AP Biology 1 Student Edition

Page 1


AP BIOLOGY 1 Student Edition Meet the writing team

Tracey

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

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.

Senior Author

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 Kumbuka, the 16-year-old silverback western lowland gorilla (Gorilla gorilla gorilla) from London Zoo. The western lowland gorilla is one of two subspecies of western gorilla occupying montane, primary and secondary forests and lowland swamps in central Africa. Wild populations are threatened by the bushmeat trade, habitat loss, and low fertility. Image: Gary Brookshaw ŠBrookshaw Photography

Richard

Founder & CEO

Thanks to: The staff at BIOZONE, including Mike Campbell and Holly Coon for design and graphics support, Paolo Curray and Malaki Toleafoa for IT support, Allan Young and Arahi Hippolite for office handling and logistics, and the BIOZONE sales team. Special thanks to Jason Crean and Cherylann Hollinger for their advice and lively discussion throughout. Second edition 2017

ISBN: 978-1-927309-62-9 Copyright Š 2017 Richard Allan Published by BIOZONE International Ltd Printed by Walsworth Print Group www.wpcdirect.com

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

BIOZONE Corporation USA and Canada FREE phone: 1-855-246-4555 FREE fax: 1-855-935-3555 Email: sales@thebiozone.com Web: www.thebiozone.com

AP Biology 1 2E 2017.indb 2

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 workbook may not be re-sold. The conditions of sale specifically prohibit the photocopying of exercises, worksheets, and diagrams from this workbook for any reason.

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Contents

Using the Student Edition ....................................vi Using the Tab System ...................................... viii AP Biology Course Guide .................................. ix AP Biology: A Concept Map ................................ x Using BIOZONE'S Weblinks and Biolinks ......... xii

Science Practices Essential Skills for AP Biology

Essential knowledge .......................................... 1 How Do We Do Science? ................................... 2 Systems and Systems Models............................ 4 Accuracy and Precision ...................................... 6 Variables and Data.............................................. 8 Planning a Quantitative Investigation.................. 9 Working with Numbers ..................................... 11 Fractions, Percentages, and Ratios ................. 12 Logs and Exponents......................................... 13 Properties of Geometric Shapes ...................... 14 Practicing with Data ......................................... 15 Apparatus and Measurement............................ 16 Using Table and Graphs ................................... 17 Constructing Graphs ........................................ 18 Interpreting Line Graphs................................... 19 Correlation or Causation?................................. 20 Mean, Median, Mode........................................ 21 Spread of Data.................................................. 23 Reliability of the Mean ...................................... 24 Interpreting Sample Variability.......................... 26 Which Test to Use?........................................... 28

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

EK 2.A., 3.A. The Biochemistry of Life

Essential knowledge ........................................ 29

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

CODES:

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The Biochemical Nature of the Cell................... The Role of Water............................................. The Properties of Water.................................... Organic Molecules ........................................... Building an Organism........................................ Biochemical Tests.............................................. Nucleotides....................................................... Nucleic Acids..................................................... Amino Acids ..................................................... Protein Structure............................................... Protein Shape is Related to Function ............... Comparing Fibrous and Globular Proteins ....... Modification of Proteins..................................... Lipids ................................................................ Phospholipids ................................................... Carbohydrate Chemistry................................... Condensation and Hydrolysis of Sugars .......... Colorimetry ....................................................... Polysaccharides ............................................... Cellulose and Starch......................................... Cell Sizes ......................................................... Limitations to Cell Size ..................................... Investigating the Effect of Cell Size .................. KEY TERMS AND IDEAS: Did You Get It? .......

Activity is marked:

to be done

30 31 32 33 34 35 36 37 39 40 41 42 44 45 47 48 49 50 51 53 54 55 56 58

EK 2.B., 4.A. Cell Structure and Processes

Essential knowledge ........................................ 59

45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76

Optical Microscopes ......................................... 61 Preparing a Slide ............................................. 63 Staining a Slide ................................................ 64 Measuring and Counting Using a Microscope . 65 Calculating Linear Magnification ...................... 66 Making Biological Drawings.............................. 67 Prokaryotic Cells............................................... 69 Animal Cells ..................................................... 70 Identifying Structures in an Animal Cell ........... 71 Plant Cells ........................................................ 72 Identifying Structures in a Plant Cell ................ 73 The Cell's Cytoskeleton .................................... 74 Cell Walls.......................................................... 75 Cell Structures and Organelles ........................ 76 The Structure of Membranes ........................... 78 How Do We Know? Membrane Structure......... 80 Cell Processes ................................................. 81 Diffusion ........................................................... 83 Factors affecting Membrane Permeability......... 84 Investigating Transport Across the Membrane.. 86 Investigating Diffusion ...................................... 87 Osmosis ........................................................... 88 Water Relations in Plants ................................. 89 Making Dilutions ............................................... 91 Estimating Osmolarity of Cells.......................... 92 Active Transport................................................. 93 Ion Pumps......................................................... 94 Disturbances to Ion Transport .......................... 95 Membranes and the Export of Proteins ........... 96 Endocytosis....................................................... 98 Active and Passive Transport Summary ........... 99 KEY TERMS AND IDEAS: Did You Get It? ...... 100

EK 2.A., 3.A. Cellular Communication

Essential knowledge ..................................... 101 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91

Signals and Signal Transduction..................... 103 The Nature of Signaling Molecules ................ 105 Communication Among Unicellular Organisms.. 106 Types of Cell Signaling.................................... 108 Cell to Cell Communication............................. 110 Local Regulators............................................. 111 Action of Insulin .............................................. 112 Hormones are Regulated by Negative Feedback ....................................................... 113 Types of Signal Transduction (Overview) ....... 114 Signal Transduction Using Second Messengers........................................ 116 Signal Transduction Involving Protein Modification ........................................ 117 Cell Signaling and DNA Repair ...................... 118 Effect of Blocking Signals ............................... 119 KEY TERMS AND IDEAS: Did You Get It? ..... 121 Synoptic Questions ........................................ 122

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Contents

EK 3.A., 4.A. DNA and RNA

Essential Knowledge....................................... 126 92 Genomes, Genes, and Alleles ........................ 128 93 Prokaryotic Chromosomes ............................. 129 94 Plasmid DNA .................................................. 130 95 Eukaryotic Chromosome Structure ................ 131 96 Does DNA Really Carry the Code? ................. 133 97 The Evidence for DNA Structure .................... 135 98 RNA Molecules .............................................. 136 99 RNAi can Control Gene Expression ............... 137 100 Creating a DNA Molecule ............................... 138 101 DNA Replication ............................................. 141 102 Enzyme Control of DNA Replication .............. 143 103 Meselson and Stahl's Experiment................... 144 104 What is Gene Expression? ............................. 146 105 The Genetic Code........................................... 147 106 Transcription is the First Step in Gene Expression ............................................ 148 107 mRNA Processing in Eukaryotes.................... 150 108 Translation ...................................................... 151 109 Gene Expression Summary............................ 153 110 Protein Activity Determines Phenotype........... 154 111 What is Genetic Engineering? ....................... 155 112 Making Recombinant DNA.............................. 156 113 New Tools: Gene Editing With CRISPR........... 158 114 Polymerase Chain Reaction ........................... 159 115 Gel Electrophoresis ........................................ 161 116 Preparing a Gene for Cloning ......................... 163 117 Gene Cloning ................................................. 164 118 Aseptic Technique and Streak Plating............. 166 119 Testing for Transformation .............................. 168 120 What is Transgenesis? ................................... 169 121 Vectors for Transgenesis ................................ 170 122 Applications of Genetic Manipulation ............. 171 123 Using Recombinant Plasmids in Industry ...... 172 124 Engineering for Improved Nutrition ................ 174 125 Production of Insulin ....................................... 176 126 Food for the Masses ....................................... 178 127 KEY TERMS AND IDEAS: Did You Get It? ..... 180

EK 3.A. Chromosomes and Cell Division

Essential Knowledge ..................................... 181 128 Cell Division ................................................... 129 The Eukaryotic Cell Cycle.............................. 130 Regulation of the Cell Cycle ........................... 131 Defective Gene Regulation in Cancer ............ 132 Mitosis ............................................................ 133 Mitosis and Cytokinesis .................................. 134 Modeling Mitosis ............................................ 135 The Effect of Environment on Mitosis ............. 136 Meiosis ........................................................... 137 Meiosis and Variation...................................... 138 Modeling Meiosis ........................................... 139 Mitosis vs Meiosis .......................................... 140 Crossing Over Problems ................................ 141 KEY TERMS AND IDEAS: Did You Get It? ..... CODES:

Activity is marked:

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182 183 184 186 188 189 191 192 193 194 195 197 198 199

EK 3.A. The Chromosomal Basis of Inheritance

Essential Knowledge ..................................... 200

142 Alleles.............................................................. 143 Mendel's Pea Plant Experiments ................... 144 Mendel's Laws of Inheritance ......................... 145 Basic Genetic Crosses ................................... 146 Monohybrid Cross .......................................... 147 Codominance ................................................. 148 Codominance in Multiple Allele Systems........ 149 Incomplete Dominance .................................. 150 Lethal Alleles .................................................. 151 Problems Involving Monohybrid Inheritance .. 152 Dihybrid Cross ................................................ 153 Inheritance of Linked Genes .......................... 154 Recombination and Dihybrid Inheritance ....... 155 Detecting Linkage in Dihybrid inheritance....... 156 Using the Chi-Squared Test in Genetics ........ 157 Chi-Squared Exercise in Genetics ................. 158 Problems Involving Dihybrid Inheritance ........ 159 Sex Linkage ................................................... 160 Inheritance Patterns ....................................... 161 Pedigree Analysis ........................................... 162 Hunting for a Gene ......................................... 163 Genetic Counseling ........................................ 164 Polygenes ...................................................... 165 Sex Determination .......................................... 166 Sex Limited Characteristics ............................ 167 Nonnuclear Inheritance .................................. 168 KEY TERMS AND IDEAS: Did You Get It? .....

201 202 203 204 205 206 207 209 210 211 212 213 215 217 218 219 220 221 223 224 226 227 229 231 232 233 234

EK 2.E., 3.B, 4.A Regulation of Gene Expression Essential Knowledge ..................................... 235

169 Cellular Differentiation .................................... 170 Gene Expression and Development................ 171 Homeotic Genes and Development................ 172 The Timing of Development............................ 173 Factors Regulating Seed Germination............ 174 Mutations and Development............................ 175 Gene Transplantation Experiments................. 176 miRNA and Development............................... 177 Apoptosis: Programmed Cell Death................ 178 Structural and Regulatory Genes.................... 179 Gene Regulation in Prokaryotes..................... 180 Eukaryotic Gene Structure and Regulation .... 181 KEY TERMS AND IDEAS: Did You Get It? .....

237 238 239 241 242 243 244 245 246 247 248 250 252

EK 3.C., 4.C. Sources of Variation

Essential Knowledge ...................................... 253

182 Sources of Genetic Variation .......................... 183 Gene-Environment Interactions ..................... 184 Using the Student's t Test ............................... 185 Quantitative Investigation of Variation ............ 186 Mutagens ....................................................... 187 Mutations ........................................................ 188 Beneficial Mutations .......................................

255 257 259 261 263 264 266

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Contents

189 Types of Gene Mutations ............................... 267 190 Inherited Metabolic Disorders ........................ 270 191 Sickle Cell Mutation ........................................ 271 192 Chromosome Mutations.................................. 272 193 Gene Duplication ............................................ 273 194 Non-disjunction can Produce Aneuploidies .... 274 195 Aneuploidy in Sex Chromosomes .................. 275 196 Polyploidy as a Source of Variation................. 276 197 The Genetic Basis of Resistance in Bacteria... 278 198 Replication in Bacteriophages ....................... 279 199 Consequences of Lysogeny ........................... 280 200 Replication in Animal Viruses ......................... 281 201 Antigenic Variability in Influenzavirus ............. 283 202 HIV Evolves Rapidly ....................................... 284 203 KEY TERMS AND IDEAS: Did You Get It? ..... 285 204 Synoptic Questions ........................................ 286

EK 1.A. Genetic Change in Populations Essential Knowledge ...................................... 289 205 A Pictorial History of Evolutionary Thought .... 291 206 Darwin's Theory .............................................. 293 207 Adaptation and Fitness .................................. 294 208 Sexual Selection ............................................ 296 209 Gene Pools and Evolution .............................. 297 210 Gene Pool Exercise ......................................... 299 211 Changes in a Gene Pool ................................. 301 212 Calculating Allele Frequencies in Populations... 302 213 Analysis of a Squirrel Gene Pool .................... 304 214 Types of Natural Selection ............................. 306 215 Stabilizing Selection For Human Birth Weight .. 307 216 Selection for Skin Color in Humans ............... 309 217 Heterozygous Advantage ............................... 310 218 Directional Selection in Moths ........................ 312 219 Insecticide Resistance.................................... 313 220 The Evolution of Antibiotic Resistance............ 314 221 The Founder Effect ......................................... 315 222 Population Bottlenecks ................................... 317 223 Genetic Drift ................................................... 319 224 Artificial Selection in Animals ......................... 320 225 Selection in Livestock ..................................... 321 226 Selection and Population Change................... 322 227 Artificial Selection in Crop Plants ................... 323 228 Breeding Modern Wheat ................................ 325 229 Selection in Fast Plants .................................. 326 230 KEY TERMS AND IDEAS: Did You Get It? ..... 328

231 The Evidence for Evolution ............................ 232 Fossils ............................................................ 233 Methods of Dating Fossils .............................. 234 Interpreting the Fossil Record ........................ 235 Chronometric Dating....................................... 236 The Evolution of Horses ................................. 237 Homologous Structures ................................. 238 Vestigial Structures ........................................

CODES:

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239 Homologous Proteins...................................... 240 Molecular Clocks ............................................ 241 Homologous DNA Sequences ....................... 242 Gene Duplication and Evolution ..................... 243 Evolution of Novel Forms ............................... 244 Analyzing Data from Extant Populations ........ 245 The Role of Master Genes ............................. 246 Biogeographical Evidence .............................. 247 Oceanic Island Colonizers ............................. 248 Continental Drift and Evolution ....................... 249 KEY TERMS AND IDEAS: Did You Get It? .....

330 331 333 334 336 338 339 340

341 343 344 345 346 347 348 350 351 354 358

EK 1.B. The Relatedness of Organisms

Essential Knowledge ..................................... 359

250 Descent and Common Ancestry .................... 251 The Origin of Eukaryotes ............................... 252 What is a Phylogenetic Tree? ......................... 253 The Phylogeny of Animals ............................. 254 Constructing Phylogenies Using Cladistics .... 255 Why are Birds Dinosaurs? ............................. 256 Constructing a Cladogram .............................

360 362 363 365 366 368 369

EK 1.C. Speciation and Extinction

Essential Knowledge ..................................... 370

257 Divergence is an Evolutionary Pattern ........... 258 What is a Species? ........................................ 259 Prezygotic Isolating Mechanisms ................... 260 Postzygotic Isolating Mechanisms ................. 261 Allopatric Speciation ...................................... 262 Sympatric Speciation ..................................... 263 Speciation in Action ........................................ 264 Rate of Evolutionary Change ......................... 265 Chloroquine Resistance in Protozoa .............. 266 Drug Resistance in HIV .................................. 267 Evolution in E. coli .......................................... 268 Adaptive Radiation in Mammals ..................... 269 Adaptive Radiation in Ratites ......................... 270 Convergent Evolution ..................................... 271 Coevolution .................................................... 272 Extinction ........................................................ 273 Causes of Mass Extinctions ........................... 274 The Sixth Extinction ....................................... 275 KEY TERMS AND IDEAS: Did You Get It? .....

EK 1.A. Evidence for Biological Evolution

Essential Knowledge ..................................... 329

371 372 373 376 377 379 380 382 383 384 385 386 388 390 392 394 395 396 398

EK 1.D. The Origin of Living Systems

Essential Knowledge ..................................... 399

276 The Origin of Life on Earth ............................. 277 An RNA World ................................................ 278 Prebiotic Experiments .................................... 279 Landmarks in Earth's History ......................... 280 KEY TERMS AND IDEAS: Did You Get It? ..... 281 Synoptic Questions ........................................

400 402 403 404 406 407

Image credits and data for birthweights.................. 410 Index ...................................................................... 411 when completed

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vi

Using The Student Edition Activities make up most of this book. These are usually presented as short instructional sequences allowing you to build a deeper understanding of core concepts and content and the science practices that accompany them as you progress through each chapter. Each activity is accompanied by questions or specific tasks for you to complete.

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

Activity pages

Did you get it?

Synoptic questions

Describes a checklist of essential knowledge and the associated activities.

These make up the bulk of the book. Each covers one concept and leads in with a key idea capturing the main focus of the page.

Tests your knowledge and understanding of key terms and ideas in the chapter.

These conclude each major section of work and can be used as a formal assessment of the content in the preceding chapters.

345

242 Gene Duplication and Evolution Key Idea: Gene duplication can provide new genes for natural selection to act on. Gene duplication can potentially create more genetic information, which can then be influenced by selection pressures. Studies have shown that gene duplication has occurred in virtually every species, often multiple times. The duplication of a gene frees one of those genes to develop a

new function, while the original continues as it was. Duplicate genes tend to accumulate mutations faster than a single copy of a gene as there is a "backup" gene able to perform the original function. Mutations that cause the duplicated gene to perform a new beneficial function are likely to be retained and lead to evolution. Entire genomes can also be duplicated, a somewhat common occurrence in plants.

Experimental evidence of gene duplication enhancing fitness

A

When Susumu Ohno first expressed the idea that gene duplication could be an important evolutionary force, he argued that the gene was duplicated first, then subjected to selection pressures. However, this required that a duplicate gene should be able to "wait around" for a selection pressure to present itself, something unlikely under natural selection.

Original gene A with primary and secondary roles

A

Ohno's model was later modified to one in which the gene could already perform a secondary function (as the result of a mutation) and duplication then allowed that function to be enhanced.

B

Gene A continues with original roles.

Gene B accumulates mutations that enhance the secondary role.

A

This was demonstrated by Dan Anderson and John Roth in an experiment with the bacteria Salmonella enterica. They grew a strain missing a gene for expressing the amino acid tryptophan. A second gene with a different primary role also had a weak ability to perform the missing gene's role. The researchers encouraged S. enterica. to duplicate the second gene. This copy gathered multiple mutations and, after 3000 generations, was able to perform the complete role of the missing gene, manufacturing tryptophan.

B Gene B takes over gene A's secondary role as its primary role.

Hemoglobin: the role of gene duplication and mutation Ancestral globin gene

Millions of years ago

800

400

α

α

200

β α

ζ

α

γ

β

ζ

ψζ ψα2 ψα1 α2 α1 ψθ

ε

ψβ

δβ β

β-globin gene family is found on chromosome 11

407

281 Synoptic Questions 1. The average birth weight for babies born in the US is 3.5 kg but the range of typical birth weights is quite narrow. This is an example of:

2. Circle the graph below which shows directional selection: (a)

(b)

A

The preserved remains or traces of a past organism.

B

Method that gives an absolute time since the material was made plus or minus a standard error. Techniques include radioactive dating and electron spin resonance.

(a) Directional selection

evolution

C

The science determining the relative order of past events, without necessarily determining their absolute age

(b) Disruptive selection

fossil

D

The evolutionary history or genealogy of a group of organisms represented as a ‘tree’ showing descent of new species from the ancestral one.

half-life

E

The time taken for a quantity to reduce to half the original amount.

phylogenetic tree

F

A branch of geology concerned with the order of position of strata and their relationship to the geological time line.

radioisotopic decay

G

The change in heritable characteristics of biological population over many successive generations.

(a) Mutations

(a) Mutation

relative dating

H

The distribution of plants and animals over geographical areas.

(b) Homologous structures

(b) Extinction

I

The spontaneous breakdown of an atomic nucleus resulting in the release of matter and energy from the nucleus and the production of a new atom.

(c) Vestigial structures

(c) Convergent evolution

(d) Analogous structures

(d) Adaptive radiation

biogeography chronometric dating

stratigraphy

2. Comparing fossils is a useful tool in determining relatedness between organisms, but even if fossils did not exist (no extinct organisms were fossilized) how would we still be able to tell all life on Earth evolved from a common ancestor?

(c)

(c) Stabilizing selection (d) None of the above 3. The diagrams (far right) show the forelimbs of a whale, bird, and human. These structures share many of the same bones. These type of structures are called:

4. The structures pictured right arose through:

Human

Bird

Whale

5. The statements (A)-(D) below describe scenarios in hypothetical populations: A A forest fire burns through an area containing black and gray squirrels. It kills 70% of the black squirrels and 30% of the gray squirrels. B Dark colored mice are able to avoid being eaten by owls more easily than light colored mice. C Deer from a population become isolated from another population because of a huge slip blocking a hilltop pass.

3. Compare and contrast DNA hybridization and DNA sequence comparison as methods for generating phylogenies:

D A plant nursery is infected with a fungal infection that affects the leaves of plants. Plants with thick leaf cuticles are more likely to survive. In which of the scenarios is genetic drift likely to have a significant effect on allele frequencies? (a) A

(b) B and D

(c) A and C

(d) All of the above

6. A river changes its course separating a population of snails into two. They can no longer interact and over time the two populations become separate species. Identify the type of speciation involved?

f The third event was the transposition (change in position) of these genes to different chromosomes (3). Further duplications and mutations produced the wide family of globin proteins found in humans (4).

0 α-globin gene family is found on chromosome 16

249 KEY TERMS AND IDEAS: Did You Get It? 1. Match each term to its definition, as identified by its preceding letter code.

f The globin proteins are a family of ironcontaining oxygen binding proteins. The evolution of human hemoglobin genes has involved both duplication and mutation. f Duplication of an ancestral globin gene (1), followed by an accumulation of mutations to both copies (2) produce the a- and b-hemoglobin genes. This was a significant event because it allowed for the formation of the large mutliunit haemoglobin protein from the different subunits (a- and b-globins).

β

358

(a) Sympatric speciation

f Myoglobin, another globin gene, was also formed by duplication, mutation, and transposition. Myoglobin has an oxygen storage role in muscle and the gene is found on chromosome 22.

4. Insects are extremely adaptable and have a wide range of body forms. Consider the wing structure of the insects below:

(b) Adaptive radiation

(c) Allopatric speciation

(d) Artificial selection

7. The diagram below right represents part of a sedimentary rock profile from two states in the USA. Identify which statement is incorrect: South Carolina

(a) Species C is the youngest (b) Species B was not found in Virginia

Fish species A

A. Forewings

(c) Species A is the youngest

D. Hindwings

Fly

(b) What age of rocks are missing from the profile in Virginia?

(b) What does the homology of these structures indicate?

©2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

CONNECT

CONNECT

AP1

AP1

250

193

WEB

242

TEST

KNOW

(a) What sort of dating is involved here?

Beetle

(a) Use the letters to identify the wing structures that are homologous on the images above:

©2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

Fish species A

Fish species B

8. Again, with reference to the diagram right:

C. Halteres E. Wing covers

Butterfly

©2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

Fish species C

Middle Eocene

B. Hindwings

2. Outline how a number of different globin proteins have been produced from a single ancestral globin gene:

Virginia

(d) The order of fossils makes it possible to correlate rock layers of the same age

Early Eocene

1. What evidence is there that gene duplication can lead to enhanced gene function?

Fish species C

TEST

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vii Understanding the activity coding system and making use of the online material identified will enable you to get the most out of this book. 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 program of independent and group-based work. Look out for these features and know how to use them: The chapter introduction provides a summary of the essential knowledge and skills required for the topic, phrased as a set of learning tasks. Use the check boxes to identify and mark off the points as you complete them. Activities to support required AP investigations are identified by a blue flag. A list of key terms for the chapter is also provided, from which you can construct your own glossary.

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

184

Chromosomes and Cell Division

Enduring Understanding

3.A

130 Regulation of the Cell Cycle Key Idea: Regulatory checkpoints are built into the cell cycle to ensure that the cell is ready to proceed from one phase to the next. The failure of these systems can lead to cancer. Cell checkpoints give cells a way to ensure that all cellular processes have been completed correctly before entering the next phase. There are three checkpoints in the cell cycle.

cancer

3.A.2 In eukaryotes, heritable information is passed on via the cell cycle and mitosis or meiosis and fertilization Essential knowledge

cell cycle checkpoint chromatid

(a) The cell cycle is a complex set of stages that is regulated via checkpoints, which determine the ultimate fate of the cell

crossing over

c 1

Outline the events in interphase to include growth (G1), DNA synthesis (S), and preparation for mitosis (G2).

c 2

Explain how the cell cycle is regulated by checkpoints. Describe the internal and external signals that provide stop and go signs at checkpoints (e.g. MPF). Explain how cancer results from disruptions to the cell cycle controls.

fertilization

c

G1

PR-7

c 3

Explain how cyclins and cyclin-dependent kinases control the cell cycle

c 4

Understand that mitosis alternates with interphase in the cell cycle

c 5

Describe how specialized cells may enter a stage where they no longer divide, but reenter the cell cycle when given certain cues. Understand that nondividing cells may exit the cell cycle or remain static at a particular stage.

c

PR-7

cytokinesis diploid DNA replication

G2 gamete haploid homologous chromosomes independent assortment

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

Activity number

Interphase

129 130 131

G2 checkpoint

130 129

Investigate the effect of environmental factors on mitosis [part 2].

129 130

135

Pass this checkpoint if: • Cell is large enough. • Chromosomes have been successfully duplicated.

Mitosis

131

Compare karyotypes of normal and cancerous cells [part 3].

At each checkpoint, a set of conditions determines whether or not the cell will continue into the next phase. Cancer can result when the pathways regulating the checkpoints fail. Non-dividing cells enter a resting phase (G0), where they may remain for a few days or up to several years. Under specific conditions, they may re-enter the cell cycle.

Checkpoints during the cell cycle

G1 checkpoint

Key terms

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.

Resting phase (G0) • Cells may exit the cell cycle in response to chemical cues. • Cells in G0 may be quiescent, differentiated, or senescent (aged). • Quiescent (waiting) cells may reenter the cycle in response to chemical cues.

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

M phase meiosis mitosis S phase Cancerous cells

spindle Afunguy

(b) Mitosis passes a complete genome from the parent cell to daughter cells Know that mitosis occurs after DNA replication and state why.

c 2

Know that mitosis is followed by cytokinesis and state the result of this.

c 3

Describe the roles of mitosis in the life cycle of organisms.

c 4

Describe mitosis as a continuous process and identify the order of events and the main features of each stage.

c

PR-7

129 128 139 132 133

Model chromosome behavior and the events in mitosis [part 1].

Skin cancer (melanoma). The cancer cells grow more rapidly than the normal skin cells because normal cell regulation checkpoints are ignored. This is why the cancerous cells sit higher than the normal cells and can rapidly spread (metastasize).

Many fully differentiated (specialized) cells, e.g. neurons (above), exit the cell cycle permanently and stay in G0. These cells continue their functional role in the body, but do not proliferate. Senescent cells have accumulated mutations, lose function, and die.

136

c 1

Explain how meiosis ensures that gametes each receive one complete haploid set of chromosomes (1n).

c 2

Describe the behavior of homologous chromosomes in prophase and metaphase of meiosis I. Explain the significance of the random orientation of chromosome pairs on the spindle axis [also see 3.A.3.b.1].

136 137 138

c 3

Describe the events during anaphase of meiosis I and explain their significance.

136 137

c 4

Describe crossing over in prophase of meiosis I and explain its significance.

137 140

c 5

Explain how fertilization unites genetically dissimilar gametes, increasing genetic 128 136 variation in populations, and restoring the diploid number of chromosomes. 137

c

PR-7

2. In terms of the cell cycle and the resting phase (G0), distinguish between the behavior of fully differentiated cells, such as neurons, and cells that are quiescent, such as B memory cells

138 WEB

KNOW

A TASK CODE identifies the type of activity. For example, is it primarily information-based (KNOW) or does the content prepare for or support one of the 13 AP Investigations (PRAC)? A full list of codes is given on the following page but the codes are relatively self explanatory.

B memory cell

Most lymphocytes in human blood are in the resting G0 phase and remain there unless they are stimulated by specific antigens to reenter the cell cycle via G1. G0 phase cells are not completely dormant, continuing to carry out essential cell functions in reduced form.

1. Explain the importance of cell cycle checkpoints:

(c) Meiosis, a reduction division, followed by fertilization ensures genetic diversity in sexually reproducing organisms

Model chromosome behavior and the events in meiosis [part 4].

Normal cells

NCI

129

c 1

130

CONNECT

CONNECT

AP1

AP1

88

176

WEB tabs 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, 3-D model, or quiz. Bookmark the Weblinks page (see page ix) and visit it frequently as you progress through the book.

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CONNECT tabs identify related concepts and content across the entire AP program. The tab identifies the relevant activity number in either AP1 or AP2.

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(a) 21 mL:

(b) 48 mL:

(c) 9 mL:

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

viii

Using the Tab System (b) A 10 mL pipette delivers a measured volume of 9.98 mL:

The tab system is a useful way to quickly identify science practices and connected ideas across the AP program. They also indicate the type of task involved and show whether or not the activity is supported online. ff The CONNECT tabs indicate concept and content connections across Big Ideas and their Enduring Understandings. The connections made some of the important you can yourboth own using the concept map0.02 provided on yet pages x-xi.percentage (c)are The pipettes used ones, in (a)butand (b) make above under-delivered mL, the a particularthe Science Practice is emphasized in the activity. A guide to the picture ff The PRACTICES picture Usecodes this indicate data tothat describe effect of volume on percentage error: codes and the science practice they represent is provided below. There may be more than one or none.

errors

ff The WEBLINKS code is always the same as the activity number on which it is cited. On visiting the WEBLINKS page (opposite), find the number and it will correspond to a 3D model or one or more external websites providing an explanatory video or animation. Occasionally, a WEBLINK may provide a link to a spreadsheet model (provided by BIOZONE).

WEB

KNOW

CONNECT

38

11

AP1

CONNECT

68 AP1

CONNECT

64

PRACTICES

PRACTICES Picture codes indicate where one of the 7 science practices is emphasized. See guide below.

AP2

TASK CODES

WEBLINKS

CONNECT

These identify the nature of the activity

Bookmark the weblinks page: www.thebiozone.com/weblink/ AP1-9629 Access the external URL for the activity by clicking the link.

Concept and content connections are made to other activities within AP1 or in AP2. Use these to reinforce connections between Big Ideas and their Enduring Understandings across all topics.

COMP = comprehension of text DATA = data handling and interpretation KNOW = content you need to know PRAC = supports one of the 13 AP Biology investigations REFER = reference - use for information SKILL = supporting a mathematical or practical skill TEST = test your understanding

Recognizing the Science Practices Codes PRACTICES

SCIENCE PRACTICE 1

PRACTICES

PRACTICES

SCIENCE PRACTICE 2

PRACTICES

Use mathematics appropriately, including justifying the use of mathematical routines, applying mathematical routines, and making numerical estimates.

PRACTICES

SCIENCE PRACTICE 3 Engage in scientific questioning to extend thinking or to guide investigations, including posing, refining, and evaluating scientific questions

PRACTICES

SCIENCE PRACTICE 5

Perform data analysis and evaluation of evidence, including analyzing data to identify patterns or relationships and evaluating evidence provided by data in relation to a particular question.

Use representations and models to communicate scientific phenomena and solve scientific problems. Includes creating, describing, refining, and using representations and models of natural or man-made phenomena and systems.

SCIENCE PRACTICE 6 Work with scientific explanations and theories, including justifying claims with evidence, constructing explanations and making claims and predictions about natural phenomena.

PRACTICES

SCIENCE PRACTICE 7 Connect and relate knowledge across various scales, concepts, and representations in and across domains. Includes connecting phenomena and models across scales such as time, size, and complexity, and describing how enduring understandings and/or big ideas are connected.

SCIENCE PRACTICE 4 Plan and implement data collection strategies appropriate to a particular scientific question. Includes posing, refining, and evaluating scientific questions.

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Using BIOZONE's Weblinks and Biolinks WEBLINKS is an online resource compiled by BIOZONE to enhance or extend the content provided in the activities, largely though explanatory animations and short videos. WEBLINKS also provides information relevant to research projects or as a starting point for group work. WEBLINKS comprises external URLS as well as 3D models provided by BIOZONE. External sites have been selected for their suitability and accuracy and are regularly checked. Any errata for the student book or model answers are also posted on the WEBLINKS page.

www.thebiozone.com/weblink/AP1-9629 This WEBLINKS page provides links to external websites and 3D models supporting the activities. The external links are distinct from those provided in the BIOLINKS area of BIOZONE's web site. For the most part, they are narrowly focused 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

Activity number and title in the book

Hyperlink to the external website

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

BIOLINKS is a database of useful sites, organized into broad subject areas. Use it to locate information on specific topics or find out more about an area you are interested in Access the BIOLINKS database directly from the homepage of BIOZONE's website.

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AP Biology: Concepts and Connections This map shows the structure of the AP Biology program, showing the four Big Ideas, their Enduring Understandings in blue ovals and some of the connections between them. You can draw in your own connections and add labels with keys words to show how different parts of the program relate to each other. The white clouds with blue text indicate content in AP Biology 1. The gray clouds with gray text indicate content covered in AP Biology 2.

BIG IDEA 1 The process of evolution drives the unity and diversity of life

1.A. Change in the genetic makeup of a population over time is evolution

1.B. Organisms are linked by lines of descent from common ancestry

1.C. Life continues to evolve within a changing environment

1.D. The origin of living systems is explained by natural processes

Reactions of life occurred in a suitable medium

Monomers as buildng blocks

Variation is subject to natural selection

Sexual reproduction and variation

Variation is subject to natural selection

Viral replication and life cycles

3.A. Heritable information provides for continuity of life

3.B. Expression of genetic information involves cellular and molecular mechanisms

Metabolic pathways

Communication is adaptive

Signal transduction pathways

3.C. The processing of genetic information is imperfect and is a source of genetic variation

3.D. Cells communicate by generating, transmitting, and receiving chemical signals

Signal transduction pathways

3.E. Transmission of information results in changes within and between biological systems

DNA and RNA

BIG IDEA 3 Living systems, store, retrieve, transmit, and respond to information essential to life’s processes

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BIG IDEA 2 Biological systems use free energy and molecular building blocks to grow, reproduce, and maintain homeostasis

2.A. Growth, reproduction, and maintenance of the organization of living systems require free energy and matter

2.B. Growth, reproduction, and dynamic homeostasis require that cells create and maintain internal environments.

2.C. Organisms use feedback mechanisms to regulate growth and reproduction, and maintain dynamic homeostasis

2.D. Growth and dynamic homeostasis of a biological system are influenced by changes inthe system’s environment

2.E. Many processes involved in growth, reproduction, and homeostasis include temporal regulation and coordination

Signal transduction pathways

Properties of membranes

Ecosystem stability and change

Mitochondria Human impact on ecosystems

Chloroplasts

Compartments in cells

Species interactions

Food chains and webs Molecular diversity and range of function

Energy allocation

Antibodies and immune response Species interactions

Molecular diversity and range of function

4.A. Interactions within biological systems lead to complex properties

Compartments in cells

4.B. Competition and cooperation are important aspects of biological systems

4.C. Naturally occurring diversity among and between components within biological systems affects interactions with the environment

BIG IDEA 4 Biological systems interact, and these systems and their interactions possess complex properties

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xii

AP Biology Guide The AP biology program is organized into four underlying big ideas. The guide below lists the enduring understandings for Big Idea 1: The process of evolution drives the diversity and unity of life 1A: Change in the genetic makeup of a population over time is evolution 1.A.1

Natural selection is a major mechanism of evolution

1.A.2

Natural selection acts on phenotypic variations in populations

1.A.3

Evolutionary change is also driven by random processes

1.A.4

Biological evolution is supported by scientific evidence from many disciplines

1.B.1 1.B.2

Phylogenetic trees and cladograms are graphical models of evolutionary history

Speciation and extinction have occurred throughout the Earth's history

1.C.2

Speciation may occur when populations become reproductively isolated

1.C.3

Populations continue to evolve

Evidence for Biological Evolution

The Relatedness of Organisms

Hypotheses about the natural origin of life

1.D.2

Scientific evidence from different disciplines supports models of life's origin

Speciation & Extinction

The Origin of Living Systems

2A: Growth, reproduction and maintenance of the organization of living systems require free energy and matter 2.A.1

All living systems require energy

Energy in Living Systems, Homeostasis & Energy Allocation

2.A.2

Organisms capture and store free energy for use in biological processes

Energy in Living Systems, Energy Flow & Nutrient Cycles

2.A.3

Energy exchange maintains life processes

The Biochemistry of Life

2B: Growth, reproduction and dynamic homeostasis require that cells create and maintain internal environments that are different from their external environments Cell membranes are selectively permeable

2.B.2

Movement of molecules across membranes maintains growth and homeostasis

2.B.3

Cell Structure and Processes

3.A.2

In eukaryotes, heritable information is passed on via the cell cycle and mitosis or meiosis plus fertilization

Chromosomes & Cell Division

3.A.3

The chromosomal basis of inheritance gives an understanding of transmission of genes from parent to offspring

Chromosomes & Cell Division, The Chromosomal Basis of Inheritance

3.A.4

The inheritance pattern of many traits is not explained by Mendelian genetics

The Chromosomal Basis of Inheritance

3B: Expression of genetic information involves cellular and molecular mechanisms 3.B.1

Gene regulation results in differential gene expression and cell specialization

3.B.2

Signals mediate gene expression

3.C.1 3.C.2 3.C.3

3.D.1  3.D.2 3.D.3 3.D.4

2.C.1

Organisms used feedback mechanisms to maintain internal environments and respond to external environmental change

2.C.2

Organisms respond to change in their external environments

Homeostasis & Energy Allocation

Genotype changes can alter phenotype Processes that increase genetic variation Viral replication and genetic variation

Sources of Variation

Commonalities in cell communication Signaling by direct contact or chemicals Signal transduction pathways Changes to signal transduction pathways

Cellular Communication

3.E.1

Communicating information with others

Timing & Coordination

3.E.2

Nervous systems and responses

Communicating & Responding

Big Idea 4: Biological systems interact, and these systems and their interactions possess complex properties 4A: Interactions within biological systems lead to complex properties 4.A.1

Properties of a molecule are determined by its molecular construction

The Biochemistry of Life

4.A.2

The structure and function of subcellular components, and their interactions, provide essential cellular processes

Cell Structure and Processes

4.A.3

Gene expression results in specialization of cells, tissues and organs

Regulation of Gene Expression

4.A.4

Organisms exhibit complex properties due to interactions between their parts

Plant Structure & Adaptation, Comparing Animal Systems, Interactions in Physiological Systems

4.A.5

Communities are composed of populations that interact in complex ways

Populations and Communities

4.A.6

Movement of matter and energy

Populations & Communities, Energy Flow & Nutrient Cycles, The Diversity and Stability of Ecosystems

4B: Competition and cooperation are important aspects of biological systems

2D: Growth & dynamic homeostasis are influenced by changes in the environment Biotic and abiotic factors affect biological systems

Regulation of Gene Expression

3E: Transmission of information results in changes within and between systems

Internal membranes in eukaryotic cells partition the cell into specialized regions

2C: Organisms use feedback mechanisms to regulate growth and reproduction, and to maintain dynamic homeostasis

2.D.1

DNA and RNA

3D: Cells communicate by generating, transmitting and receiving chemical signals

Big Idea 2: Biological systems utilize free energy and molecular building blocks to grow, to reproduce and to maintain dynamic homeostasis

2.B.1

DNA, and in some cases RNA, is the primary source of heritable information

3C: Processing of genetic information is imperfect and a source of genetic variation

1D: The origin of living systems is explained by natural processes 1.D.1

3.A.1 Genetic Change in Populations

1C: Life continues to evolve within a changing environment 1.C.1

Big Idea 3: Living systems store, retrieve, transmit & respond to information essential to life processes 3A: Heritable information provides for continuity of life

1B: Organisms are linked by lines of descent from common ancestry Organisms share many conserved core processes and features that have evolved

each big idea, and identifies where the material is located in AP Biology 1 (blue) or AP Biology 2 (black).

4.B.1

Interactions between molecules affect their structure and function

Enzymes & Metabolism

4.B.2

Cooperative interactions within organisms promote efficiency

Plant Structure & Adaptation, Comparing Animal Systems, Interactions in Physiological Systems

4.B.3

Population interactions influence species distribution and abundance

Populations & Communities, The Diversity & Stability of Ecosystems

4.B.4

Ecosystem distribution changes over time

The Diversity & Stability of Ecosystems

The Nature of Ecosystems

2.D.2

Homeostatic mechanisms reflect both common ancestry and divergence due to adaptation in different environments

2.D.3

Biological systems are affected by disruptions to their dynamic homeostasis

Homeostasis & Energy Allocation, Plant Structure & Adaptation, Comparing Animal Systems, Interactions in Physiological Systems

2.D.4

Plants and animals have chemical defenses against infections

Internal Defense, Plant Structure & Adaptation

2E: Many biological processes involved in growth, reproduction & dynamic homeostasis include temporal regulation & coordination

4C: Naturally occurring diversity among and between components within biological systems affects interactions with the environment

2.E.1

Timing and coordination of events are regulated and necessary for development

Regulation of Gene Expression

4.C.1

Variation in molecular units provides cells with a wider range of functions

Internal Defense

2.E.2

Multiple mechanisms regulate timing & coordination of physiological events

Timing & Coordination

4.C.2

Environmental factors influence the expression of the genotype

The Chromosomal Basis of Inheritance

2.E.3

Timing and coordination are regulated and are important in natural selection

Communicating & Responding

4.C.3 4.C.4

Variation in populations affects dynamics Diversity may influence ecosystem stability

The Diversity & Stability of Ecosystems

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Essential Skills for AP Biology

Key terms

Inquiry is the basis of science

accuracy

Essential knowledge [points also covered in activities throughout AP1&2]

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chi-squared control

controlled variable

dependent variable experiment

Inquiry begins with observation

c 1

Describe the role of inquiry-based investigations in science. Show, through your work, your understanding of science as a non-linear process. [SP3]

c 2

Use a variety of methods to answer questions you raise as a result of observation. These include field and laboratory-based investigations, simulations and models, and data analysis. [SP3] [SP4]

c 3

Use scientific models to illustrate biological processes and concepts, communicate information, make predictions, and describe systems. [SP1]

graph

hypothesis

Activity number

independent variable

1

4 5

2

mean

median

model (scientific) observation percentage precision

prediction rate

PASCO

scientific method

Communication

statistical test

Essential knowledge [points also covered in activities throughout AP1&2]

table

trend

variable

Activity number

The results of investigations must be communicated to peers to have value

c 1

Demonstrate an ability to communicate the findings of your investigations through oral and written presentations, including lab reports, and through graphs and/or contributions to online resources. [SP3] [SP4]

c 2

Use a lab notebook or portfolio to organize your work and provide a record of ideas, methods, results, further questions, and references. [SP3] [SP4]

Quantitative skills

Essential knowledge [points also covered in activities throughout AP1&2]

15

5

Activity number

Quantitative reasoning is an essential part of inquiry in biology

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Demonstrate ability to use basic mathematical skills to collect data. These include making counts and measurements. Distinguish between accuracy and precision and explain their importance when collecting quantitative data. [SP2]

c 2

Demonstrate appropriate application of mathematical routines to data., e.g. determining mean and median, calculating rates and percentages, and chisquare analysis. [SP2]

6 - 10

c 3

Use tables or spreadsheets to organize different types of data, including any calculated values (e.g. means and standard deviation). [SP2] [SP5]

12 16

c 4

Construct graphs for different types of data, including that collected during your investigations. Indicate error in calculated values as appropriate. [SP2] [SP5]

12 - 17

c 5

Use descriptive statistics (e.g. mean and standard deviation) and simple statistical tests (e.g. chi-squared, Student's t test, correlation) to analyze your data and test your hypothesis. [SP2] [SP5]

16 - 19

c 6

Develop and use quantitative models (e.g. spreadsheet models) to analyze biological phenomena (e.g. population growth) and predict outcomes. [SP1]

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

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

EXPLORE AND DISCOVER

Ask questions

Find inspiration

Share ideas & information

Make observations

Explore the literature

TEST IDEAS Gathering data

Make hypotheses

Expected results

Actual results

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

…support a hypothesis

RESULTS AND BENEFITS Develop technologies

…oppose a hypothesis

Solve problems

…change assumptions

…suggest a new hypothesis

Build knowledge

Satisfy curiosity

Inform society

ANALYSIS AND FEEDBACK Peer review

Repeat investigation

Discussion with peers

Publication

New questions

Theory building

Citation and references

Citation and reference by authors

Author

All scientific work acknowledges sources of information through citation and a list of references. Citations support the statements made in the text in context, and all citations are then listed alphabetically, or identified and referenced sequentially by number. Internet sites are dated and site author acknowledged.

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Citation and reference by numbers

Year Title Publication Volume and pages The style you choose is not as important as being consistent, thorough, and honest about drawing on other people's work. All the information needed to locate the reference should be included (above).

Thorough and accurate citation and referencing shows you have explored the topic, have evidence to support your work, and you are not taking credit for work that is not your own. Each publication sets its own particular referencing style and these can vary widely. In your own work, it is most important to be consistent.

1. What is the role of citation and correct referencing when reporting on scientific investigations?

2. Study the diagram and write a paragraph on the scientific process and the role of surprising results in the progression of science. Staple it to this page. At the end of your course, reexamine what you wrote. Have your ideas changed? WEB

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Observations, questions, and hypotheses

PR E V ON IEW LY

Observation is the beginning of any scientific investigation. Often the best investigations are based on a series of fortuitous or specific observations. For example, in 1765 Edward Jenner developed the first vaccination for smallpox after hearing that milkmaids who contracted cowpox (a harmless disease) never got smallpox. After observing a phenomenon, questions must be asked: What causes the phenomenon? Is it linked to other observations? Can it be manipulated? Questions and observations lead to a hypothesis that can be tested by using a repeatable method. For every hypothesis, there is a corresponding null hypothesis, i.e. a hypothesis of no difference or no effect. Creating a null hypothesis enables a hypothesis to be tested in a meaningful way using statistical tests. If the results of an experiment are statistically significant, the null hypothesis can be rejected. If a hypothesis is accepted, anyone should be able to test the predictions with the same methods and get a similar result each time. Example: Two observations were made, as described below and used to produce a hypothesis:

Observation 1: Some caterpillar species are brightly colored and appear to be conspicuous to predators (e.g. insectivorous birds). Predators appear to avoid these species. These caterpillars are often found in groups, rather than being solitary.

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

Assumptions

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

Hypothesis: Bright color patterns might signal to predators that the caterpillars are distasteful. The corresponding null hypothesis would be there is no difference in palatability between the bright and cryptically colored caterpillars.

ff Insectivorous birds have color vision.

ff Caterpillars that look bright or cryptic to us, also appear that way to insectivorous birds.

ff Insectivorous birds can learn about the palatability of prey by tasting them.

3. Based on the hypothesis above, generate a prediction about the behavior of insectivorous birds towards caterpillars:

4. During a routine preparation of bacterial colonies on agar plates, a laboratory assistant noticed that the colonies left overnight on the side of a bench near a heating unit grew faster than those left on the opposite side of the bench. The assistant decided to test this observation by experiment: (a) State a hypothesis for the investigation:

(b) Generate a prediction based on the hypothesis:

(c) Formulate a possible design for the experiment to test the observation:

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Systems and Models

Key Idea: Systems are assemblages of interrelated components working together. Models can be mathematical or visual representations of these systems. An example of a system is our eight-planet solar system. Each of the planet's orbits represents a single component of the system. The driving force of the system is gravity from the

PR E V ON IEW LY

Sun. Modeling systems helps to understand how they work. A model is a representation of an object or system that shares important characteristics with the object or system being studied. A model does not necessarily have to incorporate all the characteristics or be fully accurate to be useful. It depends in the level of understanding required.

Barfooz CC 3.0

Open systems are able to exchange matter, energy and information with their surroundings. This causes them to be constantly changing, although the overall processes and outcomes remain relatively constant. Open systems are the most common type in natural systems. Examples include ecosystems, living organisms, and the ocean.

Models are extremely important when trying to understand how a system operates. Models are useful for breaking complex systems or organizations down into manageable parts and often only part of a system is modelled at a time. As understanding of the system progresses, more and more data can be built into the model so that it more closely represents the real world system or object. A common example is the use of models to represent atoms. The three illustrations (right) become more complex from left to right.

Closed systems exchange energy with their surroundings, but not matter. Closed systems are uncommon on Earth, although the cycling of certain materials, such as water and nitrogen, approximates them. The Earth itself is essentially a closed system. It receives energy from the Sun but exchanges virtually no matter with the universe (apart from the occasional meteorite).

Isolated systems exchange no energy, information or matter with their surroundings. No such systems are known to exist (with the possible exception of the entire universe). Some natural systems approximate isolated systems, at least for certain lengths of time. The solar system is essentially isolated, as is the Milky Way galaxy if gravity from nearby stars or galaxies is ignored.

-

-

-

+

+ + + +

-

Atomic model showing position of charge

-

-

-

Atomic model showing 2D electron orbitals

+ + + +

Atomic model showing 3D electron clouds

Feedback loops

Time

No Cla t fo ssr r o Us om e

Change

Change

A feedback loop is a system in which the output is fed back into the system as input.

Time

Negative feedback loops

Positive feedback loops

Negative feedback loops are probably the most common in natural systems and counteract change in an equilibrium. As one component of the equilibrium changes, it causes changes in a second component, which returns the first component to its original state. Examples include predator-prey cycles, population changes in K-selected species, and the regulation of body systems in living organisms.

These are less common in natural systems than negative feedback loops because they accelerate or exaggerate change and are therefore destabilizing. A change in one part of the system causes change in a second part which increases the magnitude of the first. Examples include exponential population growth, certain possible aspects of global warming, and some physiological systems in animals, with the intent of achieving a purpose, e.g. childbirth.

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1. Identify each of the following as either an open, closed, or isolated system: (a) Reef ecosystem: (e) Solar system: (f) Digestive system:

(c) Earth:

(g) A National Park:

(d) Biosphere:

(h) A large lake:

PR E V ON IEW LY

(b) Nitrogen cycle:

2

2. Identify each of the following examples as a negative or positive feedback loop: (a) Alarm leading to panic:

(d) x = (x+1):

(b) Temperature regulation:

(e) Predator prey oscillation:

(c) Hormonal changes during childbirth:

(f) Exponential population growth:

3. Explain why the water cycle approximates a closed system:

4. Explain why there are no known isolated systems:

5. (a) Explain why there are generally few positive feedback loops in the physiological systems of living organisms:

(b) Explain the purpose of positive feedback in physiological systems:

No Cla t fo ssr r o Us om e

6. Explain why models are never 100% accurate representations of the system being studied:

7. Discuss the advantages and disadvantages of using models to explain a system:

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

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

Accurate but imprecise

The measurements are all close to the true value but quite spread apart. Analogy: The arrows are all close to the bullseye.

Precise but inaccurate

The measurements are all clustered close together but not close to the true value. Analogy: The arrows are all clustered close together but not near the bullseye.

Inaccurate and imprecise

The measurements are all far apart and not close to the true value. Analogy: The arrows are spread around the target.

Accurate and precise

The measurements are all close to the true value and also clustered close together. Analogy: The arrows are clustered close together near the bullseye.

Increasing precision

The accuracy of a measurement refers to how close the measured (or derived) value is to the true value. The precision of a measurement relates to its repeatability. In most laboratory work, we usually have no reason to suspect a piece of equipment is giving inaccurate measurements (is biased), so making precise measures is usually the most important consideration. We can test the precision of our measurements by taking repeated measurements from individual samples.

Population studies present us with an additional problem. When a researcher makes measurements of some variable in a study (e.g. fish length), they are usually trying to obtain an estimate of the true value for a parameter of interest (e.g. the mean size, therefore age, of fish). Populations are variable, so we can more accurately estimate a population parameter if we take a large number of random samples from the population.

RA

Increasing accuracy

A digital device such as this pH meter (above left) will deliver precise measurements, but its accuracy will depend on correct calibration. The precision of measurements taken with instruments such as callipers (above) will depend on the skill of the operator.

1. Distinguish between accuracy and precision:

No Cla t fo ssr r o Us om e

2. Describe why it is important to take measurements that are both accurate and precise:

3. A researcher is trying to determine the temperature at which an enzyme becomes denatured. Their temperature probe is incorrectly calibrated. Discuss how this might affect the accuracy and precision of the data collected:

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7 Reducing error

The accuracy of a measurement can be increased by increasing the number of measurements taken. For example, the accuracy of the mean mass of individuals in a population can be increased by increasing the number of individuals measured (i.e. increasing the sample size).

Sometimes reducing error requires taking more measurements over a longer period of time. For example, hypothetical waves breaking on a shore do so with a relatively regular frequency. Recording the time between one wave breaking and the next depends when the wave is defined as breaking. This may be difficult to determine precisely for each individual wave and the waves may be breaking too quickly to allow enough time to elapse for recordings to be made accurately (especially if the timer is being started and stopped by a person).

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Accuracy and equipment

2

In many cases, the accuracy of the measuring equipment needs to be taken into account. For example, electronic balances may give readings to one or more decimal places based on their accuracy (laboratory balances may read to a hundred thousandths of a gram).

The table below illustrates the difference between a balance weighing to 0.1 of a gram and 0.01 of a gram Sample

1

2

3

4

5

Mean

Mass (g) (2 s.f)

1.1

1.2

1.4

1.2

1.3

1.2

Mass (g) (3 s.f)

1.12

1.23

1.44

1.19

1.28

1.25

The difference in mean mass is slight (just 0.05 g) but over larger samples or larger masses the differences can add up.

The table above also shows the importance of significant figures (s.f.). The actual numerical mean for the second row is 1.252. However because we are measuring to three significant figures we cannot be sure of the final number thus the answer must be given in the same significant figures as the measurements.

To increase the accuracy of measuring the period between each wave, it is best to record the time for a number of waves to break (e.g. 10) and divide by that number to obtain the period between each wave. This has the effect of allowing for slight variations in the period and reduces the total error in the measurement. Example: Actual wave period: 5.0 seconds Accuracy of timer (i.e. reaction speed) 0.3 seconds

Measurements of individual periods (seconds): 5.4, 5.7, 5.7, 5.8, 4.5, 4.6, 5.7, 5.8, 5.1, 5.3, Mean: 5.4 In each measurement above, the error is about 0.3 seconds producing an error of up to 6.7% (0.3 á 4.5 x 100) of the recorded value of a wave period.

If the time recorded for ten waves to break was 51.1 seconds, then the time for one wave to break is 5.1 seconds. The error is spread over the whole 51.1 seconds (0.3 á 51.1) and thus is much smaller at just 0.6% of the wave period.

4. The period of a pendulum is based on the length of the pendulum and the mass at its end. Two students measure the time it takes for a pendulum to swing back and forth (its period). Student A measures three individual swings and calculates a mean value. Student B measures three sets of ten swings and calculates a mean. Each student measures the accuracy of the timer as 0.2 seconds. The results are shown below: Student A

Student B

Time for swing (s)

Set

Time for ten swings (s)

1

2.7

1

20.3

2

2.1

2

20.1

3

2.5

3

19.8

Mean (1 swing)

Mean (10 swings) 1 swing

(a) Calculate the mean for each student's results and the time for one swing for student B.

(b) Explain why student B's results are more accurate than student A's:

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5. In a class of 20 students, the individual heights of the students in cm are: 135, 139, 141, 146, 147, 149, 156, 151, 158, 155, 156, 159, 161, 167, 162, 163, 161, 172, 171, 170.

(a) Calculate the mean height of the students:

(b) A person takes a sample of five of the students: 139, 151, 162, 172, 170. Calculate the mean of the sample and comment on its accuracy:

(c) A second person takes a sample of ten of the students: 135, 146, 147, 156, 155, 156, 161, 167, 162, 170. Calculate the mean of the sample and comment on its accuracy:

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

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

Types of data

Quantitative (interval or ratio)

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

Qualitative (nominal)

Ranked (ordinal)

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

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

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

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

Discontinuous or discrete data:

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

Continuous

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

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

Continuous data:

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

WBS

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

B: Eggs per nest

A: Skin color

(a) Skin color:

(c) Tree trunk diameter:

C: Tree trunk diameter

(b) Number of eggs per nest:

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

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

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

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Planning a Quantitative Investigation

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

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

Carrying out your practical work

Preparation

Execution and recording

Analysis and reporting

Familiarize yourself with the equipment and its set it up. Calibrate equipment if necessary to give accurate measurements. Read through the methods and identify key stages and how long they will take.

Know how you will take your measurements and how often. Record your results systematically as you go in a log book. You could record results a handwritten table or in a spreadsheet. If using a data logger, data will be logged.

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

Identifying variables

Experimental controls

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

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

Dependent variable

• Measured during the investigation.

Independent variable

Controlled variables

• Factors that are kept the same or controlled.

• List these in the method, as appropriate to your own investigation.

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Independent variable

• Set by the experimenter.

• Recorded on the graph's x axis.

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Dependent variable

• Recorded on the y axis of the graph.

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

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Investigation: effect of light on rate of photosynthesis Background

20

25

30

35

40

No direct light

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

Distance from direct light source (cm)

The method

ff 6 x 1.0 g of Cabomba stems were placed into each of 6 test-tubes filled with 10 mL room temperature solution containing 0.2 molL-1 sodium hydrogen carbonate (to supply carbon dioxide).

ff Test tubes were placed at distances (20, 25, 30, 35, 40 cm) from a 60W light source (light intensity reduces with distance at a predictable rate). One test tube was not exposed to the light source. ff Before recording, the Cabomba stems were left to acclimatize to the new light level for 5 minutes. The bubbles emerging from the stem were counted for a period of three minutes at each distance.

1.0 g Cabomba

0.2 molL-1 NaHCO3

6CO2 + 12H2O

Oxygen bubbles

Light

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

C6H12O6 + 6O2 + 6H2O

1. Write a suitable aim for this experiment:

2. Write a possible hypothesis for this experiment:

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

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

(c) Name the unit for the independent variable:

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

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

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

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

5. Which tube is the control for this experiment?

6. Identify two assumptions being made about this system: (a)

(b)

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

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

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Working with Numbers

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

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

Decimal and standard form

Conversion factors and expressing units

Decimal form (also called ordinary form) is the longhand way of writing a number (e.g. 15,000,000). Very large or very small numbers can take up too much space if written in decimal form and are often expressed in a condensed standard form. For example, 15,000,000 is written as 1.5 x 107 in standard form. In standard form a number is always written as A x 10n, where A is a number between 1 and 10, and n (the exponent) indicates how many places to move the decimal point. n can be positive or negative.

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

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

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

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

0. 0 0 1 0 1 = 1.01 x

10-3

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

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

2. Add the two numbers together:

4.

The value of a variable must be written with its units where possible. SI units or their derivations should be used in recording measurements: volume in cm3, dm3, or liter (L), mass in kilograms (kg) or grams (g), length in meters (m), time in seconds (s). For example the rate of oxygen consumption would be expressed: 3

Oxygen consumption ( cm g

–1

s

–1

)

Estimates

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

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

Time Cumulative gas (Minute) produced ( cm3)

Rate of reaction (cm3 min-1)

0

0

0

2

34

17

4

42

4*

6

48

3

8

50

1

10

50

0

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

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

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

5. 43.2 x 1044:

6. 3.4 x 72 ÷ 15: 7. 658 ÷ 22:

Probability

Probability is how likely something is to happen. It is an important part of biology. Its uses include calculating the statistical significance of a difference between means or the probability of an event occurring.

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Rates

AP1 1 Essential Skills for AP Biology 2017.indd 11

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

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

3. Convert to standard form:

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

The probability of an event ranges from 0 to 1. The sum of all probabilities equals 1.

Product rule: for independent events A and B the probability (P) of A and B occurring is P(A) x P(B). E.g. the probability two children born one after the other both being male is 0.5 x 0.5 = 0.25.

Sum rule: For mutually exclusive events Y and Z the probability that one will occur (Y or Z) is P(Y) + P(Z). E.g. in an Aa x Aa cross the probability a person will have a dominant phenotype = 0.25 + 0.5 = 0.75.

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

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

Fractions

Ratios

Percentages

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

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

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

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

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

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

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

• Ratios do not require units.

• Ratios are usually expressed as a : b.

• Allows meaningful comparison between different samples.

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

• Useful to monitor change (e.g. % increase from one year to the next).

882 inflated

299 constricted

Volume of food coloring (cm3)

Volume of water (cm3)

Concentration of solution ( %)

10

0

100

8

2

80

6

4

60

4

6

40

2

8

20

0

10

0

Pea pod shape: Ratio = 2.95 : 1

495 round yellow

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

158 round green

152 wrinkled yellow

55 wrinkled green

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

Example: Calculating phenotype ratios in Mendelian genetics

Example: Producing standards for a calibration curve.

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

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

2. Simplify the following fractions:

(a) 3/9:

(b) 84/90:

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

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Interphase

140

Prophase

70

Telophase

15

Metaphase

10

Anaphase

5

Total

240

No. of cells calculated

4800

(c) 11/121:

Athlete

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

No. of cells counted

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Cell cycle stage

Normal weight Overweight Obese

Body mass (kg)

Lean body mass (kg)

50

38

56

41

65

46

80

48

95

52

% lean body mass

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

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

Power function

Exponential function

Log transformations

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

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

A log transformation has the effect of normalizing data and making very large numbers easier to work with. Biological data often have a positive skew so log transformations can be very useful.

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

Basal metabolic rate

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

BMR = aMb

Body mass

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

• The log of a number is the exponent to which a fixed value (the base) is raised to get that number. So log10 (1000) = 3 because 103 = 1000.

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

• Exponential growth is easy to identify because the curve has a J-shape appearance due to its increasing steepness over time. It grows more rapidly than a power function

• Examples of exponential growth include the growth of microorganisms in an unlimiting growth environment.

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

• Log transformations are useful for data where there is an exponential increase in numbers (e.g. cell growth). In these cases the log transformed data will plot as a straight line.

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

Log 10 cell numbers

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

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

Cell numbers

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

Time

Time

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

Example: Yeast cell growth plotted on logarithmic scale.

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

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

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

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

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Properties of Geometric Shapes materials into a cell? The cells of organisms, and sometimes the organisms themselves, are often rather regular shapes, so their physical properties (e.g. cell volume or surface area) can be calculated (or approximated) using the simple formulae applicable to standard geometric shapes.

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Key Idea: Circumference, surface area, and volume are useful calculations that can be applied in biological situations. Biology often requires you to evaluate the effect of a physical property, such as cell volume, on function. For example, how does surface area to volume ratio influence the transport of

r

r

r

h

h

w

w

The circumference is the linear distance around the edge of a circle or sphere and is given by the formula 2πr

r = radius

l

l = length

w = width

h = height

w = 3.14

Sphere

Cube

Rectangular prism

Cylinder

Staphylococcus bacterial cell

Kidney tubule cell

Intestinal epithelial cell

Axon of neuron

Surface area: The sum of all areas of all shapes covering an object's surface.

4wr2

6w2

2(lh + lw + hw)

(2πr2) + (2πrh)

Volume: The amount that a 3-dimensional shape can hold.

4/3πr3

w3

lwh

πr2h

Biological example.

1. For a sphere with a radius of 2 cm, calculate the:

(a) Circumference: (b) Surface area:

(c) Volume: 2. For a rectangular prism with the dimensions l = 3 mm, w = 0.3 mm, and h = 2 mm calculate the:

(a) Surface area: (b) Volume: 3. For a cylinder with a radius of 4.9 cm and height of 11 cm, calculate the:

(a) Surface area: (b) Volume: 4. Find the height of a rectangular prism with a volume of 48 cm3, a length of 4 cm, and a width of 2.5 cm:

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5. Find the radius of a cylinder with a volume of 27 cm3 and a height of 3 cm:

6. A spherical bacterium with a radius of 0.2 µm divides in two. Each new cell has a radius that is 80% of the original cell. (a) Calculate the surface area of the 'parent' bacterial cell: (b) Calculate the volume of the 'parent' bacterial cell:

(c) Calculate the surface area of each new cell:

(d) Calculate the volume of each new cell:

(e) Which cell has the greatest surface area to volume ratio: PRACTICES

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10 Practicing with Data

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

(a) Photosynthetic rate at different light intensities

(b) Plant water loss using a bubble potometer

Light intensity (%)

Average time for leaf disc to float (min)

Reciprocal of time* (min-1)

Time (min)

Pipette arm reading (cm3)

Plant water loss (cm3 min-1)

100

15

0.067

0

9.0

50

25

5

8.0

0.2

25

50

10

7.2

11

93

15

6.2

6

187

20

4.9

* Reciprocal of time gives a crude measure of rate.

(c) Incidence of cyanogenic clover in different areas

Clover plant type

Frost free area

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

Frost prone area

Number

%

Number

Cyanogenic

124

78

26

Acyanogenic

35

Total

159

%

Totals

Size class (mm)

Frequency

Relative frequency (%)

0-50

7

2.6

50-99

23

100-149

59

150-199

98

200-249

50

250-299

30

300-349

3

Total

270

115

Clover

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

(a) 8970

(b) 0.046

(c) 1,467,851

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

(b) 0.0031 x 10-2

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

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

Dietary fiber? Sugars?

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

(c) 6.2 x 104

Chilli beans nutrition facts Serving size 1 cup (253 g)

Amount per serving Total fat

– Saturated fat

Total carbohydrate

AP Biology 1 2E 2017.indb 15

% composition

8g 3g

22 g

– Dietary fiber

9g

– Sugars

4g

Protein

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25 g

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

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

Selecting the correct equipment

Recognizing potential sources of error

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

Percentage errors

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

A cuvette (left) is a small clear tube designed to hold spectrophotometer samples. Inaccurate readings occur when:

x 100

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

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

x 100

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

• The cuvette is dirty or scratched (light is absorbed giving a falsely high reading). • Some cuvettes have a frosted side to aid alignment. If the cuvette is aligned incorrectly, the frosted side absorbs light, giving a false reading.

• Not enough sample is in the cuvette and the beam passes over, rather than through the sample, giving a lower absorbance reading.

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

(a) 21 mL:

(b) 48 mL:

(c) 9 mL:

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2. Calculate the percentage error for the following situations (show your working):

(a) A 1 mL pipette delivers a measured volume of 0.98 mL:

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

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

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

Presenting data in tables

Presenting data in graphs

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

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

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

Mean rate of internode growth (mm day-1)

Mean internode length (mm)

Control

50

0.60

32.3

Hormone 1

46

1.52

41.6

Hormone 2

98

0.82

38.4

Hormone 3

85

2.06

50.2

1.1

Yield (absorbance at 550 nm)

Treatment

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

1.0

0.9 0.8 0.7 0.6

Sensitive strain Resistant strain

0.5

0

0

1

2

3

4

5

Antibiotic concentration (g m-3)

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

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

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

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

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

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

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

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

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

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13 Constructing Graphs

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plotting of points. Before representing data graphically, it is important to identify the kind of data you have. Common graphs include scatter plots and line graphs (for continuous data), and bar charts (for categorical data). For continuous data with calculated means, points can be connected. On scatter plots, a line of best fit is often drawn.

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Key Idea: Graphs are useful for visually displaying numerical data, trends, and relationships between variables. Graphs are an excellent way to summarize trends in data or relationships between different variables. Presenting graphs properly demands attention to a few basic details, including correct orientation and labeling of the axes, and accurate

Growth rate in peas at different temperatures

Line graphs are used when one variable (the independent variable) affects another, the dependent variable. Important features include: • The data must be continuous for both variables. The independent variable is often time or experimental treatment. The dependent variable is usually the biological response.

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

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

Mean growth rate (mm day –1)

Guidelines for line graphs

0.9

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

0.8

0.7

0.6 0

10

8

12

14

16

18

20

22

24

Temperature (°C)

Body length vs brood size in Daphnia

A scatter graph is a common way to display continuous data where there is a relationship between two interdependent variables. • The data must be continuous for both variables.

• There is no independent (manipulated) variable, but the variables are often correlated, i.e. they vary together in a predictable way. • Useful for determining the relationship between two variables.

Number of eggs in brood

Guidelines for scatter graphs

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

80

Line of best fit

60 40

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

20 0

0

1

2

3

Body length (mm)

4

Distribution of mass of shrimp in a population

Guidelines for histograms

15 10

5

Bar graphs are appropriate for data that are non-numerical and discrete (categorical) for one variable. There are no dependent or independent variables. Important features include: • Data for one variable are discontinuous, non-numerical categories (e.g. place, species), so the bars do not touch.

• Multiple sets of data can be displayed side by side for direct comparison (e.g. males and females in the same age group).

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

45-49.9

40-44.9

35-39.9

30-34.9

25-29.9

20-24.9

15-19.9

Mass (g) Size of various woodlands in Britain

Cwm Clydach

20

Burnham Beeches

450

Scords Wood

• Data values may be entered on or above the bars if you wish.

Mass (g)

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Guidelines for bar graphs

10-14.9

0

0-4.9

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

20

5-9.9

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

25

Frequency

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

30

350

Wyre Forest

500

Yarner Wood

400

Wistmans Wood

4

0

100

200

300

400

500

600

Area of woodland (Hectares)

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14 Interpreting Line Graphs of a straight line and tells us about the relationship between x and y (how fast y is changing relative to x). For a straight line, the rate of change of y relative to x is always constant. A line may have a positive, negative, or zero slope.

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Key Idea: The equation for a straight line is y = mx + c. A line may have a positive, negative, or zero slope. The equation for a linear (straight) line on a graph is y = mx + c. The equation can be used to calculate the gradient (slope)

Measuring gradients and intercepts y

The equation for a straight line is written as:

y = mx + c

6

Where:

5

y = the y-axis value

4

m = the slope (or gradient)

3

x = the x-axis value

2

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

1

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

5

11

Determining "m" and "c"

1

To find "c" just find where the line crosses the y-axis. To find m: 1.

Choose any two points on the line.

2.

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

3.

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

4.

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

2

3

4

5

6

7

8

9

10

11

x

For the example above: c=1 m = 0.45 (5 á11)

Once c and m have been determined you can choose any value for x and find the corresponding value for y. For example, when x = 9, the equation would be: y = 9 x 0.45 + 1 y = 5.05

change in x

Interpreting gradients

a

b

c

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

0

0

1

2

3

4

5

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

0

0

1

3. For the graph (right):

(a) Identify the value of c:

(b) Calculate the value of m:

(c) Determine y if x = 2:

(d) Describe the slope of the line:

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4

5

0

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

1. State the gradient for graphs a, b, and c (above): (a) 2. For a straight line y = 3x + 2, (a) Identify the value of c:

2

1

2

3

4

5

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

(b)

(c)

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(b) Determine y if x = 4:

3 2

1

1

2

3

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

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

Correlation does not imply causation

Drawing the line of best fit

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

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

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

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

25 000

300

Autism Organic food sales

20 000

200

15 000 10 000

100

5000 0

1998

2000

2002

2004

2006

2008

0

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

Too steep

Number of individuals diagnosed with autism x1000

Organic food sales ($ millions)

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

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

Good fit

Too shallow

Year

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

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

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

(b) Describe the results:

Handspan vs foot length in adults

350

250

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

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300

150

200

250

300

Handspan (mm)

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

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

Variation in data

25 20

Frequency

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

A: Normal distribution

15 10 5

Type of variable sampled

x

0

Mass (g)

The shape of the distribution when the data are plotted

Ranked

Qualitative

25

B: Skewed distribution

20

Mode

Mode

Frequency

Quantitative (continuous or discontinuous)

15 10

Negative skew: the left tail is longer

5 0

Mass (g)

Symmetrical peak

Skewed peak or outliers present

Two peaks (bimodal)

Mean Median

Median

Modes

25

C: Bimodal (two peaks)

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

Frequency

20 15 10 5 0

Mass (g)

Statistic

Definition and use

Method of calculation

When NOT to calculate a mean:

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

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

Median

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

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

Mode

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

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

Range

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

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

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

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Mean

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

• DO NOT calculate a mean of ratios (e.g. percentages) for several groups of different sizes. Go back to the raw values and recalculate.

• DO NOT calculate a mean when the measurement scale is not linear, e.g. pH units are not measured on a linear scale.

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22 Case study: height of swimmers

Total of data entries = 5221 = 180 cm Number of entries 29

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

Height of swimmers (in rank order)

174 175 175 175 176 176 176 176 176 177

177 177 178 178 178 178 180 180 180 181

Height (cm)

Mean Tally

Total

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

Raw data: Height (cm)

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

178 180 180 178 176

177 181 185 186 188

188 178 185 176 180

176 178 175 185 186

186 176 189 177 177

175 175 174 176

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

Case study: fern reproduction

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

Total of data entries = 1641 = 66 sori Number of entries

25

Mean

Raw data: Number of sori per frond 60 70 69 64

64 63 59 63

62 70 70 64

68 70 66

66 63 61

63 62 70

Fern spores

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

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

Ladybird mass (mg) 10.1 8.0 6.7 9.8 6.2

8.2 8.8 7.7 8.8 8.8

7.7 7.8 8.8 8.9 8.4

Number of sori per frond (in rank order)

Sori per frond

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

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

71

Tally

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

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

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

Standard deviation

25

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

Normal distribution

Frequency

20 15

68%

10

2.5%

2.5%

5

95%

0

x-2s

x -1s

x

Size class

x +1s

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

x+2s

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

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

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

Frequency

Frequency

Calculating s Standard deviation is easily calculated using a spreadsheet.

2.5%

2.5%

68%

68%

2.5%

2.5%

95%

x -1s

x

x +1s

x+2s

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

Birth weights (kg)

3.740 3.830 3.530 3.095 1.560 3.910 4.180 3.570 3.150 3.400 3.380 2.660 3.840 3.630 3.810

2.640 2.980 3.350 3.780 3.260 3.800 4.170 4.400 3.770 3.825 3.130 3.400 3.260 3.220 3.135

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AP Biology 1 2E 2017.indb 23

3.090 3.830 3.840 4.710 4.050 4.560 3.380 3.690 1.495 3.260

x-2s

s=

∑(x – x̄)2 n–1

x -1s

x

x +1s

x+2s

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

1. Two data sets have the same mean. The standard deviation of the first data set is much larger than the standard deviation of the second data set. What does this tell you about the spread of data around the central measure for each set?

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x-2s

95%

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

(a) Calculate the mean for the data:

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

(c) State the mean ± 1s:

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

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

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18 Reliability of the Mean

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reason. However, it is does allow you calculate the 95% confidence interval (95% CI). By the end of this activity you should be able to: • Enter data and calculate descriptive statistics using a spreadsheet program such as Microsoft Excel®. You can follow this procedure for any set of data. • Calculate standard error and 95% CIs (and confidence limits) for sample data and plot these data with error bars. • Interpret the graphically presented data and reach tentative conclusions about the findings of an investigation.

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Key Idea: The 95% confidence limit gives a measure of the reliability of the sample mean as an estimate of the true value of the mean of the population. You have already seen how to use the standard deviation (s) to quantify the spread or dispersion in your data. Usually, you will also want to know how good your sample mean (x̄ ) is as an estimate of the true population mean (µ). This can be indicated by the standard error of the mean (or just standard error or SE). SE is often used as an error measurement simply because it is small, rather than for any good statistical

Reliability of the sample mean

When we take measurements from samples of a larger population, we are using those samples as indicators of the trends in the whole population. Therefore, when we calculate a sample mean, it is useful to know how close that value is to the true population mean (µ). This is not merely an academic exercise; it will enable you to make inferences about the aspect of the population in which you are interested. For this reason, statistics based on samples and used to estimate population parameters are called inferential statistics.

Sample

The standard error (SE)

Ladybird population

The standard error (SE) is simple to calculate and is usually a small value. Standard error is given by:

When we measure a particular attribute from a sample of a larger population and calculate a mean for that attribute, we can calculate how closely our sample mean (the statistic) is to the true population mean for that attribute (the parameter). Example: If we calculated the mean number of spots from a sample of six ladybird beetles, how reliable is this statistic as an indicator of the mean number of spots in the whole population? We can find out by calculating the 95% confidence interval.

S

SE =

n

where s = the standard deviation, and n = sample size.

Standard errors are sometimes plotted as error bars on graphs, but it is more meaningful to plot the 95% confidence intervals (see box below). All calculations are easily made using a spreadsheet (see following pages).

The 95% confidence interval

df

P

0.05

Do not be alarmed by this calculation; once you have calculated the value of the SE, it is a simple matter to multiply this value by the value of t at P = 0.05 (from the t table) for the appropriate degrees of freedom (df) for your sample (n – 1).

