daughter) DNA.

4 3' 3'5' 5'

replicationparentEnzymesDNAunzipDNAatthefork.Enzymes

The two strands are joined by base pairing Enzymes unwind parent DNA double helix

acts as a

The enzymes can work in only one direction and the strands are anti-parallel, so one strand is made in fragments that are later joined by other enzymes.

1. How are the new strands of DNA lengthened?

4. Describe three activities carried out by enzymes during DNA replication:

‘parent’Original add free nucleotides to the exposed bases on the template.

to match

DNA base pairing rule G pairs with C A pairs with T

2. What rule ensures that the two new DNA strands are identical to the original strand?

f The DNA strands can only be replicated in one direction, so one strand has to be copied in short segments, which are joined together later.

3. Why does one strand of DNA need to be copied in segments?

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Parent strand of DNA template nucleotides new DNA strand.

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New ‘daughter’ DNA

‘parent’Original DNA

(c)(b)(a)

Enzymes are involved at every step of DNA replication. They unzip the parent DNA, add the free nucleotides to the 3’ end of each single strand, join DNA fragments, and check and correct the new DNA strands.

Stages in DNA replication

New ‘daughter’ DNA G TAC

for the

2 3

Parent DNA is made up of two anti-parallel strands coiled into a double helix.

Two new double-stranded DNA molecules

f During DNA replication, new nucleotides (the units that make up the DNA molecule) are added at a region called the replication fork. The replication fork moves along the chromosome as replication progresses.

Nucleotide symbols

f This whole process occurs simultaneously for each chromosome of a cell and the entire process is tightly controlled by enzymes.

Key Idea: DNA replication is achieved by enzymes attaching new nucleotides to the growing DNA strand at the replication fork.

f Nucleotides are added in by complementary base-pairing: Nucleotide A is always paired with nucleotide T. Nucleotide C is always paired with nucleotide G.

Key Idea: The eukaryotic cell cycle can be divided into phases, although the process is continuous. Specific cellular events occur in each phase.

During interphase the cell increases in size, carries out its normal activities, and replicates its DNA in preparation for cell division. Interphase is not a stage in mitosis.

1. Briefly outline what occurs during the following phases of the cell cycle:

During interphase, the cell grows and acquires the materials needed to undergo mitosis. It also prepares the nuclear material for separation by replicating it.

f The first gap phase (G1).

Cytokinesis: Cytoplasm divides and the two cells separate. It is distinct from mitosis.

InterphaseDNAgrow

(b) Mitosis:

Mitosis Nuclear:division

During mitosis the chromosomes are separated. Mitosis is a highly organised process and the cell must pass checkpoints before it proceeds to the next phase.

f The S-phase (S).

Mitosis and cytokinesis (M-phase)

TheProhibitedEukaryotic Cell Cycle

Interphase

S (synthesis) phase: DNA replication, the chromosomes are duplicated.

Cells spend most of their time in interphase. Interphase is divided into three stages:

Mitosis and cytokinesis occur during M-phase. During mitosis, the cell nucleus (containing the replicated DNA) divides in two equal parts. Cytokinesis occurs at the end of M-phase. During cytokinesis the cell cytoplasm divides, and two new daughter cells are produced.

(a) Interphase:

The life cycle of a eukaryotic cell is called the cell cycle. The cell cycle can be divided into two broad phases; interphase and M phase. Specific activities occur in each phase.

First gap phase (G1): Cell increases in size and makes the mRNA and proteins needed for DNA replication.

Second gap phase (G2): Rapid cell growth and protein synthesis. Cell prepares for mitosis.

Mitosis grow replicate and prepare for division

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During interphase the nuclear material is unwound. As mitosis approaches, the nuclear material begins to reorganise in readiness for nuclear division.

f The second gap phase (G2).

Checkpoints during the cell cycle

One of the most important proteins in the regulation of the cell cycle is the protein produced by the gene p53. Mutations to the p53 gene are found in about 50% of cancers. p53

cellInstituteCancerNational

InterphaseMitosis

G2 Checkpoint:

Pass this checkpoint if: Cell is large enough.

Checkpoints in the Cell Cycle

Pass this checkpoint if: Cell is large enough.

cycle checkpoints:

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The cell cycle checkpoints are controlled by enzymes and regulatory proteins. Mutations to genes that encode these enzymes and proteins can lead to cancer. However, many genes are involved so many mutations are needed to develop cancer.

Pass this checkpoint if:  All chromosomes are attached to the mitotic spindle.

There are three regulatory points in the cell cycle called checkpoints. At each checkpoint, a set of conditions determines whether or not the cell will continue into the next phase. For example, cell size is important in regulating whether or not the cell can pass through the G1 checkpoint.

Cell has enough nutrients.

The product of the gene BRCA1 has been linked to DNA repair and may be involved in the metaphase checkpoint. Mutations to this gene and another gene called BRCA2 are found in about 10% of all breast cancers and 15% of ovarian cancers.

Chromosomes have been successfully duplicated.

1. Explain the importance cell

How could the failure of a cell cycle checkpoint lead to cancer?

Signals from other cells have been received.

Cancer

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G1 checkpoint

Key Idea: Regulatory checkpoints are built into the cell cycle to ensure that the cell is ready to proceed from one phase to the next. The failure of these systems can lead to cancer.

Metaphase checkpoint

InagloryBrocken

f Asexual reproduction: Some unicellular eukaryotes (such as yeasts) and some multicellular organisms (e.g. Hydra) reproduce asexually by mitotic division. Replacement reproduction:

Growth

Fertilised egg

Key Idea: Mitosis has three primary functions: growth of the organism, replacement of damaged or old cells, and asexual reproduction (in some organisms).

Some simple eukaryotic organisms reproduce asexually by mitosis. Yeasts (such as baker's yeast, used in baking) can reproduce by budding. The parent cell buds to form a daughter cell (right). The daughter cell continues to grow, and eventually separates from the parent cell.

(zygote) Embryo Adult boneBroken limbsDamaged

Mitotic cell division has three purposes:

1. Use examples to explain the role of mitosis in: (a) Growth of an organism: (b)

Matthias Zepper

f Growth: Multicellular organisms grow from a single fertilised cell into a mature organism. Depending on the organism, the mature form may consist of several thousand to several trillion cells. These cells that form the building blocks of the body are called somatic cells.

f Repair: Damaged and old cells are replaced with new cells.

Asexual reproduction

of damaged cells: (c) Asexual

Jpbarrass

Multicellular organisms develop from a single fertilised egg (zygote) and grow by increasing in cell numbers. Cells complete a cell cycle, in which the cell copies its DNA and then divides to produce two identical cells. During the period of growth, the production of new cells is faster than the death of old ones. Organisms, such as the 12 day old mouse embryo (above, middle), grow by increasing their total cell number and the cell become specialised as part of development. Cell growth is highly regulated and once the mouse reaches its adult size (above, right), physical growth stops and the number of cell deaths equals the number of new cells produced.

cellParentcellDaughter

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Repair

FunctionsProhibited of Mitosis

Mitosis is vital in the repair and replacement of damaged cells. When you break a bone or graze your skin, new cells are generated to repair the damage. Some organisms, like this sea star (above right) are able to generate new limbs if they are broken off.

(a) How many daughter cells are created:

(b) How many chromosomes does each daughter cell have?

occursreplicationDNA

f Mitosis results in the separation of the nuclear material and division of the cell. It does not result in a change of chromosome number.

KNOW 72

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(b) Plants:

Key Idea: Mitosis is an important part of the eukaryotic cell cycle in which the replicated chromosomes are separated and the cell divides, producing two new cells.

f In plants, mitosis takes place in the meristems. The meristems are regions of growth (where new cells are produced), such as the tips of roots and shoots.

The growing tip (meristem, M) is the site of mitosis in this plant root. The root cap below the meristem protects the dividing cells.

2. Where does mitosis take place in:

Mitosis produces identical daughter cells

f In animals, mitosis takes place in the somatic (body) cells. Somatic cells are any cell of the body except sperm and egg cells.

At any one time, only a small proportion of the cells in an organism will be undergoing mitosis. The majority of the cells will be in interphase.

f Mitosis is one of the shortest stages of the cell cycle. When a cell is not undergoing mitosis, it is said to be in interphase.

Parent cell 2N = 4

(a) Animals:

Mitosis is a stage in the cell cycle

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f M-phase (mitosis and cytokinesis) is the part of the cell cycle in which the parent cell divides in two to produce two genetically identical daughter cells (right).

Mitosis in onion cells (DIC light micrograph)

Daughter cell, 2N = 4Daughter cell, 2N = 4

RCN

Mitosis

1. Briefly outline the events in mitosis:

3. A cell with 10 chromosomes undergoes mitosis.

The cell divides forming two identical daughter cells. The chromosome number remains the same as the parent cell.

f Plant cells lack centrioles, and the spindle is organised by structures associated with the plasma membrane. In plant cells, cytokinesis involves formation of a cell plate in the middle of the cell. This will form a new cell wall.

Division of the cytoplasm. When cytokinesis is complete, there are two separate daughter cells.

f Mitosis is continuous, but is divided into stages for easier reference (1-6 below). Enzymes are critical at key stages. The example below illustrates the cell cycle in an animal cell.

1. What must occur before mitosis takes place?

Interphase refers to events between mitosis. The cell replicates its DNA and prepares for mitosis. The centrosome (containing two centrioles) also divides.

Some spindle fibres attach to and organise the chromosomes on the equator of the cell. Some spindle fibres span the cell.

The animal cell cycle and stages of mitosis

f In animal cells, centrioles (located in the centrosome), form the spindle. During cytokinesis (division of the cytoplasm) a constriction forms dividing the cell in two. Cytokinesis is part of M-phase, but it is distinct from mitosis.

Interphase

Other spindle fibres lengthen, pushing the poles apart and causing the cell to elongate.

DNA condenses into chromosomes. The nuclear membrane breaks down. The centrosomes migrate to the poles.

Anaphase

Centrosome containing centrioles (forms spindle)

LateCytokinesisTelophaseanaphase

Chromosomes appear as two chromatids held together at the centromere. The centrioles begin to form the spindle (made up of microtubules and proteins).

Key Idea: Mitosis is a continuous process, but it is divided into stages to help identify and describe what is occurring.

The cell cycle and stages of mitosis

Early prophase

Spindle fibres attached to chromatids shorten, pulling the chromatids apart. Spindle shortening is catalysed by enzymes.

Nucleus Nuclear membrane

Two new nuclei form. A furrow forms across the midline of the parent cell, pinching it in two.

Centrosomes move to opposite poles

Late prophase

Homologous pair of replicated chromosomes Metaphase Spindle

Mitosis and Cytokinesis

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3. formingplate (onion)

Cleavage furrow Plant

A ring of microtubules assembles in the middle of the cell, next to the plasma membrane, constricting it to form a cleavage furrow. In an energy-using process, the cleavage furrow moves inwards, forming a region of separation where the two cells will separate.

Cytokinesis

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HealthofDepartmentStateYorkNewCenter-dsworth Animal cell Constriction Cleavagemicrotubulesbyfurrow

In plant cells (below right), cytokinesis (division of the cytoplasm) involves construction of a cell plate (a precursor of the new cell wall) in the middle of the cell. The cell wall materials are delivered by vesicles derived from the Golgi. The vesicles join together to become the plasma membranes of the new cell surfaces. Animal cell cytokinesis (below left) begins shortly after the sister chromatids have separated in anaphase of mitosis.

Summarise what happens in each of the following phases: (a) Prophase: (b) Metaphase: (c) Anaphase: (d) Telophase:

(a) What is the purpose of cytokinesis? (b) Describe the differences between cytokinesis in an animal cell and a plant cell: Cytokinesis in an animal cell Cytokinesis in a plant cell Cell

Key Idea: Using pipe cleaners to model the stages of mitosis helps to visualise and understand the process and its end result.

6. Photo 4 shows the completion of mitosis.

5. In the space below, draw what happens next:

(b) On the photo, draw the centrosomes in their correct positions.

Photo 1 Photo 2 A B C

(b) How many chromosomes are in each cell?

Photo 4

(b) On photo 2, draw the new position of the centrosomes:

Photo 3

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1. (a) Photo 1 represents a cell in interphase before mitosis begins. The circular structures are the centrosomes. Name the labelled structures:

4. (a) What is happening in photo 3?

f Students used pipe cleaners and wool to model mitosis in an animal cell. Four chromosomes were used for simplicity (2N = 4). Images of their work are displayed below.

3. In the space below, draw what happens next, and give an explanation: A

(b)C:B:A: Why are there two copies of the centrosomes?

(a) How many cells are formed?

2. (a) What is happening to the structure labelled A in photo 2?

Adenosine triphosphate (ATP). The hydrolysis of the terminal phosphate provides the energy for cellular reactions.

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Putting it all Together:

1. In this diagram, fill in the rectangles with the process and the ovals with the substance used or produced in the process. The start molecules, carbon dioxide and water, have been provided to help you begin:

Metabolic reactions occur continuously and are frequently linked together in pathways. In plants, the ultimate source of energy for these reactions is sunlight, which is used to fix carbon and produce glucose. In animals, the energy comes from glucose directly, via the food that is eaten. Glucose is used in catabolic pathways that transfer the energy from the covalent bonds in glucose to ATP. ATP is then used to power cellular processes such as cell division. Each metabolic reaction is controlled by enzymes. The specific form and function of the cell determines, in part, how the metabolic reactions proceed.

Key Idea: Metabolic reactions do not occur in isolation. The reactions happen in a continuous series of reactions. Metabolism!

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2. Highlight or circle the processes in the diagram that occur in both plant and animal cells:

TEST 76

Light Alcohol or lactic acid H2O HCO22O (b) (c) (m) (k) (i) (l) (j) (a) A 2 CoA (f)

4. Complete the diagram below by naming organelle and a metabolic process that occurs there:

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ProcessesOrganelle: that occur here: 2)1)

ProcessOrganelle:that occurs here:

Equation for process:

Glycolysis is catalysed by a series of __________ reactions. It is an _____________ pathway (requiring no oxygen). The product of glycolysis is then __________ in the link reaction, __________ and the electron transport chain.

3. Use the word lists to complete the paragraphs below:

Enzyme DNA molecule

(n) (o)

(a) Word list: ATP, carbon, energy, glucose, photosynthesis. transforms light _________ into chemical energy by producing _______ and NADPH. These provide energy and reducing power to fix inorganic ____________ to form ___________.

(b) Word list: Anaerobic, enzyme, oxidised, Krebs cycle

ProcessOrganelle:that occurs here:

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Plant cell

G1 G2 M S

Equation for process:

(c) Word list: Cell cycle, DNA, enzymes, mitosis, replication. DNA _________________ occurs in the cell nucleus. It is controlled by _______________. The parent _____ unwinds and is copied to create two new ______ molecules. This occurs in the S phase of the ________ ________, before ___________.

The cell cycle and mitosis

HINT: Describe steps in DNA replication including the base pairing rule. Include a diagram of DNA replication.

REVISE 78

HINT: Describe the stages in the cell cycle and the events of mitosis. Note the differences in cytokinesis between animal cells and plant cells.

DNA replication

What You Know So Far: Cell Division56

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• a labelled diagram to suppor t your answer Cell

• the units that make up DNA

• the process of replication

1. What is the purpose of DNA replication?

2. Explain how the parent DNA is copied by semi-conser vative replication (you may use extra paper if required). Your answer should include:

DNA replication occurs prior to mitosis and cell division. It produces new DNA by semi-conservative replication.

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NCEA Style Question:

Count and record the number of cells in the image which are undergoing mitosis and those that are in interphase Estimate the amount of time a cell spends in each phase. StageNo. of cells% of cellstotal timeEstimatedinstage 100

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A chromatid contains half original (parent) ... ...is made up of two chromatids.

C The physical process of cell division, which divides the cytoplasm of a parental cell into two daughter cells.

The light micrograph (right) shows a section of cells in an onion root tip. These cells have a cell cycle of approximately 24 hours. The cells can be seen to be in various stages of the cell cycle. By counting the number of cells in the various stages it is possible to calculate how long the cell spends in each stage of the cycle.

E A term to describe the DNA replication process in which each double stranded DNA molecule contains one original strand and one new strand.

2. Match the statements in the table below to form complete sentences, then put the sentences in order to make a coherent paragraph about DNA replication and its role:

Replication is tightly controlled... ...to correct any mistakes. After replication, the chromosome... ...and half new (daughter) DNA. DNA replication... ...during mitosis

3. DNA replication occurs during the S (synthesis) phase of the cell cycle

1. Match each term to its definition, as identified by its preceding letter code

F The process of nuclear division in cells.

DNA replication is the process by which the DNA molecule... ...by enzymes.

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A The process of producing two identical copies of DNA from one original DNA molecule

D An organised structure of protein and DNA found in the nucleus of eukaryotic cells. It contains most of the cell's genetic material containing most of the DNA.

The enzymes also proofread the DNA during replication......is required before mitosis can occur.

B The sequence of events that a cell completes prior to and including its division.

InterphaseMitosisTotal

The chromatids separate... ...is copied to produce two identical DNA strands.

Onion root tip cells

cell

Write the complete paragraph here: semi-conservativemitosisDNAcytokinesischromosomecyclereplication

c E Demonstrate comprehensive understanding of genetic variation and change: Link biological ideas about genetic variation and change. The discussion may involve justifying, relating, evaluating, comparing and contrasting, or analysing.

Select biological ideas and processes from...

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c i Mutation as a source of new alleles. 60 62 64

c i Natural selection (differential survival and reproduction of individuals as a result of differences in phenotype). 87 92

crosscodominanceallele(genetic cross) crossing over dihybrid variationnaturalflow)migrationgeneticgenefoundereffectbottleneckGenetraitrecombinationrecessivephenotypemutationinheritancemonohybridmeiosislocuslinkedlethalassortmentindependentdominanceincompletehomozygousheterozygousgenotypegeneticdominantinheritanceallelevariationallelegenesallelepoolseffectpooldrift(=geneselection

Biological ideas and processes relating to factors affecting allele frequencies within a gene pool numberActivity

c 2 Factors that cause changes to the allele frequency in a gene pool. 87 - 95

c v The effect of crossing over and linked genes on dihybrid inheritance. 80 81 82

Select biological ideas and processes from...

c A Demonstrate understanding of genetic variation and change: Define and use annotated diagrams or models to describe genetic variation and change. Describe characteristics of, or provide an account of, genetic variation and change.

c 1 Sources of variation within a gene pool. 59 - 82

Achievement criteria for achieved, merit, and excellence

Genetic variation and change

A population's gene pool consists of all the alleles (genetic variants) present. Mutation is the source of all new alleles. Processes in gene pools, including natural selection and genetic drift, result in allele frequencies changing over time.

Achievement criteria and explanatory notes

c ii Independent assortment, segregation, and crossing over during meiosis. 65 69

c iii Monohybrid inheritance to show the effect of codominance, incomplete dominance, lethal alleles, and multiple alleles. 70 78

Sources

Key terms of variation

Genetic variation and change involves the following concepts...

c iv Dihybrid inheritance with complete dominance. 79 82

Kristian Peters

c ii Migration (transfer of alleles or genes from one population to another). 87

c iii Genetic drift (random fluctuations in the numbers of gene variants in a population). 93 - 95

Explanatory notes: Genetic variation and change numberActivity

c M Demonstrate in-depth understanding of genetic variation and change: Provide reasons as to how or why genetic variation and change occurs.

Biological ideas and processes relating to sources of variation within a gene pool numberActivity

Achievement Standard 2.5

c Define the terms dominant allele, recessive allele, lethal allele, codominant alleles, incomplete dominance, and multiple alleles.

c Define genetic drift and explain its consequences to allele frequencies in populations. Using examples, explain when genetic drift is most important.

Sources of variation within a gene pool

Activities 59 86

c Solve monohybrid crosses involving simple dominant-recessive inheritance, codominance, incomplete dominance, and lethal alleles. Describe the resulting phenotype and genotype ratios in each case.

Marc King

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Jeff Podos

c Define genotype, phenotype, monohybrid cross, dihybrid cross, F1 generation, F2 generation and demonstrate correct use of these terms in studies of inheritance.

c Define the term gene pool and allele frequency. Understand that a change in the allele frequencies of a population over time is evolution.

c Define the founder effect. Using examples, describe the genetic and evolutionary consequences of the founder effect, including the greater impact of genetic drift in founder populations.

c Describe the main events occurring in meiosis, including the reduction division and its significance, the segregation of alleles, crossing over and independent assortment, and second division.

c Describe the process of natural selection and explain its role in sorting variation and establishing adaptive genotypes. Using examples, explain how natural selection tends to reduce genetic variation within a gene pool and increase genetic differences between populations.

c Recall the sources of variation in gene pools as a result of mutation and sexual reproduction.

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

c Explain what is meant by mutation and describe mutation as the source of all new alleles.

c Identify both genetic and environmental causes of variation. Describe examples of discontinuous and continuous variation in populations and their origin.

c Define the terms allele, locus, heterozygous, and homozygous in relation to chromosomes. Demonstrate your understanding of how these terms are used in inheritance.

c List factors that a cause change in allele frequencies in a gene pool, including mutation, natural selection, gene flow (migration), and genetic drift.

Factors causing changes to allele frequencies in gene pools

c How gene flow (migration) may affect allele frequencies. Explain how gene flow tends to reduce the genetic differences between populations.

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

c Solve problems involving dihybrid crosses of unlinked, autosomal genes for two independent traits showing complete dominance.

c Define the bottleneck effect (also called genetic bottleneck or population bottleneck). Using examples, describe the genetic and evolutionary consequences of the bottleneck effect, including the greater impact of genetic drift in populations that have been through a bottleneck.

c Describe Mendel’s principles of inheritance and explain their importance to our understanding of heredity and evolution.

Activities 87 98

c Explain recombination as the exchange of alleles between homologous chromosomes as a result of crossing over.

c Explain how meiosis and fertilisation contribute to variation in the offspring.

c Describe and explain the effect of crossing over and linked genes on dihybrid inheritance.

What you need to know for this Achievement Standard

Jeff Podos

(a) Heterozygous:

3. What is a homologous pair of chromosomes?

1. Define the following ter ms describing the allele combinations of a gene in a diploid sexually reproducing organism: (a) Heterozygous:

Paternal chromosome originating from the sperm of this person's father.

1 2 2 3 4 3 4 5 1 5 BAcBac

(b) Homozygous dominant:

Homologous chromosomes

Maternal chromosome originating from the egg of this person's mother.

When both chromosomes have identical copies of the dominant allele for gene B the organism is homozygous dominant for that gene.

Key Idea: Eukaryotes generally have paired chromosomes. Each chromosome contains many genes, and each gene may have a number of versions called alleles.

The diagram above shows the complete chromosome complement for a hypothetical organism. It has a total diploid (2N) number of ten chromosomes, as five, nearly identical pairs (each pair is numbered).

When both chromosomes have identical copies of the recessive allele for gene C the organism is said to be homozygous recessive for that gene.

f The pairs are called homologues or homologous pairs. Each homologue carries an identical collection of genes, but the version of the gene (the allele) from each parent may differ. This diagram shows the position on the same chromosome of three different genes (A, B and C) controlling three different traits (features).

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(b) Homozygous dominant: (c) Homozygous recessive:

Having two different versions of gene A is a heterozygous condition. Only the dominant allele (A) will be expressed.

2. For a gene given the symbol ‘A’, identify the alleles present in an organism that is:

f In most sexually reproducing organisms, the cells of the body (not gametic cells) contain two complete sets of chromosomes (called diploid) and the chromosomes are found in pairs. Each parent contributes one chromosome to the pair.

Genes occupying the same position or locus on a chromosome code for the same characteristic (e.g. ear lobe shape).

2. What factors determine the phenotype?

(a) Mutation:

3. How could two individuals with the same genotype have a different phenotype?

Sexual reproduction

Mutation: Substitute T instead of C

Sources of Variation

Genotype

Genes interact to influence the phenotypic result.

Key Idea: Variation may come from changes to the genetic material (mutation), through sexual reproduction, and as a result of the effects of the environment.

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KNOW 84

Environmental factors influence expression of the genotype. The external environment includes physical factors (e.g. temperature) or biotic factors (e.g. competition). The internal environment (e.g. hormones) and chemical tags and markers on the DNA may also affect genotype expression.

Photocopying Prohibited

(b) Genotype:

Mutations are changes to the DNA Changes to the DNA modify existing genes. Mutations can create new alleles.

Environmental factors

OriginalMutantDNADNA

The genetic make up of the individual is its genotype. It determines the genetic potential of the individual.

Sexual reproduction involves meiosis and mate selection Sexual reproduction rearranges and reshuffles the genetic material into new combinations. Fertilisation unites dissimilar gametes to produce more variation.

The genetic instructions for creating the individual (the genotype) may be modified by the environment during and after development to produce the phenotype (the physical expression of the modified genetic information).

Phenotype

60 LINK 61 WEB 60

Mutations

1. Define the following ter ms:

Why is variation important?

Variation by sexual reproduction by asexual

f Variation refers to the diversity of phenotypes or genotypes within a population or species. Variation arises through mutation (the source of new alleles) and sexual reproduction (which shuffles alleles into new combinations).

(b) Why is variation important in a changing environment?

f Variation provides the raw material for adaptation and so enables populations respond to changes in their environment. Very gradual environmental changes, e.g. mountain building, allow sufficient time for populations (even those that reproduce asexually) to acquire the variability to adapt. However rapid environmental changes, e.g. the emergence of a new strain of disease, demand a more rapid response.

