Se AP BIOLOGY Nichoel D'Alessio a _ GET A HIGHER SCORE IN LESS TIME - Acomplete AP Biology course in a concise, time-saving format it - Targeted review covers only material that will actually be tested - Strategies for answering every type of question - Free online practice exam pinpoints your strengths and weaknesses GREEN “APs oregisteed trademark fhe Coleg Bord wich ws atinaved inthe prodction EDITION | of and dos nat endorse this product,REA: THE TEST PREP AP TEACHERS RECOMMEND AP BIOLOGY CRASH COURSE By Michael B’Alessio Watchung Hills Regional High School Warren, New Jersey Research & Education Association Visit our website at: www.rea.comPlanet Friendly Publishing ‘v Made in the United States | At REA we're committed to producing books in an v Printed on Recycled Paper | earth-friendly manner and to helping our customers Text: 10% Cover: 10% | make greener choices. “eam mors: www-greenesion.r9_| Manufacturing books in the United States ensures compliance with strict environmental laws and eliminates the need for international freight ship- ping, a major contributor to global air pollution, ‘And printing on recycled paper helps minimize our consumption of trees, water and fossil fuels. This book was printed on paper made with 10% post-consumer waste. According to Environmen- tal Defense's Paper Calculator, by using this innovative paper instead of conventional papers, we achieved the following environmental benefits: Trees Saved: 4 * Air Emissions Eliminated: 649 pounds Water Saved: 638 gallons * Solid Waste Eliminated: 192 pounds For more information on our environmental practices, please visit us online at www.rea.com/green Research & Education Association 61 Ethel Road West Piscataway, New Jersey 08854 E-mail: info@rea.com AP BIOLOGY CRASH COURSE Copyright © 2011, 2010 by Research & Education Association, Inc. All rights reserved. No part of this book may be reproduced in any form without permission of the publisher. Printed in the United States of America Library of Congress Control Number 2010940972 ISBN-13: 978-0-7386-0662-0 ISBN-10: 0-7386-0662-6 = REA® is a registered trademark of Research & Education Association, Inc. uoAP BIOLOGY CRASH COURSE Begin Reading Table of Contents Copyright Page Access Your ExamAP BIOLOGY CRASH COURSE TABLE or CONTENTS About This Book. About Our Author. Acknowledgments [Gy INTRODUCTION UE Chapter 1: Keys for Success on the AP Biology Exam, FyVy] MOLECULES AND CELLS: CHEMISTRY OF LIFE, A CELLS, AND CELLULAR ENERGETICS Chapter 2: Water... Chapter 3: Organic Molecules in Organisms... Chapter 4: Free Energy Changes.... Chapter 5: Enzymes... Chapter 6: Prokaryotic and Eukaryotic Cells Chapter 7: Membranes.... Chapter 8: Subcellular Organization ..... Chapter 9: Cell Cycle and Regulation... Chapter 10: Coupled Reactions..... Chapter 11: Fermentation and Cellular Respiration .. Chapter 12: Photosynthesis.AP Crash Course [yy] HEREDITY AND EVOLUTION: fal HEREDITY, MOLECULAR GENETICS AND EVOLUTIONARY BIOLOGY PART Chapter 13: Chapter 14: Chapter 15: Chapter 16: Chapter 17: Chapter 18: Chapter 19: Chapter 20: Chapter 21: Chapter 22: Chapter 23: Meiosis and Gametogenesis ... Eukaryotic Chromosomes..... Inheritance Patterns... RNA and DNA Structure and Functiot Gene Regulation... Mutation... Viral Structure and Replication, Bacterial Gene Regulation... Nucleic Acid Technology and Applications........ 91 Early Evolution of Life ... Evidence for Evolution .. Mechanisms of Evolutio1 ORGANISMS AND POPULATION: DIVERSITY OF ORGANISMS, STRUCTURE AND FUNCTION OF PLANTS AND ANIMALS, ECOLOGY Chapter 24: Evolutionary Patterns... Ww Chapter 25: Survey of the Diversity of Life . 115 Chapter 26: Phylogenetic Classification... 121 Chapter 27: Evolutionary Relationships ... 2125 Chapter 28: Plants and Animals—Reproduction, Growth, and Development... Chapter 29: Plants—Structural, Physiological, and Behavioral Adaptations... Chapter 30: Animals—Structural, Physiological, and Behavioral Adaptations..... p35) Chapter 31: Plants—Response to the Environment............ 147Table of Contents al i Chapter 32: Chapter 33: Chapter 34: Chapter 35: Chapter 36: THE EXAM Chapter 37: Chapter 38: Chapter 39: Biosphere ... ar Population Dynamics. Communities and Ecosystems... Ecology—Behavior . Ecology—Global Issues Major AP Biology Themes and Their Relationship to the Exam The 12 AP Biology Labs . 175 Essay Writing—Exemplars, Data Analysis /Graphing Techniques, Setting Up an Experiment... .. 203About This Book HE asout THis Book REA's AP Biology Crash Course is the first book of its kind for the last- minute studier or any AP student who wants a quick refresher on the course. The Crash Course is based upon a careful analysis of the AP Biology Course Description outline and actual AP test questions. Written by an AP teacher, our easy-to-read format gives students a crash course in Biology. The targeted review chapters prepare stu- dents for the exam by focusing on the important topics tested on the AP Biology exam. Unlike other test preps, REA’s AP Biology Crash Course gives you areview specifically focused on what you really need to study in order to ace the exam. The review chapters offer you a concise way to learn all the important facts, terms, and biological processes before the exam. The introduction discusses the keys for success and shows you strate- gies to help you build your overall point score. Parts Two, Three, and Four are made up of our review chapters. Each chapter presents the essential information you need to know about Biology. Part Five focuses exclusively on the format of the AP Biology exam. Each chapter in this section explains a specific aspect of the test, including Major AP Biology Themes and Their Relationship to the Test, the 12 AP Biology Labs, and Essay Writing (exemplars, data analysis/graphing techniques, setting up an experiment). No matter how or when you prepare for the AP Biology exam, REA’s Crash Course will show you how to study efficiently and strategically, so you can boost your score! To check your test readiness for the AP Biology exam, either before or after studying this Crash Course, take our FREE online practice exam. To access your free practice exam, visit www.rea.com/crashcourse and follow the on-screen instructions. This true-to-format test features automatic scoring, detailed explanations of all answers, and will help you identify your strengths and weaknesses so you'll be ready on exam day! Good luck on your AP Biology exam! vilAP Crash Course ABOUT OUR AUTHOR Michael D’Alessio earned his BS in Biology from Seton Hall Univer- sity, South Orange, New Jersey, and his MS in Biomedical Sciences from the University of Medicine and Dentistry of New Jersey. In 2004, he earned his Executive Master of Arts in Educational Leadership from Seton Hall University. Mr. D‘Alessio has had an extensive career teaching all levels of math- ematics and science, including AP Biology, chemistry, physics, algebra and geometry. Most of Mr. D’Alessio’s AP Biology students received scores of 4 or 5 on the AP Biology exam. Currently, Mr. D’Alessio serves as the Supervisor of the Mathematics and Business Department at Watchung Hills Regional High School in Warren, New Jersey, overseeing a department of 30 teachers. In addi- tion, Mr. D’Alessio participates in numerous SAT workshops around the country, preparing students for the mathematics portion of the test. In 2003, Mr. D’Alessio received the Governor's Teacher of the Year recognition for Watchung Hills Regional High School. In 2004, Mr. D’Alessio received a Certificate of Recognition of Excellence in Sci- ence Teaching from Sigma Xi, the Scientific Research Society of Rut- gers University and in 2005, he was voted National Honor Society Teacher of the Year by the students of Watchung Hills. Mr. D’Alessio would like to thank Larry Krieger for encouraging him to write this AP Biology Crash Course and Dr. Jeffrey Charney for his invaluable help in proofreading and editing the manuscript. ACKNOWLEDGMENTS In addition to our author, we would like to thank Larry B. Kling, Vice President, Editorial, for his overall guidance, which brought this pub- lication to completion; Pam Weston, Publisher, for setting the qual- ity standards for production integrity and managing the publication to completion; Diane Goldschmidt, Senior Editor, for editorial proj- ect management; Alice Leonard, Michael Reynolds, and Kathleen Casey, Senior Editors, for preflight editorial review; Rachel DiMatteo, Graphic Artist, for page design; and Weymouth Design for designing our cover. We also extend our thanks to Rebekah Warner for technically editing the manuscript, Ellen Gong for proofreading, and Kathy Caratozzolo of Caragraphics for typesetting this edition. viiiPART I: INTRODUCTIONKeys for Success on the AP Biology Exam AP Biology textbooks are very thick and contain thousands of bio- logical facts and concepts. if the AP Biology exam contained all of these facts, the challenge to earn a good score on the exam would be daunting. Studying for this exam requires the student to be a pragmatic learner who can delineate the important tested material from the material that is “interesting to know.” This book will help you become more pragmatic in your studying and streamline your chances to score a 4 or 5 on this exam. The keys to success on the exam include knowing the following: 1. The Content of the Advanced Placement Biology Examination The Advanced Placement Biology curriculum is based on the content of an introductory biology course taught at the col- lege level. The topics taught in the class reflect major themes that are presented in a number of college-level textbooks. The curriculum of the course also includes 12 thematic-based laboratories. The underpinnings of these labs are based on common biological tenets that are taught in the course. In order to succeed on the exam, students need to master the basic concepts of biology and apply these concepts to vari- ous situations in a traditional test format. The make-up of the exam is based on the following percentages in three topics of understanding: |. Molecules and Cells, 25% Il. Heredity and Evolution, 25% Ill, Organisms and Populations, 50% 3 JaydeydChapter 1 The AP Biology examination is 3 hours in length, consisting of an 80-minute, 100 multiple-choice-question section and a 90-minute, free-response section with 4 essays. You are given a 10-minute reading period in which you may read the essay questions and outline a format to respond to the questions. Part |. 80 minutes, 100 multiple-choice questions = 60% of the grade The number of multiple-choice questions from each topic is based on the percentages above. For example, expect the fol- lowing: 25 multiple-choice questions pertaining to Molecules and Cells 25 multiple-choice questions pertaining to Heredity and Evolution 50 multiple-choice questions pertaining to Organisms and Populations Part Il. 90 minutes, 4 essay free-response section = 40% of the grade One essay will be from Topic |, Molecules and Cells One essay will be from Topic Il, Heredity and Evolution Two essays will be from Topic Ill, Organisms and Populations One or more of these essays could require you to interpret sci- entific data from one of the 12 required AP laboratories. Each free response is weighted equally, and you must answer them in essay form (outlines are not acceptable). Process- vs. Content-Based Questions Content-based questions are simply designed in order to test your ability to know and recall facts about a particular topic. For example, below is a multiple-choice question on cell res- piration: In cell respiration, glucose is converted to pyruvate in which of the following metabolic pathways? (A) Gluconeogenesis (B) Krebs cycleKeys for Success on the AP Biology Exam (©) Glycolysis (D) Light Dependent Reaction (E) Oxidative Phosphorylation The correct answer is (C). Glycolysis is a ten-step anaerobic pathway that converts glucose (6 carbon sugar) to 2 pyruvate molecules (3 carbon atoms each). The good news is that content-based questions make up a good portion of the AP exam. It is essential to know these ba- sic ideas when confronted by this type of question. This book will give you the content you need to know in a Crash Course manner. Process-based questions are designed for you to apply the content you have learned to a situation. You should not be intimidated by these types of questions, because knowing the content goes a long way in helping you determine the correct answer. The questions tend to be in the laboratory/ experimental-based questioning part of the exam. It is crucial for you to know what should have been interpreted and con- cluded in the 12 AP labs. Below is an example of a process- based question: A biologist prepares an in vitro sample of the activity of the enzyme protease, which promotes the hydrolysis of proteins to amino acids. Three flasks containing 20 milliliters of 5% hemoglobin (a protein) in water are prepared with the addition at time zero of each of the substances indicated in the diagram below. I mL protease Tmtwater mC botted provease / \ i i. \ / \ fi \ / \ / 596 hemoglobin / s%hemagiabin \, / 5% hemoglobin \, \ ¢ s ) FLASK A FLASKB FLASK C 5Chapter 1 In the experiment above, which flask(s) would you ex- pect hemoglobin molecules to maintain a higher-level protein structure? (A) Flask A (B) Flask B (C) Flask C (D) Flasks A and B (E)_ Flasks B and C The correct answer is (E). Water is the control, and boiled pro- tease will have lost its enzymatic activity because of enzymatic denaturation. The content to apply to correctly answer this process-based question is the definition of a controlled experi- ment and physical properties of enzymes. The good news is that process-based questions make up a smaller portion of the AP exam when compared to content- based items. This Crash Course will help you know the con- tent needed for making the process-based questions easier to handle. Expect to see multiple-choice questions in the following formats. Type |. Traditional multiple-choice questions with choices (A) through (E). There is only one correct answer! Which of the following is MOST inclusive when clas- sifying organisms using taxonomy? (A) Domain (B) Kingdom (C) Phylum (D) Class (E) Order The answer is (A). Domain includes the most organisms within a particular group.Keys for Success on the AP Biology Exam Type Il. Heading multiple-choice questions with choices (A) through (E). There is only one correct answer for each ques- tion. However, the answers could be used in a set of questions once, more than once, or not at all. Questions 70-74 (A) Immune System (B) Nervous System (© Digestive System (D) Circulatory System (E) Excretory System 70. Involves blood that is oxygenated 7 Enzyme called amylase is an important part of this system 72. Involved in allergic attacks 73. Basic unit is called the nephron 74. Peristalsis is a key type of action Answers 70. D 71. 72. A 73. E 74. C How Is the Test Scored? Section I. Total scores on the multiple-choice section of the exam are based on the number of questions answered cor- rectly. Points are not deducted for incorrect answers and no points are awarded for unanswered questions. Multiple-Choice Raw Score Number Right x .90 = Weighted Score for Section | 7Chapter 1 Section Il. The free-response section is scored by multiply- ing the number of raw points earned on each essay by 1.50. x 1.50 = (possible points out of 10) Free-Response Question 1 x 1.50 = (possible points out of 10) Free-Response Question 2 x 1.50 = (possible points ‘out of 10) Free-Response Question 3 x 1.50 = Free-Response Question 4 ___ (possible points J out of 10) Combined Free-Response Score = Composite Score = (MC Weighted Score) + (Combined Free-Response Score) Composite Score Range* | AP Grade 95-150 5 81-94 4 69-80 3 58-68 & 0-57 1 * From the 2009 released exam. Composite score ranges change year to year. On the low end you can score 95 out of a possible 150 points (63%) and STILL earn a 5 on the test. To earn a 4 you need 81 points out of a possible 150, or just 54%!!! This Crash Course will help you achieve the highest grade possible on this test.Keys for Success on the AP Biology Exam 4, What Is the Breakdown of AP Biology Grades Across the United States? % of Students Earning Examination Grade of Year 5 4 3 2 1 2006 19.6% | 20.3% | 21.2% | 23.3% | 15.6% 2007 19.3% | 20.3% | 21.2% | 23.2% | 15.9% 2008 | 18.6% | 15.6% | 16.1% | 15.2% | 34.6% (Data obtained from College Board Student Grade Distribution Reports, 2006-2008.) The data above indicates that about one out of five of the stu- dents who take the AP Biology Examination earned a 5, with a staggering 34.6% of students earning a 1 in 2008. In 2008, there were 154,000 test takers for AP Biology, which was an all-time high. This Crash Course is tailored for all of the stu- dents represented in the data and will help you earn a grade that will allow you to receive college credit for biology. 5. Using College Board and REA Materials to Supplement Your Crash Course Your Crash Course contains everything you need to know to score a4 or 5. You should, however, supplement it with materi- als provided by the College Board and other resources. The AP Biology Course Description Booklet, and the 1999, 2002, and 2008 released AP Biology exams, can all be ordered from the College Board’s Online Store (http://store.collegeboard.com). In addition, the College Board’s AP Central site (http://apcentral. collegeboard.com) contains a wealth of materials, including essay questions with exemplars and rubrics. And finally, REA’s AP Biology (7th Edition) contains excellent narrative chapters that supplement this Crash Course material. 9——— PART II: MOLECULES AND CELLS: Chemistry of Life, Cells, and Cellular EnergeticsWater Chemistry of Life—Water A. The Amazing Properties of the Water Molecule 1. Water is a polar molecule; this means it is a molecule whose ends have opposite charges (hydrogen being positive and oxygen being negative). . Hydrogen bonding—the ability of hydrogen to interact with the elements fluorine, oxygen, and nitrogen. This is not a covalent bond. . For every 1 water molecule, 4 more molecules can hydrogen- bond with the original. .. Most abundant molecule in living organisms. .. Water exhibits hydrogen bonding leading to the following properties. i. Cohesion—the ability of water molecules to stick together. Example: the movement of water against gravity on plant surfaces. ii, Adhesion—the ability of water to adhere to molecules other than water. Example: water moving through a hose and adhering to the inside of the hose. Surface Tension—the measure of how difficult it is to break the surface of water. Example: the high surface tension of water allows a water strider (skater) to walk on the surface of a pond with minimal effort. iv. Specific Heat—the amount of energy (heat) it takes to raise or lower the temperature of one gram of a substance 1 degree Celsius. Water has a high specific heat; therefore, large bodies of water do not evaporate easily. 13 JaydeydChapter 2 v. Evaporative Cooling—since water has a high heat of vaporization, once water leaves a surface, the surface cools down. Example: the sweat of an athlete assists in maintaining homeostasis during a vigorous workout. B. Dissociation of Water (or the shifting of hydrogen from one water molecule to another) 1, 2H,O <> H,O* (hydronium ion) + OH" (hydroxide ion) 2. The counterbalance of hydronium and hydroxide results with a pH of 7.0 for water. 3. Acid (biological definition)—a molecule that increases hydronium concentration. Example: HCl > Ht + Cl 4, Base (biological definition)—a molecule that increases hydroxide concentration. Example: NaOH - Na* + OH- 5. pH scale—ranges from 0-14, with 7 being neutral. Each increment of 1 on the pH scale is a ten-fold change. 6. The deleterious effects of acid rain—production of carbon dioxide and sulfur oxides from industrial manufacturing leads to the production of acidic conditions (H,CO, and H,SO,). Acid rain affects ecosystems such as lakes and forests. Molecules and Cells—Inquiry about the amazing properties of water has been expanded in AP Biology essays. Water is a recurring theme found in all aspects of molecules and cells.Organic Molecules in Organisms Chemistry of Life—Organic Molecules in Organisms A. Organic Chemistry—the study of carbon-containing compounds 1. Carbon is the building block of the four major macromolecules: carbohydrates, lipids, proteins, and nucleic acids. . Tetra-valence or the ability to form 4 covalent bonds is a hallmark of the carbon atom. B. Functional Groups—groups of accessory elements that attach to molecules that give them a different structure, and thus a different function 1. Hydroxyl Group—hydrogen bonded to oxygen attached to a carbon skeleton. Molecules with hydroxyl groups are considered alcohols. —oH HYDROXYL . Carbonyl Group—carbon joined to oxygen via a double bond. If the carbonyl group is on the end of a carbon skeleton, it is called an aldehyde; otherwise the compound is called a ketone. 