Biological Science, 6th Edition Scott Freeman Solution Manual

Type:

Solution Manual

Resource:

Biological Science

Edition: Author(s):

6th Edition Scott Freeman Kim Quillin Lizabeth A. Allison Michael Black Greg Podgorski Emily Taylor Jeff Carmichael

For exam and test prep, contact excellentessaywriters@gmail.com

INSTRUCTOR GUIDE

Melissa Reedy UNIVERSITY OF ILLINOIS AT URBANA

Brian Bagatto, PhD UNIVERSITY OF AKRON

Biological Science SIXTH EDITION

Kim Quillin Lizabeth Allison Michael Black Greg Podgorski Emily Taylor Jeff Carmichael

Contents

CHAPTER 1

Biology and the Tree of Life

1

CHAPTER 2

Water and Carbon: The Chemical Basis of Life

9

CHAPTER 3

Protein Structure and Function

23

CHAPTER 4

Nucleic Acids and the RNA World

29

CHAPTER 5

An Introduction to Carbohydrates

37

CHAPTER 6

Lipids, Membranes, and the First Cells

47

CHAPTER 7

Inside the Cell

56

CHAPTER 8

Energy and Enzymes: An Introduction to Metabolic Pathways

70

Cellular Respiration and Fermentation

78

CHAPTER 9

CHAPTER 10 Photosynthesis

90

CHAPTER 11 Cell-Cell Interactions

102

CHAPTER 12 The Cell Cycle

109

CHAPTER 13 Meiosis

120

CHAPTER 14 Mendel and the Gene

128

CHAPTER 15 DNA and the Gene: Synthesis and Repair

141

CHAPTER 16 How Genes Work

148

CHAPTER 17 Transcription, RNA Processing, and Translation

155

CHAPTER 18 Control of Gene Expression in Bacteria

170

CHAPTER 19 Control of Gene Expression in Eukaryotes

177

CHAPTER 20 The Molecular Revolution: Biotechnology and Beyond

189

CHAPTER 21 Genes, Development, and Evolution

201

CHAPTER 22 Evolution by Natural Selection

214

CHAPTER 23 Evolutionary Processes

223

CHAPTER 24 Speciation

233

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CHAPTER 25 Phylogenies and the History of Life

241

CHAPTER 26 Bacteria and Archaea Protists Green Algae and

248

CHAPTER 27 Land Plants Fungi An Introduction to Animals

259

CHAPTER 28 Protostome

Animals Deuterostome Animals

270

CHAPTER 29 Viruses Plant Form and Function Water and

284

CHAPTER 30 Sugar Transport in Plants Plant Nutrition Plant

295

CHAPTER 31 Sensory Systems, Signals, and Responses Plant

306

CHAPTER 32 Reproduction and Development Animal Form

316

CHAPTER 33 and Function Water and Electrolyte Balance in

327

CHAPTER 34 Animals Animal Nutrition Gas Exchange and

338

CHAPTER 35 Circulation Animal Nervous Systems Animal

349

CHAPTER 36 Sensory Systems Animal Movement Chemical

360

CHAPTER 37 Signals in Animals Animal Reproduction and CHAPTER 38 Development The Immune System in Animals CHAPTER 39 An Introduction to Ecology Behavioral Ecology CHAPTER 40 Population

Ecology

Community

Ecology

CHAPTER 41 Ecosystems and Global Ecology Biodiversity and CHAPTER 42 Conservation Biology CHAPTER 43

384 396 407 417 429 441 459

CHAPTER 44

471

CHAPTER 45

481

CHAPTER 46

494

CHAPTER 47

510

CHAPTER 48

521

CHAPTER 49

534

CHAPTER 50

542

CHAPTER 51

551

CHAPTER 52

561

CHAPTER 53

571

CHAPTER 54

iv

369

CONTENTS

CHAPTER

1

Biology and the Tree of Life

Learning Objectives: Students should be able to...

• Describe the five fundamental characteristics of life.

• Describe the three unifying ideas of biology.

• Explain how biologists use predictions of the theory of evolution and genetic sequence analysis to construct a tree of life.

• Describe key steps in the scientific process.

Lecture Outline I. What Does It Mean to Say That Something Is Alive?

A. All living organisms share five fundamental traits: 1. Organisms are made up of membrane-bound cells.

2. Organisms can reproduce.

3. Populations of organisms evolve.

4. Organisms have hereditary information encoded in genes.

5. Organisms acquire and use energy.

II. Life Is Cellular A. Are all organisms made of cells?

1. Cells were first described and identified in cork tissue (Hooke, 1665) and in water and a variety of living tissues (van Leeuwenhoek). (Fig. 1.1) 2. Scientists have examined thousands of plant and animal samples and have concluded that all organisms are composed of cells.

3. A cell is an organized compartment bounded by a thin, flexible plasma membrane and contains concentrated chemicals in an aqueous solution.

4. Most chemical reactions important to life occur inside cells.

5. Cells reproduce via cell division.

B. Where do cells come from?

1. Two components of theories: observable pattern and the mechanism or process that creates the pattern 2. Spontaneous generation hypothesis (Virchow) versus all-cells-from-cells (Pasteur) hypothesis a. Pasteur’s experiment supported the all-cells-from-cells hypothesis. (Fig. 1.2)

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3. All individuals in a population of organisms are related by ancestry.

4. All cells in a multicellular organism are descended from the same ancestral cell.

C. Statement of the cell theory: All organisms are made of cells, and all cells come from preexisting cells.

1. Theories are explanations for phenomena or observations supported by evidence.

2. Chemical evolution: Biologists have evidence that life arose from non-life early in life’s history through this process.

D. Life replicates through cell division.

1. New cells arise when preexisting cells split.

2. All species are connected by common ancestry was also a founding idea published in the same year as all-cells-from-cells hypothesis.

III. Life Evolves A. What is evolution?

1. Species are related by common ancestry (pattern component).

2. The characteristics of a species can be modified from generation to generation (process component). a. Darwin and Wallace (1858) proposed that this happens by natural selection (descent with modification, Darwin).

3. Evolution is a change in heritable characteristics of a population over time.

4. Population is a group of individuals of the same species living in the same area at the same time.

B. What is natural selection?

1. Natural selection occurs whenever two conditions are met: a. Individuals within a population vary in characteristics that are heritable. b. Certain heritable traits help individuals survive and produce offspring.

2. How do these two conditions lead to evolution? a. If certain heritable traits help individuals produce more offspring, then those traits become more common (more frequent) in the population over time.

3. Natural selection acts on individuals, but evolutionary change affects populations as a whole.

4. Speciation: In recent decades, natural selection has caused populations of one species to diverge and form new species.

5. Fitness is the ability to produce offspring.

6. Adaptation is a trait that increases an individual’s fitness in a particular environment.

IV. Life Processes Information A. Chromosome Theory of Inheritance (1902): Inside cells, hereditary or genetic information is encoded in genes, the units located on chromosomes.

1. Chromosomes consist of a molecule of deoxyribonucleic acid (DNA).

B. The Central Dogma (DNA  RNA  Protein) (Fig. 1.5) 1. Watson/Crick double-stranded helix (Fig. 1.4) 2. DNA encodes a message in the sequence of its building blocks: A, T, C, and G.

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3. A pairs with T, C pairs with G. 4. DNA is made into RNA (ribonucleic acid), messenger RNA is read to make a protein.

C. Life requires energy.

1. Two fundamental nutritional needs are required by organisms: ATP (adenosine triphosphate) and obtaining molecules that can be used as building blocks for the synthesis of complex functions of the cell.

V. The Tree of Life A. Evolution leads to speciation, the generation of new species.

1. This implies that all species come from preexisting species and that their ancestry can be traced back to a single common ancestor.

2. Therefore, we should be able to reconstruct the tree of life—a family tree of all organisms.

B. How can we use molecules to understand the tree of life?

1. Evolutionary relatedness should be reflected in molecular similarities.

2. Woese and colleagues proposed using RNA to assess the relatedness (phylogeny) of all living groups of organisms.

C. What does it mean to analyze genetic variation?

1. Looking at genetic variation means looking at the DNA or RNA of that organism and its close relationship to related species. (Making Models 1.1 and plant/algae example) 2. The goal in analysis is to then produce a diagram, a phylogenetic tree, that describes those organisms being compared. (Fig. 1.6) D. How is the tree of life estimated from genetic data?

1. Construction of a phylogenetic tree is done using computer programs to find the arrangement of branches that is most consistent with the similarities and differences observed in the genetic data.

2. The tree of life contains a diverse array of species with a main node that extends to the last universal common ancestor (LUCA). a. There are three major lineages in the tree of life: Bacteria, Archaea (barely known prior to this analysis), and Eukarya. (Fig. 1.7) b. Fungi are more closely related to animals than to plants. c. The data did not support older models, such as the five-kingdom model.

E. The tree of life is a work in progress. a. Students should understand that the lines in the tree of life, specifically the length of the lines between species, reflect the variation in nucleotide sequences between species. (BioSkills 13) F. How should we name branches on the tree of life? (BioSkills 15) 1. The three major lineages of life⎯Bacteria, Archaea, and Eukarya⎯are called domains.

2. A phylum (plural: phyla) is a major lineage within one of the three domains.

3. Several (or more) distinct phyla per domain are each represented by distinct major branches on the tree of life.

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4. Each species has a unique, formal, two-part (Genus species) scientific name. This system was developed by Linnaeus (1735) and is still used today.

VI. Doing Biology A. What is the nature of science?

1. Biologists test the predictions made by alternative hypotheses, either by making observations or by setting up carefully designed experiments.

2. Biologists explore only those types of questions that can be tested by collecting data from the observable world.

3. Are science and religion compatible? a. Science explores the what (patterns) and how (processes) questions of life. b. Religion explores why we exist and how we should live.

B. Why do giraffes have long necks? An introduction to hypothesis testing 1. State the hypothesis as precisely as possible, and list the predictions that it makes.

(Example: Giraffes have long necks for food competition.) a. Remember to search for alternative hypotheses, too. (Example: Giraffes have long necks to compete for mates.)

2. Design an observational or experimental study that can test the predictions. (Where do giraffes feed? Prediction for food competition would be high in trees.)

3. Interpret the results. a. If the predictions are accurate, the hypothesis is supported. b. If the predictions are not accurate, the hypothesis as originally stated is not correct.

(Fig. 1.8a) It may need to be modified or discarded altogether in favor of an alternative hypothesis.

4. Example: Why do giraffes have long necks? a. Students should be able to state whether the giraffe study was an observational or experimental study. Students should also be able to develop hypotheses, generate ideas for studies, and interpret hypothetical data for similar questions, such as, “Why do zebras have stripes?”

C. Let’s look at an introduction to experimental design.

1. Experiments allow researchers to test the effect of a single, well-defined factor on a particular phenomenon.

2. Example: Navigation by Saharan desert ants—How do ants find their way back to their nest? (Fig. 1.9) a. Hypothesis: Ants know the distance back based on number of steps taken and stride length. b. Variable tested—stride length (stubs, normal, and stilts) (Figs. 1.10, 1.11) 3. Important features of experimental design: a. It is critical to include control groups to rule out the effect of other factors. (Control with normal ants to ensure that manipulating ants doesn’t alter ability to attain an accurate distance.) b. Control groups and experimental groups are exposed to exactly the same conditions. c. Repeating the test is essential; larger sample sizes reduce the effect of “noise” in the data.

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4. Students should be able to explain: (1) What you would conclude if in the first test the normal individual had not walked 10 m on the return trip before looking for the nest; and (2) What you would conclude if the stilts and stumps ants had not navigated normally during the second test.

Chapter Vocabulary This vocabulary list may be used in several ways, such as to: • Add or modify terms to customize the list to your lectures.

• Remind yourself about concepts to emphasize or clarify during lectures.

• Easily determine which terms or topics are discussed in a chapter, and why.

• Distribute the list to students to use as a study aid for exams.

To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are also important to the current chapter. It also includes terms other than those highlighted in bold type in the chapter text. adaptation Archaea ATP (adenosine triphosphate) cell cell theory central dogma chemical evolution chromosome chromosome theory of control group DNA domain double helix energy eukaryote evolution experiment

experimental study fitness genes genus (s.), genera (pl.) heritable trait hypothesis hypothesis testing inheritance kingdom LUCA microscope natural selection null hypothesis observational study organism phylogenetic tree phylogeny phylum

population prediction prokaryote protein replication RNA scientific name speciation species species spontaneous generation spontaneous generation taxon (s.), taxa (pl.) taxonomy theory tree of life

Lecture Activities Most instructors of introductory college biology classes struggle to find a balance between covering the large volume of material required and engaging the students. These lecture activities are suggestions for ways to engage the students while reinforcing a concept that is foundational to their understanding of the subject at hand. We recognize that class sizes for introductory courses vary depending on the size of the institution. For this reason, we frequently provide different suggestions for both large and small class sizes. Sometimes, the activity ideas are completely different; at other times, we include suggestions on how to modify an activity to make it work with a large class. Following each set of lecture activities is a discussion idea that may help you to engage the students in a simpler way that takes up less class time than an activity does. .

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Developing Hypotheses Estimated duration of activity: 5 to 15 minutes (depending on the depth of discussion) Present students with a set of observations (or ask them to make some observations), and instruct them to develop one or more hypotheses that might explain the observed phenomena.

The observations do not necessarily have to be biological. The goal is first to show students that they make observations, hypotheses, and predictions every day and then to get them to think about how a scientist would formally approach a question to test the hypothesis.

Ideas for observations: • How are students seated in the classroom—clustered, evenly distributed, or some other way? An analogy could be made to population distributions in the wild. Ecologists study these types of population distributions to learn more about a community.

• Why do zebras have stripes? Use the example in the book for this exercise instead of lecturing on it.

• Do antibacterial soaps really work? Have students discuss how they could test this idea.

• Do low-carbohydrate diets put individuals at higher risk for heart disease?

This would be a nice example for discussing why most biological questions have to be answered by using experimental species other than humans.

Ask students what observations they might make and what experiments they might do to test these hypotheses. For an experiment, students must determine the controls they would need in order to achieve valid results.

Practicing BioSkills: Reading Graphs (BioSkills 2) Estimated duration of activity: 10 to 15 minutes (depending on whether time is allotted to have students draw their own versions of the graphs) One of the challenges for introductory students is learning how to read graphs. Figure 1.8a (the giraffe feeding data) shows an interesting type of graph that is essentially a modified histogram. Figure 1.9 (the ant experiment) shows a box-and-whiskers plot, a type of graph commonly used to show behavioral data. This graph will be new to most students. You can use these two graphs to give students practice at interpreting graphs and to drive home the point that the same data can be displayed in many different ways.

First, explain how to read both types of graphs, using these steps outlined in BioSkills 2: 1. What do the axes represent?

2. What do the marks on the graph⎯the lines and boxes⎯represent?

Students may benefit from the following tips: • Note that the giraffe graph shows percentages of total bites taken.

• The ant experiment uses box-and-whiskers plots, interpreted as follows: − The box always encloses the middle 50 percent of the data (25th to 75th percentile). − The line inside the box is always the median, which divides the data into a top half and a bottom half. − In this case, the whiskers enclose 80 percent of the data (10th to 90th percentile). − Often, outliers are shown as isolated dots, but not in this case.

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3. What is the overall trend or message? After students understand how to read both graphs, ask them to make a brief sketch of what each of these graphs would have looked like if the data had been shown in a simple bar graph.

(For the giraffe data, students may use the mode⎯the peak of the line⎯for the height of the bars in their bar graphs. For the ant data, tell students to use the median line⎯the line inside the box⎯for the height of the bar.) Ask students: Which graph is simpler to read? Which gives a “truer” picture of the overall data? Which is a better way to show the data, and why?

Remind students that they can refer to BioSkills 3 for further practice and for examples of other types of graphs. Encourage students by suggesting that reading and understanding graphs will be included on classroom exams or quizzes.

Practicing Taxonomy Estimated duration of activity: 10 minutes To illustrate how biologists classify organisms, take a few minutes to have your students build a classification system of the students in the class. Ask them to start with the largest classification and work toward more specific distinctions. The criteria they suggest may include some of the following: • Appearance: This criterion is analogous to biologists using structural characteristics to classify organisms.

• Academic affiliations (like college major): This criterion is loosely analogous to classifying organisms by looking at the functions of specific appendages or by looking at their ecological roles (terrestrial predator, marine reef-builder, etc.).

• Ethnicity: This criterion is loosely analogous to classifying organisms based on their evolutionary relationships.

As you are diagramming the classifications the students are making, you may draw them using a tree or cladogram, explaining that the base category for this example is a trait they all have in common, whereas the base of a phylogenetic tree is a presumed common origin.

Asking Good Scientific Questions Estimated duration of activity: 10 to 15 minutes Begin with a discussion on the elements of high-quality scientific questions. Introduce the idea that although scientists may begin their inquiry with general questions, they must focus on one or more specific questions in order to make any progress.

1. Have students discuss the following idea in small groups (2 to 4 students): What are the elements of a good scientific question?

2. After a few minutes, ask some groups to report to the class. Guide the class discussion toward sound principles such as these: • Good scientific questions imply specific experiments or observations; in other words, • •

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they can be tested.

Good questions are focused.

Good questions may suggest specific hypotheses.

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3. Present a list of potential scientific questions (see below). Have the groups rate the questions in terms of their utility for conducting scientific research. Have them provide the reasons for their ratings.

Suggested questions: (Note: Some of these examples contain details with which the students may not be familiar. Explain to students that it is not necessary to know all the details in order to be able to judge the quality of the question.)

• How many species of fish are there in a certain lake?

• Do embryonic cells differentiate into specialized cells (e.g., liver cells) by the inactivation of genes or by the physical removal of them from chromosomes?

• What determines the vertical distribution of barnacles on rocks in the intertidal zone?

• Can infections with the HIV virus be treated with drugs that inhibit the viral enzyme • known as reverse transcriptase?

• What makes cells become cancerous?

•

How do steroid hormones stimulate lymphocytes to die?

• Is the anticancer gene known as p53 activated in response to DNA damage initiated by radiation?

• Can animals distinguish colors?

• Are Agkistrodon piscivorus and Crotalus atrox so closely related that they should be • classified in the same genus?

• Does smoking cause cancer?

• What carbon-based compounds are present on Mars?

Do the retinas of rats contain the specialized cells (cones) responsible for color vision in humans?

Do all viruses contain DNA?

Discussion Idea: Virchow and Darwin In 1858, both Virchow and Darwin introduced their ideas on biological ancestry. How do you think these ideas challenged the popular thought of the day? How are the two ideas related to each other, and how have they shaped the field of biology since then? Why is one still socially controversial while the other one is not?

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CHAPTER

2

Water and Carbon: The Chemical Basis of Life

Learning Objectives: Students should be able to...

• Describe the structure of an atom and the role of electrons in forming bonds and interactions between atoms that form molecules.

• Explain how the structure of water molecules relates to water’s unusual properties and to its importance to life.

• Define a chemical reaction and the role of energy and entropy in determining if a reaction •

will be spontaneous or not.

Describe the chemical evolution theory and evaluate the two models proposed to explain

• the process component.

Describe the six major functional groups attached to carbon atoms, their structural formulas, and their basic characteristics.

Lecture Outline • Scientific explanation of how life on Earth began: 1. Pattern component—Both small molecules and complex carbon-containing molecules are required for life.

2. Process component—Simple chemical compounds combined to make more complex chemicals and compounds that eventually led to the first life-form (evolution).

3. This entity became the first life-form from which all life has descended.

• Switch from chemical evolution to biological evolution, which is governed by natural selection.

I. Atoms, Ions, and Molecules: The Building Blocks of Chemical Evolution A. Four atoms make up 96 percent of every organism: 1. Carbon, hydrogen, oxygen, and nitrogen 2. Physical structure of these atoms (C, H, O, and N) affects the function of the molecules that they form.

3. Small molecules formed (water, carbon dioxide, etc.) are building blocks for more complex structures required for chemical evolution and, ultimately, life.

B. Basic atomic structure 1. Structure of an atom (Fig. 2.1a) a. Nucleus .

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(1) Protons: large, in nucleus, positive charge (i) The number of protons gives an atom its chemical identity. (ii) Number of protons does not vary = atomic number. (Fig. 2.2) (2) Neutrons: large, in nucleus, no charge (i) The number of neutrons does not affect the atom’s chemical identity but affects its mass. (ii) Number of neutrons can vary in an element, and these are called isotopes.

(Example: 14C has 6 protons and 8 neutrons and is considered an unstable isotope that decays into 14N when one neutron decays to a proton making 7 protons and 7 neutrons.) (iii) Number of protons + number of neutrons = atomic mass (based on average of all naturally occurring isotopes of an atom). b. Orbitals (1) Electrons: travel in orbital, have negligible mass and a negative charge (2) Most of an atom’s volume is empty space found in the orbitals.

(3) Atoms that have the same number of electrons as protons have no charge (electrically neutral).

2. Measuring mass of electrons, protons, and neutrons a. Dalton—mass of one proton or one neutron equals approximately one Dalton; electrons are so small they are considered to have no mass. b. Carbon has 6 protons and 6 neutrons and an atomic mass of ≈12 daltons (12C). c. If carbon had 6 protons and 7 neutrons, it would have an atomic mass of 13 (13C); 12C and 13C are isotopes.

3. Movement of electrons around the nucleus of an atom a. Electrons move around the nucleus in orbitals. b. Each orbital has two electrons. c. Orbitals are organized into shells numbered 1, 2, 3, etc. d. Orbitals are more distant from the nucleus as the shell number increases. e. Each shell has a specific number of orbitals (a shell with one orbital can thus hold two electrons, while a shell with four orbitals can hold eight electrons). f. Electrons fill the shells closest to the nucleus first before filling any outer shells.

4. The outermost shell is called the valence shell and contains valence electrons. a. Electrons in the outermost shell of the highlighted atoms in Figure 2.3 have unpaired electrons (at least one) in their valence shell. b. Number of unpaired electrons = atom’s valence. (Example: Carbon has four valence electrons.) c. An atom is most stable when its valence shell is filled with electrons. d. Filling outer shells can be accomplished by formation of chemical bonds that bind atoms together. e. Sharing electrons between atoms results in the formation of a covalent bond.

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C. How does covalent bonding hold molecules together?

1. How does a hydrogen atom, with one unpaired valence electron, become stable? The outer shell can only hold a total of two electrons. a. By pairing together two hydrogen atoms, each atom can share its electron with the other atom’s valence shell. b. By sharing, both outer shells are now full. c. Sharing of electrons between two hydrogens results in H–H or H2, a molecule of hydrogen. (Fig. 2.4) d. A molecule results when covalent bonds are formed by sharing electrons between atoms (two or more). e. Covalent bonds result from the attraction between protons in the atom’s nucleus and the surrounding electrons; the attraction must be stronger than the repulsion of positively charged nuclei to each other or repulsion of electrons by one another to make a covalent bond.

2. Equal sharing of electrons between atoms results in a covalent bond that is nonpolar.

Examples: H−H bonds and C−H bonds. (Fig. 2.5a) 3. Unequal sharing of electrons (if the electrons are pulled closer to one of the atom’s nuclei) results in a covalent bond that is polar. (Fig. 2.5b) a. The tendency of an atom to hold electrons tightly is referred to as electronegativity and results from the number of protons in the nucleus and the distance of the shells from the nucleus. b. Atoms with increasing electronegativity are observed on the periodic table when moving from left to right in any given row (more electrons). c. Atoms decrease in electronegativity on the periodic table going down any column (increasing distance from nucleus). d. Polar covalent bonds result in partial charges (δ) on certain parts of the molecule.

(Example: water) (Fig. 2.5b) 4. Single bonds share one pair of electrons, double bonds share two pairs, and triple bonds share three pairs. (Fig. 2.8b, c) D. Ionic bonding, ions, and the electron-sharing continuum 1. In an ionic bond, the electron(s) is lost entirely from one atom and donated to the other atom, usually resulting in a filled outer shell for each atom. (Fig. 2.6a) 2. The donor atom carries an extra (+) charge (cation); the recipient atom carries one less (–) charge (anion). (Example: sodium chloride) (Fig. 2.6b, c) 3. A sodium chloride molecule, where the sodium ions and the chloride ions pack together tightly, results from the strong attraction between the ions that create the salt crystals.

4. The degree of sharing of electrons exists on a continuum; equal sharing of covalent nonpolar bonds, unequal sharing of polar covalent bonds, and complete sharing of ionic bonds. (Fig. 2.7) E. Some simple molecules formed from C, H, N, and O 1. There are four atoms required for most of life molecules.

2. Valence shells of each show how many covalent bonds each atom can form.

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a. H = 1 b. O = 2 c. N = 3 d. C = 4 3. Molecules form from single, double, and triple bonds between atoms. (Fig. 2.8) a. Methane—CH4 (single bonds) b. Ammonia—NH3 (single bonds) c. Water—H2O (single bonds) d. CO2 (double bonds) e. N2 (triple bonds) 4. Students should be able to draw arrows between the atoms in each molecule shown in Figure 2.8 to indicate the relative position of the shared electrons. If the electrons are equally shared, then students should draw a double-headed arrow.

F. The geometry of simple molecules 1. The shape of molecules depends on the atoms bound to one another and the shapes of the electron orbitals for each bound atom.

2. CO2 and N2 are linear.

3. Methane (CH4) is tetrahedral, with the electron pairs in each C–H orbital trying to get as far away from one another as possible. (Fig. 2.9a) 4. Water is a bent two-dimensional structure. (Fig. 2.9b) a. Four orbitals are in oxygen’s outer valence shell. b. Two orbitals have pairs from the oxygen atom. c. Two orbitals have mixed shared pairs from oxygen and hydrogen. d. All four orbitals repel each other.

G. Representing molecules (Fig. 2.10) 1. Molecular formula shows types and numbers of atoms but nothing about shape.

2. Structural formula display which atoms are bound to one another and what types of bonds bind the atoms together; geometry is also displayed.

3. Ball-and-stick models show a 3-dimensional representation of molecules and the relative size and shape of the overall structure.

4. Space-filling model is the most accurate 3-dimensional spatial depiction of a molecule.

5. Specific colors in models denote particular atoms, with black representing carbon, blue representing oxygen, and red representing nitrogen.

6. Building blocks for chemical evolution existed in volcanic gases, in deep-sea hydrothermal vents, and in the atmospheres of nearby planets.

7. Chemical evolution hypothesis suggests that these building blocks combined into more complex molecules in an aqueous (water-based) environment that facilitated the reactions necessary for this evolution.

II. Properties of Water and the Early Oceans A. Life is based on water.

1. Life arose in an aqueous environment.

2. Every cell is approximately 75 percent water (volume), making it the most abundant molecule in living organisms. (Fig. 2.11) 12

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3. Humans can survive weeks without eating, but only three to four days without drinking water.

B. Water is an efficient solvent—an agent for dissolving substances into a solution.

1. O–H bonds are polar due to high electronegativity of oxygen and display partial positive and negative charges.

2. H2O is a bent molecule because the partial negative charge on oxygen points away from the partial positive charges on the hydrogen atoms.

3. Weak electrical interactions between two water molecules allow the partially negative charge of the oxygen atom of water to be attracted to the partially positive charges on a hydrogen atom on an adjoining water molecule, forming a hydrogen bond.

(Fig. 2.12) 4. Students should understand why the shape of the water molecule results in hydrogen bonding that is stronger than if the water molecule were to be linear.

5. Students should be able to (a) draw a fictional version of Figure 2.12b that shows water as a linear (not bent) molecule with partial charges on the oxygen and hydrogen atoms and (b) explain why electrostatic attractions between such water molecules would be much weaker as a result.

6. Hydrogen bonding in water allows for most charged and polar molecules to form hydrogen bonds and dissolve in water (hydrophilic).

7. Uncharged and nonpolar molecules do not interact with water and are called hydrophobic. (Fig. 2.13) 8. Hydrogen bonds are not as strong as covalent or ionic bonds, but they are critical to life because of the sheer number of them that form between water and hydrophilic molecules.

C. What properties are correlated with water’s structure?

1. Cohesion is the attraction between like molecules. (Fig. 2.14a) a. The meniscus in a tube of water displays water at the bottom of the meniscus that is lower than the sides because of the attraction of water molecules to one another (pulling one another down).

2. Adhesion is the attraction between unlike molecules; the sides of the meniscus with a tube of water results from cohesion between the water molecules and the tube’s walls, pulling the water at the sides up toward the tube.

3. Water displays surface tension due to cohesion; hydrogen bonding between water molecules or with other hydrophilic substances resists forces that would increase surface area, creating an “elastic membrane.” (Fig. 2.14b) 4. Water is denser as a liquid than a solid (there are more water molecules in a given volume of liquid water than the same volume of solid water, or ice). a. Number of hydrogen bonds per water molecule in ice is four, which is more than in liquid water; as such, water molecules in the liquid phase can pack more densely together. b. Ice is less dense than water and floats. Ice stays on top of the water, providing a layer of insulation for the water below; if ice did not float, it would fall to the bottom of lakes and oceans and would never melt. (Fig. 2.15) 5. Water has a high capacity to absorb energy; this means that as heat hits water molecules, it takes a lot of heat to break the hydrogen bonds of the water molecules .

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before those molecules can begin moving more quickly (heat becomes kinetic energy to move the water molecules faster); specific heat is the amount of energy required to raise the temperature of 1 gram of a substance 1˚C. (Table 2.1) a. The amount of hydrogen bonding of a molecule correlates with the molecule’s specific heat; polar molecules display higher specific heat values than nonpolar molecules because it requires more energy to break the increased number of hydrogen bonds. b. Water has a high heat of vaporization because of hydrogen bonding, which prevents the conversion of water from liquid to gas; heat of vaporization is the amount of heat required to change 1 gram of a liquid substance to gas. c. Evaporation of water from your body, when you sweat or put water on your body when it is hot, removes heat as it changes from the liquid to gas phase.

6. The protection of forming molecules in Earth’s ocean during chemical evolution from the high energy of the sun’s radiation is critical to the evolution of higher-order molecules.

7. Table 2.2 summarizes the key properties of water.

D. The role of water in acid–base reactions 1. Water molecules continuously undergo a chemical reaction (when one substance is combined with others or broken down into another substance; or when atoms undergo rearrangements, usually with chemical bonds being broken and made).

– a. Water can undergo dissociation: H 2O  H ++ OH . b. The dissociation of water is completely reversible, denoted by the double arrows. c. Water dissociates into hydrogen (H+) ions and hydroxide ions (OH–); but protons (H+) do not exist by themselves and instead associate with a water molecule to form a hydronium ion (HO+3) in the following reaction: H2O + H 2O  H3O++ OH . – 2. Acids donate protons during a chemical reaction; bases accept protons; water acts as both an acid and a base; however, it is a weak base with very few water molecules donating protons to make hydronium ions.