USDA

5. How could a favourable mutation spread through different bacterial lineages?

4. (a) What is variation and how does it arise in sexually reproducing organisms?

f Variation in species that reproduce asexually is generated by mutation and sometimes (as in bacteria) by gene transfers.

Some asexually reproducing organisms (e.g. bacteria) are able to exchange genes occasionally. Bacteria exchange genes with other bacteria during a process called conjugation (shown by the thicker blue line). This allows mutations arising in one lineage to be passed to another.

Variation

DEFDEFDEFDEFreproductionFDEClones Mutation D Mutation E Mutation F DDDDDDABABC AB AB ABAB ABCABCABBCMutation A Mutation B Variation MutationrecombiningfromallelesCA PREVIEWONLYNotforClassroomUseNotforClassroomUse

Aphids can reproduce both sexually and asexually. Females hatch in spring and give birth to clones. Many generations are produced asexually. Just before autumn, the aphids reproduce sexually. The males and females mate and the females produce eggs which hatch in spring. This increases variability in the next generation.

During meiosis, alleles are recombined in new combinations. Some combinations of alleles may be better suited to a particular environment than others. This variability is produced without the need for mutation. Beneficial mutations in separate lineages can be quickly combined through sexual reproduction.

f Variation is important in providing adaptable defences against disease. Species with adaptations to survive a disease flourish. Those without, die out. It is thought that sexual reproduction is an adaptation to increase the chances that any one of the offspring produced will have the allele combinations that enable them to survive a disease.

These diagrams show how three beneficial mutations could be combined through sexual or asexual reproduction.

f Examples of quantitative traits include height in humans for any given age group, skin colour range for any particular population, length of leaves in plants, grain yield in corn, growth rate in young mammals, and milk production in cattle. Most quantitative traits are influenced by environmental factors.

f A phenotype is an observable characteristic of an individual (e.g. eye colour). Individuals show particular variants of phenotypic characters called traits, e.g. blue eye colour.

Quantitative traits

Growth rate in piglets

Measureable phenotypic range

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Key Idea: Genetic variation may be continuous as a result of quantitative traits (e.g. leaf length) or it may be discontinuous as a result of qualitative traits (e.g. AB blood type).

Three genes (call them A,B, and C) largely account for most of the skin colour differences in human populations. Shades of skin colour range from very dark to very pale, with most individuals being somewhat intermediate. An individual with no dominant alleles lacks dark pigment (aabbcc). Full pigmentation (black) requires six dominant alleles (AABBCC).

Albinism is qualitative trait

f Quantitative traits are measurable phenotypes that depend on the actions of many genes and the environment. These traits vary over a range, producing a large number of phenotypic variants.

Low High Frequency

KNOW 86

Milk production in cattle

Phenotypes exhibit variation

f Traits can be quantitative (they show measureable variation) or qualitative (they show variation that falls into non-numerical descriptive categories).

Grain yield in corn

f Within any particular population, a quantitative trait has a continuous number of variants, so any value along a continuum is possible. This can be displayed as a histogram of trait value against its frequency. Individuals fall somewhere on a normal distribution curve of the phenotypic range (right).

Examples Genetic Variation

1. Using examples, explain the difference between continuous and discontinuous variation:

f Other examples of qualitative traits include:

(a) Wool production in sheep:

(f) Flower colour in snapdragons:

2. (a) Why do quantitative traits show continuous variation?

f Qualitative traits are determined by only one gene or two genes. As a result, there are only a few phenotypic variations possible and individuals fall into discrete categories.

1614121086420

(a) Does this represent a quantitative or qualitative trait?

(b) Hand span in humans:

the comb shape in chickens (right) and the flower colour in snapdragons (left) are qualitative traits. They are either one or the other, but not in-between.

Length of leaf (mm)

Qualitative traits

f For example, the ABO blood typing in humans recognises four discontinuous traits determined by three codominant alleles: A, B, AB or O. Individuals fall into discrete categories (right). Variation in qualitative traits can be displayed in a bar graph.

3. Identify each of the following phenotypic traits as continuous (quantitative) or discontinuous (qualitative):

(b) Why do qualitative traits show discontinuous variation?

• Comb shape in poultry (below, right). Birds have one of four phenotypes depending on which combination of four alleles they inherit. The dash (missing allele) indicates that the allele may be recessive or dominant.

CRCR CWCW Single comb rrpp Walnut comb R_P_ Pea comb rrP_ Rose comb R_pp Frequency of blood types in New Zealand ABABO (%)Frequency 50403020100 Blood type leavesofNumber

• Albinism is the result of the inheritance of recessive alleles for melanin production. Those with the albino phenotype lack melanin pigment in the eyes, skin, and hair.

(e) Body weight in mice:

4. The graph on the right shows leaf length in ivy.

answer:Both

Leaf length in ivy 3045607590

15 105

• Flower colour in snapdragons (below) is also a qualitative trait determined by two alleles (red and white) The alleles show incomplete dominance and the heterozygote (CRCW) exhibits an intermediate phenotype between the two homozygotes.

(b) Explain your

(d) Albinism in mammals:

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f Mutations can arise naturally through copying errors during DNA replication or the can be induced by chemical or physical agents called mutagens

Most mutations are harmful. This is because changes to the DNA sequence of a gene may change the amino acid encoded by the gene. The alteration may change the way the protein folds and prevent it from carrying out its usual biological function. Individuals with the mutation for albinism (above) are more prone to many types of cancers and may have impaired vision.

Beneficial mutations

KNOW 88

Mutated sequenceNormal sequence to

Arg GGC GCCMutates

sickle cell mutation is caused by a single mutation resulting in the production of an abnormal haemoglobin molecule. Under low oxygen conditions, the red blood cells take on a sickle shape.

Silent mutations

The Role of Mutations in Populations

Sometimes a mutation can be beneficial. The mutation may result in a more efficient protein or produce an entirely different protein that can improve the survival of the organism. In viruses (e.g. Influenzavirus above) genes coding for the glycoprotein spikes (arrowed) are constantly mutating, producing new strains that avoid detection by the host's immune system.

Spikes

GC A G T C Arg

Photocopying Prohibited

1. (a) Define the term mutation: (b) How do mutations introduce new alleles into a population? 2. Distinguish between harmful, beneficial, and silent mutations:The

Normal RBC

CDC

cc3.0collegeOpenStax PREVIEWONLYNotforClassroomUseNotforClassroomUse

Key Idea: Changes to the DNA sequence are called mutations. Mutations are the ultimate source of new genetic information, i.e. new alleles.

f Mutations are changes to the DNA sequence. An altered DNA sequence changes the gene sequence and results in a new allele. If the mutation is inherited it can be passed on to future generations.

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f A mutation may affect a single gene (gene mutation) or large blocks of genes (chromosomal mutation).

Harmful mutations

A silent mutation encodes the same amino acid (therefore the same protein) even though the DNA sequence is changed. This occurs because several codons in the genetic code may code for the same amino acid so the phenotype is unaffected. However, there is evidence that silent mutations may affect the efficiency of protein translation and may be important if selection pressures on that allele change.

Sickled RBC

f Beneficial mutations spread through a gene pool. Harmful mutations are usually eliminated from the gene pool because their effects on the phenotype are detrimental to survival and reproduction.

The code is read in 3-base sequences called codons. One codon codes for one amino acid. GlyGly DNA sequence formed acid sequence)

2. Why has the connexin 26 mutation not been eliminated from the population?

f The mutation (shown below) involves the deletion of the 35th nucleotide base. It produces a shortened polypeptide chain and results in a protein that does not function correctly.

The deletion creates a recessive allele. A person is deaf when they carry two recessive alleles.

f Recall that mutations can produce new alleles. A single mutation to the gene coding for a protein called connexin 26 produces a new allele.

The image above shows a single connexin 26 protein. Connexin 26 is found in many areas of the body including the inner ear.

1. (a) Identify the type of mutation that produces the connexin 26 mutation:

f If a person inherits two copies of the mutation (one from each parent) they suffer from a form of genetic hearing loss called NSRD.

Mutation causes a short polypeptide chain

Polypeptide formed codes for stopsequence AT T TG GG G G GGG Gly Val Stop

Six connexin 26 proteins join together. They form a connection called a gap junction between cells which allows molecules, ions and electrical impulses to directly pass between cells. If these proteins are not made correctly, they cannot form the final structure.

A A

(amino

Leu

Mutations Can Produce New Alleles

TGA

Val Asn Normal

(b) What phenotype does the connexin 26 mutation produce?

Connexin polypeptide26chains

The most common mutation in this gene is deletion of the 35th base (guanine (G)). Deleting the guanine changes the code and results in a short chain, which cannot function.

Polypeptide

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CC 28 42 Leu

f Carriers (individuals with only one copy of the mutated gene) are unaffected.

The gene that codes for the protein connexin 26 is on chromosome 13.

28 42 A AT T TGGG G G GGG CC 2

Mutated

Key Idea: A mutation to the gene coding for the connexin 26 protein produces a new allele. The phenotype for the new allele is congenital deafness.

f Only gametic mutations will be inherited. Somatic mutations are not inherited but may affect an organism in its lifetime (e.g. a cancer).

f The red delicious apple (right) is a chimaera (an organism with a mixture of mutated and normal cells). A mutation occurred in the part of the flower that developed into the fleshy part of the apple. The seeds are unaffected by the mutation, so it is not inherited.

(goldphenotypeMutantcolour) (redphenotypeNormalcolour) 1. Distinguish between somatic and gametic mutations: 2. Why are only gametic mutations inherited? LINK 60 WEB 64 LINK 62 Gametic mutation Gametic mutation to mutationcarriesorganismEntirespermthe Half of thegametesthecarrymutationGametesOrganismgametesParentalEmbryoofoffspring Somatic mutation mutationSomaticNoneofthegametesthecarrymutationPatchofaffectedarea cellSperm cellEgg PREVIEWONLYNotforClassroomUseNotforClassroomUse

Somatic and Gametic Mutations

Key Idea: Gametic mutations affect egg and sper m cells, these mutations are inherited. Somatic mutations affect body cells, they are not inherited.

f Gametic cells are the reproductive (sex) cells of an organism (the egg and sperm). Mutations occurring in these cells are called germline or gametic mutations

f Somatic cells (body cells) are all the remaining cells. Mutations to these cells are called somatic mutations

KNOW 90

chromosomesHomologous pair up. The pairs are separated. An chromosomes.completeTheygametesHaploidaretimelineTheeachreplicatedcellintermediateformswithacopyofchromosome.chromosomesupagain.Thisthechromatidsseparated.(N)form.haveonesetof

The process of meiosis

f DNA replication precedes meiosis. If genetic mistakes (mutations) occur here, they will be passed on (inherited by the offspring).

Meiosis produces variation

Crossing over has occurred.

Key Idea: Meiosis is a reduction division that produces haploid cells for the purposes of sexual reproduction.

f Meiosis involves a single chromosomal duplication followed by two successive nuclear divisions, and it halves the diploid chromosome number.

Chromatid

2. Describe how meiosis produces variation in the gametes:

Diploid (2N) cell has two complete sets of chromosomes, one from the mother, one from the father. N is the haploid number of chromosomes. Each chromosome is replicated.

1. (a) What is the purpose of meiosis?

f Meiosis is a special type of cell division necessary for the production of sex cells (gametes) for the purpose of sexual reproduction.

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f During meiosis, a process called crossing over may occur when homologous chromosomes may exchange genes. This further adds to the variation in the gametes.

What is meiosis?

(b) Where does meiosis take place?

Meiosis

f Meiosis is an important way of introducing genetic variation. The assortment of chromosomes into the gametes (the proportion from the father or mother) is random and can produce a huge number of possible chromosome combinations.

f Recall that each sister chromatid is one half of a replicated chromosome. When they are separated, they are once again called chromosomes.

f An overview of meiosis is shown on the right. Meiosis occurs in the sex organs of plants and animals.

KNOW 92 Meiosis and Variation66

Possible gametes with crossing over (parental types and recombinants) crossingover

chromatidsSisterCentromere chromatidsNon-sister No

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Meiosis produces variation

Crossing over and recombination

1. (a) What is crossing over? (b) How does crossing over increase the variation in the gametes (and hence the offspring)? PREVIEWONLYNotforClassroomUseNotforClassroomUse

chromatidsNon-sister Crossingover chromosomesRecombined chromatidsRecombined

f Meiosis creates genetic variation in the gametes as alleles are reshuffled into different combinations. This variation arises through crossing over and independent assortment.

f Because of meiosis, siblings with the same biological parents can all appear very different, although there is often a family resemblance.

f Crossing over refers to the mutual exchange of pieces of chromosome (and their genes) between homologous chromosomes.

Key Idea: Meiosis produces variation. Independent assortment and crossing over are two important ways of introducing variation into the gametes formed during meiosis.

f Independent assortment is the random alignment and distribution of homologous chromosomes to the gametes.

f When the replicated chromosomes are paired during the first stage of meiosis, non-sister chromatids may become entangled and segments may be exchanged in a process called crossing over

f Crossing over results in the recombination of alleles (variations of the same gene) producing greater variation in the offspring than would otherwise occur.

Possible gametes with no crossing over (only parental types)

f Chromosomes replicate during interphase, before meiosis, to produce replicated chromosomes with sister chromatids held together at the centromere (see below).

(b) How does it increase genetic variation in the gametes?

2. (a) What is independent assortment?

Independent assortment

Gamete 3 Gamete 4

f Independent assortment is the random alignment and distribution of chromosomes during meiosis. It is an important mechanism for producing variation in gametes.

(b) Which genes have been exchanged between the homologous chromosomes?

f For the example right, there are two chromosome pairs. The number of possible allele combinations in the gametes is 22 = 4 (only two possible combinations are shown).

f This results in the production of 2x different possible combinations (where x is the number of chromosome pairs).

(a) Draw the gene sequences for the four chromatids (above), after crossing over has occurred at crossover point 2:

3. (a) Draw the other two gamete combinations not shown in the diagram above:

(b) For each of the following 2N chromosome numbers, calculate the number of possible gamete combinations: i. 8: ii. 24: iii 64:

4. Homologous chromosomes are shown below, with possible crossover points for a chromatid marked with numbered arrows.

IntermediateOocytecellsEggsGenotypeAaBb a Ab Ab aB a BBbb AAaa AA bb BB aa b A b A aB B a B a MeiosischromosomesReplicated 1 23 4 5 6 7 89chromosomesHomologous Chromatid 1 Chromatid 2 Chromatid 3 Chromatid 4 a bcdefg h ijklmn o p a bcdefg h ijklmn o p A BCDEFG A BCDEFG H IJKLMN O P H IJKLMN O P 4321 PREVIEWONLYNotforClassroomUseNotforClassroomUse

f The law of independent assortment states that the alleles for separate traits are passed independently of one another from parents to offspring. In other words, the allele a gamete receives for one gene does not influence the allele received for another gene.

Each of your somatic cells contain 46 chromosomes, 23 maternal chromosomes and 23 paternal chromosomes. Therefore, you have 23 homologous (same) pairs. For simplicity, the number of chromosomes studied in this exercise has been reduced to four (two homologous pairs). To study the effect of crossing over on genetic variability, you will look at the inheritance of two of your own traits: the ability to tongue roll and handedness. This activity will take 25-45 minutes.

HandednessTraitPhenotypeGenotypeTonguerolling 10 chromosomePaternal chromosomePaternal ChromosomenumberHomologouspair Homologouspair initialsYour chromosomeMaternalYourgenotype 210 2 LB LBLB LB tT r r 10 LB t 10 LB t10 LB T 10 LB T 2LB r 2LB r 2LB r 2LB r

Background

Chromosome # Phenotype Genotype

Tongue roller TT, Tt roller handed RR, Rr Left handed

tt 2 Right

1. Collect four ice-blocks sticks. These represent four chromosomes. Colour two sticks blue or mark them with a P. These are the paternal chromosomes. The plain sticks are the maternal chromosomes. Write your initial on each of the four sticks. Label each chromosome with their chromosome number (below).

BEFORE YOU START THE SIMULATION: Partner up with a classmate. Your gametes will combine with theirs (fertilisation) at the end of the activity to produce a child. Decide who will be the female, and who will be the male. You will need to work with this person again at step 6.

3. Simulate prophase I by lining the duplicated chromosome pair with their homologous pair (below). For each chromosome number, you will have four sticks touching side-by-side (A). At this stage crossing over occurs. Simulate this by swapping sticky dots from adjoining homologues (B).

Label four sticky dots with the alleles for each of your phenotypic traits, and stick each onto the appropriate chromosome. For example, if you are heterozygous for tongue rolling, the sticky dots with have the alleles T and t, and they will be placed on chromosome 10. If you are left handed, the alleles will be r and r and be placed on chromosome 2 (below).

(B)(A) LB10 t LB10 t LB10 T LB10 T LB2 r LB2 r LB2 r LB2 r LB2LB10 t LB10 T LB10 T LB2 rr LB2 r LB2 r LB10 t Key Idea: We can simulate crossing over, gamete production, and the inheritance of alleles during meiosis using ice-block sticks to represent chromosomes. Modelling Meiosis67 LINK 66 WEB 67 PREVIEWONLYNotforClassroomUseNotforClassroomUse

10 Non-tongue

Record your phenotype and genotype for each trait in the table (right). If you have a dominant trait, you will not know if you are heterozygous or homozygous for that trait, so you can choose either genotype for this activity.

Photocopying ProhibitedKNOW 94

2. Randomly drop the chromosomes onto a table. This represents a cell in either the testes or ovaries. Duplicate your chromosomes (to simulate DNA replication) by adding four more identical ice-block sticks to the table (below). This represents interphase

4. Randomly align the homologous chromosome pairs to simulate alignment on the metaphase plate (as occurs in metaphase I). Simulate anaphase I by separating chromosome pairs. For each group of four sticks, two are pulled to each pole.

5. Telophase I: Two intermediate cells are formed. If you have been random in the previous step, each intermediate cell will contain a mixture of maternal and paternal chromosomes. This is the end of meiosis 1 Now that meiosis 1 is completed, your cells need to undergo meiosis 2. Carry out prophase II, metaphase II, anaphase II, and telophase II. Remember, there is no crossing over in meiosis II. At the end of the process each intermediate cell will have produced two haploid gametes (below).

LB10 t LB10 T LB10 T LB2 r LB2 r LB2 r LB2 r LB10 t LB10 T LB2 r LB2 r LB10 t LB10 t LB10 T LB10 T LB2 r LB2 r LB2 r LB2 r LB10 t 10LB T LB2 r LB2 r 10LB t LB10 t LB2 r LB10 T LB2 r 10LB T LB2 r LB2 r 10LB t chromosomeMaternal chromosomeMaternal chromosomePaternal Intermediate

Haploid (N) gamete Intermediate

TongueHandednessTraitPhenotypeGenotyperolling PREVIEWONLYNotforClassroomUseNotforClassroomUse

6. Pair up with the partner you chose at the beginning of the exercise to carry out fertilisation. Randomly select one sperm and one egg cell. The unsuccessful gametes can be removed from the table. Combine the chromosomes of the successful gametes. You have created a child! Fill in the following chart to describe your child's genotype and phenotype for tongue rolling and handedness. cell 1 cell 2

Key Idea: Genetic information is inher ited from parents in units called genes. Mendel's laws of inheritance govern how these genes are passed on to the offspring.

White Purple Parent Generationplants1Generation 2 The offspring are inbred (self pollinated) X X Meiosis EggEggEggEgg

Homologous pair of chromosomes, each has a copy of the gene on it (A or a)

This diagram shows two genes (A and B) that code for different traits. Each of these genes is represented twice, one copy (allele) on each of two homologous chromosomes. The genes A and B are located on different chromosomes and, because of this, they will be inherited independently of each other, i.e. the gametes may contain any combination of the parental alleles.

Characteristics of both parents are passed on to the next generation as discrete entities (genes).

KNOW 96

Photocopying Prohibited

Law of segregation

Allele pairs separate independently during gamete formation, and traits are passed on to offspring independently of one another (this is only true for genes on separate chromosomes).

Particulate inheritance

These gametes are eggs (ova) and sperm cells. The allele in the gamete will be passed on to the offspring.

2. An oocyte (egg producing cell) had the following allele combination: Tt Gg and produced the following combination of alleles in the egg cells: TG, Tg, tG, tg. Which law of inheritance applies to this scenario?

Mendel’s Laws of Inheritance68 LINK 66 WEB 68 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Law of independent assortment

1. State the property of genetic inheritance which applies to the following scenario: The crossing of a tall and dwarf pea plant produces all tall pea plants in the first generation, which when crossed, produce both tall and dwarf pea plants in the second generation.

This model explained many observations that could not be explained by the idea of blending inheritance, which was universally accepted prior to Mendel's work. The trait for flower colour (right) appears to take on the appearance of only one parent plant in the first generation, but reappears in later generations.

Oocyte AaAa a aAA aaEggsAA BB b b Oocyte Genotype:AaBbAA aa BBb Intermediatebcell aBIntermediatecella BAAb b AAb b aB a B

During meiosis, the two members of any pair of alleles segregate unchanged and are passed into different gametes.

f Traits are particular variants of a phenotypic (observed physical) character. For example, a phenotype is eye colour, a trait is blue eye colour.

Some of the best known experiments in phenotypes are the experiments carried out by Gregor Mendel (right) on pea plants. During one of the experiments (shown below) he noticed how traits expressed in one generation disappeared in the second generation, but reappeared in the third generation. In his experiments Mendel used true breeding plants. When self-crossed, true breeding organisms produce offspring with the same phenotypes as the parents.

Traits are inherited

• Height (tall or short)

Mendel was able to explain his observations in the following way:

RR RR Rr RRRr Rr Rr Rr RrRr RrRr RrRr Rr rr rr rr

Key Idea: Traits are controlled by genes, and can be inherited and passed from one generation to the next. A trait may be caused by a dominant or recessive allele.

In this diagram R (round) is dominant to r (wrinkled).

• Pod shape (inflated or constricted)

f Traits are controlled by alleles (variations of a gene).Traits are inherited and the alleles controlling a trait may be dominant or recessive (although degrees of dominance are also possible).

f Traits are determined by a unit, which passes unchanged from parent to offspring (we now know that these units are genes).

f Traits may not physically appear in an individual, but the units (genes) for them can still be passed to its offspring.

Mendel studied seven phenotypic characters of the pea plant:

f In contrast, a recessive trait will only be expressed when an individual has two copies of the recessive allele.

• Position of the flowers on the stem (axial or terminal)

How can this be explained?

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P1 F1 F2 Wrinkled Round

f Dominant alleles produce a dominant trait. A dominant trait overrides a recessive trait. Only one copy of the dominant allele needs to be present for the dominant trait to be expressed.

One of the experiments crossed true breeding round seed plants with true breeding wrinkled seed thousands of seeds produced, all were round, none were wrinkled.

• Seed shape (round of wrinkled)

• Pod colour (green or yellow)

• Seed colour (yellow or green)

Mendel's experiments

f Each individual inherits one unit (gene) for each trait from each parent (each individual has two units).

• Flower colour (violet or white)

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Mendel then crossed the F1 offspring together. The wrinkled seed reappeared in the next generation. He saw similar results with all the other phenotypic characters he studied.

Outplants.ofthe

© 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited 98 Dwarf Trait Possible phenotypes Results Dominance Ratio shapeSeed Recessive:Dominant: colourSeed RecessiveDominant: colourPod RecessiveDominant: positionFlower RecessiveDominant: shapePod RecessiveDominant: lengthStem RecessiveDominant: Wrinkled Round Green Yellow Green Yellow Axial TerminalInflatedConstricted RoundWrinkled TOTAL YellowGreen TOTAL YellowGreen TOTAL TerminalAxial TOTAL InflatedConstricted TOTAL DwarfTall TOTAL 54741850 7324 60222001 8023 152428 580 207651 858 882299 1181 277787 1064 WrinkledRound 2.96 : 1 1. Define a trait: 2. Describe the difference between the expression of dominant and recessive traits: 3. In Mendel's pea experiments on the previous page: (a) What was the ratio of smooth seeds to wrinkled seeds in the F2 generation? (b) Why did the wrinkled seed trait not appear in the F1 generation? 4. Mendel examined seven phenotypic traits. Some of his results from crossing heterozygous plants are tabulated below. The numbers in the results column represent how many offspring had those phenotypic features. (a) Study the results for each of the six experiments below. Determine which of the two phenotypes is dominant, and which is the recessive. Place your answers in the spaces in the dominance column in the table below. (b) Calculate the ratio of dominant phenotypes to recessive phenotypes (to two decimal places). The first one has been done for you (5474 ÷ 1850 = 2.96). Place your answers in the spaces provided in the table below: PREVIEWONLYNotforClassroomUseNotforClassroomUse

PP Homozygous purple X Homozygous white pp F1Genotypes: All Pp Phenotypes: All purple X 75% Pp- purple 25% pp- white Monohybrid cross F1 Monohybrid cross F2 PPppPp Heterozygous purple pPPpgametesMale gametesFemale Gametes:Parents Parents A true-breeding organism is homozygous for the gene involved. The F1 (first filial generation) offspring of a cross between two true breeding parent plants are all purple (Pp). A cross between the F1 offspring (Pp x Pp) would yield a 3:1 ratio in the F2 of purple (PP, Pp, Pp) to white (pp). Pp Heterozygous purple © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited KNOW 99 1. Study the diagrams above and explain why white flower colour does not appear in the F1 generation but reappears in the F2 generation: 2. Complete the crosses below: The Monohybrid Cross70 Pp Heterozygous purpleX Homozygous white pp Monohybrid cross ParentsGametes: Monohybrid cross PP Homozygous purpleX Heterozygous purple PpParentsMalegametes gametesFemale Key Idea: The outcome of a cross depends on the parental genotypes. A true breeding parent is homozygous for the gene involved. WEB 70 LINK 71 PREVIEWONLYNotforClassroomUseNotforClassroomUse

(a)Genotype (b)Phenotypefrequency:frequency: 50% Ee, 50% ee 50% yellow-brown, 50% ebony

A: (a)Genotype (b)Phenotypefrequency:frequency: 100% Ee 100% yellow-brown Cross

test cross Female gametes gametes Female gametes gametes

The basis of a

Parent 2 Homozygous recessive genotype (no dominant traits)

Photocopying ProhibitedKNOW 100 ???? X abab

(but

f The principle is simple. The individual with the unknown genotype is bred with a homozygous recessive individual for the trait(s) of interest. The homozygous recessive can produce only one type of allele (recessive), so the phenotypes of the offspring will reveal the genotype of the unknown parent (below). The test cross can be used to determine the genotype of single genes or multiple genes.