1 2 3 4 5

12.71 4.303 3.182 2.776 2.571

For example: where the SE = 0.6 and the sample size is 10, the calculation of the 95% CI is:

6 7 8 9 10

2.447 2.365 2.306 2.262 2.228

20 30 40 60 120 >120

2.086 2.042 2.021 2.000 1.980 1.960

95% CI = 0.6 x 2.262 =

1.36

Part of the t table is given to the right for P = 0.05. Note that, as the sample becomes very large, the value of t becomes smaller. For very large samples, t is fixed at 1.96, so the 95% CI is slightly less than twice the SE All these statistics, including a plot of the data with Y error bars, can be calculated using a program such as Microsoft Excel®.

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Relationship of Y against X (± 95% confidence intervals, n = 10)

Maximum value of t at this level of P

Confidence limits are given by the X ± 95%CI

Trendline

Small 95% CI

Mean

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95% CI = SE x t P(n-1)

Critical values of Student’s t distribution at P = 0.05.

Dependent variable (biological response)

SE is required to calculate the 95% confidence interval (CI) of the mean. This is given by:

Value of t at n–1 = 9

Large 95% CI

Range of the independent variable

Plotting your confidence intervals

Once you have calculated the 95% CI for the means in your data set, you can plot them as error bars on your graph. Note that the 95% confidence limits are given by the value of the mean ± 95%CI. A 95% confidence limit (i.e. P = 0.05) tells you that, on average, 95 times out of 100, the limits will contain the true population mean.

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Research has indicated that different pastures have different susceptibility to infestation by clover root weevil (left). Armed with this knowledge, two students reasoned that the most susceptible grass type would have the greatest weevil population. The students chose five pasture types, and recorded the number of weevil larvae in each pasture type at six sample sites (sample area 1 m2). Their results are presented in the table below.

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The clover root weevil (Sitona lepidus) is a pest of white clover pastures. The adults feed on clover leaves, while the larvae feed on clover nodules and roots, causing root loss and a reduction in nitrogen fixation.

Clover root weevil

Environment

Sample

Perennial ryegrass

Fescue

White clover

Red clover

Chicory

1

42

42

48

42

45

2

45

46

54

46

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3

41

38

48

45

45

4

42

41

52

42

38

5

49

45

49

44

40

6

43

44

52

44

47

Number of weevils

1. Complete the table below by calculating the mean, standard deviation, standard error, and 95% confidence interval (95% CI) for each of the grass environments. Perennial ryegrass

Fescue

White clover

Red clover

Chicory

Mean

Standard deviation Standard error 95% CI

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2. Select the appropriate graph format and plot the means for each of the grass environments below. Include bars to show the 95% confidence intervals.

3. Study your plot and decide if there are any significant differences between the abundance of clover root weevils in the five environments. Write a conclusion for the investigation below:

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19 Interpreting Sample Variability these data sets, the results would (incorrectly) suggest that they were alike. The assumptions we make about a population will be affected by what the sample data tell us. This is why it is important that sample data are unbiased (e.g. collected by random sampling) and that the sample set is as large as practicable. This exercise will help to illustrate this principle.

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Key Idea: The sampling method can affect the results of the study, especially if it has an unknown bias. The standard deviation (s) gives a simple measure of the spread or dispersion in data. It is usually preferred over the 2 variance (s ) because it is expressed in the original units. Two data sets could have the same mean, but very different values of dispersion. If we simply used the mean to compare

Random sampling, sample size, and dispersion in data

Complete sample set n = 689 (random)

Look at this data set and then complete the exercise to calculate the same statistics from each of two smaller data sets (tabulated right) drawn from the same population. This exercise shows how random sampling, large sample size, and sampling bias affect our statistical assessment of variation in a population.

Length of year zero perch

x-2s

x -1s

x

x+2s

x +1s

50 40 30 20

10

Descriptive statistics

69

65

61

57

53

49

45

41

37

0

33

Length in mm

MEAN 48 MEDIAN 47 MODE 45 VARIANCE 61.03 STANDARD DEVIATION 7.81

Length in mm

Freq

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68

1 0 0 0 0 0 0 0 0 0 2 0 0 3 2 1 3 0 0 0 0 1 0 2 0 0 1 3 0 0 0 0 1 0 3 2 2 0 0 0 0 0 2 1

30

689

Small sample set n = 50 (bias) Length in mm

Freq

46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67

1 0 0 1 0 0 1 1 1 1 0 2 2 4 1 0 8 10 13 2 0 2

50

The person gathering this set of data was biased towards selecting larger fish because the mesh size on the net was too large to retain small fish This population was sampled randomly to obtain this data set

This column records the number of fish of each size Number of fish in the sample

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1 0 0 0 0 0 0 2 3 3 4 5 10 23 22 33 39 41 41 36 49 32 14 32 27 25 24 17 18 27 21 20 11 18 16 22 13 8 10 5 7 2 3 3 1 0 1

29

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

Sample size and sampling bias can both affect the information we obtain when we sample a population. In this exercise you will calculate some descriptive statistics for some sample data. The complete set of sample data we are working with comprises 689 length measurements of year zero (young of the year) perch (column left). Basic descriptive statistics for the data have been calculated for you below and the frequency histogram has also been plotted.

25

Freq

Frequency

Length in mm

Small sample set n = 30 (random)

1. For the complete data set (n = 689) calculate the percentage of data falling within: (a) ± one standard deviation of the mean:

(b) ± two standard deviations of the mean:

(c) Explain what this information tells you about the distribution of year zero perch from this site:

2. Give another reason why you might reach the same conclusion about the distribution:

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PRACTICES

PRACTICES

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Enter the data values in separate cells under an appropriate descriptor

Calculating descriptive statistics using Excel® You can use Microsoft Excel® or other similar spreadsheet programme to easily calculate descriptive statistics for sample data.

Ignore this WEIGHT column. Sometimes the data we are interested in is part of larger data set.

In this first example, the smaller data set (n = 30) is shown as it would appear on an Excel® spreadsheet, ready for the calculations to be made. Use this guide to enter your data into a spreadsheet and calculate the descriptive statistics as described.

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The variables being measured. Both length and weight were measured, but here we are working with only the length data.

Type in the name of the statistic

The cells for the calculations below are B2 to B31

Excel ® will calculate. This gives you a reference for the row of values.

Type the formula into the cell beside its label. When you press return, the cell will contain the calculated value.

=COUNT(B2:B31) =AVERAGE(B2:B31) =MEDIAN(B2:B31) =MODE(B2:B31) =VAR(B2:B31) =STDEV(B2:B31)

When using formulae in Excel®, = indicates that a formula follows. The cursor will become active and you will be able to select the cells containing the data you are interested in, or you can type the location of the data using the format shown. The data in this case are located in the cells B2 through to B31 (B2:B31). 3. For this set of data, use a spreadsheet to calculate:

(a) Mean:

(b) Median:

(c) Mode:

(d) Sample variance:

(e) Standard deviation:

Staple the spreadsheet into your workbook.

4. Repeat the calculations for the second small set of sample data (n = 50) on the previous page. Again, calculate the statistics as indicated below and staple the spreadsheet into your workbook: (a) Mean: (b) Median: (c) Mode:

(d) Variance:

(e) Standard deviation:

5. On a separate sheet, plot frequency histograms for each of the two small data sets. Label them n = 30 and n = 50. Staple them into your workbook. If you are proficient in Excel ® and you have the "Data Analysis" plug in loaded, you can use Excel® to plot the histograms for you once you have entered the data. 6. Compare the descriptive statistics you calculated for each data set with reference to the following: (a) How close the median and mean to each other in each sample set:

(b) The size of the standard deviation in each case:

(c) How close each small of the sample sets resembles the large sample set of 689 values:

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7. (a) Compare the two frequency histograms you have plotted for the two smaller sample data sets:

(b) Why do you think two histograms look so different?

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20 Which Test to Use?

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28

significance of your investigation. Never under-estimate the value of plotting your data, even at a very early stage. This will help you decide on the best type of data analysis. Sometimes, statistical analysis may not be required.

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Key Idea: How your data is analyzed depends on the type of data you have collected. Plotting your initial data can help you to decide what statistical analysis to carry out. Data analysis provides information on the biological

Regression

Finding how one factor affects another

Trend

Plot a scatter graph

Non-Linear: The data do not plot in a straight line (i.e. curved). Example: oxygen consumption at different temperatures.

Normal data

Testing for a relationship between variables

Spearman correlation coefficient

Non-normal data

What kind of test?

More than two groups of data

Testing for a difference between groups (e.g. habitats or treatments)

START HERE

Example: Frequency of occurrence of different species at two sites.

ANOVA (Analysis of variance) Example: Survival of weevils in different pasture types.

Same individuals

Paired t-test

Example: Comparison of ratios of arm to leg length in chimpanzees and gorillas.

Normal data

Difference

Measurements or counts

Pearson correlation coefficient

Example: Wing length vs tail length in birds.

Testing for a correlation

Calculate mean and 95% confidence intervals from replicates

Linear: The data plot in a straight line (uncommon biologically). Example: clutch size vs body size in Daphnia.

Plot a bar graph

Unpaired t-test

Two groups of data

What kind of data are you recording?

Different individuals

Example: Suitability of clay pots and plastic pots for plant growth.

Mann-Whitney U-test

Non-normal data

Data must be ranked in order of increasing size. Example: Size of fruit from a plant species grown in two different habitats.

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Frequencies

(counts only, not measurements)

Comparing observed counts to an expected count

Chi-squared for goodness of fit Example: An expected genetic ratio, or preference for different habitats.

Test for significance of observed vs expected

Covered in AP Biology 1 & 2

Covered in the Teacher’s Digital Edition and as spreadsheets via WEBLINKS

Covered in AP Biology 1

Not required

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Testing an association between groups of counts

PRACTICES

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Chi-squared for independence Example: Association of one plant with another in an area.

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The Biochemistry of Life

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Enduring Understanding

2.A 4.A

amino acid

2.A.3 Organisms must exchange matter with the environment to grow, reproduce, and maintain organization Essential knowledge

amphipathic

carbohydrate

dehydration synthesis (=condensation) denaturation DNA

hydrolysis

(a) Molecules and atoms from the environment are necessary to build new molecules c

1

Explain how carbon moves from the environment and how it is used to build new molecules and in storage and cell formation in organisms.

21 24 25

c

2

Describe how nitrogen and phosphorus move from the environment to organisms and how they are used in building new molecules in organisms.

21 24 25

c

3

Explain how living systems depend on the properties of water that arise from its polarity and hydrogen bonding. Include reference to cohesion, adhesion, thermal conductivity, high specific heat capacity, heat of vaporization, and heat of fusion, and role as a universal solvent.

hydrophilic

hydrophobic lipid

monomer

monosaccharide

c

1

Describe how the SA:V changes as cells increase in size and how this constrains cell size. Using examples, explain how the cells and structures of organisms enable adequate exchanges of materials and energy with the environment.

41 42

c

2

Understand that the surface area of the plasma membrane must be large enough to supply the needs of the cell. Explain why smaller cells have a more favorable SA:V ratio for the exchange of materials with the environment.

42 43

nucleotide

organic molecule phospholipid polar

polymer

polypeptide

polysaccharide

primary structure protein

PR-4

c

Activity number

(a) The structure and function of polymers are derived from the way their monomers are assembled c

1

Describe the structure of nucleic acids, including reference to the components of individual nucleotides and how the nucleotides are organized into polymers. Describe the structural and functional differences between DNA and RNA.

27 28

c

2

Explain how the specific order of amino acids in polypeptide (the primary structure) interacts with the environment to determine the overall shape of the protein molecule and its function. Explain how the properties of the amino acid R groups and their interactions determine final protein structure and function.

29 - 33

c

3

Describe the non-polar nature of a typical lipid (e.g. a triacyl glycerol) and explain how phospholipids differ in having polar and non-polar regions. Explain how differences in fatty acid saturation determine lipid structure and function.

34 35

c

4

Explain how carbohydrates are constructed from sugar monomers by dehydration synthesis (condensation) and how the structures and bonding determine the properties and functions of the polymer. To illustrate this, compare the structure of glucose polymers such as cellulose, starch, and glycogen.

secondary structure sugar

tertiary structure

triglyceride (= triacyl glycerol)

43

Essential knowledge

RNA

surface area to volume ratio

Investigate factors affecting diffusion rates in model cells [Procedure 1].

4.A.1 The subcomponents of biological molecules and their sequence determine the properties of that molecule

quaternary structure saturated fatty acid

22 23

(b) Surface area to volume ratios affect ability to make exchanges with the environment

non-polar

nucleic acid

Activity number

unsaturated fatty acid

SKILL

c

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Key terms

Use simple biochemical tests to distinguish sugars, proteins, and lipids.

36 - 40

26

(b) Directionality influences structure and function of the polymer

AP Biology 1 2E 2017.indb 29

c

1

Nucleic acids have directionality, defined by the 3' and a 5' carbons of the sugar in the nucleotide. Explain how the directionality of the molecule determines the direction of nucleotide addition during DNA synthesis and transcription (5'3').

27 28

c

2

Describe how proteins have an amino (NH2) end and a carboxyl (COOH) end. Explain how this determines the primary structure of a protein as a linear sequence of amino acids joined by peptide bonds.

29 30

c

3

Explain how the nature of the bonding between sugar monomers determines their orientation in the carbohydrate, which then determines 2° structure.

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

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

Chloroplasts in plant cells

Plant epidermis

ist

ia

n

Lo

uis

a

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

Kr

H

t Pe

a ow

e rs

rd

D ar

tmout College

Centrioles

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

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

Animal cell

Plant cell

Plant cell wall

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

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

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

(a) Carbohydrates:

(b) Lipids:

(c) Proteins:

(e) Inorganic ions:

(d) Nucleic acids:

(f) Water:

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Chloroplast membranes

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

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Chromosome

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

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

H

H

O

O

O

Hydrogen bonds

H

δ+

H

O

δ+

Small +ve charges

Na+

H

H

Intermolecular bonds between water and Small other polar molecules or ions are important for -ve biological systems. Inorganic ions may havecharge a positive or negative charge (e.g sodium ion is positive, chloride ion is negative). The O chargedSmall +ve charge water molecule is attracted to the Hcharged H ion and surrounds it (right). This formation of intermolecular bonds between water and the ions is what keeps the ions dissolved in water. Polar molecules such as amino acids and Water molecule carbohydrates also dissolve readily in water. Formula: H2O

δ-

H

Small –ve charge

A water molecule is polar, meaning it has a positively and a negatively charged region. In water, oxygen has a slight negative charge and each of the hydrogens have a slight positive charge. Water molecules have a weak attraction for each other, forming large numbers of weak hydrogen bonds with other water molecules (far right).

H

Water forms hydrogen bonds

Oxygen is attracted to the Na+

Cl–

Hydrogen is attracted to the Cl-

Water surrounding a negative ion (Cl-)

Water surrounding a positive ion (Na+)

The importance of water

Leptospira bacterium

The metabolic reactions carried out by all organisms depend on dissolved reactants (solutes) coming into contact. Water provides the medium for metabolic reactions. Water can also act as an acid (donating H+) or a base (receiving H+) in chemical reactions.

Water provides an aquatic environment for organisms to live in. Ice is less dense than water and floats, insulating the underlying water and maintaining the aquatic habitat. A lot of energy is needed for water to change state, so water has a buffering effect on climate.

Water is colorless, with a high transmission of visible light. Light penetrates aquatic environments, allowing photosynthesis to continue at depth. Water also has a high heat capacity, absorbing and releasing energy slowly. This means large bodies of water are thermally stable.

1. The diagram at the top of the page shows a positive sodium ion and a negative chloride ion surrounded by water molecules. On the diagram, draw on the charge of the water molecules.

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2. Explain the formation of hydrogen bonds between water and other polar molecules:

3. Explain how the dipole structure of water gives it a central role in metabolic processes:

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23 The Properties of Water come about because of its polarity and its ability to form hydrogen bonds with other polar molecules. Water's physical and chemical properties are essential for sustaining life.

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Key Idea: Water's chemical properties influence its physical properties and account for its central role in life's processes. Water's cohesive, adhesive, thermal, and solvent properties

Cohesive properties

Thermal properties

Water molecules are cohesive, they stick together because hydrogen bonds form between water molecules. Cohesion allows water to form drops and allows the development of surface tension. Example: The cohesive and adhesive properties of water allow it to be transported as an unbroken column through the xylem of plants.

ff Water has the highest heat capacity of all liquids, so it takes a lot of energy before it will change temperature. As a result, water heats up and cools down slowly, so large bodies of water maintain a relatively stable temperature. ff Water is liquid at room temperature and has a high boiling point because a lot of energy is needed to break the hydrogen bonds. The liquid environment supports life and metabolic processes.

Adhesive properties

Water is attracted to other molecules because it forms hydrogen bonds with other polar molecules. Example: The adhesion of water molecules to the sides of a capillary tube is responsible for a meniscus (the curved upper surface of a liquid in a tube).

ff Water has a high latent heat of vaporization, meaning it takes a lot of energy to transform it from the liquid to the gas phase. In sweating, the energy is provided by the body, so sweat has a cooling effect.

Solvent properties

Other substances dissolve in water because water's dipolar nature allows it to surround other charged molecules and prevent them from clumping together. Example: Mineral transport through a plant.

Reef corals secrete calcium carbonate to form a hard skeleton

Water is known as the universal solvent, because many substances will dissolve in it. In natural waters, dissolved minerals, such as calcium, are available to aquatic organisms, e.g. corals. Blood plasma is about 92% water and transports many water-soluble substances, including ions, glucose, and amino acids, around the body.

Many aquatic organisms depend on a thermally stable environment

Sweating cools by evaporation

Water has a high heat capacity, meaning it takes a lot of energy to heat it up and a lot of energy is lost when it cools. This means that large water bodies will maintain a relatively stable temperature, even when the air temperature fluctuates. It also allows temperature sensitive metabolic reactions to take place within organisms.

Water's high latent heat of vaporization means that a lot of energy is needed for water to change state (e.g from liquid to gas). Sweating is therefore an effective cooling mechanism. As the water in sweat evaporates from the skin's surface, heat from the body is transferred to the air, cooling the body.

2. How does water act as a coolant during sweating?

3. (a) What is the significance of water's solvent properties within organisms?

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1. What is the biological significance of water having a high heat capacity?

(b) What is the significance of water's solvent properties in natural water bodies?

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24 Organic Molecules Key Idea: Organic molecules are those with carbon-hydrogen bonds. They make up most of the chemicals found in living organisms and can be portrayed as formulae or models. Molecular biology is a branch of science that studies the molecular basis of biological activity. All life is based around carbon, which is able to combine with many other elements to form a large number of carbon-based (or organic) molecules.

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Specific groups of atoms, called functional groups, attach to a C-H core and determine the specific chemical properties of the molecule. The organic macromolecules (large complex molecules) that make up living things can be grouped into four classes: carbohydrates, lipids, proteins, and nucleic acids. The diagram (bottom) illustrates some of the common ways in which organic molecules are portrayed.

H

C

H

H H

A carbon atom (above) has four electrons that are available to form up to four covalent bonds with other atoms. A covalent bond forms when two atoms share a pair of electrons. The number of covalent bonds formed between atoms in a molecule determines the shape and chemical properties of the molecule.

O

C

C

O

Methanal (molecular formula CH2O) is a simple organic molecule. A carbon (C) atom bonds with two hydrogen (H) atoms and an oxygen (O) atom. In the structural formula (blue box), the bonds between atoms are represented by lines. Covalent bonds are very strong, so the molecules formed are very stable.

Organic macromolecule

Structural unit

Elements

Carbohydrates

Sugar monomer

C, H, O

Proteins

Amino acid

C, H, O, N, S

Lipids

Not applicable

C, H, O

Nucleic acids

Nucleotide

C, H, O, N, P

The most common elements found in organic molecules are carbon, hydrogen, and oxygen, but organic molecules may also contain other elements, such as nitrogen, phosphorus, and sulfur. Most organic macromolecules are built up of one type of repeating unit or 'building block', except lipids, which are quite diverse in structure.

Portraying organic molecules

The numbers next to the H O carbon atoms are used 1C for identification when the 2 molecule changes shape H C OH HO

3

H

H

4

OH

C6H12O6

H

5

OH

Glucose

H

C C C

6

C

OH

6 5

H

CH2OH

6

C

O

H

C

OH

4

HO

3

C

H

H

2

H

C

4

1

5

OH

C

1

2

3

OH

H

Molecular formula

Structural formula Glucose (straight form)

Structural formula α-glucose (ring form)

The molecular formula expresses the number of atoms in a molecule, but does not convey its structure. This is indicated by the structural formula.

Ball and stick model Glucose

Space filling model β-D-glucose

A ball and stick model shows the arrangement of bonds while a space filling model gives a more realistic appearance of a molecule.

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1. Study the table above and state the three main elements that make up the structure of organic molecules:

2. Name two other elements that are also frequently part of organic molecules:

3. (a) On the diagram of the carbon atom top left, mark with arrows the electrons that are available to form covalent bonds with other atoms.

(b) State how many covalent bonds a carbon atom can form with neighboring atoms:

4. Distinguish between molecular and structural formulae for a given molecule:

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25 Building an Organism contain many smaller components, such as inorganic ions and water. In order to carry out life processes, organisms must obtain nutrients from the environment. In animals, this is achieved by the consumption of other organisms. In plants, nutrients are obtained from the soil (via roots) or from the atmosphere.

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Key Idea: Living organisms are made up of different macromolecules and inorganic compounds. Living organisms are very complex biological structures, containing many different components. The major components are the four macromolecules: proteins, nucleic acids, carbohydrates, and lipids. In addition, organisms also

OXYGEN Source: atmosphere Use: cellular respiration, incorporated in to macromolecules

PHOSPHORUS Source: food Use: lipids, nucleic acids

CARBON Source: food Use: carbohydrates, proteins, lipids, nucleic acids

CARBON Source: atmosphere Use: carbohydrates, proteins, lipids, nucleic acids

Mnolf

VM

In plants, carbon is stored as starch in special organelles called amyloplasts.

R

C

VM

Inorganic ions are important for the structure and metabolism of all living organisms. An ion is simply an atom (or group of atoms) that has gained or lost one or more electrons. Many of these ions are water soluble. Some of the inorganic ions required by organisms and examples of their biological roles are described in this table (right). A deficiency in any of these ions can result in specific deficiency disorders.

NITROGEN Source: soil Use: proteins, nucleic acids

Ion

Name

Example of biological roles

Ca2+

Calcium

Component of bones and teeth, required for muscle contraction

Mg2+

Magnesium

Component of chlorophyll, role in energy metabolism

Fe2+

Iron (II)

Component of hemoglobin and cytochromes

NO3-

Nitrate

Component of amino acids

Na+

Sodium

Component of extracellular fluid and the need for nerve function

K+

Potassium

Important intracellular ion, needed for heart and nerve function

Cl-

Chloride

Component of extracellular fluid in multicellular organisms

1. State the main source of carbon, phosphorus, and nitrogen for animals: 2. (a) State the main source of carbon for plants:

PHOSPHORUS Source: soil Use: lipids, nucleic acids

(b) State the main source of phosphorus and nitrogen for plants:

3. List the four main macromolecule components of living organisms:

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Glycogen in muscle

In animals, carbon is stored as glycogen.

NITROGEN Source: food Use: proteins, nucleic acids

4. Explain why carbon is so important for building the molecular components of an organism:

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26 Biochemical Tests carbohydrates (sugars and starch). However, they cannot be used directly to determine absolute concentrations or distinguish between different molecules of the same type (e.g. different sugars in a mixed solution).

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Key Idea: Qualitative biochemical tests detect the presence of a specific molecule in food. Qualitative biochemical tests can be used to detect the presence of molecules such as lipids, proteins, or

Simple food tests

A qualitative test for reducing sugar

To determine whether this muffin contains any reducing sugars (e.g. glucose), the Benedict's test for reducing sugar is carried out.

Proteins: The biuret test Reagent:

Biuret solution.

Procedure:

A sample is added to biuret solution and gently heated.

Positive result:

Solution turns from blue to lilac.

The muffin is placed in a blender with some water and mixed until it forms an homogenous (uniform) mixture.

Starch: The iodine test Reagent:

Iodine.

Procedure:

Iodine solution is added to the sample.

Positive result:

Blue-black staining occurs.

Lipids: The emulsion test Reagent:

Ethanol.

Procedure:

The sample is shaken with ethanol. After settling, the liquid portion is distilled and mixed with water.

Positive result:

The solution turns into a cloudy-white emulsion of suspended lipid molecules.

Sugars: The Benedict's test Reagent:

Benedict’s solution.

Procedure:

Non reducing sugars: The sample is boiled with dilute hydrochloric acid (acid hydrolysis), then cooled and neutralized. A test for reducing sugars is then performed. Reducing sugar: Benedict’s solution is added, and the sample is placed in a water bath.

Positive result:

Solution turns from blue to orange to red-brown.

2-3 mL of the muffin mixture is placed into a test tube with 1 mL of Benedict's solution. The tubes are heated for 4 -10 minutes.

The intensity of the color depends on the concentration of glucose present in the sample. The darker the color, the more glucose is present. A colorimetric analysis enables the amount of glucose present to be quantified (see the following activity).

Negative result: solution remains blue

None Trace Clear blue Green

Moderate High Orange Brick red

1. Explain why lipids must be mixed in ethanol before they will form an emulsion in water:

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2. Explain why the emulsion of lipids, ethanol, and water appears cloudy:

3. What is the purpose of the acid hydrolysis step when testing for non-reducing sugars with Benedict's reagent?

4. What are the limitations of qualitative tests such as those described above?

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36 OH

P O CH 2

N

NH

27 Nucleotides O

H N

H

OH

N

H

OH

N

N

Nucleotide structure

N

H

NH

N

NH2

A

N

P O CH H 2

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O

N

H

OH

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O three components:Na base, a sugar, and a phosphate group. Key Idea: is H Nup nucleicNacids. A nucleotide N H Nucleotides make H Nucleotides may contain one of five bases. The combination made up ofHa base,NH aH2sugar, andONa phosphate group. Phosphate: Links N of bases in the nucleotides making up DNA or RNA stores Nucleotides areNthe building blocks of Hthe nucleic acids DNA neighboring sugars. H OOH H Base: One of four types and RNA, which are involved in the transmission of inheritedH the information controlling the cell's activity. The bases in Guanine Adenine H possible. The base carries H Phosphate Sugar DNA are the same as RNA except that thymine (T) in DNA is information. Nucleotide derivatives, Base such as H ATP and GTP, the coded genetic replaced with uracil (U) in RNA. are involved in energy transfers in cells. A nucleotide has message in a nucleic acid. OH

Pyrimidines Phosphate

O

NH2

CH3

H N

H

H N

H

N

O

O

H

N

H

Sugar

H

N

O

H

Phosphate Base

OHribose Sugar: One of two types: Phosphate groups are represented in RNA and deoxyribose in DNA. by circles. Along with the pentose HO P OH sugar they form the "backbone" of O the DNA or RNA molecule.

N

H

N

N

N

H

NH2

N

Phosphate: Links neighboring sugars.

H

Guanine

Sugar: One of two types: ribose in RNA and deoxyribose in DNA.

Adenine

Purines are double ringed bases. Both DNA and RNA contain the purines adenine (A) and guanine (G).

H N

O

CH2OH

Nucleotide derivatives OH

CH3

A

H

N Adenine

H

Thymine ATP

NH2

O

H

N

H

H

H N

H

O

N

O

Cytosine Ribose

H

Sugars

CH2OH

H

Uracil

H

H

H

OH

N

O

H

OH

O

OH

O

H

3 phosphate HO P groups OH

H

H

OH

H

deoxyribose

H

OH

ribose

Nucleotides contain one of two different sorts of sugars. Deoxyribose sugar is only found in DNA. Ribose sugar is found in RNA.

2 phosphates

A

Formation of a nucleotide

Nucleotide formation

Inorganic phosphate

ADP+ Pi

Condensation

Ribose

ATP is a nucleotide derivative used to provide chemical energy for metabolism. It consists of an adenine linked to a ribose sugar and 3 phosphate groups. Energy is made available when a phosphate group is transferred to a target molecule. Other nucleoside triphosphates (NTPs) have similar roles.

A

H2 O

In nucleotide formation, a phosphoric acid and a base are bonded to a sugar molecule by condensation reactions in which water is given off (also called dehydration synthesis). The reverse reaction is hydrolysis.

(b) In RNA:

2. Name the sugar present: (a) In DNA:

A

H2 O

1. List the nucleotide bases present: (a) In DNA:

(water removed)

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Adenine

(b) In RNA:

3. (a) What molecules would you label with a fluorescent tag if you wanted to label only the RNA in a cell and not the DNA?

(b) What molecules would you label with a fluorescent tag if you wanted to label only the DNA in a cell and not the RNA?

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H

H

N

O

O

OH

H

A

Base: One of five bases possible. The base carries the coded genetic message in a nucleic acid.

NH2

N

H

OH

O

deoxyribose

Purines

H

H

H

Thymine Cytosine Uracil Pyrimidines are single ringed bases. DNA contains the pyrimidines cytosine (C) and thymine (T). RNA contains the pyrimidines cytosine and uracil (U).

H N

CH2OH

N

O

H

H

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28 Nucleic Acids Key Idea: Nucleic acids are macromolecules made up of long chains of nucleotides, which store and transmit genetic information. DNA and RNA are nucleic acids. DNA and RNA are nucleic acids involved in the transmission of inherited information. Nucleic acids have the capacity to store the information that controls cellular activity. The central nucleic acid is called deoxyribonucleic acid (DNA). Ribonucleic acids (RNA) are involved in ‘reading’ and

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translating the information in DNA. Messenger RNAs carry the code for the amino acids in proteins, tRNAs move amino acids to the ribosome binding sites, and rRNA forms part of ribosome responsible for catalyzing the formation of a polypeptide. All nucleic acids are made up of nucleotides linked to form chains or strands. The strands vary in the nucleotide base sequence. It is this sequence that provides the coding for the amino acids that make up proteins.

RNA molecule

DNA molecule

DNA molecule

RNA molecule

U

In RNA, uracil replaces thymine in the code.

C

A

Ribose sugar

Ribonucleic acid (RNA) comprises a single strand of nucleotides linked together. Although it is single stranded, it is often found folded back on itself, with complementary bases joined by hydrogen bonds.

C

G

G

T

A

C

G

Deoxyribose sugar

Hydrogen bonds hold the two strands together. Only certain bases can pair.

T

A

DNA molecule

Symbolic representation

Space filling model

Deoxyribonucleic acid (DNA) comprises a double strand of nucleotides linked together. It is shown unwound in the symbolic representation (above left). The DNA molecule takes on a double helix shape as shown in the space filling model above right.

Double-stranded DNA

5'

3'

ff The DNA backbone is made up of alternating phosphate and sugar molecules, giving the DNA molecule an asymmetrical structure.

ff The way the pairs of bases come together to form hydrogen bonds is determined by the number of bonds they can form and the configuration of the bases.

A

G

ff The asymmetrical structure gives a DNA strand direction. Each strand runs in the opposite direction to the other.

ff The ends of a DNA strand are labeled 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).

C

T

A

3'

G

C

T

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The double-helix structure of DNA is like a ladder twisted into a corkscrew shape around its longitudinal axis. It is ‘unwound’ here to show the relationships between the bases.

5'

1. Label the following parts on the diagram of the double-stranded DNA molecule above: (a) Deoxyribose (b) Phosphate (c) Hydrogen bonds (d) Purine bases (e) Pyrimidine bases

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Nucleotides are joined by condensation polymerization

A condensation reaction joins Formation of a two molecules together with the loss of a water molecule dinucleotide (dehydration synthesis). In the formation of nucleic acids, nucleotides are joined together into polymers through a condensation reaction between the phosphate of one nucleotide and the sugar of another. Water is released. Because of the way they are formed, nucleic acids are called condensation polymers.

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T

H2O

5

4

C

1

3

2

New nucleotides added to this end.

The carbon atoms on the pentose sugar are labeled one to five. During DNA replication (when new DNA is made) new nucleotides are added to the 3' end (the third carbon) of the existing nucleotide chain. It is therefore said DNA replication works in the 5' to 3' direction.

2. (a) Explain the base-pairing rule that applies in double-stranded DNA:

(b) How is the base-pairing rule for mRNA different?

(c) What is the purpose of the hydrogen bonds in double-stranded DNA?

3. Briefly describe the roles of RNA:

4. How can simple nucleotide units combine to store genetic information?

5. (a) What makes the DNA strands asymmetrical?

(b) What are the differences between the 5' and 3' ends of a DNA strand?

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de

6. Complete the following table summarizing the differences between DNA and RNA molecules:

DNA

Sugar present

Bases present

Number of strands Relative length

RNA

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29 Amino Acids Amino acids are the basic units from which proteins are made. Twenty amino acids commonly occur in proteins and they can be linked in many different ways by peptide bonds to form a huge variety of polypeptides. Proteins are made up of one or more polypeptide molecules.

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Key Idea: Amino acids join together in a linear chain by condensation reactions to form polypeptides. The sequence of amino acids in a protein is defined by a gene and encoded in the genetic code. In the presence of water, they can be broken apart by hydrolysis into their constituent amino acids.

The structure and properties of amino acids

Chemically variable 'R' group

R

C

C

NH2

Amine group

Carboxyl group

OH

H

Carbon atom

O

NH 2 CH2 CH2 CH2 CH22

SH

CH2

COOH CH 2

Hydrogen atom

All amino acids have a common structure (above), but the R group is different in each kind of amino acid (right). The property of the R group determines how it will interact with other amino acids and ultimately determines how the amino acid chain folds up into a functional protein. For example, the hydrophobic R groups of soluble proteins are folded into the protein's interior, while the hydrophilic groups are arranged on the outside.

Cysteine

Lysine

Aspartic acid

This 'R' group can form disulfide bridges with other cysteines to create cross linkages in a polypeptide chain.

This 'R' group gives the amino acid an alkaline property.

This 'R' group gives the amino acid an acidic property.

Condensation and hydrolysis reactions

Two amino acids

R1

H2N

R2

O

H2N

C

C

OH

H

C

C

OH

H

Condensation

Hydrolysis

Two amino acids are joined to form a dipeptide with the release of a water molecule.

H2N

O

When a dipeptide is split, a water molecule provides a hydrogen and a hydroxyl group.

R1

O

H

R2

C

C

N

C

H

H

Amino acids are linked by peptide bonds to form long polypeptide chains of up to several thousand amino acids. Peptide bonds form between the carboxyl group of one amino acid and the amine group of another (left). Water is formed as a result of this bond formation. The sequence of amino acids in a polypeptide is called the primary structure and is determined by the order of nucleotides in DNA and mRNA. The linking of amino acids to form a polypeptide occurs on ribosomes. Once released from the ribosome, a polypeptide will fold into a secondary structure determined by the composition and position of the amino acids making up the chain.

A polypeptide chain

O

C

+ H2O

OH

Dipeptide

Peptide bond

Peptide bond

Peptide bond

Peptide bond

Peptide bond

Peptide bond

1. (a) What makes each of the amino acids in proteins unique? (b) What is the primary structure of a protein?

(c) What determines the primary structure?

(d) How do the sequence and composition of amino acids in a protein influence how a protein folds up?

2. (a) What type of bond joins neighboring amino acids together?

(b) How is this bond formed?

(d) How are di- and polypeptides broken down?

(c) Circle this bond in the dipeptide above:

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30 Protein Structure

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Key Idea: The sequence and type of amino acids in a protein determine its three-dimensional shape and function. Proteins are large, complex macromolecules, built up from a linear sequence of repeating units called amino acids. Proteins account for more than 50% of the dry weight of most cells and are important in virtually every cellular process. The

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various properties of the amino acids, which are conferred by the different R groups, determine how the polypeptide chain folds up. This three dimensional tertiary structure gives a protein its specific chemical properties. If a protein loses this precise structure (through denaturation), it is usually unable to carry out its biological function.

Primary (1°) structure (amino acid sequence)

Phe

Glu

Tyr

Ser

Iso

Met

Ala

Ala

Peptide bond

Ser

Glu

1. Describe the main features in the formation of each part of a protein's structure: (a) Primary structure:

Amino acid

Hundreds of amino acids are linked by peptide bonds to form polypeptide chains. The attractive and repulsive charges on the amino acids determine the higher levels of organization in the protein and its biological function.

Secondary (2°) structure (a-helix or b-pleated sheet)

Secondary (2°) structure is maintained by hydrogen bonds between neighboring CO and NH groups. The hydrogen bonds are individually weak but collectively strong.

a-helix

Hydrogen bonds

(b) Secondary structure:

(c) Tertiary structure:

(d) Quaternary structure:

Amino acid chain

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)

a-helix

Tertiary (3°) structure is maintained by more distant interactions such as disulfide bridges between cysteine amino acids, ionic bonds, and hydrophobic interactions.

Aspartic acid Ionic bond Lysine

Disulfide bond

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.

Alpha chain

Some complex proteins are only functional when as a group of polypeptide chains. Hemoglobin has a 4° structure made up of two alpha and two beta polypeptide chains, each enclosing a complex iron-containing prosthetic (or heme) group.

Prosthetic (heme) group

Beta chain

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. WEB

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2. How are proteins built up into a functional structure?

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Quaternary (4°) structure

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31 Protein Shape is Related to Function 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.

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Key Idea: The three dimensional shape of a protein reflects its role. When a protein is denatured, it loses its functionality. As we have seen, a protein may consist of one polypeptide chain, or several polypeptide chains linked together. Hydrogen bonds between amino acids cause the polypeptide chain to form its secondary structure, either an a-helix or a

The shape of a protein reflects its biological role

H2N

Active site formed by the precise configuration of the protein

b sheets

Glu

Arg

Asn

Glu

Tyr

Gln

Val

Leu

b chain

Leu

Val

Ser

H2N

S

COOH

Asn

Phe

Leu

Cys

Gly

S

Gly

Cys

His

His

S

Gly

a helix

Ala

Cys

Gly

Phe

Val

Leu

S

Ile

Tyr

Val

Asn

Phe

Gln

Glu

Cys

Thr

S

Glu

Tyr

a chain

Leu

Thr

Cys

Gln

Pro

Tyr

Ser

S

Ile

Cys

Leu

Ser

Lys

Amylase

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 catalyze reactions. They are specific to their substrates and their tertiary structure creates an active site where the substrate can bind and the reaction can occur. The specificity of the active site is determined by the interactions of amino acid R groups. Denaturation alters the active site and causes a loss of function.

Many proteins, e.g. insulin and hemoglobin, 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 stabilized by disulfide bridges between neighboring 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 stabilize 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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32 Comparing Fibrous and Globular Proteins proteins are spherical and somewhat soluble forming colloids in water (e.g. enzymes). Fibrous proteins have an elongated structure and are not water soluble. Some, such as keratin, are even insoluble in organic solvents. Fibrous proteins provide stiffness and rigidity to the more fluid components of cells and tissues and have important structural roles.

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Key Idea: The very different structure and properties of globular and fibrous proteins reflect their contrasting roles. Proteins can be classified according to structure or function. Globular and fibrous proteins form two of the main broad structural groups of proteins (the others being membrane proteins and disordered proteins such as casein). Globular

Variable regions

Globular proteins

Globular shape is a function of their tertiary structure. Some proteins (e.g. actin and tubulin) globular and soluble as monomers, but polymerize to form long, stiff fibers. Properties

• Easily water soluble

• Tertiary structure critical to function

• Polypeptide chains folded into a spherical shape

Function

Constant regions

• Catalytic, e.g. enzymes

• Regulatory, e.g. hormones (insulin)

• Transport, e.g. hemoglobin, serum albumin

IgG2, a common antibody (immunoglobulin) in human serum. The darker domains are the constant regions of the molecule, whereas the lighter colored domains are the variable regions and dictate antibody binding specificity.

• Protective, e.g. immunoglobulins (antibodies) • Structural, e.g. actin monomers, tubulin monomers (cytoskeletal elements)

Hemoglobin

RuBisCO

Zephyris cc 3.0

Insulin

Heme group

Insulin is a peptide hormone involved in the regulation of blood glucose. Insulin is composed of two peptide chains (the A chain and the B chain) linked together by two disulfide bonds.

Hemoglobin is an oxygen-transporting protein found in vertebrate red blood cells. One hemoglobin molecule consists of four polypeptide chains (two identical a chains and two identical b chains). Each contains an ironcontaining heme group, which binds oxygen.

RuBisCo is a large multi-unit enzyme found in green plants and catalyzes the first step of carbon fixation in the Calvin cycle. It consists of 8 large (L) and 8 small (S) subunits arranged as 4 dimers. RuBisCO is the most abundant protein on Earth.

1. How are globular proteins involved in the functioning of organisms? Use examples to help illustrate your answer:

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2. Using an example, explain how the shape and properties of a globular protein relate to its functional role:

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Fibrous proteins

• Water insoluble and not easily denatured

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Properties

• Very tough physically; may be supple or stretchy

• Parallel polypeptide chains in long fibers or sheets

Function

• Structural role in cells and organisms e.g. collagen. • Contractile e.g. actin polymers

Many collagen molecules form fibrils and the fibrils group together to form fibers.

Covalent cross links between the collagen molecules

Hydrogen bond

Collagen (above) is the main protein in the connective tissues of animals. It can be mineralized to varying degrees (e.g. with calcium and phosphorus) to provide different amounts of rigidity and compliance (as in bone, ligaments, tendons, cartilage, and skin).

Glycine

Keratins are found in hair, nails, claws (above),horn, hooves, wool, feathers, and the outer layers of the skin. They fall into two classes (a and b keratins) defined on their secondary structures (a-helices as above or b-sheets). Keratins are characterized by a high sulfur content, with large numbers of disulfide bridges between cysteine residues. These form permanent, thermally stable cross links.

The collagen molecule consists of three polypeptides wound into a helical ‘rope’. Every third amino acid in each polypeptide is a glycine (Gly) where hydrogen bonding holds the three strands together. The collagen molecules self assemble into fibrils of many molecules held together by covalent cross linkages between hydroxylysine and lysine residues (above). Bundles of fibrils form the fibers found in connective tissue.

Elastin is another fibrous connective tissue protein which has elastic properties that allow tissues to resume their shape after stretching. Skin, arteries, lungs, and bladder all contain elastin. Elastin has many hydrophobic amino acids such as glycine and proline, which form mobile hydrophobic regions flanked by stable crosslinks between lysine residues.

3. How are fibrous proteins involved in the functioning of organisms? Use examples to help illustrate your answer:

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4. Using an example, explain how the shape and properties of a fibrous protein relate to its functional role:

5. What common feature contributes to the strength and stability of collagen, keratin, and elastin?

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33 Modifications of Proteins position and orientate the glycoprotein in the membrane, guide a protein to its final destination, or help in cell-to-cell recognition and cell signaling. Other proteins may have fatty acids added to them in the rER to form lipoproteins. These modified proteins transport lipids in the plasma between various organs in the body (e.g. gut, liver, and adipose tissue). Other common post-translational modifications include degradation, cleavage, and phosphorylation (below).

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Key Idea: The modification of proteins allows the cell to specify their use and final destination. Proteins may be modified after they have been produced by post translational modification. Two important modifications involve adding carbohydrates or lipids to the protein. Glycoproteins are formed by adding carbohydrates to proteins as they pass through the rough endoplasmic reticulum (rER) and Golgi. The carbohydrates may help

Cleaving: Polypeptide chains may be cleaved to give smaller chains, which then fold or join to give the functional protein. An example is human insulin which is transcribed as one long polypeptide chain before being cleaved in two places to form two shorter chains that form the functional protein.

Glycosylation (adding carbohydrate groups): This is used to add an ID tag to the protein that will allow the cell to recognize its use and where it is to be transported (2a). The resulting glycoprotein may be used in the cell membrane or secreted. The carbohydrate tag may help position the glycoprotein within the membrane (2b).

P

P

Phosphorylation (the addition of phosphate groups) takes place in the Golgi. It may contribute to the protein's three dimensional structure or help with cell signaling.

P

P

Lipid attachment: Proteins may have lipids attached to them which anchor the protein to the plasma membrane.

Degradation: Some polypeptide chains may be tagged for degradation when they are no longer useful and their amino acids reused in the formation of other proteins.

(b) Why are these changes necessary?

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1. (a) Describe some of the modifications that polypeptide chains undergo before becoming functional proteins:

2. Why might the orientation of a protein in the plasma membrane be important?

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34 Lipids Key Idea: Lipids are non-polar, hydrophobic organic molecules, which have many important biological functions. Fatty acids are the building blocks of more complex lipids. Lipids are organic compounds which are mostly nonpolar (have no overall charge) and hydrophobic, so they do not readily dissolve in water. Lipids include fats, waxes, sterols,

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and phospholipids. Fatty acids are a major component of neutral fats and phospholipids. Most fatty acids consist of an even number of carbon atoms, with hydrogen bound along the length of the chain. The carboxyl group (–COOH) at one end makes them an acid. They are generally classified as saturated or unsaturated fatty acids (below).

Triglycerides (triacylglycerols)

Triglycerides are formed by condensation

Triglycerides form when glycerol bonds with three fatty acids. Glycerol is an alcohol containing three carbons. Each of these carbons is bonded to a hydroxyl (-OH) group. When glycerol bonds with the fatty acid, an ester bond is formed and water is released. Three separate condensation (or dehydration synthesis) reactions are involved in producing a triglyceride.

O

H2C

O O

HC

O

H2C

O

9

12

15

Fatty acids

H H

H H C O C O

Triglyceride: an example of a neutral fat

H H

C O C O

H H

HO HO

CO CO

CH CH CH22 CH33

C O C O H H Glycerol

H H

HO HO

CO CO

CH CH CH22 CH33

O

Glycerol

Ester bond

H H

Neutral fats and oils are the most abundant lipids in living things. They make up the fats and oils found in plants and animals. Neutral fats and oils consist of a glycerol attached to one (mono-), two (di-) or three (tri-) fatty acids by ester bonds.

H H

HO HO

CO CO

CH CH CH22 CH33

Fatty acids

Condensation

Esterification: A condensation reaction of an alcohol (e.g. glycerol) with an acid (e.g. fatty acid) to produce an ester and water. In the diagram right, the ester bonds are indicated by thicker blue lines.

Lipolysis: The breakdown of lipids. It involves hydrolysis of triglycerides into glycerol molecules and free fatty acids.

Hydrolysis

H H

H H C C

O O

CO CO

CH CH CH22 CH33

H H

C C

O O

CO CO

CH CH CH22 CH33

H H

C C H H

O O

CO CO

CH CH CH22 CH33

O H O H H H O H O H H H O H O H H H

Triglyceride

Water

Saturated and unsaturated fatty acids

Fatty acids are carboxylic acids with long hydrocarbon chains. They are classed as either saturated or unsaturated. Saturated fatty acids contain the maximum number of hydrogen atoms. Unsaturated fatty acids contain some doublebonds between carbon atoms and are not fully saturated with hydrogens. A chain with only one double bond is called monounsaturated, whereas a chain with two or more double bonds is called polyunsaturated. O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

O C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

Palmitic acid

Butter

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H

O

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

O C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

--

H

H

Linoleic acid

Structural formulae and space-filling models for a saturated fatty acid, palmitic acid (left) and an unsaturated fatty acid, linoleic acid (right). The arrows indicate double bonded carbon atoms that are not fully saturated with hydrogens.

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--

Lipids containing a high proportion of saturated fatty acids tend to be solids at room temperature (e.g. butter, left). Lipids with a high proportion of unsaturated fatty acids are oils and tend to be liquid at room temperature (e.g. olive oil, right). This is because the unsaturation causes kinks in the straight chains so that the fatty acid chains do not pack closely together.

Oil

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46 Why are lipids high energy?

Biological functions of lipids

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Lipids have a high proportion of hydrogen present in the fatty acid chains. When the molecule is metabolized, the chemical energy is released. Being so reduced and anhydrous (without water), they are an economical way to store fuel reserves, and provide more than twice as much energy as the same quantity of carbohydrate. In addition, respiration (oxidation) of stored fat (e.g. in a camel's hump) produces carbon dioxide and water, making stored fat a source of water.

WMU

Plasma membrane

Waxes and oils secreted onto surfaces provide waterproofing in plants and animals.

TG

Phospholipids form the structure of cellular membranes in eukaryotes and prokaryotes.

Oxidation of lipids provides large amounts of ATP and metabolic water.

Fat absorbs shocks. Organs that are prone to bumps and shocks (e.g. kidneys) are cushioned with a relatively thick layer of fat.

Stored lipids provide fuel but also insulation. Increased body fat levels in winter reduce heat losses to the environment.

1. Identify the main components (a-c) of the symbolic triglyceride right: (a)

a

(b)

b

(c)

c

2. Why do lipids have such a high energy content?

3. (a) Distinguish between saturated and unsaturated fatty acids:

(b) Relate the properties of a neutral fat to the type of fatty acid present:

4. (a) Describe what happens during the esterification (condensation) process to produce a triglyceride:

(b) Describe what happens when a triglyceride is hydrolyzed:

5. Discuss the biological role of lipids:

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35 Phospholipids bilayers in aqueous solutions and are the main component of all cellular membranes. The fatty acid tails can be saturated (forming straight chains) or unsaturated (kinked chains). The level of phospholipids with saturated or unsaturated tails affects the fluidity of the phospholipid bilayer.

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Key Idea: Phospholipids are modified triglycerides. They are important components of cellular membranes. Phospholipids are similar in structure to a triglyceride except that a phosphate group replaces one of the fatty acids attached to the glycerol. Phospholipids naturally form

Phospholipids

Phospholipids and membranes

Phospholipids consist of a glycerol attached to two fatty acid chains and a phosphate (PO43-) group. The phosphate end of the molecule is attracted to water (it is hydrophilic) while the fatty acid end is repelled (hydrophobic). The hydrophobic ends turn inwards in the membrane to form a phospholipid bilayer.

The amphipathic (having hydrophobic and hydrophilic ends) nature of phospholipids means that when in water they spontaneously form bilayers. This bilayer structure forms the outer boundary of cells or organelles. Modifications to the different hydrophobic ends of the phospholipids cause the bilayer to change its behavior. The greater the number of double bonds in the hydrophobic tails, the greater the fluidity of the membrane.

+

Hydrophilic head

CH2

N (CH3)3

CH2 O

P

O

–

O

O

CH

O

O

C

O C

CH2

O

Membrane containing only phospholipids with saturated fatty acid tails.

Hydrophobic tails

H2C

Membrane containing phospholipids with unsaturated fatty acid tails. The fact that the phospholipids do not stack neatly together produces a more fluid membrane.

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1. (a) Relate the structure of phospholipids to their chemical properties and their functional role in cellular membranes:

(b) Suggest how the cell membrane structure of an Arctic fish might differ from that of tropical fish species:

2. Explain why phospholipid bilayers containing many phospholipids with unsaturated tails are particularly fluid:

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36 Carbohydrate Chemistry

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48

Monosaccharide polymers form the major component of most plants (as cellulose). Monosaccharides are important as a primary energy source for cellular metabolism. Carbohydrates have the general formula Cx(H2O)y, where x and y are variable numbers (often but not always the same).

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Key Idea: Monosaccharides are the building blocks for larger carbohydrates. They can exist as isomers. Sugars (monosaccharides and disaccharides) play a central role in cells, providing energy and joining together to form carbohydrate macromolecules, such as starch and glycogen.

Monosaccharides

Ribose: a pentose monosaccharide

Monosaccharides are single-sugar molecules and include glucose (grape sugar and blood sugar) and fructose (honey and fruit juices). They are used as a primary energy source for fuelling cell metabolism.

HOCH2

Examples monosaccharide structures Singleofsugars (monosaccharides) Triose

Pentose

Hexose

C C C

e.g. ribose, deoxyribose

e.g. glucose, fructose, galactose

H

H

H

C

C

OH

OH

H

Ribose is a pentose (5 carbon) monosaccharide which can form a ring structure (left). Ribose is a component of the nucleic acid ribonucleic acid (RNA).

Glucose isomers CH2OH

CH2OH

C

H

C

OH

O

H

H

OH

H

C

C

H

OH

C

H

C

C

OH

HO

OH

H

C

C

H

Îą-glucose

O

H

HO

C

H

OH

β -glucose

Isomers are compounds with the same chemical formula (same types and numbers of atoms) but different arrangements of atoms. The different arrangement of the atoms means that each isomer has different properties.

Molecules such as glucose can have many different isomers (e.g. a and b glucose, above) including straight and ring forms.

Rufino Uribe

e.g. glyceraldehyde

C

C

They can be joined together to form disaccharides (two monomers) and polysaccharides (many monomers). Monosaccharides can be classified by the number of carbon atoms they contain. Some important monosaccharides are the hexoses (6 carbons) and the pentoses (5 carbons). The most common arrangements found in sugars are hexose (6 sided) or pentose (5 sided) rings (below). The commonly occurring monosaccharides contain between three and seven carbon atoms in their carbon chains and, of these, the 6C hexose sugars occur most frequently. All monosaccharides are reducing sugars (they can participate in reduction reactions).

OH

O

Glucose is a versatile molecule. It provides energy to power cellular reactions, can form energy storage molecules such as glycogen, or it can be used to build structural molecules.

Plants make their glucose via the process of photosynthesis. Animals and other heterotrophic organisms obtain their glucose by consuming plants or other organisms.

Fructose, often called fruit sugar, is a simple monosaccharide. It is often derived from sugar cane (above). Both fructose and glucose can be directly absorbed into the bloodstream.

(a)

(b)

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1. Describe the two major functions of monosaccharides:

2. Describe the structural differences between the ring forms of glucose and ribose:

3. Using glucose as an example, define the term isomer and state its importance:

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37 Condensation and Hydrolysis of Sugars

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CH2OH

Key Idea: Dehydration synthesis reactions (also called condensations) join monosaccharides together to H form C disaccharides and polysaccharides. Hydrolysis reactions split disaccharides and polysaccharides into smaller molecules. OH Monosaccharide monomers can be linked together by condensation reactions to produce disaccharides and

CH2OH

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Cpolysaccharides by dehydration Csynthesis.O The reverse O H HO into H sugars down Hreaction, hydrolysis, breaks compound H C monosaccharides. C their constituent DisaccharidesC (doubleH H OH OH sugars) are produced when OH H are HO two monosaccharides Cjoined together. C C are formed C Different disaccharides by joining H

OH β -glucose

Condensation and Hydrolysis Reactions

Disaccharides

Monosaccharides can combine to form compound sugars in what is called a condensation reaction. Compound sugars can be sugars (monosaccharides) broken down bySingle hydrolysis to simple monosaccharides.

Disaccharides (below) are double-sugar molecules and 2OH as energy sources and as CH 2OH blocks for areCH used building larger molecules. They are important in human nutrition C C O and are foundOin milk (lactose), H table sugar (sucrose), and H H Double sugars (disaccharides) malt H (maltose). H C C C H H OH OH Examples OH OH OH sucrose, C C C C

H

Triose

C C C

Two monoPentose saccharides

OH

OH

H

C

C

OH

C

O

CH2OH

C

H

C

OH

H

H

OH β -glucose

Glycosidic bond

H

C

C

H

OH

CH2OH C

H

C

OH

OH

OH

H

C

C

H

α -glucose

CH2OH

C

OH

Glycosidic bond

O

H

lactose, maltose, OH cellobiose

C

OH

OH

Maltose is composed of two a-glucose molecules. Germinating seeds contain maltose because the plant breaks down their starch stores to use it for food.

B

Maltose

C

H

α -glucose

A

H

O

H

C

OH

H

C

Disaccharide + water O

H

α -glucose α -glucose The type of disaccharide formed depends on the monomers involved and whether they are in their a- or b- form. Only a few disaccharides (e.g. lactose) are classified as reducing sugars. Some common disaccharidesCH areOH described below. CH2OH 2 Maltose C C O O H Lactose, a milk sugar,H H H H is made up of b-glucose C C C C + b-galactose. Milk H OH H OH OH OH contains 2-8% lactose O C C C by weight. It is C the primary carbohydrate H H OH OH Glycosidic bond source for suckling Disaccharide + water mammalian infants.

H

+

H2O

O

OH

fructose, galactose When a disaccharide is split, as in digestion, a water molecule is used CH2OH as a source of hydrogen C a hydroxyl O group. The and reaction is catalyzed HO by H specific enzymes. C H OH H

HO

H OH α-glucose

H

e.g. glucose, Hydrolysis reaction

deoxyribose Two monosaccharides are joined together to form a disaccharide CH2OHwith the release of a water molecule (also C called dehydration Osynthesis). H A net energy input is required H H H for C the reaction to proceed. C C OH

C

Hexose

e.g. glyceraldehyde e.g. ribose, Condensation reaction

oes

together different combinations of monosaccharides (below).

H OH α-glucose

C

H

H

C

CH2OH

C

OH

H

C

C

H

OH Glycosidic bond

O

O

H

OH

H

C

C

H

OH

H

C

OH

Sucrose (table sugar) is a simple sugar derived from plants such as sugar cane, sugar beet, or maple sap. It is composed of an a-glucose molecule and a b-fructose molecule.

1. Explain briefly how disaccharide sugars are formed and broken down:

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Disaccharide + water

+ water

2. On the diagram above, name the reaction occurring at points A and B and name the product that is formed:

3. On the lactose, maltose, and sucrose molecules (above right), circle the two monomers on each molecule. ©2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

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38 Colorimetry to produce a color. After a set time the samples are placed in a colorimeter, which measures the solution's absorbance at a specific wavelength. A dilution series can be used to produce a calibration curve, which can then be used to quantify that substance in samples of unknown concentration. The example below is measuring glucose in urine samples.

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Key Idea: Colorimetry determines the concentration of colored compounds in solution. Colorimetry is a quantitative technique used to determine the concentration of a specific substance in a solution. A specific reagent (e.g. Benedict's, which detects glucose) is added to the test solution where it reacts with the substance of interest

Colorimetric analysis of glucose

Aim

To find the concentration of glucose in three unknown urine samples.

Prepare glucose standards

Method

A calibration curve was prepared using prepared artificial 'urine' containing known concentrations of glucose (right). A Benedict's test was performed on each and the absorbance of the resulting solution was measured using a colorimeter. The Benedict's test was then performed on unknown urine samples and the absorbance of the resulting solutions measured and compared to the calibration curve.

0

2

4

6

8

10

Concentration of glucose (mg dL-1)

Benedict's test for reducing sugars

Benedict’s solution is added, and the sample is placed in a water bath at 90°C or heated over a Bunsen burner.

Produce the calibration curve

Cool and filter samples as required. Using a red filter, measure the absorbance (at 735 nm) for each of the known dilutions and use these values to produce a calibration curve for glucose.

Results:

Absorbance (arbitary units)

Sample 1 absorbance: 0.10 Sample 2 absorbance: 0.05 Sample 3 absorbance: 0.82

Negative test (clear blue)

Positive test (red precipitate)

1.2 1.0 0.8 0.6 0.4 0.2 0.0

0

36

72

108

144

180

Concentration of glucose (mg dL ) -1

1. Why is Benedict's reagent added to the samples?

(a) Sample 1:

(b) Sample 2:

(c) Sample 3:

3. What is the purpose of the 0 mg dL-1 tube?

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2. Use the calibration curve provided to estimate the glucose content of the urine samples:

4. What would you do if the absorbance values you obtained for most of your 'unknowns' were outside the range of your calibration curve?

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39 Polysaccharides common polysaccharides (cellulose, starch, and glycogen) contain only glucose, but their properties are very different. These differences are a function of the glucose isomer involved and the types of linkages joining the monomers. Different polysaccharides can thus be a source of readily available glucose or a structural material that resists digestion.

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Key Idea: Polysaccharides consist of many monosaccharides joined by condensation. Their functional properties depend on composition and monosaccharide isomer involved. Polysaccharides are macromolecules consisting of straight or branched chains of many monosaccharides. They can consist of one or more types of monosaccharides. The most

Cellulose

The structure of polysaccharides (also called complex carbohydrates) can be compared using molecular visualization software

Public Domain

Cellulose is a structural material found in the cell walls of plants. It is made up of unbranched chains of b-glucose molecules held together by b-1,4 glycosidic links. As many as 10,000 glucose molecules may be linked together to form a straight chain. Parallel chains become cross-linked with hydrogen bonds and form bundles of 60-70 molecules called microfibrils. Cellulose microfibrils are very strong and are a major structural component of plants, e.g. as the cell wall. Few organisms can break the b-linkages so cellulose is an ideal structural material.

Cellulose

Cotton fibers contain more than 90% cellulose fiber.

Starch

Starch granules

BF

Starch is also a polymer of glucose, but it is made up of long chains of a-glucose molecules linked together. It contains a mixture of 25-30% amylose (unbranched chains linked by a-1,4 glycosidic bonds) and 70-75% amylopectin (branched chains with a-1, 6 glycosidic bonds every 24-30 glucose units). Starch is an energy storage molecule in plants and is found concentrated in insoluble starch granules within specialized plastids called amyloplasts in plant cells (see photo, right). Starch can be easily hydrolyzed by enzymes to soluble sugars when required.

Amylose

Starch granules in a plant cell (TEM).

Glycogen, like starch, is a branched polysaccharide. It is chemically similar to amylopectin, being composed of a-glucose molecules, but there are more a-1,6 glycosidic links mixed with a-1,4 links. This makes it more highly branched and more water-soluble than starch. Glycogen is a storage compound in animal tissues and is found mainly in liver and muscle cells (photo, right). It is readily hydrolyzed by enzymes to form glucose making it an ideal energy storage molecule for active animals.

J. Miquel, D. Vilavella, Z.widerski, V. V. Shimalov and J. Torres

Glycogen

Glycogen

Chitin

Chitin is a tough modified polysaccharide made up of chains of N-acetylglucosamine, a derivative of glucose. It is chemically similar to cellulose but each glucose has an amine group (–NH2–) attached. The addition of the amine groups allows for stronger hydrogen bonding to occur than in cellulose. This makes chitin very strong. After cellulose, chitin is the second most abundant carbohydrate in nature. It is found in the cell walls of fungi and is the main component of the exoskeleton of insects (right) and other arthropods. Š2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

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Glycogen (G) in the spermatozoa of a flatworm. M1, M2=mitochondria, N=nucleus.

A single N-acetylglucosamine molecule

Chitinous insect exoskeleton.

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1. (a) What is a polysaccharide?

(b) What do all polysaccharides have in common?

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2. Suggest why polysaccharides are such a good source of energy?

3. Contrast the properties of the polysaccharides starch, cellulose, and glycogen and relate these to their roles in the cell. Use the symbolic forms in the diagram right to help construct your explanation. Symbolic form of cellulose 4

1

Symbolic form of amylopectin 1

Symbolic form of cellulose 4 6

4

1

6

Symbolic form form of of amylopectin glycogen Symbolic 1

4

6

6

Symbolic form of glycogen Symbolic form of chitin

NHCOCH 3 3

4. Amylopectin is very similar in structure to glycogen but is less soluble. Explain why:o

NHCOCH 3 3

6

o

o

6

NHCOCH 3 3

6 6

5. (a) The symbolic structure of chitin is given below right. Which polysaccharide is it most similar to structurally? Symbolic form of chitin

(b) Describe the major structural difference between chitin and the polysaccharide you named in (a):

o

NHCOCH 3 3

NHCOCH 3 3

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6

6

o

o

NHCOCH 3 3

6 6

(c) Explain what functional advantage this structural difference provides to chitin:

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40 Cellulose and Starch Key Idea: Starch and cellulose are important polysaccharides in plants. Starch is a storage carbohydrate made up of two α-glucose polymers, amylose and amylopectin. Cellulose is a b-glucose polymer which forms the plant cell wall.

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Glucose monomers can be linked in condensation reactions to form large structural and energy storage polysaccharides. The glucose isomer involved and the type of glycosidic linkage determines the properties of the molecule.

Plant cell

Starch is manufactured and stored in amyloplasts (left), non-pigmented storage organelles within plant cells. Starch consists of two types of molecules: the linear and helical amylose and the branched amylopectin.

Plant cells are surrounded by a cell wall made from cellulose microfibrils. They provide the cell with strength and rigidity.

Amylopectin makes up 70-75% of starch

Amylose makes up 25-30% of starch

The microfibrils (below) consist of between 40-70 cellulose chains joined by hydrogen bonds.

a-glucose monomer

Cellulose

O

O

Cellulose is an unbranched polymer of β-glucose molecules bonded by extremely stable β-1, 4 glycosidic bonds. The unbranched structure of cellulose produces parallel chains which become cross linked with hydrogen bonds to form strong microfibrils.

O

O

β-glucose monomer

O

O

O

a-1, 6 linkage creates branching

O

O

O

O

O

O

O

O

O

O

O

O

O

a-1, 4 glycosidic bond

β-1, 4 glycosidic bond

Amylose is made from many thousands of α-glucose monomers. It is a linear molecule, which forms a helix as a result of the angle of the α-1, 4 glycosidic bonds. Every turn of the amylose helix requires six α-glucose molecules. Amylose forms 25-30% of the structure of starch.

Amylopectin consists of the same -1, 4 linked glucose monomers as amylose with occasional -1,6 glycosidic bonds which provide branching points around every 2430 glucose residues. This branching allows many millions of glucose molecules to be stored in a compact form.

1. (a) Where is starch stored in plants?

(b) Where is cellulose found in plants?

2. Compare and contrast the structure of amylose and amylopectin:

3. Account for the differences in structure between cellulose and starch:

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41 Cell Sizes a variety of cell types, including a multicellular microscopic animal for comparison. For each of these images, note the scale and relate this to the type of microscopy used.

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

Parenchyma cell of flowering plant

Unit of length (international system)

Human white blood cell

Eukaryotic cells

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

Meters

Equivalent

1 meter (m)

1m

= 1000 millimeters

1 millimeter (mm)

10

-3

m

= 1000 micrometers

10

-6

m

= 1000 nanometers

10

-9

m

= 1000 picometers

1 micrometer (µm) 1 nanometer (nm)

Prokaryotic cells

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

Unit

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

Nerve cell

50 µm

C: Paramecium is a protozoan commonly found in ponds.

CDC

3 µm

5 µm

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

A: A human red blood cell (erythrocyte). They transport oxygen around the body.

n

10 µm

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

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

100 µm

ney Too

C C 3. 0

CDC

RCN

c

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

50 µm

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

µm

mm (d) Salmonella:

µm

mm

(b) Giardia:

µm

mm

(e) Nucleus:

µm

mm

(c) Paramecium:

µm

mm

(f) Chloroplast:

µm

mm

2. Mark and label the pictured examples above (A-E) on the scale below according to their size:

0.1 nm

1 nm WEB

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10 nm

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100 nm

1 mm

10 mm

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

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

Cell size and functional efficiency

Solving the size problem

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

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

Oxygen

Carbon dioxide

r

Nitrogenous wastes

Ellipsoid V = 2 cm3 SA= 8.8 cm2

Sphere V = 2 cm3 SA= 7.65 cm2

Disc shaped ellipsoid V = 2 cm3 SA= 14.98 cm2

Food

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

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

White blood cell

Skeletal muscle cells

Red blood cell

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

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

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

(a) 2 µm:

(c) 10 µm:

(b) 5 µm:

(d) 30 µm:

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

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

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

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

(b)

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43 Investigating the Effect of 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 cells with a large surface area relative to their volume. When an object (e.g. a cell) is small it has a large surface area in comparison to its volume. Diffusion is an effective way to transport materials (e.g. gases) into and out of the cell. As an object becomes larger, its surface area compared to its

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volume is smaller and diffusion is no longer an effective way to transport materials to and from the inside. In this activity you will design an experiment to demonstrate the effect of surface area: volume ratios on diffusion in model cells. Think about how you will plan your investigation and analyze your data to obtain meaningful results. This will help you to make valid conclusions about your findings.

Background information

Oxygen, water, cellular waste, and many nutrients are transported into and out of cells by diffusion. However, at a certain surface area to volume ratio, diffusion becomes inefficient. In this activity you will create model cells of varying sizes from agar and use them to test the relationship between cell size and rate or efficiency of diffusion.

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

Region of no color change

ff Phenolphthalein is an acid/base indicator and turns a pink in the presence of a base.

Region of color change

ff As the NaOH diffuses into the agar, the phenolphthalein changes to a pink color and thus indicates how far into the agar block the NaOH has diffused (right). ff By cutting an agar block into cubes of various sizes, it is possible to investigate the effect of cell size on diffusion.

A phenolphthalein-infused agar cube after exposure to NaOH.

Equipment list

Timer

Paper towel

Glass beaker

Scalpel

Laboratory tongs

Ruler

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Agar blocks infused with phenolphthalein

Sodium hydroxide (NaOH) solution

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Based on the aim of this experiment "To investigate the effect of cell size on diffusion in a model cell", the background information provided, and the equipment list provided, design your own experiment using the questions below to guide you.

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1. State an hypothesis for your experiment:

2. Write your method as step by step instructions:

4. Write your conclusions here:

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3. In the space below draw a table to record your results. Remember to record cell volume and include space to work out how much diffusion has occurred.

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44 KEY TERMS AND IDEAS: Did You Get It?

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

carbohydrates condensation denaturation

A The loss of a protein's three dimensional functional structure is called this.

B The splitting of a molecule into smaller components by addition of a water molecule.

C Organic compounds, usually linear polymers, made of amino acids linked together by peptide bonds. D The building blocks of proteins. E

Molecules with carbon-hydrogen bonds. They make up most of the chemicals found in living organisms.

F

Organic molecules consisting only of carbon, hydrogen and oxygen that serve as structural components in cells and as energy sources.

hydrolysis lipids

monomer

G A large molecule made up of linked repeating monomer subunits.

organic molecules

H A class of organic compounds with an oily, greasy, or waxy consistency. Important as energy storage molecules and as components of cellular membranes.

polymer

I

A general term for a reaction in which water is released.

J

Simple molecule that can combine with others of the same kind to form a polymer.

proteins

2. The structure on the right represents a phospholipid bilayer.

(a) What does label A represent?

(b) What does label B represent?

A

B

(c) Explain how the properties of the phospholipid molecule result in the bilayer structure of membranes:

3. The organic molecule on the right is hemoglobin.

(b) Which order of structure does the molecule on the right represent?

(c) What factors could cause this molecule to lose its shape?

(d) What would a loss of shape do to the functionality of this molecule?

Zephyris cc 3.0

(a) What class of organic molecules does it belong to? Explain how you decided this?

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4. (a) What general reaction combines two molecules to form a larger molecule? (b) What general reaction cleaves a larger molecule by the addition of water?

(c) Describe what happens to water in each of the reactions described above:

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Cell Structure and Processes

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Enduring Understanding

2.B 4.A

Key terms

SKILL Basic skills when studying cells and tissues

active transport

Practical skills and knowledge

amphipathic

Light microscopy enables us to examine the structure of cells

aquaporin

carrier protein cell wall

channel protein chlorophyll

chloroplast

concentration gradient cotransport diffusion

endocytosis

endoplasmic reticulum eukaryote

Activity number

c

SKILL

Prepare and stain a tissue sample for viewing with an optical microscope. 45 46 47 Use a light microscope to locate specimen material and focus images.

c

SKILL

Demonstrate an ability to measure and count objects under a microscope.

48

c

SKILL

Calculate the magnification of drawings and the size of cell structures in drawings and micrographs. Produce a scientific drawing from observation.

49 50

2.B.1

Cell membranes are selectively permeable

Essential knowledge

Activity number

(a) Cell membranes separate the internal and external environments of the cell c 1

Describe how the plasma membrane separates the internal environment of the cell from its external environment and explain the importance of this.

59

(b) Selective permeability is a consequence of membrane structure

c 1

Describe the structure of cell membranes with reference to phospholipids, embedded proteins, cholesterol, glycoproteins, and glycolipids.

59 60

facilitated diffusion

c 2

Describe the amphipathic nature of phospholipids and the consequences of this.

59 60

Golgi

c 3

Describe nature of embedded (integral) membrane proteins including the significance of hydrophilic and hydrophobic regions in the protein.

59 60

c 4

Describe how molecules of different size and charge pass through the 59 61 62 membrane. Include reference to the transport of small, uncharged polar molecules (e.g. H2O and ethanol) and small non-polar molecules (such as gases), as well as large polar molecules (e.g. glucose) and ions. Describe the role of aquaporins in increasing the rate at which water crosses the membrane.

c

SKILL

Design an experiment to investigate factors affecting permeability of the membrane and its ability to act as a selective barrier to molecules.

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c

PR-4

Use a model of a living cell to investigate the diffusion of substances and the movement of water across cellular membranes [Procedure 2].

65

c

PR-4

Design and conduct an investigation to measure the water potential of plant cells [Procedure 3].

exocytosis

integral protein ion pump isotonic

lysosome

mitochondrion

optical microscope osmosis

passive transport

plasma membrane plasmolysis prokaryote

68 69

ribosome

(c) Cell walls provide a structural boundary and a permeability barrier for some substances

roughER

c 1

Describe the structure and location of the plant cell wall. Explain the role of the plant cell wall in providing structural support and limiting the volume of the cell.

c 2

Describe the structure and roles of cell walls in prokaryotes and fungi. Explain how the structure of each differs from the structure of plant cell walls.

57

Growth and dynamic homeostasis are maintained by the constant movement of molecules across membranes

Activity number

smoothER stain

turgor

vacuole

water potential

2.B.2

Essential knowledge

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selectively permeable

(a) Passive transport does not require input of energy

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c 1

Define passive transport and explain its role in the import of resources (e.g. nutrients) and export of wastes (e.g. carbon dioxide).

c 2

Using examples, explain the role of membrane proteins in facilitated diffusion of 59 61 64 charged and polar molecules through a membrane.

c 3

Explain the terms hypotonic, hypertonic, and isotonic with reference to the external environment of cells.

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(b) Active transport requires free energy to move molecules against their concentration gradient Define active transport. Explain the role integral membrane proteins in moving molecules and/or ions across the membrane in establishing and maintaining concentration gradients.

70

c 2

Understand that active transport is dependent on membrane proteins.

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c 1

(c) Endocytosis and exocytosis move large molecules from the external environment to the internal environment and vice versa respectively.

c 1

Explain the process of exocytosis and describe its role.

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c 2

Explain the process of endocytosis and describe its role.

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2.B.3

Eukaryotic cells maintain internal membranes

Activity number

Essential knowledge

(a) Internal membranes facilitate cellular processes

Kristian Peters

c 1

58 61

Describe how internal membranes partition the cell into specialized regions and explain how this maximizes the efficiency of cellular processes.

(b) Membranes and membrane-bound organelles localize processes within the cell

c 1

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Using examples, explain how compartmentalization allows specific metabolic processes and enzymatic reactions to be localized.

(c) Archaea and Bacteria lack internal membranes and organelles and have a cell wall Contrast prokaryotic and eukaryotic cells with respect to their internal structure and cell wall.

51 52 54

4.A.2 The structure and function of subcellular components, and their interactions, provide essential cellular processes

Activity number

c 1

Essential knowledge and learning objectives

(a) Ribosomes are small universal structures c 1

Describe the structure of ribosomes and explain how the subunits interact to become the site of protein synthesis.

51 - 55 61

NIH

(b) Endoplasmic reticulum occurs in two forms c 1

Describe the structure and role of rough endoplasmic reticulum.

54 61 73

c 2

Describe the structure and role of smooth endoplasmic reticulum.

52 54 61

(c) The Golgi is a membrane-bound structure consisting of flattened sacs c 1

Describe the structure and roles of the Golgi, including reference to production of lysosomes and synthesis and packaging of small molecules for transport.

52 54 61 73

c 1

Describe the structure of a mitochondrion. Explain how the structure allows compartmentalization within the organelle and relate this to its function.

c 2

Contrast the properties of the outer and inner membranes of the mitochondrion.

c 3

Relate the structure and properties of the cristae to their role in ATP generation.

52 54 61 61

(e) Lysosomes carry out intracellular digestion in a variety of ways

c 1

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Vossman CC 3.0

(d) Mitochondria specialize in energy capture and transformation

Describe the structure and roles of lysosomes, including their roles in intracellular digestion, apoptosis, and recycling of materials.

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(f) Vacuoles have different roles in plant and animal cells

c 1

Describe the structure of vacuoles and compare and contrast their size and roles in plant and animal cells.

52 54 61

(g) Chloroplasts are specialized organelles found in algae and green plants

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c 1

Explain the relationship between structure and function in the chloroplast.

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c 2

Describe the location and role of chlorophyll pigments, including chlorophyll a.