3. Strong acids readily give up a proton, such as hydrochloric acid (HCl): HCl + H2O – HO+ + Cl. 3 4. Strong bases readily accept protons, such as sodium hydroxide (NaOH): NaOH(aq) → Na+ + OH–and OH– + HO+3 2 H2O. Determining the concentration of protons: a. There is no way to count the actual number of protons. b. A mole is 6.022×1023. c. A mole of a substance is equal to its molecular weight in grams; example of water = 1+1+16 = 18 approximately; thus, 18 g of water would be approximately 6.02×1023 molecules, or 1 mole. d. Substances dissolved in water have a concentration based on molarity; number of moles of a substance in 1 L of water is molarity; 1 mole of protons (H+) in a liter of water would be a 1 molar solution of protons. A 1 L sample of pure water has 1.0×10–7M H+.

5. The pH of a solution determines whether it is acidic or basic. a. Concentration of protons is so small, chemists and biologists use the logarithmic notation, pH = –log [H+]; solving the equation for [H+] = antilog (–pH) = 10–pH. (Fig. 2.16) 14

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b. Solutions with proton concentrations > 1×10–7 M are acidic (pH<7); solutions with proton concentrations <1×10–7 M are basic (pH>7). c. Students should understand that a change of pH from 7 to 5 represents a solution that has 100-fold more protons and is 100 times more acidic. d. Students should be able to calculate the concentration of protons in a solution that has a pH of 8.5. e. Pure water is a standard on the pH scale, and living cells have a pH of approximately 7 (neither acidic or basic, but neutral). f. Living organisms are sensitive to changes in pH. g. Compounds that minimize changes in pH are buffers (weak acids); buffers protect cells and maintain homeostasis. h. Buffers are usually weak acids, willing to give up protons in solution; when many protons are present, acids are regenerated.

– i. Carbonic acid dissociates to bicarbonate and a proton: CH 2O3 CHO3 + H. + j. If carbonic acid and bicarbonate are in equal concentrations, they act as a buffer; when protons are incrementally increased, they combine with bicarbonate and form carbonic acid and the pH does not change; when protons are slightly decreased, carbonic acid dissociates and the pH does not change. k. Students should be able to predict what would happen to the concentration of bicarbonate ions if a base like sodium hydroxide (NaOH) were added to the solution of carbonic acid.

III. Chemical Reactions, Energy, and Chemical Evolution A. Theory of chemical evolution suggests that simple molecules combined to create complex organic (carbon-containing) molecules—proteins, nucleic acids, sugars, and lipids.

1. Environments where chemical evolution occurred a. Atmosphere—gases ejected from volcanoes (CO2, N2) with water vapor, H2, and CO. b. Deep-sea hydrothermal vents—rich in CO2 and H2 and minerals such as nickel and iron.

B. How do chemical reactions happen?

1. In a chemical reaction, reactants are converted into products. a. Most common reaction in gases from volcanoes is water and CO2 to produce carbonic acid (acid rain). b. Balanced equation: CO2 (g) + H2O (l)  CH2O3 (aq) demonstrating the conservation of mass in closed system.

2. Most reactions are reversible, denoted by .

3. Chemical equilibrium: a. Rate of forward reaction equals rate of reverse reaction; equilibrium is dynamic but stable; quantities of reactants and products are not necessarily equal. b. Equilibrium can be disturbed by adding or removing reactant or product or by altering the temperature.

4. Reactions that absorb heat are endothermic; reactions that release heat are exothermic.

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C. What is energy?

1. Energy is the capacity to do work or supply heat.

2. Examples of different forms of energy: a. Potential energy = stored energy (Fig. 2.17) (1) The potential energy in chemical bonds is called chemical energy and derives from electrons in outer shells trying to occupy the shell closest to the nucleus of the atom possible; an electron in an outer shell far from the nucleus has more potential energy than one close to the nucleus. b. Kinetic energy = energy of motion (1) The kinetic energy of molecular motion is called thermal energy.

(2) Temperature equals the amount of thermal energy possessed by a molecule (cold objects have slowly moving molecules; hot objects have quickly moving molecules).

(3) Heat—the transfer of energy between two molecules when they meet.

3. First law of thermodynamics: Energy is not created or destroyed; it just changes form or is transferred. a. Chemical evolution resulted from the transfer of energy during the early Earth years, with energy input from radiation, electricity in lightning, and energy stored in chemical bonds.

D. What makes chemical reactions spontaneous?

1. Spontaneous chemical reactions are those that proceed on their own without any added energy. a. Reactions tend to be spontaneous if the products are less ordered than the reactants; example is explosion of nitroglycerin into CO2, N2, and H2O with the release of heat.

(1) Amount of disorder is called entropy.

(2) Entropy increases when products of reaction are more disordered than reactants.

(3) Second law of thermodynamics states that entropy increases in a closed system. b. Reactions tend to be spontaneous if the products have a lower potential energy than reactants.

(1) If electrons are held more tightly in products than in the reactants, then the product has lower potential energy.

(2) Equal sharing of electrons in molecules such as H2 or O2 results in higher potential energy than in H2O, where the electronegative oxygen holds the electrons tighter than in O2.

(3) In the reaction of H2+O2 → H2O, entropy decreases, but the potential energy of the products is much lower than the reactants, and a large drop in potential energy as heat results in the reaction occurring spontaneously. (Fig. 2.18) (4) Spontaneous processes result in lower potential energy, increased disorder, or both. (Fig. 2.19)

IV. Model Systems for Investigating Chemical Evolution A. To probe the kinds of reactions that have set chemical evolution in motion, researchers focus on small molecules that were present on early Earth.

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1. Prebiotic soup model—molecules were synthesized from gases in atmosphere or on meteorites; these molecules ended up in the oceans upon condensation; additional reactions resulted in more complex molecules.

2. Surface metabolism model—surface gases contacted minerals near deep-sea ocean vents and resulted in reactions generating more complex molecules.

C. Early origin-of-life experiments 1. Stanley Miller, a graduate student, designed prebiotic soup experiment by simulating early Earth’s atmosphere and conditions. (Fig. 2.20) a. Large flask with “mini atmosphere” containing CH2, NH3, and H2 b. Connected to small flask containing boiling water (ocean)—added water vapor c. Collected water vapor that cooled and condensed and sent it back into boiling water (mimicking rain) d. Nothing happened without the addition of energy as electrical charges (simulating lightning bolts). e. Solution turned pink to dark red and cloudy due to formation of hydrogen cyanide and formaldehyde.

D. Recent origin-of-life experiments 1. Miller experiment criticized for including CH4 and NH3 2. Synthesis of molecules (precursors of life) using light energy 3. Synthesis of formaldehyde from carbon dioxide and hydrogen (CO2 (g) + 2 H2 (g) → CH2O (g) + H2O (g)) may have played a large role in chemical evolution. a. Heating formaldehyde can result in higher-order molecules being formed, such as sugars (requiring a large input of energy). Early Earth was probably bombarded with high-energy photons, which can break molecules apart to form highly reactive free radicals. (Fig. 2.21) b. Ozone was not present in appreciable amounts in early atmosphere and did not protect Earth from high-energy photons. c. Free radicals resulting from high-energy photons hitting molecules are thought to be responsible for some of the key reactions in early chemical evolution; they contain unpaired electrons in their outermost shells, making them highly reactive.

4. Experiments modeling conditions on early Earth indicate that formaldehyde (H2CO) would have formed.

5. Experiments also showed that HCN (hydrogen cyanide) would have formed—another precursor for life. a. Thus, key intermediates in forming larger organic molecules would have formed and rained down into oceans. (Fig. 2.22a) 6. Chemical energy and chemical evolution a. During chemical evolution, the energy in sunlight was converted to chemical energy (the potential energy stored in chemical bonds).

7. This allowed for larger, more organized molecules to be formed from smaller, simpler ones.

8. Hydrothermal vents a. Surface metabolism model to account for the potential dilution of formaldehyde and hydrogen cyanide in Earth’s early oceans

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(1) Reactants recruited to defined space—reactive minerals present on walls of deep-sea vent chimneys (Fig. 2.22b) (2) Minerals play a critical role in rate of reactions as catalysts.

(3) Catalysts influence only the rate of reactions and does not provide energy for reactions.

(4) Example reaction: Carbon dioxide and hydrogen react to synthesize acetic acid (2 CO2 (aq) + 4 H2 (aq) → CH3COOH (aq) + 2 H2O (l)). (i) Spontaneous—driven by energy in reactants (5) Acetic acid can form under the conditions present at hydrothermal vents.

(6) Acetic acid is important in the synthesis of Acetyl CoA (important throughout the tree of life).

(7) Data suggests that minerals from thermal vents are the original source of catalysts used in modern reactions.

(8) Precursors for synthesis of nucleotides (molecules of inheritance) also found in experiments mimicking the surface metabolism conditions

V. The Importance of Organic Molecules A. Carbon atoms provide the structural framework for important compounds for life (excluding water).

1. Carbon atoms can be linked together in long chains such as octane in gasoline.

(Fig. 2.23a) 2. Carbon atoms can be linked together in rings, such as in the sugar glucose.

(Fig. 2.23b) B. Functional groups 1. Carbon atoms create the framework for a molecule’s overall shape.

2. Addition of H-, N-, O-, P- —or S-containing groups, which are functional groups—to a carbon skeleton imparts a variety of chemical reactivities to carbon molecules.

(Table 2.3) 3. There are six major functional groups: a. Amino groups function as bases, attracting protons in solution. b. Carboxyl groups function as acids, dropping protons in solution.

(1) Important amino- and carboxyl-containing molecules are amino acids.

(2) NOTE: More than one functional group may occur on any molecule.

(3) Covalent bonds can form between amino- and carboxyl-groups.

(4) Amino- and carboxyl-groups participate in H–bonding. c. Carbonyl groups are found on aldehydes and ketone molecules. d. Hydroxyl groups are highly soluble in water and act as weak acids; hydroxyl groups are polar, tend to form H–bonds, and are soluble in water. e. Phosphate groups have two negative charges and can affect the shape of the molecule or a molecule to which they are transferred; large amounts of chemical energy are stored in phosphate bonds. f. Sulfhydryl groups can link two molecules via disulfide bonds.

4. Functional groups attached to a carbon framework decide how a molecule is going to “behave.” When examining new compounds, do the following:

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a. Look at the overall shape based on carbon atoms. b. Note where covalent bonds localize and where polarities exist. c. Identify functional groups to predict functions associated with the molecule.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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ΔH

electron

meniscus

ΔS acetaldehyde acid acid–base reaction acidity adhesion alcohol aldehyde alkalinity amino group anion aqueous atom atomic number atomic weight ball-and-stick model base buffer carbon carbonyl group carboxyl group carboxylic acid catalyst cation chemical bond chemical energy chemical equilibrium chemical evolution chemical reaction cohesion concentration covalent bond dalton disulfide bond double bond

electron shell electronegativity element endergonic endothermic energy entropy evolution exergonic exothermic first law of thermodynamics formaldehyde free radicals functional groups Gibbs free energy heat heat of vaporization homeostasis hydrogen hydrogen bond hydrogen ion hydrophilic hydrophobic hydrophobic interactions hydroxide ion hydroxyl group inorganic molecules ion ionic bond isotope ketone kinetic energy LUCA mass number mechanical energy

molarity mole molecular formula molecular weight molecule neutron nitrogen nonpolar covalent bond nonspontaneous chemical reactions nucleus orbital organic compounds organic molecules organic phosphate oxygen ozone pH pH scale phosphate group photons polar covalent bond polarity potential energy prebiotic soup model product proton radioactive isotope reactant second law of thermodynamics single bond solute solution solvent space-filling model

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specific heat spontaneous chemical reactions structural formula sulfhydryl group

surface metabolism model surface tension system temperature thermal energy

triple bond valence valence electron valence shell

Lecture Activities Student-Led Concept Illustrations Estimated duration of activity: A few minutes as you lecture Often, getting students involved is as simple as having them illustrate a phenomenon in their own terms. The following questions can be presented to students, allowing them to explain these concepts in terms that their peers can understand.

Types of chemical bonds: After describing hydrogen, covalent, and ionic bonds, ask the students to come up with nonbiological analogies to these types of bonds. You can start their thinking by comparing the molecules in covalent bonds to partners in a three-legged race.

Another chemical-bond analogy that students tend to identify with is different levels of interpersonal relationships: Hydrogen bonds are analogous to acquaintances, ionic bonds are analogous to dating relationships, and covalent bonds are analogous to marriages. Ask the students to explain why each type of bond can be represented by each type of interpersonal relationship.

The properties of water: After explaining water’s unique properties, ask students to explain the following phenomena: • A leaf can land on the water’s surface, but a rock sinks.

• There is less variation between day and night temperatures at the beach than between day and night temperatures in the desert.

• Salt dissolves in water, but gasoline does not.

• All mammals that live in hot environments keep themselves cool using only two mechanisms: sweating and/or panting.

Energy: After introducing different forms of energy, ask students what sort of energy transformation is occurring in the following common daily activities. (Once they have the idea, you may challenge them to come up with additional examples.)

• Rubbing hands together to get warm (mechanical energy to thermal energy) • Hearing (sound energy to mechanical energy; sound waves cause our eardrums to vibrate) • A plant growing (electromagnetic energy to chemical energy; photosynthesis) • A person riding a bicycle (chemical energy to mechanical energy)

Practicing BioSkills: Using Logarithms (BioSkills 5) Estimated duration of activity: 15 minutes Many introductory students are not familiar with the concepts of logarithms, scientific notation, and molarity. The pH scale is a good opportunity to review all three concepts.

First, model for students how to convert a concentration of hydrogen ions (e.g., 0.001 mole of hydrogen ions per liter) to scientific notation (1 × 10–3) and then to pH (3). Ask students 20

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why there is a 3 in the first number (i.e., in the number 0.001, the 1 is three places away from the decimal point). Then write the formal pH equation: pH = –log[H+] It may be useful to “translate” this mathematical equation into English, such as: “Log is a formal way of asking, What’s the exponent?”

When students understand the basic concepts of logarithms and scientific notation, have pairs of students work on these questions: • What is the pH of a solution that has 0.01 mole of hydrogen ions per liter?

• What is the pH of a solution that has 0.0000001 mole of hydrogen ions per liter?

• A solution has a pH of 4. What is the molarity of H+ ions?

• As the pH gets higher, does the concentration of hydrogen ions get lower or higher?

• A solution changes from pH 1 to pH 2. How much did the hydrogen-ion concentration change?

Allow 5 to 10 minutes, and then poll the class about their results, review the correct answers, and correct any misconceptions. Remind students that they can consult BioSkills 5 for further practice with logarithms.

Concept-Processing Pairs Estimated duration of activity: 10 to 20 minutes (depending on the level of student understanding and the number of items used) This activity has two parts. Students are divided into pairs. One member of each pair is designated A and the other B. Part 1, Paired review: Give students 3 to 5 minutes to clarify with each other the basic concepts of the chapter, such as: • Types of chemical bonds • Polarity • Properties of water • Solubility Part 2, Elaborative questions: Have students take turns asking each other questions that test understanding. Designate who will attempt to answer the first question (i.e., pick either student A or student B to start; this saves time for the student pairs with too much inertia to decide for themselves who will start). The number of questions asked depends on the length of time allotted for the activity. Make sure adequate time is provided for each question.

Students will need time to process their answers, hear your optimal answer, and then discuss their understanding. The purpose of the exercise is to help students recognize and repair misconceptions and holes in their understanding.

A typical agenda for this activity: 1 to 2 minutes: Student A explains the answer to the question (as best as s/he can).

1 to 2 minutes: Student B assesses and amends the answer (as best as s/he can).

1 minute: Instructor shares the optimal answer with the class.

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1 to 2 minutes: Student pairs discuss their answer relative to the answer.

1 to 3 minutes: Instructor entertains questions from the class.

Repeat the process with another question.

Sample questions: • A molecule of octane contains about six times the mass of a molecule of water. Nevertheless, if one pours liquid octane onto liquid water, the octane floats on top of the water.

How can this be?

• A salt crystal dropped into a beaker of water becomes smaller and eventually seems to disappear. However, the same salt crystal remains intact at the bottom of a beaker of octane. Explain.

• A beaker of water is allowed to sit underneath another, larger beaker placed upside down over it (thereby trapping air above the water). After a long period of time, a few air molecules are found in the water and a few water molecules are present in the air above the beaker. However, the air and the water mostly remain separated. Why? • (Follow-up to the preceding question) Air contains mostly nitrogen (N2) and oxygen (O2).

Explain how the structures of these two gases contribute to the observation in the preceding question.

• One liter of water is mixed with 0.2 L of table salt. After the salt is dissolved, the new volume is slightly greater than 1 L but much less than 1.2 L. Why?

Discussion Idea: Calories and Body Temperature Students like to relate what they are learning to the processes that occur in their own bodies.

While discussing Gibbs free energy and the laws of thermodynamics, you can relate these concepts to nutrition and body temperature. Begin by asking students these two questions: • Why do you heat up when you exercise?

• Why is exercising said to “burn calories”?

Give students two hints: (1) In the field of nutrition, the calorie is a unit of energy; that is, if one serving of a food contains 50 “calories,” that food will release 50 kilocalories of energy when broken down by cells. (2) In any spontaneous reaction, at least some of the energy is converted into heat and is “lost” (cannot be captured in chemical bonds). With those two hints, students should be able to reach the following conclusions: • The oxidation of food releases energy that is captured by cells in the bonds of ATP; that is, chemical energy in food molecules is converted into chemical energy in ATP molecules.

Some energy is lost as heat.

• When you exercise, the energy in the bonds of ATP is used to fuel muscle contraction.

Again, some of this energy is lost as heat.

• The more molecules of ATP a muscle uses, the more energy is lost as heat. Muscles rapidly heat up. They also rapidly “run out” of ATP molecules and soon need more food molecules to produce more molecules of ATP. (Students may be interested to learn that at maximum contraction, a muscle will run out of ATP molecules in just a few seconds.

This is why athletes can sustain a maximum contraction⎯such as in Olympic weight lifting⎯for only a few seconds.)

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CHAPTER

3

Protein Structure and Function

Learning Objectives: Students should be able to...

• Describe how the structure of an amino acid affects its function and how amino acids polymerize to form polypeptides.

• Describe the four levels of protein structure and predict how changes to one level may affect the structure of others.

• Provide examples to show how protein folding is directly related to function.

• Describe the various roles of proteins in living systems.

Lecture Outline A. What type of molecule was responsible for the origin of life?

1. Stanley Miller and later researchers repeatedly discovered early molecules containing amino acids.

2. Amino acids found in meteorites and in experiments that approximate environment of interstellar space.

I. Amino Acids and Their Polymerization A. The structure of amino acids 1. Most proteins are composed of only 20 different amino acids.

2. All amino acids have the same core structure: (Fig. 3.1) a. A central carbon with four bonds, attached to these four components: (Fig. 3.1) (1) A hydrogen atom (H) (2) A basic amino group, NH2 (3) An acidic carboxyl group, COOH (4) An R-group, also called a side chain 3. At +pH 7, NH attracts pro ton to form NH3 and carboxyl acts as an acid (COO) – 2 losing a proton easily 4. Importance of charges a. Amino acids stay in solution b. Affect on reactivity B. The nature of side chains 1. The side chain makes each of the 20 amino acids unique. (Fig. 3.2) 2. Side chains can have functional groups that affect reactivity.

3. Polarity affects solubility. Polar side chains are hydrophilic, and nonpolar side chains are hydrophobic. .

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4. Side chains containing carboxyl, sulfhydryl, hydroxyl, or amino groups tend to be both soluble and reactive.

5. Types of amino acids grouped as: charged (acid/base), uncharged polar, nonpolar 6. Students should be able to use Figure 3.2 to order the following amino acids from most hydrophilic to most hydrophobic: valine, aspartate, asparagine, and tyrosine.

Explain why you have chosen this order.

C. How do amino acids link to form proteins?

1. Polymerization is the process of linking monomers (subunits) to form a polymer (a long chain of subunits). (Fig. 3.3) a. A protein is a macromolecule that consists of linked amino acid monomers. b. Polymerization always requires an input of energy.

2. Could polymerization occur in the energy-rich environment of early Earth? a. Polymerization occurs via condensation (dehydration) reactions, in which a molecule of water is lost and energy must be added. (Fig. 3.4a) b. Hydrolysis breaks monomers by adding water, with a release of energy. (Fig. 3.4b) c. In most solutions, hydrolysis predominates (polymers break down). d. Polymerization can happen spontaneously in some situations, such as on clay particles and near undersea volcanoes. This may have occurred on ancient Earth.

3. The peptide bond a. A peptide bond is the C−N bond between the carboxyl group of one amino acid and the amino group of the next amino acid. (Fig. 3.5) b. Peptide bonds are unusually stable and help provide a structural framework or “backbone” for the entire molecule. (Fig. 3.6a) (1) The side chains extend out from the backbone.

(2) The backbone has directionality, with a free amino group at one end and a free carboxyl group at the other end. (Fig. 3.6b) (3) The backbone is flexible. (Fig. 3.7)

II. What Do Proteins Look Like?

A. Protein structure 1. Proteins differ in size, shape, and the chemistry of the amino acids. (Fig. 3.8) 2. A protein’s structure has four basic levels of organization. (Table 3.1) B. Primary structure is the amino acid sequence.

1. Every protein has a unique sequence of amino acids.

2. The sequence of amino acids dictates all higher levels of protein structure due to the nature of the R-groups present.

3. Even small changes in the amino acid sequence can result in significant changes in overall protein function. (Example: hemoglobin and sickle-cell anemia.) (Fig. 3.9) C. Secondary structure emerges when parts of the polypeptide backbone interact.

1. Each peptide bond is polar because the amino group has a slight positive charge and the carboxyl group has a slight negative charge.

2. The carboxyl portion of one peptide bond can form a hydrogen bond with the amino portion of another peptide bond. (Fig. 3.10a) 3. The hydrogen bonding can result in two different structures: (Fig. 3.10b, c) 24

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a. α-helix b. β-pleated sheet c. Which one forms, if either, is determined by the primary structure.

D. Tertiary structure is due to side-chain interactions.

1. The secondary structure brings R-groups in close proximity to one another.

2. Depending on their identity, R-groups can interact in the following ways: (Fig. 3.11) a. Hydrogen bonds can stabilize interactions between two polar R-groups. b. Hydrophobic interactions cause hydrophobic side chains to coalesce into a globular mass. c. van der Waals interactions stabilize hydrophobic side chains once they are near each other. d. Covalent bonds called disulfide bridges can occur between the sulfur-containing R-groups of two cysteine residues. e. Ionic bonds can form between oppositely charged R-groups.

E. Quaternary structure arises when different polypeptides interact.

1. Many proteins are actually multiple polypeptide chains (subunits) folded together.

2. The subunits are held together by the same interactions between R-groups that stabilize tertiary structure. The difference is that the interactions are between R-groups on separate polypeptide chains. (Table 3.1) 3. Examples: The Cro protein is a dimer (two polypeptide subunits); hemoglobin is a tetramer (four polypeptide subunits). (Fig. 3.12)

III. Folding and function

1. Most elements of protein structure are based on folding of polypeptide chains.

2. Folding happens spontaneously because the folded molecule is more stable energetically than the unfolded molecule.

A. Normal folding is crucial to function.

1. If a protein is unfolded, or denatured (such as a ribonuclease), it is unable to function normally. (Fig. 3.13) 2. Folding is often facilitated by molecular chaperones (e.g., heat-shock proteins).

B. Protein shape is flexible.

1. Even though proteins have a characteristic folded shape, most are flexible and dynamic, not rigid and static.

2. Protein folding is often regulated. a. Controlling when or where a protein is folded will regulate the protein’s activity.

3. Misfolding can be “infectious.” a. Prions are an unusual class of infectious agents that are improperly folded proteins. b. Example: PrP and spongiform encephalopathy (Fig. 3.14)

IV. Protein Functions Are as Diverse as Protein Structures A. Proteins are crucial to most tasks required for cells to exist.

1. Catalysis—Many proteins are specialized to catalyze, or speed up, a reaction. A protein that functions as a catalyst is called an enzyme.

2. Defense—Antibodies and complement proteins attack and destroy viruses and bacteria that cause disease. .

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3. Movement—Motor proteins and contractile proteins are responsible for moving the cell itself.

4. Signaling—Proteins are involved in carrying and receiving signals from cell to cell inside the body.

5. Structure—Structural proteins make up body components such as fingernails and hair and define the shapes of individual cells.

6. Transport—Proteins allow particular molecules to enter and exit cells and carry them throughout the body.

B. Why are enzymes good catalysts?

1. Enzymes have an active site where reactants (substrates) bind. a. The active site is usually a cleft or cavity on a globular enzyme.

2. How do enzymes and substrates interact to catalyze a reaction? a. An enzyme binds specifically to a particular reactant or set of reactants (lock-andkey model). b. Enzymes often flex or change shape when they bind to a substrate (a flexible “lock”). (Fig. 3.15) C. Did life arise from a self-replicating enzyme?

1. Amino acids were likely abundant during chemical evolution.

2. To achieve the attributes of life, proteins would need to possess information, replicate, and evolve. The information in proteins cannot be used as a template for their own replication and therefore cannot evolve on their own.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. α-helix β-pleated sheet active site amino acids catalyze chemical evolution condensation reaction covalent bonding C-terminus dehydration reaction denaturation dimer disulfide bonds/bridges electrically charged enzyme

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functional group: amino group carboxyl group sulfhydryl group hydroxyl group hydrolysis hydrophilic hydrophobic hydrophobic interactions hyrdrogen bonding ionic bonding lock-and-key model macromolecular machines macromolecule macromolecule

INSTRUCTOR GUIDE FOR BIOLOGICAL SCIENCE, 6e

molecular chaperone monomer nonpolar N-terminus oligopeptide peptide bond polar polymer polymerization polypeptide prebiotic soup primary structure prion proteins quaternary structure

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renaturation residue R-group scientific theory

tetramer van der Waals interactions

secondary structure side chain substrates tertiary structure

Lecture Activities Visualizing Protein Structure Estimated duration of activity: 5 to 10 minutes Many students have a difficult time conceptualizing protein structure and how it is related to protein function. Fortunately, if your classroom has a web-enabled computer projection system, many bioinformatics computer tools now allow us to display 3-D protein structure in class. Browse the websites listed below for some interesting proteins to show in class.

Note especially the University of Massachusetts hemoglobin tutorial, which allows you to rotate the molecule with the mouse. This website can display both normal and sickle-cell hemoglobin, and it has an excellent explanation of how a single amino acid change can cause sickle-cell hemoglobin to aggregate in clumps.

Any other protein you have discussed in class can likely be found in the Protein Database or one of the other sites listed below. Note also that the Protein Database has a “Molecule of the Month” feature that students often find entertaining.

Websites: University of Massachusetts Hemoglobin Tutorial: www.umass.edu/molvis/tutorials/hemoglobin/ Online Macromolecular Museum (OMM): www.clunet.edu/BioDev/omm/gallery.htm Protein Database (PDB): www.rcsb.org/pdb Davidson College Molecular Biology: www.bio.davidson.edu/courses/RasMol.html Note: Most of these websites display protein structure using JMol, which runs in both Mac OSX (Safari 2.0 or higher) and Windows (Firefox is recommended). Some websites may use an older, molecular-display software called CHIME. If necessary, you can download a free CHIME plug-in at www.symyx.com.

Discussion Idea: Prions and Mad Cow Disease Mad cow disease has been in the news repeatedly during the past decade. However, few people really understand the disease and what causes it. This chapter has a short explanation of prions, but students may find it helpful and interesting to discuss prions in more depth and explore how they are related to diseases (in both animals and humans). Here are some suggested issues to discuss, with Internet resources for you to read before the discussion: • What is a prion? How does it “infect” and cause disease? www.hepatitis-central.com/hcv/whatis/prion.html

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• What is mad cow disease? www.cdc.gov/ncidod/dvrd/bse/ • What human diseases are caused by prions? omim.org/entry/176640 www.microbiologybytes.com/virology/Prions.html

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CHAPTER

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Nucleic Acids and the RNA World

Learning Objectives: Students should be able to...

• Describe the structure of a nucleotide and explain how nucleotides polymerize to form nucleic acids.

• Describe the secondary structure of DNA and explain how it allows organisms to store and copy information.

• Compare and contrast RNA and DNA in their structure and function.

• Use experimental evidence to evaluate the hypothesis that RNA was the first “living molecule.”

Lecture Outline A. Life began when chemical evolution led to the product of a molecule that could replicate.

1. RNA World Hypothesis proposes that there was a stage in the evolution of life when RNA stored genetic information and catalyzed its own replication.

2. The point in time when molecules reproduced, chemical evolution ended and biological evolution began.

I. What Is a Nucleic Acid?

A. What is a nucleic acid made of?

1. A nucleic acid is a polymer made up of monomers called nucleotides.

2. One nucleotide consists of a phosphate group, a sugar, and a nitrogenous base.

(Fig. 4.1a) 3. The sugar can be either ribose or deoxyribose. (Fig. 4.1b) a. Ribonucleotides contain the sugar ribose; deoxyribonucleotides contain the sugar deoxyribose. b. On the 2' carbon, ribose has an −OH group; deoxyribose has an −H. 4. The nitrogenous bases (Fig. 4.1c) a. There are two classes of nitrogenous bases: purines and pyrimidines.

(1) Adenine (A) and guanine (G) are purines.

(2) Cytosine (C), uracil (U), and thymine (T) are pyrimidines. b. Ribonucleotides can have the bases A, G, C, and U. c. Deoxyribonucleotides can have the bases A, G, C, and T. 5. Students should be able to diagram a ribonucleotide and a deoxyribonucleotide.

Use a ball for the phosphate group, a pentagon to represent the nitrogenous base.

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Students should also be able to label the 2', 3', and 5' carbons on the sugar molecule, and add the atoms that are bonded to each.

B. Could chemical evolution result in the production of nucleotides?

1. (Stanley) Miller-like simulations have shown that nitrogenous bases and sugars can be synthesized under conditions that mimic prebiotic soup.

2. Recent work in deep sea thermal vents have found minerals preferentially bind to ribose.

3. Experiments to simulate early Earth have yet to synthesize complete nucleotides.

C. How do nucleotides polymerize to form nucleic acids?

1. Nucleic acids form when nucleotides polymerize. a. The polymerization occurs via a condensation reaction that forms a phosphodiester linkage, linking the 5' carbon of one nucleotide to the 3' carbon of the sugar of the next nucleotide. (Fig. 4.2) b. A polymer of ribonucleotides is ribonucleic acid (RNA); a polymer of deoxyribonucleotides is deoxyribonucleic acid (DNA).

2. DNA and RNA strands are directional. (Fig. 4.3) a. One end has an unlinked 5' carbon with a free phosphate group. b. The other end has an unlinked 3' carbon with a free −OH group where new nucleotides can be added. c. The sequence of bases of a strand of DNA or RNA is conventionally written from the 5' to the 3' end.

3. Polymerization requires an energy source. a. The free energy of the nucleotides is first raised by the addition of two phosphate groups to each nucleotide. (Fig. 4.4) b. This raises the potential energy of the substrate molecules enough to make the polymerization reaction possible. c. The potential energy is stored in the bonds between the phosphate (Fig. 4.4a). In nucleic acid, these are activated nucleotides.

