AP Biology Premium Prep Book 2025 PDF

Acknowledgments I wish to thank: – My husband, Austin, for his love and support – My children, Jamie and Didi, for making every day a great one – My editor, Samantha Karasik, for her guidance and valuable input – The evaluators of this book, who made great recommendations – The students and teache...

Acknowledgments I wish to thank: – My husband, Austin, for his love and support – My children, Jamie and Didi, for making every day a great one – My editor, Samantha Karasik, for her guidance and valuable input – The evaluators of this book, who made great recommendations – The students and teachers I have worked with during my career—I have learned so much from all of you AP® is a registered trademark of the College Board, which is not affiliated with Barron’s and was not involved in the production of, and does not endorse, this product. © Copyright 2024, 2023, 2022 by Kaplan North America, LLC, d/b/a Barron’s Educational Series All rights reserved under International and Pan-American Copyright Conventions. By payment of the required fees, you have been granted the non-exclusive, non-transferable right to access and read the text of this eBook on screen. No part of this text may be reproduced, transmitted, downloaded, decompiled, reverse engineered, or stored in or introduced into any information storage and retrieval system, including but not limited to generative artificial intelligence (gen AI) systems and machine learning systems, in any form or by any means, whether electronic or mechanical, now known or hereinafter invented, without the express written permission of the publisher. Published by Kaplan North America, LLC, d/b/a Barron’s Educational Series 1515 West Cypress Creek Road Fort Lauderdale, Florida 33309 www.barronseduc.com ISBN: 978-1-5062-9167-3 About the Author Mary Wuerth has taught AP Biology for more than 20 years at Tamalpais High School in Mill Valley, California. She earned her B.S. degree in Biochemistry at UCLA and her M.S. degree in Biological Sciences at Clemson University. In addition to having taught Biology at the College of Marin, she has taught teens and teachers in a variety of environments, under various circumstances, including as a visiting teacher and a remote teacher prior to, and during, the COVID-19 pandemic. Mary has written items for the AP Biology exam, served as a Table Leader for the scoring of the free-response questions on the AP Biology exam, and served as chair for the Test Development Committee for the SAT Subject Test in Biology. She has also been presenting AP Biology workshops to new and experienced teachers around the world since 1999. Mary is a winner of the Presidential Award for Excellence in Math and Science Teaching and has received national awards for the use of technology in the classroom. She was selected as one of 25 Lead Teachers for WGBHTV’s Evolution Project and is currently serving as an HHMI Biointeractive Ambassador. Table of Contents How to Use This Book Barron’s Essential 5 ABOUT THE EXAM 1 Introduction Exam Format Tips for Section I: Multiple-Choice Questions Tips for Section II: Free-Response Questions Scoring of the AP Biology Exam Suggested Study Plans 2 Statistics in AP Biology Overview What Is a Null Hypothesis? Chi-Square Test Descriptive Statistics Practice Questions Answer Explanations UNIT 1: CHEMISTRY OF LIFE 3 Water Overview Water and the Importance of Hydrogen Bonds Practice Questions Answer Explanations 4 Macromolecules Overview Biological Macromolecules Protein Structure Nucleic Acids Practice Questions Answer Explanations UNIT 2: CELL STRUCTURE AND FUNCTION 5 Cell Organelles, Membranes, and Transport Overview Cell Organelles and Their Functions Endosymbiosis Hypothesis The Advantages of Compartmentalization The Importance of Surface Area to Volume Ratios Structure of the Plasma Membrane What Can (and Cannot) Cross the Plasma Membrane Passive Transport Active Transport Practice Questions Answer Explanations 6 Movement of Water in Cells Overview Water Potential Osmolarity and Its Regulation Practice Questions Answer Explanations UNIT 3: CELLULAR ENERGETICS 7 Enzymes Overview Enzyme Structure and Function Environmental Factors that Affect Enzyme Function Activation Energy in Chemical Reactions  Energy and Metabolism/Coupled Reactions Practice Questions Answer Explanations 8 Photosynthesis Overview Light-Dependent Reactions Light-Independent Reactions (The Calvin Cycle) Practice Questions Answer Explanations 9 Cellular Respiration Overview Glycolysis Oxidation of Pyruvate Krebs Cycle (Citric Acid Cycle) Oxidative Phosphorylation Fermentation Practice Questions Answer Explanations UNIT 4: CELL COMMUNICATION AND CELL CYCLE 10 Cell Communication and Signaling Overview Types of Cell Signaling Signal Transduction Disruptions in Signal Transduction Pathways Feedback Mechanisms Practice Questions Answer Explanations 11 The Cell Cycle Overview Phases of the Cell Cycle Regulation of the Cell Cycle, Cancer, and Apoptosis Practice Questions Answer Explanations  UNIT 5: HEREDITY 12 Meiosis and Genetic Diversity Overview How Meiosis Works How Meiosis Generates Genetic Diversity Practice Questions Answer Explanations 13 Mendelian Genetics and Probability Overview Mendelian Genetics Probability in Genetics Problems Practice Questions Answer Explanations 14 Non-Mendelian Genetics Overview Linked Genes Multiple Gene Inheritance Nonnuclear Inheritance Phenotype = Genotype + Environment Practice Questions Answer Explanations UNIT 6: GENE EXPRESSION AND REGULATION 15 DNA, RNA, and DNA Replication Overview Structure of DNA and RNA DNA Replication Practice Questions Answer Explanations 16 Transcription and Translation Overview Transcription in Prokaryotes vs. Transcription in Eukaryotes  Translation Flow of Information from the Nucleus to the Cell Membrane Practice Questions Answer Explanations 17 Regulation and Mutations Overview Regulation of Gene Expression in Prokaryotes Regulation of Gene Expression in Eukaryotes Gene Expression Helps Cells Specialize Mutations Practice Questions Answer Explanations 18 Biotechnology Overview Bacterial Transformation Gel Electrophoresis Polymerase Chain Reaction (PCR) CRISPR-Cas9 Practice Questions Answer Explanations UNIT 7: NATURAL SELECTION 19 Types of Selection Overview Evidence of Evolution Natural Selection Artificial Selection Sexual Selection Practice Questions Answer Explanations 20 Population Genetics Overview Population Genetics and Genetic Drift Hardy-Weinberg Equilibrium  Practice Questions Answer Explanations 21 Phylogeny, Speciation, and Extinction Overview Phylogeny and Common Ancestry Speciation Extinction Modern-Day Examples of Continuing Evolution Practice Questions Answer Explanations UNIT 8: ECOLOGY 22 The Basics of Ecology Overview How Organisms Respond to Changes in the Environment Energy Flow Through Ecosystems Practice Questions Answer Explanations 23 Population Ecology, Community Ecology, and Biodiversity Overview Population Ecology K-Selected vs. r-Selected Populations Community Ecology and Simpson’s Diversity Index Biodiversity Practice Questions Answer Explanations LAB REVIEW 24 Labs Overview Lab 1: Artificial Selection Lab 2: Hardy-Weinberg Lab 3: BLAST  Lab 4: Diffusion and Osmosis Lab 5: Photosynthesis Lab 6: Cellular Respiration Lab 7: Mitosis and Meiosis Lab 8: Bacterial Transformation Lab 9: Restriction Enzyme Analysis of DNA Lab 10: Energy Dynamics Lab 11: Transpiration Lab 12: Fruit Fly Behavior Lab 13: Enzyme Activity Practice Questions Answer Explanations PRACTICE TESTS Practice Test 1 Section I: Multiple-Choice Section II: Free-Response Answer Explanations Practice Test 2 Section I: Multiple-Choice Section II: Free-Response Answer Explanations APPENDICES A Frequently Used Formulas and Equations B Choosing the Right Graph in AP Biology Visit Barron’s Online Learning Hub for more full-length practice tests. How to Use This Book This book provides comprehensive review and extensive practice for the latest AP Biology course and exam. About the Exam Start with Chapter 1, which summarizes the Big Ideas and the eight units of this course, provides an exam overview, and discusses scoring. Review the tips for answering each question type. Consult the suggested study plans to map out your test prep. Next, review Chapter 2, which covers the statistical tests and descriptive statistics you need for this exam. Don’t worry if math isn’t your strongest skill—this chapter will teach you simple calculations needed for test day. Review and Practice Study Chapters 3 through 23, which are organized according to the eight units of AP Biology. The most complex material is divided into multiple chapters to provide you with more manageable chunks of content. Every chapter includes Learning Objectives that will be covered, a review of each topic, dozens of figures that illustrate key concepts, and a set of multiple- choice and short and long free-response practice questions (with detailed answer explanations) to check your progress. Then, review Chapter 24, which focuses on labs. While no specific labs are required for this course, the curriculum emphasizes inquiry-based labs that require you to make hypotheses, evaluate data, make predictions, and justify your conclusions with evidence. This chapter covers 13 common labs that allow you to refine your skills in applying these science practices. Practice Tests This book concludes with two full-length practice tests that mirror the actual exam in format, content, and level of difficulty. Each test is followed by detailed answers and explanations for all questions. Online Practice There are also four additional full-length practice tests online. You may take these tests in practice (untimed) mode or in timed mode. All questions are answered and explained. For Students Whether you are using this book at the start of the school year or in the weeks leading up to the exam, this book will provide you with the support you need to maximize your score. Try to answer as many questions as you can before checking the explanations to determine which topics you know well and which chapters you need to review further. After studying the test- taking tips, reviewing each topic, and completing every practice question and test, you can (and will) achieve success on the AP Biology exam. For Teachers This book is fully aligned with the eight units and exam format outlined in the latest AP Biology Course and Exam Description. You can use this book as a resource in the classroom, or you can assign chapters as supplemental reading or practice questions as homework or test material.  