0 0 I I) c c aS x H ALDEHYDE KETONE 15 JaydeydChapter 3 . Carboxyl Group—carbon is doubled bonded to an oxygen and a hydroxyl (carboxyl is a combination of carbonyl and hydroxyl) 0 | ¢ as oH CARBOXYLIC ACID . Amino Group—a nitrogen atom bonded to two hydrogen atoms and one carbon. Amines are organic molecules that contain an amino group. Ni, ) .. Phosphate Group—a phosphate ion covalently attached to a carbon skeleton. Phosphate groups have a tremendous amount of energy associated with them. PHOSPHATE C. Biologically Important Molecules 1 Macromolecules—large molecules that fall into four categories: carbohydrates, lipids, proteins, and nucleic acids (see the molecular genetics chapter). . Polymer—a long molecule consisting of similar molecules held together by covalent bonds. (Poly = many) . Monomers—the repetitive molecules of a polymer. (Mono = one) . The synthesis of polymers takes place through a reaction called dehydration synthesis (condensation reaction). In this reaction water is lost.Organic Molecules in Organisms » = | Ny i Ho— 4 5. The breakdown of polymers takes place through a reaction called hydrolysis. In this reaction water is added across the covalent bond. (Hydro = water, lysis = break) HO. H | 2 Ho HO -—H D. Carbohydrates 1. Definition: sugar and the polymers of sugar. Ending with suffix -ose. 45-65% of dietary allowance. 4 calories per gram. Excess carbohydrates can be converted to fat. 2. Monosaccharide—single sugars with a formula of CH,O. 3. Most important monosaccharide is glucose. C,H,,O,. —CH,0H Hon of 09H A : \ VA on H a Ou OH ‘Alpha (cx) Glucose Beta (B) Glucose Difference is hydroxyl on carbon 1. 17Chapter 3 18 4. Disaccharides—two monosaccharides linked via dehydration synthesis. 5. Glycosidic bond is the covalent bond formed between monosaccharides to form di- and polysaccharides. . Maltose—glucose/glucose disaccharide. . Sucrose—glucose/fructose disaccharide. . Lactose—glucose/galactose disaccharide. . Polysaccharides—macromolecules with 100-1000 of monomer subunits. i. Storage Polysaccharide—starch for plants. Glycogen for animals. ii. Structural Polysaccharide—cellulose for plants (indigestible for humans). Chitin found in arthropods and fungi. COND . Lipids (Fats/Triglycerides, Phospholipids, Steroids) 1. The only class of large molecules that are not considered polymers. 2. Lipids are hydrophobic—fear of water. 3. Used for insulation and buoyancy in marine and Arctic animals. 4. Major component of membranes. 5. fat or triglyceride—glycerol and 3 fatty acids. 9 Calories per gram, thus making it the most energy rich of the biologically important compounds. 10-35% of dietary allowance. ii, Too much fat causes a build-up in the arteries called atherosclerosis. iii, Fatty Acids—long chains of carbon attached to a carboxyl group. iv. Saturated Fatty Acids—no double bonds between carbon atoms. Example: Animal fat—lard or butter (solids at room temperature).Organic Molecules in Organisms v. Unsaturated Fatty Acid—has some double bonds between carbon atoms, Example: Plant and fish fat and many vegetable oils (liquids at room temperature). ° 6. Phospholipids—are lipids but unlike triglycerides only have 2 fatty acids. i, Glycerol + phosphate + 2 fatty acids = a phospholipid. ii, Head region contains the glycerol and phosphate and is hydrophilic (attracted to water). iii, Tail region contains the fatty acids (1 saturated and 1 unsaturated) and is hydrophobic. iv, Phospholipids are the basic units of biological membranes. } Hydrophilic Head ydrophobic Tail 7. Steroids—tipids that consist of 4 fused rings. i. Structure of the precursor lipid cholesterol. is H3C CH cH Ho ii. Many steroid hormones in animals are produced from cholesterol. 19Chapter 3 20 F. Proteins 1 Molecules that perform structural, catalytic, signaling, defense, and transport duties within the cell. Make up about 50% of the dry weight of a cell. 4 calories per gram. 10-35% of dietary allowance. . Proteins are made up of monomer units called amino acids. An amino acid consists of an amino group, a carboxyl group, and up to 20 different side chain groups (20 different amino acids) that give each amino acid unique physical and chemical properties (different structure, thus a different function). R R= side chain . Dipeptide is formed when two amino acids are linked via dehydration synthesis. . Peptide bond is the covalent bond formed between amino acids. . When many amino acids come together via dehydration synthesis, a polypeptide is formed. Polymers of amino acids are called polypeptides, and when one or more polypeptides fold in specific conformations, a protein is formed. . A protein conformation (shape) depends on 4 levels of organization. i, Primary Structure—unique sequence of amino acids. ii. Secondary Structure—the interaction of hydrogen bonds along the peptide chain forming an alpha (a) helix or beta (A) pleated sheet. Tertiary Structure—the interaction between the « helix and f pleated sheet. Disulfide bridges, hydrophobic interaction, hydrogen bonding, and Van der Waals forces allow the structure to form. iv. Quaternary Structure—the interaction of 2 or more polypeptide chains forming a multi-subunit protein. Examples include DNA polymerase, collagen, and hemoglobin.Organic Molecules in Organisms v. Denaturation—an inactive form of a protein (back to primary sequence) can take place with changes to pH, salt concentration, temperature, or exposure to toxic compounds. '. Protein homology—refers to the fact that certain proteins have similar structures that are found in more than one species that share a common ancestor. These proteins share a high degree of amino acid correspondence. An example is the cytochrome c protein found in the electron transport chain. The data below indicates humans and rhesus monkeys are closer on the evolutionary chain than humans and tuna. Number of Amino Acid Differences Species When Compared to Humans Human 0 Rhesus monkey 1 Kangaroo 10 Snapping turtle 15 Bullfrog 18 Tuna 21 Molecules and Cells—Biologically important molecules are great examples of the “structure and function” theme. Be sure to use these examples in essay questions to gain critical points. 21Free Energy Changes Chemistry of Life—Free Energy Changes A. Metabolism—the totality of all chemical reactions that occur within an organism. B. Anabolic Reactions—sets of reactions that consume energy to make molecules. For example, the synthesis of DNA or protein is anabolic. C. Catabolic Reactions—sets of reactions that produce energy in the breakdown of molecules. For example, the breakdown of glucose into carbon dioxide and water with the production of ATP. D. Energy—capacity to do work. E. 1* Jaw of thermodynamics—energy can neither be created nor destroyed but can change from one form to another. Example: Plants convert light energy from the sun to make chemical energy in the form of glucose. F. 2 law of thermodynamics—every energy transfer increases the entropy of the universe (disorder). G. Free Energy—the total amount of energy in a system (a cell) that can be tapped to do work. Not all energy transfers are 100%. H. Exergonic Reaction—a reaction that produces a net release of free energy. AB > A+B + Energy 23 JaydeydChapter 4 24 Endergonic Reaction—a reaction that absorbs free energy from its surroundings. A+B 4+ Energy > AB Energy Coupling—using the products of an exergonic reaction to help run an endergonic reaction. The molecule that is essential for coupling of reactions and cellular work is adenosine triphosphate (ATP). Adenine Phosphate Why Groups i Arh a eee NH OH OH Ribose When the terminal (last) bond of phosphate of ATP is broken (via hydrolysis), energy is released (exergonic). Adenosine diphosphate (ADP) is formed. Hydrolysis of ATP Po PH PonFree Energy Changes The ATP cycle is the process in which ATP is hydrolyzed and ADP is phosphorylated (addition of a phosphate group) to reform ATP. AP Phosphorylation Hydrolysis ADP + Pi SZ Molecules and Cells—ATP is used in a plethora of biological situations including active transport, muscle movement, DNA replication, and cell respiration/photosynthesis. 25Enzymes Chemistry of Life—Enzymes A. 99.9% of enzymes are proteins in structure. 0.1% are ribozymes made of RNA (found in rRNA). B. Enzymes are catalysts or biological chemical agents that speed up a reaction without being consumed or produced by the reaction. Most enzymes end with the suffix -ase. C. An enzyme works by lowering the activation energy (E,) of a reaction. Activation energy is defined as the amount of input energy needed to start a chemical reaction. ENERGY PROFILE FOR ENZYME TRANSITION STATE __— ACTIVATION ENERGY E, > ENZYME LOWERS E, FREE ENERGY ——> \ Propucts PROGRESS OF THE REACTION ——> D. The enzyme combines with the substrate or molecule that the enzyme will work on. When the enzyme and substrate 27 JaydeydChapter 5 28 are joined, a catalytic reaction takes place, forming a product. The enzyme can be recycled and used for later reactions. . Active site of the enzyme is the region that the substrate binds. -. The catalytic cycle of an enzyme: Enzyme + Substrate <> Enzyme — Substrate Complex <> Enzyme + Product Induced Fit Model of Catalysis—the enzyme's active site and substrate bind to each other, “inducing” a very slight change in the structure of the enzyme, which promotes the reaction. . Environmental factors that disrupt the activity of enzymes include: 1. Each enzyme has an optimal temperature. If the temperature is too high, the enzyme will denature (3-D shape will be lost). If the temperature is too low, the enzyme will slow down. 2. Each enzyme has an optimal pH. If the pH is too high or too low, the enzyme will denature. 3. Cofactor—non-organic molecules that augment enzyme activity. Examples: iron, magnesium, and copper. 4. Coenzyme—organic molecule that augments enzyme activity. Example: vitamins |. Allosteric Site and Regulation—a portion of the enzyme in which a regulator will bind. This is not the same as the active site. The effect on the enzyme may be stimulatory or inhibitory. Feedback Inhibition—when the end product of a biological pathway inhibits the activity of an enzyme involved in that pathway. Cooperativity—different from allosteric regulation because the enzyme contains multiple subunits. When a regulator binds to one subunit, the other subunits “follow course” and a conformational change in the enzyme’s shape takes place. This has the effect of amplifying the activity of the enzyme.Enzymes IMPORTANT ENZYMES TO KNOW Enzyme Lipase Protease DNAse RNAse Kinase Carboxylase Decarboxylase Dehydrogenase Substrate Lipids Proteins DNA RNA Any molecule Any molecule Any molecule Any molecule Action Breaks down lipids Breaks down protein Breaks down DNA Breaks down RNA Transfers phosphate groups Adds carbon dioxide to create an additional -COOH group Removes carbon dioxide Adds electrons and hydrogen Molecules and Cells—Knowing a list of enzymes and their substrates is helpful in AP essays. Correctly naming an enzyme and its substrate will translate to points on your essay. 29Prokaryotic and Eukaryotic Cells Cells—Prokaryotic and Eukaryotic Cells A. Cytology—“cyto” meaning cell, “logy” meaning study of—is the branch of biology that studies the structure and function of cells. Microscopes are the instruments that allow the study of the structure of a cell. He w Light Microscope—based on light passing through a two-lens system. Ocular lenses = 10x magnification, objectives lenses = 4x, 10x, 40x, or 100x magnification. Total magnification = ocular x objective. Can view prokaryotic and eukaryotic cells from the light microscope, but not a virus. .. Magnification—ability to enlarge the image. . Resolution—ability to provide clarity to the image. . Electron Microscope—uses a beam of electrons to provide greater magnification and resolution. i. Transmission Electron Microscope—used to view the internal components of cells (organelles). ii, Scanning Electron Microscope—used to view the surface of cells. . Cell Fractionation is the ability to use centrifuges (ultracentrifuges) to spin at different high speeds to “fractionate” (split apart) cells. i. Homogenate (tear apart)—2 layers (pellet and supernatant) — supernatant contains mitochondria and/or chloroplasts — supernatant contains cell membrane and internal membranes — supernatant contains ribosomes. B. Prokaryotic CelH—unicellular bacteria. ile 2. Have no nucleus!!! Generally considered the first form of life and were most likely anaerobic. 31 JaydeydChapter 6 32 3. Nucleoid region—“devoid” of a true nucleus, but had DNA floating in the cytoplasm. No nuclear membrane. 4. Cytoplasm—region between the nucleus and the plasma membrane. 5. Cytosol—semi-fluid within the cell. 6. Generally 1 to 10 micrometers in size. 7. Have a cell membrane—regulates transport and is very selective in permeability. 8. Have a cell wall—protective layer external to the cell membrane. The cell wall does not contain phospholipids or transmembrane proteins. 9. Contains ribosomes—location of protein synthesis. 10. Capsule—ties outside of the cell wall, made of carbohydrate, and is pathogenic (disease causing). C. Eukaryotic Cell—animal cell or plant cell; members are all multicellular. There are unicellular eukaryotic cells (fungi). 1. Does contain a nucleus. 2. Nuclear envelope—double membrane made of phospholipids and proteins. Maximum protection of the DNA in the nucleus. 3. Nuclear pores—holes in the envelope that allow transport into and out of the nucleus. 4. Nuclear lamina—set of proteins that give the nucleus structure. 5. Chromatin—DNA and proteins combine within the nucleus. Coiled DNA around proteins allows maximum occupancy of the nucleus by DNA. 6. Chromosomes—coiled chromatin that contains genes. 7. Nucleolus—inside the nucleus and the location of where ribosomes are synthesized. Stains dark using cell-staining procedures. 8. Ribosomes—found outside of the nucleus and are the location of the synthesis of proteins. (Common between both prokaryotes and eukaryotes.)Prokaryotic and Eukaryotic Cells Organelles Found in Animal Cells but Not in Plant Cells Centrioles Lysosomes Cilia Flagella Intermediate Filaments Organelles Found in Plant Cells but Not in Animal Cells Cell Wall Chloroplast Central Vacuole Plasmodesmata Test & Tip 6CO, + 6H,O + ATP . Both involve oxidation-reduction reactions: i, Loss of electrons is oxidation (glucose to carbon dioxide). Gain of electrons is reduction (oxygen to water). ili. Electrons = Energy . In cell respiration there are two “electron and hydrogen carrier molecules” that help carry the energy from glucose to the mitochondria where it will be harnessed —NADH (Nicotinamide Adenine Dinucleotide) and FADH, (Flavin Adenine Dinucleotide). i. NAD* = oxidized form, NADH = reduced form ji. FAD = oxidized form, FADH, = reduced form B. The Four Parts of Cellular Respiration 1. Glycolysis— “splitting of sugar” i, Takes place in the cytoplasm and is an anerobic process. Oldest metabolic pathway since it is found in prokaryotes, which were the first organisms on earth. iii, The 6 carbon sugar, glucose, is broken down into two 3-carbon molecules called pyruvate. This is just a re- arrangement of the carbons from glucose. iv. Ten-step metabolic reaction that initially invests ATP that will yield a “small” energy payoff. |49 JaydeydChapter 11 50 v. NAD* is reduced to NADH. vi. End products of glycolysis are 2 pyruvate molecules, 2 NADH molecules, and 2 ATP via substrate level phosphorylation (2 of each molecule because of the splitting of the glucose molecule in step 4). Substrate level phosphorylation—formation of ATP via a direct transfer of phosphate from a donor molecule to ADP. viii, There is no loss of CO, (decarboxylation) in glycolysis. vii. . Shuttle Step—“process of converting pyruvate to Acetyl CoA.” i. Takes place in the mitochondria. ii, Pyruvate is decarboxylated to form acetate a 2-carbon compound. Acetate then has a coenzyme A group attached to it to form Acetyl Coenyzme A (Acetyl-CoA). NADH is made as a by-product via a reduction reaction. Pyrvuate (3 carbons) 225, acetyl-CoA (2 carbons) + 2C0 +NADH . Krebs Cycle—“generation of NADH and FADH,,” i. Takes place in the mitochondrial matrix. ii, Eight-step metabolic reaction that produces the majority of NADH, FADH,, and CO, (waste product) for cellular respiration. Key intermediate to know: Oxaloacetate (OAA); the 2-carbon fragment from Acetyl Co-A is added to OAA to make citrate. Electron Transport Chain Coupled to Oxidative Phosphorylation i. The inner mitochondrial membrane or cristae is the site where Electron Transport Chain (ETC) proteins are found. i. Proteins in the ETC accept electrons from the electron carriers NADH and FADH,. The cytochrome complex is a series of proteins in the ETC (evolutionary, conserved through time). Electrons get passed down the ETC via oxidation- reduction reactions until they meet the final electron acceptor molecule, oxygen (O,), to form HO. iv, In the ETC, no direct ATP is made; it must be coupled to Oxidative Phosphorylation via chemiosmosis (or the diffusion of H* ions across the membrane). Electron Transport Chain A255, oy dative PhosphorylationFermentation and Cellular Respiration v. As NADH and FADH, are oxidized, H* inside the mitochondrial matrix is transported to the intermembrane space. A proton motive force (or storing of energy) is created across the membrane due to this transport. Once the H* is in the intermembrane space, it “leaks” back through the membrane by using an enzyme called ATP synthase and ATP is produced. vi. Electrons that passed via NADH produce 3 ATP, while electrons from FADH, produce 2 ATP. Cellular Respiration Scorecard Direct ATP from Substrate-Level NADH FADH, Step Phosphorylation | Produced | Produced Glycolysis 2 2 0 Shuttle Step 0 af 0 Krebs Cycle 2 6 2 *Could be FADH, depending on the type of shuttle: 1 NADH = 3 ATP; 1 FAHD, = 2 ATP ante roe “ONADH x 3ATP = 30ATP —> seme < 2FADH2 X 2ATP = 4ATP THAN C. Fermentation le Lactic Acid Fermentation—" anaerobic process in which only glycolysis takes place.” i. Takes place in prokaryotes and in humans. ii, Pyruvate is converted to lactic acid producing only 2 ATP and 2 NADH. iii, Prokaryotes can only use fermentation (glycolysis) because they have no mitochondria that are required for the other parts of cellular respiration. iv. In humans this anaerobic process (lack of oxygen to the muscles cells) causes the phenomenon of “the burn” in muscles cells. Vigorous exercise can cause lactic acid build-up and therefore cause muscle pain and fatigue. |51Chapter 11 cps Giuase 2 2Pywate +2a1? 2NAD* 2NADH 2Lactate <—— 2 Pyrwate 2. Alcoholic Fermentation—“ anaerobic process in which only glycolysis takes place.” i. Takes place in fungi, such as yeast, and is used in the beer and wine industry. . Pyruvate is converted to acetylaldehyde and then ethanol, producing only 2 ATP, 2NADH, and 2 CO,. Fungi have mitochondria, but in anaerobic conditions carry out this process. Glucose 2", > pyruvate +241° 2NAD* 2NADH 2€thanol <———— 2 Acetyladehyde \ = Te St esa Molecules and Cells—Don't stress out to learn the names of Tip/ all the enzymes involved in cellular respiration. AP questions on this topic are designed for you to understand the “big picture” of cell respiration. 52Photosynthesis Cellular Energetics—Photosynthesis A. Autotrophs/Heterotrophs 1. 