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c 3

Describe how the organization of the chloroplast supports its function.

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

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

(a)

Typical compound light microscope

RCN

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

Stoma in leaf epidermis

(e) (f)

(b)

(g) (h)

(c)

What is magnification?

(d)

Magnification refers to the number of times larger an object appears compared to its actual size. Magnification is calculated as follows:

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

Objective lens power

X

Eyepiece lens power

Knob for the adjustment of the microscope on the arm

(i)

What is resolution?

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

(j)

(k) (l)

RCN

High resolution

Attached light source (not always present)

Drosophila

(m)

Dissecting microscope

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

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Low resolution

Dissecting microscopes are a special type of binocular microscope used for observations at low total magnification (X4 to X50), where a large working distance between the objectives and stage is required. A dissecting microscope has two separate lens systems, one for each eye. Such microscopes produce a 3-D view of the specimen and are sometimes called stereo microscopes for this reason.

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

EII

JDG

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62

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

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

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

2. Determine the magnification of a microscope using:

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

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

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

4. What type of microscope would you use to:

(a) Count stream invertebrates in a sample:

(b) Observe cells in mitosis:

5. (a) Distinguish between magnification and resolution:

(b) Explain the benefits of a higher resolution:

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

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

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

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

Focus and center the specimen using the medium objective lens. Focus firstly with the coarse focus knob, then with the fine focus knob (if needed). Turn on the light source.

Focus and center the specimen using the low objective lens. Focus firstly with the coarse focus knob, then with the fine focus knob.

Focus the eyepieces to adjust your view.

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

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

These photos KP

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

Preparing a specimen Upper epidermis peeled away

Upper epidermis

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

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

Mounting a specimen

Viewing

Mounted needle

Mounting fluid

Coverslip

Specimen

Microscope slide

Mounting: The thin layer is placed in the center of a clean glass microscope slide and covered with a drop of mounting liquid (e.g. water, glycerol, or stain). A coverslip is placed on top using a mounted needle to support and lower it gently over the specimen. This avoids including air in the mount.

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

2. What is the purpose of the coverslip?

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

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

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

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47 Staining a Slide Key Idea: Staining material to be viewed under a microscope can make it easier to distinguish particular cell structures. Stains and dyes can be used to highlight specific components or structures. Most stains are non-viable, and are used on dead specimens, but harmless viable stains can be applied

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to living material. Stains contain chemicals that interact with molecules in the cell. Some stains bind to a particular molecule making it easier to see where those molecules are. Others cause a change in a target molecule, which changes their color, making them more visible.

Stain

Final color

Used for

Iodine solution

blue-black

Starch

Crystal violet

purple

Gram staining

Aniline sulfate

yellow

lignin

Methylene blue

blue

Nuclei

Hematoxylin and eosin (H&E)

H=dark blue/violet E=red/pink

H=Nuclei E=Proteins

Mnolf

Some commonly used stains

Iodine stain

Iodine stains starch-containing organelles, such as potato amyloplasts, blue-black.

Red

Pink

Mnolf

PloS

Lung tissue

CDC: Dr Lucille K. Georg

Blue

H&E stain is one of the most common histological stains. Nuclei stain dark blue, whereas proteins, extracellular material, and red blood cells stain pink or red.

Vital (viable) stains do not immediately harm living cells. Trypan blue is a vital stain that stains dead cells blue but is excluded by live cells. It is also used to study fungal hyphae.

How to apply a simple stain

Methylene blue is a common temporary stain for animal cells, such as these cheek cells. It stains DNA and so makes the nuclei more visible.

1

2

If a specimen is already mounted, a drop of stain can be placed at one end of the coverslip and drawn through using filter paper (below). Water can be drawn through in the same way to remove excess stain.

Irrigation

Specimen

Coverslip

Filter paper

1. What is the main purpose of using a stain?

2. What is the difference between a viable and non-viable stain?

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The light micrographs 1 and 2 (above) illustrate how the use of a stain can enhance certain structures. The left image (1) is unstained and only the cell wall is easily visible. Adding iodine (2) makes the cell wall and nuclei stand out.

3. Identify a stain that would be appropriate for distinguishing each of the following:

(a) Live vs dead cells:

(c) Lignin in a plant root section:

(b) Red blood cells in a tissue preparation:

(d) Nuclei in cheek cells:

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48 Measuring and Counting Using a Microscope Two common pieces of equipment are the graticule and the hemocytometer. A graticule can be used to measure the size of an object whereas a hemocytometer is used to count the number of cells in a set area or volume.

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Key Idea: Graticules make it possible to measure cell size. Hemocytometers are used to count the number of cells. Measuring and counting objects to be viewed under a microscope requires precisely marked measuring equipment.

Measuring cell size

A graticule is a scale placed in the eyepiece of a microscope. It is usually about 1 mm long and divided into 100 equal units. A graticule is used in combination with a stage micrometer to work out the size of an object being viewed. The stage micrometer is a slide with a scale that is exactly 1 mm long and also divided into 100 divisions (so that each division is 0.01 mm) and is placed on the microscope stage. The stage micrometer allows the graticule to be calibrated so that a precise scale can be calculated at each magnification.

Eyepiece

0

10

20

30

40

50

60

70

80

90 100

Graticule

0.01 mm

x400

The scale on the graticule is lined up with the stage micrometer. The number of graticule divisions between the divisions of the stage micrometer can then be read off. In the example right, each division of the stage micrometer is equal to four large divisions of the graticule. Each large division of the graticule is therefore 2.5 x 10-3 mm at 400x magnification.

0

10 20

30

40

50

60

70

80 90 100

View through eyepiece

Stage micrometer

Counting cells

1 mm

Counting grid at 100x

Jeffery M. Vinocur CC 3.0

11 mm mm

1 mm

+

+

Microscopes can be used as a tool to count cells or other small objects (e.g pollen grains). By counting the number of cells in a known area, the total number of cells in a larger area can be calculated. A hemocytometer is commonly used to count cells viewed with a light microscope. It is a simple slide with precisely etched lines forming a grid and was developed for counting blood cells. There are a number of types of hemocytometer, including the Improved Neubauer, shown below. The slide holds a coverslip 0.1 mm above the surface of the grid, allowing volume to be calculated. The central counting grid is divided into 25 large squares, each of which is further divided into 16 squares.

1. A student using the graticule scale shown at the top of this page found a cell to be 56 divisions wide. Calculate the width of the cell in mm and in mm:

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2. A second student grew yeast cells in 5 cm3 of nutrient solution. The student used the hemocytometer shown above to count the number of yeast cells each day for 3 days. (a) Calculate the area and volume of the grid shown in blue: Area: Volume: (b) The student counted yeast cells in the central blue grid. Complete the table below based on the counts obtained: Number of cells counted

Cells in 5 cm3

Day 1

Day 2

Day 3

4

9

17

3. A botanist wished to know the number of pollen grains produced per anther by a flower with eight anthers. She cut the anthers and placed them in 3 cm3 of distilled water, shaking the mix vigorously. Using a hemocytometer she counted 6 grains in the large central counting grid (1 x 1 mm). Calculate the total number of pollen grains produced per anther:

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

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

Calculating linear magnification: a worked example

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

1

1.0 mm

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

2

The magnification of the scale line is 10 (10 mm / 1 mm)

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

Microscopy equations

measured size of the object

1. Magnification =

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

3

2. Actual object size =

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

actual size of the object size of image

magnification

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

x 140

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

0.5 mm

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

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

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

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

ff Biological drawings require you to pay attention to detail. It is very important that you draw what you actually see, and not what you think you should see. ff Biological drawings should include as much detail as you need to distinguish different structures and types of tissue, but avoid unnecessary detail which can make your drawing confusing.

ff Attention should be given to the symmetry and proportions of your specimen. Accurate labeling, a statement of magnification or scale, the view (section type), and type of stain used (if applicable) should all be noted on your drawing. ff Some key points for making good biological drawing are described on the example below. The drawing of Drosophila (right) is good but lacks the information required to make it a good biological drawing.

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

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

Daphnia

Antenna

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

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

Eye

Heart

Brood chamber with eggs

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

Abdominal legs

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

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

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Gut

All parts of your drawing must be labeled accurately.

Use simple, narrow lines to make your drawings.

Scale 1 mm

Labeling lines should be drawn with a ruler and should not cross over other label lines. Try to use only vertical or horizontal lines although this is not always possible..

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Annotated diagrams Root tranverse section from Ranunculus

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

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Root hairs: increase surface area for water absorption

Epidermal cell: outer layer of cells

Parenchyma cells: make up the bulk of the soft parts of plants.

Xylem: conducts water from the roots to the stem and leaves.

Scale

Phloem: tramsports soluble compunds throughout the plant.

0.05 mm

Plan diagrams

Light micrograph of a transverse section through a sunflower stem

Plan diagrams are drawings made of samples viewed under a microscope at low or medium power. They are used to show the distribution of the different tissue types in a sample without any cellular detail. The tissues are identified, but no detail about the cells within them is included.

EII

The example here shows a plan diagram produced after viewing a light micrograph of a transverse section through a dicot stem.

Epidermis Phloem

Vascular bundle

Vascular cambium Xylem

Sclerenchyma (fiber cap)

Pith (parenchyma cells)

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Cortex of collenchyma cells

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51 Prokaryotic Cells Key Idea: Prokaryotic cells lack many of the features of eukaryotic cells, including membrane-bound organelles. Bacterial (prokaryotic) cells are much smaller than eukaryotic cells and lack many eukaryotic features, such as a distinct

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nucleus and membrane-bound cellular organelles. The cell wall is an important feature. It is a complex, multi-layered structure and has a role in the organism's ability to cause disease. A generalized prokaryote, E. coli, is shown below.

E. coli structure

The cell wall lies outside the plasma membrane. It gives the cell shape, prevents rupture, and serves as an anchorage point for flagella. It is composed of a carbohydrate macromolecule called peptidoglycan, with variable amounts of lipopolysaccharide and lipoprotein.

Plasma membrane is similar in composition to eukaryotic membranes, although less rigid.

Cytoplasmic inclusions include aggregations of storage compounds, e.g. glycogen, fatty acids, sulfur, or phosphorus.

70S ribosomes are free in the cytoplasm.

Fimbriae are hairlike structures. They are shorter, straighter, and thinner than flagella and used for attachment, not movement.

Cytoplasm

Prokaryotes may have small accessory chromosomes called plasmids. These often carry genes for antibiotic resistance and may be exchanged with other bacterial cells.

Nucleoid region (pale)

The circular chromosome occurs within a region called the nucleoid. It is not enclosed in a membrane. The DNA is 'naked' meaning it is not associated with proteins.

Flagella (sing. flagellum) are used for locomotion. They are anchored in the plasma membrane. There may be one or more flagella.

Some bacteria, including E. coli, have a polysaccharide capsule outside the cell wall. The capsule contributes to its ability to cause disease.

Fimbriae

Dividing cells

A spiral shape is one of four bacterial shapes (the others being rods, commas, and spheres). These Campylobacter cells also have flagella.

Escherichia coli is a rod-shaped bacterium, common in the human gut. The fimbriae surrounding the cell are used to adhere to the intestinal wall.

Bacteria usually divide by binary fission. During this process, DNA is copied and the cell splits into two cells, as in these round (cocci) cells.

NASA

Halobacterium

CDC

Flagellum

Prokaryotes include the Archaea and Bacteria. The Archaea were first grouped with Bacteria but are now classified in their own domain on the basis of their unique characteristics.

1. Describe three features distinguishing prokaryotic cells from eukaryotic cells: (a)

(b)

2. (a) Describe the function of flagella in bacteria:

(b) Explain how fimbriae differ structurally and functionally from flagella:

3. Describe the location and general composition of the bacterial cell wall:

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(c)

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

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

Lysosome: A sac bounded by a single membrane. They are pinched off from the Golgi apparatus and contain and transport enzymes that break down food and foreign matter. Lysosomes show little internal structure but often contain fragments of material being broken down. Specialized lysosomes are generally absent from plant cells.

Generalized animal cell

Smooth endoplasmic reticulum: ER without ribosomes. It is a site for lipid and carbohydrate metabolism, including hormone synthesis.

Nuclear pore: A hole in the nuclear membrane allowing the nucleus to communicate with the rest of the cell.

Nucleolus: A dense, solid structure composed of crystalline protein and nucleic acid. They are involved in ribosome synthesis.

Tight junctions: Join cells together in the formation of tissues. Nuclear membrane: Double layered

Ribosomes: These small structures may be free in the cytoplasm or associated with the endoplasmic reticulum (ER). Ribosomes in animal cells are 80S ribosomes

Nucleus

Cytoplasm

Plasma (cell surface) membrane

Rough endoplasmic reticulum: A site of protein synthesis.The rough ER also synthesizes new membranes, growing in place by adding proteins and phospholipids.

Centrioles: Structures within a centrosome associated with nuclear division. They are composed of microtubules, but appear as small, featureless particles, 0.25 µm diameter, under a light microscope. They are absent in higher plant cells and some protists.

Golgi apparatus (20-200 nm): A series of flattened, disc-shaped sacs, stacked one on top of the other and connected with the ER. The Golgi stores, modifies, and packages proteins. It ‘tags’ proteins so that they go to their correct destination.

Mitochondrion (pl. mitochondria): An organelle bounded by a double membrane system. The number in a cell depends on its metabolic activity.

1. What is the difference in ribosomes between prokaryotic and eukaryotic cells?

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

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

A

White blood cells and red blood cells (blood smear)

C

Photos: EII

(b)

Neurons (nerve cells) in the spinal cord

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(c)

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53 Identifying Structures in an Animal Cell When viewing TEMs, the cellular organelles may have quite different appearances depending on whether they are in transverse or longitudinal section.

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

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

(b)

(c)

(d)

(e)

(f)

(g)

(h)

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

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

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

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

(b) It is metabolically very active:

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

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54 Plant Cells cytoplasm, which is itself enclosed by the plasma membrane. Plant cells are enclosed in a cellulose cell wall, which gives them a regular, uniform appearance. The cell wall protects the cell, maintains its shape, and prevents excessive water uptake. It provides rigidity to plant structures but permits the free passage of materials into and out of the cell.

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

Mitochondrion: 1.5 µm X 2–8 µm. They are the cell's energy transformers, converting chemical energy into ATP. Mitochondria contain 70S ribosomes.

Generalized plant cell

Starch granule: Carbohydrate stored in amyloplasts (plastids specialized for storage). Plastids are not found in animal cells. Non-photosynthetic plastids are usually used for storage.

Chloroplast

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

Large central vacuole: usually filled with an aqueous solution of ions. Vacuoles are prominent in plants and function in storage, waste disposal, and regulation of cell turgor. The plant cell vacuole also helps increase cell size during growth.

Chloroplast: Specialized plastids, 2 µm x 5 µm, containing the green pigment chlorophyll. They contain dense stacks of membranes (grana) within a colorless fluid which is much like cytosol. They are the sites for photosynthesis and occur mainly in leaves. Chloroplasts contain 70S ribosomes.

Nuclear pore: 100 nm diameter Nuclear membrane: a double layered structure.

Nucleus: A conspicuous organelle 5 µm diameter.

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

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

Endoplasmic reticulum (ER): Comprises a network of tubes and flattened sacs. ER is continuous with the nuclear membrane and may be smooth or have attached ribosomes (rough ER). While it is extensive, its membranous structure makes it difficult to resolve with a light microscope.

Nucleolus

Ribosomes: 80S ribosomes. These small (20 nm) structures manufacture proteins. They may be free in the cytoplasm or associated with the surface of the endoplasmic reticulum. Golgi apparatus

P

Alison Roberts

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

Cytoplasm: A watery solution containing dissolved substances, enzymes, and the cell organelles and structures. The site of translation in the cell.

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

(b) What are its roles?

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

3. (a) Distinguish between the cytoplasmic ribosomes and the ribosomes found in chloroplasts and mitochondria:

(b) What might this suggest about the origins of these organelles?

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55 Identifying Structures in a Plant Cell Key Idea: The position and appearance of the organelles in an electron micrograph can be used to identify them.

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1. Study the diagrams on the other pages in this chapter to familiarise yourself with the structures found in eukaryotic cells. Identify the 11 structures in the cell below using the following word list: cytoplasm, smooth endoplasmic reticulum, mitochondrion, starch granule, chromosome, nucleus, vacuole, plasma membrane, cell wall, chloroplast, nuclear membrane

(a)

(b)

(c)

(d) (e) (f)

(g) (h) (i)

(j)

(k)

BF

TEM

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

(b) Describe what cytoplasm is made up of:

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

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

(a)

(b)

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56 The Cell's Cytoskeleton

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Plasma membrane

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Key Idea: The cytoskeleton is a complex structure of tubules and fibers. It resists tension and so provides structural support to maintain the cell's shape. The cell's cytoplasm is not a fluid filled space; it contains a complex network of fibers called the cytoskeleton. The cytoskeleton is made up of three proteinaceous elements: microfilaments, intermediate filaments, and microtubules. Each has a distinct size, structure, and protein composition, and a specific role in cytoskeletal function. Cilia and flagella are made up of microtubules and for this reason they are considered to be part of the cytoskeleton. The elements of the cytoskeleton are dynamic; they move and change to alter the cell's shape, move materials within the cell, and move the cell itself. This movement is achieved through the action of motor proteins, which transport material by 'walking' along cytoskeletal 'tracks', hydrolyzing ATP at each step. It is an active transport process.

Rough ER

Microtubule

Nucleus

Intermediate filament Microfilament

Intermediate filaments

Microfilaments

7 nm

Microtubules

8-12 nm

25 nm

Actin subunit

Microfilaments

Intermediate filaments

Microtubules

Protein subunits

Actin

Fibrous proteins, e.g. keratin

a and b tubulin dimers

Structure

Two intertwined strands

Fibers wound into thicker cables

Hollow tubes

Functions

• Maintain cell shape • Motility (pseudopodia) • Contraction (muscle) • Cytokinesis of cell division

• Maintain cell shape • Anchor nucleus and organelles

• Maintain cell shape • Motility (cilia and flagella) • Move chromosomes (spindle) • Move organelles

Central pair

Intermediate filaments can be composed of a number of different fibrous proteins and are defined by their size rather than composition. The protein subunits are wound into cables around 10 nm in diameter. Intermediate filaments form a dense network within and projecting from the nucleus, helping to anchor it in place.

1. Describe what all components of the cytoskeleton have in common:

2. Explain the importance of the cytoskeleton being a dynamic structure:

Microtubule doublet

Microtubules are the largest cytoskeletal components and grow or shrink in length as tubulin subunits are added or subtracted from one end. The are involved in movement of material within the cell and in moving the cell itself. This EM shows a cilia from Chlamydomonas, with the 9+2 arrangement of microtubular doublets.

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Microfilaments are long polymers of the protein actin. Microfilaments can grow and shrink as actin subunits are added or taken away from either end. Networks of microfilaments form a matrix that helps to define the cell's shape. Actin microfilaments are also involved in cell division (during cytokinesis) and in muscle contraction.

NIH

Y tambe

Actin microfilaments in mouse embryo cells

Dartmouth Electron Micrscopy Facility

Intermediate filaments surrounding nucleus

3. Explain how the presence of a cytoskeleton could aid in directing the movement of materials within the cell:

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57 Cell Walls Key Idea: Cell walls give structure to cells. They are found in bacteria, fungi, and plants. Cell walls are structural components of cells external to the plasma membrane. They give the cell support and protection, preventing over expansion when the pressure inside a cell

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rises. In bacteria, cell walls also contribute to virulence so they are important in bacterial pathogenicity. Cell walls come in a variety of molecular structures. A major component of bacterial cell walls is peptidoglycan, whereas the major component of plant cell walls is cellulose.

Gram-negative bacteria

Gram-positive bacteria

Lipopolysaccharide

Teichoic acid

Porin

Lipoteichoic acid

Outer membrane

Periplasmic space

Lipoprotein

Peptidoglycan

Peptidoglycan

Periplasmic space

Lipoprotein

Plasma membrane

Plasma membrane

Bacteria can be divided into two major groups, Gram-positive or Gram-negative based on how they appear when treated with certain stains. This is determined by wall structure. Gram negative bacteria have an outer membrane over a thin layer of peptidoglycan. The Archaeans also lack peptidoglycan in their cell wall.

Gram-positive bacteria have a thick layer of peptidoglycan beyond the plasma membrane. Peptidoglycan is made up of polysaccharide chains cross-linked by various peptides. The gram positive bacterial cell wall is a major target in the development of antibiotics.

Fungi

Plant

Mannoproteins

Cellulose

Pectin

Hemicellulose

β-glucan

Chitin

Plant cell walls comprise three major elements: cellulose, hemicellulose, and pectin. Hemicellulose links the cellulose into a matrix, which is embedded with pectin. The cell wall provides the cell with support. Internal pressure (turgor) pushes against the wall and produces a rigid cell that can (together with other cells) support the mass and structure of the plant.

Lipoprotein

Plasma membrane

Fungal cell walls are a unique structure of chitin, b-glucans (b-D-glucose polysaccharides), and mannoproteins (proteins with mannose sugar attached). As in plants, the cell wall offers the fungus support and structure. Antifungal medicines target the chitin in the fungal cell wall because its uniqueness to fungi.

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Plasma membrane

1. What is the main difference between the cell walls of Gram-negative and Gram-positive bacteria?

2. Explain why antibiotics targeting the bacterial cell wall will not affect the people taking them:

3. Describe two functions of cell walls:

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

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Key Idea: Each type of organelle in a cell has a specific role. Not all cell types contain every type of organelle. The diagram below provides spaces for you to summarize 1. (a) Name this organelle:

2. (a) Name this organelle:

(b) Structure and location:

(b) Structure and location:

(c) Function:

(c) Function:

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

3. (a) Name this organelle:

(b) Structure and location:

1

(c) Function:

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

2

5

3

4. (a) Name this organelle:

(b) Structure and location:

(c) Function:

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

4

6

Plant cell

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5. (a) Name this organelle:

6. (a) Name this organelle:

(b) Structure and location:

(b) Structure and location:

(c) Function:

(c) Function:

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

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Plasma membrane

Nucleus Golgi

Chloroplast

Animal cell

Plant cell

Leaf section

Leaf

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Ribosome

DNA

0.1 nm

1 nm

10 nm

Mitochondrion

100 nm

1 mm

SEM TEM

Viewing range

Light microscope

10 mm

7. (a) Name this organelle:

7

(b) Structure and location:

(c) Function:

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

9

8

Animal cell

8. (a) Name this organelle:

9. (a) Name this organelle:

(b) Structure and location:

(b) Structure and location:

(c) Function:

(c) Function:

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

(d) This organelle is found (circle the correct answer): only in plant cells / only in animal cells / in both plant and animal cells

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10. Use the scale at the top of the page and the information on previous activities to identify which of the organelles (1-9) can be seen through a light microscope:

11. Identify which of the organelles (1-9) require a TEM (transmission electron microscope) to be seen:

12. Identify one other structure in the plant cell not labeled opposite and describe its function:

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59 The Structure of Membranes

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cell's contents and regulates many of the cell's activities. Importantly, it controls what enters and leaves the cell by the use of carrier and channel proteins.

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Key Idea: A cellular membrane is made of a phospholipid bilayer with proteins of different sorts embedded in it. The cell surface (or plasma) membrane encloses the

Simple membrane structure

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.

CO2

Phospholipids naturally form a bilayer.

Phosphate head is hydrophilic

Fatty acid tail is hydrophobic

H2O

Glucose

Na+

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

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

Water molecules pass between the phospholipid molecules by osmosis.

What can cross a lipid bilayer?

Gases

Hydrophobic molecules

Large polar molecules

Glucose

O2

Charged molecules

Na+

H2O

Benzene

CO2

Small polar molecules

Cl-

Ca2+

Small uncharged molecules can diffuse easily through the membrane.

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

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Polar molecules are small enough to diffuse through. Aquaporins increase rate of water movement.

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Ethanol

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

Ions can be transported across the membrane by ion channels (passive) or ion pumps (active).

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What do proteins in the cell surface membrane really look like?

Vossman cc 3.0

A2-33 cc 3.0

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The structure of membrane proteins enables them to perform their particular function in transport, cell signaling, or cell recognition. The proteins are integral to the membrane, and often have parts of their structure projecting from both internal and external sides of the membrane.

Aquaporins are a special type of channel protein that speed up the passage of water molecules across the membrane. Their tertiary structure creates a pore through the center of the protein through which molecules can pass.

The GLUT1 glucose transporter is a carrier protein that facilitates the transport of glucose across the cell surface membranes of mammalian cells. It enables glucose to be transported into cells at a rate high to supply the cell's energy needs (50,000x >0).

The receptor (darker) is bound to intracellular G protein (lighter).

Extracellular

Intracellular

G-protein coupled receptors are proteins involved in signaling pathways. A signal molecule binds to the receptor protein outside the cell to trigger a reaction involving intracellular G protein. The receptor in this example binds to adrenaline.

Cell surface antigens provide an identifiable cell signature so that the body can distinguish between its own cells and foreign molecules. They are often glycoproteins. The image above shows how the antigens project from the membrane.

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

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

3. Identify the molecule(s) that:

(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:

(a) Aquaporins:

(b) GLUT1 protein:

(c) G protein:

(d) Cell surface antigens:

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4. Describe the role of the following proteins in the plasma membrane:

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60 How Do We Know? Membrane Structure

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Key Idea: The freeze-fracture technique for preparing and viewing cellular membranes has provided evidence to support the fluid mosaic model of the plasma membrane.

Cellular membranes play many extremely important roles in cells and understanding their structure is central to understanding cellular function. Moreover, understanding the structure and function of membrane proteins is essential to understanding cellular transport processes, and cell recognition and signaling. Cellular membranes are far too small to be seen clearly using light microscopy, and certainly any detail is impossible to resolve. Since early last century, scientists have known that membranes were composed of a lipid bilayer with associated proteins. The original model of membrane structure, proposed by Davson and Danielli, was the unit membrane (a lipid bilayer coated with protein). This model was later modified by Singer and Nicolson after the discovery that the protein molecules were embedded within the bilayer rather than coating the outside. But how did they find out just how these Cleaving the membrane molecules were organized? The answers were provided with electron microscopy, and one technique in particular – freeze fracture. As the name implies, freeze fracture, at its very simplest level, is the freezing of a cell and then fracturing it so the inner surface of the membrane can be seen using electron microscopy. Membranes are composed of two layers of phospholipids held together by weak intermolecular bonds. These split apart during fracture. The procedure involves several steps:

Cells are immersed in chemicals that alter the strength of the internal and external regions of the plasma membrane and immobilize any mobile macromolecules.

The cells are passed through a series of glycerol solutions of increasing concentration. This protects the cells from bursting when they are frozen. The cells are mounted on gold supports and frozen using liquid propane.

Razor blade

-150oC. A

The cells are fractured in a helium-vented vacuum at razor blade cooled to -170o C acts as both a cold trap for water and the fracturing instrument.

Proteins leave bumps and holes in the membrane when it is cleaved

The surface of the fractured cells may be evaporated a little to produce some relief on the surface (known as etching) so that a three-dimensional effect occurs. For viewing under an electron microscope (EM), a replica of the cells is made by coating them with gold or platinum to ~3 nm thick. A layer of carbon around 30 nm thick is used to provide contrast and stability for the replica. The samples are then raised to room temperature and placed into distilled water or digestive enzymes, which separates the replica from the sample. The replica is then rinsed in distilled water before it is ready for viewing.

The freeze fracture technique provided the necessary supporting evidence for the current fluid mosaic model of membrane structure. When cleaved, proteins in the membrane left impressions that showed they were embedded into the membrane and not a continuous layer on the outside as earlier models proposed.

50 nm

Photo: Louisa Howard and Chuck Daghlian, Dartmouth College

1. Explain how freeze-fracture studies provided evidence for our current model of membrane structure:

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2. The Davson and Danielli model of membrane structure was the unit membrane; a phospholipid bilayer with a protein coat. Explain how the freeze-fracture studies showed this model to be flawed:

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

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Key Idea: Cells perform the processes essential to life. Each cellular organelle carries out one or more processes that contribute to the functioning of the cell as a whole. The cell can be compared to a factory with an assembly line. Organelles in the cell provide the equivalent of the power supply, assembly line, packaging department, repair and maintenance, transport system, and the control center. The sum total of all the processes occurring in a cell is known

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, vacuoles 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.

Cytosis

Secretion

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, vacuoles. Solids or fluids can be engulfed to bring them into the cell (endocytosis) or the plasma membrane can fuse with the Golgi secretory vesicles to expel material from the cell (exocytosis). In animal cells, vacuoles may be involved as part of cytosis.

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).

Plant cell

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Chloroplast

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Membranes allow compartmentalization of reactions and processes

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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Membranes play an important role in separating regions within the cell (and within organelles) where particular reactions occur. Specific enzymes are therefore often located in particular organelles. The reaction rate is controlled by controlling the rate at which substrates enter the organelle and therefore the availability of the raw materials required for the reactions.

The Golgi apparatus is a specialized membrane-bound organelle which compartmentalizes 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. Identify two examples of intracellular membranes and describe their functions: (a)

(b)

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3. Explain how compartmentalization within the cell is achieved and how it contributes to functional efficiency:

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

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

What is diffusion?

Factors affecting the rate of diffusion

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

High concentration

Low concentration

Concentration gradient

Concentration gradient

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

The distance moved

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

The surface area involved

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

Barriers to diffusion

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

Temperature

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

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

Glucose

Lipid soluble solutes

Inorganic ion

Channel protein

Carrier protein

Carrier-mediated facilitated diffusion

Channel-mediated facilitated diffusion

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

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

Channels (hydrophilic pores) in the membrane allow inorganic ions to pass through the membrane. Example: K+ ions exiting nerve cells to restore resting potential.

1. What is diffusion?

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Simple diffusion

2. What do the three types of diffusion described above all have in common?

3. How does facilitated diffusion differ from simple diffusion?

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

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

The aim and hypothesis

To investigate the effect of ethanol concentration on membrane permeability. The students hypothesized that the amount of pigment leaking from the beetroot cubes would increase with increasing ethanol concentration.

Beetroot cubes

Method for determining effect of ethanol concentration on membrane permeability

Background

Raw beetroot was cut into uniform cubes using a cork borer with a 4 mm internal diameter. The cubes were trimmed to 20 mm lengths and placed in a beaker of distilled water for 30 minutes. The following ethanol concentrations were prepared using serial dilution: 0, 6.25, 12.5, 25, 50, and 100%.

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 color. 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.

Eighteen clean test tubes were divided into six groups of three and labeled with one of the six ethanol concentrations. Three cm3 of the appropriate ethanol solution was placed into each test tube. A beetroot cube (dried by blotting) was added to each test tube. The test tubes were covered with parafilm (plastic paraffin film with a paper backing) and left at room temperature. After one hour the beetroot cubes were removed and the absorbance measured at 477 nm. Results are tabulated, below.

Absorbance of beetroot samples at varying ethanol concentrations

Ethanol concentration (%)

Sample 1

Sample 2

Sample 3

0

0.014

0.038

0.038

6.25

0.009

0.015

0.023

12.5

0.010

0.041

0.018

25

0.067

0.064

0.116

50

0.945

1.100

0.731

100

1.269

1.376

0.907

Absorbance at 477 nm

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. Complete the table above by calculating the mean absorbance for each ethanol concentration:

3. What is absorbance measuring and why is it increasing with increasing ethanol concentration?

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4. What was the purpose of the 0% ethanol solution in the experiment described opposite?

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

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

6. (a) Plot a line graph of ethanol concentration against mean absorbance on the grid (right):

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

7. Some students wanted to find out how temperature affected membrane permeability. They prepared the beetroot cubes the same way as in the previous experiment.

(a) Write a hypothesis for their experiment:

(b) List an appropriate range of temperatures for the experiment:

(c) Write the method for the experiment:

(d) Make a prediction about the results based on your understanding of cellular processes:

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64 Investigating Transport Across Membranes readily diffuse across the membrane, the molecules will enter the RBCs by diffusing down their concentration gradient. This will draw water into the RBCs (by osmosis) and they will burst (hemolyze). When the RBCs burst, the cellular material settles out of suspension and the solution becomes clear. By using a spectrophotometer to measure the rate at which the solution becomes clear, it is possible to determine the rate at which the molecules are crossing the plasma membrane.

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Key Idea: The rate of diffusion of molecules through the plasma membrane can be determined by measuring the change in light absorbance as a solution of red blood cells hemolyzes. How a cell behaves when suspended in a solution depends on whether or not the molecules or ions in the solution can cross the plasma membrane. If red blood cells (RBCs) are suspended in a concentrated solution of molecules that can

The aim

0.3 mol L-1 solution

To investigate how the size and membrane solubility of molecules affects the rate of diffusion across the plasma membrane.

The method

Diffusing molecule

ff 0.3 mol L-1 solutions of glucose, sucrose, urea, and glycerol were prepared (this concentration is greater than the cell internal concentration). A blank solution of distilled water was also prepared. Molecular weights (MW) are as follows: glucose (MW 180), sucrose (MW 342), urea (MW 60), and glycerol (MW 92).

Water molecule

ff Both urea and glycerol readily diffuse across the plasma membrane. Glucose is transported across the membrane by a carrier protein.

ff 3 mL of each solution was mixed with 0.1 mL of a sheep RBC suspension and added to cuvettes. The cuvettes were placed into a spectrophotometer and absorbance measured over 15 minutes. The results are plotted below:

Hypertonic solution

Absorbance vs time for sheep RBCs

0.7 0.6

Membrane-penetrating molecule crosses membrane, increasing the cell's internal solute concentration and drawing water in by osmosis.

Absorbance

0.5 0.4

Glycerol Glucose Sucrose Urea

0.3

Cell bursting

0.2 0.1

Linda Scott et al 1993 Dept. Bio. Hartwick College, NY.

0

0

5 10 20 30 60 120

240

320

480

600

720

Increased cellular volume causes the cell to burst (hemolysis). The cellular material settles out.

840

Time (seconds)

1. (a) Which molecule crosses the membrane the fastest?

(b) Which molecule appears to be unable to cross the plasma membrane?

(c) List the molecules in order of their ability to cross the plasma membrane (fastest to slowest):

2. (a) What is the largest molecule used in the experiment?

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(b) What is the smallest molecule used in the experiment?

(c) How does size affect the rate at which molecules can cross the plasma membrane?

3. Why don't the RBCs in the glucose solution hemolyze even though glucose is transported across the membrane?

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65 Investigating Diffusion tubing determine the size of the molecules that can pass through. The experiment described below demonstrates how glucose will diffuse down its concentration gradient from a high glucose concentration to a low glucose concentration.

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Key Idea: Dialysis tubing can be used to model the diffusion of glucose down its concentration gradient. Diffusion through a partially permeable membrane can be modeled using dialysis tubing. The pores of the dialysis

The aim

To demonstrate diffusion through a selectively permeable membrane.

Hypothesis

If there is no glucose outside the dialysis tubing, then glucose will diffuse down its concentration gradient from the dialysis tubing into the distilled water until the glucose concentrations are equal.

Dialysis tubing (partially-permeable membrane)

Distilled water

Solution containing starch and glucose

Background

Dialysis tubing acts as a partially (or selectively) permeable membrane. It comes in many pore sizes and only allows molecules smaller than the size of the pore to pass through.

Lugol's indicator contains iodine, and turns blue/ black in the presence of starch. The presence of glucose can be tested using a glucose dipstick test. If glucose is present, the indicator window will change color. The color change can be compared against a reference to determine the concentration of glucose present.

Method

Dialysis tubing was filled with 5 cm3 each of a 1% starch solution and a 10% glucose solution. A 1 cm3 sample was removed and tested for the presence of starch using Lugol's indicator, and glucose using a glucose dipstick.

The dialysis tubing was tied, and the outside of the tubing washed with distilled water to remove any starch or glucose that spilled on to the outer surface during filling. The tubing was placed into a beaker of distilled water.

After 30 minutes, the solution inside the dialysis tubing and the distilled water were tested for the presence of starch and glucose.

1. Why was it important to wash the dialysis tubing before placing it into the beaker of distilled water?

2. What part of a cell does the dialysis tubing represent?

Dialysis tubing start

Beaker start

Dialysis tubing end

Beaker end

Starch

++

-

++

­-

Glucose

++

-

+

+

S Starch

G Glucose

3. The results for the experiment are tabulated right.

(a) In the spaces provided (below, right) draw the distribution of starch and glucose at the start and at the end of the experiment. Use the symbols shown under the table to represent starch and glucose: (b) Describe why glucose has moved across the partially permeable membrane during the experiment:

Dialysis tubing start

Dialysis tubing end

(c) Why was there no starch present in the beaker at the end of the experiment?

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Beaker start

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Beaker end

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66 Osmosis as the plasma membrane, allows some molecules, but not others, to pass through. Water molecules can diffuse directly through the lipid bilayer, but movement is aided by specific protein channels called aquaporins. There is a net movement of water molecules until an equilibrium is reached and net movement is then zero. Osmosis is a passive process and does not require any energy input.

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Key Idea: Osmosis is the term describing the diffusion of water molecules down their concentration gradient across a partially permeable membrane. The diffusion of water down its concentration gradient across a partially permeable membrane is called osmosis and it is the principal mechanism by which water moves in and out of living cells. A partially permeable membrane, such

Demonstrating osmosis

Osmotic potential

Osmosis can be demonstrated using dialysis tubing in a simple experiment (described below). Dialysis tubing, like all cellular membranes, is a partially permeable membrane.

Osmotic potential is a term often used when studying animal cells. The presence of solutes (dissolved substances) in a solution increases the tendency of water to move into that solution. This tendency is called the osmotic potential or osmotic pressure. The greater a solution's concentration (i.e. the more total dissolved solutes it contains) the greater the osmotic potential.

A sucrose solution (high solute concentration) is placed into dialysis tubing, and the tubing is placed into a beaker of water (low solute concentration). The difference in concentration of sucrose (solute) between the two solutions creates an osmotic gradient. Water moves by osmosis into the sucrose solution and the volume of the sucrose solution inside the dialysis tubing increases.

Describing solutions

The dialysis tubing acts as a partially permeable membrane, allowing water to pass freely, while keeping the sucrose inside the dialysis tubing. Glass capillary tube

Dialysis tubing (partially permeable membrane)

Sucrose molecule

Water molecule

Zephyris

Dialysis tubing containing sucrose solution

Water movements in cells, particularly plant cells, are often explained in terms of water potential (see next activity). But you will often see other terms used to compare solutions of different solute concentration, especially in animal biology: Isotonic solution: Having the same solute concentration relative to another solution (e.g. the cell's contents). Hypotonic solution: Having a lower solute concentration relative to another solution. Hypertonic solution: Having a higher solute concentration relative to another solution.

Water

Net water movement

The red blood cells above were placed into a hypertonic solution. As a result, the cells have lost water and have begun to shrink, losing their usual discoid shape.

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1. What is osmosis?

2. (a) In the blue box on the diagram above, draw an arrow to show the direction of net water movement.

(b) Why did water move in this direction?

3. What would happen to the height of the water in the capillary tube if the sucrose concentration was increased?

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67 Water Movement in Plant Cells any particular direction can be calculated on the basis of the water potential of the cell sap relative to its surrounding environment. The use of water potential to express the water relations of plant cells is used in preference to osmotic potential and osmotic pressure although these terms are still frequently used in areas of animal physiology and medicine.

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Key Idea: Water potential explains the tendency of water to move from one region to another by osmosis. Water molecules moves to regions of lower water potential. The water potential of a solution (denoted by ψ) is the term given to the tendency for water molecules to enter or leave a solution by osmosis. The tendency for water to move in

Water potential and water movement

Less negative Ψs

More negative Ψs

Less negative Ψ

More negative Ψ

Hypotonic

Hypertonic

Loses water by osmosis

Gains water by osmosis

Water molecule

Solute molecule cannot pass through the membrane

Partially permeable membrane

The pressure potential (Ψp) The pressure potential is the hydrostatic pressure to which water is subjected (e.g. by a plant cell wall). The pressure potential is usually positive and is zero when cells are in equilibrium. It is sometimes called turgor or wall pressure.

The solute potential (Ψs) The solute potential is a measure of the reduction in water potential due to the presence of solute molecules. It is the negative component of water potential, sometimes referred to as the osmotic potential or osmotic pressure.

Water moves towards more negative Ψs until the concentration of water molecules equalizes

As water molecules move around some collide with the plasma membrane and create pressure on the membrane called water potential (ψ).The greater the movement of water molecules, the higher their water potential. The presence of solutes (e.g. sucrose) lowers water potential because the solutes restrict the movement of water molecules. Pure water has the highest water potential (zero). Dissolving any solute in water lowers the water potential (makes it more negative).

Water always diffuses from regions of less negative to more negative water potential. Water potential is determined by two components: the solute potential, ψs (of the cell sap) and the pressure potential, ψp, expressed by:

ψcell = ψs + ψp

The closer a value is to zero, the higher its water potential.

1. What is the water potential of pure water?

(a)

(b)

A ψs = –400 kPa ψp = 300 kPa

ψ for side A:

ψ for side B: Direction:

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B ψs = –500 kPa ψp = 300 kPa

A ψs = –500 kPa ψp = 100 kPa

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2. The diagrams below show three hypothetical situations where adjacent cells have different water potentials. For each pair (a)-(c) calculate ψ for each side and describe the net direction of water flow (AB, BA or no net movement): (c)

B ψs = –600 kPa ψp = 100 kPa

A ψs = –600 kPa ψp = 200 kPa

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B ψs = –500 kPa ψp = 300 kPa

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90 is irreversible. The diagram below shows two situations: when the external water potential is less negative than the cell and when it is more negative than the cell. When the external water potential is the same as that of the cell, there is no net movement of water.

Plasmolysis in a plant cell

Turgor in a plant cell

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When the contents of a plant cell push against the cell wall they create turgor (tightness) which provides support for the plant body. When cells lose water, there is a loss of turgor and the plant wilts. Complete loss of turgor from a cell is called plasmolysis and

Salt solution

Ψ = –600

Water

Pure water

Ψ=0

Water

Water

Cell wall is freely permeable to water molecules.

The Ψs is due to the solute concentration of the cytoplasm.

Cell wall bulges outward

Water

Cytoplasm takes on water, putting pressure on the plasma membrane and cell wall. Ψp rises, offsetting Ψs at full turgor.

Cytoplasm

Plasma membrane

Water

Water

Water

When external water potential is more negative than the water potential of the cell (Ψcell = Ψs + Ψp), water leaves the cell and, because the cell wall is rigid, the plasma membrane shrinks away from the cell wall. This process is termed plasmolysis and the cell becomes flaccid (Ψp = 0). Full plasmolysis is irreversible; the cell cannot recover by taking up water.

Water

When the external water potential is less negative than the Ψcell, water enters the cell. A pressure potential is generated when sufficient water has been taken up to cause the cell contents to press against the cell wall. Ψp rises progressively until it offsets Ψs. Water uptake stops when the Ψcell = 0. The rigid cell wall prevents cell rupture. Cells in this state are turgid.

3. What is the effect of dissolved solutes on water potential?

4. Why don't plant cells burst when water enters them?

5. (a) Distinguish between plasmolysis and turgor:

(b) Describe the state of the plant in the photo on the right and explain your reasoning:

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6. (a) Explain the role of pressure potential in generating cell turgor in plants:

(b) Explain the purpose of cell turgor to plants:

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68 Making Dilutions Simple dilutions are often used to produce calibration curves. A serial dilution is a stepwise dilution that quickly amplifies the dilution factor. Serial dilutions are useful when you require a volume or amount that is too small to measure accurately, or when you need to quickly dilute a solution that is very concentrated to begin with (e.g. bacterial cells in a solution).

Simple dilution

Serial dilution

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Key Idea: Dilution reduces the concentration of a stock solution by a known factor. A dilution reduces the concentration of a solution by a known value. Simple dilutions are based on ratios, and involve taking a volume of stock solution and adding it to an appropriate volume of solvent to achieve the desired dilution.

1 cm3

1 cm3 stock solution into 9 cm3 of solvent is a 1 to 10 dilution (10% of original concentration).

5 cm3 stock solution into 5 cm3 of solvent is a 1 to 1 dilution (50% of original concentration).

Stock solution

1 cm3

1 cm3 of original culture

9 cm3 of nutrient broth in each tube

1:10

1 cm3

1:100

1 cm3

1 cm3

Thick growth

1 cm3

Isolated colonies

Serial dilution is often used in microbiology to produce a plate of countable bacterial colonies.

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A microbial culture, grown in a broth, is serially diluted (above). The diluted culture is plated onto agar (left). this technique is called dilution plating.

C1 = initial concentration of stock solution V1 = initial volume of stock solution C2= final concentration required V2= final volume required

V1 = (C2 x V2) / C1

1:10 000

1:1000

The following equation is used to calculate the volume needed to make a simple dilution: C1 x V1 = C2 x V2

You will always know three of the values, so by rearranging the equation you can determine what volume of stock solution is needed to achieve the desired final concentration.

1 cm3

When a culture has been sufficiently diluted, the colonies are discrete and can easily be counted. Each colony arises from a single cell.

(a) 0.75 mol L-1:

(b) 0.50 mol L-1:

(c) 0.25 mol L-1:

2. (a) Use the equation below to calculate the solute potential (ψs) of the solutions in (1) and also of the 1.00 mol L-1 solution (the solutions were at 22°C):

0.75 mol L-1:

0.25 mol L-1:

0.50 mol L-1:

1.00 mol L-1:

ψs = -iCRT

i = ionization constant (for sucrose, this is 1) C = molar concentration R = pressure constant = 8.31 L kPa K−1mol−1 T = temperature (°K) = 273 + °C of solution.

(b) Plot sucrose concentration vs solute potential on the grid.

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1. A student had a 1.00 mol L-1 stock solution of sucrose. Calculate the dilutions required to produce 5 cm3 of sucrose solution at the following concentrations:

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69 Estimating Osmolarity of Cells tissue into a series of solutions of known concentration and observing if the tissue loses (hypertonic solution) or gains (hypotonic solution) water. The solution in which the tissue remains unchanged indicates the osmolarity of the tissue.

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Key Idea: A cell placed in a hypotonic solution will gain water while a cell placed in a hypertonic solution will lose water. The osmolarity (which is directly proportional to the solute potential) of a cell or tissue can be estimated by placing the

The aim

To determine the solute potential of potatoes by placing potato cubes in varying solutions of sucrose, C12 H22O11 (table sugar).

The method

Potato cubes

1. Complete the table (right) by calculating the total mass of the potato cubes, the total change in mass, and the total % change in mass for all the sucrose concentrations:

Fifteen identical 1.5 cm3 cubes of potato where cut and weighed in grams to two decimal places. Five solutions of sucrose were prepared in the following range (in mol dm-3): 0.00, 0.25, 0.50, 0.75, 1.00. Three potato cubes were placed in each solution, at 22°C, for two hours, stirring every 15 minutes. The cubes were then retrieved, patted dry on blotting paper and weighed again.

The results

2. Use the grid below to draw a line graph of the sucrose concentration vs total percentage change in mass:

[Sucrose] 0.00 mol L-1

Potato sample

Initial mass (I) (g)

Final mass (F) (g)

1

5.11

6.00

2

5.15

6.07

3

5.20

5.15

1

6.01

4.98

2

6.07

5.95

3

7.10

7.00

1

6.12

5.10

Total

Change (C) (F-I) (g)

% change (C/I x 100)

[Sucrose] 0.25 mol

L-1

Total

Change (C) (F-I) (g)

% change (C/I x 100)

[Sucrose] 0.50 mol

L-1

2

7.03

6.01

3

5.11

5.03

1

5.03

3.96

2

7.10

4.90

3

7.03

5.13

Total

Change (C) (F-I) (g)

% change (C/I x 100)

[Sucrose] 0.75 mol L-1

Total

(b) Use the calibration curve (previous page) to determine the solute potential (ψs) of your potato (in kPa):

(c) What is the pressure potential (ψp) of the potato cells at equilibrium?

Change (C) (F-I) (g)

% change (C/I x 100)

[Sucrose] 1.00 mol

Total

(d) Use the equation ψ = ψs + ψp to determine the water potential of the potato cells at equilibrium:

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3. (a) Use this graph to estimate the osmolarity of the potato (the point where there is no change in mass):

L-1

1

5.00

4.03

2

5.04

3.95

3

6.10

5.02

Change (C) (F-I) (g)

% change (C/I x 100)

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70 Active Transport Key Idea: Active transport uses energy to transport molecules against their concentration gradient across a partially permeable membrane. Active transport is the movement of molecules (or ions) from

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regions of low concentration to regions of high concentration across a cellular membrane by a transport protein. Active transport needs energy to proceed because molecules are being moved against their concentration gradient.

ff The energy for active transport comes from ATP (adenosine triphosphate). Energy is released when ATP is hydrolyzed (water is added) forming ADP (adenosine diphosphate) and inorganic phosphate (Pi).

Active

Active

ff Transport (carrier) proteins in the membrane are used to actively transport molecules from one side of the membrane to the other (below).

Passive

A ball falling is a passive process (it requires no energy input). Replacing the ball requires active energy input.

ff Active transport can be used to move molecules into and out of a cell.

ff Active transport can be either primary or secondary. Primary active transport directly uses ATP for the energy to transport molecules. In secondary active transport, energy is stored in a concentration gradient. The transport of one molecule is coupled to the movement of another down its concentration gradient, ATP is not directly involved in the transport process.

It requires energy to actively move an object across a physical barrier.

Sometimes the energy of a passively moving object can be used to actively move another. For example, a falling ball can be used to catapult another (left).

Active transport

1

ATP binds to a transport protein.

2

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

3

ATP is hydrolyzed and the energy released is used to transport the molecule or ion across the membrane.

4

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

Transport protein

High molecule concentration

ATP

ATP

H2O

ADP

H

P

OH

Low molecule concentration

Molecule to be transported

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

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1. What is active transport?

3. What is the difference between primary active transport and secondary active transport?

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71 Ion Pumps Key Idea: Ion pumps are transmembrane proteins that use energy to move ions and molecules across a membrane against their concentration gradient. Sometimes molecules or ions are needed in concentrations that diffusion alone cannot supply to the cell, or they cannot diffuse through the plasma membrane. In this case ion

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pumps move ions (and some molecules) across the plasma membrane. The sodium-potassium pump (below) is found in almost all animal cells and is common in plant cells also. The concentration gradient created by ion pumps is often coupled to the transport of other molecules such as glucose across the membrane.

Proton pump

Sodium-potassium pump

Na+

H+

H+

(the sodium-glucose symport) Na+

Na+

Na+

H+

H+

Cotransport

(the Na+/K+ ATPase)

Extracellular fluid or lumen

Potassium ion

K+ binding site

Na+

K+

Na+

Diffusion of sodium ions

+ + + + +

Glucose

Na+

Plasma membrane

– – – – –

ATP

H+

Carrier protein

Carrier protein

Na + binding site

ATP

Na+

Cell cytoplasm

K+

K+

Sodium ion

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

Na+

Proton pumps

Sodium-potassium pump

Cotransport (coupled transport)

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, the energy for moving the H+ comes from electrons, and the flow of H+ back across the membrane drives ATP synthesis via the membrane-bound enzyme ATP synthase.

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

A gradient in sodium ions drives the active transport of glucose in intestinal epithelial cells. The specific transport protein couples the return of Na+ down its concentration gradient to the transport of glucose into the intestinal epithelial cell. Glucose diffuses from the epithelial cells and is transported away in the blood. A low intracellular concentration of Na+ (and therefore the concentration gradient) is maintained by a sodium-potassium pump.

1. Why is ATP required for membrane pump systems to operate?

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

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

(c) What happens to the glucose that is transported into the intestinal epithelial cells?

3. Describe two consequences of the extracellular accumulation of sodium ions:

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72 Disturbances to Ion Transport cell, as water moves by osmosis from areas of low solute concentration to high solute concentration. Disturbances to the correct flow of ions can affect this balance. In the example of cholera infection below, transport channels for chloride ions out of the cell are permanently opened, causing an incorrect balance of ions and resulting in a massive loss of electrolytes and water from the cells.

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Key Idea: Disturbing the regulation of the movement of ions in and out of cells also has the effect of changing the amount of water entering and leaving a cell. Regulation of ion transport across a plasma membrane is important for maintaining the correct balance of ions inside and outside the cell. Maintaining this balance is in turn an important way of regulating the amount of water in a

Mechanism of cholera induced dehydration Blood vessel

Intestinal epithelial cell

Mechanism of oral rehydration salts

Gut lumen

Cholera toxin

Intestinal epithelial cell

Blood vessel

A B

Membrane receptors bind toxin

ATP

A

Glucose

Na-glucose symport

Cl-

cAMP

Cl-

CFTR channel

Na+

Na+

H2O

Cl-

H 2O

Cholera toxin contains two subunits (A and B), which bind to receptors on the cell membrane. Subunit A enters the cell and activates pathways that cause the production of cAMP from ATP. cAMP opens the Cystic Fibrosis Transmembrane conductance Regulator (CFTR) channel, causing the loss of chloride ions (Cl-) into the gut lumen. This results in a negative charge within the lumen. Sodium follows down its electrochemical gradient (the electrical potential and a difference in the chemical concentration across a membrane) while water follows down its osmotic gradient. These are replaced from the blood, causing a drop in blood volume.

Cholera is treated in the first instance with the use of oral rehydration salts, rather than antibiotics. Glucose and sodium enter the gut lumen in high concentrations and are transported into the cell by the sodium-glucose symport (ion pump). The influx of sodium causes a positive charge within the cell and results in Cl- moving down its electrochemical gradient and back into the cell. Water follows down its osmotic gradient. Water can then reenter the blood to restore blood volume (rehydration). This effect can occur rapidly, but oral rehydration must continue until the patient's immune system is able to eliminate the cholera bacteria.

1. How does the cholera toxin cause a loss of water into the gut lumen?

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2. Oral rehydration salts are used to treat cholera. They contain high concentrations of glucose and sodium. Explain how these are able to replace the water and electrolytes lost from a cell:

3. Why was it important to understand ion transport in the intestine when devising a treatment for cholera?

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73 Membranes and the Export of Proteins

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functional. This modification takes place in the rough endoplasmic reticulum (rER). From the rER, proteins are transported to the Golgi where the protein is further modified before packaged and shipped to its final destination.

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Key Idea: The synthesis, packaging and movement of macromolecules inside the cell involves coordination between several membrane-bound organelles. Many proteins need to be modified in order to become

1

3

As it enters the cisternal space inside the ER, it folds up into its correct 3-dimensional shape.

Ribosomes on the surface of the endoplasmic reticulum (ER) translate mRNA into a polypeptide chain.

4

Ribosome

2

The chain is threaded through the ER membrane into the cisternal space, possibly through a pore.

6

These vesicles are received by the Golgi apparatus which further modifies, processes, and packages the proteins. Proteins move through the Golgi stack from one side of the organelle to the other, undergoing modification by different enzymes along the way. They are eventually shipped to the cell's surface, where they can be exported from the cell by exocytosis.

5

Transport vesicle

Most proteins destined for secretion are glycoproteins (i.e. they are proteins with carbohydrates added to them). Attachment of carbohydrate to the protein by enzymes (glycosylation) occurs in the rER.

Proteins destined for secretion leave the ER wrapped in transport vesicles which bud off from the outer region of the ER.

Transport vesicle

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1. Explain how ribosomes, endoplasmic reticulum, transport vesicles, and the Golgi interact to enable cellular secretion:

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Exocytosis

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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 nerve cells) are specialized to manufacture products, such as proteins, and then export them from the cell to elsewhere in the body or outside it.

Exocytosis (and its counterpart endocytosis) require energy because they involve movement of cytoskeletal proteins.

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.

Dartmouth college

Nerve cell

NT

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).

EII

Nerve cell

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

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

2. (a) What is the purpose of exocytosis?

(b) How does it occur and what cellular components are involved?

3. Describe two examples of the purpose of exocytosis in cells: (a)

(b)

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74 Endocytosis membrane folds around a substance to transport it across the plasma membrane into the cell. The ability of cells to do this is a function of the flexibility of the plasma membrane.

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Key Idea: Endocytosis is an active transport process in which the cell engulfs material and draws it in. Endocytosis is a type of active transport in which the plasma Material (solids or fluids) that are to be brought into the cell are engulfed by an infolding of the plasma membrane.

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 or receptormediated. 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 cells of the liver, and some plant cells.

1. What is the purpose on endocytosis?

2. Is endocytosis active or passive transport?

(a) Phagocytosis:

(b) Receptor mediated endocytosis:

(c) Pinocytosis:

4. Explain how the plasma membrane can form a vesicle:

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3. Describe the following types of endocytosis:

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75 Active and Passive Transport Summary The diagram below summarizes the movement of material in and out of a cell. Use the information to complete the activity.

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

Cytoskeleton

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

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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76 KEY TERMS: Did You Get It?

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100

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1. Match each term to its definition, as identified by its preceding letter code. active transport carrier protein

A A partially permeable phospholipid bilayer forming the boundary of all cells.

B Movement of substances across a biological membrane without energy expenditure. C The passive movement of molecules from high to low concentration.

concentration gradient

D A type of passive transport facilitated by transport proteins.

diffusion

E

A membrane-bound protein involved in the transport of a specific molecule across the membrane either by active transport or facilitated diffusion.

endocytosis

F

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

facilitated diffusion

G Chemical used to enhance the contrast or highlight specific parts of a cell or tissue to be viewed under a microscope.

ion pump

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

osmosis

I

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

J

The gradual difference in the concentration of solutes in a solution between two regions. In biology, this usually results from unequal distribution of ions across a membrane.

passive transport

plasma membrane

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

stain

2. The diagrams below depict what happens when a red blood cell is placed into three solutions with differing concentrations of solutes. Describe the tonicity of the solution (in relation to the cell) and describe what is happening: A

B

C

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

Particle with diameter of 20 nm

After one hour:

Container of water at 20° C

Partially permeable membrane with pores of 10 nm.

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Soluble particles placed in at high concentration

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Cellular Communication

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Enduring Understanding

3.D

Key terms

cell signaling cyclic AMP cytokine

hormone ligand

local regulator morphogen

neurotransmitter pheromone

phosphorylation cascade protein kinase

3.D.1 Cell communication processes share common features that reflect a shared evolutionary history

Activity number

Essential knowledge

(a) Communication involves transduction of signals c 1

Explain what is meant by communication and how it involves transduction of stimulatory or inhibitory signals from other cells, organisms, or the environment.

77 78

(b) Signal transduction processes are under strong selective pressure

c 1

Explain why there is strong selective pressure for signal transduction processes to be correct and appropriate.

77

(c) Signal transduction pathways influence responses to the environment in single-celled organisms

c 1

quorum sensing receptor

second messenger

Use examples to show how signal transduction pathways determine responses to the environment in bacteria. Examples include: • Quorum sensing • Cell movements in response to external signals • Use of pheromones to trigger reproduction and development.

79

signal molecule

signal transduction

(d) Signal transduction pathways coordinate activities in multicellular organisms

c 1

Use examples to show how signal transduction pathways coordinate activities within the individual cells that support the functioning of the whole organism. Examples include: • Epinephrine stimulation of glycogen breakdown in mammals • Temperature determination of sex in some vertebrates • DNA repair

Essential knowledge

(a) Cells can communicate by cell-to-cell contact c 1

Activity number

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3.D.2 Cells communicate with each other in a variety of ways

78 85 88

Use examples to explain how cells communicate by cell-to-cell contact. Examples include: • Antigen presentation in immune cells • Cell-to-cell transport of material between plant cells via plasmodesmata.

80 81 82

(b) Cells communicate over short distances using local regulators

c 1

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Explain how cells use local regulators to communicate with target cells in the vicinity of the emitting cell. Examples of local regulators include: • Neurotransmitters and cytokines in vertebrates • Quorum sensing in bacteria • Signal molecules in the plant immune response • Morphogens in embryonic development

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(c) Signals released by one cell type can target another cell type some distance away c 1

Describe the characteristic features of endocrine signaling, e.g. in a mammal. 80 83 - 85 Explain how the signal is transmitted and why only target cells respond. Examples include: • Polar (hydrophilic) hormones: insulin and human growth hormone • Non-polar (hydrophobic) hormones: testosterone, estrogen, thyroid hormones

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Insulin

3.D.3 Signal transduction pathways link signal reception with cellular response

Activity number

Essential knowledge

(a) Signaling begins with a receptor protein recognizing a ligand c 1

Define receptor and ligand. Understand that different receptors recognize 77 78 80 different chemical messengers in a one-to-one relationship. Know that peptides, small molecules, or proteins can act as chemical messengers.

c 2

Explain the process of signal transduction including reference to ligand binding and shape change in the receptor protein. Examples could include: • G-protein-linked receptors • Ligand gated channels • Receptor tyrosine kinases.

PBP with bound bomykol

85

(b) In signal transduction, the signal is converted to a cellular response

c 1

Explain how signaling cascades relay signals from receptors to cell targets, often with amplification of the signal. Describe the result.

77 85

c 2

Using an example, explain the role of second messengers in signal cascades. Examples could include: • Ligand-gated ion channels • Second messengers such as cyclic AMP (cAMP) and cyclic GMP (cGMP) • Calcium ions (Ca2+) • Inositol triphosphate (IP3)

85 86

c 3

Explain signal transduction pathways involving: i) Protein modifications such as methylation ii) Phosphorylation cascades involving protein kinases

85 87

p53

Chk2

3.D.4 Changes in signal transduction pathways can alter cellular response

Activity number

Essential knowledge

(a) Signal transduction can be blocked or defective

Use examples to show how conditions where signal transduction is blocked or defective can be deleterious, preventative, or prophylactic. Examples include: • Diabetes, heart disease, autoimmune disease, cancer, or cholera • Effects of neurotoxins, poisons, and pesticides • Drugs such as hypertensive drugs, antihistamines, and birth control drugs

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77 Signals and Signal Transduction a cascade and usually involves phosphorylation (charging) of a number of molecules in a sequence. The type of response varies and may include a change in metabolism (activating a pathway), gene expression (to produce a specific protein), or membrane permeability (to allow entry of specific molecules). Signal transduction processes are generally under strong selection pressure during evolution because failure can result in death of the organism. The basic process, which is shared across evolutionary lines of descent, is outlined below.

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Key Idea: Signal transduction is the conversion of an external signal to a functional change within the cell through a series of biochemical reactions. Signal transduction is the process by which molecular signals are transmitted from outside the cell to inside, bringing about a cellular response. The transduction involves an external signal molecule binding to a receptor and triggering a series of biochemical reactions, which lead to a specific cellular response. The series of biochemical reactions is often called

An overview of signal transduction

Cytoplasm

Plasma membrane

Extracellular fluid

1. Reception

2. Transduction

3. Response

The cellular response within the target cell is activated

Receptor

Many molecules are involved in a signal transduction pathway

Signal transduction can be broken into three main steps:

Signal molecule

ff Reception: An extracellular signal molecule binds to its receptor on a target cell. ff Transduction: The activated receptor triggers a chain of biochemical events within the cell. Many different enzymes are involved, and the entire reaction is often called a signaling cascade. ff Response: The signal cascade results in a specific cellular response.

Cytokines are signaling molecules

ff Cytokines are a category of small signaling proteins with roles in regulating immunity, inflammation, and the formation and differentiation of blood cells (hematopoiesis).

ff The unique sequence of cytokine expression and combination of cytokines (along with other control processes such as methylation) determine blood cell fate during differentiation.

Multipotential hematopoietic stem cell IL 3 O

O

Myeloid progenitor

Lymphoid progenitor

IL 1 IL 2 IL 4 IL 6 IL 7

Myeloblast

Megakaryocyte

IL 3 O

B cell

1. (a) What are cytokines?

IL 3 IL 6 O

IL 3 IL 5 O

IL 3 IL 6 O

T cell

Erythrocyte (red blood cell)

Thrombocytes

Basophil

Eosinophil

Neutrophil

(b) What is their role in hematopoiesis?

(c) Explain how so many different cell types are generated from the multipotent hematopoietic stem cell:

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IL 3 IL 6 O

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ff Interleukins (IL) are common cytokines in hematopoiesis (right). Other cytokines (O), including tumor necrosis factors and growth factors, are also important.

Multipotent cells can give rise to any of the cell types of a particular tissue (e.g. blood).

IL 1 IL 3 IL 6 O

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What is a receptor?

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In order for a chemical signal to have an effect on a cell it must bind to a receptor. Cell receptors are protein molecules, and are very specific about the type of chemical signal they bind. Each receptor is linked to a specific biochemical pathway. Once the chemical signal has been bound a message is passed to the target cell causing a cellular response. The specificity of the receptor or a particular signal molecule prevents cells wasting resources and energy on producing cellular products that are not needed.

The binding sites of cell receptors are specific only to certain ligands (signal molecules). This stops them reacting to every signal the cell encounters. Receptors generally fall into two main categories:

Extracellular domain

Outside of cell

Intracellular receptors

Cell membrane

Intracellular receptors (also called cytoplasmic receptors) are located within the cell cytoplasm and they bind ligands that are able to cross the plasma membrane.

Extracellular receptors

Extracellular receptors (also called transmembrane receptors) span the cell membrane and bind ligands that cannot cross the plasma membrane. They have an extracellular domain outside the cell, and an intracellular domain within the cell cytosol. A diagram of an extracellular receptor is shown on the right.

Cell cytosol

Intracellular domain

Structure of a transmembrane receptor

2. Identify the three stages of signal transduction and describe what occurs at each stage:

(a)

(b)

(c)

3. (a) Why doesn't every cell respond to a signal molecule?

(b) Why is it important the cells don't respond to every signal molecule produced by an organism?

5. In the space below label the sequence (1-3) in the correct order:

Cellular response

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4. Why would signal transduction pathways be under strong selection pressure?

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78 The Nature of Signaling Molecules others (such as neurotransmitters) take effect rapidly, but the effect is short lived as the chemical is quickly broken down. Chemical signals can affect cells within an organism or, in the case of microbes, can be used to communicate with nearby cells. Pheromones are chemical signals secreted into the environment. They may travel over long distances to influence members of the same species.

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Key Idea: Signaling molecules are widespread in nature. The effect they have is highly varied and they may act over short or long distances to cause a response. Signaling molecules bind to specific receptors to cause a response in a target cell. There are a huge number of different signal molecules, and each has a specific effect. Some (such as hormones) tend to be slow acting and long lasting, while

Hormones help animals prepare for and adjust to seasonal changes

Gibberellins break dormancy in seeds

Plant hormones have important roles in plant growth and development (e.g. stem elongation, breaking dormancy, and fruit fall). Plant hormones are transported around the plant by the plant's vascular tissue. Tiny channels called plasmodesmata allow small signal molecules to be transported between plant cells, resulting in a cellular response.

Neurotransmitters are chemicals that carry signals between nerve cells or between a nerve cell and another type of cell such as a muscle or gland. Neurotransmitters act on the cell immediately next to it. They are released into a synapse (gap between the cells) and bind to receptors on the receiving cell. The response is rapid and short lived.

In mammals, hormones are secreted by endocrine glands (e.g. the pituitary) and carried in the blood to target cells. Hormones are very potent, and effective at low concentrations. Hormonal responses tend to be slow (because it takes time for the signal to reach its target) and generally long lasting because they induce metabolic changes.

USDA

A nerve cell

Lymphocytes produce cytokines

Staphylococcus aureus bacteria

Cytokines are a large group of peptides and small proteins involved in coordinating the response of cells in the immune system both within their immediate vicinity or over large distances. Cytokines are produced by a wide range of cells, including immune cells and endothelial cells. They include interferons, interleukins, and tumor necrosis factors.

Stallion exposing pheromone receptors

Communication between bacterial cells occurs by the release (and detection) of specific signaling molecules. This type of communication is used to monitor extracellular conditions, ensure the microbe is in a nutrient-rich environment so it can grow and reproduce, and to coordinate the release of toxins against a host.

Pheromones are chemical signals released into the external environment, and are common in social insects and mammals. Pheromones have different roles (e.g. aggression, aggregation, reproduction, territoriality) but all act to generate a specific response in members of the same species. In mammals, pheromones are used to signal sexual receptivity and attract mates.

2. What role do plasmodesmata play in cell signaling between plant cells?

3. (a) Identify a type of long lasting chemical signal: (b) Identify a type of short lasting chemical signal: 4. How do pheromones differ from the other types of chemical signals?

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1. Compare how hormones are transported in plants and animals:

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79 Communication Between Unicellular Organisms Key Idea: Bacteria gather information about their environment and synchronize their responses to environmental change by the use of signaling molecules. Unicellular organisms (e.g. bacteria) gather information about their environment and respond to it appropriately by using signaling pathways triggered by chemicals in the

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environment. These chemicals can relay information about how close a food source is, how many of the same bacteria are nearby (quorum sensing), or the location of a favorable environment. Chemical substances produced by unicellular organisms can also be used to communicate between individual cells and influence their cellular processes.

What is quorum sensing?

ff Quorum sensing is a process of cell to cell communication between bacterial cells. It involves extracellular signaling molecules called autoinducers. ff The bacteria share information about cell density in their environment and then alter their cellular response appropriately (e.g. through changes in gene expression).

ff Quorum sensing allows the bacterial population to act together to produce a coordinated response. A critical number (quorum) is required for the action to be beneficial. The quorum sensing response helps to ensure that valuable resources and energy are not expended unnecessarily and ensure that the effect on a host will be maximized. ff Many physiological activities are regulated by quorum sensing such as symbiosis, virulence, motility, antibiotic production, and biofilm formation.

CDC

Staphylococcus aureus forming a biofilm.

Bioluminescence in bacteria

ff Chemical communication between bacterial cells was first investigated in 1970 by Nealson and Woodland, after the observation that the luminescent bacterium Aliivibrio fischeri only luminesces when the population reaches a certain density. ff Luminescent bacteria produce light as the result of a chemical reaction during which chemical energy is converted to light energy.

Inactive LuxR

More autoinducer is produced

DNA

3 When the level of autoinducer is high enough, it activates the protein LuxR, which in turn stimulates the expression of the genes responsible for producing the proteins involved in luminescence.

Genes for luminescence

2

1

The mechanism of luminescence is controlled by a messenger molecule called an autoinducer, which travels between bacterial cells as a local regulator.

Note how the luminescence pathway is only activated when the level of autoinducer reaches a critical value, indicating the density of bacteria is high.

1. Explain what is meant by quorum sensing and describe its biological role:

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Proteins involved in luminescence

2. Briefly outline the process of cell to cell signaling resulting in luminescence in bacteria:

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Control of chemotaxis in bacteria

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2

Chemotaxis is the movement or orientation of an organism along a chemical concentration gradient either toward or away from the chemical stimulus. Chemotaxis in bacteria is controlled by signal transduction, which affects the direction of flagella rotation. Clockwise rotation of the flagellum causes tumbling and is the native condition (occurs without receptor stimulation) in bacteria. Tumbling occurs when the molecule CheA activates the molecule CheY, which attaches to the flagellum motor and causes its clockwise rotation.

When an attractant molecule (a chemical the bacterium wants to move towards) is encountered, the action of CheA is inhibited so that it can't activate CheY. Flagellum rotation is anticlockwise, causing the bacterium to swim straight (a run). After a second, the bacterium becomes insensitive to the attractant molecule and resumes tumbling. Encountering another attractant molecule causes it to start another run. In this way bacteria can move towards an attractant source, as a greater concentration of attractant molecules causes more runs and less tumbling. The protein CheZ is also able to directly inhibit CheY and stop it binding to the motor complex.

Anticlockwise rotation = directional run

Receptor

Clockwise rotation = random tumble

CheA

CheZ (-)

Attractant molecule (-)

Che complex

Flagellum

CheY

Flagellum motor

3. (a) Explain how the autoinducer molecule in luminescent bacteria signals when to luminesce:

(b) Explain how this enables the bacterium to detect the population density:

(c) How might this information be of survival advantage to the bacterium?

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4. Explain how the Che molecules allow a bacterium to move towards or away from a chemical source:

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the cell and bring about a response. Cells that possess the receptors to bind a particular signal molecule are called target cells for that signal molecule. If a cell does not have the specific receptor, then it is unaffected by the chemical signal. Chemical signals can be classified based on how far they travel to cause an effect. Some signals have a localized effect, while others can influence cells (or organisms) a considerable distance away.

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Key Idea: Cells use signals (chemical messengers) to communicate and to gather information about, and respond to, changes in their cellular environment. In order to communicate and respond to changes in their environment, cells must be able to send, receive, and process signals. Most signals are chemicals (signal molecules or ligands) and many different types exist. In order for a signal to have an effect on a cell it must be able to bind to T-cell

Signal molecules

Autocrine signaing Cells can produce and react to their own signals. This type of signaling is important during growth and development and in the functioning of the immune system. Example: In vertebrates, the presence of a foreign antibody causes T-cells to produce a growth factor to stimulate their own production. The increased number of T-cells helps to fight the infection.

Receptor

Cell-to-cell communication Cell-to-cell communication involves cells interacting directly with one another. There are two forms: 1) communication via special channels between adjacent cells and 2) two cells bind to and communicate with each other because they have complementary proteins on their surfaces.

Antigen presenting cell

Example: Plasmodesmata are microscopic channels that run through the cell wall of adjacent plant cells. Signal molecules can pass through to the next cell.

Helper T-cell

T-cell

Antigen presenting cell

Synapse

Signaling by local regulators Some cell signaling occurs between cells that are close together. The signal molecule binds to receptors on a nearby cell causing a response. Both neurotransmitters (signaling molecules in the nervous system) and cytokines (small molecules produced by a range of different cells) are involved in this type of local regulation.

Postsynaptic cell

Signal molecule (neurotransmitter)

Example: Neurotransmitters released from a nerve cell travel across the synapse (gap) to another cell to cause a response.

Synapse

Endocrine signaling A signal is carried in the bloodstream to target cells, often some distance away.

Endocrine cell

Target cells

Endocrine signaling may involve hormones (released from the cells of endocrine glands) or cytokines as the signaling molecule although cytokines are also important in local regulation and circulate in more variable concentrations than hormones.

Blood vessel

Signal molecule

Receptor

Insulin WEB

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Presynaptic cell

Example: In the immune system, antigen presenting cells present antigens to helper T-cells for destruction.

Example: Insulin from the pancreas stimulates the cellular uptake of glucose.

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Signaling between organisms by pheromones Some organisms produce chemical signals that are secreted externally to cause a response in other organisms. The signal molecules are called pheromones, and they affect the physiology or behavior of members of the same species.

OH

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Pheromone (bombykol) secreted by the female silk moth (Bombyx mori) is detected by the male's antennae causing his wings to flutter.

PBP

B

Pheromone Binding Protein (PBP) with bound bomykol (B)

Example: The pheromone bombykol is released by female silk moths and is detected in very low concentrations by males. The pheromone binding protein on the male's antennae and binds the bombykol, triggering a wing fluttering response. He uses the chemical concentration gradient to locate his mate.

1. Describe the following types of cell signaling:

(a) Autocrine signaling:

(b) Cell-to-cell communication:

(c) Local regulation:

(d) Endocrine signaling:

(e) Communication by pheromones:

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2. Identify the components that all types of communication by signaling have in common:

3. Endocrine signaling and communication by pheromones both work over long distances. How do they differ?

4. Cytokines and hormones are both involved in endocrine signaling. How do they differ?

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81 Cell-to-Cell Communication

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cells allowing for communication to take place. In animal cells these are called gap junctions and in plant cells they are plasmodesmata (sing. plasmodesma). A second form of cell-to-cell communication occurs when a receptor of one cell interacts directly with the receptor of a second cell. This type of cell signaling is important in embryonic development and the immune response.

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Key Idea: Some cell signaling involves direct interaction between cells, which may even be permanently adjoined. Cell-to-cell signaling requires close contact between adjacent cells. It is also known as contact dependent signaling. There are several mechanisms by which cell-to-cell communication can occur. Plant and animal cells may have connecting microscopic channels running between adjacent

Cell-to-cell communication in animal cells

Antigen presenting cell

Cell-to-cell communication in plant cells

Plasma membrane

Middle lamella

Cell wall

Direct cell-to-cell contact occurring

Cytokines

Presented antigen

Smooth ER

Plasmodesma

Helper T-cell

Cell A

Helper T-cells are activated by direct cell-to-cell signaling with antigen presenting cells. In the immune system, helper T-cells and antigen presenting cells (APCs) carry out cell-to-cell communication through direct physical contact with each other. APCs, such as dendritic cells and macrophages, ingest antigens, process them, and present them on the cell surface where they are recognized by helper T-cells. This presentation (along with cytokines released from the APC) stimulates the production of killer T-cells and B-cells and helps an organism to destroy foreign antigens.

