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NCEA LEVEL 2 BIOLOGY EXTERNALS Meet the writing team

Tracey

Senior Author

Kent

Tracey Greenwood I have been writing resources for students since 1993. I have a Ph.D in biology, specialising in lake ecology and I have taught both graduate and undergraduate biology. Kent Pryor I have a BSc from Massey University majoring in zoology and ecology and taught secondary school biology and chemistry for 9 years before joining BIOZONE as an author in 2009.

Author

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

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

Cover photograph 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. PHOTO: James T. Reardon

Richard

Founder & CEO

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

ISBN: 978-1-927309-51-3 Copyright Š 2017 Richard Allan Published by BIOZONE International Ltd Printed by Wickcliffe Solutions www.wickliffe.co.nz

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

BIOZONE International Ltd. P.O. Box 5002 Hamilton 3242, New Zealand Telephone: (07) 856 8104 Fax: (07) 856 9243 Email: sales@biozone.co.nz Website: www.BIOZONE.co.nz

www.BIOZONE.co.nz

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


Contents

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

NCEA Style Question: Cell Structure and Transport ............................ 32 KEY TERMS AND IDEAS: Cell Structure and Transport.............................. 34 Reactions in Cells............................................. 35 Enzymes .......................................................... 36 Enzymes and Activation Energy....................... 37 How Enzymes Work.......................................... 38 Cofactors and Inhibitors Affect Enzyme Activity.. 39 Enzyme Reaction Rates .................................. 40 Investigating Catalase Activity ......................... 41

29 30 31 32 33 34 35 36 37 38 39 40

ATP .................................................................. 42 Energy Transformations in Cells ...................... 43 Cellular Respiration: Inputs and Outputs ......... 44 Steps in Cellular Respiration ........................... 46 Measuring Respiration ..................................... 48 Anaerobic Metabolism in Plants and Fungi ..... 49 Anaerobic Metabolism in Animals..................... 50 Leaf Structure and Photosynthesis .................. 51 The Structure of Chloroplasts .......................... 53 Photosynthesis: Inputs and Outputs...............  .. 54 Steps in Photosynthesis .................................. 56 Factors Affecting Photosynthesis ..................... 57

CODES:

The Structure of Plant Cells................................ 3 Identifying Structures in Plant Cells.................... 5 The Structure of Animal Cells............................. 6 Identifying Structures in Animal Cells................. 8 Structure Relates to Function............................. 9 Cell Processes.................................................. 11 Identifying Organelles in Micrographs ............. 13 Cell Structures and Organelles ........................ 15 The Structure of Membranes ........................... 16 Diffusion in Cells............................................... 18 Osmosis in Cells .............................................. 20 Water Relations in Plants.................................. 21 Diffusion and Cell Size ..................................... 22 Temperature and Membrane Permeability ....... 24 Active Transport ............................................... 25 Ion Pumps ........................................................ 26 Cytosis ............................................................. 28 Active and Passive Transport Summary .......... 30 What You Know So Far: Cell Structure and Transport ............................ 31

Activity is marked:

to be done

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

Investigating Photosynthetic Rate .................... 59 Modelling Photosynthesis and Cellular Respiration ....................................................... 60 What You Know So Far: Energy Transformations ................................... 63 NCEA Style Question: Photosynthesis............ .. 64 NCEA Style Question: Cellular Respiration....... 65 KEY TERMS AND IDEAS: Energy Transformations..................................... 66 DNA Replication .............................................. 67 Details of DNA Replication .............................. 68 The Eukaryotic Cell Cycle ................................ 69 Checkpoints in the Cell Cycle .......................... 70 Functions of Mitosis .......................................... 71 Mitosis .............................................................. 72 Mitosis and Cytokinesis ................................... 73 Modelling Mitosis ............................................. 75 Putting it all Together: Metabolism! .................. 76 What You Know So Far: Cell Division ............... 78 NCEA Style Question: Cell Division ................. 79 KEY TERMS AND IDEAS: Cell Division .......... 80

AS  2.5 Genetic Variation and Change

Achievement criteria and explanatory notes ..... 81

59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83

Alleles .............................................................. Sources of Variation ......................................... Examples of Genetic Variation ......................... The Role of Mutations in Populations .............. Mutations Can Produce New Alleles ...............

83 84 86 88 89

Somatic and Gametic Mutations ...................... 90 Meiosis ............................................................. 91 Meiosis and Variation ....................................... 92 Modelling Meiosis ............................................ 94 Mendel's Laws of Inheritance........................... 96 Dominant and Recessive Traits ........................ 97 The Monohybrid Cross .................................... 99 The Test Cross................................................ 100 Practising Monohybrid Crosses...................... 101 Codominance in Alleles ................................. 102 Codominance in Multiple Allele Systems ....... 103 Incomplete Dominance.................................... 105 Lethal Alleles................................................... 106 Problems Involving Monohybrid Inheritance.... 107 Dihybrid Cross................................................. 108 Inheritance of Linked Genes............................ 110 Recombination and Dihybrid Inheritance......... 112 Problems Involving Dihybrid Inheritance.......... 114 What You Know So Far: Sources of Variation.. 115 NCEA Style Question: Sources of Variation..... 116

when completed


Contents 84 85 86 87 88 89 90 91 92 93 94 95

NCEA Style Question: Inheritance of Alleles... 118 KEY TERMS AND IDEAS: Sources of Variation......................................... 120 Processes in Gene Pools................................. 121

How Natural Selection Works.......................... 123 Types of Natural Selection............................... 125 Modelling Natural Selection............................. 126 Gene Pool Model............................................. 127 Natural Selection in Pocket Mice..................... 129 The Founder Effect.......................................... 131 Population Bottlenecks.................................... 133 Genetic Drift..................................................... 135 What You Know So Far: Changes in Gene Pools................................... 137 96 NCEA Style Question: Changes in Gene Pools................................... 138 97 KEY TERMS AND IDEAS: Changes in Gene Pools................................... 140

122 A Case Study: Cystic Fibrosis.......................... 178 123 Skin Cancer: A Non-Inherited Mutation........... 180 124 Environment and Phenotype........................... 181 125 Genes and Environment Interact..................... 184 126 What You Know So Far: Genotype and Phenotype................................ 186 127 NCEA Style Question: Genotype and Phenotype................................ 187 128 KEY TERMS AND IDEAS: Genotype and Phenotype................................ 189 APPENDIX 1: Hints For Drawing Graphs........ 190 APPENDIX 2: Questioning Terms.................... 191 PHOTO CREDITS........................................... 191

INDEX ............................................................ 192

AS 2.7 Gene Expression

Achievement criteria and explanatory notes ... 141

98 The Role of DNA in Cells................................. 143 99 The Structure of Chromosomes....................... 144 100 Nucleotides...................................................... 145 101 DNA and RNA.................................................. 146 102 Constructing a DNA Model.............................. 148 103 Genes Code for Proteins................................. 152 104 Transcription.................................................... 153 105 What is the Genetic Code?.............................. 154 106 Cracking the Genetic Code.............................. 156 107 Translation....................................................... 157 108 Protein Synthesis Summary............................ 159 109 Amino Acids Make Up Proteins....................... 160 110 The Structure of Proteins................................. 161 111 Protein Shape is related to Function................ 162 112 Globular and Fibrous Proteins......................... 163 113 What You Know So Far: Nucleic Acids and Proteins.............................. 164 114 NCEA Style Question: Nucleic Acids and Proteins.............................. 165 115 KEY TERMS AND IDEAS: Nucleic Acids and Proteins.............................. 167 116 Metabolic Pathways ........................................ 168 117 Interrupting Metabolic Pathways...................... 170 118 Influences on Phenotype................................. 172 119 Mutagens......................................................... 173 120 Mutation, Genotype, and Phenotype............... 174 121 Effect of Substitution Mutation on Phenotype.. 177

CODES:

Activity is marked:

to be done

when completed


v

Using This Resource 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 outline of the chapter structure below will help you to navigate through the material in each chapter. Sections within a chapter share the same structure. They correspond to natural topic breaks within the Achievement Standard.

Introduction

Activities

Review

Test

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

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

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

• NCEA style questions conclude clusters of related activities • These enable you to practise your NCEA exam skills

16

9

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

The Structure of Membranes

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. 78 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.

56 What You Know So Far: Cell Division

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

Summarise you know about this topic so far under the Carrier proteins permit thewhat passage headings of specific molecules byprovided. facilitatedYou can draw diagrams or mind maps, or writetransport. short notes diffusion or active Ion to organise your thoughts in preparation for style essay question that follows. Use the points pumps are a the typeNCEA of carrier protein.

CO2

in the introduction and the hints provided to help you:

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

DNA replication head HINT:Phosphate Steps in DNA replication including base is hydrophilic matching. Include a diagram of DNA replication. (water loving)

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.

79

57 NCEA Style Question: Cell Division

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?

80

58 KEY TERMS AND IDEAS: Cell Division 1. Match each term to its definition, as identified by its preceding letter code.

Fatty acid tail is hydrophobic (water hating)

H2O

Glucose

cell cycle 2. Explain how the parent DNA is copied by semi-conservative replication (you may use extra paper if required). Your answer should include: chromosome • the units that make up DNA • the process of replication • a labelled diagram to support your answer cytokinesis

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.

DNA replication

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

A The process of producing two identical copies of DNA from one original DNA molecule.

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

mitosis

Water molecules pass freely between the phospholipid molecules by osmosis.

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

semi-conservative

1. Describe the fluid mosaic model of membrane structure:

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.

F The process of nuclear division in cells.

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:

2. (a) What are the two main roles of the plasma membrane?

(b) How is its structure related to these roles?

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:

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

KNOW

9

LINK

6

LINK

14

© 1994-2016 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

3. DNA replication occurs during the S (synthesis) phase of the cell cycle.

REVISE

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Onion root tip 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. 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. Stage

© 1994-2016 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

Interphase Mitosis Total

TEST

© 1994-2017 BIOZONE International

Photocopying Prohibited

No. of cells

% of total cells

Estimated time in stage

TEST 100

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vi

Understanding the activity coding system and making use of the online material identified will enable you to get the most out of this resource. The chapter content is structured to build knowledge and skills but this structure does not necessarily represent a strict order of treatment. Be guided by your teacher, who will assign activities as part of a wider programme of independent and group-based work. Look out for these features and know how to use them: The activities form most of this book. They are numbered sequentially and each has a task code identifying the skill emphasised. Each activity has a short introduction with a key idea identifying the main message of the page. Most of the information is associated with pictures and diagrams, and your understanding of the content is reviewed through the questions. Some of the activities involve modelling and group work.

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.

16

Life processes at the cellular level

Achievement Standard

2.4

9

Free response questions allow you to use the information provided to answer questions about the content of the activity, either directly or by applying the same principles to a new situation. In some cases, an activity will assume understanding of prior content.

The Structure of Membranes

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.

Key terms

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.

Cell structure eukaryotic organelle Movement of materials active transport diffusion endocytosis exocytosis ion pump osmosis partially permeable passive transport plasma membrane surface area: volume ratio Enzymes and energy active site aerobic anaerobic ATP Calvin cycle catalyst cellular respiration chloroplast denaturation electron transport chain enzyme fermentation glycolysis Krebs cycle light dependent phase light independent phase metabolic pathway metabolism mitochondrion photosynthesis Cell division cell cycle chromosome cytokinesis DNA replication mitosis semi-conservative

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

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.

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 the cellular level.

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.

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

CO2

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

Phosphate head is hydrophilic (water loving)

Fatty acid tail is hydrophobic (water hating)

H2O

Glucose

EII

Kristian Peters

Explanatory notes: Life processes at the cellular level Life processes for plant and animal cells to include the following...

Activity number

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 - 55

c

i

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

c

ii

Movement of materials, including diffusion, osmosis, and active transport.

c

iii

Enzyme activity (although specific names of enzymes is not required).

c

iv

Factors affecting the process (e.g. photosynthesis, cellular respiration, cell transport processes, enzyme activity, cell division).

c

v

Details of the processes as they relate to the overall functioning of the cell (although the names of specific stages are not required).

Biological ideas relating to life processes

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

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

Water molecules pass freely between the phospholipid molecules by osmosis.

1. Describe the fluid mosaic model of membrane structure:

Activity number

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

1-9

2. (a) What are the two main roles of the plasma membrane?

9 - 18 22 - 28 14 26 - 28 33 40 41 53

(b) How is its structure related to these roles?

30 - 32 38 -39 47 48 51 54

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

KNOW

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

9

LINK

6

LINK

14

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.

Š 1994-2017 BIOZONE International

Photocopying Prohibited

Š 1994-2016 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

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


vii

Using the Tab System

2. (a) What are the two main roles of the plasma membrane?

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 In structure the example below for theroles? activity (b) defined. How is its related to these "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.

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

Activities are coded

KNOW

KNOW = content you need to know

9

LINK

LINK

14

6

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

Link

Weblinks

Bookmark the weblinks page: www.biozone.co.nz/weblink/ NZL2E-9513 Access the external URL for the activity by clicking the link

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

www.biozone.co.nz/weblink/NZL2E-9513 This WEBLINKS page provides links to external websites with supporting information for the activities. These sites are distinct from those provided in the BIOLINKS area of BIOZONE's web site. For the most part, they are narrowly focussed animations and video clips directly relevant to some aspect of the activity on which they are cited. They provide great support to help your understanding of basic concepts.

Chapter in the book

Hyperlink to the external website page. Activity in the book

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

Š 1994-2017 BIOZONE International

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viii

How to Scaffold an NCEA Style Answer The external NCEA exams require you to demonstrate your understanding of a particular concept by providing a written paragraph or essay. 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 mark in these questions you need to lay out your answer in a clear and logical way, so that the examiner can easily see how you have demonstrated your understanding of the topic. The difference between you obtaining an achievement, merit, or excellence grade depends on how well you demonstrated your understanding of a concept. • Defining, drawing, annotating, or giving a description of a process is achievement level only. • Explaining how a process works, why it works, and how it changes to it may affect an outcome is merit level. • Linking biological ideas, comparing and contrasting, analysing, or justifying ideas is excellence level. The following example question shows how an answer can be built up from a simple definition, through explanation, to comparisons and linking of ideas.

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:

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

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. 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:

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

How point mutations cause reading frame shifts.

ATT GGC CGC TTA TTT GTG original strand aa 1 aa2 aa3 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 GCG CTT ATT TGT G Insert G between CC cause aa1 aa2 aa10 aa11 aa 12 aa13 frame shift to the right 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). ATT GGC AGC TTA TTT GTG Substitution of second C for A aa 1 aa2 aa14 aa4 aa5 aa6

Link protein structure to amino acid sequence.

Definition of mutation provided. Point mutation defined.

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

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The effect of reading frame shifts on the amino acid sequence


Life processes at the cellular level

PR E V ON IEW LY

No Cla t fo ssr r o Us om e

Achievement Standard

2.4

Key terms

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.

Cell structure eukaryotic organelle

Enzymes and energy active site aerobic anaerobic ATP Calvin cycle catalyst cellular respiration chloroplast denaturation electron transport chain enzyme fermentation glycolysis Krebs cycle light dependent phase light independent phase metabolic pathway metabolism mitochondrion photosynthesis Cell division cell cycle chromosome cytokinesis DNA replication mitosis semi-conservative

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.

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

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.

EII

Kristian Peters

Explanatory notes: Life processes at the cellular level Life processes for plant and animal cells to include the following...

Activity number

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 - 55

Biological ideas relating to life processes

Activity number

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

i

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

c

ii

Movement of materials, including diffusion, osmosis, and active transport.

c

iii

Enzyme activity (although specific names of enzymes is not required).

c

iv

Factors affecting the process (e.g. photosynthesis, cellular respiration, cell transport processes, enzyme activity, cell division).

c

v

Details of the processes as they relate to the overall functioning of the cell (although the names of specific stages are not required).

c

1 - 9

No Cla t fo ssr r o Us om e

Movement of materials active transport diffusion endocytosis exocytosis ion pump osmosis partially permeable passive transport plasma membrane surface area: volume ratio

Achievement criteria and explanatory notes

9 - 18

22 - 28

14 26 - 28 33 40 41 50

30 - 32 38 -39 47 48 52 53


PR E V ON IEW LY

No Cla t fo ssr r o Us om e

What you need to know for this Achievement Standard Structure of plant and animal cells Activities 1 - 9, 19 - 21

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

c

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

c

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.

c

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

Jeff Podos

Movement of materials Activities 10 - 21

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

c

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

c

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

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.

c

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

c

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

Mnolf

Enzymes and energy transformations Activities 22 - 46

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

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

c

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

c

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

c

Describe and explain enzyme denaturation.

c

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

c

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

c

Describe ATP production by fermentation when oxygen is absent.

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

Describe factors that affect the rate of photosynthesis.

Cell division Activities 47 - 58

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

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c

c

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

c

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

c

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

c

Describe the role of mitosis in organisms.

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?


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The Structure of Plant Cells

Key Idea: Plant cells are eukaryotic cells. They share many features in common with animal cells, but

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they also have several unique features.

Amyloplast

Specialised plastids which synthesise and store starch.

Mitochondrion

Chloroplast

Convert chemical energy into ATP.

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

Cellulose cell wall

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.

Endoplasmic reticulum (ER)

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).

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.

Nuclear pore

Nuclear membrane

Nucleus contains most of a cell's DNA.

Plasma membrane

Tonoplast: A special membrane surrounding the vacuole.

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

Nucleolus

Ribosomes

Manufacture proteins.

Cytoplasm

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

l

Golgi apparatus

Cell cytoskeleton

Stores, modifies, and packages proteins.

Features of a plant cell

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

ffPresence of a membrane-bound nucleus.

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

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

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:

ffCellulose cell wall ffChloroplast ffAmyloplast

ffLarge vacuole (often centrally located)

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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.

Mnolf

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.

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

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

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

(b) What are its roles?

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) Why are they found there?

6. Briefly describe the functions of the following:

(a) Chloroplasts:

(b) Plasma membrane:

(c) Nucleus:

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5. What is the function of cellulose and lignin?

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Identifying Structures in Plant Cells

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

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identify them.

(a)

(b)

(c)

(d) (e) (f)

(g)

(h)

TEM

(i)

BF

(j)

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.

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

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

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

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

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

(b)

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The Structure of Animal Cells

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

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also have several unique features.

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

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

Mitochondria

Oval-shaped organelles bounded by a double membrane.

Lysosome

These contain enzymes that break down foreign material.

Smooth endoplasmic reticulum (smooth ER) An interconnected network of membranes. A site for lipid and carbohydrate metabolism, including hormone synthesis.

Plasma membrane

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

Rough endoplasmic reticulum (rough ER)

Nuclear pore

n

A hole in the nuclear membrane. It allows communication between the nucleus and the rest of the cell.

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

Nucleus

Ribosomes

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

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

Golgi apparatus

Centrioles

Microtubular structures associated with nuclear division.

Cytoplasm

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

Features of an animal cell

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

ffPresence of a membrane-bound nucleus.

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

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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.

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:

ffCentrioles

ffIrregular shape

ffLack of a cell wall

ffLack of chloroplasts

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7 SEM: Skin cells

SEM: Egg cell

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SEM: Blood cells

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.

Nerve cell

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

A

A:

B:

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

Neurones (nerve cells) in the spinal cord

(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.

B

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White blood cells and red blood cells (blood smear)

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

(a) Microvilli:

(b) Ribosomes:

(c) Golgi apparatus:

(d) Rough endoplasmic reticulum (rER):

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4. Briefly describe the function of the following:


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Identifying Structures in Animal Cells

Key Idea: The number and types of organelles present in a cell can reflect the cell's function.

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Organelles seen in an electron micrograph can vary in appearance depending on position.

(a)

(b)

(c)

(d)

(e)

(f)

(g) (h)

TEM

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.

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

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

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

4. The cell above contains a large amount of rough ER. What does this tell you about the cell?

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Structure Relates to Function

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Key Idea: Eukaryotic cells come in a wide range of shapes and sizes. Each cell type has specific

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features that enable it to fulfil its particular functional role in the organism.

Examples of specialised plant cells

Cell type: Palisade mesophyll cell Function: Photosynthesis

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

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

B

Vacuole

Sperm cells

Uneven thickening

Chloroplast

Open pore

Nucleus

Cell wall lacks waterproof cuticle

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

Companion cell

Phloem parenchyma cell

Vacuole High surface area

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

Lignin

Perforations

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

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

Elongated, lignified (strengthened) cells. End of cell perforated to allow movement of water.

D

Pollen grain

Elongated cell. Chloroplasts located near outer edge of cell to gather light.

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

Cell type: Xylem vessel Function: Transports water and minerals

Tube cell nucleus

E

Perforated cell end allows movement of sap.

F

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

(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.

(b)

(c)

(d)

(e)

(f)

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Examples of specialised animal cells Cell type: Retinal cell Function: Detects and responds to light

A

B

Cell type: Sperm cell Function: Transfers genetic material to egg cell

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Cell type: White blood cell (neutrophil) Function: Internal defence Nucleus

Nerve endings

Receptor membranes with light sensitive pigments

C

Head

Tail

Midpiece

Bacteria

Large spheroid cell. Able to change shape and exhibit amoeba-like behaviour. Engulfs and destroys foreign material. Cell type: Ciliated epithelial cells Function: Moves mucus and trapped debris out of airways

D

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

Cell type: Red blood cell (erythrocyte) Function: Carries oxygen in the blood and offloads it at the tissues

E

Cilia

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

Contains many haemoglobin molecules to carry oxygen No nucleus and few organelles

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

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

Cell type: Muscle cell (fibre) Function: Contracts to create movement

F

Nerve fibre

Myofibrils of muscle fibre 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:

(a) 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.

(b)

(c)

(e)

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Cell Processes

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Key Idea: Cells perform the processes essential to life. Each cellular organelle carries out one or

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more processes that contribute to the functioning of the cell as a whole.

ffThe 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.

ffTo 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.

Processes in an animal cell

Protein synthesis

Transport in and out of the cell

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

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

Autolysis

Organelle involved: lysosome. Destroys unwanted cell organelles and foreign material.

Cell division

Organelles involved: nucleus, centrioles. 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.

Secretion

Cytosis

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

Organelle involved: plasma membrane. 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).

Organelles involved: Golgi apparatus, plasma membrane. 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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Cellular respiration

Plant cells carry out photosynthesis

Photosynthesis (plant cell)

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

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Chloroplast

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Membranes in cells

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

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

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.

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):

(a) Secretion: (b) Respiration: (c) Endocytosis: (d) Protein synthesis:

(e) Photosynthesis:

(f) Cell division:

(g) Autolysis:

(h) Transport in/out of cell:

2. In what way is a cell like an assembly line in a factory?

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

4. Identify two examples of intracellular membranes and describe their functions: (a)

(b)

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Identifying Organelles in Micrographs

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Key Idea: Cellular organelles can be identified in electron micrographs by their specific features.

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.