4. Used by autotrophs or “self feeders”—organisms that create their own organic molecules or food. Autotrophs are also known as producers. . Plants are photoautotrophs—organisms that create their own organic molecules or food by using light energy. . Heterotrophs—organisms that cannot create their own organic molecules or food. Heterotrophs are consumers. Photosynthesis takes place in the chloroplast of the plant. B. The two parts of photosynthesis 1. Light Dependent Reaction—“process in which NADPH, ATP, and O, are produced.” i. Accessory molecules to know in the light-dependent reaction: * Chlorophyll a—absorbs red light (680 nm) and purple light (400 nm). * Chlorophyll b—absorbs yellow light (650 nm) and blue light (450 nm). * Carotenoids—absorbs at various wavelengths and protects the chlorophyll molecules. |. The accessory molecules absorb photons of light and become “excited” when their electrons gain energy. Photosystems—light-gathering centers that contain chlorophyll and accessory molecules. © Photosystem II—P680—named such because light is absorbed best at 680 nm. |53 JaydeydChapter 12 * Photosystem I—P700—named such because light is absorbed best at 700 nm. iv. Route of electrons though non-cyclic electron flow of the light dependent reaction. Photosystem IP 680 absorbs light and elections ae excited. ¥ Electrons in P680 are now “boosted toa higher level and must be replaced. Water is split via light (photolysis) and the electrons are replenished: [MOLECULAR oxygen is produced. Y Tl Electrons pass down an electron transport chain similar tothe one found in cellular respiration. Chemiosmasis takes place and ATP is produced. Electrons are passed to Photosystem |-P700 where they are again boosted toa higher level. Electrons are then passed down a second election transport chain that produces NADPH. FINAL PRODUCTS OF LIGHT REACTION 0, via photolysis AP via ETC NADPH via ETC v. Route of electrons through cyclic electron flow of the light- dependent reaction.Photosynthesis * This reaction takes place by electrons cycling back from P700 through the first ETC. This has the net effect of producing ATP in greater amounts. No NADPH or O, is produced. This reaction takes place because the Calvin Cycle uses more ATP per mole than NADPH per mole, and hence replenishes the used ATP. 7 2. Dark Reaction/Calvin Cycle/Light Independent Reaction— “process in which NADPH and ATP are used to make organic compounds such as glucose.” i. Uses the products of the light reaction (ATP and NADPH) to produce glucose. ii. Takes place in the stroma of the chloroplast. Carbon Fixation Reduction 0, isfxed to ribulose ‘ATP and NADPH are biphosphate(RuB?) used to produce via tubisco to form alycetaldehyde 3-phosphoglycerate. 3-phosphate (63) Regeneration uBPis remade 5556 Chapter 12 C. Comparison Chart Process Breakdown of glucose synthesis of glucose O, is released O, is consumed Chemiosmosis co, is released CO, is consumed/ fixed ATP is produced ATP is consumed Pyruvate as intermediate NADH produced NADPH produced Takes Place in Cellular Respiration Yes No No Yes—ETC and Oxidative Phosphorylation Yes—ETC Yes—Shuttle Step and Krebs Cycle No Yes—Glycolysis, Krebs Cycle, ETC and Oxidative Phosphorylation Yes—Glycolysis initial investment Yes—Glycolysis Yes—Glycolysis, shuttle step, Krebs cycle No Takes Place in Photosynthesis No Yes—Calvin Cycle Yes—light- dependent reaction No Yes—ETC No Yes—Calvin Cycle Yes—light- dependent reaction Yes—Calvin Cycle No No Yes—light- dependent reactionPhotosynthesis D, Alternatives to Photosynthesis 1. Photorespiration i. i ra C, plants make up roughly 90% of the plants on earth and fix carbon into a 3-phosphoglycerate (3 carbon compound). |. Normal photosynthetic processes take place within C, plants. i. When the day is hot and/or dry, stomata are closed; therefore, less carbon dioxide enters the cell for the Dark Reaction. Rubisco will fix oxygen rather than carbon dioxide through a process called photorespiration. Photorespiration will produce no ATP and the Dark Reaction will not fix carbon, thus decreasing glucose output. Plants Plants that prefer an alternative mode of carbon fixation that forms a four-carbon compound first rather than a three-carbon compound. . Leaf anatomy is the main reason why C, plants have alternative carbon fixation. Bundle-sheath cells are found around the vein of the leaf, and the mesophyll cells are found between the bundle- sheath cells and the leaf surface. @ Fixed to 4 carbon Mesophyll Decarboxylation and Calvin Cycle Bundle-sheath 57Chapter 12 58 iv. C, plants minimize photorespiration because the enzyme Rubisco is not as available to fix oxygen. A secondary enzyme called PEP carboxylase fixes the carbon dioxide to a four-carbon compound and has no affinity for oxygen. C, plants fix carbon dioxide more efficiently than C, plants. . Crassulacean Acid Metabolism (CAM) Plants i. Based on arid conditions and plants that are succulent (water-storing). ii, At night, CAM plants open their stomata and during the day close their stomata. iii, Closing the stomata during the day halts the drying out of the plant, but also prevents carbon dioxide from entering. iv. Opening the stomata during the night allows the plant to take in carbon dioxide where it will be stored in the form of organic acids. Molecules and Cells—Some similarities between cellular respiration and photosynthesis are always tested on the AP Biology exam. Included are the following: ATP production, electron transport use, compartmentalization between chloroplast and mitochondria, hydrogen and electron acceptor molecules, such as NADH, FADH,, and NADPH.———PART III: HEREDITY AND EVOLUTION: Heredity, Molecular Genetics, and Evolutionary Biology2nediploid ell Meiosis and Gametogenesis Heredity—Meiosis and Gametogenesis A. Genetic Terms 1, Genes—heredity units made up of DNA. 2. Locus—a gene’s specific location on the chromosome. 3. Asexual Reproduction—a form of reproduction not requiring meiosis or fertilization. Only passes a copy of genes to its progeny. A type of reproduction in which there is no variation in genetic make-up. Bacteria reproduce via asexual reproduction. 4. Clone—an individual that arises from asexual reproduction. 5. Sexual Reproduction—a type of reproduction that involves variation because two parents give rise to their progeny. Major evolutionary advantage because of genetic variation. 6. Somatic Cell—a body cell that is diploid. 7. Gamete—a sex cell such as egg or sperm. Life cles of Animats and Plants n-haploid cell sperm oregg Fertilization JaydeydChapter 13 62 B. Meiosis—Key Points 1 2. Meiosis, like mitosis, is preceded by replication of chromosomes. Unlike mitosis, meiosis has two consecutive cell divisions called Meiosis | and II. . Division results in four daughter cells that are genetically different and are haploid. . Tetrads are formed or the pairing of homologous chromosomes via synapsis. Chiasmata or the site of crossing over/exchange of genetic material is formed in Prophase | of Meiosis |. ._ In Metaphase | of Meiosis | homologous chromosomes pair with each other at the metaphase plate. . Homologous chromosomes separate in Anaphase | of Meiosis | and sister chromatids stay together. . Meiosis Il has no further chromosomal replications and looks just like Mitosis. . Chromosomes will independently assort from each other producing variations. . Crossing over in Prophase | is another mechanism for variation from meiosis.Meiosis and Gametogenesis C. Comparison of Mitosis and Meiosis Event DNA Replication Homologous Chromosomes Sister Chromatid Separation Divisions Cells Produced Crossing Over Mitosis Occurs during interphase Align one after another on metaphase plate Anaphase 1 2 Diploid— genetically identical Does not occur Meiosis Occurs during interphase Pair with each other during metaphase |. Align one after another on metaphase plate during metaphase II Meiosis II Anaphase II 2 4 Haploid—genetically different Meiosis | Prophase | Molecules and Cells—Every AP Biology test-taker should know the main differences between meiosis and mitosis. Mitosis produces diploid identical cells that have no genetic variation. Meiosis produces gametes (haploid) that are genetically different because of crossing over in Prophase | of meiosis. Similarly, know the stages of mitosis and meiosis and special structures that are formed. 63Eukaryotic Chromosomes Heredity—Eukaryotic Chromosomes A. Structure and Function of Eukaryotic Chromosome Part Genes Chromatids Centromere Chromatin Kinetochore Nucleosomes Telomeres Structure Made up of the nucleic acid DNA Two replicated chromosomes that are held together at the centromere DNA region found near the middle (not always) chromosome DNA and protein combination Proteins Histone proteins and DNA Ends of DNA Remember: Eukaryotic DNA is linear, meaning it has definite ends. Most eukaryotic organisms are diploid. Fungi, such as yeast, can exist as haploid or diploid. Function Will be transcribed onto mRNA Will be translated for proteins Allows proper segregation of chromosome during meiosis and mitosis Hold chromatids together to form a chromosome. Aids in packaging DNA, DNA replication, and expression of proteins Allows for the attachment of the mitotic spindle to the centromere Aids in packaging of DNA Protection against the destruction of the DNA from nucleases 65 JaydeydChapter 14 66 B. Advantages to Chromosomes 1. Genetic variation—crossing over, independent assortment. 2. Gene regulation 3. Allows for a large number of genes to be expressed. 4. . Diploid—allows one to have two copies of genes as a backup mechanism. 5. Linkage of genes to be inherited together. . Structure and Function of Prokaryotic Chromosome as a Comparison 1. Circular in shape and much smaller than eukaryotic chromosome. 2. Genes are arranged in operons—one promoter controlling many genes. . Transcription and translation are coupled processes. . Plasmids are prevalent—extra chromosomal pieces of DNA that carry antibiotic resistance. They are not part of the chromosome. Autonomously replicating. 5. One origin of replication. 6. No histone proteins to condense, but DNA is supercoiled. RwInheritance Patterns Heredity—Inheritance Patterns A. Inheritance Terms 1. 2. ONAWSD Characteristic—a heritable feature such as hair color (phenotype). Trait—a variant of a characteristic. Example: red or blond hair color. . Allele—alternative form of a gene, such as tall (7) plants are dominant to short plants (t). . Dominant allele—the allele that is fully expressed. . Recessive allele—the allele that is not expressed. . Genotype—the genetic makeup of an organism. . Phenotype—organism’s appearance. . Gregor Mendel—father of genetics who worked on genetic crosses of pea plants. . Law of Segregation—separation of alleles into gametes. B. Genetic Crosses 1. Monohybrid cross—a cross that tracks the inheritance pattern of a single character. Example: In pea plants, tall (7) plants are dominant to short plants (t). There are three allelic combinations: » TT—homozygous dominant (considered true breeding)/Tall » Tt—heterozygous or hybrid/Tall » tt—homozygous recessive (considered true breeding)/Short 67 JaydeydChapter 15 Cross two true breeding plants of tall and short P (parental) Punnett Square Generation-true TT x tt breeding parents Tt Tt tall and short ee Tt Tt F; (fist) Tex Tt Generation- Punnett Square hybrids > IT It F, (second filial) TT, Tt, Tt, tt Tt tt Generation- — |-_——p> 3:1 ti ee 3:1 tall to short phenotypically 2. Test cross—A cross that determines whether the dominant parent is homozygous dominant or heterozygous. Always cross the dominant parent to a homozygous recessive. Assume black (B) is dominant to white (b) for cat coat color. x Black parent could be BB or Bb. White parent is bb. > If BB x bb, all progeny will be black carriers. > If Bb x bb, 2 of the progeny are black and ¥% are white. ~ 68Inheritance Patterns 3. . Law of Independent Assortment—is observed with dihybrid crosses or crosses between two different characters. Alleles assort independently from each other; therefore, dominant can combine with recessive. Example: In pea plants, tall (T) plants are dominant to short plants (t). Green leaf (G) is dominant to yellow leaf (g). Cross two true breeding plants of tall green and short yellow P (parental) TIGG x ttgg Generation-true Gametes Produced breeding parents talland short a TG and tg Punnett Square F; (fist fila) Gaepnin a Tt6g TTGG | TTGg | TTGg | TTgg 166 | Ttgg | Tt6G | TtGg TtGg | TtGg | TtGg | Ttgg F, (second filial) Generation. =) L_____ ttGG | ttGg | ttGg | ttgg 9:3:3:1 ratio 9 tall, green: 3 tall, yellow 3 short, green: 1 short, yellow C. Using the Laws of Probability in Genetics le The law of multiplication—independent events are governed by this; therefore, for genes that are linked the law of multiplication cannot be followed. i, Assume the following cross: AaBbCc x AabbCC. What are the chances of the following progeny? a) AabbCC b) aabbCc ©) AAbbCC Answer: Perform each individual monohybrid cross and use the law of multiplication. Aa x Aa = 1/2 Aa, 1/4 aa, 1/4 AA Bb x bb = 1/2 Bb, 1/2 bb Cc x CC = 1/2 Cc, 1/2 CC 69Chapter 15 a) AabbCC = 1/2 x 1/2 x 1/2 b) aabbCc = 1/4 x 1/2 x 1/2 c) AAbbCC = 1/4 x 1/2 x 1/2 = 1/16 i, Assume the following genotype: AaBBCcddEeFf. How many different gametes are possible? Answer: Determine how many different gametes are possible for each set of alleles. Aa = 2 (either A or a) (only B) (either C orc) | 2x1x2x1x2x2= (only d) 16 different gametes (either E or e) Ff =2 (either F or f) D. Non-Mendelian Genetics—genetics that do not follow the inheritance patterns of Mendel’s initial pea plant experiments. 1. Incomplete dominance—the phenotype of the offspring has an appearance that is between that of both parents. This is not a blending hypothesis. The dominant allele is not fully expressed. Snapdragons P (parental) it Punnett Square (era ie | Red (CC) x White (cc) al breeding parents Cc Cc red and white = ie Cc F, (first filial) Geneation- tybrdsal pink a naad ‘unnett square i» cc Cc Cc cc F, (second filial) Generation- —- |__________ 12:1 ratio 1:2:1 red to pink to white 70Inheritance Patterns 2. Codominance—both alleles are expressed at the same time. > MN Blood system (M and N are blood group antigens found on the cell surface of a red blood cell). » There are three allelic combinations: * MM—homozygous dominant (only produce M antigen on cell surface). * MN—heterozygous (produce M and N antigens on cell surface). * NN—homozygous recessive (only produce N antigen on cell surface). 3. Multiple Alleles—many different alleles can control the expression of a character. i. ABO Blood System—carbohydrate antigens found on the cell surface. Antigen on Cell Surface | Antibodies Geno- of Red Present in type Phenotype Blood Cell Blood ii O blood type None Anti A, Anti B For | Ablood type B Anti A an Por i) B blood type A Anti B te AB blood type AB None Example ABO Cross: Assume that a child has type B blood and the father was type A. What are the possible genotypes of the mother? Answer: Child could be /* /’ or /* i and the father had to be i, The mother could be either /* Por /° i or * F, If the father was 4 no matter what genotype the mother is, a type B child could not be produced. 4. Pleiotropy—one gene causes multiple different phenotypic effects on an organism. Human Disease PKU (phenylketouria) Phenotypes Mental Retardation Hair Loss Skin Pigmentation 71Chapter 15 72 5. Epistasis—one gene affecting the expression of another gene. F, offspring phenotypic ratio is usually 9:3:4. Polygenic inheritance—two or more genes affecting one phenotype. Examples include skin color and cancer, which is the most common polygenic inherited disorder. Polygenic inheritance leads to a bell curve distribution of phenotypes. E. Pedigree Analysis—a visual depiction of inheritance patterns in multiple family generations. ne Basic Rules i. If two affected people have an unaffected child, it must be a dominant pedigree: [A] is the dominant mutant allele and [a] is the recessive allele. Both parents are Aa (hybrid carriers) and the unaffected child is aa. ii. If two unaffected people have an affected child, it is a recessive pedigree: [A] is the dominant allele and [a] is the recessive allele. Both parents are Aa (hybrid carriers) and the affected child is aa. iii, If every affected person has an affected parent, it is a dominant pedigree (no skipping of generations). iv. Dominant traits never skip generations, while recessive traits can skip. v. Squares are male, vi. Circles are females. O vii. Mating is indicated by the connection with a line. — viii, Filled-in circles or squares indicate affected person. ix. Sex-linked dominant—all females descending from the affected males have the disease. x. Sex-linked recessive—no male carriers possible and skips generations. xi. Autosomal recessive—carriers are present, so skips generations. 50% males and females affected.Inheritance Patterns xii, Autosomal dominant—no carriers or skipping of generations. 50% males and females affected. F. Autosomal Genetic Disorders Recessive Inherited Disorder—absence or malfunction of protein Must receive both non-functional copies from parents; therefore, affected individual is homozygous recessive (aa). Disease Albinism Cystic Fibrosis Tay-Sachs Sickle cell disease Outcome Lack of pigment in the skin, eyes, and hair. May lead to skin cancers. Defective or absent chloride channel protein in membranes, causing a build- up of mucus in lungs. Person is prone to bacterial infections. Defective or absent lipase enzyme in brain. Predominant in Jewish population. Defective hemoglobin protein. Mostly affects the African-American population. Dominant Inherited Disorder—absence or malfunction of protein Must receive at least one non-functional copy from one parent; therefore, affected individual is heterozygous (Aa) or homozygous dominant (AA). Disease Achondroplasia Huntington’s Disease Outcome Dwarfism Degenerative breakdown of the nervous system. G. Linked Genes 1. Thomas Hunt Morgan performed genetic crosses with the fruit fly (Drosophila melanogaster) 73Chapter 15 2. Terms used in fruit fly crosses. i. Wild Type—most common phenotype in the population. ii. Mutants—alternative phenotypes to the wild type. 3. Morgan performed the following dihybrid mating: > Example: In fruit flies, gray (g*) body color is dominant to black body color (g). Normal wings (w*) are dominant to dumpy wings (w). » Cross a double heterozygote to a double recessive (g*g w'w x ggww). » Expected phenotypes of 1000 offspring would be: 250 wild type (gray normal)/parental phenotype 250 black dumpy/parental phenotype 250 gray dumpy/recombinant phenotype 250 black normal/recombinant phenotype » Observed phenotypes of 1000 offspring were: 450 wild type (gray normal) /parental phenotype 450 black dumpy/parental phenotype 50 gray dumpy/recombinant phenotype 50 black normal/recombinant phenotype The high number of observed parental phenotypes indicated that the genes for body color and wings were linked to each other. Linked genes are on the same chromosome and are very close to each other. Linked genes are inherited together and recombination between the genes is very low. » Calculation of Recombination Frequency or the measure of genetic linkage between 2 genes (also called map units). Recombination Frequency = eotirecomblnants x 100 total offspring Using the data above: Recombination Frequency = a x 100 = 10% 1000 Only 10% of the time will there be recombination between the genes for body type and wings. 4. Genetic Maps i, Recombination Frequency allows you to create genetic maps that estimate the distance between genes. 74Inheritance Patterns Assume the following Recombination Frequencies. Determine the genetic map for genes W,X,Y, Z W-Y, 7 map units W-X, 26 map units W-Z, 24 map units Y-X, 19 map units Y-Z, 31 map units Answer: X --- 5. Sex-linked genes Genes that are carried on the X-chromosome. Females carry two X chromosomes, XX. iii, Males carry 1X and 1Y, XY. iv. Inheritance patterns of sex-linked genes: 1. A father will always transmit the sex-linked trait to his daughter. His son receives the Y, and does not inherit the trait. 2. Only females can be carriers of sex-linked traits. Therefore, a carrier female who mates with a normal male transmits the mutant allele to half her sons and half her daughters. 3. Examples of sex-linked traits include hemophilia and muscular dystrophy. Barr body—one of the female’s X chromosomes is randomly inactivated in order to have the same gene dosage as males for sex chromosomes. The chromosome tends to look smaller in physical structure. Example of the phenotypic output of X-inactivation are Calico-colored cats. 6. Chromosomal Mutations Non-disjunction of chromosomes 1. If non-disjunction takes place in Meiosis |, all 4 cells will be aneuploid, or have abnormal chromosome numbers. 2 cells will be monosomic (n-1) and 2 cells will be trisomic (n+1). 2. If non-disjunction takes place in Meiosis II, 2 cells will be normal (n), 1 cell (n-1) monosomic, and 1 cell trisomic (n+1). 