Plasmodesmata allow adjacent plant cells to communicate directly with each other. Each cell has 103-105 plasmodesmata, laid down at the time of development. They allow signal molecules, such as transcription factors and regulatory RNAs, to pass between neighboring cells. Intercellular communication allows the plant to coordinate tissue or organ responses to environmental stress or as required for growth and development. An increase in calcium has shown to constrict the opening of the plasmodesmata, and provides a way to regulate the transport of signal molecules between cells.

Plasmodesmata

Alison Roberts

Killer T-cell

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B-cell

Cell B

1. Compare direct cell-to-cell communication between animal cells of the immune system and adjacent plant cells:

2. How could plants regulate the level of cell signaling occurring through plasmodesmata?

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82 Local Regulators number of cells in the immediate vicinity of the emitting cell can be stimulated, resulting in a specific cellular response. This communication allows neighboring cells to coordinate their activities (e.g. coordinating cell specialization during development and coordinating behavior in bacteria). Local regulation pathways are evident in all organisms.

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Key Idea: Signaling by local regulation occurs between cells located in the close proximity to one another. A signal molecule is released from one cell and travels to another causing a cellular response. During local regulation, the signal molecule travels only a short distance from the secreting cell to the target cell. A

Local regulation between animal cells

Local regulation between plant cells

ff Local regulation across synapses occurs in the nervous system and takes place between a neuron (nerve cell) and another cell (e.g. another neuron or a muscle cell).

In plants, local regulation occurs via a process called transcellular transport. Signaling molecules are transferred between cells over a short distance, but without the use of plasmodesmata. The mechanism of exchange may include import and export mechanisms such as cytosis or the use of specialist receptor and transport proteins located in cell membranes. This mechanism is used by plants to launch an immune response to disease.

ff In this 'synaptic signaling', an electrical impulse travels along a long conducting process of the neuron called an axon. It eventually reaches the synapse (the gap between the two cells) where it triggers the release of chemical signal molecules called neurotransmitters from the axon terminals. ff The neurotransmitter travels across the synapse to affect the target cell. The effect can increase a response (excitatory) or decrease a response (inhibitory) depending on the cell and neurotransmitter involved. Chemical changes in the cell bring about a range of specific cellular responses (e.g. opening ion channels, causing muscle contraction, and regulation of heart rate and blood pressure).

Example: Salicylic acid is produced in response to pathogens (disease-causing organisms). It results in the activation of a number of genes involved in defensive behaviors which may:

ff Close stomata (to prevent more pathogens entering). ff Produce proteins with antimicrobial activity. ff Target and destroy non-plant nucleic acids. Cell wall

Axon of neuron (stimulating cell)

Cell A

Plasma membrane

Axon terminal

Exocytosis

Neurotransmitter released from vesicle by exocytosis

Synapse

Endocytosis

ell tc e g ar Mem brane of t

Cell B

The neurotransmitter is broken down or reabsorbed after diffusing across the synapse. This prevents continued response by the target cell.

Cell response

Signal molecules

1. Contrast the two mechanisms of transcellular transport in plants:

Cell response

Permease/ transporter

Receptor

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Receptor

2. Describe a similarity in cell signaling involving local regulation in plants and animals:

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83 Action of Insulin tightly regulated and under strong selection (evolutionary) pressure. If insulin is lacking or cells fail to respond to it, blood glucose remains elevated while the body's cells themselves are starved of fuel. Insulin circulates in the blood where it binds to protein kinase receptors on the surface of cells and triggers a signal cascade that results in activation of the membrane transporters that bring glucose into the cell.

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Key Idea: Activation of the insulin receptor by insulin causes a signal cascade that results in cellular glucose uptake. Insulin is a peptide hormone secreted by the pancreas. It is involved in regulating blood glucose levels by promoting the uptake of glucose by cells. Malfunctions in the signaling pathways for insulin production and reception have serious physiological consequences (including death) so these are

Extracellular environment

Bound insulin

Glucose uptake pathway

1 Two molecules of insulin must bind

to the extracellular domain of the insulin receptor to activate it.

Glucose

Unbound insulin

Insulin receptors are made up of four subunits: 2 α-subunits and 2 β-subunits.

Glut4 glucose transporter

2 Once the insulin is bound,

phosphate groups are added to the receptor in a process called autophosphorylation.

Active molecules

Inactive molecules

5 Glut4 glucose transporters

insert into the membrane allowing the uptake of glucose.

Glut4 secretory vesicle

3 The autophosphorylation of the

4 The cascade sequence results in

receptor begins a signal cascade, in which several other proteins are phosphorylated in sequence. Each can activate many other proteins.

the activation of many Glut4 secretory vesicles, which produce the Glut4 glucose transporters.

Intracellular environment

1. (a) What type of signaling does this example represent (circle one): autocrine / local regulation / endocrine / pheromone

(b) Explain why you chose this answer:

2. Why must blood glucose levels be tightly regulated?

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3. Describe the process by which insulin signaling causes the uptake of glucose in to cells:

4. How does the signal cascade increase the response of the insulin receptor?

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84 Hormones are Regulated by Negative Feedback the production of a protein called insulin-like growth factor (IGF-1). GH levels are regulated by negative feedback (a regulatory mechanism in which a stimulus input causes an opposite output in order to maintain a steady state). GH has a critical role in human development and malfunctions in its regulation can result in a number of serious conditions including growth deficiencies (low GH) and abnormal tissue growth resulting in tumors or giantism (excessive GH).

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Key Idea: Growth hormone is required for normal growth and development. Its levels are tightly regulated through negative feedback mechanisms to prevent growth disorders. Growth hormone (GH) is a peptide hormone that stimulates growth, cell reproduction, and cell regeneration through endocrine signaling. GH is released from the pituitary gland at the base of the brain and circulates in the blood where it binds to two adjacent receptors on liver cells and stimulates

The effects and regulation of growth hormone

ff Growth hormone (GH) is released in response to GHRH (growth hormone releasing hormone) from the hypothalamus of the brain.

ff GH acts both directly and indirectly to affect metabolic activities associated with growth.

--

Bone

GHRH Hypothalamus

+

Anterior pituitary

ff GH directly stimulates metabolism of fat, but its major role is to stimulate the liver and other tissues to secrete IGF-1 (Insulin-like Growth Factor 1) and through this stimulate bone and muscle growth.

GH secretion is regulated via negative feedback.

ff High levels of IGF1 suppress GHRH secretion (and therefore GH secretion from the anterior pituitary).

Stimulates bone growth

Fat

+

+

GH

Directly stimulates utilization of fat

IGF-1

Liver

ff High levels of IGF1 stimulate the release of growth hormone inhibiting hormone (GHIH or somatostatin) from the hypothalamus. GHIH suppresses GH secretion (this path is not shown).

IGF-1

Stimulates muscle growth

+

Muscle

Growth hormone

1. Describe the metabolic effects of growth hormone:

3. (a) In the space on the right, construct a negative feedback pathway to show how GHIH regulates GH secretion:

(b) Predict the effects of low GH levels during infancy:

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2. Describe the two main mechanisms through which the secretion of growth hormone is regulated:

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85 Types of Signal Transduction

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signal molecule does not have to pass across the plasma membrane to cause a cellular response. Most cell receptors are extracellular receptors. Intracellular receptors bind signal molecules that have passed into the cell directly across the plasma membrane. Intracellular receptors may be located in the cytoplasm or on the nucleus.

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Key Idea: The majority of cell signals bind to extracellular receptors to exert their effect. However some cell signals are able to pass through the plasma membrane and bind directly to intracellular receptors within the cell to exert their effect. Cell receptors fall into two broad classes. Extracellular receptors bind signal molecules outside of the cell. The

Hydrophilic signal molecules are received by extracellular receptors

Hydrophilic signal molecules such as epinephrine are water soluble and cannot cross the plasma membrane. Epinephrine accelerates heart rate and is involved in the fight or flight response. Extracellular fluid

Hydrophobic signal molecules are received by intracellular receptors

Estrogen is the primary female sex hormone. It is involved in the development and maintenance of female characteristics. Extracellular fluid

The first messenger (signal molecule) binds to the receptor protein

Signal molecule

Enzyme

Receptor protein

Plasma membrane

Plasma membrane

Lipid soluble signal molecule passes across the plasma membrane

Protein subunit from the receptor protein activates the enzyme

Second messenger

Signal binds to receptor to form a receptor/signal complex

Active enzyme produces a second messenger

Receptor/signal complex acts as a transcription factor, binding to DNA to begin transciption

The second messenger triggers a cascade of phosphorylation events leading to a cellular response

DNA

Nucleus

P

Cell response

Cell response

Cytoplasm

Cytoplasm

Hydrophilic signal molecules cannot cross the plasma membrane and must exert their effect by interacting with an extracellular receptor. Hydrophilic signals include water soluble hormones such as epinephrine and insulin. The signal molecule is the first messenger. When it binds, the extracellular receptor is activated, triggering a sequence of biochemical reactions, including activation of a second messenger. As a consequence, the original signal is amplified, bringing about a cellular response. This pathway is given in more detail in the next activity. WEB

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The protein product produced alters the cell’s activity

Hydrophobic hormones, such as steroids (e.g. estrogen), diffuse freely across the plasma membrane and into the cytoplasm of target cells. Once inside the cell, they bind to intracellular receptors in the cytoplasm to form a receptorsignal complex. The complex moves to the cell nucleus where it binds directly to the DNA and acts as a transcription factor, resulting in the transcription of a one or more specific genes. Concentrations of the different gene products (proteins) change as a result (a phenotypic change).

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1. Describe the differences between an intracellular receptor and an extracellular receptor:

2. What must a signal molecule do in order to activate a receptor?

3. In terms of their ability to cross the plasma membrane, describe the difference between a hydrophobic signal molecule and a hydrophilic signal molecule:

4. (a) Outline the process when signal transduction occurs via an extracellular receptor:

(b) Describe the differences between a first messenger and a second messenger:

5. Outline the process when signal transduction occurs via an intracellular receptor:

6. The diagram on the right represents a cell signaling process.

(a) Does this diagram represent an extracellular or intracellular signaling process? Explain your answer:

Plasma membrane

A

(b) What type of receptor is B? (c) What does A represent? (d) Would A be hydrophobic or hydrophilic? Explain your answer:

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Cell response

B

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86 Signal Transduction Using Second Messengers

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original signal, so many molecules are activated. Common second messengers include cyclic AMP (cAMP), Ca2+, or inositol triphosphate. Signal transduction pathways involving phosphorylation cascades, such as the one below for epinephrine, involve activation of protein kinases. In the example below the end result is that glycogen is broken down into glucose monomers.

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Key Idea: Second messengers relay signals received at receptors on the cell surface to target molecules in the cytoplasm or nucleus to bring about a cellular response. Second messengers relay a signal received by an extracellular receptor to a target within the cell to bring about a specific response such as enzyme activation. The process involves a signal cascade, which amplifies the

Second messenger activation of a signal cascade

When activated by G protein, the enzyme adenylate cyclase catalyzes the synthesis of cyclic AMP (cAMP).

The hormone (epinephrine) is the first messenger. It operates through cAMP as a second messenger.

Hormone binds to receptor

cAMP activates protein kinase A, beginning a phosphorylation cascade which amplifies the original signal and brings about a cellular response.

Plasma membrane of target cell

1 molecule

G-protein linked receptor (β adrenergic receptor)

Adenylate cyclase

α

cAMP

γ

GDP

ATP

α

P

ADP

Inactive phosphorylase kinase

Inactivated G protein is trimeric (has three parts)

GTP

Activated G protein

Activated protein kinase A

ATP

β

α

ATP

When a ligand binds to the Gprotein linked receptor, GDP is exchanged for GTP on the Gprotein α subunit, activating it.

Activated phosphorylase kinase

ADP

A kinase transfers phosphate groups from ATP to specific substrates (phosphorylation).

Activated glycogen phosphorylase

P

Inactive glycogen phosphorylase

ATP

Cellular response 108 molecules

ADP

Glucose-1-phosphate

P

Glycogen

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1. Explain the significance of signal amplification:

2. Explain why each molecule in the cAMP signal cascade is phosphorylated:

3. Briefly summarize the main steps in the signal transduction process described above:

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87 Signal Transduction Involving Protein Modification Key Idea: Signal transduction pathways can be regulated by the addition of a methyl group (-CH3) to proteins involved in the signaling pathway. Protein methylation involves the addition of a methyl group (-CH3) to certain amino acids in a protein sequence. This action is carried out by a specific group of enzymes called

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methyltransferases. Protein methylation influences a number of functions within a cell including regulation of some signal transduction pathways. For example, the signal transduction pathway responsible for the run and tumble chemotaxis response in bacteria is regulated by the level of methylation of the Methyl-accepting Chemotaxis Protein (MCP).

3

1. Briefly describe how methylation of MCP proteins influences the signal transduction pathway responsible for flagellar rotation in bacteria:

The degree of methylation of MCP causes the run or tumble response

A2-33 CC3.0

Methyl-accepting Chemotaxis Proteins (MCP) are a group of transmembrane chemoreceptor proteins found in bacteria. MCPs (right) allow bacteria to detect concentrations of attractants (nutrients) and repellents (toxins) in their environment. The degree of methylation of the MCPs reflects the current environmental conditions. It sets off a signal transduction pathway which determines whether the bacteria should initiate a run or tumble response.

ff Attractants cause a decrease in the level of methylation of the MCPs and cause a run response towards the attractant.

ff Increasing repellent concentration results in an increased level of MCP methylation and initiates the tumble response, allowing the bacterium to reorientate itself in a new direction to find a more favorable environment. The default state for flagella rotation is clockwise (tumble). The toggle between the two states is maintained through a signaling pathway involving MCP and the Che molecules. MCP interacts with CheW/CheA, which in turn transmit signals to CheY/CheZ and CheB/CheR to bring about a response.

2. The two diagrams A and B below represent MCP protein from a bacterium at two different times. A

B

ff CheW/CheA generate receptor signals.

Methylated

ff CheY/CheZ control the spin direction of the flagellar motor. CheY promotes clockwise rotation and CheZ prevents this.

Non-methylated

ff CheB/CheR regulate the MCP methylation state by adding or removing methyl groups (adaptation to environment). Ligand binding

Attractants

Transmembrane MCP

Repellents

Cell membrane

A:

CheR

E

E

CheW

CheB

C

CheA

O

CheW

C

O

O-

CheA

O CH3

B:

Sensory adaptation

Anticlockwise CheZ=run

Clockwise CheY=tumble

Transmembrane receptors (MCPs) exist in two states (methylated and non-methylated) and this determines the rotational direction of the flagella. E=glutamic acid (amino acid residue).

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(a) Predict the environmental conditions at each different time:

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Transmembrane MCP

(b) Predict whether the bacterium would be in a run phase or a tumble phase:

A: B:

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88 Cell Signaling and DNA Repair encoded in the genome. DNA damage as a result of ionizing radiation is repaired via a signal cascade beginning with the autophosphorylation of the protein kinase ATM. ATM is a highly conserved protein and it is a key component in DNA repair. Depending on the extent of the damage, the signal cascade results in repair of the DNA, arrest of cell cycle, or apoptosis (programmed cell death).

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Key Idea: DNA repair is an essential cellular process involving a signal cascade initiated by DNA damage. DNA can acquire errors, either through copying errors during replication or as a result of environmental agents such as ionizing radiation. Repairing DNA damage is an essential process that guards against structural damage that would hamper the cell's ability to correctly transcribe the information

Initiating DNA repair

DNA damage

Au to

ph os p

hor

ylat

Ionizing radiation (UV light)

ion of

A signal cascade or signal transduction pathway requires an initiator. This begins a cascade of chemical reactions within the cell that activates the required molecules and transmits the signal to the correct area of the cell. Often the signal is a molecule, such as a hormone, that is produced elsewhere in the organism and is transported to the cell. In the case of DNA damage by ionizing radiation, it is the radiation itself that activates the signal cascade.

ATM

Phosphorylated ATM

The initial signal is passed through a series of messenger molecules. Some of these will take part in the final response while others may simply pass on the signal to other molecules.

p53 protein binds DNA, which in turn stimulates another gene to produce a protein called p21. When p21 is complexed with Chk2 the cell cannot pass through to the next stage of cell division.

P

Phosphorylated Chk2

Chk2

P

Phosphorylated p53

p53

Chk2 (a checkpoint kinase)

P

DNA repair

p53

p21

Apoptosis

p21

The signal cascade results in activating molecules that participate in DNA repair, arrest the cell cycle, or bring about apoptosis (programmed cell death). DNA repair and cell cycle arrest are immediate responses to DNA damage and protect a damaged cell and to promote its recovery. The activation of apoptosis or senescence occurs later in time. These delayed responses eliminate a damaged cell.

Cell cycle arrest

(b) Explain why a signal cascade can be useful in certain cell processes:

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1. (a) Describe how DNA repair is signalled in cells:

2. Use the example of DNA damage to explain why there is a strong selection pressure for the correct functioning of signal transduction pathways:

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89 Effect of Blocking Signals of a pathway to operate properly would cause significant detrimental effects. Many therapeutic drugs focus on blocking the effects of overactive pathways or restoring signaling pathways when they fail. The negative effects of many toxins are caused by their actions on signal transduction pathways.

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Key Idea: The failure of correct cell signaling can lead to many serious diseases, including cancer. Almost as soon as biologists realized the importance of signal transduction pathways in the body, their importance in the development of disease became obvious. Failure Signal molecule

Signals, responses, and inhibitors

Normal cell and signaling pathway

Universal inhibitor

Diseased cell and signaling pathway

Selective inhibitor

Receptor B

Membrane receptors

Unregulated pathway

Single blocked pathway

Blocked pathways

Molecule B

Regulated pathway

Molecule A

Receptor A

When choosing a drug to correct a transduction pathway, it is important to understand how the pathway operates. The signaling pathways of diseased cells (e.g. tumor cells) are often overactive (e.g. overexpressing receptors for growth factors). In the example above, receptor A signals the production of more receptor A via the messenger molecule A. Receptor B initiates a signal cascade activating molecule B, which inactivates molecule A and so regulates receptor A's production. The diseased cell fails to make molecule B and so cannot regulate receptor A's production.

Drug selection also involves the question of how many receptors are affected by a particular drug. Universal signal transduction inhibitors affect diseased cells by blocking many signal pathways, but also affect healthy cells, and so can cause damage outside the target area. Selective signal transduction inhibitors target a single overactive pathway, thus causing no damage outside the target area. However the specificity of these drugs may mean many different types are required to treat a disease effectively.

The Wednesday Island CC 3.0

Public domain

Psoriasis results from the hyper-proliferation of keratinocytes (epidermal cells) which is driven by EGFR (epidermal growth factor receptor) causing scaly patches or plaques on the skin.

1. (a) Explain the link between signal transduction and disease:

(b) Explain why understanding the link is important:

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Type II (adult onset) diabetes is caused by the failure of cells to respond to signaling by the hormone insulin, which results in the failure of cells to take up glucose.

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Breast cancer (like all cancers) is the result of cells dividing out of control. In breast cancer, the HER2 receptor protein has been identified as a major cause.

Finger prick test for diabetics

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Pesticides and signals

Action of antihistamine

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Many pesticides work by affecting the signaling of nerves cells by either blocking uptake of signaling molecules or facilitating the uptake of far greater amounts than normal.

Hayfever and similar allergic reactions occur when mast cells produce the chemical histamine as part of an immune response to foreign particles such as pollen. Histamine triggers the inflammatory response and results in swelling, red eyes, and a runny nose.

Cholinesterase inhibitors work by blocking the enzyme responsible for breaking down acetylcholine after it has transmitted a message across a synapse. As a result, the receiving neuron continues to fire, causing over-stimulation and death.

Neonicotinoid insecticides (NNIs) mimic the action of acetylcholine in synapses. This again causes the over-stimulation of the nerve cells and results in death. Recent evidence links use of NNIs to persistent negative effects on reproduction in bees.

Antihistamines block the histamine receptor of cells and so block the signal for the inflammatory response. Mast cell

Histamine

Pyrethrins act on the sodium channels that move sodium across the neuron membrane, preventing them from closing and causing over-stimulation and death.

Organochlorines inhibit the GABA receptor, which itself has an inhibitory function in nerve cells. The result is over-stimulation of the nervous system and death.

Antihistamine

Histamine receptor

Chitin is a major part of an insect's exoskeleton. Pesticides block the chitin synthesis pathway, preventing the insect from growing a new exoskeleton and molting (so it can't grow), eventually killing the insect.

2. (a) Describe the action of a universal signal transduction inhibitor:

(b) Describe the action of a selective transduction inhibitor:

(c) Explain the role of these drugs in controlling a disease:

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3. (a) Describe the general function of insecticides that act on the nervous system:

(b) Explain why pesticides based on cholinesterase inhibitors are also toxic to humans while pesticides based on chitin synthesis inhibitors are not:

4. Explain how antihistamines prevent hayfever:

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90 KEY TERMS AND IDEAS: Did You Get It?

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

cell signaling

A A molecule that relays signals from receptors on the cell surface to target molecules inside the cell.

cyclic AMP

B These chemical signals are secreted externally by one organism and cause a response in other organisms.

hormone ligand

C A signal molecule that is carried in the bloodstream to target cells, often a considerable distance away. D A molecule that binds to another.

pheromones

E

The process by which cells use signals (chemical messengers) to communicate and to gather information about, and respond to, changes in their cellular environment.

phosphorylation cascade

F

A second messenger important in many biological processes.

protein kinase

quorum sensing

G The process by which an external cell signal binds to a receptor and the receptor triggers a series of biochemical reactions in the cell to cause a specific response. H A process of cell to cell communication between bacterial cells. I

A sequence of events where one enzyme phosphorylates another, causing a chain reaction leading to the phosphorylation of many proteins.

J

Specific protein molecules that signal molecules bind to.

receptor

second messenger signal molecule

K A molecule that relays signals from receptors on the cell surface to target molecules inside the cell. L

signal transduction

A kinase enzyme that modifies other proteins by chemically adding phosphate groups to them (phosphorylation).

2. Choose the correct answer from the choices below:

3. Choose the correct answer from the choices below:

Hormones are relatively long lived signals that travel through the bloodstream. This type of signaling is called:

(a) Autocrine signaling (b) Cell-to-cell communication (c) Local regulation (d) Endocrine signaling (e) Pheromone signaling

(a) Hormones (b) Neurotransmitters (c) Pheromones (d) Cytokines

A

(b) What prevents the other two signal molecules from binding to this receptor?

B

C

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4. (a) The molecules labelled A-C are signaling molecules. Identify the signal molecule that will bind to the receptor shown:

What type of signal molecule is used in synaptic signaling?

Receptor

(c) Why is it important that not all cells react to every signal molecule?

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91 Synoptic Questions 1. Name the process illustrated in the diagram on the right: (a) Condensation (dehydration synthesis)

(b) Denaturation

(c) Hydrolysis

(d) Lysis

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2. Name the opposite reaction to the one illustrated in the diagram on the right:

(a) Condensation (dehydration synthesis)

(b) Denaturation

(c) Hydrolysis

(d) Lysis

3. Identify the molecule depicted in the diagram below:

R

NH2

C

C

H

O

+ H2O

4. Which statement is untrue?

(a) Polysaccharides are made up of many monosaccharides.

(b) Nucleotides are the building blocks of DNA and RNA.

(c) Fatty acids are the building blocks of proteins.

(d) Phospholipids are an important component of cellular membranes.

O

OH

(a) Monosaccharide

(b) Lipid

(c) Amino acid

(d) Triglyceride

5. Identify the type of solution that the red blood cell below has been placed into:

(a) Isotonic

(b) Hypertonic

(c) Hypotonic

(d) None of the above

6. The diagram right shows a plant cell in plasmolysis. This has occurred because:

(a) The cell membrane has ruptured

(b) Water has left the cell

(c) Water has entered the cell

(d) The cell wall has collapsed

8. The organelle on the right is the:

(a) Nucleus

(a) Site of protein production

(b) Chloroplast

(b) Site of cellular respiration

(c) Golgi body

(c) Site of photosynthesis

(d) Mitochondrion

(d) Site of protein modification

9. The image of a macrophage (type of white blood cell) right was measured at 20 mm when viewed at 1000 X magnification. The actual size of the cell is:

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7. Identify the organelle on the right:

10. Which of these statements about carrier-mediated diffusion is untrue?

(a) It enables the transport of lipid-insoluble molecules across the membrane

(b) 20 µm

(b) It is a passive process

(c) 200 µm

(d) 2000 µm

(c) It moves molecules against their concentration gradient

(d) It is important in the transport of glucose into red blood cells

(a) 2 mm

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123

Dialysis tubing (selectively-permeable membrane)

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11. A solution of starch and glucose was placed into a dialysis bag and then suspended in a beaker of water (right). Which of the following descriptions best describes the contents of the beaker after two hours.

(a) The beaker contains distilled water only.

(b) The beaker contains water, starch, and glucose.

(c) The beaker contains water and glucose.

(d) The beaker contains water and starch.

12. The dark lines on the photograph right show cytoplasmic connections between two plant cells. These are called:

(a) Tight junctions

(b) Gap junctions

(c) Adhesions

(d) Plasmodesmata

14. Signal transduction pathways may be regulated by:

(a) Second messengers

(b) Transcription factors

(c) Protein methylation

(d) All of the above

Solution containing starch and glucose Distilled water

13. Describe the purpose of the structures identified in the photograph

15. Are these structures present in animal cells?

16. The drawing right shows two cells of differing water potential. Calculate the water potential of each cell:

(a) The ψ of cell A = (b) The ψ of cell B = (c) Net water movement is (identify the correct answer): i) From cell A to cell B ii) From cell B to cell A iii) Net water movement = 0

A

B

ψs = -300 kPa ψp = 400 kPa

ψs = -500 kPa ψp = 300 kPa

17. For the protein collagen, pictured right, describe how the structure of the molecule relates to its functional role in providing skin strength and elasticity:

18. For the enzyme human salivary alpha amylase, pictured right, describe how the structure of the molecule relates to its functional role in hydrolyzing alpha-linked polysaccharides:

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Collagen

Alpha-amylase

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19. The molecule pictured right is cellulose. (a) To what class of organic molecules does it belong?

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(b) Where is this molecule found?

(c) What is the monomer from which it is made?

(d) Describe features of this molecule that relate to its functional role?

20. Cells use signals (chemical messengers) to communicate and to gather information about, and respond to, changes in their cellular environment. The diagram below shows a form of cell signaling.

Cellulose

Extracellular fluid

A

(a) Describe the basis of cell signaling in terms of a receptor, cell signaling molecule, and a target cell:

Plasma membrane

Cytoplasm

B

(b) Why is it important that not all cells respond to every cell signal?

C

(c) On the diagram, what type of signal molecule is A? Explain your reasoning:

(d) What is the structure labelled B?

(e) What role is structure B playing at point C on the diagram?

(f) What is happening at point D on the diagram?

(g) In general terms, describe the effect of this on the cell:

D

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21. Plant and animal cells are eukaryotic cells, and share several common features. They also have features unique to each cell type. Annotate the following diagrams to compare and contrast the features of plant and animal cells and their functions. You may use extra paper if required. In your answer you should: • Identify each cell type • Identify and describe features common and unique to each.

Cell A

Cell B

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22. Describe the characteristics of the plasma membrane and discuss how substances are transported across it. Include simple diffusion, facilitated diffusion, osmosis, and active transport. You may use extra paper if required.

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DNA and RNA

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Enduring Understanding

3.A 4.A

Key terms adenine

anticodon

base-pairing rule coding strand cytosine DNA

double-helix exons

fibrous protein

3.A.1 DNA and RNA are the sources of heritable information Essential knowledge

(a) Genetic information is transmitted from one generation to the next c 1

Explain how genetic information is stored in and passed to subsequent generations through DNA, or sometimes RNA.

92 98 100

c 2

Contrast chromosome structure in eukaryotes and non-eukaryotes.

93 95

c 3

Describe the nature of plasmid DNA in prokaryotes, eukaryotes, and viruses.

94

c 4

Describe the historical investigations that showed that DNA is the carrier of genetic information.

96

c 5

Describe the semi-conservative replication of DNA and explain how it provides continuity of hereditary information. Identify the role of enzymes the process.

101  -103

c 6

Describe an alternate flow of information, from RNA to DNA, in retroviruses, including the role of reverse transcriptase.

106 116

gene

gene expression

Activity number

genetic code genome

globular protein guanine

guanine cap

hydrogen bonding introns

nucleic acids nucleotides

peptide bond polypeptide

primary transcript protein

proteome purine

pyrimidine

(b) DNA and RNA have similarities and differences that define function

c 1

Describe the structure of a nucleotide and explain how nucleotides are joined to form linear DNA and RNA molecules with specific properties.

98 100

c 2

Identify differences and similarities between DNA and RNA.

98 100

c 3

Describe the base pairing rule and its universal and conserved nature. Distinguish between purines and pyrimidines and explain the significance of the differences.

98 100

c 4

Relate RNA structure to function. Describe the structure and role of mRNA, tRNA, rRNA, and RNAi.

ribosome

RNA (mRNA, rRNA, tRNA) semi-conservative DNA replication template strand thymine

transcription translation uracil

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98 99

(c) Genetic information flows from gene to protein

c 1

Describe transcription of DNA by RNA polymerase to produce mRNA.

104 106

c 2

Describe modifications to the primary mRNA transcript in eukaryotes, including addition of caps and tails and removal of introns.

104 107

c 3

Know that translation of mRNA occurs on ribosomes in the cytoplasm.

104 108

c 4

Know that transcription and translation are coupled processes in prokaryotes. 104 - 106 Describe steps in translation (initiation, elongation, termination). Describe how the 108 109 mRNA is read by ribosomes and the role of tRNAs and start and stop codons.

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(d) Phenotypes are determined through protein activities Explain how the activities of proteins determine the phenotype of an organism. Choose illustrative examples from the following:

110

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c 1

Yikrazuul cc 3.0

• The

role of enzymatic reactions.

• The

regulation of transport processes in cells and organisms.

• The

synthesis or degradation of a protein or proteins.

(e) Genetic engineering techniques can manipulate the heritable information

c 1

Explain how techniques in genetic engineering (aka genetic manipulation) can manipulate heritable information (DNA or, in some cases, RNA).

111

Choose from the following examples of genetic engineering techniques:

c

i Electrophoresis as a technique for identifying DNA fragments.

115

c

ii Plasmid-based transformation of a genome.

117

c

iii Use of restriction enzymes to manipulate and analyze DNA.

c

iv Use of polymerase chain reaction (PCR) to amplify DNA fragments.

c

PR-8

Investigate bacterial transformation using plasmids.

119

c

PR-9

Use restriction enzymes and gel electrophoresis to analyze DNA.

115

Yikrazuul cc 3.0

112 113 114

(f) Illustrative examples of products of genetic engineering include… c

i Genetically modified foods (e.g. rice, soy) or enzymes for food production.

124 - 126

c

ii Transgenic organisms, including their use as biofactories.

120 - 122

c

iii Cloned animals, including cloned transgenics.

122

c

iv Pharmaceuticals such as human insulin or factor X.

125

4.A.1 The subcomponents and sequence of biological molecules determine their properties

Activity number

Essential knowledge

(a) The structure and function of polymers are derived from the way their monomers are assembled c 1

Recall the structure of nucleotide monomers and how these are joined to produce nucleic acids (DNA and RNA), which differ in structure and function.

100

(b) Directionality influences the structure and function of the polymer

Describe the directionality of DNA and RNA, determined by the 3' and 5' carbons of the sugar in the nucleotide. Recall the importance of this during DNA synthesis and transcription.

98 100 101 106

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c 1

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92 Genomes, Genes, and Alleles Every cell in an individual has a complete copy of the genome. Within the genome are sections of DNA, called genes, which code for proteins. Collectively, genes determine what an organism looks like (its traits). Eukaryotes have two copies of each gene (one inherited from each parent), so it is possible for one individual to have two different versions of a gene. These different versions are called alleles.

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Key Idea: A genome is the entire haploid amount of genetic material, including all the genes, of a cell or organism. Eukaryotes can have different versions of a gene (alleles) because they have two copies of each gene. The genome refers to all the genetic material in one haploid set of chromosomes. The genome contains all of the information the organism needs to function and reproduce.

The location and size of the genome varies between organisms Protein coat

Human papillomavirus (HPV)

NIH

Agrobacterium

Human: a eukaryote

MSU

Genome size

Small

Large

Number of genes

Few

Many

The viral genome is contained within the virus's outer protein coat. Viral genomes are typically small and highly variable. They can consist of single stranded or double stranded DNA or RNA and contain only a small number of genes.

In bacteria (prokaryotes) most of the DNA is located within a single circular chromosome, which makes them haploid (i.e. one allele) for most genes. Many bacteria also have small accessory chromosomes called plasmids, which carry genes for special functions such as antibiotic resistance and substrate metabolism.

In eukaryotes, most of the DNA is located inside the cell's nucleus. A small amount resides in the mitochondria and chloroplasts (in plants). The DNA is arranged into linear chromosomes and most eukaryotes are diploid, with two sets of chromosomes, one from each parent.

The HPV genome consists of a double stranded circular DNA molecule ~8000 bp long.

The Agrobacterium genome is 5.7 Mb long and consists of a linear chromosome and two plasmids, one of which enables it to infect plants.

The human genome is ~3000 Mb long in 23 chromosomes. The diploid number is 46 chromosomes.

Genome size is often expressed as the number of base pairs. The unit most often used to show the size of a genome is the megabase (Mb). Note: 1 megabase = 1 million base pairs. The image right is of the φX174 bacteriophage, a virus that infects bacterial cells. Its entire genome is only 5375 base pairs long (0.005375 Mb) and it contains only nine genes, coding for nine different proteins. At least 2000 times this amount of DNA would be found in a single bacterial cell. Half a million times the quantity of DNA would be found in the genome of a single human cell.

Spikes on protein coat

Model of φX174 bacteriophage

Fdardel cc3.0

Measuring genomes

(a) Genome:

(b) Gene:

(c) Allele:

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1. Define the following terms:

2. Describe the general trend for genome size and gene number for viruses, bacteria, and eukaryotic organisms:

3. Explain why an individual eukaryote can have different versions of a gene (allele) but viruses and bacteria do not:

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93 Prokaryotic Chromosomes double stranded DNA, attached to the plasma membrane and located in a nucleoid region, which is in direct contact with the cytoplasm. As well as the bacterial chromosome, bacteria often contain small circular, double-stranded DNA molecules called plasmids. Plasmids are independent of the main bacterial chromosome and usually contain 5-100 genes that are not crucial to cell survival under normal conditions.

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Key Idea: Prokaryote DNA is packaged as one single chromosome that is not associated with protein. DNA is a universal carrier of genetic information but it is packaged differently in prokaryotic and eukaryotic cells. Unlike eukaryotic chromosomes, the prokaryotic chromosome is not enclosed in a nuclear membrane and is not associated with protein. It is a single circular (rather than linear) molecule of

The prokaryotic genome

ff In contrast to eukaryotes, prokaryotic DNA consists almost entirely of protein coding genes and their regulatory sequences. It was the study of prokaryotic genomes that gave rise to the one gene-one protein hypothesis, which still holds true for bacteria.

Proteins associated with the plasma membrane carry out DNA replication and segregate the new chromosome to a daughter cell in cell division.

Single circular chromosome is attached to the plasma membrane and not associated with proteins.

Plasmids occur in the cytoplasm. Plasmids replicate independently of the main chromosome.

Cytoplasm

ff The chromosomal DNA is located in a nucleoid region. It is not enclosed in a membrane. In actively growing cells, the nucleoid may take up as much as 20% of the cell's volume.

ff Most bacteria have a single, circular chromosome. This makes them haploid for most genes, unless copies are located on plasmids (small circular auxiliary DNA strands).

Nucleoid

1 Îźm

Conjugative cell

Recipient

Super-coiled DNA

Sex pilus (pl. pili)

Wiki CC3.0

Linear DNA

Plasmid DNA

Plasmids vary in size from 1000 base pairs (bp) to hundreds of thousands of base pairs. In bacteria, they play an important role in providing extra genetic material that confers properties such as antibiotic resistance. Plasmids can be transferred between bacterial cells by a process of plasmid transfer called conjugation (right)

Conjugation: Special conjugative plasmids contain transfer genes, which enable conjugation and the transfer of genetic information between bacterial cells. The transfer occurs via special sex pili, which form a bridge-like structure between the donor and the recipient. Conjugation provides an alternative route for acquiring genes.

1. State three important ways in which prokaryote chromosomes differ from eukaryotic chromosomes:

(b)

(c)

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(a)

2. Explain the consequences to protein synthesis of the prokaryotic chromosome being free in the cytoplasm:

3. Most of the bacterial genome comprises protein coding genes and their regulatory sequences. What is the consequence of this to the relative sizes of bacterial and eukaryotic genomes:

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94 Plasmid DNA important genes, such as those for the production of toxins that eliminate prokaryote competitors. Plasmids are less common in eukaryotes but some species, such as the yeast Saccharomyces, do have them. The genetic material from viruses may form plasmid-like structures called episomes once they have infected a cell.

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Key Idea: Plasmids are small circular pieces of DNA that are not associated with the chromosomal DNA. Prokaryotes store most of their genetic information in one large chromosome. However, a small percentage can be found as independently replicating, circular, extra-chromosomal pieces of DNA known as plasmids. Plasmids may carry

Spliced gene for a valuable protein

Introduce to bacterium

Features of a plasmid: • Small in size (usually not bigger than 1000-2000 kb) • Forms circular loops • Extra-chromosomal • Self replicating

Grow in culture

Genetically modified organism, e.g. E. coli

Gene product expressed in vat culture

The properties of plasmids have enabled them to be used as tools to introduce novel genetic material into organisms. A gene is spliced into a plasmid and the plasmid inserted into a recipient organism (e.g. a bacterium). The organism will then produce the gene product as part of its normal metabolism. This technique has enabled the industrial-scale production of valuable gene products, such as human insulin, from GMOs. The pLW1043 plasmid

The make-up of plasmids

T

T

C

Cytokinin production (cell proliferation)

The Ti plasmid Auxin production

T Trimethoprim resistance

V Vancomycin resistance resistance D Disinfectant pLW1043 plasmid D S Streptomycin D Disinfectant resistanceresistance

pLW1043 plasmid

P

P

PD Penicillin resistance S Streptomycin

V

Border of transfer region (25 bp)

Border of transfer region (25 bp)

Catabolism of opines (used as a nitrogen and energy source by Agrobacterium)

Origin of replication

Virulence region A - H

C Conjugative ability

C T Trimethoprim resistance V Vancomycin resistance

Opine synthesis

Transfer DNA region (transferred to host cell) (20,000 bp)

Genes for Genes for C Conjugative ability

S

V

resistance

P Penicillin resistance

S

Plasmids often carry genes for beneficial traits, including antibiotic resistance and the ability to use new substrates. The ability of bacteria to exchange plasmids has contributed to the spread of antibiotic resistance. The pLW1043 plasmid (above) from the superbug Staphylococcus aureus (MRSA) carries several genes for antibiotic resistance, acquired progressively over time.

The bacterium Agrobacterium tumefaciens often contains the Ti (tumor inducing) plasmid. This plasmid is able to transfer genes into plant cells and cause disease. Several regions on the plasmid (left) help it to infect plants. The plasmid is just over 200,000 bp long and contains 196 genes. The mapping of its genes has made it of great importance in creating genetically modified plants.

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1. What is a plasmid?

2. Explain how a plasmid can convey a survival advantage to bacteria under certain conditions:

3. (a) Why are plasmids (such as the Ti plasmid) useful to genetic engineers?

(b) Into which region of the Ti plasmid would you insert a gene in order for it to be transferred into a host plant cell?

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95 Eukaryotic Chromosome Structure DNA efficiently so that it can fit into the nucleus. Prior to cell division, the chromatin is at its most compact, forming the condensed metaphase chromosomes that can be seen with a light microscope.

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Key Idea: A chromosome consists of DNA complexed with proteins to form a highly organized, tightly coiled structure. The DNA in eukaryotes is complexed with histone proteins to form chromatin. The histones assist in packaging the

EII

SEM

The packaging of chromatin

Chromatin is the combination of DNA and proteins that make up the contents of the cell nucleus. Chromatin structure is based on successive levels of DNA packing. Histone proteins are responsible for packing the DNA into a compact form. Without them, the DNA could not fit into the nucleus. Five types of histone proteins form a complex with DNA, in a way that resembles “beads on a string”. These beads, or nucleosomes, form the basic unit of DNA packing.

2 nm

DNA molecule

A cluster of human chromosomes seen during metaphase of cell division. Individual chromatids (arrowed) are difficult to discern on these double chromatid chromosomes.

Each bead has two molecules of each of four types of histone (H2A, H2B, H3, and H4)

Chromatid

DNA

10 nm

Nucleosomes

Chromatid

Centromere Chromatin fibers

Chromosome TEM

The nucleosome bead consists of DNA wrapped around a protein core

A fifth histone, H1, attaches near the bead and organizes the next level of packing.

Banding

Histone H1 helps the beaded string to coil to form a chromatin fiber roughly 30 nm thick.

Human chromosome 3

A human chromosome from a dividing white blood cell (above left). Note the compact organization of the chromatin in the two chromatids. The LM photograph (above right) shows the banding visible on human chromosome 3.

30 nm chromatin fiber

30 nm

The 30 nm fiber organized by H1 forms loops called looped domains

The looped domains are attached to a scaffold of non-histone protein.

300 nm

BF

TEM

Looped domains of the 30 nm fiber

In non-dividing cells, chromosomes exist as single-armed structures. They are not visible as coiled structures, but are 'unwound' to make the genes accessible for transcription (above).

Looped domains coil and fold

Looped domains

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The looped domains coil and fold....

700 nm

EII

1400 nm

The evidence for the existence of looped domains comes from the study of giant lampbrush chromosomes in amphibian oocytes (above). Under electron microscopy, the lateral loops of the DNA-protein complex appear brushlike. ©2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

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Metaphase chromosome

...making the chromatin even more compact and producing the characteristic metaphase chromosome.

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132 Exons and introns

Only about 2% of the DNA in humans codes for proteins. Much of the rest codes for RNA (e.g. tRNA) or has a regulatory function, e.g. the centromere or telomeres. In general, the genomes of more complex organisms contain much more of this so-called "non protein-coding" DNA. Even within protein coding sequences, parts of the DNA are excised from the primary transcript to create the mRNA that codes for the protein to be translated.

Most protein-coding genes in eukaryotic DNA are not continuous. The protein-coding regions (exons) are interrupted by nonprotein-coding regions called introns. Introns are edited out of the protein-coding sequence prior to translation (protein synthesis). After processing, the introns may go on to serve a regulatory function.

Alex Popovkin

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Non-coding regions

The bladderwort, Utricularia gibba, has just 3% non-coding DNA.

98% of the DNA in humans is non-coding DNA.

To help you remember, exons are expressed and introns are in-between the protein-coding regions.

Exons: protein-coding regions

DNA

Intron

Intron: edited out during protein synthesis

Intron

1. Where is the DNA located in eukaryotes?

2. Why does DNA need to be tightly packaged?

3. How do histone proteins help in the coiling of DNA?

4. Explain the significance of the following terms used to describe the structure of chromosomes:

(a) DNA:

(b) Chromatin:

(c) Nucleosome:

(d) Chromatid:

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5. What is the difference between an exon and an intron?

6. Each human cell has about a 1 metre length of DNA in its nucleus. Discuss the mechanisms by which this DNA is packaged into the nucleus and organized in such a way that it does not get ripped apart during cell division:

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96 Does DNA Really Carry the Code? function suggested they could account for the many traits we see in organisms. Two early experiments, one by Griffith and another by Avery, MacLeod, and McCarty, provided important information about how traits could be passed on and what cellular material was responsible. The experiments involved strains of the bacterium Streptococcus pneumoniae. The S strain is pathogenic (causes disease). The R strain is harmless. Later experiments by Hershey and Chase helped confirm DNA as the genetic material.

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Key Idea: The progressive studies of several scientists helped to establish DNA as the heritable material responsible for the characteristics we see in organisms. Many years before Watson and Crick discovered the structure of DNA, biologists had deduced, through experimentation, that DNA carried the information that was responsible for the heritable traits we see in organisms. Prior to the 1940s, it was thought that proteins carried the code. Little was known about nucleic acids and the variety of protein structure and

Griffith (1928)

ff Griffith found that when he mixed heat-killed pathogenic bacteria with living harmless cells, some of the living cells became pathogenic. Moreover, the newly acquired trait of pathogenicity was inherited by all descendants of the transformed bacteria. He concluded that the living R cells had been transformed into pathogenic cells by a heritable substance from the dead S cells.

Living R cells (no capsule, harmless)

Mouse lives

Living S cells (capsule, pathogenic)

Mouse dies

Heat-killed S cells

Mouse lives

Heat-killed S cells and living R cells

Mouse dies Blood sample contained living S cells that could reproduce to yield more S cells.

Avery-MacLeod-McCarty (1944)

ff What was the unknown transformation factor in Griffith's experiment? Avery designed an experiment to determine if it was RNA, DNA, or protein. He broke open the heat-killed pathogenic cells and treated samples with agents that inactivated either protein, DNA, or RNA. He then tested the samples for their ability to transform harmless bacteria.

Treat sample with enzyme that destroys RNA, protein, or DNA.

Add the treated samples to cultures of R bacterial cells

RNase

Protease

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S and R strains

S and R strains

Heat-killed S cells, remove lipids and carbohydrates, homogenize and filter

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DNase

R strain only

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Hershey and Chase (1952)

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ff The work of Hershey and Chase in 1952 followed the work of Avery and his colleagues and was instrumental in the acceptance of DNA as the hereditary material. Hershey and Chase worked on viruses that infect bacteria called phages. Phages are composed only of DNA and protein. When they infect, they inject their DNA into the bacteria, leaving their protein coat outside.

ff Hershey and Chase had two batches of phage. Batch 1 phage were grown with radioactively labeled sulfur, which was incorporated into the phage protein coat. Batch 2 phage were grown with radioactively labeled phosphorus, which was incorporated into the phage DNA. The phage were mixed with bacteria, which they then infected. When the bacterial cells were separated from the liquid, Hershey and Chase studied the liquid and the bacteria cells to determine where the radioactivity was.

ff Hershey and Chase were able to show that DNA is the only material transferred directly from the phage to the bacteria when the bacteria are infected by the viruses.

Protein

The radioactivity is in the supernatant

Radioactive protein capsule (35S)

Power

Speed

BBT Inc.

Radioactive DNA (32P)

DNA

The radioactivity is in the pellet

Power

Speed

BBT Inc.

Blender separates phage outside the bacteria from the cells and their contents.

Cells and phage separated using centrifugation. Cells and their DNA form the pellet.

1. Griffith did not predict transformation in his experiment. What results was he expecting? Explain:

2. (a) What did Avery's experimental results show?

(b) How did Avery's experiment build on Griffith's findings?

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3. (a) How did the Hershey-Chase experiment provide evidence that nucleic acids, not protein are the hereditary material?

(b) What assumptions were made about the role of DNA in viruses and bacteria and the role of DNA in eukaryotes?

(c) How would the results of the experiment have differed if proteins carried the genetic information?

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97 The Evidence for DNA Structure in different areas many years to determine DNA's structure. In particular, four scientists, Watson, Crick, Franklin, and Wilkins are recognized as having made significant contributions in determining the structure of DNA. Once the structure of DNA was known scientists could determine how it was replicated.

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Key Idea: Many scientists contributed to the discovery of DNA's structure. Once the structure of DNA was known, it immediately suggested a mechanism for its replication. DNA is easily extracted and isolated from cells. This was first done in 1869, but it took the work of many scientists working Late 1940s Linus Pauling determined by x-ray crystallography that proteins have a helical structure.

1949 Chargaff’s rules are announced: DNA contains equal proportions of bases A and T and G and C.

1951 Rosalind Franklin studied the structure of DNA using x-ray crystallography at King's College. Maurice Wilkins, also at King's College, was also using x-ray crystallography to study DNA. Wilkins and Franklin did not get on well.

1951 James Watson and Francis Crick build their first DNA model - a three stranded helix, with bases to the outside and phosphate groups to the inside. Franklin points out that their model is incorrect and is not consistent with the data.

30 January 1953 Wilkins shows Watson and Crick the "photo 51" without Franklins' approval or knowledge. It provided the structural information they needed to finalize their model, completed on 7 March 1953.

May 1952 Franklin produces "photo 51", showing DNA is a helix. She was working on a less hydrated form of DNA and did not return to the photo again until 1953.

16 April 1958 Franklin dies at age 37 of ovarian cancer. She was never nominated for a Nobel Prize.

1962 Watson, Crick, and Wilkins win the Nobel Prize in Physiology or Medicine. Franklin was not acknowledged.

Discovering the structure of DNA ... a story of collaboration and friction

Although Watson and Crick are often credited with discovering DNA's structure, the contributions of many scientists were important and personal conflicts and internal politics probably prevented the structure being determined earlier. Professional friction between Franklin and Wilkins meant that they worked independently of each other. Watson and Crick analyzed Franklin's results without her knowledge or consent and Watson himself recalls that he tended to dismiss her. Only later did he acknowledge her considerable contribution. Franklin did not receive the Nobel prize, which cannot be awarded posthumously.

A

A

Photo 51

Wilkins and Franklin were both crystallographers. Their X-ray diffraction patterns of DNA provided measurements of different parts of the molecule and the position of different groups of atoms. Franklin's X-ray image (photo 51) gave Watson and Crick the necessary information to produce their model of DNA. As soon as Franklin saw it, she readily accepted it as correct.

1. What made Watson and Crick realize that DNA was a double helix?

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Distinct parts of the famous "photo 51" X-ray diffraction image (recreated in the illustration right) indicate specific qualities of the DNA. The X pattern indicates a helix, but Watson and Crick realized that the apparent gaps in the X (labeled A) were due to the repeating pattern of a double helix. The diamond shapes (in blue) indicate the helix is continuous and of constant dimensions and that the sugar-phosphate backbone is on the outside of the helix. The distance between the dark horizontal bands allows the calculation of the length of one full turn of the helix.

2. Do you regard the discovery of DNA's structure as a collaboration between scientists working in related fields? Explain:

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98 RNA Molecules

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136

mRNA) where the information can be converted to useful materials (via rRNA and tRNA). RNA is also involved in regulating gene expression by regulating mRNA and can form catalytic structures called ribozymes which can act in similar (but more limited) ways to enzymes.

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Key Idea: RNA plays vital roles in transcribing and translating DNA, forming messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). RNA (ribonucleic acid) is involved in processes in the cell that relate to the nuclear material (DNA and RNA). It is involved in conveying information from the DNA to the cytoplasm (via

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.

Dicer catalyzes the cleavage of dsRNA to siRNAs and miRNAs.

Ribosomal RNA (above) forms ribosomes from two separate ribosomal components (the large and small subunits) and assembles amino acids into a polypeptide chain. Ribozymes within the ribosome help with its function of producing polypeptide chains.

Opabinia regalis, cc 3.0

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.

RNAs contain selfcomplementary sequences that allow parts of the RNA to pair with itself to form short helices joined by H bonds.

RNA interference (RNAi) regulates gene expression through miRNAs (microRNA) and siRNAs (short interfering RNA), which bind to specific mRNA sequences, causing them to be cleaved. RNAi is important in gene regulation during development and in defense against viruses, which often use double-stranded RNA as an infectious vector.

(a)

(b)

(c)

(d)

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1. Briefly describe four specific roles of RNA:

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99 RNAi Can Silence Gene Expression sequences, causing them to be cleaved. RNAi is important in regulating gene expression during development and in defense against viruses, which often use double-stranded RNA as an infectious vector. Regulation of translation is achieved by destroying specific mRNA targets using short RNA lengths, which may be exogenous (short interfering RNAs) or endogenous (microRNAs). Mechanisms of RNA interference (RNAi) are illustrated below.

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Key Idea: RNA interference (RNAi) regulates gene expression through miRNAs and siRNAs, which act to silence genes. RNA plays vital roles in transcribing and translating DNA, forming messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). RNA is also involved in modifying mRNA after transcription and regulating translation. RNA interference (RNAi) regulates gene expression through miRNAs and siRNAs, which bind to specific mRNA

MicroRNA (miRNA) is formed from hairpin loops of nonprotein-coding RNA copied from endogenous (within the cell) DNA. Cell nucleus

Short interfering RNA (siRNA) is formed from exogenous dsRNA, such as the genetic material from a virus.

Endogenous DNA

Exogenous (outside the cell) double stranded RNA (dsRNA)

dsRNA

Dicer cuts the dsRNA into siRNA. These are short lengths of dsRNA with a two bp overhang on the 3' end.

siRNA

Hairpin loops of dsRNA

The dicer enzyme cuts the hairpin loops into short lengths of miRNA.

RISC incorporates siRNA to form a complex in the same way as for miRNA. It forms ss siRNA, which can bind to target mRNA.

RISC

The RNA induced silencing complex (RISC) is a multiprotein complex that incorporates the miRNA and mediates its unwinding to a form a single stranded miRNA.

mRNA

siRNA

miRNA

RISC

mRNA

The RISC complex binds to complementary sequences of target mRNA, cleaves it, and so stops its translation. Up to 30% of human proteincoding genes may be regulated by miRNAs.

Cleaved mRNA

miRNA

The cleaved mRNA is recognized by the cell as abnormal, and is destroyed. This prevents it from being translated.

Cleaved mRNA

2. How does RNAi regulate gene expression?

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1. Distinguish between siRNA and miRNA:

3. Under-expression of microRNAs is associated with tumor formation (uncontrolled cell growth). Based on your understanding of the role of microRNAs, account for this association:

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100 Creating a DNA Molecule bases. These factors cause the nucleotides to join together in a predictable way, referred to as the base pairing rule. The strands of the DNA are antiparallel (they run in opposite directions) and there are 10 base pairs per 360° turn of the helix. The activity below will guide you through constructing a three dimensional model of DNA.

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Key Idea: Nucleotides pair according to the base pairing rule. There are ten base pairs per turn of the DNA double helix. DNA is made up of structures called nucleotides. Two primary factors control the way in which these nucleotide building blocks are linked together: the available space within the DNA double helix and the hydrogen-bonding capability of the

DNA base pairing rule

Adenine

is always attracted to Thymine A T

Thymine

is always attracted to Adenine

T A

Cytosine

is always attracted to Guanine

C G

Guanine

is always attracted to Cytosine G C

1. Cut out page 139. Cut out the template strand. Dark black lines should be cut. Fold on the blue dotted lines so that the gray surfaces are facing (a valley fold).

2. Cut out the complementary strand. The first base (G) is already in position as a guide. Again fold on the blue line so that the blue surfaces are facing each other. 3. (a) To help you place the remaining bases in the correct order on the complementary strand, fill in the table below.

Complementary strand Template strand Cytosine (C)

Guanine (G)

Guanine (G) Thymine (T) Adenine (A)

Thymine (T) Adenine (A)

Thymine (T) Thymine (T)

Cytosine (C) Guanine (G)

A finished model

(b) Cut out the bases and slot them into the slots on the complementary strand using the order in the table above. Use short lengths of tape to fix them in position. Make sure the blue surfaces are facing and the base is in the same orientation as the guide (G).

4. Line up the first base pairs (C and G) and stick them together with tape. Note that the bases are facing in opposite directions.

6. What does antiparallel mean?

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5. Continue sticking base pairs together, working your way around the helix, to complete the DNA molecule.

7. Record how your model is similar to a DNA molecule and any differences you have noticed:

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PRACTICES

PRACTICES

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G

A

T

C

A

Template strand

A

G

A

A

T

T

A

T

Bases

G

C

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139

Bases

T T

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Complementary strand

G

C

C

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140

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The page has been deliberately left blank

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101 DNA Replication Key Idea: Semi conservative DNA replication produces two identical copies of DNA, each containing half original material and half new material. Before a cell can divide, it must double its DNA. It does this by a process called DNA replication. This process ensures that each resulting cell receives a complete set of genes from the

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original cell. After the DNA has replicated, each chromosome is made up of two chromatids, joined at the centromere. The two chromatids will become separated during cell division to form two separate chromosomes. During DNA replication, nucleotides are added at the replication fork. Enzymes are responsible for all of the key events.

Step 1 Unwinding the DNA molecule

3'

A normal chromosome consists of an unreplicated DNA molecule. Before cell division, this long molecule of double stranded DNA must be replicated.

Helicase at the replication fork

es

is

DNA polymerase

io

n

DNA polymerase catalyzes the condensation reaction that joins adjacent nucleotides. The strand is synthesized in a 5' to 3' direction, with the polymerase moving 3' to 5' along the strand it is reading. Thus the nucleotides are assembled in a continuous fashion on one strand but in short fragments on the other strand. These fragments are later joined by an enzyme to form one continuous length.

Free nucleotides are used to construct the new DNA strand.

th

The formation of new DNA is carried out mostly by an enzyme complex called DNA polymerase.

Temporary break allows the strand to swivel

yn

Step 2 Making new DNA strands

Single-armed chromosome as found in a non-dividing cell.

of s

For this to happen, it is first untwisted and separated (unzipped) at high speed at its replication fork by an enzyme called helicase. Another enzyme relieves the strain that this generates by cutting, winding and rejoining the DNA strands.

5'

D ire ct

Step 3 Rewinding the DNA molecule

Each of the two new double-helix DNA molecules has one strand of the original DNA (dark gray and white) and one strand that is newly synthesized (blue). The two DNA molecules rewind into their double-helix shape again.

Each of the newly formed DNA molecules create a chromatid.

3'

5'

3'

The two new strands of DNA coil into a double helix

Some viruses, called retroviruses, store their genetic material as RNA and use the enzyme reverse transcriptase to copy it to DNA when they infect a cell.

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DNA replication is semi-conservative, with each new double helix containing one old (parent) strand and one newly synthesized (daughter) strand. The new chromosome has twice as much DNA as a non-replicated chromosome. The two chromatids will become separated in the cell division process to form two separate chromosomes.

5'

Replicated chromosome ready for cell division.

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1. What is the purpose of DNA replication?

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2. Summarize the three main steps involved in DNA replication: (a)

(b)

(c)

3. For a cell with 22 chromosomes, state how many chromatids would exist following DNA replication: 4. State the percentage of DNA in each daughter cell that is new and the percentage that is original:

5. What does it mean when we say DNA replication is semi-conservative?

6. How are the new strands of DNA lengthened during replication?

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

8. Why does one strand of DNA need to be copied in short fragments?

9. 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 or meiosis 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 DNA.

DNA replication...

...during mitosis.

The chromatids separate...

...is copied to produce two identical DNA strands.

A chromatid contains half original ...

...is made up of two chromatids.

Write the complete paragraph here:

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The enzymes also proofread the DNA during replication...

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102 Enzyme Control of DNA Replication Key Idea: The process of DNA replication is controlled by many different enzymes. DNA replication involves many enzyme-controlled steps. They are shown below as separate, but many of the enzymes are clustered together as enzyme complexes. As the DNA

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is replicated, enzymes ‘proof-read’ it and correct mistakes. The polymerase enzyme can only work in one direction, so that one new strand is constructed as a continuous length (the leading strand) while the other new strand (the lagging strand) is made in short segments to be later joined together.

5'

3'

Double strand of original (parental) DNA

Overall direction of replication

DNA replication occurs during interphase of the cell cycle at an astounding rate. As many as 4000 nucleotides per second are replicated. This explains how bacterial cells, with as many as 4 million nucleotides, can complete a cell cycle in about 20 minutes. Note that the nucleotides are present as deoxynucleoside triphosphates. When hydrolyzed, these provide the energy for incorporating the nucleotide into the strand.

1

2

Gyrase: Relieves tension on the DNA strand ahead of the replication fork.

Helicase: Unwinds and separates the double stranded DNA molecule 3

Swivel point

DNA primase is an RNA polymerase which synthesizes a short RNA primer, which is later removed.

4 Single-strand binding proteins: Prevent single strands from rewinding and protect them from cleavage.

DNA polymerase III adds nucleotides in the 5' to 3' direction so the leading strand is synthesized continuously in this direction

5

DNA polymerase III: Extends RNA primer with short lengths of complementary DNA 6

RNA primers

Parental strand provides a 'template' for the new strand's synthesis

DNA polymerase I: Digests RNA primer and replaces it with DNA

Replication fork

D

n

io

ct

ire

The lagging strand is formed in fragments, 1000-2000 nucleotides long. These Okazaki fragments, are later joined together.

DNA ligase: Joins neighboring fragments together

D

ire

ct

io

n

of

sy

nt

he

si

s

7

of

3'

5'

nt

sy

si

he

5'

s

3'

1. Describe the general role of enzymes in DNA replication:

(a) Helicase:

(b) DNA polymerase I:

(c) DNA polymerase III:

(d) Ligase:

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2. State the specific role of each of the following enzymes in DNA replication:

3. Determine the time it would take for a bacterium to replicate its DNA (see note in diagram above):

4. How is the energy for incorporating the nucleotides into the strand provided?

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103 Meselson and Stahl's Experiment Key Idea: Meselson and Stahl devised an experiment that showed DNA replication is semi-conservative. Three models were proposed to explain how DNA replicated. Watson and Crick proposed the semi-conservative model in which each DNA strand served as a template, forming a new DNA molecule that was half old and half new DNA. The

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conservative model proposed that the original DNA served as a complete template so that the resulting DNA was completely new. The dispersive model proposed that the two new DNA molecules had part new and part old DNA interspersed throughout them. Meselson and Stahl's experiment confirmed that DNA replication is semi-conservative.

Meselson and Stahl's experiment

E. coli were grown for several generations in a medium containing a heavy nitrogen isotope (15N). Once all the bacterial DNA contained 15N, the bacteria were transferred to a medium containing a light nitrogen isotope (14N). DNA newly synthesized in the 14N medium would contain 14N, old DNA would contain 15N.

1

E. coli

3

2

Sample

14N solution (NH4Cl)

15N

DNA extraction

14N

Excess in solution

solution

E. coli were grown in a nutrient solution containing 15N. After 14 generations, all the bacterial DNA contained 15N. A sample was removed. This was generation 0.

CsCl solution

Generation 0 was added to a solution with excess 14N (as NH4Cl). During replication, new DNA would incorporate 14N and be 'lighter' than the original DNA (which had only 15N).

Each generation (~ 20 minutes), a sample was taken and treated to release the DNA. The DNA was placed in a CsCl solution which provided a density gradient for separating the DNA.

Models for DNA replication

5

4

Extracted DNA

Extracted DNA in CsCl solution

Generation

0

1

2

1000

0

RPM

2000

1000

0

Timer

RPM

2000

On/Off

Speed

Samples were spun in a high speed ultracentrifuge at 140,000 g for 20 h. Heavier 15N DNA moved closer to the bottom of the test tube than light 14N DNA or 14N/ 15N intermediate DNA.

Timer

All the DNA in the generation 0 sample moved to the bottom of the test tube. All the DNA in the generation 1 sample moved to an intermediate position. At generation 2, half the DNA was at the intermediate position and half was near the top of the test tube. In subsequent generations, more DNA was near the top and less was in the intermediate position.

1. Describe each of the DNA replication models:

(a) Conservative:

(b) Semi-conservative:

(c) Dispersive:

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Conservative

Semiconservative

Dispersive

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145 Conservative Generation 0

Generation 1

Generation 2

Conservative

Semi-conservative

Dispersive

Heavy

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2. Three models of the mechanism of DNA replication are shown below. In the blue boxes right and below write down the percentage of heavy (15N 15N), intermediate (14N 15N), and light (14N 14N) DNA for each generation of each model.

Intermediate

2

Light

1

0

Generation

DNA containing 14N

DNA containing 15N

% Generation 0

Semiconservative

% Generation 1

% Generation 2

Dispersive

Generation 1

Generation 2

% Generation 0

% Generation 1

% Generation 2

Generation 0

Generation 1

Generation 2

% Generation 0

% Generation 1

% Generation 2

Light

Light

Intermediate

Intermediate

Heavy

Heavy

Generation 0

I

H

50% Light (L) DNA + 50% intermediate 14N/15N DNA. 14N

2

1

0

100% Intermediate (I) 15N /14N DNA

(a) Compare the models above with the results gained by Meselson and Stahl to decide which of the three DNA replication models is supported by the data:

(b) Was Watson and Crick's proposal correct?

4. Identify the replication model that fits the following data: 100% of generation 0 is "heavy DNA", 50% of generation 1 is "heavy" and 50% is 100% Heavy (H) "light", and 25% of generation 2 is "heavy" and 75% is "light": 15 N DNA

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3. The results from Meselson's and Stahl's are shown graphically left. All of the generation 1 DNA contained one light strand (14N) and one heavy strand (15N) to produce an intermediate density. At generation 2, 50% of the DNA was light and 50% was intermediate DNA.

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104 What is Gene Expression? protein. Eukaryotic genes include non-protein coding regions called introns. These regions of intronic DNA must be edited out before the mRNA is translated by the ribosomes. Transcription of the genes and editing that primary transcript to form the mature mRNA occurs in the nucleus. Translation of the protein by the ribosomes occurs in the cytoplasm.

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Key Idea: Genes are sections of DNA that code for proteins. Genes are expressed when they are transcribed into messenger RNA (mRNA) and then translated into a protein. Gene expression is the process by which the information in a gene is used to synthesize a protein. It involves transcription of the DNA into mRNA and translation of the mRNA into

A summary of eukaryotic gene expression

mRNA

TRANSLATION

Nucleus

EDITING

TRANSCRIPTION

Ribosome

Nuclear pore

Primary transcript

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.

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.

Because prokaryotes do not have a nucleus, translation is coupled to transcription. This means translation can happen while the mRNA is still being synthesized. Also, because there are no introns in prokaryotic DNA, there is no need to edit the primary mRNA.

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?

(iii)

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(i) (ii)

3. The photograph right shows an SEM of a giant polytene chromosome. These chromosomes are common in the larval stages of flies and form as a result of repeated cycles of DNA replication without cell division. This creates many copies of genes. Within these chromosomes, visible 'puffs' indicate regions where there is active transcription of the genes.

(a) What is the consequence of active transcription in a polytene chromosome?

(b) Why might this be useful in a larval insect?

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105 The Genetic Code Key Idea: The genetic code is the set of rules by which the genetic information in DNA or mRNA is translated into proteins. The genetic information for the assembly of amino acids is stored as three-base sequence. These three letter codes on mRNA are called codons. Each codon represents one of 20 amino acids used to make proteins. The code is effectively

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universal, being the same in all living things (with a few minor exceptions). The genetic code is summarized in a mRNAamino acid table, which identifies the amino acid encoded by each mRNA codon. The code is degenerate, meaning there may be more than one codon for each amino acid. Most of this degeneracy is in the third nucleotide of a codon.

Amino acid

Ala

Alanine

Arg

Codons that code for this amino acid

No.

4

Leu

Leucine

Arginine

Lys

Lysine

Asn

Asparagine

Met

Methionine

Asp

Aspartic acid

Phe

Phenylalanine

Cys

Cysteine

Pro

Proline

Gln

Glutamine

Ser

Serine

Glu

Glutamic acid

Thr

Threonine

Gly

Glycine

Trp

Tryptophan

His

Histidine

Tyr

Tyrosine

Ile

Isoleucine

Val

Valine

GCU, GCC, GCA, GCG

Codons that code for this amino acid

Amino acid

No.

1. Use the mRNA-amino acid table (below) to list in the table above all the codons that code for each of the amino acids and the number of different codons that can code for each amino acid (the first amino acid has been done for you). 2. (a) How many amino acids could be coded for if a codon consisted of just two bases?

(b) Why is this number of bases inadequate to code for the 20 amino acids required to make proteins?

3. Describe the consequence of the degeneracy of the genetic code to the likely effect of a change to one base in a triplet:

Read second letter here

U

C on the left row, A on the top column, G on the right row CAG is Gln (glutamine)

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C

U

UUU UUC UUA UUG

Phe Phe Leu Leu

UCU UCC UCA UCG

C

CUU CUC CUA CUG

Leu Leu Leu Leu

CCU CCC CCA CCG

A

AUU AUC AUA AUG

Ile Ile Ile Met

ACU ACC ACA ACG

G

GUU GUC GUA GUG

Val Val Val Val

GCU GCC GCA GCG

A

G

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Read first letter here

Example: Determine CAG

Read third letter here

Second letter

Ser Ser Ser Ser

UAU UAC UAA UAG

Tyr Tyr STOP STOP

UGU UGC UGA UGG

Cys Cys STOP Trp

U C A G

Pro Pro Pro Pro

CAU CAC CAA CAG

His His Gln Gln

CGU CGC CGA CGG

Arg Arg Arg Arg

U C A G

Thr Thr Thr Thr

AAU AAC AAA AAG

Asn Asn Lys Lys

AGU AGC AGA AGG

Ser Ser Arg Arg

U C A G

Ala Ala Ala Ala

GAU GAC GAA GAG

Asp Asp Glu Glu

GGU GGC GGA GGG

Gly Gly Gly Gly

U C A G

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Third letter

How to read the table: The table on the right is used to 'decode' the genetic code as a sequence of amino acids in a polypeptide chain, from a given mRNA sequence. To work out which amino acid is coded for by a codon (triplet of bases) look for the first letter of the codon in the row label on the left hand side. Then look for the column that intersects the same row from above that matches the second base. Finally, locate the third base in the codon by looking along the row from the right hand end that matches your codon.

First letter

mRNA-amino acid table

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106 Transcription is the First Step in Gene Expression protein-coding portion 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 must first be removed and the remaining exons spliced together to form the mature mRNA before the gene can be translated into a protein. This editing process also occurs in the nucleus.

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Key Idea: Transcription is the first step of gene expression. It involves the enzyme RNA polymerase rewriting the information into a primary RNA transcript. In eukaryotes, transcription takes place in the nucleus. Transcription is the first stage of gene expression. It takes place in the nucleus and is carried out by the enzyme RNA polymerase, which rewrites the DNA into a primary RNA transcript using a single template strand of DNA. The

Transcription is carried out by RNA polymerase (RNAP) Coding (sense) strand of DNA

5'

3'

RNA polymerase (RNAP) adds nucleotides to the 3' end so the strand is synthesized 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 synthesized RNA strand is complementary to the template strand

RNA polymerase dissociates at the terminator region

3'

5'

The primary RNA transcript is edited to form the mature mRNA and then passes to the cytoplasm where the nucleotide sequence is translated into a polypeptide.

Translation will begin at the start codon AUG

5'

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 colored pen to mark the beginning and end of the protein-coding region being transcribed.

2. (a) In which direction is the RNA strand synthesized?

(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? WEB

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149 Comparing gene expression in prokaryotes and eukaryotes

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In both prokaryotes and eukaryotes, genes are transcribed by RNA polymerase and translated by ribosomes. However, there are some important differences. In eukaryotes, primary RNA must be edited and processed before passing from the nucleus to the cytoplasm. In prokaryotes, there is no nucleus and ribosomes can begin translating a gene while it is still being transcribed.

DNA

Nucleus

DNA

RNAP

Transcription

5'

mRNA

RNAP

Cytoplasm

Transcription and processing 3'

mRNA

Transport

5'

Translation by ribosomes

3'

Translation by ribosomes

5'

Growing polypeptide

Prokaryote

Eukaryote

4. For the following triplets on a DNA template strand, state the codon sequence for the mRNA that would be synthesized: Triplets on the DNA:

T A C

Codons on the mRNA: Triplets on the DNA:

T A C

T A G

C C G

C G A

T T T

A A G

C C T

A T A

A A A

Codons on the mRNA: 5. What is the significance of the promoter and terminator regions on the DNA?

6. Why might a cell employ several RNA polymerases to produce multiple RNA transcripts of a gene at any one time?

7. Based on the diagram above, describe two differences between gene expression in prokaryotes and eukaryotes:

(b)

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(a)

8. Based on the diagram above, how is gene expression in prokaryotes and eukaryotes similar?

9. What benefit might there be to the way in which prokaryotes can begin translation while a gene is still being transcribed?

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107 mRNA Processing in Eukaryotes cytoplasm. Modifications to the 5' and 3' ends of the transcript enable the mRNA to exit the nucleus and remain stable long enough to be translated. Other post transcriptional modifications remove non-protein coding intronic DNA and splice exons in different combinations to produce different protein end products.

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Key Idea: Primary mRNA molecules are modified after transcription so that the mRNA can exit the nucleus. Post transcriptional modification also enables the cell to produce a wide variety of proteins from a smaller number of genes. Once a gene is transcribed, the primary transcript is modified to produce the mRNA strand that will be translated in the

Primary RNA is modified by the addition of caps and tails

ff After transcription, both ends of the primary RNA are modified by enzymes to create 'caps' and 'tails' (below). These modifications protect the RNA from degradation and help its transport through the nuclear pore. Primary RNA

5' modification (capping) A guanine nucleotide cap is added to the 5' end of the primary transcript to protect it from degradation during transport from the nucleus to the cytoplasm.

3' modification (poly-A tails) Multiple adenosine nucleotides are added to the primary transcript. These poly-A tails aid nuclear export, translation, and stability of the mRNA.

Post transcriptional modification

ff Human DNA contains 25,000 genes, but produces up to one million different proteins. Each gene must therefore produce more than one protein. This is achieved through both post-transcriptional modification of the mRNA as well as post translational modifications, such as glycosylation and addition of phosphates.

ff Primary RNA contains both protein coding exons and non-protein coding introns. Introns are usually removed after transcription and may be processed to create regulatory elements such as microRNAs. The exons are then spliced together ready to be translated. However, there are many alternative ways to splice the exons and these alternatives create variations in the translated proteins. The most common method of alternative splicing involves exon skipping, in which not all exons are spliced 1 (below). Other splicing Exon 2 Exonvariants. 3 Exon 4 Exon 5 into the finalExon mRNA options create further

Exon 1

Exon 2

Intron

Exon 3

Exon 4

Intron

Exon 5

Intron

Intron

Splicing

1

1

2

2

3

3

4

4

5

5

1

1

2

2

4

4

5

5

1

1

2

2

3

3

5

5

Three splicing alternatives creates three different proteins

2. (a) What happens to the intronic sequences in DNA after transcription?

(b) What is one possible fate for these introns?

3. How can so many proteins be produced from so few genes?

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1. What is the purpose of the caps and tail on mRNA?

4. If a human produces 1 million proteins, but human DNA codes for only 25,000 genes, on average how many proteins are produced per gene?

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108 Translation molecules, providing the catalytic environment for the linkage of amino acids delivered by transfer RNA (tRNA) molecules. Protein synthesis begins at the start codon and, as the ribosome wobbles along the mRNA strand, the polypeptide chain elongates. On reaching a stop codon, the ribosome subunits dissociate from the mRNA, releasing the protein.

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Key Idea: Translation is the final stage of gene expression in which ribosomes read the mRNA and decode (translate) it to synthesize a protein. This occurs in the cytoplasm. In eukaryotes, translation occurs in the cytoplasm either at free ribosomes or ribosomes on the rough endoplasmic reticulum. Ribosomes translate the code carried in the mRNA

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

Anticodon is a 3-base sequence complementary to the codon on mRNA.

subunit Functional Small ribosome

1. Describe the structure of a ribosome:

2. What is the role of each of the following components in translation?

(a) Ribosome:

(b) tRNA:

(c) Amino acids:

(d) Start codon:

(e) Stop codon:

3. There are many different types of tRNA molecules, each with a different anticodon (HINT: see the mRNA table).

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Small ubunit

Ribosome structure

(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:

Codons on the mRNA:

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U A C

U A G

C C G

C G A

U U U

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152 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

Charged Arg-tRNA enters the ribosome A (acceptor) site. The amino acid is added to the growing polypeptide chain.

The ribosome P (peptidyl) site carries the growing polypeptide chain.

Met

Thr

Phe

Unloaded Thr-tRNA leaves the ribosome E (exit) site

Charging Lys-tRNA

Charged Val-tRNA

Val

Arg

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

3'

5'

mRNA

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).

START

3'

Direction of protein synthesis

4. Describe the events occurring during translation:

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5'

STOP

5. Many ribosomes can work on one strand of mRNA at a time (a polyribosome system). What would this achieve?

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109 Gene Expression 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: (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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(a) Structure A:

3. Describe two factors that would determine whether or not a particular protein is produced in the cell: (a)

(b)

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110 Protein Activity Determines Phenotype

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different type of protein. These differences in the proteins affect how each type carries out its task (e.g fast, slow, or not at all) and so produces differences in metabolic pathways or biological structures that will be expressed as different phenotypes (e.g. differences in eye color.)