1

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

(b) Describe the function of this organelle:

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2. (a) Name the circled organelle:

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

(c) Describe the function of this organelle:

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

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

(c) Describe the function of this organelle:

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

(c) Name the dark ‘blobs’ attached to the organelle:

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

BF

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

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

(c) Describe the function of this organelle:

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Cell Structures and Organelles

Key Idea: Cells contain a variety of organelles that carry out specialised functions within the cell. Not

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all cell types contain every type of organelle.

(a) Name:

Plasma membrane

Location: Surrounds the cell Function: Encloses cell contents and regulates

Lipid bilayer: a double layer of phospholipids

movement of substances into and out of cell.

(b) Name:

Small subunit

Protein

Large subunit

Location:

Function:

(c) Name:

Location:

Function:

Smooth & rough endoplasmic reticulum Location: Penetrates the whole cytoplasm (d) Name:

Function of smooth ER:

Function of rough ER:

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Plant cell

Transport pathway

Ribosomes

Rough

Smooth

Flattened membrane sacs

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Budding vesicles

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(e) Name:

Golgi apparatus

Location: Cytoplasm associated with smooth ER

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Secretory vesicle budding off the trans face

Function:

Cisternae

Transfer vesicles enter from the smooth endoplasmic reticulum

Middle lamella

(f) Name:

Cellulose cell wall

Location:

Function:

Pectins

Hemicelluloses

Cellulose fibres

(g) Name:

Double membrane

Pores

Location:

Function:

Genetic material

Stacks of membranes

Nucleolus

Outer membrane Inner membrane

(h) Name:

Location: Within the cytoplasm

Stroma (fluid)

Outer membrane

Lamellae

Inner membrane Matrix (fluid)

(i) Name:

Location:

Function:

Folded inner membranes

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Function:


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The Structure of Membranes

Key Idea: The plasma membrane encloses the cell and regulates the entry and exit of substances into

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the cell. It is a phospholipid bilayer with embedded proteins moving freely within it.

ffThe 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.

ffThe 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.

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.

CO2

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

Phosphate head is hydrophilic (water loving)

Fatty acid tail is hydrophobic (water hating)

H2O

Glucose

Na+

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

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

Water molecules pass freely between the phospholipid molecules by osmosis.

2. (a) What are the two main roles of the plasma membrane?

(b) How is its structure related to these roles?

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1. Describe the fluid mosaic model of membrane structure:

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

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Hydrophobic molecules

Small polar molecules

Large polar molecules

Charged molecules

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Gases

What can cross a lipid bilayer?

CO2

Glucose

O2

Small uncharged molecules can diffuse easily through the membrane.

Lipid soluble molecules diffuse into and out of the membrane unimpeded.

Na+

H2O

Benzene

Cl-

Ca2+

Ethanol

Small polar molecules are small enough to diffuse through. Aquaporins (water channels) increase rate of water movement.

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

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

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

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

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

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

(c) Must be transported across the membrane by carrier proteins:

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

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

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10 Diffusion in Cells

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Key Idea: Diffusion is the movement of molecules from high to low concentration. Diffusion through

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plasma membranes can be facilitated by various transport proteins.

What is diffusion?

Factors affecting the rate of diffusion

ffDiffusion 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.

Concentration gradient

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

The distance moved

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

The surface area involved

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

ffMost 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.

High concentration

Barriers to diffusion

Low concentration

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

Concentration gradient

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

Temperature

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

Glucose

Lipid soluble solutes

Inorganic ion

Simple diffusion

Facilitated diffusion by carriers

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

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.

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Channel protein

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Carrier protein

Facilitated diffusion by channels 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.

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1. What is diffusion?

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

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

4. Consider the two diagrams below. For each, draw in the appropriate box what you would expect to see after one hour. Particle with diameter of 5 nm

Particle with diameter of 20 nm

After one hour:

Soluble particles placed in at high concentration

Container of water at 20° C

Partially permeable membrane with pores of 10 nm.

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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:


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11 Osmosis in Cells Key Idea: Osmosis is the diffusion of water molecules from a lower solute concentration to a higher

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solute concentration across a partially permeable membrane.

Osmosis using dialysis tubing

Osmotic potential

ffAny solution of water and a solute (e.g. glucose)

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.

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).

ffA 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.

Glass capillary tube

Cells and tonicity

Dialysis tubing (partially permeable membrane)

Dialysis tubing containing glucose solution

Glucose molecule

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

Water molecule

ffHypertonic solutions have a

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

ffHypotonic solutions have a lower

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

ffWhen two solutions have the

same solute concentration they are called isotonic.

Water

Net water movement

1. In the blue box below the diagram above, draw an arrow to show the direction of the net water movement.

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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?

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12 Water Relations in Plants Key Idea: Plant cells in a hypertonic solution lose water and undergo plasmolysis. In a hypotonic

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solution they gain water, creating the turgor pressure that helps support the plant.

Osmosis and tonicity

ffA 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

Wilted plant (cells have lost turgor)

Plant cells are turgid

Cell in pure water (hypotonic) Water

Water

Cell wall is freely permeable to water.

Water

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

Cell contents more dilute than the external environment

Cell contents less dilute than the external environment

Cytoplasm

Plasma membrane

Water

Water

Cell wall bulges outward but prevents cell rupture

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

Water

Water

Water

Plasmolysis in a plant cell

Turgor in a plant cell

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.

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).

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

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

(c) A plant cell in an isotonic solution:

2. What is the turgor pressure in a fully plasmolysed cell? 3. Why does a plant with flaccid cells wilt?

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1. Identify the outcome of the following situations:

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13 Diffusion and Cell Size

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Key Idea: Diffusion is less efficient in cells with a small surface area relative to their volume than in

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ones with a large surface area relative to their volume.

Diffusion in organisms of different sizes

Single-celled organisms

Multicellular organisms

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

Multicellular organisms (e.g. plants and animals) are often 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.

Carbon dioxide

Oxygen

Food

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

Wastes

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

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

2 cm cube

3 cm cube

4 cm cube

5 cm cube

Cube size

2 cm cube

3 cm cube

4 cm cube

5 cm cube

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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.

Surface area

Volume

2 x 2 x 6 = 24 cm2

2 x 2 x 2 = 8 cm3

(2 cm x 2 cm x 6 sides)

(height x width x depth)

Surface area to volume ratio

24 to 8 = 3:1

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

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

5. The diffusion of molecules into a cell can be modelled by using agar cubes infused with phenolphthalein indicator and soaked in sodium hydroxide (NaOH). Phenolphthalein turns 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.

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

Cube 1

Cube

1

2

3

2 cm

Cube 2

NaOH solution

1 cm

4 cm

Cube 3

Region of no colour change

Region of colour change

2. Volume not pink (cm3)

3. Diffused volume (1. – 2. ) (cm3) 4. Percentage diffusion

Cubes shown to same scale

(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:

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Agar cubes infused with phenolphthalein

1. Total volume (cm3)

6. Explain why a 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:

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14 Temperature and Membrane Permeability Key Idea: High temperatures can disrupt the structure of cellular membranes and alter their

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permeability, making them leaky.

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

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.

Beetroot cubes

Experimental method

Raw beetroot was cut into uniform cubes using a cork borer with a 4 mm internal diameter. The cubes were trimmed to 20 mm lengths and placed in a beaker of distilled water for 30 minutes.

The aim and hypothesis

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.

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.

Absorbance of beetroot samples at varying temperatures

Temperature (°C)

Absorbance at 530 nm

Observation

Sample 1

Sample 2

Sample 3

0

No colour

0

0.007

0.004

20

Very pale pink

0.027

0.022

0.018

40

Very pale pink

0.096

0.114

0.114

60

Pink

0.580

0.524

0.509

90

Red

3

3

3

Mean

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

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

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

(c) Explain why this effect occurs:

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15 Active Transport Key Idea: Active transport is a process that uses energy to transport molecules across the plasma

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membrane against their concentration gradient. The energy comes from ATP.

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

Active

Active

ffThe 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.

Passive

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

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

ffThe 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).

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.

ATP and active transport

1

ATP binds to a transport protein.

2

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

3

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

4

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

Transport protein

ATP

ADP

H2O

P

Molecule to be transported

1. Define active transport:

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

H

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ATP

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

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16 Ion Pumps Key Idea: Ion pumps are transmembrane proteins that directly or indirectly use energy to move ions

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and molecules across a plasma membrane against their concentration gradient.

ffMembrane 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.

ffThe transport of molecules against their concentration gradient requires an input of energy. This is called active transport. ffSpecial membrane proteins, called ion pumps, directly or indirectly use energy to transport ions across the membrane against a concentration gradient.

ffThe 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).

Sodium-potassium pump

Extracellular fluid Extracellular fluid or gut orlumen lumenofof gut

Cotransport

(the sodium-glucose symport) Na+

Na+

Na+

Na+

K+ binding site

Na+

K+

Na+

Diffusion of sodium ions

+ + + + + + + + +

Glucose

Na+

Plasma membrane

– – – – – – – – –

Carrier protein

Na + binding site

ATP

Na+

K+

K+

Cell cytoplasm

3 Na+ are pumped out of the cell for every 2 K+ pumped in

Na+

Cotransport (coupled transport)

The Na+/K+ pump is a protein in the membrane that uses energy in the form of ATP to exchange sodium ions (Na+) for potassium ions (K+) across the membrane. The unequal balance of Na+ and K+ across the membrane creates 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.

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.

1. What is an ion pump?

2. Explain why ion pumps require energy to carry out their functional role:

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Sodium-potassium (Na+/K+) Pump

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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.

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The proton (hydrogen) pump and its role in photosynthesis and respiration Proton pump

Extracellular fluid Hydrogen ion H+

H+

H+

Stroma

ATP synthase

H+

++++++

ADP

Hydrogen pump

H+

Plasma membrane

Thylakoid membrane

– – – – –

ATP or electrons

ATP

H+

Carrier protein

Cell cytoplasm

Chlorophyll

H+

Chlorophyll

Thylakoid space

H+

Proton pumps in the thylakoid membranes of chloroplasts are powered by electrons passed to them by chlorophyll.

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

(a) Proton pump:

(b) Sodium-potassium pump:

(c) Sodium-glucose symport:

4. (a) Explain what is meant by cotransport:

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

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

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:

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17 Cytosis

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Key Idea: Endocytosis and exocytosis are active transport processes. Endocytosis involves the cell

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engulfing material. Exocytosis involves the cell expelling material.

ffMost 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.

ffCytosis 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.

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

ffExocytosis 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).

Exocytosis

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. 3

Plasma membrane

2

1

From Golgi apparatus

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

Vesicle fuses with the plasma membrane.

Vesicle from the Golgi carrying molecules for export moves to the perimeter of the cell.

NT

Nerve cell

Golgi apparatus forming vesicles

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).

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

1. What is the purpose of exocytosis?

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Nerve cell

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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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Endocytosis

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

Exocytosis and endocytosis require energy because they involve movement of cytoskeletal proteins.

Plasma membrane

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

Vesicle buds inwards from the plasma membrane

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.

HIV particle

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).

Dartmouth College

CDC

Dartmouth College

Receptors and pit beginning to form

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.

2. Describe two examples of exocytosis in cells:

3. Describe the following types of endocytosis:

(a) Phagocytosis:

(b) Receptor mediated endocytosis:

(c) Pinocytosis:

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18 Active and Passive Transport Summary Key Idea: Cells move materials into and out of the cell by either passive transport, which does not

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use energy, or by active transport which requires energy, usually as ATP.

A

C

B

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.

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

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.

e.g. Cl –

CO2

D

O2

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

H2O

E

Golgi

rER

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.

Nucleus

F

Na+

K+

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.

sER

G

Plasma membrane

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

Cytoskeleton

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.

(a) Uptake of extracellular fluid by liver cells:

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

(c) Movement of water into the cell:

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

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

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

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

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19 What You Know So Far: Cell Structure & Transport Active and passive transport processes

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

HINT: Include reference to membrane proteins and their roles.

Cell structure and function

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

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

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Organelle structure and function

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20 NCEA Style Question: Cell Structure & Transport

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

In your answer you should:

• discuss features common to plant and animal cells • discuss features unique to plant and animal cells • briefly describe and contrast the role of the plasma membrane and cell wall

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TEST

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

(a) Osmosis:

(b) Active transport:

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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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21 KEY TERMS AND IDEAS: Cell Structure & Transport

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

active transport

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

adenosine triphosphate (ATP)

B

diffusion

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

endocytosis

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

ion pump

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

organelle

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

osmosis

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

passive transport

I

Movement of substances down a concentration gradient without energy expenditure.

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

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

plasma membrane

2. (a) Identify organelle 1:

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

(c) Identify organelle 2:

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

1

2

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. (a)

(b)

Cells are the basic...

...such as photosynthesis or respiration.

A cell is enclosed by a plasma membrane...

...hydrophilic head and a hydrophobic tail.

A phospholipid is made up of a...

...units of life.

Proteins are embedded...

...in the plasma membrane.

Eukaryotic cells contain different types of organelle...

...made of a phospholipid bilayer with embedded proteins

Each organelle has a specific function in the cell...

...some of which are composed of membranes.

Transport of molecules though the plasma membrane...

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

Active transport requires the input of energy...

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

Passive transport involves the movement of molecules from...

...can be active or passive.

Simple diffusion can occur...

...directly across the lipid bilayer of the membrane.

Facilitated diffusion involves proteins in the plasma membrane...

...which help molecules or ions to move through.

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

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...whereas passive transport does not.

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22 Reactions in Cells Key Idea: Anabolic reactions build complex molecules and structures from simpler ones. Catabolic

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reactions break down larger molecules into smaller molecules.

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

ffThere are two categories of metabolic reactions: anabolic and catabolic.

Anabolic reactions

Catabolic reactions

ffAnabolic reactions are reactions that result in the

ffCatabolic reactions are reactions that break

down large molecules into smaller components.

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.

ffCatabolic reactions involve a net release of

energy. They are called exergonic reactions.

ffCatabolic reactions are the opposite of anabolic

ffAnabolic reactions need a net input of energy to

reactions.

proceed. They are called endergonic reactions.

ffThe energy released from catabolic reactions

can be used to drive other metabolic processes.

Energy

Many small molecules

One large molecule

Energy

Many small molecules

One large molecule

Plants carry out photosynthesis in organelles called chloroplasts (left). Photosynthesis is an anabolic process because it converts carbon dioxide and water into glucose. Energy from the sun is required to drive photosynthesis.

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).

2. (a) What is a catabolic reaction?

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1. What is an anabolic reaction?

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

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

(a) Protein synthesis:

(c) Digestion:

(b) ATP conversion to ADP:

(d) DNA synthesis:

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23 Enzymes

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

ffMost 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).

ffEnzymes 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.

ffEnzymes 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.

Enzymes can be intracellular or extracellular

The 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

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

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. Example: Catalase. 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.

complex tertiary structure.

The substrate is the chemical that an enzyme acts on. A specific enzyme acts on a specific substrate. Extremes of temperature or pH can alter the enzyme's active site and lead to loss of function. This is called denaturation.

Substrates collide with an enzyme's active site

Incorrect reactant orientation = no reaction

X

Enzyme orientates the reactants making the reaction more likely.

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

Catalase

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. 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.

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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.

2H2O + O2

2H2O2

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

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

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24 Enzymes and Activation Energy Key Idea: Enzymes lower the activation energy required for a reaction to proceed. This helps speed

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up the reaction rate.

ffChemical 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).

ffAny 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.

ffAn 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.

Transition state (unstable)

High

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

Amount of energy stored in the chemicals

Energy barrier

Low

Ea

High energy

Ea

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

Reactants

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

Product

Low energy

Start

Finish

Direction of reaction

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

(c) How do enzymes lower the activation energy for a reaction?

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1. (a) What is the transition state?

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

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

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25 How Enzymes Work Key Idea: Enzymes catalyse reactions by providing a reaction site for a substrate. The model that

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describes the behaviour of enzymes the best is the induced fit model.

ffAn 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).

ffThe 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.

Induced fit model of enzyme activity Substrate molecules

1

Enzyme

2

Enzyme

Active site

Two substrate molecules are drawn into the active site of the enzyme.

2b

3

Enzyme changes shape

Enzyme

Enzyme

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.

End product released

The end product is released and the enzyme returns to its previous shape.

Enzymes in anabolic and catabolic reactions

ffComplex 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. Substrates Substrates

Substrates drawn into the active site.

Substrate drawn into the active site.

Product is released

Enzyme Enzyme

Enzyme catalyses the formation of bonds.

Product Product

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.

Enzyme catalyses the breaking of bonds.

Substrate Substrate

Products are released

Enzyme Enzyme

Products Products

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.

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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?:

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

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26 Cofactors and Inhibitors Affect Enzyme Activity Key Idea: Some enzymes require cofactors to function. Enzyme activity can be reduced or stopped by

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inhibitors.

Some enzymes need cofactors Substrate molecules

Substrate molecules bind to the active site

Cofactor

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.

Enzyme

Active site cannot receive substrate until cofactor binds

Cl- ion

Ca2+ion

a-amylase

Cofactor completes the active site and the enzyme is functional.

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 cofactors.

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).

Zinc ion

Carbonic anhydrase

Inhibitors slow or stop enzyme activity

Hg

Enzyme

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

Enzyme changes shape

Enzyme

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 drug.

Non-competitive inhibitor binds outside the active site and substrate cannot bind.

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?

2. Distinguish between competitive and non-competitive inhibition:

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Drugs

Penicillin targets cell wall synthesis

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Inhibitor, e.g. mercury

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27 Enzyme Reaction Rates Key Idea: Enzymes have a narrow range of conditions in which they work optimally. Outside this range,

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activity decreases and the enzyme may lose its structure.

Rate of reaction

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

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

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

With ample substrate

Enzyme concentration

Rate of reaction

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

With fixed amount of enzyme

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

Substrate concentration

Rate of reaction

No inhibitor

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

Competitive inhibitor

(a) What is the general effect of an enzyme inhibitor:

Non-competitive inhibitor

(b) How can the effect of a competitive inhibitor be overcome?

Too cold to operate

0

10

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.

Enzyme denatures

20

30

40

50

60

Enzyme activity

Temperature (°C)

Urease

Pepsin

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

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Enzyme activity

Substrate concentration

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

(c) What is the effect of denaturation on the enzyme's active site?

Trypsin

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

1 2 Acid

3

6

5

7

pH WEB

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8 9 10 Alkaline

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

Pepsin:

Trypsin:

Urease:

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

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28 Investigating Catalase Activity Key Idea: The effect of increasing the amount of enzyme in a reaction can be measured indirectly by

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measuring the volume of reaction products.

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

Timed for 5 minutes.

Oxygen released by the reaction

Hypothesis and prediction Reaction rate increases with amount of enzyme present, so a greater mass of potato (therefore more enzyme) will produce a greater reaction rate. Method 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: Mass of potato (g)

Volume oxygen (cm3) (5 minutes) Test 1

Test 2

Test 3

1

6

5

6

2

10

9

9

3

14

15

15

4

21

20

20

5

24

23

25

Mean

Water in the 50 cm3 cylinder is displaced by the oxygen.

Tube transfers released oxygen

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

Potato cubes + excess H2O2

Mean rate of O2 production (cm3 min-1)

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

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

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

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

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

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

(b) Explain this result:

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29 ATP Key Idea: ATP transfers energy to where it is needed in the cell. Hydrolysis of the phosphate group

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releases energy, which can be used to do work.

Structure of an ATP molecule

Adenosine triphosphate (ATP)

Adenine

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

• A purine base (adenine) • A pentose sugar (ribose) • Three phosphate groups.

ffATP 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).

How does ATP provide energy? ffThe 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).

ffWhen 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).

Phosphate groups

Ribose

Adenine + ribose = adenosine

P

Adenosine

P

P

Adenosine

Adenosine triphosphate (ATP)

+

Pi

O

H

H

H

A Reactants A and B have low energy

A

P

Adenosine diphosphate (ADP)

+

Water

P

B Reactant A combines with Pi raising its energy

A

Pi

O

P

+

H

C The energy gained by A lets it react with B

A

B

Pi

B

B

Pi

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.

1. What are the three components of ATP?

(b) Where is the energy stored in ATP?

(c) What products are formed during hydrolysis of ATP?

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

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

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30 Energy Transformations in Cells Key Idea: Photosynthesis uses energy from the sun to produce glucose. Glucose breakdown

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produces ATP, which is used by all cells to provide the energy for metabolism.

ffDuring 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.

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

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

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

Light energy

Oxygen

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

Other uses of glucose

Glucose *

Photosynthesis

Oxygen

Fuel

ADP + Pi

Pi

Carbon dioxide + water

Cellular respiration

ATP

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

A photosynthetic plant cell

Cellular respiration is a chemical process in which the step-wise breakdown of glucose provides the energy to form high energy ATP from ADP and inorganic phosphate (Pi).

Water

Heat energy

Carbon dioxide

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

2. What is the immediate source of energy for reforming ATP from ADP? 3. (a) What is the ultimate source of energy for plants?

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

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

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

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31 Cellular Respiration: Inputs and Outputs Key Idea: During cellular respiration, the energy in glucose is transferred to ATP in a series of enzyme

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controlled steps.

ffCellular 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.

ffGlycolysis 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.

Cytoplasm

Mitochondrion outer membrane

Mitochondrion

Matrix: enzymes for the Krebs cycle reside here.

NADH + FADH

Glucose

Glycolysis

Pyruvate

Link reaction

2 ATP

tyl Ace me y z n coe A

Krebs cycle

NA

DH

2

Electron transport chain

Cristae: folded inner membrane. Enzymes for the electron transport chain reside here

2 ATP

34 ATP

The general equation for cellular respiration

C6H12O6 + 6O2

6CO2 + 6H2O + energy

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

(a) Glycolysis:

(b) The link reaction: (c) Krebs cycle reactions:

(d) Electron transport chain:

3. What is the total number of ATP produced from one glucose molecule? 4. Describe three functions of glycolysis in cellular respiration:

(i)

(ii) (iii)

5. Describe two functions of the Krebs cycle in cellular respiration:

(i) (ii)

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

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How does cellular respiration provide energy? ffA molecule's energy is contained in the electrons within the molecule's chemical bonds. During a chemical

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reaction, energy (e.g. heat) can break the bonds of the reactants.

ffWhen 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.

ffGlucose 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.

A model for ATP production and energy transfer from glucose

A model for ATP use in the muscles

Chemical potential energy

Glucose

Carbohydrates in muscles

2 ATP

ADP

Glycolysis

Pi

Glucose

Link reaction

Energy transfer

2 ATP

Krebs cycle

34 ATP

Electron transport chain Oxygen

Water

Muscle contraction

ATP

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:

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:

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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:

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32 Steps in Cellular Respiration

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Key Idea: There are four major steps to cellular respiration. The majority of ATP is produced in the

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electron transport chain. Enzymes catalyse each reaction of cellular respiration.

ffATP is produced during cellular respiration by both substrate level phosophorylation and oxidative phosphorylation. ffSubstrate 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.

ffOxidative 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.

Glycolysis

Glycolysis

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

Glucose

2NADH

ffThe electrons are transferred to NAD+ to form the

coenzyme NADH. NADH carries hydrogens to the electron transport chain.

2 ATP

ffGlycolysis 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).