BARRON’S ESSENTIAL 5 As you review the content in this book to work toward earning that 5 on your AP Biology exam, here are five things that you MUST know above everything else: Everything in AP Biology connects to the four Big Ideas. The AP Biology course and exam are structured around four Big Ideas: 1. Evolution 2. Energetics 3. Information Storage and Transmission 4. Systems Interactions As you review the eight units of AP Biology, notice how the Big Ideas are interwoven throughout multiple units. For example, Big Idea 3 (Information Storage and Transmission) is covered in five units. This idea is first introduced in Unit 1 during the discussion of nucleic acids. This idea reappears in many topics of Unit 4 (such as cell communication, signal transduction, and the cell cycle) and in several topics of Unit 5 (including Mendelian genetics and non- Mendelian genetics). Big Idea 3 is also included in every topic of Unit 6 (gene expression and regulation). Finally, this concept is covered in Unit 8, when discussing ecological responses to the environment. As seen from this example, the Big Ideas are the foundation of this course and exam, and it is important to recognize how each of these four concepts connects with multiple units and topics. Learn how to explain concepts and how to analyze visual representations (Science Practices 1 and 2). You must be able to: Describe and explain biological concepts in applied situations Construct a visual model (a diagram or a graph) of the characteristics of a biological system Given a visual representation of a biological system, describe the interactions and relationships between the components of the system Know how to ask testable questions and how to design experiments and methods to test those questions (Science Practice 3). You must be able to: State the null hypothesis and design an experiment to test it State the alternative hypothesis and design an experiment to test it Identify the components of an experiment, including appropriate controls, the independent and dependent variables, and experimental constants Use experimental data to evaluate a hypothesis Design a follow-up experiment Show that you can represent data in an appropriate graph and that you can accurately describe that data (Science Practice 4). You must be familiar with the uses of the following types of graphs: Line graph Bar graph Histogram Pie chart  Scatterplot Box and whisker plot Graphs with two y-axes (dual y) Demonstrate that you can use statistical tests to analyze data and that you know how to use the results of these statistical tests to support or reject claims and hypotheses (Science Practices 5 and 6). You should be familiar with the following calculations and statistical tests: Means Rate calculations Ratios and percentages 95% confidence intervals (error bars) Chi-square About the Exam  1 Introduction Before beginning your review, it is important to understand the guiding principles and units that make up the AP Biology course and exam. The AP Biology curriculum focuses on four Big Ideas: Big Idea 1: Evolution—The process of evolution drives the diversity and unity of life. Big Idea 2: Energetics—Biological systems use energy and molecular building blocks to grow, reproduce, and maintain dynamic homeostasis. Big Idea 3: Information Storage and Transmission—Living systems store, retrieve, transmit, and respond to information essential to life processes. Big Idea 4: Systems Interactions—Biological systems interact, and these systems and their interactions exhibit complex properties. These Big Ideas are the overarching themes covered in the eight units of content that make up the AP Biology course and exam. The review chapters that follow this introduction are all grouped according to these eight units, so you can test which units you are strongest in and which you may want to study more closely. Table 1.1 lists each of these eight units and the approximate percentage of questions that will be devoted to each unit on the AP Biology exam. Table 1.1 AP Biology Units Unit % of Questions 1—Chemistry of Life 8–11% 2—Cell Structure and Function 10–13% 3—Cellular Energetics 12–16% 4—Cell Communication and Cell Cycle 10–15% 5—Heredity 8–11% 6—Gene Expression and Regulation 12–16% 7—Natural Selection 13–20% 8—Ecology 10–15% Exam Format You will have three hours total to complete the AP Biology exam, which consists of the two sections outlined in Table 1.2. Table 1.2 AP Biology Exam Format Section I Section II Question Type Multiple-Choice Free-Response Number of Questions 60 2 Long Free-Response 4 Short Free-Response Time 90 minutes 90 minutes % of Overall Score 50% 50% Section I: Multiple-Choice You will have 90 minutes to complete 60 multiple-choice questions, which will make up 50% of your overall score. Each question will have four possible answer choices, and you need to select the choice that best answers the question. Some questions (in both Section I and Section II of the exam) may require you to use your math skills and the AP Biology Equations and Formulas sheet, which will be provided to you on test day.  Many of the multiple-choice questions will require you to evaluate data in tables, graphs, or diagrams. Thus, to prepare yourself for Section I, practice analyzing and interpreting as many different types of tables and graphs as possible. A wide variety of tables and graphs are incorporated throughout this book to provide you with as much practice with them as possible. TIP Get familiar with the AP Biology Equations and Formulas sheet. You do NOT need to memorize any of those formulas, but you DO need to know when to use them and how to apply them. Section II: Free-Response You will have 90 minutes to complete six free-response questions, two of which are long free-response questions and four of which are short free- response questions. Each of the six free-response questions will consist of four parts. All six free-response questions combined will make up 50% of your overall exam score. Long Free-Response Questions Questions 1 and 2 of Section II will be the long free-response questions. Each will be worth between 8 and 10 points, for a total of 18 points between both questions. Both will likely involve interpreting and evaluating experimental results. Question 1 may ask you to evaluate data presented in a table or graph, while question 2 may ask you to construct a graph using the appropriate confidence intervals. Short Free-Response Questions Questions 3, 4, 5, and 6 of Section II will be the short free-response questions. Each of these questions will be worth 4 points.  Question 3 may describe an experimental scenario. You may be asked to identify the parts of the experiment (such as any controls, the independent variable, and the dependent variable), predict results, and justify your predictions. You may also be asked to describe the biological processes covered in the experiment. Question 4 is typically a conceptual analysis question. In this question, you may be asked to describe and explain a biological process. Given a disruption in the process, you must predict how that disruption will affect the process and justify your prediction with evidence. Question 5 may ask you to analyze a model or visual representation of a biological concept. You may be presented with a diagram and asked to describe the characteristics of the process represented in this model. Then, you may be asked to explain the relationships between the different parts of the model and relate or apply the model to a larger biological concept. Question 6 may ask you to analyze data. You may see data in a graph or table, and you might be asked to describe the data and to use the data presented to evaluate a claim. Finally, you may be asked to explain how the data presented relates to a larger biological concept. Tips for Section I: Multiple-Choice Questions Do NOT skip over the scenarios and/or diagrams presented in the stem of the question. A stem that contains a description of a scenario and/or a diagram or graph will precede many of the multiple-choice questions. In a testing situation where time is limited, students are sometimes tempted to save time by skipping over the stem and proceeding directly to the question. Don’t do this! Often, taking just 30 seconds to read over the data or scenario presented will make it easier to answer the question or questions that follow it. The scenario presented in the stem of the question often will have important background information that will help you answer the question. If you are presented with a graph, note the variables shown on each axis and their units, and try to detect any patterns in the data. In data tables or charts, note the column  headings and their units, and observe any trends or patterns in the data. Do NOT be afraid of organisms or genes you may not have heard of before. There are so many great examples of organisms, genes, and ecosystems that apply to the content of the AP Biology course, and no teacher or textbook can mention all of them. Any example that is not explicitly included in the AP Biology Course and Exam Description will be described in enough detail in the question so that you will have enough background information to answer the question. Therefore, don’t worry if you see a question about the CYP6M2 gene in Anopheles gambiae and you’ve never heard of either before! The stem of the question will tell you what you need to know about that gene and organism (for example, that the CYP6M2 gene confers insecticide resistance to Anopheles mosquitoes), so all you need to do is apply your knowledge and skills to that background information to find the correct answer. Do NOT be tempted by the “distractors.” Incorrect answer choices are called distractors. As you read each question, cover the answer choices with a piece of paper or your hand. Before you reveal the answer choices, think of the characteristics that a good answer to the question at hand will contain. Then, reveal the answer choices and choose the answer that best fits the characteristics you know a good answer will have. It is often easier to focus your brain on finding the best answer rather than trying to eliminate each of the distractors. DO pace yourself. You will have 90 minutes to answer 60 multiple- choice questions. If it is taking you more than two minutes to answer a question, move on to the next question and go back to that question later. Just make sure to skip the bubble in the answer sheet for each question you skip so that the following answers are filled in in the correct bubbles. DO answer every question. There is no guessing penalty on the AP Biology exam. If you leave a question blank, you are guaranteed to not earn points for that question, so answer every question, even if you have to guess. Never leave a question blank on the AP Biology exam! Reserve the last two or three minutes of the time allotted for  Section I to check that you have answered all of the questions and have not left any questions blank. Tips for Section II: Free-Response Questions Do NOT leave any questions blank. Even if you think you don’t know how to answer the question, reread the question to see what terms in the question you do know something about. Then, use those terms as the basis for your answer, keeping in mind the task verbs in the question. As in Section I, if you leave a question blank, you are guaranteed to not earn points on that question, but if you write something, you may earn some points that could make the difference between a score of a 3, 4, or 5. Never give up—remember, you CAN do this! Do NOT make any contradictory statements. For example, if you state that the function of the mitochondria is to generate energy for the cell (a correct statement) but then later in your response state that the function of the mitochondria is also to perform photosynthesis (an incorrect statement), you have made two contradictory statements. Thus, you will