B A homozygous recessive female (ee) with an ebony body is crossed with a heterozygous male (Ee).

2. 50% of the resulting progeny have red eyes, and 50% have brown eyes. What is the genotype of the male Drosophila?

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Cross B:

f It is not always possible to determine an organism's genotype by its appearance because gene expression is complicated by patterns of dominance and by gene interactions. The test cross was developed by Mendel as a way to establish the genotype of an organism with the dominant phenotype for a particular trait.

Parent 1 Unknown genotype with dominant traits)

A. A homozygous recessive female (ee) with an ebony body is crossed with a homozyogous dominant male (EE).

The common fruit fly (Drosophila melanogaster) is often used to illustrate basic principles of inheritance because it has several genetic markers whose phenotypes are easily identified. Once such phenotype is body colour. Wild type (normal) Drosophila have yellow-brown bodies. The allele for yellow-brown body colour (E) is dominant. The allele for an ebony coloured body (e) is recessive. The test crosses below show the possible outcomes for an individual with homozygous and heterozygous alleles for ebony body colour.

1. In Drosophila, the allele for brown eyes (b) is recessive, while the red eye allele (B) is dominant. Describe how you would set up and carry out a test cross to determine the genotype of a male who has red eyes:

Key Idea: If an individual's genotype is unknown, it can be determined using a test cross.

The Test Cross

Monohybrid crosses can be used to determine the genotype and phenotype outcomes for coat colour in guinea pigs. Complete the crosses below by determining the gametes and phenotype and genotype ratios of the F1. The F1 is the first filial generation of offspring of distinctly different parental types. The first cross has been completed.

5. (a) Genotype frequency: (b) Phenotype frequency: whiteHeterozygous

7. (a) Genotype frequency: (b) Phenotype frequency:

allele? Homozygous whiteHomozygous black BBXbb bb BB BbBb Bb BbOffspringGametes(F1)

bb X bb

4. (a) Genotype frequency: (b) Phenotype frequency:

Homozygous

Key Idea: A monohybrid cross studies the inheritance pattern of one gene. The offspring of these crosses occur in predictable ratios.

Photocopying Prohibited

100% Bb 100%

6. (a) Genotype frequency: (b) Phenotype frequency: whiteHomozygous white

4. (a) Genotype frequency: (b) Phenotype frequency:

Practising Monohybrid Crosses

Parents Heterozygous blackHeterozygous black BbXBb

(g) Which is the recessive

BBXBB Homozygous blackHomozygous black

3. (a) Genotype frequency: (b) Phenotype frequency:

(e) Which is the dominant allele?

2. (a) Genotype frequency: (b) Phenotype frequency:

1. (a) Genotype frequency: (b) Phenotype frequency: black

(f) Which coat colour is recessive?

black Bb X bb

KNOW 101

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Homozygous

5. (a) Genotype frequency: (b) Phenotype frequency:

(d) Which coat colour is dominant?

1. A white bull is mated with a roan cow (right):

Red Roan Red Roan

Codominance is an inheritance pattern in which both alleles in a heterozygote contribute to the phenotype and both alleles are independently and equally expressed. Examples of codominance include:

3. What is the phenotypic ratio for a codominant cross of two heterozygous parents (e.g. a cross between two roan cattle):

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(a) Fill in the spaces to show the genotypes and phenotypes for parents and calves:

(b) Which bull serviced the cows? red or roan (delete one)

? Red bull White cow GametesParents Offspring Roan Roan Roan Roan CRCR CWCW CR CR CW CW CRCW CRCW CCRCW RCW

Roan cowUnknown bull

(b) What is the phenotypic ratio for this cross?

(a) Fill in the spaces (right) to show the genotype and phenotype for parents and calves.

Roan cow

f Human AB blood group, which is the result of two alleles: A and B, both being equally expressed.

KNOW 102

Codominance in Alleles

Key Idea: In codominance, neither allele is recessive and both alleles are equally and independently expressed in the heterozygote.

White bull

f Roan colouration in horses and cattle. Reddish coat colour is equally dominant with white. Animals that have both alleles have coats that are a mix of red and white hairs (roan). The red hairs and white hairs are expressed equally and independently (not blended to produce pink). In the Shorthorn cattle breed, white Shorthorn parents always produce calves with white coats. Red parents always produce red calves. However, a cross between red and white parents produces roan coats.

2. A farmer has only roan cattle on his farm. He suspects that one of the neighbours' bulls may have jumped the fence to mate with his cows earlier in the year because half the calves born were red and half were roan. One neighbour has a red bull, the other has a roan.

f The codominant alleles A and B each encode a different enzyme that adds a different, specific sugar to the basic sugar molecule. The recessive allele O produces a non-functioning enzyme. Blood group A and B antigens react with antibodies in the blood of people with incompatible blood types so must be matched for transfusion.

X group:Blood AB group:Blood AB group:Blood A group:Blood A group:Blood B group:Blood Ogroup:Blood Ogroup:Blood AB XgenotypesChildren'sfertilisationsPossibleGametesgenotypesParentalgroupsBlood Cross 2Cross 1 XX Cross 4Cross 3 ABABAA BB ABABAB AA BOAB ABABAOABAB OOOO genotypesChildren'sfertilisationsPossibleGametesgenotypesParentalgroupsBlood Blood (phenotype)group genotypesPossible FrequencyinNZ A 40%AA AO O 49%OO B 9% AB 2% Key

f If a person has the AO allele combination, their blood group is A, because O is recessive to A. AA also produces blood group A. Idea: The human ABO blood group system is a multiple allele system. Alleles A and B are codominant and allele O is recessive. The possible phenotypes are A, B, AB, and O.

Codominance in Multiple Allele Systems74

2. Below are four crosses possible between couples of various blood group types. The first example has been completed for you. Complete the genotype and phenotype for the other three crosses below:

f On the surface of red blood cells there are sugars called the ABO antigens, corresponding to the ABO locus on chromosome 9. These antigens are made by enzymes that link particular combinations of sugars together.

1. Use the information above to complete the table for the possible genotypes for blood group B and group AB.

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(b) One child having blood group O:

104

(a) If the man has a blood group O and the woman has a blood group A, could the child be his son? Use the diagram on the right to illustrate the genotypes of the three people involved.

B AB Child 1 Child 2 B AB Male Female Marriage Pedigree chart key X groupsBloodgenotypesChildren’sfertilisationsPossibleGametesgenotypesParental groupBlood groupBloodA O PREVIEWONLYNotforClassroomUseNotforClassroomUse

3. A woman is heterozygous for blood group A and the man has blood group O

X BChild’sGametesgenotypesParentalgenotypeBlood group groupBlood groupBloodA O

(b) State with reasons whether the man can be correct in his claim:

Determine the probability of:

(d) One child having blood group AB:

(a) Give the genotypes of each parent (fill in spaces on the diagram on the right).

(b) Explain why child 1 has two possible genotypes and child 2 has only one possible genotype:

5. The pedigree chart on the right shows the phenotypes of two children. Their father is blood group B and their mother is blood group AB.

(a) State the sex of the children:

4. In a court case involving a paternity dispute (i.e. who is the father of a child) a man claims that a male child (blood group B) born to a woman is his son and wants custody. The woman claims that he is not the father.

Snapdragons also show incomplete dominance in their flower colours. True breeding snap dragons produce red (CR) or white flowers (Cr). When red and white-flowered parent plants are crossed, offspring have pink flowers. If the offspring (F1) are then crossed, all three phenotypes (red, pink, and white) are produced in the F2

Photocopying Prohibited KNOW 105

Example 1

Incomplete Dominance

Key Idea: In incomplete dominance, the action of one allele does not completely mask the action of the other. Neither is dominant and an intermediate phenotype is produced.

3. A second breeder wanted to prove that pink snapdragon flowers were heterozygous. Briefly explain how this could be done:

2. What parental phenotypes would you choose if you wanted to produce snapdragons that were only pink or white (i.e. no red)? Use the Punnett square to help you.

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f In incomplete dominance, the action of one allele does not completely mask the action of the other and neither is dominant. Heterozygous offspring are intermediate in phenotype between the contrasting parental phenotypes.

(b) The F1 plants were crossed. When the F2 plants flowered, the flowers were produced as follows: crimson: 40, orange-red: 87, yellow: 42 State the genotype of each flower colour and the phenotypic ratio:

f In crosses involving incomplete dominance, the phenotype and genotype ratios are identical.

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1. (a) A crimson four o'clock and a yellow four o'clock were crossed. Complete the Punnett square (below) to show the F1 genotype.

Example 2

flowerRed flowerWhite ParentsGametesOffspringCRCR Pink CRCr Pink CRCr Pink CRCr Pink CCRCrrCrCrCRCR Cr Gametes from male femalefromGametes Gametes from male femalefromGametes flowerWhite PhenotypeParentsGametesfertilisationsPossibleOffspringflowerCrimsonMagentaMagentaMagentaMagenta CRCW CRCW CRCW CRCW CWCWCRCR CW CWCR CR

Four o'clocks (Mirabilis jalapa) (below) have flower colours that show incomplete dominance. True breeding four o'clocks produce crimson (CR), yellow (Cr), or white (CW) flowers. Crimson flowers crossed with yellow flowers produce orange-red flowers, while crimson flowers crossed with white flowers produce magenta (reddish-pink) flowers.

(b) State the phenotype ratio of Manx to normal cats and explain why it is not the expected 3:1 ratio:

2. Huntington disease (HD) is caused by an autosomal dominant mutation in either of the alleles of the gene Huntingtin. Suggest why HD persists in the human population when it is caused by a lethal, dominant allele:

(a) Complete the Punnett square for the cross:

Lethal Alleles

f Some lethal alleles are lethal in both the homozygous dominant and heterozygous conditions. Other lethal alleles are lethal only in the homozygous condition (either dominant or recessive). Furthermore, lethal alleles may take effect at different stages in development, e.g. symptoms of Huntington disease usually appear after 30 years of age. A

AYA AYA AY A Non-yellow¼AAYellow

½ AY GametesA AAY

Normal cat

Cats possess a gene for producing a tail. The tailless Manx phenotype in cats is produced by an allele that is lethal in the homozygous state. The Manx allele ML severely interferes with normal spinal development. In heterozygotes (MLM), this results in the absence of a tail. In MLML homozygotes, the double dose of the gene produces an extremely abnormal embryo, which does not survive.

Yellow mouse mouse Yellow

KNOW 106

When Lucien Cuenot investigated inheritance of coat colour in yellow mice in 1905, he reported a peculiar pattern. When he mated two yellow mice, about 2/3 of their offspring were yellow, and 1/3 were non-yellow (a 2:1 ratio). This was a departure from the expected Mendelian ratio of 3:1.

Y fertilisationsPossible

A test cross of the yellow offspring showed that all the yellow mice were heterozygous. No homozygous dominant yellow mice were produced because they had two copies of a lethal allele (Y). The Y allele is a mutation of the wild type agouti gene (A).

Manx cats are born without a tail

f Lethal alleles are usually a result of mutations in essential genes. They may result in death of an organism because an essential protein is not produced.

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Normal cats are born with a tail Manx cat

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Gametes from male

femalefromGametes

Yellow

ISBN: 978-1-927309-51-3

1. In Manx cats, the allele for taillessness (ML) is incompletely dominant over the recessive allele for normal tail (M). Tailless Manx cats are heterozygous (MLM) and carry a recessive allele for normal tail. Normal tailed cats are MM. A cross between two Manx (tailless) cats, produces two Manx to every one normal tailed cat (not a regular 3 to 1 ratio).

Key Idea: Lethal alleles are mutations of a gene that produce a non-functional gene product which may affect an organism's survival.

Photocopying Prohibited

5. In a dispute over parentage, the mother of a child with blood group O identifies a male with blood group A as the father. The mother is blood group B. On a separate sheet of paper draw Punnett squares to show possible genotype/phenotype outcomes to determine if the male is the father and the reasons (if any) for further dispute:

(b) Both parents are albino and have only albino children:

(a) Both parents have nor mal phenotypes; some of their children are albino and others are unaffected:

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3. A recessive allele, a, is responsible for albinism, an inability to produce or deposit melanin in tissues. Humans and a variety of other animals can exhibit this phenotype. In each of the following cases, determine the possible genotypes of the mother and father, and of their children:

Key Idea: Monohybrid crosses involve the inheritance of a single gene. The problems below will give you an oppor tunity to practise solving Mendelian monohybrid crosses.

4. Chickens with shor tened wings and legs are called creepers. When creepers are mated to normal birds, they produce creepers and normals with equal frequency. When creepers are mated to creepers they produce two creepers to one normal. Crosses between normal birds produce only normal progeny. Explain these results:

1. A dominant gene (W) produces wire-haired texture in dogs and its recessive allele (w) produces smooth hair. A group of heterozygous wire-haired individuals are crossed and their progeny are then test-crossed. Determine the expected genotypic and phenotypic ratios among the test cross progeny:

Problems Involving Monohybrid Inheritance

2. In sheep, black wool is due to a recessive allele (b) and white wool to its dominant allele (B). A white ram is crossed to a white ewe. Both animals carry the black allele (b). They produce a white ram lamb, which is then back crossed to the female parent (a back cross is a cross with a parent). Determine the probability of the back cross offspring being black:

Photocopying ProhibitedKNOW 108 The offspring can be arranged in groups with similar phenotypes 9 black, short hair 3 black, long hair 3 white, short hair 1 white, long hair PhenotypeGenotype Bbll Offspring (F2) BBLlBbLLBbLl BBLlBbLl BbLl BBLL BBll Bbll BbLLBbLl bbLLbbLl bbLlbbll Female gametes gametesMale BL BlbL bl BL Bl bL bl Homozygous black, short hair Homozygous white, long hair Parents Gametes(P) fertilisationsPossiblefertilisationsPossibleOffspring (F1) X BbLl BbLlX bl blbl BLbl BLBL BL bbll bbll1 bbLlbbLL21 BbllBBll21 BbLlBBLlBbLLBBLL4221

total of 9 offspring with one of 4 different genotypes can produce black, short hair

total of 3 offspring with one of 2 different genotypes can produce black, long hair

Only

1 offspring of a given genotype can produce white, long hair BBLL

Idea: A dihybrid cross studies the inheritance

Each of the 16 animals represents the possible zygotes formed by different combinations of gametes coming together at fertilisation.

When these genes are on separate chromosomes, the offspring of these crosses occur in predictable ratios. Dihybrid Cross78 LINK 79 WEB 78 LINK 80 PREVIEWONLYNotforClassroomUseNotforClassroomUse

two

BIOZONE International

F2 offspring: The F1 were mated together. Each individual from the F1 is able to produce four different kinds of gamete. Using a Punnett square (left), it is possible to determine the expected genotype and phenotype ratios in the F2 The notation F2 is only used for the offspring produced by crossing F1 heterozygotes.

Parents: The notation P, is only used for a cross between true breeding (homozygous) parents. In this case black hair (B) and short hair (L) are dominant.The genes are on separate chromosomes (the alleles assort independently).

F1 offspring: The first generation offspring are all the same genotype. The notation F1 is used for the hybrid offspring of genetically distinct parents.

total of 3 offspring with one of 2 different genotypes can produce white, short hair

Gametes: Each parent is homozygous for both traits so only one type of gamete is produced from each parent.

Key pattern of genes.

© 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited 109 Female gametes gametesMale XbbLL Bbl l Homozygous white, short hairBlack, long hair Offspring fertilisationsPossiblle OffspringGametesParents fertilisationsPossiblle Female gametes gametesMale BL BlbL bl bL bL bl bl GametesParents X BLBlbLblbLbLblbl BbLl bbLl Heterozygous black, short hairWhite, short hair The dihybrid cross on the right has been partly worked out for you. You must 1.determine:Thegenotype and phenotype for each animal (write your answers in its dotted outline). 2. Genotype ratio of the offspring: 3. Phenotype ratio of the offspring: For the dihybrid cross on the right, 1.determine:Gametes produced by each parent (write these in the circles).

2. The genotype and phenotype for each animal (write your answers in its dotted outline).

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3. Genotype ratio of the offspring: Phenotype ratio of the offspring:

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f Note that in the hypothetical example above there are only two possible genotype outcomes, both the same as the parent type.

AaBb

White body, white eyes aabb

Genes are linked when they are found on the same chromosomes. In this case A (body colour) and B (eye colour) are linked.

f Linkage is indicated in crosses when a greater proportion of the offspring from a cross are of the parental type (than would be expected if the alleles were assorting independently and on separate chromosomes).

chromosomereplicatedEach gametes.identicaltwoproduces here.showniseachofOne

White body, white eyes

GametesOffspringX

f Linked genes tend to be inherited together and so fewer genetic combinations of their alleles are possible, which reduces the variety of offspring that can be produced (contrast this with recombination).

beforeChromosomesreplicationChromosomesafterreplication

aAB baBA ababbaAB

KNOW 110

Dark body, blue eyes

f If the genes in question had been assorting independently and on separate chromosomes, there would have been more genetic variation in the gametes (and in the offspring i.e. four outcomes instead of two).

Overview of

Only two kinds of genotype combinations are possible. They are they same as the parent genotype.

Possible Offspring

MeiosislinkageB AAB baab a a b b baab AaBb AaBb aabb aabb AB ba ba ba abba baabb

Parent 1 (2N) Parent 2 (2N)

Dark body, blue eyes

Key Idea: Linked genes are genes found on the same chromosome and tend to be inherited together. Linkage reduces the genetic var iation in the offspring.

Wildfemaletype Mutantmale

4. If the female had been homozygous for the dominant wild type alleles (CuCu EbEb), state:

OffspringLinkageParent

(a) The genotype(s) of the F1:

cueb

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Straight wing Grey body

2. Complete the linkage diagram (right) by adding the gametes in the ovals and offspring genotypes in the rectangles.

Gametes from female fly (N) Gametes from male fly (N)

Gametes from male

Contact Newbyte Educational Software for details of their superb Drosophila Genetics software package which includes coverage of linkage and recombination. Drosophila images © Newbyte Educational Software.

5. A second pair of Drosophila are mated. The female genotype is Vgvg EbEb (straight wings, grey body), while the male genotype is vgvg ebeb (vestigial wings, ebony body). Assuming the genes are linked, carry out the cross and list the genotypes and phenotypes of the offspring. Note vg = vestigial (no) wings. Use the Punnett square to help you:

The genes for wing shape and body colour are linked (they are on the same chromosome) cueb cueb

Curled wing Ebony bodyPhenotype

The genotype(s) of the F1:

An example of linked genes in Drosophila

3. List the possible genotypes in the offspring (right) if genes Cu and Eb had been on separate chromosomes:

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Sex of offspring is irrelevant in this case

1. What is linkage and what is its effect on the inheritance of genes?

The phenotype(s) of the F1:

111

In the example shown on the right, wild type (the typical form found in nature) alleles are dominant and are given an upper case symbol of the phenotype caused by mutant recessive alleles (Cu or Eb). This symbology used for Drosophila departs from the convention of using the dominant gene to provide the symbol. This convention is used because there are many mutant alternative phenotypes to the wild type.

Cucu Ebeb cucu ebebGenotype

An example of linked genes in Drosophila

(b) The phenotype(s) of the F1:

Meiosis

CuEb

femalefromGametes

f During crossing over, alleles on homologous chromosomes can be exchanged and new associations of alleles are formed in the gametes.

These two offspring exhibit unexpected allele combinations. They can only arise if one of the parent's chromosomes has undergone crossing over. body, blue eyes body, body, white eyes Idea: Recombination is the exchange of alleles between homologous chromosomes as a result of crossing over. Recombination increases the genetic variation in the offspring.

Dark

AaBbDark body, blue eyes White

These two offspring exhibit allele combinations that are expected as a result of independent assortment during meiosis. Also called parental types. offspring

blue eyes Dark body, white eyes White

f The offspring formed show combinations of characteristics that are different from the parents and are known as recombinants. The appearance of recombinants in the offspring indicates recombination.

aabb White body, white eyes ChromosomesbeforeChromosomesreplicationafterreplicationGametesOffspringXaAB baabbaAB Parent 1 (2N) Parent 2 (2N) Overview of MeiosisrecombinationB AAb Baab AaBb aaBb b AB Ab a b b a aabb baaB ba ba ba ba ba a a b b baab Because homozygous,individualthisis if these genes cross over there is no change to the allele combination. Crossing over has chromosomesbetweenoccurredthese Ba Aabb baAb Key

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Non-recombinant offspring

f The proportion of recombinants in the offspring can be used to calculate the frequency of recombination. The frequency is fairly constant for any given pair of alleles.

Recombinant

Wildfemaletype cueb CuEbcueb cueb

Non-recombinantoffspring Recombinantoffspring

Recombination produces variation

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Cucu StraightEbebwingGreybody

Crossing over has occurred, giving four types of gametes

3. How does recombination increase the amount of genetic variation in offspring?

2. Complete the recombination diagram right, adding the gametes in the ovals and offspring genotypes and phenotypes in the boxes:

OffspringLinkageParent

If genes remain linked, the possible combinations in the gametes remains limited. Crossing over and recombination increase the variation in the offspring. In humans, even without crossing over, there are approximately (223)2 or 70 trillion genetically different zygotes that could form for every couple.

Photocopying Prohibited

(c) What does the low number of recombinants tell you about how often the alleles cross over and their relative positions on the chromosome?

(b) Give their genotypes:

Meiosis

Sex of offspring is irrelevant in this case

(a) Which phenotypes were the recombinants?

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In the female parent, crossing over occurs between genes for wing shape and body colour

Curled wing Ebony bodyPhenotype

An example of recombination in Drosophila

Gametes from female fly (N) Gametes from male fly (N)

Mutantmale

cucu ebebGenotype

An example of recombination

Taking crossing over and recombination into account produces (423)2 or 5000 trillion trillion genetically different zygotes for every couple.

1. Describe the effect of recombination on the inheritance of genes:

5. A third pair of Drosophila are mated. The female was Vgvg Yy (straight wing, grey body), while the male was vgvg yy (vestigial wing, yellow body). The numbers of offspring produced were: 84 straight wing, grey body, 78 vestigial wing, yellow body, 16 vestigial wing, grey body, and 18 straight wing, yellow body.

The cross (right) uses the same genotypes as the previous activity but, in this case, crossing over occurs between the alleles in a linkage group in one parent. The notation used is the same.

4. A second pair of Drosophila are mated. The female is Cucu YY (straight wing, grey body) and the male is Cucu yy (straight wing, yellow body). Assuming recombination, perform the cross and list offspring genotypes and phenotypes:

Only one type of gamete is produced in this case

Phenotyperatio:ratio:

2. In humans, two genes affecting the appearance of the hands are the gene for thumb hyperextension (curving) and the gene for mid-digit hair. The allele for curved thumb, H, is dominant to the allele for straight thumb, h. The allele for mid digit hair, M, is dominant to that for an absence of hair, m

(a) Give all the genotypes of individuals who are able to cur ve their thumbs, but have no mid-digit hair:

Genotype

(b) Complete the Punnett square (right) to show the possible genotypes from a cross between two individuals heterozygous for both alleles:

Genotype

(a) Determine which alleles responsible for the phenotypes observed are dominant and which are recessive:

3. A plant with orange-striped flowers was cultivated from seeds. The plant was self-pollinated and the F1 progeny appeared in the following ratios: 89 orange with stripes, 29 yellow with stripes, 32 orange without stripes, 9 yellow without stripes. Call orange O and stripes S.

Problems Inheritance

Phenotyperatio:ratio:

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1. In cats, the following alleles are present for coat characteristics: black (B), brown (b), short (L). long (l), tabby (T), blotched tabby (tb). Use the information to complete the dihybrid crosses below:

(d) What is the probability one of the offspring would have mid-digit hair?

Key Idea: Dihybrid crosses involve the inheritance of two genes. The problems below will give you an opportunity to practise solving dihybrid crosses.