2 Pyruvate

Link reaction

Acetyl coenzyme A

ffThe 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.

2NADH

2CO2

Krebs cycle

ffAcetyl coenzyme A is attached to a 4C

intermediate and coenzyme A is released.

2 ATP

ffThe 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.

6NADH

Krebs cycle

2FADH2

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

ff1 ATP is made by substrate level phosphorylation and 2 CO2 are produced per turn.

Electron transport chain (ETC)

ffElectrons carried by NADH and FADH2 are

passed to a series of electron carrier enzymes embedded in the inner membrane of the mitochondria.

Electron transport chain H+

ffThe 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.

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

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4CO2

NADH

H+

H+

H+

34 ATP (via ETC)

e–

NAD+

FADH2

FAD+

H2O

2H+ +1/2O2

H+

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1. (a) What is substrate level phosphorylation?

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

2. (a) What is oxidative phosphorylation?

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

3. Which parts of cellular respiration produce CO2?

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?

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

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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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33 Measuring Respiration Key Idea: Oxygen consumption and carbon dioxide production in respiring organisms can be

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measured with a respirometer.

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. Scale

Screw clip

Coloured bubble

Capillary tube

Measuring respiration with a simple respirometer ffSoda lime or potassium hydroxide is placed into the

chamber to absorb any CO2 produced during respiration. The respirometer measures O2 consumption.

ffRespiring 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).

ffThe 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.

ffChanges in temperature or atmospheric pressure

Perforated metal cage Germinating seeds

Soda lime (or KOH) pellets (CO2 absorbant)

may change the readings and give a false measure of respiration when using a simple respirometer.

ffDifferential 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.

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. Clamp stand

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

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

(c) Describe the results in your plot:

See Appendix

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

3. Why would it have been better to use a differential respirometer?

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Time (minutes)

Distance bubble moved (mm)

Rate (mm min-1)

0

0

_

5

25

5

10

65

15

95

20

130

25

160

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34 Anaerobic Metabolism in Plants and Fungi Key Idea: Alcoholic fermentation is an anaerobic process in plants and fungi, which passes electrons

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directly to the final electron acceptor ethanal.

Alcoholic fermentation

ffOrganisms can generate ATP when oxygen is absent

by using a molecule other than oxygen as the terminal electron acceptor for the pathway.

ffIn 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.

Glucose C6H12O6

2 x pyruvate CH3COCOO–

2 NAD+

ffYeasts 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.

2 ATP yield

2 ADP

Ethanol CH3CH2OH

ffThe root cells of plants also use fermentation as a

2 NADH + 2H+

E

Ethanal CH3CHO

E

Alcoholic fermentation (higher plants, yeast)

CO2

pathway when oxygen is unavailable but the ethanol must eventually be converted back to respiratory intermediates and respired aerobically.

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.

Yeasts 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.

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.

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

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

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

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

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35 Anaerobic Metabolism in Animals Key Idea: Lactic acid fermentation is an anaerobic process that passes electrons directly to pyruvate

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as the final electron acceptor. It occurs in the muscles of mammals.

Lactic acid fermentation

2 ADP

ffMammalian 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+.

2 x pyruvate CH3COCOOH

Glucose C6H12O6

2 NAD+

ffThe 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'.

ffImportantly, 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.

2 ATP yield

2 NADH + H+

pyruvate Lactic acid E CH3COCOOH CH3CHOHCOO- +H+ + + NAD+ NADH +H+ Lactic acid fermentation (animal tissues)

ffLactate 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.

ffDuring moderate exercise, lactate released from

Wintec

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.

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

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

(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:

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

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36 Leaf Structure and Photosynthesis Key Idea: Leaves are the main site for photosynthesis in plants. Leaf structure aids the

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photosynthetic process.

ffThe 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.

ffThe 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.

Dicot leaf structure

Cuticle forms a barrier to the diffusion of gases

Upper epidermis (lacks chloroplasts)

Palisade mesophyll cell with chloroplasts

E

Mp

Spongy mesophyll cell with chloroplasts

O2

Substomatal air space

CO2

Lower epidermis

CO2

Ms

Leaf vein (xylem and phloem)

O2

Entry and exit of gases CO2 through the stomata

O2

Net gas exchanges in a photosynthesising dicot leaf

Guard cell

Stoma (pore)

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

ffRespiring 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).

ffWhen 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.

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

(b) How is this movement regulated?

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

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

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.

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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.

Sun Sun plant plant

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

Intense Intense light light

Palisade Palisade mesophyll mesophyll layer layer often often 2-32-3 cells cells thick thick

Shade Shade plant plant

A shade A shade leaf leaf can can absorb absorb thethe light light available available at at lower lower light light intensities. intensities. If exposed If exposed to to high high light, light, most most would would pass pass through. through. Palisade Palisade mesophyll mesophyll layer layer only only 1 cell 1 cell thick thick Low Low light light intensity intensity

Thick Thick leaves leaves

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

Thin Thin leaves leaves

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

Sun Sun leaves leaves

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

Shade Shade leaves leaves

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.

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:

Sunny

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

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

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

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

Shady

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Wet

Dry

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37 The Structure of Chloroplasts Key Idea: The complex membranous structure of chloroplasts creates compartments in which the

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separate stages of photosynthesis can occur.

ffPhotosynthesis takes place in disk-shaped organelles called

chloroplasts (4-6 Âľm 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.

Chloroplasts

ffThis structure creates distinct regions in which the separate

stages of photosynthesis occur. One stage depends on light and the other doesn't.

Kristian Peters

Cell wall

ffPigments on the thylakoid membranes called chlorophylls

capture light energy by absorbing light of specific wavelengths. Chlorophylls reflect green light, giving leaves their green colour.

A mesophyll leaf cell contains 50-100 chloroplasts!

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

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

Liquid stroma

Grana (sing. granum)

Starch granule Lipid droplet

Chloroplast is enclosed by an inner and outer membrane

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

1. Label the transmission electron microscope (TEM) image of a chloroplast below:

(d)

(b)

(e)

(c) (c)

(f)

Image: Dartmouth College

2. What does chlorophyll do?

3. How do plants maximise the amount of light that can be absorbed?

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(a)

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38 Photosynthesis: Inputs and Outputs Key Idea: Photosynthesis transforms the energy in light into chemical energy. It uses carbon dioxide

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and water and produces glucose and oxygen.

ffPhotosynthesis 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.

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

Light dependent phase (LDP):

Light independent phase (LIP):

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.

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

Sun

phase does not need light to proceed.

Chloroplast outer membrane

ligh

t

LDP

H2O (waste)

LIP

O2

ATP

LDP

Triose phosphate (C3H7O6P)

NADPH

Converted to

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

CO2

CO2 from the air provides raw materials for glucose production.

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.

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

Photosynthesis can be summarised in the equation: Light

6CO2 + 12H2O

C6H12O6 + 6O2 + 6H2O

Chlorophyll

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

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

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

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

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

3. Explain why most photosynthetic organisms are green:

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

Raw materials

(a)

ADP

ATP

Light dependent phase

Solar energy

(f)

(c)

Light independent phase

Process:

(g) Location:

(h) Process:

NADPH

Main product

(e)

(i) Location:

NADP+

(b)

By-products

(d)

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:

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

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

From these results, what would you conclude about the source of the oxygen in:

(a) The carbohydrate produced?

(b) The oxygen released?

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

(a) The light dependent phase of photosynthesis:

(b) The light independent phase of photosynthesis:

8. What is the function of each of the following in photosynthesis:

(a) ATP:

(b) NADPH:

(c) Light:

(d) Chlorophyll:

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7. Describe what happens during:


39 Steps in Photosynthesis

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56

Key Idea: Photosynthesis involves two linked phases. The light dependent phase captures light

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energy and stores it as ATP. The light independent phase uses ATP to fix carbon.

Light dependent phase

Stroma

Cytochrome complex

ffThe 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.

Photosystem II

ffPhotosystem 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).

H+

Light energy

ADP

NADP+

e-

2H2O

NADPH

e-

H+

4H+ + O2

H+

Photosystem I

ATP synthase

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

Thylakoid space

Light independent phase (Calvin cycle)

3 x CO2

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

E

ffThe 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.

ffThree CO2 molecules are needed for the production of

3 x 5C RuBP

3 x 6C

Carbon fixation

3 ADP 3 ATP

6 x 3C GP

6 ATP

Regeneration

one TP molecule. Two TP molecules are used to form one glucose molecule. Stages of the Calvin cycle

ATP

Reduction

6 ADP

3 x 5C

Carbon fixation: The enzyme (E) RuBisCo joins CO2 with RuBP to form glycerate 3-phosphate (GP). Reduction: ATP driven reactions form intermediates then NADPH driven reactions form triose phosphate (TP).

12 NADPH

5 x 3C

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

6 x 3C

1 x 3C

12NADP+

TP

Glucose and other products

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

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

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

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

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40 Factors Affecting Photosynthesis Key Idea: The two main factors that affect the rate of photosynthesis are the availability of light and

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carbon dioxide.

ffThe 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.

ffThe 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. The effect of different limiting factors on the rate of photosynthesis in cucumber plants 280

A: Light intensity vs photosynthetic rate

Rate of photosynthesis (mm3 CO2 cm-2 h-1)

Rate of photosynthesis (mm3 CO2 cm-2 h-1)

90 80 70 60 50 40

1

2 3 4 5 6 Units of light intensity (arbitrary scale)

7

B: Light intensity, CO2, and temperature vs photosynthetic rate

High CO2 at 30oC

240 200

High CO2 at 20oC

160 120

Low CO2 at 30oC

80 40

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

Low CO2 at 20oC

1

2 3 4 5 6 Units of light intensity (arbitrary scale)

7

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.

ffManipulating 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.

ffTemperature, carbon dioxide (CO2) concentration, and

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

(a) An increase in CO2 concentration:

(b) An increase in light intensity:

(c) A decrease in temperature:

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Glasshouse environments can artificially boost CO2 levels

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.

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58

High wind increases water loss from stomata

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.

Plants have adaptations to reduce water loss from stomata

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.

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

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

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

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

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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:

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41 Investigating Photosynthetic Rate Key Idea: Measuring the production of oxygen in provides a simple means of measuring the rate of

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photosynthesis.

Experimental set up

Background

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. Aim

Oxygen bubbles

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

Test tube with NaHCO3 solution

Hypothesis

Inverted funnel

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

Beaker with NaHCO3 solution at 20°C

Method

Cabomba stem

ff0.8-1.0 g of Cabomba stem were weighed. The stem

was cut and inverted to ensure a free flow of oxygen.

ffThe stem was placed into a beaker filled with a 20°C

See Appendix

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.

ffThe 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).

ffBefore 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.

The results

Bubbles counted in three minutes

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

Bubbles per minute

1. Complete the table (left) by calculating the rate of oxygen production (bubbles of oxygen gas per minute): 2. Use the data to draw a graph on the grid above of the bubbles produced per minute vs light intensity:

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Light intensity in lx (distance)

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:

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?

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42 Modelling Photosynthesis and Cellular Respiration Key Idea: Modelling photosynthesis and cellular respiration using paper cut outs will help you better

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understand the net chemical changes occurring.

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

ffIn 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.

Glucose

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

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.

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

Carbon

Hydrogen

Water

Carbon dioxide

Oxygen

2. Write the equation for photosynthesis here:

(a) State the starting reactants in photosynthesis: (b) State the total number of atoms of each type needed to make the starting reactants:

Carbon:

Hydrogen:

Oxygen:

(c) Use the atoms you have cut out to make the starting reactants in photosynthesis.

(d) State the end products of photosynthesis:

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

Carbon:

Hydrogen:

Oxygen:

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

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

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

3. Write the equation for cellular respiration here:

(a) State the starting reactants in cellular respiration: (b) State the total number of atoms of each type needed to make the starting reactants:

Carbon:

Hydrogen:

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Oxygen:

(c) Use the atoms you have cut out to make the starting reactants in cellular respiration.

(d) State the end products of cellular respiration:

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

Carbon:

Hydrogen:

Oxygen:

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

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

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Sunlight energy

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+

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+

ATP


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43 What You Know So Far: Energy Transformations Photosynthesis

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

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

How enzymes work

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

Cellular respiration

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

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

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Factors affecting enzyme activity

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44 NCEA Style Question: Photosynthesis

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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: • the raw material(s) for photosynthesis and the final product(s) • the location of photosynthesis in plant cells • an outline of the light dependent and light independent reactions. • the role of enzymes in photosynthesis • factors that affect the rate of photosynthesis

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TEST

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45 NCEA Style Question: Cellular Respiration

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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: Your discussion should include: • the purpose of aerobic cellular respiration • the anaerobic generation of ATP • 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 • the role of enzymes in these processes • situations in which aerobic or anaerobic pathways are used.

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46 KEY TERMS AND IDEAS: Energy Transformations 1. Complete the diagram of cellular respiration below by filling in the boxes: (b)

(c)

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(a)

No. ATP

Glycolysis

Input

(f)

Waste

No. ATP

No. ATP

(d)

(e)

Link reaction

CO2

(g)

Waste

Waste

(h)

2. (a) Write the process of photosynthesis as: A word equation: A chemical equation:

(b) Where does photosynthesis occur?

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

4. In the image below, part 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:

Foil

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

electron transport chain Krebs cycle

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White

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.

B The phase in photosynthesis when light energy is converted to chemical energy.

C The phase in photosynthesis in which chemical energy is used for the synthesis of

light dependent phase

carbohydrate. Also called the Calvin cycle.

D Chain of enzyme-based redox reactions, which passes electrons from high to

light independent phase

TEST

low redox potentials. The energy released is used to pump protons across a membrane and produce ATP.

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47 DNA Replication Key Idea: Before a cell can divide, the DNA must be copied. DNA replication produces two identical

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copies of DNA. A copy goes to each new daughter cell.

ffBefore 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.

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

DNA replication duplicates chromosomes DNA replication creates a chromosome with two identical chromatids

Replicated chromosome consists of two chromatids joined at the centromere. Centromere links sister chromatids

ffAfter 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.

ffDNA 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.

Chromatid

Parent chromosome

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

DNA replication is semi-conservative 3'

1. What is the purpose of DNA replication?

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

5'

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

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

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

Original ‘parent’ DNA

New ‘daughter’ DNA

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

(b) What is the purpose of the centromere?

4. Explain what semi-conservative replication means:

3'

5'

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

3'

DNA that will become one chromatid

5'

DNA that will become the other chromatid

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

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48 Details of DNA Replication

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Key Idea: DNA replication is achieved by enzymes attaching new nucleotides to the growing DNA

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strand at the replication fork.

Stages in DNA replication

ffDuring 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.

The two strands are joined by base pairing

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

ffNucleotides are added in by complementary

base-pairing: Nucleotide A is always paired with nucleotide T. Nucleotide C is always paired with nucleotide G.

1

ffThe DNA strands can only be replicated in one

Enzymes unwind parent DNA double helix

direction, so one strand has to be copied in short segments, which are joined together later.

ffThis whole process occurs simultaneously

for each chromosome of a cell and the entire process is tightly controlled by enzymes.

1. How are the new strands of DNA lengthened?

5'

3'

Parent strand of DNA acts as a template to match nucleotides for the new DNA strand.

Enzymes unzip parent DNA at the replication fork.

2

Original ‘parent’ DNA

5'

3'

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

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

3

Enzymes add free nucleotides to the exposed bases on the template.

Nucleotide symbols G

C

A

T

New ‘daughter’ DNA

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.

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

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

(b)

Original ‘parent’ DNA

New ‘daughter’ DNA

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.

(c)

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Two new double-stranded DNA molecules

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49 The Eukaryotic Cell Cycle Key Idea: The eukaryotic cell cycle can be divided into phases, although the process is continuous.

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Specific cellular events occur in each phase.

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.

Interphase

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

Interphase

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

replicate DNA

ffThe first gap phase (G1).

ffThe S-phase (S).

grow

ffThe second gap phase (G2).

grow and prepare for division

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.

Mitosis

Mitosis and cytokinesis (M-phase)

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

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

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

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

Mitosis: Nuclear division

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

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

(a) Interphase:

(b) Mitosis:

(c) Cytokinesis:

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

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50 Checkpoints in the Cell Cycle Key Idea: Regulatory checkpoints are built into the cell cycle to ensure that the cell is ready to

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proceed from one phase to the next. The failure of these systems can lead to cancer.

Checkpoints during the cell cycle

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.

G2 Checkpoint: Pass this checkpoint if: Cell is large enough. Chromosomes have been successfully duplicated.

Interphase

G1 checkpoint Pass this checkpoint if: Cell is large enough. Cell has enough nutrients. Signals from other cells have been received.

Mitosis

p53

Cancer cell

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.

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.

1. Explain the importance of cell cycle checkpoints:

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

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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.

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National Cancer Institute

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

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51 Functions of Mitosis Key Idea: Mitosis has three primary functions: growth of the organism, replacement of damaged or old

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cells, and asexual reproduction (in some organisms).

Mitotic cell division has three purposes:

ffGrowth: 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.

ffRepair: Damaged and old cells are replaced with new cells.

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

Fertilised egg (zygote)

Embryo

Adult

Matthias Zepper

Asexual reproduction

Growth

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.

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.

Parent cell

Daughter cell

Jpbarrass

Broken bone

Brocken Inaglory

Damaged limbs

Repair

1. Use examples to explain the role of mitosis in:

(a) Growth of an organism:

(b) Replacement of damaged cells:

(c) Asexual reproduction:

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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.

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52 Mitosis Key Idea: Mitosis is an important part of the eukaryotic cell cycle in which the replicated

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chromosomes are separated and the cell divides, producing two new cells.

Mitosis is a stage in the cell cycle

ffM-phase (mitosis and cytokinesis) is the part of the cell cycle

Mitosis produces identical daughter cells

in which the parent cell divides in two to produce two genetically identical daughter cells (right).

ffMitosis results in the separation of the nuclear material and division

Parent cell 2N = 4

of the cell. It does not result in a change of chromosome number.

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

ffIn animals, mitosis takes place in the somatic (body) cells. Somatic

DNA replication occurs

cells are any cell of the body except sperm and egg cells.

ffIn plants, mitosis takes place in the meristems. The meristems are

RCN

regions of growth (where new cells are produced), such as the tips of roots and shoots.

M

Mitosis in onion cells (DIC light micrograph)

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.

Daughter cell, 2N = 4

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

Daughter cell, 2N = 4

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

2. Where does mitosis take place in:

(a) Animals:

(b) Plants:

3. A cell with 10 chromosomes undergoes mitosis.

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1. Briefly outline the events in mitosis:

(a) How many daughter cells are created:

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

(c) The genetic material of the daughter cells is the same as / different to the parent cell (delete one). WEB

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53 Mitosis and Cytokinesis Key Idea: Mitosis is a continuous process, but it is divided into stages to help identify and describe

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what is occurring.

The cell cycle and stages of mitosis

ffMitosis 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.

ffIn 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.

ffPlant 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.

The animal cell cycle and stages of mitosis

Interphase

Nucleus

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

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

Early prophase

Nuclear membrane

Centrosomes move to opposite poles

Centrosome containing centrioles (forms spindle)

Cytokinesis

Late prophase

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

1

2

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

Homologous pair of replicated chromosomes

Telophase

3

Metaphase

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

5 Late anaphase

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

Spindle

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4

Anaphase

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

1. What must occur before mitosis takes place?

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

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Cytokinesis

dsworth Center- New York State Department of Health

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In plant cells (below right), cytokinesis (division of the cytoplasm) involves construction of a cell plate (a precursor of the new cell wall) in the middle of the cell. The cell wall materials are delivered by vesicles derived from the Golgi. The vesicles join together to become the plasma membranes of the new cell surfaces. Animal cell cytokinesis (below left) begins shortly after the sister chromatids have separated in anaphase of mitosis. A ring of microtubules assembles in the middle of the cell, next to the plasma membrane, constricting it to form a cleavage furrow. In an energy-using process, the cleavage furrow moves inwards, forming a region of separation where the two cells will separate.

Cleavage furrow

Constriction by microtubules

Animal cell

Plant (onion) cells

Cleavage furrow

Cytokinesis in an animal cell

Cell plate forming

Cytokinesis in a plant cell

2. Summarise what happens in each of the following phases:

(a) Prophase:

(b) Metaphase:

(c) Anaphase:

(d) Telophase:

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3. (a) What is the purpose of cytokinesis?

(b) Describe the differences between cytokinesis in an animal cell and a plant cell:

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54 Modelling Mitosis Key Idea: Using pipe cleaners to model the stages of mitosis helps to visualise and understand the

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process and its end result.

ffStudents 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.

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

Photo 1

A:

A

B:

C

C:

B

(b) Why are there two copies of the centrosomes?

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

Photo 2

A

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

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

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

Photo 3

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

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

Photo 4

6. Photo 4 shows the completion of mitosis.

(a) How many cells are formed?

(b) How many chromosomes are in each cell? Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

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55 Putting it all Together: Metabolism!

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

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of reactions.

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.

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

Light

CO2

(b)

(a)

H2O

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:

(c)

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

(d)

Alcohol or lactic acid

(f)

(e)

(j)

(g)

(i)

(h)

(l)

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CoA

(m)

(k)

TEST

H2O

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3. Use the word lists to complete the paragraphs below: (a) Word list: ATP, carbon, energy, glucose, photosynthesis.

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______________ 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.

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.

(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 ___________.

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

Organelle:

Processes that occur here:

Process that occurs here:

1)

Equation for process:

2)

Organelle:

Process that occurs here:

Plant cell

Equation for process:

Organelle:

Process that occurs here:

DNA molecule

(o)

M

(n)

S

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G1

Enzyme

G2

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56 What You Know So Far: Cell Division

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The cell cycle and mitosis

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

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

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

REVISE

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57 NCEA Style Question: Cell Division

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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?

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

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58 KEY TERMS AND IDEAS: Cell Division

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

A The process of producing two identical copies of DNA from one original DNA molecule.

chromosome

B The sequence of events that a cell completes prior to and including its division.

cytokinesis

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

DNA replication

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.

mitosis

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

semi-conservative

F The process of nuclear division in cells.

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: ...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:

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. Stage

No. of cells

% of total cells

Interphase Mitosis Total

TEST

100

Estimated time in stage

Onion root tip cells

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The enzymes also proofread the DNA during replication...

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Genetic variation and change

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Achievement Standard

2.5

Key terms

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.

Sources of variation allele codominance

Achievement criteria and explanatory notes

cross (genetic cross)

Achievement criteria for achieved, merit, and excellence

crossing over

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

M

Demonstrate in-depth understanding of genetic variation and change: Provide reasons as to how or why genetic variation and change occurs.

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.

dihybrid inheritance dominant allele

genetic variation genotype

heterozygous homozygous incomplete dominance

independent assortment lethal allele

linked genes locus

meiosis

Kristian Peters

monohybrid inheritance

Explanatory notes: Genetic variation and change

mutation

Genetic variation and change involves the following concepts...

phenotype

Activity number

recessive allele

c

1

Sources of variation within a gene pool.

59 - 82

recombination

c

2

Factors that cause changes to the allele frequency in a gene pool.