not earn any points for either of those statements. DO plan your approach to Section II. Take the first 5–10 minutes allotted for Section II to “read and rank.” Read all six free-response questions, and then place the number 1 next to the question you think will be the easiest for you, the number 2 next to the next easiest question, and so on. You do not have to answer the questions in the order they appear in the test. Sometimes the easiest free- response questions are at the end of this section, and if you get hung up on a more challenging question that appears earlier, you may never get to the easier questions you are likely to earn points on. DO read each question carefully. Read each question carefully at least two times. Each time you read the question, circle or underline key words, especially any bolded words (which are the action or task verbs), any numbers, or any words like and or or (which  indicate whether all or some of the items mentioned need to be addressed). DO pace yourself. You will have 90 minutes to complete all six free- response questions. Some of the free-response questions will require less time; others will require more time. Here is a suggested time plan for Section II: First 5–10 minutes for “read and rank” (see above) 20 minutes for each of the two long free-response questions (Questions 1 and 2) for a total of 40 minutes 5–10 minutes for each of the short free-response questions (Questions 3, 4, 5, and 6) for a total of 20–40 minutes DO write legibly. This may seem obvious, but if your answer is unclear or unreadable, the AP reader cannot award you points for it. Use a black ballpoint pen to write your answer. If you make a mistake, just cross it out with a single strikethrough—any more than that is unnecessary. If your handwriting is particularly difficult to read, consider writing on every other line in the test booklet. Don’t worry about running out of pages—the test booklet usually contains more blank pages than are typically needed, and the test proctor is required to give you extra pages if you do run out of paper in the test booklet. DO label your graphs completely with units. If a question asks you to construct a graph, always make sure the axes are labeled clearly with the appropriate units. A unitless graph will not earn points. Use consistent scaling on your axes, and give your graph a title. DO label the parts of your answer appropriately. This makes it easier for the reader who scores your exam to award you points. However, if you happen to answer Part A of a question in the section you labeled Part B, the reader will still award you points for it. DO use complete sentences. As per the instructions for Section II, use complete sentences in your answers. You will not be awarded points for bulleted lists. If you use a drawing in your answer, make sure to also describe it in complete sentences. DO ATP (Address the Prompt). Do not waste time writing an introductory paragraph, a thesis statement, or a concluding paragraph. Do not restate the question—the reader knows what the  question is! While you need to be clear in your writing, you are not being evaluated on your ability to write a well-constructed essay, as you might be in an AP English course. You ARE being evaluated on your knowledge of biology. Make sure you understand the question prompt and what it is asking you to do. Then, reread your answer to make sure you addressed all of the task verbs in the question and did not make any contradictory statements. DO pay attention to the task verbs! Pay attention to these action verbs, which are typically bolded in the long and short free-response questions, as these words indicate what the question requires you to provide in your response. Some of the most frequently used task verbs are the following: Predict—state what you think will happen if a change is made in a system or process Justify—give evidence to support your prediction Make a claim—make a statement based on the available data or evidence Support a claim—give evidence to defend a claim Describe—note the characteristics of something Explain—state “why” or “how” something happens (Note: This is more demanding than describing.) Identify—provide the information that is asked for (Note: This is less demanding than describing.) Calculate—perform the requested calculation, and ALWAYS show your work and your units! Construct—make a graph (show units!) or a diagram that illustrates data or a relationship Determine—make a conclusion based on evidence State—give a null hypothesis or an alternative hypothesis that is supported by data/evidence Evaluate—assess the validity or accuracy of a claim or hypothesis Scoring of the AP Biology Exam The AP Biology exam is scored on a scale from 1 to 5, with 5 being the highest possible score. Table 1.3 describes each score. Table 1.3 AP Biology Exam Scores AP Exam Score Recommendation 5 Extremely Well Qualified 4 Well Qualified 3 Qualified 2 Possibly Qualified 1 No Recommendation Scores of 3 or above may earn you college credit or allow you to skip introductory courses and take more advanced courses earlier in your college career. Policies regarding credit for AP exam scores vary widely between schools and may even vary between majors at the same school. Always check with the college or university you plan to attend to find out the latest information. Fifty percent of your total score is based on your performance on Section I (the multiple-choice section), and the other 50% of your total score is based on your performance on Section II (the free-response section). For this reason, it is very important to practice answering all question types (multiple-choice, short free-response, and long free- response)—that is why you will see all of these types of questions at the end of every chapter of this book. Suggested Study Plans The following are suggested study plans depending on how much time is left until test day. If there’s a lot of time left before the exam, read through all of the chapters in this book, answer all of the practice questions, and complete all the practice tests. If time is limited, refer to these study plans to skip to the areas that you may want to study further. Follow what works best for you and your schedule. Remember, by reviewing and practicing with this book, you are already taking the first step toward achieving success on the AP Biology exam! Six Weeks Until the Exam Start by taking all of Practice Test 1. Once you’ve completed Practice Test 1, review the answer explanations and use the Self-Analysis Chart for Section I to determine what your strengths are and to diagnose the four units where you need the most improvement. Read through the chapters that cover those four units, and answer all the practice questions in those chapters. Reread the preceding tips for Section I and Section II. Review Chapter 2, which focuses on statistics (since 95% confidence intervals, the null hypothesis, and the chi-square test are key tools used to evaluate experimental results in AP Biology). Take all of Practice Test 2. Once you’ve completed Practice Test 2, review the answer explanations and use the Self-Analysis Chart for Section I to determine where you’ve improved and what two units you’re still having trouble with. Reread the chapters related to those two units, and answer all the practice questions in those chapters. Review Chapter 24, which focuses on the lab component of the course. Answer all the practice questions at the end of this chapter, and review the answer explanations for any questions you may have answered incorrectly. Revisit the preceding tips for Section I and Section II one last time so that those reminders are fresh in your mind for test day. Two Weeks Until the Exam Complete all of Practice Test 1. Once you’ve completed Practice Test 1, review the answer explanations and use the Self-Analysis Chart for Section I to  determine what your strengths are and to diagnose the three units where you need the most improvement. Read through the chapters that cover those three units, and answer all the practice questions in those chapters. Reread the preceding tips for Section I and Section II. Review Chapter 2, which focuses on statistics. Complete all of Practice Test 2. Once you’ve completed Practice Test 2, review the answer explanations and use the Self-Analysis Chart for Section I to determine where you’ve improved and the one unit you’re still having trouble with. Reread the chapters related to that unit, and answer all the practice questions in those chapters. Review Chapter 24, which focuses on the lab component of the course. Answer all the practice questions at the end of this chapter, and review the answer explanations for any questions you may have answered incorrectly. One Week Until the Exam Complete all of Practice Test 1. Once you’ve completed Practice Test 1, review the answer explanations and use the Self-Analysis Chart for Section I to determine what your strengths are and to diagnose the two units where you need the most improvement. Read the chapters related to those two units, and answer all the practice questions in those chapters. Reread the preceding tips for Section I and Section II. Review Chapter 2, which focuses on statistics. Complete all of Practice Test 2. Once you’ve completed Practice Test 2, review the answer explanations and use the Self-Analysis Chart for Section I to determine where you’ve improved and the one unit you’re still having trouble with.  Reread the chapters related to that unit, and answer all the practice questions in those chapters. Review Chapter 24, which focuses on the lab component of the course. Answer all the practice questions at the end of this chapter, and review the answer explanations for any questions you may have answered incorrectly. The Day Before the Exam Complete just Section I of one of the two practice tests in this book. Once you’ve finished Section I, review the answer explanations and use the Self-Analysis Chart to diagnose the one unit where you need the most improvement. Skim through the chapters related to that unit, and answer all the practice questions in those chapters. Reread the preceding tips for Section I and Section II. Review Chapter 2, which focuses on statistics. Review Chapter 24, which focuses on the lab component of the course. 2 Statistics in AP Biology Learning Objectives In this chapter, you will learn: What Is a Null Hypothesis? Chi-Square Test Descriptive Statistics Overview Scientists make hypotheses and then design experiments to test these hypotheses. Data are gathered during these experiments and then analyzed. Scientists use these analyses to draw conclusions about the data. An important tool in data analysis is statistics. Statistical tests are used to evaluate hypotheses. Descriptive statistics describe data sets. This chapter will review some of the statistical tests and descriptive statistics you need to understand for the AP Biology course and exam. What Is a Null Hypothesis? The null hypothesis (H0) states that there is no statistically significant difference between two groups in an experiment. For example, a student designs an experiment to see if plants watered with bottled water will exhibit more growth than plants watered with tap water. The null hypothesis for this experiment would be that there will be no statistically significant difference in plant growth between the plants watered with bottled water and the plants watered with tap water.  