4. Could nucleic acids have formed in the absence of cellular enzymes? a. During chemical evolution, activated nucleotides may have polymerized on the surfaces of tiny mineral particles. b. In one experiment, researchers added activated nucleotides to tiny particles and were able to create RNA molecules up to 50 nucleotides long. c. If nucleotides managed to form in the prebiotic soup, they could then probably have polymerized to form DNA and RNA.

II. DNA Structure and Function A. DNA’s primary structure 1. DNA has a sugar-phosphate backbone with four types of nitrogenous bases.

2. DNA’s primary structure is the sequence of nitrogenous bases.

B. What is the nature of DNA’s secondary structure?

1. The discovery of DNA’s secondary structure by Watson and Crick (1953) was one of the greatest scientific breakthroughs of the twentieth century.

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a. Chemists already knew that DNA must have a sugar-phosphate backbone. b. Chargaff found that nucleotides seem to be “paired” in some way: (1) The total numbers of purines and pyrimidines are equal in any given nucleic acid molecule.

(2) The amount of adenine is always equal to the amount of thymine, and cytosine is always equal to guanine. c. Franklin’s and Wilkins’s X-ray crystallography studies suggested that DNA had a helix, with a regular, repeating structure, with certain spacing.

3. DNA strands form an antiparallel double helix. a. Antiparallel strands: By building models, Watson and Crick deduced that the two DNA strands must run in opposite directions (antiparallel) with the nitrogenous bases in the middle. (Fig. 4.5) b. Complementary base pairing: To maintain the correct distance between the two strands, a purine must always pair with a pyrimidine. This allows hydrogen bonds to form between A and T and between G and C. (Fig. 4.5a, b) c. The patterns of hydrogen bonding form when an opposite strand is flipped 180 degrees. Strand 1 is 5'  3' and Strand 2 is 3' 5' making the strands anti-parallel d. Double helix is formed from the hydrogen bonds causing the backbone to twist, with a major and minor groove. (Fig. 4.6) 4. Students should be able to explain why complementary base pairing would not be possible if two DNA strands were aligned in parallel fashion instead of the antiparallel alignment shown in Figure 4.6 a, b, c.

5. Helical turns occur every 10 bases (Franklin/Wilkens data) allowing base pairs to line up in a way that makes hydrogen bonding possible.

6. Carbon-nitrogen ring structure is nonpolar causing the twist into a helix to minimize contact of hydrophobic rings and surround water.

7. DNA secondary structure a. Outside of helix are two grooves: major and minor b. Grooves allow access for proteins to bid DNA (Making Models 4.2 Draw Simplified Versions of DNA) C. The tertiary structure of DNA 1. Supercoils form when DNA is wound tightly.

D. DNA functions as an information-containing molecule.

1. In all cells studied, DNA carries the information required for the organism’s growth and reproduction.

2. This information is contained in the sequence of nucleotides.

3. The double helix allows the information to be copied, by using one strand as a template for the synthesis of a complementary strand. (Fig. 4.7) a. Steps for replication: strand separation, base pairing, polymerization E. The DNA double helix is a stable structure.

1. Stability in symmetry, phosphodiester linkages, hydrogen bonding, hydrophobic interactions, and a few functional groups exposed limiting reactibility.

2. Function is consistent among cells.

3. DNA is inept at catalysis; its stability and orderliness make it nonreactive. .

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4. Because of its inability to act as a catalyst and self replicate, DNA has no support for being the first life form as standalone.

III. RNA Structure and Function A. Structurally, RNA differs from DNA. (Table 4.1) 1. Primary structure a. Like DNA, RNA has a sugar-phosphate backbone with phosphodiester linkages between nucleotides, with a sequence of four nitrogenous bases. b. Unlike DNA, RNA contains ribose instead of deoxyribose. This makes RNA more reactive and less stable than DNA. c. Unlike DNA, RNA contains uracil instead of thymine. Uracil pairs with adenine.

2. Secondary structure a. Unlike DNA, a single RNA strand tends to fold and form complementary base pairs with itself, resulting in structures such as hairpin loops. (Fig. 4.8) 3. Tertiary structure a. Unlike DNA, RNA molecules can have tertiary structure and quaternary structure, resulting in very different shapes and chemical properties. (Table 4.1) B. RNA’s versatility 1. RNA is not as good at information storage as DNA or as good at catalysis as protein, but it can do “a little of everything.”

2. RNA has recently been found to perform many diverse and important functions in cells.

C. RNA can function as a catalytic molecule.

1. Due to RNA’s structural complexity, RNA is capable of stabilizing a few transition states and can catalyze some chemical reactions.

2. RNA catalysts are called ribozymes.

3. Altman and Cech (1989 Nobel Prize) found that ribozymes catalyze important reactions in living cells, such as the formation of phosphodiester bonds. (Fig. 4.9) 4. If a ribozyme could catalyze its own replication, it could make a copy of itself. Such a ribozyme could have been the first living entity.

IV. In Search of the First Life-Form A. The first living molecule had to do two things: 1. Provide a template that can be copied. (Fig. 4.10) 2. Catalyze polymerization reactions that link monomers into a copy of that template.

3. RNA is the only molecule capable of both processes. Therefore, most researchers think that the first life-form was made of RNA. B. How biologists study the RNA world 1. Johnston and Bartel’s research a. Their goal: To produce, from scratch, an RNA molecule that can catalyze the addition of nucleotides to an existing RNA strand b. Their strategy: (1) Synthesize many RNA molecules.

(2) Select those that can catalyze the addition of the most nucleotides.

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(3) Copy the selected RNA molecules in a way that introduces “mutations.” c. Their results: After 18 rounds of selection, a ribozyme evolved that was very effective at adding ribonucleotides to a growing RNA strand. d. Other scientists have produced other ribozymes capable of catalyzing many of the key reactions needed for life.

C. The RNA world may have sparked the evolution of life.

1. The RNA world likely preceded protein enzymes.

2. The evolution of protein enzymes provided the means for catalyzing reactions necessary for life to emerge in a cellular form.

3. Following this event, three of the five fundamental characteristics would be in place: a. Information—stored in nucleic acids b. Replication—enzymes and possibly ribosomes replicating nucleic acids c. Evolution—random changes in the synthesis of proteins and selective advantages resulting from some of these changes

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. deoxyribonucleotide 3' carbon deoxyribose 5' carbon double helix activated nucleotides guanine (G) adenine (A) hairpin adenosine triphosphate major groove (ATP) minor groove antiparallel nitrogenous basebase complementary nucleic pairingacid nucleotide complementary strand phosphate groupstrand complementary phosphodiester linkage cytosine (C) phosphorylation deoxyribonucleic acid polymerization (DNA)

purine pyrimidine ribonucleic acid (RNA) ribonucleotide ribose ribozyme RNA world hypothesis stem-and-loop structure sugar sugar-phosphate backbone template strand thymine (T) uracil (U) X-ray crystallography

Lecture Activities Practicing BioSkills: Reading and Making Visual Models (BioSkills 12) Estimated duration of activity: 15 minutes Drawing a concept map can help students review and grasp the many new terms and concepts associated with nucleic acids. If the class is not familiar with concept maps, you may model the idea by starting the map yourself and soliciting more input from students as the map progresses. Students can also create concept maps in small groups and then compare their results. Concept maps can also be made into a game, such as with two teams that alternate

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sending people up to the board to add a new term or a new arrow. (Example: A team gains a point if the team can add a correct element to the map within 10 seconds.) Refer students to BioSkills 12 for more information.

Suggested list of starting terms: DNA, RNA, nucleotide, phosphate group, ribose, deoxyribose, nitrogenous base, purine, pyrimidine, catalysis, information storage.

Computer Activity: Visualizing the Structure of DNA (This exercise requires that the classroom be equipped with a web-enabled computer projection system.)

Estimated duration of activity: 10 minutes Many students have a difficult time conceptualizing nucleic acid structure and how it is related to function. Fortunately, many bioinformatics computer tools now allow us to actually visualize molecular structure. The company MDL has designed a program called CHIME that you can use to view the 3-D image of any molecule. You can download a free copy of the CHIME plug-in by visiting www.mdli.com. In CHIME, you can show the students the 3-D conformation of DNA and use the mouse to rotate the molecules in three dimensions. This is an invaluable classroom tool for demonstrating the double helix, the sugar-phosphate backbone, base pairs and their hydrogen bonds, antiparallelism, the genetic code, and other features. The websites listed below have user-friendly tutorials to help teachers demonstrate various aspects of nucleic acid structure (download the CHIME plug-in before using these websites): • www.umass.edu/molvis/tutorials/dna/ • www.bio.davidson.edu/courses/genomics/jmol/longDNAframe.html • www.callutheran.edu/BioDev/omm/gallery.htm

Critical-Thinking Problems: Nucleic Acids Estimated duration of activity: 10 to 15 minutes For this activity, divide students into groups of three. Assign groups the problems suggested below. Each member of the group is responsible for one of the three problems. Students are then given 3 minutes to determine individually the answer to their assigned problem (with the textbook and notes closed). Student A then presents the solution to his/her problem, with an explanation justifying the answer. Student B expresses an opinion validating or rejecting student A’s answer (also without reference to notes or the text). Student C (using notes and text) evaluates the responses from A and B and moderates any differences between the two.

The roles are then rotated as student B presents his/her answer to the second item, student C responds, and student A evaluates. The roles are rotated one last time for the third problem.

Suggested problems: • A DNA molecule composed of normal nucleotides (blue) is incubated with the appropriate enzymes to create a copy. During the copying process, radioactive nucleotides (red) are made available for the reaction. Which of the three diagrams in the figure on the next page correctly illustrates the two DNA copies? Why?

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• The diagram below illustrates a portion of a DNA molecule. The diagram contains three flaws. What are they? What are the appropriate corrections?

5'

3'

A

C

A

A

G

C

T

T

G

C

G C

C

T

A

C

T

G

T

T

C

C

A

A

C

G

C

G

A

T

G

G

5'

3'

• A DNA molecule contains 16 percent adenine nucleotides. What percentage of the DNA molecule is represented by guanine nucleotides?

Discussion Idea: History of the Double-Helix Discovery Biology students learn about the double-helix model of DNA structure from their earliest days of science education, so they often don’t realize how exciting Watson and Crick’s 1953 publication was. They may not know that earlier, reputable scientists proposed other models for the structure of DNA. Your students may benefit from a discussion of the models that were proposed in the early 1950s before Watson and Crick published their famous Letter to Nature.

Note also that Watson and Crick’s letter is very short (one page) and is fairly accessi- ble for introductory students. (For instance, most students will be able to grasp the implica- tions of the famous last sentence.) The letter can be displayed in a single PowerPoint slide and can also be used to introduce the concepts of primary literature and peer-reviewed journals.

Finally, Watson’s book The Double Helix provides excellent fodder for class discussions (or extra-credit assignments) about how scientists think and how scientific discoveries happen.

Consult the following websites for further information: • www.exploratorium.edu/origins/coldspring/ideas/ This site provides an annotated copy of Watson and Crick’s Letter to Nature, with comprehensive information about the history of the search for the structure of DNA. • osulibrary.oregonstate.edu/specialcollections/coll/pauling/dna/index.html This site is a comprehensive overview of the “race for DNA.” It details all of the major players but focuses on Linus Pauling’s contributions.

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• osulibrary.oregonstate.edu/specialcollections/coll/pauling/index.html This link will take you to a scanned copy of Linus Pauling’s paper, which suggests a triple-helix structure of DNA. • www.nature.com/nature/dna50/archive.html This is Nature’s archival tribute to the 50th anniversary of the discovery of the double-helix structure of DNA. It contains several of the hallmark papers published in 1953 by Watson, Crick, Franklin, and Wilkins.

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CHAPTER

5

An Introduction to Carbohydrates

Learning Objectives: Students should be able to...

• Define the chemical nature of monosaccharides and provide three ways in which they can differ from one another.

• Compare and contrast the linkages between monosaccharides with those in other macromolecule.

• Compare and contrast the structure and function of starch, glycogen, cellulose, chitin, and •

peptidoglycan.

Use examples to describe the three major functions that carbohydrates perform in cells.

Lecture Outline What is a carbohydrate?

• Monosaccharides are one sugar; oligosaccharides are polymers of several sugars; polysaccharides have many sugars.

• All carbohydrates are variations of (CH2O)n, where n is the number of “carbon-hydrate” groups.

• Carbohydrates contain a carbonyl group and several hydroxyl groups.

• Named based on molecular formula (CH2O)n, with n the number of carbon-hydrates (n = 3 up to 1000).

1. Not carbon bound to water (hydrate) 2. Carbonyl C=O and hydroxyls (–OH) with many C–H bonds

I. Sugars as Monomers A. Required during early chemical evolution to make ribose for nucleotides 1. Carry chemical energy used by cells 2. Required for synthesis of complex compounds B. What distinguishes one monosaccharide from another?

1. Monosaccharides (simple sugars) are the monomers of carbohydrates. (Fig. 5.1) 2. The carbonyl group can be at the end (aldose) or in the middle of the chain (ketose).

3. Reactive groups include carbonyls and hydroxyls and allow carbohydrates to participate in chemical reactions.

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4. Monosaccharides can have three, five, or six carbons (trioses, pentoses, and hexoses, respectively), numbered consecutively starting with the one closet to the carbonyl group.

5. The spatial arrangement of their −OH groups distinguishes one molecule from another.

(Fig. 5.2) a. Glucose and galactose have the same molecular formula (C6H12O6), but their structures differ by the placement of a hydroxyl group. b. Functions of galactose and glucose differ based on structure, with glucose providing the carbons for biosynthetic reactions and power for energy; galactose must be converted to glucose to have these functions. c. The importance of structure is based on the interactions between molecules requiring a particular shape.

6. Carbohydrates with five or more carbons form rings in aqueous solutions. (Fig. 5.3) a. Bond required for ring structure forms between carbon with carbonyl and a carbon with a hydroxyl. b. Glucose C-1 carbon bonds with oxygen of the C-5 hydroxyl; hydrogen from C-5 transferred to C-1 carbonyl. c. Newly formed hydroxyl can be above the plane of the ring (beta) or below the plane of the ring structure (alpha) (NOTE: based on Hayworth structure).

7. Each monosaccharide has a unique structure and function.

C. Can monosaccharides form by chemical evolution?

1. Most monosaccharides are readily synthesized in conditions that mimic those predicted for early Earth. a. Heating of formaldehyde resulted to form almost all pentoses and hexoses.

2. A three-carbon ketose and other related sugars were also found on a meteorite in Australia in 1969.

3. Speculation that sugars are synthesized on dust particles and other debris in space and fell on early Earth.

4. Minerals on deep-sea thermal vents also allow for sugar synthesis.

5. Students should understand the differences between simple sugars and be able to draw a three-, five-, and six-carbon sugar illustrating those differences.

II. The Structure of Polysaccharides A. Polysaccharides 1. Linking together of simple sugars into chains (complex carbohydrates) a. Chain length varies from small oligomers or oligosaccharides to very large polymers called polysaccharides. b. Two sugars linked together make a disaccharide.

2. Structure and function of linked chains of monomers depend on the structure of monomers and how they are linked together. (Fig. 5.4) a. Example is maltose, which is composed of two glucose molecules and lactose, which links together glucose and galactose. (Fig. 5.5a, b) 3. Condensation reaction between monosaccharides at hydroxyl groups results in a glycosidic linkage. (Fig. 5.5) a. Hydrolysis cleaves glycosidic linkages.

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4. Glycosidic linkages are similar to peptide bonds of amino acids and phosphodiester bonds between nucleic acids in that they hold monomers together. a. Peptide bonds and phosphodiester bonds always form at the same locations between monomers. b. Glycosidic linkages can vary because of the presence of multiple hydroxyl groups. c. This results in a large variety of polysaccharides. d. α-1,4- and β-1,4-glycosidic linkages are two of the most common linkages with different geometries (see above); form between C-1 and C-4 carbons.

5. Structure and function of these α-1,4- and β-1,4-glycosidic linkages varies dramatically. a. Storage polysaccharides use one type of glycosidic linkage α-1,4-glycosidic linkage (starch, glycogen). b. Structural polysaccharides use β-1,4-glycosidic linkages (cellulose, chitin and peptidoglycan).

B. Starch: a storage polysaccharide in plants 1. Starch consists of α-glucose monomers joined by α-glycosidic linkages. (Table 5.1) 2. Linkages result in a coiled helix structure.

3. Starch is a mixture of unbranched amylose containing only α-1,4-glycoside linkages and branched amylopectin with glycosidic linkages between C-1 carbon of glucose with C-6 carbon on another strand.

4. Branches occur in amylopectin once in every 30 monomers.

C. Glycogen: a highly branched storage polysaccharide in animals (Table 5.1) 1. Glycogen is stored in liver and muscles.

2. Like starch, glycogen consists of α-glucose monomers joined by α-linkages once in every 10 monomers.

D. Cellulose: a structural polysaccharide in plants (Table 5.1) 1. Many cells are surrounded by cell walls; a protective sheet outside the membrane; plants, bacteria, fungi and other cells have cell walls composed of polysaccharides.

2. Cellulose is a major component of plant cell walls.

3. Cellulose consists of β-glucose monomers joined by β-1,4-glycosidic linkages.

4. Every other glucose is flipped compared to the previous monomer. a. Flipping every other monomer results in a linear molecule (not helical). b. Linear molecule has H bonds between adjacent parallel strands.

5. The structure provided due to flipping every other monomer results in cellulose fibers being straight and strong, requirements for cell structural support.

E. Chitin: a structural polysaccharide in fungi and animals (Table 5.1) 1. Chitin stiffens the cell walls of fungi and some algae and is an important component of the exoskeletons of insects and crustaceans.

2. Chitin has β-1,4-glycosidic linkages between monosaccharides called N-acetylglucosamine (NAG).

3. Like cellulose, every other chitin monomer is flipped, allowing for the formation of H bonds and making a strong, tough sheet resulting in stiffness and protection.

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F. Peptidoglycan: a structural polysaccharide in bacteria (Table 5.1) 1. Most bacteria have cell walls composed of peptidoglycan (a polysaccharide).

2. It has two different alternating monosaccharides joined by β-1,4-glycosidic linkages.

3. It has a short chain of amino acids attached to every other monomer, allowing for the formation of peptide bonds between adjacent strands.

4. The peptide bonds play a similar role to the H-bonds of chitin and cellulose by providing structural polysaccharides that are long, straight, parallel strands bound to each other, making these polysaccharides strong and stable.

G. Polysaccharides and chemical evolution 1. Cellulose and chitin are two of the most abundant organic compounds on Earth today.

2. But polysaccharides probably played little or no role in the origin of life because: a. Monosaccharides probably could not polymerize under the conditions present on early Earth.

(1) Polymerization requires protein enzymes (no enzymatic RNAs have been identified to synthesize monosaccharides). b. Polysaccharides are not known to catalyze any reactions; they lack structural complexity required for catalysis. c. The monomers of polysaccharides cannot do complementary pairing and thus have no way of storing information, like nucleotides.

III. What Do Carbohydrates Do? A. Sugars are used as starting materials for many other complex molecules.

1. Nucleotides start with ribose or deoxyribose.

2. Amino acids often start from a carbon chain of a sugar.

B. Carbohydrates can provide structural support 1. Cellulose, chitin, and peptidoglycan form strong fibers that cross-link to form tough sheets that provide strength and elasticity.

2. These polysaccharides are resistant to degradation because most organisms do not have enzymes that can break β-1,4-glycosidic linkages. a. Water is excluded from β-1,4-glycosidic linkages, making them insoluble, and hydrolysis of the fibers is difficult. b. Dietary fiber from cellulose that is eaten creates a spongelike mass that moves through the intestinal tract, preventing constipation.

C. The role of carbohydrates in cell identity (Fig. 5.6) 1. There is enormous structural diversity in some complex carbohydrates based on monomers and the bonds they can form.

2. Polysaccharides act as identification badges on plasma membranes surrounding cells.

3. A protein with one or more carbohydrate groups covalently bound is a glycoprotein, and the carbohydrates of glycoproteins are always on the outside of the cell.

4. Glycoproteins are key molecules for cell recognition and cell signaling.

5. Each cell carries a set of glycoproteins on its cell surface, identifying it as part of that body and identifying its tissue type (nervous, muscle, etc.).

6. The diversity of monosaccharides allows for a virtually unlimited amount of unique glycoproteins to exist, so that each species can have a unique identity.

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7. Wasserman et al., examined binding of the sperm to the egg during fertilization and demonstrated that sperm only recognize eggs of the same species. (Fig. 5.6) a. Competitive binding assay added purified egg-surface glycoproteins to sperm and blocked binding of sperm to eggs. b. Purified carbohydrates but not proteins caused the block in binding.

D. Carbohydrates and energy storage 1. Carbohydrates store sunlight as chemical energy using the process of photosynthesis. a. Photosynthesis can be summarized as: CO2 + H2O + sunlight → (CH2O)n + O2. b. Electrons in reactants have unequally shared electrons and therefore low potential energy (C=O and C–O have electrons held tightly). (Fig. 5.8) c. C–H bonds in product have equal sharing of electrons and are therefore of higher potential energy than the unequally shared electrons in the reactants. d. Even C–C bonds have equally shared electrons and relatively high potential energy compared to C–O or C=O bonds.

(1) Fats and carbohydrates store fuel, but fat has twice the amount of energy per gram compared to carbohydrates.

2. Enzymes hydrolyze carbohydrates to release glucose. a. Starch and glycogen are good storage molecules because they contain α-glycosidic linkages, which are easily broken down by enzymes.

(1) Phosphorylase catalyzes the hydrolysis of glycogen in many cells.

(2) Amylase catalyzes the breakdown of α-glycosidic linkages in starch and is found in saliva as well as the pancreas and intestine.

3. Energy stored in glucose is used to make ATP. a. The free energy in the bonds of carbohydrates is used to transfer an inorganic phosphate to adenosine diphosphate (ADP), synthesizing ATP. b. Ten calories in a LifeSaver can make 2 × 1023 molecules of ATP; however, an average human’s need for energy would use this amount of ATP in just over a minute.

4. Carbohydrates are the water behind the dam, and the ATP is the energy generated when the water (carbohydrates) are released.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. adenosine triphosphate (ATP) aldehyde aldose amylase amylopectin amylose carbohydrate .

carbonyl group cell signaling cell wall cell-cell recognition cellulose chitin complex carbohydrate disaccharide

galactose glucose (α and β) glycogen glycoprotein glycosidic linkages (α and β) hexose hydroxyl group CHAPTER 5 An Introduction to Carbohydrates

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ketone ketose linear form monosaccharide N-acetylglucosamine (NAG)

oligosaccharide pentose peptidoglycan phosphorylase photosynthesis polysaccharide

ribose ring form simple sugar starch sugar triose

Lecture Activities Artificial Sweeteners Estimated duration of activity: 15 minutes The use of artificial sweeteners is ubiquitous in our society. However, many of us do not know what the artificial sweeteners really are or why they taste sweet but have no dietary calories. Your students may find it interesting to compare the tastes and molecular structures of sucrose, aspartame, saccharin, and sucralose.

To conduct the taste test, have a TA or student volunteer pass out two or three different samples to several students (if the class is small enough, all students may get samples). Label the samples A, B, C, and so on. Have the students record their observations about the taste, appearance, texture, and other features. (Not all of the students participating have to taste all of the samples.) At the end of the taste test, you can create a class-wide data chart to get an overall comparison. While the students taste the sweeteners, discuss the structural differences between the molecules and ask students to make hypotheses as to why the artificial sweeteners have no dietary caloric value.

Sucrose: A naturally occurring disaccharide composed of glucose and fructose, commonly known as table sugar. It has dietary caloric value because the human digestive system can break it down into its monosaccharide parts, and those monosaccharides can be metabolized to produce ATP. Aspartame: Commonly known as Nutrasweet, derived from the amino acids aspartatic acid and phenylalanine. It is 180 times sweeter than sugar, so less of it can be added to foods to achieve the same degree of sweetness. It is broken down in the digestive tract, but because so little of it is added to foods to make them sweet, the added caloric value is extremely low.

Saccharin: Commonly known as Sweet-n-Low, derived from toluene, 300 times sweeter than sugar.

Sucralose: Commonly known as Splenda, 600 times sweeter than sugar. It is made from sugar but is not digestible, so it adds no dietary calories to the foods that contain it. The FDA has recently approved it for use in any food.

The following website has molecular structures and further information on all of these sweeteners (requires CHIME plug-in): www.elmhurst.edu/~chm/vchembook.

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High-Fructose Corn Syrup Estimated duration of activity: 10 minutes The fructose debate: This debate involves a homework assignment for students and an in-class discussion/debate on the health effects of high-fructose corn syrup (HFCS) versus sucrose (table sugar) in the American diet. Background: HFCS is about 50 percent fructose and 50 percent glucose, both in monosaccharide form. It therefore contains the same monomers as sucrose (table sugar), but sucrose is a disaccharide made up of one glucose molecule and one fructose molecule. There is considerable debate about whether HFCS is nutritionally worse than sucrose. Since the 1970s, HFCS has found its way into almost every processed food that requires a sweetener. Food manufacturers prefer HFCS because it is cheaper than sucrose, is more soluble, and has a longer shelf life. Moreover, it gives foods an attractive consistency, mixes well with other flavorings, retains moisture, and resists bacterial growth. It is widely used in soft drinks and baked products. Many nutritional scientists have done research indicating that HFCS intake is correlated with adult-onset diabetes and obesity. The HFCS manufacturers vehemently deny such claims, saying that HFCS is really not all that different from sucrose. Activity: Divide a large class into groups using the first letters of students’ last names. Have A–M research and defend HFCS, and instruct N–Z to research the potential health risks and be prepared to argue against HFCS. You will act as the moderator of the debate, making sure that students take the time to compare and contrast HFCS and sucrose in terms of their structures, activity in the digestive system, and ultimate implications for health. You will likely find that once students do a bit of research for themselves on any given topic, they become invested and retain much more of the basic information relating to that topic. This can also be a good opportunity to introduce the concepts of correlational versus experimental studies.

Misconceptions about Carbohydrates Estimated duration of activity: 10 to 15 minutes For this activity, divide class into groups of 3 to 5 students. The purpose of this exercise is for students to identify as many misconceptions as possible in their understanding of issues relating to carbohydrates. Introduce the activity by explaining the importance of identifying and rectifying misconceptions during the learning process. There will be an informal competition to see which group can identify the greatest number of misconceptions in the time allotted. It is critical to establish a safe and fun environment to help students overcome the natural reluctance to have one’s misconceptions exposed publicly.

Each group member is assigned a question from those listed on the next page. Group members are first given a few minutes to formulate answers to their assigned question. This part of the activity is done individually without notes or text. Group members then take turns sharing their responses. During the sharing, the other members of the group should have their notes and textbooks open to monitor the responses and search for misconceptions. At the end of each group member’s answer, the misconceptions are revealed, discussed, and repaired by group members. Students should be made aware that misconceptions can be discovered at any time, including during the discussion following the answer to each question.

At the end of the allotted time, ask for short reports from the groups. These reports may take any of the following forms:

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• Identifying which group found the largest number of misconceptions • Asking what was the most important misconception each group discovered • Asking how many groups discovered this same misconception • Asking what was the most significant thing each student learned from this exercise Sample questions: • Suppose a cell were capable of using the molecules shown in Figure 5.1 as a source of energy. Rank the molecules in order of the amount of potential energy available. Give the basis for your ranking.

H H

C

H O

H

C

H H

H O

C

O

O

O

C

H

H H

C

O

H

H

• Explain why our bodies can digest starch but not cellulose. Draw a picture to illustrate the molecular basis for your answer.

• If 100 sugar molecules were joined into a polysaccharide to form a structural sheet, which type of sheet would be the strongest: cellulose, peptidoglycan, or chitin? Draw a picture to explain the molecular basis for your answer.

• Why does a gram of fat provide many more calories of energy than a gram of carbohydrate? Draw a picture to illustrate the molecular basis for your answer.

• Embryonic liver cells can be removed from a developing mouse fetus and then dispersed on a cell culture dish. Over time, the cells adhere to each other to form small aggregates.

Suppose that, prior to dispersal of the cells on the culture dish, they were treated with an enzyme that hydrolyzes the chemical bonds between sugar monomers. What would be likely to occur? Why? • Lactose and cellulose both have sugar monomers with a β-1,4-glycosidic linkage. Explain why our bodies can digest lactose but not cellulose. Draw a picture to illustrate the molecular basis for your answer.

Discussion Idea: Glucosamine This chapter discusses N-acetylglucosamine as a structural component of chitin in the exoskeletons of insects and crustaceans. Students may recognize glucosamine as a dietary supplement marketed to people with arthritis and joint pain. Students may find it interesting to talk about how glucosamine and other polysaccharides are important structural molecules in cartilage and how taking a glucosamine supplement can presumably alleviate some of the discomfort from arthritis by promoting the reformation and repair of damaged cartilage. It is interesting that the bulk of glucosamine on the market today is isolated from crab, lobster, and shrimp shells.

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Getting the “Big Picture”: Concept Maps of Cellular Energy Estimated duration of activity: 15 minutes This chapter contains several key points related to cellular energy, including the first mention of photosynthesis and cellular respiration. You can reinforce these basic concepts by having students draw their own concept maps of cellular energy terms. After they have produced their own concept maps, have them compare their maps with the “Big Picture⎯Energy” concept map from the textbook and see how many terms they already know.

Here is a suggested list of starting terms: monosaccharide, polysaccharide, glucose, glycogen, starch, sunlight, photosynthesis, oxygen, carbon dioxide, water, cellular respiration, ATP, energy.

Discussion Idea: So What’s Wrong with Oreos, Anyway?

The recent talk of the evils of trans fats has led to the vilification of America’s favorite sandwich cookie. The debate became so heated that Kraft has eliminated trans fats from Oreo cookies. The FDA recently required that food manufacturers list the trans-fat content on all food ingredient labels. Shortly afterward, New York City passed a law limiting trans-fat content in restaurant food to 0.5 gram per serving. This “trans-fat ban” led most major restaurant chains, such as Starbucks and McDonald’s, to eliminate trans fats from their products nationally rather than develop separate recipes just for New York City. Thus, the New York City ban resulted in one of the most sudden dietary changes that has ever occurred in the United States.

Ideas for class discussion: • Did we have enough evidence about trans fats to justify banning them? (That is, is one study enough?) As further food for thought: Some natural trans fats may actually be beneficial for health.

• The New York City law effectively spread immediately to the entire country. Should other regions have been given the chance to make their own decisions?

• Should people have the right to eat anything they wish, whether or not it is unhealthy?

Or should there be laws that require people to eat healthy foods?

Background information on trans fats • What are trans fats? Trans fats are triglycerides in which one of the double bonds in the fatty acid has been flipped from the normal cis configuration, which produces the “kink” in the fatty acid, to a rare trans configuration. The trans configuration gives the fatty acid an unusual straight, nonkinked double bond.