87 - 95

trait

Biological ideas and processes relating to sources of variation within a gene pool

Activity number

Gene pools bottleneck effect

c

i

Mutation as a source of new alleles.

founder effect

c

ii

Independent assortment, segregation, and crossing over during meiosis.

65  - 69

gene pool

c

iii

Monohybrid inheritance to show the effect of codominance, incomplete dominance, lethal alleles, and multiple alleles.

70 - 78

c

iv

Dihybrid inheritance with complete dominance.

79 82

c

v

The effect of crossing over and linked genes on dihybrid inheritance.

Select biological ideas and processes from...

migration (=gene flow) natural selection variation

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genetic drift

60 62 - 64

Biological ideas and processes relating to factors affecting allele frequencies within a gene pool

80 81 82

Activity number

Select biological ideas and processes from...

i

Natural selection (differential survival and reproduction of individuals as a result of differences in phenotype).

c

ii

Migration (transfer of alleles or genes from one population to another).

c

iii

Genetic drift (random fluctuations in the numbers of gene variants in a population).

c

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What you need to know for this Achievement Standard Sources of variation within a gene pool Activities 59 - 86

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

Marc King

Jeff Podos

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

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

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

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 Mendel’s principles of inheritance and explain their importance to our understanding of heredity and evolution.

c

Define the terms dominant allele, recessive allele, lethal allele, codominant alleles, incomplete dominance, and multiple alleles.

c

Define genotype, phenotype, monohybrid cross, dihybrid cross, F1 generation, F2 generation and demonstrate correct use of these terms in studies of inheritance.

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.

c

Solve problems involving dihybrid crosses of unlinked, autosomal genes for two independent traits showing complete dominance.

c

Describe and explain the effect of crossing over and linked genes on dihybrid inheritance.

Factors causing changes to allele frequencies in gene pools Activities 87 - 98

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

Jeff Podos

Define the term gene pool and allele frequency. Understand that a change in the allele frequencies of a population over time is evolution.

c

Recall the sources of variation in gene pools as a result of mutation and sexual reproduction.

c

List factors that a cause change in allele frequencies in a gene pool, including mutation, natural selection, gene flow (migration), and genetic drift.

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

How gene flow (migration) may affect allele frequencies. Explain how gene flow tends to reduce the genetic differences between populations.

c

Define genetic drift and explain its consequences to allele frequencies in populations. Using examples, explain when genetic drift is most important.

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

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.

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59 Alleles Key Idea: Eukaryotes generally have paired chromosomes. Each chromosome contains many genes,

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and each gene may have a number of versions called alleles.

Homologous chromosomes

ffIn 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.

3

4

5

4

ffThe 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). Having two different versions of gene A is a heterozygous condition. Only the dominant allele (A) will be expressed.

When both chromosomes have identical copies of the dominant allele for gene B the organism is homozygous dominant for that gene. When both chromosomes have identical copies of the recessive allele for gene C the organism is said to be homozygous recessive for that gene.

1

3

1

2

5

2

a

A

B

B

c

c

Maternal chromosome originating from the egg of this person's mother.

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).

Genes occupying the same position or locus on a chromosome code for the same characteristic (e.g. ear lobe shape). Paternal chromosome originating from the sperm of this person's father.

(a) Heterozygous:

(b) Homozygous dominant:

(c) Homozygous recessive:

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1. Define the following terms describing the allele combinations of a gene in a diploid sexually reproducing organism:

2. For a gene given the symbol ‘A’, identify the alleles present in an organism that is:

(a) Heterozygous:

(b) Homozygous dominant:

3. What is a homologous pair of chromosomes?

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(c) Homozygous recessive:

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60 Sources of Variation Key Idea: Variation may come from changes to the genetic material (mutation), through sexual

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reproduction, and as a result of the effects of the environment.

Mutations

Sexual reproduction

Mutations are changes to the DNA Changes to the DNA modify existing genes. Mutations can create new alleles.

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.

Mutation: Substitute T instead of C

Original DNA Mutant DNA

Genotype

Phenotype

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).

The genetic make up of the individual is its genotype. It determines the genetic potential of the individual.

Genes interact to influence the phenotypic result.

Environmental factors

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.

(a) Mutation:

(b) Genotype:

(c) Phenotype:

2. What factors determine the phenotype?

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1. Define the following terms:

3. How could two individuals with the same genotype have a different phenotype?

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Why is variation important?

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ffVariation refers to the diversity of phenotypes or genotypes

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within a population or species. Variation arises through mutation (the source of new alleles) and sexual reproduction (which shuffles alleles into new combinations).

ffVariation 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.

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.

ffVariation in species that reproduce asexually is generated by mutation and sometimes (as in bacteria) by gene transfers.

USDA

ffVariation is important in providing adaptable defences against

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.

These diagrams show how three beneficial mutations could be combined through sexual or asexual reproduction. Variation by sexual reproduction

Variation by asexual reproduction

Variation from recombining alleles

A

B

Clones

Mutation D

D

Mutation B

Mutation A

D

AB

AB

AB

AB

AB

ABC

F

Mutation C

ABC

D

DE

D

Mutation E

C

AB

D

Mutation F

ABC

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.

D

DEF

DEF

DEF

DEF

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.

(b) Why is variation important in a changing environment?

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4. (a) What is variation and how does it arise in sexually reproducing organisms?

5. How could a favourable mutation spread through different bacterial lineages?

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61 Examples of Genetic Variation

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86

Key Idea: Genetic variation may be continuous as a result of quantitative traits (e.g. leaf length) or it

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may be discontinuous as a result of qualitative traits (e.g. AB blood type).

Phenotypes exhibit variation

ffA 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.

ffTraits can be quantitative (they show measureable variation) or qualitative (they show variation that falls into non-numerical descriptive categories).

Albinism is qualitative trait

Quantitative traits

ffQuantitative traits are measurable phenotypes

ffWithin 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).

Frequency

that depend on the actions of many genes and the environment. These traits vary over a range, producing a large number of phenotypic variants.

ffExamples 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.

Low

Measureable phenotypic range

High

Grain yield in corn

Growth rate in piglets

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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).

Milk production in cattle

1. Using examples, explain the difference between continuous and discontinuous variation:

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Qualitative traits

ffQualitative traits are determined by only one gene or two genes.

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As a result, there are only a few phenotypic variations possible and individuals fall into discrete categories.

Frequency of blood types in New Zealand

40

Frequency (%)

ffFor 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.

ffOther examples of qualitative traits include:

• 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.

30 20 10

• 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.

0

A

B AB Blood type

O

• 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.

Both 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.

Single comb rrpp

Walnut comb R_P_

Pea comb rrP_

Rose comb R_pp

CRCR

CWCW

2. (a) Why do quantitative traits show continuous variation?

(b) Why do qualitative traits show discontinuous variation?

3. Identify each of the following phenotypic traits as continuous (quantitative) or discontinuous (qualitative): (a) Wool production in sheep: (d) Albinism in mammals: (b) Hand span in humans:

(e) Body weight in mice:

(c) Blood groups in humans:

(f) Flower colour in snapdragons:

4. The graph on the right shows leaf length in ivy.

(a) Does this represent a quantitative or qualitative trait?

(b) Explain your answer:

Leaf length in ivy

16

Number of leaves

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14 12 10 8 6 4 2 0

15

30

45

60

75

Length of leaf (mm)

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62 The Role of Mutations in Populations Key Idea: Changes to the DNA sequence are called mutations. Mutations are the ultimate source of

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new genetic information, i.e. new alleles.

ffMutations 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.

Normal RBC

Harmful mutations are usually eliminated from the gene pool because their effects on the phenotype are detrimental to survival and reproduction.

Sickled RBC

ffA mutation may affect a single gene (gene mutation) or large blocks of genes (chromosomal mutation).

ffMutations can arise naturally through copying errors during DNA replication or the can be induced by chemical or physical agents called mutagens.

OpenStax college cc3.0

ffBeneficial mutations spread through a gene pool.

The 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.

Arg

Spikes

G C T

Arg

Mutates to

CDC

C G A

G C C C G G

Normal sequence

Mutated sequence

Harmful mutations

Beneficial mutations

Silent mutations

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.

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.

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.

1. (a) Define the term mutation:

(b) How do mutations introduce new alleles into a population?

2. Distinguish between harmful, beneficial, and silent mutations:

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63 Mutations Can Produce New Alleles Key Idea: A mutation to the gene coding for the connexin 26 protein produces a new allele. The

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phenotype for the new allele is congenital deafness.

ffRecall that mutations can produce new alleles. A single

mutation to the gene coding for a protein called connexin 26 produces a new allele.

ffIf a person inherits two copies of the mutation (one from

each parent) they suffer from a form of genetic hearing loss called NSRD.

ffCarriers (individuals with only one copy of the mutated

The image above shows a single connexin 26 protein. Connexin 26 is found in many areas of the body including the inner ear.

gene) are unaffected.

ffThe 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.

Polypeptide formed (amino acid sequence)

Leu

2

1

Gly

Gly

Val

The code is read in 3-base sequences called codons. One codon codes for one amino acid.

42

28

The gene that codes for the protein connexin 26 is on chromosome 13.

Asn

C T G G G G G G T G T G A A C

Normal DNA sequence

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.

Mutation causes a short polypeptide chain

Polypeptide formed

3

Connexin 26 polypeptide chains

Leu

Gly

Val

28

Stop

42

G

C T G G G G G T G T G A A C A

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.

Mutated sequence

4

TGA codes for stop

The deletion creates a recessive allele. A person is deaf when they carry two recessive alleles.

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1. (a) Identify the type of mutation that produces the connexin 26 mutation:

(b) What phenotype does the connexin 26 mutation produce?

(c) How does the mutation stop connexin 26 from carrying out its biological role?

2. Why has the connexin 26 mutation not been eliminated from the population?

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64 Somatic and Gametic Mutations

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90

Key Idea: Gametic mutations affect egg and sperm cells, these mutations are inherited. Somatic

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mutations affect body cells, they are not inherited.

ffGametic cells are the reproductive (sex) cells of an organism (the

egg and sperm). Mutations occurring in these cells are called germline or gametic mutations.

ffSomatic cells (body cells) are all the remaining cells. Mutations to these cells are called somatic mutations.

ffOnly gametic mutations will be inherited. Somatic mutations are not inherited but may affect an organism in its lifetime (e.g. a cancer).

Mutant phenotype (gold colour)

Normal phenotype (red colour)

ffThe 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.

Somatic mutation

Sperm cell

Egg cell

Gametic mutation

Parental gametes

Gametic mutation to sperm

Embryo

Somatic mutation

Organism

None of the gametes carry the mutation

Gametes of offspring

1. Distinguish between somatic and gametic mutations:

2. Why are only gametic mutations inherited?

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Entire organism carries the mutation

Half of the gametes carry the mutation

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Patch of affected area

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65 Meiosis Key Idea: Meiosis is a reduction division that produces haploid cells for the purposes of sexual

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reproduction.

What is meiosis?

The process of meiosis

ffMeiosis is a special type of

cell division necessary for the production of sex cells (gametes) for the purpose of sexual reproduction.

ffDNA replication precedes meiosis. If genetic mistakes (mutations) occur here, they will be passed on (inherited by the offspring).

ffMeiosis involves a single

chromosomal duplication followed by two successive nuclear divisions, and it halves the diploid chromosome number.

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.

Crossing over has occurred.

Homologous chromosomes pair up. The pairs are separated.

ffAn overview of meiosis is shown

on the right. Meiosis occurs in the sex organs of plants and animals.

Meiosis produces variation

An intermediate cell forms with a replicated copy of each chromosome.

ffDuring meiosis, a process called

crossing over may occur when homologous chromosomes may exchange genes. This further adds to the variation in the gametes.

ffMeiosis 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.

ffRecall that each sister chromatid is

one half of a replicated chromosome. When they are separated, they are once again called chromosomes.

Chromatid

The chromosomes line up again. This time the chromatids are separated.

Haploid (N) gametes form. They have one complete set of chromosomes.

(b) Where does meiosis take place?

(c) Why are the gametes produced by meiosis haploid?

2. Describe how meiosis produces variation in the gametes:

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1. (a) What is the purpose of meiosis?

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66 Meiosis and Variation Key Idea: Meiosis produces variation. Independent assortment and crossing over are two important

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ways of introducing variation into the gametes formed during meiosis.

Meiosis produces variation

ffMeiosis creates genetic variation in the gametes as alleles are reshuffled into different combinations. This variation arises through crossing over and independent assortment.

ffBecause of meiosis, siblings with the same biological

parents can all appear very different, although there is often a family resemblance.

ffCrossing over refers to the mutual exchange of pieces

of chromosome (and their genes) between homologous chromosomes.

ffIndependent assortment is the random alignment and

distribution of homologous chromosomes to the gametes.

Crossing over and recombination

ffChromosomes replicate during interphase, before meiosis, to produce replicated chromosomes with sister chromatids held together at the centromere (see below).

ffWhen 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.

ffCrossing over results in the recombination of alleles (variations of the same gene) producing greater variation in the offspring than would otherwise occur. No crossing over

Crossing over

Sister chromatids

Recombined chromosomes

Recombined chromatids

Non-sister chromatids

Centromere

Non-sister chromatids

1. (a) What is crossing over?

Possible gametes with crossing over (parental types and recombinants)

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Possible gametes with no crossing over (only parental types)

(b) How does crossing over increase the variation in the gametes (and hence the offspring)?

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Independent assortment ffIndependent assortment is the random

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Oocyte

alignment and distribution of chromosomes during meiosis. It is an important mechanism for producing variation in gametes.

A

a

Replicated chromosomes

a

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A

B

ffThe law of independent assortment states

B

b

b

Meiosis

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.

A

Genotype AaBb

A

b

a

Intermediate cells

b

ffThis results in the production of 2x different

B

a

possible combinations (where x is the number of chromosome pairs).

A

A

a

B

a

a

Eggs

a

ffFor 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).

b

B

B

aB

aB

b

Ab

Ab

2. (a) What is independent assortment?

(b) How does it increase genetic variation in the gametes?

3. (a) Draw the other two gamete combinations not shown in the diagram above:

Gamete 4

Gamete 3

(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.

h i

j

k l

m n o p

Chromatid 1

a b c d e f g

h i

j

k l

m n o p

Chromatid 2

1

2

3

4

7

8

9

A B C D E F G

H I

J

K L M N O P

Chromatid 3

A B C D E F G

H I

J

K L M N O P

Chromatid 4

(a) Draw the gene sequences for the four chromatids (above), after crossing over has occurred at crossover point 2: (b) Which genes have been exchanged between the homologous chromosomes?

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Homologous chromosomes

a b c d e f g

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4


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67 Modelling Meiosis Key Idea: We can simulate crossing over, gamete production, and the inheritance of alleles during

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meiosis using ice-block sticks to represent chromosomes.

Chromosome # Phenotype

10 10 2 2

Trait

r

T

LB 2

LB 2

r

r

LB

LB

10

10

2

2

Your initials

T

t

t

LB LB LB LB

Homologous pair

10 10 10 10

t

T

t

LB LB LB LB 10 10 10 10

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r

r

r

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T

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t

LB

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t

LB

LB

r

T

10

(A)

(B)

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LB

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).

Paternal chromosome

LB

Homologous pair

10

r

Maternal chromosome

Paternal chromosome

Chromosome number

r

LB

Your genotype

t

10

T

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.

LB

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).

Tongue rolling

10

Genotype

LB

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).

Phenotype

TT, Tt tt RR, Rr rr

Handedness

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.

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.

Genotype

Tongue roller Non-tongue roller Right handed Left handed

2

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.

2

Background

LB LB LB LB 2

2

2

2

r

r

r

r

LB LB LB LB 2

2

2

2

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4. Randomly align the homologous chromosome pairs to simulate alignment on the metaphase plate (as occurs in metaphase I). Simulate anaphase I by separating chromosome pairs. For each group of four sticks, two are pulled to each pole. t

T

t

t

T

t

T

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T

LB LB

LB LB

LB LB

LB LB

10 10

10 10

10 10

10 10

r

r

r

LB LB 2

r

r

LB LB

2

2

r

r

LB LB

2

2

r

LB LB

2

2

2

5. Telophase I: Two intermediate cells are formed. If you have been random in the previous step, each intermediate cell will contain a mixture of maternal and paternal chromosomes. This is the end of meiosis 1. Now that meiosis 1 is completed, your cells need to undergo meiosis 2. Carry out prophase II, metaphase II, anaphase II, and telophase II. Remember, there is no crossing over in meiosis II. At the end of the process each intermediate cell will have produced two haploid gametes (below). Intermediate cell 1

t

Intermediate cell 2

t

T

T

r

r

10

Maternal chromosome

10

LB

Maternal chromosome

r

Paternal chromosome

LB

r

LB 2

LB 2

LB 2

LB 2

LB LB 10 10

Haploid (N) gamete

t

t

r

r

r

T

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10

LB 2

LB 2

LB 2

10

LB

LB

T

r

LB

10

LB

10

LB 2

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.

Trait

Handedness

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Phenotype

Genotype

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68 Mendel’s Laws of Inheritance Key Idea: Genetic information is inherited from parents in units called genes. Mendel's laws of

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inheritance govern how these genes are passed on to the offspring.

Particulate inheritance

Characteristics of both parents are passed on to the next generation as discrete entities (genes).

White

Purple

X

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.

Law of segregation

During meiosis, the two members of any pair of alleles segregate unchanged and are passed into different gametes.

Generation 1

X

The offspring are inbred (self pollinated) Generation 2

Homologous pair of chromosomes, each has a copy of the gene on it (A or a)

A A

a

a

Oocyte

Meiosis

These gametes are eggs (ova) and sperm cells. The allele in the gamete will be passed on to the offspring.

A

A

Egg

a

Egg

a

Egg

Law of independent assortment

Egg

Oocyte

A A

Intermediate cell

B b

Intermediate cell

b

a

Genotype: AaBb

b

b

a

B

a

A

b

Ab

a

B

A A

A

a

b

Ab

Eggs

B

a

B

B

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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).

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.

Parent plants

aB

aB

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.

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? WEB

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69 Dominant and Recessive Traits Key Idea: Traits are controlled by genes, and can be inherited and passed from one generation to the

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next. A trait may be caused by a dominant or recessive allele.

Traits are inherited

ffTraits are particular variants of a phenotypic (observed physical) character. For example, a phenotype is eye colour, a trait is blue eye colour.

ffTraits 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).

ffDominant 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.

ffIn contrast, a recessive trait will only be expressed when an individual has two copies of the recessive allele.

Mendel's experiments

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.

Mendel studied seven phenotypic characters of the pea plant: • Flower colour (violet or white) • Pod colour (green or yellow)

• Height (tall or short)

• Position of the flowers on the stem (axial or terminal) • Pod shape (inflated or constricted) • Seed shape (round of wrinkled) • Seed colour (yellow or green)

rr

RR

Wrinkled

F1

Rr

Rr

Rr

One of the experiments crossed true breeding round seed plants with true breeding wrinkled seed plants.

Round

Rr

Rr

Rr

F2

rr

RR

Rr

Rr

Rr

rr

RR

How can this be explained?

Mendel was able to explain his observations in the following way:

Rr

Rr

Rr

Out of the thousands of seeds produced, all were round, none were wrinkled.

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.

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P1

In this diagram R (round) is dominant to r (wrinkled).

ffTraits are determined by a unit, which passes unchanged from parent to offspring (we now know that these units are genes).

ffEach individual inherits one unit (gene) for each trait from each parent (each individual has two units).

ffTraits may not physically appear in an individual, but the units (genes) for them can still be passed to its offspring. © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

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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 á 1850 = 2.96). Place your answers in the spaces provided in the table below:

Possible phenotypes

Results

Seed shape

Wrinkled

Inflated

Stem length

Tall

Round

Recessive:

Wrinkled

Green Yellow TOTAL

2001 6022 8023

Dominant:

Green Yellow TOTAL

428 152 580

Dominant:

Axial Terminal TOTAL

651 207 858

Dominant:

Terminal

Pod shape

Constricted

Dominant:

2.96 : 1

Recessive

Recessive

Yellow

Flower position

Axial

1850 5474 7324

Yellow

Pod colour

Green

Ratio

Wrinkled Round TOTAL

Round

Seed colour

Green

Dominance

Dwarf

Constricted Inflated TOTAL

299 882 1181

Tall Dwarf TOTAL

787 277 1064

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Trait

Recessive

Dominant:

Recessive

Dominant:

Recessive

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70 The Monohybrid Cross Key Idea: The outcome of a cross depends on the parental genotypes. A true breeding parent is

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homozygous for the gene involved.

Monohybrid cross F1

Homozygous purple

Parents

Gametes:

PP

P

Monohybrid cross F2

Homozygous white

X

P

Heterozygous purple

Parents

pp

Heterozygous purple

X

Pp

p

Pp

p

Male gametes

P

P

p

p

F1

Genotypes: All Pp

Female gametes

Phenotypes: All purple

75% Pp - purple 25% pp - white

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).

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:

Monohybrid cross

Parents

Pp

X

Gametes:

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Homozygous white

Homozygous purple

Parents

pp

Male gametes

PP

Heterozygous purple

X

Pp

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Heterozygous purple

Monohybrid cross

Female gametes

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71 The Test Cross

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Key Idea: If an individual's genotype is unknown, it can be determined using a test cross.

ffIt 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.

ffThe 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 unknown parent (below). The test cross can be used to The basisofofthe a test cross determine the genotype of single genes or multiple genes. Parent 1 Unknown genotype (but with dominant traits)

?

?

?

?

Parent 2 Homozygous recessive genotype (no dominant traits)

X

a

b

a

b

The common fruit fly (Drosophila melanogaster) is often used to illustrate basic principles of inheritance because it has several genetic markers whose phenotypes are easily identified. Once such phenotype is body colour. Wild type (normal) Drosophila have 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. A. A homozygous recessive female (ee) with an ebony body is crossed with a homozyogous dominant male (EE).

B. A homozygous recessive female (ee) with an ebony body is crossed with a heterozygous male (Ee).

Female gametes

Male gametes

Female gametes

Male gametes

Cross A:

(a) Genotype frequency:

(b) Phenotype frequency:

100% Ee

100% yellow-brown

Cross B:

(a) Genotype frequency:

(b) Phenotype frequency:

50% Ee, 50% ee

50% yellow-brown, 50% ebony

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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:

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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72 Practising Monohybrid Crosses Key Idea: A monohybrid cross studies the inheritance pattern of one gene. The offspring of these

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crosses occur in predictable ratios.

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. Homozygous white

bb

b

Homozygous black

BB

Homozygous black

BB

X

Homozygous black

b

Parents

BB

X

B

Gametes

B

5. (a) 3. (a) Genotype Genotypefrequency: frequency:

Bb

Bb

Bb

Bb

100% Bb

1. (a) Genotype frequency:

(b) Phenotype frequency:

100% black

(b) Phenotype frequency: (b) Phenotype frequency:

Homozygous white

Offspring (F1)

bb

Heterozygous black

Bb

X

(c) Write a definition for the F1 generation:

(d) Which coat colour is dominant?

(e) Which is the dominant allele?