Here’s another example: You want to test if dogs prefer dog food brand A over dog food brand B. The null hypothesis would be that there will be no statistically significant difference between the number of dogs choosing dog food brand A and the number of dogs choosing dog food brand B. If 100 dogs are presented with both dog food brand A and dog food brand B, the null hypothesis would predict that 50 dogs would choose brand A and 50 dogs would choose brand B. Chi-Square Test The chi-square test is a statistical test that is used to compare the observed results to the expected results in the experiment. In AP Biology, the chi- square test is used to evaluate the null hypothesis, and it is often used in genetics problems, in the lab on mitosis (Lab 7), and in the lab on animal behavior (Lab 12). It is important to note that the chi-square test is used to compare primary or raw data, such as the number of items in each category of data. The chi-square test should not be used to compare processed data, such as percentages or means. For example, it would be appropriate to use the chi-square test if you were comparing the number of purple flowers and the number of white flowers that resulted from a genetic cross. It would not be appropriate to use the chi- square test to compare the percentage of purple flowers to the percentage of white flowers. Consider the experiment that measured whether dogs prefer dog food brand A or dog food brand B. If there are 100 dogs, the null hypothesis would predict that 50 dogs would choose brand A and 50 dogs would choose brand B. These are the expected results. If the experiment was carried out and 45 dogs chose brand A and 55 dogs chose brand B, those are the observed results. The chi-square test could be used to evaluate the null hypothesis that there will be no statistically significant difference between the expected results and the observed results of the experiment. The steps of the chi-square test are as follows: TIP You do NOT need to memorize any of the formulas reviewed in this chapter—they are included on the AP Biology Equations and Formulas sheet, which will be supplied to you on test day. For quick reference, you can review those formulas in Appendix A of this book. 1. Calculate the chi-square value. The formula for chi-square is: The symbol means “summation.” This means you need to do this calculation for each category of data (brand A and brand B) and then add the values. Using the observed and expected values (45 observed and 50 expected for brand A; 55 observed and 50 expected for brand B): 2. Determine the number of degrees of freedom (df) in the experiment. The number of degrees of freedom in an experiment is defined as the number of possible outcomes in the experiment minus 1. In this experiment, there are two possible outcomes, brand A or brand B, so the df is 2 – 1 = 1. 3. Using the degrees of freedom and the p-value, find the critical value in the chi-square table. The p-value is defined as the probability that the observed data would be produced by random chance alone. For biology, the typical p-value used is 0.05. Using the df of 1 calculated above and a p-value of 0.05, the critical value is 3.84 according to the chi-square table that follows. Chi-Square Table  4. Compare your calculated chi-square value to the critical value. The calculated chi-square value from the first step (1) is less than the critical value (3.84). 5. Based on the comparison, decide whether to reject the null hypothesis or whether you fail to reject the null hypothesis. If the calculated chi- square value is less than or equal to the critical value, fail to reject the null hypothesis. If the calculated chi-square value is greater than the critical value, reject the null hypothesis. In this example, since the calculated chi-square value is less than the critical value, you would fail to reject the null hypothesis that there is no statistically significant difference between the observed and expected data. This does not mean the null hypothesis is proven—it just means you cannot reject the null hypothesis. In other words, the null hypothesis cannot be ruled out. Here is another example: A coin has heads on one side and tails on the other side. If the coin is flipped 40 times, you would expect the coin to come up heads 20 times and come up tails 20 times. These are the expected results. The coin is flipped 40 times, and the coin comes up heads 12 times and it comes up tails 28 times. Those are the observed results. The null hypothesis is that there is no statistically significant difference between the observed and expected numbers of heads and tails. To evaluate this null hypothesis, here are the steps involved: 1. Calculate the chi-square value. 2. Determine the number of degrees of freedom (df) in the experiment. There are two possible outcomes in this experiment (heads or tails), so the df is 2 – 1 = 1.  3. Using a df of 1 and a p-value of 0.05, the critical value is 3.84 according to the chi-square table. 4. Compare your calculated chi-square value to the critical value. The calculated chi-square value (6.40) is greater than the critical value (3.84). 5. Decide whether to reject the null hypothesis or whether you fail to reject the null hypothesis. In this example, you should reject the null hypothesis because the calculated chi-square value is greater than the critical value. Since the null hypothesis is rejected, you can then come up with an alternative hypothesis—for example, perhaps this is a “trick coin”! Descriptive Statistics Descriptive statistics are used to describe data sets. There are descriptive statistics that describe the center of a data set, and other descriptive statistics describe the amount of variability or spread of a data set. The mean and median can be used to characterize the center of a data set. Thus, the mean and median are sometimes referred to as measures of the central tendency of a data set. The mean, or average, of a data set is calculated with the following formula: In the above formula: n = sample size (the number of data points in the data set) = summation (add all members of the data set, starting with the first member of the set, and continue to the nth member of the data set) xi = the ith member of the data set  To calculate the mean, add the values of all the data points in the data set, and then divide that by the sample size. (Multiplying by gives the same result as dividing by n.) Here’s how you would calculate the means for data set A and data set B: Data Set A: 1, 2, 3, 4, 5; n = 5 Data Set B: 3, 3, 3, 3, 3; n = 5 Notice that data set A looks very different from data set B, but they both have the same mean. Means can be distorted or skewed by extreme values in the data set. These two data sets illustrate how looking at just the mean is sometimes not enough to accurately describe the data set. The median is the midpoint of the data set. To find the median, place the members of the data set in numerical order from lowest to highest value. The middle of the data set is the median. If there are an even number of data points in the set, calculate the mean of the two numbers in the middle of the data set to find the median. Here’s how you would find the median for data set A and data set B: Data Set A: 1, 2, 3, 4, 5 Since the members of the data set are already arranged from lowest to highest, look for the middle value, which in this case is 3. Data Set B: 3, 3, 3, 3, 3 Since the members of this data set are all the same, the median is 3. Again, even though data sets A and B are very different, they have the same median.  Extremes in a data set do not affect the median, but they do affect the mean of the data set. For example, add the numbers 0 and 100 to data set A and data set B to form new data sets C and D, respectively. This changes the means greatly from the original calculations, but the medians do not change: Data Set C: 0, 1, 2, 3, 4, 5, 100 Data Set D: 0, 3, 3, 3, 3, 3, 100 Using the mean and median alone are often not enough to accurately describe a data set. It is also important to use descriptive statistics that describe the spread of a data set. Standard deviation and standard error of the mean can be used to describe how spread out the data points are. You will not be required to calculate standard deviation or standard error of the mean on the AP Biology exam. However, it is important to understand what standard deviation and standard error of the mean can tell you about a data set. You also must be able to use standard error of the mean to construct 95% confidence intervals. Standard deviation (s) averages how far each data point is from the mean of the data set. The formula for standard deviation is: Here’s how you would calculate the standard deviation for data set A and data set B: Data Set A: 1, 2, 3, 4, 5; and n = 5 Data Set B: 3, 3, 3, 3, 3; and n = 5 If a data set is more spread out, the standard deviation will be larger. The less spread out the data points, the smaller the standard deviation. The standard error of the mean ( ) is another measure of how spread out a data set is. Each time you repeat an experiment, random chance will lead to slightly different means. If an experiment is repeated multiple times and a mean is calculated for each experiment, the standard error of the mean predicts the distribution of the means of those repeated experiments. The prediction would be that 95% of the means would fall within two standard errors of the mean (above or below the mean of the original experiment). If the standard error of the mean is large, repeating the experiment will result in a larger range of means than if the standard error of the mean was small. Data sets with smaller standard errors of the mean are considered more accurate. Standard error of the mean is calculated with the following formula: In the above formula: s = standard deviation n = sample size Notice that as the sample size (n) increases, the standard error of the mean decreases. You may already understand that an experiment performed with a larger sample size is considered more reliable than an experiment performed with a smaller sample size. Standard error of the mean gives a mathematical reason for why experiments with larger sample sizes are typically more reliable. Here’s how you would calculate the standard error of the mean for data set A and data set B: Data Set A: 1, 2, 3, 4, 5; s = 1.58 and n = 5 Data Set B: 3, 3, 3, 3, 3; s = 0 and n = 5 Standard error of the mean can be used to create a type of error bar on a graph called a 95% confidence interval (95% CI). A 95% confidence interval does NOT mean you are 95% confident in your data! What a 95% confidence interval does mean is that if you repeated an experiment 100 times and calculated the mean of the data you collected each time, the mean would fall within the 95% confidence interval 95 of those 100 times. In other words, you would expect your mean to fall within the 95% confidence interval 95% of the time, but 5% of the time you would expect the mean to be outside of the 95% confidence interval. To construct a 95% confidence interval, you need to know the upper limit of the interval and the lower limit of the interval. The upper limit of the 95% confidence interval is found by starting with the mean and then adding two times the standard error of the mean: To find the lower limit of the 95% confidence interval, start with the mean and then subtract two times the standard error of the mean: Here’s an example to practice working with 95% confidence intervals. An AP Biology class grows “Fast Plants” (Brassica rapa) and records the height of each plant on day 28 of growth. The mean plant height and standard error of the mean are calculated. The 10 tallest plants are cross- pollinated and produce seeds. These seeds are harvested and planted to form a second generation of plants. The height of each plant in this second generation is measured, and again the mean plant height and standard error of the mean are calculated. The data are shown in the following table: Mean Plant Height on Day 28 (mm) Generation 1 158.4 8.8 Generation 2 203.1 9.6 Construct a graph of the mean plant height for each generation, showing 95% confidence intervals. First, find the upper and lower limits for the 95% confidence intervals for each generation. For Generation 1: Upper limit of 95% CI = 158.4 + 2(8.8) = 176.0 mm Lower limit of 95% CI = 158.4 – 2(8.8) = 140.8 mm For Generation 2: Upper limit of 95% CI = 203.1 + 2(9.6) = 222.3 mm Lower limit of 95% CI = 203.1 – 2(9.6) = 183.9 mm Graphing the means for each generation with the 95% confidence intervals leads to Figure 2.1.  