• How are trans fats produced, and why? Unsaturated plant oils are cheaper than saturated animal fats. However, oils do not have the proper consistency or “mouth feel” for qualitybaked goods and certain other products⎯like the white cream filling in Oreos. So food manufacturers alter the oils by adding hydrogen atoms. This process breaks the double bonds, converts the oils to saturated fats, and makes them semisolid at room temperature.

They are also able to retain their flavor and freshness longer. (They are then called “partially hydrogenated oils.”) However, we now know that during this process, a small proportion of the fatty-acid molecules do not become normal hydrogenated oils as intended but instead end up with a trans double bond.

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• What evidence links trans fats to health? According to a study released in 1997, consumption of trans fats equivalent to just 2 percent of total caloric intake—or about 40 calories of trans fats per day—was strongly linked with doubling the risk of heart disease. This is the single largest dietary/lifestyle risk factor for heart disease ever discovered. In other words, eating 1 teaspoon of trans-fat margarine is worse (in terms of heart disease risk) than eating twice that amount of saturated fats, consuming lots of salt, or never exercising (Hu et al. 1997).

• How do trans fats do their damage? It is still not clear. Two clues are: − Membrane integrity: Trans fats are incorporated into cell membranes, decreasing fluidity and compromising cell membrane function. − Free radical damage: Trans fats are known to increase the amount of free radicals (damaging molecules that are missing electrons) in the bloodstream.

For further information, see this website: www.sciencenews.org/articles/20030719/food.asp

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CHAPTER

6

Lipids, Membranes, and the First Cells

Learning Objectives: Students should be able to...

• Describe the molecular structure of lipids, their physical properties, and the roles they play in cells.

• Use the properties of phospholipids to explain how they spontaneously form selectively permeable membranes.

• Predict how gradients can affect the movement of water and solutes across a membrane.

• Compare and contrast how membrane proteins help substances cross a cell membrane via passive and active transport.

Lecture Outline A cell membrane (or plasma membrane) separates life from nonlife.

• The formation of the cell membrane was an important milestone in the history of life.

• A cell membrane performs two essential tasks: • It acts as a selective barrier, letting desired molecules in and keeping undesired ones out.

• It speeds up reactions by concentrating reactants in the cell.

I. Lipid Structure and Function A. What are lipids?

1. A lipid is any molecule that is hydrophobic (will not dissolve in water). a. Most lipids are hydrophobic because they contain many C−H bonds, which are nonpolar. (Fig. 6.1a) 2. There are many classes of lipids with highly different chemical structures.

B. How does bond saturation affect hydrocarbon structure?

1. A fatty acid is a simple lipid consisting of a hydrocarbon chain bonded to a carboxyl (–COOH) functional group. (Fig. 6.1b) 2. Fatty acids typically contain 14–20 carbon atoms.

3. The type of bond between carbons in hydrocarbon chains lead to the structure/function of the lipid.

4. Hydrocarbon chains that consist of only single bonds between the carbons are called saturated.

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5. If one or more double bonds exist in the hydrocarbon chains, they are unsaturated.

6. Fatty acids have unbranched hydrocarbon chains. Unsaturated hydrocarbons contain carbon-carbon double bonds. Saturated hydrocarbons do not. Some double bonds produce a kink. (Fig. 6.1b) 7. The fluidity of lipids depends on the length and saturation of their hydrocarbon tails.

(Fig. 6.2 a, b, c, d) C. A look at three types of lipids found in cells 1. A fat (triacylglycerol, triglyceride) is a glycerol with three fatty acids. (Fig. 6.3) a. Each fatty acid is linked with an ester linkage to one of the carbons of the glycerol. b. Fats are not polymers because the fatty acids are not linked in a chain.

2. Steroids have a nonpolar four-ring structure with a polar side group. Example: cholesterol. (Fig. 6.4a) 3. Phospholipids have a polar “head” and a nonpolar “tail.” (Fig. 6.4b) a. The head is a three-carbon glycerol with a phosphate group (PO–4). b. The tail is two fatty acids.

D. How membrane lipids interact with water 1. Functions of lipids include storing chemical energy, pigments that capture or respond to sunlight, signaling, waterproofing, vitamins, and the cell membrane.

2. Membrane lipids have a polar, hydrophilic region. a. Head region: contains negatively charged phosphate attached to polar group b. Carbon tails: nonpolar and hydrophobic 3. Water interactions a. With head groups: charges and polar bonds interact when placed in solution. b. With hydrophobic tail: water cannot form hydrogen bonds with tail.

E. The structures of membrane lipids 1. Lipids that can form membranes are amphipathic, with a polar hydrophilic end and a nonpolar hydrophobic end.

2. Students should be able to look back at Figure 6.1b and explain why fatty acids are also amphipathic.

F. Were lipids present during chemical evolution?

1. Along with amino acids, nucleic acids, and carbohydrates, evidence indicates that lipids were present during chemical evolution.

II. Phospholipid Bilayers A. Phospholipids in water 1. In water, phospholipids can spontaneously (without energy input) form two structures.

2. The polar heads interact with water; the nonpolar tails are driven together.

3. Phospholipids in water form either micelles or bilayers. a. Micelles are tiny droplets created when the hydrophilic heads are on the outside and the nonpolar tails are on the inside. (Fig. 6.5a) b. Lipid bilayers are two sheets back to back, with the nonpolar tails together.

(Fig. 6.5b)

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B. Artificial membranes as an experimental system 1. Researchers use liposomes (small balls of bilayers) and planar bilayers (flat sheets) to study the behavior of lipid bilayers. (Figs. 6.6, 6.7) C. Selective permeability of lipid bilayers (Fig. 6.8) 1. Small nonpolar molecules cross membranes quickly. Example: O2 2. Large molecules or charged molecules (including all ions) cross membranes slowly, if at all. Examples: K+, Cl–, glucose, and sucrose 3. Some small polar molecules can cross membranes rapidly if they are uncharged.

Example: H2O 4. Students should be able to predict where amino acids and nucleotides would be placed in Figure 6.8 and explain their reasoning.

D. How does lipid structure affect membrane permeability?

1. Bond saturation and hydrocarbon chain length affect membrane fluidity and permeability. (Fig. 6.9) a. C=C double bonds in a fatty acid form a kink in a carbon chain. b. Unsaturated lipids have one or more double bonds and display kinks. c. Saturated lipids have no double bonds and are straight.

2. Cholesterol reduces membrane permeability. (Fig. 6.10) a. Because the steroid rings in cholesterol are bulky, adding cholesterol fills gaps that would otherwise be present in the hydrophobic section of the membrane.

E. How does temperature affect the fluidity and permeability of membranes?

1. The fluid nature of the membrane allows phospholipids to move laterally within the bilayer.

2. Phospholipids move side to side readily but rarely “flip.” (Fig. 6.11) 3. At lower temperatures, hydrophobic tails pack more closely; fluidity decreases, permeability decreases, and membranes will solidify if the temperature is low enough.

III. How Substances Move across Lipid Bilayers: Diffusion and Osmosis A. Diffusion: a spontaneous movement from high to low concentration 1. Molecules and ions have thermal energy and are in constant motion.

2. This produces a net movement from regions of high concentration to regions of low concentration. Analogy: blue and white billiard balls 3. The concentration gradient is a difference in solute concentrations.

4. At equilibrium, molecules and ions are randomly distributed with equal concentrations everywhere. (They continue to move, but there is no net movement in any one direction.) (Fig. 6.12) 5. Diffusion is a spontaneous process because it increases entropy (second law of thermodynamics).

6. Students should be able to predict how a difference in temperature across a membrane would affect the concentration of a solute at equilibrium.

B. Osmosis: a diffusion of water 1. Osmosis occurs whenever two solutions are separated by a membrane that is permeable to water but not to some (or all) of the solutes.

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2. Water molecules show a net movement from high to low water concentration⎯that is, from low to high solute concentration. (Fig. 6.13) 3. Osmosis can cause shrinking or swelling of membrane-bound vesicles (liposomes or cells). (Fig. 6.14) a. Hypertonic solution (higher solute concentration outside than inside): water moves out; vesicle shrinks. b. Hypotonic solution (lower solute concentration outside than inside): water moves in; vesicle expands and may burst. c. Isotonic solution (same solute concentration outside and inside): no net movement of water; vesicle does not change.

C. Membranes and chemical evolution 1. The first lipid bilayers likely provided a container for replicating RNA. 2. Protocells are simple vesicle-like structures that harbor nucleic acids, and these are viewed as possible intermediates in the evolution of the cell. (Fig. 6.15)

IV. Proteins Alter Membrane Structure and Function A. What are membrane proteins like?

1. Membrane proteins are amphipathic, with both nonpolar and polar amino acids.

(Fig. 6.16) a. Polar amino acids “stick out” into the aqueous solution on either side. b. Nonpolar amino acids are stabilized in the interior of the bilayer and keep the protein in the membrane.

2. Secondary and tertiary protein structures can form a channel or pore.

B. Development of the fluid-mosaic model 1. Davson and Danielli (1935) proposed that cell membranes are like a sandwich, with an inner bilayer coated by a layer of proteins on either side. (Fig. 6.17a) 2. Singer and Nicolson (1972) proposed the fluid-mosaic model. a. Some proteins may span the membrane. (Fig. 6.17b) b. The model was confirmed by freeze-fracture electron microscopy. (Fig. 6.18) (BioSkills 9) 3. Current understanding a. The fluid-mosaic model is largely correct. b. Some proteins span the membrane entirely (transmembrane proteins), while others are on just one side (peripheral membrane proteins). c. These proteins strongly affect membrane permeability.

C. Systems for studying membrane proteins 1. Membrane proteins can be isolated using amphipathic detergents that break up the bilayer and solubilize the membrane proteins. (Fig. 6.19) 2. The different solubilized proteins can then be isolated for further study.

D. Channel proteins facilitate diffusion.

1. Ion channels a. Ion channels are proteins that form a pore in the membrane which allows specific ions or molecules to pass passively through.

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b. The ions move in response to the electrochemical gradient⎯the combined effects of the concentration gradient and electrical (charge) gradient. (Fig. 6.20) 2. Students should be able to add an arrow to Figure 6.20, indicating the electrochemical gradient for chloride ions.

3. Is an ion channel involved in cystic fibrosis? a. Cystic fibrosis transmembrane conductance regulator (CFTR) (1) Researchers were able to find the gene that is defective in cystic fibrosis and produce the protein (CFTR) that this gene should make.

(2) Insertion of this protein into lipid bilayers produced a flow of electric current, indicating that ions were crossing the bilayer. (Fig. 6.21) (3) CFTR is now known to be a chloride channel.

4. Protein structure determines channel selectivity. a. A channel protein is selective, allowing only a certain ion or molecule to pass through. Example: aquaporins and water (Fig. 6.22) 5. Movement through many membrane channels is regulated. a. Ion channels are gated channels, meaning that they open or close in response to a particular signal. Example: K+ channel (Fig. 6.23) b. Once a gated channel is open, movement through it is passive.

6. Ion channels allow facilitated diffusion, a form of passive transport. a. Facilitated diffusion is passive transport of substances that otherwise could not cross the membrane readily on their own. b. Passive transport is transport powered only by diffusion along an electrochemical gradient.

E. Carrier proteins facilitate diffusion.

1. Facilitated diffusion can also occur via carrier proteins, which change shape during the process.

2. Unlike channel proteins, carrier proteins do not form a “pore.”

3. The search for a glucose transporter a. Observation: Red blood cell “ghosts” are much more permeable to glucose than pure phospholipid bilayers. b. Researchers isolated the glucose transporter protein (GLUT-1) from red blood cell membranes and placed them in liposomes to study them.

4. How does GLUT-1 work? (Fig. 6.24) a. It is thought that GLUT-1 changes shape when glucose binds to it, similar to the shape change of a catalytic enzyme. The change in shape propels the glucose to the interior.

5. Practice modeling transport across a membrane. (Making Models 6.1) F. Pumps perform active transport.

1. In active transport, ions and molecules are moved against their concentration gradient (from low to high concentration).

2. Active transport requires energy input, usually via phosphorylation from ATP. a. Example: sodium-potassium pump (Fig. 6.25, steps 1-8) b. For each ATP used, three Na+ are transported out and two K+ in.

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3. Secondary active transport (co-transport) a. Protein pumps can set up electrochemical gradients. b. The electrochemical gradient can then be used to power the transport of a different molecule against its concentration gradient. c. Example: a co-transporter in your gut uses the Na+ gradient created by Na+/K+-ATPases to import glucose against its chemical gradient.

G. Plasma membranes define the intracellular environment.

1. The lipid bilayer and the membrane proteins create an internal cellular environment that is very different from the extracellular environment. (Fig. 6.26)

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. active transport amphipathic aquaporin ATP carrier protein cell membrane CFTR channel protein cholesterol cholesterol concentration gradient cotransport cystic fibrosis detergent diffusion electrochemical gradient equilibrium ester linkage facilitated diffusion fat fatty acid fluid-mosaic model freeze-fracture electron microscopy

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gated channel gel electrophoresis glucose transporter (GLUT-1) glycerol hydrocarbon hydrocarbon tail hydrophilic hydrophobic hypertonic hypotonic integral membrane protein ion channel isotonic lipid lipid bilayer liposome membrane fluidity micelle nonpolar oil osmosis passive transport peripheral membrane protein permeability

INSTRUCTOR GUIDE FOR BIOLOGICAL SCIENCE, 6e

phospholipid planar bilayer plasma membrane polar polar head group protocell pumps red blood cell “ghost” saturated lipid scanning electron microscope (SEM) secondary active transport selective permeability sodium-potassium pump solutes steroid transmembrane protein transport protein triacylglycerol triglyceride unsaturated lipid vesicle wax

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Lecture Activities Protein Architects Estimated duration of activity: 20 minutes In this activity, students will apply what they have learned about protein structure, membrane structure, and membrane transport to design a transmembrane protein that will be able to perform a certain function⎯for example, to facilitate the diffusion of Na+ but not Cl–, or to actively transport glucose into the cell.

Divide the class into teams of 2 to 5 students (depending on class size). Tell them they are biological architects and have been asked to submit a protein design that will perform one of the above functions. They must draw their designed protein and indicate what type of amino acids (with some examples) make up certain domains of the protein. Sample answers are given below. Give the teams 5 to 10 minutes for their design, and then have each team present its design to the class. Other students can analyze the design and assess whether it would actually perform its function. You may want to remind students of some of the structural features of the nonprotein molecules involved in this exercise (Na+, Cl–, glucose, etc.).

Correct answers should include the following information: • A transmembrane protein that will facilitate the diffusion of Na+ but not Cl –: The protein should traverse the membrane, with a cylindrical 3-D structure. To attract Na+, amino acids should be hydrophobic (Ala, Val, Leu, Iso, Pro, etc.) on the outside of the cylinder, hydrophilic (Ser, Thr, Gln, and Asn) on the ends in contact with the aqueous environment, and negatively charged (Glu and Asp) on the inside of the cylinder.

• A transmembrane protein that will actively transport glucose into the cell: The protein should traverse the membrane and be globular (not a cylinder). Amino acids should be hydrophobic (Ala, Val, Leu, Iso, Pro, etc.) where they are in touch with the inside of the membrane and hydrophilic (Ser, Thr, Gln, and Asn) where they are in contact with the aqueous environment. The protein should have a binding site for glucose on the extracellular surface, resembling the active site of an enzyme. Inside the binding site, there should be amino acids with polar side chains (Ser, Thr, etc.) that can form a hydrogen bond with glucose. Students should indicate that when glucose is bound to this binding site, the protein changes conformation to move the glucose molecule through the membrane and into the cell.

Permability Puzzle Estimated duration of activity: 5 minutes Consider a hypothetical experiment testing the permeability of various molecules in two different experimental systems. System 1 is artificial liposomes, prepared by agitating pure phospholipid bilayers in water. System 2 is red blood cell (RBC) “ghosts,” prepared by rupturing RBCs in a hypotonic solution and then resealing the holes in the membrane once all the cell contents have been expelled.

Write the following chart on the board but with the numbers missing, or hand out the chart to each student or student group:

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Molecule

Properties

V

Small, nonpolar

5

5

W

Large, polar

2

3

X

Small, charged

3

4

Y

Small, polar

4

2

Z

Large, charged

1

1

Transport/Permeability Transport/ in Artificial Liposomes Permeability RBC Ghosts

Ask the students to fill in the column for artificial liposomes with numbers representing the relative speed at which they think each of the molecules would pass through the artificial liposome (5 = fastest, 1 = slowest). Their answers should match the numbers given above.

Discuss and clarify. (Most students should be able to place 5 correctly and possibly 1, but they may need hints to place 2, 3, and 4.)

Now fill in the column for RBC ghosts yourself, showing the students the permeabilities.

Ask students to develop a hypothesis to explain the difference between the numbers in the two columns. Answer: The membrane of the RBC ghost contains transporter proteins for compounds W and X that facilitate their diffusion. The artificial liposomes, in this experiment, have no proteins in the membrane.

Discussion Idea: Too Much Water Show students this news headline from March 2003: “Fraternity Pledge Dies from Drinking Too Much Water in a Hazing Event.” Then ask students: How can too much water kill somebody?

The answer is that drinking too much water results in a condition called hyponatremia, or water intoxication. As a person drinks water, it passes through the digestive system to the blood and increases the volume of blood. As the water content of blood increases, the salt concentration in blood decreases. Because blood plasma (the aqueous part of the blood) is continuous with extracellular fluid, this increase in water content surrounding the cells of the body can cause edema. Likewise, the decrease in salt concentrations in body tissues can have serious implications for nerve function, kidney function, and so on. The most common symptoms of water intoxication are headache, nausea, vomiting, irritability, and seizures. These symptoms result from brain swelling. All other body tissues seem able to tolerate swelling, but since the brain is confined by the skull, swelling results in tissue damage that can lead to death.

Practicing BioSkills: Reading Graphs (BioSkills 2) Estimated duration of activity: 10 minutes Figure 6.10 summarizes an experiment on the effects of cholesterol and temperature on membrane permeability. Since this graph is one of the first graphs presented in the textbook, students may benefit from instruction in how to read graphs. Remind students that BioSkills 2 offers a quick, easy tutorial in reading graphs. You can then walk students through the steps in reading a graph:

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1. What do the axes represent?

2. What do the lines represent?

3. What is the overall trend or message?

Finally, encourage students to think about why the data are presented in this particular way and what other graph types could have been used.

Questions for discussion: • Why did the author present the data with three lines on one graph rather than three graphs, each with one line?

• Is it clear how many measurements were actually taken⎯that is, how many data points were used to generate each line? Why or why not? Is this a problem?

• Could the authors have used a scatterplot, pie chart, or bar graph?

If time permits, conduct another exercise, which will take an additional 5 minutes. Using an overhead or whiteboard with the original graph displayed, add dots to all three lines at exactly 10°C and 20°C. Now tell students to suppose that these six dots represent the only data gathered in the entire experiment. Ask them if they still think the line graph is an appropriate way to show the data. Then ask them to make a brief sketch showing what the same data would look like if they were displayed in a bar graph. (Students may need a hint about how bars can be grouped together to represent categories.) Discuss briefly, emphasizing that (a) the same data can often be displayed in several different ways, and (b) the best choice is the graph that is simplest, clearest, and most “honest”⎯that is, the least likely to mislead the reader about what the data actually are.

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CHAPTER

7

Inside the Cell

Learning Objectives: Students should be able to...

• Describe the structure and function of prokaroytic cell components.

• Describe the structure and function of eukaryotic cell components.

• Explain how organelles are functionally integrated and how they vary among specialized cells in a multicellular eukaryotic organism.

• Define molecular “zip codes” and explain how they function in protein transport.

• • Make predictions or interpret data from experiments used to track the movement of proteins in the cell.

Describe the structure and function of the cytoskeletal filaments and associated motor proteins.

Lecture Outline I. Cell Theory A. All cells are derived from pre-existing cells. All life forms contain the following characteristics: 1. Proteins 2. Nucleic acids 3. Carbohydrates 4. Plasma membrane B. The two fundamental cell types based on morphology are eukaryotes and prokaryotes.

1. Eukaryotes have a membrane bound nucleus.

2. Prokaryotes have no nucleus.

C. Phylogentically, we have three domains of life: 1. Bacteria 2. Archaea 3. Eukarya

II. Bacterial and Archaeal Cell Structures and Their Functions A. A revolutionary new view (Fig. 7.1) 1. Bacterial cells are highly organized, with an array of distinctive structures found among millions of species. (BioSkills 9)

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B. Prokaryotic cell structures: a parts list 1. The chromosome is organized in a nucleoid. a. Bacterial DNA is inside a chromosome, which contains genes. b. In order to fit within the nucleoid, the chromosome is supercoiled. (Fig. 7.2) c. Some bacterial cells have additional DNA molecules called plasmids that carry genes which help cells adapt to unusual circumstances.

2. Ribosomes manufacture proteins. a. Ribosomes are composed of large and small subunits. b. Ribosome size and function in bacteria and Archaea are similar, but the primary structures of RNA and protein differ.

3. Photosynthetic species have internal membrane complexes. (Fig. 7.3) 4. Organelles perform specialized functions. a. Recent research indicate the presence of internal compartments in bacteria that qualify as organelles. b. These organelles have specialized tasks related to the storage of ions, enzymes, and minerals.

5. The cytoskeleton structures the cell interior. a. Current research reveals a more complex cytoskeletal network with protein fibers and filaments that aid in cell division and cell shape.

6. The plasma membrane separates life from nonlife. a. Bacterial membranes consist of fatty acids bound to glycerol. b. Archaeal membranes consist of isoprenoid chains bound to glycerol, making it more stable for extreme environments.

7. The cell wall forms a protective “exoskeleton.” a. Tough fibrous layer surrounds the plasma membrane, giving it shape and rigidity. b. Molecularly bacterial cell walls contain peptidoglycan while Archaeal walls do not.

8. External structures enable movement and attachment. a. Flagella and fimbriae commonly found on bacterial surfaces b. Flagella used to propel the cell through water c. Fimbriae used to establish infection by serving as the “glue” to the surface of tissues 9. Prokarytic cell components (Fig. 7.5).

III. Eukaryotic Cell Structures and Their Functions A. The benefits of organelles 1. Eukaryotic cells are compartmentalized by membrane-bound organelles. a. Compartmentalization offers two key advantages: (1) Incompatible chemical reactions can be separated.

(2) Chemical reactions can become more efficient.

2. Prokaryotic versus eukaryotic cells a. Eukaryotic cells are usually larger than prokaryotic cells. b. Eukaryotic chromosomes are inside a membrane-bound nucleus. c. Eukaryotic cells have a larger number of distinct organelles.

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B. Eukaryotic cell structures: a parts list (Fig. 7.6) (Table 7.1) 1. Nucleus (Fig. 7.7) a. The nucleus is the largest organelle in the cell. b. It is enclosed by a double membrane called the nuclear envelope. c. It has a nuclear envelope supported by the nuclear lamina. d. It contains the genetic information of the cell, DNA. e. It contains the nucleolus, the site of rRNA synthesis and ribosome assembly.

2. Ribosomes (Fig. 7.8) a. Ribosomes are found in the cytosol. b. They are molecular machines that synthesize proteins.

3. Endoplasmic reticulum (Fig. 7.9) a. The endoplasmic reticulum (ER) is a network of membrane-bound sacs and tubules. b. Although a single structure, the ER has two distinct regions: (1) The rough ER has ribosomes bound to the cytoplasmic surface on which secreted and transmembrane proteins are made. It contains enzymes in the lumen that fold and modify proteins after they are synthesized on bound ribosomes.

(2) The smooth endoplasmic reticulum is a portion of the ER that does not have ribosomes attached. It contains enzymes involved in the synthesis of lipids, detoxifying harmful substances and storing calcium ions.

4. Golgi apparatus (Fig. 7.10) a. The Golgi apparatus consists of flattened membrane sacs called cisternae. b. It has polarity: a cis side that receives proteins from the rough ER, and a trans side that ships proteins to their final destination.

5. Lysosomes (Fig. 7.11) a. Lysosomes are involved in solid-waste processing and material storage. b. Lysosomes have a proton pump in their membrane that keeps the lumen acidic. c. They contain digestive enzymes, called acid hydrolases, that work best in acidic conditions to hydrolyze macromolecules. d. Golgi/ER/Lysosomes jointly form the endomembrane system. The endomembrane system is for producing, processing, and transporting proteins and lipids in eukaryotic cells.

6. Vacuoles (in plants and fungi) (Fig. 7.12) a. Vacuoles occupy most of the volume of plant cells. b. They act as a storage depot for water, ions, and sometimes proteins. c. They may contain noxious substances that protect the plant from predators.

7. Peroxisomes (Fig. 7.13) a. Peroxisomes are centers for oxidation reactions. b. They generate hydrogen peroxide as a by-product of these reactions. c. They contain an enzyme, catalase, which converts hydrogen peroxide into water and oxygen. d. Plants have special peroxisomes, called glyoxosomes, which convert the products of photosynthesis into a storable sugar.

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8. Mitochondria (Fig. 7.14) a. Mitochondria are the sites of ATP synthesis. b. They have two membranes: (1) The outer membrane is smooth.

(2) The inner membrane contains folds called cristae.

(3) The mitochondrial matrix is inside the inner membrane. c. Mitochondria contain some of their own DNA, mitochondrial DNA (mtDNA). d. They make their own ribosomes. e. They can divide independently of cell division.

9. Chloroplasts (Fig. 7.15) a. Chloroplasts convert the energy in sunlight into the chemical energy in sugar. b. They have two membranes: (1) The outer membrane is smooth.

(2) The inner membrane is also smooth.

(3) The inner membrane is filled with a series of flattened sacs, called thylakoids, which are stacked in grana.

(4) Thylakoids are surrounded by stroma, the fluid filling the inner membrane. c. Chloroplasts have some of their own DNA and can divide independently of nuclear division. d. Since mitochondria and chloroplast grow and divide independently of cell division, it is believed both were once free-living bacteria engulfed by ancestors and then developed a mutually beneficial relationship. This theory is called the endosymbiosis theory.

10. Cytoskeleton a. The cytoskeleton is a system of protein filaments. b. It provides structural support to the cell. c. It facilitates many types of cellular movement.

11. The eukaryotic cell wall a. Algae, fungi, and plant cells have cell walls. b. Cell walls are composed of carbohydrate rods and fibers running through a stiff polysaccharide matrix.

IV. Putting the Parts into a Whole A. Structure and function at the whole-cell level 1. Examples (Fig. 7.16) a. Pancreatic cells are specialized for secreting digestive enzymes. They are packed with ER and Golgi apparatus, the organelles that manufacture secreted proteins.

(Fig. 7.16a) b. Testis cells are specialized for making testosterone, a steroid hormone. They have an extensive smooth ER network. (Fig. 7.16b) c. Leaf cells are specialized for absorbing light and manufacturing sugars. They contain a large number of chloroplasts. (Fig. 7.16c) d. Numerous mitochondria are present in cardiac tissue due to the constant need of ATP. (Fig. 7.16d) .

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B. The dynamic cell 1. Differential centrifugation allows us to isolate and study the functions of individual organelles (BioSkills 7).

2. Studies involving green fluorescent protein and confocal microscopy allow us to study how all of the organelles in a cell act together (BioSkills 9).

3. Both techniques have limitations. Differential centrifugation splits cells into parts that are analyzed independently. Electron microscopy give a fixed snapshot of the cell or organism being observed. The limitation is that both are dynamic.

V. Cell Systems I: Nuclear Transport A. Structure and function of the nuclear envelope 1. The nucleus is enclosed by a double membrane called the nuclear envelope.

(Fig. 7.17) 2. Nuclear pores penetrate the nuclear envelope and connect the cytoplasm to the nucleoplasm. a. Nuclear pores consist of 30 different proteins called the nuclear pore complex.

3. Molecules travel into and out of the nucleus through the nuclear pores. a. rRNA and ribosomal subunits are made in the nucleolus but are functional in the cytoplasm. b. mRNA and tRNA are made in the nucleus but function in the cytoplasm. c. Nucleotide triphosphates—enzymes involved in DNA replication, transcription, and so forth—are made in the cytoplasm but function in the nucleus.

B. How do molecules enter the nucleus?

1. Size is not the sole factor in selective transport across the nuclear envelope.

2. This observation led to the hypothesis that the nuclear pore complex serves as a gate to control passage through the envelope.

3. Experiments using a protein called nucleoplasmin (involved in chromosome assembly) revealed the nature of the nuclear localization signal. (Fig. 7.18) a. What portion of the molecule targets it to the nucleus?

(1) Experimental design: Researchers used enzymes called proteases to separate the tail region of nucleoplasm from the head region. They labeled each with a radioactive atom and injected them into different living cells.

(2) Results: The researchers found that only the tail regions were transported into the nucleus.

(3) Conclusion: The tail region contains the nuclear transport “zip code” tag. b. Further experiments demonstrated that a 17-amino-acid stretch of the nucleoplasm in the tail region made up the nuclear localization signal. Follow-up studies showed that other proteins have nuclear localization signals that interact with proteins that deliver molecules into or out of the nucleus.

VI. Cell Systems II: The Endomembrane System Manufactures, Ships, and Recycles Cargo A. Studying the pathway through the endomembrane system (Fig. 7.19) 1. The secretory pathway hypothesis proposed that proteins intended for secretion from the cell are synthesized and processed in a series of steps: ER  Vacuole  Golgi  Cell Surface.

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B. Tracking protein movement via pulse-chase assay 1. George Palade studied pancreatic cells that secreted a large volume of proteins in order to understand the role of the endomembrane system in the production and processing of secreted proteins.

2. Hypothesis: The rough ER and Golgi apparatus form an endomembrane system that directs the production and secretion of specific proteins.

3. Experimental design: Palade used the pulse-chase technique to label newly synthesized proteins and visualize them over time. a. Palade exposed cells to a brief “pulse” of radioactively labeled amino acid (leucine). b. Then he gave the cells a large amount of unlabeled amino acid (the “chase”). c. After varying amounts of time, he killed the cells and examined them under the electron microscope (see BioSkills 6 and 9).

C. Results of the pulse-chase 1. These experiments demonstrated that the proteins pass from the rough ER to the Golgi apparatus to secretory granules and are ultimately secreted outside the cell. (Fig. 7.20) B. Entering the endomembrane system: the signal hypothesis 1. Günter Blobel hypothesized that proteins destined to be secreted have a signal in the first few amino acids that function as an address tag directing them to the ER. 2. Researchers found that when secreted proteins are synthesized in a test tube without the ER, they are about 20 amino acids longer than the same protein secreted by cells. a. When the researchers added the ER to their test tubes, the synthesized proteins were the same length as the endogenous proteins. b. The sequence of 20 amino acids is located on the N-terminus (the beginning) of the protein. It has been named the ER signal sequence.

3. Recent studies have documented the mechanisms for receiving this send-to-ER signal and inserting the protein into the rough ER. (Fig. 7.21) a. Protein synthesis begins on a free ribosome in the cytosol, and it synthesizes the ER signal sequence. b. The signal sequence binds to a signal recognition particle (SRP)—a complex of RNA and a protein. c. The ribosome, signal sequence, and SRP move to the ER membrane, where it binds to a receptor. d. After the binding to the rough ER, the synthesis of the polypeptide is completed, the signal sequence is removed, and there are two possibilities for the polypeptide: (1) The polypeptide may be transferred to the lumen of the ER so that it can be processed and ultimately packaged into an organelle or secreted from the cell.