(f) Which coat colour is recessive?

4. (a) Genotype Genotypefrequency: frequency: 6. (a)

(g) Which is the recessive allele?

Phenotype frequency: (b) (b) Phenotype frequency:

Bb

Heterozygous black

X

4. 2. (a) (a) Genotype Genotypefrequency: frequency:

(b) Phenotype frequency: (b) Phenotype frequency: Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

Homozygous white

Bb

bb

Homozygous white

bb

X

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Heterozygous black

5. (a) Genotype Genotypefrequency: frequency: 7. (a)

(b) Phenotype frequency: frequency: (b) Phenotype

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73 Codominance in Alleles Key Idea: In codominance, neither allele is recessive and both alleles are equally and independently

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expressed in the heterozygote.

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:

Red bull

White cow

CRCR

CWCW

ffHuman AB blood group, which is the result of two

Parents

alleles: A and B, both being equally expressed.

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.

1. A white bull is mated with a roan cow (right):

(a) Fill in the spaces to show the genotypes and phenotypes for parents and calves:

(b) What is the phenotypic ratio for this cross?

(c) How could a cattle farmer control the breeding so that the herd ultimately consisted of only red cattle?

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. (a) Fill in the spaces (right) to show the genotype and phenotype for parents and calves.

(b) Which bull serviced the cows? red or roan (delete one)

3. What is the phenotypic ratio for a codominant cross of two heterozygous parents (e.g. a cross between two roan cattle):

CR

CR

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CW

Offspring

CRCW

CRCW

CRCW

CRCW

Roan

Roan

Roan

Roan

White bull

Roan cow

Roan cow

Unknown bull

?

Red WEB

CW

Gametes

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ffRoan colouration in horses and cattle. Reddish

Roan

Red

Roan

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74 Codominance in Multiple Allele Systems Key Idea: The human ABO blood group system is a multiple allele system. Alleles A and B are

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codominant and allele O is recessive. The possible phenotypes are A, B, AB, and O.

ffOn 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.

Blood group (phenotype)

Possible genotypes

O

OO AA AO

ffThe 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.

A B

ffIf a person has the AO allele combination, their blood group is A,

Frequency in NZ

49% 40% 9%

AB

because O is recessive to A. AA also produces blood group A.

2%

1. Use the information above to complete the table for the possible genotypes for blood group B and group AB.

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:

Blood group: AB

Parental genotypes

X

AB

A

Gametes

Blood group: AB

Cross 1

B

Blood group: O

OO

AB

A

Blood group: O

Cross 2

X

OO

B

Possible fertilisations

Blood groups

AA

AB

AB

BB

A

AB

AB

B

Blood group: AB

Parental genotypes

Blood group: A

Cross 3

AB

Gametes

Possible fertilisations Children's genotypes Blood groups

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X

AO

Blood group: A

Blood group: B

Cross 4

AA

BO

X

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Children's genotypes

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104 Blood group

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3. A woman is heterozygous for blood group A and the man has blood group O. (a) Give the genotypes of each parent (fill in spaces on the diagram on the right).

O

X

Parental genotypes

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Blood group

A

Determine the probability of:

(b) One child having blood group O:

(c) One child having blood group A:

(d) One child having blood group AB:

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.

(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) State with reasons whether the man can be correct in his claim:

Gametes

Possible fertilisations Children’s genotypes Blood groups

Blood group

Blood group

A

X

Parental genotypes

Gametes

Child’s genotype

Blood group

B

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.

O

Pedigree chart key

Male

(a) State the sex of the children:

Female

Marriage

(b) Explain why child 1 has two possible genotypes and child 2 has only one possible genotype:

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B

AB

B

AB

Child 1

Child 2

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75 Incomplete Dominance Key Idea: In incomplete dominance, the action of one allele does not completely mask the action of

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the other. Neither is dominant and an intermediate phenotype is produced.

ffIn 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. ffIn crosses involving incomplete dominance, the phenotype and genotype ratios are identical.

Example 1

Example 2

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.

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.

White flower

Crimson flower

Parents

CWCW

CRCR

Gametes

CR

Red flower

CR

CW

White flower

CRCR

CW

CR

CrCr

CR

Cr

Parents

Gametes

Cr

Possible fertilisations

Offspring

Offspring

C C R

Phenotype

C C

W

R

Magenta

W

Magenta

C C R

W

Magenta

C C R

W

Magenta

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.

C C

C C R

R

r

Pink

C C

Pink

R

r

Pink

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. Gametes from male

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r

Gametes from female

Gametes from female

(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:

R

Pink

Gametes from male

C C

r

3. A second breeder wanted to prove that pink snapdragon flowers were heterozygous. Briefly explain how this could be done:

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76 Lethal Alleles Key Idea: Lethal alleles are mutations of a gene that produce a non-functional gene product which

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may affect an organism's survival.

ffLethal 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.

ffSome 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.

AYA

AY

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).

A YA

A

Gametes

AY

A

Possible fertilisations

Dead

¼ AYAY

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

Yellow mouse

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.

Yellow

Yellow

½ AYA

Non-yellow

¼ AA

Manx cats are born without a tail

Normal cats are born with a tail

Normal cat

Manx cat

(ML)

(a) Complete the Punnett square for the cross:

(b) State the phenotype ratio of Manx to normal cats and explain why it is not the expected 3:1 ratio:

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Gametes from male

Gametes from female

1. In Manx cats, the allele for taillessness 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).

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:

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77 Problems Involving Monohybrid Inheritance Key Idea: Monohybrid crosses involve the inheritance of a single gene. The problems below will give

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you an opportunity to practise solving Mendelian monohybrid crosses.

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:

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:

3. A recessive allele, a, is responsible for albinism, an inability to produce or deposit melanin in tissues. Humans and a variety of other animals can exhibit this phenotype. In each of the following cases, determine the possible genotypes of the mother and father, and of their children:

(a) Both parents have normal phenotypes; some of their children are albino and others are unaffected:

(b) Both parents are albino and have only albino children:

(c) The woman is unaffected, the man is albino, and they have an albino child and three unaffected children:

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4. Chickens with shortened 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:

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:

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78 Dihybrid Cross

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Key Idea: A dihybrid cross studies the inheritance pattern of two genes. When these genes are on

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separate chromosomes, the offspring of these crosses occur in predictable ratios.

BBLL

Parents (P)

BL

Gametes

BL

BL

Homozygous white, long hair

bbll

X

BL

bl

bl

bl

bl

Possible fertilisations

Offspring (F1)

bL

BL

BBLL

BBLl

BbLL

BbLl

Bl

BBLl

BBll

BbLl

Bbll

bL

BbLL

BbLl

bbLL

bbLl

bl

BbLl

Bbll

bbLl

bbll

BBLL BbLL BBLl BbLl

1 BBll 2 Bbll

1 bbLL 2 bbLl

1 bbll

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bl

Bl

The offspring can be arranged in groups with similar phenotypes

Genotype 1 2 2 4

BbLl

BL

Possible fertilisations

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A total of 9 offspring with one of 4 different genotypes can produce black, short hair

A total of 3 offspring with one of 2 different genotypes can produce black, long hair

A total of 3 offspring with one of 2 different genotypes can produce white, short hair Only 1 offspring of a given genotype can produce white, long hair

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Gametes: Each parent is homozygous for both traits so only one type of gamete is produced from each parent.

F1 offspring: The first generation offspring are all the same genotype. The notation F1 is used for the hybrid offspring of genetically distinct parents.

Female gametes

Offspring (F2)

Male gametes

X

BbLl

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).

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.

Each of the 16 animals represents the possible zygotes formed by different combinations of gametes coming together at fertilisation.

Phenotype

9 black, short hair

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Homozygous black, short hair

3 black, long hair

3 white, short hair

1 white, long hair

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Cross No. 1

White, short hair

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The dihybrid cross on the right has been partly worked out for you. You must determine:

Heterozygous black, short hair

1. The genotype and phenotype for each animal (write your answers in its dotted outline).

2. Genotype ratio of the offspring:

BbLl

Parents

Gametes

BL

Bl

bL

bl

bbLl

bL

bL

bl

bl

Female gametes

Offspring

Possiblle fertilisations

X

BL

Bl

bl

bL

3. Phenotype ratio of the offspring:

Male gametes

bL bL bl

bl

Cross No. 2

For the dihybrid cross on the right, determine:

Homozygous white, short hair

Parents

1. 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).

Gametes

3. Genotype ratio of the offspring:

Offspring

b b LL

X

Black, long hair

Bb l l

Female gametes

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4. Phenotype ratio of the offspring:

Male gametes

Possiblle fertilisations


79 Inheritance of Linked Genes

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Key Idea: Linked genes are genes found on the same chromosome and tend to be inherited together.

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Linkage reduces the genetic variation in the offspring.

Overview of linkage

Parent 1 (2N)

Parent 2 (2N)

Dark body, blue eyes AaBb

A

a

B

b

White body, white eyes aabb

Chromosomes before replication

Genes are linked when they are found on the same chromosomes. In this case A (body colour) and B (eye colour) are linked.

a

a

b

b

a

a

a

a

b

b

b

b

Chromosomes after replication

A

A

a

a

X

B

B

b

b

Each replicated chromosome produces two identical gametes. One of each is shown here.

Meiosis

A

a

B

b

Gametes

A

a

A

a

B

b

B

b

AaBb

AaBb

Dark body, blue eyes

Offspring

a

a

b

b

a

a

a

a

b

b

b

b

aabb

aabb

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White body, white eyes Possible Offspring Only two kinds of genotype combinations are possible. They are they same as the parent genotype.

ffLinked 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).

ffLinkage 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).

ffIf 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).

ffNote that in the hypothetical example above there are only two possible genotype outcomes, both the same as the parent type.

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An example of linked genes in Drosophila

An example of linked genes in Drosophila

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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.

1. What is linkage and what is its effect on the inheritance of genes?

The genes for wing shape and body colour are linked (they are on the same chromosome)

Wild type female

Mutant male

Phenotype

Straight wing Grey body

Curled wing Ebony body

Genotype

Cucu Ebeb

cucu ebeb

Parent

Cu

Eb

cu

eb

cu

eb

cu

eb

Linkage

Meiosis

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)

3. List the possible genotypes in the offspring (right) if genes Cu and Eb had been on separate chromosomes:

4. If the female had been homozygous for the dominant wild type alleles (CuCu EbEb), state:

(a) The genotype(s) of the F1:

(b) The phenotype(s) of the F1:

Offspring

Sex of offspring is irrelevant in this case

The genotype(s) of the F1:

The phenotype(s) of the F1:

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Gametes from male

Gametes from female

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:

Contact Newbyte Educational Software for details of their superb Drosophila Genetics software package which includes coverage of linkage and recombination. Drosophila images Š Newbyte Educational Software.


80 Recombination and Dihybrid Inheritance

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Key Idea: Recombination is the exchange of alleles between homologous chromosomes as a result

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of crossing over. Recombination increases the genetic variation in the offspring.

Overview of recombination

Parent 1 (2N)

Parent 2 (2N)

Dark body, blue eyes AaBb

A

a

B

b

White body, white eyes aabb

Chromosomes before replication

a

a

b

b

Chromosomes after replication

Crossing over has occurred between these chromosomes

A

A

a

a

B

b

B

b

X

a

a

a

b

b

b

Because this individual is homozygous, if a these genes cross over there is no b change to the allele combination.

Meiosis

A B

A

a

b

a

Gametes

b

B

A

a

a

a

B

b

b

b

a

a

a

a

b

b

b

b

A

a

a

a

b

b

B

b

Offspring

AaBb Dark body, blue eyes

Aabb Dark body, white eyes

aabb White body, white eyes

aaBb White body, blue eyes

Recombinant offspring

These two offspring exhibit unexpected allele combinations. They can only arise if one of the parent's chromosomes has undergone crossing over.

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Non-recombinant offspring

These two offspring exhibit allele combinations that are expected as a result of independent assortment during meiosis. Also called parental types.

ffDuring crossing over, alleles on homologous chromosomes can be exchanged and new associations of alleles are formed in the gametes.

ffThe 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.

ffThe 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.

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An example of recombination in Drosophila An example of recombination In the female parent, crossing over occurs between genes for wing shape and body colour

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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.

Recombination produces variation 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. 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:

Wild type female

Mutant male

Phenotype

Straight wing Grey body

Curled wing Ebony body

Genotype

Cucu Ebeb

cucu ebeb

Parent

Cu

Eb

cu

eb

cu

eb

cu

eb

Linkage

Meiosis

Gametes from female fly (N) Crossing over has occurred, giving four types of gametes

Gametes from male fly (N) Only one type of gamete is produced in this case

2. Complete the recombination diagram right, adding the gametes in the ovals and offspring genotypes and phenotypes in the boxes:

3. How does recombination increase the amount of genetic variation in offspring?

Offspring

Non-recombinant offspring

Recombinant offspring

Sex of offspring is irrelevant in this case

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.

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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:

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.

(a) Which phenotypes were the recombinants?

(b) Give their genotypes:

(c) What does the low number of recombinants tell you about how often the alleles cross over and their relative positions on the chromosome?

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81 Problems Involving Dihybrid Inheritance Key Idea: Dihybrid crosses involve the inheritance of two genes. The problems below will give you an

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opportunity to practise solving dihybrid crosses.

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:

(a) A black short haired (BBLl) male is crossed with a black long haired (Bbll) female. Determine the genotypic and phenotypic ratios of the offspring:

Genotype ratio:

Phenotype ratio:

(b) A tabby, short haired male (TtbLl) is crossed with a blotched tabby, short haired (tbtbLl) female. Determine ratios of the offspring:

Genotype ratio:

Phenotype 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 curve their thumbs, but have no mid-digit hair:

(b) Complete the Punnett square (right) to show the possible genotypes from a cross between two individuals heterozygous for both alleles:

(c) State the phenotype ratios of the F1 progeny:

(d) What is the probability one of the offspring would have mid-digit hair?

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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.

(a) Determine which alleles responsible for the phenotypes observed are dominant and which are recessive:

(b) Determine the genotype of the original plant with orange striped flowers:

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82 What You Know So Far: Sources of Variation Meiosis and variation

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

HINT: Explain how independent assortment and crossing over during meiosis produces variation.

Mutation as a source of new alleles

HINT: Define the term allele and explain how mutations produce new alleles.

Dihybrid inheritance

HINT: Include definitions and expected genotype and phenotype ratios for monohybrid inheritance patterns, including crosses involving codominance, incomplete dominance, and lethal alleles.

HINT: Explain dihybrid inheritance and explain how linkage and recombination affect variation.

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Monohybrid inheritance

REVISE


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83 NCEA Style Question: Sources of Variation

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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 definition for allele, heterozygote, and homozygote A description of mutation and why some detrimental alleles are retained within a population while others are not 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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• • • •

TEST

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2. Meiosis is a special type of cell division. It produces gametes (sex cells) for sexual reproduction. (a) Discuss how meiosis produces genetic variation. Your answer should include:

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• A description of meiosis and the type of cells produced • An explanation of how genetic variation is increased through segregation, independent assortment, and crossing over during meiosis • An explanation of why the daughter cells are genetically different to the parent cells

You may use diagrams to illustrate your answer and more paper if you need it.

(b) Explain linkage and discuss its effect on variation in the offspring, contrasting it with recombination:

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84 NCEA Style Question: Inheritance of Alleles

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1. Snapdragons have red, pink, or white flowers. Roan coats in Shorthorn cattle is a mix of red and white hairs.

(a) Define monohybrid cross:

(b) Use the inheritance patterns of codominance and incomplete dominance to explain these results (you may use extra paper if required):

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?

(b) What sort of cross is this? Justify your answer:

(c) Is there evidence of crossing over in the offspring?

(d) Explain your answer:

(e) Determine the genotypes of the offspring:

TEST

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Straight wing: Cucu Curled wing: cucu Grey body: Ebeb Ebony body: ebeb

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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 Relevant genotypes and phenotypes The use of a Punnett squares and the correct use of the allele notation given 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.

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85 KEY TERMS AND IDEAS: Sources of Variation

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

A Observable characteristics in an organism.

codominance

B Allele that will only be expressed in the absence of the dominant allele is called this.

dihybrid cross

C A genetic cross involving the inheritance of two genes.

dominant

genotype

D Possessing two different alleles of a particular gene, one inherited from each parent. E Sequences of DNA occupying the same gene locus (position) on different, but homologous, chromosomes.

heterozygous

F The process of double nuclear division (reduction division) to produce four nuclei, each containing half the original number of chromosomes (haploid).

incomplete dominance

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.

meiosis

H Exchange of alleles between homologous chromosomes as a result of crossing over.

mutation

phenotype recessive

recombination

I Allele that is expressed irrespective of the other allele is called this.

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. K The allele combination of an organism.

L An inheritance pattern in which two different alleles for a trait are equally and independently expressed.

2. The allele for wrinkled seeds in pea plants is considered recessive because:

3. True breeding individuals:

A It is not expressed in the F2 generation

B It is not expressed in the heterozygote.

B Always have the same coat or flower colour.

C Individuals who have the allele are less likely to pass genes on to the next generation.

C Are homozygous for a particular allele.

D Round peas are smaller than wrinkled peas.

D Always produce heterozygous offspring.

A Only ever breed with others of the same kind.

4. Use lines to match the statements in the table below to form complete sentences: Mutations are the ultimate …

… of a gene.

Alleles are variations …

… dominant or recessive.

A person carrying two of the same alleles (one on each homologous chromosome) …

… for the gene, they are heterozygous.

If the person carries two different alleles …

… expressed if it is homozygous.

… is said to be homozygous.

Alleles may be …

… source of new alleles.

A dominant allele …

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… is always expressed whether it is homozygous or heterozygous.

A recessive allele is only …

5. Complete the following crosses for pea plants, including genotype and phenotype ratios: Round seeds (R) are dominant to wrinkled seeds (r):

(a) RR x Rr:

TEST

(b) Rr x rr

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86 Processes in Gene Pools Key Idea: The proportions of alleles in a gene pool can be altered by processes that increase or

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decrease genetic variation.

ffA 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.

ffThe 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.

ffFour 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). Immigration: New genes may be introduced.

aa

Mutations: Changes to the DNA sequence can create new alleles.

Aa

AA

Emigration: Genes may be lost.

Aa

A'A

AA

AA

Aa

Aa

Aa

AA

aa

Aa

Aa

aa

AA

Aa

aa

aa

Aa

AA

AA

Aa

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.

Deme 1

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.

Geographical barriers (e.g. mountains or rivers) isolate the gene pool and prevent regular gene flow between populations.

Aa

Aa

aa

Deme 2

Aa

aa

Aa

Aa

Aa

Aa

aa

Aa

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. Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

AA

Aa

Aa

aa

aa

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Aa

aa

aa

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. LINK

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122

1. Define the following terms: (a) Gene pool:

(b) Allele frequency:

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2. For each phase in the gene pool below, fill in the tables provided as follows (some have been done for you): (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). (c) For each of the above, work out the frequencies as percentages (bottom row of table):

Allele frequency =

No. counted alleles á Total no. of alleles x 100

Phase 1: Initial gene pool

Black

Black A

aa

Aa

AA

a

AA

No.

27

7

%

54

28

Aa

aa

Pale

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

Allele types Allele combinations

Phase 2: Natural selection

Two pale individuals died. Their alleles are removed from the gene pool.

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).

AA

Aa

These individuals (surrounded by small white arrows) are not counted for allele frequencies; they are dead! A

a

AA

Aa

Aa

aa

Aa

Aa

aa

Aa

Aa

AA

AA

Aa

AA

AA

No. No.

aa

Aa

Aa

AA

aa

aa

AA

Aa

Aa

aa

Aa

Aa

%%

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. Individuals coming into the gene pool (AA) are counted for allele frequencies, but those leaving (aa) are not. A

No. No. %%

a

AA

Aa

AA

Aa

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Phase 3: Immigration and emigration

This individual is leaving the population, removing its alleles from the gene pool.

aa

Aa

AA

Aa

Aa

Aa

AA

aa

AA

Aa

aa

AA

Aa

Aa

AA

AA

Aa

Aa

aa

Aa

Aa

AA

Aa

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87 How Natural Selection Works Key Idea: Evolution by natural selection describes how organisms that are better adapted to their

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environment survive to produce a greater number of offspring.

What is evolution?

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.

Darwin's theory of evolution by natural selection 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.

Overproduction

Variation

Populations produce too many young: many must die

Individuals show variation: some variations more favourable than others

We can demonstrate the basic principles of evolution using the analogy of a 'population' of M&M's sweets.

#1

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.

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.

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.

Natural selection

Natural selection favours the individuals best suited to the environment at the time

The blue sweets become more common...

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.

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#2

Inherited

Variations are inherited: the best suited variants leave more offspring

#3

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. Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

The variations (both favourable and unfavourable) are passed on to offspring. Each generation will contain proportionally more descendents of individuals with favourable characters.

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124 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.

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Variation, selection, and population change

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).

Red

Brown mottled

Red 2 spot

2. Selective predation: Brown mottled beetles are eaten by birds but red ones are avoided.

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.

1. What produces the genetic variation in populations? 2. (a) Define evolution:

(b) Identify the four principles of evolution by natural selection as proposed by Darwin:

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3. Using your answer 2(b) as a basis, explain how the genetic make-up of a population can change over time:

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88 Types of Natural Selection Key Idea: Natural selection is the differential survival of favourable phenotypes and their associated

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genotypes. It is an important cause of change in gene pools.

ffNatural selection is the differential survival of favourable

phenotypes, produced by their particular combinations of alleles.

Types of selection

ffStabilising selection: Selects against phenotypic extremes, reducing phenotypic variation in a population.

ffAs 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.

ffDirectional selection: The adaptive

phenotype shifts in one direction, favouring phenotypes at one extreme and establishing a new phenotypic norm.

ffOver time, natural selection may lead to a permanent change in the genetic makeup of a population.

ffDisruptive selection: Favours two

phenotypic extremes at the expense of intermediate forms. It creates two divergent phenotypic norms.

ffNatural 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.

The effects of different types of selection on a population

Stabilising selection

Number of individuals

Number of individuals

Number of individuals

Phenotypic range

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Phenotypic range

Phenotypic range

Phenotypic range

Number of individuals

Number of individuals

Phenotypic range

Distribution after many generations

Diversification of habitat or resources

Phenotypic range

Number of individuals

Number of individuals

Phenotypic range

Distribution after a few generations

Disruptive selection

Changing climate or resources

Number of individuals

Initial distribution

Stable climate and resources

Number of individuals

Typical environment

Directional selection

Phenotypic range

Phenotypic range

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?

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89 Modelling Natural Selection

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Key Idea: Natural selection acts on phenotypes. Those individuals better suited to an environment will

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leave more offspring in the next generation. This can be illustrated using a simple model.

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.

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). 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.

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.

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).

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. 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.

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. 8. Repeat steps 4 to 7 for generation 2, and 3 more generations (5 generations in total or more if you wish).

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?

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?

1

Black

Number

Proportion

2

Number

Proportion

3

Number

Proportion

4

Number

Proportion

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Proportion WEB

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White

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90 Gene Pool Model

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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.