(a) A black shor t haired (BBLl) male is crossed with a black long haired (Bbll) female. Determine the genotypic and phenotypic ratios of the offspring:

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(b) A tabby, short haired male (TtbLl) is crossed with a blotched tabby, short haired (tbtbLl) female. Determine ratios of the offspring:

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(b) Determine the genotype of the original plant with orange striped flowers:

Involving Dihybrid

Dihybrid inheritance

What You Know So Far: Sources of Variation

Monohybrid inheritance

HINT: Define the term allele and explain how mutations produce new alleles.

Meiosis and variation

HINT: Explain dihybrid inheritance and explain how linkage and recombination affect variation.

HINT: Explain how independent assortment and crossing over during meiosis produces variation.

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HINT: Include definitions and expected genotype and phenotype ratios for monohybrid inheritance patterns, including crosses involving codominance, incomplete dominance, and lethal alleles.

Mutation as a source of new alleles

NCEA Style Question: Sources of Variation

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TEST 116

1. Sickle cell disease is caused by a mutation to the gene encoding the oxygen-carrying protein haemoglobin (Hb). The mutation causes an alteration in the phenotype of the red blood cells containing Hb and they take on a sickle shape. Heterozygotes are carriers of the sickle cell mutation, but do not have the disease. People with two copies of the allele have sickle cell disease. Sickle cell disease causes many medical complications and often results in early death.

Discuss how mutations can lead to phenotypic variation within a population. Your answer should include:

• A description of mutation and why some detrimental alleles are retained within a population while others are not

• A definition for allele, heterozygote, and homozygote

• State whether sickle cell disease is inherited in a dominant or recessive pattern and justify your answer

• Discuss why the sickle cell mutation remains in the population

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• An explanation of why the daughter cells are genetically different to the parent cells

• An explanation of how genetic variation is increased through segregation, independent assortment, and crossing over during meiosis

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(b) Explain linkage and discuss its effect on variation in the offspring, contrasting it with recombination:

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2. Meiosis is a special type of cell division. It produces gametes (sex cells) for sexual reproduction.

You may use diagrams to illustrate your answer and more paper if you need it.

• A description of meiosis and the type of cells produced

(a) Discuss how meiosis produces genetic variation. Your answer should include:

NCEA Style Question: Inheritance of Alleles

2. A Drosophila male with genotype Cucu Ebeb (straight wing, grey body) is crossed with a female with genotype cucu ebeb (curled wing, ebony body). The phenotypes of the F1 were recorded and the percentage of each type calculated. The percentages were straight wings, grey body 45%, curled wings, ebony body 43%, straight wings, ebony body 6%, and curled wings grey body 6%.

(a) What are the phenotypic characteristics being investigated here?

(d) Explain your answer:

1. Snapdragons have red, pink, or white flowers. Roan coats in Shorthorn cattle is a mix of red and white hairs.

(e) Determine the genotypes of the offspring:

TEST 118

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(a) Define monohybrid cross:

Straight wing: Cucu Grey body: Ebeb Curled wing: cucu Ebony body: ebeb

(b) Use the inheritance patterns of codominance and incomplete dominance to explain these results (you may use extra paper if required):

(b) What sort of cross is this? Justify your answer:

• The use of a Punnett squares and the correct use of the allele notation given

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• Relevant genotypes and phenotypes

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3. In guinea pigs, black hair (B) is dominant over white hair (b) and short hair (L) is dominant over long hair (l). Breeders use true breeding parents in order to produce consistent and reliable coat colour in guinea pigs. A breeder has a male guinea pig she knows to be BBLL and a female she knows to be bbll. She wants to use them to begin a breeding programme to produce a true breeding white, short haired guinea pig. In your answer you should include:

• A definition of true breeding

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• An explanation of how the desired pure breeding animal can be produced

• An explanation of how the animal's ability to breed true could be proved.

… of a dominantgene.or recessive.

… for the gene, they are heterozygous. … is said to be homozygous. … expressed if it is homozygous. … source of new alleles. is always expressed whether it is homozygous or heterozygous.

1.recombinationrecessivephenotypemutationmeiosisincompleteheterozygousgenotypedominantdihybridcodominanceallelescrossdominanceTestyourvocabulary

A dominant allele … A recessive allele is only …

H Exchange of alleles between homologous chromosomes as a result of crossing over.

4. Use lines to match the statements in the table below to form complete sentences:

(b) Rr x rr

C A genetic cross involving the inheritance of two genes.

G A change to the DNA sequence of an organism. This may be a deletion, insertion, duplication, inversion or translocation of DNA in a gene or chromosome.

E Sequences of DNA occupying the same gene locus (position) on different, but homologous, chromosomes.

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F The process of double nuclear division (reduction division) to produce four nuclei, each containing half the original number of chromosomes (haploid).

K The allele combination of an organism.

L An inheritance pattern in which two different alleles for a trait are equally and independently expressed.

A It is not expressed in the F2 generation

B It is not expressed in the heterozygote.

I Allele that is expressed irrespective of the other allele is called this

3. True breeding individuals:

A Only ever breed with others of the same kind.

Mutations are the ultimate … Alleles are variations …

5. Complete the following crosses for pea plants, including genotype and phenotype ratios: Round seeds (R) are dominant to wrinkled seeds (r):

A Observable characteristics in an organism.

2. The allele for wrinkled seeds in pea plants is considered recessive because:

A person carrying two of the same alleles (one on each homologous chromosome) …

KEY TERMS AND IDEAS: Sources of Variation

C Are homozygous for a particular allele

TEST 120

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

B Allele that will only be expressed in the absence of the dominant allele is called this.

D Always produce heterozygous offspring.

J An inheritance pattern in which one allele for a trait is not completely expressed over its paired allele and neither shows dominance. An intermediate phenotype is produced.

D Round peas are smaller than wrinkled peas.

B Always have the same coat or flower colour.

If the person carries two different alleles … Alleles may be …

D Possessing two different alleles of a particular gene, one inherited from each parent.

C Individuals who have the allele are less likely to pass genes on to the next generation.

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(a) RR x Rr:

Mate choice (non-random mating): Individuals do not select their mate randomly but may seek out particular phenotypes, increasing the frequency of the associated alleles in the population.

f The proportion of each allele within a gene pool is called the allele frequency. Allele frequencies in populations can vary as a result of the events occurring the gene pool. Changes to the allele frequency over time is called evolution.

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Geographical barriers (e.g. mountains or rivers) isolate the gene pool and prevent regular gene flow between populations.

Processes in Gene Pools

Immigration: New genes may be introduced.

f A gene pool is the collection of all the alleles in a population. The term deme is sometimes used to refer to a local population of individuals that interbreed freely and share a distinct gene pool.

Deme 1

f Four processes act to cause genetic change in populations: mutation, gene flow migration (gene flow), natural selection, and genetic drift. These processes are summarised in the diagram below (definitions in blue).

Deme 2

Emigration: Genes may be lost.

86 AA aa A'A AA AA AA AA aa aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa AA aa aa aa aa aa Aaaa Aa Aa Aa Aa Aa Aa Aa Aa Aa AA aa AA aa aa

Mutations: Changes to the DNA sequence can create new alleles.

Gene flow: The exchange of alleles between gene pools as a result of migration. Gene flow is a source of new genetic variation and tends to reduce differences between populations that have accumulated because of natural selection or genetic drift.

Genetic drift: Random changes to the allele frequencies of populations due to chance events. Genetic drift has a relatively greater effect on small populations and can be an important process in their evolution.

Key Idea: The proportions of alleles in a gene pool can be altered by processes that increase or decrease genetic variation.

Natural selection: The differential survival of favourable phenotypes (unfavourable allele combinations have lower survival or reproductive success). Natural selection accumulates and maintains favourable allele combinations. It reduces genetic diversity within the gene pool and increases differences between populations.

Photocopying Prohibited 122 AA Aa Aa Aa Aa Aa Aa Aa AaAa aa Aa Aa Aa Aa AA AA AA AA AA AA aaAA aa AAAaaa No.% Aa AA aa AA AA AA AA AA AA aa aa aa aa Aa Aa AA aa AAAaaa No% Aa 5427 287 Black Pale Allele typesAllele combinations Black Aa Aa Aa Aa Aa AaAa Aa Aa Aa Aa Aa Aa aa AA AA AA aa aa aa Aa Aa Aa Aa AaAa Aa Aa Aa AAAaaa No.% Aa AA AA AA Aa Aa Aa aa Aa AA

This individual is leaving the population, removing its alleles from the gene pool.

(a) Count the number of A and a alleles separately. Enter the count into the top row of the table (left hand columns).

(b) Count the number of each type of allele combination (AA, Aa and aa) in the gene pool. Enter the count into the top row of the table (right hand columns).

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In the same gene pool at a later time there was a change in the allele frequencies. This was due to the loss of certain allele combinations due to natural selection. Some of those with genotype 'aa' were eliminated (poor fitness).

These individuals (surrounded by small white arrows) are not counted for allele frequencies; they are dead!

Two pale individuals died. Their alleles are removed from the gene pool.

Individuals coming into the gene pool (AA) are counted for allele frequencies, but those leaving (aa) are not.

Allele frequency = No. counted alleles ÷ Total no. of alleles x 100

This individual is entering the population and will add its alleles to the gene pool.

This particular kind of beetle exhibits wandering behaviour. The allele frequencies change again due to the introduction and departure of individual beetles, each carrying certain allele combinations.

(b) Allele frequency:

2. For each phase in the gene pool below, fill in the tables provided as follows (some have been done for you):

1. Define the following terms:

(a) Gene pool:

%No.%No.

Phase 3: Immigration and emigration

Phase 2: Natural selection

Phase 1: Initial gene pool

Evolution is the change in the genetic makeup of a population (the allele frequencies) over time. Evolution is the consequence of the interaction between four factors: (1) The potential for populations to increase in numbers, (2) genetic variation as a result of mutation and sexual reproduction, (3) competition for resources, and (4) proliferation of individuals with better survival and reproduction.

Natural selection is the term for the mechanism by which better adapted organisms survive to produce a greater number of viable offspring. This has the effect of increasing their proportion in the population so that they become more common. This is the basis of Darwin's theory of evolution by natural selection.

We can demonstrate the basic principles of evolution using the analogy of a 'population' of M&M's sweets.

The blue sweets common...

Key Idea: Evolution by natural selection describes how organisms that are better adapted to their environment survive to produce a greater number of offspring.

become more

Populations generally produce more offspring than are needed to replace the parents. Natural populations normally maintain constant numbers. A certain number will die without reproducing.

What is evolution?

Natural selection favours the individuals best suited to the environment at the time

The variations (both favourable and unfavourable) are passed on to offspring. Each generation will contain proportionally more descendents of individuals with favourable characters.

Populations produce too many young: many must die

Individuals in the population compete for limited resources. Those with favourable variations will be more likely to survive. Relatively more of those without favourable variations will die.

Individuals in a population have different phenotypes and therefore, genotypes. Some traits are better suited to the environment, and individuals with these have better survival and reproductive success.

Eventually, you are left with a bag of blue M&M's. Your selective preference for the other colours changed the make-up of the M&M's population. This is the basic principle of selection that drives evolution in natural populations.

Variations are inherited: the best suited variants leave more offspring

In a bag of M&M's, there are many colours, which represents the variation in a population. As you and a friend eat through the bag of sweets, you both leave the blue ones, which you both dislike, and return them to bag.

Darwin's theory of evolution by natural selection is outlined below. It is widely accepted by the scientific community and it is one of founding principles of modern science.

Inherited

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Individuals show variation: some variations more favourable than others

Natural selection

How Natural Selection Works

Darwin's theory of evolution by naturalVariationselection

Overproduction

Natural populations, like the ladybird population above, show genetic variation. This is a result of mutation (which creates new alleles) and sexual reproduction (which produces new combinations of alleles). Some variants are more suited to the environment of the time than others. These variants will leave more offspring, as described for the hypothetical population (right).

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2. (a) Define evolution:

3. Using your answer 2(b) as a basis, explain how the genetic make-up of a population can change over time:

Variation, selection, and population change

1. What produces the genetic variation in populations?

Red Brown mottled Red 2 spot

(b) Identify the four principles of evolution by natural selection as proposed by Darwin:

1. Variation through mutation and sexual reproduction: In a population of brown beetles, mutations independently produce red colouration and 2 spot marking on the wings. The individuals in the population compete for limited resources.

2. Selective predation: Brown mottled beetles are eaten by birds but red ones are avoided.

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3. Change in the genetics of the population: Red beetles have better survival and fitness and become more numerous with each generation. Brown beetles have poor fitness and become rare.

f Natural selection is a dynamic process because it is always linked to the suitability of particular phenotypes in the environment of the time. A change in the environment may favour the proliferation of once unfavourable phenotypes.

Key Idea: Natural selection is the differential survival of favourable phenotypes and their associated genotypes. It is an important cause of change in gene pools.

1. What is the effect on the phenotypic mean (the average phenotype) under the following selection patterns? (a) Stabilising selection: (b) Directional selection: (c) Disruptive selection: 2. Why is natural selection a dynamic process?

Types of Natural Selection

f Disruptive selection: Favours two phenotypic extremes at the expense of intermediate forms. It creates two divergent phenotypic norms.

Types of selection

The effects of different types of selection on a population

f Stabilising selection: Selects against phenotypic extremes, reducing phenotypic variation in a population.

f Natural selection is the differential survival of favourable phenotypes, produced by their particular combinations of alleles.

Stabilising selection Disruptive selectionDirectional selection environmentTypicaldistributionInitial Stable climate and resources Changing climate or resourcesDiversification of habitat or resources generationsafterDistributiongenerationsafterDistributionafewmany individualsofNumber individualsofNumber individualsofNumber individualsofNumber individualsofNumber individualsofNumber Phenotypic range Phenotypic range Phenotypic range Phenotypic range Phenotypic range Phenotypic range Phenotypic range Phenotypic range Phenotypic range individualsofNumber individualsofNumber individualsofNumber WEB 88 LINK 87 LINK 91 PREVIEWONLYNotforClassroomUseNotforClassroomUse

f Over time, natural selection may lead to a permanent change in the genetic makeup of a population.

f As a result of natural selection, organisms with phenotypes most suited to the prevailing environment are more likely to survive and breed successfully than those with less suited phenotypes. Favourable phenotypes will become relatively more numerous and unfavourable phenotypes will become less common.

f Directional selection: The adaptive phenotype shifts in one direction, favouring phenotypes at one extreme and establishing a new phenotypic norm.

6. Count the remaining phenotype colours and calculate the proportions of each phenotype. Record them in the table below in the proportions row of generation 2. Use the formula: Proportion = number of coloured squares ÷ total number of squares remaining. For example: for one student doing this activity: proportion of white after predation = 10/30 = 0.33.

Modelling Natural Selection

7. Before the next round of selection, the population must be rebuilt to its original total number using the newly calculated proportions of colours and the second half of the squares from step 3. Use the following formula to calculate the number of each colour: number of coloured squares required = proportion x number of squares in original population (42 if you are by yourself, 84 with a partner). For example: for one student doing this activity: 0.33 x 42 = 13.9 = 14 (you can't have a fraction of a phenotype). Therefore in generation 2 there should be 14 white squares. Do this for all phenotypes using the spare colours to make up the numbers if needed. Record the numbers in the numbers row of generation 2. Place generation 2 into the bag.

10. Now repeat the whole activity using a white sheet background instead of the black sheet. What do you notice about the proportions this time? White

1. Cut out the squares on the following pages and record the number of black, grey, and white squares. Work out the proportion of each phenotype in the population (e.g. 0.33 black 0.34 grey, 0.33 white) and place these values in the table below. This represents your starting population (you can combine populations with a partner to increase the population size for more reliable results).

1 ProportionNumber 2 ProportionNumber 3 ProportionNumber 4 ProportionNumber 5 ProportionNumber WEB 89 LINK 87 PREVIEWONLYNotforClassroomUseNotforClassroomUse

2. For the first half of the activity you will also need a black sheet of paper or material that will act as the environment (A3 is a good size). For the second half of the activity you will need a white sheet of paper.

4. Now take the squares out of the bag and randomly distribute them over the sheet of black paper (this works best if your partner does this while you aren't looking).

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Key Idea: Natural selection acts on phenotypes. Those individuals better suited to an environment will leave more offspring in the next generation. This can be illustrated using a simple model.

8. Repeat steps 4 to 7 for generation 2, and 3 more generations (5 generations in total or more if you wish).

Generation Black Grey

3. Place 14 each of the black, grey, and white squares in a bag and shake them up to mix them. Keep the others for making up population proportions later.

PRAC 126

5. For 20 seconds, pick up the squares that stand out (are obvious) on the black paper. These squares represent snails in the population that have been preyed on and killed (you are acting the part of a predator). Place them to one side and pick up the remaining squares. These represent the population that survive to reproduce.

Natural selection can be modelled in a simple activity based on predation. You can carry out the following activity by yourself, or work with a partner to increase the size of the population. The black, grey, and white squares opposite represent the phenotypes of a population. Cut them out and follow the instructions below to model natural selection. You will also need a sheet of white paper and a sheet of black paper.

9. On separate graph paper, draw a line graph of the proportions of each colour over the five generations. Which colours have increased, which have decreased?

© 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited PRAC 127 Gene Pool Model90 The set of all the versions of all the genes in a population (its genetic make-up) is called the gene pool. Cut out the squares below and use them to model the events described in Modelling Natural Selection LINK 89 PREVIEWONLYNotforClassroomUseNotforClassroomUse

2. Using the data in the table above and the grids below and on the facing page, draw column graphs of the percent reflectance of the mice coats and the rocks at each of the 14 collection sites.

1. (a) What is the genotype(s) of the dark coloured mice?

Natural Selection in Pocket Mice

f The coat colour of the Arizona rock pocket mice is controlled by the Mc1r gene (a gene that in mammals is commonly associated with the production of the pigment melanin). Homozygous dominant (DD) and heterozygous mice (Dd) have dark coats, while homozygous recessive mice (dd) have light coats. Coat colour of mice in New Mexico is not related to the Mc1r gene.

(b) What is the genotype of the light coloured mice?

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f 107 rock pocket mice from 14 sites were collected and their coat colour and the rock colour they were found on were recorded by measuring the percentage of light reflected from their coat (low percentage reflectance equals a dark coat). The data are presented right:

Key Idea: The need to blend into their surroundings to avoid predation is an important selection pressure acting on the coat colour of rock pocket mice.

Rock pocket mice are found in the deserts of southwestern United States and northern Mexico. They are nocturnal, foraging at night for seeds, while avoiding owls (their main predator). During the day they shelter from the desert heat in their burrows. The coat colour of the mice varies from light brown to very dark brown. The usual environment of the mice is sandy desert, but lava flows in some regions have created a dark environment. This new landscape and the mice that live there present an excellent study in natural selection.

Percent reflectance (%) Site Rock type (V volcanic) Mice coat Rock KNZ V 4 10.5 ARM V 4 9 CAR V 4 10 MEX V 5 10.5 TUM V 5 27 PIN V 5.5 11 AFT 6 30 AVR 6.5 26 WHT 8 42 BLK V 8.5 15 FRA 9 39 TIN 9 39 TUL 9.5 25 POR 12 34.5

(d) How might these exceptions have occurred?

5. The rock pocket mice populations in Arizona use a different genetic mechanism to control coat colour than the New Mexico populations. What does this tell you about the evolution of the genetic mechanism for coat colour?

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3. (a) What do you notice about the reflectance of the rock pocket mice coat colour and the reflectance of the rocks they were found on?

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4. What type of selection appears to operating in each of the environments (dark and light rock)? Explain:

(b) Suggest a cause for the pattern in 3(a). How do the phenotypes of the mice affect where the mice live?

Some individuals from the mainland population are carried at random to the offshore island by natural forces such as strong winds.

AA aa aa aa aa aa aa AA AA AA AA AA AA AA aa AA AA AA AA AA AA AA Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa Aa AaAa Aa Aa Aa Aa Aa Aa Aa AaAa Aa Aa

(a) Count the phenotype numbers for the two populations (i.e. the number of black and pale beetles).

1. Compare the mainland population to the island population (use the spaces in the tables below):

2. How are the allele frequencies of the two populations different?

This population may not have the same allele frequencies as the mainland population.

(b) Count the allele numbers for the two populations: the number of dominant alleles (A) and recessive alleles (a). Calculate these as a percentage of the total number of alleles for each population.

Allele frequencies frequenciesPhenotype numbersActual Calculate% BlackPale Allele A Allele a Total Allele frequencies frequenciesPhenotype numbersActual Calculate% BlackPale Allele A Allele a Total AAAaaa AAAaaa Mainland population Colonising island population

Key Idea: The founder effect is the loss of genetic variation when a new colony is formed from a small number of individuals. The founder population may evolve differently to the parent population.

f It may be possible for certain alleles to be missing altogether from the isolated population (the allele has been lost).

The Founder Effect92 WEB 92 LINK 94 PREVIEWONLYNotforClassroomUseNotforClassroomUse

populationIsland

3. Describe how organisms could become isolated from a parent population:

f Occasionally, a small number of individuals may become isolated from their original large population, e.g. by dispersal, chance events, or as a result of changing geography. The small colonising, or founder, population is unlikely to have a representative sample of the population's alleles and its genetic diversity will be reduced.

f As a consequence of this founder effect, the evolution of the colonising population is likely to differ from that of the parent population. The colonising population will be more susceptible to genetic drift and the selection pressures for the colonisers in the new environment may be very different to those experienced by the parent population.

populationMainland

sclae)to(notRoad ©

7. Identify the colony in block A that appears to be isolated from the rest of the block itself:

6. Both the MDH-1 alleles fully operative enzymes. Suggest why the frequencies of the alleles have become significantly different.

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Microgeographic isolation in garden snails

The snails were found in several colonies in each block. Allele frequencies for the gene MDH-1 (alleles A and a) were obtained and compared. Statistical analysis of the allele frequencies of the two populations showed them to be significantly different (P << 0.05). Note: A Mann-Whitney U test was used in this instance. It is similar to a Student's t test, but does not assume a normal distribution of data (it is non-parametric)

produce

1 1 3 3 2 2 4 4 5 5 6 6 7 7 14 13 13 8 8 15 10 10 9 9 12 1211 11

132 4.

Colony123456789101112131415 MDH-1 A % 393936423947324244424450505875 MDH-1 a % MDH-1 A % 81617568706170605861545447 MDH-1 a %ABlockBlockB 19753,No.29,VolEvolution,Source: PREVIEWONLYNotforClassroomUseNotforClassroomUse

Block A Block B Building 1994-2017 BIOZONE ISBN: 978-1-927309-51-3 Prohibited Complete the table above by filling in the frequencies of the MDH-1 a allele: Suggest why these snail populations are effectively geographically isolated:

Snail colony (circle size is proportional to colony size).

International

The European garden snail (Cornu aspersum, formerly Helix aspersa) is widely distributed throughout the world, both naturally and by human introduction. However because of its relatively slow locomotion and need for moist environments it can be limited in its habitat and this can lead to regional variation. The study below illustrates an investigation carried out on two snail populations in the city of Bryan,Texas. The snail populations covered two adjacent city blocks surrounded by tarmac roads.

Population numbers are drastically reduced. Much of the genetic diversity is lost.

Time

f Before increasing in size again, the population is also more susceptible to the effects of genetic drift and may be vulnerable to the detrimental effects of inbreeding.

Population numbers increase again, but the larger population has low genetic diversity.

Large,

diversegeneticallypopulation.

Population Bottlenecks

Bottleneck

f A population crash may not select against one phenotype. It can affect all phenotypes equally although, for disease losses, it may be particular allele combinations that survive. Large scale catastrophes, such as fire or volcanic eruption, are examples of non-selective events.

The effect of population bottlenecks on genetic diversity

f A population (or genetic) bottleneck is a sharp reduction in the size of a population (a population crash). Causes include natural disasters (earthquake, flood, fire, drought), disease, predation, climatic change, and human activity.

1. Define the term population bottleneck: 2. Explain how a population bottleneck can decrease genetic diversity in a population: 3. What events might cause a population bottleneck? WEB 93 LINK 94 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Cheetahs are a species that has experienced a population bottleneck. Cheetahs nearly became extinct at the end of the last ice age. Today there are fewer than 8,000 surviving. The entire population exhibits very little genetic diversity and this lack of genetic diversity, threatens cheetah survival.

numbersPopulation HighLow diversityGeneticHighLow

f Following bottleneck events, the small number of individuals contributing to the gene pool may not have a representative sample of the alleles in the pre-catastrophe population. The population may recover, but genetic diversity is lost.

Key Idea: Population bottlenecks occur when population numbers and diversity fall dramatically. Although a population's numbers may recover, its genetic diversity may not.

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5. Study the graph above and suggest why (based on unique DNA fragments) the South Island robin is less likely to be endangered than the Chatham Island black robin:

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(b) How has this led to increased susceptibility to disease, specifically infectious cancer?

Low allelic diversity for MHC is implicated in the spread of devil facial tumour disease (DFTD). DFTD is a contagious cancer that appeared in Tasmanian devil populations in the mid 1990s and has resulted in the loss of 80% of the devil population. The cancerous cells are transmitted when the devils fight. Ordinarily this foreign material would be recognised and destroyed by the immune system. In Tasmanian devils, the immune diversity is so low that tumours can spread without invoking an immune response. However, recent evidence shows that some populations are developing immunity to DFTD. This may originate in individuals with MHC alleles distinctly different from the susceptible individuals.