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Key Idea: The ability of a protein to perform its task affects the phenotype of the individual. Proteins are the link between genotype (the alleles a person carries) and phenotype (what they look like). For a particular gene there may be multiple alleles, all coding for a slightly

Examples of phenotype related to protein activity

Enzymatic activity

Proteins involved in synthesis

The gene PON1 codes for the enzyme paraoxonase in humans. This enzyme acts on multiple substrates including the chemical paraoxon. Paraoxon is produced in the body from the organophosphate parathion, found in some insecticides. Paraoxon inhibits the transmission of nervous signals.

The pigment melanin gives a dark color to the skin and fur in mammals. The synthesis of melanin is regulated by the enzyme tyrosinase as part of the metabolism of the amino acid phenylalanine. Ordinarily tyrosinase has a optimum temperature similar to body temperature (37°C).

PON1 has the alleles R and Q. The R alleles encodes an enzyme that is able to degrade paraoxon ten times faster than the enzyme encoded by the Q allele. The two enzymes differ by just one amino acid. The R allele encodes the amino acid arginine at position 192 whereas the Q allele encodes glutamine.

In various mammals, including Himalayan rabbits and Siamese cats, a mutation in the enzyme causes it to operate at a much lower temperature than normal. This causes the animal to lack pigment in the fur as the enzyme is inactive. However, the animal's extremities (e.g. ears, nose, paws) usually experience cooler temperatures than the core body temperature and so the enzyme is active there, producing dark patches on the fur, a phenomenon called color-pointing.

People homozygous for the R allele (RR) display a phenotype with high resistance to pesticides containing paraoxon, whereas people homozygous for the Q allele are much more sensitive.

Color-pointing on extremities

MgtE

Degradation of proteins

Transport by proteins

Protein degradation is an important mechanism for regulating cellular processes. Mis-folded, faulty proteins, or proteins that have served their purpose are broken down by protein complexes called proteasomes. Various proteasomes are involved in regulating cell division, DNA repair, and regulation of membrane receptors, as well as many other processes. Failure of the degradation system can lead to disease phenotypes including plaques, cancers, and neurodegenerative disorders.

Proteins play an important role in transporting ions in and out of cells. Mg2+ is needed by various cellular enzymes and is transported by the membrane protein SLC41A1. It is homologous to the prokaryote magnesium transporter MgtE. Mutations that alter the structure of this protein often cause it to be unable to transport Mg2+. This can cause various diseases including kidney disease.

A2-33 cc.3.0

Paraoxonase

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1. Explain how the activity of a protein affects a phenotype:

2. Describe the structural difference between the Q and R proteins encoded by PON1 and the effect of this difference:

3. Explain how color-pointing occurs in Siamese cats:

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111 What is Genetic Engineering? aims to produce improved or novel organisms with specific desirable traits. Genetic engineering has wide applications in food technology, industry, agriculture, environmental clean up, pharmaceutical production, and vaccine development. Organisms that have had their DNA altered are called genetically modified organisms (or GMOs).

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Key Idea: DNA manipulation alters an organism's DNA either by adding new DNA or editing the existing DNA. DNA manipulation (genetic engineering) involves the direct manipulation of an organism's genome using biotechnology. This can be achieved by introducing new DNA into an organism or by editing its existing DNA. DNA manipulation

How are genetically modified organisms produced?

Foreign gene is inserted into host DNA

Existing gene is altered

Host DNA

Gene is deleted or deactivated

Host DNA

Host DNA

Add a foreign gene

Alter an existing gene

Delete or ‘turn off’ a gene

A novel (foreign) gene is inserted from another species. This will enable the GMO to express the trait encoded by the new gene. Organisms genetically altered in this way are referred to as transgenic.

An existing gene may be altered to make it express at a higher level (e.g. growth hormone) or in a different way (in tissue that would not normally express it). The technique may provide a way to fix a malfunctioning gene.

An existing gene may be deleted or deactivated (switched off) to prevent the expression of a trait (e.g. the deactivation of the ripening gene in tomatoes produced the Flavr-Savr tomato).

NIH

Tumor

Human insulin, used to treat diabetic patients, is now produced using transgenic bacteria and yeast.

New gene editing technologies, such as CRISPR, are being explored to treat breast cancer (above) and sickle cell disease.

Manipulating gene action is one way in which to control processes such as ripening in fruit.

1. (a) What is DNA manipulation?

(b) Using examples, discuss the ways in which an organism may be genetically modified (to produce a GMO):

2. Describe some of the applications of DNA manipulation:

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112 Making Recombinant DNA

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made of the same building blocks (nucleotides). rDNA allows a gene from one organism to be moved into, and expressed in, a different organism. Two important tools are used to create rDNA. Endonucleases (such as restriction enzymes or the CRISPR-Cas9 system) cut the DNA and the enzyme DNA ligase is used to join the sections of DNA together.

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Key Idea: Recombinant DNA (rDNA) is produced by first isolating (or synthesizing) a DNA sequence, then inserting it into the genome of a different organism. Recombinant DNA (rDNA) is produced by combining genetic material from two or more different sources. The production of rDNA is possible because the DNA of every organism is

What are restriction enzymes?

1

2

3

A restriction enzyme is an enzyme that cuts a double-stranded DNA molecule at a specific recognition site (a specific DNA sequence). There are many different types of restriction enzymes, each has a unique recognition site.

Restriction enzyme cuts here

Recognition site

G A AT T C

G A AT T C

C T TA A G

C T TA A G

cut

cut

cut

DNA

Some restriction enzymes produce DNA fragments with two sticky ends (right). A sticky end has exposed nucleotide bases at each end. DNA cut in such a way is able to be joined to other DNA with matching sticky ends. Such joins are specific to their recognition sites.

Some restriction enzymes produce a DNA fragment with two blunt ends (ends with no exposed nucleotide bases). The piece it is removed from is also left with blunt ends. DNA cut in such a way can be joined to any other blunt end fragment. Unlike sticky ends, blunt end joins are non-specific because there are no sticky ends to act as specific recognition sites.

Recognition site

G

A AT T

C T TA A

C

G

G

Fragment

DNA fragment with two sticky ends

A AT T C

G

G

A

G

C T TA A

Sticky end

Recognition site

The fragments of DNA produced by the restriction enzymes are mixed with ethidium bromide, a molecule that fluoresces under UV light. The DNA fragments are then placed on an electrophoresis gel to separate the different lengths of DNA.

C T TA

Sticky end

Restriction enzyme cuts here

CDC

A AT T C

Recognition site cut

CCCGGG

CCCGGG

DNA

GGGCCC GGGCCC Once the DNA fragments are separated, The solution containing the DNA is cut cut the gel is placed on a UV viewing platform. centrifuged at high speed to separate The area of the gel containing the DNA out the DNA. Centrifugation works by fragments of the correct length is cut molecules of different densities. Theout cutand by thisseparating type of restriction placed in a solution that dissolves enzyme the gel. leavesOnce isolated, the DNA can be spliced into no overhang This releases the DNA into the solution. another DNA molecule. CCC GGG CCC GGG

1. What is the purpose of restriction enzymes in making recombinant DNA?

2. Distinguish between sticky end and blunt end fragments:

DNA fragment with two blunt ends

3. Why is it useful to have many different kinds of restriction enzymes?

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CCC

GGG

CCC

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GGG

GGG

CCC

CCC

GGG

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157 NOTE: This other end of the foreign DNA is attracted to the remaining sticky end of the plasmid

G

TA A C T

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Creating a recombinant DNA plasmid

1

2

3

4

Two pieces of DNA are cut by the same restriction enzyme (they will produce fragments with matching sticky ends).

Fragments with matching sticky ends can be joined by basepairing. This process is called annealing. This allows DNA fragments from different sources to be joined.

The two different DNA fragments are attracted to each other by weak hydrogen bonds

Plasmid DNA fragment

A AT T C

G

G

C

T

T A

C T TA A

Foreign DNA fragment

A

Detail of restriction site

G

A A T T

C

TA A G C T

The fragments of DNA are joined together by the enzyme DNA ligase, producing a molecule of recombinant DNA.

The joined fragments will usually form either a linear or a circular molecule, as shown here (right) as recombinant plasmid DNA.

G

Restriction sites on the fragments are attracted by base pairing only

Gap in DNA molecule’s ‘backbone’

Plasmid DNA fragment G

C

T

Foreign DNA fragment

A A T T C

T A

A

G

DNA ligase

Detail of restriction site

G

A A T T

C

TA A G C T

Recombinant plasmid DNA

25kartika

pGLO is a plasmid engineered to contain Green Fluorescent Protein (gfp). pGLO has been used to create fluorescent organisms, including the bacteria above (bright patches on agar plates).

Fragments linked permanently by DNA ligase

No break in DNA molecule

The fragments are joined by the enzyme DNA ligase

G

C

T

A A T T C

T A

A

G

(a) Annealing:

(b) DNA ligase:

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4. Explain in your own words the two main steps in the process of joining two DNA fragments together:

5. Explain why ligation can be considered the reverse of the restriction digestion process:

6. Why can recombinant DNA be expressed in any kind of organism, even if it contains DNA from another species?

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113 New Tools: Gene Editing with CRISPR for CRISPR to work: an RNA guide that locates and binds to the target piece of DNA and the Cas9 endonuclease that unwinds and cuts the DNA. The technology has potential applications in correcting mutations responsible for disease, switching faulty genes off, adding new genes to an organism, or studying the effect of specific genes. It represents a major advance because it allows more precise and efficient gene editing at much lower cost than ever before.

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Key Idea: CRISPR is a complex comprising Cas9 endonuclease and sgRNA. The CRISPR complex cuts DNA at very specific sequences and can be used to edit genes. CRISPR-Cas9 (shortened to CRISPR and pronounced crisper) is an endonuclease complex occurring naturally in bacteria, which use it to edit the DNA of invading viruses. CRISPR is able to target specific stretches of DNA and edit it at very precise locations. Two key components are required

Single guide RNA (sgRNA) is a short synthetic RNA sequence designed to guide Cas9 to the site of interest (e.g. a faulty gene sequence). It contains a nucleotide section which is complementary to the DNA of interest.

Cas9 is guided to the target site by sgRNA. Cas9 unwinds the DNA and cuts both strands at a specific point.

Cutting point

Target DNA sequence

5'

3' 5'

5'

3'

The PAM sequence (NGG)* lies directly downstream of the target sequence on the non-target DNA strand. Recognition of PAM by Cas9 destabilizes the DNA allowing the sgRNA to be inserted. Cas9 will not function if PAM is absent. *N can be any nucleotide

5'

5'

3' 5'

3'

5'

5' The cut DNA can be repaired using one of the following methods:

Gene knock in "gene editing"

A new DNA sequence is inserted into the DNA break. For example allows a faulty gene sequence can be replaced with the correct sequence to restore normal gene function.

Gene knock out "gene silencing"

As the cell's normal repair process mend the broken DNA, errors occur resulting in the insertion or deletion of nucleotide bases. The resulting frame-shift mutation changes the way the nucleotide sequence is read, either disabling gene function or producing a STOP signal. This technique can be used to silence a faulty gene.

1. What are the roles of the following in CRISPR gene editing:

(a) Cas9:

2. Outline two ways CRISPR can be used to edit genes:

3. What benefits are offered by CRISPR technology?

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(b) sgRNA:

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114 Polymerase Chain Reaction chain reaction) is a technique for reproducing large quantities of DNA in the laboratory from an original sample. For this reason, it is often called DNA amplification. The technique is outlined below for a single cycle of replication. Subsequent cycles replicate DNA at an exponential rate, so PCR can produce billions of copies of DNA in only a few hours.

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Key Idea: PCR uses a polymerase enzyme to copy a DNA sample, producing billions of copies in a few hours. Many procedures in DNA technology, e.g. DNA sequencing and profiling, require substantial amounts of DNA yet, very often, only small amounts are obtainable (e.g. DNA from a crime scene or from an extinct organism). PCR (polymerase

DNA polymerase: A thermally stable form of the enzyme is used (e.g. Taq polymerase). . This is extracted from thermophilic bacteria.

A single cycle of PCR

A DNA sample (called target DNA) is obtained. It is denatured (DNA strands are separated) by heating at 98oC for 5 minutes.

Primer annealed

Nucleotides

Primer moving into position

Direction of synthesis

The sample is cooled to 60oC. Primers are annealed (bonded) to each DNA strand. In PCR, the primers are short strands of DNA; they provide the starting sequence for DNA extension.

Free nucleotides and the enzyme DNA polymerase are added. DNA polymerase binds to the primers andsynthesises complementary strands of DNA, using the free nucleotides.

After one cycle, there are now two copies of the original DNA.

Repeat cycle of heating and cooling until enough copies of the target DNA have been produced

Repeat for about 25 cycles

Loading tray Prepared samples in PCR tubes are placed in the loading tray and the lid is closed.

Temperature control Inside the machine are heating and refrigeration mechanisms to rapidly change the temperature.

Dispensing pipette Pipettes with disposable tips are used to dispense DNA samples into the PCR tubes.

Thermal cycler

Amplification of DNA can be carried out with machines called thermal cyclers. Once a DNA sample has been prepared, the amount of DNA can be increased billions of times in just a few hours.

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DNA quantitation The amount of DNA in a sample can be determined by placing a known volume in this quantitation machine. For many genetic engineering processes, a minimum amount of DNA is required.

RA

Controls The control panel allows a number of different PCR programmes to be stored in the machine’s memory. Carrying out a PCR run usually just involves starting one of the stored programmes.

Reducing contamination

The PCR process will amplify all the DNA within the sample including unwanted DNA from contamination. Therefore great care must be taken not to contaminate the sample with unwanted DNA (e.g from microbes in the environment, from dirty equipment, or from the researcher). Contamination is reduced by following strict protocols. The researcher must make sure that they are wearing appropriate clothing (hair net, gloves, coat) to stop their DNA contaminating the sample. In addition clean work surfaces, sterile solutions, and use of disposable equipment (e.g. pipettes and tubes) will help reduce contamination.

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1. What is the purpose of PCR?

2. Describe how the polymerase chain reaction works:

3. Describe two situations where only very small DNA samples may be available for sampling and PCR could be used: (a)

(b)

4. After only two cycles of replication, four copies of the double-stranded DNA exist. Calculate how much a DNA sample will have increased after:

(a) 10 cycles:

(b) 25 cycles:

5. The risk of contamination in the preparation for PCR is considerable.

(a) Describe the effect of having a single molecule of unwanted DNA in the sample prior to PCR:

(b) Describe two possible sources of DNA contamination in preparing a PCR sample:

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Source 1:

Source 2: (c) Describe two precautions that could be taken to reduce the risk of DNA contamination: Precaution 1:

Precaution 2:

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115 Gel Electrophoresis the gel depends primarily on their size and the strength of the electric field. The gel they move through is full of pores (holes). Smaller DNA molecules move through the pores more quickly than larger ones. At the end of the process, the DNA molecules can be stained and visualized as a series of bands. Each band contains DNA molecules of a particular size. The bands furthest from the start of the gel contain the smallest DNA fragments.

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Key Idea: Gel electrophoresis is used to separate DNA fragments on the basis of size. Gel electrophoresis is a tool used to isolate DNA of interest for further study. It is also used for DNA profiling (comparing individuals based on their unique DNA banding profiles). DNA has an overall negative charge, so when an electrical current is run through a gel, the DNA moves towards the positive electrode. The rate at which the DNA molecules move through

DNA solutions: Mixtures of different sizes of DNA fragments are loaded in each well in the gel.

(-ve)

(-ve)

C

DNA markers, a mixture of DNA molecules with known molecular weights (size) are often run in one lane. They are used to estimate the sizes of the DNA fragments in the sample lanes. The figures below are hypothetical markers (bp = base pairs).

G

(-ve)

(-ve)

A

T

DNA is negatively charged because the phosphates (blue) that form part of the backbone of a DNA molecule have a negative charge.

5 lanes

Negative electrode (–)

Wells: Holes are made in the gel with a comb, acting as a reservoir for the DNA solution.

DNA fragments move: The gel matrix acts as a sieve for the negatively charged DNA molecules as they move towards the positive terminal. Small fragments move easily through the matrix, whereas large fragments don't. As DNA molecules migrate through the gel, large fragments will lag behind small fragments. As the process continues, the separation between larger and smaller fragments increases.

Large fragments

50,000 bp 20,000 bp 10,000 bp 5000 bp 2500 bp

Small fragments

1000 bp

500 bp

Tray: The gel is poured into this tray and allowed to set.

Positive electrode (+)

Gel: A gel is prepared, which will act as a support for separation of the fragments of DNA. The gel is a jelly-like material, called agarose.

Steps in the process of gel electrophoresis of DNA

1. The gel is placed in an electrophoresis chamber and the chamber is filled with buffer, covering the gel. This allows the electric current from electrodes at either end of the gel to flow through the gel. 2. DNA samples are mixed with a “loading dye” to make the DNA sample visible. The dye also contains glycerol or sucrose to make the DNA sample heavy so that it will sink to the bottom of the well. 3. The gel is covered, electrodes are attached to a power supply and turned on. 4. When the dye marker has moved through the gel, the current is turned off and the gel is removed from the tray. 5. DNA molecules are made visible by staining the gel with methylene blue or ethidium bromide which binds to DNA and will fluoresce in UV light. 6. The band or bands of interest are cut from the gel and dissolved in chemicals to release the DNA. This DNA can then be studied in more detail (e.g. its nucleotide sequence can be determined).

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1. What is the purpose of gel electrophoresis?

2. Describe the two forces that control the speed at which fragments pass through the gel: (a) (b)

3. Why do the smallest fragments travel through the gel the fastest?

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Recognition sites for selected restriction enzymes Source

Recognition sites

DNA fragments for gel electrophoresis are produced by restriction digestion of DNA using restriction enzymes. Restriction enzymes are produced by bacteria as a method of eliminating foreign DNA. About 3000 different restriction enzymes have been isolated. Around 600 are commonly used in laboratories.

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Enzyme

EcoRI

Escherichia coli RY13

GAATTC

HaeIII

Haemophilus aegyptius

GGCC

HindIII

Haemophilus influenzae Rd

AAGCTT

Hpal

Haemophilus parainfluenzae

GTTAAC

HpaII

Haemophilus parainfluenzae

CCGG

MboI

Moraxella bovis

GATC

TaqI

Thermus aquaticus

TCGA

Restriction enzymes are named according to the species they were first isolated from, followed by a number to distinguish different enzymes isolated from the same organism.

4. (a) A scientist uses HpaII to cut a length of DNA. State the recognition site for HpaII:

(b) Circle where on the DNA sequence below HpaII would cut the following DNA sequence:

GTTAGGCCCGGCTAGCTTGACCAGTCCCGGGTCACAGTCTCTGACCCGGCTTTAGACACACTCCGGTTACTACCG 5. In 1988, the disease BLAD (Bovine Leukocyte Adhesion Deficiency) specific to Holstein cattle, was affecting the U.S. dairy industry. The disease was traced to the bull Osborndale Ivanhoe. The disease is recessive and two alleles are needed for its expression. The disease is caused by two mutations of the CD18 gene. One of the affected regions is shown below. The DNA of three individuals is shown: an unaffected individual, a carrier, and an affected individual.

Unaffected

Carrier

Affected

GTGACCTTCCGGAGGGCCAAGGGCTACCCCATCGGCCTGTACTACCTGATGGACCTCT

Allele 1

GTGACCTTCCGGAGGGCCAAGGGCTACCCCATCGGCCTGTACTACCTGATGGACCTCT

Allele 2

GTGACCTTCCGGAGGGCCAAGGGCTACCCCATCGACCTGTACTACCTGATGGACCTCT

Allele 1

GTGACCTTCCGGAGGGCCAAGGGCTACCCCATCGGCCTGTACTACCTGATGGACCTCT

Allele 2

GTGACCTTCCGGAGGGCCAAGGGCTACCCCATCGACCTGTACTACCTGATGGACCTCT

Allele 1

GTGACCTTCCGGAGGGCCAAGGGCTACCCCATCGACCTGTACTACCTGATGGACCTCT

Allele 2

(a) Use the restriction enzymes TaqI, HaeII, and HpaII to "cut" the sequences. Write the length of the segments produced in the spaces below (NOTE: TaqI cuts between T and C. HaeII cuts between G and C. HpaII cuts between C and C:

Taq1: Unaffected:

Carrier:

Affected:

HaeIII: Unaffected:

Carrier:

Affected:

HpaII: Unaffected:

Carrier:

Affected:

(b) On the gels below draw in the bands that would be seen for each individual for each restriction enzyme:

10 bp

40 bp

60 bp

Ladder

Unaffected Carrier Affected Ladder

TaqI

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20 bp

Unaffected Carrier Affected Ladder

HaeIII

Unaffected Carrier Affected

HpaII

(c) Decide which restriction enzyme(s) would be useful for identifying carriers of BLAD:

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116 Preparing a Gene for Cloning between introns and exons (coding sequences). The solution is to engineer a eukaryotic gene that can be transcribed and translated (expressed) by a prokaryote. This is achieved using the retroviral enzyme reverse transcriptase, which copies the mature mRNA (containing exons only) to produce a complementary strand of DNA. This task is important in both in vivo gene cloning and ex vivo gene cloning (by PCR) because it creates a gene ready for amplification.

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Key Idea: Eukaryotic genes contain introns, which must be removed before a gene can be inserted into a prokaryotic cell for in vivo cloning. The presence of introns (sequences of a gene's DNA that do not code for proteins) presents a problem when preparing a eukaryotic gene for insertion into a prokaryotic cell. Prokaryotes are the most commonly used organism in large scale culture of gene products but do not distinguish 1

Double stranded DNA of a gene from a eukaryotic organism (e.g. human) containing introns.

Intron

DNA

Intron

Intron

Intron

Double stranded molecule of genomic DNA

Exon

Exon Exon

Exon

Exon

Exon

Transcription

2 Transcription creates a

primary RNA molecule as a part of the cell's normal gene expression.

Intron

Primary RNA

Exons are joined together

3 The introns are removed by restriction

enzymes to form a mature mRNA (now excluding the introns) that codes for the production of a single protein.

synthesizes a single stranded DNA molecule complementary to the mRNA. In retroviruses, this enzyme makes a DNA strand from the viral ssRNA.

6 The second DNA strand is made

by using the first as a template, and adding the enzyme DNA polymerase.

• In cases of in-vivo cloning, it makes the gene shorter and therefore easier to insert into plasmids

Reverse transcription

the cell and purified.

5 Reverse transcriptase is added which

Why remove the introns?

mRNA

4 The mRNA is extracted from

• It means that large amounts of noncoding DNA are not made by PCR.

mRNA DNA

DNA strand being synthesized by reverse transcriptase.

DNA DNA

Introns

Introns are removed

• In cases of in-vivo cloning, it allows the bacterial enzymes to properly translate the human gene from the reassembled DNA.

Completed artificial gene consisting of a double stranded molecule of complementary DNA (cDNA).

1. What is the role of restriction enzymes in preparing a clone?

(b) What is the role of reverse transcriptase in this process?

3. What is the normal role of reverse transcriptase?

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2. (a) Why are introns removed before cloning a gene?

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117 Gene Cloning

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technique is known as gene cloning and it is widely used to produce valuable commodities at low cost. To be useful, all vectors must be able to replicate inside their host organism, they must have one or more sites at which a restriction enzyme can cut, and they must have some kind of genetic marker that allows them to be identified. Replica plating is often used to identify organisms of interest.

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Key Idea: Recombinant plasmids will be taken up by bacterial cells, which will then go on to multiply the gene of interest or produce its protein product. Recombinant DNA techniques can be used to insert a gene into a vector such as a plasmid. The recombinant vector can then be used to transmit the gene to another organism (such as E. coli) where it is replicated with the host DNA. This Human cell

Bacterial cell. e.g. E. coli

Chromosome

DNA in chromosome

A gene of interest (DNA fragment) is isolated and prepared by removal of introns (non-protein coding regions).

Human gene

Sticky end

Both the human DNA and the plasmid are treated with the same restriction enzyme to produce identical sticky ends.

Sticky end

A commercially available plasmid engineered to contain certain restriction sites is used in the cloning process.

Restriction enzyme recognition sequence

Tetracyclineresistance gene

Plasmid

Plasmid vector

Ampicillinresistance gene

Promoter and terminator sequences added

The restriction enzyme cuts the plasmid DNA at its single recognition sequence, disrupting the tetracycline resistance gene.

Sticky ends

Human gene

gfp as a gene marker

Increasingly there are concerns about the spread of antibiotic resistant genes in the environment, so more often today, another gene acts as a marker for transformation instead of an antibiotic resistance gene. The gene for Green Fluorescent Protein (gfp above), isolated from the jellyfish Aequorea victoria, has become well established as a marker for gene expression in the recombinant organism. The gfp gene is recombined with the gene of interest and transformed cells can then be detected by the presence of the fluorescent product (cells with gfp present glow green under fluorescent light).

Agar plate with bacterial colonies. Only some have the plasmid with the human gene.

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Recombinant DNA molecule

The DNA fragments are mixed together and the enzyme DNA ligase is used to anneal (join) the complementary sticky ends.

The recombinant plasmid is introduced into a bacterial cell by adding the DNA to a bacterial culture. Under the right conditions, some bacteria will take up the plasmid from solution by the process of transformation. These bacteria are transgenic organisms.

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Bacterial plasmids are commonly used vectors because they are easy to manipulate, their restriction sites are well known, and they are readily taken up by cells in culture.

Replica plating can be used to identify colonies of bacteria that carry the recombinant plasmid. Recombinant bacteria are resistant to ampicillin but sensitive to tetracycline. These colonies can be isolated and grown in culture.

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Replica plating identifies colonies with desirable qualities Four colonies grow on the ampicillin plate.

Two colonies are TR, so do not contain the insulin gene.

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After gene cloning, it is important to be able to identify the colonies in which transformation has occurred. This is achieved by replica plating.

TetracyclineTetracyclineReplica plating transfers colonies from a resistance master gene resistance gene plate to test plates enriched with specific Plasmid Plasmid nutrients or antibiotics. The original A pattern of A Master plate colonies (spatial arrangement) is maintained during the transfer. Growth (or lack of) onAmpicillinthe test Ampicillinresistance gene plates can be used to identify colonies ofresistance interest gene (e.g. colonies containing the insulin gene). In the example (right) colonies are tested for their susceptibility to the antibiotic tetracycline. Those New gene gene New with ampicillin resistance but no tetracycline resistance contain the insulin gene (plasmid B). The insulin gene has interrupted the tetracycline gene, so they are sensitive to tetracycline.

Plasmid Plasmid A A

Ampicillin resistance (AR): YES Tetracycline resistance (TR): YES

Plasmid Plasmid B B

Plates are incubated at 30°C and scored for growth

Master plate media

Media containing ampicillin

Media containing tetracycline

Orientation marker

Sterile velvet is pressed against the colonies on the master plate.

The colonies are transferred to different media.

YES NO

1. Explain how a human gene is removed from a chromosome and placed into a plasmid:

2. (a) What is the purpose of replica plating?

(b) Suggest why the colonies are replated onto the master media:

(c) In the replica plating example above, explain how the colonies with the recombinant plasmids are identified:

(d) What can you say about the colony that did not grow on the ampicillin plate?

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3. Explain why the gfp marker is a more desirable gene marker than genes for antibiotic resistance:

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118 Aseptic Technique and Streak Plating show confluent growth (growth as a continuous sheet), while the area at the end of the streak will show individual colonies. Isolated colonies can be removed using aseptic techniques, and transferred to a sterile medium. After incubation, assuming aseptic techniques have been used, all organisms in the new culture will be descendants of the same organism (i.e. a pure culture). The organism can then be identified and studied (e.g. for sensitivity to particular antibiotics).

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Key Idea: Streak plating is a simple method to isolate individual bacterial colonies for further study. Aseptic technique is the best way of avoiding contamination The most common way of separating bacterial cells on the agar surface is the streak plate method. This method dilutes the sample by mechanical means. Contamination is minimized using a procedure called aseptic technique. After incubation, the area at the beginning of the streak pattern will The streaking starts here. Streaks are made in the order indicated by the numbers on the plate. The first streak is made from the initial bacterial mixture.

In each streak, the loop picks up bacteria from the previous series, diluting the number of cells each time.

Individual colonies (arising from one cell) should be obtained here. These can be removed and then cultured separately.

After incubation

Latex gloves ensure no contamination from either bacteria or fungi on the hands.

BBT Inc.

.cnI TB B

The inoculating loop is sterilised with flame and alcohol after each streak. It is cooled before a new streak is made.

The lid of the petri dish (not shown) is lifted only enough to allow the loop inside. It is replaced after each streak.

GT

Rough colonies on blood agar

Colonies become visible when approximately 10 to 100 million bacterial cells are present. Note the well-isolated colonies in the photo above. A single colony may be removed for further investigation.

A swab containing a single strain of bacteria is used to inoculate additional nutrient plates to produce pure cultures of bacteria.

Smooth colonies on bicarbonate agar

CDC

Bacillus anthracis

To test purity, a sample of a culture can be grown on a selective medium that promotes the growth of a single species. A selective medium may contain a nutrient specific to a particular species.

1. What is the purpose of streak plating?

3. Why is the lid only partially removed during streaking?

4. (a) How would you know if your streak plating had been effective?

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2. Outline the process of streak plating:

(b) What could you do to test that all your colonies were the same species?

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Aseptic technique

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Aseptic technique is a fundamental skill in microbiology as it prevents unwanted microorganisms contaminating a laboratory culture. The technique involves the use of heat (a flame) and sometimes alcohol to sterilize the tools used to transfer a microbial inoculum to the growth medium.

1

Always wear gloves when working with microbes. Wipe the work surface down with a disinfectant such as ethanol.

Exposure of the culture media to the environment is limited to reduce the risk of contamination from the environment. For this reason, the lid of an agar plate or screw cap of a liquid broth are only partially opened for as little time as possible to inoculate the media. Aseptic technique also minimizes the risk of microbes being released into the environment. This is especially important when dealing with pathogenic microbes. The example provided on the right shows the inoculation of an agar plate using aseptic technique.

Sources of contamination

Sources of contamination include: ff Airborne microbes

Hold the inoculating loop in the flame until it glows red hot. Remove the lid from the culture broth and pass the neck of the bottle through the flame.

ff Contamination from the researcher's body

ff Dirty (unsterilized) equipment or bench top ff Contaminated culture media

Dip the cool inoculating loop into the broth. Flame the neck of the bottle again and replace the lid.

The environment contains many microbes that could potentially contaminate an inoculum if correct aseptic technique is not followed. The agar plate above was left exposed in a laboratory for one hour and then incubated. Many different types of microbes have grown on it.

4 Raise the lid of the plate just enough to allow the loop to streak the plate. Streak the surface of the media. Seal the plate with tape and incubate upside down.

5. Identify three sources of contamination and how contamination from them can be minimized: (a)

(c)

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(b)

6. Why is it important to use aseptic technique when growing microbial cultures?

7. What would happen if you did not cool the inoculation loop before you dipped it into the culture broth?

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119 Testing for Transformation first observed and studied by Frederick Griffith in 1928. Transformation efficiency is a measure of the ability of bacteria to take up extracellular DNA. Bacteria that are able to do this are termed competent. Transformation is commonly used in genetic engineering to insert novel genes into bacteria in order to produce proteins on an industrial scale.

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Key Idea: Bacteria are able to obtain new genetic information by taking up genetic material from the environment. This process called transformation. Bacteria may obtain new genetic material via mutation, obtain it from other bacteria, or by taking up naked genetic material from the environment. This process of transformation was

Aim

Method for transforming E. coli

To investigate the efficiency of transformation in E. coli bacteria when mixed with a plasmid containing a variant of GFP (green fluorescent protein) gene and the ampicillin resistance gene.

250 mL of ice cold CaCl2 was transferred to two microcentrifuge tubes using a sterile transfer pipette. One tube was labeled +plasmid, the other was labeled -plasmid. Both tubes were placed on ice. A starter colony was transferred to each tube from an agar plate of E. coli using a sterile inoculation loop. The tubes were inverted several times to ensure mixing, then returned to the ice.

Background

GFP is a common marker gene that is used to indicate bacterial colonies that have acquired a target plasmid. GFP glows under florescent light. The variant of this gene also causes the bacterial colonies to turn yellow-green in ordinary white light (E. coli colonies normally have a whitish appearance). GFP is normally a preferred alternative to using a gene for resistance to ampicillin (an antibiotic) as it reduces the risk of antibiotic resistance spreading in bacteria. In this example, both genes are being used as only transformed bacterial colonies are being counted and the ampicillin resistance makes this simpler because untransformed colonies can be eliminated by using ampicillin-containing agar plates. Colonies can be assumed to be derived from individual bacterial cells, so the number of colonies relates directly to the number of cells originally on the agar plate.

10 mL of 0.005 mg mL-1 solution of plasmid was transferred by sterile pipette to the tube labeled +plasmid. Both tubes were then incubated on ice for 10 minutes. The tubes were then placed in a water bath at 42°C for 50 seconds to heat shock the bacteria, after which the tubes were returned to ice for two minutes. Each tube then had 250 mL of nutrient broth added and were incubated at room temperature for ten minutes.

Two plates containing nutrient agar and two plates containing nutrient agar and ampicillin were prepared. 100 mL of -plasmid was transferred to one of each type of plate and streaked using a sterile inoculating loop. The same was done with the +plasmid. The plates were then covered and placed in an incubator at 37°C for 24 hours. The number of colonies on the plate containing the +plasmid and ampicillin agar were counted and recorded.

Results

The four agar plates are shown below:

Plate 1: No ampicillin -plasmid

Transformed E.coli

Plate 2: No ampicillin +plasmid

Untransformed E.coli

Plate 3: Ampicillin -plasmid

Plate 4: Ampicillin +plasmid

1. (a) Determine the mass of plasmid pipetted into the microcentrifuge tubes. Use the formula mass (mg) (of plasmid) = concentration (mg mL-1) x volume (mL):

(b) Determine the fraction of this amount spread on the plate (volume spread on plate ÷ total volume in tube):

(c) Determine the mass of plasmid spread on the plate (answer 1. (a) x answer 1. (b)):

(d) Calculate the transformation efficiency (transformants mg-1) using the number of colonies ÷ mass of DNA spread:

2. What is the purpose of the -plasmid tube?

3. Account for the differences between plate 1 and 2:

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4. How has plate 4 made counting the transformed colonies easier compared to plate 2?

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120 What is Transgenesis? Transgenesis allows direct modification of a genome to introduce traits to an organism in which they are not naturally present. To be successful, the transgenes must be transmitted to the offspring. The genes are inserted using vectors or by direct insertion of the DNA. The first successful transgenic animal was a mouse, produced using DNA microinjection.

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Key Idea: Transgenesis is the insertion of a gene from one species into another, so its protein product is expressed in the second species and the gene is transferred to its offspring. Transgenesis is the insertion of a gene from one species into another that does not normally carry the gene. An organism modified in this way is called a transgenic organism.

Creating transgenic mice using pronuclear injection

Pronuclear injection

A gene that has been transferred into another organism is called a transgene. Genes can be introduced directly into an animal cell by microinjection. Multiple copies of the transgene are injected via a glass micropipette into a recently fertilized egg cell, which is then transferred to a surrogate mother. Transgenic livestock are produced in this way but the process is inefficient: only 2-3% of eggs give rise to transgenic animals and only a proportion of these animals express the transgene adequately. For this reason, successful transgenics are often cloned. Micropipette injects gene

Egg cell

This example outlines a successful experiment that used DNA microinjection technology to produce the world’s first transgenic animal.

2b Micropipette injects rat growth hormone gene into a fertilized egg.

3b

Transformed egg is cultured to an embryo, then implanted in a surrogate mother.

4b

Mouse B Weight 44 g

1 Two eggs are removed

from a single female mouse and are fertilized artificially in a test tube.

Mouse A Weight 29 g

4a

2a This fertilized

egg is unaltered.

Egg nucleus

Blunt holding pipette

Normal egg is cultured to

3a an embryo, then implanted

The two mice above are siblings, but mouse B is a transgenic organism. A rat growth hormone gene was introduced into its genome.

in a surrogate mother.

1. (a) What is transgenesis?

(b) How can transgenesis be used to produce organisms with desirable traits?

(c) What are the limitations of transgenesis?

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2. Briefly describe how a transgenic mouse can be produced using pronuclear injection:

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121 Vectors for Transgenesis requires a vector (carrier), such as a virus, liposome, or plasmid. The vector (with its foreign gene) must also be delivered somehow to the host cell. This may occur easily, as in bacteria, or may require tools such as gene guns.

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Key Idea: Several different carriers, called vectors, can be used to introduce a gene into a cell. There are advantages and disadvantages associated with each type of vector. The transfer of a gene to the cell of another organism usually

Viral vectors

Liposome vectors

Plasmid vectors

Retrovirus

Recombinant plasmids contain DNA from one or more other organisms

Adenovirus

ff Viruses are well known for their ability to insert DNA into a host cell. For this reason they have become a favored tool in transgenesis.

ff Different types of viruses integrate their DNA into the host in different ways. This allows scientists to control where and for how long the new DNA is expressed in the host. However, the size of the piece of DNA that can be transferred is limited to about 8 kb. ff Integration of the DNA into the host DNA can cause unexpected side effects depending on where in the host's chromosome the DNA inserts itself.

Lipid bilayer

Novel gene

ff Liposomes are spherical bodies of lipid bilayer.

ff Plasmids are circular lengths of DNA up to 1000 kb long (1 kb = 1000 bp).

ff They can be quite large and targeted to specific types of cells by placing specific receptors on their surfaces.

ff Recombinant plasmids are frequently used to produce transgenic organisms, especially bacteria. The bacteria may be the final target for the recombinant DNA (e.g. transgenic E. coli producing insulin) or it can be used as a vector to transfer the DNA to a different host (e.g. Agrobacterium tumefaciens is used to transfer the Ti plasmid to plants).

ff Liposomes can carry plasmids 20 kb or more.

ff They do not trigger immune responses when used in gene therapy, but are less efficient than viruses at transferring the plasmid into a target cell.

ff In gene therapy, plasmids by themselves, as naked DNA, are unstable and not particularly efficient at integrating DNA into a target cell.

Transformation is the direct uptake of foreign DNA and is common in bacteria. Recombinant DNA plasmids are mixed with bacteria and the bacteria that take up the DNA are used.

Transduction is the transfer of DNA into a bacterium by a virus. Bacteriophages (viruses that infect bacteria) are commonly used to integrate recombinant DNA into a target bacterium.

Transfection is the deliberate, often non-viral, introduction of foreign DNA into a cell. There are numerous methods including electroporation and the use of the gene gun (above).

Electroporation cuvettes

Zephyris cc 3.0

BioRad/RA

EII

Dr Graham Beards, cc 3.0

Transferring the DNA

Electroporation is a method in which an electric field is applied to cells, causing the plasma membrane to become more permeable. This allows DNA to cross the plasma membrane.

2. (a) Why are viruses often the preferred vector?

(b) Identify two problems with using viral vectors for DNA transfer:

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1. What is the role of a vector in transgenesis?

3. What type of vector would be most suitable for transferring a 400 kb length of DNA to a plant? WEB

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122 The Applications of Genetic Manipulation used GMOs, with applications ranging from pharmaceutical production and vaccine development to environmental clean-up and rehabilitation. Crop plants are also commonly genetically modified because they are easily propagated and the potential gains are great. However their use is controversial because transgenes are easily spread between plant species.

USDA

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Key Idea: GMOs are used in the food industry, agriculture, horticulture, medicine, and environmental practices. Techniques for genetic manipulation are now widely applied throughout modern biotechnology: in food and enzyme technology, in industry and medicine, and in agriculture and horticulture. Microorganisms are among the most widely

Extending shelf life: The shelf life of soy oil has been extended by genetic modification of its fatty acid profile, reducing its tendency to oxidize. In GM soy, the genes for the enzymes that convert the fatty acids into easily oxidized forms are silenced.

Pest or herbicide resistance: Plants can be engineered to carry and express genes for insect toxins or herbicide resistance. Pest resistant crops do not require spraying and herbicide resistance allows the grower to control weeds without damaging the crops.

1. Suggest one economic advantage of extending shelf life of products such as soy oil:

2. Suggest one disadvantage of engineering crop plants to be herbicide resistant:

Pronuclear injection

David Wells

3. How is cloning technology used in the livestock industry?

4. What advantages could be gained by developing a GE crop that produces more protein?

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Transgenic animals are difficult and costly to produce, so cloning technology is often used to reproduce successful transgenics more quickly. Clones can be produced using a technique called somatic cell nuclear transfer (SCNT). An egg cell with its chromosomes removed is fused with a transgenic somatic cell. After artificially activating this single cell embryo so that it divides like a normal embryo, it is grown for 7 days till it reaches the blastocyst stage. Viable embryos are then transplanted into recipient mothers (e.g. the Blonde d'Aquitaine calves above). SCNT has also been used to clone and preserve rare breeds.

Biofactories: Transgenic bacteria are widely used to produce desirable commodities, such as hormones or proteins. Large quantities of a product can be produced using bioreactors, e.g. injectable human insulin produced by recombinant bacteria or yeast (above).

Livestock improvement using transgenic animals: Transgenic sheep have been used to enhance wool production in flocks (above, left). The keratin protein of wool is largely made of a single amino acid, cysteine. Injecting developing sheep with the genes for the enzymes that generate cysteine produces woollier sheep. Transgenic sheep carrying the human gene for a protein, ι-1-antitrypsin, produce the protein in their milk. The antitrypsin is extracted from the milk and can be used to treat hereditary emphysema. Š2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

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5. Using animals as live biofactories to produce a valuable product (e.g. a human protein) is controversial. What welfare issues could be associated with such practices?

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Crop improvement: Gene technology is now an integral part of the development of new crop varieties. Crops can be engineered to produce higher protein or vitamin levels (e.g. golden rice) or to grow in inhospitable conditions (e.g. salty or arid land).

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123 Using Recombinant Plasmids in Industry

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The issue ff Recombinant DNA technology can be used to produce industrially important enzymes in large quantities.

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Key Idea: Inserting useful genes into bacteria to produce biofactories can solve the problem of shortages in the manufacturing and food industries.

Concept 1

Enzymes are proteins made up of amino acids. The amino acid sequence of chymosin can be determined and the mRNA coding sequence for its translation identified.

ff Chymosin (also known as rennin) is an enzyme that digests milk proteins. It is the active ingredient in rennet, a substance used by cheesemakers to clot milk into curds. ff Traditionally rennin is extracted from "chyme", i.e. the stomach secretions of suckling calves (hence its name of chymosin).

ff By the 1960s, a shortage of chymosin was limiting the volume of cheese produced. ff Enzymes from fungi were used as an alternative but were unsuitable because they caused variations in the cheese flavor.

Concept 2

Concept 3

Concept 4

Concept 5

Reverse transcriptase can be used to synthesize a DNA strand from the mRNA. This process produces DNA without the introns, which cannot be processed by bacteria.

DNA can be cut at specific sites using restriction enzymes and rejoined using DNA ligase. New genes can be inserted into selfreplicating bacterial plasmids.

Under certain conditions, bacteria are able to lose or take up plasmids from their environment. Bacteria are readily grown in vat cultures at little expense.

The protein in made by the bacteria in large quantities.

Techniques

The amino acid sequence of chymosin is first determined and the RNA codons for each amino acid identified.

Plasmid isolated from E. coli bacteria.

Initially, the gene coding for chymosin was isolated from the stomach of a milk-fed suckling calf (less than 10 days old). Now genes are produced by PCR.

Plasmid

mRNA matching the identified sequence is isolated from the stomach of young calves. Reverse transcriptase is used to transcribe mRNA into DNA. The DNA sequence can also be made synthetically once the sequence is determined. The DNA is amplified using PCR. Plasmids from E. coli bacteria are isolated and cut using restriction enzymes. The DNA sequence for chymosin is inserted using DNA ligase. Plasmids are returned to E. coli by placing the bacteria under conditions that induce them to take up plasmids.

Outcomes

The transformed bacteria are grown in vat culture. Chymosin is produced by E. coli in packets within the cell that are separated during the processing and refining stage.

The recombinant plasmid is taken up by the bacteria.

Transformed bacterial cells are grown in a vat culture

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Recombinant chymosin entered the marketplace in 1990. It established a significant market share because cheesemakers found it to be cost effective, of high quality, and in consistent supply. Most cheese is now produced using recombinant chymosin such as CHY-MAX.

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Recombinant plasmid

Restriction enzyme cuts the plasmid and DNA ligase joins the chymosin gene into the plasmid DNA.

Further applications

A large amount of processing is required to extract chymosin from E.coli. There are now a number of alternative bacteria and fungi that have been engineered to produce the enzyme. Most chymosin is now produced using the fungi Aspergillus niger and Kluyveromyces lactis. Both are produced in a similar way as that described for E. coli.

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Enzymes from GMOs are widely used in the baking industry. Maltogenic a-amylase from Bacillus subtilis bacteria is used as an anti-staling agent to prolong shelf life. Hemicellulases from B. subtilis and xylanase from the fungus Aspergillus oryzae are used to improve dough, crumb structure, and volume during the baking process.

Lipase from Aspergillus oryzae is used to process palm oil to produce substitutes for cocoa butter (above), which have similar textural qualities but lower cost.

Dual Freq

Romain Behar

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Acetolactate decarboxylase from B. subtilis is an enzyme used in the brewing industry. It reduces maturation time of the beer by by-passing a rate-limiting step.

1. Describe the main use of chymosin:

2. What was the traditional source of chymosin?

3. Summarize the key concepts that led to the development of the technique for producing chymosin:

(a) Concept 1:

(b) Concept 2:

(c) Concept 3:

(d) Concept 4:

(e) Concept 5:

4. Discuss how the gene for chymosin was isolated and how the technique could be applied to isolating other genes:

5. Describe three advantages of using chymosin produced by GM bacteria over chymosin from traditional sources:

(b)

(c)

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(a)

6. Explain why the fungus Aspergillus niger is now more commonly used to produce chymosin instead of E. coli:

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124 Engineering for Improved Nutrition The issue

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Key Idea: The use of recombinant DNA to build a new metabolic pathway has greatly increased the nutritional value of a variety of rice.

Concept 1

Rice is a staple food in many developing countries. It is grown in large quantities and is available to most of the population, but it lacks many of the essential nutrients required by the human body for healthy development. It is low in b-carotene.

ff Beta-carotene (β-carotene) is a precursor to vitamin A which is involved in many functions including vision, immunity, fetal development, and skin health.

ff Vitamin A deficiency is common in developing countries where up to 500,000 children suffer from night blindness, and death rates due to infections are high due to a lowered immune response. ff Providing enough food containing useful quantities of β-carotene is difficult and expensive in many countries.

Concept 2

Concept 3

Concept 4

Rice plants produce b-carotene but not in the edible rice endosperm. Engineering a new biosynthetic pathway would allow b-carotene to be produced in the endosperm. Genes expressing enzymes for carotene synthesis can be inserted into the rice genome.

The enzyme carotene desaturase (CRT1) in the soil bacterium Erwinia uredovora, catalyzes multiple steps in carotenoid biosynthesis. Phytoene synthase (PSY) overexpresses a colorless carotene in the daffodil plant Narcissus pseudonarcissus.

DNA can be inserted into an organism's genome using a suitable vector. Agrobacterium tumefaciens is a gall-forming bacterial plant pathogen that is commonly used to insert novel DNA into plants.

The development of golden rice

Techniques

The PSY gene from daffodils and the CRT1 gene from Erwinia uredovora are sequenced.

Ti plasmid

PSY is needed for the synthesis of a colorless carotene.

PSY

T1 CR

The tumor-inducing Ti plasmid is modified to delete the gallforming gene and insert the genes of interest. The parts of the Ti plasmid required for plant transformation are retained.

CRT1 can catalyse multiple steps in the synthesis of carotenoids. These steps require many enzymes in plants.

Recombinant plasmid

SGR1

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Agrobacterium is incubated with rice plant embryo. Transformed embryos are identified by their resistance to hygromycin.

Outcomes

Further applications

Modified plants are identified by resistance to hygromycin.

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Modified Ti plasmid is inserted into the bacterium.

The rice produced had endosperm with a distinctive yellow color. Under greenhouse conditions golden rice (SGR1) contained 1.6 µg per g of carotenoids. Levels up to five times higher were produced in the field, probably due to improved growing conditions.

Recombined plasmid is inserted into Agrobacterium. This is then mixed with rice plant embryos.

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The Ti plasmid from Agrobacterium is modified using restriction enzymes and DNA ligase to delete the gall-forming gene and insert the synthesized DNA packages. A gene for resistance to the antibiotic hygromycin is also inserted so that transformed plants can be identified later. The parts of the Ti plasmid required for plant transformation are retained.

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Gene for the enzyme phytoene synthase (PSY) extracted from the daffodil plant Narcissus pseudonarcissus

Gene for the enzyme carotene desaturase (CRT1) extracted from the soil bacterium Erwinia uredovora.

DNA sequences are synthesized into packages containing the CRT1 or PSY gene, terminator sequences, and endosperm specific promoters (these ensure expression of the gene only in the edible portion of the rice).

Further research on the action of the PSY gene identified more efficient methods for the production of β-carotene. The second generation of golden rice now contains up to 37 µg per g of carotenoids. Golden rice was the first instance where a complete biosynthetic pathway was engineered. The procedures could be applied to other food plants to increase their nutrient levels.

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Golden rice: A controversial solution

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Golden rice (top bowl) is promoted as one way to increase the beta-carotene (and thus vitamin A) intake in countries where rice is a staple part of the diet. However there are many groups that actively resist its promotion and use (e.g. Greenpeace).

Many of these anti-GM groups believe there is no need for golden rice because people can obtain enough beta-carotene by eating a variety of fruits and vegetables. They are also worried about the possibility of golden rice contaminating ordinary rice crops.

In June 2016 the The National Academies of Sciences, Engineering, and Medicine released the results of an extensive study, reporting there was no evidence to suggest GM crops were unsafe and that GM crops are as safe to eat as non-GM crops

1. Describe the basic methodology used to create golden rice:

2. Explain how scientists ensured b-carotene was produced in the endosperm:

3. What property of Agrobacterium tumefaciens makes it an ideal vector for introducing new genes into plants?

4. (a) How could this new variety of rice reduce disease in developing countries?

(b) Absorption of vitamin A requires sufficient dietary fat. Explain how this could be problematic for the targeted use of golden rice in developing countries:

5. Explain why golden rice is a controversial product:

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125 Production of Insulin The Issue

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ff Type I diabetes mellitus is a metabolic disease caused by a lack of insulin. Around 25 people in every 100,000 suffer from type I diabetes.

Key Idea: By using microorganisms to make human insulin, problematic issues of cost, allergic reactions, and ethics have been addressed.

ff It is treatable only with injections of insulin.

ff In the past, insulin was taken from the pancreases of cows and pigs and purified for human use. The method was expensive and some patients had severe allergic reactions to the foreign insulin or its contaminants.

Insulin A chain

Insulin B chain

Concept 1

Concept 2

Concept 3

DNA can be cut at specific sites using restriction enzymes and joined together using DNA ligase. Genes can be inserted into self-replicating bacterial plasmids at the point where the cuts are made.

Plasmids are small, circular pieces of DNA found in some bacteria. They usually carry genes useful to the bacterium. E. coli plasmids can carry promoters required for the transcription of genes.

Under certain conditions, bacteria are able to lose or pick up plasmids from their environment. Bacteria can be readily grown in vat cultures at little expense.

Concept 4

The DNA sequences coding for the production of the two polypeptide chains (A and B) that form human insulin can be isolated from the human genome.

Techniques

The gene is chemically synthesized as two nucleotide sequences, one for the insulin A chain and one for the insulin B chain. The two sequences are small enough to be inserted into a plasmid.

The nucleotide sequences for each insulin chain are synthesized separately and placed into separate plasmids

Plasmids are extracted from Escherichia coli. The gene for the bacterial enzyme b-galactosidase is located on the plasmid. To make the bacteria produce insulin, the insulin gene must be linked to the b-galactosidase gene, which carries a promoter for transcription.

Restriction enzymes are used to cut plasmids at the appropriate site and the A and B insulin sequences are inserted. The sequences are joined with the plasmid DNA using DNA ligase.

The recombinant plasmids are introduced into the bacterial cells

b-galactosidase + chain A

The gene is expressed as separate chains

b-galactosidase + chain B

The recombinant plasmids are inserted back into the bacteria by placing them together in a culture that favours plasmid uptake by bacteria.

The bacteria are then grown and multiplied in vats under carefully controlled growth conditions.

Outcomes

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Insulin A chain

The product consists partly of b-galactosidase, joined with either the A or B chain of insulin. The chains are extracted, purified, and mixed together. The A and B insulin chains connect via disulfide cross linkages to form the functional insulin protein. The insulin can then be made ready for injection in various formulations.

Further applications

Disulfide bond

Insulin B chain

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The techniques used to produce human insulin from genetically modified bacteria can be applied to a range of human proteins and hormones. Proteins currently being produced include human growth hormone, interferon, and factor VIII.

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production in Saccharomyces yeast

The gene for human insulin is inserted into a plasmid. The yeast plasmid is larger than that of E.coli, so the entire gene can be inserted in one piece rather than as two separate pieces.

Cleavage site

The proinsulin protein that is produced folds into a specific shape and is cleaved by the yeast's own cellular enzymes, producing the completed insulin chain.

S

S

S

S

S

By producing insulin this way, the secondary step of combining the separate protein chains is eliminated, making the refining process much simpler.

S

S

S

S

S

S

S

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Yeast cells are eukaryotic and Insulin hence are much larger than bacterial cells. This enables them to accommodate much larger plasmids and proteins within them.

Cleavage site

1. Describe the three major problems associated with the traditional method of obtaining insulin to treat diabetes: (a) (b) (c)

2. Explain the reasoning behind using E. coli to produce insulin and the benefits that GM technology has brought to diabetics:

3. Explain why, when using E. coli, the insulin gene is synthesized as two separate A and B chain nucleotide sequences:

4. Why are the synthetic nucleotide sequences (‘genes’) 'tied' to the b-galactosidase gene?

(b) Secretion and purification of the protein product:

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5. Yeast (Saccharomyces cerevisiae) is also used in the production of human insulin. Discuss the differences in the production of insulin using yeast and E. coli with respect to: (a) Insertion of the gene into the plasmid:

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126 Food for the Masses

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the Earth's land area, leaving little room to grow more crops or farm more animals. Development of new fast growing and high yield crops appears to be part of the solution, but many crops can only be grown under a narrow range of conditions or are susceptible to disease. Moreover, the farming and irrigation of some areas is difficult, costly, and can be environmentally damaging. Genetic modification of plants may help to solve some of these looming problems by producing plants that will require less intensive culture or that will grow in areas previously considered not arable.

Useful organisms

Enzymes

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Key Idea: Genetic engineering has the potential to solve many of the world's food shortage problems by producing crops with greater yields than those currently grown. Currently 1/6 of the world's population are undernourished. If trends continue, 1.5 billion people will be at risk of starvation by 2050 and, by 2100 (if global warming is taken into account), nearly half the world's population could be threatened with food shortages. The solution to the problem of food production is complicated. Most of the Earth's arable land has already been developed and currently uses 37% of

Restriction enzyme

Fungus that is able to survive dry conditions using two enzymes WA-UT1 and Ter-UT2 to facilitate water uptake.

Reverse transcriptase

Plant identified for modification

Engineering your solution

Bacterium known to thrive in dry conditions using a single enzyme DRI-X1 to enhance water absorption.

A solution to the possible future food crisis is to genetically engineer food crops so that they can maximize their growth under adverse conditions. Standard selective breeding techniques could be used to do this, but in some plants this may not be possible or feasible and it may require more time than is available. A selection of genetic tools and organisms with useful characteristics are described. Your task is to use the items shown to devise a technique to successfully create a plant that could be successfully farmed in semi-desert environments such as sub-Saharan Africa. The following page will take you through the procedure. Not all the items will need to be used.

Petri dish

DNA ligase

Plasmid

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Adenovirus

Retrovirus

Agrobacterium

Incubator

Equipment

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Liposome

Possible vectors

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1. Identify the organism you would chose as a 'donor' of drought survival genes and explain your choice:

2. Describe a process to identify and isolate the required gene(s) and identify the tools to be used:

3. Identify a vector for the transfer of the isolated gene(s) into the crop plant and explain your decision:

4. Explain how the isolated gene(s) would be integrated into the vector's genome:

5. (a) Explain how the vector will transform the identified plant:

(b) Identify the stage of development at which the plant would most easily be transformed. Explain your choice:

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6. Explain how the transformed plants could be identified:

7. Explain how a large number of plants can be grown from the few samples that have taken up the new DNA:

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127 KEY TERMS AND IDEAS: Did You Get It?

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

A reaction that is used to amplify fragments of DNA using cycles of heating and cooling.

genetic code

E

Form of intermolecular bonding between hydrogen and an electronegative atom such as oxygen.

hydrogen bonding

F

The sequence of DNA that is read during the synthesis of mRNA.

nucleic acids

G

Two-ringed organic bases, including adenine and guanine.

nucleotides

H

The DNA strand with the same base sequence as the RNA transcript produced (although with thymine replaced by uracil in mRNA).

purines

I

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.

J

The structural units of nucleic acids, DNA and RNA.

K

Single-ringed organic bases, including uracil, cytosine, and thymine.

L

A small circular piece of DNA commonly found in bacteria and often used as a vector in genetic modification.

M

A process that is used to separate different lengths of DNA by placing them in a gel matrix placed in a buffered solution through which an electric current is passed.

N

The process of inserting of a gene from one species into another that would not normally contain the gene. The gene is then expressed in the next generation.

O

Universally found macromolecules composed of chains of nucleotides. These molecules carry genetic information within cells.

P

An enzyme that is able to cut a length of DNA at a specific sequence or site.

base-pairing rule coding strand DNA

gel electrophoresis

plasmid

polymerase chain reaction pyrimidines

restriction enzyme RNA

template strand transgenesis

2. Below is a DNA sequence of sections, A, B, C, D, and E and A', B', C', D', and E'. A scientist wants to isolate sections B, C, and D as a continuous group by PCR. Primers are B and D'. Complete the PCR process, using the right hand (') strand to show how B, C, and D are isolated. The first step is done for you. Note that the process could be shown for the left hand strand also but the sequence of events would be different. A

A'

Primer

A'

B

B'

B

B'

B

B'

C

C'

C'

C

C'

D

D'

D'

D

D'

E

E'

E'

E

E'

Isolation using LH strand

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Unzipped

A'

3. Complete the following paragraph by deleting one of the words in the bracketed () pairs below:

In eukaryotes, gene expression begins with (transcription/translation) which occurs in the (cytoplasm/nucleus).

(Transcription/Translation) is the copying of the DNA code into (mRNA/tRNA). The (mRNA/tRNA) is then transported to

the (cytoplasm/nucleus) where (transcription/translation) occurs. Ribosomes attach to the (mRNA/tRNA) and help match

the codons on (mRNA/tRNA) with the anticodons on (mRNA/tRNA). The (mRNA/tRNA) transports the animo acids to the

ribosome where they are added to the growing (polypeptide/carbohydrate) chain.

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Chromosomes and Cell Division

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Enduring Understanding

3.A

Key terms cancer

3.A.2 In eukaryotes, heritable information is passed on via the cell cycle and mitosis or meiosis and fertilization Essential knowledge

cell cycle

checkpoint chromatid

(a) The cell cycle is a complex set of stages that is regulated via checkpoints, which determine the ultimate fate of the cell

crossing over

c 1

Outline the events in interphase to include growth (G1), DNA synthesis (S), and preparation for mitosis (G2).

c 2

Explain how the cell cycle is regulated by checkpoints. Describe the internal and external signals that provide stop and go signs at checkpoints (e.g. MPF). Explain how cancer results from disruptions to the cell cycle controls.

cytokinesis diploid

Activity number

DNA replication

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130 131

fertilization

c

G1

PR-7

c 3

Explain how cyclins and cyclin-dependent kinases control the cell cycle

130

c 4

Understand that mitosis alternates with interphase in the cell cycle

129

c 5

Describe how specialized cells may enter a stage where they no longer divide, but reenter the cell cycle when given certain cues. Understand that nondividing cells may exit the cell cycle or remain static at a particular stage.

c

PR-7

G2

gamete haploid

homologous chromosomes independent assortment

Compare karyotypes of normal and cancerous cells [part 3].

Investigate the effect of environmental factors on mitosis [part 2].

131

129 130

135

M phase meiosis mitosis

S phase spindle

Afunguy

(b) Mitosis passes a complete genome from the parent cell to daughter cells c 1

Know that mitosis occurs after DNA replication and state why.

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c 2

Know that mitosis is followed by cytokinesis and state the result of this.

129

c 3

Describe the roles of mitosis in the life cycle of organisms.

c 4

Describe mitosis as a continuous process and identify the order of events and the main features of each stage.

c

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Model chromosome behavior and the events in mitosis [part 1].

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(c) Meiosis, a reduction division, followed by fertilization ensures genetic diversity in sexually reproducing organisms

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c 1

Explain how meiosis ensures that gametes each receive one complete haploid set of chromosomes (1n).

c 2

Describe the behavior of homologous chromosomes in prophase and metaphase of meiosis I. Explain the significance of the random orientation of chromosome pairs on the spindle axis [also see 3.A.3.b.1].

136 137 138

c 3

Describe the events during anaphase of meiosis I and explain their significance.

136 137

c 4

Describe crossing over in prophase of meiosis I and explain its significance.

137 140

c 5

Explain how fertilization unites genetically dissimilar gametes, increasing genetic 128 136 variation in populations, and restoring the diploid number of chromosomes. 137

c

PR-7

Model chromosome behavior and the events in meiosis [part 4].

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128 Cell Division from each parent cell. Mitosis is responsible for growth and repair processes in multicellular organisms, and asexual reproduction in some eukaryotes, e.g. yeasts. Meiosis is a special type of cell division concerned with producing sex cells (gametes or haploid spores) for sexual reproduction. It occurs in the sex organs of plants and animals.

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Key Idea: New cells arise from the division of existing cells. There are two types of cell division, mitosis and meiosis. New cells are formed when existing cells divide. In eukaryotes, cell division begins with the replication of a cell's DNA followed by division of the nucleus. There are two forms of cell division. Mitosis produces two identical daughter cells

The 2N (diploid) number refers to the cells each having two whole sets of chromosomes. For a normal human embryo, all cells will have a 2N number of 46.

Female embryo 2N

Many mitotic divisions

Gametes are produced by meiosis; a special division which reduces the chromosome number to half that of a somatic cell. The 1N (haploid) number indicates a single set of chromosomes.

Female adult 2N

Meiosis

Egg 1N

Mitotic division is responsible for growth of body cells (somatic growth) to the adult size.

Male embryo 2N

Fusion of the sperm and the egg in fertilization produces a diploid zygote. This cell will give rise to a new individual through growth and differentiation.

Male adult 2N

Many mitotic divisions

Many mitotic divisions give rise to the adult. Mitosis continues throughout life for cell replacement and repair of tissues. For example blood cells are replaced at a rate of two million per second, and a layer of skin cells is constantly lost and replaced about every 28 days.

Zygote 2N

Embryo 2N

Mitosis

Adult 2N

Mitosis

Cell division and the life cycle of an organism

Meiosis

Jpbarrass

Matthias Zepper

Sperm 1N

Mitosis for repair: 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, such as sea stars, can generate new limbs if they are broken off.

Mitosis for growth: Multicellular organisms develop from a single cell. Organisms, such as this 12 day old mouse embryo, grow by increasing their cell number. Cell growth is highly regulated and once the mouse reaches its adult size, physical growth stops.

Mitosis for reproduction: Sexually reproducing organisms produce haploid gametes by meiosis, but some eukaryotes, such as these yeast cells, can reproduce asexually by budding. The parent cell buds to form a daughter cell, which grows and then separates from the parent cell.

1. (a) Where does mitosis take place in animals? (b) What are the roles of mitosis?

2. (a) Where does meiosis take place in animals?

(b) What is the purpose of meiosis?

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3. Why do gametes produced by meiosis have a haploid chromosome number?

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129 The Eukaryotic Cell Cycle around twice a day, while cells in the liver divide once a year, and those in muscle tissue do not divide at all. If any of these tissues is damaged, however, cell division increases rapidly until the damage is repaired. This variety of length in the cell cycle can be explained by the existence of regulatory mechanisms that are able to slow down or speed up the cell cycle in response to changing conditions.

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Key Idea: The eukaryotic cell cycle can be divided into phases, although the process is continuous. 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 interphase and M phase. Aspects of the cell cycle can vary enormously between cells of the same organism. For example, intestinal cells divide

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

Interphase

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

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

Interphase

ff The first gap phase (G1). ff The S-phase (S).

DNA replication

ff The second gap phase (G2).

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.

growth and preparation for cell division

growth

Mitosis nuclear division

Mitosis and cytokinesis (M-phase)

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

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

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

Mitosis: Nuclear division

During interphase the nuclear material is unwound. As mitosis approaches, the nuclear material begins to reorganize in readiness for 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 organized 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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130 Regulation of the Cell Cycle

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At each checkpoint, a set of conditions determines whether or not the cell will continue into the next phase. Cancer can result when the pathways regulating the checkpoints fail. Non-dividing cells enter a resting phase (G0), where they may remain for a few days or up to several years. Under specific conditions, they may re-enter the cell cycle.

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Key Idea: Regulatory checkpoints are built into the cell cycle to ensure that the cell is ready to proceed from one phase to the next. The failure of these systems can lead to cancer. Cell checkpoints give cells a way to ensure that all cellular processes have been completed correctly before entering the next phase. There are three checkpoints in the cell cycle.

Checkpoints during the cell cycle

G1 checkpoint

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

Interphase

G2 checkpoint

Pass this checkpoint if: • Cell is large enough. • Chromosomes have been successfully duplicated.

Mitosis

Resting phase (G0)

• Cells may exit the cell cycle in response to chemical cues. • Cells in G0 may be quiescent, differentiated, or senescent (aged). • Quiescent (waiting) cells may reenter the cycle in response to chemical cues.

Metaphase checkpoint

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

Cancerous cells

Skin cancer (melanoma). The cancer cells grow more rapidly than the normal skin cells because normal cell regulation checkpoints are ignored. This is why the cancerous cells sit higher than the normal cells and can rapidly spread (metastasize).

B memory cell

Most lymphocytes in human blood are in the resting G0 phase and remain there unless they are stimulated by specific antigens to reenter the cell cycle via G1. G0 phase cells are not completely dormant, continuing to carry out essential cell functions in reduced form.

1. Explain the importance of cell cycle checkpoints:

Many fully differentiated (specialized) cells, e.g. neurons (above), exit the cell cycle permanently and stay in G0. These cells continue their functional role in the body, but do not proliferate. Senescent cells have accumulated mutations, lose function, and die.

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NCI

Normal cells

2. In terms of the cell cycle and the resting phase (G0), distinguish between the behavior of fully differentiated cells, such as neurons, and cells that are quiescent, such as B memory cells

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185 Checkpoints, cyclins, and cancer

Experiments with the eggs of the African clawed frog (Xenopus laevis) provided evidence that a substance found in an M-phase cell could induce a G2 cell to enter M phase. The substance was called M-phase promoting factor (MPF).

Cancer can result when the cell checkpoints, such as the G1 checkpoint, fail. Signals from other cells play a part in this checkpoint. Many of these signals are growth factors (such as platelet derived growth factor) which stimulate the synthesis of cyclin. In cancerous cells, over-production of cyclin leads to uncontrolled growth. The actions of growth factors, cyclin, and CDK are described in the diagram below.

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The trigger for mitosis

Growth factors arrive from other cells

Production of cyclin without the stimulus of growth factors can lead to cancer ADP

Cytoplasm of interphase cell injected into G2 cell.

ATP

Cytoplasm of mitotic phase (M phase) cell injected into G2 cell.

Cyclin concentration increases

Spindle fiber forms

G2 cell enters M phase.

No change in G2 cell.

Other studies have shown that MPF is made up of two subunits. The first subunit is a protein kinase which activates proteins by transferring a phosphate group from ATP to the protein. The second subunit, called a cyclin, activates the first subunit (and thus the first subunit is called a cyclin-dependent kinase or CDK).CDK is constantly present in the cell, while cyclin is not.

CDK phosphorylates and activates target proteins

Cyclin activates CDK

ATP

ADP

3. Explain why the cytoplasm from a M-phase cell could induce a G2 cell to enter M phase:

4. (a) Which checkpoint ensures that replicated chromosomes will separate correctly?

(b) Why is this important?

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5. (a) Explain why signals from other cells play a part in regulating the cell cycle:

(b) How does the over-production of cyclin lead to cancer?

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131 Defective Gene Regulation and Cancer divide without any checks on their proliferation even though they are faulty. Agents capable of causing cancer are called carcinogens. Most carcinogens are also mutagens (they damage the DNA). Any one of a number of cancer-causing factors (including defective genes) may interact to disrupt the cell cycle and result in cancer.

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Key Idea: When cell checkpoints fail, unregulated cell division can occur and cancerous tumors can form. Cells that become damaged beyond repair normally undergo a controlled process called programmed cell death or apoptosis. However, cancerous cells have managed to evade this control and become immortal, continuing to

Cancer: cells out of control

Proto-oncogenes and tumor-suppressor genes

Cancerous transformation results from changes in the genes controlling normal cell growth and division. The resulting cells become immortal and no longer carry out their functional role.

Two types of gene are normally involved in controlling the cell cycle: proto-oncogenes, which start cell division and are essential for normal cell development, and tumor-suppressor genes, which switch off cell division. In their normal form, these types of gene work together, enabling the body to perform vital tasks such as repairing defective cells and replacing dead ones. Mutations in these genes can disrupt this regulation. Proto-oncogenes, through mutation, can give rise to oncogenes, which cause uncontrolled cell division. Mutations to tumor-suppressor genes initiate most human cancers. The best studied tumor-suppressor gene is p53, which encodes a protein that halts the cell cycle so that DNA can be repaired before division.

Normal cell

DNA molecule

Damaged DNA

If repairs are made, then p53 allows the cell cycle to continue.

p53

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. WEB

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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.

Proto-oncogenes Genes that turn on cell division. The mutated form or oncogene leads to unregulated cell division.

Paulo Henrique Orlandi Mourao cc 3.0

Emmanuelm cc3.0

A mutation to one or two of the controlling genes causes a benign (nonmalignant) tumor. As the number of controlling genes with mutations increases, so too does the loss of control until the cell becomes cancerous.

Tumor-suppressor genes When damage occurs, the tumor suppressor gene p53 commands other genes to bring cell division to a halt.

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If the damage is too serious to repair, p53 activates other genes to cause the cell to enter apoptosis (programmed cell death).

Blood smear from a patient with chronic myelogenous leukemia, a disease in which the granulocytic white blood cells (neutrophils, eosinophils, and basophils) proliferate out of control. The disease is associated with a recognizable chromosomal abnormality.

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Karyogram B: Individual with chronic myelogenous leukemia (male and female sex chromosomes are both shown)

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Karyogram A: Normal individual (male and female sex chromosomes are both shown)

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1. How do cancerous cells differ from normal cells?

2. Explain how the cell cycle is normally controlled, including reference to the role of tumor-suppressor genes:

3. With reference to the role of oncogenes, explain how the normal controls over the cell cycle can be lost:

4. Study the two karyograms at the top of the page and answer the following questions: (a) How does karyogram B differ from karyogram A?

(b) Suggest what has happened in the karyogram depicting chronic myelogenous leukemia:

(c) The result of the changes in karyogram B is an abnormal gene (BCR_ABL1). It promotes cell division and blocks apoptosis. Predict the effect of this on the spread of cancerous cells:

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132 Mitosis cells). Mitosis is preceded by interphase in which the cell's DNA is duplicated. The genetic material is then apportioned equally to the two daughter cells. Each daughter cell is genetically identical to the parent cell and there is no change in chromosome number.

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

Mitosis is a stage in the cell cycle

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

Mitosis produces identical daughter cells

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

Parent cell 2N = 4

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

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

DNA replication occurs

RCN

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

M

Mitosis in onion cells (DIC light micrograph)

The mitotic index (number of cells in mitosis á total number of cells) gives a measure of cell proliferation. The mitotic index is high in areas of rapid growth, such as in plant meristems.

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.

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1. Contrast the location of mitosis in plants and animals:

2. When a slide of plant meristematic tissue is examined, only a small proportion of the cells are dividing. Explain why:

3. A cell with 10 chromosomes undergoes 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).

4. What feature of mitosis is central to its role in repairing and replacing damaged tissues in an adult organism:

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133 Mitosis and Cytokinesis dividing the cell in two. Cytokinesis is part of M-phase, but it is distinct from mitosis. Plant cells lack centrioles, and the spindle is organized by equivalent 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 example below illustrates the cell cycle in an animal cell.

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Key Idea: Mitosis is a continuous process, but it is divided into stages to help identify and describe what is occurring. Although mitosis is part of a continuous cell cycle, it is divided into stages (1-6 below) for easier reference. Enzymes are involved at key stages in mitosis. In animal cells, centrioles (located in the centrosome), form the spindle. During cytokinesis (division of the cytoplasm) a constriction forms

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The animal cell cycle and stages of mitosis

Interphase

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

Nucleus

Nuclear membrane

Early prophase

1

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

Centrosomes move to opposite poles

Centrosome (forms spindle)

Late prophase

Cytokinesis

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

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

Metaphase

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

5

Late anaphase

1. What must occur before mitosis takes place?

2. (a) What is the purpose of the spindle fibers?

(b) Where do the spindle fibers originate:

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Spindle

Anaphase

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

Some spindle fibers attach to and organize the chromosomes on the equator of the cell. Some spindle fibers span the cell.

3

4

Spindle fibers attached to chromatids shorten, pulling the chromatids apart. Spindle shortening is catalyzed 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

3. Summarize what happens in each of the following phases:

(a) Prophase:

(b) Metaphase:

(c) Anaphase:

(d) Telophase:

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4. Describe the differences between cytokinesis in an animal cell and a plant cell:

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134 Modeling Mitosis to model mitosis in an animal cell. Four chromosomes were used for simplicity (2N = 4). Images of their work are displayed below.

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Key Idea: Using chenille stems to model the stages of mitosis will help you to visualize and understand the process. Students used chenille stems (pipe cleaners) and yarn

OW

1. (a) Photo 1 represents the stage before mitosis begins. The circular structures are the centrosomes. Name the labeled structures:

Photo 1

A:

B:

A

C

C:

(b) Name the stage shown in the photo:

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

B

Photo 2

Photo 3

Photo 4

Photo 5

Photo 6

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2. Study photos 2-6 below. Identify the mitosis stage shown and briefly describe what is occurring:

3. Draw the final stage of mitosis, in the box. 4. What happens next?

5. How many cells are there?

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135 Environmental Factors Influence Mitosis

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Key Idea: The effect of lectin on mitosis rates in onion plants can be tested for significance using the chi-squared test. Some environmental factors, such as presence of the carbohydrate-binding proteins lectins, have been shown to increase the rate of mitosis in blood cells. Students wanted to determine if PHA-M (a lectin) would increase the level of mitosis in the roots tip cells of onion plants. They exposed the roots of some onion plants to a 50 mg L-1 solution of PHA-M and others to a control solution (water). After an exposure period of 2 days, they prepared and stained the samples so they could count the number of cells in mitosis and interphase.

Onion root tip cells

1. Generate a null hypothesis and an alternative hypothesis for this experiment:

(a) Null hypothesis (H0):

(b) Hypothesis (HA):

Procedure

ff The combined class data is given in table 1. The percentage of cells in interphase or mitosis for the control and treated samples are presented in table 2.

ff A chi-squared test (c2) was used to determine if the null hypothesis should be accepted or rejected at P=0.05 (see table 3). Expected values (E) were calculated by applying the control percentages in table 2 (blue column) to the treated total (i.e. 0.81 x 1712 = 1386). At the end of the calculation, a chi-squared value (∑(O-E)2 / E) is generated.

ff The c2 value is compared to a critical value of c2 for the appropriate degrees of freedom (number of groups - 1). For this test it is 1 (2-1=1). For this data set, the critical value is 3.84 (at 5% probability and 1 degree of freedom). If c2 is < 3.84, the result in not significant and H0 cannot be rejected. If c2 is ≥ 3.84 H0 can be rejected in favor of HA.