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91 Natural Selection in Pocket Mice Key Idea: The need to blend into their surroundings to avoid predation is an important selection

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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.

ffThe 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.

ff107 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:

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

8.5

15

FRA

9

39

TIN

9

39

TUL

9.5

25

POR

12

34.5

BLK

V

1. (a) What is the genotype(s) of the dark coloured mice?

(b) What is the genotype of the light coloured mice?

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.

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See Appendix

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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?

(b) Suggest a cause for the pattern in 3(a). How do the phenotypes of the mice affect where the mice live?

(c) What are two exceptions to the pattern you have noticed in 3(a)?

(d) How might these exceptions have occurred?

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4. What type of selection appears to operating in each of the environments (dark and light rock)? Explain:

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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92 The Founder Effect Key Idea: The founder effect is the loss of genetic variation when a new colony is formed from a small

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number of individuals. The founder population may evolve differently to the parent population.

ffOccasionally, 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.

ffAs 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.

ffIt may be possible for certain alleles to be missing altogether from the isolated population (the allele has been lost). Aa

AA

Aa

Aa

AA

Aa

Aa

Aa

AA

Aa

aa

Aa

aa

Aa

AA

Aa

Aa

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

This population may not have the same allele frequencies as the mainland population.

AA

Aa

AA

Aa

AA

AA

Aa

aa

Aa

aa

Aa

Aa

Aa

Aa

AA

Aa

aa

Aa

AA

Aa

Aa

Island population

Mainland population

1. Compare the mainland population to the island population (use the spaces in the tables below): (a) Count the phenotype numbers for the two populations (i.e. the number of black and pale beetles). (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.

Mainland population

Actual numbers

Calculate %

Phenotype frequencies

Black

Actual numbers

Pale

Allele A

Allele A

Allele a

Allele a

Total

AA

Aa

aa

Phenotype frequencies

Allele frequencies

Total

2. How are the allele frequencies of the two populations different?

Calculate %

Black

Pale

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Allele frequencies

Colonising island population

AA

Aa

aa

3. Describe how organisms could become isolated from a parent population:

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Microgeographic isolation in garden snails

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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. 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).

Block A

Block B

1

3

5

4

2

6

Road (not to sclae)

7

1

14

13

Source: Evolution, Vol 29, No. 3, 1975

8

2

3

15

9

10

4

11

5

11

12

12

10

6

7

8

13

9

Building

Colony

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Block A

MDH-1 A %

39

39

36

42

39

47

32

42

44

42

44

50

50

58

75

Block B

Snail colony (circle size is proportional to colony size).

MDH-1 A %

81

61

75

68

70

61

70

60

58

61

54

54

47

MDH-1 a %

MDH-1 a %

4. Complete the table above by filling in the frequencies of the MDH-1 a allele:

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5. Suggest why these snail populations are effectively geographically isolated:

6. Both the MDH-1 alleles produce fully operative enzymes. Suggest why the frequencies of the alleles have become significantly different.

7. Identify the colony in block A that appears to be isolated from the rest of the block itself:

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93 Population Bottlenecks Key Idea: Population bottlenecks occur when population numbers and diversity fall dramatically.

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Although a population's numbers may recover, its genetic diversity may not.

ffA 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.

ffA 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.

ffFollowing 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.

ffBefore 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.

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.

High

High

Bottleneck

Low

Low

Genetic diversity

Population numbers

The effect of population bottlenecks on genetic diversity

Time

Large, genetically diverse population.

Population numbers are drastically reduced. Much of the genetic diversity is lost.

Population numbers increase again, but the larger population has low genetic diversity.

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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?

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Case study: Tasmanian devils 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.

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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.

60

South Island robin

50

S.L. Arden, D.M. Lambert 1997

Total number of unique [DNA] fragments

70

40 30 20

Black robin

10

0

0

100

200

300

400

500

600

Number of [DNA] fragments sampled

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).

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.

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:

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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:

6. (a) What has been the genetic consequence of bottleneck events in the Tasmanian devil population?

(b) How has this led to increased susceptibility to disease, specifically infectious cancer?

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94 Genetic Drift Key Idea: Genetic drift is the term for the random changes in allele frequency that occur in all

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populations. It has a much greater effect on the allele frequencies of small populations.

ffGenetic 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.

ffThe allele combinations in any generation are determined by what is passed on from the generation before.

ffAlleles may become lost from the gene pool altogether (frequency becomes 0%) or fixed as the only allele for the gene present (frequency becomes 100%).

ffGenetic 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. The genetic makeup (allele frequencies) of the population changes randomly over a period of time

Generation 1 A = 16 (53%) a = 14 (47%)

Generation 2 A = 15 (50%) a = 15 (50%)

Generation 3 A = 12 (41%) a = 18 (59%)

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

aa

Aa

Aa

Aa

Aa

aa

Killed

Fail to locate a mate due to low population density

Further chance events will affect allele frequencies in subsequent generations.

Fail to locate a mate due to low population density

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.

Computer simulation of genetic drift

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. 100

Breeding population = 2000

80

80

60

60

40

40

20

20

0

0

20

40

60

80

100

Generations

120

140

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.

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0

100

Breeding population = 200

Breeding population = 20

80 60

Allele lost from the gene pool

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Allele frequency ( %)

100

40 20

0

20

40

60

80

100

Generations

120

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.

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140

0

0

20

40

60

80

100

Generations

120

140

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%).

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1. (a) What is genetic drift?

(b) How does genetic drift differ from natural selection?

2. (a) What does it mean to say an allele has become fixed in the population? (b) What does it mean to say an allele has become lost from a population?

3. (a) Why is the effect of genetic drift more pronounced in small populations?

(b) Explain why genetic drift is an important process in the evolution of colonising (founder) 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. (a) Explain the likely impact of the founder effect and genetic drift on takahe populations relocated to offshore islands:

(b) Without any intervention, 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:

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95 What You Know So Far: Changes in Gene Pools Genetic bottleneck (the bottleneck effect), founder effect, and genetic drift

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

HINT: Explain these processes and their effect on allele frequencies.

Factors affecting allele frequencies in gene pools

HINT: List, include definitions, and state whether each factor increases or decreases genetic variation.

Natural selection

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HINT: Include definition and examples of types of natural selection.

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96 NCEA Style Question: Changes in Gene Pools

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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. (a) Define the term population bottleneck:

(b) Explain the nature of the population bottleneck experienced by the black robin:

(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:

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Photo: Dept of Conservation

TEST

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(a) Define the 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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Photo: Tony Wills, CC 2.5

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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 established.


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97 KEY TERMS AND IDEAS: Changes in Gene Pools

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1. Test your vocabulary by matching each term to its definition, as identified by its preceding letter code. bottleneck effect founder effect gene pool

A The process by which heritable traits become more or less common in a population through differential survival and reproduction. B The differences between individuals in a population as a result of genes and environment.

C The sum total of all alleles of all breeding individuals in a population at any one time.

genetic drift

D An evolutionary event in which a significant proportion of a population's alleles are lost.

migration

E The movement of individuals into and out of a population. Also called gene flow.

natural selection

F The loss of genetic variation when a new colony is formed by a very small number of individuals from a larger population.

variation

G The change in allele frequency in a population as a result of chance events. The effect is proportionally larger in small populations.

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: (a) Directional selection:

(b) Stabilising selection:

Environment: Stable climate and resources / steady trend in one direction / diversification in habitat or resources

(c) Disruptive selection:

Environment: Stable climate 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:

100

80

60

40

Population 2000

20

A

0

0

(b) What has happened at point A on the diagram?

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Environment: Stable climate and resources / steady trend in one direction / diversification in habitat or resources

Allele frequency (%)

20

40

Population 200 Population 20

60 80 Generations

100

120

140

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Gene expression

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Achievement Standard

2.7

Key terms

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.

Nucleic acids

base pairing rule DNA

Achievement criteria and explanatory notes

double helix

Achievement criteria for achieved, merit, and excellence

gene expression

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

M

Demonstrate in-depth understanding of gene expression: Provide a reason as to how or why biological processes affect gene expression.

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.

hydrogen bond nucleic acid nucleotide RNA

Genetic code and protein synthesis codon

degeneracy

globular protein fibrous protein mRNA

Explanatory notes: Gene expression

polypeptide

Gene expression involves a selection from ...

protein

redundancy

c

1a

rRNA

c

b

transcription

c

translation

triplet code tRNA

99  -  103

Nucleic acid structure.

The nature of the genetic code.

106 107

2

Proteins and their significance.

110 -113

c

3

Protein synthesis.

104 - 109

c

4

The determination of phenotype via metabolic pathways.

117 118

c

5

Effect of environment on genotype through mutations.

119 - 125

c

6

Effect of environment on expression of phenotype.

126 127

Nucleic acid structure and the genetic code

Activity number

Metabolic pathways deletion (mutation) environment

Select biological ideas and processes from...

c

i

Molecular components of nucleic acids and their role in carrying the genetic code.

genotype

c

ii

Nature of the genetic code including triplets, codons, and anticodons.

c

iii

Redundancy due to code degeneracy.

metabolic pathway mutagen

Significance of proteins

mutation

Select biological ideas and processes from...

phenotype

substitution (mutation)

99 - 104

106 - 108

106

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enzyme

insertion (mutation)

Activity number

Activity number

c

i

Proteins as the products of gene expression: DNA  mRNA  polypeptide.

104 

c

ii

Identification of one gene  one polypeptide relationship.

104

c

iii

Significance of proteins with reference to their structural and catalytic roles.

Protein synthesis

110 - 113 Activity number

Select biological ideas and processes from...

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

ii

The role of enzymes in controlling the process (enzyme names not required).

105 108


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Determination of phenotype via metabolic pathways

Activity number

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Select biological ideas and processes from...

c

i

Biochemical reactions are catalysed by specific enzymes and every enzyme is encoded by a specific gene(s).

117 - 118

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.

117 - 118

c

iii

Phenotype is determined by the presence, absence, or amount of metabolic products. 117 - 118

Effect of environment on expression of phenotype

Activity number

Select biological ideas and processes from...

120

i

Mutagens (recognise specific mutagens but not their effect at the molecular level).

c

ii

The potential effect on genotype and phenotype of mutations at the gene level.

121 - 124

c

iii

Ideas and processes relating to the effect of environment on phenotype involve how environmental factors may change phenotype without changing genotype.

119 125 126

c

What you need to know for this Achievement Standard Nucleic acids and the genetic code Activities 99 - 104, 106 - 107, 114 - 116

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

c

Recognise DNA and RNA as nucleic acids made from nucleotide monomers. Describe their biological roles.

c

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.

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

Describe and explain the main features of the genetic code, including: • the 4-letter alphabet and 3-letter triplet code (codon) of base sequences. • the non-overlapping, linear nature of the code which is read in one direction from start to finish. • The degeneracy of the code and how this leads to redundancy.

Proteins and protein synthesis Activities 104 - 116

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

Outline flow of information from DNA  mRNA  polypeptide or protein. Describe the one gene-one polypeptide relationship.

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

Identify transcription and translation as the two stages of gene expression.

c

Explain transcription in eukaryotes, including the role of enzymes in the process.

c

Explain translation in eukaryotes, including the role of mRNA, tRNA, ribosomes, and enzymes.

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c

Metabolic pathways, mutation, and phenotype Activities 117 - 128

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

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

Use examples to explain how phenotype is determined via metabolic pathways.

c

Use examples to explain how mutagens can change genotype through mutation.

c

Describe types of gene and chromosome mutations and outline their general effect on phenotype.

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).


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98 The Role of DNA in Cells Key Idea: A cell's genetic information is called DNA. In eukaryotic cells, DNA is located in the cell's

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nucleus.

About DNA

ffDNA stands for deoxyribonucleic acid.

ffDNA is called the blueprint for

life because it contains all of the information an organism needs to develop, function, and reproduce.

ffDNA stores and transmits genetic information.

ffDNA is found in every cell of all living organisms.

ffDNA 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.

DNA

Nucleus

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.

In eukaryotes, 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.

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.

(b) Where is most of the DNA found in eukaryotes?

(c) What does DNA do?

2. (a) How is DNA organised in eukaryotes?

(b) Why does DNA have to be tightly packaged?

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1. (a) What does DNA stand for?

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99 The Structure of Chromosomes Key Idea: A chromosome is a single long molecule of DNA coiled around histone proteins.

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Chromosomes contain protein-coding regions called genes.

ffIn 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.

ffEach 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.

In eukaryotes, chromosomes are formed from the coiling of chromatin.

Chromosome

The chromatin is coiled up, reducing its length.

Chromatids (2)

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 string.

Chromatin consists of a complex of DNA and 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.

Histone proteins

Gene (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. DNA

Start sequence

Exon Intron

Stop sequence

(b) What is the role of histones in this process?

2. Identify three parts of the gene that do not appear in the final mRNA:

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1. (a) Suggest a purpose for DNA coiling:

3. What has to happen before a gene can be expressed (made into a protein)?

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100 Nucleotides Key Idea: Nucleotides are the building blocks of the nucleic acids, DNA and RNA. A nucleotide has

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three components; a base, a sugar, and a phosphate group.

The structure of a nucleotide

Nucleotide bases

Nucleotides are the building blocks of nucleic acids (DNA and RNA). Nucleotides have three parts to their structure (see diagrams below):

Five different kinds of nitrogen bases are found in nucleotides. These are: Adenine (A)

ffA nitrogen containing base

Guanine (G)

ffA five carbon sugar

Cytosine (C)

ffA phosphate group

Thymine (T) Uracil (U)

DNA contains adenine, guanine, cytosine, and thymine.

Symbolic form of a nucleotide

RNA also contains adenine, guanine, and cytosine, but uracil (U) is present instead of thymine.

(showing positions of the 5 C atoms on the sugar)

Purines: (two-ring bases)

Phosphate group

5

1

Guanine

Adenine

Sugar

2

3

G

A

Base

4

Pyrimidines:

(single-ring bases)

Chemical structure of a nucleotide

OH

N

H

OH

P O CH 2

N

O

O

H

H

OH

T

Cytosine

Thymine (DNA only)

H

H

Uracil (RNA only)

Sugars

N

N

H

Phosphate

NH 2

U

C

Nucleotides contain one of two different sorts of sugars. Deoxyribose sugar is only found in DNA. Ribose sugar is found in RNA.

H

Sugar

Base

OH

H

OH

Deoxyribose sugar (found in DNA)

2. List the nucleotide bases present:

(a) In DNA:

(b) In RNA:

3. Name the sugar present: (a) In DNA: Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

Ribose sugar (found in RNA)

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1. What are the three components of a nucleotide?

OH

(b) In RNA:

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101 DNA and RNA

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146

Key Idea: DNA and RNA are nucleic acids made up of long chains of nucleotides, which store and

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transmit genetic information. DNA is double-stranded. RNA is single-stranded.

DNA

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.

5'

3'

Carbon no. 3

Carbon no. 5

C

T

A

G

A

ffThe DNA backbone is made up of alternating

Hydrogen bond

G

C

T

phosphate and sugar molecules.

ffEach "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).

3'

5'

ffEach 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.

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.

G

T

ffThe 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).

RNA

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:

H2O

In RNA, uracil (U) replaces thymine in the code.

G

A

C

U

ffMessenger RNA (mRNA) ffTransfer RNA (tRNA)

The presence of two OH groups on the ribose sugar stops it from forming a double stranded helix.

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 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?

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ffRibosomal RNA (rRNA)

C

A

G

G

T

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?

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100 102

T

A

C

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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.

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.

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.

4. What is the purpose of the hydrogen bonds in double-stranded DNA?

5. Briefly 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?

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7. Complete the following table summarising the differences between DNA and RNA molecules:

DNA

Sugar present

Bases present

Number of strands

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RNA


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102 Constructing a DNA Model Key Idea: Nucleotides pair together in a specific way called the base pairing rule. In DNA, Adenine

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always pairs with thymine, and cytosine always pairs with guanine.

ffDNA are double stranded molecules that contain genetic information. Each strand is made up of nucleotides.

ffThe 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.

DNA base pairing rule

Chargaff's rules

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.

Adenine

always pairs with

Thymine

Thymine

always pairs with

Adenine T

A

Cytosine

always pairs with

Guanine C

G

Guanine

always pairs with

Cytosine G

C

A

T

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. 2. Place one of each of the four kinds of nucleotide on their correct spaces below:

Place a cut-out symbol for thymine here

Place a cut-out symbol for cytosine here

Thymine

Cytosine

Place a cut-out symbol for adenine here

Place a cut-out symbol for guanine here

Adenine

Guanine

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3. Identify and label each of the following features on the adenine nucleotide immediately above: phosphate, sugar, base, hydrogen bonds.

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. 5. Create the complementary strand of DNA by using the base pairing rule above. Note that the nucleotides have to be arranged upside down.

6. It is not normally possible for base mismatches to occur. Describe two factors that prevent a mismatch from occurring:

Factor 1:

Factor 2:

7. Once you have checked that the arrangement is correct, you may glue, paste or tape these nucleotides in place. WEB

PRAC

102

LINK

47

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100 101

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Cut

Cut

Cut

P

S

S

S

S

Cut

P

P

S

S

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P

P

P

Adenine

Guanine

Adenine

Guanine

Adenine

Guanine

P

P

P

P

P

P

S

S

S

S

S

S

Thymine

Cytosine

Thymine

Cytosine

Thymine

Cytosine

P

P

P

P

P

P

S

S

S

S

S

S

Adenine

Guanine

Adenine

Guanine

Adenine

Guanine

P

P

P

P

P

P

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S

S

S

S

S

Thymine

Cytosine

Thymine

Cytosine

Thymine

Cytosine

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149

Nucleotides

Tear out this page and separate each of the 24 nucleotides by cutting along the columns and rows (see arrows indicating the cutting points).

Cut

Cut

Cut

Cut


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150

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This page is left blank deliberately

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151

DNA molecule

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P

Adenine

S

Thymine S

Thymine

P

Put the named nucleotides on the left hand side to create the template strand

Put the matching complementary nucleotides opposite the template strand

Cytosine

Adenine

Adenine

Guanine

Thymine

Cytosine

Guanine

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Thymine


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103 Genes Code for Proteins Key Idea: Genes are sections of DNA that code for proteins. Genes are expressed when they are

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transcribed into messenger RNA (mRNA) and then translated into a protein.

ffA gene is a section of DNA that codes for a protein (or some other functional mRNA product).

ffGene 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).

ffEukaryotic 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.

Nucleus

TRANSLATION

EDITING

TRANSCRIPTION

mRNA

Ribosome

Nuclear pore

mRNA

Amino acids are linked together at the ribosome to form the protein encoded by mRNA.

Cytoplasm

The primary transcript is edited. The non-protein coding introns are removed and modifications are made to help the mRNA exit the nucleus.

Primary transcript

DNA

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.

1. What is a gene?

2. (a) What does gene expression mean?

(b) What are the three stages in gene expression in eukaryotes and what happens in each stage?

(i)

(ii)

(iii)

3. (a) Where does transcription occur in eukaryotes?

(b) Where does translation occur in eukaryotes? LINK

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104 Transcription Key Idea: Transcription is the first step of gene expression. In eukaryotes, transcription takes place in

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the nucleus and is carried out by RNA polymerase.

ffTranscription 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.

ffThe 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. Coding (sense) strand of DNA

3'

RNA polymerase (RNAP) adds nucleotides to the 3' end so the strand is synthesised in a 5' to 3' direction.

mRNA nucleotides. Free nucleotides are used to construct the RNA strand

3'

Template (antisense) strand of DNA stores the information that is transcribed into mRNA

Direction of transcription

RNA polymerase binds at the upstream promoter region

Newly synthesised RNA strand is complementary to the template strand

RNA polymerase dissociates at the terminator region

3'

5'

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.

Translation will begin at the start codon AUG

Several RNA polymerases may transcribe the same gene at any one time, allowing a high rate of mRNA synthesis.

1. (a) Name the enzyme responsible for transcribing the DNA:

(b) What strand of DNA does this enzyme use?

(c) The code on this strand is the [ same as / complementary to ] the RNA being formed (circle correct answer).

(d) Which nucleotide base replaces thymine in mRNA?

(e) On the diagram, use a coloured pen to mark the beginning and end of the protein-coding region being transcribed.

2. (a) In which direction is the RNA strand synthesised?

(b) Explain why this is the case:

3. (a) Why is AUG called the start codon?

(b) What would the three letter code be on the DNA coding strand? Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

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5'

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105 What is the Genetic Code? Key Idea: Each mRNA codon codes for a specific amino acid. This activity will help you practise

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using the amino acid table to determine the amino acids encoded by the genetic code. Read second letter here

The mRNA - amino acid table

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.

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: i

Find the first letter of the codon in the row on the left hand side of the table. AUG is the start codon.

ii Find the column that intersects that row from the top, second letter, row. iii Locate the third base in the codon by looking along the row on the right hand side that matches your codon. e.g. GAU codes for Asp (aspartic acid)

UUU UUC UUA UUG

Phe Phe Leu Leu

UCU UCC UCA UCG

Ser Ser Ser Ser

UAU UAC UAA UAG

UGU UGC STOP UGA STOP UGG

CUU CUC CUA CUG

Leu Leu Leu Leu

CCU CCC CCA CCG

Pro Pro Pro Pro

CAU CAC CAA CAG

His His Gln Gln

CGU CGC CGA CGG

Arg Arg Arg Arg

AUU AUC AUA AUG

Ile Ile Ile Met

ACU ACC ACA ACG

Thr Thr Thr Thr

AAU AAC AAA AAG

Asn Asn Lys Lys

AGU AGC AGA AGG

Ser Ser Arg Arg

GUU GUC GUA GUG

Val Val Val Val

GCU GCC GCA GCG

Ala Ala Ala Ala

GAU GAC GAA GAG

Asp Asp Glu Glu

GGU GGC GGA GGG

Gly Gly Gly Gly

Read first letter here

Tyr Tyr

Cys Cys

STOP

Trp

U

Read third letter here

1. (a) Use the base-pairing rule for to create the complementary strand for the DNA template strand shown below.

(b) For the same DNA template strand, then determine 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. Template strand

DNA

T A C C C A A T G G A C T C C C A T T A T G C C C G T G A A A T C

Complementary strand (this is the DNA coding strand)

Gene expression

Template strand

DNA

T A C C C A A T G G A C T C C C A T T A T G C C C G T G A A A T C

mRNA

Translation

Amino acids

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Transcription

2. What do you notice about the sequence on the DNA coding strand and the mRNA strand?

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Redundancy and degeneracy

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Redundancy and degeneracy are important concepts in understanding the genetic code.

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ffRedundancy is when several situations code for or control the actions of one specific thing. ffDegeneracy is when a particular output can be produced by several different pathways.

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.

Flight computers

Control lines

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 normally.

Degeneracy is seen 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).

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.

CCU

CCU

CCC

CCC

Pro

Pro

CCA

CCA

CCG

CCG

Salivary amylase (above) is structurally different to pancreatic amylase, but has the same function.

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.