(2) It may be integrated into the rough ER membrane so that it can be processed and shipped to the plasma membrane. e. What happens to proteins when they are in the rough ER? (1) Proteins are folded into their tertiary and quaternary structures with the help of chaperone proteins.

(2) Other enzymes modify proteins by adding carbohydrate groups (glycosylation).

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C. Moving from the ER to the Golgi apparatus 1. Palade and colleagues hypothesized that vesicles bud off of the ER, move through the cytoplasm to the nearby Golgi apparatus, fuse with the cis face of the Golgi apparatus, and dump their contents inside.

2. Studies using electron microscopy combined with differential centrifugation confirmed this hypothesis and demonstrated that distinct types of vesicles shuttle materials through the Golgi.

D. What happens inside the Golgi apparatus?

1. Recent studies have shown that the cisternae are constantly regenerated by the fusion of vesicles with the cis face and disposed of as vesicles bud off the trans face.

2. The function of the Golgi is well understood. a. Cisternae have a suite of enzymes at each stage of maturation, which catalyze different parts of the glycosylation reactions. b. Existing sugar groups on proteins are modified, phosphorylated, or even removed, while new sugar groups may be added.

E. How do proteins reach their destinations?

1. Once proteins are manufactured and processed through the ER and Golgi, they must be sent to the appropriate location or organelle within the cell.

2. Studies involving lysosomal enzymes reveal that these proteins are tagged with specific carbohydrate groups (mannose-6- phosphate) acting as molecular zip codes that target the proteins to the vesicles destined for the lysosome.

3. Proteins to be secreted are packaged into vesicles destined for the plasma membrane, where they will fuse and dump their contents via a process called exocytosis.

(Fig. 7.22) 4. Proteins synthesized in the cytoplasm also have zip codes, directing them to mitochondria, chloroplasts, or other destinations. a. The arrays of proteins produced in a cell are sorted via distinctive zip codes. b. These molecular addresses allow proteins to be shipped to the compartment where they function.

F. Recycling material in the lysosome (Fig. 7.23) 1. Receptor-mediated endocytosis begins when macromolecules bind to receptors.

2. Phagocytosis is the uptake of small particles or cells.

3. Autophagy begins when damaged organelles are delivereed to a lysosme for recycling.

VII. Cell Systems III: The Dynamic Cytoskeleton A. Actin filaments (Table 7.2) 1. Actin filaments are long, fibrous polymers of globular protein called actin. a. Actin resembles two strands twisted together. b. Actin has polarity—a plus end and a minus end; microfilaments grow at the plus end.

2. Microfilaments form a matrix that helps define the cell’s shape.

3. Actin filaments are also involved in several types of cell movement that are facilitated by the action of myosin binding to actin filaments, causing them to slide across one another. (Fig. 7.24)

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a. Cytokinesis: Actin filaments form a ring beneath the plasma membrane of a dividing cell; contraction of this ring pinches the cell in two. b. Cytoplasmic streaming: Actin filaments sliding across one another direct the movement of cytoplasm in plant and fungal cells. c. Cell crawling: Extension of actin filaments in one part of the cell forms pseudopodia that move the cell in a given direction.

B. Intermediate filaments (Table 7.2) 1. Many types of intermediate filaments exist, each consisting of different types of protein, although they are similar in size and structure.

2. Intermediate filaments (nuclear lamins) form a dense network on the inside of the nuclear envelope that maintains the shape of the nucleus.

3. Intermediate filaments also form a flexible skeleton between the nucleus and the rest of the cell that helps the nucleus stay in place.

C. Microtubules (Table 7.2) 1. Microtubules are composed of tubulin dimers that polymerize to form a hollow tube that has polarity (a plus end and a minus end).

2. Microtubules are essential for mitosis. a. Microtubules move and separate chromosomes during mitosis. b. Centrosomes (made up of centrioles in animals) or microtubule organizing centers (in plants) organize the microtubules used in mitosis. (Fig. 7.25) 3. Microtubules serve as tracks for vesicle transport. (Fig. 7.26) a. Ronald Vale and colleagues studied how vesicles move throughout the cell by using the extruded cytoplasm of a squid giant axon. They used this cell because: (1) It is very large and easy to manipulate.

(2) Large numbers of vesicles are transported down the very long length of the cell.

(3) They could squeeze cytoplasm out of the cell and still observe vesicle transport without the plasma membrane in the way.

4. Motor proteins pull vesicles along the tracks. a. Vale went on to identify the means by which a vesicle moves along a microtubule.

(1) Kinesin is a motor protein that contains three major regions: a head region with two globular pieces, a tail, and a stalk that connects the head to the tail.

(2) The globular pieces on the head contain microtubule binding sites as well as ATP binding sites; the tail region can bind the vesicle. (Fig. 7.27) (3) When the globular pieces bind and release ATP, they undergo a conformational change.

(4) The two globular head regions alternate between binding ATP and binding to the microtubule.

(5) This alternation results in the two globular head segments actually acting as “feet” as they “walk” down the microtubule. (Fig. 7.27) D. Flagella and cilia: moving the entire cell (Fig. 7.28) 1. How are cilia and flagella constructed? (Fig. 7.29) a. Cilia and flagella have nine pairs of microtubules that surround two individual microtubules. This “9 + 2” structure is called an axoneme. (Fig. 7.29) b. The axoneme attaches to the cell at the basal body, a structure identical to a centriole. .

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c. Electron microscopic studies revealed that the nine microtubule pairs are connected by motor proteins called dynein, and the center microtubules are connected to the nine doublets by protein spokes. (Fig. 7.29) 2. What provides the force required for movement? a. Dynein hydrolyzes ATP and changes shape, thus “walking” along in a manner similar to that of kinesin. b. This walking causes the microtubules within the cilia or flagella to slide past one another. c. However, this sliding is limited because the whole structure is anchored to the center microtubules by the protein spokes. d. This constrained movement results in the bending of the cilia or flagella, thus creating movement. (Fig. 7.29)

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to thecurrent chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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actin filaments

ER signal sequence

nuclear localization

Archaea autophagy autophagy axoneme basal body cell crawling cell wall centrioles centrosome chloroplast chromosome cilia/cilium cisternae cristae cytokinesis cytoplasm cytoplasmic cytoskelton cytosol cytosol differential centrifugation dimers dynein early endosome endocytosis endoplasmic reticulum endosymbiosis theory

eukaryote exocytosis fimbria flagella gene glycoprotein glycosylation glyoxysomes Golgi apparatus grana intermediate filaments kinesin late endosome lumen lysosomes messenger RNA (mRNA) microfilaments microtubule microtubules mitochondria mitochondrial DNA (mtDNA) mitochondrial matrix morphology motor protein nuclear envelope nuclear lamina nuclear lamins

nuclear pore complex nuclear pores nucleoid nucleolus nucleus organelles organizing center perixosomes phagocytosis phagocytosis phylogeny plasma membrane plasmids prokaryote pulse-chase experiment receptor-mediated endocytosis ribosomal RNA (rRNA) ribosomes rough endoplasmic reticulum

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(RER) signal recognition particle smooth endoplasmic reticulum streaming stroma thylakoids vacuole vacuoles .

Lecture Activities Case Study: Asbestosis Estimated duration of activity: 15+ minutes Case studies are an excellent way to increase student interest in biology. This exercise may be done completely in class (if class or a discussion section can meet in a computer lab), or it may be started in class and finished as an out-of-class assignment. It may also be assigned as homework and discussed during the next class session.

This case study guides students to investigate the important roles that lysosomes play in human cells and to understand the consequences of lysosomal dysfunction.

Procedure: 1. After dividing the class into groups, hand out the first part of the case study; ask them to brainstorm and make a list of questions about the case.

2. Have each group come up with one or two questions that will be assigned to each group member for part 1 (see the following part descriptions).

3. After the lists are complete (this should take about 10 minutes), give the groups part 2; have students add to their lists and revise their questions from part 1.

4. Repeat for parts 3 and 4. By the end of the class period, each group should have formulated a list of questions that can be divided among the group members.

5 Group members must then find the answers to their questions. As information sources, encourage them to use the Internet (reputable, referenced sources only), their textbook, other textbooks, scientific or medical journals, and so on. You may want to have them bring their information to the next class, allowing groups to meet again to finish the assignment.

6. Alternatively, students may report to each other by e-mail (or on a class website) and then write individual responses to the assignment.

7. As a third alternative, you may choose to do this exercise quickly as a class. In this case, you may provide all the answers to the brainstorming group questions. This approach allows the students to reach a conclusion by the end of class but does lose the self-directed research aspect of the exercise.

Part descriptions: • Part 1: Chuck hasn’t been feeling well lately. He is often out of breath, his chest feels tight, and he isn’t eating much. But, worst of all, he is constantly coughing up mucus.

• Part 2: Chuck has worked for American/Atlantic Construction Company for 25 years.

They are renovation specialists who restore old office buildings. Chuck is on the team that removes old, asbestos-containing insulation in the buildings, replacing it with asbestos-free insulation. Chuck likes his job, but he hates the protective suits he must wear while removing the old insulation. He much preferred the early years with the company, when they didn’t have to wear all that stuff! The company says the suits are protection against asbestos. Chuck doesn’t understand the big fuss about asbestos. He can’t see it, so how could it possibly hurt him?

• Part 3: Chuck goes to the doctor and describes his symptoms. The doctor listens to his chest and says he can hear a crackling sound when Chuck breathes. He also has Chuck

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breathe into an instrument and then comments that Chuck has very low tidal volume. Last, the doctor orders chest X-rays and, after reviewing them, tells Chuck that he has found bilateral calcified plaques in the lower quadrant of the lungs as well as several irregular opacities in the same regions.

• Part 4: You are a healthcare representative for the union Chuck belongs to. Chuck is overwhelmed by medical bills and is contacting you to see whether he has a valid case to claim this condition on the company’s worker’s compensation program. Chuck has not yet gotten a diagnosis from his doctor, but he hopes that you can help him figure out what is going on and what caused this problem. He has heard that you are really nice and might be able to explain to him what is happening. Compose an e-mail to Chuck covering these questions: What is wrong with Chuck? Which specific organ is affected by this ailment?

What is happening in those cells? What caused this ailment? What are the treatment and prognosis for this ailment? Do you think this is a valid claim for worker’s compensation?

Do you think the company was at fault, or was Chuck?

Internet resources: www.mayoclinic.com/health/asbestosis/DS00482 www.nlm.nih.gov/medlineplus/ency/article/000118.htm

Think, Pair, and Share Estimated duration of activity: 20 to 25 minutes Toward the end of Chapter 7, pose the following question to the students to Think about in preparation for the next class. In the next class, have them form groups, or Pair with each other, and discuss their thoughts on the question. Then, ask each group to Share their thoughts with the entire class. The question is: Eukaryotic cells are generally about 1,000 times the size of prokaryotic cells. What are the structures and mechanisms that evolved in eukaryotic cells that enabled them to become so much larger than prokaryotic cells? Tell the students there are at least two major differences between the two types of cells that account for their size difference.

Here are the answers that should come out by the end of the discussion: 1. Prokaryotes have only one membrane surrounding each cell. This limits the size of the cell because the cell’s internal volume cannot be so large that the force of diffusion is unable to move a molecule from the outside of the cell to the center of the cell. If the distance from the cell surface to the center of the cell becomes too large, molecules will not be able to diffuse into the center of the cell because diffusion is a local force that cannot move molecules great distances.

2. Eukaryotic cells, in contrast, have evolved many internal membrane-bound organelles.

All of those membranes count as part of the total surface area of the cell. Therefore, the distance from one membrane to another is never so great as to prevent the force of diffusion from moving a molecule.

3. The first two answers are part of the one major difference the students need to identify.

The other difference is how molecules are moved around in the eukaryotic cell—packaged in vesicles traveling on the fibers of the cytoskeleton. This is directed motion, powered by the hydrolysis of ATP. Thus, the expenditure of energy allows the cell to overcome the limitations of the force of diffusion.

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Exercise objectives: • Review the relevant portions of Chapter 7 that cover the internal structure of prokaryotic and eukaryotic cells.

• Compare and contrast those structures.

• Come up with ideas that address the question.

• Participate in class discussions that further the entire class’s understanding of the topic.

Procedure: 1. Ask students to review Chapter 7 and compare eukaryotic cell structures with prokaryotic cell structures.

2. Ask students to think about the question and then write down their responses, issues, and additional questions that are relevant.

3. Ask students to bring their written ideas to class and then form groups to discuss their ideas. At the end of the class period, collect each student’s written ideas. The written ideas should be evaluated in a formative way, as should anything collected from this exercise.

The evaluation should be just a check mark to indicate that the exercise was done. Provide students with helpful comments to consider for the next time this type of exercise is attempted.

4. The groups must discuss each person’s idea(s), refine the ideas, and then write down the answers the group is most comfortable with. Collect these ideas for a formative assessment. The discussion should be limited to 10 to 15 minutes.

5. The groups should then informally share their ideas with the rest of the class. After each group has spoken, say something like, “What does everybody think about the other groups’ ideas? Are any of the ideas really good? Are there any ideas we can develop further?”

Survey on the Think, Pair, and Share Activity Name (optional): __________________ Course/section #: __________________ Date of the discussion: ______________ 1. How much time did you spend on your own thinking about the question for discussion?

2. How much time did you spend discussing the question with your group?

3. How well did the group discussion go for you? Did you feel you made your opinion known? Did you feel your opinion was carefully considered by the group?

4. How well did you think the class discussion went?

5. Overall, how much did you get out of this exercise?

6. Any other comments?

In-Class Demonstration: Pulse-Chase Experiment Estimated duration of activity: 5 minutes for lecturing, then 3 to 5 minutes to explain the demonstration This activity is rather elementary, but it will give the students a fun way to visualize how a pulse-chase experiment works. Because so many students are visual learners, role-playing

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demonstrations are often a fun and visual way to help them understand dynamic biological processes. Even the brightest students will at least appreciate the break from lecture.

After or during an explanation of the pulse-chase experiment and its usefulness in visualizing molecule movement in a cell, perform the following demonstration.

Purpose: Tell students they are going to create a pathway that you will be able to define without knowing it ahead of time (even the students won’t know the pathway because it is created as the demonstration progresses) and without watching the pathway directly.

Instructions to students: Tell students that when they are handed a piece of paper (differentcolored sticky notes work well), they must discreetly hand it to a nearby student of their choosing. The main rule is that each student must always hand every note they are given to the same nearby student. They must pass their note along only when they receive a new note.

Instruct them to end the “pathway” after a reasonable number of students have become involved; the self-designated “end” person may accumulate the sticky notes. Instruct students to temporarily stop passing when you say “Data collection.” At that time, only the people holding pink notes should hold them up. You can then say “Resume” to indicate that they can start passing notes again.

Logistics: Start the demonstration by handing a student in the front row a yellow note (it works better when a student or a teaching assistant does this, so the instructor can turn his/her back or simply lecture without paying attention to where the notes are going). Ask the students to wait several seconds before passing the note along. After they have given their note away, keep handing that first student yellow notes and have him/her pass them along.

After the tenth note or so, begin handing the first student pink notes. After handing out five or six pink notes, switch back to yellow. (The pink is your “pulse” and the yellow is your “chase.”) When you “collect data,” record the position of the students holding pink papers.

After several cycles of data collection, you should be able to map the pathway that the students have created.

Analogies: The classroom represents the interior of the cell, and the students represent an organelle system (the endomembrane system or the cytoskeleton, for example). The sticky notes represent molecules: The pink notes are labeled molecules that form the pulse, and the yellow notes are the chase of unlabeled molecules. The data-collection periods represent the method researchers use to “stop” different samples in order to visualize where the label is at different times after the pulse has been initiated.

Discussion Idea After discussing protein synthesis and modification through the endomembrane system, ask your students the following questions: How are proteins sorted into vesicles? How are those vesicles targeted to the appropriate destination? Then ask students to develop one or more hypotheses, and ask them what experiments they might perform to test their hypotheses.

Describe the experimental design performed by Lippincott-Schwartz and colleagues as described in their 1997 paper in Nature (389: 81–85). Describe each experiment and review each data figure, pointing out the pertinent points. As you look at each figure, ask a student to summarize each experiment and record the data on the board. When all of the data have

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been recorded, ask the students to summarize the major findings of the paper. Then use their conclusions to segue into discussing Figure 7.22, the current model for how proteins are sorted into distinct vesicles and targeted to the appropriate destination.

Note: This activity will work best in a classroom outfitted with computer projection and an Internet connection, or in a computer lab.

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CHAPTER

8

Energy and Enzymes: An Introduction to Metabolic Pathways

Learning Objectives: Students should be able to...

•

Explain how energy is created or used up in chemical reactions.

•

Describe how chemical reactions will continue until all the reactants are exhausted.

•

Discuss enzymes, which provide the energy required to make endergonic reactions occur.

•

Explain why exothermic reactions are always spontaneous and endothermic reactions are always nonspontaneous.

• • •

Explain why breaking the bonds between phosphates in ATP is exergonic.

Describe redox reactions and explain how hydrogen atoms are transferred between molecules.

Explain how, if a reaction is spontaneous, it will occur quickly.

Lecture Outline I. What Happens to Energy in Chemical Reactions?

A. Two types of energy: 1. Potential energy = stored energy Examples include: gravitational, electrical, chemical bonds, and chemical gradients 2. Kinetic energy = energy of motion Examples include: sound, thermal, electricity, and electromagnetic radiation 3. Energy transformation in a waterfall (Fig. 8.1) a. A water droplet at the top of a waterfall has a defined amount of potential energy. b. As it falls, some of its potential energy is converted to kinetic energy. c. When it hits the bottom, its potential energy is lower. The change in potential energy has been transformed into equal amounts of kinetic.

B. Chemical reactions involve energy transformations.

1. Kinetic energy is typically in the form of potential/thermal energy stored in chemical bonds. The amount of potential energy is based on the position of the shared electrons relative to the nuclei of the bonded atoms. (Fig. 8.2) 2. First law of thermodynamics: Energy is not created or destroyed; it just changes form.

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3. The total energy in a molecule is referred to as its enthalpy (H).

4. Example of change in enthalpy (ΔH) illustrated by the bombardier beetle (Fig. 8.3). a. When provoked, the bombardier beetle can produce scalding hot liquid. b. A large change in enthalpy occurs when hydrogen peroxide is broken down into water and oxygen (2 H2O2  2 H2O + O2). c. The heat from the reaction also increases the temperature. When a reaction releases heat, it is exothermic and ΔH is negative.

5. If heat is taken up, ΔH is positive and the products of the reaction have higher potential energy than the reactants.

6. Entropy (S) or the amount of disorder also changes during a chemical reaction. a. When products are less ordered than the reactants were, entropy is positive (ΔS positive) b. Second law of thermodynamics states that entropy always increases in a system that includes the reactants and surroundings.

7. Spontaneous chemical reactions are those that proceed on their own without any added energy. a. Reactions tend to be spontaneous if either or both of these conditions are met: (1) The products have lower potential energy than the reactants (negative ΔH; exothermic reaction). (Fig. 8.3) (2) The product molecules are less ordered than the reactant molecules (positive ΔS; increase in entropy).

8. The Gibbs free-energy change, ΔG, summarizes the combined effects of changes in heat and disorder. a. ΔG = ΔH – TΔS. b. If ΔG is negative, the reaction is exergonic and spontaneous. c. If ΔG is positive, the reaction is endergonic and not spontaneous. d. If ΔG = 0, the reaction is at equilibrium.

9. Students should be able to explain why the same reaction can be nonspontaneous at low temperature but spontaneous at high temperature.

10. Students should be able to explain why some exothermic reactions are nonspontaneous.

B. Temperature and concentration affect reaction rates.

1. Higher concentrations and temperatures tend to speed up chemical reactions. (Fig. 8.4) and (BioSkills 3)

II. Nonspontaneous Reactions May Be Driven Using Chemical Energy A. Redox reactions transfer energy via electrons.

1. Energetic coupling between exergonic and endergonic reactions allows chemical energy released from one reaction to drive another. (Fig. 8.5) 2. In a reduction-oxidation or redox reaction, one compound loses an electron and another compound gains it. a. Electrons may be completely transferred, or they may shift positions within a covalent bond.

3. The electron donor becomes oxidized, and the electron acceptor becomes reduced.

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4. The reduced compounds have a high potential energy and act as electron donors to oxidized molecules that can receive that energy.

5. The electron that is transferred is usually accompanied by a proton so that the reduced atom gains a hydrogen atom.

B. An example of redox in action.

1. The oxidation of glucose releases energy: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + 686 kcal. (Fig. 8.6) a. Glucose is the reduced molecule with a high potential energy that acts as the electron donor. b. Oxygen is the oxidized molecule that ultimately acts as the electron acceptor.

C. Another approach to understanding redox. (Fig. 8.7) 1. During redox reactions, electrons may be transferred from an atom in one molecule (electron donor) to an atom in a different molecule (electron acceptor).

2. Following hydrogen atoms, reduction adds H and oxidation removes H. a. FAD is reduced by 2 electrons (and 2 protons) to form FADH2, which readily donates these high-energy electrons to other molecules (considered the electron carrier). b. NAD+ is another common electron carrier that is reduced by 2 electrons to form NADH; however, only 1 proton is carried.

D. ATP transfers energy via phosphate groups.

1. ATP (adenosine triphosphate) has a lot of potential energy.

2. Its three phosphate groups contain four negative charges that repel one another but are confined to a small space. (Fig. 8.8a) 3. When ATP is hydrolyzed, the terminal phosphate is released in a highly exergonic reaction. (Fig. 8.8b) 4. ATP hydrolysis releases free energy. a. The energy released from the hydrolysis of ATP is used to make things happen in a cell. b. The phosphate released from ATP is transferred to other molecules (like proteins) via a phosphorylation reaction.

5. How does ATP drive endergonic reactions? a. Phosphorylation activates substrates or enzymes, giving them a higher free energy. b. This higher free energy transforms endergonic reactions into exergonic reactions that can now proceed. c. Therefore, phosphorylation couples endergonic reactions to exergonic reactions.

(Fig. 8.9) d. Students should be able to compare and contrast how energy is transferred via redox reactions and ATP hydrolysis.

III. How Enzymes Work A. Enzymes help reactions clear two hurdles: 1. Enzymes bring substrates together in a precise orientation making reactions more likely. Enzymes are catalysts.

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a. An enzyme binds specifically to a particular reactant or set of reactants (lock-andkey model). b. Enzymes often flex or change shape when they bind to a substrate (a flexible “lock”). This change in shape is called induced fit. (Fig. 8.10) c. After the binding, interactions with R-groups at the active site stabilize the transition state. (The transition state is the “key.”)

2. Enzymes lower the activation energy.

3. Spontaneous reactions have an overall decrease in free energy but need an investment of energy (activation energy) to get going.

4. Activation energy is the energy necessary to reach the transition state. a. Reactions happen only when reactants can reach the transition state, an intermediate state of new and old bonds. (Fig. 8.11) b. Reactants can reach the transition state on their own if they have enough kinetic energy.

5. Enzymes are catalysts, meaning that they lower activation energy. (Fig. 8.12) a. A catalyst is any substance that lowers the activation energy of a reaction and thus increases the rate of the reaction. b. A catalyst is not consumed in the reaction and does not affect the overall energy change (∆G). c. Proteins that catalyze reactions are called enzymes. d. Most of the important reactions in biology would proceed at imperceptible rates, or not at all, without enzymes.

6. Enzyme catalysis is a three-step process: (Fig. 8.13) a. Initiation: The enzyme precisely orients the substrates as they interact with R-groups in the active site via hydrogen bonds, ionic bonds, and others. b. Transition-state facilitation: The interaction between substrate and active site is maximized as the transition state forms. c. Termination: The products of the reaction have a low affinity for the active site, and they are released.

7. Students should be able to complete the following sentences: 1) Enzymes speed reaction rates by ______ and lowering activation energy. 2) Activation energies drop because enzymes destabilize bonds in the substrates, forming the ______. 3) Enzyme specificity is a function of the active site’s shape and the chemical properties of the _____ at the active site. 4) In enzymes, as in many molecules, function follows from _____.

B. What limits the rate of catalysis?

1. The rate of an enzyme-catalyzed reaction increases with increasing substrate concentration, up to a point. (Fig. 8.14) 2. After all the active sites are occupied, further increases in substrate concentration have no effect.

C. Do enzymes work alone?

1. To function properly, some enzymes need additional atoms or molecules that are not part of their primary structure:

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a. Cofactors—metal ions (such as Zn2+, Mg2+, or Fe2+) that reversibly interact with enzymes b. Coenzymes—organic molecules that reversibly interact with enzymes, such as the electron carriers NADH or FADH2 c. Prosthetic groups—non-amino acid atoms or molecules that are permanently attached to proteins (e.g., retinal and many vitamins)

IV. What Factors Affect Enzyme Function?

A. Enzymes are optimized for particular environments.

1. Any condition (such as temperature and pH) that alters an enzyme’s structure alters its function.

2. Temperature affects the kinetic energy of the substrates and enzyme and can directly affect enzyme structure. Example: chitinase (Fig. 8.15a) 3. pH often affects an enzyme’s structure and function. Example: chitinase (Fig. 8.15b) B. Most enzymes are regulated.

1. Regulating enzymes via noncovalent interactions does not permanently affect the enzymes’ primary structure.

2. Reversible modifications (noncovalent) affect enzyme function in two ways: a. Competitive inhibitors bind to the active site, physically blocking substrate binding.

(Fig. 8.16a) b. Allosteric modulators indirectly affect the active site by binding elsewhere on the enzyme and changing its overall shape. (Fig. 8.16b) 3. Regulating enzymes via covalent modifications occurs when the enzyme is altered by a chemical change in its primary structure. a. The most common modification is the addition of one or more phosphate groups.

(Fig. 8.17)

V. Enzymes Can Work Together in Metabolic Pathways A. Metabolic pathways are regulated.

1. Each of the molecules of life is built by a series of reactions, each catalyzed by a different enzyme. These multistep processes are called metabolic pathways.

2. Students should be able to predict how inactivating the enzyme catalyzing the second reaction in a 3-reaction sequence would affect the concentrations of the products before and after this particular reaction.

3. When an enzyme pathway is inhibited by the product of the reaction sequence, feedback inhibition occurs. (Fig. 8.18) B. Metabolic pathways evolve.

1. The retro-evolution hypothesis was developed by Norman Horowitz. If an original substrate were depleted, natural selection would next favor the evolution of a new enzyme to make the product from other existing molecules. (Fig. 8.19) 2. Patchwork evolution refers to how early enzymes catalyzing some pathways were recruited to new pathways. For example, bioremediation of pentachlorophenol by bacteria.

3. Pathways that break down molecules for sources of energy and carbon building blocks are called catabolic pathways, while those that use energy and carbon building blocks to synthesize molecules are called anabolic pathways.

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Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

(NAD+)/(NADH) activation energy active site adenosine triphosphate (ATP) allosteric regulation anabolic pathways anabolic pathways bioremediation catabolic pathways catabolic pathways catalysis catalysts coenzymes cofactors competitive inhibition covalent modification electron acceptor

electron carrier electron donor endergonic endothermic energetic coupling enthalpy entropy enzymes exergonic exothermic feedback inhibition first law of thermodynamics flavin adenine dicnucleotide (FAD)/(FADH2) Gibbs free-energy induced fit kilocalorie

kinetic energy metabolic pathways nicotinamide adenine dinucleotide oxidation phosphorylation potential energy prosthetic groups redox reactions reduction reduction-oxidation reactions second law of thermodynamics Standard free energy change substrate transition state

Lecture Activities Student-Led Concept Illustrations Estimated duration of activity: A few minutes as you lecture Often, getting students involved is as simple as having them illustrate a phenomenon in their own terms. The following questions can be presented to students, allowing them to explain these concepts in terms that their peers can understand.

Energy: After introducing different forms of energy, ask students what sort of energy transformation is occurring in the following common daily activities. (Once they have the idea, you may challenge them to come up with additional examples.)

• Rubbing hands together to get warm (mechanical energy to thermal energy) • Hearing (sound energy to mechanical energy; sound waves cause our eardrums to vibrate) • A plant growing (electromagnetic energy to chemical energy; photosynthesis) • A person riding a bicycle (chemical energy to mechanical energy)

Discussion Idea: Calories and Body Temperature Students like to relate what they learn to the processes that occur in their own bodies. While discussing Gibbs free energy and the laws of thermodynamics, you can relate these concepts to nutrition and body temperature. Begin by asking students these two questions: • Why do you heat up when you exercise? .

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• Why is exercising said to “burn calories”? Give students two hints: (1) In the field of nutrition, the calorie is a unit of energy; that is, if one serving of a food contains 50 “calories,” that food will release 50 kilocalories of energy when broken down by cells. (2) In any spontaneous reaction, at least some of the energy is converted into heat and is “lost” (cannot be captured in chemical bonds). With those two hints, students should be able to reach the following conclusions: • The oxidation of food releases energy captured by cells in the bonds of ATP; that is, chemical energy in food molecules is converted into chemical energy in ATP molecules. Some energy is lost as heat.

• When you exercise, the energy in the bonds of ATP is used to fuel muscle contraction.

Again, some of this energy is lost as heat.

• The more molecules of ATP a muscle uses, the more energy is lost as heat. Muscles rapidly heat up. They also rapidly “run out” of ATP molecules and soon need more food molecules to produce more molecules of ATP. (Students may be interested to learn that at maximum contraction, a muscle will run out of ATP molecules in just a few seconds. This is why athletes can sustain a maximum contraction, such as in Olympic weight lifting, for only a few seconds.)

Evaluating Data Estimated duration of activity: 10–15 minutes For this activity, students may work individually, in pairs, or in small groups. Another variation that works quite well is to have students work individually and then share their answers with others. At the end of the activity, allow time for general class discussion and present the optimal answers.

Provide students with Figure 8.1 below:

Explain either verbally or with written instructions that the graph represents an experiment in which the rate of a biochemical reaction is measured at a constant enzyme concentration with

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varied concentrations of substrate. The experiment is repeated with the same enzyme at different pH values.

Have students refer to Figure 8.2 (below) to complete the following items:

1. This drawing illustrates part of the experiment. The test tube on the left contains enzyme but no substrate at pH 8. The next tube contains enzyme with the highest concentration of substrate shown in the graph in Figure 8.1, also at pH 8. For the two remaining test tubes, diagram the experimental conditions relating to points A and B on the graph in Figure 8.1.

2. Which of the pH values tested in this experiment is most ideal for the action of the enzyme? How can you tell?

3. Figure 8.3 (below) is a diagram that illustrates a more detailed view of the interaction between individual molecules of this enzyme and the substrate at pH 7. Draw how the same protein might interact with the substrate at pH 8 and at pH 6.