75Chapter 15 76 vi. vii. viii. Polyploidy—having more than two chromosome sets such as triploid (3N). Deletion—entire fragment of chromosome is lost. Duplication—entire fragment of chromosome is duplicated on the chromosome. Inversion—part of the chromosome reverses its orientation. Translocation—one part of a chromosome is attached to another part of a different chromosome. Down Syndrome or Trisomy 21—presence of all or part of an extra chromosome 21. XXY—Klinefelter syndrome. Phenotypically male, but are sterile and have reduced testes size and enlarged breasts. XO—Turner syndrome. Phenotypically female, but have a lack of development of secondary female characteristics. Fragile X—abnormal X chromosome, which causes mental retardation and autism. Heredity and Evolution—You may have to perform simple Mendelian crosses that are either monohydrid or dihydrid. This may require you to work backwards from data to find the genotypes of parents.RNA and DNA Structure and Function Molecular Genetics—RNA and DNA Structure and Function A. Deoxyribonucleic Acid (DNA)—Genetic material B. Ribonucleic Acid (RNA) 1. 2. 3 types: mRNA (messenger RNA), rRNA (ribosomal RNA), tRNA (transfer RNA) All 3 are involved in the expression of genes into proteins. C. Nucleotides 1. Nucleic acids are made of nucleotides. A nucleotide is composed of a 5 carbon sugar, a nitrogenous base, and a phosphate group. .. Two types of nitrogenous bases: purines (6-member ring fused to a 5-member ring) and pyrimidines (6-member ring). Purines Pyrimidines Adenine Thymine (not found in RNA) Guanine Cytosine Uracil (only found in RNA) . Two types of sugars: deoxyribose for DNA and ribose for RNA. |. Phosphate group (see chapter on Molecules and Cells). . Nucleoside is a nitrogenous base and sugar without the phosphate. 77 JaydeydChapter 16 78 Discovery of DNA as the Genetic Material A. Transformation Experiments of Griffith, Avery, McCarty, MacLeod. 1. Smooth (contains capsule) living Streptococcus pneumonia injected into live mouse, resulted in a dead mouse. 2. Rough (no capsule) living Streptococcus pneumonia injected into live mouse, resulted in a healthy mouse. 3. Heat-killed smooth (capsule destroyed) Streptococcus pneumonia injected into live mouse, resulted in a healthy mouse. 4. Heat-killed smooth (contains capsule) mixed with living rough (no capsule) Streptococcus pneumonia injected into live mouse, resulted in a dead mouse. 5. Interpretation of the experiment was that the DNA from the heat-killed smooth cells “transformed” the rough cells into smooth cells that killed the mouse. Transforming agent was DNA. . Hershey-Chase Experiment 1. Worked with T2 bacteriophage or a virus that infects bacteria. 2. Bacteriophage were radioactively labeled with P32 (DNA) or $35 (protein coat of bacteriophage). 3. When separate experiments were completed, it was found that bacteria contain the radioactively labeled P32 DNA of the bacteriophage. 4. Interpretation of the experiment was that the bacteriophage injected their DNA into the host bacterium in order to produce progeny phage, indicating DNA as the genetic material.RNA and DNA Structure and Function C. Watson and Crick le 2. DNA ase pair Suga phosphat backbone James Watson and Francis Crick used X-ray crystallography experiments to show that DNA is a double helix. Data of the experiment also indicated that the double helix had a sugar phosphate backbone with nitrogenous bases pairing interior to the double helix. . Base pairing rules of purines and pyrimidines were established (also known as Chargaff’s rule). i, Adenine (purine) pairs with thymine (pyrimidine). 2 hydrogen bonds for base pairing. ii, Guanine (purine) pairs with cytosine (pyrimidine). 3 hydrogen bonds for base pairing. Hydrogen bonds indicated by dashed lines. N a oor MHN oF Ni ‘Adenine Thymine Q----------- 22 == - Ni NH-~----->7 >>> wa < 3 NHN Wnt --- === A NH Guanine Gtosine 79Chapter 16 D. Meselson-Stahl 1. Experiment indicated that replication of DNA is semi- conservative, or one old strand is used for the synthesis (template) of a new strand. 2. Experiment showed that both heavy and light nitrogen would be incorporated into the daughter DNA during the first round of DNA replication. In the second round of replication, daughter strands would only have light nitrogen since the heavy nitrogen was removed. Banding patterns indicated a semi-conservative model is favored over conservative or dispersive. \ Heredity and Evolution—Every AP Biology test-taker should Te lay know the main differences between DNA and RNA. DNA est BARS consists of A, T, C, G as nitrogenous bases, deoxyribose as 5 Tip carbon sugar, and phosphate. RNA consists of A, U, G, C as nitrogenous bases, ribose as 5 carbon sugar, and phosphate. Structurally DNA is a double-stranded helix, while RNA is single-stranded. DNA Replication 1. Where does DNA replication begin? i. Origins of replication (Ori) or specific DNA sequences where replication will commence. In prokaryotes there is one origin for replication while in eukaryotes there are thousands of origins for replication. ii. DNA replication proceeds in both directions on the strand, forming what is called a replication fork. 2. Elongation i. Elongation at the replication fork is catalyzed by an enzyme called DNA polymerase. 3p eating sand 7 v4 3° Lagging strand . DNA PolymeraseRNA and DNA Structure and Function . Nucleotides are the substrate for DNA polymerase and ATP is hydrolyzed by the enzyme. Strands have to be primed by an enzyme called RNA primase. This enzyme adds a small strand of RNA for DNA polymerase to hook to and begin the synthesis of anew DNA strand. The RNA primer is removed later in replication and filled in with DNA. iv. The two strands of DNA are anti-parallel or 5’ phosphate to 3’ hydroxyl in one direction and at the opposite end is 3’ hydroxyl to 5’ phosphate. v. DNA polymerase only elongates the strand from 5’ to 3’. - 5 phosphate me 3 hydroxyl 5 5; phosphate vi. Leading Strand—the daughter strand that is synthesized into the replication fork. This strand is synthesized in a continuous fashion. Lagging Strand—the daughter strand that is synthesized away from the replication fork. This strand is synthesized in a discontinuous fashion or in fragments called Okazaki fragments. viii. DNA ligase is the enzyme that will take a 5’ phosphate and 3’ hydroxyl and link them together, helping join the Okazaki fragments into a single strand. vii. 81Chapter 16 82 Major Enzymes and Proteins in DNA Replication Enzyme DNA helicase Single-stranded binding proteins DNA primase DNA polymerase DNA ligase Substrate Double- stranded DNA Single- stranded DNA Single- stranded DNA Single- stranded DNA Single- stranded DNA Action Opens up the DNA strand for replication Binds single-stranded DNA and keeps replication fork open Lays down a RNA primer on single-stranded DNA for DNA polymerase to hook up with Adds the complementary base to the daughter strand using the parental template. Follows base pairing rules; adenine with thymine, guanine with cytosine Links a 5’ phosphate with a 3’ hydroxyl on the lagging strandGene Regulation Molecular Genetics—Gene Regulation A. One Gene-One Enzyme Hypothesis of Beadle and Tatum 1. 2. Used auxotrophs (nutritional mutants) of Neurospora crassa. Wild type can grow on a minimal media while several other class mutants were not able to grow on minimal media unless supplemented. . Each mutant had a specific metabolic deficiency because the gene that produced the required enzyme was mutated. B. Transcription 1. Synthesis of single-stranded mRNA (messenger RNA) from DNA. 2. Nuclear event. . Major enzyme involved is RNA polymerase, and runs in a 5’ to 3’ direction. . RNA polymerase binds to a promoter or sequence in the DNA that allows for the initiation of transcription. Part of the promoter has a sequence of 5’TATAAA3’ called TATA box. . Eukaryotic cell transcription factors along with RNA polymerase promote transcription. .. Complementary base pair in RNA is different from that found in DNA. The base pairing is adenine with uracil and guanine with cytosine. . Termination of transcription requires a sequence in the DNA called the terminator. 83 JaydeydChapter 17 8. mRNA in eukaryotes is modified after it is synthesized from RNA polymerase. A poly A tail and 5’ 7-methylguanosine cap are added to aid in the stability of the message. 9. RNA splicing is a process in which nonprotein coding portions of pre-mRNA called introns are excised, while coding portions called exons are spliced together to form a fully functional mRNA molecule. Protein complex called the Spliceosome catalyzes this cut-and-paste mechanism. C. Translation 1. Synthesis of protein from mRNA in the 5’ to 3’ direction. 2. Takes place in the cytoplasm and the rough ER. 3. Requires all 3 RNA molecules (mRNA, tRNA, and rRNA) and codons (sets of 3 nucleotide RNA bases that code for amino acids). 4, As mRNA threads itself through the ribosome, initiation of the process begins with a location of the start codon AUG (calls for the amino acid methionine). 5. Once the start codon is found, a t-RNA molecule containing the appropriate amino acid is brought to the ribosome. The tRNA molecule binds to the codon via an anticodon complementary sequence which is located on tRNA. Once the binding happens, the entire ribosome translocates down another 3 bases and reads another codon sequence, where another tRNA brings in the appropriate amino acid. A peptide bond between amino acids is formed via an enzymatic reaction promoted by the rRNA portion of the ribosome. 6. Termination of translation happens when one of the following codons is read: UAA, UGA, UAG. 84Molecular Genetics—Mutation Mutation A. Mutation—changes in the genetic material. 1. Two types of point mutations: Base pair substitution and base pair insertions or deletions. 2. Mutagens—Physical or chemical agents that promote mutations. Examples include UV light, X-rays, and chemicals such as bromine or chlorine. Name of Mutation Base Pair Substitution Silent Mutation Base Pair Substitution Missense Mutation Example Wild Type mRNA AUG-CCU-UAC Protein MET-PRO-TYR Mutant mRNA AUG-CCG-UAC Protein MET-PRO-TYR Wild Type mRNA AUG-CCU-UAC Protein MET-PRO-TYR Mutant mRNA AUG-CCU-UGC Protein MET-PRO-CYS Consequence No change in protein sequence because the genetic code is redundant. Change in protein sequence because the UGC calls for a different amino acid. Most likely will change the structure, thus the function of protein. (continued) 85 JaydeydChapter 18 Name of Mutation Base Pair Substitution Nonsense Mutation Insertion or Deletion Single Base Pair Insertion or Deletion Three Base Pair Example Wild Type mRNA AUG-CCU-UAC Protein MET-PRO-TYR. Mutant mRNA AUG-CCU-UAA Protein MET-PRO-STOP Wild Type mRNA AUG-CCU-UAC Protein MET-PRO-TYR Mutant mRNA AUG-CCA-UUA-C Protein MET-PRO-LEU Wild Type mRNA AUG-CCU-UAC Protein MET-PRO-TYR Mutant mRNA AUG-CCU-UAC-UUU Protein MET-PRO-TYR- PHE Consequence Premature stop codon put into protein. Protein is usually nonfunctional. Change in protein sequence because of the shifting of the sequence by the addition of a base. Similarly can happen if one base is deleted. Protein is usually nonfunctional. Addition or loss of amino acid due to insertion or deletion of 3 bases. Protein will most likely retain some normal function. 86Viral Structure and Replication, Bacterial Gene Regulation Viral Structure and Replication, Bacterial Gene Regulation A. Genome Various types of genomes including the following: double- stranded DNA, single-stranded DNA, double-stranded RNA, or single-stranded RNA. B. Capsids and Envelopes 1. 2. Protein coat that protects the viral genome is called a capsid. Viral envelopes—extra protection layer that surrounds the capsid. C. Viral Reproduction 1. Host cells—the virus will invade and eventually live off of the host by taking over the metabolic machinery (parasitic). Viruses cannot reproduce independently. . Interferon—class of drugs produced by cells in response to viral infection. . Virus attaches to the host via cell surface receptors and injects its DNA. .. Lytic Cycle—type of viral reproduction that eventually kills the host. . Lysogenic Cycle—virus will replicate its genome, but does not kill the host. Formation of prophage, or incorporation of the viral DNA into the host chromosome. 87 JaydeydChapter 19 88 . Human Immunodeficiency Virus or HIV, the causative agent of Acquired Immune Deficiency Syndrome (AIDS) 1. Retrovirus—RNA virus that uses an enzyme called reverse transcriptase to synthesize DNA from an RNA strand. 2. Infects T4 helper cells. . Prion—protein infectious particle or misfolded protein that converts other normal proteins into a mutant form. Causative agent for “mad cow disease.” Bacterial Gene Regulation A. Plasmid—autonomous piece of self-replicating DNA. B. Genetic Recombination in Bacteria 1. Transformation—bacteria uptaking DNA from an outside source. This is the crux of the Griffith, Avery, McCarty, MacLeod experiment outlined in Chapter 16. 2. Transduction—bacteriophages acting as vectors by moving bacterial genes from one bacterial cell and transferring to another. 3. Conjugation—formation of sex pilus between two bacteria that allows them to share genes with each other. 4. Transposons—transposable genetic elements or jumping genes. Genes being moved from one location to another in the cell’s genome. C. The Lactose Operon 1. Operon—sets of genes under the control of promoter. 2. Lactose Operon—operon required for the utilization of lactose as a carbon source. i. Structural genes—lacZ, lacY, lacA ii, Regulatory genes—lacl (repressor) 3. All bacterial gene regulation is done through the interaction of repressor proteins with specific DNA sequences called operators (found in the promoter region of a gene).Viral Structure and Replication, Bacterial Gene Regulation 4. Repressor proteins bind to the operator sequence, acting as a blockade to stop RNA polymerase from binding to the promoter. 5. When the inducer is present, it will bind to the repressor, causing it to fall off the operator allowing RNA polymerase to transcribe the structural genes of the pathway. 6. The interplay of the inducer and repressor is the “hallmark” of bacterial gene regulation. Repressor bound to operator Operon in off position; no lactose present tact Promoter Operator lock lacy lack | Repressor RNA Polymerase Repressor not bound to operator ‘Operon in on position; lactose present fac\ Promoter Operator | lacZ | foc | tack RNA oe Polymerase wow Gene | Product Action lacl Repressor Shuts operon off lacZ Beta-galactosidase | Breaks down lactose lacY Permease Allows lactose in cell lacA Transacetylase Modifies lactose breakdown Heredity and Evolution—The classic example of gene expression is the Lactose Operon. It is a good example to compare and contrast with eukaryotic gene expression in essays. |s9Nucleic Acid Technology and Applications DNA Cloning 1. Restriction Enzymes—enyzmes that cut DNA at particular sequences called restriction sites. Example: Restriction site ACTGGA— ed A CIGGA— ———— —Toacct— Tone TF 2. Recombinant DNA—combining DNA sequences that would not normally occur together to form one piece of DNA. The enzyme DNA ligase is added to seal the strands together. CIGGA~ TGACC be A GG > GACC - DNALigase ut DNA Foreign DNA Recombinant DNA 3. Cloning Vector—original plasmid that is used to carry foreign DNA into a cell and replicate there. fsa DNA with ONA cue) = | a igen lacZ genes restriction site. fu foreign DNA [seme Into bacterial hast, eS Plate on media containing ¥-gal and ampcilin as Bacterial cone thats white and grows on plates clone containing recombinant DNA plasmid (See explanation on next page.) 91 JaydeydChapter 20 92 . When the restriction enzyme is added to plasmid, lac Z is destroyed and non-functional. The fac Z gene produces the enzyme -galactosidase, which breaks down the sugar X-gal causing the colony to appear blue. If the /ac Z gene product is not made, the colony appears white; if the gene is functional the colony appears blue. . Ampicillin resistance gene allows bacteria to grow in the presence of the antibiotic ampicillin. . Media is selective for clones that have the ampicillin resistance gene, and differential for blue or white colonies. . Colony that is growing on plate (ampicillin resistance) and white are correct clones carrying recombinant DNA. DNA Gel Electrophoresis Ne 2. w DNA is placed in gel made of a polysaccharide called agarose or acrylamide (used for smaller fragments). Migration of DNA is based on size differential of DNA fragments. An electric field is passed through DNA molecules and the molecules will travel toward the positive end (cathode) due to negative charge of phosphate on DNA. . Larger molecules travel slower, smaller molecules travel faster. 4. Marker DNA of a standard size is used to approximate the size of unknown molecules. Marker is measured in kilobase pairs. . Visualization of DNA is done by staining gel with ethidium bromide, which increases the visual difference between DNA and gel.Nucleic Acid Technology and Applications Lane 1 lane? lane 3 lane 4 Lane 1— Marker DNA/Standard Size Lane 2— 2 bands roughly 9 kb and 3 kb Lane 3~ 1 band roughly 1.8kb Lane 4— 3 bands roughly 12 kb, 7 kb, and 1 kb Electrophoresis can be used for DNA and protein identification, isolation of types of DNA or protein, calculating the size of fragments (DNA and protein), crime scene investigation, and genetic testing. Polymerase Chain Reaction Ub 2. A method to take a small amount of DNA and amplify the amount. Based on progressive heating and cooling of DNA strands with the addition of primers and DNA polymerase. 93Chapter 20 94 DNA Fingerprinting 1. A technique used by forensic scientists to help determine the DNA of individuals. 2. The DNA of humans is highly homologous. There are sequences called Short Tandem Repeats (STR). These repeats vary in length and size for each human, and therefore, can be used as identifying factors of humans. 3. STRs can be visualized using DNA gel electrophoresis. Heredity and Evolution—Fully understanding the cloning process is considered a major concept in the AP Biology curriculum. You should understand how restriction enzymes and vectors are used in tandem to construct a recombinant plasmid.Early Evolution of Life Early Evolution of Life A. Earth and Early Life 1. 2. Earth is most likely around 5 billion years old. Anaerobic prokaryotes emerged approximately 4 billion years ago and represent the first origin of life. . Earliest living organisms were unicellular, had a genetic code, and were able to evolve and reproduce. . Prokaryotes diverged into two types—bacteria and archaea— about 2.5 billion years ago. .. Oxygen accumulated in the atmosphere approximately 2.5 million years ago as a result of photosynthetic bacteria. . Eukaryotes emerged 2 billion years ago via the Endosymbiotic Theory—larger prokaryotic species engulfed smaller ones that continued to live and function within the larger host cell. Eventually, the mitochondria and chloroplasts were the two organelles created from these smaller cells, resulting in a eukaryotic cell. . Prior to 500 million years ago, life was confined to aquatic environments. Plants eventually found a foothold on earth (root system) via a symbiotic relationship with fungi. B. The Early Atmosphere 1. Primitive earth was thought to have the following atmospheric molecules—water (H,O), methane (CH,), hydrogen (H,), and ammonia (NH,). No oxygen. . Miller-Urey Experiment—tested the Oparin-Haldane model that the atmosphere on primitive earth was the precursor for the synthesis of organic molecules. |95 JaydeydChapter 21 96 i, The primitive atmosphere components when given electric spark eventually formed crude organic molecules including sugars, lipids, amino acids, and nucleic acids. ii, Heterotrophic Hypothesis—first forms of life were prokaryotic heterotrophs that used non-biologically produced organic matter, via the Oparin-Haldane model, as their carbon source. Still no oxygen produced. C. RNA—The First Genetic Material 1. RNA was most likely a self-replicating molecule found in aquatic environments. 2. RNA can self-catalyze its synthesis via its ribozyme (RNA enzyme) capability. 3. Eventually, DNA became the genetic material, because of its stability over RNA and ability to correct mutations. D. Current Models for the Origins of Prokaryotic and Eukaryotic Cells 3-Domain Model Domain Archaea Dom: Bacteria Domain Eukarya \ Indicates Archaea and Eukarya have commonalities Universal Ancestor 1. Carl Woese developed a domain level that is more inclusive than the Kingdom system. 2. Universal Ancestor—primordial organism that was unicellular, anaerobic, prokaryotic, used ATP for energy and DNA as the genetic material. 