Case study: The Chatham Island black robin

Many native New Zealand birds have passed through bottleneck events, mostly due to the arrival of humans and loss of habitat. The Chatham Island (or black) robin (Petroica traversi) is an extreme example. When first described in 1872, it was common on the Chatham Islands. By 1938, it was extinct on all the islands except for one population of about 60 on Little Mangere Island. By the early 1970s, the black robin population had dropped to just 18. By 1980 it was five. Thanks to recovery efforts, the population is now close to 250. All of these are descended from one female, Old Blue. Analysis of the black robin's DNA shows it to have among the lowest genetic diversity of any reported bird species. The black robin is listed as critically endangered.

4. Populations of endangered species have often experienced historic genetic bottlenecks. Explain how genetic bottlenecks might affect the ability of an endangered species to recover and maintain healthy populations:

Tasmanian devils were once found throughout mainland Australia, but are now restricted to Tasmania. Genetic evidence suggests that the devils went through two population crashes, one 30,000 years ago and another 3000 years ago. Further modern declines (1850 to 1950) occurred as a result of trapping and disease. These historic population crashes are likely to be responsible for the very low diversity of MHC genes in devils. MHC genes are important in immunity and the body's self recognition system.

Case study: devils

Tasmanian

6. (a) What has been the genetic consequence of bottleneck events in the Tasmanian devil population?

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DNA profiling reveals that almost all the black robin's genetic variation is found after sampling just a few birds (unlike the related South Island robin).

ofnumberTotal fragments[DNA]unique Number of [DNA] fragments sampled 0 706050100203040 100 200 300400500600 Black robin South Island robin 1997LambertD.M.Arden,S.L.

Generation 2

Small breeding population Fluctuations are more severe in smaller breeding populations because random changes in a few alleles cause a greater percentage change in allele frequencies

Further chance events will affect allele frequencies in subsequent generations.

Breeding population = 2000

Aa Aa Aa Aa Aa Aa Aa Aa Aa

A = 12 (41%)

Large breeding population Fluctuations are minimal because the large numbers buffer the population against random loss of alleles. On average, losses for each allele type are similar in frequency and little change occurs.

Computer simulation of genetic drift

f The allele combinations in any generation are determined by what is passed on from the generation before.

0 20 40 60 80 100%)(frequencyAllele 100806040200 120140 0 20 40 60 80 100 100806040200 120140 0 20 40 60 80 100 100806040200 120140

Generations

The diagram above shows the gene pool of a hypothetical small population over three generations. As a result of chance events, not all individuals contribute alleles to the next generation. With the random loss of the alleles carried by these individuals, the allele frequency changes from one generation to the next. The change in frequency is directionless as there is no selection pressure. The allele combinations for each successive generation are determined by how many alleles of each type are passed on from the preceding one.

AA AA Aa Aa Aa Aa Aa Aa Aa Aa AA AA aa aa aa AA AA AA aa aa aa

Fail to locate a mate due to low population density

A = 15 (50%)

Generation 1

f Genetic drift is the change in allele frequencies in a population as a result of random events. The change in allele frequency is not due to selection pressures but to chance.

Genetic Drift

f Alleles may become lost from the gene pool altogether (frequency becomes 0%) or fixed as the only allele for the gene present (frequency becomes 100%).

AA AA aa aa aa aa aa AaAa Aa Aa Aa Aa Aa Aa

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Key Idea: Genetic drift is the term for the random changes in allele frequency that occur in all populations. It has a much greater effect on the allele frequencies of small populations.

The genetic makeup (allele frequencies) of the population changes randomly over a period of time

Fail to locate a mate due to low population density

Killed

Generations

Very small breeding population Fluctuations in very small breeding populations are so extreme that the allele can become fixed (frequency of 100%) or lost from the gene pool altogether (frequency of 0%).

Breeding population = 20

A = 16 (53%) a = 14 (47%)

Generation 3 a = 18 (59%)

f Genetic drift is the evolutionary equivalent of sampling error, so its effects are much greater in small populations and alleles may be fixed or lost relatively quickly. Small populations can arise as a result of founder effect or genetic bottlenecks. The loss of alleles may be more detrimental in a small population if its genetic diversity is already low.

a = 15 (50%)

Breeding population = 200

Generations Allele lost from the gene pool

The changes in allele frequencies in computer simulations of random genetic drift for different sized populations (2000, 200, or 20) are shown below. Each simulation was run for 140 generations.

(b) Explain why genetic drift is an impor tant process in the evolution of colonising (founder) populations:

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(b) What does it mean to say an allele has become lost from a population?

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3. (a) Why is the effect of genetic drift more pronounced in small populations?

4. Many native New Zealand bird species have suffered population bottlenecks due to loss of habitat or predation. The flightless takahe (Porphyrio hochstetteri) experienced a severe bottleneck associated with climate change and hunting by early Polynesians and were already in low numbers when Europeans arrived in New Zealand. Translocation of takahe to five, predator-free islands has helped to save the species from extinction, but the genetic diversity in these populations is low. Genetic analyses (2005, 2008) have indicated that island populations of takahe have significantly less genetic variation than the main Fiordland population, as well as significantly different allele frequencies, with some alleles now fixed. In addition, it is estimated that island takahe have lost 7.5% of allelic diversity since the founder populations were established.

2. (a) What does it mean to say an allele has become fixed in the population?

(a) Explain the likely impact of the founder effect and genetic drift on takahe populations relocated to offshore islands:

(b) Without any inter vention, the island population is projected to maintain only 76% of the original founding genetic diversity in the next 100 years. Suggest how humans could manage the population to limit this loss of genetic diversity and discuss the implications for the long term resilience of the island populations:

(b) How does genetic drift differ from natural selection?

136

1. (a) What is genetic drift?

HINT: List, include definitions, and state whether each factor increases or decreases genetic variation.

What You Know So Far: Changes in Gene Pools

Factors affecting allele frequencies in gene pools

HINT: Explain these processes and their effect on allele frequencies.

Natural selection

HINT: Include definition and examples of types of natural selection.

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Genetic bottleneck (the bottleneck effect), founder effect, and genetic drift

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1. The Chatham Island (or black) robin (right) was once common on the Chatham Islands, but became extinct through most of its range in the 1930s as a result of predation by introduced cats and rats. The population fell to just 5 birds in 1980 and included a single breeding pair. Through captive breeding, the population is now around 250 birds. All of the birds are descended from one female, Old Blue. Its very small population shows very low genetic diversity and there is very high relatedness between the birds. The black robin is listed as critically endangered.

(c) Explain the effect of the bottleneck on the genetic diversity of the black robin population and discuss the likely implications for the persistence of this critically endangered species:

TEST 138

Pools96

ISBN: 978-1-927309-51-3

ConservationofDeptPhoto:

(a) Define the term population bottleneck:

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(b) Explain the nature of the population bottleneck experienced by the black robin:

NCEA Style Question: Changes in Gene

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term founder effect:

(b) Explain why genetic drift is likely to be important in the evolution of the weta population on Matiu-Somes Island:

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2. The Wellington tree weta (right) was introduced to MatiuSomes Island in Wellington Harbour from Mana Island as part of an ecological restoration programme. A total of 59 weta were introduced over 2 years (1996-1997). The population is now well (a)established.Definethe

139

TEST 140

bottleneck effect

F The loss of genetic variation when a new colony is formed by a very small number of individuals from a larger population.

E The movement of individuals into and out of a population. Also called gene flow.

KEY TERMS AND IDEAS: Changes in Gene Pools

(a) Directional Environment:selection:Stableclimate and resources / steady trend in one direction / diversification in habitat or resources

3. The graph on the right shows the effect of genetic drift on the frequency of an allele (A) in populations of three different sizes.

(a) Describe how the impact of genetic drift varies depending on population size:

(b) Stabilising Environment:selection:Stable climate and resources / steady trend in one direction / diversification in habitat or resources

B The differences between individuals in a population as a result of genes and environment.

G The change in allele frequency in a population as a result of chance events. The effect is proportionally larger in small populations.

40 40 60 60 Generations80 (%)frequencyAllele 80 100 100 120140 Population 2000 Population 200 Population 20 A PREVIEWONLYNotforClassroomUseNotforClassroomUse

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

2 Describe the characteristics of each of the following types of selection and circle the type of environment in which each is most likely to operate:

(c) Disruptive Environment:selection:Stable climate and resources / steady trend in one direction / diversification in habitat or resources

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D An evolutionar y event in which a significant proportion of a population's alleles are lost.

A The process by which heritable traits become more or less common in a population through differential survival and reproduction.

(b) What has happened at point A on the diagram? 0 20

C The sum total of all alleles of all breeding individuals in a population at any one time

founder effect gene variationnaturalmigrationgeneticpooldriftselection

c M Demonstrate in-depth understanding of gene expression: Provide a reason as to how or why biological processes affect gene expression.

Nucleic acids base pairing rule doubleDNA helix gene substitutionphenotypemutationmutagenmetabolicinsertiongenotypeenzymeenvironmentdeletionMetabolictRNAtriplettranslationtranscriptionrRNAredundancyproteinpolypeptidemRNAfibrousglobulardegeneracycodonproteinGeneticRNAnucleotidenucleichydrogenexpressionbondacidcodeandsynthesisproteinproteincodepathways(mutation)(mutation)pathway(mutation)

c A Demonstrate understanding of gene expression: Define and use annotated diagrams or models to explain gene expression. Give characteristics of, or provide an account of, gene expression.

c i Molecular components of nucleic acids and their role in carrying the genetic code. 99 104

Gene expression

c i Proteins as the products of gene expression: DNA  mRNA  polypeptide. 104

c iii Redundancy due to code degeneracy. 106

Protein synthesis numberActivity

c 5 Effect of environment on genotype through mutations. 119 - 125 c 6 Effect of environment on expression of phenotype. 126 127

Achievement Standard 2.7

c 4 The determination of phenotype via metabolic pathways. 117 118

Select biological ideas and processes from...

c ii Identification of one gene  one polypeptide relationship. 104

Significance of proteins numberActivity

c 1a Nucleic acid structure. 99 - 103

Select biological ideas and processes from...

c ii Nature of the genetic code including triplets, codons, and anticodons. 106 - 108

c ii The role of enzymes in controlling the process (enzyme names not required). 105 108

c i The role of the DNA sequence in determining the structure of a protein and how that protein is produced by transcription and translation. 104 110

c E Demonstrate comprehensive understanding of gene expression: Link biological ideas and processes about gene expression. The discussion may involve justifying, relating, evaluating, comparing and contrasting, or analysing.

Select biological ideas and processes from...

Key terms

c 3 Protein synthesis. 104 - 109

c b The nature of the genetic code. 106 107

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c iii Significance of proteins with reference to their structural and catalytic roles. 110 113

Nucleic acid structure and the genetic code numberActivity

Gene expression involves a selection from ...

Achievement criteria and explanatory notes

c 2 Proteins and their significance. 110 -113

The genetic code is the set of rules by which information encoded in genes is translated into proteins by cells. Catalytic proteins, called enzymes, regulate metabolic pathways and are involved in determining phenotype. Mutation and environmental effects can affect phenotype.

Explanatory notes: Gene expression numberActivity

Achievement criteria for achieved, merit, and excellence

• the 4-letter alphabet and 3-letter triplet code (codon) of base sequences.

117

c Identify transcription and translation as the two stages of gene expression.

Select biological ideas and processes from...

Proteins and protein synthesis

Activities 104 116

c Outline flow of information from DNA  mRNA  polypeptide or protein. Describe the one gene-one polypeptide relationship.

c Use examples to explain how phenotype is determined via metabolic pathways.

Metabolic pathways, mutation, and phenotype

Describe the structure of a nucleotide. State the role of the five-carbon sugar (deoxyribose or ribose), phosphate, and nitrogenous bases in its structure and function.

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

c Describe the double-helix model of DNA. Describe complementary base pairing in DNA and demonstrate an understanding of the base-pairing rule.

c Contrast the structure and roles of RNA and DNA.

c ii The potential effect on genotype and phenotype of mutations at the gene level. 121 - 124

c Use examples to explain how the environment can affect the phenotype without any alteration to genotype. You can include reference to the physical (external) environment and the effect of chemical signatures and markers that control the expression of genes (epigenetic factors).

c Describe the basic structure of amino acids and proteins. Describe the structural roles of fibrous proteins and the catalytic roles of globular proteins.

c Describe types of gene and chromosome mutations and outline their general effect on phenotype.

c Describe and explain the main features of the genetic code, including:

Select biological ideas and processes from...

Determination of phenotype via metabolic pathways numberActivity

c iii Phenotype is determined by the presence, absence, or amount of metabolic products. 117 - 118

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c Explain translation in eukaryotes, including the role of mRNA, tRNA, ribosomes, and enzymes.

119

c Explain how, in a metabolic pathway, the end product of one reaction can be the substrate for the next in a series of linked reactions. Each reaction is catalysed by an enzyme.

c Explain transcription in eukaryotes, including the role of enzymes in the process.

• The degeneracy of the code and how this leads to redundancy.

117

Recognise DNA and RNA as nucleic acids made from nucleotide monomers. Describe their biological roles.

Nucleic acids and the genetic code

Activities 99 104, 106 107, 114 116

c Use examples to explain how mutagens can change genotype through mutation.

Activities 117 128

• the non-overlapping, linear nature of the code which is read in one direction from start to finish.

c i Mutagens (recognise specific mutagens but not their effect at the molecular level). 120

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

c iii Ideas and processes relating to the effect of environment on phenotype involve how environmental factors may change phenotype without changing genotype. 126125

c ii Biochemical reactions do not occur in isolation but form part of a chain reaction in which the product of one reaction becomes the substrate of the next. 118

What you need to know for this Achievement Standard

Effect of environment on expression of phenotype numberActivity

c i Biochemical reactions are catalysed by specific enzymes and every enzyme is encoded by a specific gene(s). 118

(b) Where is most of the DNA found in eukar yotes?

DNA contains the instructions an organism needs to develop, function, and reproduce. Small differences in the DNA within a species cause the differences in appearance we see between individuals.

Nucleus

most of the cell's DNA is located in the nucleus, associated with proteins that help package it up. A small amount is located in mitochondria and in the chloroplasts of plants.

f DNA is called the blueprint for life because it contains all of the information an organism needs to develop, function, and reproduce.

f DNA stores and transmits genetic information.

The DNA in eukaryotes is packaged into chromosomes (above). Each chromosome consists of a DNA molecule and packaging proteins. The number and appearance of chromosomes is specific to the species.

f DNA is found in every cell of all living organisms.

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packaged?Ineukaryotes,

f DNA stands for deoxyribonucleic acid.

About DNA

The Role of DNA in Cells

2. (a) How is DNA organised in eukaryotes?

f DNA has a double-helix structure (right). If the DNA in a single human cell was unwound, it would be more than two metres long! The long DNA molecules are packed with protein in an organised way so that they can fit into the nucleus.

1. (a) What does DNA stand for?

DNA

Key Idea: A cell's genetic information is called DNA. In eukaryotic cells, DNA is located in the cell's nucleus.

(b) Why does DNA have to be tightly

The Structure of Chromosomes

Chromatin consists of a complex of DNA and histone proteins

Genestring.(protein

The chromatin is coiled up, reducing its length.

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DNA

3. What has to happen before a gene can be expressed (made into a protein)?

coding sequence). Genes on a chromosome can only be expressed when the DNA is unwound. This allows the enzymes involved to access the DNA.

Start sequence

f In eukaryotes, DNA is complexed with proteins to form chromatin. The proteins in the chromatin are responsible for packaging the chromatin into discrete linear structures called chromosomes. The extent of packaging changes during the life cycle of the cell, with the chromosomes becoming visible during cell division.

f Each chromosome includes many DNA sequences called genes. A eukaryotic gene codes for protein but also includes regulatory regions which are involved in expression of the gene into the protein.

ExonIntron

Stop sequence

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Key Idea: A chromosome is a single long molecule of DNA coiled around histone proteins. Chromosomes contain protein-coding regions called genes.

(b) What is the role of histones in this process?

1. (a) Suggest a purpose for DNA coiling:

Histone proteins

A gene includes start and stop sequences that control the gene's expression as well as some sections, called introns, which do not code for protein. Only the proteincoding sequences (exons) appear in the final mRNA, which is read by ribosomes to make the protein.

Nucleosome (eight histone proteins wrapped in DNA). The nucleosome is the basic unit of DNA packing and acts as a spool around which the DNA winds. The nucleosomes have the appearance of beads on a

2. Identify three parts of the gene that do not appear in the final mRNA:

Chromatids (2)

In eukaryotes, chromosomes are formed from the coiling of chromatin

Chromosome

OHOHOH H

Adenine Guanine A G

Nucleotides100 1 23 4 5 LINK 101 LINK 102 WEB 100 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Five different kinds of nitrogen bases are found in nucleotides. These are:

Uracil (U)

Pyrimidines:Purines:

Cytosine (C)

Cytosine only) Uracil only)

1. In RNA:

CH 2 H HOHH H H N N N NNH 2 P O O OH PhosphateOH Sugar Base O H H Chemical structure of a nucleotide

Nucleotide bases

Ribose

The structure of a nucleotide

f A five carbon sugar

form of a nucleotide (showing positions of the 5 C atoms on the sugar) groupPhosphate Sugar Base

Nucleotides are the building blocks of nucleic acids (DNA and RNA). Nucleotides have three parts to their structure (see diagrams below):

What are the three components of a nucleotide? 2. List the nucleotide bases present: (a) In DNA: (b) In RNA: 3. Name the sugar present: (a) In DNA: (b)

DNA contains adenine, guanine, cytosine, and RNAthymine.also contains adenine, guanine, and cytosine, but uracil (U) is present instead of thymine.

U (two-ring (single-ringbases)bases)

f A phosphate group

C Thymine (DNA

f A nitrogen containing base

Sugars

sugar

Nucleotides contain one of two different sorts of sugars. Deoxyribose sugar is only found in DNA. Ribose sugar is found in RNA.

(RNA

Symbolic

Key Idea: Nucleotides are the building blocks of the nucleic acids, DNA and RNA. A nucleotide has three components; a base, a sugar, and a phosphate group.

Adenine (A)

Deoxyribose (found in DNA) sugar (found in RNA)

Thymine (T)

Guanine (G)

Nucleotides are linked together by a condensation reaction (left). It is called a condensation reaction because water is produced. The link occurs between the phosphate group of one nucleotide and the sugar group of another.

101

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When long chains of nucleotides are joined together, they form nucleic acids. Deoxyribonucleic acid (DNA) is a nucleic acid.

DNA consists of a two strands of nucleotides linked together, forming a double helix. A double helix is like a ladder twisted into a corkscrew shape around its length. It is ‘unwound’ in the diagram on the right to show how the bases pair up.

f Transfer RNA (tRNA)

3. If you wanted to use a radioactive or fluorescent tag to label only the DNA in a cell and not the RNA, what compound(s) would you label? A C G C T G

DNA

In RNA, uracil (U) replaces thymine in the code. A C UG

1. The diagram on the right shows a double-stranded DNA molecule. Label the following: (a) Sugar group (d) Purine bases (b) Phosphate group (e) Pyrimidine bases (c) Hydrogen bonds (f) 5' and 3' ends

f Messenger RNA (mRNA)

f Each "rung" of the DNA molecule is made up of two nitrogen bases, joined together by hydrogen bonds between the bases (a hydrogen bond is a bond between hydrogen and an electronegative atom such as oxygen).

RNA

2. If you wanted to use a radioactive or fluorescent tag to label only the RNA in a cell and not the DNA, what compound(s) would you label?

f The ends of a DNA strand are labelled the 5' (five prime) and 3' (three prime) ends. The 5' end has a terminal phosphate group (off carbon 5), the 3' end has a terminal hydroxyl group (off carbon 3).

f The DNA backbone is made up of alternating phosphate and sugar molecules.

f Each DNA strand has a direction. Each single strand runs in the opposite direction to the other. This gives the DNA molecule an asymmetrical (uneven) structure.

bondHydrogen T A C G A C T G5'3' 3'5' T G H2O Carbon no. 3 no.Carbon5 LINK 100 LINK 102 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Ribonucleic acid (RNA) is a type of nucleic acid. Like DNA, the nucleotides are linked together through condensation reactions. RNA is single stranded. Its functions include protein synthesis and cell regulation. There are 3 types of RNA:

DNA and RNA

f Ribosomal RNA (rRNA)

The presence of two OH groups on the ribose sugar stops it from forming a double stranded helix.

Key Idea: DNA and RNA are nucleic acids made up of long chains of nucleotides, which store and transmit genetic information. DNA is double-stranded. RNA is single-stranded.

RNAs are involved in decoding the genetic information in DNA, as messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). Some RNAs can also act as enzymes involved in regulating gene expression.

4. What is the purpose of the hydrogen

describe the roles of RNA: (a) mRNA: (b) tRNA: (b) rRNA: 6 (a) Why do the DNA strands have an asymmetrical structure? (b) What are the differences between the 5' and 3' ends of a DNA strand? 7. Complete the following table summarising the differences between DNA and RNA molecules: DNA RNA Sugar present Bases Numberpresentofstrands PREVIEWONLYNotforClassroomUseNotforClassroomUse

5 Briefly

Transfer RNA (above) carries amino acids to the growing polypeptide chain. One end of the tRNA carries the genetic code in a three-nucleotide sequence called the anticodon. The amino acid links to the 3' end of the tRNA.

Ribosomal RNA (above) forms ribosomes from two separate ribosomal components (the large and small subunits) and assembles amino acids into a polypeptide chain.

Messenger RNA (above) is transcribed (written) from DNA. It carries a copy of the genetic instructions from the DNA to ribosomes in the cytoplasm, where it is translated into a polypeptide chain.

bonds in double-stranded DNA?

1. Cut out each of the nucleotides on page 149 by cutting along the columns and rows (see arrows indicating cutting points). Although drawn as geometric shapes, these symbols represent chemical structures.

4. Create one strand of the DNA molecule by placing the 9 correct 'cut out' nucleotides in the labelled spaces on page 151 (DNA molecule). Make sure these are the right way up (with the P on the left) and are aligned with the left hand edge of each box. Begin with thymine and end with guanine.

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6. It is not normally possible for base mismatches to occur. Describe two factors that prevent a mismatch from occurring:

Place a cut-out symbol for thymine here

Key Idea: Nucleotides pair together in a specific way called the base pair ing rule. In DNA, Adenine always pairs with thymine, and cytosine always pairs with guanine. Model

7. Once you have checked that the arrangement is correct, you may glue, paste or tape these nucleotides in place.

Factor 1:

Place a cut-out symbol for cytosine here

Place a cut-out symbol for guanine here

Place a cut-out symbol for adenine here

Before Watson and Crick described the structure of DNA, an Austrian chemist called Chargaff analysed the base composition of DNA from a number of organisms. He found that the base composition varies between species but that within a species the percentage of A and T bases are equal and the percentage of G and C bases are equal. Validation of Chargaff's rules was the basis of Watson and Crick's base pairs in the DNA double helix model.

Constructing a DNA

Adenine

3. Identify and label each of the following features on the adenine nucleotide immediately above: phosphate, sugar, base, hydrogen bonds.

Cytosine always pairs with Guanine C G

Thymine

Chargaff's rules

f DNA are double stranded molecules that contain genetic information. Each strand is made up of nucleotides.

102 LINK 47 WEB 102 LINK 101 LINK 100

Thymine always pairs with Adenine T A

2. Place one of each of the four kinds of nucleotide on their correct spaces below:

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Factor 2:

Guanine always pairs with Cytosine G C

DNA base pairing rule

GuanineCytosine

5. Create the complementary strand of DNA by using the base pairing rule above. Note that the nucleotides have to be arranged upside down.

f The chemical properties of each nucleotide mean it can only bind with a one other type of nucleotide. This is called the base pairing rule, and is explained in the table below, right. The exercise on the following pages is designed to help you learn and remember this rule.

Adenine always pairs with Thymine A T

149 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited Nucleotides P Cytosine S P Cytosine S P Cytosine S S P Adenine S P Adenine S P Adenine S P Guanine S P Thymine S P Thymine S P Thymine S P Guanine S P GuanineCutCutCut Cut Cut Cut Cut Cut P Cytosine S P Cytosine S P Cytosine S S P Adenine S P Adenine S P Adenine S P Thymine S P Thymine S P Thymine S P Guanine S P Guanine S P Guanine Tear out this page and separate each of the 24 nucleotides by cutting along the columns and rows (see arrows indicating the cutting points). PREVIEWONLYNotforClassroomUseNotforClassroomUse

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This page is left blank deliberately

151 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited GuanineCytosineThymineThymineGuanineAdenineAdenineCytosine S P Thymine S P Adenine Thymine Put the nucleotidesnamedon the left hand side to create the template strand Put the matching complementary nucleotides opposite the template strand DNA molecule PREVIEWONLYNotforClassroomUseNotforClassroomUse

The primary transcript is edited. The non-protein coding introns are removed and modifications are made to help the mRNA exit the nucleus.

Nucleus LINK 104 LINK 107 PREVIEWONLYNotforClassroomUseNotforClassroomUse

In the nucleus, the gene is rewritten into a single stranded primary RNA transcript, using one strand of DNA as a template. RNA polymerase catalyses this process.

2. (a) What does gene expression mean? (b) What are the three stages in gene expression in eukaryotes and what happens in each stage?