Figure 2.1 Mean Plant Height on Day 28 for Generations 1 and 2 Now that you know how to construct graphs with 95% confidence intervals, it is important to understand what they can tell us when comparing data sets. In the previous example, the 95% confidence intervals for generations 1 and 2 do not overlap. If the 95% confidence intervals for two sets of data do not overlap, it is likely there is a statistically significant difference between the two groups. In that example, there is likely a statistically significant difference between the mean plant heights on day 28 for generations 1 and 2. However, if the 95% confidence intervals do overlap, the data are inconclusive, and it is not possible to say whether or not there is a significant difference between the groups. The experiment would need to be repeated. Here is another example that illustrates this point. On field trips to a salt marsh, an AP Biology class counted the number of crustaceans in a 100 cm2 quadrant. Three different locations in the marsh were visited, and multiple quadrants were counted at each location. The mean number of crustaceans per quadrant and the standard error of the mean were calculated from the data at each location. The data are shown in the following table.  Mean Number of Standard Error of Location Crustaceans per Quadrant the Mean A 8 1.5 B 11 0.9 C 14 1.1 The data from each location were plotted with 95% confidence intervals (calculated as follows). Figure 2.2 shows the graph of this data. Location Upper limit of 95% CI = 8 + 2(1.5) = 11 A: Lower limit of 95% CI = 8 – 2(1.5) = 5 Location Upper limit of 95% CI = 11 + 2(0.9) = 12.8 B: Lower limit of 95% CI = 11 – 2(0.9) = 9.2 Location Upper limit of 95% CI = 14 + 2(1.1) = 16.2 C: Lower limit of 95% CI = 14 – 2(1.1) = 11.8 Figure 2.2 Mean Number of Crustaceans per 100 cm2 in Three Locations in the Salt Marsh Which of the two locations are most likely to have a statistically significant difference in the number of crustaceans found per 100 cm2? The answer is that locations A and C would most likely have statistically significant differences because the 95% confidence intervals for locations A and C do not overlap. The 95% confidence intervals for locations A and B do overlap, so it is not possible to make any conclusions about statistically significant differences between those two locations. Similarly, the 95% confidence intervals for locations B and C overlap, so it is also not possible to make any conclusions about statistically significant differences between locations B and C. Practice Questions Multiple-Choice 1. Which of the following can be used to describe the center of a data set and is NOT affected by extreme values in the data set? (A) mean (B) median (C) standard deviation (D) standard error of the mean 2. Data Set A consists of {10, 15, 15, 20, 25}. Data Set B consists of {15, 15, 17, 19, 19}. Which of the following descriptive statistics is the same for both data set A and data set B? (A) mean (B) median (C) standard deviation (D) standard error of the mean 3. The chi-square test is appropriate to compare which of the following types of data? (A) means (B) percentages (C) processed data (D) raw data 4. In a dihybrid cross of two organisms (AaBb × AaBb), four different phenotypes of offspring can be produced. How many degrees of freedom would there be? (A) 1 (B) 2 (C) 3 (D) 4 5. In pea plants, the tall (T) allele is dominant to the dwarf (t) allele. Two heterozygous (Tt) pea plants are crossed and produce 400 offspring. Three hundred offspring are expected to be tall, and 100 are expected  to be short. There are 290 tall offspring observed and 110 dwarf offspring observed. Calculate the chi-square value from this data (using a p-value of 0.05), and state whether there is likely a statistically significant difference between the observed and expected data. (A) The chi-square value is 0.67. There is a statistically significant difference between the observed and expected data. (B) The chi-square value is 0.67. There is NOT a statistically significant difference between the observed and expected data. (C) The chi-square value is 1.33. There is a statistically significant difference between the observed and expected data. (D) The chi-square value is 1.33. There is NOT a statistically significant difference between the observed and expected data. 6. A student wanted to see if isopods preferred banana or watermelon as a food source. Twenty isopods were placed in the center of a choice chamber with two compartments. Banana was placed in one compartment, and watermelon was placed in the second compartment. After 15 minutes, the number of isopods in each compartment was counted. Data are shown in the table. Number of Isopods in Type of Food in Compartment Compartment After 15 Minutes Banana 6 Watermelon 14 The null hypothesis was that the isopods would have no food preference and would be found in equal numbers in both compartments. Calculate the chi-square value (using a p-value of 0.05), and determine if there was a statistically significant difference between the observed and expected data in this experiment. (A) The chi-square value is 0.80. There is a statistically significant difference between the observed and expected data.  (B) The chi-square value is 0.80. There is NOT a statistically significant difference between the observed and expected data. (C) The chi-square value is 3.20. There is a statistically significant difference between the observed and expected data. (D) The chi-square value is 3.20. There is NOT a statistically significant difference between the observed and expected data. 7. Data Set X and Data Set Y have the same standard deviations. Data Set X has a sample size of 10, while Data Set Y has a sample size of 50. How will their standard errors of the mean compare? (A) Data Set X will have a larger standard error of the mean than Data Set Y. (B) Data Set Y will have a larger standard error of the mean than Data Set X. (C) Both data sets will have the same standard error of the mean. (D) The difference in the standard errors of the mean cannot be determined from the information given. Questions 8 and 9 The heights of oak trees in four different locations were measured. The mean height and the standard error of the means were calculated for each location and are shown in the table. Mean Height of Oak Standard Error of Location Trees (m) the Mean Oakville 20.1 2.5 Sacramento 16.4 1.6 San Rafael 28.7 4.3 Oak Valley 34.1 2.0 8. Which location showed the greatest variability in the heights of the oak trees?  (A) Oakville (B) Sacramento (C) San Rafael (D) Oak Valley 9. Which of the following two locations are likely to have a statistically significant difference between the heights of their oak trees? (A) Oakville and Sacramento (B) Oakville and Oak Valley (C) San Rafael and Oak Valley (D) San Rafael and Oakville 10. Which of the following correctly describes how to calculate a 95% confidence interval? (A) Mean ± 1(Standard Deviation) (B) Mean ± 1(Standard Error of the Mean) (C) Mean ± 2(Standard Deviation) (D) Mean ± 2(Standard Error of the Mean) Short Free-Response 11. Some chemicals are known to increase the frequency of chromosome breakage in dividing cells. Thirty Petri dishes with dividing cells were treated with either lead (10 Petri dishes), cadmium (10 Petri dishes), or control solutions (10 Petri dishes). Forty-eight hours after treatment, each Petri dish was examined, and the percentage of cells that showed chromosome breakage in each Petri dish was calculated. The mean percentage of cells that showed chromosome breakage and the standard error of the mean for each treatment were calculated and are shown in the table.  Mean Percentage of Standard Error of the Cells That Showed Mean Chromosome Breakage Control 5.5 0.05 Lead 24.3 5.2 Cadmium 46.1 9.1 Part A (i) Describe which treatment produced the greatest variability in the data. Part B (i) Describe which treatment was least likely to result in chromosome breakage. Part C A student claims that exposure to cadmium is more likely to result in chromosome breakage than is exposure to lead. (i) Evaluate this claim using the data provided. Part D (i) Identify the independent variable in this experiment. (ii) Identify the dependent variable in this experiment. 12. In tobacco plants, green (G) is dominant to albino (g). A heterozygous green (Gg) tobacco plant is crossed with an albino (gg) tobacco plant, and 100 offspring are produced. Fifty offspring are expected to be green (Gg), and 50 offspring are expected to be albino (gg). However, when the offspring are counted, 60 green plants and 40 albino plants are observed. Part A (i) Identify the degrees of freedom in this experiment.  Part B (i) Calculate the chi-square value for this data. Part C (i) Make a claim about whether or not the observed data are likely significantly different statistically from the expected data. Use a p-value of 0.05. Part D (i) Justify your claim from Part C with your knowledge of the chi- square test. Long Free-Response 13. Organic gardeners will sometimes use ladybugs as a method of reducing aphid populations in their gardens. A garden infested with aphids is divided into three sections. In the first section, no treatment (to reduce the aphid populations) is applied. In the second section, ladybugs are introduced to reduce the aphid populations. In the third section, a chemical pesticide is used to reduce the aphid populations. One week after the treatments are applied, 10 measurements of the density of the aphid populations (in aphids per square decimeter) are taken in each section of the garden. The means and the 95% confidence intervals for each section of the garden are shown in the graph.  