3. Bacteria and archaea are both at the prokaryotic level, but archaea live in more harsh conditions than bacteria.4. Early Evolution of Life Types of Archaea Methanogens—produce methane and are used to convert waste to CH,. Acidophiles—thrive under highly acidic conditions. | Halophiles—thrive under high salt conditions. | Thermophiles—thrive under high temperature conditions. When comparing archaea to eukarya, they have much in common with each other (more than bacteria). Examples include several types of RNA polymerase, introns, ribosome structure, RNA sequence, and non-response to antibiotics. E. Early Prokaryotic Characteristics io 2. Origination—around 4 billion years ago Anaerobic fermentation (glycolysis)\—primary metabolic pathway . Used organic compounds made via Oparin-Haldane Model— Heterotrophic Hypothesis |. Eventually chemoautotrophs were selected because of their ability to make energy when the environment had been depleted for ATP. . Cyanobacteria (blue-green algae)—photosynthetic bacteria that were able to produce oxygen. F. Early Eukaryotic Characteristics wRWnNo . Earliest eukaryotic descendents of prokaryotes were Protists. . Diverse unicellular organisms. . Motile, having cilia and flagella. . Can be photoautotrophs or heterotrophs. . Endosymbiotic Theory—Larger prokaryotic species engulfed smaller ones that continued to live and function within the larger host cell. Eventually, the mitochondria and chloroplasts were the two organelles created from these smaller cells, resulting in a eukaryotic cell. Developed by Lynn Margulis. 97Chapter 21 98 Evidence for Endosymbiotic Theory Mitochondria and chloroplasts have DNA similar to prokaryotic DNA (sequence and shape are circular). Mitochondria and chloroplasts are double membrane and double bound, similar to prokaryotes. Mitochondria and chloroplasts are similar in size to prokaryotes. Mitochondria and chloroplasts have their own ribosomes like prokaryotes. Mitochondria and chloroplasts reproduce via binary fission like prokaryotes. & 4 Heredity and Evolution—Being able to clearly identify the characteristics of early prokaryotes has been a hallmark of all AP Biology tests.Evidence for Evolution Evidence for Evolution A. Early Theories 1. In the 1700s and 1800s, the biological sciences were defined in terms of natural theology, rather than scientific data and extrapolation. Several scientists began to use data to debunk natural theology as a means for explaining scientific findings. . Carolus Linnaeus—founder of taxonomy or the classification of organisms based on anatomical similarities and differences. Developed the binomial system or Genus species, ex. Homo erectus. . Georges Cuvier—developed paleontology or the study of fossils found in sedimentary rock. Coined the term catastrophism—strata in sedimentary rock was reflective of a catastrophic event that took place. . James Hutton—used the geologic concept of gradualism— major change in geology is a slow and ongoing process. . Charles Lyel—used the geologic concept of uniformitarianism—that the earth was shaped entirely by slow-moving forces still in operation today, acting over a very long period of time . Thomas Malthus—indicated that populations could grow so rapidly, they could outgrow their food supplies. . Jean Baptiste Lamarck—developed Lamarckism—traits can be inherited from one generation to the next via the process of “the inheritance of acquired characteristics.” For example, a giraffe’s neck gets longer with each successive generation or a good baseball player can pass his traits to his children. There is no evidence, however, to prove this theory that |99 JaydeydChapter 22 in “acquired characteristic” can be passed to the next generation. Another name for this is use and disuse; if a trait is used it will be passed down to the next, but, if not used, it will be discarded and not passed along. His theory was still considered visionary because of its emphasis on adaptation to the environment. B. Descent with Modification—The Darwinian World 1. English scientist Charles Darwin voyages on the HMS Beagle to explore the South American coastline. 2. On the Galapagos Islands he found 12 different types of finches that were similar, but each had unique characteristics. 3. Darwin made a link between the origin of a new species and the environment in which these species reside. 4. The theory of natural selection is coined—reproductive success of an organism depends on its ability to adapt to the environment in which it resides. For example, several of the finches in the Galapagos Islands adapted their beak structure in order to find food. 5. Postulates of natural selection— i. If the environment cannot support the individuals who occupy it, then most of the offspring will perish. ii, Survival of individuals within a population will depend on the genetic background of the individual. Individuals with the best traits will survive. Another name for this is the “survival of the fittest.” |. Over time, the fittest organism will survive and therefore changes in the population will take place to benefit the reproduction of the population. The result of natural selection is the adaption of populations to their environment, thus giving them a competitive advantage to survive. iv. Below is a classic example of natural selection (hypothetical data). The graph indicates that as time changes, the average beak length of a finch changes depending on the season. During a dry season the average beak length gets slightly larger, giving the finches 100Evidence for Evolution a better advantage to transverse terrain and outcompete other birds for seeds that are less abundant in a wet season. A larger beak indicates a competitive advantage and survival of the fittest. Dry seasons are 1950 and 1980; wet seasons are 1960, 1970, 1990, and 2000. eS Beak Length (cm) PTT 1940 ©1950-1960» «1970 «1980» 1990» 20002010 Year C. Evidence for Evolution le Biogeography—study of organisms and how they relate to the environment. Organisms may be unique to certain geographies that are present in one particular location. Hence, the organism has adapted to live in that environment. . Fossils—help indicate the progression of organisms from simple to complex. For example, transitional fossils are fossils of animals that display a trait that helped the organism attain a competitive advantage. At one time whales had limb-like fossils indicating they may have been land dwellers at one time. . Comparative Anatomy—study of anatomical similarities between organisms. Homologous structures—structures with organisms that indicate a common ancestor. For example, a human arm, cat leg, whale flipper, and bat wing all have a similar structure, but different functions. Vestigial organs— remnants of structures that were at one time important for ancestral organisms. |. Comparative Embryology—comparing the embryonic development of one organism to another. 101Chapter 22 5. Molecular Biology—used in the study of evolution by looking at homology in DNA and protein sequences. See chart in Chapter 3 on cytochrome proteins. This is the best evidence that can be used when studying the relationships of species. Heredity and Evolutlon—Darwinian biology permeates all aspects of biology. Knowing the evidence of evolution and early theories of evolution can make up a series of questions on the AP Biology exam. 102Mechanisms of Evolution Mechanisms of Evolution A. Population Genetics 1. 2. Population Genetics—study of genetic variation within a population of individuals. Population—a group of individuals that belong to the same species. . Gene Pool—the total sum of genes within a population at a given time. . Hardy-Weinberg—study of the gene pool of a non-evolving population. . Hardy-Weinberg can only be met if the following five conditions are met: No mutation . No gene flow or genetic variation iii, A very large population sample iv. No natural selection v. Random mating .. Mathematical Relationship of Hardy-Weinberg—equation that can help you determine whether a population is in Hardy-Weinberg equilibrium i, _p =represents the frequency of the dominant allele = represents the frequency of the recessive allele iii, Assume a population of 500 pea plants in which green is dominant to yellow. 103 JaydeydChapter 23 A = green, a= yellow Phenotype Green Green Genotype AA Aa Number of pea 320 160 plants (total = 500) Genotypic 320/500 = | 160/500 = frequencies 0.64 AA 0.32 Aa Number of alleles 320x2= 160A& in gene pool 640A 160a Allelic frequencies | 800/1000= | 200/1000 = 0.8A 0.2a p = frequency | q = frequency of A=0.8 ofa=0.2 Yellow aa 20 20/500 = 0.04 aa 40a >» p+q=1 (combined gene frequency must be 100%) > p?+2pq+q?=1 * p?= frequency of AA genotype = 0.8 x 0.8 = 0.64 = 64% * 2pq = frequency of Aa genotype = 2 x 0.8 x 0.2 = 0.32 = 32% * q?= frequency of aa genotype = 0.2 x 0.2 = 0.04 = 4% 7. Sample AP Biology Problem Using Hardy-Weinberg (Know this concept . . . it is always on AP tests): Assume that in a population of insects, body color is being studied. 36% represent the orange color which is recessive to 64% which represent the black dominant phenotype. each successive generation maintains thealll frequency, the population is said tobe in Hardy-Weinberg equilibrium. 1) Determine the allelic frequencies. 2) Determine the genotypic frequencies. © The recessive phenotype is key to this problem because the dominant represents both AA and Aa. However, recessive is ONLY represented by aa. Use logic that q’ =aa, therefore the square root of .36 or q = Since p+ q=1, p +0.6 =1, then p = 0.4. 104 0.6.Mechanisms of Evolution * Allelic frequencies are A = 0.4, a = 0.6. * Genotypic frequencies follow the equation p?+2pq+q?=1. © p?=0.47 =0.16 = 16% (AA or homozygous dominant) — Black phenotype 2pq=2 x 0.6 x 0.4 = 0.48 = 48% (Aa or heterozygous) > Black phenotype 0.6? = 0.36 = 36% (aa or homozygous recessive) > Orange phenotype 16 + 48 + 36 = 100 q B. Microevolution Up Microevolution—defined as evolution on small scale from one generation to the next. i. Genetic Drift—changes in the gene pool due to chance because of a small population. The small population directly contrasts the large population needed to maintain Hardy-Weinberg equilibrium. |. Bottleneck Effect—changes in the gene pool due to some type of disaster or massive hunting that inhibits a portion of the population from reproducing. The small population directly contrasts the large population needed to maintain Hardy-Weinberg equilibrium. |. Founder Effect—a new colony is formed by a few members of a population. The smaller the sample size, the less the genetic makeup of the population. The small population directly contrasts with the large population needed to maintain Hardy-Weinberg equilibrium. iv. Gene Flow—transfer of alleles from one population to another through migration. The gametes of fertile offspring mix within a population, providing genetic variation. Genetic variation directly contrasts the no- gene-flow postulate needed to maintain Hardy-Weinberg equilibrium. v. Mutation—a change in the genetic makeup of an organism at the DNA level. Mutation directly contrasts the no mutation postulate needed to maintain Hardy- Weinberg equilibrium. ‘i, Non-random Mating—individuals mating with those in close vicinity. Non-random mating directly contrasts with vi |105Chapter 23 the random mating postulate needed to maintain Hardy- Weinberg equilibrium. vii. Natural Selection—reproductive success of organisms depends on their ability to adapt to the environment in which they reside. Natural selection directly contrasts with the no natural selection postulate needed to maintain Hardy-Weinberg equilibrium. C. Modes of Natural Selection 1. Natural Selection will favor the allelic frequency in three ways. Below is an example of a bell curve or normal distribution population. Three types of selections can take place that shift the bell curve to different frequencies (hence, evolution is taking place). Original frequency of individuals shows a normal“bell-curve” distribution Frequency of Individuals Phenotype Ht Stabizing Selection —Extreme phenotypes are removed and more common phenotypes are selected Frequengy of Individuals Phenotype 106Mechanisms of Evolution Directional Selection —One of the extreme phenotypesis selected Frequency of Individuals Diversitying Selection —Both of ‘the extreme phenotypes are selected Frequency of Individuals Phenotype D. Macroevolution 1. Macroevolution—evolution that forms new taxonomic groups. 2. Speciation—the origin of new species (a population of individuals who can mate with each other and produce viable offspring). 3. Barriers to Speciation (Pre-zygotic)—before the zygote is formed. i. Habitat Isolation—two species may live in the same area, but have a different habitat. One may be tree dwelling, while another may be terrestrial. ii. Behavioral Isolation—two species may display totally different behaviors that do not attract each other for mating. iii. Temporal Isolation—two species may breed at different times of the day or different seasons. iv. Mechanical Isolation—two different species may be anatomically different. 107Chapter 23 v. Gametic Isolation—two different species produce gametes that will not fuse to form a zygote. 4. Barriers to Speciation (Post-zygotic)—after the zygote is formed. i, Reduced Hybrid Viability—zygote is formed but the zygote does not live. ii, Reduced Hybrid Fertility—hybrid offspring are born and mature, but the hybrid cannot produce viable offspring. E. Speciation—How Does It Occur? 1. Allopatric Speciation—populations are separated by geographical isolation, thus a new species can be formed following adaption to new surroundings. i. Adaptive Radiation—Evolution of a large number of species from a common ancestor. Examples of this are the finches Darwin found on the Galapagos Islands. 2. Sympatric Speciation—populations are not separated by geographical isolation, but a new species is formed amongst the parent populations. i. Autopolyploidy—meiotic error causes a species to have more than two sets of chromosomes. Contribution is from one species. ii, Allopolyploidy—polyploidy is a result of two different species. F. Speciation—How Fast Does It Occur? 1. Gradualism—species are produced by slow evolution of intermediate species. 2. Punctuated Equilibrium—speciation happens quickly at first and then is followed by small changes over a long period of time. Heredity and Evolutlon—Do you like math? Let's hope so jase because there may be easy mathematical calculations on Test sec6 % the AP Biology test. For example, you should know how to Tip EY ~ appropriately use the Hardy-Weinberg equation: p + q = 1, p? + 2pq + q’ = 1. Also make sure you understand how to use the chi-square equation—don’t memorize it. If needed, it will <7]. 2 be given to you: X? = 5 108———PART IV: ORGANISMS AND POPULATION: Diversity of Organisms, Structure and Function of Plants and Animals, EcologyEvolutionary Patterns Jaydeyd Evolutionary Patterns of Animals A. Body Plans of Animals 1. Phylum Porifera—represent sponges. Lack true tissues and do not have nervous, digestive, or circulatory systems. Many Porifera are sessile or attached to some type of environment. 2. Phylum Cnidaria—represent hydra, jellyfish, and sea anemones. These organisms have radial symmetry. Radial symmetry indicates the organism has a top (dorsal end) and a bottom (ventral), but no left or right. Bilateral symmetry— organism has dorsal, ventral, anterior (head), and posterior (tail). Phyla included are Platyhelminths (flatworms), Rotifera (rotifers), Nematoda (nematodes), Mollusca (clams, snails), Annelida (segmented worms), Arthropoda (crustaceans, insects, spiders), Echinodermata (sea urchins), and Chordata (vertebrates). i. Cephalization—all bilateral organisms have this process associated with them. Anterior region of organism contains sensory paraphernalia. This will allow for the development of the central nervous system (brain and extending nerve cord). Bilateral organisms need to move, crawl, burrow, or even walk, which indicates the salient need for a CNS. B. Germ Layers 1. Gastrulation—the process of layering the embryo into three germ layer regions (triploblastic). Phylum Cnidaria are diploblastic; all others are triplobastic. 111Chapter 24 112 i, Ectoderm germ layer—outer part (ecto-) of the germ layer and produces central nervous system, the lens of the eye, ganglia and nerves, the epidermis, hair, and mammary glands. ii. Endoderm germ layer—innermost part (endo-) of the germ layer and produces the stomach, the colon, the liver, the pancreas, the urinary bladder, the intestines, the lining of the urethra, the epithelial parts of the trachea, the lungs, the pharynx, the thyroid, and the parathyroid. Mesoderm layer—middlemost part (meso-) of the germ layer and produces skeletal muscle, the skeleton, the dermis of the skin, connective tissue, the urogenital system, the heart, blood, and the spleen. C. Body Cavities Ne Coelom—a body cavity completely lined with mesodermic tissue allowing the organs to grow independent of a body wall (thus allowing regulation of one-organ system not affecting another). Having a coelom also allows for larger body size and cushioning and protecting internal body parts. A coelom is a true evolutionary advantage for an organism. i. Coelomates have a true coelom. Phyla include Mollusca, Annelida, Arthropoda, Echinodermata, and Chordata. |. Acoelomates completely lack a coelom. Phylum includes Platyhelminthes. Pseudocoelom—a body cavity that is only partially lined by mesodermic tissue. Phyla include Rotifera and Nematoda. D. The Coelomate Dichotomy ls Protostomes—coelomate phylum that includes Mollusca, Annelida, amd Arthropoda. i. Spiral Cleavage and Determinate Cleavage of Embryonic Cell—causes small cells to be adjacent to larger cells at the 8-cell stage. Determinate cleavage at this point indicates the cells at this stage form an inviable embryo if separated from one another.Evolutionary Patterns ii, Coelom Formation—coelom forms from splits in the mesoderm. iii. Blastopore (opening of archenteron or primitive gut system)—mouth develops from blastopore 2. Deuterostome—coelomate phylum that includes Echinodermata and Chordata. i. Radial Cleavage and Indeterminate Cleavage of Embryonic Cell—causes all cells to be aligned with each other at the 8-cell stage. Indeterminate cleavage at this point indicates the cells at this stage form a viable embryo if separated from one another. ii, Coelom Formation—coelom forms from folds of the archenteron. iii, Blastopore (opening of archenteron or primitive gut system)—anus develops from blastopore Organisms and Population—The topic of evolutionary patterns of animals eludes many students during studying. This topic is worth 2-3 questions on the AP Biology test. Be sure you know it! 113,Survey of the Jaydeyd Diversity of Life Survey of the Diversity of Life A. Representative Organisms of Domain Bacteria Kingdom Monera 1. 2. 3. 4. Unicelluar, prokaryotic, with no nucleus. No membrane-bound organelles Archaea—“ancient bacteria.” See chapter 21. Eubacteria—true bacteria. i. Have specialized cell wall made up of peptidoglycan (protein and sugar). ii. Gram-negative bacteria stain pink because of complex cell wall. iii. Gram-positive bacteria stain purple because of less complex cell wall. iv. Decomposers—feed on dead or decaying organisms. v. Nitrogen fixation vi. Coccus—spherical shape. vii, Bacillus—rod shape. viii. Spirillum—spiral shape ix. Endospore—protective layer formed by bacteria that protects them from harsh conditions such as heat, pressure, or lack of food. . Common Eubacteria i. Escherichia coli or E.coli—intestinal pathogen. ii. Streptococcus pneumoniae—causative agent of bacterial pneumonia. iii. Bacillus anthracis—soil dwelling and causative agent of anthrax. . How Eubacteria Gain Energy i. Photoautotroph—use photosynthesis or light energy to make carbon sources. |115Chapter 25 i. Chemoautotroph—need carbon dioxide for carbon source, but inorganic molecules for energy. Photoheterotroph—use light energy to make ATP, but need carbon from an outside source. iv. Chemoheterotroph—consume organic molecules for energy and carbon. B. Representative Organisms of Domain Eukarya Kingdom Protista 1. Most diverse of all eukaryotes. 2. Unicelluar or multicelluar. 3. Protozoans—animal-like protists. i, Amoeba ii, Algae 4. Unicellular algae—plant-like protists. i. Euglena ii. Dinoflagellates iii, Diatoms 5. Motion classification i. Amoeba—move by a false foot or pseudopodia. Ciliates—move via cilia. iii. Flagellates—move via flagella. C. Representative Organisms of Domain Eukarya Kingdom Fungi 1. Yeast (unicellular) or multicellular. 2. Saprobes—derive their nutrition from nonliving organic material. 3. Lichens—symbiotic relationship between fungus and cyanobacterium. 