Table 1: Class data

Table 2: Percentage of cells in interphase or mitosis

Number of cells

% of cells

Interphase

Mitosis

Total

Control

1292

302

1598

Treated

1406

306

1712

Control

Treated

Interphase cells

81

82

Mitotic cells

19

18

Table 3: Calculation of chi-squared value

Observed (O)

Expected (E)

Interphase cells

1406

1386

Mitosis cells

306

325

(O - E)

(O - E)2

(O -E)2 / E

2. Calculate the chi-squared value by completing table 3. c2 =

3. (a) Should the null hypothesis be accepted or rejected?

(b) Is the result you would have expected? Explain:

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∑ (O -E)2 / E

4. Suggest further investigations to verify the results presented here. Use more paper and attach here if you wish:

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136 Meiosis of the diploid chromosome number. Meiosis occurs in the sex organs of animals and the sporangia of plants. If genetic mistakes (gene and chromosome mutations) occur here, they will be passed on to the offspring (they will be inherited).

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Key Idea: Meiosis is a special type of cell division. It produces sex cells (gametes) for the purpose of sexual reproduction. Meiosis involves a single chromosomal duplication followed by two successive nuclear divisions, and results in a halving

Paternal chromosome

Interphase 2N

When a cell is not dividing (interphase) the chromosomes are not visible, but the DNA is being replicated. The cell shown in the diagram (left) is 2N, where N is the number of copies of chromosomes in the nucleus. N = one copy of each chromosome (haploid). 2N = two copies of each chromosome (diploid).

Maternal chromosome

Meiosis I

Meiosis starts here

Homologous chromosomes pair up: Prior to cell division, the chromosomes condense into visible structures. Replicated chromosomes appear as two sister chromatids held together at the centromere. Homologous chromosomes pair up (synapsis). Crossing over may occur at this time.

Prophase I

(Reduction division)

The first division separates the homologous chromosomes into two intermediate cells.

Independent assortment: Homologous pairs line up in the middle of the cell independently of each other. This results in paternal and maternal chromosomes assorting independently into the gametes.

Metaphase I

Anaphase I

Homologous pairs separate.

Telophase I

Intermediate daughter cells form.

Intermediate cells

Prophase II

Spindle apparatus forms. Chromosomes migrate towards the metaphase plate.

Meiosis II

Metaphase II

(Mitotic division)

Anaphase II

The second division is merely a mitotic one in nature, where the chromatids are pulled apart, but the number of chromosomes remains the same. This allows large numbers of gametes to be produced.

Chromosomes line up on the metaphase plate.

Sister chromatids (now individual chromosomes) separate.

Separate gametes are produced

N

Telophase II

N

N

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N

1. Describe the behavior of the chromosomes in the first division of meiosis:

2. Describe the behavior of the chromosomes in the second division of meiosis:

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137 Meiosis and Variation

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194

chromosomes and results in the recombination of alleles in the gametes. In addition, during meiosis, chromosomes are distributed randomly to the gametes, a process known as independent assortment. These features of meiosis, together with the random fusion of gametes at fertilization, result in genetically variable offspring.

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Key Idea: Independent assortment and crossing over during meiosis introduce variation into the gametes. The events during meiosis (crossing over and independent assortment of chromosomes) create variation in the gametes produced. Crossing over involves the mutual exchange of alleles between non-sister chromatids of homologous

Crossing over and recombination

Chromosomes replicate during interphase, before meiosis, to produce replicated chromosomes with sister chromatids held together at the centromere (see below). When the replicated chromosomes are paired during the first stage of meiosis, non-sister chromatids may become entangled and segments may be exchanged in a process called crossing over. Crossing over results in the recombination of alleles (variations of the same gene) producing greater variation in the offspring than would otherwise occur. No crossing over

Sister chromatids

Crossing over

Recombined chromosomes

Non-sister chromatids

Non-sister chromatids

Centromere

Non-sister chromatids

Possible gametes with no crossing over

Possible gametes with crossing over

Independent assortment

ff Independent assortment is the random alignment and distribution of chromosomes during meiosis. It is an important mechanism for producing variation in gametes.

or

ff The law of independent assortment states that the alleles for separate traits are passed independently of one another from parents to offspring. In other words, the allele a gamete receives for one gene does not influence the allele received for another gene. ff This results in the production of 2x different possible combinations (where x is the number of chromosome pairs).

1

2

3

4

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ff For the example right, there are two chromosome pairs. The number of possible allele combinations in the gametes is 22 = 4.

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

2. How does independent assortment increase genetic variation in gametes?

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138 Modeling Meiosis over increases genetic variability. This is demonstrated by studying how two of your own alleles are inherited by the child produced at the completion of the activity. Completing this activity will help you to visualize and understand meiosis. It will take 25-45 minutes.

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Key Idea: We can simulate crossing over, gamete production, and the inheritance of alleles during meiosis using ice-block sticks to represent chromosomes. This practical activity simulates the production of gametes (sperm and eggs) by meiosis and shows you how crossing

Background

Dominant: Tongue roller

Each of your somatic cells contain 46 chromosomes. You received 23 chromosomes from your mother (maternal chromosomes), and 23 chromosomes from your father (paternal chromosomes). Therefore, you have 23 homologous (same) pairs. For simplicity, the number of chromosomes studied in this exercise has been reduced to four (two homologous pairs). To study the effect of crossing over on genetic variability, you will look at the inheritance of two of your own traits: the ability to tongue roll and handedness. Chromosome #

Phenotype

Genotype

Tongue roller Non-tongue roller Right handed Left handed

TT, Tt tt RR, Rr rr

10 10 2 2

Recessive: Non-roller

Trait

Record your phenotype and genotype for each trait in the table (right). NOTE: If you have a dominant trait, you will not know if you are heterozygous or homozygous for that trait, so you can choose either genotype for this activity.

Label four sticky dots with the alleles for each of your phenotypic traits, and stick it onto the appropriate chromosome. For example, if you are heterozygous for tongue rolling, the sticky dots with have the alleles T and t, and they will be placed on chromosome 10. If you are left handed, the alleles will be r and r and be placed on chromosome 2 (right).

2. Randomly drop the chromosomes onto a table. This represents a cell in either the testes or ovaries. Duplicate your chromosomes (to simulate DNA replication) by adding four more identical ice-block sticks to the table (below). This represents interphase.

Recessive: Left hand

Phenotype

Genotype

Handedness

Tongue rolling

BEFORE YOU START THE SIMULATION: Partner up with a classmate. Your gametes will combine with theirs (fertilization) at the end of the activity to produce a child. Decide who will be the female, and who will be the male. You will need to work with this person again at step 6. 1. Collect four ice-blocks sticks. These represent four chromosomes. Color two sticks blue or mark them with a P. These are the paternal chromosomes. The plain sticks are the maternal chromosomes. Write your initial on each of the four sticks. Label each chromosome with their chromosome number (right).

Dominant: Right hand

Your genotype

t

T

Paternal chromosome

r

r

Paternal chromosome

Maternal chromosome

Your initials

LB 10

LB

LB

LB

10

2

2

Homologous pair

Chromosome number

Homologous pair

3. Simulate prophase I by lining the duplicated chromosome pair with their homologous pair (below). For each chromosome number, you will have four sticks touching side-by-side (A). At this stage crossing over occurs. Simulate this by swapping sticky dots from adjoining homologs (B). (A)

T

T

t

t

r

r

r

r

r

2

T

LB

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LB

LB LB LB LB

r

2

r

LB 2

LB 2

r

10 10 10 10

T

LB LB LB LB 2

2

2

2

(B)

LB

LB

10

10

LB L B 10 1 0

t

t

T

t

T

t

LB LB LB LB 10 10 10 10

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r

r

r

r

LB LB LB LB 2

2

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2

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

T

t

T

t

t

T

t

T

LB LB

LB LB

LB LB

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 2

t

Intermediate cell 1

t

T

T

r

r

r

Paternal chromosome

LB

r

LB 2

Haploid (N) gamete

10

Maternal chromosome

10

LB

Maternal chromosome

LB 2

LB 2

t

t

r

r

r

T

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10

LB 2

LB 2

LB 2

10

LB

LB

T

r

LB 2

LB LB 10 10

LB

10

LB

10

LB 2

6. Pair up with the partner you chose at the beginning of the exercise to carry out fertilization. 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

Tongue rolling

Phenotype

Genotype

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139 Mitosis vs Meiosis purposes and outcomes. Mitosis is the simpler of the two and produces two identical daughter cells from each parent cell. Mitosis is responsible for growth and repair processes in multicellular organisms and reproduction in single-celled and asexual eukaryotes. Meiosis involves a reduction division in which haploid gametes are produced for the purposes of sexual reproduction. Fusion of haploid gametes in fertilization restores the diploid cell number in the zygote.

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Key Idea: Mitosis produces two daughter cells genetically identical to the parent cell. Meiosis produces four daughter cells that contain half the genetic information of the parent  cell. Cell division is fundamental to all life, as cells arise only by the division of existing cells. All types of cell division begin with replication of the cell's DNA. In eukaryotes, this is followed by division of the nucleus. There are two forms of nuclear division: mitosis and meiosis, and they have quite different

Meiosis

2N

2N

Genetic material can be exchanged between chromosomes in meiosis I

Homologous chromosomes line up in pairs (bivalents) at the equatorial plate

Cell division

Meiosis II: 'Mitotic' division

Homologous chromosomes line up at the equatorial plate

Cell division

Cell division

2N

Meiosis I: Reduction division

Mitosis

N

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1. Explain how mitosis conserves chromosome number while meiosis reduces the number from diploid to haploid:

2. Describe a fundamental difference between the first and second divisions of meiosis:

3. How does meiosis introduce genetic variability into gametes and offspring (following gamete fusion in fertilization)?

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140 Crossing Over Problems Key Idea: Crossing over can occur in multiple places in chromosomes, producing a large amount of genetic variation. The diagram below shows a pair of homologous chromosomes about to undergo crossing over during meiosis I. There are known crossover points along the length of the chromatids

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(same on all four chromatids shown in the diagram). In the prepared spaces below, draw the gene sequences after crossing over has occurred on three unrelated and separate occasions (it would be useful to use different colored pens to represent the genes from the two different chromosomes).

Homologous chromosomes

a b c d e f g

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

6

5

7

8

9

Possible known crossover points on the chromatid

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

1. Crossing over occurs at a single point between the chromosomes above.

1

(a) Draw the gene sequences for the four chromatids (on the right), after crossing over has occurred at crossover point: 2

2 3

(b) Which genes have been exchanged with those on its homologue (neighbor chromosome)?

4

2. Crossing over occurs at two points between the chromosomes above.

(a) Draw the gene sequences for the four chromatids (on the right), after crossing over has occurred between crossover points: 6 and 7.

1 2 3

(b) Which genes have been exchanged with those on its homologue (neighbor chromosome)?

4

3. Crossing over occurs at four points between the chromosomes above.

(a) Draw the gene sequences for the four chromatids (on the right), after crossing over has occurred between crossover points: 1 and 3, and 5 and 7. (b) Which genes have been exchanged with those on its homologue (neighbor chromosome)?

1 2

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3

4

4. What would be the genetic consequences if there was no crossing over between chromatids during meiosis?

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141 KEY TERMS AND IDEAS: Did You Get It?

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

cell cycle

cell division checkpoint

crossing over

A The phase of a cell cycle involving nuclear division in which the replicated chromosomes in a cell nucleus are separated into two identical sets.

B The process of double nuclear division (reduction division) to produce four nuclei, each containing half the original number of chromosomes (haploid).

C Control mechanisms within the cell cycle to ensure that certain prerequisites have been meet by the cell before it proceeds to the next stage of the cell cycle. D The stage in the cell cycle between divisions.

cytokinesis

independent assortment

E

The exchange of genetic material between two non-sister chromatids of homologous chromosomes during meiosis.

F

Process by which a parent cell divides into two or more daughter cells.

interphase

G The ordered sequence of events that occur in a cell leading to its division into two daughter cells.

meiosis

H The random alignment and distribution of chromosomes during meiosis.

mitosis

I

The part of the cell cycle in which division of the cytoplasm occurs.

2. (a) Label the cell cycle right with the following labels: G1, G2, M, S, cytokinesis, G1 checkpoint, G2 checkpoint, metaphase checkpoint.

(b) Briefly describe what happens in each of the following phases:

G1:

C

G2:

M phase:

S phase:

3. (a) Identify the process occurring in the circled chromosomes right:

(b) Does this occur during mitosis or meiosis?

(c) At what specific stage of your answer in (b) does this occur?

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4. (a) A region of plant meristem (growing region) was examined on a slide under a microscope. It was found that 65 cells were in the process of mitosis and 216 were in interphase. What is the mitotic index of the plant meristem? 0.12

(b) What does the plot right show about the rate of mitosis in plant roots?

Mitotic index

0.10 0.08 0.06 0.04 0.02 0

0.0

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0.5 1.0 1.5 Distance from root cap (mm)

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The Chromosomal Basis of Inheritance

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Enduring Understanding

3.A

Key terms allele

3.A.3 The chromosomal basis of inheritance helps us understand how genes are transmitted from parent to offspring

Activity number

Essential knowledge

chi-squared test codominance

(a) Rules of probability can be applied to analyze the inheritance of single gene traits

dihybrid cross

c 1

genetic disorder

Show how the rules of probability can be applied to Mendelian crosses to determine the expected phenotypes and genotypes of offspring.

6 142 145

incomplete dominance

(b) Segregation and independent assortment of chromosomes result in genetic variation

independent assortment

c 1

Explain how the rules of segregation and independent assortment are applied to genes on different chromosomes.

c 2

Define linkage and describe its consequences. Know that the probability that linked genes will segregate as a unit is a function of the distance between them.

c 3

Explain how the pattern of inheritance (monohybrid, dihybrid, linked, or sexlinked) can often be predicted from data providing parent genotype/phenotype and/or offspring phenotypes/genotypes.

145 146 151 - 155 158 - 160

c

SKILL

Use the chi-square test to analyze and explain observed vs predicted results of genetic crosses.

156 157

lethal allele linkage

Mendel's laws

Mendelian inheritance monohybrid cross multiple alleles multiple genes (=polygenes)

non-disjunction non-nuclear inheritance

144 145 153

(c) Some genetic disorders can be attributed to the inheritance of single gene traits or chromosomal changes

c 1

recombination

Use examples to explain how genetic disorders in humans are the result of inheriting single gene traits or chromosomal changes arising through nondisjunction. Examples include sickle cell disease, Huntington's disease, X-linked color blindness, trisomy 21 and Klinefelter syndrome.

rules of probability

(d) Ethical, social, and medical issues surround human genetic disorders

sex determination

c 1

sex linked gene X-linkage

Discuss some of the ethical, social, and medical issues associated with genetic disorders in humans. You could include reference to reproductive issues and issues around ownership of genetic information and privacy.

3.A.4 The inheritance patterns of many traits cannot be explained by simple Mendelian genetics

160 - 162

162 163

Activity number

Essential knowledge

(a) Many traits are the product of multiple genes and/or physiological processes

Explain how you can identify departures from Mendelian patterns of inheritance by quantitative analysis of observed phenotypic ratios. Inheritance patterns may involve codominance, incomplete dominance, lethal alleles, and multiple genes.

147 - 150 164

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c 1

(b) Some traits are determined by genes on sex chromosomes c 1

Explain how the sex chromosomes determine sex and influence the inheritance of sex linked characteristics in mammals and flies. To help your explanation: • Explain why, in mammals and flies, there are few genes on the Y chromosome. • Describe how sex is determined in mammals and flies. • Explain why X-linked recessive traits are always expressed in males. • Describe sex-limited traits in mammals.

165 166

(c) Some traits result from nonnuclear inheritance

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c 1

Explain why the inheritance of chloroplast and mitochondrial DNA does not follow simple Mendelian rules.

167

c 2

Explain why mitochondrial-determined traits are maternally inherited.

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142 Alleles of chromosomes, one set from each parent. The equivalent chromosomes that form a pair are termed homologues. They carry equivalent sets of genes, but there is the potential for different versions of a gene (alleles) to exist in a population.

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Key Idea: Eukaryotes generally have paired chromosomes. Each chromosome contains many genes and each gene may have a number of versions called alleles. Sexually reproducing organisms usually have paired sets

Homologous chromosomes In sexually reproducing organisms, most cells have a homologous pair of chromosomes (one coming from each parent). This diagram shows the position of three different genes on the same chromosome that control three different traits (A, B and C).

3

4

5

2

Chromosomes are formed from DNA and proteins. DNA tightly winds around special proteins to form the chromosome.

Having two different versions (alleles) of gene A is a heterozygous condition. Only the dominant allele (A) will be expressed. Alleles differ by only a few bases.

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.

3

a

1

4

A

1

5

2

B

B

c

c

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

This diagram shows the complete chromosome complement for a hypothetical organism. It has a total of ten chromosomes, as five, nearly identical pairs (each pair is numbered). Each parent contributes one chromosome to the pair. The pairs are called homologues or homologous pairs. Each homologue carries an identical assortment of genes, but the version of the gene (the allele) from each parent may differ.

A gene is the unit of heredity. Genes occupying the same locus or position on a chromosome code for the same trait (e.g. dimpled chin). Paternal chromosome originating from the sperm of this individual's father.

1. Define the following terms used to describe the allele combinations in the genotype for a given gene:

(a) Heterozygous:

(b) Homozygous dominant:

(c) Homozygous recessive:

2. For a gene given the symbol ‘A’, name the alleles present in an organism that is identified as: (b) Homozygous dominant:

3. What is a homologous pair of chromosomes?

4. Discuss the significance of genes existing as alleles:

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(c) Homozygous recessive:

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(a) Heterozygous:

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143 Mendel’s Pea Plant Experiments

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Key Idea: Many genes produce phenotypic traits that are inherited in predictable ratios, as shown by Mendel's pea experiments. Gregor Mendel (1822-84), right, was an Austrian monk who carried out the pioneering studies of inheritance. Mendel bred pea plants to study the inheritance patterns of a number of traits (specific characteristics). He showed that characters could be masked in one generation but could reappear in later generations and proposed that inheritance involved the transmission of discrete units of inheritance from one generation to the next. At the time the mechanism of inheritance was unknown, but further research has provided an accepted mechanism and we know these units of inheritance are genes. The entire genetic makeup of an organism is its genotype.

Mendel examined six traits and found that they were inherited in predictable ratios, depending on the phenotypes (the physical appearance) of the parents. Some of his results from crossing heterozygous plants are tabulated below. The numbers in the results column represent how many offspring had those traits.

1. 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. 2. Calculate the ratio of dominant phenotypes to recessive phenotypes (to two decimal places). The first one has been done for you (5474 ÷ 1850 = 2.96). Place your answers in the spaces provided in the table below: Possible phenotypes

Seed shape

Wrinkled

Round

Seed color

Green

Axial

Constricted

Stem length

Inflated

Round

Recessive:

Wrinkled

Round

5474

TOTAL

7324

Green

2001

Yellow

6022

TOTAL

8023

Green

428

Yellow

152

TOTAL

580

Axial

651

Terminal

207

TOTAL

858

Recessive

Constricted

299

Dominant:

Inflated

882

TOTAL

1181

Terminal

Pod shape

Dominant:

1850

Yellow

Flower position

Dominance

Wrinkled

Yellow

Pod color

Green

Results

Tall

787

Dwarf

277

TOTAL

1064

Ratio

2.96 : 1

Dominant:

Recessive

Dominant:

Recessive

Dominant:

Recessive

Dominant:

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Trait

Recessive

3. Mendel’s experiments identified that two heterozygous parents should produce offspring in the ratio of three times as many dominant offspring to those showing the recessive phenotype.

(a) Which three of Mendel’s experiments provided ratios closest to the theoretical 3:1 ratio?

(b) Suggest why these results deviated less from the theoretical ratio than the others:

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

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Mendel's laws of inheritance were based on his observations and govern how genes are passed to the offspring.

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

X

White

Purple

This model explained many observations that could not be explained by the idea of blending inheritance, which was universally accepted prior to this theory. The trait for flower color (right) appears to take on the appearance of only one parent plant in the first generation, but reappears in later generations.

The offspring are inbred (self-pollinated)

Generation 2

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

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

a

a

Oocyte

A

NOTE: This diagram has been simplified, omitting the stage where the second chromatid is produced for each chromosome.

This diagram shows are 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.

A A

Meiosis

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

Allele pairs separate independently during gamete formation, and traits are passed on to offspring independently of one another (this is only true for unlinked genes).

Generation 1

X

Law of segregation

Law of independent assortment

Parent plants

A

Egg

a

Egg

a

Egg

Egg

Oocyte

Intermediate cell

A A a

B B b

A A

b

A

Intermediate cell

a

a

Genotype: AaBb

b

A

b

Ab

b

B B

a

Eggs

b

Ab

a

a

B

aB

B

aB

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1. State the property of genetic inheritance that allows parent pea plants of different flower color to give rise to flowers of a single color in the first generation, with both parental flower colors reappearing in the following generation:

2. The oocyte is the egg producing cell in the ovary of an animal. In the diagram illustrating the law of segregation above:

(a) State the genotype for the oocyte (adult organism):

(c) State how many different kinds of gamete can be produced by this oocyte:

(b) State the genotype of each of the four gametes:

3. The diagram illustrating the law of independent assortment (above) shows only one possible result of the random sorting of the chromosomes to produce: Ab and aB in the gametes. (a) List another possible combination of genes (on the chromosomes) ending up in gametes from the same oocyte:

(b) How many different gene combinations are possible for the oocyte?

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145 Basic Genetic Crosses

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describes the offspring of a cross between true-breeding (homozygous) parents. A back cross is a cross between an offspring and one of its parents. If the back cross is to a homozygous recessive, it can be used as a test cross.

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Key Idea: The outcome of a cross depends on the parental genotypes and can be determined using Punnett squares. Examine the diagrams below on monohybrid (one gene) and dihybrid (two gene) inheritance. The F1 generation

Monohybrid cross F1

Homozygous purple

Parents:

PP

P

Gametes:

Monohybrid cross F2

Heterozygous purple

Homozygous white

X

p

P

Parents:

Heterozygous purple

X

pp

Pp

Pp

p

P

P

Male gametes:

Female gametes:

p

p

F1

Genotypes: All

Pp

Phenotypes: All purple

75% Pp - purple 25% pp - white

A true-breeding organism is homozygous for the gene involved. The F1 offspring of a cross between two true breeding parent plants are all purple (Pp).

Dihybrid cross

A dihybrid cross studies the inheritance patterns of two genes. In pea seeds, yellow color (Y) is dominant to green (y) and round shape (R) is dominant to wrinkled (r). Each true breeding parental plant has matching alleles for each of these characters (YYRR or yyrr). F1 offspring will all have the same genotype and phenotype (yellow-round: YyRr).

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).

Homozygous yellow-round

Homozygous green-wrinkled

Parents:

Gametes:

X

YR

yr

YyRr

F1 all yellow-round

2. In the boxes below, use fractions to indicate the numbers of each phenotype produced from this cross.

YR

Yellow-round

Male gametes

Green-round

Yellow-wrinkled Green-wrinkled

3. Express these numbers as a ratio:

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Yr

yR yr

for the F2

Female gametes

Offspring (F2) Possible fertilizations

YyRr

YR

Yr

yR

yr

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1. Fill in the Punnett square (below right) to show the genotypes of the F2 generation.

X

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146 Monohybrid Cross allele (b), produces white. Each parent can produce two types of gamete by meiosis. Determine the genotype and phenotype frequencies for the crosses below. For questions 3 and 4, also determine the gametes produced by each parent (write these in the circles) and offspring genotypes and phenotypes (write these inside the offspring shapes).

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Key Idea: A monohybrid cross studies the inheritance pattern of one gene. The offspring of these crosses occur in predictable ratios. In this activity, you will examine six types of matings possible for a pair of alleles governing coat color in guinea pigs. A dominant allele (B) produces black hair and its recessive Homozygous white

X

bb

b

Heterozygous black

Homozygous black

b

Parents

BB

B

B

Gametes

X

Bb

B

Homozygous black

BB

b

B

BB

Bb

B

Possible fertilisations

Bb

Bb

Bb

Bb

100% Bb Phenotype frequency: 100% black

1. (a) Genotype frequency: (b)

Heterozygous black

Bb

Offspring (F1)

Heterozygous black

X

Bb

3. (a) Genotype frequency:

(b) Phenotype frequency:

BB

Bb

2. (a) Genotype frequency:

(b) Phenotype frequency: Homozygous white

bb

Heterozygous black

X

Bb

4. (a) Genotype frequency:

(b) Phenotype frequency:

5. Describe the back cross that produces 100% black guinea pigs:

(b) Explain the result:

(c) What is this type of cross called?

(d) Explain why it is diagnostic:

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6. A white guinea pig and a black guinea pig are crossed. All of the guinea pigs that are born are white. (a) What is the genotype of the black guinea pig?

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147 Codominance are independently and equally expressed. Examples include the human blood group AB and certain coat colors in horses and cattle. Reddish coat color is equally dominant with white. Animals that have both alleles have coats that are roan (both red and white hairs are present).

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Key Idea: In inheritance involving codominant alleles, neither allele is recessive and both alleles are equally and independently expressed in the heterozygote. Codominance is an inheritance pattern in which both alleles in a heterozygote contribute to the phenotype and both alleles Red bull

White cow

CRCR

CWCW

Parents

CR

CRCW

CR

Gametes

CRCW

CRCW

CW

CW

A roan shorthorn heifer

CRCW

Offspring

Roan

Roan

Roan

Roan

1. Explain how codominance of alleles can result in offspring with a phenotype that is different from either parent:

In the shorthorn cattle breed, coat color is inherited. White shorthorn parents always produce calves with white coats. Red parents always produce red calves. However, when a red parent mates with a white one, the calves have a coat color that is different from either parent; a mixture of red and white hairs, called roan. Use the example (left) to help you to solve the problems below. White bull

Roan cow

2. 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 phenotype ratio for this cross?

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

Roancow cow Roan

Unknown Unknown bullbull

(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)

4. Describe the classical phenotypic ratio for a codominant gene resulting from the cross of two heterozygous parents (e.g. a cross between two roan cattle):

?

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3. A farmer has only roan cattle on his farm. He suspects that one of the neighbors' 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 neighbor has a red bull, the other has a roan.

Red

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Roan

Red

Roan

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148 Codominance in Multiple Allele Systems enzyme that adds a different, specific sugar to the basic sugar molecule. Because alleles A and B are codominant they are expressed equally. The recessive allele O produces a non-functioning enzyme and is unable to make any changes to the basic sugar molecule. Blood group A and B antigens react with antibodies in the blood of people with incompatible blood types so must be matched for transfusion.

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Key Idea: The human ABO blood group system is a multiple allele system involving the codominant alleles A and B and the recessive allele O. On the surface of red blood cells there are sugars called the ABO antigens. These antigens are made by enzymes that link particular combinations of sugars (antigens) together. The codominant alleles A and B each encode a different ff Recessive allele: O produces a non-functioning protein ff Dominant allele:

A

produces an enzyme which forms A antigen

ff Dominant allele:

B

produces an enzyme which forms B antigen

If a person has the AO allele combination then their blood group will be group A. The presence of the recessive allele (O) has no effect on the blood group in the presence of a dominant allele. AA also produces blood group A. 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 (phenotype)

Possible genotypes

O A

White

Black

Native American

OO

45%

49%

79%

AA AO

40%

27%

16%

B

11%

20%

4%

AB

4%

4%

1%

* Frequency is based on North American population

Source: www.kcom.edu/faculty/chamberlain/Website/MSTUART/Lect13.htm

Blood group: O

Blood group: O

Cross 2

OO

AB

A

Frequency*

X

OO

B

Possible fertilizations

Blood groups

AA

AB

AB

BB

A

AB

AB

B

Blood group: AB

Parental genotypes

Gametes

Possible fertilizations Children's genotypes Blood groups

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Blood group: A

Cross 3

AB

X

AO

Blood group: A

Blood group: B

Cross 4

AA

X

BO

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Children's genotypes

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208 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 fertilizations Children’s genotypes

Blood groups

Blood group

Blood group

A

O

X

Parental genotypes

Gametes

Child’s genotype

Blood group

B

Pedigree chart key

Male

Female

Marriage

(a) State the sex of the children:

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

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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.

B

AB

B

AB

Child 1

Child 2

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149 Incomplete Dominance homozygous parental phenotypes. In crosses involving incomplete dominance the phenotype and genotype ratios are identical. Examples of incomplete dominance includes flower color in snapdragons (Antirrhinum) and four o'clocks (Mirabilis) (below). In this type of inheritance the phenotype of the offspring results from the partial influence of both alleles.

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Key Idea: Incomplete dominance refers to the situation where the action of one allele does not completely mask the action of the other and neither allele has dominant control over the trait. In incomplete dominance the heterozygous offspring are intermediate in phenotype between the contrasting Red flower

CR CR

Pure breeding snapdragons produce red or white flowers (left). When red and white-flowered parent plants are crossed a pinkflowered offspring is produced. If the offspring (F1 generation) are then crossed together all three phenotypes (red, pink, and white) are produced in the F2 generation.

White flower

CWCW

Parents

Gametes

CR

CW

CW

Offspring

CR CW

CR CW

CR CW

CR CW

Pink

Pink

Pink

Pink

Khalid Mahmood

CR

Four o'clocks (above) are also known to have flower colors controlled by incompletely dominate alleles. Pure breeding four o'clocks produce crimson, yellow or white flowers. Crimson flowers crossed with yellow flowers produced reddish-orange flowers, while crimson flowers crossed with white flowers produce magenta (reddish-pink) flowers.

2. A plant breeder wanted to produce snapdragons for sale that were only pink or white (i.e. no red). Determine the phenotypes of the two parents necessary to produce these desired offspring. Use the Punnett square (right) to help you:

Gametes from female

1. Explain how incomplete dominance of alleles differs from complete dominance:

Gametes from male

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

(b) State the phenotype ratio:

Magenta flower

Gametes

Possible fertilizations

Offspring

Phenotypes

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3. Another plant breeder crossed two four o'clocks, known to have its flower color controlled by a gene which possesses incomplete dominant alleles. Pollen from a magenta flowered plant was placed on the stigma of a crimson flowered plant.

Crimson flower

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150 Lethal Alleles 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.

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Key Idea: Lethal alleles are mutations of a gene that produce a non-functional gene product which may affect an organism's survival. Lethal alleles are usually a result of mutations in essential genes. They may result in death of an organism because an essential protein is not produced. Some lethal

AYA

AY

AYA

A

Gametes

AY

A

Possible fertilizations

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 color 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. 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).

Yellow

Yellow

½ AYA

Non-yellow ¼ AA

Manx cats are born without a tail

Normal cats are born with a tail

Normal cat

(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:

Gametes from male

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1. In Manx cats, the allele for taillessness (ML) is incompletely dominant over the recessive allele for normal tail (M). Tailless Manx cats are heterozygous (MLM) and carry a recessive allele for normal tail. Normal tailed cats are MM. A cross between two Manx (tailless) cats, produces two Manx to every one normal tailed cat (not a regular 3 to 1 ratio).

Gametes from female

Manx cat

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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151 Problems Involving Monohybrid Inheritance alleles involved are associated with various phenotypic traits controlled by a single gene. The problems are to give you practise in problem solving using Mendelian genetics.

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Key Idea: For monohybrid crosses involving autosomal unlinked genes, the offspring appear in predictable ratios. The following problems involve Mendelian crosses. The

1. A dominant gene (W) produces wire-haired texture in dogs; its recessive allele (w) produces smooth hair. A group of heterozygous wire-haired individuals are crossed and their F1 progeny are then test-crossed. Determine the expected genotypic and phenotypic ratios among the test cross progeny:

2. In sheep, black wool is due to a recessive allele (b) and white wool to its dominant allele (B). A white ram is crossed to a white ewe. Both animals carry the black allele (b). They produce a white ram lamb, which is then back crossed to the female parent. Determine the probability of the back cross offspring being black:

3. A homozygous recessive allele, aa, is responsible for albinism. Humans can exhibit this phenotype. In each of the following cases, determine the possible genotypes of the mother and father, and their children:

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

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

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

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. 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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4. Two mothers give birth to sons at a busy hospital. The son of the first couple has hemophilia, a recessive, X-linked disease. Neither parent from couple #1 has the disease. The second couple has an unaffected son, despite the fact that the father has hemophilia. The two couples challenge the hospital in court, claiming their babies must have been swapped at birth. You must advise as to whether or not the sons could have been swapped. What would you say?

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152 Dihybrid Cross

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chromosomes and are sorted independently of each other during meiosis. The two genes in the example below are on separate chromosomes and control two unrelated characteristics, hair color and coat length. Black (B) and short (L) are dominant to white and long.

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Key Idea: A dihybrid cross studies the inheritance pattern of two genes. In crosses involving unlinked autosomal genes, the offspring occur in predictable ratios. There are four types of gamete produced in a cross involving two genes, where the genes are carried on separate Homozygous black, short hair

BBLL

Parents (P)

BL

Gametes

BL

BL

Homozygous white, long hair

bbll

X

BL

Parents: The notation P is used for a cross between true breeding (homozygous) parents.

bl

bl

bl

bl

Gametes: Only one type of gamete is produced from each parent (although they will produce four gametes from each oocyte or spermatocyte). This is because each parent is homozygous for both traits.

Possible fertilizations

Offspring (F1)

BbLl

Female gametes

Offspring (F2)

Possible fertilizations

X

BbLl

F1 offspring: There is only one kind of gamete from each parent, therefore only one kind of offspring produced in the first generation. The notation F1 is used to denote the heterozygous offspring of a cross between two true breeding parents.

BL

Bl

bL

bl

Male gametes

BL Bl

F2 offspring: The F1 were mated with each other (selfed). Each individual from the F1 is able to produce four different kinds of gamete. Using a grid called a Punnett square (left), it is possible to determine the expected genotype and phenotype ratios in the F2 offspring. The notation F2 is used to denote the offspring produced by crossing F1 heterozygotes.

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

bL

bl

BBLL BbLL BBLl BbLl

The offspring can be arranged in groups with similar phenotypes:

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

BBll Bbll

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

bbll

9 black, short hair

3 black, long hair

3 white, short hair

bbLL bbLl

Phenotype

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Genotype

A total of 3 offspring with one of 2 different genotypes can produce white, short hair

Only 1 offspring of a given genotype can produce white, long hair

1 white, long hair

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153 Inheritance of Linked Genes chromosome. In genetic crosses, linkage is indicated when a greater proportion of the offspring from a cross are of the parental type (than would be expected if the alleles were on separate chromosomes and assorting independently). Linkage reduces the genetic variation that can be produced in the offspring.

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Key Idea: Linked genes are genes found on the same chromosome and tend to be inherited together. Linkage reduces the genetic variation in the offspring. Genes are linked when they are on the same chromosome. Linked genes tend to be inherited together and the extent of crossing over depends on how close together they are on the Parent 1 (2N)

Parent 2 (2N)

Overview of linkage

Dark body, blue eyes AaBb

White body, white eyes aabb

Chromosomes before replication

A

a

B

b

Genes are linked when they are found on the same chromosomes. In this case A (body color) and B (eye color) are linked.

a

a

b

b

a

a

a

a

b

b

b

b

Chromosomes after replication

A

A

a

a

B

B

b

b

X

Each replicated chromosome produces two identical gametes. One of each is shown here.

Meiosis

A

a

B

b

Gametes

A

a

A

a

B

b

B

b

Offspring

AaBb AaBb Dark body, blue eyes

a

a

b

b

a

a

a

a

b

b

b

b

aabb aabb White body, white eyes

Possible offspring

1. What is the effect of linkage on the inheritance of genes?

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Only two kinds of genotype combinations are possible. They are they same as the parent genotype.

2. Explain how linkage decreases the amount of genetic variation in the offspring:

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An example of linked genes in Drosophila

In the example shown left, wild type alleles are dominant and are given an upper case symbol of the mutant phenotype (Cu or Eb). This notation used for Drosophila departs from the convention of using the dominant gene to provide the symbol. This is necessary because there are many mutant alternative phenotypes to the wild type (e.g. curled and vestigial wings). A lower case symbol of the wild type (e.g. ss for straight wing) would not indicate the mutant phenotype involved.

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The genes for wing shape and body color are linked (they are on the same chromosome)

Drosophila and linked genes

Parent

Phenotype Genotype Linkage

Wild type female

Mutant male

Straight wing Gray body

Curled wing Ebony body

Cucu Ebeb

cucu ebeb

Cu

Eb

cu

eb

cu

eb Meiosis

cu

eb

Gametes from female fly (N)

Drosophila melanogaster is known as a model organism. Model organisms are used to study particular biological phenomena, such as mutation. Drosophila melanogaster is particularly useful because it produces such a wide range of heritable mutations. Its short reproduction cycle, high offspring production, and low maintenance make it ideal for studying in the lab.

Gametes from male fly (N)

W

KTBN

Sex of offspring is irrelevant in this case

Drosophila melanogaster examples showing variations in eye and body color. The wild type is marked with a w in the photo above.

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.

3. Complete the linkage diagram above by adding the gametes in the ovals and offspring genotypes in the rectangles.

4. (a) List the possible genotypes in the offspring (above, left) if genes Cu and Eb had been on separate chromosomes:

(b) If the female Drosophila had been homozygous for the dominant wild type alleles (CuCu EbEb), state: The genotype(s) of the F1: The phenotype(s) of the F1:

The genotype(s) of the F1:

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5. A second pair of Drosophila are mated. The female genotype is Vgvg EbEb (straight wings, gray 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:

The phenotype(s) of the F1:

6. Explain why Drosophila are often used as model organisms in the study of genetics:

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154 Recombination and Dihybrid Inheritance In contrast to linkage, recombination increases genetic variation in the offspring. Recombination between the alleles of parental linkage groups is indicated by the appearance of non-parental types in the offspring, although not in the numbers that would be expected had the alleles been on separate chromosomes (independent assortment).

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Key Idea: Recombination is the exchange of alleles between homologous chromosomes as a result of crossing over. Recombination increases the genetic variation in the offspring. The alleles of parental linkage groups separate and new associations of alleles are formed in the gametes. Offspring formed from these gametes are called recombinants and show combinations of characteristics not seen in the parents.

Overview of recombination

Parent 1 (2N)

Parent 2 (2N)

Dark body, blue eyes AaBb

White body, white eyes aabb

A

a

B

b

Chromosomes before replication

a

a

b

b

Chromosomes after replication

Crossing over has occurred between these chromosomes

A

A

a

a

a

a

a

b

b

b

Because this individual is homozygous, if a these genes cross over there is no change to the allele b combination.

X

B

b

B

b

Meiosis

A

A

b

B

a

a

Gametes

b

B

A

a

a

a

B

b

b

b

a

a

a

a

b

b

b

b

A

a

a

a

b

b

B

b

Offspring

aabb

White body, white eyes

Aabb

aaBb

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AaBb

Dark body, blue eyes

Dark body, white eyes

White body, blue eyes

Non-recombinant offspring

Recombinant offspring

These two offspring show allele combinations that are expected as a result of independent assortment during meiosis. Also called parental types.

These two offspring show unexpected allele combinations. They can only arise if one of the parent's chromosomes has undergone crossing over.

1. Describe the effect of recombination on the inheritance of genes:

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An example of recombination

The cross (left) 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 symbology used is the same.

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In the female parent, crossing over occurs between the linked genes for wing shape and body color Parent

Phenotype Genotype Linkage

Recombination produces variation

Wild type female

Mutant male

Straight wing Gray body

Curled wing Ebony body

Cucu Ebeb

cucu ebeb

Cu

Eb

cu

eb

cu

eb

cu

eb

If crossing over does not occur, the possible combinations in the gametes is 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 at least (423)2 or 5000 trillion trillion genetically different zygotes for every couple.

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

Family members may resemble each other, but they'll never be identical (except for identical twins).

Using recombination

Analyzing recombination gave geneticists a way to map the genes on a chromosome. Crossing over is less likely to occur between genes that are close together on a chromosome than between genes that are far apart. By counting the number of offspring of each phenotype, the frequency of recombination can be calculated. The higher the frequency of recombination between two genes, the further apart they must be on the chromosome.

Non-recombinant offspring

y

w

v m

r

0

1

31 34

58

Recombinant offspring

The sex of the 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 Š Newbyte Educational Software.

Distances of more than 50 map units show genes that assort independently

Map of the X chromosome of Drosophila, showing the relative distances between five different genes (in map units).

2. Complete the recombination diagram above, adding the gametes in the ovals and offspring genotypes and phenotypes in the rectangles:

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3. Explain how recombination increases the amount of genetic variation in offspring:

4. Explain why it is not possible to have a recombination frequency of greater than 50% (half recombinant progeny):

5. A second pair of Drosophila are mated. The female is Cucu YY (straight wing, gray body), while the male is Cucu yy (straight wing, yellow body). Assuming recombination, perform the cross and list the offspring genotypes and phenotypes:

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155 Detecting Linkage in Dihybrid Crosses Key Idea: Linkage between genes can be detected by observing the phenotypic ratios in the offspring. Shortly after the rediscovery of Mendel's work early in the 20th century, it became apparent that his ratios of 9:3:3:1 for heterozygous dihybrid crosses did not always hold true.

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Experiments on sweet peas by William Bateson and Reginald Punnett, and on Drosophila by Thomas Hunt Morgan, showed that there appeared to be some kind of coupling between genes. This coupling, which we now know to be linkage, did not follow any genetic relationship known at the time.

Sweet pea cross

Pedigree for nail-patella syndrome

I

Red flowers, round pollen (ppll)

OO

P

X

II

Purple flowers, long pollen (PPLL)

OO BO OO BO OO BO

AO BO BO

BO BO

AO BO

BO OO AB

III

F1

X

Purple flowers, long pollen (PpLl)

BO

Individual with nail-patella syndome Blood types OO, BO, AO, AB

Bateson and Punnett studied sweet peas in which purple flowers (P) are dominant to red (p), and long pollen grains (L) are dominant to round (l). If these genes were unlinked, the outcome of an cross between two heterozygous sweet peas should have been a 9:3:3:1 ratio.

Observed

Expected

AO

+

Linked genes can be detected by pedigree analysis. The diagram above shows the pedigree for the inheritance of nailpatella syndrome, which results in small, poorly developed nails and kneecaps in affected people. Other body parts such as elbows, chest, and hips can also be affected. The nailpatella syndrome gene is linked to the ABO blood group locus.

Purple flowers, long pollen (PpLl)

Table 1: Sweet pea cross results

BO

1. Fill in the missing numbers in the expected column of Table 1, remembering that a 9:3:3:1 ratio is expected:

Purple long (P_L_)

284

2. (a) Fill in the missing numbers in the expected column of Table 2, remembering that a 1:1:1:1 ratio is expected:

Purple round (P_ll)

21

Red long (ppL_)

21

Red round (ppll)

55

Total

381

(b) Add the gamete type (parental/recombinant) to the gamete type column in Table 2: (c) What type of cross did Morgan perform here?

381

3. (a) Use the pedigree chart below to determine if nailpatella syndrome is dominant or recessive, giving reasons for your choice: Morgan performed experiments to investigate linked genes in Drosophila. He crossed a heterozygous red-eyed normal-winged (Prpr Vgvg) fly with a homozygous purple-eyed vestigial-winged (prpr vgvg) fly. The table (below) shows the outcome of the cross.

X

Purple eyed vestigial winged (prpr vgvg)

(b) What evidence is there that nail-patella syndrome is linked to the ABO blood group locus?

(c) Suggest a likely reason why individual III-3 is not affected despite carrying the B allele:

Table 2: Drosophila cross results Observed

Expected

Gamete type

Prpr Vgvg

1339

710

Parental

prpr Vgvg

152

Prpr vgvg

154

prpr vgvg

1195

Total

2840

Genotype

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Red eyed normal winged (Prpr Vgvg)

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156 Using the Chi-Squared Test in Genetics

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Key Idea: The chi-squared test for goodness of fit (χ2) can be used for testing the outcome of dihybrid crosses against an expected (predicted) Mendelian ratio. When using the chi-squared test, the null hypothesis predicts the ratio of offspring of different phenotypes according

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to the expected Mendelian ratio for the cross, assuming independent assortment of alleles (no linkage). Significant departures from the predicted Mendelian ratio indicate linkage of the alleles in question. Raw counts should be used and a large sample size is required for the test to be valid.

Ebony body, vestigial wing, male

Ebony body, long wing, female

Gray body, vestigial wing, female

In a Drosophila genetics experiment, two individuals were crossed (the details of the cross are not relevant here). The predicted Mendelian ratios for the offspring of this cross were 1:1:1:1 for each of the four following phenotypes: gray body-long wing, gray bodyvestigial wing, ebony body-long wing, ebony body-vestigial wing. The observed results of the cross were not exactly as predicted. The following numbers for each phenotype were observed in the offspring of the cross: Observed results of the example Drosophila cross

Gray body, long wing Gray body, vestigial wing

Gray body, long wing, male

Step 1: Calculate the expected value (E) In this case, this is the sum of the observed values divided by the number of categories (see note below)

Using c2 in Mendelian genetics

98 88

Ebony body, long wing 102 Ebony body, vestigial wing 112

Using c2, the probability of this result being consistent with a 1:1:1:1 ratio could be tested. Worked example as follows:

400 4

= 100

Step 2: Calculate O – E

The difference between the observed and expected values is calculated as a measure of the deviation from a predicted result. Since some deviations are negative, they are all squared to give positive results. This step is usually performed as part of a tabulation (right, darker blue column).

(O – E)2

E

O–E

Gray, long wing

98

100

–2

4

0.04

Gray, vestigial wing

88

100

–12

144

1.44

Ebony, long wing

102

100

2

4

0.04

Ebony, vestigial wing

112

100

12

144

1.44

Total = 400

Where:

E

O = the observed result E = the expected result ∑ = sum of

The calculated χ2 value is given at the bottom right of the last column in the tabulation.

Step 4: Calculating degrees of freedom The probability that any particular χ2 value could be exceeded by chance depends on the number of degrees of freedom. This is simply one less that the total number of categories (this is the number that could vary independently without affecting the last value). In this case 4-1 = 3

(O – E)2 E

O

Step 3: Calculate the value of χ2 χ2 =

(O – E)2

Category

χ2

∑ = 2.96

Step 5a: Using the χ2 table On the χ2 table (part reproduced in Table 1 below) with 3 degrees of freedom, the calculated value for χ2 of 2.96 corresponds to a probability of between 0.2 and 0.5 (see arrow). This means that by chance alone a χ2 value of 2.96 could be expected between 20% and 50% of the time. Step 5b: Using the χ2 table The probability of between 0.2 and 0.5 is higher than the 0.05 value which is generally regarded as significant. The null hypothesis cannot be rejected and we have no reason to believe that the observed results differ significantly from the expected (at P = 0.05).

Footnote: Many Mendelian crosses involve ratios other than 1:1. For these, calculation of the expected values is not simply a division of the total by the number of categories. Instead, the total must be apportioned according to the ratio. For example, for a total of 400 as above, in a predicted 9:3:3:1 ratio, the total count must be divided by 16 (9+3+3+1) and the expected values will be 225: 75: 75: 25 in each category.

Level of probability (P)

Degrees of freedom

0.98

0.95

0.80

1

0.001

0.004

2

0.040

0.103

3

0.185

4

5

0.50

0.20

0.10

0.064

0.455 χ2 1.64

2.71

0.466

1.386

3.22

0.352

4.61

1.005

2.366

4.64

0.429

0.711

6.25

1.649

3.357

5.99

0.752

0.145

7.78

2.343

4.351

7.29

9.24

Do not reject H0

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Table 1: Critical values of χ2 at the different levels of probability. By convention, the critical probability for rejecting the null hypothesis (H0) is 5%. If the test statistic is less that the tabulated critical value for P = 0.05 we cannot reject H0 and the result is not significant. If the test statistic is greater than the tabulated value for P = 0.05 we reject H0 in favor of the alternative hypothesis.

0.05

0.02

0.01

0.001

3.84

5.41

6.64

10.83

5.99

7.82

9.21

13.82

7.82

9.84

11.35

16.27

9.49

11.67

13.28

18.47

11.07

13.39

15.09

20.52

Reject H0

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157 Chi-Squared Exercise in Genetics A worked example illustrating the use of the chi-squared test for a genetic cross is provided on the previous page.

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Key Idea: The following problems examine the use of the chisquared (χ2) test in genetics.

1. In a tomato plant experiment, two heterozygous individuals were crossed (the details of the cross are not relevant here). The predicted Mendelian ratios for the offspring of this cross were 9:3:3:1 for each of the four following phenotypes: purple stemjagged leaf edge, purple stem-smooth leaf edge, green stem-jagged leaf edge, green stem-smooth leaf edge. The observed results of the cross were not exactly as predicted. The numbers of offspring with each phenotype are provided below: Observed results of the tomato plant cross 12 9

Purple stem-jagged leaf edge Purple stem-smooth leaf edge

Green stem-jagged leaf edge Green stem-smooth leaf edge

8 0

(a) State your null hypothesis for this investigation (H0):

(b) State the alternative hypothesis (HA):

2. Use the chi-squared (χ2) test to determine if the differences observed between the phenotypes are significant. The table of critical values of χ2 at different P values is provided on the previous page. (a) Enter the observed values (number of individuals) and complete the table to calculate the χ2 value: O

Category

E

O—E

(O — E)2 (O — E)2 E

(b) Calculate χ2 value using the equation: χ2 =

Purple stem, jagged leaf

(O – E)2

χ2 =

E

Purple stem, smooth leaf

(c) Calculate the degrees of freedom:

Green stem, jagged leaf

(d) Using the χ2 table, state the P value corresponding to your calculated χ2 value:

Green stem, smooth leaf

(e) State your decision: (circle one) reject H0 / do not reject H0

3. Students carried out a pea plant experiment, where two heterozygous individuals were crossed. The predicted Mendelian ratios for the offspring were 9:3:3:1 for each of the four following phenotypes: round-yellow seed, round-green seed, wrinkled-yellow seed, wrinkled-green seed. Round-yellow seed 441 Round-green seed 159

Use a separate piece of paper to complete the following:

(a) State the null and alternative hypotheses (H0 and HA). (b) Calculate the χ2 value.

Wrinkled-yellow seed 143 Wrinkled-green seed 57

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The observed results were as follows:

(c) Calculate the degrees of freedom and state the P value corresponding to your calculated χ2 value.

(d) State whether or not you reject your null hypothesis: reject H0 / do not reject H0 (circle one)

4. Comment on the whether the χ2 values obtained above are similar. Suggest a reason for any difference:

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158 Problems Involving Dihybrid Inheritance predictable ratio. This activity will allow you to test your understanding of dihybrid inheritance.

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Key Idea: Dihybrid inheritance involves two genes. For autosomal unlinked genes, the offspring appear in

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. 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. (a) Determine the genotype of the original plant with orange striped flowers:

(b) A seed was chosen at random from the F1 progeny of the selfed plant with orange striped flowers. The seed was grown and crossed with a plant grown from a randomly chosen seed from the plant with yellow non-striped flowers. What is the probability that a yellow striped flower will be produced?

(a) State the genotypes:

1: Parent

Parent 2:

Offspring:

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3. In rabbits, spotted coat S is dominant to solid color s, while for coat color, black B is dominant to brown b. A brown spotted rabbit is mated with a solid black one and all the offspring are black spotted (the genes are not linked).

(b) Use the Punnett square to show the outcome of a cross between the F1:

(c) Using ratios, state the phenotypes of the F2 generation:

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159 Sex Linkage sex (the heterogametic sex). In humans, recessive sex linked genes cause a number of heritable disorders in males, e.g. hemophilia. Women who have a recessive allele are said to be carriers. One of the gene loci controlling coat color in cats is sex-linked. The two alleles, red and non-red (or black), are found only on the X-chromosome.

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Key Idea: Many genes on the X chromosome do not have a match on the Y chromosome. In males, a recessive allele cannot therefore be masked by a dominant allele. Sex linkage occurs when a gene is located on a sex chromosome (usually the X). The result of this is that the character encoded by the gene is usually seen only in one

Allele types Xo = Non-red (=black)

XO = Red

Genotypes XoXo, XoY

Phenotypes = Black coated female, male

XOXO, XOY

= Orange coated female, male

XOXo

= Tortoiseshell (intermingled black and orange in fur) in female cats only

1. An owner of a cat is thinking of mating her black female cat with an orange male cat. Before she does this, she would like to know what possible coat colors could result from such a cross. Use the symbols above to fill in the diagram on the right. Summarize the possible genotypes and phenotypes of the kittens in the tables below.

Genotypes

Parent cats

Black female

X

Orange male

Phenotypes

Male kittens

Gametes

Possible fertilizations (kittens)

Female kittens

Tortoiseshell female

Describe the father cat's:

(a) Genotype:

(b) Phenotype:

Gametes

Possible fertilizations (kittens)

2 orange females

3. The owner of another cat, a black female, also wants to know which cat fathered her two tortoiseshell female and two black male kittens. Use the symbols above to fill in the diagram on the right. Show the possible fertilizations by placing appropriate arrows.

Describe the father cat's:

(a) Genotype:

(b) Phenotype:

(c) Was it the same male cat that fathered both this litter and the one above?

YES / NO (delete one)

Black female

Gametes

Possible fertilizations (kittens)

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1 black male

1 tortoiseshell female

2 orange males

Unknown male

Parent cats

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X

X

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2. A female tortoiseshell cat mated with an unknown male cat in the neighborhood and has given birth to a litter of six kittens. The owner of this female cat wants to know what the appearance and the genotype of the father was of these kittens. Use the symbols above to fill in the diagram on the right. Also show the possible fertilizations by placing appropriate arrows.

Unknown male

Parent cats

1 tortoiseshell female

1 black male

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Dominant allele in humans A rare form of rickets in humans is determined by a dominant allele of a gene on the X chromosome (it is not found on the Y chromosome). This condition is not successfully treated with vitamin D therapy. The allele types, genotypes, and phenotypes are as follows: Allele types

Genotypes

XR = affected by rickets XRXR, XRX

X = normal

Phenotypes

= Affected female

XRY

= Affected male

XX, XY

= Normal female, male

As a genetic counselor you are presented with a married couple where one of them has a family history of this disease. The husband is affected by this disease and the wife is normal. The couple, who are thinking of starting a family, would like to know what their chances are of having a child born with this condition. They would also like to know what the probabilities are of having an affected boy or affected girl. Use the symbols above to complete the diagram right and determine the probabilities stated below (expressed as a proportion or percentage). 4. Determine the probability of having:

(a) Affected children:

(b) An affected girl:

(c) An affected boy:

Another couple with a family history of the same disease also come in to see you to obtain genetic counselling. In this case the husband is normal and the wife is affected. The wife's father was not affected by this disease. Determine what their chances are of having a child born with this condition. They would also like to know what the probabilities are of having an affected boy or affected girl. Use the symbols above to complete the diagram right and determine the probabilities stated below (expressed as a proportion or percentage).

Affected husband

Normal wife

Parents

X

Gametes

Possible fertilizations

Children

Affected wife (whose father was normal)

Parents

Normal husband

X

Gametes

5. Determine the probability of having: (a) Affected children:

(b) An affected girl:

(c) An affected boy:

Possible fertilizations

Children

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6. Describing examples other than those above, discuss the role of sex linkage in the inheritance of genetic disorders:

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160 Inheritance Patterns of inheritance patterns in humans: autosomal recessive, autosomal dominant, sex linked recessive, and sex linked dominant inheritance.

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Key Idea: Sex-linked traits and autosomal traits have different inheritance patterns. Complete the following monohybrid crosses for different types

1. Inheritance of autosomal recessive traits Example: Albinism Albinism (lack of pigment in hair, eyes and skin) is inherited as an autosomal recessive allele (not sex-linked). Using the codes: PP (normal) Pp (carrier) pp (albino) (a) Enter the parent phenotypes and complete the Punnett square for a cross between two carrier genotypes. (b) Give the ratios for the phenotypes from this cross. Phenotype ratios:

Female parent phenotype:

Male parent phenotype:

P

Phenotype ratios:

sperm

p

Female parent phenotype:

Male parent phenotype:

eggs

w

sperm

w

Female parent phenotype:

Inheritance of hemophilia is sex linked. Males with the recessive (hemophilia) allele, are affected. Females can be carriers.

Using the codes: XX (normal female) XXh (carrier female) XhXh (hemophiliac female) XY (normal male) XhY (hemophiliac male) (a) Enter the parent phenotypes and complete the Punnett square for a cross between a normal male and a carrier female.

W

W

3. Inheritance of sex linked recessive traits Example: Hemophilia

p

P

2. Inheritance of autosomal dominant traits Example: Woolly hair Woolly hair is inherited as an autosomal dominant allele. Each affected individual will have at least one affected parent. Using the codes: WW (woolly hair) Ww (woolly hair, heterozygous) ww (normal hair) (a) Enter the parent phenotypes and complete the Punnett square for a cross between two heterozygous individuals. (b) Give the ratios for the phenotypes from this cross.

eggs

(b) Give the ratios for the phenotypes from this cross.

Male parent phenotype:

X

eggs

Xh

X

sperm

Y

4. Inheritance of sex linked dominant traits Example: Sex linked form of rickets A rare form of rickets is inherited on the X chromosome.

Using the codes: XX (normal female); XY (normal male) XRX (affected heterozygote female) XRXR (affected female) XRY (affected male) (a) Enter the parent phenotypes and complete the Punnett square for a cross between an affected male and heterozygous female. (b) Give the ratios for the phenotypes from this cross. Phenotype ratios: Š2017 BIOZONE International ISBN: 978-1-927309-62-9 Photocopying Prohibited

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Phenotype ratios:

Female parent phenotype:

Male parent phenotype:

XR

eggs

X

XR

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2

3

4

5

6

III

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2 the inheritance 3 6 7 make used1to study of4 genetic5 disorders and it possible to follow the genetic history of an individual.

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Key Idea: Pedigree charts are a way to graphically illustrate inheritance patterns over a number of generations. They are

Pedigree charts

Key to symbol s

A pedigree chart is a diagram that shows the occurrence and appearance of a particular gene or trait from one generation to the next. In humans, pedigree charts are often used to analyze the inheritance of heritable conditions. In domestic animals, pedigree charts are often used to trace the inheritance of characteristics in selective breeding programmes for horses and dogs.

Normal female

Sex unknown

Normal male

Died in infancy

Affected female (expresses allele of interest)

Identical twins

Non-identical twins

Affected male (expresses allele of interest)

Pedigree charts use symbols to indicate an individual's particular traits. The key (right) explains the meaning of the symbols. Particular individuals are identified by their generation number and their order number in that generation. For example, II-6 is the sixth person in the second row. The arrow indicates the person through whom the pedigree was discovered (i.e. who reported the condition).

I, II, III

Carrier (heterozygote)

I

If the chart on the right were illustrating a human family tree, it would represent three generations: grandparents (I-1 and I-2) with three sons and one daughter. Two of the sons (II-3 and II-4) are identical twins, but did not marry or have any children. The other son (II-1) married and had a daughter and another child (sex unknown). The daughter (II-5) married and had two sons and two daughters (plus a child that died in infancy).

Generations

1, 2, 3

1

Children (in birth order)

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For the particular trait being studied, the grandfather was expressing the phenotype (showing the trait) and the grandmother was a carrier. One of their sons and one of their daughters also show the trait, together with one of their granddaughters (arrow).

3

4

5

6

7

Key to symbol s

The pedigree of lactose intolerance

Normal female Sex unknown Lactose intolerance is the inability to digest the milk sugar lactose. It occurs because some people do not produce lactase, the enzyme needed to break down lactose. The pedigree chart below was one of the original studies to determine the inheritance pattern Died in infancy Normaltolerant male parents can produce a lactose of lactose intolerance. Researchers concluded that because two lactose intolerant child, lactose intolerance must be a recessively inherited condition (it needs two copies of the gene for lactose intolerance to show up). Affected female (expresses Identical twins allele of interest)

?

?

I

Non-identical twins

Affected male (expresses 1 allele of interest) 2

II

?

?

1

2

I, II, III

Carrier (heterozygote)

?

1, 2, 3

4

3

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Children (in birth order)

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Lactose tolerant male

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Lactose intolerant male

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Lactose tolerant female

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Lactose intolerant female

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? Lactose tolerance unknown

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(a) Write the genotype for each of the individuals on the chart using the following letter codes: PP normal skin color; Pnormal, but unknown if homozygous; Pp carrier; pp albino.

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Albinism in humans

1. Autosomal recessive traits Albinos lack pigment in the hair, skin and eyes. This is an autosomal recessive trait.

II

1

(b) Why must the parents (II-3) and (II-4) be carriers of a recessive allele:

(a) Write the genotype for each of the individuals on the chart using the codes: XY normal male; XhY affected male; XX normal female; XhX female carrier; XhXh affected female:

3

4

5

III

1

2. Sex linked recessive traits Hemophilia is a disease where blood clotting is affected. A person can die from a simple bruise (which is internal bleeding). The clotting factor gene is carried on the X chromosome.

2

2

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Hemophilia in humans

I

1

2

3

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(b) Why can males never be carriers?

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3. Autosomal dominant traits An unusual trait found in some humans is woolly hair (not to be confused with curly hair). Each affected individual will have at least one affected parent.

2

3

4

Woolly hair in humans

(a) Write the genotype for each of the individuals on the chart using the following letter codes: WW woolly hair; Ww woolly hair (heterozygous); W- woolly hair, but unknown if homozygous; ww normal hair

I

(b) Describe a feature of this inheritance pattern that suggests the trait is the result of a dominant allele:

II

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2

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4. Sex linked dominant traits A rare form of rickets is inherited on the X chromosome. All daughters of affected males will be affected. More females than males will show the trait.

I

II

1

III

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A rare form of rickets in humans

(b) Why will more females than males be affected?

5. Using the examples on this these two pages, make up your own set of guidelines for interpreting pedigree charts. How do you distinguish an autosomal inheritance pattern from an X-linked one? What are the features of autosomal recessive inheritance? Of autosomal dominant? Of X-linked dominant traits and X-linked recessive traits. Attach your summary to this page.

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(a) Write the genotype for each of the individuals on the chart using the following letter codes: XY normal male; XRY affected male; XX normal female; XR– female (unknown if homozygous); XRXR affected female.

2

1

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162 Hunting for a Gene

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Key Idea: Huntington's disease is caused by a repeating section of DNA. The longer the repeating pattern, the earlier the disease tends to appear and the worse its symptoms. Huntington's disease (HD) is a genetic neuro-degenerative disease that normally does not affect people until about the age of 40. Its symptoms usually appear first as a shaking of the hands and an awkward gait. Later manifestations of the disease include serious loss of muscle control and mental function, often ending in dementia and ultimately death. All humans have the huntingtin (HTT) gene, which in its normal state produces a protein with roles in gene transcription, synaptic transmission, and brain cell survival. The mutant gene (mHTT) causes changes to and death of the cells of the cerebrum, the hippocampus, and cerebellum, resulting in the atrophy (reduction) of brain matter. The gene was discovered by Nancy Wexler in 1983 after ten years of research working with cell samples and family histories of more than 10,000 people from the town of San Luis in Venezuela, where around 1% of the population have the disease (compared to about 0.01% in the rest of the world). Ten years later the exact location of the gene on the chromosome 4 was discovered.

New research has shown that the mHTT gene activates an enzyme called JNK3, which is expressed only in the neurons and causes a drop in nerve cell activity. While a person is young and still growing, the neurons can compensate for the accumulation of JNK3. However, when people get older and neuron growth stops, the effects of JNK3 become greater and the physical signs of HD become apparent. Because of mHTT's dominance, an affected person has a 50% chance of having offspring who are also affected. Genetic testing for the disease is relatively easy now that the genetic cause of the disease is known. While locating and counting the CAG repeats does not give a date for the occurrence of HD, it does provide some understanding of the chances of passing on the disease.

NYWTS

The identification of the HD gene began by looking for a gene probe that would bind to the DNA of people who had HD, and not to those who didn't. Eventually a marker for HD, called G8, was found. The next step was to find which chromosome carried the marker and where on the chromosome it was. The researchers hybridized human cells with those of mice so that each cell contained only one human chromosome, a different chromosome in each cell. The hybrid cell with chromosome 4 was the one with the G8 marker. They then found a marker that overlapped G8 and then another marker that overlapped that marker. By repeating this many times, a map of the genes on chromosome 4 was produced. The researchers then sequenced the genes and found people who had HD had one gene that was considerably longer than people who did not have HD, and the increase in length was caused by the repetition of the base sequence CAG. The HD mutation (mHTT) is called a trinucleotide repeat expansion. In the case of mHTT, the base sequence CAG is repeated multiple times on the short arm of chromosome 4. The normal number of CAG repeats is between 6 and 30. The mHTT gene causes the repeat number to be 35 or more and the size of the repeat often increases from generation to generation, with the severity of the disease increasing with the number of repeats. Individuals who have 27 to 35 CAG repeats in the HTT gene do not develop Huntington disease, but they are at risk of having children who will develop the disorder. The mutant allele, mHTT, is also dominant, so those who are homozygous or heterozygous for the allele are both at risk of developing HD.

American singer-songwriter and folk musician Woody Guthrie died from complications of HD

1. Describe the physical effects of Huntington's disease:

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2. Describe how the mHTT gene was discovered:

3. Discuss the cause of Huntington's disease and its pattern of increasing severity with each generation:

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Increasingly, there are DNA tests for the identification of specific defective genes. People usually consider genetic counseling if they have a family history of a genetic disorder, or if a routine prenatal screening test yields an unexpected result. While screening for many genetic disorders is now recommended, the use of presymptomatic tests for adultonset disorders, such as Alzheimer's, is still controversial.

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Key Idea: Genetic counseling provides an analysis of the risk of producing offspring with known gene defects within a family. Genetic Counselors identify families at risk, investigate the problem present in the family, interpret information about the disorder, analyze inheritance patterns and risks of recurrence, and review available options with the family.

Genetic testing may involve biochemical tests for gene products such as enzymes and other proteins, microscopic examination of stained or fluorescent chromosomes, or examination of the DNA molecule itself. Various types of genetic tests are performed for various reasons, including:

Autosomal recessive conditions

Common inherited disorders caused by recessive alleles on autosomes. Recessive conditions are evident only in homozygous recessive genotypes.

Cystic fibrosis: Malfunction of the pancreas and other glands; thick mucus leads to pneumonia and emphysema. Death usually occurs in childhood. CF is the most frequent lethal genetic disorder in childhood (about 1 case in 3700 live births).

Maple syrup urine disease: Mental and physical retardation produced by a block in amino acid metabolism. Isoleucine in the urine produces the characteristic odor.

Carrier screening

?

Tay-Sachs disease: A lipid storage disease which causes progressive developmental paralysis, mental deterioration, and blindness. Death usually occurs by three years of age. Autosomal dominant conditions Inherited disorders caused by dominant alleles on autosomes. Dominant conditions are evident both in heterozygotes and in homozygous dominant individuals.

Huntington's disease: Involuntary movements of the face and limbs with later general mental deterioration. The beginning of symptoms is highly variable, but occurs usually between 30 to 40 years of age.

Identifying unaffected individuals who carry one copy of a gene for a disease that requires two copies for the disease to be expressed.

Preimplantation genetic diagnosis

Screens for genetic flaws in embryos used for in vitro fertilization. The results of the analysis are used to select mutation-free embryos.

Prenatal diagnostic testing

Tests for chromosomal abnormalities such as Down syndrome.

Newborn screening

Newborn babies are screened for a variety of enzyme-based disorders.

Presymptomatic testing

Testing before symptoms are apparent is important for estimating the risk of developing adult-onset disorders, including Huntington’s disease, cancers, and Alzheimer's disease.

Down karyotype

Auditory test

About half of the cases of childhood deafness Genetic counseling provides information to are the result of an autosomal recessive disorder. families who have members with birth defects Early identification of the problem prepares or genetic disorders, and to families who may families and allows early appropriate treatment. be at risk for a variety of inherited conditions.

Cytogenetics Dept., Waikato Hospital

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163 Genetic Counseling

Most pregnant women in developed countries will have a prenatal test to detect chromosomal abnormalities such as trisomy 21 ( Down syndrome).

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1. Outline the benefits of carrier screening to a couple with a family history of a genetic disorder:

2. (a) Suggest why Huntington's disease persists in the human population when it is caused by a lethal, dominant allele:

(b) How could presymptomatic genetic testing change this?

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Understanding how genes influence one's health and possible medical future raises many beneficial options and ethical issues. For example, a person diagnosed with a high risk of developing a treatable genetic disorder can plan for its occurrence or make lifestyle changes to delay its onset. A person diagnosed at high risk for an untreatable disorder may face years of anxiety over something that may not even occur. Moreover, how will health insurers react? Will they deny insurance cover or charge exorbitantly for cover?

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Benefits and ethical issues arising from genetic testing

Medical benefits

ff Improved diagnosis of disease and predisposition to disease by genetic testing.

ff Better identification of disease carriers, through genetic testing.

Non-medical benefits

Possible ethical issues

ff Greater knowledge of family relationships through genetic testing, e.g. paternity testing in family courts. ff Advances forensic science through analysis of DNA at crime scenes.

ff Better drugs can be designed using ff Improved knowledge of the knowledge of protein structure (from evolutionary relationships between gene sequence information) rather than humans and other organisms, which by trial and error. will help to develop better, more ff Greater possibility of successfully accurate classification systems. using gene therapy to correct genetic disorders.

Couples can already have a limited range of genetic tests to determine the risk of having offspring with some diseasecausing mutations.

When DNA sequences are available for humans and their ancestors, comparative analysis may provide clues about human evolution.

ff It is unclear whether third parties, e.g. health insurers, have rights to genetic test results.

ff If treatment is unavailable for a disease, genetic knowledge about it may have no use. ff Genetic tests are costly, and there is no easy answer as to who should pay for them. ff Genetic information is hereditary so knowledge of an individual’s own genome has implications for members of their family.

Legislation is needed to ensure that there is no discrimination on the basis of genetic information, e.g. at work or for health insurance.

3. Huntington's disease can be tested for by looking for the number of CAG repeats on the HTT gene. Explain why someone with a family history of the disease may or may not want to be tested for their own disease risk:

(a) Medical:

(b) Non-medical:

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4. Describe two possible benefits of genetic testing:

5. Dave has a family history of a specific heart disease. He has a DNA test and finds he has a small chance of developing the disease, but if he lives a healthy lifestyle it is highly unlikely to develop and need treatment. Later, he buys health insurance and, after having a medical exam in which the doctor pronounces him in perfect health with no pre-existing medical conditions, he ticks the box "no pre-existing conditions" on the insurance forms but informs the insurance company of his family history of heart disease. The insurance company agrees to insure him and for many years Dave has no medical problems other than impacted wisdom teeth, which the insurance pays to have removed. Then, 15 years later, still in apparently perfect health, Dave suddenly develops heart disease and needs treatment. His lodges a claim with his medical insurance. The insurance company investigates and obtains a copy of Dave's DNA test showing he has a chance of developing the disease. They deny his claim on the grounds that he failed to disclose a pre-existing condition. Dave argues he had no condition at the time and told them about his family history.

As a class discuss the ethical and medical issues surrounding a case like this. Who is in the right? Should the insurance company pay out or should Dave have included the DNA test results in his insurance application?

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164 Polygenes skin pigment melanin in humans is controlled by at least three genes. The amount of melanin produced is directly proportional to the number of dominant alleles for either gene (from 0 to 6).

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Key Idea: Many phenotypes are affected by multiple genes. Some phenotypes (e.g. kernel color in maize and skin color in humans) are determined by more than one gene and show continuous variation in a population. The production of the

Very pale

A light-skinned person A dark-skinned person

Light

Medium light

Medium

Medium dark

Dark

Black

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2

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0

1. Complete the Punnett square for the F2 generation (below) by entering the genotypes and the number of dark alleles resulting from a cross between two individuals of intermediate skin color. Color-code the offspring appropriately for easy reference.

Parental generation

X

(a) How many of the 64 possible offspring of this cross will have darker skin than their parents:

(b) How many genotypes are possible for this type of gene interaction:

There are seven shades skin color ranging from very dark to very pale, with most individual being somewhat intermediate in skin color. No dominant allele results in a lack of dark pigment (aabbcc). Full pigmentation (black) requires six dominant alleles (AABBCC).

Black (AABBCC)

Pale (aabbcc)

2. Explain why we see many more than seven shades of skin color in reality:

Medium (AaBbCc)

F2 generation (AaBbCc X AaBbCc) GAMETES

ABC

ABc

AbC

Abc

aBC

aBc

abC

abc

ABC

AbC Abc

aBC aBc

abC abc

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3. Discuss the differences between continuous and discontinuous variation, giving examples to illustrate your answer:

4. From a sample of no less than 30 adults, collect data for one continuous variable (e.g. height, weight, shoe size, hand span). Record and tabulate your results in the space below, and then plot a frequency histogram on the grid below:

Raw data

Tally chart (frequency table)

Frequency

Variable:

(a) Calculate the following for your data and attach your working. Mean:

Mode:

Standard deviation:

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Median:

(b) Describe the pattern of distribution shown by the graph, giving a reason for your answer:

(c) What is the genetic basis of this distribution?

(d) What is the importance of a large sample size when gathering data relating to a continuous variable?

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165 Sex Determination sex because each somatic cell has one X and one Y chromosome. Sex determination is based on the presence or absence of the Y chromosome. Without a Y, an individual will be female. In mammals, the male is always the heterogametic sex, but this is not necessarily the case in other taxa. In birds and butterflies, the female is the heterogametic sex, and in some insects the male is simply X whereas the female is XX.

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Key Idea: The sex of an individual is usually determined by sex chromosomes, but can be influenced by the environment. The determination of sex in an organism is controlled in most cases by the sex chromosomes provided by each parent. These have evolved to regulate the ratios of males and femalesSex produced and preservein thehumans genetic differences determination between the sexes. In humans, males are the heterogametic

Sex determination in humans

Y-bearing

X-bearing

sperm cell

sperm cell

X-bearing oocyte

X-bearing oocyte

XY sex determination

Female: XX Male: XY Examples: Humans (and all mammals) and some dioecious plants (having separate male and female plants) such as kiwifruit. In humans, the female is the homogametic sex and has two similar sex chromosomes (XX), whereas the male is the heterogametic sex with two unlike chromosomes (XY). The primary sexual characteristics are determined by genes on the X chromosome. Females must have 2X chromosomes. Maleness is determined by the presence of the Y.

Parents

Female

Male

XX

XY

Gametes

Possible fertilizations Offspring

XY: Male XY: male

XX: Female XX: female

Sex determination in Drosophila

Sex:

Temperature dependent sex determination

In Drosophila, maleness is determined by the ratio of X chromosomes to autosomes, not the presence of the Y chromosome. Two X chromosomes in a diploid cell produce a female fly, but one X chromosome in a diploid cell produces a male fly. The Y chromosome is not involved in determining sex, but does contain genes involved in the production of sperm in adult males. Thus XO in Drosophila is a sterile male, while in mammals it is a sterile female.

In some vertebrate species, mostly reptiles, sex is determined by the temperature at which the eggs are incubated. In turtles, males are produced at lower incubation temperatures than females (22°C-27°C as opposed to 30°C). The hormone testosterone may be converted to estradiol at higher temperatures to produce females, but how temperature triggers gene expression and the pathway for genetic sex determination is poorly understood.

1. (a) Complete the diagram above, to show the resulting gametes, genotype and sex of the offspring: (b) Determine the probability of a conception producing a male child:

(c) Determine the probability of a second conception producing a female child:

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2. Explain what determines the sex of the offspring at the moment of conception in humans:

3. Explain why many genes on the X chromosome in males will be expressed regardless of their dominance status:

4. Estimate the ratio of males to females produced when a clutch of turtle eggs is incubated at 28.5°C:

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166 Sex Limited Characteristics hormone levels in males and females. Sex influenced traits are traits that may be displayed in both genders but are more developed or pronounced in one.

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Key Idea: Some phenotypic traits are displayed only in one sex, or differently between sexes. Some genes are only expressed in one gender and are called sex limited. This can be due to responses to the different

Sex limited traits

Sex limited traits are commonly associated with secondary sexual characteristics. Lactation in mammals is an example. The secondary development of the breast tissue in females during puberty and preparation for milk production during pregnancy are both the result of the hormones estrogen and progesterone, produced by the ovaries.