Cytoplasm transcriptPrimarymRNA

Nuclear

EDITING

TRANSLATION TRANSCRIPTION

Ribosome pore

f Gene expression is the process by which the information in a gene is used to synthesise the gene product. Typically this is illustrated by the one gene-one protein model which involves transcription of the DNA into mRNA and translation of the mRNA into protein (shown below).

f A gene is a section of DNA that codes for a protein (or some other functional mRNA product).

Amino acids are linked together at the ribosome to form the protein encoded by mRNA.

mRNA

f Eukaryotic genes include non-protein coding regions called introns. These regions of intronic DNA must be edited out before the mRNA is translated by the ribosomes. Transcription of the genes and editing that primary transcript to form the mature mRNA occurs in the nucleus. Translation of the protein by the ribosomes occurs in the cytoplasm.

(b) Where does translation occur in eukaryotes?

(iii)(ii)(i)

3. (a) Where does transcription occur in eukar yotes?

Key Idea: Genes are sections of DNA that code for proteins. Genes are expressed when they are transcribed into messenger RNA (mRNA) and then translated into a protein.

1. What is a gene?

DNA

Key Idea: Transcription is the first step of gene expression. In eukaryotes, transcription takes place in the nucleus and is carr ied out by RNA polymerase.

Coding (sense) strand of DNA

RNA polymerase (RNAP) adds nucleotides to the 3' end so the strand is synthesised in a 5' to 3' direction.

RNA promoterbindspolymeraseattheupstreamregion RNA terminatordissociatespolymeraseattheregion 3' 3' 3' 5' 5'

Several RNA polymerases may transcribe the same gene at any one time, allowing a high rate of mRNA synthesis.

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f Transcription is the first stage of gene expression. In eukaryotes, it occurs in the nucleus. It is catalysed by the enzyme RNA polymerase, which rewrites the DNA into a primary RNA transcript using a single template strand of DNA.

f The protein-coding part of a gene is bounded by an upstream start (promoter) region and a downstream terminator region. These regions control transcription by telling RNA polymerase where to start and stop transcription. In eukaryotes, non protein-coding sections called introns are removed. The remaining exons are spliced together to form the mature mRNA before the gene can be translated into a protein. This editing process also occurs in the nucleus.

104

(b) What strand of DNA does this enzyme use?

Direction transcriptionof

codonbeginTranslationstrandwillatthestartAUG

TranscriptionProhibited

3. (a) Why is AUG called the star t codon?

2. (a) In which direction is the RNA strand synthesised?

Recall that the primary RNA transcript is edited to form the mature mRNA, which passes to the cytoplasm where the nucleotide sequence is translated into a polypeptide.

(d) Which nucleotide base replaces thymine in mRNA?

Newly synthesised RNA strand is complementary to the template

(e) On the diagram, use a coloured pen to mark the beginning and end of the protein-coding region being transcribed.

(b) Explain why this is the case:

(b) What would the three letter code be on the DNA coding strand?

Template (antisense) strand of DNA stores the information that is transcribed into mRNA

mRNA nucleotides. Free nucleotides are used to construct the RNA strand

1. (a) Name the enzyme responsible for transcribing the DNA:

Gene expression

The table on the right is used to ‘decode' the genetic code. It shows which amino acid each mRNA codon codes for. There are 64 different codons possible, 61 code for amino acids, and three are stop codons.

ii Find the column that intersects that row from the top, second letter, row.

i Find the first letter of the codon in the row on the left hand side of the table. AUG is the start codon.

iii Locate the third base in the codon by looking along the row on the right hand side that matches your codon.

The mRNA - amino acid table

e.g. GAU codes for Asp (aspartic acid)

Amino acid names are written as three letter abbreviations (e.g. Ser = serine). To work out which amino acid a codon codes for, carry out the following steps:

What is the Genetic Code? Idea:

Key

strand

2. What do you notice about the sequence on the DNA coding strand and the mRNA strand? UUU Phe UUC Phe UUA Leu UUG Leu UCU Ser UCC Ser UCA Ser UCG Ser UAU Tyr UAC Tyr ACU Thr ACC Thr ACA Thr ACG Thr GCU Ala GCC Ala GCA Ala GCG Ala CCU Pro CCC Pro CCA Pro CCG Pro GUU Val GUC Val GUA Val GUG Val CUU Leu CUC Leu CUA Leu CUG Leu AUU Ile AUC Ile AUA Ile AUG Met Read second letter here Read first letter here Read third letter here U UAA STOP UAG STOP CAU His CAC His CAA Gln CAG Gln AAU Asn AAC Asn AAA Lys AAG Lys GAU Asp GAC Asp GAA Glu GAG Glu UGU Cys UGC Cys UGA STOP UGG Trp CGU Arg CGC Arg CGA Arg CGG Arg AGU Ser AGC Ser AGA Arg AGG Arg GGU Gly GGC Gly GGA Gly GGG Gly

1. (a) Use the base-pairing rule for to create the complementary strand for the DNA template strand shown below.

mRNA TranscriptionTranslation DNAacidsAmino TA C CCAATGGACTCCCA TTA T GCCCGTGAA AT C DNA TA C CCAATGGACTCCCA TTA T GCCCGTGAA AT C Complementary strand (this is the DNA coding strand) Template strand

Template

(b) For the same DNA template strand, then deter mine the mRNA sequence and use the mRNA amino acid table to determine the amino acid sequence. Note that in mRNA, uracil (U) replaces thymine (T) and pairs with adenine.

Each mRNA codon codes for a specific amino acid. This activity will help you practise using the amino acid table to determine the amino acids encoded by the genetic code. 105 LINK 103 WEB 105 LINK 107 LINK 106 PREVIEWONLYNotforClassroomUseNotforClassroomUse

(b) Glycine (Gly) is encoded in how many ways?

(b) The codons that encode aspartic acid (Asp):

6. (a) Arginine (Arg) is encoded in how many ways?

f Degeneracy is when a particular output can be produced by several different pathways.

FlightControlcomputerslines

Degeneracynormally.isseen

The degeneracy of the genetic code creates redundancy, so that several codons code for the same amino acid (e.g. CCU, CCC, CCA, and CCG code for proline). Note that although there is redundancy, there is no ambiguity - none of the codons encodes any other amino acid.

Redundancy and degeneracy are important concepts in understanding the genetic code.

Salivary amylase (above) is structurally different to pancreatic amylase, but has the same function.

in the production of the enzymes salivary and pancreatic amylase. Salivary amylase breaks down carbohydrates in the mouth, whereas pancreatic amylase does so in the small intestine. The enzymes are encoded by different genes (AMY1A and AMY2A) but have the same functional role (right).

Modern aircraft (left) have multiple redundant features for safety. Often there are three or four flight computers linked independently to the flight surfaces and other input/output devices. If one computer or control line fails the others can continue to fly the plane

f Redundancy is when several situations code for or control the actions of one specific thing.

CCGCCACCCCCU Pro CCGCCACCCCCUPro PREVIEWONLYNotforClassroomUseNotforClassroomUse

5. Identify the following:

Examples of redundancy and degeneracy are illustrated below. In modern aircraft redundant features add safety by making sure if one system fails others will ensure a smooth, safe flight. Degeneracy can be seen in proteins when different proteins have the same function.

3. Explain how degeneracy adds "safety" to the coding of protein chains:

(a) The codons that encode valine (Val):

The genetic code shows degeneracy. This means that a number of 3 base combinations specify one amino acid. The codons for the same amino acid often differ by only a single letter (often the second or third). For example, proline is encoded by four different codons.

Redundancy and degeneracy

4. The genetic code shows redundancy but no ambiguity. What does this mean and why is it important?

2314

(43) 4 x 4 x 464 amino acids

DNase was added to destroy bacterial DNA so there was no template for mRNA to be made. Cell free E. coli extract.

mRNA was added an amino acid was produced. The codon UUU produced the amino phenylalanineacid(Phe).

Once it was discovered that DNA carries the genetic code needed to produce proteins, the race was on to "crack the code" and find out how it worked.

The triplet code was accepted once scientists confirmed that some amino acids have multiple codes.

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The first step was to find out how many nucleotide bases code for an amino acid. Scientists knew that there were four nucleotide bases in mRNA and 20 amino acids in proteins. Simple mathematics (below) showed that a one or two base code did not produce enough amino acids, but a triplet code produced more amino acids than were found in proteins.

Double

Radio-labelled amino acids and a synthetic mRNA containingstrandonly uracil (U) were added.

UUUU

(b) What would it have been difficult to crack the code if no DNase was added?

1. (a) How many types of nucleotide bases are there in mRNA?

OncemRNA).the

LINK 105 WEB 106

A cell free E. coli extract was produced for their experiment by rupturing the bacterial cells to release the cytoplasm. The extract had all the components needed to make proteins (except

Key Idea: Scientists used mathematics and scientific experiments to unlock the genetic code. A series of three nucleotides, called a triplet, codes for a single amino acid.

106

The genetic code

How was the genetic code cracked?

Once the triplet code was discovered, the next step was to find out which amino acid each codon produced. Two scientists, Marshall Nirenberg and Heinrich Matthaei, developed an experiment (below) to crack the code.

3. (a) Why was DNase added to the cell free E. coli extract?

A triplet (three nucleotide bases) codes for a single amino acid. The triplet code on mRNA is called a codon.

Triple

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Number of bases in the code Working Number of amino acids produced

(41)4 4 amino acids

Phe Phe Phe Phe

Cracking the Genetic Code

(b) How many types of amino acids are there in proteins?

(42) 4 x 416 amino acids

U UUUUUUU

Single

2. A triplet code could potentially produce 64 amino acids. Why are only 20 amino acids produced?

Over the next few years, similar experiments were carried out using different combinations of nucleotides until all of the codes were known.

Ribosomes are made up of a complex of ribosomal RNA (rRNA) and ribosomal proteins. These small cellular structures direct the catalytic steps required for protein synthesis and have specific regions that accommodate transfer RNA (tRNA) molecules loaded with amino acids.

(c) Determine the mRNA codons and the amino acid sequence for the following tRNA anticodons: tRNA anticodons: UAC U AG CCG CGA UUU Codons on the mRNA:

(a) Ribosome:

(a) How many different tRNA types are there, each with a unique anticodon?

3. There are many different types of tRNA molecules, each with a different anticodon (HINT: see the mRNA table).

Amino acids encoded:

(e) Stop codon:

SmallLargesubunitsubunit

Anticodon is a 3-base sequence complementary to the codon on mRNA.

Ribosome structure

subunitLarge Functional ribosome

(d) Start codon:

tRNA molecules are RNA molecules, about 80 nucleotides long, which transfer amino acids to the ribosome as directed by the codons in the mRNA. Each tRNA has a 3-base anticodon, which is complementary to a mRNA codon. There is a different tRNA molecule for each possible codon and, because of the degeneracy of the genetic code, there may be up to six different tRNAs carrying the same amino acid.

2. What is the role of each of the following components in translation?

Key Idea: Translation is the final stage of gene expression in which ribosomes read the mRNA and decode (translate) it to synthesise a protein. This occurs in the cytoplasm.

(b) Explain your answer:

Translation107

tRNA structure

Amino acid attachment site. Enzymes attach the tRNAs to their specific amino acids.

1. Describe the structure of a ribosome:

f In eukaryotes, translation occurs in the cytoplasm either at free ribosomes or ribosomes on the rough endoplasmic reticulum. Ribosomes translate the code carried in the mRNA molecules, providing the catalytic environment for the linkage of amino acids delivered by transfer RNA (tRNA) molecules.

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f Protein synthesis begins at the start codon and, as the ribosome wobbles along the mRNA strand, the polypeptide chain elongates. On reaching a stop codon, the ribosome subunits dissociate from the mRNA, releasing the protein.

Ribosomes exist as two separate sub-units (below) until they are attracted to a binding site on the mRNA molecule, when they come together around the mRNA strand.

(b) tRNA:

Arg Met-tRNAUnloaded Val

The ribosome P (peptidyl) site carries the growing polypeptide chain.

158

tRNA molecules match amino acids with the appropriate codon on mRNA. As defined by the genetic code, the anticodon specifies which amino acid the tRNA carries. The tRNA delivers its amino acid to the ribosome, where enzymes join the amino acids to form a polypeptide chain. During translation the ribosome "wobbles" along the mRNA molecule joining amino acids together. Enzymes and energy are involved in charging the tRNA molecules (attaching them to their amino acid) and elongating the peptide chain.

Lys-tRNACharging Val-tRNACharged 3'

The polypeptide chain grows as more amino acids are added STOP

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mRNA

tRNA molecules deliver amino acids to ribosomes

Protein synthesis begins when the ribosome reads the start codon (AUG). This codes for methionine (Met) which may be removed after transation depending on the protein.

The polypeptide chain continues to grow as more amino acids are added.

Charged Arg-tRNA enters the ribosome A (acceptor) site The amino acid is added to the growing polypeptide

Chargedchain.tRNAs

enter at the A site except for the first amino acid methionine (Met), which enters at the P site to begin the process.

Start codon

5. Many ribosomes can work on one strand of mRNA at a time (a polyribosome system). What would this achieve?

Protein synthesis stops when a stop codon is reached (UGA, UAA, or UAC). The ribosome falls off the mRNA and the polypeptide is released.

4. Describe the events occurring during translation:

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Lys

Thr-tRNAUnloaded leaves the ribosome E (exit) site

5'STARTDirection of protein synthesis

PheThrMet

Ribosome (only large subunit shown)

(a) Process 1: (b) Process 2: (c) Process 3: (d) Process 4: (e) Process 5: (f) Process 6: (g) Process (h) Process Process

1. Briefly describe each of the numbered processes in the diagram above:

(f) Structure (g) Structure (h) Structure Structure Structure

G: (c) Structure C:

3. Describe two factors that would determine whether or not a particular protein is produced in the cell:

8: (i)

This diagram provides a visual overview of gene expression. It combines information from the previous activities. Each of the major steps in the process are numbered, whereas structures are identified with letters.

(a) Structure

2. Identify each of the structures marked with a letter and write their names below in the spaces provided:

F: (b) Structure B:

(b)(a) Thr 3'5' Lys Met Phe Phe ThrArg Met Tyr Tyr TyrAsn Phe Thr Arg Met Arg Val Val Cys Lys Lys Thr Ala 5' 3' Protein Synthesis Summary108

H: (d) Structure D: (i)

Gene #1 Gene #2 LINK 103 LINK 104 LINK 107 WEB 108 PREVIEWONLYNotforClassroomUseNotforClassroomUse

I: (e) Structure E: (j)

This 'R' group gives the amino acid an acidic property.

A Polypeptide Chain

2. (a) Name the type of bond that links amino acids together:

(b) What makes each of the 20 amino acids found in proteins unique?

Aspartic acid

NHAspartic2CHCHCOOH

Structure of an amino acid

Aspartic acid

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NHCysteine2COOHCCH2SHH

NH 2 COOHCHCHCHC22CH22HNH2

atomHydrogen Aminegroup Carbonatom C

NH 2 COOHCHCHCHC22CH22HNH2

(b) What is the name given to many amino acids joined together by this bond?

1. (a) Describe the structure of an amino acid:

f Twenty amino acids occur commonly in proteins. Nonprotein amino acids have roles as intermediates in metabolic reactions.

PeptidebondPeptidebond Peptidebond Peptidebond Peptidebond LINK 103 WEB 109 LINK 110 LINK 107 PREVIEWONLYNotforClassroomUseNotforClassroomUse

NH 2 COOH

This 'R' group can form disulfide bridges with other cysteines to create cross linkages in a polypeptide chain.

CHCH COOH2

CHCHCHC22CH22HNH2

The 'R' group varies in chemical make-up with each type of amino acid.

109

NHCysteine2COOHCCH2SHH

CHCH COOH2

Lysine NH 2 COOH

NH2 C CarboxylOHO group makes the molecule behave like a weak acid.

Key Idea: Amino acids are the basic units of proteins. They have a common structure but contain a variable "R" group, which gives each of them unique properties.

Lysine

NH 2 COOH

f Peptide bonds form during condensation reactions in which the carboxyl group from one amino acid reacts with the amine group of another. Water is produced as a by-product.

Amino acids are the basic units from which proteins are made. Plants manufacture all the amino acids they require from simpler molecules, but animals must obtain a certain number of ready-made amino acids called essential amino acids. These must be taken in with the diet.

A polypeptide chain

This 'R' group gives the amino acid an alkaline property.

NHCysteine2COOHCCH2SHH

f Amino acids are linked together by peptide bonds to form long chains of up to several thousand amino acids. These are called polypeptide chains. The order of amino acids in a polypeptide is directed by the order of nucleotides in DNA (and thus mRNA).

3. What is the function of a disulfide bridge?

f All amino acids have a common structure. The only difference between the different types lies with the 'R' group in the general formula. This group is variable, which means that it is different in each kind of amino acid (right and below).

Amino Acids Make up Proteins

Some complex proteins are only functional when as a group of polypeptide chains. Haemoglobin has a 4° structure made up of two alpha and two beta polypeptide chains, each enclosing a complex iron-containing prosthetic (or haem) group.

The

Tertiary (3°) structure is maintained by more distant interactions such as disulfide bridges between cysteine amino acids, ionic bonds, and hydrophobic interactions.

A protein's 3° structure is the three-dimensional shape formed when the 2° structure folds up and more distant parts of the polypeptide chains interact.

Secondary (2°) structure (a-helix or b pleated sheet)

Hydrogen bonds Amino acid chain

Primary (1°) structure (amino acid sequence)

(b) Secondary structure:

Polypeptide chains fold into a secondary (2°) structure based on H bonding. The coiled a-helix and b-pleated sheet are common 2° structures. Most globular proteins contain regions of both 2° configurations.

Ionic AsparticLysinebondacid a-helix bondDisulfide Beta chain Alpha(haem)Prostheticchaingroup LINK 109 LINK 111 LINK 112 WEB 110 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Tertiary (3°) structure (folding of the 2° structure)

Quaternary (4°) structure

1. Describe the main features that aid the formation of each part of a protein's structure:

b sheet-pleated a-helix

Hundreds of amino acids are linked together by peptide bonds to form polypeptide chains. The attractive and repulsive charges on the amino acids determines how the protein will fold up.

f The folding of a protein into its functional form creates a three dimensional structure. It is this tertiary structure that gives a protein its unique chemical properties. If a protein loses this precise structure (i.e. the protein is denatured), it is usually unable to carry out its biological function.

f Proteins are large, complex macromolecules, built up from a linear sequence of amino acids. Proteins are molecules of central importance in the chemistry of life. They account for more than 50% of the dry weight of most cells, and they are important in virtually every cellular process.

Key Idea: The sequence and type of animo acids in a protein determines that protein's threedimensional shape and its function. Structure Proteins

A protein's 4° structure describes the arrangement and position of each of the subunits in a multiunit protein. The shape is maintained by the same sorts of interactions as those involved in 3° structure.

(a) Primary structure:

110 Peptide bond Amino acid Phe Glu Glu Tyr Ser Ser Ser Iso Met MetAla andneighbouringbondsbyis(2°)SecondarystructuremaintainedhydrogenbetweenCONHgroups.

2. Which two structures can act as fully functional proteins or enzymes?

Enzymes

2. Why do channel proteins often fold non-polar the

Channel proteins

f A protein may consist of one polypeptide chain, or several polypeptide chains linked together. Hydrogen bonds between amino acids cause the polypeptide chain to form its secondary structure, either an a-helix or a b-pleated sheet. The interaction between R groups causes a polypeptide to fold into its tertiary structure, a three dimensional shape held by ionic bonds and disulfide bridges (bonds formed between sulfur containing amino acids). If bonds are broken (through denaturation), the protein loses its tertiary structure, and its functionality.

111

KNOW 162 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited H2N H2N COOH COOH Ser Pro Arg Glu Glu Thr Thr Tyr Tyr His HisHis Cys Cys Gly Gly Gly Gln Ala Lys Val Val Val Leu Leu Leu Leu Asn Val Glu Glu Ile Ile Leu SerThr Ser Tyr Tyr Gly Cys Cys CysCysGln Gln Leu Asn Asn Phe PhePhe S S S S S S a chain b chain

Protein denaturation

1. Using the example of

Cooked (denatured) egg white

b sheets a helix

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insulin, explain how interactions between R groups stabilise the protein's functional structure:

Key Idea: The three dimensional shape of a protein reflects its role. When a protein is denatured, it loses its functionality.

with

channel's exterior and polar R groups to its interior?

Proteins that fold to form channels in the plasma membrane present non-polar R groups to the membrane and polar R groups to the inside of the channel. Hydrophilic molecules and ions are then able to pass through these channels into the interior of the cell. Ion channels are found in nearly all cells and many organelles.

Enzymes are globular proteins that catalyse specific reactions. Enzymes that are folded to present polar R groups at the active site will be specific for polar substances. Non-polar active sites will be specific for non-polar substances. Alteration of the active site by extremes of temperature or pH cause a loss of function.

AmylasesiteActive

3. Why does denaturation often result in the loss of protein functionality?

When the chemical bonds holding a protein together become broken the protein can no longer hold its three dimensional shape. This process is called denaturation, and the protein usually loses its ability to carry out its biological function. There are many causes of denaturation including exposure to heat or pH outside of the protein's optimum range. The main protein in egg white is albumin. It has a clear, thick fluid appearance in a raw egg (right). Heat (cooking) denatures the albumin protein and it becomes insoluble, clumping together to form a thick white substance (far right).

R groups to

Raw (native) egg white

Sub-unit proteins

Many proteins, e.g. insulin and haemoglobin, consist of two or more sub-units in a complex quaternary structure, often in association with a metal ion. Active insulin is formed by two polypeptide chains stabilised by disulfide bridges between neighbouring cysteines. Insulin stimulates glucose uptake by cells.

Protein Shape is Related to Function

Globular proteins

HydrogenGlycinebond

Their properties include:

Human insulin is a relatively small globular protein involved in regulating blood glucose. It has two polypeptide subunits.

Their functions include:

• Transpor t, e.g. haemoglobin

Collagen (above) consists of three helical polypeptides wound around each other to form a ‘rope’ and held together by hydrogen bonds. Many collagen molecules form fibrils and the fibrils group together to form larger fibres.

• Structural role in cells and organisms, e.g. collagen in connective tissue, cartilage, bones, tendons, and blood vessel walls.

• Regulatory, e.g. hormones (insulin)

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Bonds between different polypeptide chains help hold globular proteins in their required shape.

• Easily water soluble

• Defensive, e.g. antibodies

112 LINK 110 LINK 111

Fibrous proteins

• Polypeptide chains folded into a roughly spherical shape

• Catalytic, e.g. enzymes

2. How does the shape of a fibrous protein relate to its functional role?

Globular proteins have a roughly spherical shape and are soluble in water. Many globular proteins are enzymes and have a catalytic role in regulating metabolic pathways.

• Tertiar y structure is critical to function

Covalent cross links between the collagen molecules.

• Contractile in the cytoskeleton and muscle

Fibrous proteins are fibre-like and not water soluble. They are often made up of repeating units and give stiffness and rigidity to the fluid components of cells and tissues. Their roles are structural and contractile.

• Physically ver y tough. They are supple or stretchy

• Parallel polypeptide chains in long fibres or sheets

3. How does the shape of a catalytic protein (enzyme) relate to its functional role?

Their properties include:

• Water insoluble

RuBisCo is a large multiunit enzyme. It catalyses the first step of carbon fixation in photosynthesis. It consists of 8 large and 8 small subunits and is the most abundant protein on Earth.

1. Contrast the properties of globular and structural proteins:

Their functions include:

Key Idea: Globular proteins have mainly catalytic and regulatory functions, while fibrous proteins have a mainly structural function.

Globular and Fibrous Proteins

Proteins

The structure of DNA and RNA

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Gene expression

113

HINT: You should define and explain transcription and translation including key molecules and events.

HINT: Include reference to amino acids, bonds, protein structure, and denaturation.

HINT: Include the features of each molecule, their roles in the cell, and the base pairing rule.

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REVISE 164

What You Know So Far: Nucleic Acids and Proteins

(a) Use the information to complete the table below (you only need one codon per amino acid): (codingDNA strand) (templateDNA strand) AminomRNA acids Met Pro Glu Asn Stop

NCEA Style Question: Nucleic Acids and Proteins114 LeuLeuPhePhe SerSerSerSer TyrTyr ProProProProThrThrThrThrAlaAlaAlaAlaMetIleIleIleLeuLeuValValValValLeuLeu Second letter letterrdThiFirstletter USTOPSTOPHisHisGlnGlnAsnAsnLysLysAspAspGluGlu GlyGlyGlyGlyArgArgSerSerArgArgArgArgTrpCysCysSTOP PREVIEWONLYNotforClassroomUseNotforClassroomUse

(b) Explain how the nature of the genetic code protects against the effect of point mutations in the DNA sequence:

1. The table below lists the amino acids and their codons:

166 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited 2. Discuss the production of a functional protein via translation. In your discussion, include the role(s) of: • Ribosomes • tRNA • Codons (including start and stop codons) and anticodons PREVIEWONLYNotforClassroomUseNotforClassroomUse

4. Complete the following paragraph by deleting one of the words in the bracketed () pairs below: In eukaryotes, gene expression begins with (transcription/translation) which occurs in the (cytoplasm/nucleus). (Transcription/Translation) is the copying of the DNA code into (mRNA/tRNA). The (mRNA/tRNA) is then transported to the (cytoplasm/nucleus) where (transcription/translation) occurs. Ribosomes attach to the (mRNA/tRNA) and help match the codons on (mRNA/tRNA) with the anticodons on (mRNA/tRNA). The (mRNA/tRNA) transports the animo acids to the ribosome where they are added to the growing (polypeptide/carbohydrate) chain.