4. Produce antibiotics that work against prokaryotic cells. The best example of this is Penicillium chrysogenum, or fungus that produces the antibiotic penicillin. D. Nine Major Animal Phyla 1. Phylum Porifera i. Sponges. ii, Invertebrates that have a porous body, thus they are filter feeders. 116Survey of the Diversity of Life ii, Anchored to a stratum. /, Radial symmetry. . Phylum Cnidaria i. Hydra, sea anemone, jellyfish (tentacle-containing organisms). ii. Invertebrates that have a true gastrovascular cavity (mouth). iii. Radial symmetry. . Phylum Platyhelminthes (planaria) i. Flatworms. ii, Invertebrates that a have a true gastrovascular cavity (mouth). iii, Bilateral symmetry. iv. Simple nervous system and first organisms to have cephalization. . Phylum Nemotoda i. Nemotodes or round worms (parasitic). ii, Invertebrates with a complete digestive tract (mouth and anus). iii. Bilateral symmetry. . Phylum Annelida i, Invertebrates. Segmented worms such as the earthworm or leech. iii, Bilateral symmetry. . Phylum Mollusca (mollusks) i. Very diverse phylum of invertebrates. ii. Clams, snails, squid, octopi, scallops. iii, Major parts of all mollusks a. Visceral mass—organ-containing system b. Head-foot region—head for sensory function, and foot for motor function. c. Mantle—secretes the shell. d. Radula—sharp tongue for eating. . Phylum Arthropoda i. Phylum containing the most number of animals because of insect class. ii, Invertebrates iii, Major parts of all arthropods. a. They have a so-called exoskeleton or hard outer covering that supports and protects muscles. 117Chapter 25 b. Jointed appendages. c. Segmented body. 8. Phylum Echinodermata i. Invertebrates that include sea stars and sea urchins. ii, Radial symmetry. iii. Spiny skin, an endoskeleton, and a water vascular system for food and waste transportation. 9. Phylum Chordata i. Most chordates are vertebrates, but not all. ii. Major parts of all chordates. a. Notochord that supports the back. b. Dorsal hollow nerve cord. c. Pharyngeal gill slits for feeding. d. Post-anal tail. E. Chordate Classes 1. Class Osteichthyes i. Bony fish ii, Gill formation for breathing iii. Operculum—flap above gill that allows the fish to breathe without moving. 2. Class Amphibia i. First terrestrial vertebrates Limb development for movement |. Cold-blooded iv. Lay eggs in water v. Gills replaced by lungs 3. Class Reptilia i. Scales for waterproofing. ii, Terrestrial amniotic egg. There are four extra embryonic membranes that are adaptations of the terrestrial egg: chorion contributes to the formation of the placenta; amnion sac cushions the embryo; yolk sac provides nourishment for the egg; and allantois collects waste materials and exchanges gases from the embryo. iii. Ectothermic—warm bodies via heat from environment. 4. Class Aves i. Wings, feather, and hollow light bones for flying. 118Survey of the Diversity of Life 5. Class Mammalia i. Warm blooded and four-chambered heart. ii. Endothermic—warm blooded. iii, Hair and mammary glands. iv. Development of offspring within mother prior to live birth. F. Major Plant Divisions 1. Division Bryophyta—Non-vascular seedless plants. i. Moss, liverworts, hornworts. ii. Grow in moist conditions because of lack of vascular system. iii, Flagellated sperm that require water so they can swim to the egg. iv. Gametophyte is dominant life cycle. 2. Division Pteridophyta—Vascular seedless plants. i, Ferns Spores born on underside of leave. iii, Sporophyte is dominant life cycle. 3. Phylum Coniferae—vascular seed plants i. Called gymnosperms. ii, Douglas Fir, Cedar, and other woody plants. iii. Cone-bearing plants. iv. Sporophyte is dominant life cycle 4, Phylum Anthophyta—vascular seed plant. i. Called angiosperms or flowering plants. ii, Formation of flower allowed for co-evolution with other organisms. iii. Sporophyte is dominant life cycle was a major evolutionary adaption for birth to happen ~ 4 Organisms and Population—The terrestrial amniotic egg \ CG outside of water. 119Jaydeyd Phylogenetic Classification Diversity of Organisms—Phylogenetic Classification A. Taxonomy—Classification of Organisms 1, Developed by Carolus Linnaeus and based on similarity of organisms. 2. Hallmark is binomial nomenclature or genus species (Homo Sapien). Kingdom (most inclusive) Phylum cass Order Family Genus Species, B. Major Kingdoms 1. Animalia i, Animals are multicellular with organ systems. ii. Cells have organelles including a nucleus, but no chloroplasts or cell walls. 121Chapter 26 Move with the aid of cilia, flagella, or muscular organs. iv, Heterotrophic—must ingest other organisms for sustenance. 2. Plantae i, Plants are multicellular with organelles including chloroplasts and cell walls. ii. Nutrients are acquired by photosynthesis (autotrophic). 3. Protista i. Protists are single-celled organisms. Move by cilia, flagella, or by amoeboid mechanisms. iii, There is usually no cell wall, although some forms may have a cell wall. iv. May have chloroplasts. v. Nutrients are acquired by photosynthesis, ingestion of other organisms, or both (autotrophic and heterotrophic). 4. Fungi i. Fungi are multicellular (except yeast, which are unicellular), with a cell wall but no chloroplasts. ii, Contain filamentous structures called hyphae for growth. iii. Can be haploid and/or diploid (some have stages of both). iv. Can reproduce asexually via spore formation. v. Saprobic. 5. Monera i, Monerans are single-celled, contain a cell wall, have no chloroplasts or other organelles. ii, Nucleoid. iii, Haploid. iv. Include Archaea and Eubacteria. C. Domain System—developed in 1990 by Carl Woese and places prokaryotes into 2 domains. Domains are more inclusive than kingdoms (see Chapter 21). 1. Domain Bacteria i. Kingdom Monera 2. Domain Archaea i, Kingdom Monera 122Phylogenetic Classification 3. Domain Eukarya i. Kingdom Protista ii. Kingdom Fungi iii, Kingdom Plantae iv. Kingdom Animalia Organisms and Population—The domain system of classification is fairly new when compared to the kingdom system. Even though the domain system is more inclusive, test questions can be derived from both forms of classification. 123Evolutionary Relationships Jaydeyd Diversity of Organisms—Evolutionary Relationships A. Sedimentary Rocks 1. Fossil record—strata that are formed are based on time. Top strata contained the most recent fossil record, while bottom strata contain older fossils. B. Molecular Biology 1. 2. Protein comparison—homology of the primary amino acid structure. DNA and RNA comparisons i. DNA-DNA hybridization—measuring the extent of hydrogen bonding between DNA of two different sources. ii, Restriction Enzyme Mapping—restriction enzymes recognize and cut sequences of DNA. A pattern can be observed using DNA gel electrophoresis. iii, DNA sequence analysis—comparing nucleotide sequence. C. Cladistics 1. A clade is a group of organisms that includes an ancestor and all descendents of that ancestor. Cladistics is based on molecular techniques (DNA and RNA sequences) rather than morphology or form and structure of animals. 125Chapter 27 Species B Shedding of tall Enlarged cranium Bi-pedalism > Species B and C are more closely related to each other than to species A. > All species are generated from an ancestor species with bi- pedalism. » All species retain traits from the ancestor but have evolved to gain some specific trait through time. Organism | Bi-pedal | Large Cranium | Tail Loss Species A x Species B x Species C x x x 126Plants and Animals— Reproduction, Growth, and Development Jaydeyd Reproductive Process of All Land Plants A. Alternation of Generations 1, Two main multi-cellular forms called gametophyte and sporophyte. 2. Gametophyte is haploid and produces the egg and sperm. 3. Sporophyte is diploid formed by the fusion of egg and sperm. 4. Meiosis produces the spore that will eventually give rise to sporophytes via mitosis. 5. Fertilization of male and female gametophyte produce a sporophyte. { (muttictilar) Mitosis * Meiosis Tygote “ar | (> > an otal 127Chapter 28 The Rise of Land Plants A. Green Algae 1. Gave rise to land plants. 2. Most live in fresh water, but there are some marine (not a land plant). 3. No vascular tissue or root system. B. Bryophytes 1. Moss, liverworts, hornworts. 2. None vascular and seedless. 3. Have no root system, but are anchored via rhizoids or non- vascular containing cells. 4. The dominant stage during the lifecycle of bryophytes is the gametophyte, i. Antheridia—produces the male gametophyte Archegonia—produces the female gametophyte iii. Fertilization takes place in the archegonia. Water droplets are required to transport male gametophyte to archegonia. iv. Sporangium produces spores that will eventually produce the mature male or female gametophyte. C. Tracheophytes 1, 4 major characteristics i. Protective layer that surrounds the gametes. ii, Multicellular embryos. iii, Cuticles or waxy layer that covers all parts above the root system. iv. Vascular system (xylem and phloem) D. Pteridophytes (most basic tracheophytes) 1, Ferns and horse tails. 2. Vascular and seedless. 3. The dominant stage during the life cycle of pteridophytes is the sporophyte. 128Plants and Animals—Reproduction, Growth, and Development i, Antheridia—produces the male gametophyte. ii, Archegonia—produces the female gametophyte. 4, Water droplets are required to transport male gametophyte to archegonia. 5. Sporangia are found on the underside of the leaf and produce spores that will undergo fertilization. E, The Seed 1. Allows for protection. 2. Seed = sporophyte embryo along with a food source. 3. Seed can remain dormant but viable until favorable conditions are met. 4. Dispersal via wind or animals. F. Gymnosperm—conifers or cone-bearing plants such as pine. 1. The dominant stage during the life cycle of gymnosperms is the sporophyte. 2. Ovule—structure containing the eggs that are produced via meiosis. 3. Pollen Cone—pollen grains (haploid) are produced via meiosis. 4. Ovulate Cone—contains two ovules. 5. Pollen Grain—2 male gametophytes formed via pollination land on ovulate cone. One will be destroyed while the other will wait at least one year before fertilization takes place. 6. Once a new embryo (only 1 of the eggs is fertilized) is produced (sporophyte), a seed coat will be produced and the unfertilized female gametophyte will become the food reserve, G. Angiosperm—flowering plant. 1. The dominant stage during the life cycle of angiosperms is the sporophyte. 2. Reproductive structure of angiosperms is the flower. i. Sepal—enclose and protect the flower before it buds. ii, Petal—bright-colored structure that attracts insects for pollination. 129Chapter 28 Stamen—male reproductive organ. Carpel (pistal)—female reproductive organ. v. Anther—part of the stamen where pollen is produced via meiosis. i. Stigma—part of the carpel; receives the pollen. |. Style—part of the carpel that leads to the ovary. Ovary—part of the carpel where the ovule is encased. 3. Fruit—a mature ovary. i. After fertilization, the ovary thickens to aid in protection. ii, Fruits aid in dispersing seeds because some are edible by other species. Once eaten, the seed’s protective coat will not break down and the organism will pass the seed through feces at a distant location from where it was eaten. The feces also contain a plethora of fertilizer that will aid in germination. 4. Double fertilization and the endosperm i, In contrast to gymnosperms where one of the pollen grains is destroyed, both pollen grains fertilize one egg in angiosperms. This is called double fertilization. i. The triploid nucleus will continually divide giving rise to a rich food reserve called the endosperm. iii. Cotyledon or seed leaves are produced. Monocots produce 1 seed leaf while dicots produce 2 seed leaves. Organisms and Population—Be sure to know the four main plant groups: bryophytes, seedless vascular plants, gymnosperms, and angiosperms. The gametophyte is the dominant generation of bryophytes, while the sporophyte is dominant in seedless vascular plants. In seed plants, such as gymnosperms and angiosperms, the seed replaces the spore 4s the main means of dispersing offspring. Animal Reproduction A. Various Types of Reproductive Patterns 1. Asexual reproduction—no genetic diversity since all genes come from one parent. No fusion of egg and sperm. i, Budding—outgrowths from a parent form and pinch off to live independently. 130Plants and Animals—Reproduction, Growth, and Development ii. Binary Fission—a type of cell division by which prokaryotes reproduce; each daughter cell receives a single parental chromosome. iii, Fragmentation—breaking of a body piece that will form an adult via regeneration of body parts. 2. Sexual reproduction—genetic diversity since genes will be inherited from both parents. Fusion of egg and sperm. . Parthenogenesis—egg develops without being fertilized. 4, Hermaphroditism—having both male and female reproductive organs. w B. Spermatogenesis (happens in gonads) 1. Continuous throughout the life of a male. 2. 4 viable sperm produced via meiosis in the testes. C. Oogenesis (happens in ovary) 1. After first meiotic division cytokinesis is unequal with the secondary oocyte receiving all of the cytoplasm from the oogonium. 2. At birth the ovary contains all cells that will develop into the egg. 3. Only 1 viable ovum is produced with 3 non-viable polar bodies. D. The Reproductive Cycle of the Human Female 1, Menstrual cycle or changes in the uterus or female reproductive organ. 2. 28-day cycle in which the destruction and regeneration of the uterine lining (endometrium) occurs. i. Menstrual Phase (day 0-5)—menstruation (bleeding due to destruction of endometrium). |. Proliferative Phase (day 6-14)—regeneration of endometrium. Secretory Phase (day 15-28)—endometrium becomes more vascularized and is ready for implantation of embryo. If the embryo is not implanted, the entire menstrual cycle will happen again. 131Chapter 28 132 3. Ovarian Cycle—parallels the menstrual cycle. i. Follicular Phase (day 0-13)—egg cell enlarges in a follicle. ii. Ovulation (day 14)—oocyte is released and pregnancy can take place. iii. Luteal Phase (days 15-30)—the corpus luteum is formed. Astructure that grows on the surface of the ovary where a mature egg was released at ovulation. The corpus luteum produces progesterone in preparing the body for pregnancy. 4. Important hormones i. Gonadotropin-releasing hormone—aids in the release of FSH and LH. ii, Luteinizing hormone (LH)—stimulates ovaries. iii. Follicle-stimulating hormone (FSH)—stimulates production of ova (ovulation) and corpus luteum. iv. Estrogen—stimulates uterine lining growth. v. Progesterone—stimulates uterine lining growth. Animal Development A. Fertilization 1. Process of egg and sperm fusion to make a zygote, 2. Activation of egg that will lead to embryonic development. B. After Fertilization 1. Cleavage—a series of rapid cell divisions to form blastomeres (blastula) or small cells with their own nucleus. 2. Cleavage is the succession of rapid cell divisions without growth during early embryonic development which converts the zygote into a ball of cells. 3. When the blastomere reaches the 16-32 cell stage, it is called a morula. 4. The morula will then become a gastrula by gastrulation. The gastrula will form the three embryonic tissues of the ectoderm, endoderm, and mesoderm.Plants—Structural, Physiological, and Behavioral Adaptations Jaydeyd Structure and Function of Plants—Structural, Physiological, and Behavioral Adaptations A. Diversity of Angiosperms 1. Monocots and dicots Part Monocot Dicot Embyro One cotyledon Two cotyledons (di) (mono) Leaf Venation | Parallel Net like Stems Complex Ring arrangement arrangement of of vascular bundles vascular bundles Roots Fibrous root Taproot Flowers Flowers in Flowers in multiples multiples of 3 of 4or5 B. Cell Types 1. Parenchyma cells—perform the metabolic processes of cells. 2. Collenchyma cells—support the plant. 3. Sclerenchyma cells—rigid cells that are found in areas where the plant is no longer growing. C. Tissue Types 1. Vascular-xylem tissue i, Type of vascular tissue that transports water and dissolved minerals from the roots up the plant. |133Chapter 29 134 ii, Tracheids and vessel elements—dead cells that conduct water and minerals. 2. Vascular-phloem tissue i. Type of vascular tissue that transports food from the leaves to the roots of the plant. li. Sieve-tube members—live cells, but have no organelles. Main function is to transport sucrose. Companion cells—next to sieve-tube member and provide all the metabolic resources for the sieve tube members. Xylem works upward; Phloem works downward! 3. Dermal Tissue—protects the plant. i. Cuticle—waxy coat that helps the plant retain water. 4. Ground Tissue—occupies space between vascular and dermal tissues i. Mostly made up of parenchyma cells. D. Plant Growth 1. Apical meristems or primary growth—located at the tips of roots or shoot buds and contain the cells undergoing mitosis for vertical or expansive cell growth. 2. Lateral meristems or secondary growth—located through the length of the shoot system and roots and is considered outward horizontal growth (increases plant’s diameter). Organisms and Population—Memorize! Memorize! Memorize! Do not lose points because you forgot the properties of plant cells and tissues.Animals—Structural, Physiological, and Behavioral Adaptations Jaydeyd Structure and Function of Animals—Structural, Physiological, and Behavioral Adaptations A. Mammalian Digestive System 1. Oral Cavity (only carbohydrates are broken down) i, Mouth secretes salivary amylase which breaks down starch. Chewing or mechanical digestion is also carried out. Bolus or ball of food is formed. ii. Pharynx, or back of the throat, has a structure called the epiglottis that blocks food from going down the windpipe or trachea. Esophagus—food tube that conducts bolus down to the stomach via a smooth muscle contraction called peristalsis. 2. Stomach (only protein is broken down) i. Gastric Juice—a digestive fluid with pH of about 2 that aids digestion. ii, Pepsin—a protease secreted in an inactive form called pepsinogen until food is present in the stomach. iii, Acid Chyme—food and gastric juice that is processed in the stomach. 3. Small Intestine and Accessory Organs (all 3 macromolecules are broken down) i, Organ that digests most food and absorbs it into the blood. ii, Duodenum—first part of the small intestine where digestion takes place. Pancreatic enzymes are sent to the small intestine to aid in digestion. These enzymes are protease, amylase, and lipase. Bile from the liver (stored in the gallbladder) |135Chapter 30 emulsifies or opens up the fat for lipase to break it down. Fat is broken down for the first time in the small intestine. iv. Microvilli of the small intestine increase the surface area and allow for absorption of nutrients. 4, Large Intestine or Colon (No Digestion) i. Main purpose is for reabsorption of water. ii. Feces are created and eliminated through the rectum or end of the large intestine. B. Circulatory System 1. Open Circulatory System—blood mixes with internal organs directly. Insects, arthropods, and mollusks have an open circulatory system. 2. Closed Circulatory System—blood is confined to blood vessels that lead to the organs. Earthworms, octopi, and vertebrates have a closed circulatory system. 3. Fish—one ventricle, one atrium with gill capillaries allowing for gas exchange. 4. Amphibian—one ventricle and two atrium; have lung and skin capillaries for gas exchange. Have double circulation or oxygen-rich blood going to the organs and oxygen deficient blood returning to the right atrium. 5. Mammal—two ventricles and two atrium, with lung capillaries for gas exchange. Also makes use of double circulation. 6. Flow of Blood in Mammalian Heart i, Deoxygenated blood from the vena cava enters the right atrium and passes through the right atrioventricular valve (tricuspid valve) into the right ventricle. From the right ventricle it travels out of the heart via the pulmonary artery where it becomes oxygenated at the lungs (CO, is exchanged for O,). Oxygenated blood then travels via the pulmonary vein to the left atrium and through the left atrioventricular valve (mitral or bicuspid valve) into the left ventricle. Oxygenated blood then leaves the left ventricle of the heart via the aorta to the organs of the body. 7. Beating of the Heart i. Cardiac muscle transfers an electrical signal via the following: Sinoatrial (SA) node or “pacemaker” of the 136Animals—Structural, Physiological, and Behavioral Adaptations 8. heart (top of right atrium) generates an electrical impulse that is relayed to the atrioventricular (AV) node located between the right atrium and right ventricle. The signal then transfers to the bundle branches and Purkinje fibers at the heart’s apex. Blood Pressure i. The force of blood against a blood vessel wall is a measure of blood pressure. ii. Systolic pressure is peak pressure in the arteries, which occurs when the ventricles are contracting. Diastolic pressure is minimum pressure in the arteries, which occurs when the ventricles are filled with blood. The ratio of systolic and diastolic pressure is the measurement of blood pressure (systolic pressure/diastolic pressure). C. Respiratory System Us 2. 