Horned sheep

The appearance of horns on Dorset sheep was first studied by Arkell and Davenport in 1912. Both sexes in Dorset sheep are horned. In the Suffolk breed, both sexes are hornless. In a cross of these two breeds, F1 males have horns and F1 females are hornless. If F1 individuals are crossed, F2 male offspring occur in a 3:1 ratio of horned to hornless rams, while the female ratio is the opposite. Horned male

Parental cross Hornless female

Traits such as this become important in the breeding of dairy herds. Bulls may carry genes for high milk production and quality and crossing them with cows that carry similar genes is important in increasing milk production and quality in the offspring. DNA tests are able to identify these genes at birth, allowing farmers to save on the huge cost of raising bulls to breeding age to see if they sire good milk-producing offspring.

Sex influenced traits

Two well known examples of sex influenced traits are male pattern baldness in humans and the appearance of horns in Dorset sheep.

H+

H+

H

H+ H

H+ H

H

H+ H

H+ H

Horned male

1. Describe the difference between a sex limited trait and a sex influenced trait:

Hornless female

F1 cross

H+

H

H+ H+ H+ H+ H H+ H

H

HH

Male pattern baldness

Male pattern baldness (androgenic alopecia) is often cited as an example of a sex influenced trait which acts as a dominant trait in males and as a recessive trait in females. Certainly, it is more pronounced in males, being less severe and occurring later in life in females. However, it appears to be controlled by many genes, rather than one, and it is still not clear how these genes interact.

3. (a) On the first Punnett square above right, use a blue pen to circle the genotype of the F1 generation that would produce horned males and red pen to circle the genotype of horned females:

(b) Repeat this for the F2 generation on the second Punnett square:

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Research in 2005 confirmed that the recessive allele of the androgen receptor gene (found on the X chromosome) played a role in pattern baldness. In this case, males require just one recessive allele while females require two. However, many other genes and their products are involved. Levels of the enzyme 5-alpha reductase are higher in males with pattern baldness. Drugs that interfere with this enzyme greatly reduce the progress of pattern baldness. Other genes on the Y chromosome and on chromosome 3 also appear to play a role. Sons whose fathers have hair loss are 2.5 times more likely to experience it than sons with fathers with no hair loss.

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2. Why is the bull's genotype important when breeding high quality milk producing cows?

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167 Non-Nuclear Inheritance organelle's function, e.g. mitochondrial DNA contains genes for producing some of the proteins associated with the electron transport chain. Because mitochondria and chloroplasts are distributed randomly in gametes, traits determined by their genes do not follow simple Mendelian rules.

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Key Idea: Not all genes are encoded within nuclear DNA. Both mitochondria and chloroplasts contain their own plasmid-like copies of DNA, (denoted mtDNA and cpDNA respectively). Mitochondrial and chloroplast DNA replicates independently of the main genome and contains genes essential for that

Mitochondrial and chloroplast DNA

Using mtDNA and cpDNA

In most multicellular organisms, the mtDNA and cpDNA are inherited from the maternal line. Mitochondria in the cytoplasm are randomly divided between dividing primary oocytes so that the final egg cell has a random selection of mitochondria from the original cell. Mitochondria are numerous in sperm cells, providing the energy to power the tail. However, they do not enter the egg cell, or are destroyed by enzymes within it.

Because mitochondria and chloroplasts are passed through the maternal line they can be used to trace maternal lineage. Mutations occur in mtDNA at a rate of fewer than one per 100 people. This means that the mtDNA of any one person is probably the same as their direct maternal ancestor back many generations. The same applies for plant cpDNA and mtDNA. In humans, this concept has been used to trace our mitochondrial ancestor, a single female from Africa. This does not mean this person is the ancestor of all humans, rather that no other lineages produced daughters through which the mtDNA was passed to our generation.

Mitochondrion

Primary oocyte

Generation 1 2 3

Dividing egg cell

4 5 6

Polar body

Experimental evidence for maternal inheritance of cpDNA

Randomly assigned mitochondrion

Sperm cell

Mitochondria

In 1909, Carl Correns, a German geneticist, noticed that the branches of the four o'clock (Mirabilis jalapa) developed leaves that could be green, white, or variegated (green and white). After performing a series of crosses, Correns noticed that seeds from branches with solid green leaves grew into plants with solid green leaves, even if the pollen was from a branch with white or variegated leaves. Seeds from branches with variegated leaves produced plants with all three leaf types. Again, where the pollen came from did not matter. In addition, the appearance of these three types was never in any predictable Mendelian ratio. Correns concluded that the trait was being passed via the maternal line. We now know a defect in the cpDNA of the four o'clock causes the white color and these, like mitochondria, are passed randomly to the offspring.

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1. (a) Why are mtDNA and cpDNA only passed along the maternal line?

(b) Why do the inheritance patterns of mtDNA and cpDNA not follow predictable Mendelian rules?

2. (a) In which generation of the mtDNA diagram (blue box above) did a mutation in the mtDNA occur?

(b) How can this be used to help trace the lineage?

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168 KEY TERMS AND IDEAS: Did You Get It?

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

allele

A Observable characteristics in an organism.

dominant

B Possessing two different alleles for a gene, one inherited from each parent.

heterozygous

C One of a number of alternative versions of a gene

homozygous

D Possessing two identical alleles of a gene, one inherited from each parent.

phenotype recessive

E

Allele that expresses its trait irrespective of the other allele.

F

An allele that will only express its trait in the absence of the dominant allele.

2. A plant breeder crossed a group of true breeding red flowers with a group of true breeding purple flowers and collected the seeds. When they were grown it was found that all the plants had flowers that where wine colored. When these plants were crossed and the seeds collected and grown it was found that of 124 plants, 30 had red flowers and 33 had purple flowers and 61 has wine colored flowers.

(a) Complete the Punnett squares below. Use the alleles FR and FP for the alleles for the red and purple flowers:

Parental cross

F1 cross

(b) What kind of inheritance pattern is shown here?

3. (a) A second plant breeder used a plant known to produce yellow or orange flowers and long or short petals. Orange (O) is dominant to yellow (o) and long petals (L) is dominant to short petals (l). The breeder started with true breeding orange long petaled flowers and crossed them with true breeding yellow short petal flowers to produce the F1. The breeder then crossed the F1 plants with the intention of producing yellow long petaled flowers. Use the Punnett square to show the F1 cross. The genes are not linked:

(b) Out of 245 seeds produce in the F1 cross, how many will grow to produce yellow long petalled flowers?

(c) How many of these will be pure breeding and how could the breeder find out which these are?

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4. The pedigree below shows the occurrence of a genetic disorder in a family. Analyze the pedigree and decide with reasons what type of inheritance pattern is shown:

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Regulation of Gene Expression

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Enduring Understanding

2.E 3.B, 4.A.

activator

Timing and coordination of specific events are necessary for the normal development of an organism

apoptosis (=PCD)

Essential knowledge

Key terms

cellular differentiation enhancer sequence gene expression gene induction

gene repression

2.E.1

(a) Cell differentiation results from the expression of genes for tissue-specific proteins c 1

Hox gene inducer

microRNA operon

promoter

regulatory DNA sequence

regulatory protein

Explain how cellular differentiation involves differential gene expression.

169 170

(b) Induction of transcription factors during development results in genes being expressed in sequence

homeotic gene

regulatory gene

Activity number

c 1

Describe the role of homeotic genes (including Hox genes) in development.

c 2

Explain how embryonic induction during development results in the correct timing of events.

c 3

Explain how temperature and the water availability determine seed germination.

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c 4

Explain how mutations can result in abnormal development.

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c 5

Explain how genetic transplantation experiments support the link between gene expression and normal development.

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c 6

Describe the role of genetic regulation by microRNAs in development and control of cellular functions.

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171

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(c) Apoptosis plays a role in normal development and differentiation c 1

repressor

Use examples to show understanding of how programmed cell death (apoptosis) is involved in normal development and differentiation.

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structural gene

terminator sequence transcription

transcription factor

3.B.1

Gene regulation results in differential gene expression, leading to cell specialization

Activity number

Essential knowledge

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(a) DNA regulatory sequences, regulatory genes, and small regulatory RNAs are involved in gene expression c 1

Explain what is meant by a regulatory DNA sequence and explain how they control transcription. Examples include promoters, terminators, and enhancers.

c 2

Define regulatory genes and explain their role.

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(b) Positive and negative controls regulate gene expression in bacteria and viruses

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c 1

Explain how the presence of an inducer can switch on the expression of genes.

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c 2

Explain how the presence of a repressor can inhibit the expression of genes.

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c 3

Explain how repressors and inducers bring about their effects.

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c 4

Explain how regulatory proteins can inhibit gene expression by binding to DNA and blocking transcription (negative control).

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c 5

Explain how regulatory proteins can stimulate gene expression by binding to DNA and stimulating transcription (positive control) or inactivating repressors.

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c 6

Explain how and why certain genes are continuously expressed.

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(c) In eukaryotes, gene expression is complex and control involves the action of regulatory genes, regulatory elements, and transcription factors c 1

Explain how transcription factors bind to specific DNA sequences and/or other regulatory proteins.

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c 2

Describe the role of activators and repressors in eukaryotic gene expression.

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c 3

Explain how control of transcription determines if and how much of a gene product will be produced at any one time.

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(d) Gene regulation accounts for some phenotypic differences between organisms c 1

3.B.2

Explain how differences in gene regulation can lead to phenotypic difference in genetically similar organisms.

A variety of intercellular and intracellular signal transmissions mediate gene expression

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Activity number

Essential knowledge (content coverage in other chapters also indicated)

(a) Signal transmission within and between cells mediates gene expression c 1

Using an example, explain how the transmission of signals (chemical signal molecules) brings about gene expression. Examples include: • Regulation of gene expression by cytokines, which enable cell division, proliferation, and differentiation, e.g. retinal cells, immune cells. • Mating pheromones in yeast triggering mating gene expression. • Levels of cyclic AMP regulating metabolic gene expression in bacteria. • Ethylene levels causing changes in enzyme production and fruit ripening. • The effect of gibberellin on seed germination.

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(b) Signal transmission within and between cells mediates cell function c 1

Using an example, explain how the transmission of signals (chemical signal molecules) within and between cells mediates cell function. Examples include: • The role of morphogens in stimulating cell differentiation and development. • Changes in p53 activity resulting in cancer. • HOX genes and their role in development in animals.

4.A.3 Interactions between external stimuli and regulated gene expression result in specialization of cells, tissues, and organs

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Essential knowledge (content coverage in other chapters also indicated)

(a) Differentiation in development is due to external and internal cues that trigger gene regulation by proteins that bind to DNA [also 3.B.1, 3.B.2] c 1

Explain how differentiation of cells during development is due to internal and external cues that trigger gene regulation by transcription factors.

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(b) Structural and functional divergence of cells in development is due to expression of genes specific to a particular tissue or organ type c 1

Explain how cells become different (specialized for different roles) as a result of them expressing genes specific to a particular tissue. [also 3.B.1, 3.B.2]

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(c) Environmental stimuli can affect gene expression in a mature cell

Explain how stimuli in the environment can affect gene expression, e.g. induction of the lac operon in E. coli. [also 3.B.1, 3.B.2]

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c 1

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169 Cellular Differentiation giving rise to the three germ layers and, subsequently, the specialized cells that make up the tissues and organs of the body. This process by which more specialized cells develop from more generalized ones is called cellular differentiation and it is achieved through selectively switching genes on and off in particular sequences. As a cell proceeds along its developmental pathway, its choices become more limited. Once fully differentiated, it can not turn into another cell type.

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Key Idea: A zygote divides and produces all the cell types in the body by cellular differentiation. Specific patterns of gene switching determine what cell type develops. Multicellular organisms consist of many different types of cells, each specialized to carry out a particular function. A zygote and its first few divisions are totipotent and can differentiate to form any cell type in the body. During development, these cells divide and follow different developmental pathways,

Kidney

Lung

Muscle

Urinary bladder

Endoderm

Germline

As stem cells differentiate, different genes are switched on and off and the cell becomes increasingly committed to its fate.

Circulatory system

Mesoderm

Gut

Germline cells (cells that produce gametes) are set aside early in development.

Blood and immune cells

Bones and other connective tissue

Expression of retinal specific gene  retinal cell

Ectoderm

RPE

The retina is the light sensitive tissue layer of the eye. The retina comprises several different cell types, all originating from retinal precursor cells. Many transcription factors control retinal cell fate during embryogenesis. Improper regulation of the signaling pathways can cause eyes to form in the wrong place or improper eye development.

Expression of neuronspecific gene  nerve cell

Nerve fiber

Signaling controls cell differentiation

Skin cells

Photoreceptors

Signal molecules initiate signal transduction pathways that activate transcription factors that control the differentiation process.

Glands

Inner limiting membrane

The zygote (fertilized egg) is totipotent. It has all the information stored in the chromosomes to make a complete new individual.

Rod cell

Cone cell

Ganglion cell

Bipolar cell

Amacrine cell

Horizontal cell

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1. Multicellular organisms consist of many different cell types. Explain how it is possible for these all to arise from a single fertilized egg (zygote):

2. What role do transcription factors play in cellular differentiation of the retina?

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170 Gene Expression and Development normal development of an organism from a zygote to an adult. It is achieved by switching the transcription of specific genes on and off in response to the activity of transcription factors and the signaling molecules with which they interact. The precise nature of this regulation is crucial in embryogenesis and at specific stages during an organism's life, e.g. during molting in insects or puberty in humans.

Control over genes

Control over developmental processes

Many insect larvae have large polytene chromosomes (chromosomes that have undergone repeated rounds of DNA replication without cell division so that many sister chromatids remain synapsed together). They contain many copies of the same gene, so have a high level of gene expression.

Development is the process of progressive change through the lifetime of an organism. Signal transmission mediates gene expression to ensure the correct cell type is produced, or the correct process is undertaken, at the right time.

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Key Idea: Differential gene expression has a crucial role in development. Developmental processes are tightly regulated by transcription factors and chemical signals, which switch gene expression on and off in specific sequences. The controlled expression of different genes at different times and at different locations in the organisms is called differential gene expression and it is responsible for the

For example, Chironomus (bloodworm) larvae produce large amounts of saliva to aid digestion of detritus. As they molt and grow, salivary gland development must keep pace with their increased food requirements. The molting hormone, ecdysone, acts as the signal to turn on the gene regulating the development of the salivary glands. The regions of the chromosome with this gene puff out when the genes are being transcribed (below).

In plants, fruit ripening is coordinated by the gaseous hormone ethylene. Ethylene is a signal transduction molecule. It binds to specific receptors and sets off a signal cascade, switching on the genes associated with ripening. Translation of these genes produces the many enzymes involved in the ripening process.

Many physical and chemical changes occur during ripening. For example, when a banana ripens (below) the long chain starch molecules are broken down by enzymes into smaller sucrose molecules. The progressive loss of starch makes the flesh softer, and the increase in sucrose makes the banana sweeter.

Stephen Moore

Bloodworm larva

EII

Increasing level of ripeness

Chromosome puff

1. Explain why the timing and location of gene expression during development is so important:

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Ethylene Firmness

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Immature green Fruit development stage

Ethylene (nl ng-1 FW)

Firmness (kg cm-2)

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3. Study the graph (right) and describe the relationship between ethylene levels and ripening:

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2. Using an example, explain why genes might be upregulated (show higher activity) or down regulated (show reduced activity) at different times during the development of an organism:

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Overripe

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171 Homeotic Genes and Development structures (morphogenesis) during embryonic development. Homeotic genes are highly conserved (have changed very little over time) and regulate patterns of morphogenesis is a wide range of organisms including plants, animals, and fungi. Differences in the regulation of homeotic genes leads to the many different forms we see in multicellular organisms.

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Key Idea: Homeotic genes control the development of body segments or specific structures during embryogenesis. Their control is achieved through regulating the activity of the transcription factors that control gene expression. Homeotic genes (or homeobox genes) are a large family of similar genes that direct the formation of many body

What are homeotic genes?

ff Homeotic genes are master genes, meaning they regulate the expression of other genes by switching them on or off (right) via transcription factors. This is analogous to a director directing scenes in a play. If the actors are called to play a scene in the wrong order, the play will be presented differently. ff Homeotic genes contain a highly conserved (unchanging) sequence called the homeobox. The homeobox is about 180 base pairs long and encodes a string of about 60 amino acids called a homeodomain that can bind to DNA.

ff Homeodomain-containing proteins act as transcription factors and so regulate transcription of other genes. A homeobox gene is not a specific gene. It is a large, ancient group of genes that all contain the homeobox sequence.

Master gene …

ff In humans, there are about 235 homeobox genes, but homeobox genes are found in all eukaryotes, including plants, animals, and fungi. The genes themselves may be different, but the homeobox sequence itself hasn't changed much at all throughout evolution.

… switches other genes on or off in a specific sequence

Hox genes are found in animals

Hox genes are a special group of homeobox genes that are found only in animals. Hox genes control the development of the back and front parts of the animal body. The same genes (or homologous ones) are present in essentially all animals, including humans. Drosophila embryo

Head

Thorax

Abdomen

MADS genes control flower development in plants

MADS-box genes are highly conserved sequences found in nearly all eukaryotes, but their abundance and diversity is highest in plants.

In plants, MADS-box genes called MIKCc genes encode transcription factors controlling the development of the flower organs (sepals, petals, carpels, and stamens). The development of specific parts of the flower is determined by differential activity in the MIKCc genes, which in turn produces different combinations of transcription factors. Each combination of transcription factors recognizes different DNA sites, activating or repressing the expression of specific genes by interacting with the DNA. This results in the development of the different flower parts. DNA

AP3

AG SEP

PI

SEP AG

AP1 SEP Petal

?

SepalAP1 AP1 Sepal

The Hox genes are located on a single chromosome in Drosophila, and on four separate chromosomes in mice. The shading indicates where in the body the genes are expressed.

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?

Carpel

PI

AP3

SEP AG

Stamen

Transcription factors produced by expression of the regulatory MADS-box genes

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240 Mice

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Shifting Hox expression

Chickens

Geese

Snakes

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Huge diversity in morphology in organisms within and across phyla could have arisen through small changes in the genes controlling development.

Differences in neck length in vertebrates provide a good example of how changes in gene expression can bring about changes in appearance (morphology). Different vertebrate species have different numbers of neck vertebrae (denoted by the black ovals on the diagram). In all cases, the boundary between neck and trunk vertebrae is marked by expression of the Hox c6 gene (c6 denotes the sixth cervical or neck vertebra) but the position varies in each animal relative to the overall body. The forelimb (arrow) arises at this boundary in all four-legged vertebrates. In snakes, the boundary is shifted forward to the base of the skull and no limbs develop. As a result of these differences in expression, mice have a short neck, geese a long neck, and snakes, no neck at all.

Hox c6

Hox c6

Hox c6

1. (a) What are homeotic genes?

(b) Describe the function of the homeodomain:

2. (a) What are Hox genes?

(b) What is the general function of Hox genes?

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3. Explain the role of MADS-box genes in the development of plant flowers:

4. Using an example, discuss how changes in gene expression can bring about changes in morphology:

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172 The Timing of Development Key Idea: Embryonic development is regulated by signal molecules that affect gene expression. The differentiation of cells is controlled by cell signalling, either by producing a gradient of molecules through the embryo or by signalling a specific cell that then signals other

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cells in sequence. Cell signals regulate transcription factors that in turn control the differential gene expression that shapes embryos. This process, called embryonic induction, takes hours to days, while the time a cell can respond to an inducing signal is strictly limited.

Development in Drosophila

Morphogen concentration and cell fate

Concentration of morphogen

Threshold 3

The bicoid gene is important in Drosophila development. After fertilization, bicoid mRNA from the mother is passed to the egg where it is translated into bicoid protein. Bicoid protein forms a concentration gradient in the developing embryo, which determine where the anterior (front) and posterior (rear) develop. Embryo

Threshold 2

Threshold 1

Decreasing bicoid concentration

In a region of high bicoid concentration, the head and anterior structures develop.

Cell Cell fate 2 fate 1

Morphogens are signal molecules that govern patterns of tissue development. Morphogens induce cells in different positions to adopt different fates by diffusing from an area of production through a field of cells. Bicoid protein (right) is an example of a morphogen.

Normal embryo

If bicoid is absent from the egg, the embryo develops two posterior ends and no head. Development stops.

Development in Caenorhabditis elegans

Apoptotic cell deaths

The death of these cells is controlled by the genes ced-3 and ced-4, which are themselves regulated by gene ced-9. There are three waves of apoptosis. The first removes 113 cells, the second removes 18 more cells, and the third removes half of the developing egg cells (oocytes). This process is so tightly regulated that cell death events occur at the same time and same place in all C. elegans individuals.

100

Embryonic cell deaths (first wave, 113 cells)

80 60 40

Larval cell deaths (second wave, 18 cells)

20

0

20

2. Describe the pattern of apoptosis in the development of C. elegans:

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60

80

120

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Hours after fertilization

(b) Explain the purpose of the bicoid protein in Drosophila:

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Germ-cell deaths (third wave)

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1. (a) What is a morphogen?

G Lettre and M Hengartner, Nature.com 2007

The development of C. elegans never varies. Exactly 1090 cells are generated to form an adult hermaphrodite. Of these 1090 cells exactly 131 cells undergo programmed cell death (apoptosis).

Apoptosis during development in C. elegans

120

The nematode worm, Caenorhabditis elegans, is a model organism for developmental studies as its simple body plan and short life cycle makes it easy to study.

Bicoid mutant

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Cell fate 3

In low bicoid concentration, the anus and posterior structures develop.

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173 Factors Regulating Seed Germination

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affect germination rate. Water is needed to swell and break the seed coat, and also to activate cellular metabolism. One important metabolic activity is the breakdown of stored starch into simple sugars by the a-amylase enzyme. The hormone gibberellic acid is activated during germination and acts as a signal molecule to initiate the production of a-amylase.

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Key Idea: Seed germination is affected by temperature and water availability. Gibberellic acid activates a-amylase which in turn hydrolyzes starch to provide energy. Seed germination depends on several environmental factors including adequate soil water levels and suitable temperatures. Variations in one or both of these factors can

Gibberellic acid activation of a-amylase

GA

GA receptor

Cell of the aleurone layer

α-amylase

Myb

RNAP

DNA

Gibberellic acid (GA) binds to a receptor on the plasma membrane.

mRNA

mRNA

Myb protein binds to DNA and activates transcription of a-amylase enzyme.

GA receptor signals for the production of a transcription factor called Myb protein.

How temperature affects germination and a-amylase activity

a-amylase is produced and hydrolyzes the starch in the seeds into simple sugars.

1. What role does water play in seed germination?

a-amylase activity vs temperature

1.56

a-amylase activity

1.60

2. How does temperature affect a-amylase activity?

1.50 1.40 1.30

1.26

1.20 1.10

1.09

Ching, 1975, Plant Physiology

10

20 Temperature °C

3. Study the graph relating to germination vs temperature:

30

(a) Which plant(s) have the best germination success?

(b) Would you choose Zea mays if you were planting crops in environmental conditions less than 20°C? Explain:

Above: the synthesis and activity of a-amylase is temperature dependent, peaking between 20°C and 30°C (there is some variation between plants).

Percentage germination vs temperature in crop plants Wheat strain A

Wheat strain C

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Wheat strain B

40

Zea mays

20

4. Describe the role of gibberellic acid in a-amylase production:

Mendeny, 2007, World Journal Ag.

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Temperature oC

Above: the effect of temperature on the germination of seedlings varies between plants. However, there is often a peak of germination activity at an optimal temperature.

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Germination (%)

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174 Mutations and Development changes) the genetic message can be altered and a different gene product (protein) will be produced. If the protein is critical to development, it can disrupt normal development. Gene regulation can be studied by investigating the effect of particular mutations on the development of an individual.

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Key Idea: Abnormal expression of Hox genes in the model organism Drosophila melanogaster results in abnormal development of the embryo. Genes play a critical role in regulating normal development. However, if a gene becomes mutated (its DNA sequence

Eight Hox genes control development in Drosophila

Drosophila have eight different Hox genes to control body development. They are clustered into two complexes, the antennapedia complex (ANT-C) consists of five genes and the bithorax complex (BX-C) consists of three genes (right).

PhiLiP public domain

Mutations to highly conserved Hox genes can drastically alter the body plan of affected individuals and abnormal forms may develop. Drosophila melanogaster (the fruit fly) is a model organism and has been extensively studied to determine how gene expression can affect developmental processes. The general principles learned by studying Drosophila can be applied to understanding the development of other organisms.

BX-C

ANT-C

lab

pb

Dfd Scr Antp

Ubx

Abd-A

Abd-B

ff The antennapedia (Antp) gene directs the development of the second pair of legs and so is normally highly expressed in the second thoracic segment. It also allows, although doesn't directly activate, wing formation ff Expression of the Antp gene in other regions of the body can result in mutant strains of Drosophila. Antp expression in the head region (where it is usually not expressed at all) results in legs instead of antennae growing (photo, right). This type of mutation is called a gain-of-function mutation and is dominant (a leg will always form if this gene is expressed).

Legs

ff Another mutant form of Drosophila is the recessive loss-offunction Antp mutant. In these individuals the Antp gene is not expressed in the second thoracic segment, and antennae form instead of legs.

Toony cc3.0

ff The name antennapedia simply means antenna-foot.

An antennapedia mutant Drosophila

1. Describe the role of the Antp gene in Drosophila development:

2. (a) Where is the Antp gene usually expressed in a normal Drosophila individual? (b) Where is the Antp gene expressed in an antennapedia mutant Drosophila?

(c) What is the effect of the mutation in the antennapedia mutant?

(d) What happens if the Antp gene is not expressed in the second thoracic segment?

3. Why is the antennapedia mutation considered to be dominant?

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4. In the third thoracic segment, Ubx directs the development of a pair of legs and a pair of halteres (very reduced wings used for balance). It operates largely by suppressing wing formation. Predict the effect in Ubx loss-of-function mutants:

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175 Gene Transplantation Experiments processes. One technique used is gene transplantation. Gene transplantation involves inserting a gene from one cell type into another or from one organism into another. Researchers can study how expression of the gene affects the developmental process in the recipient.

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Key Idea: Genes can be transplanted between cells or individuals and still express functional protein products. These studies provide insight into control of development. Studying gene expression allows researchers to understand the role of gene expression in normal developmental

Genes and the Fate of Cells

As an embryo develops, populations of undifferentiated cells pass a point where they become committed to becoming a certain type of cell. An understanding of what causes this was developed after gene experiments by Harold Weintraub were carried out (below). Myoblasts are cells that are committed to becoming muscle cells. They contain regulatory proteins that control their developmental fate.

A variety of mRNAs coding for various proteins were isolated from myoblasts.

mRNA was converted into DNA using reverse transcriptase.

Promoters were attached to the DNA to ensure it was transcribed later. Promoter

Transplanting eye genes into blind cavefish

The Mexican tetra (Astyanax mexicanus), also called the blind cavefish, is a freshwater fish that can develop two distinct forms. Cavefish living in areas with light have functioning eyes, whereas those living in caves (no light) do not. Both forms can interbreed.

The eyes of both forms begin development in the early embryo stage, but development stops soon after in the blind form. Researchers thought that a gene in the eye cells might be regulating the eye's development. To test this they took a lens from an eyed Astyanax and transplanted it into a blind embryo. The eye in the blind embryo then developed normally, showing the cells in the lens were producing the substances required for eye development. This also showed that the cells in the eye of blind cavefish embryos are able to develop into correctly functioning eyes if they are given the correct signals at the correct time.

When two different and geographically isolated populations of blind cavefish were bred together, the offspring developed functioning eyes. This was because eye development is controlled by numerous genes. Each isolated population of cavefish developed blindness through mutations to different genes. The hybrid offspring gained a working version of each gene, one from each parent, enabling their eyes to develop normally.

Fibroblast

One of the cell lines developed into a muscle cell. The gene responsible is now called MyoD (myoblast determination).

Ltshears

A single new gene was introduced to a single fibroblast cell (connective tissue cell).

Blind cavefish

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1. (a) Why do myoblasts form muscle cells?

(b) Explain why transferring the MyoD gene into non-muscle cells caused them to develop into muscle cells:

2. What do the results of the cavefish gene transplantation experiment tell us about genes and normal development?

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176 miRNA and Development miRNAs play important roles in gene regulation through interactions with mRNA. This is achieved through three mechanisms: 1) cleavage of mRNA, 2) mRNA destabilization, and 3) reduction in the efficiency of mRNA translation so that less protein product is produced. Many studies of embryos have shown that miRNA activity is required for normal development in multicellular eukaryotes.

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Key Idea: The role of miRNA in development processes can be studied in zebrafish. Absence of the dicer enzyme involved in miRNA processing results in abnormal morphology. MicroRNAs (miRNA) are small, highly conserved RNA molecules found in plants and animals and some DNA based viruses. They are around 22 nucleotide bases in length and are non-coding RNA in that they are not translated into protein.

Wildtype

Recall that the dicer enzyme cuts the hairpin loops of dsRNA into short lengths of miRNA. Making the dicer enzyme nonfunctional prevents miRNA being produced. Thus, researchers can study how its absence affects developmental processes.

MZdicer

MZdicer + miR430

Gastrulation

Zebrafish (Danio rerio) embryos contain maternal dicer enzyme. When maternal dicer is removed, the embryos continue to develop normally for 24 hours. This indicates that miRNA activity is not important in regulating early development.

Adapted from Schier, A.F and Giraldex, A.J MicroRNA function and mechanism: Insights from zebra fish Cold Spring Harbor Symposia on Quantitative Biology, 2006

Evidence for the role of miRNA in development

Brain (30 hours)

After the initial normal development, the dicer mutant embryos showed many physical defects and died between five and eight days after fertilization. This indicates that miRNA is critical for controlling embryonic development at this stage.

This information is also supported by expression studies which show that in zebrafish most miRNAs are not expressed early on, but are highly expressed in specific tissues at later stages of development. This also suggests that miRNAs are involved in cellular differentiation.

KEY: ff Wildtype: individuals with normal dicer activity

ff MZdicer: individuals with all dicer activity removed

ff MZdicer +miR430: individuals with all dicer activity removed, but injected with miRNA430 at the onset of zygotic transcription (around the 500 cell stage).

Zebrafish (Danio rerio)

The results above show gastrulation and brain development in the three study groups. Wildtype and MZdicer +miR430 individuals developed normally. MZdicer individuals showed developmental defects and died on day five. The arrows on the images of the 30 hour brain indicate the midbrain-hindbrain boundary.

1. What evidence is there that miRNA is not important in early embryo development (up to 24 hours)?

2. Why was it important to remove all dicer enzyme when studying the effect of miRNA on embryonic development?

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3. (a) Compare the wildtype and MZdicer development results (above right):

(b) What effect did injecting miR430 have on the MZdicer+miR430 individuals?

(c) What do these results reveal about the role of miRNA on developmental processes?

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177 Apoptosis: Programmed Cell Death adult cell numbers and is a defense against damaged or dangerous cells, such as virus-infected cells and cells with DNA damage. Apoptosis also has a role in “sculpting” embryonic tissue during development, e.g. in the formation of digits in developing embryos and resorption of the larval tail during amphibian metamorphosis.

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Key Idea: Apoptosis is a controlled cell suicide that maintains cell numbers and sculpts body parts during development. It is a tightly regulated process. Apoptosis, also called programmed cell death (PCD), is a natural and necessary mechanism in multicellular organisms to trigger the death of a cell. Apoptosis helps to maintain

An overview of apoptosis

ff Apoptosis is a controlled process of cell suicide. It occurs in response to specific cell signals and involves an orderly series of biochemical events. ff The cell and its nucleus shrink and there is an orderly dissection of chromatin by endonucleases.

ff Death is finalized by a rapid engulfment of the dying cell by phagocytosis. This safely disposes of the remains of the cell.

Ed Uthman

The cell shrinks and forms blebs. The nucleus degrades and the chromatin forms clumps.

Blebs

Cells that are stressed, damaged, or triggered by signal molecules begin apoptosis.

Nucleus

Organelle

The nucleus collapses but some organelles are not affected. The cell sends signals to attract macrophages.

The nucleus breaks up into spheres, the DNA breaks up, and the cell fragments.

The cell breaks into numerous apoptotic bodies. Macrophages remove them by phagocytosis.

Apoptotic body

In humans, the mesoderm tissue initially formed between the fingers and toes is removed by apoptosis. Forty one days after fertilization (top), the digits of the hands and feet are webbed, appearing paddle-like. Apoptosis selectively destroys this superfluous webbing, sculpting them into digits, which can be seen in later stages of development (above).

(b) Describe its roles in multicellular organisms:

2. (a) How is apoptosis triggered?

(b) Predict a possible consequence of this mechanism failing:

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1. (a) What is apoptosis?

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178 Structural and Regulatory Genes function. Regulatory genes code for proteins and other small molecules such as microRNAs involved in controlling the expression of structural genes, although they may be some distance away from them. Within a section of DNA, expressed structural genes are enclosed on either side by untranslated regions (UTRs). UTRs contain regulatory sequences, such as the promoter, that control transcription. How the structural and regulatory elements are arranged differs between prokaryotes and eukaryotes (below).

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Key Idea: Structural genes code for all proteins except for regulatory proteins. Regulatory genes code for the molecules involved in controlling the expression of structural genes. Genes are sections of DNA that code for proteins or other mRNA products. They are divided into two broad groups, structural genes and regulatory genes. Structural genes code for any protein product other than a regulatory protein. The proteins encoded by structural genes are diverse and have roles in maintaining a cell's structure and

Prokaryotic gene structure

ff In prokaryotes, several structural genes with related functions are grouped together between UTRs (below). These groupings of structural genes and their regulatory elements are called operons. The UTR upstream of the structural genes contains a regulatory sequence to initiate transcription of the structural genes. The downstream UTR stops transcription of the genes. ff In prokaryotes, the entire transcribed mRNA sequence for the structural genes is translated into proteins (there is no gene editing). Upstream of the operon, there is also a regulatory gene, which encodes a regulatory protein (not shown). Regulatory sequence

DNA

mRNA

Structural genes

Regulatory sequence

UTR

Gene 1

Gene 2

Gene 3

UTR

UTR

Gene 1

Gene 2

Gene 3

UTR

Protein 1

Protein 2

Protein 3

Eukaryotic gene structure

ff In eukaryotes, transcription of structural genes is also under the control of regulatory sequences. However, only one structural gene is enclosed by UTRs and there is no 'bulk control' of a structural gene sequence as seen in prokaryotes.

ff It is important to remember that before the primary RNA transcript is translated in eukaryotes, the non-protein coding introns are removed (only the protein-coding exons form the mature mRNA for translation).

DNA

UTR

Exon

Intron

Exon

Intron

Exon

Intron

Exon

UTR

Introns are removed leaving only the exons to be translated

mRNA

UTR

UTR

Gene 1

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Protein 1

1. (a) What is the difference between a regulatory gene and a structural gene?

(b) What are UTRs and what is their role?

2. How does gene expression differ in eukaryotes and prokaryotes?

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179 Gene Regulation in Prokaryotes

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248

gene can produce a repressor molecule that can bind to the operator and block the transcription of the structural genes. The presence or absence of a functional repressor molecule switches the structural genes on or off and controls the metabolic pathway. Two mechanisms operate in this model: gene induction and gene repression. In gene induction (below), genes are switched on when an inducer binds to the repressor and deactivates it. In gene repression (opposite), genes are normally on but will be switched off when the endproduct of the metabolic pathway is present in excess.

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Key Idea: Prokaryotic genes are organized as operons. An operon is a cluster of several structural genes under the control of the same regulatory genes and sequences. The lac operon (lactose operon) is an operon required for the transport and metabolism of lactose in many bacteria including Escherichia coli. It is often used to study the control of gene expression in prokaryotes. The operon model applies only to prokaryotes because eukaryotic genes are not found as operons. Transcription of the structural genes in an operon is controlled by the promoter and the operator. The regulator

Prokaryotic genes occur as operons

A number of structural genes encoding the enzymes for a metabolic pathway are under the control of the same regulatory elements. Regulatory gene encodes repressor

Regulatory sequences in the untranslated region (UTR)

DNA Regulator gene

Promoter

Operator

Structural genes: translated region

Structural gene 1

Structural gene 2

Structural gene 3

OPERON

The operon consists of the structural genes and the promoter and operator sites

The regulator is some distance from the operon. It codes for the repressor that prevents the expression of specific genes.

Structural genes. At least one structural gene is present in an operon but usually there are more. The lac operon in E. coli has three. Structural genes code for the synthesis of enzymes in a metabolic pathway.

The promoter site is where the RNA polymerase enzyme first attaches itself to the DNA to begin synthesis of the mRNA.

The operator is an ‘on-off’ switch that controls RNA polymerase’s access to the structural genes. It is the repressor binding site.

The lac operon: an inducible operon

Repressor is active and the genes are normally switched off

Glucose is the preferred substrate for E. coli. The disaccharide lactose is uncommon so E. coli transcribes the genes for using it only when it is present.

An active repressor molecule binds to the operator site, switching the gene off.

RNAP

RNA polymerase cannot bind to the promoter

DNA Regulator gene

Genes are not transcribed

Promoter

Operator

lacZ

lacY

lacA

Lactose switches the lac operon on

Presence of lactose results in inactivation of the repressor so the genes can be transcribed er

Induc

The inducer binds to the repressor altering its shape. It can no longer bind to operator site and the gene is switched on.

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When lactose is available, some of it is converted into the inducer allolactose.

Genes are transcribed

DNA Regulator gene

Promoter

Operator

lacZ

lacY

lacA

With the operator site free, RNA polymerase binds to the promoter and the genes for lactose metabolism are transcribed

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The trp operon: a repressible operon

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Transcription is normally on and must be switched off When the effector (tryptophan) is in high concentration, some binds to the trp repressor, activating it.

The combined repressor and effector bind to the operator and prevent transcription of the structural genes encoding tryptophan synthetase, the enzyme required for tryptophan synthesis.

Tryptophan repressor bound to operator.

Genes are not transcribed

Trp repressor

DNA Regulator gene

Promoter

Operator

With the operator site occupied, RNAP cannot bind to the promoter

trpE

trpD

trpC

trpB

trpA

Five structural genes encode the multiunit enzyme tryptophan synthetase

In E. coli, the enzyme tryptophan synthetase produces the amino acid tryptophan. Tryptophan is an important amino acid so the genes for producing tryptophan synthetase are normally switched on. When tryptophan is present in excess, some of it acts as an effector (or co-repressor), activating the repressor and preventing transcription of the structural genes.

1. Outline the role of each of the following components of an operon:

(a) Promoter:

(b) Operator:

(c) Structural genes:

2. What is the role of the repressor molecule in operon function?

3. Summarize the function of the lac operon by completing the following (delete the wrong answer in each choice):

(a) When lactose is absent, the repressor is active / inactive, RNA polymerase can / cannot bind, and the structural genes are / are not transcribed.

(b) When lactose is present, the repressor is active / inactive, RNA polymerase can / cannot bind, and the structural genes are / are not transcribed.

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4. (a) Why is the lac operon usually switched off in E.coli?

(b) What is the advantage in having an inducible enzyme system that is regulated by the presence of a substrate?

(c) Suggest when it would not be an advantage to have an inducible system for metabolism of a substrate:

5. How does the lac operon differ from the trp operon and explain reasons for the difference:

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180 Eukaryotic Gene Structure and Regulation sequences ensure that transcription begins and ends at the correct points and that the necessary sequences are present to create the mature mRNA that can be exported from the nucleus. The promoter region is the binding site for regulatory proteins called transcription factors and it contains several highly conserved regions (sequences that have remained unchanged throughout evolution). Transcription's dependence on sequence recognition and transcription factors provides close control over the expression of genes.

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Key Idea: Eukaryotic genes include both translated and untranslated regions, including control sequences. Regulatory proteins called transcription factors regulate gene expression by ensuring the necessary elements come together. A eukaryotic gene includes more than just the region of exonic DNA that is translated into protein. It also contains introns, regulatory untranslated regions (UTRs), a promoter to which the RNA polymerase binds, and a terminator region that signals the stop point of transcription. The control

The structure of eukaryotic genes

A gene contains both translated and untranslated regions, including control sequences

Upstream control elements

5'

CCAAT TATA box box

5' UTR

Enhancers

DNA coding strand

Poly-A tail signal sequence within the 3' UTR

Transcription start (+1) 25-35 base pairs downstream of the TATA box. Exon

Intron

Exon

Intron

Exon

3' UTR

Terminator sequence signals STOP transcription

Promoter region RNA polymerase recognises and binds

Highly conserved sequences within eukaryotic genes include the TATA box within the promoter and parts of the 5' and 3' UTR. These sequences must have very important functions in gene regulation.

5'

5' UTR

Exon

Transcribed and translated

Intron

Exon

Intron

Exon

3' UTR

3'

Primary transcript

Introns are removed to create the final mRNA

Transcribed and transported to the cytoplasm but not translated

Transcribed and removed before transport to the cytoplasm

3'

Translation start codon (AUG)

5' cap

5' UTR

Translation stop codon (UAG, UAA, or UGA)

Poly A tail 3' UTR ~AAA~AAA~AAA

Mature mRNA

(a) Promoter region:

(b) Terminator sequence:

(c) Transcription start signal:

(d) AUG codon:

(e) UAA, UAG, and UGA codons:

2. What happens to each of the following regions of a eukaryotic gene?

(a) Exons:

(b) Introns:

(c) 5' and 3' UTR: WEB

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1. Outline the role of each of the following regions of a eukaryotic gene:

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RNA polymerase can only transcribe genes in the presence of transcription factors Assembly of the transcription initiation complex

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ff In eukaryotes, RNA polymerase cannot initiate the transcription of structural genes alone. It requires the presence of transcription factors. Transcription factors are encoded by regulatory genes and have a role in creating an initiation complex for transcription.

E n hancer

ff Transcription factors bind to distinct regions of the DNA, including the promoter and upstream enhancers, and act as a guide to indicate to RNA polymerase where transcription should start.

ff The TATA binding protein is a subunit of a multiunit general transcription factor. It is the first to bind to DNA, recruiting other transcription factors to form a transcription initiation complex. Once bound to the promoter sequence, the transcription factors capture RNA polymerase, which can then begin transcription.

Activator proteins bind to enhancer region

General transcription factors, including TATA-binding protein UTR

TATA box

Transcription start signal

RNA polymerase

RNA polymerase binds and transcription begins

DNA

DNA hairpin loop

RNA synthesis

TBP

General transcription factors including the TATA binding protein (TBP) bound to DNA

RNA polymerase binds to the promoter region and begins transcription

Transcription intitiation complex

3. Why would a gene contain regions that are transcribed but not translated?

4. (a) What is a transcription factor?

(b) What sort of genes encode transcription factors?

(c) How are transcription factors involved in the regulation of gene expression?

5. (a) What does it mean to say a DNA sequence is highly conserved?

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(b) Some of the most highly conserved regions of genes include untranslated sequences. Why do you think this is?

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181 KEY TERMS AND IDEAS: Did You Get It?

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

apoptosis

A

Master genes responsible for the development of specific structures during embryonic development.

cellular differentiation

B

A plant hormone involved in triggering seed germination.

C

Short, highly conserved sections of RNA with an important role in regulating gene expression.

D

A controlled process of cell suicide, also called programmed cell death.

E

The method of transplanting genes between organisms. This technique can be used to study developmental processes.

F

These genes are only found in animals and control the development of the front and back regions of the body.

G

The process by which a cell becomes more specialized.

gibberellic acid

gene transplantation homeotic genes Hox genes miRNA

2. Webbing of the toes and fingers in early human fetal development is normal. At 16 weeks a process occurs to remove the webbing.

(a) What is this process called?

(b) Did this process occur normally in the development of the person shown right?

(c) What is the purpose of this process in adults?

3. Study the diagram below, and answer the following questions: Regulatory sequence

Regulatory sequence

Gene 1

Gene 2

Gene 3

Protein 1

Protein 2

Protein 3

(a) What is occurring in the diagram?

(b) Is this process occurring in a eukaryote or prokaryote?

(c) Explain your decision in 3 (b):

UTR

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UTR

Structural genes

4. Distinguish between a structural gene, a regulatory gene, and a regulatory sequence and give an example of each:

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Sources of Variation

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Enduring Understanding

3.C 4.C

Key terms aneuploidy

bacteriophage

3.C.1 Changes in genotype can result in changes in phenotype Essential knowledge

conjugation

(a) Alterations in a DNA sequence can lead to changes in phenotype

crossing over

c 1

fertilization

gene duplication genetic variation genotype

heterozygote advantage

Using examples, explain how DNA mutations can be positive (beneficial), negative (harmful), or neutral based on how they affect the resulting nucleic acid or protein and the phenotype that results.

Activity number

182 186 - 192

(b) Errors in DNA replication or DNA repair, as well as external factors, can cause random mutations in DNA

c 1

Explain how spontaneous and induced mutations occur. Explain how the effect of a mutation (detrimental, beneficial, or neutral) depends on the environmental context. Recognize mutations as the source of all new alleles.

186 187

hybrid vigor

(c) Errors in mitosis or meiosis can result in changes in phenotype

independent assortment

c 1

Explain how changes in chromosome number can result in new phenotypes. Include reference to sterility in triploids and increased vigor of polyploids.

194 - 196

c 2

Using examples, explain how changes in chromosome number can cause human disorders with limitations to development. Examples include trisomy 21 and Turner syndrome (XO).

194 195

lysogenic cycle lytic cycle mutagen mutation

natural selection non-disjunction phenotype plasticity

polyploidy

sexual reproduction transduction

transformation translocation transposition transposons

A. Bolzer, G. Kreth, I. Solovei, D. Koehler, K. Saracoglu, C. Fauth, S. Müller, R. Eils, C. Cremer, M.R. Speicher, T. Cremer-Bolzer et al., (2005) Three-Dimensional Maps of All Chromosomes in Human Male Fibroblast Nuclei and Prometaphase Rosettes. PLoS Biol 3(5): e157 DOI: 10.1371/journal.pbio.0030157, Figure 7a. CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=1371900

(d) Changes in genotype may produce phenotypes that, under natural selection, have greater fitness in the prevailing environment

c 1

Use examples to show how selection for phenotypes that have higher fitness in a particular selective environment can produce evolutionary change. Include: • Mutations for antibiotic resistance in bacteria [also 1.A.2.d] • Mutations for pesticide resistance in insects [see 1.A.2.c] • Sickle cell mutation and heterozygote advantage [also 1.A.2.c]

Essential knowledge

197 220 219 191 217 Activity number

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3.C.2 Multiple processes increase genetic variation in biological systems

188

(a) DNA replication and repair are imperfect and this increases variation c 1

Describe how mutations can arise as a result of error in DNA replication and failure of repair mechanisms.

186 187

(b) Prokaryotes can acquire genetic information through horizontal gene transmission

c 1

Explain how genetic information can be acquired horizontally in prokaryotes through transformation, transduction, conjugation, and transposition.

197

(c) Sexual reproduction in eukaryotes increases genetic variation

c 1

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Describe how the processes involved in gamete formation and fertilization increase genetic variation in the offspring of sexually reproducing organisms. Explain why these processes are conserved and what this means.

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Activity number

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3.C.3 Viral replication results in genetic variation, and viral infection can introduce genetic variation into hosts Essential knowledge

NHGRI

(a) Various mechanisms during viral replication can generate genetic variation

198 200 201 202

c 1

Describe the highly efficient replicative abilities of viruses and explain how these allow viruses to evolve rapidly and acquire new phenotypes.

c 2

Describe the lytic cycle in viruses and explain how it allows one virus to produce many progeny simultaneously (component assembly model).

198

c 3

Explain how viral replication allows for mutations to occur through the usual host pathways.

200

c 4

Describe replication in RNA viruses. Explain how their lack of replication errorchecking results in high mutation rates.

200

c 5

Describe how related viruses can combine/recombine genetic information if they infect the same host cell. Explain how this increases genetic variation.

201

c 6

Using HIV as an example, describe how the rapid evolution of a virus within the host contributes to pathogenicity.

202

(b) The reproductive cycles of viruses facilitate the transfer of genetic information

c 1

Using examples, explain how viruses transmit DNA or RNA when they infect a host cell. Examples could include transduction in bacteria and the movement of transposons (a type of mobile genetic element) present in incoming DNA.

197 200

c 2

Explain how some viruses (e.g. phage) are able to integrate into the host genome to form a latent (or lysogenic) infection. Use examples to describe how these lysogenic infections can result in new properties in the host.

198 199

4.C.1 Variation in molecular units provides cells with a wider range of functions

Activity number

Essential knowledge

(a) Variations within molecular classes provide cells and organisms with a wider range of functions c 1

Explain how variation in molecules or their units provides a wider range of useful functions. Examples include different types of MHC antigens, antibodies, phospholipids in membranes, hemoglobin, or chlorophylls. [also 2.B.1, 4.A.1]

189

(b) Multiple copies of alleles or genes can provide new phenotypes

c 1

Using an example, explain the basis of heterozygote advantage. [also 3.C.1.d]

c 2

Using examples, explain how gene duplication can allow for the evolution of a new property. Examples include the antifreeze gene in fish, leaf digesting enzyme in colobine monkeys, and hemoglobins in vertebrates.

4.C.2 Environmental factors influence expression of the genotype Essential knowledge (content coverage in other chapters also indicated)

191

192 193

Activity number

(a) Environmental factors directly and indirectly influence many traits

Using examples, explain how environmental factors influence the expression of the genotype, directly (e.g. lactose and induction of the lac operon) or indirectly (e.g. diet and disease risk). Other examples include height and weight in humans, flower color and soil pH, seasonal fur color in Arctic animals, density of plant hairs as a function of herbivory, sex determination in reptiles, UV and melanin production in mammals, and presence of the opposite mating type on pheromone production in yeast. [also 3.B. 2.a]

182 183 also 179

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c 1

(b) Adaptation to the local environment reflects a flexible response of the genome

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c 1

Using examples, explain how an organism's adaptation to local environment reflects the ability of its genome to respond in a flexible way (termed plasticity). Examples include darker fur in cooler extremities in mammals and changes to the timing of flowering in response to climate changes.

c

SKILL

Use the Student's t test to determine the significance of phenotypic differences between populations in different environments.

182 183

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182 Sources of Genetic Variation In sexually reproducing organisms, variation in genotype accounts for most of the phenotypic variation we see. Sexual reproduction (meiosis and random fertilization) creates new allele combinations in the offspring and these many variants are tested in the prevailing environment for their ability to survive and leave their own offspring (their fitness). Thus, the variation generated by sexual reproduction provides raw material for selecting the best suited phenotypes. This process is called natural selection.

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Key Idea: Sexually reproducing organisms show variation in their phenotype (appearance) as a result of mutation and the processes involved in meiosis and fertilization. Both genes and environment contribute to the final phenotype. Variation refers to the diversity of genotypes (allele combinations) and phenotypes (appearances) in a population. Phenotypic variation is a feature of sexually reproducing populations and is the product of both genotype and environmental effects on the expression of that genotype.

Mutations

Sexual reproduction

Mutations are changes to the DNA

Sexual reproduction involves meiosis and fertilization Sexual reproduction rearranges and reshuffles the genetic material into new combinations. Fertilization unites dissimilar gametes to produce more variation.

Mutation modifies existing genes to form new alleles. Not all mutations have a phenotypic effect because of the degeneracy of the genetic code. These silent mutations are neither harmful nor beneficial, so may escape selection pressure until conditions change. Mutation: Substitute T for C

Mutated DNA

Don Horne

Original DNA

Genotype

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

Gene interactions and epigenetics

Environmental factors

Dominant, recessive, codominant, and multiple alleles, epigenetic modifications (such as methylation), and interactions between genes, combine in their effects.

The environment can influence the expression of the genotype. Both the external environment (e.g. physical and biotic factors) and internal environment (e.g. hormones) can be important.

Phenotype

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The phenotype is the physical appearance of an individual. An individual’s phenotype is the result of the interaction of genetic and environmental factors during its lifetime. Gene expression can be influenced by both the internal and external environment during and after development.

1. (a) What is the basis of the genetic variation of sexually reproducing organisms?

(b) How does the environment contribute to the phenotype we see:

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256 What does variation look like?

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ff Phenotypic variation can be either continuous, with a large number of phenotypic variants approximating a bell shaped (normal) curve or discontinuous, with only a limited number of phenotypic variants in the population.

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ff Phenotypic characteristics showing discontinuous variation are determined by only one or two genes, e.g. flower color.

ff Phenotypic characteristics showing continuous variation are determined by a large number of genes and are often heavily influenced by environment, e.g. milk production influenced by diet.

Quantitative traits are characterized by continuous variation, with individuals falling somewhere on a normal distribution curve of the phenotypic range. Typical examples include, grain yield in corn (above left), milk production in cattle (center), and growth in pigs (above, right). Quantitative traits are determined by genes at many loci (polygenic) but most are also influenced by environmental factors.

Single comb rrpp

Walnut comb R_P_

Pea comb rrP_

Rose comb R_pp

Comb shape in poultry is a qualitative trait and 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.

Albinism (above) 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.

CWCW

CRCR

Flower color in snapdragons (right) 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

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

(b) Hand span in humans:

(e) Body weight in mice:

(c) Blood groups in humans:

(f) Flower color in snapdragons:

3. (a) Distinguish between continuous and discontinuous variation in phenotype:

(b) Explain the genetic basis of each type:

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4. Suggest why quantitative traits are more likely to be influenced by the environment than qualitative traits:

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between a person with an good diet and a person with an inadequate diet (the effect of which is mostly formed before the age of eighteen), but there are also other effects such as the effect of wider lifestyles on the development of cancer, especially on those predisposed to it through their genes.

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Key Idea: The external or internal environment can have a large effect on the development of an organism. The development of an organism is influenced by a combination of environmental factors and its genome. This may produce obvious effects such as differences in height

Cancers are caused by a complex interaction of the environment, genes, and epigenetic factors. The relatively new field of epigenetics (meaning "on top of genetics") is highlighting the role of gene-environment interactions in gene expression. Epigenetics alters the way the genes are read and expressed but not the genes themselves. Epigenetic processes include DNA methylation (in which methyl groups are added to cysteine bases in DNA) and histone modification. There is increasing evidence that lifestyle, including diets high in sugars, high alcohol consumption, obesity, and lack of exercise, cause epigenetic changes to the DNA encoding microRNAs. MicroRNAs have roles in regulating various genes. An example is the microRNA miR-182, which has a role in regulating the BRCA1 gene. BRCA1 is a tumor suppressor gene and produces a protein that helps repair DNA. Epigenetic modifications increase the activity of miR-182 which in turn decreases BRCA1 activity, which can lead to breast cancer. Snowshoe hares are well known for their change of coat color as the seasons change. During fall, the hares molt and replace their brown coat with a white coat. This again molts in spring and is replaced by a brown coat. This change in coat color is triggered by changes in the photoperiod (day length) and regulated by hormonal changes. Little is known about the genes involved in the color change but a 2014 study showed at least 568 genes were differentially expressed between the white coat stage and the brown coat stage, while another 186 were specifically involved in the molt from white to brown. This transition from white to brown and back helps with camouflaging the hare and thus reducing predation. Because the change is based on the photoperiod and not temperature some scientists have expressed concerns over the effect of global warming and a mismatch of the environment and the timing of the color change.

Normal tissue

Cancerous mass arrowed

Breast cancer mammogram

NIH

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183 Gene-Environment Interactions

Saccharomyces cerevisiae has two mating types, a and a (both haploid). The presence of pheromones from either type can induce the other to express genes that produce proteins that inhibit the cell cycle. The two cells fuse to form a diploid cell. This grows and, under starvation, begins meiosis, producing four haploid spores. Under the right conditions these germinate and undergo mitotic growth. a-factor receptor

α-factor receptor

a

α

α-factor produced by α cell

Arrest of cell cycle

a-factor produced by a cell

a

α

a- factor docks with a receptor

a

α

Cell fusion

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Winter morph

Diploid cell

- nutrients

+ nutrients

Meiosis

Walter Siegmund CC 3.0

Mitotic growth

Summer morph

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Spores

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Acacia xanthophloea

Acacia seyal

3.0

Thorn or leaf length (cm)

Thorn or leaf length (cm)

4.5

4.0

3.5

3.0

2.5

2.0

1.5

Thorn length

Thorn length

Leaf length

Leaf length

0-50

50-100

100 -150

150-200

Fewer giraffes More giraffes Distance from research camp (m)

0-50

50-100

100 -150

150-200

Fewer giraffes More giraffes Distance from research camp (m)

Thorns as induced defenses: experimental A.V. Milewski, Truman P. Young, and Dere k Madden, School for Field Studies, 16 Broadway, Beverly, MA, 01915, USA 1990

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Plants respond to herbivores in a variety of ways including increasing trichome (leaf hair) density with increasing herbivory and increasing the concentration of distasteful chemicals in their leaves. Observations of Acacia trees in Kenya noted those that were regularly browsed had significantly longer thorns than unbrowsed trees. Moreover, trees that were regularly pruned also produced longer thorns. Together with this, leaves were longer at the tops of the trees (above 5 m) than lower down where they were shorter than the thorns, presumably to avoid browse damage by giraffes and other herbivores.

1. What is epigenetics and what role does it play in determining phenotype?

2. Using an example, explain how lifestyle factors and epigenetic changes can lead to cancers:

3. Explain the cause of a snowshoe hare's color change and its survival advantages:

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4. How do pheromones from an 'a' yeast cell affect the behavior of an 'a' yeast cell?

5. (a) Describe the effect of herbivore browsing on the phenotype of Acacia trees:

(b) Why would this growth pattern occur only after browsing damage?

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184 Using the Student’s t Test Key Idea: Differences between two populations can be tested for significance using the Student's t test. How could you test if observed phenotypic differences between two populations were significant or simply due to chance? The Student's t test is commonly used in this situation to compare two sample means, e.g. means for a

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measured characteristic between two populations. It is a simple test for distinguishing differences between samples and is robust even when sample sizes are small. An example outlining the steps in the test is given below. It compares data for a treatment and a control from a hypothetical experiment (the units are not relevant in this case, only the values).

Steps in performing a Student’s t test

1 Calculate summary statistics for the two data sets Control (A)

Treatment (B)

6.6

6.3

5.5

7.2

6.8

6.5

5.8

7.1

6.1

7.5

5.9

7.3

nA = 6, x̄ A = 6.12, sA = 0.496

nB = 6, x̄ B = 6.98, sB = 0.475

1. (a) In an experiment, data values were obtained from four plants in experimental conditions and three plants in control conditions. The mean values for each data set (control and experimental conditions) were calculated. The t value was calculated to be 2.16. The null hypothesis was: "The plants in the control and experimental conditions are not different”. State whether the calculated t value supports the null hypothesis or its alternative (consult t table below):

nA and nB are the number of values in the first and second data sets respectively (these do not need to be the same). x̄ is the mean.

s is the standard deviation (a measure of scatter in the data).

(b) The experiment was repeated, but this time using 6 control and 6 “experimental” plants. The new t value was 2.54. State whether the calculated t value supports the null hypothesis or its alternative now:

2 Set up and state your null hypothesis (H0)

H0: there is no treatment effect. The differences in the data sets are the result of chance and they are not really different. The alternative hypothesis is that there is a treatment effect and the two sets of data are truly different.

2. Explain what you understand by statistical significance:

3 Decide if your test is one or two tailed

4 Calculate the t statistic

For our sample data above the calculated value of t is –3.09. The degrees of freedom (df) are n1 + n2 – 2 = 10.

Calculation of the t value uses the variance which is simply the square of the standard deviation (s2). You may compute t using a spreadsheet but manual computation is not difficult (see opposite). It does not matter if the calculated t value is a positive or negative (the sign is irrelevant).

The absolute value of the t statistic (3.09) well exceeds the critical value for P = 0.05 at 10 degrees of freedom.

We can reject H0 and conclude that the means are different at the 5% level of significance. If the calculated absolute value of t had been less than 2.23, we could not have rejected H0.

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Table of critical values of t at different levels of P. Degrees of freedom

Level of Probability

0.05

0.01

0.001

1 2 3 4 5

12.71 4.303 3.182 2.776 2.571

63.66 9.925 5.841 4.604 4.032

636.6 31.60 12.92 8.610 6.869

6 7 8 9 10

2.447 2.365 2.306 2.262 2.228

3.707 3.499 3.355 3.250 3.169

5.959 5.408 5.041 4.781 4.587

2.131 2.120 2.110 2.101 2.093 2.086

2.947 2.921 2.898 2.878 2.861 2.845

4.073 4.015 3.965 3.922 3.883 3.850

2.060 2.042 2.021 2.009 2.000 1.984

2.787 2.750 2.704 2.678 2.660 2.626

3.725 3.646 3.551 3.496 3.460 3.390

15 16 17 18 19 20

25 30 40 50 60 100

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A one-tailed test looks for a difference only in one particular direction. A two-tailed test looks for any difference (+ or –). This tells you what section of the t table to consult. Most biological tests are two-tailed. Very few are one-tailed.

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3. The table below presents data for heart rate (beats per minute) in samples of ten males and females from a population. (a) Complete the calculations to perform the t test for these two samples. The steps are outlined in the right hand column. x – x̄ (deviation from the mean)

(x – x̄ )2 (deviation from mean)2

Step 1: Summary statistics Tabulate the data as shown in the table (left). Calculate the mean and give the n value for each data set. Compute the standard deviation if you wish (these have been done for you).

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x (bpm)

Male

Female

Male

Female

Male

Female

70

69

-2.3

1

5.29

1

74

62

1.7

-6

2.89

36

x̄ A = 72.3

80

75

nA = 10

73

66

sA = 5.87

75

68

82

57

62

61

69

84

70

61

68

77

nA = 10

nB = 10

The number of samples in each data set

Males

The sum of each column is called the sum of squares

x̄ B = 68.0 nB = 10

sB = 8.47

Step 2: State your null hypothesis

∑ (x – x̄ )2

(b) The variance for males:

Females

The variance for females:

∑ (x –x̄ )2

s2A =

s2B =

Step 3: Test is one tailed / two tailed (delete one) Step 4: Calculating t

4a: Calculate sums of squares Complete the computations outlined in the table left. The sum of each of the final two columns (left) is called the sum of squares.

4b: Calculate the variances Calculate the variance (s2) for each data set. This is the sum of squares ÷ by n – 1 (number of samples in each data set – 1). In this case the n values are the same, but they need not be.

(c) The difference between the means for males and females

(x̄ A – x̄ B) =

(d) t (calculated) =

s2A = ∑(x – x̄ )2(A) nA – 1

s2B = ∑(x – x̄ )2(B) nB – 1

4c: Differences between the means

Calculate the difference between the means (x̄ A – x̄ B)

4d: Calculate t t

(e) Determine the degrees of freedom (d.f.)

d.f. (nA + nB – 2) =

(critical value)

s2A + s2B nA

nB

4e: Determine the degrees of freedom

(f) P =

t

(x̄ A – x̄ B)

=

(g) Your decision is:

Degrees of freedom (d.f.) = nA + nB – 2 where nA and nB are the number of counts in each of populations A and B.

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=

Step 5: Consult the t table Consult the t-tables (opposite) for the critical t value at the appropriate degrees of freedom and the acceptable probability level (e.g. P = 0.05).

5a: Make your decision

Make your decision whether or not to reject H0. If tcalc is large enough you may be able to reject H0 at a lower P value (e.g. 0.001), increasing confidence in the alternative hypothesis.

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185 Quantitative Investigation of Variation It has white flowers and distinctive leaves with three (or occasionally four) leaflets. The leaves are held on petioles that can be 150 mm or more long if left undisturbed. In pasture that is regularly grazed petiole length can be shorter.

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Key Idea: The Student's t test can be used to test the significance of differences between populations for a variable phenotypic character. White clover (Trifolium repens) is a common pasture plant.

Two paddocks containing white clover were grazed by cattle under different regimes during the peak growing season (late winter to early summer). Paddock A was grazed for one day every week whereas paddock B was grazed for one day every four weeks. At the end of the trial, quadrats were used to select random samples of clover and the lengths of the petioles were measured to evaluate the effect of grazing on morphology (in this case petiole length). The results are shown below. Use the Student's t test to determine the significance of the differences between the two populations (grazing regimes). The calculation steps are given in the blue boxes. Steps for calculating the summary statistics with a calculator are in the grey boxes. x (length / mm)

x – x̄ (deviation from the mean)

Leaflet

Petiole

(x – x̄ )2 (deviation from mean)2

Clover leaf

Paddock A

Paddock B

Paddock A

Paddock B

Paddock A

Paddock B

83

30

40.2

-77.5

1616.04

6006.25

70

87

27.2

-20.5

739.84

420.25

32

48

61

92

70

54

45

33

28

135

34

60

* These can be calculated using the variance equation or a standard scientific calculator (see below).

37

81

Step 2: State your null hypothesis

20

139

25

90

30

78

31

125

35

174

80

167

22

184

62

80

35

125

25

163

44

197

30

116

Step 1: Summary statistics

Tabulate the data as shown in the first 2 columns of the table (left). Calculate the mean and give the n value for each data set. Compute the standard deviation if you wish.

Popn A x̄ A = nA = sA* =

Step 3: Test is one tailed / two tailed (delete one) Summary statistics on a calculator

∑ (x – x̄ )2

Most standard scientific calculators will be able to provide you with the number of sample entrants (n), the mean (x̄ ) and the standard deviation (s) once you have entered the data. The procedure shown below is for a Casio fx-82 calculator, a standard classroom calculator. In most Casio models the procedure is similar. Consult the calculator's manual if necessary.

∑ (x –x̄ )2

Input the data on a calculator

Set the calculator to SD mode:

MODE

Step 2

2

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CLR

Enter the data for paddock A

83

Clear the memory:

SHIFT

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Step 3

1

=

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The sum of each column is called the sum of squares

Step 1

Popn B x̄ B = nB = sB* =

DT

70

32

DT

DT

Repeat this procedure for paddock B after you have retrieved all the summary statistics and calculated the variance for paddock A (see next page).

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Step 4: Calculating t

Retrieving the summary statistics SHIFT

1

3

=

(calculates n)

SHIFT

2

1

=

(calculates x̄ )

SHIFT

2

3

=

(calculates s)

3

x2

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4a: Calculate sums of squares Complete the computations outlined in the table left. The sum of each of the final two columns (left) is called the sum of squares.

4b: Calculate the variances Calculate the variance (s2) for each data set. This is the sum of squares ÷ by n – 1 (number of samples in each data set – 1). In this case the n values are the same, but they need not be. s2A = ∑(x – x̄ )2(A) nA – 1

s2B = ∑(x – x̄ )2(B) nB – 1

Calculate the variance SHIFT

2

=

4c: Differences between the means

Calculate the difference between the means (x̄ A – x̄ B)

1. The variance for population A: s2A =

4d: Calculate t

t

=

The variance for population B: s2B =

(x̄ A – x̄ B)

2. The difference between the population means:

s2

2 A+ s B

nA

nB

(x̄ A – x̄ B) =

3. (a) Calculate t:

4e: Determine the degrees of freedom

Degrees of freedom (d.f.) = nA + nB – 2 where nA and nB are the number of counts in each of populations A and B.

Step 5: Consult the t table Consult the t-tables in the previous activity for the critical t value at the appropriate degrees of freedom and probability level (e.g. P = 0.05).

5a: Make your decision Make your decision whether or not to reject H0. If tcalc is large enough you may be able to reject H0 at a lower P value (e.g. 0.001), increasing confidence in the alternative hypothesis.

(b) t (calculated) =

4. Determine the degrees of freedom (d.f.)

d.f. (nA + nB – 2) =

5. P =

t

(critical value)

=

7. Write a conclusion for the investigation:

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6. Your decision is:

8. To further the investigation, it was decided to find out if the regular grazing affected the rate of dry matter increase in the clover. Suggest a way in which this could done:

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186 Mutagens low, but this rate can be increased by environmental factors such as ionizing radiation and mutagenic chemicals (e.g. benzene). In some cases a mutation may trigger the onset of cancer, if the normal controls over gene regulation and expression are disrupted.

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Key Idea: Mutations disrupt the normal DNA sequence and can be caused by a range of external factors. Any factor that causes a mutation is called a mutagen Mutations occur in all organisms spontaneously. The natural rate at which a gene will undergo a change is normally very

Mutagen and effect

1. Describe examples of environmental factors that induce mutations under the following headings:

(a) Radiation:

(b) Chemical agents:

Ionizing radiation

Nuclear radiation from nuclear fallout or exposure to radioisotopes. Ultraviolet radiation from the sun and tanning lamps. X-rays and gamma rays from medical diagnosis and treatment. Ionizing radiation is associated with the development of cancers, e.g. thyroid cancers, and skin cancer (from high exposure to ultraviolet). Fair skinned people at low latitudes are at risk from ultraviolet radiation. Safer equipment has considerably reduced the risks to those working with ionizing radiation (e.g. radiographers).

Viruses and microorganisms

Some viruses integrate into the human chromosome, upsetting genes and triggering cancers. Examples include hepatitis B virus (liver cancer), HIV (Kaposi’s sarcoma), and Epstein-Barr virus (Burkitt’s lymphoma, Hodgkin’s disease), and HPV (left) which is implicated in cervical cancer. Aflatoxins produced by the fungus Aspergillus flavus are potent inducers of liver cancer. Those at higher risk of viral infections include intravenous drug users and those with unsafe sex practices

HIH

2. Explain how mutagens cause mutations:

Poisons and irritants

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. Those most at risk include workers in the chemicals industries, including the glue, paint, rubber, resin, and leather industries, petrol pump attendants, and those in the coal and other mining industries.

3. Explain how cancer can be caused by mutagens:

Photo left: Firefighters and those involved in environmental clean-up of toxic spills are at high risk of exposure to mutagens.

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Diet, alcohol and tobacco smoke

Diets high in fat, especially those containing burned or fatty, highly preserved meat, slow the passage of food through the gut giving time for mutagenic irritants to form in the lower bowel. High alcohol intake increases the risk of some cancers and increases susceptibility to tobacco-smoking related cancers. Tobacco tar is one of the most damaging constituents of tobacco smoke Tobacco tars contain at least 17 known carcinogens (cancer inducing mutagens) that cause chronic irritation of the gas exchange system and cause cancer in smokers.

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187 Mutations mutations). Heritable mutations that are beneficial (provide an advantage) may be passed on and become established within the population. Some mutations are silent, meaning they have no phenotypic effect. These may be carried in the genome and not subject to selection pressure as long as the environment stays the same (i.e. they are neutral). However, a change in the environment can alter selection pressure and may result in the mutation being beneficial or harmful. The effect of a mutation will always be determined by the selection pressures on the population at the time.

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Key Idea: Mutations are changes to an organism's DNA. Beneficial mutations may spread within a population but harmful mutations are not usually retained. A mutation is a permanent change to the DNA sequence of an organism. Mutations allow for new genetic material to arise and be tested within the current environmental conditions. Most mutations are harmful because they reduce fitness. They remain at low levels within a population or they are eliminated altogether. Heritable mutations occur when the gamete producing cells are affected (gametic

Some mutations are retained, others are eliminated

If the mutation is harmful (reduces fitness) in the current environment it is selected against and is usually eliminated from the population.

Original amino acid sequence

Individual with the mutated protein

Mutated amino acid sequence

There is evidence that some 'silent' changes affect mRNA stability and transcription, even though they do not change codon information. In these cases, the mutations are not neutral.

This mutation results in a different amino acid being added to the peptide chain. It changes the protein made.

A mutation occurs. Mutations may arise through errors in DNA replication or from environmental factors (e.g. UV radiation).

A silent mutation is a DNA sequence change that has no phenotypic effect. This may be because the change occurs outside a protein-coding region (in introns), because there is no change in the amino acid (due to code degeneracy) or because an amino acid has been replaced by one with the same properties. Heritable silent mutations may be carried without effect and may only be subject to selection pressure when environmental conditions change.

If the mutation is beneficial (increases fitness) and it is heritable (occurs in the gametes) it is selected for and retained in the population. It may become more common over several generations.

1. What is a mutation?

3. (a) What is a silent mutation?

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2. Why are some mutations retained within a population and others eliminated?

(b) What is the potential advantage of a silent mutation being retained within a population?

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ff Gametic cells are the reproductive (sex) cells of an organism (the egg and sperm). Mutations occurring in these cells are called germ-line or gametic mutations.

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ff Somatic cells (body cells) are all the remaining cells. Mutations to these cells are called somatic mutations. ff Only 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 color)

Normal phenotype (red color)

ff The red delicious apple (right) is a chimera (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

Patch of affected area

Organism

None of the gametes carry the mutation

Gametes of offspring

Entire organism carries the mutation

Half of the gametes carry the mutation

5. Why are only gametic mutations inherited?

6. What is a chimera?

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4. Distinguish between somatic and gametic mutations:

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188 Beneficial Mutations are heritable, they can spread through the population. Some beneficial mutations are not very common in the human population. This is because the mutations have only been in existence for a relatively short time, so the mutations have not yet had time to become widespread.

Apolipoprotein A1-Milano mutation

Lactose tolerance mutation

Lactose free milk allows lactose intolerant adults to consume milk without experiencing unpleasant side effects

The village of Limone, Italy

ff Mutation: Apolipoprotein A1-Milano (a mutation to the apolipoprotein A1 protein).

ff Mutation: Lactose tolerance/lactose persistence.

ff Effect: Helps remove cholesterol from the blood by transporting it to the liver. The mutation causes a change to one amino acid and increases the protein's effectiveness at transporting cholesterol by ten times. ff Benefit: Reduces incidence of heart disease by reducing plaque build up in the arteries (atherosclerosis).

ff Origin: The mutation can be traced back to Limone, Italy, in 1644.

ff Effect: Continued production of lactase enzyme in adults allows the milk sugar lactose (found in dairy products) to be digested. ff Benefit: Adults can digest lactose and gain the nutritional benefits from consuming dairy products. The ability to digest lactose is lost as the young mammal is weaned and lactase production declines.

ff Origin: Lactose tolerance first evolved in cattle or camel-raising populations in Northern Europe, East Africa, and the Middle East around 10,000 years ago.

Until 1932, the only way to reach the town was over steep mountains or across the lake by boat. Limone

Italy

Lake Garda

10%

Brescia

Verona

90%

Adapted from Leonardi, M., et.al J. Int. Dairy J. 22, 88–97 (2012).

it:Utente:Cits

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Key Idea: Beneficial mutations increase the fitness of the organisms that possess them, but they are relatively rare. A beneficial mutation is one that provides a selective advantage by increasing an individual's fitness. Beneficial mutations are rare relative to harmful mutations, but if they

The ability to digest lactose remains highest in populations with a long history of consuming natural milk products. For example, 95% of people of Northern European descent are lactose tolerant. In contrast only 5-10% of people from East Asia can digest lactose. Adults without lactose tolerance have adverse reactions to dairy products including abdominal cramps, diarrhoea, and vomiting.

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1. Explain why many beneficial mutations have not spread through the entire human population:

2. (a) Why would the ability to digest lactose in adults have first developed in cattle-raising populations?

(b) What is the advantage of being able to digest lactose as an adult?

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189 Types of Gene Mutations mRNA transcribed. A point mutation may not alter the amino acid sequence because the degeneracy of the genetic code means that more than one codon can code for the same amino acid. Such mutations are often called silent because the change is not recorded in the amino acid sequence. Mutations that cause a change in the amino acid sequence may be harmful because they alter protein function, but may occasionally be beneficial in the right selective environment.

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Key Idea: Gene mutations are localized changes to the DNA base sequence. In most cases this will cause a change in phenotype. Gene mutations are small, localized 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. Point mutations can occur by substitution, insertion, or deletion of bases and alter the

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

Gly

Glu

GCT TGT TTA TGC TCT GGC CAT CTT CAC CAA AAT

Cys

Ala

S

Arg

His

H2N

1

S

Leu

Cys

GTT TTT GCT CTG TAT CTG GTT TGT GGT GAA CGT GGG TTT TTT TAT ACT CCC AAA ACT TGT

Gly

Leu

Phe

31

Phe

His

Ser

Cys

Gly

S

Thr

Tyr

S

Pro

Lys

Thr

Cys COOH

31 Skin tone with normal active pigment

Substitution mutation

ff In a substitution mutation a single base is substituted with another. Some substitutions may still code for the same amino acid (silent mutation) because of code degeneracy, but they may also result in a codon that codes for a different amino acid.

ff In the example below left, a substitution mutation has altered the 31st triplet TGT to TCT. This results in the amino acid serine appearing where cystine should be (called a missense substitution). This could affect this protein’s function. ff In the example below right, the change from TGT to TGA at the 19th triplet produces a triplet that translates as a stop codon. When the mRNA is translated, translation ends prematurely, leaving a shortened protein. This is a nonsense substitution.