3. Explain how degeneracy adds "safety" to the coding of protein chains:

5. Identify the following:

(a) The codons that encode valine (Val):

(b) The codons that encode aspartic acid (Asp):

6. (a) Arginine (Arg) is encoded in how many ways?

(b) Glycine (Gly) is encoded in how many ways?

(c) Which amino acid(s) are encoded in only one way? Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

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4. The genetic code shows redundancy but no ambiguity. What does this mean and why is it important?


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106 Cracking the Genetic Code Key Idea: Scientists used mathematics and scientific experiments to unlock the genetic code. A series

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of three nucleotides, called a triplet, codes for a single amino acid.

The genetic code

How was the genetic code cracked?

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.

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.

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.

The triplet code was accepted once scientists confirmed that some amino acids have multiple codes.

Number of bases in the code

Working

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 mRNA).

1 DNase was added

to destroy bacterial DNA so there was no template for mRNA to be made.

Number of amino acids produced

2 Radio-labelled amino acids and a synthetic mRNA strand containing only uracil (U) were added.

U U U U U U U U U U U U

Single (41)

4

4 amino acids

Double (42)

4x4

16 amino acids

Triple (43)

4x4x4

64 amino acids

Cell free E. coli extract.

Phe Phe Phe Phe

A triplet (three nucleotide bases) codes for a single amino acid. The triplet code on mRNA is called a codon.

3 Once the mRNA was

added an amino acid was produced. The codon UUU produced the amino acid phenylalanine (Phe).

4 Over the next few years, similar

experiments were carried out using different combinations of nucleotides until all of the codes were known.

1. (a) How many types of nucleotide bases are there in mRNA? (b) How many types of amino acids are there in proteins?

(c) Why did scientists reject a one or two base code when trying to work out the genetic code?

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2. A triplet code could potentially produce 64 amino acids. Why are only 20 amino acids produced?

3. (a) Why was DNase added to the cell free E. coli extract?

(b) What would it have been difficult to crack the code if no DNase was added?

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107 Translation Key Idea: Translation is the final stage of gene expression in which ribosomes read the mRNA and

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decode (translate) it to synthesise a protein. This occurs in the cytoplasm.

ffIn 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.

ffProtein 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.

Ribosome structure

tRNA structure

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.

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.

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.

Amino acid attachment site. Enzymes attach the tRNAs to their specific amino acids.

Large subunit

Large subunit

Functional ribosome

Small subunit

Anticodon is a 3-base sequence complementary to the codon on mRNA.

1. Describe the structure of a ribosome:

(a) Ribosome:

(b) tRNA:

(c) Amino acids:

(d) Start codon:

(e) Stop codon:

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2. What is the role of each of the following components in translation?

3. There are many different types of tRNA molecules, each with a different anticodon (HINT: see the mRNA table).

(a) How many different tRNA types are there, each with a unique anticodon?

(b) Explain your answer:

(c) Determine the mRNA codons and the amino acid sequence for the following tRNA anticodons:

tRNA anticodons:

U A C

Codons on the mRNA:

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U A G

C C G

C G A

U U U

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tRNA molecules deliver amino acids to ribosomes

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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. Unloaded Met-tRNA

Lys

The ribosome P (peptidyl) site carries the growing polypeptide chain.

Charged Arg-tRNA enters the ribosome A (acceptor) site. The amino acid is added to the growing polypeptide chain.

Met

Thr

Phe

Unloaded Thr-tRNA leaves the ribosome E (exit) site

Charged Val-tRNA

Arg

Val

Charged 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

mRNA

Charging Lys-tRNA

5'

3'

Ribosome (only large subunit shown)

The polypeptide chain grows as more amino acids are added The polypeptide chain continues to grow as more amino acids are added.

Protein synthesis stops when a stop codon is reached (UGA, UAA, or UAC). The ribosome falls off the mRNA and the polypeptide is released.

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.

5'

Direction of protein synthesis

4. Describe the events occurring during translation:

STOP

3'

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START

5. Many ribosomes can work on one strand of mRNA at a time (a polyribosome system). What would this achieve?

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108 Protein Synthesis Summary 3'

Thr

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5'

Phe

Cys

Asn

Lys

Met

Lys

Tyr

Arg

Val

Met

Tyr

Thr

Phe

Arg

Val

Lys

Tyr

Met

Arg

Thr

Thr

Ala

Phe

Gene #2

5'

3'

Gene #1

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.

1. Briefly describe each of the numbered processes in the diagram above:

(a) Process 1:

(b) Process 2:

(c) Process 3:

(d) Process 4:

(e) Process 5:

(f) Process 6:

(g) Process 7:

(h) Process 8: (i) Process 9: 2. Identify each of the structures marked with a letter and write their names below in the spaces provided: (a) Structure A:

(f) Structure F:

(b) Structure B:

(g) Structure G:

(c) Structure C:

(h) Structure H:

(d) Structure D:

(i) Structure I:

(e) Structure E:

(j) Structure J:

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3. Describe two factors that would determine whether or not a particular protein is produced in the cell: (a)

(b)

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109 Amino Acids Make up Proteins Key Idea: Amino acids are the basic units of proteins. They have a common structure but contain a

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variable "R" group, which gives each of them unique properties.

Structure of an amino acid

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.

The 'R' group varies in chemical make-up with each type of amino acid.

ffAll amino acids have a common structure. The only

Carbon atom

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).

Amine group

ffTwenty amino acids occur commonly in proteins. Nonprotein amino acids have roles as intermediates in metabolic reactions.

This 'R' group gives the amino acid an alkaline property.

SH

CH2

NH 2

SH

CH2

C NHCOOH C 2 H

NH 2 CH2 CH2 CH2 CH2

H

Cysteine Cysteine

H

H

Lysine

Lysine

NH2

C

Hydrogen atom

H

C

COOH

Carboxyl group makes the molecule behave like a weak acid.

This 'R' group can form disulfide bridges NH 2with other cysteinesCH to 2create cross linkages in CHa2 polypeptide chain. CH2 CH2

CH 2

CH2

CH 2

NH 2 C NHCOOH COOH C 2 H

SH

COOH

NH 2

COOH

H

Aspartic acid Aspartic acid

ffAmino acids are linked together by peptide bonds

O

OH

This 'R' group gives the amino acid an acidic property.

NH 2 CH2 CH2 CH2 CH2

NH 2 C NHCOOH C 2 COOH

R

C

COOH

NH 2

C

H

H

Cysteine

Lysine

COOH

A polypeptide chain

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).

A Polypeptide Chain

ffPeptide 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.

Peptide bond

Peptide bond

Peptide bond

Peptide bond

Peptide bond

(b) What makes each of the 20 amino acids found in proteins unique?

2. (a) Name the type of bond that links amino acids together:

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1. (a) Describe the structure of an amino acid:

(b) What is the name given to many amino acids joined together by this bond?

(c) How is this bond formed?

3. What is the function of a disulfide bridge?

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C

C NH 2

C

H

Aspar


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110 The Structure of Proteins Key Idea: The sequence and type of animo acids in a protein determines that protein's three-

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dimensional shape and its function.

ffProteins 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.

ffThe 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.

Secondary (2°) structure (a-helix or b pleated sheet)

Primary (1°) structure (amino acid sequence)

Phe

Glu

Tyr

Ser

Iso

Met

Ala

Ser

Peptide bond

Ser

Glu

Met

Secondary (2°) structure is maintained by hydrogen bonds between neighbouring CO and NH groups.

Amino acid

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.

a-helix

Amino acid chain

Hydrogen bonds

b-pleated sheet

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.

Tertiary (3°) structure (folding of the 2° structure)

Quaternary (4°) structure

Alpha chain

a-helix

Beta chain

Prosthetic (haem) group

Aspartic acid Ionic bond

Disulfide bond

Lysine

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.

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.

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.

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Tertiary (3°) structure is maintained by more distant interactions such as disulfide bridges between cysteine amino acids, ionic bonds, and hydrophobic interactions.

1. Describe the main features that aid the formation of each part of a protein's structure: (a) Primary structure:

(b) Secondary structure:

(c) Tertiary structure:

(d) Quaternary structure:

2. Which two structures can act as fully functional proteins or enzymes? © 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

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111 Protein Shape is Related to Function

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162

Key Idea: The three dimensional shape of a protein reflects its role. When a protein is denatured, it

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loses its functionality.

ffA 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. H2N

b sheets

Ala

Asn

Glu

Tyr

Active site

Gln

Val

Leu

Leu

Val

Glu

Cys

Gly

Arg

Ser

H2N

S

COOH

Asn

Phe

Leu

Cys

Gly

S

Gly

Cys

S

Ile

Tyr

Val

Asn

Phe

Glu

Gln

a chain

Leu

Thr

Cys

Cys

Thr

S

Glu

Tyr

Gln

Pro

Amylase

His

b chain

His

S

Gly

a helix

Phe

Val

Leu

Tyr

Ser

S

Ile

Cys

Leu

Ser

Lys

Thr

COOH

Channel proteins

Enzymes

Sub-unit proteins

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.

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 denaturation

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).

Cooked (denatured) egg white

Raw (native) egg white

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1. Using the example of insulin, explain how interactions between R groups stabilise the protein's functional structure:

2. Why do channel proteins often fold with non-polar R groups to the channel's exterior and polar R groups to its interior?

3. Why does denaturation often result in the loss of protein functionality?

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112 Globular and Fibrous Proteins Key Idea: Globular proteins have mainly catalytic and regulatory functions, while fibrous proteins have

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a mainly structural function.

Globular proteins

Fibrous proteins

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.

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.

Their properties include:

• Easily water soluble • Tertiary structure is critical to function • Polypeptide chains folded into a roughly spherical shape

Their properties include:

• Water insoluble • Physically very tough. They are supple or stretchy • Parallel polypeptide chains in long fibres or sheets

Their functions include:

Their functions include:

• • • •

• Structural role in cells and organisms, e.g. collagen in connective tissue, cartilage, bones, tendons, and blood vessel walls. • Contractile in the cytoskeleton and muscle

Regulatory, e.g. hormones (insulin) Catalytic, e.g. enzymes Transport, e.g. haemoglobin Defensive, e.g. antibodies Bonds between different polypeptide chains help hold globular proteins in their required shape.

Hydrogen bond

Glycine

Covalent cross links between the collagen molecules.

Human insulin is a relatively small globular protein involved in regulating blood glucose. It has two polypeptide subunits.

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.

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.

2. How does the shape of a fibrous protein relate to its functional role?

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1. Contrast the properties of globular and structural proteins:

3. How does the shape of a catalytic protein (enzyme) relate to its functional role?

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113 What You Know So Far: Nucleic Acids and Proteins

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164

Gene expression

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

HINT: You should define and explain transcription and translation including key molecules and events.

The structure of DNA and RNA

HINT: Include the features of each molecule, their roles in the cell, and the base pairing rule.

Proteins

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HINT: Include reference to amino acids, bonds, protein structure, and denaturation.

REVISE

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114 NCEA Style Question: Nucleic Acids and Proteins

Second letter Tyr Tyr

Phe Phe Leu Leu

Ser Ser Ser Ser

STOP STOP

Trp

Leu Leu Leu Leu

Pro Pro Pro Pro

His His Gln Gln

Arg Arg Arg Arg

Ile Ile Ile Met

Thr Thr Thr Thr

Asn Asn Lys Lys

Ser Ser Arg Arg

Val Val Val Val

Ala Ala Ala Ala

Asp Asp Glu Glu

Gly Gly Gly Gly

Cys Cys

STOP

Third letter

First letter

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1. The table below lists the amino acids and their codons:

U

(a) Use the information to complete the table below (you only need one codon per amino acid):

DNA (coding strand)

DNA (template strand) mRNA

Amino acids

Pro

Glu

Asn

Stop

(b) Explain how the nature of the genetic code protects against the effect of point mutations in the DNA sequence:

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Met

TEST


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2. Discuss the production of a functional protein via translation. In your discussion, include the role(s) of:

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• Ribosomes • tRNA • Codons (including start and stop codons) and anticodons

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115 KEY TERMS AND IDEAS: Nucleic Acids and Proteins

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

anticodon

base-pairing rule coding

strand

A Single stranded nucleic acid that consists of nucleotides containing ribose sugar.

B A set of rules by which information encoded in DNA or mRNA is translated into proteins. C The rule governing the pairing of complementary bases in DNA. D The structural units of nucleic acids, DNA and RNA.

DNA

E Intermolecular bond between hydrogen and an electronegative atom such as oxygen.

gene expression

F The process by which genetic information is used to produce a functional gene product.

genetic code

G The sequence of DNA that is read during the synthesis of mRNA.

hydrogen bond

H Universally found macromolecules composed of chains of nucleotides. These molecules carry genetic information within cells.

nucleic acids

I

nucleotides

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.

RNA

template strand

The DNA strand with the same base sequence as the RNA transcript produced (although with thymine replaced by uracil in mRNA).

K The region of a transfer RNA with a sequence of three bases that is complementary to a codon in the messenger RNA.

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.

DNA (template strand):

G A A A C C C T T A C A T A T C G T G C T

mRNA:

Amino acids: 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

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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.

5. Decide if the nucleotide shown right is from DNA or RNA. Explain your choice:

OH

P O CH 2

N

N

H

H

H

OH

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NH 2

N

O

O

Phosphate

N

H

OH

Sugar

H

H

H

Base

TEST

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116 Metabolic Pathways Key Idea: A metabolic pathway is a series of linked biochemical reactions. Specific enzymes catalyse

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each step of a metabolic pathway.

ffMetabolic pathways are linked biochemical reactions that occur within living organisms to maintain life.

ffEnzymes catalyse each step of a

metabolic pathway and, in turn, each enzyme is coded for by specific genes.

ffMetabolic 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.

Fred the Oyster CC 4.0

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.

A simplified metabolic pathway

ffThe pathway shown below involves just two enzymes, two substrates, and one end product. It illustrates that: 1: Biochemical reactions are catalysed by specific enzymes which are in turn encoded by specific genes.

2: Biochemical reactions occur in a chain of reactions in which the product of one reaction forms the substrate for the next reaction. Gene B

Gene A

Expression of gene B produces enzyme B

Expression of gene A produces enzyme A

Enzyme A

Starting substrate

Enzyme B

Enzyme A catalyses a reaction to chemically alter the starting substrate into an intermediate.

Intermediate

Enzyme B catalyses a reaction to chemically alter the intermediate into the end product.

End product

2. (a) What is the role of enzymes in metabolic pathways?

(b) Describe the two general ways that metabolic pathways are regulated:

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1. Define the term metabolic pathway:

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Cyclic pathways

ffSome metabolic pathways flow in a cycle. The product from one step in the cycle is the substrate for the next step

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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. Ammonia (toxic) NH3 + CO2

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.

2 ATP E

Carbamyl phosphate

Citrulline

E

Ornithine

E

The urea cycle

E

oxaloacetate

UREA (waste product)

E

E

Arginine

Aspartate

The Krebs Cycle

Îą-ketoglutarate

Malate

E

E

Argininosuccinate

E

E

Fumarate

E

Acetyl CoA E Citrate

Fumarate

E

Succinate

Step is catalysed by an enzyme

3. (a) Describe the features of a cyclic metabolic pathway:

(b) Explain the role of enzymes in these metabolic pathways:

4. (a) What molecule remains when urea is given off as a waste product in the urea cycle?

(b) What is the role of this molecule in the cycle?

5. What two molecules combine to form the first substrate in the Krebs cycle?

6. Identify a product from one cycle that could act as a substrate in the other:

7. Feedback inhibition occurs when the product of a metabolic pathway inhibits the processing of the original substrate.

Enzyme 1

Product C

Intermediate

Inhibits

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Original substrate

Enzyme 2

(a) Using the example above explain what would happen if there was a build up of product C?

(b) Explain how feedback inhibition provides a simple way to regulate metabolic pathways:

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117 Interrupting Metabolic Pathways

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170

Key Idea: Metabolic pathways are interrupted when one or more of the enzymes catalysing the

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biochemical reactions involved are not expressed correctly.

ffEach metabolic pathway is a series of biochemical reactions in which each step relies on the completion of the previous step.

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.

ffAn 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.

Substrate 1

Enzyme 1

Enzyme 2

Substrate 2

Substrate 3

Enzyme 3

Product

The metabolism of phenylalanine Protein

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.

Phenylkeronuria

Phenylalanine

Phenylalanine hydroxylase

Thyroxine

a series of enzymes

Faulty enzyme causes:

Congenital hypothyroidism

This in turn causes:

Faulty enzyme causes buildup of:

Tyrosinase

Tyrosine

Phenylpryuvic acid

Melanin

Faulty enzyme causes:

Transaminase

Albinism

Hydroxyphenylpyruvic acid

Faulty enzyme causes:

Ty

Homogentisic acid

Homogentisic acid oxidase

Carbon dioxide and water

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Faulty enzyme causes:

Maleylacetoacetic acid

Tyrosinosis

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Hydroxyphenylpyruvic acid oxidase

a

Alkaptonuria

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Photo: msdonna

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Guthrie card

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.

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.

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 function).

1. Identify four end products of the normal metabolism of phenylalanine:

2. In the metabolic pathway of the metabolism of phenylalanine, identify the faulty enzyme that results in:

(a) Albinism:

(b) Phenylketonuria:

(c) Tyrosinosis:

(d) Alkaptonuria:

3. Why do people with phenylketonuria have light skin?

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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:

(a) Albinism:

(b) Phenylketonuria:

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118 Influences on Phenotype Key Idea: An organism's phenotype is influenced by the effects of the environment during and after

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development, even though the genotype remains unaffected.

ffThe 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).

ffEven identical twins have minor differences in their appearance due to epigenetic and environmental factors such as diet and intrauterine environment. Genes, together with epigenetic and environmental factors determine the unique phenotype that is produced. Genotype

Sources of variation in organisms

Epigenetics

Alleles

Sexual reproduction

Phenotype

Single nucleotide variations

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).

Chemical tags and markers that regulate how the DNA is expressed

Mutations

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.

Competition

Pathogens

Predators

Nutrition

Drugs

Toxins

Physical environment

Environment

1. (a) What are some sources of genetically induced variation?

(b) What are some sources of environmentally induced variation?

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2. Explain why genetically identical twins are not always phenotypically identical:

3. Give an example in which an organism's environment produces a marked phenotypic change:

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119 Mutagens

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Key Idea: Mutagens are any chemical or physical agent that causes a change in the DNA sequence.

The natural 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:

1. Describe examples of environmental factors that induce mutations under the following headings:

(a) Radiation:

(b) Chemical agents:

Ionising radiation

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.

Viruses and microorganisms

Some viruses integrate into the human chromosome, upsetting genes and triggering cancers. Examples include hepatitis B virus (liver cancer), Hodgkinâ&#x20AC;&#x2122;s disease, and HPV (right) which is implicated in cervical cancer. Aflatoxins produced by the fungus Aspergillus flavus are potent inducers of liver cancer.

2. How do mutagens cause mutations?

Poisons and irritants

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 mutagens).

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3. Explain why people such as radiographers, or even dentists have at a higher risk of exposure to cancer causing radiation:

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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.

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120 Mutation, Genotype, and Phenotype

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174

Key Idea: Gene mutations are localised changes to the DNA base sequence. In most cases this will

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cause a change in phenotype.

ffGene 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.

ffPoint mutations can occur by substitution, insertion, or deletion of bases and alter the mRNA transcribed.

ffA 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.

ffMutations that cause a change in the amino acid sequence are usually harmful because they alter protein function.

Normal sequence

Leu

Ala

Tyr

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.

Phe

Val

Leu

Asn

Gln

Val

1

Glu

GCT TGT TTA TGC TCT GGC CAT CTT CAC CAA AAT

Gly

Cys

Ala

S

Arg

His

H2N

1

S

Leu

Cys

GTT TTT GCT CTG TAT CTG GTT TGT GGT GAA CGT

Gly

Leu

Phe

GGG TTT TTT TAT ACT CCC AAA ACT TGT 31

Phe

Ser

Cys

His

Gly

S

Tyr

Thr

S

Pro

Lys

Thr

Cys COOH

31 Skin tone with normal

active pigment

Substitution mutation ffIn 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.

ffIn 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 proteinâ&#x20AC;&#x2122;s function.

ffIn 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.

Missense substitution Nonsense substitution: Substitute A instead of T

Mutant DNA

Leu

Phe

Val

Gln

Val

Glu

Gly

Leu

Cys

Val

S

S

His

STOPH2N Ala

Leu

Cys

Leu

Cys

Ser

Phe

Tyr

Thr

Gly

31

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LINK

LINK

Lys

Thr

Ser

LINK

mRNA

120 121 122 123

Asn

Gln

Val

19

Ala

His

H2N

Cys

Leu

Cys

Ser

Leu Missense substitutio Substitute C instead o His

Gly

Mutant DNA Abnormal pigment is not active. The mRNA person has light skin.

Mutant DNA

Amino acids

WEB

Val

Nonsense substitution: Substitute A instead of T

Abnormal pigment is not active. The person has light skin.

Thr

Phe

Leu

COOH

Lys

Ser

Amino acids

Ala

Tyr

His

Pro

COOH

Leu

mRNA

Asn

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Leu

Gly

Phe

Nonsense substitution

Mutant DNA

Ala

Tyr

mRNA

Amino acidsArg

Missense substitution: Substitute C instead of G

Leu

Val

STOP

Amino acids

Lys

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Thr

S


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Silent mutations

Leu

Ala

Tyr

Phe Val

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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.

Leu

19

Glu

Gly

Asn

Gln

Val

Cys

Ala

S

Arg

Gly

Leu

Phe

Ser

His

Gly

S

Tyr

Thr

S

Pro

Lys

Cys COOH

Thr

mRNA

Val

Cys

Phe

Mutant DNA

Leu

Leu

Cys

Silent mutation: Substitute C instead of T

Amino acids

His

H2N

S

31

Skin tone with normal active pigment

Cys

Insertion mutation

Leu

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.

Val

Gly

Ala

S

Gly

Leu

Phe

Phe

Tyr

Thr

Ala

Deletion mutation Amino

Leu

Gln

Asn

Cys

Ser

His

Gly

27

Tyr

Thr

Ala

Gln

Leu

Asn

COOH

Leu

Insertion of G

mRNA

His

H2N

S

Abnormal pigment is not active. The person has light skin.

Reading frame shift results in a new sequence of amino Mutant acids. The protein is unlikely to have any biological activity. DNA Deletion of G

Tyr Thr Ala Gln Asn Leu The deletion of an nucleotide is called a deletion mutation. acids Mutant It also causes a reading frame shift. In this example the DNA nucleotide G is deleted after the 17th DNA triplet. This leads Reading frame shift results in a new sequence of amino to a reading frame shift that affects the rest of the protein. acids. The protein is unlikely to have any biological activity. mRNA This could lead to nonsense.

Leu

Ala

Tyr

17

Phe

Val

Leu

Asn

Gln

Leu Phe Deletion of G

Val

Val

Asn

22

Asn

Val

Leu

?

Leu

Phe

Val

Val

Asn

His

H2N

Leu

Cys

?

?