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CHAPTER

9

Cellular Respiration and Fermentation

Learning Objectives: Students should be able to...

• Evaluate the claim that glycolysis, pyruvate processing, and the citric acid cycle represent the central metabolic pathway in cells.

• Summarize the two phases of glycolysis, including the inputs and outputs and how it is regulated.

• Describe the three steps of pyruvate processing and how inputs and outputs regulate •

enzyme activity.

Explain what happens to the redox state of the carbons from the original glucose molecule

• during the citric acid cycle.

Use the analogy of a hydroelectric dam to describe how electron transport is coupled to • oxidative phosphorylation.

Compare and contrast the inputs, outputs, and energetic yield of fermentation and cellular respiration.

Lecture Outline I. An Overview of Cellular Respiration A. In cells, glucose is oxidized through a long series of controlled steps; its energy is then harnessed to fuel the synthesis of ATP. (Fig. 9.1) B. What happens when glucose is oxidized?

1. Glycolysis: One molecule of glucose is broken down into two molecules of pyruvate, two ATP are produced, and NAD+ is reduced to NADH. (Fig. 9.2) 2. Pyruvate processing: Pyruvate is processed to form acetyl CoA, during which another NADH is produced. (Fig. 9.2) 3. Citric acid cycle: Acetyl CoA is oxidized to two CO2 molecules. During this cycle, more ATP and NADH are produced as well as FADH2. (Fig. 9.2) 4. Electron transport and chemiosmosis: Electrons from NADH and FADH2 move through a series of proteins called the electron transport chain. The potential energy released during these reactions is used to create a proton gradient, which is used to make ATP. (Fig. 9.2)

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C. Cellular respiration plays a central role in metabolism.

1. Catabolic pathways break down a variety of molecules such as carbohydrates, fats, and proteins. (Fig. 9.3) 2. Catabolic intermediates are used in anabolic pathways. a. About half of the amino acids can be synthesized from molecules siphoned from the citric acid cycle. b. Acetyl CoA is the starting point for fatty-acid synthesis. c. Intermediates in glycolysis can be used to make nucleotides. d. If ATP is abundant, pyruvate and lactate can be used to synthesize glucose.

3. Regulating these reactions is the key to the metabolic activity of the cell and to homeostasis. (Fig. 9.4) D. Tips for drawing models of cellular respiration (Making Models 9.1)

II. Glycolysis: Oxidizing Glucose to Pyruvate A. Glycolysis is a sequence of 10 reactions. (Fig. 9.5) (Table 9.1) 1. Initially, energy input is needed to phosphorylate (energize) glucose. a. ATP is used to phosphorylate glucose into glucose-6-phosphate. b. An enzyme rearranges this substance into fructose-6-phosphate. c. ATP is used again to add another phosphate, forming fructose-1,6-bisphosphate.

2. The next set of reactions yields an energy payoff. a. The sixth reaction results in the reduction of two NAD+ molecules to yield two NADH molecules. b. The enzymes in reactions 7 and 10 remove phosphates placed on the intermediates in the energy-input phase and use them to phosphorylate ADP, forming ATP. This process is called substrate-level phosphorylation. (Fig. 9.6) c. For each molecule of glucose processed, the net yields are 2 NADH, 2 ATP, and 2 pyruvate molecules.

B. How is glycolysis regulated?

1. High concentrations of ATP regulate glycolysis by inhibiting the glycolytic enzyme phosphofructokinase; the enzyme catalyzes the synthesis of fructose-1,6-biphosphate from fructose-6-phosphate.

2. Regulation of glycolysis occurs at the “committed step” and is an example of feedback inhibition. (Fig. 9.5) a. This is a committed step in the glycolytic pathway because fructose-1,6bisphosphate can be used only in the glycolytic reactions. b. Before this step, the process can be halted and the intermediates used elsewhere in the cell. c. When an enzyme is inhibited by a product of the pathway in which it is involved (like PFK inhibition by ATP), the event is called feedback inhibition. d. Regulation of the glycolytic pathway by ATP allows the cell to conserve glucose when ATP is plentiful.

3. Determination of the three-dimensional structure of the enzyme revealed two binding sites for ATP. (Fig. 9.7) a. One ATP binding site is in the active site, where ATP is used as a reactant.

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b. The other ATP binding site is a regulatory site that, when occupied, results in a conformational change that in turn results in inhibition of the enzyme. c. Phosphofructokinase is an allosteric enzyme.

4. Students should be able to explain the importance of the two ATP binding sites on phosphofructokinase. They should also be able to predict the consequences if the regulatory site had higher affinity for ATP than the active site.

III. Processing Pyruvate to Acetyl CoA A. The pyruvate generated from glycolysis moves into the mitochondrial matrix via active transport. (Fig. 9.8) B. Pyruvate reacts with coenzyme A to form acetyl CoA. 1. When coenzyme A is acetylated, the acetyl group becomes activated and can easily be transferred to another molecule.

2. The enzyme pyruvate dehydrogenase catalyzes the reaction to form acetyl CoA. (Fig. 9.9) a. During this reaction, an NADH is produced. b. One of the carbons in pyruvate is oxidized to form carbon dioxide. c. The two remaining carbons are transferred to coenzyme A. C. Pyruvate dehydrogenase is regulated by phosphorylation.

1. Phosphorylation occurs when ATP, NADH, or acetyl CoA concentrations are high.

2. High levels of AMP, NAD, or coenzyme A speed up the catalytic rate of pyruvate dehydrogenase.

3. Pyruvate dehydrogenase is under both positive- and negative-feedback control, making it a key regulatory point in glucose oxidation.

IV. The Citric Acid Cycle: Oxidizing Acetyl CoA to CO

2

A. Discovery of the citric acid cycle 1. Early researchers observed that eight small carboxylic acids are oxidized, presumably during glucose metabolism. a. When researchers added these molecules to cells, the rate of respiration increased. b. All eight of these molecules could be recovered later, implying that they are not used but recycled.

2. Hans Krebs hypothesized that this pathway was cyclical and linked to the processing of pyruvate produced by glycolysis. a. Krebs and W. A. Johnson found that pyruvate could react with the end product of the pathway, oxaloacetate, to produce the first carboxylic acid in the pathway, citrate. (Fig. 9.10) (Table 9.2) b. Krebs concluded that during this pathway, pyruvate is oxidized to carbon dioxide.

B. How is the citric acid cycle regulated? (Fig. 9.11) 1. The enzyme complex that converts pyruvate to acetyl CoA is regulated. a. This complex is phosphorylated and inhibited when levels of ATP are high. b. High levels of NADH or acetyl CoA also inhibit the activity of this complex. c. In contrast, high levels of AMP, NAD+, and CoA enhance the activity level of this enzyme complex.

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2. High levels of ATP also regulate the enzyme that catalyzes the formation of citrate from oxaloacetate and acetyl CoA. 3. Regulation of both glycolysis and the Krebs cycle allow cells to carefully match the rate at which they use glucose to their energy requirements.

C. What happens to the NADH and FADH2?

1. For each molecule of glucose oxidized, 10 NADH are produced, 2 FADH2 are produced, 4 ATP are produced, and 6 carbon dioxide molecules are released. (Fig. 9.12) 2. ATP is used to fuel endergonic reactions, and the carbon dioxide is released as waste.

3. The NADH and FADH are used to reduce oxygen to form water. The oxidized NAD+ 2 and FAD are recycled.

4. Figure 9.13 summarizes the free-energy changes that take place.

V. Electron Transport and Chemiosmosis: Building a Proton Gradient to Produce ATP A. The ETC 1. Four of the ETC components are proteins, anchored to the inner mitochondrial membrane, that contain chemical groups such as flavin, iron-sulfur, heme, or copper, where the reduction-oxidation reactions take place.

2. One of the components is a lipid-soluble coenzyme called ubiquinone (Q), which can move through the membrane and shuttle electrons from one protein complex to the other.

3. The electron carriers NADH and FADH2 donate electrons to the ETC. a. The ETC components are organized from least oxidized to most oxidized. b. NADH donates electrons to a flavin-containing protein at the top of the chain. c. FADH2 donates electrons to an iron-sulfur complex that donates them to Q, effectively skipping a step in the ETC. d. Oxygen is highly oxidized and is the final electron acceptor in many cell types.

4. Organization of the ETC a. Figure 9.14 shows how electrons step down in potential energy from the electron carriers NADH and FADH2. (Table 9.3) b. Students should be able to use Figure 9.14 to identify the ETC electron acceptor with the highest redox potential and the acceptor with the lowest redox potential.

5. The movement of electrons through the ETC actively transports protons from the matrix, across the membrane, and into the intermembrane space. (Fig. 9.15) a. In complexes I and IV, the protons presumably pass directly through the complex of electron carriers. b. In complex III, Q shuttles protons and electrons. When it donates the electrons to complex IV, it releases the protons into the intermembrane space. c. Protons released into the intermembrane space contribute to the proton-motive force. d. Cytochrome c and Q shuttle the electrons from one complex to the other.

B. The discovery of ATP synthase 1. Efraim Racker noticed that inner mitochondrial membrane had many large proteins studded along the surface. (Fig. 9.16)

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2. He also found that the stalks and knobs protruding from these proteins could hydrolyze ATP, forming ADP. 3. By removing and replacing them, he discovered the protein used in the synthesis of ATP: ATP synthase.

C. The chemiosmosis hypothesis 1. Peter Mitchell hypothesized that the ETC indirectly provides energy for ATP production by producing a proton gradient (chemiosmosis). a. The ETC uses the energy from reduction-oxidation reactions to fuel the pumping of protons up a concentration gradient from the matrix to the inner membrane space. b. This creates an energy-rich, proton-motive force. c. An enzyme in the inner membrane then uses the proton-motive force to synthesize ATP. 2. Several researchers performed experiments that supported the chemiosmotic hypothesis. a. One experiment utilized laboratory-synthesized vesicles to demonstrate that a proton gradient can fuel ATP synthesis. (Fig. 9.17) (1) The researchers synthesized membranes that contained ATP synthase from mitochondria and bacteriorhodopsin, a protein that pumped protons across a membrane when exposed to light.

(2) They illuminated these synthetic vesicles while in a solution of ADP and inorganic phosphate and observed ATP production.

(3) These studies demonstrated that NADH and the ETC were not necessary for ATP synthesis when something else was present to create the proton-motive force.

3. Chemiosmosis can be compared to a hydroelectric dam. a. The hydroelectric pumps are analogous to the electron transport chain. b. The turbines are analogous to ATP synthase.

4. Students should be able to explain why ATP production is characterized as indirect and explain the relationship among glucose oxidation, the proton gradient, and ATP synthase.

5. The proton-motive force couples electron transport to ATP synthesis. (Fig. 9.18) 6. Hydrothermal vents may have played a role in chemical evolution of electrochemical gradients.

D. Organisms use a diversity of electron acceptors.

1. Figure 9.19 summarizes glucose oxidation and cellular respiration.

2. About 29 ATP are made per glucose: 25 by ATP synthase via oxidative phosphorylation and 4 via substrate-level phosphorylation.

3. Species that depend on oxygen as an electron acceptor for the ETC use aerobic respiration and are called aerobic organisms.

4. Cells that depend on electron acceptors other than oxygen are using anaerobic respiration.

5. Aerobic respiration is much more efficient than anaerobic respiration.

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VI. Fermentation A. Fermentation allows glycolysis to continue. (Fig. 9.20) 1. High levels of NADH inhibit the enzyme complex that converts pyruvate into acetyl CoA. 2. As a result, pyruvate builds up and is made available as a substrate for the enzymes of fermentation.

3. These enzymes use NADH to reduce pyruvate, producing NAD+ that can be used for glycolysis.

B. Many different fermentation pathways exist.

1. The reduction of pyruvate results in the production of lactate or ethanol, depending on the set of enzymes involved. a. Muscle cells contain enzymes that convert pyruvate to lactic acid. (Fig 9.21a) b. Some bacteria and yeast contain enzymes that convert pyruvate into ethanol.

(Fig. 9.21b) C. Fermentation as an alternative to cellular respiration.

1. Fermentation is extremely inefficient. a. It produces only 2 ATP molecules per glucose processed. b. In contrast, aerobic respiration produces about 29 ATP per glucose processed.

2. Organisms that can perform both aerobic and anaerobic respiration are called facultative anaerobes.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. acetyl CoA aerobic respiration alcohol fermentation anabolic pathways anaerobic respiration ATP synthase carboxylic acids catabolic pathways cellular respiration chemiosmosis citric acid cycle (TCA) coenzyme A (CoA) coenzyme Q cristae cytochrome c

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dehydrogenase electrochemical gradient electron carrier electron transport chain (ETC) facultative anaerobes FAD FADH2 feedback inhibition glucose glycolysis guanosine triphosphate (GTP) homeostasis Kreb Cycle

lactic acid fermentation mitochondrial matrix NADH nicotinamide adenine dinucleotide (NAD) oxidative phosphorylation phosphofructokinase phosphorylation proton-motive force pyruvate pyruvate processing redox potential regulatory site substrate-level ubiquinone

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Lecture Activities A Case Study about Dinitrophenol (DNP) Estimated duration of activity: 15+ minutes Case studies are an excellent way to increase student interest in biology. This exercise could be done completely in class (if the class or discussion section can meet in a computer lab), or it can be started in class and finished as an out-of-class assignment. It can also be assigned as homework and discussed during the next class session.

This case study allows students to investigate the concept of chemiosmosis and the consequences of oxidative uncoupling.

One day during winter break, Mary was talking with her grandmother. While reminiscing about her college years in the 1930s, the grandmother recalled a sad memory. When she was a junior in college, she had a good friend named Barbara who became very sick and died during the spring semester. Like many young women, Barbara had been preoccupied with her weight. She was planning to work as a model the following summer and wanted to lose weight before then. A new miracle diet pill had just come on the market, and Barbara began taking it. Figuring that more was better, she took more than the recommended dose of pills.

Before long, Barbara began experiencing blurry vision and a constant high fever that she couldn’t shake. Barbara eventually died, and her death was ultimately blamed on the diet pills. The pills were later taken off the market because they caused cataracts and sometimes death in many people who were taking them.

Mary asked her grandmother why the pills made people die, but her grandmother didn’t know. All she remembered was that they were called Dinistriso and had something called DNP in them.

Help Mary figure out why these diet pills were killing people. What is DNP, and why was it in a diet pill? Why did it cause the symptoms that Barbara was experiencing, and why did it eventually kill her? What is DNP (legally) used for today? Why does it work for that purpose?

This case study could also lead into a discussion of brown fat and naturally occurring oxidative uncouplers in babies and bears.

A Concept Map for the Chapter Estimated duration of activity: 20–30 minutes The material in this chapter is often overwhelming for students. They tend to get lost in the details and lose sight of how it all relates. After students read this chapter, present a helpful review by having your students draw a large concept map that links everything together, from glycolysis to ATP synthesis.

Steps in making a concept map: 1. Draw a cutaway diagram showing a single (very large) mitochondrion, reserving about half the board to represent cytoplasm. (If you have enough board space, dedicate half to all of it for this diagram. If you do not have enough board space, bring in a large piece of mural paper or several pieces of poster board.)

2. Ask a student to come to the board. By following his classmates’ directions, have him begin with a glucose molecule in the cytoplasm. Instruct the student to use generic lines and circles for the pathways, simply to illustrate the big picture.

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3. After you have finished each section of material (glycolysis, Krebs, ETC), ask a student to go to the concept map and fill in the details you have emphasized. For example, it is probably important to include arrows showing where ATP is used and where it is made, where NAD+ is used, and where NADH is made. As the students link the pathways together, they can draw arrows to show how the pathways interact. For example, the NADH molecules made during glycolysis and the Krebs cycle donate electrons to the electron transport chain.

4. Once the class has drawn the entire respiratory chain, have a student draw an X through oxygen and then discuss what happens when going back through the chain to initiate a switch to fermentation. Then ask them to add fermentation to the diagram. Last, have the student draw an X through glucose and discuss where fats and proteins can be metabolized and enter the chain. You can also use the diagram to highlight the major points of regulation.

Note: This idea will also work for a large class. It may not be practical to have students come to the board, but you could draw and fill in the concept map, or have a teaching assistant do it. The key is to get the students involved; let them direct the activity.

Learning about Metabolism through Investigations of Metabolic Diseases Estimated duration of activity: two class periods (potentially extended to a major writing assignment) It is common for students to “not see the forest for the trees” when studying metabolic pathways. Even if the material is well organized and well presented in the textbook, students still fail to fully grasp why they are studying metabolic pathways because they are overwhelmed by the details. Unfortunately, instructors frequently do not have the time to go much beyond teaching the details and specific concepts involved with each pathway, simply because there is so much to cover.

This assignment is designed to get students to learn about metabolism from a disease-state point of reference. If they learn about what goes wrong as a result of metabolic disorders, they can better appreciate the importance of metabolism to the basic functions of organisms.

You may want to give students a list of diseases to pick from for their project to ensure that the projects will be productive. Also, give students at least one good source they can easily access, such as a website from the National Institutes of Health. Alternatively, the project may be entirely student designed.

Possible topics: 1. What are the symptoms of the disorder? How has it been diagnosed in the past? How is it treated currently? Do other animals suffer from this disorder?

2. Which pathway is specifically involved in the disorder? The students should thoroughly describe how the pathway in question works normally, and then present research showing how the pathway is dysfunctional in the disease state.

3. How prevalent is the disorder in the population? Is there some benefit to the disorder, such as the relationship of sickle-cell anemia to malaria?

4. What current research is going on regarding this disorder? Have modern genetics and molecular biology influenced the diagnosis and treatment of the disorder? Is a genetics .

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screening test available, or is such a test in the works? Are there any genetics-based treatments for the disease, or are such treatments being researched?

Procedure: 1. Each group of four students should pick a disorder to cover from a list provided by the instructor, or each group should pick a disorder on its own.

2. Each member of the group must pick a question from the possible topics listed on the previous page. One of the questions for each group must be: Which pathway is specifically involved in the disorder?

3. Each member should do a little research on the group’s question, just enough to be able to present 2 minutes of information to the class.

4. At the next class period, each group should give a 10-minute presentation, during which each group member speaks for about 2 minutes. Presentations should just be “chalk talks,” to avoid making this into a formal presentation requiring more preparation than the project is worth.

Going Beyond Estimated duration of activity: significant time outside of class, 15-minute presentation per group The preceding activity can be turned into a major assignment if you want to emphasize learning about metabolism through disease states. A major assignment could be done as follows.

Procedure: 1. Each group should pick a disorder to cover from a list provided by the instructor, or each group should pick a disorder on its own.

2. Each group should do some preliminary research on the disorder, with a firm deadline set for this research to be completed.

3. The group should equally divide the questions among its members from the possible topics listed on the previous page. If the group comes up with other questions to answer, they should assign those equally as well. If there are not enough questions to evenly divide, a few questions can be assigned more than once in the group.

4. Each group member must write a five-page paper on the group’s topic and submit it to the group by an agreed-upon date.

5. The group will then evaluate each paper and recommend revisions as necessary. The revisions should address all concerns related to the paper’s format, grammar, and sentence structure as well as the quality, accuracy, and organization of the topic presented.

6. On the due date for the assignment, each group member must submit their five-page paper to the instructor, and the group must give a 15-minute overview presentation of the entire project. Each group member must participate in the presentation. Also, when the paper is turned in, each group should include a summary of the changes each group member was asked to make, along with proof that the changes were appropriately addressed.

Requirements: 1. Each paper must be five full pages long. The paper must include a title page and a reference page, and those two pages do not count as part of the five-page requirement.

There will be severe penalties for short papers.

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2. Margins for the paper must be 0.8 inches all around. The font must be Times New Roman, 12 pt. Students should use double spacing but not make an extra return between headings or paragraphs.

3. There must be citations in the body of the work; students must include at least one citation per paragraph, but as many citations as needed should be used.

4. Citations and references must follow the American Psychological Association guidelines.

Grading: 1. For each group, the instructor will evaluate each paper, average the scores for the papers, and assign that score to each individual in the group. A few extra points may be awarded to papers that are clearly of higher quality than the other papers in the group. Thus, it is essential that step 3 of the procedure be carefully and thoroughly carried out in each group.

2. Each group member will also get a common grade for the group’s overview presentation.

Student-Performed In-Class Demonstration: The Electron Transport Chain (ETC), Chemiosmosis, and ATP Synthesis Estimated duration of activity: 5–8 minutes This activity is rather elementary, but it will provide the students a fun way to visualize how the ETC is tied to ATP synthesis. Because so many students are visual learners, role-playing demonstrations are often an interesting, visual way to help them understand dynamic biological processes. Even the brightest students will at least appreciate the break from lecture.

Purpose: To have students “act out” the ETC and chemiosmotic hypothesis to help them visualize and understand how it works.

Materials: For this demonstration, you will need the following materials (add strings to the signs so students can hang them around their necks): Signs with the following labels on one side: • Complex I (or NADH-Q reductase) • Ubiquinone (or Q) • Succinate-Q reductase • Complex II (or cytochrome reductase) • Cytochrome c • Complex III (or cytochrome oxidase) • ATP synthase • ADP + Pi fi ATP Double-sided signs with the following labels: • NAD+ on one side; NADH on the other side • FADH on one side; FADH2 on the other side • O2 on one side; H2O on the other side Several small signs with H+ on them A ball or other object labeled as an electron (e–)

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Procedure: 1. Have student volunteers come to the front of the class (preferably in front of a board) to act as parts of the ETC, ATP synthase, NADH (should be given the e–), FADH2, or oxygen.

Hand out signs, and have the students decide on the correct order in which to stand.

2. Designate yourself as Madam or Master Matrix, and remind students that they are in the inner mitochondrial membrane. Designate the board as the intermembrane space. As Madam or Master Matrix, you will supply the system with the molecules it needs to proceed.

3. Ask the class to direct the scene. They should tell NADH to give the electron to complex I. As complex I passes the electron, the class should tell you to give her protons. She should then turn around and tape them on the board. The class should direct the passage of the electron through the chain and the pumping of protons through the appropriate complexes.

As O2 accepts the electron, the sign should be turned to reveal H2O (water) as the product.

As ATP synthase allows the diffusion of protons back to the matrix (moving them from the board back to you), the protons should be given the ADP reaction sign to display.

4. Repeat the demonstration with FADH2.

Highlights: It is important that you stop the demonstration at crucial points to emphasize that a proton-motive force is created, that FADH2 results in the pumping of fewer protons and thus the generation of fewer molecules of ATP, that the proteins undergo conformation changes as they transfer electrons and protons, that Q and cytochrome c should be the only components moving laterally through the membrane, and so on.

Discussion Topics Teenagers and young adults are very interested in health-related issues such as weight loss and exercise physiology. To the average student, the material in this chapter may seem bogged down in details. However, a good discussion of how cellular respiration relates to topics in which they are invested will really increase their interest level.

Weight loss: Many people today are hooked on so-called fat-free or low-fat foods. We believe that if something is fat free, we can eat more of it and still maintain our current weight—or even lose weight. We may even think this means we can lose weight without exercising. We are often baffled to find that, after ignoring the serving size shown on the packages of these foods, we gain weight. If “fat is stored as fat” and we are not eating any fat, why are we getting fatter? This issue is a good synthesis question for students after participating in a discussion of anabolic pathways.

Answer: The students should come to realize that when ATP (energy) is abundant, cellular respiration slows down; and that intermediates of glycolysis and the Krebs cycle can be enzymatically converted into fatty acids and glycerol and packaged into adipose tissue (even if the original energy source was glucose).

This activity can also lead into a nice discussion of why fats are so rich in energy and why proteins are the “last resort” for energy.

Exercise physiology: Many people who exercise are familiar with the muscle “burn” associated with lactic acid. Many of us even know that the burn is caused by lactic acid, but we do not know what lactic acid is or why it causes the burn. Open your discussion of

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fermentation by asking students what lactic acid is, why it is an issue in intense exercise, whether it is good or bad for the cells, and why it causes a burning sensation.

Answer: Lactic acid is the by-product of fermentation in our muscle cells. During vigorous exercise, our muscle cells are rapidly using and making ATP. It is not long before the oxygen in the muscle is used for oxidative phosphorylation faster than our blood can deliver it, so oxygen stores are depleted. Muscle cells have the enzymes used for lactic acid fermentation, and for short periods of time they can generate ATP via glycolysis in the absence of oxygen.

Lactic acid is not good for cells; it is actually toxic in high levels. The “lactic acid burn” is not caused by high levels of lactate in your blood. Subjects who were given high levels of sodium lactate did not suffer the burning sensation. These symptoms result from the drop in the blood’s pH level that results from increasing acid content. An increase in any acid would cause the same sensation.

This activity can also lead into a nice discussion contrasting brain and muscle cells.

Because brain cells do not have the enzymes for lactic acid fermentation, they can survive only a short time without oxygen.

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CHAPTER

10

Photosynthesis

Learning Objectives: Students should be able to...

• Sumarize how the light-capturing reactions and the Calvin cycle transform light energy to chemical energy.

• Explain how light energy is captured and used to drive an endergonic redox reaction.

• Describe how ATP and NADPH are produced during the light-capturing reactions of • photosynthesis.

• Describe the three phases of the Calvin cycle.

Compare and contrast the roles of the Calvin cycle and the C4 cycle.

Lecture Outline I. Photosynthesis Harnesses Sunlight to Make Carbohydrates A. What is photosynthesis?

1. Photosynthesis is the process by which autotrophs (e.g., plants) convert sunlight into chemical energy. (Fig. 10.1) 2. Overall reaction: 6 CO2 + 12 H2O + light → C6H12O6 + 6 O2 + 6 H2O 3. Nonphotosynthetic organisms are called heterotrophs.

B. Photosynthesis: two linked sets of reactions 1. Cornelius van Niel studied photosynthesis in purple sulfur bacteria. a. He found that CO2 and H2S do not combine directly during photosynthesis. b. He also found that the oxygen atoms in CO2 are not released as oxygen gas.

2. Experiments using isolated chloroplasts and heavy isotopes of oxygen confirmed that the oxygen atoms released during photosynthesis come from water.

3. Melvin Calvin found that radioactively labeled carbon dioxide is incorporated into plant carbohydrates, even in the dark.

4. Therefore, photosynthesis consists of two independent reactions: the light-dependent reactions that generate oxygen from water and the Calvin cycle that generates sugar from carbon dioxide. a. The light-dependent reactions split water, make ATP, and reduce the electron carrier NADPH. b. The Calvin cycle (light-independent reactions) uses the energy in ATP and NADPH to reduce CO2 to carbohydrate. (Fig. 10.2) 5. To keep track of the main events of photosynthesis, drawing flowcharts can help with the complex processes. (Making Models 10.1) 90

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C. Photosynthesis occurs in chloroplasts. (Fig. 10.3) 1. Photosynthesis occurs in leaf cells in organelles called chloroplasts.

2. Structure of a chloroplast a. Chloroplasts have two organelle membranes. b. The inside of the chloroplast is filled with membrane sacs called thylakoids.

(1) A stack of thylakoids is a granum (plural: grana).

(2) The fluid-filled space around the grana is the stroma.

(3) Thylakoids have a considerable amount of the blue-green absorbing pigment chlorophyll in their membranes.

II. How Do Pigments Capture Light Energy?

A. Light is a type of electromagnetic radiation, a form of energy 1. Light behaves like a wave. a. Visible light ranges in wavelength from 400 to 710 nm within the electromagnetic spectrum. (Fig. 10.4) b. Light with shorter wavelength has more energy than light of longer wavelength.

2. Light behaves like a particle. a. Light is packaged in discrete units called photons. b. Each photon of a particular wavelength has a specific amount of energy.

B. Photosynthetic pigments absorb light.

1. When a photon of light hits a pigment, one of three things can happen: a. The photon can be absorbed. b. The photon can be transmitted. c. The photon can be reflected.

2. Most pigments absorb specific wavelengths of light.

3. The wavelengths that are not absorbed by a pigment are transmitted or reflected, thus giving the pigment that color.

4. Different pigments absorb different wavelengths of light. a. This hypothesis can be tested by using thin-layer chromatography. (Fig. 10.5) (BioSkills 6) (1) Leaf extract is spotted at the bottom of a piece of filter paper, which is placed in a solvent that wicks up the paper, carrying the pigments with it.

(2) The pigment molecules are carried at different rates, depending on their size and solubility in the solvent.

(3) The pigment spots can be cut out and placed in a spectrophotometer to measure the light wavelength that each absorbs. (BioSkills 8) b. Using this approach, biologists can plot absorption spectrums that illustrate which wavelengths a given pigment will absorb and reflect.

(1) Chlorophylls a and b absorb red and blue wavelengths, and they transmit green.

(2) Carotenoids absorb blue and green wavelengths, and they transmit yellow, orange, or red.

5. Which of these wavelengths drive photosynthesis? a. Experiments using photosynthetic algae allowed biologists to plot an action spectrum. (Fig. 10.6) .

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b. An action spectrum shows which wavelengths are effective at facilitating photosynthesis. c. Action spectra show that blue and red wavelengths are the most effective at driving photosynthesis. Since chlorophylls absorb these wavelengths, these data are consistent with chlorophylls being the main photosynthetic pigments. (Fig. 10.7) 6. Which part of a pigment absorbs light? a. Chlorophylls have a long isoprenoid “tail” and a “head” consisting of a large ring structure with a magnesium atom in the middle. (Fig. 10.8) b. The head is where the light is absorbed.

7. What is the role of carotenoids and other accessory pigments? a. Carotenoids (xanthophylls) are common accessory pigments found in chloroplasts. b. They are responsible for leaf color in the autumn when chlorophyll begins to degenerate. c. They absorb photons belonging to the blue and green part of the spectrum and pass the energy on to chlorophyll, thus functioning to extend the range of wavelengths that can drive photosynthesis. d. They also function to protect chlorophyll from free radical damage. When carotenoids are absent, chlorophyll is destroyed and the plant dies.

C. When light is absorbed, electrons enter an excited state.

1. When a photon hits chlorophyll, the energy is transferred to an electron in the head region.

2. This energy causes that electron to move to a higher orbital, one with higher potential energy (because it is farther from the atom’s nucleus). (Fig. 10.9) 3. Once the electron is in the higher orbital, there are several ways it can return to its resting state. a. If the electron simply falls back to its original position, the excess energy is released as light (fluorescence) and heat. (Fig. 10.10) 4. The antenna complex a. When the energy is passed from one chlorophyll molecule to the next, an electron is excited in response. b. As energy is transmitted, the original electron falls back into its original orbital. c. The antenna complex acts to transmit the energy from specific wavelengths of light to the receiver—a special chlorophyll molecule called the reaction center. Energy in the electron is transferred to nearby pigment via resonance. d. The antenna complex and the reaction center form a photosystem.