Part A (i) Based on the graph, identify the section of the garden with the least variability in aphid density one week after treatment. (ii) Based on the graph, identify the section of the garden with the most variability in aphid density one week after treatment. Part B (i) Analyze the data shown in the graph to determine which sections of the garden are most likely to have a statistically significant difference in aphid density. Part C In this experiment, 10 measurements (n = 10) of aphid density were taken in each section of the garden. (i) If 50 measurements (n = 50) were taken, make a prediction about the effect, if any, this would have on the 95% confidence intervals. Part D (i) Using your knowledge of how 95% confidence intervals are calculated, justify your prediction from Part C. Answer Explanations Multiple-Choice 1. (B) The median is the midpoint of a data set and is not as affected by extreme values in a data set in the same way that the mean is. Extreme values in a data set skew the mean, so choice (A) is incorrect. The mean is used to calculate standard deviation, so standard deviation would be affected by extreme values in a data set. Thus, choice (C) is incorrect. Choice (D) is incorrect because standard error of the mean is also affected by extreme values in a data set. 2. (A) The mean for both data sets is 17. Choice (B) is incorrect because the median for data set A is 15, while the median for data set B is 17. The spread for data set A (25 – 10 = 15) is larger than the spread for data set B (19 – 15 = 4), so the standard deviation and the standard error of the means for those data sets will likely be different. Therefore, choices (C) and (D) are incorrect. 3. (D) The chi-square test is most appropriately used on raw data, such as the numbers of individuals in each category. The values used in the chi-square table have been calculated by statisticians and are designed to be used with raw data. Processed data, such as means and percentages, cannot be used with the chi-square test, making choices (A), (B), and (C) incorrect. 4. (C) Since there are four different phenotypes that could be produced, there are 3 degrees of freedom. Degrees of freedom = number of possible outcomes in the experiment – 1 = 4 – 1 = 3. 5. (D). Using the observed and expected values in the problem:. There are two possible outcomes in the experiment (tall or dwarf), so the degrees of freedom = 2 – 1 = 1. Using the p-value of 0.05 and the chi-square table, the  critical value is 3.84. The calculated chi-square value from the data is less than the critical value, so there is likely no statistically significant difference between the observed data and the expected data. Choices (A) and (B) are incorrect because they have incorrect calculations of the chi-square data. Choice (C) is incorrect because even though the chi-square value from the data is calculated correctly, the wrong conclusion is drawn from the result. 6. (D) The chi-square =. There are two possible outcomes in the experiment (banana or watermelon), so there is one degree of freedom. Using the p-value of 0.05 and the chi- square table, the critical value is 3.84. The calculated chi-square value is less than the critical value, so there is likely not a statistically significant difference between the observed and expected data. Choices (A), (B), and (C) are incorrect because they present the wrong chi-square value and/or the wrong conclusion about the data. 7. (A) Standard error of the mean is calculated by dividing the standard deviation by the square root of the sample size. If the standard deviation is the same for both data sets, the standard error of the mean will decrease when the sample size increases. Since both data sets have the same standard deviation, the data set with the smaller sample size (Data Set X, which has n = 10) will have a larger standard error of the mean than Data Set Y, which has a larger sample size (n = 50). 8. (C) The standard error of the mean is a way to measure the variability or spread of a data set. The larger the standard error of the mean, the greater the variability in the data set. Since data collected in San Rafael has the highest standard error of the mean, San Rafael has the greatest variability in the data. 9. (B) The upper limit of the 95% confidence interval for data from Oakville (25.1) is less than the lower limit of the 95% confidence  interval for data from Oak Valley (30.1). So their 95% confidence intervals do not overlap and there is likely a significant difference between the oak tree heights in those two locations. Choice (A) is incorrect because the 95% confidence interval for the data from Sacramento overlaps with the 95% confidence interval for the data from Oakville (the upper limit for Sacramento of 19.6 is greater than the lower limit for Oakville of 15.1). Similarly, the 95% confidence intervals for data from San Rafael and Oak Valley overlap (the upper limit for San Rafael of 37.3 is greater than the lower limit for Oak Valley of 30.1), so choice (C) is incorrect. Choice (D) is incorrect because the 95% confidence intervals for data from San Rafael and Oakville overlap (the upper limit for Oakville of 25.1 is greater than the lower limit for San Rafael of 20.1). 10. (D) The upper limit of a 95% confidence interval is determined by adding two times the standard error of the mean to the mean of the data set, which is represented by choice (D). The lower limit of the 95% confidence interval is determined by subtracting two times the standard error of the mean from the mean of the data set. Choices (A), (B), and (C) are incorrect because they do not represent the upper limit or the lower limit of a 95% confidence interval. Short Free-Response 11. (A-i) The treatment with cadmium resulted in the greatest variability in the data because it has the greatest standard error of the mean. (B-i) The control treatment was the least likely to result in chromosome breakage. Its mean percentage of cells that showed chromosome breakage was far lower than the mean for the other two treatments. (C-i) The upper limit of the 95% confidence interval for lead (24.3 + 2(5.2) = 34.7) is greater than the lower limit of the 95% confidence interval for cadmium (46.1 – 2(9.1) = 27.9), so the 95% confidence intervals for those two treatments  overlap. When the 95% confidence intervals overlap, the data are inconclusive and it is not possible to claim that there is a statistically significant difference between the groups. The data provided do not support the student’s claim that exposure to cadmium is more likely to result in chromosome breakage than exposure to lead. (D-i) The independent variable is the type of treatment (control, lead, or cadmium). (D-ii) The dependent variable is the mean percentage of cells that showed chromosome breakage. 12. (A-i) There are two possible outcomes in this experiment (green or albino), so the number of degrees of freedom (df) = 2 – 1 = 1. (B-i) (C-i) The observed data are likely statistically significantly different from the expected data. (D-i) With one degree of freedom and using a p-value of 0.05, the critical value from the chi-square table is 3.84. Since the calculated chi-square value (4) is greater than the critical value, there is likely a statistically significant difference between the observed and expected values. Long Free-Response 13. (A-i) The section of the garden with the least variability in aphid density is the section treated with chemical pesticide because its 95% confidence interval is the smallest. (A-ii) The section with the most variability is the section treated with ladybugs because its 95% confidence interval is the largest. (B-i) The 95% confidence interval for the section that received no treatment does not overlap with the 95% confidence intervals for the ladybug section or the chemical pesticide section. Thus, the section that received no treatment is least like the others and is most likely to have a statistically significant difference in aphid density from the ladybug-and chemical pesticide–treated sections. (C-i) If 50 measurements were taken in each section, the 95% confidence intervals would be smaller. (D-i) To calculate 95% confidence intervals, add or subtract two times the standard error of the mean from the mean of the data set. The standard error of the mean is inversely proportional to the sample size (if the sample size increases, the standard error of the mean decreases). So if the sample size, n, increased to 50 for all three sections, the standard errors of the mean for each section would likely decrease, and the 95% confidence intervals would be smaller.  UNIT 1 Chemistry of Life 3 Water Learning Objectives In this chapter, you will learn: Water and the Importance of Hydrogen Bonds pH Overview Living organisms contain more water than any other compound. The environment of most living organisms is dominated by water. Understanding water and its properties is key for the study of life on Earth. This chapter will review water’s unique properties and how these properties affect living organisms. Water and the Importance of Hydrogen Bonds Water is a polar molecule. Its polarity allows it to form hydrogen bonds. These hydrogen bonds give water properties that are essential to life on Earth. Water contains covalent bonds (shared electrons) between the oxygen and hydrogen atoms. The element oxygen has a high electronegativity (ability to attract electrons), while the element hydrogen has a lower electronegativity. Because of this electronegativity difference, the electrons in the covalent bond between oxygen and hydrogen are unequally shared, with the electrons spending more time around the oxygen atom. This results in a polar covalent bond, with a partial negative charge around the oxygen atom and a partial positive charge around the hydrogen atoms, as shown in Figure 3.1. Figure 3.1 Polarity of Water As a result, the partial negative charge on an oxygen atom in one water molecule is attracted to the partial positive charge on a hydrogen atom in another water molecule, resulting in a hydrogen bond. This causes water molecules to be attracted to one another, as shown in Figure 3.2. Figure 3.2 Hydrogen Bonds Between Water Molecules Since water molecules can form hydrogen bonds, they have properties that help sustain life on Earth, including: Exhibiting cohesive and adhesive behavior: Water molecules are “sticky.” They are attracted to other water molecules and to other polar molecules. This is what gives water its unique properties like  water’s high surface tension and its ability to climb up the xylem in plants through capillary action. (See Figure 3.3.) Figure 3.3 Capillary Action Having a high specific heat: As a result of water’s ability to form hydrogen bonds, more energy is required to separate water molecules during phase changes, giving water a high specific heat. When a person sweats, the water in the sweat on the skin absorbs heat from the person’s body as the water/sweat evaporates, having a cooling effect on the person’s body temperature. Moderating climate: Since water has a high heat capacity, it can absorb and release large amounts of energy. This stabilizes climates in locations near large bodies of water. (See Figure 3.4.)  