Missense substitution

Nonsense substitution

Missense substitution: Substitute C instead of G

Abnormal pigment is Nonsense substitution: not active. The person Substitute A instead of T has light skin. Leu

Ala

mRNA

Phe

Tyr

mRNA

Mutant DNA

Val

Leu

Amino acids

Leu

Glu

Gly

Val Val

Asn

STOP

Cys

Ala

S

Arg

Leu

Cys

Ser

Tyr

Thr

Amino acids

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Gln

19

Leu

Leu

Cys

His Missense substitution: Substitute C instead of G

Gly

mRNA

Leu

Val

STOP

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AP2

AP2

AP2

AP1

AP1

AP1

120

Ser

Abnormal pigment is not active. The person Amino has light skin. Lys acids

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His

H2N

Mutant DNA

mRNA

31

Asn

Cys

Mutant DNA

Thr

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Leu

Nonsense substitution: Substitute A instead of T

Ser

Phe

Ala

His

Lys

COOH

Ser

COOH

Gly

Pro

Ala

Val

Phe

Phe

Thr

Leu

Cys

Gly

Lys

His

H2N

S

Amino acids

Gln

Leu

Tyr

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15

242

105

29

Thr

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Silent mutations

Leu

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.

Ala

Tyr

Phe

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Val

Leu

19

Glu

Gly

Ala

S

Gly

Leu

Leu

Cys

Phe

Phe

Ser

His

Gly

S

Tyr

Thr

S

Pro

Lys

mRNA

Val

His

H2N

S

Cys

Mutant DNA

Leu

Gln

Val

Cys

Arg

Silent mutation: Substitute C instead of T

Amino acids

Asn

Cys COOH

Thr

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 (or gain) of function.

Val

mRNA

COOH

Phe

Ala

Gln

Asn

Tyr

Cys

Leu

Leu

Cys

Ser

His

Gly

Gln

Ala

Thr

The new sequence produces a new shape in the pigment that makes it darker. This is advantageous in high UV environments.

Deletion of G

Leu

Ala

Tyr

17

Phe

Val

Leu

Asn

Gln

Tyr

Leu Phe Deletion of G

Val

Val

Asn

22

Asn

Val

?

Leu

?

Leu

Phe

Val

Val

Asn

Reading frame shift results in a new sequence of amino acids. The protein is unlikely to have any biological activity.

His

H2N

Leu

Cys

?

Tyr

Val

Ala

?

mRNA

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Phe

Reading frame shift results in a new sequence of amino Mutant acids. The protein is unlikely to have any biological activity. DNA

Amino acids

Ala

Asn

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. Amino acids

His

H2N

S

Leu

Reading frame shift results in a new sequence of amino Mutant acids. The protein is unlikely to have any biological activity. DNA

Deletion mutation Amino

Leu

27

Insertion of G

mRNA

Gly

S

Phe

Thr

Gln

Cys

Gly

Tyr

Asn

Val

Arg

Mutant DNA

Amino acids

Phe

Leu

Glu

Insertion of G

Ala

Tyr

?

?

?

?

?

?

Cys

Ser

His

Gly

COOH

31 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 147 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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190 Inherited Metabolic Disorders mutations in single genes, although most are uncommon. The three genetic diseases described below occur with relatively high frequency and are the result of recessive, dominant, and codominant allele mutations respectively.

PR E V ON IEW LY

Key Idea: Many genetic diseases in humans are the result of mutations to recessive alleles, but some are also caused by dominant or codominant alleles. There are more than 6000 human diseases attributed to Cystic fibrosis (CF)

Huntington disease (HD)

Sickle cell anemia

NYWTS

Dr Graham Beards cc 3.0

Sickled cells

Cystic fibrosis is traditionally treated with physical therapy to clear mucus from the airways.

American singer-songwriter and folk musician Woody Guthrie died from complications of HD

In a person heterozygous for the sickle cell allele, only some of the red blood cells are deformed.

Incidence: Varies with populations: United States: 1 in 1000 (0.1%). Asians in England: 1 in 10,000 European descent: 1 in 20-28 are carriers. Gene type: Autosomal recessive. The most common mutation is DF508, which accounts for around 70% of all defective CF genes. The mutation is a deletion of three bases spanning the 507/508th triplets. The net effect is the loss of the amino acid phenylalanine from the CFTR protein, which normally regulates chloride transport in cells. It is non-functional as a result. Gene location: Chromosome 7

Incidence: An uncommon disease affecting 3-7 per 100,000 people of European descent. Less common in other ethnicities (e.g. Chinese, Japanese, African). Gene type: Autosomal dominant mutation of the HTT gene caused by a trinucleotide repeat expansion on the short arm of chromosome 4. In the mutation (mHTT), the number of CAG repeats increases from the normal 6-30 to 36-125. The severity of the disease increases with the number of repeats. The repeats result in an abnormally long version of the huntingtin protein. Gene location: Chromosome 4

Incidence: Occurs most commonly in people of African descent. West Africans: 1% (10-45% are carriers). West Indians: 0.5%. Gene type: Autosomal mutation involving substitution of a single nucleotide in the HBB gene that codes for the beta chain of hemoglobin. The allele is codominant. The substitution causes a change in a single amino acid. The mutated hemoglobin behaves differently when deprived of oxygen, causing distortion of the red blood cells, anemia, and circulatory problems. Gene location: Chromosome 11

p

q

CFTR

Symptoms: Disruption of all glands including pancreas, bronchial glands (chronic lung infections), and sweat glands (high salt content becomes depleted). Inheritance: Autosomal recessive pattern. Affected people are homozygous recessive for the mutation. Heterozygotes have largely no symptoms and there is some evidence that they are less susceptible to cholera than people without the mutation.

HTT

p

HBB

q

Symptoms: The long huntingtin protein is cut into smaller toxic fragments, which accumulate in nerve cells and eventually kill them. The disease becomes apparent in mid-adulthood, with jerky, involuntary movements and loss of memory, reasoning, and personality. Inheritance: Autosomal dominance pattern. Affected people may be homozygous or heterozygous for the mutant allele.

p

q

Symptoms: Sickling of the red blood cells, which are removed from circulation, anemia, pain, damage to tissues and organs. Inheritance: Autosomal codominance pattern. People who are homozygous for the mutant allele have sickle cell disease. Heterozygotes (carriers) are only mildly affected and show greater resistance to malaria than people without the mutation.

(a) Sickle cell anemia: Gene name:

Chromosome:

(b) Cystic fibrosis:

Gene name:

Chromosome:

(c) Huntington disease: Gene name:

Chromosome:

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1. For each of genetic disorder below, indicate the following:

Mutation type:

Mutation type:

Mutation type:

2. Explain why mHTT, which is dominant and lethal, does not disappear from the population:

3. Suggest why the sickle cell mutation has been maintained in populations, despite being lethal:

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191 Sickle Cell Mutation with a deformed sickle appearance and a reduced capacity to carry oxygen. Many aspects of metabolism are also affected. The mutation is codominant (both alleles equally expressed), and people heterozygous for the mutation (carriers) have enough functional hemoglobin and suffer only minor effects.

PR E V ON IEW LY

Key Idea: The substitution of one nucleotide from T to A results in sickle cell disease. The mutation is codominant. Sickle cell disease is an inherited blood disorder caused by a gene mutation (Hbs), which produces a faulty beta (β) chain hemoglobin (Hb) protein. This in turn produces red blood cells Normal red blood cells

Each red blood cell (RBC) contains about 270 million hemoglobin molecules. In their normal state, the red blood cells have a flattened disc shape which allows them to squeeze through capillaries to offload their oxygen to tissues.

The HBB Gene The gene coding for the β-chain of hemoglobin is on chromosome 11 and consists of 438 bases.

p

HBB gene

Normal hemoglobin produces normal red blood cells

Mutant hemoglobin produces sickle-shaped red blood cells

Sickle cells

The mutated form of hemoglobin has reduced solubility and precipitates when deprived of oxygen. This deforms the red blood cells giving them a rigid sickle shape, which prevents their movement through capillaries.

Each hemoglobin molecule is made up of two α-chains and two β-chains linked together

The 438 nucleotides produce a protein made up of 146 amino acids

Sickle cell anaemia

The sickled RBCs are removed from the circulation leading to anemia. Their rigid shape blocks small vessels and leads to widespread tissue and organ damage.

β-chain hemoglobin

First base

Sickle cell and malaria

Normal base: T Substituted base: A

The sickle cell mutation (HbS) is lethal in the homozygote (two mutated alleles present) but heterozygotes (one mutated allele) are much less susceptible to malaria than unaffected people (two normal alleles). This is because the malarial parasite cannot infect the deformed blood cells. A high frequency of the mutation is present in many regions where malaria is endemic (present in the population all the time). The ability of a heterozygote to confer an adaptive advantage like this is called heterozygous advantage.

q

DNA

Code corresponding to the 1st amino acid

This sequence is the beginning of the DNA template strand for a normal β-chain of hemoglobin (excluding start sequence TAC). The sickle cell mutation involves the substitution of one base for another in the HBB gene, causing one amino acid to be altered. This new amino acid is hydrophobic rather than hydrophilic, which makes the Hb collapse in on itself when deprived of oxygen.

1. (a) Explain the genetic cause of sickle cell disease:

(b) How does the sickle cell mutation result in the symptoms of the disease?

(c) Explain why heterozygotes (carriers) suffer only minor effects:

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2. Briefly explain why there is a high frequency of the sickle cell mutation in populations where malaria is endemic:

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192 Chromosome Mutations or sequence of whole sets of genes on the chromosome (represented by letters below). Translocations may sometimes involve the fusion of whole chromosomes, thereby reducing the chromosome number of an organism. This is thought to be an important evolutionary mechanism by which instant speciation can occur.

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Key Idea: Large scale mutations occurring during meiosis can fundamentally change chromosome structure. Chromosome mutations (also called block mutations) involve the rearrangement of whole blocks of genes (involving many bases), rather than individual bases within a gene. They commonly occur during meiosis and they alter the number

Deletion

Break

Step 1

Inversion

Break

A B C D E F G H

A B C D E F G H

Segment is lost

G H

A B

M N O P Q R S T

A B F E D C G H

Step 3

1 2 3 4

A B C D E F G H

A B C D E F

5 6 7 8 9 0

Step 1

M N O P Q R S T

Fragment joins on to homologous chromosome

1 2 3 4 G H

5 6 7 8 9 0

G H

M N O P Q

M N O P Q

A B C D E F

M N O P Q

F

M N O P Q

Segment removed

Segments join

A B C D E F 1 2 3 4

A B C D E F

A B C D E F

Break

A B C D E

Step 2

M N O P Q R S T

Segment removed

Step 3

M N O P Q R S T

Duplication

Break

Step 2

M N O P Q R S T

The middle piece of the chromosome falls out and rotates through 180° and then rejoins. There is no loss of genetic material. The genes will be in a reverse order for this segment of the chromosome.

Translocation

Step 1

G H

Segment rejoins

A break may occur at two points on the chromosome and the middle piece of the chromosome falls out. The two ends then rejoin to form a chromosome deficient in some genes. Alternatively, the end of a chromosome may break off and is lost.

Non-homologous chromosomes pair during meiosis

Segment rotates 180 °

F

A B

M N O P Q R S T

E

Step 2

M N O P Q R S T

Chromosome rejoins

Step 3

Genes

D

G H

Break

C

A B

Step 1

M N O P Q R S T

C D E F

Step 2

Break

Genes

5 6 7 8 9 0

Step 3

A B C D E A B C D E F

M N O P Q

F

M N O P Q

M N O P Q R S T

In translocation mutations, a group of genes moves between different chromosomes. The large chromosome (white) and the small chromosome (blue) are not homologous. A piece of one chromosome breaks off and joins to the other. When the chromosomes are passed to gametes, some gametes will receive extra genes, while some will be deficient.

A segment is lost from one chromosome and is added to its homologue. In this diagram, the darker chromosome on the bottom is the 'donor' of the duplicated piece of chromosome. The chromosome with the segment removed is deficient in genes. Some gametes will receive double the genes while others will have no genes for the affected segment.

1. Which of the chromosome mutations above results in a loss of genetic information?

Original sequence

Mutated sequence

(a) Inversion:

A B C D E F G H M N O P Q R S T

1

(b) Translocation:

2

3

4

5

6

7

8

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2. For each of the chromosome (block) mutations below, write the new gene sequence after the mutation has occurred:

9

0

3. Which type of block mutation is likely to be the least damaging to the organism? Explain your answer:

4. Why do translocations sometimes reduce the total number of chromosomes?

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193 Gene Duplication only those providing an advantage become established. Once established, the gene may evolve to have a completely different role from the original gene. Gene duplication is very common and has occurred in virtually every species and often more than once. Many crop plants (e.g. kiwifruit) have duplicated not just one gene, but their entire genome. In humans, 38% of our genes have been duplicated.

PR E V ON IEW LY

Key Idea: Gene duplication can provide new copies of genes and thus increase variation. The duplication of whole segments of DNA containing genes is called gene duplication. Gene duplication is important because it provides new variation that can be acted upon by selection pressure, leading to evolution. Like mutations, not all gene duplications become fixed in a population and

ff Unless two copies of the same gene provide an advantage, one of the duplicated genes may develop a new function while the other copy continues on with its original function.

Before duplication

Gene containing region

After duplication

Caenorhabditis elegans (nematode): 49% genes duplicated

Duplicated region

ff In the case of genes that produce proteins with a tendency or ability to perform two functions, there may be adaptive conflict, in which its ability to perform one function compromises its ability to perform another. Gene duplication solves this problem by allowing natural selection to act on the genes so that they follow different evolutionary paths.

Drosophila melanogaster (fruit fly): 41% genes duplicated

Saccharomyces cerevisiae (yeast): 30% genes duplicated

Gene duplication in Antarctic fish

Gene duplication in colobine monkeys

Fish living in the near freezing waters of the Antarctic must have a way of ensuring their blood remains ice free. In many species, this is done by producing proteins with antifreeze properties. There are four major antifreeze proteins used by fish (called AFP types I - IV). The gene for the protein AFP III, found in Antarctic eelpout, is very similar to the gene that produces sialic acid synthase (SAS) (also found in humans).

Gene duplication in colobine monkeys has enabled the production of enzymes that optimally perform similar functions in different body environments. The primary food source of colobines, unlike most other primates, is leaves. The leaves are fermented in the gut by bacteria, which are then digested with the assistance of an enzyme produced by RNase genes. In colobines there are two forms of the RNase genes, RNase1 and RNase1B, while in other primates there is only RNase1. The optimal pH for the enzyme RNase1 and RNase1B are 7.4 and 6.3 respectively. In colobine monkeys, the pH of the digestive system is 6-7, but in other primates it is 7.4-8. RNase1B is six times more efficient at degrading RNA in the gut of colobines than RNase1. RNase1 is also expressed in cells outside the digestive system where it degrades double stranded RNA and may assist in defense against viral infection. RNase1B is 300 times less efficient at this function.

Molecular studies have found that a slight modification to the SAS gene causes the production and secretion of AFP III protein. Importantly, the SAS gene product shows ice binding capability. It appears that duplication of the SAS gene produced a new gene that was selected for its ice binding capabilities and diverged to become the AFP III gene in Antarctic eelpout. AFP III protein and SAS protein have similar structures and can be modified to have similar functions. This provides evidence for the likelihood of diversification of function after gene duplication.

ART G CC 2.0

ff Sometimes having two genes with the same role (functionality) can be an advantage and both genes retain their original function. For example, when there is a high demand for a particular protein.

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1. Explain how gene duplication can result in the evolution of a new gene function:

2. What is the evidence that the AFPIII protein found in Antarctic eel pout may have evolved through a duplication of the SAS gene?

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194 Non-Disjunction can Produce Aneuploidies two of the same type of chromosome and the other gamete receives no copy. This error is known as non-disjunction and it results in abnormal numbers of chromosomes in the gametes. The union of an aberrant and a normal gamete at fertilization produces offspring with an abnormal chromosome number. This condition is known as aneuploidy.

PR E V ON IEW LY

Key Idea: Non-disjunction during meiosis results in incorrect apportioning of chromosomes to the gametes. In meiosis, chromosomes are usually distributed to daughter cells without error. Occasionally, homologous chromosomes fail to separate properly in meiosis I, or sister chromatids fail to separate in meiosis II. In these cases, one gamete receives Non-disjunction in meiosis I

Meiosis I

24

N+1

Non-disjunction in meiosis II

Meiosis II

N

23

23

N

23

22

N-1

24

N+1

Meiosis I

24

24

N+1

46

Meiosis II

46

22

Non-disjunctions show a maternal age effect (frequency increases with maternal age).

23

22

N-1

22

N-1

Number of chromosomes

Gametes

Gametes

Datura stramonium

Down syndrome is the most common of the human aneuploidies. The incidence rate in humans is about 1 in 800 births for women aged 30 to 31 years, with a maternal age effect (the rate increases rapidly with maternal age). Nearly all cases (approximately 95%) result from non-disjunction of chromosome 21 during meiosis. When this happens, a gamete (most commonly the oocyte) ends up with 24 rather than 23 chromosomes, and fertilization produces a trisomic offspring. Right: A karyogram for an individual with trisomy 21. The chromosomes are circled.

The plant Datura stramonium has 12 sets of chromosomes. There are 12 known aneuploids, each trisomic for a different chromosome. Interestingly each aneuploid has its own variety of seed pod shape, ranging from buckling (trisomy 3) to cocklebur (trisomy 6) and spinach (trisomy 10). All the aneuploids survive to be viable adult plants indicating plants are better able to accommodate genetic shuffling.

Photo: Waikato Hospital

Agnieszka Kwiecie, license: CC-BY 3.0

Down Syndrome (Trisomy 21)

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1. Describe the consequences of non-disjunction during meiosis:

2. Explain why non-disjunction in meiosis I results in a higher proportion of faulty gametes than non-disjunction in meiosis II:

3. What is the maternal age effect and what are its consequences?

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195 Aneuploidy in Sex Chromosomes fertility. The Y chromosome carries the SRY gene, which triggers male development in the fetus. In females, only one X chromosome is ever fully active in a cell while the other is inactive except for a few essential genes. Thus the loss or gain of an X chromosome results in the under or over production of certain proteins, resulting in abnormal development.

PR E V ON IEW LY

Key Idea: Nondisjunction of the sex chromosomes (X and Y) during meiosis may result in impaired development or spontaneous abortion of the fetus. The X and Y chromosomes carry genes associated with sex, so the loss of gain of one of these chromosomes can have severe effects on sexual development, including low Faulty egg production

Faulty sperm production

Faulty meiosis during egg cell production can result in egg cells with 0 to up to 4 X chromosomes (nondisjunction in meiosis I and II).

XX

XY

X

XX

Y

X

Y

Y

XX

XX

XY

X

XY

Egg cell

X

Faulty meiosis during sperm cell production can result in an uneven distribution of X and Y chromosomes.

X

Sperm cell

XX

X

XY XY

X

X

X

Non-viable

XXX

XXY

XO

YO

___

(1)

(2)

(3)

(4)

(5)

___

___

___

(6)

(7)

(8)

Turner syndrome

Klinefelter syndrome

Turner syndrome results from having only one sex chromosome (XO). The individual is female, with physical features that typically include a webbed neck and reduced stature. Although of average intelligence, individuals may have difficulty with spatial memory. The individual is usually infertile and will not develop secondary sexual characteristics without hormone treatment.

Klinefelter syndrome results from at least one extra X chromosome (XXY, XXXY). Individuals are male and adults with Klinefelter syndrome are usually taller than average. Individuals are infertile and there may be a feminine distribution of fat tissue around the body. Intelligence varies from low to average.

1. Identify the sex chromosomes in each of the unlabeled embryos (above, right): 2. Identify which fetuses (1-8) are Turner or Klinefelter syndrome:

(a) Turner: (b) Klinefelter: 3. Why is the YO configuration (above) non-viable (i.e. there is no embryonic development)?

A: Chromosome configuration:

Sex of individual (M/F):

B: Chromosome configuration:

Sex of individual (M/F):

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4. For the karyotypes A and B, below, circle the sex chromosomes and state:

Syndrome:

Syndrome:

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196 Polyploidy as a Source of Variation

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276

common in plants, and has been important in the evolution and speciation of flowering plants (angiosperms). Polyploidy in plants produces a species that is reproductively isolated from the “parent” species, and results in instant speciation. Allopolyploidy (involving different species) and autopolyploidy (involving the same species) are both recognized.

PR E V ON IEW LY

Key Idea: Polyploidy is a condition in which a cell or organism has three or more times the haploid chromosome number. Polyploidy cells or organisms contain more than two haploid number of chromosomes (i.e. 3N or more). It arises when chromosomes fail to correctly separate during mitosis or meiosis (non-disjunction). Polyploidy is rare in animals, but

Autopolyploidy

Polyploids that arise within a species are called autopolyploids (the extra chromosomes come from another organism of the same species). Autopolyploidy occurs when chromosomes fail to separate during meiosis or when the cell fails to divide after the chromatids have separated. If a diploid gamete fuses with a haploid gamete, a triploid is formed. Triploids are generally unstable and sterile. However, if two diploid gametes fuse, the resulting tetraploid can be fertile.

Same species

Same species

AA

AA

AA

A

AA

AA

Normal haploid gamete

AA

Diploid gametes

AA

Diploid gamete

AAA

Sterile hybrid

AAAA

Triploid 3N

Fertile hybrid

Tetraploid 4N

Potatoes (left) are autopolyploids. They have a number of ploidy levels, based on a haploid number of 12, ranging from diploid (2n=24) to hexaploid (6n=72). Cultivated potato varieties are tetraploid (4n=48).

Allopolyploidy

Species A

Species B

AA

Infertile hybrid

BB

A

AB

B

Infertile hybrid reproduces asexually.

AABB

Honeyhuyue cc 3.0

AB

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Non-disjunction doubles the chromosome number in the hybrid.

AB

AABB

Many crops are allopolyploids, including wheat, rice, and modern brassicas (above).

Haploid gametes

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Allopolyploidy occurs when two species interbreed to produce a new hybrid with chromosomes from each of the parent species. The hybrid is infertile because the chromosomes cannot pair up. However, mitotic non-disjunction in the sterile hybrid can double the chromosome number and produce homologues, which can pair up during meiosis. Self-fertilization may then produce a viable, fertile hybrid. Many commercial plant varieties are allopolyploids. They show greater heterozygosity and hybrid vigor than autopolyploids.

Species C

Union of gametes from this hybrid produces a new species of interbreeding plants: a fertile allopolyploid.

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277 Advantages of polyploidy

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ff The high frequency of polyploidy in plants indicates that polyploidy provides an adaptive advantage. Often this advantage is the result of hybrid vigor, where the hybrid shows improvements over the parents (e.g. by being larger or growing more vigorously). The increase in heterozygosity (heterozygous for a gene) reduces the frequency of (expressed) recessive mutations and also contributes to hybrid vigor. ff Polyploidy results in gene redundancy and provides opportunities to diversify gene function. Extra copies of the gene not required for its original function can be adapted for use in a different way. This can provide an evolutionary advantage. Many polyploids show novel variation or morphologies relative to their parental species.

Common wheat 6N = 42

Tobacco 4N = 48

Banana 3N = 27

Boysenberry 7N = 49

Strawberry 8N = 56

Kiwifruit 6N = 174

Polyploids can be induced

ff New plant varieties can be made by inducing non-disjunction with chemicals. The induction of polyploidy is a common technique to overcome hybrid sterility during plant breeding. ff Chemicals such as colchicine (right) and N2O gas inhibit spindle fibre formation and stop the separation of chromosomes during mitosis.

ff In plants, seeds or seedlings are soaked in a solution of a spindle inhibiting chemical. The resulting plants will likely develop as polyploids. These can then be propagated and crossed (if fertile) to produce a new variety of plant.

Colchicine

ff Seedless banana and watermelon fruits are produced on triploid plants, which cannot produce fertile gametes (therefore no seeds). ff Non-disjunction is induced in a diploid to produce a tetraploid, which is crossed with a normal diploid to produce the seedless triploid hybrid.

1. Explain how polyploidy can result in instant genetic isolation?

2. (a) What advantages do polyploid organisms often have over the parent species?

(b) Explain the origin of these advantages:

4. (a) How does non-disjunction result in polyploidy?

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3. Distinguish between autopolyploidy and allopolyploidy:

(b) Using an example, explain how deliberate induction of non-disjunction can be used in producing crop varieties:

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197 The Genetic Basis of Resistance in Bacteria

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Key Idea: Bacteria are able to acquire new genetic information by mutation, conjugation, transduction, transposition, and transformation. Bacteria are well known for their ability to become resistant to antibiotics. This can be achieved through a random mutation in a gene that gives the bacteria resistance. Mutations can then be passed on to the next generation during binary fission (vertical gene transfer). However, even though bacteria have short generation times and can produce

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enormous numbers in a very short time, acquiring resistance in this way is inefficient and does not allow the resistance to spread to other bacterial populations (including other strains). A more important way for acquiring resistance in bacteria is by horizontal gene transfer (HGT). This is the transfer of genes between unrelated individuals. In this way, bacteria can acquire a resistant gene from another strain of bacterium or from the environment, allowing the gene to more quickly spread through many bacterial populations.

Methods by which bacteria acquire resistance

Spontaneous and induced resistance

Mutation caused by radiation, chemicals, or transcription error.

Conjugation

Plasmid is transferred via a sex pilus between the bacteria

Mutated gene codes for antibiotic resistance

This bacterium has plasmids that give resistance to antibiotics 1 and 2.

Plasmid giving resistance to antibiotic 1

3. How would a transposon increase variation?

Gene for antibiotic resistance

A viral vector, which has picked up an antibiotic resistance gene from one bacterium, transfers it to another.

Transduction

2. Discuss how HGT contributes to the rapid spread of drug resistance in bacteria:

Plasmid giving resistance to antibiotic 2

The gene is integrated into the genome of the bacterial cell where it confers resistance.

4. How do viruses contribute to the acquisition of new genetic information in bacteria?

The naked DNA is taken in and integrated into the bacterial genome.

5. Which of the methods by which bacteria acquire new genetic information would be useful in genetic engineering. Explain:

Transposition

DNA contains elements called transposons

Transposons are able to change positions in the DNA producing new sequences that may confer antibiotic resistance.

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Transformation

Naked DNA containing a gene for antibiotic resistance is taken up by a bacterium.

Transposon

KNOW

1. Which of the methods in the blue panel left show horizontal gene transfer?

Transposon

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198 Replication in Bacteriophages is said to be latent. The virus may be transduced into becoming active again, entering the lytic cycle and utilizing the host’s cellular mechanisms to produce new virions. The lytic cycle results in death of the host cell through cell lysis. Animal viruses follows a similar pattern to bacteriophage multiplication but there are notable differences. Animal viruses have different mechanisms by which they enter host cells and, once inside the cell, the production of new virions is different. This is partly because of differences in host cell structure and metabolism and partly because the structure of animal viruses themselves is very variable.

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Key Idea: Viruses infect living cells, commanding the metabolism of the host cell and producing new viral particles. Viruses must infect a host cell to reproduce. They do so by using the host's metabolism to produce new virus particles. In bacteriophages or phages (viruses that infect bacterial cells) this process may not immediately follow infection. Instead, the virus may integrate its nucleic acid into the host cell’s DNA, forming a provirus or prophage. This type of cycle, called lysogenic or temperate, does not kill the host cell outright. Instead, the host cell is occupied by the virus and used to replicate the viral genes. During this time, the viral infection

1 The phage attaches itself to a specific host cell, and inserts its nucleic acid and some enzymes into the bacterium.

Life cycle of a lysogenic bacteriophage, l-phage

7 Prophages can be induced

(by mutagens) to excise from the bacterial chromosome and enter the lytic cycle.

Lytic cycle

Lysogenic cycle

The lytic cycle results in cell lysis and death of the host.

Bacteria may develop new properties as a result of the integration of the phage DNA.

4 The host cell bursts

2 Phage DNA circularizes

and new phages emerge to infect new cells.

and enters the lytic cycle or lysogenic cycle.

6 Lysogenic bacterium

reproduces normally.

Prophage

3 Viral components are

produced and assembled by the host cell into virions.

5 A prophage forms when

the phage DNA integrates into the bacterial DNA.

The phage genes for directing the synthesis of new phages are repressed.

(a) The lytic cycle:

(b) The lysogenic cycle:

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1. Discuss the main stages of replication of a typical bacteriophage (e.g. l) as seen in:

2. (a) Identify the cycle during which a bacterium may acquire new properties from a virus:

(b) Explain why this can occur:

(c) Describe the implications of this ability for human health and disease:

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199 Consequences of Lysogeny dormant until induction of the lytic cycle. In some cases the genetic material remains active and confers new properties on the bacteria. This is called lysogenic conversion. In some cases, phages can enhance the virulence of the bacterial host and integrate genes that produce a toxin.

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Key Idea: Some phage genes remain active in the lysogenic cycle and result in new properties in the bacterial host. Upon entering a host cell, phages can undergo one of two life cycles (see previous activity). In most cases, once the viral material is integrated into the bacterial genome it remains

Lysogenic conversion of bacteria Phage

Gene product

Phenotype

Vibrio cholerae

CTX phage

Cholerae toxin

Cholera

Escherichia coli

Lambda phage

Shiga-like toxin

Hemorrhagic diarrhea

Clostridium botulinum

Clostridial phages

Botulinum toxin

Botulism (food poisoning)

Corynebacterium diphtheriae

Corynephage beta

Diphtheria toxin

Diphtheria

Streptococcus pyogenes

T12

Erythrogenic toxins

Scarlet fever

All images CDC

Bacterium

Vibrio cholerae (above) is associated by most people with the disease cholera, which causes severe diarrhea and dehydration. However Vibrio cholerae does not always cause disease. Infection by the CTX phage gives the bacterium its toxinogenicity (ability to produce toxin). CTX is generally only found as an integrated part of the bacterial DNA and only rarely as a free virion.

Clostridium botulinum (above) produces the botulinum toxin, the most powerful neurotoxin known. Research indicates the various toxin types produced (A-G) are inherited from a bacteriophage. This is supported by the presence of integration sites flanking the toxin gene. The toxin types A, B, E, and F are associated with disease in humans.

Streptococcus pyogenes (above) is a usually harmless bacteria. Infection by the T12 bacteriophage causes it to acquire the speA gene, which produces the speA exotoxin responsible for scarlet fever. It was first noted in 1926 that S. pyogenes could be transformed to a virulent strain but the transformation agent was not confirmed as the T12 phage until the 1980s.

(a) The bacterial host:

(b) The bacteriophage:

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1. How would the lysogenic conversion of a bacterial host by a bacteriophage be an advantage to:

2. Even though lysogenic conversion may give the host certain advantages, why could it be described as "a molecular time bomb" to the bacterial host?

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200 Replication in Animal Viruses DNA- and RNA-containing animal viruses but the methods of biosynthesis vary between these two major groups. Generally, DNA viruses replicate their DNA in the nucleus of the host cell using viral enzymes, and synthesize their capsid and other proteins in the cytoplasm using the host cell's enzymes. This is outlined below for a typical enveloped DNA virus. RNA viruses are more variable in their methods of biosynthesis. The example on the next page describes replication in the retrovirus HIV, where the virus uses its own reverse transcriptase to synthesize viral DNA and produce either latent proviruses or active, mature retroviruses.

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Key Idea: Viruses that replicate in animal cells show a range of mechanisms entering, replicating, and leaving a host cell. Animal viruses are more complex and varied in structure than the viruses that infect bacteria. Likewise, animal host cells are more diverse in structure and metabolism than bacterial cells. Consequently, animal viruses exhibit a number of different mechanisms for replicating, i.e. entering a host cell and producing and releasing new virions. Enveloped viruses bud out from the host cell, whereas those without an envelope are released by rupture of the cell membrane. Three processes (attachment, penetration, and uncoating) are shared by both

Entry of an enveloped virus into a cell

Enveloped virion such as the herpes simplex virus

1

Attachment

Digestion of the capsid releases the viral DNA, which is replicated in the host cell's nucleus using viral enzymes. Viral proteins are synthesized in the cytoplasm using the host’s enzymes.

Envelope with attachment spikes or fibers

Receptor portion of protein

Host plasma membrane engulfs the capsid

Viral envelope is discarded

Host cell surface

Virus is enclosed in a membrane

2

Penetration

3

Uncoating

RNA viruses (e.g. Ebola, above), evolve very quickly because RNA polymerases lack the error checking ability of DNA polymerases.

The nucleic acid core is uncoated and the biosynthesis of new viruses begins. Mature virions are released by budding from the host cell.

Influenzavirus

HIV virus

Retroviruses (SS RNA viruses such as HIV) also evolve very quickly. Recent studies show HIV has the fastest mutation rate of any known biological system.

Related viruses can recombine genetic material producing a new viral strain. This can happen when viruses from two different strains infect the same cell.

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Ebola virus

Once the viral particle is attached, the host cell begins to engulf the virus by endocytosis. This is the cell’s usual response to foreign particles.

CDC

NIAID https://www.flickr.com/photos/niaid/ 16436410472/in/album-72157646274075631/ CC 2.0

When a viral particle encounters the cell surface, it attaches to the receptor sites of proteins on the cell's plasma membrane.

1. Describe the purpose of the glycoprotein spikes found on some enveloped viruses:

2. (a) Explain the significance of endocytosis to the entry of an enveloped virus into an animal cell:

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282 HIV particle is attracted to CD4 receptors on a helper T cell.

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1

CD4 receptors

HIV particle fuses with the plasma membrane of the T cell and the capsid is removed by enzymes.

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2

3 Reverse transcriptase

produces viral DNA using the viral RNA is a template.

Nucleus

4 A complementary strand of DNA is formed, producing double stranded DNA.

DNA

5 The DNA is integrated into the

host’s chromosome. The viral DNA is now called a provirus. Unlike a prophage, it never comes out of the chromosome. However, it may remain as a latent infection, replicating along with the host’s DNA.

6 The viral genes are transcribed into mRNA molecules.

New virion

7

9 Budding of the new

viruses from the host cell.

Viral mRNA is translated into HIV proteins. Some mRNA also provides the genome for the next generation of viruses.

8 Assembly of the capsids around the viral genomes.

Pox virus

EII

RNA

A transposons (transposable element) is a genetic sequence that is able to copy itself and insert into or change its position in the genome. The DNA or RNA from a virus that has infected a cell acts in a similar way to transposons, leading some scientists to hypothesize that viruses may have originated from escaped transposons. Transposons are found in every living organism and make up about 50% of the human genome. Some are important in gene regulation and development. It is speculated that many of these transposons originated from viruses that inserted DNA in the human genome and subsequently became permanent sequences. In this case, transposons originated from viruses that evolved in some other way, possibly from small parasitic bacteria that slowly lost the genes needed to synthesize compounds such as ATP, ribosomes, and amino acids.

(b) State where an enveloped virus replicates its viral DNA: (c) State where an enveloped virus synthesizes its proteins:

3. (a) Describe how an HIV particle enters a host cell:

(b) Explain the role of the reverse transcriptase in the life cycle of a retrovirus:

(c) Explain the significance of the formation of a provirus:

4. Why do RNA viruses mutate rapidly?

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5. (a) HIV has a genome size of 9.2 kb. Its reported in vivo mutation rate is 4.1 x 10-3 mutations per base per cell. How many mutations would be expected in the replication of one HIV genome? (A kb is 1000 base pairs).

(b) In the study reporting the result in (a) above, it was noted that the mutation frequency was 44 times less in viral genomes derived from the host's plasma. What does this imply about the viability of most mutations?

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201 Antigenic Variability in Influenzavirus rates in influenzaviruses result in small sequential changes in the viral envelope over time. This phenomenon is called antigenic drift. The most virulent strain, Influenzavirus A, periodically also undergoes antigenic shifts as in which two or more different viral strains combine to form a new subtype. These new subtypes are associated with increased virulence.

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Key Idea: The rapid mutation rate of Influenzavirus causes its surface proteins to continually change from year to year. Influenza (flu) is a disease of the upper respiratory tract caused by Influenzavirus. Globally, flu kills up to half a million people every year and it is estimated that up 35.6 million people in the USA are affected annually. High mutation

How does the Influenzavirus change?

Influenzavirus

Three strains of Influenzavirus. (A, B, and C) affect humans, distinguished on the basis of their nuclear material. Influenzaviruses are able to combine and rearrange the 8 RNA segments of their genome, which alters the protein composition of their glycoprotein spikes. The changes make it difficult for the immune system to detect the virus.

Spikes

Influenzavirus

Antigenic drifts are small, sequential changes within a virus subtype caused by point mutations in a gene. All strains of Influenzavirus show antigenic drift but only influenza A is a public health concern because it is associated with epidemic disease. Changes due to antigenic drift mean that the influenza vaccine must be adjusted regularly to include the most recently circulating subtypes. Antigenic shift occurs when exchange of genetic material between influenza A viruses results in a new subtype with major antigenic differences from existing subtypes. The changes are large and sudden and most people lack immunity to the new subtype. Antigenic shifts are responsible for the influenza pandemics that have killed millions over the last century.

Structure of Influenzavirus

Viral strains are identified by the variation in their H and N surface antigens. Antigens are foreign proteins that alert the immune system to the presence of a pathogen. Viruses are able to combine and rearrange their RNA segments, producing new variants of their H and N glycoprotein spikes (designated by numbers. e.g. H1N2). The influenzavirus is surrounded by an envelope containing protein and lipids.

The genetic material is actually closely surrounded by protein capsomeres (these have been omitted here in order to illustrate the changes in the RNA more clearly). The neuraminidase (N) spikes help the virus to detach from the cell after infection.

H1N1

H1N2

H3N2

Influenza A virus subtypes currently circulating among humans

Type A influenza has a number of subtypes that are determined (and named) by the surface antigens hemagglutinin (H) and neuraminidase (N). Type A is harboured in wild fowl populations and periodically (every 10-40 years) undergoes an antigenic shift to produce a novel subtype.

The 2009 "swine flu"epidemic

Hemagglutinin (H) spikes allow the virus to recognise and attach to cells before attacking them.

The viral genome is contained on eight RNA segments, which enables the exchange of genes between different viral strains.

In 2009, a new strain of the H1N1 influenza subtype was identified. It appeared to be the result of a reassortment of the genes originally found in human and swine flu. The H protein on the surface was one that until 2009 had been seen only in pigs. This meant it was far more likely to avoid the human immune response, including people vaccinated against regular flu. The strain was first identified in North America but quickly spread to the rest of the world and was declared a pandemic in April 2009. Partly due to health measures taken world wide, the pandemic was declared over in August 2010.

1. The Influenzavirus is able to mutate readily and alter the composition of H and N spikes on its surface. (a) Why can the virus mutate so readily?

(b) How does this affect the ability of the immune system to recognize and respond to the virus?

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2. Why is a virus capable of antigenic shift is more dangerous to humans than a virus undergoing antigenic drift:

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202 HIV Evolves Rapidly

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284

retrovirus there is little error checking during replication by RNA polymerase. These factors contribute to HIV's ability to produce new mutations that avoid the body's immune response and make it difficult to produce a effective vaccine against infection. There is no cure and HIV infection requires a multi-drug treatment to target HIV in several ways.

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Key Idea: HIV mutates rapidly and has a short generation time, allowing it to evolve rapidly. HIV has the highest mutation rate of any known biological system and a generation time of between 1.2 and 2.6 days. A single infected cell can produce 1 x 104 new viruses per day. HIV also has a very small genome and because it is a

HIV

The human immunodeficiency virus (HIV) infects the T lymphocytes of the immune system, eventually causing AIDS, a fatal disease, which acts by impairing the body's ability to fight disease.

How does HIV evolve? HIV enters the body

Budding HIV

CDC

HIV replicates quickly. Infected cells each produce thousands of copies of HIV per day so throughout the body billions of copies are produced daily. HIV shows high genetic variability and mutates frequently. It can also combine its genetic material with other HIV viruses to form new strains. These factors have important consequences for the prevention and treatment of HIV. Any vaccine would quickly become ineffective because the virus changes so rapidly and so many different strains are present. Resistance to drugs used for treatment also arises quickly because of rapid mutation rates and short generation times. Preventing new HIV infections is critical to halting the spread of the disease.

The body's immune system responds to HIV and produces corresponding antibodies

HIV uses immune cell to manufacture new viruses

HIV mutant avoids antibody because of new surface molecules

Mutant enters new cell and replicates, producing new mutants

Anti-retroviral drugs target most HIV particles, but miss some.

Prevalence of NNRTI resistance (%)

6

4

New mutants continue to replicate in new cells.

2

0

In general, resistance to drug treatment in HIV increases as the percentage of patients receiving drug therapy increases (left). A 2012 WHO study found 0 5 10 15 20 25 30 35 40 that 10-17% of patients in Western nations had resistance to at least one anti-retroviral drug. In Africa this number was much lower, but increasing. % of people with HIV receiving the anti-retroviral drug NNRTI = non-nucleoside reverse transcriptase inhibitor

1. Describe three reasons why HIV evolves so quickly:

(b)

(c)

2. How many generations of HIV per year can occur in an infected person? 3. Explain why a combination drug therapy is used to treat HIV:

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(a)

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203 KEY TERMS AND IDEAS: Did You Get It?

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

aneuploidy

A The transfer of genetic material between bacteria via a sex pilus.

conjugation

B Phase in which the virus may integrate its nucleic acid into the host cell’s DNA. The host cell is used to replicate the viral genes.

gene duplication

C Mobile genetic elements. Sequences of DNA that are able to change their position in a genome.

lysogenic phase

D Phase in which the virus utilizes the host’s cellular mechanisms to produce new virions. This phase results in death of the host cell through cell lysis.

lytic phase

E

The duplication of any region of DNA that contains a gene. It may involve part of a chromosome that is removed an added to a homologous chromosome.

mutagen

F

The movement of genetic material for one bacteria to another via a virus released from one bacteria which then infects another.

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.

non-disjunction polyploidy

transduction transposons

H The condition of having a chromosome complement of more than 2N (e.g. 3N). I

Any chemical, influence, or object that is able to increase the mutation rate of an organism. This may be radiation, industrial or environmental chemicals, or viral infection.

J

The failure of chromosome pairs to separate properly during meiosis I or II.

K Having an extra copy of a chromosome instead of two homologous chromosomes.

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. Fill in the missing gaps using the word list provided: environment, adaptive advantage, DNA, harmful, silent, alleles, eliminated.

Mutations are changes in an organism's

and they are the source of new

population. Most mutations are

Inherited beneficial mutations provide an

that do not change the amino acid sequence are called

population but may not be subjected to selection pressures unless the

and so are

in a

from the population.

so are retained in the population. Mutations

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mutations. They are retained in the

changes.

4. (a) Circle and describe the genetic abnormality in the karyogram shown right:

(b) How would you name the condition?

(c) What caused this kind of genetic abnormality?

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204 Synoptic Questions For the following multi choice questions circle the correct answer. 2. A couple decide to have child. The father has the genotype Aa for a particular gene. The mother has the genotype aa for the same gene. The probability that the first two children will have the genotype Aa is:

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1. Identify the order of events in eukaryotic gene expression

(a) Translation, mRNA editing, transcription, protein editing.

(b) Transcription, protein editing, translation, mRNA editing.

(c) Transcription, mRNA editing, translation, protein editing.

(d) mRNA editing, transcription, protein editing. translation

(a) 1

(b) 0.5

(c) 0.25

(d) 0

3. The diagram to the right shows that the injection of M-phase cytoplasm into the cytoplasm of an egg cell triggers mitosis. What would occur if the cytoplasm of a G2 phase cell was injected into an egg cell cytoplasm?

4. Which statement about prokaryotes is not true?

(b) Prokaryotes have a single, naked, circular chromosome.

(a) DNA replication would occur.

(c) Translation can occur at the same time as transcription in prokaryotes.

(b) The cell will enter mitosis.

(c) Nothing will happen.

(d) Mitosis will stop.

(a) mRNA must be transported to the cytoplasm of the prokaryote cell before translation can begin.

(d) Prokaryote genes have no introns.

Spindle fiber forming

5. The pedigree on the right shows the inheritance of a common single gene trait. The inheritance of the trait is: (a) Autosomal dominant.

(b) Autosomal recessive.

Normal

(c) Sex linked dominant.

Affected

(d) Sex linked recessive.

6. In the early 1960s Herbert Taylor carried out an experiment on bean root tip cells. He grew them in a solution of radioactive thymidine, let them undergo one round of DNA replication and cell division then transferred them to a nonradioactive solution and let them replicate again. The chromosomes were then examined for radioactivity: Cell division

Non-radioactive

Replicated chromatids

First replication

Chromosome

Replicated chromatids

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Radioactive

Second replication

Replicated chromatids

Chromosomes

Explain the pattern of radioactivity and identify the pattern of replication supported by the results:

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7. Study the following DNA sequences: 1: GGC GTT CCG CGT AAG AGT CGT The mutation in sequence 2 is a: (a) Deletion

(b) Insertion

(c) Inversion

(d) Substitution

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2: GGC GTT CCG CCT AAG AGT CGT

8. The diagram below shows a simplified overview of DNA replication. Summarize the process by labelling the diagram according to the questions below:

(a) Label the 5' and 3' ends of all strands on the DNA being replicated.

(b) Label: (i) A parent strand (ii) A daughter strand (iii) Free nucleotides (iv) The new chromatids (v) The leading strand (vi) The lagging strand

A

(c) Draw in the positions of: DNA polymerase (i) (ii) Helicase (d) What is happening at position A?

(e) Circle the nucleotide to be added next to the leading strand and use an arrow to show where it will go.

(f) Circle the nucleotide to be added next to the lagging strand and use an arrow to show where it will go.

9. The diagrams below depict the lac operon in E. coli. The first structural gene is identified for reference.

(a) Label the promoter, operator, and regulator regions and identify the remaining two structural genes lacY and lacA:

(b) Rule a line to indicate the boundaries of the operon

(c) Which gene produces the repressor molecule?

DNA

lacZ

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(d) Draw the position of the repressor molecule and RNA polymerase when lactose is absent. Are the structural genes transcribed?

(e) Draw the position of the repressor molecule and RNA polymerase when lactose is present.

Are the structural genes transcribed?

DNA

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10. It is often said a child inherits one half of their DNA from their mother and one half of their DNA from their father. If this is true, explain why siblings are not identical:

11. A male Drosophila with genotype CucuEbeb (straight wing, grey body) is crossed with a female with genotype cucuebeb (curled wing, ebony body). The phenotypes of the F1 were recorded and the percentage of each type calculated. The percentages were straight wings, grey body 45%, curled wings, ebony body 43%, straight wings, ebony body 6%, and curled wings grey body 6%.

(a) What are the phenotypic characteristics being investigated here?

(b) What sort of cross is this? Justify your answer:

Straight wing: Cucu Grey body: Ebeb

Curled wing: cucu Ebony body: ebeb

(c) Is there evidence of crossing over in the offspring? (d) Explain your answer:

(e) Determine the genotypes of the offspring:

12. Draw a diagram to show how a change in a DNA nucleotide sequence can result in a change in the polypeptide made:

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13. Explain the connection between genetic variation in organisms and phenotypic variation in populations:

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Genetic Change in Populations

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Enduring Understanding

1.A

Key terms adaptation

1.A.1 Natural selection is a major mechanism of evolution Essential knowledge

allele frequency

artificial selection

(a) Darwin's theory explains how favorable phenotypes proliferate

evolution

c 1

fitness

founder effect gene pool

Understand that there have been many contributors to the development of evolutionary thought and that Darwin's work was preceded by that of other scientists and was continued and extended by the work of scientists after him. Describe how natural selection operates, including the role of competition for limited resources, variation, and differential survival of favorable phenotypes.

genetic drift

(b) Evolutionary fitness is measured by reproductive success

genetic equilibrium

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genetic variation

Define fitness and recognize it as a measure of individual reproductive success. Recognize sexual selection as a mode of natural selection and explain how it results in the greater reproductive success of some individuals over others.

Hardy-Weinberg equation

(c) Genetic variation and mutation play roles in natural selection

natural selection

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phenotypic variation population (genetic) bottleneck

Recall the role of mutations as the source of all new alleles. Explain the role of genetic variation in providing the raw material for natural selection.

Activity number

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(d) Environments can be relatively stable or fluctuating

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selection pressure

Use examples to explain how environmental stability or change can determine the type of selection operating and the rate at which evolutionary change occurs.

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sexual selection

Olaf Leillinger

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(e) Adaptations are genetic variations that are favored by selection

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Define adaptation and explain how adaptations are the result of selection in the prevailing environment.

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(f) Chance and random events can influence the evolution of populations

Explain how chance and random events can influence the pace and the direction of evolution. Include reference to population bottlenecks, the founder effect, and the effect of genetic drift (especially in small populations).

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(g) There are specific conditions for genetic equilibrium and these are rarely met

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Describe the conditions required for genetic equilibrium and explain why these are seldom met in real populations.

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(h) Mathematical approaches can be used to provide evidence for evolution in populations

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Explain how changes in allele frequencies can be calculated for a population and 212 213 can show evidence of evolution (change in allele frequencies over time).

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Apply mathematical methods and understanding to analyze data reflecting a change in the genetic makeup of a real or simulated population over time.

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1.A.2 Natural selection acts on phenotypic variation

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Essential knowledge

Activity number

(a) Environmental change can act as a selective mechanism on populations c 1

Using examples, explain how change in environments can alter the selective pressures on populations. Examples include melanism in peppered moth and flowering times in relation to global climate change.

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(b) Phenotypic variations are random and not directed by the environment c 1

Understand that phenotypic variation is the result of mutation and the processes involved in sexual reproduction. The environment simply selects the phenotypes that are most favorable at the time.

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(c) Some phenotypic variations significantly increase or decrease fitness

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Marc King

Describe how phenotypic variations can significantly increase or decrease the fitness of the individual or population. Examples include sickle cell disease, melanism in peppered moth, and DDT resistance in insects. Identify the role of the selective environment in each case.

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(d) Humans have an impact on variation in other species

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Using examples, explain how humans can deliberately or inadvertently alter the allele frequencies of other populations. Examples include artificial selection (animals or plants), loss of genetic diversity in crop varieties, and spread of antibiotic resistance in bacteria as a result of the overuse of antibiotics.

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Investigate the process of natural selection using artificial selection for quantitative traits in fast plants (Brassica rapa). Analyze your results graphically and statistically and discuss and present your findings.

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1.A.3 Evolutionary change is also driven by random processes Essential knowledge

Activity number

(a) Genetic drift c 1

Jeff Podos

Define genetic drift and explain why it has a disproportionate effect on small populations, e.g. founder populations or populations that have gone through a bottleneck (including endangered species).

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(b) Reduction of genetic variation within a given population can increase the difference between populations of the same species

With reference to the founder effect or population (genetic) bottlenecks, explain how the genetic variation within a population can be reduced and how this can lead to genetic divergence between population of the same species. Use examples to illustrate your explanation.

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205 A Pictorial History of Evolutionary Thought Key Idea: The modern theory of evolution was developed over many years with contributions by many scientists. Although Charles Darwin is largely credited with the development of the theory of evolution by natural selection, his ideas did not develop in isolation, but within the context of the work of others before him. The modern synthesis of evolution (below) has a long history with contributors from all

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fields of science. The diagram below summarizes just some of the important players in the story of evolutionary biology. This is not to say they were collaborators or always agreed. However, the work of many has contributed to a deeper understanding of evolutionary processes. This understanding continues to develop with the use of molecular techniques and work between scientists across many disciplines.

Find out more!

This timeline has been adapted from the University of California, Berkeley's excellent Evolution 101 website. Go to the Weblink indicated at the bottom of the page to find out more about the events and the people described.

GEOLOGY - EARTH'S HISTORY -

PALEONTOLOGY - LIFE'S HISTORY -

THE MECHANISMS OF EVOLUTION

DEVELOPMENT AND GENETICS

Modern evo-devo Stephen Jay Gould

Genetic similarities Wilson, Sarich, Sibley, & Ahlquist

1900 to present day

Endosymbiosis Lynn Margulis

Radiometric dating Clair Patterson

DNA James Watson & Francis Crick

Speciation Ernst Mayr

THE MODERN SYNTHESIS OF EVOLUTION Brought together many disciplines and showed how mutation and natural selection could produce large-scale evolutionary change. Theodosius Dobzhansky

1900

Human evolution Huxley & Dubois

The founding of population genetics Fisher, Haldane, & Sewall Wright Chromosomes and mutation Thomas Hunt Morgan

Biogeography Wallace & Wegener

Early evo-devo Ernst Haeckel

Evolution by natural selection Charles Darwin & Alfred Russel Wallace

1800s

Genes are discrete Gregor Mendel

Uniformitarianism Charles Lyell

Chromosomal basis of heredity August Weismann

Biostratigraphy William Smith

Extinctions Georges Cuvier

Old Earth and ancient life Comte de Buffon

Pre 1800

1700

The order of nature Carl Linnaeus

Fossils and the birth of palaeontology

The ecology of human populations Thomas Malthus

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Putneymark

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Developmental studies Karl Von Baer

Evolution Lamarck

Observation and natural theology Important because it addressed the question of how life works Comparative anatomy Andreas Vesalius

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The development of the modern synthesis

Wallace

Gregor Mendel (1822-1884) developed ideas of the genetic basis of inheritance. Mendel’s particulate model of inheritance was recognized decades later as providing the means by which natural selection could occur.

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From the Theodosius Dobzhansky Papers held by the APS. Used under Fair Use licence.

Charles Darwin (1809-1882) and Alfred Russel Wallace (1823-1913) jointly and independently proposed the theory of evolution by natural selection. Both amassed large amounts of supporting evidence, Darwin from his voyages aboard the Beagle and in the Galápagos Islands and Wallace from his studies in the Amazon and the Malay archipelago. Wallace wrote to Darwin of his ideas on evolution by natural selection, spurring Darwin to publish The Origin of Species.

Theodosius Dobzhansky (1900-1975) was a Ukrainian who synthesized the ideas of genetics and evolutionary biology and defined evolution as "a change in the frequency of an allele within a gene pool". Dobzhansky worked on the genetics of wild Drosophila species and was famously quoted as saying "Nothing in biology makes sense except in the light of evolution".

Ernst Mayr (1904-2005) was a German evolutionary biologist who collaborated with Dobzhansky to formulate the modern evolutionary synthesis. He worked on and defined various mechanisms of speciation and proposed the existence of rapid speciation events, which became important for later ideas about punctuated equilibrium.

The modern synthesis today

Display: Oxford University Museum of Natural History

GDallimore cc 3.0

Darwin

Ronald Fisher, JBS Haldane, and Sewall Wright founded population genetics, building sophisticated mathematical models of genetic change in populations. Their models, together with the work of others like Mayr and Dobzhansky contributed to a refinement and development of Darwin's theory into the modern synthesis. Haldane was quoted as saying that the Creator must have "an inordinate fondness for beetles".

Stephen Jay Gould (1941-2002)

Kathy Chapman online cc 3.0

After Haeckel's flawed work on embryology fell out of favor, the evolutionary study of embryos was largely abandoned for decades. However, in the 1970s, Stephen Jay Gould's work on the genetic triggers for developmental change brought studies of embryological development back into the forefront. Today evo-devo is providing some of the strongest evidence for how novel forms can rapidly arise.

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James Watson and Francis Crick's discovery of DNA's structure in 1953 revolutionized evolutionary biology. The genetic code could be understood and deciphered, and the role of mutation as the source of new alleles was realized.

In recent decades, DNA hybridization studies, DNA sequencing and protein analyses have revolutionized our understanding of phylogeny. Allan Wilson was one of a small group of pioneers in this field, using molecular approaches to understand evolutionary change and reconstruct phylogenies, including those of human ancestors.

1. Using a separate sheet, research and then write a 150 word account of the development of evolutionary thought and the importance of contributors from many scientific disciplines in shaping what became the modern synthesis. You should choose specific examples to illustrate your points of discussion.

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206 Darwin’s Theory competition for resources, and (4) proliferation of individuals with better survival and reproduction. Natural selection

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Key Idea: Evolution by natural selection describes how organisms that are better adapted to their environment survive to produce a greater number of offspring. 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)

is the term for the mechanism by which better adapted organisms survive to produce a greater number of viable offspring (they have greater fitness). 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

Overproduction

Variation

Populations produce too many young: many must die

Individuals show variation: some are more favorable than others

Populations tend to produce more offspring than are needed to replace the parents. Natural populations normally maintain constant numbers. There must therefore be a certain number that die without producing offspring.

Individuals in a population vary in their phenotype and therefore, their genotype. Some variants are better suited to the prevailing environment and have greater survival and reproductive success.

Natural selection

Natural selection favors the best suited at the time

Andrew Dunn www.andrewdunnphoto.com

The struggle for survival amongst individuals competing for limited resources will favor those with the most favorable variations. Relatively more of those without favorable variations will die.

Inherited

The variations (both favorable and unfavorable) are passed on to offspring. Each new generation will contain proportionally more descendants of individuals with favorable characters.

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The banded or grove snail, Cepaea nemoralis, is famous for the highly variable colors and banding patterns of its shell. These polymorphisms are thought to have a role in differential survival in different regions, associated with both the risk of predation and maintenance of body temperature. Dark brown grove snails are more abundant in dark woodlands, whilst snails with light yellow shells and thin banding are more commonly found in grasslands.

Variations are inherited: the best suited variants leave more offspring

1. In your own words, describe how Darwin’s theory of evolution by natural selection provides an explanation for the change in the appearance of a species over time:

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207 Adaptation and Fitness

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that do not increase fitness will not be favored and will be lost. Genetic adaptation should not be confused with physiological adjustment (acclimatization), which refers to an organism’s ability to adjust its physiology to changing conditions (e.g. a person's acclimatization to altitude). Acclimatization is a response to the need to maintain a stable internal state (homeostasis) although, of course, homeostatic mechanisms themselves are a product of evolution.

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Key Idea: An adaptation is any heritable trait that equips an organism for its functional role in the environment (its niche). An adaptation is any heritable characteristic (trait) that equips an organism for its niche, enhancing its exploitation of the environment and contributing to its survival and successful reproduction (fitness). Adaptations may be structural (morphological), physiological, or behavioral and are the result of evolution in particular environments. Traits

Mosquitoes are highly successful organisms, inhabiting almost all regions on Earth. A range of adaptations contribute to their evolutionary success.

Mosquitoes can produce large numbers of offspring. A single female can lay several batches of eggs in a year. Each batch contains 50-300 eggs.

Ben133uk cc 3.0

Adaptations in mosquitoes

Chemical receptors guide mosquitoes to food sources. Mosquitoes with human targets have carbon dioxide (CO2) receptors located in the antennae. The mosquito detects exhaled CO2 and flies towards the chemical following the concentration gradient. They also have sensors for body odor and temperature.

Jim Gathany CDC

A long proboscis allows access to nectar in flowers. In females, the sharp tube-like mouthparts allow them to pierce the host's skin and suck their blood. Only females feed on blood as it is required for egg development.

Eggs are laid in water and the larval stage (left) and pupal stage are also aquatic. Breathing occurs at the water/air boundary through spiracles located on the abdomen of the larvae. In unfavorable conditions (e.g. winter) the organisms can halt their development until conditions improve (e.g. warmer temperatures). This is called diapause, and can occur at almost every developmental stage.

Mosquito saliva contains anticoagulant, a chemical to prevent the host's blood from clotting as it is drawn out. As the mosquito feeds a special membrane forms around the blood to keep it separate from the other contents in the stomach. To cope with the high volume liquid diet, the mosquito secretes unwanted liquid (circled), leaving solid nutrients behind.

Many species have adaptations for attracting mates. A commonly cited example is the showy tail-feather display of the male peacock (right). Males with large and highly colorful tail feathers are far more likely to attract a mate and pass their genes on to the next generation then males with less impressive displays. The impressive tail display indicates a healthy bird and therefore a good mate choice for the female. This type of mate selection is called sexual section. However, there comes a point where the tail becomes detrimental to survival (i.e. individuals that cannot move quickly or cannot fold up their tail to become less conspicuous are more likely to be eaten by predators).

1. Distinguish between adaptation and acclimatization:

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Adaptations for attracting mates

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295 Ear length in rabbits and hares

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The external ears of many mammals are used as important organs to assist in thermoregulation (controlling loss and gain of body heat). The ears of rabbits and hares native to hot, dry climates, such as the jack rabbit of south-western USA and northern Mexico, are relatively very large. The Arctic hare lives in the tundra zone of Alaska, northern Canada and Greenland, and has ears that are relatively short. This reduction in the size of the extremities (ears, limbs, and noses) is typical of cold adapted species and is known as Allen's rule.

Arctic hare: Lepus arcticus

Black-tail jackrabbit: Lepus californicus

Regulation of body temperature requires a large amount of energy, and mammals exhibit a variety of structural and physiological adaptations to increase the effectiveness of this process. Heat production in any endotherm depends on body volume (heat generating metabolism) whereas the rate of heat loss depends on surface area. Increasing body size minimizes heat loss to the environment by reducing the surface area to volume ratio. Animals in colder regions therefore tend to be larger overall than those living in hot climates. This relationship is know as Bergman’s rule and it is well documented in many mammalian species. Cold adapted species also tend to have more compact bodies and shorter extremities than related species in hot climates (Allen's rule).

Drew Avery cc2.0

Body size in relation to climate

The fennec fox of the Sahara illustrates the adaptations typical of mammals living in hot climates: a small body size and lightweight fur, and long ears, legs, and nose. These features facilitate heat dissipation and reduce heat gain.

The Arctic fox shows the physical characteristics typical of cold adapted mammals: a stocky, compact body shape with small ears, short legs and nose, and dense fur. These features reduce heat loss to the environment.

2. Evaluate the mosquito's adaptations and discuss how these could have arisen through evolution by natural selection:

3 (a) Why do females often use physical characteristics to choose a mate?

(b) Using the peacock as an example, describe why overly exaggerated physical features can become detrimental to an individual's survival:

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4. Explain the nature of the relationship between the length of extremities (such as limbs and ears) and climate:

5. Explain the adaptive value of a compact body with a relatively small surface area in a colder climate:

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208 Sexual Selection

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who first introduced the concept of sexual selection, a special type of natural selection that produces anatomical and behavioral traits that affect an individual's ability to acquire mates. Biologists recognize two types: intrasexual selection (usually male-male competition) and intersexual selection (also called mate choice). One result of either type is the evolution of sexual dimorphism (differences in appearance between males and females of the same species).

Intrasexual selection

Intersexual selection

Intrasexual selection involves competition within one sex (usually males) with the winner gaining access to the opposite sex. Competition often takes place before mating, and males compete to establish dominance or secure a territory for breeding or mating. This occurs in many species of ungulates (deer, antelope, cattle) and in many birds. In deer and other ungulates, the males typically engage in highly ritualized battles with horns or antlers. The winners of these battles gain dominance over rival males and do most of the mating.

In intersexual selection (or mate choice), individuals of one sex (usually the males) advertize themselves as potential mates and members of the other sex (usually the females) choose among them. Intersexual selection results in development of exaggerated ornamentation, such as elaborate plumages. Female preference for elaborate male ornaments is well supported by both anecdotal and experimental evidence. For example, in the long-tailed widow bird (Euplectes progne), females prefer males with long tails. When tails are artificially shortened or lengthened, females still prefer males with the longest tails. They therefore select for long tails, not another trait correlated with long tails.

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Key Idea: Sexual selection is a type of natural selection in which there is differential reproductive success among individuals of the same sex (and species) because of mate choice or competition within one sex. The success of an individual is measured not only by the number of offspring it leaves, but also by the quality or likely reproductive success of those offspring. This means that it becomes important who its mate will be. It was Darwin (1871)

Mean number of nests per male

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In male-male competition for mates, ornamentation is used primarily to advertize superiority to rival males, and not to mortally wound opponents. However, injuries do occur, most often between closely matched rivals, where dominance must be tested and established through the aggressive use of their weaponry rather than mere ritual duels.

Before tail treatment

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Controls

Elongated

As shown above, there was no significant difference in breeding success between the groups before the tails were altered. When the tails were cut or lengthened, breeding success went down and up respectively relative to the unaltered controls.

2. Suggest how sexual selection results in sexual dimorphism:

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1. Explain the difference between intrasexual selection and mate choice, identifying the features associated with each:

3. Why does sexual selection fix exaggerated characteristics in a population even though they might be harmful?

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209 Gene Pools and Evolution called the allele frequency. Allele frequencies in populations can vary as a result of the events occurring in the gene pool. These changes over time are called evolution. Four microevolutionary processes contribute to genetic change in populations: mutation, gene flow, natural selection, and genetic drift. These are described below (definitions in blue).

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Key Idea: The proportions of alleles in a gene pool can be altered by the processes that increase or decrease variation. A gene pool is the collection of all the alleles in a population. The term deme is sometimes used to refer to a local population of individuals that interbreed freely and share a distinct gene pool. The proportion of each allele within a gene pool is Immigration: New genes may be introduced.

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Mutations: Changes to the DNA sequence can create new alleles.

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Emigration: Genes may be lost.

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Natural selection: The differential survival of favorable phenotypes (unfavorable allele combinations have lower survival or reproductive success). Natural selection accumulates and maintains favorable allele combinations. It reduces genetic diversity within the gene pool and increases differences between populations.

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.

Deme 2

This activity portrays two populations of a beetle species. Each beetle is a “carrier” of genetic information, represented by the alleles (A and a) for a gene that controls color and has a dominant/recessive inheritance pattern. There are normally two phenotypes: black and pale.

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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.

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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. WEB

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1. One of the important theoretical concepts in population genetics is that of genetic equilibrium, which state that "for a large, randomly mating population, allele frequencies do not change from generation to generation". If allele frequencies in a population are to remain unchanged, all of the following criteria must be met: the population must be large, there must be no mutation or gene flow, mating must be random, and there must be no natural selection. Evolution is a consequence of few if any of these conditions ever being met in natural populations. For each of the five factors (a-e) below, describe how and why each would affect the allele frequency in a gene pool. Use the diagrams to help you.

(a) Population size:

(b) Mate selection:

Factors favoring gene pool stability (no evolution)

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(b) Decrease genetic variation in populations:

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Emigration

2. Identify a factor that tends to:

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(a) Increase genetic variation in populations:

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210 Gene Pool Exercise

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topic (pages: Gene Pools and Evolution, The Founder Effect, Population Bottlenecks, and Genetic Drift).

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Cut out each of the beetles on this page and use them to model the events within a gene pool as described in this

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211 Changes in a Gene Pool Key Idea: Natural selection and gene flow as a result of migration can alter the allele frequencies in gene pools. The diagram below shows an hypothetical population of beetles undergoing changes as it is subjected to two ‘events’. The three phases represent a progression in time (i.e. the

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same gene pool, undergoing change). The beetles have two phenotypes (black and pale) determined by the amount of pigment deposited in the cuticle. The gene controlling this character is represented by two alleles A and a. Your task is to analyze the gene pool as it undergoes changes.

1. 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 ÷ Total no. of alleles x 100

Phase 1: Initial gene pool

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Aa

Aa

AA

aa

No.

Aa

aa

aa

aa

Two pale individuals died. Their alleles are removed from the gene pool.

Phase 2: Natural selection

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 a genotype of aa were eliminated (poor fitness). These individuals (marked by white arrows) are not counted for allele frequencies; they are dead!

Aa

aa

AA

Aa

Aa

Aa

AA

AA

Aa

A

a

AA

Aa

AA

AA

No. No.

AA

aa

Aa

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 behavior. 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.

AA

Aa

aa

a

AA

No. No. %%

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Aa

AA

Aa

Aa

Aa

AA

aa

AA

Aa

aa

AA

Aa

A

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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

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212 Calculating Allele Frequencies in Populations mathematical model of genetic equilibrium in a gene pool, but its main application in population genetics is in calculating allele and genotype frequencies in populations, particularly as a means of studying changes and measuring their rate.

PR E V ON IEW LY

Key Idea: The Hardy-Weinberg equation is a mathematical model used to calculate allele and genotype frequencies in populations. The Hardy-Weinberg equation provides a simple Punnett square

A

AA

A

aA

a

a

Aa

Frequency of allele combination AA in the population is represented as p2

aa

Frequency of allele combination aa in the population is represented as q2

(p + q) = 1+ q2 = 1 (p2 += q)p22 + =2pq p+2q+2 2pq

Frequency of allele combinations Frequency of allele typesFrequency Frequency of allele types of allele combinations

p = Frequency of allele A

p2 = Frequency of AA (homozygous dominant)

p = Frequency of allele A

p2

q = Frequency of allele a

2pq 2= Frequency of Aa (heterozygous)

Frequency of allele combination Aa in the population (add these together to 2pq)

q = Frequency of allele a

= Frequency of AA (homozygous dominant)

2pq = Frequency of Aa (heterozygous)

q = Frequency of aa (homozygous recessive)

q2

= Frequency of aa (homozygous recessive)

The Hardy-Weinberg equation is applied to populations with a simple genetic situation: dominant and recessive alleles controlling a single trait. The frequency of all of the dominant (A) and recessive alleles (a) equals the total genetic complement, and adds up to 1 or 100% of the alleles present (i.e. p + q = 1).

How to solve Hardy-Weinberg problems

Worked example

In most populations, the frequency of two alleles of interest is calculated from the proportion of homozygous recessives (q   2), as this is the only genotype identifiable directly from its phenotype. If only the dominant phenotype is known, q 2 may be calculated (1 – the frequency of the dominant phenotype). The following steps outline the procedure for solving a Hardy-Weinberg problem:

Among white-skinned people in the USA, approximately 70% of people can taste the chemical phenylthiocarbamide (PTC) (the dominant phenotype), while 30% are non-tasters (the recessive phenotype).

1. Examine the question to determine what piece of information you have been given about the population. In most cases, this is the percentage or frequency of the homozygous recessive phenotype q2, or the dominant phenotype p2 + 2pq (see note above).

2. The first objective is to find out the value of p or q, If this is achieved, then every other value in the equation can be determined by simple calculation.

3. Take the square root of q2 to find q.

4. Determine p by subtracting q from 1 (i.e. p = 1 – q).

5. Determine p2 by multiplying p by itself (i.e. p2 = p x p). 6. Determine 2pq by multiplying p times q times 2.

7. Check that your calculations are correct by adding up the values for p2 + q2 + 2pq (the sum should equal 1 or 100%).

(a) Homozygous recessive

Answers

phenotype(q2).

30% - provided

(b) The dominant allele (p). 45.2% (c) Homozygous tasters (p2). 20.5%

(d) Heterozygous tasters (2pq). 49.5%

Data: The frequency of the dominant phenotype (70% tasters) and recessive phenotype (30% non-tasters) are provided. Working: Recessive phenotype: q2 = 30% use 0.30 for calculation therefore: q = 0.5477 square root of 0.30 therefore: p = 0.4523 1–q=p 1 – 0.5477 = 0.4523 Use p and q in the equation (top) to solve any unknown:

Homozygous dominant p2 = 0.2046 (p x p = 0.4523 x 0.4523) Heterozygous: 2pq = 0.4953

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Remember that all calculations must be carried out using proportions, NOT PERCENTAGES!

Determine the frequency of:

1. A population of hamsters has a gene consisting of 90% M alleles (black) and 10% m alleles (gray). Mating is random. Data: Frequency of recessive allele (10% m) and dominant allele (90% M).

Recessive allele:

q

=

Dominant allele:

p

=

Recessive phenotype:

q2

=

Homozygous dominant:

p2

=

Heterozygous:

2pq

=

Determine the proportion of offspring that will be black and the proportion that will be gray (show your working):

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2. You are working with pea plants and found 36 plants out of 400 were dwarf. Data: Frequency of recessive phenotype (36 out of 400 = 9%)

q

=

Dominant allele:

p

=

Recessive phenotype:

q2

=

Homozygous dominant:

p2

=

Heterozygous:

2pq

=

3. In humans, the ability to taste the chemical phenylthiocarbamide (PTC) is inherited as a simple dominant characteristic. Suppose you found out that 360 out of 1000 college students could not taste the chemical. Data: Frequency of recessive phenotype (360 out of 1000).

Recessive allele:

q

=

Dominant allele:

p

=

Recessive phenotype:

q2

=

Homozygous dominant:

p2

=

Heterozygous:

2pq

=

Recessive allele:

q

=

Dominant allele:

p

=

Recessive phenotype:

q2

=

Homozygous dominant:

p2

=

Heterozygous:

2pq

=

Recessive allele:

q

=

Dominant allele:

p

=

Recessive phenotype:

q2

=

Homozygous dominant:

p2

=

Heterozygous:

2pq

=

Recessive allele:

q

=

Dominant allele:

p

=

Recessive phenotype:

q2

=

Homozygous dominant:

p2

=

Heterozygous:

2pq

=

PR E V ON IEW LY

Recessive allele:

(a) Calculate the frequency of the tall gene: (b) Determine the number of heterozygous pea plants:

(a) State the frequency of the gene for tasting PTC:

(b) Determine the number of heterozygous students in this population:

4. A type of deformity appears in 4% of a large herd of cattle. Assume the deformity was caused by a recessive gene. Data: Frequency of recessive phenotype (4% deformity).

(a) Calculate the percentage of the herd that are carriers of the gene:

(b) Determine the frequency of the dominant gene in this case:

Determine the proportion (%) of the population that becomes white:

6. It is known that 64% of a large population exhibit the recessive trait of a characteristic controlled by two alleles (one is dominant over the other). Data: Frequency of recessive phenotype (64%). Determine the following:

(a) The frequency of the recessive allele: (b) The percentage that are heterozygous for this trait:

(c) The percentage that exhibit the dominant trait:

(d) The percentage that are homozygous for the dominant trait:

(e) The percentage that has one or more recessive alleles:

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5. Assume you placed 50 pure bred black guinea pigs (dominant allele) with 50 albino guinea pigs (recessive allele) and allowed the population to attain genetic equilibrium (several generations have passed). Data: Frequency of recessive allele (50%) and dominant allele (50%).

7. Albinism is recessive to normal pigmentation in humans. The frequency of the albino allele was 10% in a population. Data: Frequency of recessive allele (10% albino allele).

Determine the proportion of people that you would expect to be albino:

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213 Analysis of a Squirrel Gene Pool

PR E V ON IEW LY

Key Idea: Allele frequencies for real populations can be calculated using the Hardy-Weinberg equation. Analysis of those allele frequencies can show how the population's gene pool changes over time. In Olney, Illinois, there is a unique population of albino (white) and gray squirrels. Between 1977 and 1990, students at Olney Central College carried out a study of this population. They recorded the frequency of gray and albino squirrels. The albinos displayed a mutant allele expressed as an albino phenotype only in the homozygous recessive condition. The data they collected are provided in the table below. Using the Hardy-Weinberg equation, it was possible to estimate the frequency of the normal 'wild' allele (G) providing gray fur coloring, and the frequency of the mutant albino allele (g) producing white squirrels when homozygous.

Gray squirrel, usual color form

Thanks to Dr. John Stencel, Olney Central College, Olney, Illinois, US, for providing the data for this exercise.

Albino form of gray squirrel

Population of gray and white squirrels in Olney, Illinois (1977-1990) Year

Gray

White

Total

GG

Gg

gg

Freq. of g

Freq. of G

1977

602

182

784

26.85

49.93

23.21

48.18

51.82

1978

511

172

683

24.82

50.00

25.18

50.18

49.82

1979

482

134

616

28.47

49.77

21.75

46.64

53.36

1980

489

133

622

28.90

49.72

21.38

46.24

53.76

1981

536

163

699

26.74

49.94

23.32

48.29

51.71

1982

618

151

769

31.01

49.35

19.64

44.31

55.69

1983

419

141

560

24.82

50.00

25.18

50.18

49.82

1984

378

106

484

28.30

49.79

21.90

46.80

53.20

1985

448

125

573

28.40

49.78

21.82

46.71

53.29

1986

536

155

691

27.71

49.86

22.43

47.36

52.64

1987

No data collected this year

1988

652

122

774

36.36

47.88

15.76

39.70

60.30

1989

552

146

698

29.45

49.64

20.92

45.74

54.26

1990

603

111

714

36.69

47.76

15.55

39.43

60.57

1. Graph population changes: Use the data in the first 3 columns of the table above to plot a line graph. This will show changes in the phenotypes: numbers of gray and white (albino) squirrels, as well as changes in the total population. Plot: gray, white, and total for each year:

1990

1989

1988

1987

1979

1978

1977

0

1986

100

1985

200

1984

300

1983

400

1982

500

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600

1981

(b) Describe the overall trend in total population numbers and any pattern that may exist:

700

1980

800

(a) Determine by how much (as a %) total population numbers have fluctuated over the sampling period:

Number of squ