5. The reaction center a. The excited electron is passed from the reaction center to a molecule that acts as an electron acceptor. b. This redox reaction is the point at which electromagnetic energy is converted into chemical energy. (Fig. 10.10) c. The energy released can: (1) be emitted as light (2) be given off as heat

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(3) induce resonance (4) be transferred to an electron acceptor

III. The Discovery of Photosystems I and II A. Research that led to the discovery of photosystems 1. Robert Emerson and colleagues illuminated green algae to various wavelengths of light. (Fig. 10.11) a. When the researchers illuminated a sample of green algae with far-red (713 nm) or red (653 nm) light, the photosynthetic response was about the same. b. If they combined the two (both 713 and 653 nm) types of light, then the photosynthetic response was more than double that of either wavelength alone.

2. Using this information, Robin Hill and Faye Bendall suggested that plants and algae have two distinct types of reaction centers. a. The two reaction centers were named photosystems I and II. b. When both systems are active, photosynthesis runs at its peak efficiency.

B. How does photosystem II work?

1. Pheophytin accepts electrons and transfers them to an electron transport chain (ETC). a. An electron in a chlorophyll P680 molecule is excited. b. A molecule called pheophytin takes the electron from chlorophyll.

(1) Pheophytin is structurally similar to chlorophyll but lacks a magnesium atom in its head region.

(2) Pheophytin functions as an electron acceptor. c. The electron is then passed from pheophytin through an electron transport chain.

(Fig. 10.12) (1) This electron transport chain is similar to the one found in mitochondria.

(2) It contains several quinones and cytochromes.

(3) One particular quinone, plastoquinone, is not anchored to a protein and can shuttle electrons from pheophytin to the rest of the electron transport chain.

2. The ETC sets up a proton gradient that drives ATP synthase. a. As plastoquinone passes the electron, it also pumps protons into the interior of the thylakoid, creating a proton gradient inside the thylakoid. (Fig. 10.13) b. The protons can diffuse out of the thylakoid only through ATP synthase.

(1) ATP synthase uses the proton-motive force to synthesize ATP. (2) The production of ATP by this process is called photophosphorylation.

3. Photosystem II obtains electrons by oxidizing water. a. The original chlorophyll molecule is oxidized after its excited electron is taken away. b. Water is split by photosystem II, and the electrons from water are used to reduce chlorophyll P680.

(1) This reaction is highly endergonic and possible only using energy from the sun.

4. Oxygenic photosynthesis and the oxygen atmosphere a. A photosystem that can split water is called oxygenic, whereas a photosystem that cannot oxidize water is called anoxygenic.

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5. Students should be able to make an energy flowchart that includes the antenna complex, ATP synthase, pheophytin, light, the proton gradient, an ETC, and a reaction center.

C. How does photosystem I work? (Fig. 10.14) 1. Heliobacteria produce NADPH when exposed to sunlight.

2. In heliobacteria, when the reaction center of photosystem II absorbs a photon of light, the excited electron is eventually passed to ferredoxin.

3. An enzyme called ferredoxin/NADP+ oxidoreductase transfers the electron and a proton to NADP+ to form NADPH. a. NADPH is an electron carrier like NADH. D. The Z scheme: Photosystems II and I work together. (Fig. 10.15) 1. The Z scheme follows a pathway: a. When a photon of light is absorbed by chlorophyll P680, the excited electron is passed to pheophytin. b. Pheophytin donates the electron to an electron transport chain. c. Using the energy released by the reduction-oxidation (redox) reactions occurring in the electron transfer chain, plastoquinone pumps protons into the interior of the thylakoid. d. ATP synthase uses the proton-motive force created by plastoquinone to generate ATP. e. A small protein called plastocyanin accepts the electron from the electron transport chain and transfers it to a chlorophyll P700 molecule in photosystem I. f. When a photon of light is absorbed by chlorophyll P700 in photosystem I, the excited electron is passed to ferredoxin. g. An enzyme called ferredoxin/NADP+ oxidoreductase transfers the electron and a proton to NADP+ to form NADPH. h. Students should be able to explain (1) the role of plastocyanin in linking the two photosystems and (2) where the electrons that flow through the system have their highest potential energy.

2. Understanding the enhancement effect: When both wavelengths are available at the same time, both photosystems are activated and work at their maximum rates.

3. Noncyclic electron flow between water and NADP+. a. Electrons pass from water to NADP+ through a chain of redox reactions in a linear fashion. (Fig. 10.16) 4. Cyclic electron flow recycles and drives phosphorylation. (Fig. 10.17) a. Intermittently, photosystem I donates an electron to the electron transport chain of photosystem II. b. This event results in the production of additional ATP but no NADPH. c. It occurs when extra ATP is needed.

5. Photosystems I and II are distributed differentially and occur in separate regions of thylakoid membranes within individual grana.

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6. Oxygenic photosynthesis and the evolution of Earth a. The impact of oxygen production by plants on the environment of early Earth cannot be overstated. b. Oxygen then became an electron acceptor in aerobic respiration.

IV. How Is Carbon Dioxide Reduced to Produce Sugars?

A. The Calvin cycle fixes carbon.

1. To isolate the molecules produced as carbon dioxide is reduced to sugar, Melvin Calvin fed radioactively labeled carbon dioxide to green algae.

2. Green algae were fed a pulse of 14CO2. After a specified amount of time, the cells were killed and ground up into a crude extract, and paper chromatography was used to isolate individual molecules at different time intervals. (Fig. 10.18) 3. Results: Immediately after the pulse, a three-carbon compound called 3-phosphoglycerate (3PG) containing 14C was isolated as the first intermediate product.

4. Conclusion: Carbon dioxide is combined with a compound to form 3-phosphoglycerate.

5. RuBP (ribulose biphosphate) is the initial reactant with CO2.

6. When no two-carbon compound in the system could be found, Calvin proposed that a five-carbon compound reacts with carbon dioxide and then splits in two.

7. An overview of the three phases of the Calvin cycle: (Fig. 10.19) a. Fixation phase: Carbon is fixed when CO2 combines with a five-carbon compound called ribulose bisphosphate (RuBP) to form a six-carbon molecule split to form two 3-phosphoglycerate molecules. b. Reduction phase: These 3PG molecules are phosphorylated by ATP and then reduced by NADPH. The resulting sugar is called glyceraldehyde-3-phosphate (G3P). c. Regeneration phase: Some G3P is drawn off to make glucose, and the rest is recycled to regenerate RuBP. 8. Students should be able to provide the minimum number of RuBP, ATP, and NADPH molecules that would be required to run through six complete cycles.

9. Producing sugar is highly endergonic. That energy is provided by ATP and NADPH made in the light-dependent reactions.

B. The discovery of rubisco (Fig. 10.20) 1. Arthur Weissbach assayed spinach-leaf cell extracts to identify an enzyme called rubisco.

2. Rubisco is found in all photosynthetic cells.

3. It is a slow, inefficient enzyme.

4. It catalyzes only three reactions per active site per second.

5. Plants make up for this lack of speed by synthesizing large amounts of rubisco.

6. It can catalyze two different reactions: (Fig. 10.21) a. In the presence of high concentrations of carbon dioxide, rubisco combines CO2 and RuBP to produce 3PG.

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b. In the presence of high concentrations of oxygen, rubisco combines with oxygen to form RuBP. (Fig. 10.20) (1) This reaction consumes ATP and releases CO2.

(2) This entire process is called photorespiration, and it effectively reverses photosynthesis.

C. How is photosynthesis regulated?

1. The presence of light triggers the production of proteins required for photosynthesis.

2. When sugar is abundant, the production of proteins required for photosynthesis is inhibited.

3. Rubisco is activated by regulatory molecules produced when light is available but inhibited in conditions of low CO2.

4. Low levels of phosphate stimulate the production of Calvin cycle proteins.

D. Oxygen and carbon dioxide pass through stomata.

1. Carbon dioxide enters the air spaces of a leaf through stomata. (Fig. 10.22a) a. Leaves have pairs of specialized cells on their surface called guard cells. b. Between guard cells is a pore. c. When guard cells change shape, the pore opens and CO2 can enter the leaf.

(Fig. 10.22b) 2. The CO2 then diffuses down its concentration gradient to chloroplasts.

E. Mechanisms for increasing CO2 concentration 1. Opening stomata allows carbon dioxide in but lets water out.

2. Some plants have adapted to living in hot, dry climates by developing alternative carbon fixation systems.

3. The C4 pathway a. In C4 plants, carbon fixation occurs independently of the Calvin cycle. (Fig. 10.23) b. CO2 is combined with small organic molecules by an enzyme called PEP carboxylase in mesophyll cells adjacent to the bundle-sheath cells where rubisco is active. c. When the level of CO2 is low in bundle-sheath cells, the organic acids are transported into those cells, and CO2 is liberated. (Fig. 10.24) d. This system reduces the occurrence of photorespiration by keeping levels of CO2 high in cells where rubisco is active.

4. CAM plants (Fig. 10.25) a. The combination of CO2 and small organic acids takes place in the same cells where rubisco is active. b. However, this initial reaction takes place at night when cells can open their stomata and not risk excessive water transpiration. c. The CO2 stored during the night can be used during the day for photosynthesis.

In contrast, C4 plants store CO2 in one cell for use in another.

F. What happens to the sugar produced by photosynthesis?

1. Gluconeogenesis is the production of glucose or fructose from G3P.

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2. G3P is used to make sucrose and starch. a. Starch acts as a temporary storage product. b. Starch is not water soluble, so it cannot be transported from photosynthetic cells to other areas of the plant. c. At night, the starch temporarily stored in leaf cells is broken down and used to manufacture sucrose molecules. d. The sucrose is then used by the photosynthetic cell in respiration or transported to other parts of the plant.

3. Glucose, derived from sucrose and starch, is used in glycolysis and cellular respiration to make ATP in the cells of the plant.

4. Animals ingest plants and digest the starch into glucose to use for the same purpose.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. absorption spectrum

fluorescence

photosystem II

absorption spectrum action spectrum anoxygenic antenna complex autotrophs bundle-sheath cells C3 pathway C4 pathway Calvin cycle carbon fixation carotenoids chlorophylls chloroplasts crassulacean acid

gluconeogenesis glyceraldehyde-3-phosphate (G3P) grana guard cells heterotrophs lumen mesophyll cells NADP+/NADPH noncyclic electron flow oxygenic PEP carboxylase pheophytin photons photophosphorylation photorespiration photosynthesis photosystem photosystem photosystem I

pigments plastocyanin (PC) plastoquinone (PQ) proton gradient reaction center resonance energy transfer ribulose bisphosphate (RuBP) rubisco spectrum starch stoma; stomata stroma sucrose thylakoids vascular tissue visible light wavelength Z scheme

metabolism (CAM) cyclic electron flow electromagnetic enhancement effect excitation state ferredoxin

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Lecture Activities A Case Study Investigating Photosynthesis Estimated duration of activity: 15+ minutes Case studies are an excellent way to increase students’ interest in biology. You can do this activity completely in class (if the class or discussion section can meet in a computer lab), or you can have students start it in class and finish as an out-of-class assignment. It can also be assigned as homework and discussed during the next class session.

In this case study, your students can investigate photosynthetic inhibitors and how they interfere with photosynthesis. This will give them an understanding of how photosynthesis normally works.

Heidi has become interested in gardening. She has planted a small potted garden in her dorm room, but she is distressed because her garden will not grow. She waters the plants every few days, just like the gardening books say. She has placed them near the window to get light, but her roommate, Mary, insists on keeping a translucent green curtain in the window. Mary wants to keep the curtain closed because she likes to sleep late, and she often forgets to open it when leaving for class. Heidi hasn’t made a big deal about it because it is translucent and lets some light through.

One morning, Heidi asks Mary to stop at a plant nursery to buy some fertilizer on her way home from work. Mary says she can save Heidi some money by getting some fertilizer from her dad’s storage shed when she goes home the next weekend. Mary’s dad is a corn farmer and has all the best plant products. When Mary returns from home, she brings Heidi a bit of fertilizer in a paper bag labeled Aatrex.

After fertilizing the plants once, Heidi notices that the plants seem to be dying—the leaves are turning yellow around the edges. Finally, Heidi asks Mary to drive her to a local plant nursery so that she can ask someone who works there why her plants are dying. When she arrives at the nursery, Heidi tells everything to an employee named John. John tries to hold back the laughter as he explains to Heidi why her plants are dying.

Have students explain in two or three pages what they believe John told Heidi. Address the following questions (but feel free to add more information): What is Heidi’s biggest gardening mistake? Given what you are learning in Chapter 10, why aren’t the plants growing? Why are they dying (include the gory molecular details)? What does Heidi need to do to fix the problem?

Note: Aatrex is the commercial name for atrazine, an herbicide that inhibits electron transport in PSII.

Can Planting Trees Decrease Global Warming?

Estimated duration of activity: 30–40 minutes Due to the burning of fossil fuels and other types of burnings, such as the burning of rain forests in equatorial regions, tremendous amounts of CO2 have been released into the atmosphere. This CO2 accumulates in the atmosphere, where it absorbs a great deal more of the sun’s energy than the atmosphere should normally absorb. Warming of the atmosphere due to the buildup of CO2 creates the “greenhouse” effect, which causes the temperature of Earth to rise much more than it should. This leads to global warming. The results of global warming are predicted to be catastrophic for Earth. Sea levels will rise, polar ice caps will melt, shorelines will recede—all of which will cause massive disruptions of agriculture and habitable areas on Earth.

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Many people hope that by planting more large plants such as trees, we will mitigate the effects of global warming. However, another type of pollution is destroying the ozone layer of the atmosphere. Destruction of the ozone layer could potentially seriously damage plants and therefore cancel out the positive effects of planting trees.

How would destruction of the ozone layer specifically damage plants? Do you think this damage will (1) totally wipe out the positive effects of planting large numbers of trees, (2) partially wipe out the positive effects of planting trees, or (3) not wipe out the positive effects of planting trees?

This activity has two purposes. The first is to have students thoroughly understand the relationship of various wavelengths of light to the functions of chlorophylls, carotenoids, and flavonoids. The second is to have students apply their knowledge to serious environmental problems to grasp how complex these problems have become.

Student objectives: 1. Briefly explain why planting trees would mitigate the effects of global warming.

2. Conduct some research of the scientific and popular media to discover the primary result of destroying the ozone layer.

3. Thoroughly explain the relationship of various wavelengths of light to the functions of chlorophylls, carotenoids, and flavonoids.

4. Discuss the positive effects of planting trees versus the negative effects of ozone depletion.

Then, as a group, students should come up with a reasoned answer regarding how serious the effects of ozone depletion would be in counteracting the positive effects of planting trees.

Procedure: After students complete each of the numbered steps below, ask what answers they came up with, taking care to move constantly from group to group while asking questions. Overall, instructor-led questioning should take 10 or 15 minutes. Hand out the instructions for this exercise some time before covering Chapter 10. Instruct students to do a bit of research about how the ozone layer of the atmosphere is being destroyed and to determine the primary damaging effect of depleting the ozone layer. The research does not need to be formal or in depth; consulting an online encyclopedia should suffice.

1. On the day of the exercise, have students form groups and instruct them to answer the following questions within the time frames given. For each group, have students appoint a facilitator, a notetaker, and two active researchers. The facilitator’s responsibility is to keep the group on track, keep discussions moving, settle disputes, and enforce time schedules. The notetaker takes detailed notes during the discussions and polishes the notes as needed. The active researchers are primarily responsible for finding the material they all need to learn. The facilitator and the notetaker also participate in research but may be too busy to fully participate. The students should consult the textbook for questions a−f below.

(15 minutes) a. Why would planting trees lead to increased overall photosynthesis and increased Calvin cycle activity and thus mitigate the effects of global warming? b. What overall range of light is absorbed by chlorophylls, carotenoids, and flavonoids? c. What specific wavelengths of light are absorbed by chlorophylls, carotenoids, and flavonoids?

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d. Which photons of light specifically are absorbed and cause the excitation of electrons to a higher state? e. What role does the antenna complex play in the absorption of photons and excitation of electrons? f. What is the role of flavonoids in protecting chlorophyll from certain harmful rays from the sun?

2. Identify the negative effect that results from the destruction of the ozone layer. (2 minutes) 3. Could those negative effects overwhelm the protecting effects of the flavonoids? If the flavonoids are overwhelmed, what would happen? If the flavonoids are overwhelmed, would the positive effects of planting trees be canceled out? (10 minutes) Notes for INSTRUCTORS ONLY: The following concepts and ideas should come out of this exercise.

• Increased tree planting could lead to increased production of ATP, which would be used in the Calvin cycle to fix more overall CO2 in nature and thus reduce the levels of CO2 in the atmosphere.

• The specific questions in step 1, a–f, should be answered.

• UV light is strong enough to eject electrons from chlorophyll and destroy the chlorophyll in the process. Flavonoids absorb UV light and thus protect chlorophyll.

• Depletion of the ozone layer allows increased amounts of UV light from the sun to strike Earth. This excess UV light could potentially be enough to overwhelm the protective effects of the flavonoids and allow chlorophylls to be damaged, thus stopping photosynthesis. So planting extra trees might not have much of an overall positive effect. However, it is not necessary for students to come up with that particular idea for this exercise to be successful. As long as they bring the basics together and come up with some kind of reasonable answer to the basic questions, the exercise should be deemed successful.

Internet sources for the effects of ozone depletion: http://en.wikipedia.org/wiki/Ozone_depletion www.epa.gov/ozone/science/sc_fact.html www.science.org.au/nova/004/004key.htm

Activity Idea Estimated duration of activity: 5 minutes Use the case study from the first activity, but go ahead and reveal that Aatrex is a photosynthetic poison that inhibits the electron transport chain in photosystem II. Ask students to explain exactly why this inhibitor stops carbohydrate production. This should force them to relate what is happening in the light-dependent reactions to what is happening in the lightindependent reactions. You may also ask them why they think the leaves are turning yellow (as the cells die, chlorophyll production decreases and the accessory pigments show through).

Then have students practice the scientific method while applying what they have learned about electromagnetic radiation, pigments, and photosynthesis. As you are discussing plant pigments and their absorption spectra, ask students to hypothesize about the effects of growing plants behind different-colored screens. In this discussion, you can ask them to design an experiment (complete with controls, replication, etc.) that would test their hypotheses.

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Discussion Idea Many students have a hard time motivating themselves to study plants. It’s difficult for them to understand how the details of photosynthesis are important to their life. To motivate your students, discuss with them the applicability and importance of photosynthesis.

The website http://photoscience.la.asu.edu/photosyn/study.html has great ideas for such a discussion. This essay explores the relevance of photosynthesis to energy production, agriculture, medicine, and even electronics. At the site, you will find an explanation of how the reaction center of chloroplast antennae is an important microscopic model for new technologies that use light as an energy source.

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CHAPTER

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Cell-Cell Interactions

Learning Objectives: Students should be able to...

• Describe the structure and function of both plant and animal extracellular material.

• Compare and contrast the function of cell-cell connections and cell-cell gaps.

• Explain how signal transduction pathways receive, process, and respond to information.

• Compare and contrast intercellular interactions and signaling in multicellular and unicellular organisms.

Lecture Outline I. The Cell Surface A. The structure and function of an extracellular layer 1. All extracellular layers are made of fiber composites. (Fig. 11.1) a. The layers consist of a cross-linked network of long filaments embedded in a ground substance. b. This network can withstand tension and compression.

B. The extracellular matrix in animals 1. The fibrous component of the extracellular matrix (ECM) is mostly collagen, a cablelike molecule made up of three intertwined chains and secreted by surrounding cells. (Fig. 11.2) 2. Collagen is surrounded by gelatinous polysaccharides (proteoglycans) that are also secreted by surrounding cells. (Fig. 11.2a) 3. The ground substance that surrounds collagen and other fibrous components of the ECM contain proteoglycans. (Fig.11.2b) 4. The amount and composition of the ECM vary in different tissues.

5. Cells attach to the ECM via transmembrane integrin proteins (including lamins) binding to fibrils in the ECM. (Fig. 11.3) C. The cell wall in plants 1. Primary cell walls a. Primary cell walls secreted by plant cells are made of fibrous cellulose bundled into microfibrils infiltrated with hydrophilic gelatinous pectins. (Fig. 11.4) b. Primary cell walls define the shape of the plant cell and limit the amount of water that can enter the cell (turgor pressure). c. Young plants, actively growing secrete expansins allowing for the microfibrils to slide past one another.

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2. Secondary cell walls a. Some plant cells secrete a secondary cell wall made of cellulose and lignin to provide extra support. (Fig. 11.5)

II. How Do Adjacent Cells Connect and Communicate?

A. Multicellularity 1. Multicellularity is defined by physical connections among cells, maintaining the structure and function of tissues.

B. Cell-cell attachments in multicellular eukaryotes 1. Materials and structures that bind cells together are important in tissues that form external and internal surfaces (epithelia).

2. Function as barriers between external and internal environments C. Indirect intracellular attachments 1. Plant cells are held together by a middle lamella made of gelatinous pectins.

(Fig. 11.6) 2. In animal cells, gelatinous polysaccharides fill the space between adjacent cells, whereas cytoskeleton-ECM connections hold individual cells together. (Fig. 11.3) D. Tight junctions form a seal between cells.

1. In animals, membrane proteins allow cell-cell attachments in epithelia and other tissues. (Fig. 11.7) 2. Tight junctions form a watertight seal between two animal cells, creating very effective barriers in regions like the lining of the digestive tract. (Fig. 11.8a) 3. Tight interactions pull the membranes of adjacent cells together, creating a “quilt” where the proteins “stitch” together making a watertight seal. (Fig. 11.8b) 4. Tight junctions are dynamic permitting more or less transport in response to environmental changes.

E. Desmosomes form secure adhesions.

1. Desmosomes are rivets that link the cytoskeletons of adjacent animal cells.

(Fig. 11.9a) 2. Desmosomes consist of integral membrane proteins that form bridges between proteins inside adjacent cells.

3. Desmosomes help form a continuous structural support system between cells in tissue.

(Fig 11.7) F. Intracellular adhesions are selective.

1. Work by H.V. Wilson using sponges revealed that treatment of adult sponges with chemicals made the cells separate, but under normal conditions the cells would stick together again.

2. Cells aggregate based on tissue type (selective adhesion).

G. Discovery of cell-cell adhesion proteins 1. Cell-cell adhesion takes place via interactions between membrane proteins.

(Fig. 11.10) 2. Biologists demonstrated the phenomenon of selective adhesion in experiments with animal cells. (Fig. 11.10)

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3. The attachment molecules of desmosomes—cadherins—were discovered by experiments that used antibodies to block the interaction between cadherins to cause cell dissociation.

H. Cells communicate via cell-cell gaps.

1. Gap junctions are protein-lined holes between some types of animal cells that allow for cytoplasmic communication. Example: cardiac muscle cells (Fig. 11.11a) 2. Plasmodesmata between adjacent plant cells allow for the sharing of cytoplasm and the coordination of activity between two adjacent cells. (Fig. 11.11b) 3. In plants, the plasma membrane separates most tissues into two independent corridors: (Fig. 11.12) a. Symplast—continuous network of cytoplasm connected by plasmodesmata b. Apoplast—the region outside the plasma membrane 4. Students should be able to explain what would happen if cells in a developing frog embryo were treated with a molecule that blocked a cadherin in muscle tissue.

III. How Do Distant Cells Communicate?

A. Cell-cell signaling in multicellular organisms 1. Hormones are released from one tissue and travel through the extracellular matrix or the bloodstream to distant cells.

2. Hormones are usually small molecules and include certain peptides, steroids, and gases.

B. Signal reception 1. Hormones are ligands for specific receptor proteins. a. If the hormone is lipid soluble (steroid hormones), then the receptor is in the cytoplasm. b. If the hormone is water soluble, then the receptor is an integral membrane protein on the cell surface.

2. Signal receptor proteins a. Receptors are restricted to particular cell types. b. Receptors are dynamic; the number of receptors in a cell may increase or decrease based on the amount of hormone stimulation over time (sensitization/desensitization). c. They are regulated by other molecules. Examples: beta-blockers and epinephrine receptors in the heart d. They are able to change conformation when the hormone binds.

C. Signal processing 1. Processing lipid-soluble signaling molecules a. Steroid hormones enter the cell and bind to cytoplasmic receptors that translocate to the nucleus to change gene expression. Example: testosterone (Fig. 11.13) 2. Processing lipid-insoluble signaling molecules a. Signal transduction pathways are created when water-soluble hormones bind to cell-surface receptors and convert an intercellular signal into an intracellular signal.

(Fig. 11.14)

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b. The intracellular signal can be amplified because many of the activated molecules are enzymes, which can activate many more molecules down the cascade.

3. Signal transduction via G-proteins-coupled receptors a. G proteins are associated with the cell membrane and integrate the signal from the cell surface to second messengers inside the cell. They are regulated by GTP or GDP. The steps in Figure 11.15 are: (1) A signaling molecule arrives and binds to a receptor in the plasma membrane.

(2) The receptor changes shape and activates the G protein, splitting it into two parts.

(3) One part of the split G protein activates a nearby enzyme. That enzyme produces small molecules called second messengers that initiate the biochemical change in the cell. b. Second messengers play many roles. (Table 11.1) (1) cAMP, cGMP, calcium, and diacylglycerol are examples of second messengers.

(Table 11.1) (2) Second messengers aren’t restricted to a single role or a single cell type.

(3) It is common for more than one second messenger to be involved in triggering a cell’s response to the same extracellular signal.

(4) Imagine the following movie scene: A spy arrives at a castle gate. The guard receives a note from the spy, but he cannot read the coded message. Instead, the guard turns to the queen. She reads the note and summons the commander of the guard, who sends soldiers throughout the castle to warn everyone of approaching danger. Students should be able to identify which characters in the scene correspond to the second messenger, G protein, hormone, receptor, and enzyme activated by the G protein.

4. Signal transduction via enzyme-linked receptors. The steps for the tyrosine kinase receptors in Figure 11.16: a. A hormone binds to a receptor tyrosine kinase (RTK). b. The complex changes conformation and is then phosphorylated. c. Proteins inside the cell form a bridge between an activated RTK and a peripheral membrane protein Ras (a G protein). d. When Ras is activated, it triggers the phosphorylation and activation of another protein. e. The phosphorylated protein catalyzes the phosphorylation of other proteins.

This continues, setting up a phosphorylation cascade. f. Students should be able to explain how to set up dominos to stimulate a phosphorylation cascade and state how many green, blue, and yellow dominos would be required to model the pathway if Ras and each protein kinase in the cascade were to activate 10 proteins.

D. Signal response 1. Changes in gene expression: All of the signal mechanisms mentioned above (steroid hormone binding to cytoplasmic receptor binding, second messenger cascades, phosphorylation cascades) can affect gene expression.

2. Changes in protein function: Second messenger and phosphorylation cascades result in the activation and deactivation of existing proteins in the cell. .

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E. Signal deactivation 1. Turning off the G protein: When the G protein hydrolyzes GTP to form GDP, its conformation changes and it becomes inactive.

2. Inactivating the second messenger: Enzymes in the cell can inactivate second messengers. Example: phosphodiesterase as cAMP. 3. Dephosphorylation: Phosphatases remove the phosphate groups that kinases added, thus reversing the effect of phosphorylation.

F. Cross-talk: synthesizing input from many signals (Fig. 11.17) 1. Signal transduction pathways form a network and allow cells to respond to extracellular signals in an integrated way.

2. Cross-talk may inhibit or stimulate steps in a different pathway.

3. Cross-talk helps regulate the flow of information.

IV. Signaling between Unicellular Organisms A. Responding to population density 1. In unicellular organisms, cell-cell communication is called quorum sensing.

2. The response of these signals can range from bioluminescence to attack of multiple hosts by secreting proteins to secretion of biofilms.

3. For example, quorum sensing via a G-protein-coupled receptor recruits the free-living amoebae of a cellular slime mold to aggregate. (Fig. 11.18) (BioSkills 11)

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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antibody

integrins

quorum sensing

apoplast cadherins collagen cross-talk desmosomes epithelia expansins extracellular matrix (ECM) G proteins gap junction GTP/GDP hormones

lignin MAPK microfibril multicellularity pectins pheromones phosphatase phosphorylation cascade plasmodesmata primary cell wall protein kinases proteoglycan

Ras receptor tyrosine kinases (RTKs) second messenger secondary cell wall selective adhesion signal receptor signal transduction symplast tight junction tissues turgor pressure

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Lecture Activities The Cell Wall and the Extracellular Matrix Estimated duration of activity: 20–25 minutes Most students coming into college-level general biology know little about the detailed structure of plant cell walls and even less about the extracellular matrix (ECM) of animal cells. This active learning exercise gives students primary responsibility for teaching themselves the concepts and facts associated with the structure of cell walls and the ECM. (The following group setup is something the instructor should have started at the beginning of the semester so that students are familiar with it. Note that each position in the group should constantly rotate throughout the semester.)

For each group, have students appoint a facilitator, a notetaker, and two active researchers.

The facilitator’s responsibility is to keep the group on track, keep discussions moving, settle disputes, and enforce time schedules. The notetaker takes detailed notes as the discussions go on, then polishes the notes as needed. The active researchers are primarily responsible for finding the material they all need to learn. The facilitator and the notetaker also participate in research but may be too busy to fully participate.

Questions: 1. What are the components of the plant cell wall and the ECM? What type of macromolecule is each component, and what types of subunits make up each component?

(5 minutes) 2. Of those components, which correspond to the steel rods and which correspond to the concrete in the reinforced concrete analogy? (5 minutes) 3. Why is the reinforced concrete model such a good analogy for the function of cell walls and the ECM? (5 minutes) Pick one group to report on question 1, another group to report on question 2, and another group to report on question 3. (5–10 minutes) Make sure all the components of the cell wall and the ECM are mentioned. (Note that other groups will be chosen to report in later exercises.)

A Problem of Scurvy Estimated duration of activity: 5–8 minutes Everybody likes a good story. The following is a true story of a sea captain and his discovery that fruits cure scurvy. When talking about the extracellular matrix, you may find this little history lesson a way to awaken students’ interest while giving you the opportunity to integrate information from previous chapters.

An experiment on the open seas: James Lind was a Scottish sea captain on the ship HMS Salisbury, headed toward Plymouth, England. Lind had 12 men on board who were suffering from scurvy. They complained of bleeding gums, fatigue, weak knees, and a rash. Since scurvy was a common disease on ships, Lind wanted to figure out the best course of treatment to get his sailors well and able to work. He divided the sick men into six groups of two. All 12 men were fed the same basic diet at meals, but each group were given a different supplement each day (see note 1). One group received apple cider, one group received an elixir of sulfuric acid, one group received a few tablespoons of vinegar, another got a mixture of

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herbs, another drank a quart of seawater each day, and the last group ate two oranges and a lemon each day. The results were striking. Within six days, the men who ate the oranges and lemons were ready for duty.

Question: What do we know today about scurvy, and why did the oranges and lemons cure the disease?

Answer: Scurvy results solely from a deficiency in vitamin C (ascorbic acid). Fresh fruits and vegetables contain lots of vitamin C, so that is why the oranges and lemons cured the men.