KEYProhibitedTERMS

115

base-pairinganticodon rule

RNA template strand

A Single stranded nucleic acid that consists of nucleotides containing ribose sugar.

G The sequence of DNA that is read during the synthesis of mRNA.

D The structural units of nucleic acids, DNA and RNA.

E Intermolecular bond between hydrogen and an electronegative atom such as oxygen.

J Macromolecule consisting of many millions of units containing a phosphate group, sugar and a base (A,T, C or G). Stores the genetic information of the cell.

coding strand geneDNA nucleotidesnucleichydrogengeneticexpressioncodebondacids

K The region of a transfer RNA with a sequence of three bases that is complementar y to a codon in the messenger RNA.

B A set of rules by which information encoded in DNA or mRNA is translated into proteins.

2. A grasshopper has the following percentages of nucleotides in its DNA: A = 29.3, G = 20.5, C = 20.7, T = 29.3, %GC = 41.2, %AT = 58.6. For a rat, the percentages are A = 28.6, G = 21.4, C = 20.5, T = 28.4, %GC = 42.9, %AT = 57.0. Use the base pairing rule to explain this data:

3. For the following DNA sequence on the template strand, give the mRNA sequence and then Identify the amino acids that are encoded. For this question you may consult the mRNA-amino acid table earlier in the chapter.

AminomRNA:acids:

5. Decide if the nucleotide shown right is from DNA or RNA. Explain your choice:

CH 2 H HOHH H H N N N NNH 2 P O O OH PhosphateOH Sugar Base O H H PREVIEWONLYNotforClassroomUseNotforClassroomUse

C The rule governing the pairing of complementary bases in DNA.

F The process by which genetic infor mation is used to produce a functional gene product.

H Universally found macromolecules composed of chains of nucleotides. These molecules carry genetic information within cells.

AND IDEAS: Nucleic Acids and Proteins

I The DNA strand with the same base sequence as the RNA transcript produced (although with thymine replaced by uracil in mRNA).

DNA (template strand): G AAACCCTTACATATCGTGCT

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

Key Idea: A metabolic pathway is a ser ies of linked biochemical reactions. Specific enzymes catalyse each step of a metabolic pathway.

LINK 23 WEB 116 LINK 117 LINK 32 4.0CCOystertheFred

Enzyme B catalyses a reaction to chemically alter the intermediate into the end product.

Metabolic Pathways

2: Biochemical reactions occur in a chain of reactions in which the product of one reaction forms the substrate for the next reaction.

Expression of gene A produces enzymeEnzymeA

1: Biochemical reactions are catalysed by specific enzymes which are in turn encoded by specific genes.

f Metabolic pathways are controlled by regulating the amount of enzyme present (by switching the genes encoding that enzyme on or off) or by controlling enzyme activity. Each step in a metabolic pathway is part of a sequence, where the product from one step becomes the substrate for the next.

A simplified metabolic pathway

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(b) Describe the two general ways that metabolic pathways are regulated:

2. (a) What is the role of enzymes in metabolic pathways?

f The pathway shown below involves just two enzymes, two substrates, and one end product. It illustrates that:

f Enzymes catalyse each step of a metabolic pathway and, in turn, each enzyme is coded for by specific genes

substrateStarting

1. Define the term metabolic pathway:

End product Intermediate GeneEnzymeA A Enzyme B

A catalyses a reaction to chemically alter the starting substrate into an intermediate.

116

The diagram right shows a very simplified model of some of the many metabolic pathways in humans. It shows that metabolic pathways are not isolated and that products and substrates are not limited to one specific metabolic pathway but can be used in many metabolic pathways.

f Metabolic pathways are linked biochemical reactions that occur within living organisms to maintain life.

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Expression of gene B produces enzyme B Gene B

(a) Using the example above explain what would happen if there was a build up of product C?

CitrateSuccinateMalateFumarateAcetyl

4. (a) What molecule remains when urea is given off as a waste product in the urea cycle?

(b) Explain the role of enzymes in these metabolic pathways:

CoA

The Krebs Cycle

Each turn of a metabolic cycle regenerates the starting molecule. The end product of the cycle becomes a substrate for the cycle to begin again.

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(b) What is the role of this molecule in the cycle?

E E E E E E E Step is catalysed by an enzyme E

3. (a) Describe the features of a cyclic metabolic pathway:

5. What two molecules combine to for m the first substrate in the Krebs cycle?

NH3 + CO2 2 ATP ArginineUREA(waste product) Argininosuccinate Citrulline Ornithine Ammonia (toxic) Aspartate Fumarate Carbamyl phosphate E E EEE The urea cycle α-ketoglutarate oxaloacetate

Cyclic pathways

7. Feedback inhibition occurs when the product of a metabolic pathway inhibits the processing of the original substrate.

(b) Explain how feedback inhibition provides a simple way to regulate metabolic pathways:

f Some metabolic pathways flow in a cycle. The product from one step in the cycle is the substrate for the next step and the starting molecule is regenerated with each turn of the cycle (e.g. the urea cycle and Krebs cycle below). As with all metabolic pathways, each step is catalysed by an enzyme or a group of enzymes (sometimes called an enzyme complex). You do not need to know the details of these cycles, but it is important to understand the key role of enzymes in regulating metabolism.

6. Identify a product from one cycle that could act as a substrate in the other:

f An interruption in the reaction series prevents the pathway from progressing to the end and can result in a metabolic disorder. When this happens, there might be too much of one substance or too little of another. Often phenotype is affected.

Substrate1 Substrate2 Product

117

Substrate3 3 metabolism of

Enzyme

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Interrupting Metabolic Pathways

Key Idea: Metabolic pathways are interr upted when one or more of the enzymes catalysing the biochemical reactions involved are not expressed correctly.

The

KNOW 170 © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited MelaninThyroxine PhenylkeronuriaAlbinismhypothyroidismCongenital Ty PhenylpryuvicAlkaptonuriaTyrosinosisaacidpyruvicHydroxyphenyl-hydroxylasePhenylalanineTransaminaseacidoxidaseHomogentisicacidoxidase Faulty enzyme causes buildup of: This in causes:turn Faulty enzyme causes: Faulty enzyme causes: Faulty enzymecauses:Faulty causes:enzyme Tyrosinasea series of enzymesHydroxyphenylpyruvicPhenylalanineTyrosineacidHomogentisicacid Carbondioxideandwater MaleylacetoaceticProteinacid Enzyme 1 Enzyme 2

The metabolism of the essential amino acid phenylalanine is a well studied metabolic pathway. The effect of defective enzymes at each stage has been identified. A simplified pathway is shown here, with the result of defective enzymes shown in black.

Enzyme 3 is not expressed or is nonfunctional. The reaction series cannot progress to the end product, so levels of substrate 3 build up. If substrate 3 is a toxic substance this can have serious health effects.

f Each metabolic pathway is a series of biochemical reactions in which each step relies on the completion of the previous step.

msdonnaPhoto:

(d) Alkaptonuria:

(a) Albinism:

2. In the metabolic pathway of the metabolism of phenylalanine, identify the faulty enzyme that results in:

1. Identify four end products of the normal metabolism of phenylalanine:

Newborn babies are tested for a number of genetic disorders, including PKU, soon after birth. Blood is collected from a heel prick on to a Guthrie card (above) and tested. The prognosis is good if the disease is detected early, and a low phenylalanine diet is followed throughout life. People with PKU must also take supplements to provide the amino acids that would otherwise be lacking in a low-phenylalanine diet (e.g. tyrosine, which is normally derived from phenylalanine and which is needed for brain

Melanin is the pigment that gives skin, hair, and eyes their colour. It is formed from the metabolism of the amino acid tyrosine. Lack of melanin results in albinism, a condition where there is little or no pigmentation (above). The most common cause of albinism is a faulty enzyme in the pathway that converts tyrosine into melanin.

(b) Phenylketonuria:

(a) Albinism:

3. Why do people with phenylketonuria have light skin?

4. Using the metabolism of phenylalanine as an example, discuss the role of enzymes in metabolic pathways:

5. The conditions illustrated in the diagram are due to too much or too little of a chemical in the body. For each condition listed below, state which chemical causes the problem and whether it is absent or present in excess:

function).USAF cardGuthrie PREVIEWONLYNotforClassroomUseNotforClassroomUse

Phenylketonuria (PKU) is an example of a metabolic disorder that occurs when there is an error in a metabolic pathway. Babies born with PKU are missing the enzyme needed to catalyse the first step in the pathway that metabolises the essential amino acid phenylalanine. Without the enzyme, phenylalanine cannot be converted to the next substrate, tyrosine, and it is metabolised to toxic derivatives, which cause central nervous system damage. Children with PKU tend to have lighter skin and hair than people without the disorder.

Phenotype

Genotype

f Even identical twins have minor differences in their appearance due to epigenetic and environmental factors such as diet and intrauterine environment. Genes, together with epigenetic and environmental factors determine the unique phenotype that is produced.

Alleles

markers

regulate how

f The phenotype encoded by genes is a product not only of the genes themselves, but of their internal and external environment and the variations in the way those genes are controlled (epigenetics).

118

2. Explain why genetically identical twins are not always phenotypically identical:

DNA

KNOW 172

variationsSingleSexualMutationsreproductionnucleotide Environment

1. (a) What are some sources of genetically induced variation?

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Physical

Predators CompetitionPathogensDrugsToxinsNutrition

Chemical tags and that theexpressedis

3. Give an example in which an organism's environment produces a marked phenotypic change:

(b) What are some sources of environmentally induced variation?

Epigenetics environment

The phenotype is the product of the many complex interactions between the genotype, the environment, and the chemical tags and markers that regulate the expression of the genes (epigenetic factors).

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Polyphenism is the expression of different phenotypes in a species due to environmental influences. Examples include sex determination in reptiles and changes in pigmentation in the wings of some butterfly species as the seasons change. The amount of change in a phenotype due to environmental influences is called its phenotypic plasticity. Plants often have high phenotypic plasticity because they are unable to move and so must adjust to environmental changes throughout their lives.

Key Idea: An organism's phenotype is influenced by the effects of the environment during and after development, even though the genotype

Influences on Phenotype

Sources of variation in organisms

Key Idea: Mutagens are any chemical or physical agent that causes a change in the DNA sequence

Themutagens).natural

Viruses and microorganisms

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(b)(a)headings:Radiation:Chemicalagents:

Some viruses integrate into the human chromosome, upsetting genes and triggering cancers. Examples include hepatitis B virus (liver cancer), Hodgkin’s disease, and HPV (right) which is implicated in cervical cancer. Aflatoxins produced by the fungus Aspergillus flavus are potent inducers of liver cancer.

1. Describe examples of environmental factors that induce mutations under the following

Diet, alcohol and tobacco

Diets high in fat, especially those containing burned or preserved meat are important causes of colon cancer. High alcohol and tobacco tar intake are known causes of cancer. Tobacco tars contain at least 17 known carcinogens (cancer inducing

Poisons and irritants

Ionising radiation is associated with the development of cancers, e.g. thyroid cancers, and skin cancer. It includes alpha, beta and gamma radiation from nuclear decay. UV radiation from the sun and tanning lamps. X-rays and gamma rays from medical diagnosis and treatment.

3. Explain why people such as radiographers, or even dentists have at a higher risk of exposure to cancer causing radiation:

2. How do mutagens cause mutations?

Many chemicals are mutagenic. Synthetic and natural examples include organic solvents such as benzene, asbestos, formaldehyde, tobacco tar, vinyl chlorides, coal tars, some dyes, and nitrites. Photo right: Firefighters and those involved in environmental clean-up of toxic spills are at high risk of exposure to mutagens.

rate at which a gene will undergo a change is normally very low, but this rate can be increased by chemical or physical agents called mutagens, which alter the DNA sequence. Mutagens and their effects and outlined below:

Ionising radiation

120 Lys Thr SeracidsAminomRNADNAMutant Missense substitution: Substitute C instead of G Leu Val STOPacidsAminomRNADNAMutant Nonsense substitution: Substitute A instead of T Lys Thr SeracidsAminomRNADNAMutant Missense substitution Substitute C instead of Leu Val STOPacidsAminomRNADNAMutant Nonsense substitution: Substitute A instead of T

S S S S Ser Pro Arg Glu Cys Thr Cys Tyr Tyr His Thr His Cys Gly Gly Gly GlnAlaAla Lys Val Leu Val Leu Leu Cys Leu Asn Phe PhePhe COOHH2N311

f In a substitution mutation a single base is substituted with another. Some substitutions may still code for the same amino acid (silent mutation) because of code degeneracy, but they may also result in a codon that codes for a different amino acid.

f Point mutations can occur by substitution, insertion, or deletion of bases and alter the mRNA transcribed.

The DNA sequence below is the coding strand sequence for a hypothetical pigment that affects skin tone (right). It will be used as a reference sequence against which to compare the effect of various point mutations.

Abnormal pigment is not The

f Mutations that cause a change in the amino acid sequence are usually harmful because they alter protein function.

f In the example below right, the change from TGT to TGA at the 19th triplet produces a triplet that translates as a stop codon. When the mRNA is translated, translation ends prematurely, leaving a shortened protein. This is a nonsense substitution.

active.

Key Idea: Gene mutations are localised changes to the DNA base sequence. In most cases this will cause a change in phenotype.

Normal sequence

f A point mutation may not alter the amino acid sequence because the degeneracy of the genetic code means that more than one codon can code for the same amino acid. Such mutations are often called silent because the change is not recorded in the amino acid sequence.

f In the example below left, a substitution mutation has altered the 31st triplet TGT to TCT. This results in the amino acid serine appearing where cystine should be (called a missense substitution). This could affect this protein’s function.

GCT TGT TTA TGC TCT GGC CAT CTT CAC CAA AAT GTT TTT GCT CTG TAT CTG GTT TGT GGT GAA CGT GGG TTT TTT TAT ACT CCC AAA ACT TGT 1 31 Skin tone with normal active pigment Ser Pro Arg Glu Cys Thr Ser Tyr Tyr His Thr His S S Cys Gly Gly Gly GlnAlaAla Lys Val Leu Val Leu Leu Cys Leu Asn Phe PhePhe H2N COOH 31 Abnormal pigment is not active. The person has light skin. Missense substitution

Mutation, Genotype, and Phenotype

Substitution mutation

f Gene mutations are small, localised changes in the DNA base sequence caused by a mutagen or an error during DNA replication. The changes may involve a single nucleotide (a point mutation) or a triplet.

person has light skin. COOH SerCys Tyr HisHis Gly GlnAlaAla Val Leu Val Leu Leu Cys Leu Asn Phe H2NNonsense19 substitution LINK 121 WEB 120 LINK 123 LINK 122 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Insertion mutation

acidsAminomRNADNAMutant

Tyr Thr Ala Gln Asn Leu Insertion of G

Silent mutations do not change the amino acid sequence nor the final protein because several codons may code for the same amino acid. In the example right, a substitution on the 19th triplet alters the TGT to TGC. The triplet still codes for cystine and so the pigment is still functional. Silent mutations are not always neutral as they may affect mRNA stability and transcription, even though they do not change codon information.

a new sequence of amino acids. The protein is unlikely to have any biological activity.

Reading frame shift results in a new sequence of amino acids. The protein is unlikely to have any biological activity.

Thr Ala Gln Asn Leu Insertion of G

Abnormal pigment is not active. The person has light skin.

Tyr

An insertion mutation involves the addition (insertion) of an additional base into the DNA sequence. The insertion of a single extra base displaces the bases after the insertion by one position (called a reading frame shift). In the example right, a G is inserted after the DNA triplet at position 27. This results in altered amino acids and may cause loss of function.

Deletion mutation

Deletion of G Tyr Leu Val Val AsnPhe S S Ser Ala Arg Glu Cys Thr Tyr Tyr His Asn Leu His Cys Gly Gly Gly GlnAlaAla Gln Val Leu Val Leu Leu Cys Leu Asn Phe PhePhe SCOOHH2NS S S Ser Pro Arg Glu Cys Thr Cys Tyr Tyr His Thr His Cys Gly Gly Gly GlnAlaAla Lys Val Leu Val Leu Leu Cys Leu Asn Phe PhePhe COOHH2N Ser ? ? Asn Cys ? ? Tyr ? His ? His Val ? Val Gly GlnAlaAla ? Val Leu Phe Leu Leu Cys Leu Asn Phe ? ? COOHH2N 27 17 22 3131Skin tone with normal active pigment

Abnormal pigment is not active. The person

acidsAminomRNADNAMutant

Leu Val Cys

acidsAminomRNADNAMutant

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Silent mutations

Reading frame shift results in a new sequence of amino acids. The protein is unlikely to have any biological activity.

Reading frame shift results in a new sequence of amino acids. unlikely to have any biological activity.

The deletion of an nucleotide is called a deletion mutation. It also causes a reading frame shift. In this example the nucleotide G is deleted after the 17th DNA triplet. This leads to a reading frame shift that affects the rest of the protein. This could lead to nonsense.

Reading frame shift in

Deletion of G Tyr Leu Val Val AsnPhe

acidsAminomRNADNAMutant

results

The protein is

has light skin.

Silent mutation: Substitute C instead of T

5. Why is a frame shift near the star t codon likely to have a greater impact than one near the stop codon?

3. Explain how a substitution mutation can be effectively silent:

Mutant DNA (coding strand): GTG GGT TTT TTT ATA CTC CCA AAA CTT GT?

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Mutant mRNA:

4. What is a reading frame shift?

Amino acids:

(b) Decide how affected the protein would be by this change in amino acids and justify your decision:

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7. Discuss how point mutations can affect the phenotype of an organism. Use the example of the hypothetical skin pigment to illustrate your points:

6. (a) Use the mRNA table on page 154 to fill in the missing amino acids from position 22 to 31 on the deletion mutation protein on the preceding page. The mutant DNA sequence in given below to help you:

2. Why are nonsense substitutions likely to be more damaging than missense substitutions?

1. Why are mutations that alter the amino acid sequence usually harmful?

Effect of a Substitution Mutation on Phenotype

qp geneHBB

Template DNA

Key Idea: The change of a single DNA nucleotide can change the amino acid sequence of the resulting protein, causing large scale phenotypic effects.

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f Point mutations are mutations where only one nucleotide in a DNA strand is affected. Point mutations may result from deletion of a nucleotide, insertion of an additional nucleotide, or substitution of one nucleotide for another.

The sickle cell mutation has adenine substituted at this point instead of thymine

Chromosome 11.

3. (a) Name the amino acid produced by sickle cell gene mutation: (b) Explain how this amino acid substitution results in the production of a faulty b-chain subunit:

Normal red blood cells

1. Identify the type of mutation that causes sickle cell disease:

The haemoglobin beta gene (HBB gene), is located on chromosome 11. It encodes for the b-chain subunit of the haemoglobin molecule.

f Sometimes the point mutation has no phenotypic effect (e.g. it does not result in a change in the amino acid). At other times, the mutation can disrupt the biological function of the encoded protein (e.g. the mutation changes an amino acid in a key position).

f Sickle cell disease (below) is an inherited blood disorder affecting red blood cells (RBCs). It occurs as a result of a point mutation, and produces RBCs with a deformed sickle cell appearance and a decreased ability to carry oxygen. Many aspects of metabolism are also affected.

A faulty b-chain subunit is produced because the amino acid substitution causes the b-chain to fold and behave differently.

2. Write the mRNA sequence for par t of the mutant DNA strand shown above:

Sickle cells

Red blood cells containing normal haemoglobin have a flattened disc shape.

The mutant protein, haemoglobin S is produced. Haemoglobin S changes the shape of RBCs. When the cells are exposed to low oxygen levels the RBCs become deformed into a sickle shape (right).

The mutation causes the amino acid valine to be produced instead of glutamic acid.

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The DF508 mutation of the CFTR gene on chromosome 7 is a deletion of three bases spanning the 507/508th triplets. The net effect is the loss of a single amino acid from the gene’s CFTR protein which normally regulates chloride transport in cell membranes. The mutant protein fails to do this. The template strand of the DNA containing the mutation is shown below:

strandtemplateDNA

Chloride ions build up inside the cell.

Mucus build up

Water

Key Idea: Cystic fibrosis results from a triplet deletion from the CFTR gene, causing a non-functioning protein that is unable to regulate chloride ion transport.

The DF508 mutation causes the CFTR protein to degrade rapidly, stopping it from inserting into the plasma effectively removed from the cell. protein

f The CFTR gene's protein product, the cystic fibrosis transmembrane conductance regulator, is a membrane-based protein that regulates chloride transport in cells. Over 1900 mutations of the CFTR gene have been reported, causing disease symptoms of varying severity. The D(delta)F508 mutation is particularly common and accounts for more than 70% of all defective CFTR genes. This mutation leads to an abnormal CFTR, which cannot take its proper position in the membrane (below) nor perform its transport function.

CFTR protein

The CFTR gene on chromosome 7

f Cystic fibrosis (CF) is an inherited disorder caused by a mutation of the CFTR gene. It is one of the most common lethal autosomal recessive conditions affecting people of European descent (4% of people are carriers).

This triplet codes for the 500th amino acid

The DF508 mutant form of CFTR fails to take up its position in the membrane. Its absence results in defective chloride transport and leads to the cell absorbing more water. This causes mucus-secreting glands, particularly in the lungs and pancreas, to become fibrous and produce abnormally thick mucus. CFTR is widespread throughout the body explaining why CF is a multisystem condition affecting many organs.

Chloridemembrane.ions

CFTR

Correctly

Base 1630

Normal CFTR (1480 amino acids) controls chloride ion balance in the cell

Abnormal CFTR (1479 amino acids) or little control of chloride ion balance in the cell

These three bases are deleted from the 507/508th triplets in the ΔF508 mutation CFTRgene

A Case Study: Cystic Fibrosis122 LINK 107 LINK 120 LINK 111 Water

CF patient receiving physiotherapy treatment

Cl Cl CellCellexteriorinterior

(f) Use page 154 to work out the amino acid sequence of the mutated DNA:

(d) What was the net effect of the three base deletion?

(b) Use the mRNA table on page 154 to identify the amino acids encoded by the mRNA for the mutant DNA strand:

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(a) Circle the mutation in the mutated DNA sequence:

(b) What general type of mutation is this?

(g) Explain why the mutated DNA leads to a non-functional CFTR protein:

1. (a) Write the mRNA sequence for the normal template DNA strand shown in the diagram on the opposite page:

(e) Use page 154 to work out the amino acid sequence of the nor mal DNA:

Mutated DNA: ACT TTG CAA CAG TAG AGG AAA GCC TTT GGA

(b) What effect does this have on water movement in and out of the cell?

Normal DNA: ACT TTG CAA CAG TGG AGG AAA GCC TTT GGA

(h) Recall your answer to (b). You can now give a more specific answer to the question: What type of mutation is this? Give reasons why this kind of mutation can produce non-functional proteins:

4. There are over 1900 different CFTR mutations described so far. One mutation is the W1282X mutation. This occurs at base pair 3846. The sequences below show the correct DNA sequence for the coding strand (upper line) and the mutated DNA sequence (lower line).

(b) Use the mRNA table on page 154 to identify the amino acids encoded by the mRNA for the normal DNA strand:

3. (a) Explain why the abnormal CFTR fails to transport Cl correctly:

2. (a) Rewrite the mRNA sequence for the mutated DNA strand:

3. Explain why a large number of mutations (generally) need to occur before cells become cancerous:

f Mutations can occur due to mutagens in the environment, e.g. cigarette smoke or exposure to ultraviolet light. The effect of these mutagens tends to build up over a life time so that the chance of mutation increases with age. This can be seen in the occurrence of cancers, which are more commonly seen in adults than in young people.

f It has been estimated that there might be 1012 mutations in the cells of a tumour, occurring both before and after the formation of the tumour.

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UV Light Thymine dimer of suppressortumourgene NCI

123

f A large number of repair mechanisms constantly repair mutations so many genes may need to be affected by mutations before cancer occurs.

Melanoma

f Skin cancer is commonly caused by exposure to ultraviolet light. It is an example of a non-heritable mutation, as skin cells are not involved in the production of reproductive cells (gametes).

Key Idea: Skin cancer is caused by mutations in somatic cells. These mutations are not inherited because skin cells are not involved in the production of gametes.

KNOW 180

2. Explain why skin cancer is less likely to be seen in young people than adults:

1. Discuss the difference between a non-inherited disease, such as skin cancer, and an inherited disease, such as cystic fibrosis:

f After exposure to UV light (a potent mutagen), adjacent thymine bases in DNA become cross-linked to form a 'thymine dimer'. This disrupts the normal base pairing and affects gene function. In some cases, mutations trigger the onset of cancer by disrupting the genes regulating the cell cycle.

DNA

Tests have shown that the use of sunscreen significantly reduces the production of thymine dimers. However tests also show that in unprotected skin 50% of dimers were removed after 24 hours and 75% were removed after 72 hours. A single incidence of sunburn generally won't lead to cancer because the damage is repaired but repeated cases increase the risk.