3. Gas exchange is defined as the uptake of oxygen (O,) and loss of carbon dioxide (CO,). Gills—specialized for gas exchange of aquatic organisms. Tracheal system and lungs—specialized for gas exchange of terrestrial organisms. For an insect such as a grasshopper, the tracheae opens to the outside. .. Lungs—Amphibians (frogs) are the only vertebrates that use skin along with lungs to promote gas exchange. . Flow of Air in Mammalian Lungs i. Inhaled air passes through the /arynx (upper part of the respiratory tract) into the trachea or windpipe (rings of cartilage). The trachea divides into two bronchi leading to the lungs. The bronchi branch to bronchioles which contain air sacs called alveoli. The alveoli are covered with capillaries where the gas exchange will take place. . Control of Breathing i. Medulla oblongata, lower part of the brainstem, maintains homeostasis by monitoring CO, levels. When CO, levels are high, the CO, reacts with water in the blood, dropping the pH of the blood. The medulla oblongata senses a pH drop and excess CO, is exchanged for O,.. . Hemoglobin i. Iron—containing protein of mammalian red blood cells. |137Chapter 30 D. Immune System 1. First Line of Defense (non-specific) i, Skin ii. Mucous 2. Second Line of Defense (non-specific) Inflammatory Response i. Antimicrobial Proteins 3. Third Line of Defense (specific) i, Lymphocytes (T and B cells) ii, Antibodies 4. Immune Response i. Primary Immune Response—first exposure to antigen (foreign molecule) requires 10-15 days for lymphocytes to generate the maximum response. i. Secondary Immune Response—second exposure to antigen requires less time to react to antigen. The response is greater and longer. This response is based on immunological memory that the immune system retains after the primary exposure. Secondary Immune Response: First exposure toantigen 3 10005 ig 5§ 10 SE z= | fae | seconde al toantigen : fel 0 » 0 60 Time (days) 5. Self vs. Non-self i. One of the hallmarks of the immune system is to distinguish self from non-self. Immune cells and other body cells coexist in a state known as self-tolerance. 6. Passive and Active Immunity i. Active Immunity—immunity gained by recovering from a disease. Can also be gained from vaccination (immunization). 138Animals—Structural, Physiological, and Behavioral Adaptations ii, Passive Immunity—immunity gained by one individual passing antibodies to another. 7. Antibody Structure i. Y-shaped proteins that bind antigens at specific antigenic determinants called epitopes. ii. 5 types of antibodies: » IgG—most abundant antibody, crosses the placenta and gives passive immunity to fetus. » IgA—present in body secretions such as saliva and tears. » IgM—first antibody in the primary immune response. > IgD—found on the cell surface of B cells » IgE—released in response to allergic attacks. E. Control Systems i Thermoregulation i. Body Insulation—hair, feathers, and fat reduce heat loss. ii. Vasodilation—increased blood flow to the skin, causing heat loss. iii. Vasoconstriction—decreased blood flow to the skin, reducing heat loss. iv. Cooling—sweating or panting. v. Behavorial—organism can move to a warmer or colder region. Hibernation—decreased body temperature because of decreased metabolic activity. vi. F, Urinary (Excretory) System Us Malpighian Tubules—excretory organs of insects and arthropods. 2. Mammalian Kidney—bean-shaped organ (mammals have 2). 139Chapter 30 i, Blood enters the kidney via the renal artery and leaves the kidney via the renal vein. ii. Urine leaves the kidney and enters the bladder via the ureter. |. Urine in the bladder is excreted via the urethra. . Renal medulla—inner portion of the kidney v. Renal cortex—outer portion of the kidney 3. Nephron—basic structural and functional unit of the kidney (roughly 1 million per kidney). 4. Nitrogenous Waste Products i. Ammonia—aquatic animal waste (very toxic and would be deadly for terrestrial animals). ii, Urea—mammalian and amphibian waste that breaks down to form ammonia. iii, Uric Acid—birds, snails, insects, and reptiles. 5. Production of Urine i. Glomerulus—renal artery becomes a ball of capillaries surrounded by a Bowman’s Capsule within the nephrons of the kidneys, which collects fluids (filtrate) as they pass through the glomerulus. i. The filtrate will pass through proximal tubule, loop of Henle, and distal tubule. Urine is gathered in the collecting duct, where it empties into the urinary bladder via the ureter. G. Nervous System 1, Central Nervous System (CNS)—brain and spinal cord. 2. Peripheral Nervous System (PNS)—all other nervous system structures outside the CNS. 3. Neuron—basic unit of nervous system. i. Cell Body—large portion of neuron that contains the nucleus and organelles. ii, Dendrites—communicate the nervous signal from tips of neuron to the cell body. Axon—conduit that communicates down the neuron away from the cell body. iv. Myelin Sheath—lipid-based insulation around the Schwann cells that stops the leaking of the nervous signal. 140Animals—Structural, Physiological, and Behavioral Adaptations v. Schwann Cell—chain of cells that propagates the nervous signal. vi. Node of Ranvier—space or gap between the Schwann cells. vii. Synapse—the space between the end of the axon and the target; examples of targets include muscles or other neurons. 4, Passage of a Nerve to a Reflex Action i. Signal — Sensory Receptor — Sensory Neuron > Interneuron -> Motor Neuron -> Muscle |. Presynaptic Membrane—surface of synaptic terminal that faces the synaptic cleft. Neurotransmitter (packaged in a vesicle) will be released across the cleft to a postsynaptic cell via the presynaptic membrane. Postsynaptic Membrane—surface of the cell body/axon that is on the opposite side of the synapse. Will receive the neurotransmitter and depolarization of the neuron will take place, thus propagating the signal. eft Presynaptic Cell | Postsynaptic Cell 5. Action Potential—the threshold required to be reached in order for an all or none change in the membrane potential of a neuron to occur i. All neurons have a charge associated with them because of the concentration of various ions being transported into and out of the cell. A cell at rest has a resting membrane potential of approximately -70mV. Both Na* and K* channels are closed. ii, Depolarization—reduction in the absolute value of membrane potential because Na* channels open and flood inside of the cell. K* channels are closed. |. Action Potential—if depolarization hits the threshold, an action potential will be created. This is the example of the all or none event. iv. Repolarization and hyperpolarization—decrease in membrane potential (more negative) because K* channels \141Chapter 30 open and K* leaves the cell. Na* channels are closed. Returns the membrane potential from positive (caused by depolarization) to negative. Depolarization and Acton Potential ae Theeshold i i esting Frpepolaztion Potential ail Time (msec) 6. Neurotransmitters—chemicals that are released at synapses. i, Acetylcholine—can be inhibitory or excitatory. When the target cell is a muscle, then it is usually excitatory. Acetylcholinesterase is an enzyme that destroys acetylcholine. . Norepinephrine and Epinephrine—involved in flight or fight response and generally excitatory. 7. Parts of the brain i. Medulla oblongata—controls breathing, circulatory system, and digestive system. ii. Pons—accessory portion of brain that controls breathing. ii. Cerebellum—controls movements and coordination, i.e., hand-eye coordination. iv. Cerebrum—part of the forebrain and regulates the conscious functions of the body such as thought and reasoning and perception of stimuli. v. Cerebral cortex—key role is memory. vi. Corpus callosum—facilitates communication between the right and left hemispheres of the brain. vii. Hypothalamus—regulates body temperature, hunger, and thirst. 142Animals—Structural, Physiological, and Behavioral Adaptations H. Endocrine System 1, Major Glands and Tissue Hypothalamus—inks the nervous and endocrine systems by receiving signals and propagating the endocrine response. Located in the lower half of the brain. Pituitary gland—“master gland”; many of the hormones released by the pituitary gland communicate with other glands. The pituitary is under control of the hypothalamus. Thyroid gland—located near the trachea. Parathyroid gland—located on surface of the thyroid. v. Pancreas—near the kidneys and contains specialized cells called Islets of Langerhans that secrete hormones. vi. Adrenal gland—tocated on top of the kidneys. vii. Gonads—testes and ovary. viii. Pineal gland—small peanut-shaped gland near the center of the brain. ix. Thymus—located in the upper portion of the chest cavity. Gland Hormone Action Pituitary Oxytocin Contraction of uterus; breast milk let-down Pituitary Antidiuretic hormone | Water retention in kidney Pituitary Growth Hormone Stimulates growth Pituitary Prolactin Stimulates milk production from mammary gland Pituitary Follicle Stimulating Stimulates production of Hormone ova and sperm Pituitary Luteinizing Hormone | Stimulates ovaries (continued) 143Chapter 30 Gland Pituitary Thyroid Thyroid Parathyroid Pancreas Pancreas Adrenal Gland Testes Ovaries Ovaries Pineal Hormone Thyroid Stimulating Hormone Triiodothyronine (T,) and thyroxine (T,.) Calcitonin Parathyroid Hormone Insulin Glucagon Norepinephrine and Epinephrine ‘Androgens Estrogens Progesterone Melatonin Action Stimulates the thyroid Stimulates metabolism Lowers calcium level in bloodstream; stimulates bone construction Raises blood calcium Lowers blood glucose Raises blood glucose “Fight or Flight” response; stimulates metabolism Promote sperm formation and secondary male characteristics Stimulate uterine lining and secondary female characteristics Stimulate uterine lining growth; maintains pregnancy Sleep-wake cycle and other circadian rhythms |. Sliding Filament Model of Muscle Contraction 1. Interaction of two proteins called myosin and actin. 2. Myosin binds to actin and simultaneously ATP will be hydrolyzed. 144Animals—Structural, Physiological, and Behavioral Adaptations 3. An interaction between myosin and actin causes the actin filaments to be moved (sliding) and muscle contraction to take place. 4. Ca** ions are essential ions needed for the entire process to take place. Al Test ga /. Organisms and Population—All the systems of the body Tip come together to maintain the homeostasis of the body. This is a central theme in the AP Biology curriculum. 145Plants—Response to the Environment Jaydeyd Structure and Function of Plants—Response to the Environment A. Transport of Water and Minerals in Plants 1. Water Potential 1p—combination of both solute concentrations and pressure. Water will always flow from a high water potential to a lower water potential. 2. Equation for water potential ) = Wyresssure + VPsome 3. Plasmodesmata—channels between plant cells that allow the flow of fluids. 4. Flow of water in roots begins at the root hair or extensions of the epidermal cells. i, From the root hair water will travel to the epidermis, cortex, the endodermis, and stele which contain xylem cells. ii. Casparian strip—a waxy lipid found in the endodermis that repels water and forces it into the stele. 5. Mycorrhizae—symbiotic relationship between plant roots and fungus that assists the plant in transferring water. B. Transpiration 1. Transpiration—loss of water through the plant leaf. 2. Xylem sap—water and dissolved inorganic ions that ALWAYS flow from the roots of the plant to the leaves. 3. Guttation—xylem sap that can be found as droplets on leaves. 4. Root Pressure—forcing of water up the xylem as a result of water entering the root system via osmosis. Guttation is the main result of root pressure. |147Chapter 31 5. Transpiration—Cohesion-Tension Model i. Water is taken up by the root system of the plant (transpiration). . Cohesion of water molecules to each other and the adhesion of water to the cell wall start the ascent of water up the plant (cohesion). |. The top of the plant has a lower water potential than the bottom of the plant. Water travels up the plant based on the water potential gradient. iv. As more water leaves the plant leaf via transpiration, the water potential becomes more negative (this is called tension) driving even more xylem sap up the plant. 6. Regulation of Transpiration i. Guard cells—open and close the stomata of the cell, which changes the amount of gas exchange and water loss. C. Translocation 1. Translocation—transport of food (phloem sap) in the plant via phloem. 2. Sink-Source Model i. Source Cell—cell that produces sucrose via photosynthesis. These are mesophyll cells, which are found in the leaf. i. Sink Cell—cell that will consume or store sugar. These cells are roots, stems, and fruits. iii, Sugar from the source is loaded into the sieve tube members and is transported down the phloem via pressure flow. iv. Phloem sap will reach the sink cells where sucrose will be transported into the sink. The water that assisted to transport the sucrose will be recycled to the xylem cell and move up the xylem. D. Plant Hormones 1. Auxin—stimulates stem elongation, development of fruit, and root growth. 2. Abscisic acid—inhibits growth. 148Plants—Response to the Environment 3. Ethylene—promotes the ripening of fruit. . Cytokinins—stimulated cell division, growth, and germination. 5. Gibberellic acid—stimulates the flowering of a plant and the development of fruit. * E, Tropisms 1. Phototropism—plant growing toward light. Main hormone that allows this is auxin. 2. Gravitropiim—growing in response to gravity. Positive gravitropism is exhibited by roots since they grow downwards. Shoot systems of plants exhibit negative gravitropism or growth against gravity. 3. Thigmotropism—directional growth of plants growing around structures, such as vines on walls. F, Photoperiodism—response of a plant to the length of day and night (a circadian rhythm) 1. Short-day (long-night) plant—requires a longer amount of darkness to bloom. This type of plant will bloom when the amount of darkness exceeds the critical night length. 2. Long-day (short-night) plant—requires a shorter amount of darkness to bloom. This type of plant will bloom when the amount of darkness is less than the critical night length. G. Phytochromes—pigments that plants use to detect light. 1. Phytochromes are sensitive to red and far-red light. 2. Red light promotes the growth of long-day plants, but inhibits short-day plant growth. 3. A burst of far-red light following red light can reverse the effects of the red light. Al Test $2“. Organisms and Population—The sink-source model is a Tip G) great example of structure and function with the key molecule being water. 149Biosphere Jaydeyd Ecology—Biosphere A. Abiotic Factors—non-living physical or chemical factors. Temperature, light, weather. B. Biotic Factors—living factors. Organisms in an environment. C. Levels of Ecological Organization—biosphere is more inclusive than individual. Biosphere Community = fo Individual 151Chapter 32 152 Terrestrial Biomes Biome Tropical Rain Forest Major Location South America Features * Tremendous organic diversity High rainfall Poor soil because of high rainfall Canopy or topmost layer is made of trees that block sunlight Slash and burn techniques commonly used Very low temperature range . . . . Desert Northern Africa Southwest U.S. Australia Little rainfall Cacti and other water- storing plants found (i.e., succulents) Usually hot, but there are cold deserts (i.e., Antartican Desert) * No canopy Plants are spiny for protection Savanna or the tropical grassland Central and South Africa . Large herbivores Dominant herbivores are insects Grasses, patches of trees are major plants Seasonal droughts with dry and wet seasons Warm temperature year round . . . ° Chaparral West coast of esa Known for wildfires Sloping mountains with shrubs, Coastal area with hot, dry summers and mild, rainy winters (continued)Biosphere Biome Major Location | Features Temperate | North American | ¢ Grass is major plant Grassland | Prairies * Nutrient rich soil for agricultural growth * Grazing by large animals * In the U.S., called farmland Temperate | Northeast U.S. * Deciduous—falling of Deciduous leaves prior to winter Forest ¢ Hibernation for animals because of 4 true seasons * Hickory, oak, maple are major trees Coniferous | Upper North * Cone-bearing plants such Forest or America or as pine Taiga Canadian * Cold temperatures Region * Moose, bear, deer Tundra Upper North * Permafrost or permanently America or frozen soil Canadian © Very little rainfall Region * Short growth season and little diversity * Moss and liverworts dominant plants (gymnosperms) * Too cold for reptiles or amphibians Aquatic Biomes A. Vertical Stratification 1. Upper Photic Zone—lots of light for photosynthesis. 2. Lower Photic Zone—low amount of light, therefore, little photosynthesis but some still occurs. 3. Benthic Zone (Aphotic Zone)—bottom of aquatic biomes. Benthos, or organisms that live at the bottom, feed on detritus or dead organic matter. No light penetrates. 153Chapter 32 B. Freshwater Biome 1. Littoral Zone (part of Photic Zone)—located where aquatic or root-based plants live. Close to shore. 2. Limnetic Zone (part of Photic Zone)—farther from shore where phytoplankton live. 3. Profundal Zone (part of Aphotic Zone)—deep Aphotic region. C. Wetland—any area covered with water that supports aquatic growth. Can be flooded or have soil that is permanently saturated during the growing season. ). Estuary—area where fresh and saltwater meet. . Marine Biome 1. Intertidal Zone (part of Photic zone)—where land meets water. 2. Neritic Zone (part of Photic zone)—shallow water regions. 3. Oceanic Zone (part of Photic zone)—past the continental shelf to far depths. 4. Abyssal Zone (part of Aphotic zone)—very deep benthic communities that are very cold and have total darkness, with no nutrients. No photosynthetic activity, therefore, primary productivity is low. vl Test ), Organisms and Populatlon—There is a multitude of Tip \~ information regarding biomes in this chapter. Any one of these 154 facts could be in a multiple-choice question or be used as supporting evidence for an essay.Jaydeyd Population Dynamics Ecology—Population Dynamics A. Population Density—number of people per unit area. B. Dispersion—how individuals in a population are spaced per unit area. 1. Clumped—individuals are aggregated in patches or groups. XXX XXX Example is a school of XXX XXXX fish or pride of lions. XXX 2. Random—arbitrary pattern of individuals. x x _ Example could be => a population of ants x on a sidewalk. x x ie x 155Chapter 33 3. Uniform-pattern based, or evenly spaced individuals. x x x x ng x x Example is a forest of x => trees that follow a pattern. x x x x C. Survivorship Curve—simple graph of the number of individuals alive at certain ages. A Type —Death occurs for the elderly (ie, humans) ‘Type Il—Death independent 1 ofage (Le, fish o bitds) Survivors ‘Type l—Death occurs for the young (ie, plants or sea anemone) Time D. Population Growth Models 1. Exponential Model—unlimited growth of the population because of no limitation on resources. The result is a J-shaped curve, 156Population Dynamics Population Size Number of Generations 2. Logistic Model—limited growth of the population because of limited resources. The result is an s-shaped curve. The carrying capacity or the maximum population size that a habitat can hold (defined by the letter K). Population Size Canrying Capacity Exponential ‘<—_— iowrth orLog Phase Number of Generations 3. K-selected population—have a density near their resources or their carrying capacity. For example, humans. 4. r-selected population—a population that grows fast, reproduces fast, and dies rather quickly. For example, bacteria. 