Tyr

Val

Ala

?

mRNA

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Phe

Tyr

Reading frame shift results in a new sequence of amino Mutant acids. The protein is unlikely to have any biological activity. DNA

Amino acids

Gln

Val

Cys

mRNA

Amino acids

Asn

Cys

Arg

Mutant DNA

Amino acids

Phe

Leu

Glu

Insertion of G

Ala

Tyr

?

?

?

?

?

?

Cys

Ser

His

Gly

COOH

31

Reading frame shift results in a new sequence of amino acids. The protein is unlikely to have any biological activity. Š 1994-2017 BIOZONE International ISBN: 978-1-927309-51-3 Photocopying Prohibited

Abnormal pigment is not active. The person has light skin.


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1. Why are mutations that alter the amino acid sequence usually harmful?

2. Why are nonsense substitutions likely to be more damaging than missense substitutions?

3. Explain how a substitution mutation can be effectively silent:

4. What is a reading frame shift?

5. Why is a frame shift near the start codon likely to have a greater impact than one near the stop codon?

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:

Mutant DNA (coding strand): GTG GGT TTT TTT ATA CTC CCA AAA CTT GT?

Mutant mRNA:

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:

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121 Effect of a Substitution Mutation on Phenotype Key Idea: The change of a single DNA nucleotide can change the amino acid sequence of the

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resulting protein, causing large scale phenotypic effects.

ffPoint 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.

ffSometimes 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).

ffSickle 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.

Normal red blood cells

Red blood cells containing normal haemoglobin have a flattened disc shape.

The sickle cell mutation has adenine substituted at this point instead of thymine.

HBB gene

Template DNA

p

The haemoglobin beta gene (HBB gene), is located on chromosome 11. It encodes for the b-chain subunit of the haemoglobin molecule.

The mutation causes the amino acid valine to be produced instead of glutamic acid.

q

Chromosome 11.

A faulty b-chain subunit is produced because the amino acid substitution causes the b-chain to fold and behave differently.

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).

Sickle cells

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1. Identify the type of mutation that causes sickle cell disease:

2. Write the mRNA sequence for part of the mutant DNA strand shown above:

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:

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122 A Case Study: Cystic Fibrosis

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178

Key Idea: Cystic fibrosis results from a triplet deletion from the CFTR gene, causing a non-functioning

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protein that is unable to regulate chloride ion transport.

ffCystic 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).

ffThe 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.

q

The CFTR gene on chromosome 7

CFTR protein

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:

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.

CFTR gene

Base 1630

DNA template strand

These three bases are deleted from the 507/508th triplets in the ΔF508 mutation

This triplet codes for the 500th amino acid

Normal CFTR (1480 amino acids)

Abnormal CFTR (1479 amino acids)

Correctly controls chloride ion balance in the cell

No or little control of chloride ion balance in the cell

Cell exterior

Mucus build up

Water

Cl-

The DF508 mutation causes the CFTR protein to degrade rapidly, stopping it from inserting into the plasma membrane.

Cell interior

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Chloride ions effectively removed from the cell.

CFTR protein

KNOW

CF patient receiving physiotherapy treatment

Water

Cl-

Chloride ions build up inside the cell.

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1. (a) Write the mRNA sequence for the normal template DNA strand shown in the diagram on the opposite page:

(b) Use the mRNA table on page 154 to identify the amino acids encoded by the mRNA for the normal DNA strand:

2. (a) Rewrite the mRNA sequence for the mutated DNA strand:

(b) Use the mRNA table on page 154 to identify the amino acids encoded by the mRNA for the mutant DNA strand:

(c) What amino acid did the 507th triplet encode in the normal strand and in the mutant strand: (d) What was the net effect of the three base deletion?

3. (a) Explain why the abnormal CFTR fails to transport Cl- correctly:

(b) What effect does this have on water movement in and out of the cell?

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).

Normal DNA: ACT TTG CAA CAG TGG AGG AAA GCC TTT GGA

Mutated DNA: ACT TTG CAA CAG TAG AGG AAA GCC TTT GGA

(a) Circle the mutation in the mutated DNA sequence:

(b) What general type of mutation is this?

(c) Write out the mRNA sequence for the normal DNA:

(d) Write out the mRNA sequence for the mutated DNA:

(e) Use page 154 to work out the amino acid sequence of the normal DNA:

(f) Use page 154 to work out the amino acid sequence of the mutated DNA:

(g) Explain why the mutated DNA leads to a non-functional CFTR protein:

(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:

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123 Skin Cancer: A Non-Inherited Mutation

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180

Key Idea: Skin cancer is caused by mutations in somatic cells. These mutations are not inherited

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because skin cells are not involved in the production of gametes.

ffMutations 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.

ffSkin 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).

ffIt has been estimated that there might be 1012 mutations in the cells of

UV Light

NCI

Melanoma

a tumour, occurring both before and after the formation of the tumour.

Thymine dimer

DNA of tumour suppressor gene

ffAfter 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.

ffA large number of repair mechanisms constantly repair mutations so many genes may need to be affected by mutations before cancer occurs.

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.

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1. Discuss the difference between a non-inherited disease, such as skin cancer, and an inherited disease, such as cystic fibrosis:

2. Explain why skin cancer is less likely to be seen in young people than adults:

3. Explain why a large number of mutations (generally) need to occur before cells become cancerous:

WEB

KNOW

123

LINK

53

LINK

LINK

119 120

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124 Environment and Phenotype Key Idea: An organism's phenotype is influenced by the environment in which it develops, even

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though the genotype remains unaffected.

ffEnvironmental 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.

The effect of temperature

ffThe 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).

ffColour-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 other organisms

ffThe 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. 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.

Male

Non-helmeted Daphnia

Helmeted Daphnia

Chemical signal

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ffSome organisms respond to the presence of other,

Female

1. (a) Give two examples of how temperature affects a phenotypic characteristic in an organism:

(b) Why are the darker patches of fur in colour-pointed cats and rabbits found only on the face, paws, and tail?

2. How is helmet development in Daphnia an adaptive response to environment?

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The effect of altitude Growth to genetic potential

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Severe stunting

The effect of 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.

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.

3. (a) What is a cline?

(b) What physical factors associated with altitude could affect plant phenotype?

4. Describe an example of how the chemical environment of a plant can influence phenotype:

5. Vegetable growers can produce enormous vegetables for competition. How could you improve the chance that a vegetable would reach its maximum genetic potential?

Plant species A:

Plant species B:

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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:

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183

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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:

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.

(a) Why would scientists use genetically identical organisms in gene-environment studies?

(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: Group

Identical twins raised together

88

Identical twins raised apart

69

Fraternal twins

60

Parent-child

42

Half-siblings

31

Cousins

15

Use this information to discuss the effect of genes and environment on IQ:

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% similarity in IQ


125 Genes and Environment Interact

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184

Key Idea: The environment or experiences of an individual can affect the development of subsequent

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generations.

ffStudies 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.

Mother (F0)

ffThe 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.

Fetus (F1)

Egg cells (F2)

ffAnything 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.

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. Cumulative percentage rats with breast cancer (high fat diet (HFD))

F1%

F2%

F3%

Weeks since trigger

HFD

Control

HFD

Control

HFD

Control

6

0

0

5

0

3

0

8

15

0

20

5

3

20

10

22

8

30

5

10

25

12

22

18

50

20

20

30

14

22

18

50

30

25

40

16

29

18

60

30

25

40

18

29

18

60

40

40

42

20

40

18

65

40

50

60

22

80

60

79

50

50

60

Cumulative percentage rats with breast cancer (high oestrogen diet (HOD)) F2%

F3%

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F 1%

Weeks since trigger

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

80

60

50

HOD

Control

Data source: Science News April 6, 2013

WEB

KNOW

LINK

LINK

125 118 124

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1. Use the data on the previous page to complete the graphs below. The first graph is done for you: See Appendix

% F1breast with breast cancer % F1 with cancer (HFD)(HFD)

100 100

40

40

20

20

0

0

6

100100

% rats with breast cancer

% rats with breast cancer

80 80

68 8 10 10 12 12 14 14 16 16 18 18 22 22 24 Weeks after exposure to cancer Weeks after exposure to cancer triggertrigger

High fat diet Control

60 60 40 40 20 20

High fat diet Control

60 60 40 40 20 20

0 06 6 8 8 10 10 12 12 14 14 16 16 18 18 22 22 24 24 Weeks exposure to cancer trigger Weeks afterafter exposure to cancer trigger

40

40

20

20

6

0

68 8 10 10 12 12 14 14 16 16 18 18 22 22 24 Weeks after exposure to cancer Weeks after exposure to cancer triggertrigger

80

60 40 20

0

6

8 10 12 14 16 18 22 Weeks after exposure to cancer trigger

24

100

% rats with breast cancer

% rats with breast cancer

% rats with breast cancer

80 80

% F1 with breast cancer (HFD)

60

100

0 06 6 8 8 10 10 12 12 14 14 16 16 18 18 22 22 24 24 Weeks exposure to cancer trigger Weeks afterafter exposure to cancer trigger

100100

60

0

24

% F1 with breast cancer (HFD)

80

80

60 40 20

0

6

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60

80

% rats with breast cancer

60

% rats with breast cancer

High fat diet diet 80 High fat Control Control

% rats with breast cancer

80

% rats with breast cancer

% rats with breast cancer

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100 100

8 10 12 14 16 18 22 Weeks after exposure to cancer trigger

2. (a) Which generations are affected by the original mother eating a high fat diet?

(b) Which generations are affected by the mother eating a high oestrogen diet?

(c) Which diet had the longest lasting effect?

3. What do these experiments show with respect to diet and generational effects?

24

24


126 What You Know So Far: Genotype and Phenotype

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186

Mutagens and mutations

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

HINT: Include definitions, different types of mutation, and their potential effects on phenotype and on metabolic pathways.

The effect of environment on phenotype

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HINT: Examples of how the phenotype is affected by changes in the environment.

REVISE

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127 NCEA Style Question: Genotype and Phenotype

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1. The genotype and phenotype of an organism can be affected by a variety of environmental factors. (a) Define the term mutagen. Give an example of a mutagen, its origin, and its effect:

(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).

(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).

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TEST

187


188

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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. (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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128 KEY TERMS AND IDEAS: Genotype and Phenotype

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

deletion (mutation) environment

A An agent capable of causing a change in the DNA sequence.

B The removal of a nucleotide base or bases from DNA.

enzyme

C The replacement of one nucleotide base in DNA with another.

genotype

D The genetic makeup of an organism or cell.

metabolic pathway

E A catalytic agent (usually a globular protein) in metabolic reactions.

mutagen

F Observable characteristics in an organism.

mutation

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.

phenotype

H The internal or external conditions, which may influence expression of genotype.

substitution (mutation)

I A series of enzyme controlled chemical reactions that modify an initial chemical or chemicals into an end product.

2. An original DNA sequence is shown right: GCG TGA TTT GTA GGC GCT CTG

For each of the following DNA mutations, state the type of mutation that has occurred:

(a) GCG TGT TTG TAG GCG CTC TG

(b) GCG TGA TTT GTA AGG CGC TCT G

(c) GCG TGA TTT GGA GGC GCT CTG (d) GCG TGA GTA GGC GCT CTG

3. In the following diagram label: enzyme 1, enzyme 2, end product, starting substrate, intermediate, gene 1, gene 2

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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. 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?

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TEST

189


190

Hints for Drawing Graphs

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Appendix 1

Graphs should have a concise, explanatory title.

Place the dependent variable e.g. biological response, on the vertical (Y) axis.

Place the independent variable e.g. treatment, on the horizontal (X) axis

Yield (absorbance at 550 nm)

Label both axes and provide appropriate units of measurement if necessary.

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

Overlap between these means the values not significantly different.

0.7

0.6 0.5 0.4 0.3 0.2

Sensitive strain Resistant strain

0.1

0

0

1 2 3 4 Antibiotic concentration (g m-3)

5

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.

Each axis should have an appropriate scale. Decide on the scale by finding the maximum and minimum values for each variable.

23% 17%

ffOne variable is a category ffOne variable is a count to

Use a pie graph

27%

33%

give percentage of a total

Water use key Cooling

Irrigation

Commercial

Drinking

Mangrove area in NZ harbours Herekino

ffOne variable is a category

ffOne variable is continuous data (measurements)

Whangape

Hokianga

Use a bar or column graph

Kaipara

Manakau

ffBars do not touch

Raglan

ffOne variable is continuous data (measurements)

ffOne variable is a count

6000

Weight (g)

dependent on the independent (manipulated) variable

ffA line is drawn point to point

Use a line graph

8 4 0

ffBoth variables are continuous ffThe two variables are interdependent but there is no manipulated variable

ffA line of best fit is drawn through the points

© 1994-2017 BIOZONE International

Line connecting points

12

Metabolic rate

ffThe response variable is

Temperature vs metabolic rate

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ffBoth variables are continuous

Number of eggs in brood

What type of data have you collected?

Use a histogram

2000 4000 Area (ha)

Frequency

0

0

40

Body length vs brood size

80

Use a scatter plot

10 20 30 Temperature (°C)

Line of best fit

60 40 20

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0

0

1 2 3 Body length (mm)

4


191

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Appendix 2

Questioning terms in biology

Photo credits

The following terms are often used when asking questions in examinations and assessments.

We acknowledge the generosity of those who have provided photographs for this edition: • Dartmouth College Electronic Microscope Facility • Wintec •

Interpret data to reach stated conclusions.

Annotate:

Add brief notes to a diagram, drawing or graph.

Apply:

Use an idea, equation, principle, theory, or law in a new situation.

Calculate: Find an answer using mathematical methods. Show the working unless instructed not to.

Compare: Give an account of similarities between two or more items, referring to both (or all) of them throughout.

Construct: Represent or develop in graphical form.

Contrast:

Show differences. Set in opposition.

Define:

Give the precise meaning of a word or phrase as concisely as possible.

Derive:

Manipulate a mathematical equation to give a new equation or result.

Describe: Define, name, draw annotated diagrams, give characteristics of, or an account of. Design:

Produce a plan, object, simulation or model.

Determine: Find the only possible answer.

Discuss:

Show understanding by linking ideas. Where necessary, justify, relate, evaluate, compare and contrast, or analyse.

Distinguish: Give the difference(s) between two or more items. Draw:

Represent by means of pencil lines. Add labels unless told not to do so.

Estimate:

Find an approximate value for an unknown quantity, based on the information provided and application of scientific knowledge.

Evaluate:

Assess the implications and limitations.

Explain:

Provide a reason as to how or why something occurs.

Identify:

Find an answer from a number of possibilities.

Illustrate:

Give concrete examples. Explain clearly by using comparisons or examples.

Interpret:

Comment upon, give examples, describe relationships. Describe, then evaluate.

List:

Give a sequence of answers with no elaboration.

Measure:

Find a value for a quantity.

Outline:

Give a brief account or summary. Include essential information only.

Predict:

Give an expected result.

Solve:

Obtain an answer using numerical methods.

State:

Give a specific name, value, or other answer. No supporting argument or calculation is necessary.

Suggest:

Propose a hypothesis or other possible explanation.

Summarise: Give a brief, condensed account. Include conclusions and avoid unnecessary details.

© 1994-2017 BIOZONE International

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.

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

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. 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.

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Analyse:

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Index

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192

B Base pairing rule 68, 148 Beneficial mutation 88 Beta pleated sheet 161-162 Biological catalyst 36 Black robin, genetic bottleneck 134 Blood groups, inheritance 103-104

C Calvin cycle 54, 56 Cancer 73, 179 Catabolic reaction 35, 38 Catalase activity 41 Catalyst, enzyme 36-41 Cell cycle, checkpoints 70 Cell cycle, stages of 69 Cell cytoskeleton 3 Cell division 11, 69-74, 91 Cell processes 11-12 Cell size, and diffusion 22 Cell specialisation 7, 9-10 Cell wall 3 Cell, animal 6-8 Cell, plant 3-5 Cellular energy, overview 43 Cellular organelles 3-4, 13-15 Cellular respiration 35, 43, 48, 60-61 Cellulose cell wall 3-4 Centrioles 6, 73 Channel protein 162 Chargaff's rules 148 Checkpoints, cell cycle 70 Chloroplast 3-4, 11, 53 Chromatid 144 Chromatin 144 Chromosome 144 Chromosome, homologous 83 Coding strand, of DNA 153 Codominance, of alleles 102-103 Codons 89, 154, 157 Cofactor, enzyme 39 Concentration gradient 18 Condensation reaction 147 Continuous variation 86 Cotransport 27 Crossing over 91-92 Cystic fibrosis mutation 178 Cytokinesis 69, 73-74 Cytoplasm 3, 6 Cytosine 145 Cytosis 11, 28-29 D Degeneracy, of genetic code 155 Deletion mutation 174-175 Denaturation 161-162 - of enzymes 36 Deoxyribonucleic acid 143-144, 146 Deoxyribose sugar 145 Diffusion 18, 22 Dihybrid cross 108-109, 114 Diploid 91 Directional selection 125, 129-130 Discontinuous variation 87 Disruptive selection 125

DNA 143-144, 146 DNA model, exercise 148-151 DNA, changes 88 DNA, replication of 67-68 Dominant allele 97

E Electron transport chain 44-46 Endergonic reaction 35 Endocytosis 29 Endoplasmic reticulum 3, 6 Energy, in cells 43 Environment - and phenotype 172, 181-182, 184 - and variation 84 Enzyme 162 Enzyme cofactor 39 Enzyme denaturation 36 Enzyme inhibitor 39 Enzyme reaction rate 40 Enzyme, catalase 41 Enzyme, role in metabolism 168 Enzymes 36-41 Epigenetics 172, 184 Essential amino acid 160 Eukaryotic cell 3, 6 Evolution, defined 123 Exergonic reaction 35 Exocytosis 28 Exon 144, 153 Extracellular enzyme 36

F Facilitated diffusion 18 Factors affecting gene pools 121 Factors affecting photosynthesis 57-58 Fermentation pathways 49-50 Fibrous protein 163 Fluid mosaic model 16 Founder effect 131-132

G Gametic mutation 90 Gap phase 69 Gas exchange, in photosynthesis 51 Gene 144 Gene expression 152-153, 157-158 Gene flow 121 Gene mutation 177-178 Gene pool model 127-128 Gene pools, changes in 121 Gene pools, role in evolution 121 Genetic bottleneck 133-134 Genetic code 154, 156 Genetic crosses 97-114 Genetic drift 121,135 Genetic variation, sources of 84-85 Genotype 84 - effect of mutation on 174 Geographical barrier, effect 121 Germline mutation 90 Globular protein 163 Glycolysis 44-46 Golgi apparatus 3, 6, 12 Grana 53 Guanine 145 H Haploid 91 Harmful mutation 88 Heterozygous 83 Homologous chromosome 83, 91 Homozygous 83 Hydrophilic, defined 16 Hydrophobic, defined 16 Hypertonic vs hypotonic 20-21

I Incomplete dominance, of alleles 105 Independent assortment 93 Independent assortment, law of 96 Induced fit model, enzyme activity 38 Inheritance, laws of 96 Inhibitor, enzyme 39 Insertion mutation 174-175 Interphase 69, 73 Intracellular enzyme 36 Intron 144 Ion pumps 26 Isotonic 20-21

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JK Krebs cycle 44-46

L Lactic acid fermentation 50 Leaf structure, role in photosynthesis 51-52 Lethal alleles 106 Light dependent phase 54 Light independent phase 54 Link reaction 44, 46 Linkage 110-111 Linked gene, inheritance of 110-111 Liver cell 6 Lock and key model, enzymes 38 Lysosome 6

M Mate selection 121 Meiosis 91-93 Meiosis, and variation 92-93 Meiosis, exercise 94-95 Melanin 171 Membrane permeability, factors affecting 24 Membrane proteins 16 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 121 Mitochondrion 3, 6, 12, 44-46 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 147 mRNA-amino 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 N Natural selection 121, 123-130 - model 126 - rock pocket mice 129-130 Nuclear pore 6 Nucleic acids 146-147 Nucleolus 3 Nucleosome 144 Nucleotide, base pairing rule 68, 148 Nucleotides 145 Nucleus 3, 6, 12

Pinocytosis 29 PKU 171 Plant cell 3-5 Plasma membrane 3, 6, 16-17 Plasmolysis, in plant cells 21 Plastid 4 Polypeptide chain 160 Population bottlenecks 133-134 Primary structure, of proteins 161 Primary transcript 152 Processes, cellular 11-12 Protein denaturation 161-162 Protein synthesis 11, 158-159 Protein, structure of 161 Protein, types 162-163 Proteins, enzymes 36 Proteins, membrane 16 Proton pumps 27 Purine 145 Pyrimidines 145

Q Qualitative trait 87 Quantitative trait 86 Quaternary structure, of proteins 161 R Reaction rates, of enzymes 40 Recessive allele 89, 97 Recombination 92, 112-113 Redundancy, of genetic code 155 Replication, of DNA 67-68 Respirometer 48 Respiration, measuring 48 Ribonucleic acid (RNA) 146-147 Ribose sugar 145 Ribosomes 3, 6, 157 RNA 146-147 RNA polymerase 153 rRNA 147

S S phase 69 Secondary structure, of proteins 161 Secretion, in cell 11 Segregation, law of 96 Selection 124, 129-130 Semi-conservative DNA replication 67 Sexual reproduction, variation 84-85 Shade plant 52 Sickle cell mutation 88, 176 Silent mutation 88, 175 Skin cancer mutation 180 Sodium-potassium pump 26 Somatic mutation 90, 180 Specialisation, in cells 7, 9-10 Stabilising selection 125 Stomata 51 Stroma lamellae 53 Sub-unit protein 162 Substitution mutation 174, 176 Substrate level phosphorylation 46 Sun plant 52 Surface area: volume ratio 22 T Tasmanian devil, bottleneck 134 Temperature, and membrane permeability 24 Template strand, of DNA 153 Tertiary structure, of proteins 161 Test cross 100 Thylakoid 53 Thymine 145 Tonicity 20-21 Tonoplast 3, 24 Trait 86, 97 Transcription 152-153 Translation 152, 157-158 Transport, cellular 11, 17-22, 25-30 tRNA 147, 157-158 Turgor pressure 21

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A Activation energy, of enzymes 37 Active site, of enzymes 36 Active transport 25-30 Adenine 145 Aerobic respiration 46-47 Albinism 171 Alcoholic fermentation 49 Alleles 83 - codominance of 102-103 - incomplete dominance of 105 - lethal 106 Alpha helix 161-162 Amino acid table 154 Amino acid, structure of 160 Amyloplast 3-4 Anabolic reaction 35, 38 Anaerobic metabolism 49-50 Animal cell 6-8 Anticodon 147, 158 Asexual reproduction 85 ATP 42-45 - role in active transport 30 ATP synthase 27 Autolysis 11

O Organelles, cellular 3-4, 6, 13-15 Osmosis 20-21 Osmotic potential 20 Oxidative phosphorylation 46

P Particulate inheritance 96 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 Photosynthesis 11, 35, 43, 51, 54-61 Photosynthetic rate, measuring 59 Photosystem 56

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UV Uracil 145 Vacuole 3 Variation, genetic 84-85 WXYZ Water movement 20 Zygote 71

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