Skin Cancer: A Non-Inherited Mutation

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f The sex of some animals is determined by the incubation temperature during their embryonic development. Examples include turtles, crocodiles, and the American alligator. In some species, high incubation temperatures produce males and low temperatures produce females. In other species, the opposite is true. Temperature regulated sex determination may provide an advantage by preventing inbreeding (since all siblings will tend to be of the same sex).

Female Male

f Environmental factors can modify the phenotype encoded by genes without changing the genotype. This can occur both during development and later in life. Environmental factors that affect the phenotype of plants and animals include nutrients or diet, temperature, altitude or latitude, and the presence of other organisms.

f Colour-pointing is a result of a temperature sensitive mutation to one of the melanin-producing enzymes (melanin is the pigment responsible for darker skin and hair). The mutated enzyme is thermally unstable and fails to work at normal body temperatures. However it is active at the lower temperatures found in the extremities of the body (paws, face, tail), so these areas are darker than the surrounding pale coloured hair (pointed). Colour-pointing is seen in some breeds of cats and rabbits (e.g. Siamese cats and Himalayan rabbits).

The effect of temperature

124

f The presence of other individuals of the same species may control sex determination for some animals. Some fish species, including Sandager's wrasse (right), show this characteristic. The fish live in groups consisting of a single male with attendant females and juveniles. In the presence of a male, all juvenile fish of this species grow into females. When the male dies, the dominant female will undergo physiological changes to become a male. The male and female look very different.

2. How is helmet development in Daphnia an adaptive response to environment?

The effect of other organisms

EnvironmentProhibited and Phenotype

1. (a) Give two examples of how temperature affects a phenotypic characteristic in an organism:

f Some organisms respond to the presence of other, potentially harmful, organisms by changing their body shape. Invertebrates, such as some Daphnia species, grow a helmet when invertebrate predators are present. The helmet makes Daphnia more difficult to attack and handle. Such changes are usually in response to chemicals produced by the predator (or competitor) and are common in plants as well as animals.

Helmeted Daphnia

Key Idea: An organism's phenotype is influenced by the environment in which it develops, even though the genotype remains unaffected.

(b) Why are the darker patches of fur in colour-pointed cats and rabbits found only on the face, paws, and tail?

Non-helmeted Daphnia

signalChemical

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(b) What physical factors associated with altitude could affect plant phenotype?

chemical environment

Cline

Increasing altitude can stunt the phenotype of plants with the same genotype. In some conifers, e.g. Engelmann spruce, plants at low altitude grow to their full genetic potential, but growth becomes progressively more stunted as elevation increases. Growth is gnarled and bushy at the highest, most severe sites. Gradual change in phenotype over an environmental gradient is called a cline.

4. Describe an example of how the chemical environment of a plant can influence phenotype:

Plant species A:

3. (a) What is a cline?

effect of altitude The

5. Vegetable growers can produce enormous vegetables for competition. How could you improve the chance that a vegetable would reach its maximum genetic potential?

Plant species B:

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Severe stunting

The effect of

6. Two different species of plant (A and B) were found growing together on a windswept portion of a coast, Both have a low growing (prostrate) phenotype. One of each plant type was transferred to a greenhouse where "ideal" conditions were provided to allow maximum growth. In this controlled environment, species B continued to grow in its original prostrate form, but species A changed its growing pattern and became erect in form. Identify the cause of the prostrate phenotype in each of the coastal grown plant species and explain your answer:

The chemical environment can influence the phenotype in plants and animals. The colour of hydrangea flowers varies with soil pH. Blue flowers (due to the presence of aluminium compounds in the flowers) occur in more acidic soils (pH 5.0-5.5) in which aluminium is more readily available. In less acidic soils (pH 6.0-6.5) the flowers are pink.

Half-siblings 31 Cousins 15

Identical twins raised apart 69

Identical twins raised together 88

8. Some cattle breeders on the east coast of New Zealand decided to produce a cattle breed suited to the dry east coast conditions by selecting the fastest growing cattle from each generation for breeding. A buyer from the west coast was impressed by the speed at which the cattle grew and the size they obtained and decided to try raising a herd on a west coast farm. However the buyer found subsequent generations of the cattle grew slowly and were quite small. Explain why there was such a difference in the growth and size of the cattle on the east and west coasts.

9. To examine the role of environment on phenotype, scientists try to use organisms that are genetically identical. These might include identical siblings (e.g. identical twins) or using plant cuttings.

Parent-child 42

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Group % similarity in IQ

Use this information to discuss the effect of genes and environment on IQ:

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7. With the ability of scientists now to clone animals there is often talk of people cloning their much loved pet so that they can have the same animal again after the original pet has died. Explain why it is highly unlikely that the cloned animal with be exactly like the original:

(b) Various studies on IQs have been carried out to see if cognitive ability is related to environment or genetics (the heritability of intelligence). When comparing IQs of different groups the studies have found the following:

(a) Why would scientists use genetically identical organisms in gene-environment studies?

Fraternal twins 60

Mice and environmental effects

The effect of maternal environment and diet of on later generations exposed to a breast cancer trigger (a carcinogenic chemical) was investigated in rats fed a high fat diet or a diet high in oestrogen. The length of time taken for breast cancer to develop in later generations after the trigger for breast cancer was given was recorded and compared. The data are presented below. F1 = daughters, F2 = granddaughters, F3 = great granddaughters.

f Anything the mother (F0) is exposed to will affect her and will also expose the fetus (F1). In a female fetus, egg cells (F2) develop in the ovaries, so a third generation will be exposed also. If a fourth generation (F3) proves to be affected, then there is evidence that the environmental effect is inherited.

f Studies of heredity have found that the environment or lifestyle of an ancestor can have an effect on future generations. Certain environments or diets can affect the packaging of the DNA (rather than the DNA itself), determining which genes are switched on or off and so affecting the development of the individual. These effects can sometimes be passed on to future generations. It is thought that these inherited effects may provide a rapid way to adapt to particular environmental situations.

Key Idea: The environment or experiences of an individual can affect the development of subsequent generations.

Cumulative percentage rats with breast cancer (high fat diet (HFD)) F1% F2% F3% Weekstriggersince HFD Control HFD Control HFD Control 6 0 0 5 0 3 0 8 15 0 20 5 3 20 10 22 8 30 5 10 25 12 22 18 50 20 20 30 14 22 18 50 30 25 40 16 29 18 60 30 25 40 18 29 18 60 40 40 42 20 40 18 65 40 50 60 22 80 60 79 50 50 60 Cumulative percentage rats with breast cancer (high oestrogen diet (HOD)) F1% F2% F3% Weekstriggersince HOD Control HOD Control HOD Control 6 5 0 10 0 0 0 8 10 0 10 0 15 10 10 30 15 15 20 30 20 12 38 19 30 30 40 20 14 50 22 30 40 50 20 16 50 22 30 40 50 30 18 60 35 40 40 75 40 20 60 42 50 50 80 45 22 80 55 50 50 80 60 Data source: Science News April 6, (Mother2013F0)Fetus(F1)Eggcells(F2) LINK 118 WEB 125 LINK 124 PREVIEWONLYNotforClassroomUseNotforClassroomUse

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Genes and Environment Interact

f The destruction of New York's Twin Towers on September 11, 2001, traumatised thousands of people, including 1700 pregnant women. Some of the pregnant women developed post traumatic stress disorder (PTSD). The children of PTSD mothers had much lower levels of cortisol (a stress hormone) than those whose mothers did not suffer PTSD, indicating that the maternal environment had affected the offspring. The children of women in the last three months of pregnancy at the time had the lowest cortisol levels of all. Low levels of stress hormones may seem counter-intuitive for traumatised people, but it is related to fatigue of the stress response systems.

125

100806040200 6810121416182224 cancerbreastwithrats%

High

(a) Which generations are affected by the original

Weeks after exposure to cancer triggerafter exposure to cancer trigger

Weeks after exposure to cancer trigger

100806040200 6810121416182224 cancerbreastwithrats%

are affected by the mother eating a high

diet?

100806040200 6810121416182224 cancerbreastwithrats%

% F1 with breast cancer (HFD)

Weeks after exposure to cancer trigger

100806040200 6810121416182224 cancerbreastwithrats%

% F1 with breast cancer (HFD)

diet and generational effects? See Appendix PREVIEWONLYNotforClassroomUseNotforClassroomUse

100806040200 6810121416182224 cancerbreastwithrats%

Weeks after exposure to cancer trigger

100806040200 6810121416182224 cancerbreastwithrats%

experiments show with

fat diet Control

% F1 with breast cancer (HFD)

100806040200 6810121416182224 cancerbreastwithrats% Weeks

diet had the longest lasting effect?

3. What do these respect to

High

Weeks after exposure to cancer trigger

100806040200 6810121416182224 cancerbreastwithrats%

Weeks after exposure to cancer trigger

Use the data on the previous page to complete the graphs below. The first graph is done for you:

(b) Which generations oestrogen

2. mother

Weeks after exposure to cancer trigger

Control

High fat diet Control

Weeks after exposure to cancer trigger

100806040200 6810121416182224 cancerbreastwithrats%

fat diet Control

eating a high fat diet?

High fat diet

What You Know So Far: Genotype and Phenotype

HINT: Examples of how the phenotype is affected by changes in the environment.

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HINT: Include definitions, different types of mutation, and their potential effects on phenotype and on metabolic pathways.

The effect of environment on phenotype

Mutagens and mutations

REVISE 186

NCEA Style Question: Genotype and Phenotype

(c) Discuss the effect of point mutations on genes. Use a human example to explain the genotype and phenotype changes that can occur as a result of a base substitution (use extra paper if you need to).

(a) Define the term mutagen. Give an example of a mutagen, its origin, and its effect:

1. The genotype and phenotype of an organism can be affected by a variety of environmental factors.

(b) Explain how the environment can affect an organism's phenotype without affecting the genotype. Use two examples to help you illustrate your explanation (use extra paper if you need to).

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(a) Explain what is meant by a metabolic pathway:

(b) Explain how phenylketonuria occurs and why people with the disorder must make modifications to their diet:

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2. Phenylketonuria is a metabolic disorder in which the metabolic pathway that converts the amino acid phenylalanine to tyrosine is disrupted.

G A change to the DNA sequence of an organism. This may be a deletion, insertion, duplication, inversion or translocation of DNA in a gene or chromosome.

D The genetic makeup of an organism or cell.

A An agent capable of causing a change in the DNA sequence.

For each of the following DNA mutations, state the type of mutation that has occurred:

KEYProhibitedTERMS

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

F Observable characteristics in an organism.

AND IDEAS: Genotype and Phenotype

(a) GCG TGT TTG TAG GCG CTC TG

(b) GCG TGA TTT GTA AGG CGC TCT G

(d) GCG TGA GTA GGC GCT CTG

I A series of enzyme controlled chemical reactions that modify an initial chemical or chemicals into an end product.

Compare the effect of environmental factors in bee larval development with laboratory induced effects. What does this tell us about the effect of environment on genes and phenotype?

C The replacement of one nucleotide base in DNA with another.

2. An original DNA sequence is shown right: GCG TGA TTT GTA GGC GCT CTG

3. In the following diagram label: enzyme 1, enzyme 2, end product, starting substrate, intermediate, gene 1, gene 2

128

4. All worker bees and the queen bee in a hive have the same genome, yet the queen looks and behaves very differently from the workers. Only bee larvae fed a substance called royal jelly will develop into queens. Research shows that royal jelly contains factors that silence the activation of a gene called Dnmt3, which itself silences many other genes. Studies on bee development focussed on the Dnmt3 gene. One study switched off the Dnmt3 gene in 100 bee larvae. All the larvae developed into queens. Leaving the gene switched on in larvae causes them to develop into workers.

substitutionphenotypemutationmutagenmetabolicgenotypeenzymeenvironment(mutation)pathway(mutation)

B The removal of a nucleotide base or bases from DNA.

E A catalytic agent (usually a globular protein) in metabolic reactions.

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H The internal or exter nal conditions, which may influence expression of genotype.

deletion

Hints for Drawing Graphs concentration

(g 012345m-3) 0.7 Sensitive strain Resistant strain

if verticalresponse,variablePlacenecessary.thedependente.g.biologicalonthe(Y)axis.

0.10.20.30.40.50.60nm)550at(absorbanceeldYi Antibiotic

Each axis should have an appropriate scale. Decide on the scale by finding the maximum and minimum values for each variable.

Fig. 1: Yield of two bacterial strains different 95% confidence intervals, n = 6)

Error bars may be included to show a measure of spread (e.g. standard deviation). Error bars give an indication of the reliability of the mean value. If they do not overlap between points, then these means will be significantly different.

Use a pie graph rateMetabolic Temperature (°C) 12840 010203040 Line pointsconnecting Temperature vs metabolic rate Water use key IrrigationDrinkingCommercialCooling23% 17%27% 33% broodineggsofNumber Body length (mm) 0 1 2 3 4 806040200 Line of best fit Body length vs brood size RaglanManakauKaiparaHokiangaWhangapeHerekino 0 200040006000 Mangrove area in NZ harbours AreaWeight(ha)(g)Frequency Use a scatter plot Use a histogram Use a bar or column graph Use a line graph f One variable is a category f One variable is a count to give percentage of a total f One variable is a category f One variable is continuous data (measurements) f Bars do not touch f One variable is continuous data (measurements) f One variable is a count f Both variables are continuous f The response variable is dependent on the independent (manipulated) variable f A line is drawn point to point f Both variables are continuous f The two variables are interdependent but there is no manipulated variable f A line of best fit is drawn through the points What type of data have collected?you Appendix 1 PREVIEWONLYNotforClassroomUseNotforClassroomUse

Overlap between these means the values not significantly different.both axes and provide appropriate units of measurement

Place independentthe variable e.g. treatment, on the horizontal (X) axis

Graphs should have a explanatoryconcise,title.

Label

antibiotic levels (±

Photo credits

Interpret: Comment upon, give examples, describe relationships. Describe, then evaluate.

Measure: Find a value for a quantity

State: Give a specific name, value, or other answer. No supporting argument or calculation is necessary.

Predict: Give an expected result.

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Questioning terms in biology

List: Give a sequence of answers with no elaboration.

Estimate: Find an approximate value for an unknown quantity, based on the information provided and application of scientific knowledge.

Contrast: Show differences. Set in opposition.

Distinguish: Give the difference(s) between two or more items.

Evaluate: Assess the implications and limitations.

Illustrate: Give concrete examples. Explain clearly by using comparisons or examples.

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: • Kristian Peters • Capkuckokos • Mnoff • Itayba • Jpbarrass • Matthias Zepper • Ewans love • OpenStax college • Tony Willis

Royalty free images, purchased by Biozone International Ltd, are used throughout this workbook and have been obtained from the following sources: 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.

Calculate: Find an answer using mathematical methods. Show the working unless instructed not to.

Analyse: Interpret data to reach stated conclusions.

Derive: Manipulate a mathematical equation to give a new equation or result.

Outline: Give a brief account or summar y. Include essential information only.

Compare: Give an account of similarities between two or more items, referring to both (or all) of them throughout.

Suggest: Propose a hypothesis or other possible explanation.

Define: Give the precise meaning of a word or phrase as concisely as possible.

Describe: Define, name, draw annotated diagrams, give characteristics of, or an account of.

Explain: Provide a reason as to how or why something occurs.

Appendix 2

Solve: Obtain an answer using numerical methods.

Apply: Use an idea, equation, principle, theor y, or law in a new situation.

Summarise: Give a brief, condensed account. Include conclusions and avoid unnecessary details.

Design: Produce a plan, object, simulation or model.

Determine: Find the only possible answer.

Contributors identified by coded credits are: BF: Brian Finerran (Uni. of Canterbury), CDC: Centers for Disease Control and Prevention, Atlanta, USA, DoC: Department of Conservation (NZ), EII: Education Interactive Imaging, USAF:United States Air Force USDA: United States Department of Agriculture, NCI: National Cancer Institute.

The following terms are often used when asking questions in examinations and assessments.

Annotate: Add brief notes to a diagram, drawing or graph.

Construct: Represent or develop in graphical form.

Draw: Represent by means of pencil lines. Add labels unless told not to do so.

Identify: Find an answer from a number of possibilities.

We acknowledge the generosity of those who have provided photographs for this edition: • Dartmouth College Electronic Microscope Facility • Wintec • Professor Jeff Podos, Biology Department, University of Massachusetts Amherst for his photographs of Galápagos finches • Wadsworth Centre (NYSDH) for the photo of the cell undergoing cytokinesis • Marc King for photographs of comb types in poultry • ms.donna for the photo of the albino child.

Discuss: Show understanding by linking ideas. Where necessary, justify, relate, evaluate, compare and contrast, or analyse.

Proteins, membrane 16

DNA, replication of 67-68

Skin cancer mutation 180

Genetic variation, sources of 84-85

Cell division 11, 69-74, 91

Discontinuous variation 87

Diploid

DNA model, exercise 148-151

rRNA 147

Grana Guanine53145

Stroma lamellae 53

Anabolic reaction 35, 38

Genotype 84 - effect of mutation on 174

Centrioles 6, 73

Cell, plant 3-5

C Calvin cycle 54, 56

Photosynthesis 11, 35, 43, 51, 54-61 Photosynthetic rate, measuring 59 Photosystem 56

Ion pumps 26

Pyrimidines145 145

Inhibitor, enzyme 39

Plastid Polypeptide4 chain 160

Catalyst, enzyme 36-41

Pinocytosis 29

Plasma membrane 3, 6, 16-17

T Tasmanian devil, bottleneck 134

Cell, animal 6-8

Cellular organelles 3-4, 13-15

tRNA 147, 157-158

Amyloplast 3-4

Organelles, cellular 3-4, 6, 13-15 Osmosis 20-21 Osmotic potential 20 Oxidative phosphor ylation 46

PKU 171

Qualitative trait 87

Incomplete dominance, of alleles 105

Population bottlenecks 133-134

Cell size, and diffusion 22

Cell wall 3

Animal cell 6-8

Index

Autolysis 11

Cytokinesis 69, 73-74

Fermentation pathways 49-50

Induced fit model, enzyme activity 38

DNA 143-144, 146

Primary transcript 152

Silent mutation 88, 175

Cofactor, enzyme 39

Intracellular enzyme 36

- codominance of 102-103

Plant cell 3-5

Energy, in cells 43

Enzymes Epigenetics36-41172, 184

Evolution, defined 123

Alcoholic fermentation 49

Natural selection 121, 123-130 - model 126 - rock pocket mice 129-130 Nuclear pore 6 Nucleic acids 146-147 Nucleolus Nucleosome3 144

Recombination 92, 112-113

Nucleotide, base pairing rule 68, 148 Nucleotides 145 Nucleus 3, 6, 12

B Base pairing rule 68, 148

Directional91 selection 125, 129-130

S S phase Secondary69structure, of proteins 161

Membrane171permeability, factors affecting Membrane24proteins 16

Protein, structure of 161

Founder effect 131-132

A Activation energy, of enzymes 37

Passive transport 18-22, 30 Pea plant experiments 97-98 Peptide bond 160 Phagocytosis 29 Phenotype 84, 172, 174, 181-182 - role of metabolism 170-172 Phenylalanine metabolism 170-171 Phenylketonuria 171 Phospholipid structure 16

Protein, types 162-163

Secretion, in cell 11

Gametic mutation 90

Gap phase 69

Cytosine 145

Enzyme denaturation 36

Deletion mutation 174-175

Gene mutation 177-178

Diffusion 18, 22

Globular protein 163

Sun plant 52

Interphase 69, 73

WXYZ

Proteins, enzymes 36

Active transport 25-30

Eukaryotic cell 3, 6

Beta pleated sheet 161-162

Gene pools, role in evolution 121

Linkage 110-111

Factors affecting gene pools 121

Gene flow 121

Sickle cell mutation 88, 176

Catalase activity 41

Environment-andphenotype 172, 181-182, 184 - and variation 84

Mitosis 69-75 Mitosis, model of 75 Model, DNA 148-151 Model, gene pool 127-128 Model, meiosis 94-95 Model, mitosis 75 Model, natural selection 126 Monohybrid cross 99-107 mRNA mRNA-amino147 acid table 154 Multiple alleles 103-104 Mutagen 88, 173, 180 Mutation 88-90 - and variation 84 - cystic fibrosis 178 - effect on genotype 174 - effect on phenotype 174 - in gene pools 121 - sickle cell disease 176 - types 174-176

Cell cycle, stages of 69

Lock and key model, enzymes 38 Lysosome 6

Facilitated diffusion 18

Chromatin

Dominant allele 97

Mate selection 121 Meiosis 91-93 Meiosis, and variation 92-93 Meiosis, exercise 94-95 Melanin

Tonicity Transport,TranslationTranscriptionTraitTonoplast20-213,2486,97152-153152,157-158cellular11,17-22,

RNA polymerase 153

25-30

Turgor pressure 21

Aerobic respiration 46-47

H Haploid 91

Endergonic reaction 35

Respirometer 48

Surface area: volume ratio 22

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Fibrous protein 163

Exergonic reaction 35

Cellular energy, overview 43

Catabolic reaction 35, 38

Crossing over 91-92

D Degeneracy, of genetic code 155

Electron transport chain 44-46

- role in active transpor t 30

Processes, cellular 11-12

Chromosome,Chromosome144144homologous 83

RNA 146-147

Hypertonic vs hypotonic 20-21

Cytosis 11, 28-29

Insertion mutation 174-175

Gene pools, changes in 121

Specialisation, in cells 7, 9-10 Stabilising selection 125 Stomata 51

Cell specialisation 7, 9-10

Asexual reproduction 85 ATP 42-45

Ribose sugar 145

Purine

UV Uracil Variation,Vacuole1453genetic 84-85

Chargaff's rules 148

Zygote 71

Disruptive selection 125

Isotonic 20-21

Substrate level phosphorylation 46

ATP synthase 27

Test cross 100

Factors affecting photosynthesis 57-58

Leaf structure, role in photosynthesis Lethal51-52 alleles 106

DNA, changes 88

Replication, of DNA 67-68

Intron 144

Cotransport 27

Concentration gradient 18

Light dependent phase 54 Light independent phase 54 Link reaction 44, 46

Coding strand, of DNA 153

Sub-unit protein 162

Substitution mutation 174, 176

Homologous chromosome 83, 91

Inheritance, laws of 96

Hydrophobic, defined 16

Beneficial mutation 88

Homozygous 83

Protein denaturation 161-162

Temperature, and membrane permeability 24

Genetic drift 121,135

Genetic bottleneck 133-134

Essential amino acid 160

Reaction rates, of enzymes 40

Linked gene, inheritance of 110-111 Liver cell 6

Cell cycle, checkpoints 70

Enzyme cofactor 39

Germline mutation 90

Independent assortment 93

Chromatid 144

JK Krebs cycle 44-46

Ribonucleic acid (RNA) 146-147

Chloroplast 3-4, 11, 53

Anaerobic metabolism 49-50

Ribosomes 3, 6, 157

Recessive allele 89, 97

Exocytosis 28

Golgi apparatus 3, 6, 12

Continuous variation 86

Enzyme, catalase 41

Dihybrid cross 108-109, 114

Endoplasmic reticulum 3, 6

Exon 144, Extracellular153enzyme 36

Quantitative trait 86

Sexual reproduction, variation 84-85 Shade plant 52

Gene expression 152-153, 157-158

Membrane transport 17 Membranes, roles in cells 12 Mendel, laws of inheritance 96 Mendel, pea plant experiments 97-98 Metabolic disorder 170-171 Metabolic pathway 168-171 Metabolic reaction 35, 38 Metabolism, defined 11 Metabolism, overview 76-77 Microvilli 6 Migration Mitochondrion121 3, 6, 12, 44-46

Endocytosis 29

Checkpoints, cell cycle 70

Respiration, measuring 48

Biological catalyst 36 Black robin, genetic bottleneck 134 Blood groups, inheritance 103-104

Cell cytoskeleton 3

Cell processes 11-12

Enzyme reaction rate 40

Segregation, law of 96 Selection 124, Semi-conservative129-130DNA replication 67

Cystic fibrosis mutation 178

Deoxyribose sugar 145

Plasmolysis, in plant cells 21

Adenine 145

Active site, of enzymes 36

Amino acid, structure of 160

Gene pool model 127-128

Enzyme 162

Anticodon 147, 158

Heterozygous 83

Alpha helix 161-162

Albinism 171

Water movement 20

Lactic acid fermentation 50

Cellular respiration 35, 43, 48, 60-61

Deoxyribonucleic acid 143-144, 146

Independent assortment, law of 96

- incomplete dominance of 105 - lethal 106

Template strand, of DNA 153

Hydrophilic, defined 16

Harmful mutation 88

Alleles 83

Proton pumps 27

Gas exchange, in photosynthesis 51

Amino acid table 154

Genetic code 154, 156

Cellulose cell wall 3-4

Codominance, of alleles 102-103

Genetic crosses 97-114

Denaturation 161-162 - of enzymes 36

Tertiary structure, of proteins 161

Geographical barrier, effect 121

Cancer 73, 179

Particulate inheritance 96

Enzyme, role in metabolism 168

Condensation reaction 147

Cytoplasm 3, 6

Glycolysis 44-46

Enzyme inhibitor 39

Primary structure, of proteins 161

Redundancy, of genetic code 155

Codons 89, 154, 157

Fluid mosaic model 16

Channel protein 162

Sodium-potassium pump 26 Somatic mutation 90, 180

Thylakoid 53 Thymine 145

Quaternary structure, of proteins 161

Gene 144

Protein synthesis 11, 158-159