Figure 3.4 Effects of a Large Body of Water on Climate Expanding upon freezing: Since water has the ability to form hydrogen bonds, there is more space between water molecules in the solid state than in the liquid state. As a result, ice has a lower density than that of liquid water, and thus ice floats on liquid water. This has profound consequences for organisms in ponds or lakes that freeze in the winter. The layer of ice on the surface of the lake helps protect the organisms below from temperature extremes in the atmosphere, increasing their chances of surviving the cold winter. If ice were denser than liquid water, the ice would sink, leaving the remaining water in the lake exposed and vulnerable to more freezing and increasing the likelihood that the lake would freeze solid during the winter. This would result in fewer organisms in the lake surviving the winter. Acting as a great solvent for other polar molecules and for ions: Water has a partially positive end and a partially negative end. Thus, water can readily dissolve ionic compounds (see Figure 3.5) and other polar molecules. This makes water an excellent solvent for many biological molecules.  Figure 3.5 Water Interacting with Sodium Chloride pH pH (or “power of hydrogen”) measures the concentration of H+ ions in a solution. The formula for pH is as follows: While you will not be required to calculate pH from [H+] on the AP Biology exam, you should know the following: A pH less than 7 is acidic. A pH greater than 7 is basic. A pH of 7 is neutral. Due to the negative sign in the formula for pH, a higher [H+] leads to a lower pH value and a lower [H+] leads to a higher pH value. Thus, a solution with a pH of 3 would have a higher [H+] than that of a solution with a pH of 5. Also, because the pH scale is a logarithmic scale, a pH change of one unit corresponds to a tenfold difference in H+ concentration. For example, a pH of 3 would have 10 times the H+ concentration of a pH of 4 and 100 times the concentration of a pH of 5. The pH of a water-based solution depends on how many of the water molecules are dissociated (separated into H+ ions and OH- ions) and the relative numbers of these ions. Pure water will dissociate and produce equal concentrations of H+ ions and OH- ions, resulting in a pH of 7. Acids increase the relative concentration of H+ ions in a solution, and bases increase the relative concentration of OH- ions in a solution. TIP Remember, the concentration of hydrogen ions is inversely proportional to pH. If the pH decreased, that means the [H+]  increased. Biological systems can be very sensitive to changes in pH. Buffers are crucial in maintaining relatively constant pH levels in living cells. Buffers can form acids or bases in response to changing pH levels in a cell. An example of this is the carbonic acid–bicarbonate buffering system in blood plasma. The carbon dioxide that is produced by cellular respiration reacts with water in blood plasma to produce carbonic acid (H2CO3). Carbonic acid can dissociate into bicarbonate ions (HCO3-) and hydrogen ions, as shown in the following equation: If the pH of a cell becomes too low (excess H+), the reaction will shift to the left, allowing the basic bicarbonate ions to neutralize the excess H+ ions, returning the pH to normal levels. When the pH of a cell becomes too high (excess OH-), the reaction shifts to the right, adding more H+ ions to the cell that can neutralize the excess OH-, which lowers the pH back to normal levels. Chapter 4 (“Macromolecules”) will review how changes in pH can denature proteins, changing their functions. Chapter 7 (“Enzymes”) will discuss how most enzymes have a pH at which they function optimally. Practice Questions Multiple-Choice 1. In a water molecule, hydrogen atoms are attached to oxygen atoms through which type of bond? (A) hydrogen bond (B) nonpolar covalent bond (C) polar covalent bond  (D) ionic bond 2. The attraction between the partially positive charge on a hydrogen atom on one water molecule and the partially negative charge on an oxygen atom on another water molecule is called a(n) (A) hydrogen bond. (B) nonpolar covalent bond. (C) polar covalent bond. (D) ionic bond. 3. Water’s high specific heat is due to (A) the lower density of solid ice compared to that of liquid water. (B) the amount of energy required to break the covalent bonds within a water molecule. (C) the amount of energy required to break the hydrogen bonds between water molecules. (D) the low electronegativity of oxygen atoms compared to that of hydrogen atoms. 4. Which of the following solutions has the greatest concentration of H+? (A) stomach acid with a pH of 2 (B) acetic acid with a pH of 3 (C) coffee with a pH of 5 (D) bleach with a pH of 12 5. Solution A has a pH of 4; solution B has a pH of 7. How do the [H+] in these solutions compare? (A) Solution A has 3 times the [H+] concentration of solution B. (B) Solution A has 30 times the [H+] concentration of solution B. (C) Solution A has 1,000 times the [H+] concentration of solution B. (D) Solution A has 3,000 times the [H+] concentration of solution B. 6. Coastal areas near large bodies of water tend to have more stable climates than inland areas at the same latitude. Which of the following is the property of water that best explains this difference in climate? (A) high surface tension (B) high specific heat (C) capillary action (D) density of ice 7. Small, lightweight insects can walk on the surface of water, as seen in the following figure: Which of the following is the property of water that best explains this phenomenon? (A) high surface tension (B) high specific heat (C) capillary action (D) density of ice 8. Arctic seals and walruses rely on ice floes for survival. Which of the following best explains why these ice floes exist? (A) high surface tension (B) high specific heat (C) capillary action (D) density of ice  9. Redwood trees, which are over 200 feet tall, can move water upward from their roots to other parts of the tree, despite the downward pull of gravity. Which of the following properties of water best explains this? (A) high surface tension (B) high specific heat (C) capillary action (D) density of ice 10. In hot weather, humans can cool their body temperature by sweating. Which of the following properties of water makes this possible? (A) high surface tension (B) high specific heat (C) capillary action (D) density of ice Short Free-Response 11. On hot summer days, misters will sometimes be used to cool participants at outdoor events. Part A (i) Describe the property of water that allows misters to have an effective cooling effect. Part B (i) Explain why the evaporation of water makes the participants in these events more comfortable. Part C Instead of water, nonpolar oil is spread on the skin. (i) Predict whether this would have a less effective cooling effect, a more effective cooling effect, or the exact same cooling effect as water on the skin.  Part D (i) Using what you know about the comparative properties of water and nonpolar substances, justify your prediction from Part C. 12. Refer to the following figure, which depicts water and methane. Part A (i) Describe the type of bond indicated by arrow A. Part B (i) Explain why the bond indicated by arrow A forms between water molecules. Part C (i) Would an ionic salt dissolve more readily in water or methane? Explain your reasoning. Part D Plants in arid climates often need to conserve water loss due to evaporation through the leaves of the plant. Some plant species have a waxy, nonpolar cuticle on the outer surface of their leaves. (i) A student claims that this waxy cuticle reduces water loss from the leaves. Support the student’s claim with reasoning. Long Free-Response 13. Aquatic animals produce carbon dioxide as a product of cellular respiration. Carbon dioxide combines with water to form carbonic acid (H2CO3), which releases hydrogen ions (H+) into solution. Four test tubes (containing 10 mL of water each and different numbers of aquatic snails) are prepared. pH levels were measured in each tube at the beginning of the experiment and after 20 minutes. The results are shown in the following table. Part A (ii) Explain why tubes B, C, and D all had lower pH levels after 20 minutes. Part B (i) Identify the independent variable in this experiment. (ii) Identify the dependent variable in this experiment. Part C (i) Identify the tube (B, C, or D) that contained 100 times as many H+ ions as that of tube A after 20 minutes. (ii) Explain your reasoning. Part D Aquatic plants, such as Elodea, perform cellular respiration, but they also perform photosynthesis. Photosynthesis removes carbon dioxide from the water, reducing the amount of carbonic acid. (i) Predict the effect of adding Elodea to all four tubes at the start of the experiment. (ii) Justify your prediction. Answer Explanations Multiple-Choice 1. (C) Oxygen atoms and hydrogen atoms are joined in covalent bonds within a water molecule. Because oxygen has a higher electronegativity than hydrogen does, the electrons in this bond are unequally shared, and this results in a polar covalent bond. Choice (A) is incorrect because hydrogen bonds occur between different water molecules, not within a given water molecule. Choice (B) is incorrect because nonpolar covalent bonds form between atoms with similar electronegativities. Choice (D) is incorrect because the electrons in the bonds within a water molecule are shared; they are not transferred as is the case in an ionic bond. 2. (A) Hydrogen bonds occur between the hydrogen atom of one water molecule and the oxygen atom of another water molecule. Choices (B) and (C) are incorrect because both describe intramolecular bonds, not intermolecular bonds between different molecules. Choice (D) is incorrect because there is no transfer of electrons between water molecules. 3. (C) Because of the attraction between water molecules that is the result of hydrogen bonds, more energy is required to separate water molecules. Choice (A) is incorrect because although solid ice does have a lower density than that of liquid water, this does not affect water’s specific heat. Choice (B) is incorrect because specific heat does not depend on the energy needed to break the intramolecular covalent bonds within a water molecule; rather, specific heat depends on the energy needed to break the intermolecular bonds between molecules. Choice (D) is incorrect because oxygen atoms have a higher electronegativity than that of hydrogen atoms. 4. (A) Of the choices presented, a stomach acid with a pH of 2 has the greatest concentration of H+. All the other answer choices have a  higher pH value than that of choice (A) and therefore a lower [H+]. 5. (C) pH is a logarithmic scale, so a difference of three pH units would result in a difference of 103 times the [H+] concentration. Choice (A) is incorrect because pH is not a linear scale. Choice (B) is incorrect because 103 is not 30 times. Choice (D) is incorrect because 103 is not 3,000 times. 