157Chapter 33 158 K-selected population Slow sexual maturation Long life span Large offspring Type | survivorship curve Sexual reproduction r-selected population Rapid sexual maturation Short life span Small offspring Type Ill survivorship curve Asexual reproduction . Density-dependent factors—A factor that affects population size based on the density of population. Most likely biotic factors such as food, predation, migration, or disease. . Density-independent factors—A factor that affects the population size regardless of density. Most likely abiotic factors such as weather or natural disaster. Organisms and Population—The logistic model of growth can be applied to bacterial growth as well as viral growth. It is commonly used with those examples.Communities and Ecosystems Jaydeyd Population Interaction—interactions that occur with different species living in a community (conglomerate of different species living in one location). A. Predation—predator eats the prey. One species benefits while the other does not. 1. Adaptations for predator—claws, teeth, poisons, speed, eyesight. 2. Adaptations for prey plants—thorns in plants, plant chemicals that ward off prey. 3. Adaptations for animals—cryptic coloration or camouflage, aposematic coloration or bright colors that warn, or warning noises. 4. Mimicry—prey resembles another species. i. Batesian mimicry—a harmless species mimics a species that is dangerous to the predator. ji, Mullerian mimicry—two harmful species resemble each other and create a cumulative effect against a predator. B. Parasitism—parasite lives off the host. One species benefits while the other does not (i.e., viruses, tapeworms, and mosquitoes). C. Competitive Exclusion Principle—two species cannot survive in the same ecological niche (the sum of the total abiotic and biotic factors in an ecosystem). Neither species benefits from this interaction. D. Symbiosis—means living together between a host and a symbiont. 159Chapter 34 160 1. Commensalism—One species benefits while the other is not affected. An example is a clownfish that lives within a sea anemone’s poison tentacles. The clownfish is protected and the sea anemone is neither harmed nor benefited. 2. Mutualism—Both species benefit from the interaction. Pollination between insects and plants. Fungi and algae in lichen. Rhizobium—bacteria living in the root nodules of plants supplying the organism with amino acids, while accepting carbon in the form of organic acids for respiration. iv. Bacteria living in the gut of a cow secrete the enzyme cellulase which helps break down cellulose for the cow. The bacteria have a comfortable place to live, and the cow receives nutrients from the broken-down cellulose. 3. Coevolution—the evolution of one species acting as a selective agent for the evolution of a second species. An example would be the co-evolutions of plants and insects. Ecological Succession—Changes in the Composition of a Community Over Time. A Primary Succession—Area starts off containing lifeless conditions, such as a volcanic island or barren land. Initial pioneer species, such as moss and lichen, colonize the area. Fertile soil develops and grass, shrubs, and other plants begin to grow. This process can take hundreds of years. . Secondary Succession—Area has been destroyed by a natural disaster, farming, or slash-and-burn techniques. The area returns to its initial state because the fertile soil in the area has not been removed. This process can be completed within one year.Communities and Ecosystems Ecosystems—All the Organisms Living in a Community, Including Abiotic and Biotic Factors. A. Trophic Levels—division of organisms in an ecosystem. 1. Primary Producers—photosynthetic organism such as plants and blue green algae. 2. Primary Consumers—herbivores or plant-eating organisms. 3. Secondary Consumers—carnivores that eat the primary consumers. 4. Tertiary Consumers—carnivores that eat carnivores or organisms below them. 5. Detritivores—derive their energy from dead organisms or detritus (i.e., fungi and soil microbes). Extremely helpful in recycling matter. B. Food Chain—pathway in which food is transferred from one trophic level to the next. Arrow points to the organism that is “doing” the eating. Terrestrial Food Chain ‘Marine Food Chain 161Chapter 34 162 C. Food Web—an elaborate web of organisms feeding at more than one trophic level. D. Energy Flow in Ecosystems 5 Primary Productivity—the amount of light energy converted to chemical energy in an ecosystem, by autotrophic organisms. Measured by the units of biomass or weight per unit area per unit time (g/m?/yr). 2. Gross Primary Productivity—total primary productivity. . Net Primary Productivity—the energy used by producers for cellular respiration subtracted from the gross primary productivity. . Secondary Productivity—measurement of an organism’s intake of energy that is converted to weight. . Loss of Energy from a food chain—each time energy is taken in by an organism at a particular trophic level, only 10% of that energy is converted into a new biomass for the next trophic level. This rule is called the 10% rule.Communities and Ecosystems —— 10 Tertiary consumer 100) Secondary consumer Primary consumer voy consumer 10,000 ) Primary consumer 100,000 J of ight energy 6. Pyramids of Biomass—because of the 10% rule, biomass pyramids tend to follow the same scheme and shape. Producers have the greatest biomass for the most part. E. Cycling of Elements 1. Water Cycle—cycling of water through evaporation, precipitation, percolation, and runoff. 2. Carbon Cycle—cycling of carbon through photosynthesis and cellular respiration. Burning of fossil fuels is another source of carbon dioxide in the atmosphere, as is volcanic action. 3. Nitrogen Cycle—nitrogen enters the cycle as atmospheric nitrogen or through nitrogen fixation, or the conversion of nitrogen into compounds that can be used, such as amino acids or nucleic acids. Soil bacterium called rhizobium, found in the root nodules of plants, uses nitrogen fixation. 4. Phosphorus Cycle—Phosphorus from the weathering of rocks or soil is cycled around from one organism to another. Organisms and Population—You need to understand the differences between a food web and a food chain. A food chain is exactly that—a chain of organisms that feed on each other. A food web is more complex and actually indicates the relationship among several different organisms that may feed on one another. 163Ecology—Behavior Jaydeyd Ecology—Behavior A. Innate Behavior—all organisms exhibit certain behaviors that are not learned, but are “built in.” (For example, a baby holding onto its mother’s hand when born.) Usually not changed by the environment. . Fixed Action Pattern—set of behaviors that cannot be changed. An example would be Nikolaas Tinbergen’s digger wasp experiment in which the wasp located its nest based on markings and visual cues. .. Repertoire—series of songs that can be distinguished by organisms and signal an action. Mostly done by birds for mating purposes. Learning—modifying your behavior based on specific events in a lifetime. . Habituation—loss of response to stimulus. For example, if a certain animal hears the calls of a predator, but the predator does not attack, eventually it will no longer respond to the stimulus. . Imprinting—learning a behavior at a specific time in an organism's life. An example would be Konrad Lorenz’s becoming the mother of young geese. 165Chapter 35 G. Classical Conditioning—learning associated with a reward or punishment. Pavlov’s dog experiment: when the dogs heard the sound of a bell at feeding time, they started to salivate for food. Eventually, they were conditioned to salivate anytime they heard the bell, independent of food being present or not. H. Operant Conditioning—learning by trial and error. B.F. Skinner's box experiment. |. Kinesis—changing the rate of an action in response to a stimulus. J. Taxis—movement away or toward a stimulus. K. Territoriality—defending one’s area, for example by marking the area and/or physical confrontation. L. Communication—sending information from one organism to another (or group of organisms). Chemical communication can be done via pheromones or hormones that communicate mating. A classic example is the dancing of bees to communicate the location of food, discovered by Karl von Frisch. M. Altruism—concern for the well-being of others or a group. This is most often seen in kin relationships. N. Hierarchy—certain members of the group have a higher standing over other members. As a result, the members with a higher status feed before others, and have their choice of a mate. O. Group Size—flocks, schools, herds, or prides can all be methods in which organisms organize themselves into groups. This allows for increased hunting efficiency as well as protection. However, the weaker members of the group may be subordinate to stronger ones. Communicable diseases can also have deleterious effects on the population numbers. Organisms and Population—These terms may not seem to be part of a biology course, but behavior is a key evolutionary trait that will be tested on the AP Biology exam.Jaydeyd Ecology—Global Issues Ecology—Global Issues A. Biological Magnification—toxic chemicals being increased in concentration from one trophic level to the next. Biomass from one level is created from a larger biomass from the trophic level before. Top-level consumers are mostly affected. The best-known example is the use of DDT pesticide. B. Ozone Layer—the ozone layer absorbs harmful UV light. CFCs or chlorofluorocarbons in aerosol cans and refrigeration units destroy the ozone by reducing it to oxygen. C. Greenhouse Effect—carbon dioxide emissions from the burning of fossil fuels acts as a trap of solar heat in the atmosphere. Increases in carbon dioxide warm the air and accelerate the greenhouse effect. This is thought to be the major cause of global warming. Deforestation is also a major contributor to the greenhouse effect. Organisms and Population—Environmental awareness is the hallmark to the homeostasis of biomes that are both terrestrial and marine. 167——— PART V: THE EXAMMajor AP Biology Themes and Their Relationship to the Exam The AP Biology curriculum is unified by thematic underpinnings. These underpinnings can be related to any of the topics that are presented in this AP Biology Crash Course. A good way to utilize these themes is to incorporate them into your essays. The AP Biol- ogy readers will be delighted that you were able to see the big picture of the course, rather than small isolated concepts. Below are the major themes and some conceptual examples that apply to them. Major Themes A. Science as a Process 1, The experiments of Watson and Crick and others that lead to the discovery of DNA as genetic material; 2. The experiments of Gregor Mendel that lead to the basic laws of genetics; 3. The experiments of Thomas Hunt Morgan that lead to linked genes; 4, Miller-Urey experiment and its validation of the Oparin- Haldane model of the primitive atmosphere. B. Evolution 1. Fermentation as the precursor for cellular respiration; 2. The Endosymbiotic Theory and the development of the eukaryotic cell; 3. The role of genetic variability in evolution: crossing over in meiosis, mutations, independent assortment of gametes; \171 JaydeydChapter 37 172 4. The adaptations of various animal phyla and plant divisions. C. Energy Transfer . Coupling of the four main processes of cellular respiration; . The coupling of cellular respiration and photosynthesis; . Hydrolysis of ATP in all 3 parts of the central dogma; . The utilization of energy in ecosystems. AWN D. Continuity and Change 1. The presence of carbon in all macromolecules as a means of continuity. However, the assimilation of carbon is different in carbohydrates, lipids, proteins, and nucleic acids; 2. Mitosis as a means of continuity and meiosis as a means of change; 3. The process of fertilization as a means of continuity throughout the domains of life, but the specialization of cells, tissues, and organs as a means of change after fertilization; 4. The overall structure of a biome is continuous while populations can change based on abiotic and biotic factors. E. Relationship to Structure and Function 1, The differences between the structure of DNA and RNA leading to different cellular functions; 2. The homologous structures of body parts of organisms having different functions; 3. The various structured organ systems of the human body and their individual functions; 4. The structure of the genomes of eukaryotic and prokaryotic species leading to different cellular functions. F. Regulation 1. The membrane as a means of selective permeability; 2. The function of inducers and repressors in an operon system; 3. The methods by which organisms that are ectothermic and endothermic regulate body temperature;Major AP Biology Themes and Their Relationship to the Exam 4, How various cycles such as water, carbon, and nitrogen maintain homeostatic levels. G. Interdependence in Nature 1. The global issues that can disrupt an ecosystem; 2. The use of water in moss and fern fertilization; 3. Sunlight as the driving force of photosynthesis; 4, The survival of the fittest concept that drives evolution. H. Science, Technology, and Society 1. The use of recombinant DNA techniques in all aspects of scientific inquiry; 2. Pollution in a technologically advanced society; 3. Global warming; 4. Deforestation results in overlapping ecological niches. Themes—Use the themes to your advantage. If you mention these themes in your essay, the essay scorer will surely award you more points. 173The 12 AP Biology Labs b 1 Diffusion and Osmosis Jaydeyd Presence of ars Solution Color Glucose Initial Contents Initial Final Initial | Final Bag 15% glucose | Clear Blue/ Yes | Yes & 1% starch black Beaker | H,O & IKI Yellowish | Yellowish | No Yes Exercise 1A: Diffusion The interpretation of this exercise was that starch is too large a molecule to escape through the pores of the dialysis bag. As a result, the dialysis bag turns from clear to blue/black because the IKI from the beaker diffuses through the pores and reacts with the starch (a positive test for starch). The beaker fluid stays yellowish because no starch has diffused from the dialy- sis bag. Glucose is present in the beaker because it is a smaller molecule than starch and diffuses through the dialysis bag’s pores. Residual glucose is still in the bag, thus you continue to have a positive result for the bag. 175Chapter 38 Exercise 1B: Osmosis Contents in Bag Percent Change 0.0 M Distilled Water 0.1% 0.2 M Sucrose 2.7% 0.4 M Sucrose 5.0% 0.6 M Sucrose 8.1% 0.8 M Sucrose 11.0% 1.0 M Sucrose 14.1% % Change in Mass vs. Concentration of Sucrose ooo, 140% + 12.0% + 10.0% 4 8.0% + 6.0% + % Change in Mass 4.0% + 2.0% 4 0.0% + 0 02 04 06 08 10 62.0 $$ Concentration of Sucrose (Molarity) * The interpretation of this exercise was that as the concentra- tion of the solute increased, water will diffuse into the dialysis bag (hypotonic to hypertonic), increasing the mass of the di- alysis bag. An isotonic solution is evident with distilled water 176The 12 AP Biology Labs (diffusion is happening at equal rates into and out of the di- alysis bag). Contents in Beaker Percent Change 0.0 M Distilled Water 19.0% 0.2 M Sucrose 8.0% 0.4 M Sucrose — 5.0% 0.6 M Sucrose —13.0% 0.8 M Sucrose -21.0% 1.0 M Sucrose -27.0% Exercises 1C and 1D: Water Potential % Change in Mass vs. Concentration of Sucrose 30.0% % Change in Mass Concentration of Sucrose (Molarity) ¢ The interpretation of this exercise indicates that when the line crosses the x-axis at 0.36 M (estimation) that is the concentra- tion when the potato core would be isotonic to the sucrose concentration. A net change of 0% at x = 0.36 M is the con- centration that the water potential in the potato tissue is equal to the sucrose concentration (isotonic). |177Chapter 38 * Calculation of water potential: literbar mmole Wy sit = (0) (269) (ons | (295%) = 88 bars If the calculated water potential is less than the water potential surrounding the bag, water will flow into the bag (more solute molecules inside the bag). If the calculated water potential is greater than the water potential surrounding the bag, water will flow out of the bag (less solute molecules inside the bag). Thus, water will flow from a high to low water potential. Exercise 1E: Onion Cell Plasmolysis * Interpretation of this exercise: > Onion cell in hypotonic solution will cause water to diffuse into the cell, thus creating a turgid cell. > Onion cell in hypertonic solution will cause water to dif- fuse out of the cell, thus creating a plasmolyzed cell. > Onion cell in isotonic solution will cause water to equally diffuse across the cell. The cell will be flaccid. Le) Pe see ry Exercise 2A: Test of Catalase Activity * Interpretation of this exercise: >» The enzyme is catalase with the substrate being hydro- gen peroxide (H,O,). The products released are water and oxygen. Oxygen gas is flammable gas and will reignite a glowing flint. > Boiling catalase will denature the enzyme; therefore, the enzyme will not function correctly. 178The 12 AP Biology Labs Exercises 2B, 2C, and 2D: Enzyme Catalyzed Reaction Time (seconds) 10 30 | 60 | 90 | 120 | 180 | 360 a) Base Line 36] 56] 56) S6| 56 | 5:6) 5:6 b) Final Reading 10.1 | 14.4 | 15.6 | 15.8 | 15.8 | 15.8 | 15.8 ©) Initial Reading 7.1 | 12.3 | 14.2 | 15.0 | 15.7 | 15.7 | 15.7 d) Amount of KMnO, Consumed eee | Ce e) Amount of H,0, Used 26] 3.5) 42] 48] 5.5) 5.5] 5.5 Amount of H,O, Used vs. Time a 6+ = cZ > ¥OF é44 So Hl 3 £ 1-H °°}??? 0 100 200 300, 400 Time (secs) 179Chapter 38 Lab 3 Mitosis and Meiosis Exercise 3A.1: Observing Mitosis in Plant and Animal Cells Using Prepared Slides of Onion Root Tip and Whitefish Blastula © The interpretation of this exercise: > Visually be able to draw a cell in Interphase (non-dividing portion of the cell cycle) and the 4 stages of mitosis. Interphase Prophase Metaphase / \\Anaphase Telophase The 12 AP Biology Labs /; A Exercise 3A.2: Time for Cell Replication Number of Cells Percent of | Time Field | Field | Field Total Cells | in Each 1 ae 3 | Total | Counted Stage Interphase | 500 | 600 |700 | 1800 82.0% 19 hr 41 min Prophase 50 60 70 180 8.2% Thr 58 min Metaphase | 30 40 50 120 5.5% Thr 30 min Anaphase 15 20 30 65 3.0% 43 min Telophase 8 10 12 30 1.3% 19 min 2195 |181Chapter 38 182 Anaphase, 3.0% Metaphase, 5.596 Prophase, 8.2% Telophase, 1.3% Interphase, 82.0% ° The interpretation of this exercise: >» The length of the cell cycle is roughly 24 hours with the majority of that time spent in Interphase, getting prepared for mitosis. Of the cells that one would observe in mitosis, the predominant phase is prophase. Exercise 3B.1: Meiosis * The interpretation of this exercise: > You must know the differences between meiosis and mitosis. Number of Chromosomes Number of DNA Replications Number of Divisions Number of Daughter Cells Produced Number of Chromosomes Purpose/Function Mitosis 2n-diploid 1 1 DB 2n-diploid genetically identical Growth of somatic cells Meiosis 2n-diploid il 2 4 n-haploid genetically variable Generation of gametesThe 12 AP Biology Labs Exercise 3B.2: Crossing Over during Meiosis in Sordaria fimicola * The interpretation of this exercise: >» The number of asci observed that show crossing over are called recombinants. Recombinants are progeny that do not resemble the genetic make-up of the parent. > Sordaria fimicola is a fungus that after undergoing meio- sis, another mitotic event will take place. This is unique to many fungi including Saccharomyces cerevisiae or baker's yeast. » The crossing-over event that took place is depicted below. The different combinations result in how the haploid cells arranged themselves in the ascospore. tan Meiosis Meiossit Mitosis —_ — = 7. + ue — — , — = Number of Number | Asci Showing of 4:4 | Crossing Over % Asci Sega See Total| Showing Gene Asci | Crossing Over | Distance 156 165 318 | 156/318 =.49 | 24.5 map 0.49/2 = 24.5% units 183Chapter 38 WT ee 184 Exercise 4A: Plant Pigment Chromatography * The interpretation of this exercise: » The various plant pigments that help to run the photo- synthetic machinery can be separated out based on solu- bility in the solvent being used. The molecules involved in this are chlorophyll a, chlorophyll b, carotenoids, and xanthophylls, The relationship between the distance the pigments travel in relation to the solvent is calculated by a factor called R,. R= distance pigment migrated ‘distance solvent front migrated Band Number Distance (mm) Band Color 1 10 yellow green 2 23 blue green 3 33 yellow 4 49 deep yellow 5 105 yellow Distance of solvent front 122 mm 105/122 = 0.86 | R, for carotene (yellow to yellow orange) 49/105 = 0.46 | R, for xanthophyll (yellow) 23/105 = 0.22 | R, for chlorophyll a (bright green to blue green) 10/105 = 0.095 | R, for chlorophyll b (yellow green to olive green)