6. (B) Water’s ability to form hydrogen bonds allows it to absorb and release large amounts of heat, giving water a higher specific heat than that of many other liquids. Locations near large bodies of water have more stable climates because the nearby bodies of water can absorb atmospheric heat during the day and then release it at night, leading to more stable climates. Choice (A) is incorrect because surface tension only affects the surface of the body of water and does not affect climate. Choice (C) is incorrect because capillary action describes water’s ability to climb up narrow tubes. While ice may cool limited areas, the density of ice does not explain why coastal areas have more stable climates than inland areas, so choice (D) is incorrect. 7. (A) Surface tension describes the attraction of molecules to each other on the surface of a liquid. Since water has strong hydrogen bonds, it has a higher surface tension that is sufficient enough to support the mass of very lightweight insects on its surface. Choice (B) is incorrect because specific heat is not involved in supporting the mass of an insect on the surface of water. Choice (C) is incorrect because capillary action describes water’s ability to climb up narrow tubes. While the density of ice is less than the density of liquid water, this does not explain why lightweight insects can walk on the surface of water, so choice (D) is incorrect. 8. (D) Ice has a lower density than that of liquid water, which allows ice floes to float on the ocean surface. Choice (A) is incorrect because surface tension describes the attraction between molecules of water at the surface of a liquid. Specific heat does not contribute to the creation  of ice floes, so choice (B) is incorrect. Choice (C) is incorrect because capillary action describes water’s ability to climb up narrow tubes. 9. (C) Capillary action describes water’s ability to climb up narrow tubes. The trunks of redwood trees contain many narrow tubes made of xylem cells, which allow water to travel from the roots to the rest of the tree. Choice (A) is incorrect because surface tension describes the attraction between molecules of water at the surface of a liquid. Specific heat is not involved in capillary action, so choice (B) is incorrect. Choice (D) is incorrect because the density of ice has nothing to do with water’s ability to move upward from the roots to the rest of the tree. 10. (B) Due to water’s high specific heat, more energy is required to evaporate water molecules than other liquids. Sweat on the skin absorbs heat from the body as it evaporates, lowering body temperature. Choice (A) is incorrect because surface tension describes the attraction between molecules of water at the surface of a liquid. Choice (C) is incorrect because capillary action describes water’s ability to climb up narrow tubes. Choice (D) is incorrect since ice is not involved in sweating. Short Free-Response 11. (A-i) The polarity of water molecules leads to the formation of hydrogen bonds between water molecules, so water requires more energy to evaporate than molecules that do not form hydrogen bonds. (B-i) Energy is required to break the hydrogen bonds between water molecules before those molecules can evaporate. As the hydrogen bonds are broken, heat energy is absorbed from the participant’s body, which has a cooling effect. (C-i) A nonpolar molecule would have a less effective cooling effect.  (D-i) Nonpolar molecules do not form hydrogen bonds between them, so they would require less energy to evaporate and therefore have a less effective cooling effect. 12. (A-i) Arrow A indicates a hydrogen bond between water molecules. (B-i) Oxygen has a much greater electronegativity than hydrogen does. So the electrons in the covalent bond between oxygen and hydrogen in a water molecule are not shared equally and form a polar covalent bond. This gives the oxygen atom a partially negative charge and the hydrogen atom a partially positive charge. (C-i) An ionic salt would dissolve more readily in water because the polar water molecules could form hydration shells around the ions, as shown in the following figure. Since methane is nonpolar, it could not form hydration shells around the ions. (D-i) Polar water molecules cannot cross a waxy, nonpolar cuticle layer, so less water can evaporate from leaves surrounded by a waxy, nonpolar cuticle. Long Free-Response 13. (A-i) The aquatic snails in tubes B, C, and D all produced carbon dioxide. The carbon dioxide combined with the water in the tubes to form carbonic acid, which  released H+ ions into the solution and lowered the pH. The more H+ ions there are, the lower the pH is. (B-i) The independent variable in this experiment is the number of aquatic snails in each tube. (B-ii) The dependent variable is the pH. (C-i and ii) The equation for pH (pH = –log[H+]) is a logarithmic scale, which means each pH change of 1 unit will increase the number of H+ ions by a factor of 10. A decrease of 2 pH units would indicate a 100-fold increase in H+ concentration. Since tube A has a pH of 7.0 after 20 minutes, it makes sense that tube C, with a pH of 5.0, would have 100 times the H+ concentration as that found in tube A. (D-i and ii) In tube A, the pH will increase after 20 minutes. The Elodea in tube A will remove carbon dioxide by the process of photosynthesis and therefore increase the pH. In tubes B, C, and D, which contain aquatic animals as well as Elodea, the pH will still decrease after 20 minutes since the animals are performing cellular respiration. However, since the Elodea will absorb some of the carbon dioxide produced, the decrease in pH will not be as large as it was when the tubes only contained aquatic snails. 4 Macromolecules Learning Objectives In this chapter, you will learn: Biological Macromolecules Protein Structure Nucleic Acids Overview Biological macromolecules form the basis for the structure and function of living organisms. This chapter reviews the basic reactions that form and break down these molecules. Then, the structure and function of carbohydrates, lipids, nucleic acids, and proteins are reviewed, with special attention given to the four levels of structure found in proteins. Biological Macromolecules The macromolecules necessary for life are primarily made of six elements: nitrogen, carbon, hydrogen, oxygen, phosphorus, and sulfur. Carbon is the “backbone” of these molecules. Carbon has four valence electrons and is extremely versatile in the way it can bond to other atoms. It can form single, double, or even triple bonds. Carbon can also form linear, branched, or ring-type structures. Carbon is found in all types of macromolecules.  TIP An easy mnemonic device to help you remember these elements is NCHOPS for Nitrogen, Carbon, Hydrogen, Oxygen, Phosphorus, Sulfur. Oxygen and sulfur each have six valence electrons, and each of these elements typically forms two bonds. Oxygen is found in all types of macromolecules. Sulfur is typically found in proteins. Nitrogen and phosphorus each have five valence electrons, and each of these elements typically forms three bonds. Nitrogen is found in nucleic acids and proteins. Phosphorus is found in nucleic acids and some lipids. Hydrogen has one valence electron and forms a single bond. Hydrogen is found in all types of macromolecules. In fact, hydrogen atoms are so ubiquitous that often they are not drawn in molecular structures. The structure and function of a macromolecule is determined by the types of monomers from which it is made and how the monomers are linked. TIP While you do not need to know the structures of specific molecules for the AP Biology exam, you do need to be familiar with the different types of structures and functions of macromolecules. Making and Breaking Biological Macromolecules Biological macromolecules are formed from building blocks (i.e., monomers) that are linked by dehydration synthesis to form larger molecules. These biological macromolecules are broken down by hydrolysis reactions. (See Figure 4.1.)  Figure 4.1 Dehydration Synthesis and Hydrolysis Carbohydrates Carbohydrates are polymers of sugar monomers. The types of sugars used to make the carbohydrate and how the sugars are linked determines the structure and function of the carbohydrate. The sugars may be joined in linear structures or in branched chains. Carbohydrates can be used to store energy (such as in starch or glycogen) and can also have structural functions (such as in cellulose). The type of linkages between the sugars in carbohydrates that store energy is different from the type of linkages found in carbohydrates that have a structural function. (See Figure 4.2.)  Figure 4.2 Carbohydrates Lipids Lipids are nonpolar macromolecules that function in energy storage, cell membranes, and insulation. (See Figure 4.3.) One of the building blocks of lipids is fatty acids. Fatty acids with the maximum number of C–H single bonds are called saturated, are solid at room temperature, and usually originate in animals. Fatty acids with at least one C=C double bond are called unsaturated, are liquid at room temperature, and usually originate in plants. How a lipid functions in a cell is dependent on the lipid’s saturation level. Phospholipids are extremely important in cell membranes. They are built from a glycerol molecule, two fatty acids, and a phosphate group. Because the fatty acids are nonpolar and the phosphate is polar, phospholipids are amphipathic, meaning they have both hydrophobic and hydrophilic regions.  Figure 4.3 Lipids Another class of lipids are steroids. Steroids are relatively flat, nonpolar molecules. Many steroids are formed by modifying cholesterol molecules. Examples of steroids include estradiol, testosterone, and cortisol. Nucleic Acids Nucleic acids are polymers of nucleotides and function as the carriers of genetic information. Nucleotides will be discussed in more detail later in this chapter. Proteins Proteins are polymers of amino acids. Amino acids have an amino group, a carboxylic acid group, a hydrogen atom, and a side chain (R-group) attached to a central carbon, as shown in Figure 4.4. The R-group is unique for each amino acid; it determines the amino acid’s identity and whether the amino acid will be nonpolar, polar, acidic, or basic. Proteins function in enzyme catalysis, maintaining cell structures, cell signaling, cell recognition, and more. Figure 4.4 Structure of an Amino Acid Protein Structure There are four levels of protein structure, as shown in Figure 4.5. Figure 4.5 Levels of Protein Structure Primary Structure Amino acids are joined by peptide bonds, as shown in Figure 4.6. The resulting polypeptide chains have directionality, with an amino (NH2) terminus and a carboxyl (COOH) terminus. The order of the amino acids in the polypeptide chain determines the primary structure of the protein. Figure 4.6 Peptide Bond TIP Changes in pH can disrupt hydrogen bonding and ionic interactions between amino acids in a protein, causing denaturation of the protein. This can result in a change in shape, and, consequently, reduced function of the protein. A change in the primary structure of a protein may have severe effects on the function of the protein, as seen in sickle cell disease. Just one amino acid is changed out of over 140 amino acids in sickle cell hemoglobin. Secondary Structure Once the primary structure is formed, hydrogen bonds may form between adjacent amino acids in the polypeptide chain. This drives the formation of the secondary structure of the protein. These secondary structures include alpha