Vitamin C is a reducing agent (note 2) that serves as a cofactor (note 3) for the enzyme prolyl hydroxylase (it keeps the iron atom in a reduced state). Prolyl hydroxylase is the enzyme that converts procollagen to collagen (note 4). Collagen is a major constituent of the extracellular matrix and connective tissue. A lack of collagen will result in the breakdown of the walls of capillaries, causing tissue hemorrhaging. This hemorrhaging is responsible for the bleeding gums and spots on the skin (hemorrhaging in hair follicle causes skin lesions). Inability to make collagen results in the breakdown of cartilage in joints, causing joint pain, and in the loss of dentin (connective tissue in teeth), causing severe pain while eating.

Notes for teaching: 1. This is a great place to highlight some aspects of experimental design. Although Lind did not have the luxury of a large subject population, he did establish controls so that he could draw conclusions about what was working and what was not.

2. Ask a student to explain what a reducing agent does.

3. This is a nice example of what a cofactor does and how its absence renders an enzyme incapable of function.

4. This exercise will lead right into your discussion of the extracellular matrix and the molecules that hold cells in place.

Internet resources: www.emedicine.com/derm/topic521.htm http://themedicalbiochemistrypage.org/vitamins.html

Discussion Idea—Metastasizing Cancer Cells When a tumor grows, cells can break off and migrate to the bloodstream or lymphatic system, where they circulate through the body. They eventually exit the bloodstream or lymph and integrate into a new tissue type, where they initiate the growth of another tumor. It is interesting to think of metastasis in relation to cell-cell interactions. For a cell to break away from the tumor, it has to lose or release its cell-cell and cell-extracellular matrix attachments to neighboring tumor cells. However, once it finds a new home, the migrating cell must reestablish new cell-cell connections in that tissue. It is thought that the origin of the cell type (the tissue in which the original tumor developed) will determine the tissues to which those cells can migrate and infiltrate. For instance, 70 percent of all breast cancers metastasize to bone. Cancer researchers hypothesize that the original tissue and the tissue in which secondary tumors grow have some common cell adhesion and ECM adhesion molecules. Cell-cell adhesion has become a major emphasis in cancer research in an effort to understand metastasis.

Internet resource: http://www.cancer.gov/

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The Cell Cycle

Learning Objectives: Students should be able to...

• Use experimental evidence from a pulse-chase assay to demonstrate how the four phases of the eukaryotic cell cycle were identified.

• Describe how cytoskeletal filaments and motors are used to separate chromosomes into two daughter cells.

• Predict how defects in the regulation of the M-phase-promoting factor and cell-cycle •

checkpoints would affect cells.

Explain how the G1 checkpoint is related to uncontrolled division in cancer.

Lecture Outline All cells come from preexisting cells via cell division.

Meiosis is cell division that occurs only in reproductive cells.

• The nuclei of cells generated by meiosis have half the number of chromosomes of the parent cell.

• Daughter cells (gametes) generated by meiosis are genetically different from the parent cells.

Mitosis is cell division that occurs in somatic cells.

• Mitosis generates daughter cells that are genetically identical to the parent cell.

Mitosis and meiosis are usually accompanied by cytokinesis, the division of cytoplasm into two distinct cells.

Both are responsible for 1 of the 5 attributes of life: reproduction.

I. How Do Cells Replicate?

A. Basic steps in cellular replication: 1. Copying DNA 2. Separating the copies 3. Dividing the cytoplasm to create two complete cells B. What is a chromosome?

1. Walther Flemming discovered that threadlike structures (chromosomes) in the nucleus change and move during cell division. (Fig. 12.1) a. He found that the threads (chromatids) paired before cell division, being held by cohesins. b. He also found that the pairs split during division and were isolated into new cells. .

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2. Subsequent studies found that the chromosome number ultimately remained constant during a sequence of cell divisions.

3. W. Waldeyer called the threads chromosomes.

4. Immediately before mitosis, the DNA condenses around its associated proteins (histones), resulting in a compact and condensed chromosome.

5. The region of DNA in a chromosome that codes for a particular protein/RNA is a gene.

C. Cells alternate between M phase and interphase.

1. M phase, or mitosis, is the phase of the cell cycle in which cell division occurs, when the cell separates the replicated DNA copies.

2. Interphase chromosomes uncoil into extremely long thin structures and are growing/ preparing to divide or fulfill a specialized function.

D. The discovery of S phase 1. Autoradiography is a process used to distinguish cells copying their DNA and cells that are not. (BioSkills 6) a. Procedure is done in three steps (1) Label DNA (2) Wash away radioactivity (3) Visualize with X-ray film 2. Alma Howard and Stephen Pelc used cultured cells and radioactively labeled nucleotides to study when DNA replication occurs in the cell cycle.

3. Conclusions from these experiments as well as from studies that followed: a. Because the radioactive thymidine/phosphate would be incorporated only during the replication phase, the researchers concluded that DNA replication occurs during interphase. b. The researchers called the time when DNA is synthesized the S phase.

E. The discovery of the gap phases 1. Not all interphase cells were labeled, which meant there was a “gap.” To determine where the gap(s) is/are, researchers used cultured cells. (BioSkill 11) 2. Pulse-chase approach (Chapter 7) a. Researchers added radioactively labeled thymidine to cells. A short time later, they stopped and flooded with nonradioactive thymidine. b. This labeled only those cells in S phase during the pulse. c. By taking samples at different times during the chase, researchers were able to determine how many cells were undergoing mitosis. (Fig. 12.2) d. The results led to knowledge of a gap between the end of S phase and beginning of M phase, G1 gap. e. Once the lengths of S/M phases were defined, it was determined there was a gap at the end of M and start of S. (Fig. 12.3) F. The cell cycle 1. Figure 12.3 is an overview of the cell cycle.

2. G1 allows the parent cell to grow large enough and generate enough organelles that the daughter cells will function normally.

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3. G1 gives the cell time to do its job in your body.

4. G2 allows the cell to build the protein machinery necessary for mitosis.

II. What Happens during the M Phase?

A. Chromosome structure (Fig. 12.4) 1. DNA is complexed with proteins (histones) into chromatin.

2. At the beginning of mitosis, chromatin condenses to form a compact, mobile structure. a. Each replicated chromosome is called a chromatid. b. Each set of chromatids is joined at a centromere. c. Chromatids from the same chromosome are called sister chromatids and are exact copies of one another. d. Students should be able to explain the relationships between chromosomes and genes, chromatin, and sister chromatids.

B. Events in mitosis 1. Mitosis begins when chromatin condenses to form a much more compact structure.

2. During mitosis, sister chromatids separate to form independent chromosomes. a. As a result, each daughter cell gets a copy of the genetic material on each chromosome. b. Each daughter cell ends up with exactly the same complement of genetic information.

3. Prophase (Fig. 12.5, step 2) a. Prophase is preceded by DNA replication in interphase. b. DNA condenses. c. The spindle apparatus is built.

(1) Spindles are built from microtubules, composed of α and β tubulin with +/– ends.

(2) Microtubules originate from MTOCs, polar microtubules.

(3) In animals, mitotic spindles arise from centrosomes.

3. Prometaphase (Fig. 12.5, step 3) a. The nuclear envelope breaks down. b. Spindle fibers from each centrosome attach to one of a pair of sister chromatids at the kinetochore (kinetochore microtubules). c. The centrosomes begin to move to opposite poles of the cells. d. Spindle fibers attached to kinetochores begin to move the chromosomes to the middle of the cell.

4. Metaphase (Fig. 12.5, step 4) a. The centrosomes complete their migration and reach the opposite poles. b. This results in all of the chromosomes becoming lined up with their kinetochores at the metaphase plate. c. Formation of spindle apparatus is complete. d. Spindle poles are held in place with the help of astral microtubules.

5. Anaphase (Fig. 12.5, step 5) a. Spindles attached to kinetochores begin to shorten, exerting a force on the sister chromatids that pulls them apart. .

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b. Spindle fibers not attached to kinetochores begin to push against each other via motor proteins that connect them, causing the poles of the cell to be separated even more. c. Spindle fibers continue to shorten, pulling chromatids to opposite poles. d. This process ensures that each daughter cell gets identical sets of chromosomes.

6. Telophase (Fig. 12.5, step 6) a. The nuclear envelope re-forms around the two new nuclei. b. The spindle fibers break down.

7. Students should be able to make a table with rows labeled (1) spindle apparatus, (2) nuclear envelope, and (3) chromosomes, along with columns labeled with the five phases of mitosis. They should be able to fill in the table by summarizing what happens to the three structures during each phase of mitosis. (Table 12.1) C. How do chromosomes move during anaphase?

1. Mitotic spindle forces a. Microtubules are lengthened and shortened by the addition and loss of tubulin subunits. b. Mitotic spindle shortening during anaphase is a result of the loss of tubulin subunits.

(1) Biologists studied this process using fluorescently labeled tubulin subunits to visualize spindle fibers. (Fig. 12.6) (2) They found that spindle fibers shortened at the end connected to the kinetochore.

2. Kinetochores are linked to retreating microtubule ends. a. The kinetochore is a complex of many proteins that attaches the centromere region of a chromosome. (Fig. 12.7) b. Fibers that extend from the kinetochore are tethered to a ring that surrounds the kinetochore microtubule. c. As anaphase gets underway, the plus end of the kinetochore begins to fray and disassemble. d. The result is that the chromosome is pulled to the spindle pole by depolymerization of the kinetochore microtubule.

D. Cytokinesis results in two daughter cells. (Fig. 12.5, steps 7 and 8) 1. In plants, a cell plate forms. (Fig. 12.8a) a. Vesicles from the Golgi apparatus carrying cell-wall material migrate to the middle of the cell between the two new nuclei. b. The vesicles build up and fuse, forming a new cell membrane and a cell wall dividing the two new nuclei.

2. In animals, a cleavage furrow forms between the two new nuclei. (Fig. 12.8b) a. The furrow is made from a ring of actin and myosin microfibrils that surround the inside circumference of the cell. b. The microfilaments contract, causing the cell membrane to constrict and pinch the cell in two.

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E. Bacterial cell replication 1. The mechanism of cell division in bacteria (binary fission) is similar to cytokinesis in animals. (Fig. 12.9)

III. Control of the Cell Cycle A. Control of the cell cycle varies among cell types and changes in environment.

1. The amount of time in G is usually the difference. a. In rapidly dividing cells, G1 is essentially eliminated. b. Nondividing cells are permanently stuck in G1 called G0 (cell type examples: nerve and muscle). c. Division is also in response to damage and nutrient rich/deficiencies.

2. Researchers conducted cell fusion experiments to identify the existence of a molecular cell-cycle switch in cells. (Fig. 12.10) a. Two cells from different species, and in different stages of the cell cycle, were fused.

(1) When a cell in mitosis was fused with a cell in interphase, the nucleus of the interphase cell began to enter M phase.

(2) When an S-phase cell was fused with a cell in G1, the cell in G1 began replicating its DNA. b. These experiments suggested that in the cells undergoing M and S phases, something could initiate those phases in other cells.

3. Experiments using frog eggs (oocytes) confirmed this hypothesis. a. The researchers extracted cytoplasm from frog eggs going through various stages of the cell cycle and injected it into cells at various stages of the cell cycle.

(1) When they injected cytoplasm from an egg in M phase into a cell in G2, the G2-phase eggs entered M phase.

(2) The cytoplasm from an interphase cell could not remove the G2-phase cells from their current phase. b. These experiments demonstrated there was some molecular signal in the cytoplasm that initiated mitosis. c. This factor was eventually isolated and named mitosis-promoting factor (MPF). d. MPF induces mitosis in all eukaryotic cells.

4. MPF contains a protein kinase and a cyclin. a. MPF is composed of two polypeptide subunits.

(1) One polypeptide functions as a protein kinase that can activate and inactivate other proteins by phosphorylation. This subunit is expressed at constant levels throughout the cell cycle.

(2) The other subunit is a cyclin whose concentration fluctuates throughout the cell cycle. b. When cyclin levels are high, cyclin binds with the kinase (cyclin-dependent protein kinase, or Cdk) to form active MPF. (Fig. 12.11) 5. How is MPF turned on? a. MPF activates proteins involved in chromosomal condensation. b. MPF activates microtubule-associated proteins thought to be involved in spindle formation. .

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c. MPF triggers the destruction of its own cyclin, maintaining the oscillation of cyclin concentration during the cell cycle.

6. How is MPF turned off? a. MPF is deactivated by an enzyme complex activated by the events in mitosis, thus implying negative feedback. b. Students should be able to describe what MPF does. They should also be able to explain the relationships between MPF and cyclin, Cdk, and the enzymes that phosphorylate MPF, dephosphorylate MPF, and degrade cyclin.

B. Cell-cycle checkpoints can arrest the cell cycle.

1. Different sets of cyclins and Cdks are involved at each phase shift during the cell cycle.

2. Three main switches called cell-cycle checkpoints occur during the cell cycle.

Interactions between regulatory molecules at each checkpoint “decide” whether to proceed with division. (Fig. 12.12) 3. G1 checkpoint ensures that the cell is large enough to divide and that enough nutrients are available to support the resulting daughter cells. a. Cells pass through the G1 checkpoint if the following factors have been met: (1) Size: The appropriate size must be met for daughter cells to function normally.

(2) Availability of nutrients (3) Social signals from neighboring cells (4) If DNA is damaged, a protein called p53 (a tumor-suppressor protein) activates genes that will lead to cell suicide (apoptosis).

4. G2 checkpoint ensures that DNA replication in S phase has been completed successfully.

5. Metaphase (M-phase) checkpoint ensures that all of the chromosomes are attached to the mitotic spindle by a kinetochore.

IV. Cancer: Out-of-Control Cell Division A. Properties of cancer cells 1. All cancers involve uncontrolled cell growth that results in the formation of a mass of cells called a tumor.

2. Cancer death rates have changed over time. (Fig. 12.13) 3. Cancer cells have two types of defects. a. Defects that activate the protein required for cell growth when they should not be b. Defects that prevent tumor suppressor genes from shutting down the cell cycle B. Properties of cancer cells 1. Malignant cancerous cells not only form tumors but actually leave the original site of growth and invade other organs. This movement is called metastasis. (Fig. 12.14) a. Benign tumor cells also divide rapidly but are not invasive. b. Malignant tumor cells divide and move to other tissues through lymph and blood vessels, creating secondary tumors.

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C. Cancer involves loss of cell-cycle control.

1. Social control a. Most cells leave the G1 phase because they receive a social-control signal from neighboring cells. b. Growth factors are intercellular social-control signals that can trigger cells to pass the G1 checkpoint.

(1) Growth factors trigger an increase in the expression of the cyclin involved in the G1 checkpoint.

(2) When the concentration of these cyclins increases, the cyclin molecules bind with the appropriate Cdk; the cell then moves into S phase, thus becoming committed to divide.

2. How does the G1 checkpoint work? a. Malfunctions in the regulatory protein, E2F, cause the cell to move through G1 too quickly and divide too often. b. When E2F is bound to Rb, it is in the “off ” position and can’t activate the genes required for S phase. (Fig. 12.15) c. However, the situation changes if growth factors continue to arrive. d. In this way, growth factors act as a social signal to override and pass the G1 checkpoint.

3. How do social controls and cell-cycle checkpoints fail? a. Cells can become cancerous when social controls fail. b. The G1 cyclin can be overproduced, or Rb can be defective. c. There is no single therapy or cure, since cancer is a family of diseases with a complex and highly variable molecular basis.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter. It includes terms other than those highlighted in bold type.

.

anaphase

chromatid

G2 phase

apoptosis astral microtubule benign tumor binary fission cancer cell cycle cell division cell plate cell-cycle checkpoint centrioles centromere centrosomes chemotherapy

chromatin chromosome cleavage furrow cohesins cyclin-dependent kinase (Cdk) cyclins cyclins cytokinesis daughter cell dynein embryos G1 phase

gametes gap phases gene growth factors histones interphase kinetochore kinetochore microtubule M (mitosis) phase M phase-promoting factor (MPF) malignant tumor meiosis CHAPTER 12 The Cell Cycle

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metaphase metaphase plate metastasis microtubules mitosis negative feedback oocyte p53 parent cell

polar microtubule prometaphase prophase protein kinase Rb protein replication S (synthesis) phase serum sister chromatids

social control somatic cells spindle apparatus spindle fibers telophase tumor tumor suppressor

Lecture Activities A Case Study Investigating the brca1 Gene Estimated duration of activity: 15+ minutes Case studies are an excellent way to increase students’ interest in biology. You can do this activity completely in class (if the class or discussion section can meet in a computer lab), or you can have students start it in class and finish as an out-of-class assignment. It can also be assigned as homework and discussed during the next class session.

In this case study, students can investigate the inherited form of breast cancer and the gene it is linked to, brca1. They will then better understand the role of the brca1 gene product in the cell cycle and how defects in the gene could lead to cancer.

You and your friend Sally are talking about her family. She has three older sisters and an older brother. Sally jokingly calls herself the “after thought” because all of her siblings are so much older. In fact, Sally’s sister Sarah is 18 years older and is like a second mom to her.

Sally is worried about Sarah, who has just been diagnosed with breast cancer. You are surprised that she is not more upset. When you comment on that, Sally tells you it’s only because she has become used to dealing with the breast cancer issue. Her grandma Elise died from breast cancer when Sally was very young, and Sally’s mom, Elizabeth, is currently in remission. Sally goes on to tell you that she worries about the odds being stacked against her.

She thinks there’s a good likelihood that she will also develop the disease.

You tell Sally you have heard there’s a genetic test that will inform her about her susceptibility. You don’t know much about it, but it has something to do with the brca1 gene that you heard about on the news. You are eager to help Sally learn as much as possible, so the two of you go to the library to do some research.

Have students write a two- or three-page e-mail message to Sally, including (but not limited to) answers to these questions: What is the brca1 gene, and what does it have to do with breast cancer? For what protein does the brca1 gene encode? What is the normal function (in detail) of that protein? What kind of mutation is present in the gene in the breast cancer condition, and how does that mutation affect the protein’s function? If Sally had the genetic test, what would the doctors be looking for? If they found that she had a copy of the mutated brca1 gene, what would that mean? What are the implications of the test for Sally’s life? If you were in this situation, would you choose to take the test? (This question is still relevant if you are male because these issues arise with other inherited disorders such as Huntington’s disease.)

Internet resource: http://en.wikipedia.org/wiki/BRCA1

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Exploring the Evolutionary Relationships between Mitosis and Binary Fission Estimated duration of activity: 30–40 minutes Mitosis and the rest of the cell cycle in eukaryotes, and binary fission in prokaryotes like bacteria, are somewhat similar processes. This activity gets students to compare and contrast the two processes and take an educated guess about the evolutionary relationship between them. Introduce the concept of convergent evolution to the students. This concept does not come up until a much later chapter, but it is essential to get students thinking in evolutionary terms throughout the study of biology. Convergent evolution is defined in the Glossary as “Evolution of similar traits in distantly related organisms due to adaptation to similar environments and a similar way of life.”

A good example is the comparison of an ichthyosaur to a dolphin (see Chapter 28).

Adaptation to life in the sea repeatedly produces animals with streamlined shapes and elongated jaws and little or no hair. However, these traits did not occur in the common ancestor of these two organisms; instead, adaptation produced the traits at separate times in the evolution of these two animals. In contrast, the limbs of vertebrates are traits that evolved in the ancestor of vertebrates, and bones of the limbs have evolved in many different ways in vertebrates; but such evolution is a product of descent with modification.

Students should review the stages of the eukaryotic cell cycle, including mitosis, and compare those stages to binary fission. Then they must decide whether they think the two processes are examples of convergent evolution or descent with modification. The goal is not to come up with a correct, factual answer to the question; the goal is to evaluate the evidence on hand and use analytical skills to reach a reasonable answer to the question. As a follow-up, ask the students to research the scientific literature to find out what other scientists have written about the relationship between binary fission and the eukaryotic cell cycle.

Objectives: • Describe the phases of the cell cycle, including what happens in each phase.

• Describe in detail the steps of mitosis, which is one phase of the cell cycle.

• Compare and contrast mitosis to binary fission.

• Define the concepts of convergent evolution and descent with modification.

• Discuss whether the relationship of binary fission in prokaryotes to the cell cycle of eukaryotes is convergent evolution or descent with modification.

Procedure: 1. On the day of the exercise, have students form groups and instruct them to answer the questions listed below within the time frames given. For each group, have the students appoint a facilitator, a notetaker, and two active researchers. The facilitator’s responsibility is to keep the group on track, keep discussions moving, settle disputes, and enforce the time schedules. The notetaker takes detailed notes during the discussions and polishes the notes as needed. The active researchers are primarily responsible for finding the necessary material within the textbook to answer the questions for this activity. The facilitator and the notetaker also participate in research but may be too busy to fully participate. After students complete each of the numbered steps of the procedure, ask them for the answers they came up with, taking care to move continuously from group to group while asking questions. Overall, instructor-led questioning should take about 10 to 15 minutes.

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a. What are the phases of the cell cycle? Give a brief definition of each phase and a brief description of what happens in each phase. b. Focus on the phase of mitosis. Describe what happens in each stage of mitosis in detail, including cytokinesis in both animal and plant cells. c. Look at the stages of binary fission in Figure 12.9. Compare the steps of binary fission to the phases of the cell cycle, including mitosis and cytokinesis. Note which steps of binary fission have matches with cell-cycle phases, including mitosis and cytokinesis.

Note whether the matches are strong or weak. Note which steps of binary fission have no matches with cell-cycle phases (including mitosis and cytokinesis) and which phases of the cell cycle (including mitosis and cytokinesis) have no matches in binary fission.

(15 minutes for points a–c) 2. Go to Chapter 25 and define the term, descent with modification, and then go to Chapter 28 to define the terms convergent evolution and homologous structures. (5 minutes) 3. Discuss this question: Is the relationship between binary fission in prokaryotes and the cell cycle in eukaryotes convergent evolution or descent with modification? (10–15 minutes)

Discussion Idea: Cancer Treatments How do cancer treatments work? Almost everyone knows someone who has cancer. After reading and discussing this chapter in class, students will have a better understanding of how cancer arises. However, the chapter only touches on the strategies for treating cancer. Your class can explore the cell cycle further by examining how cancer is treated. Students will find this discussion interesting because they probably do not already know the strategy behind chemotherapy and radiation. You may also want to explore and discuss new cancer drugs.

Radiation Questions about radiation • What kind of radiation is used in radiation therapy?

• How does the radiation get to the tumor?

• How does the radiation “kill” the tumor?

• What are the side effects?

• What are the advantages and disadvantages of radiation therapy?

During radiation therapy, X-rays are directed at the tumor site. The X-rays react with the water in the cells to create reactive oxygen species that disrupt DNA structure and thus disrupt replication and ultimately cell division. This treatment slows tumor growth but can also damage normal cells. However, because the cancer cells are dividing faster than the normal cells, they are more susceptible to the damage and less able to be repaired. If the treatment is given in small spurts over several days, the damage accumulates in the cancer cells faster than in the normal cells.

Chemotherapy Questions about chemotherapy • What kinds of chemicals are used in chemotherapy?

• How do the drugs get to the tumor?

• How do the drugs “kill” the tumor?

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• What are the side effects?

• What are the advantages and disadvantages of chemotherapy?

Chemotherapy is the use of drugs to fight cancer. These drugs are administered by mouth or by injection, depending on the type of cancer, the stage, and the type of drug used. See the links below for more general information. http://www.mayoclinic.com/health/chemotherapy/MY00536 http://www.nlm.nih.gov/medlineplus/ency/article/002324.htm

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CHAPTER

13

Meiosis

Learning Objectives: Students should be able to...

• Distinguish: haploid vs. diploid; haploid cell vs. gamete; chromatid vs. chromosome; sister chromatids vs. homologous chromosomes; mitosis vs. meiosis.

• Explain the relationship between chromosome number and the DNA content of a cell as it proceeds through meiosis.

• Describe the movements of chromosomes from mother to daughter cells in meioisis I and • • •

meiosis II. Explain when and how meiosis reduces ploidy.

Explain how meiosis and outcrossing create genetically distinct cells.

Discuss ways mistakes are made during meioisis and list some consequences of these

• errors.

Explain why sexual reproduction is widespread in spite of its obvious disadvantages.

Lecture Outline • In fertilization, a sperm from a male joins an egg from a female.

• Sperm and eggs are gametes.

• Gametes have half the number of chromosomes as the rest of the cells in that organism.

• Meiosis is the process whereby gametes are made.

• When nuclei of a sperm and an egg fuse to form a new individual, the original chromosome number is restored.

• Useful vocabulary for understanding this process is found in Table 13.1.

I. How Does Meiosis Occur? (Fig. 13.4) A. Chromosomes come in distinct sizes and shapes.

1. Nettie Stevens observed the karyotype of fruit fly (Drosophila melanogaster) cells.

(Fig. 13.1), (BioSkills 11) a. Karyotype is the number and type of each chromosome in a cell. b. Fruit fly body cells contain eight chromosomes but only five distinct types, distinguished by their size and shape. c. Three chromosome types in fruit flies are found in pairs; the other two (X and Y, the sex chromosomes) do not always match. (Fig. 13.1) (1) Homologs, or homologous chromosomes, are the same size and shape.

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(2) Males have one X and one Y chromosome; females have a pair of X chromosomes.

2. Subsequent work by others confirmed that chromosomes differ in size and shape but occur in homologous pairs that carry slightly different versions of the same genes, called alleles. (Fig. 13.2) B. The concept of ploidy 1. Diploid organisms have two alleles for each gene: one on each of the homologous chromosomes.

2. Haploid organisms do not contain homologous chromosomes and have only one allele for each gene.

3. The number of chromosomes of distinct type and structure in a cell is called the haploid number (n) of that species.

4. A cell’s ploidy is the number of sets of chromosomes present. (Making Models 13.1) a. Students should be able to state the haploid number, ploidy, and total number of chromosomes present in a male Drosophila. b. A maternal chromosome comes from the mother, and the paternal chromosome comes from the father.

C. An overview of meiosis 1. Prior to meiosis, the identical copies, sister chromatids, are joined at the centromere.

(Figs. 13.3, 13.4) 2. A replicated chromosome is a single chromosome made up of two sister chromatids. a. Students should be able to draw the same chromosome in unreplicated and replicated states, label the sister chromatids, indicate the number of DNA molecules present in each structure, and explain why both structures represent a single chromosome.

3. Meiosis is two cell divisions. (Fig. 13.4) a. During meiosis I, the homologs separate and go to different daughter cells.

(1) The daughter cells are haploid after meiosis I. They contain half the genetic information of a diploid cell.

(2) Each chromosome still has two sister chromatids. b. During meiosis II, the chromatids separate and go to different cells.

(1) The cells produced by meiosis II have one copy for each chromosome type; they are haploid. c. Meiosis produces haploid cells that go on to develop into gametes through gametogenesis. d. Chromosome movement during meiosis I and II is coordinated by microtubules of the spindle apparatus that attach to kinetochores at the centromeres.

4. Meiosis I is a reduction division. a. The chromosome number is reduced by half in the gametes. b. When gametes fuse in fertilization, a diploid zygote containing two sets of chromosomes results. (Fig. 13.5) c. Each zygote receives one set of its chromosomes from its mother and one set from its father.

5. Meiosis in the context of an organism’s lifecycle (Fig. 13.6) .

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a. Meiosis in a diploid organism results in haploid gametes that fuse during fertilization to form a diploid zygote. b. Students should be able to predict how many DNA molecules will be present in gametes of a Drosophila (a diploid organism that has eight replicated chromosomes in each cell entering meiosis).

D. The phases of meiosis I (Figs. 13.7, 13.8) 1. Early prophase I: synapsis a. Replicated chromosomes pair up with their homolog (synapsis), the resulting structure is called a bivalent. b. Chromatids from different homologs are referred to as nonsister chromatids.

2. Late prophase I a. Nuclear envelope breaks down and microtubule fibers from opposing poles separate the homologous pairs. b. Chiasmata form between nonsister chromatids, one to several per homologous pair. c. Nonsister chromatids in the tetrad can exchange DNA at each chiasma. This process is called crossing over; it creates chromosomes that have regions of both maternal and paternal origin.

3. Meiosis I: metaphase I a. Homologous pairs migrate to the center (metaphase plate) of the cell. b. Each tetrad is positioned randomly without regard to which side of the metaphase plate the paternal or maternal homolog is on.

4. Meiosis I: anaphase I and telophase I a. Homologous pairs are pulled to opposite poles of the cell using microtubules. b. Each replicated chromosome (pair of sister chromatids) moves independently of every other replicated chromosome.

5. Meiosis I: a recap a. When meiosis I is complete, each daughter cell is haploid, with one full set of chromosomes, but each chromosome consists of two chromatids.

E. The phases of meiosis II (Fig. 13.7) 1. Meiosis II: prophase II a. Each chromosome, consisting of two chromatids attached at the centromere, migrates to the center of each daughter cell.

2. Meiosis II: metaphase II a. Chromosomes line up at the metaphase plate.

3. Meiosis II: anaphase II a. Sister chromatids separate and are pulled to opposite poles of the cells.

4. Meiosis II: telophase II a. The cytoplasm separates again, forming a total of four haploid daughter cells.

5. Summary of meiosis a. Meiosis occurs in sexually reproducing organisms. b. Meiosis reduces the chromosome number by half, from 2n to n.

F. A closer look at synapsis and crossing over (Fig. 13.8).

1. Crossing over can occur at many locations along the length of paired homologs.

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2. At a chiasmas, nonsister chromatids are attached to each other. When they break, corresponding segments of maternal and paternal chromosomes are exchanged.

3. Steps of synapsis and crossing over a. Chromosomes condense, and chromatids are held by cohesins. b. Homologs pair in synapsis and are held by proteins called synaptonemal complex. c. One or more breaks are made in the DNA, leading to crossover. d. In late prophase I, the synatonemal complex disassembles.

G. Mitosis vs. meiosis (Table 13.2); (Fig. 13.9) 1. Homologous chromosomes pair in meiosis but not in mitosis.

2. Meiosis reduces the number of chromosomes; mitosis does not.

3. Students should be able to describe the consequences for meiosis if homologs do not pair.

H. A closer look at synapsis and crossing over (Fig. 13.8) 1. During synapsis, two replicated homologs align along their length.

2. When homologs synapse, they form a synaptonemal complex.

3. Crossing over happens along the length of the paired homologs. a. Nonsister chromatids are broken at the same point and attached to each other, so that a portion of the chromatid is exchanged. b. As a result, segments of maternal and paternal chromosomes are swapped.

II. Meiosis Promotes Genetic Variation A. Sexual reproduction versus asexual reproduction 1. Sexual reproduction is preceded by meiosis and results in offspring with a combination of both maternal and paternal traits.

2. Asexual reproduction is based on mitosis and results in offspring identical to the parent (clones).

B. Chromosomes and heredity 1. Chromosomes carry the hereditary material.

2. A gene is a chromosome segment that contains the instructions for an inherited trait.

3. Every chromosome has a linear array of many genes.

4. Offspring produced by sexual reproduction are genetically different from one another and unlike their mother and father.

C. The role of independent assortment (Fig. 13.10) 1. During metaphase of meiosis I, each homologous pair of chromosomes lines up at the plane of cell division randomly with respect to which side of the plane the paternal or maternal member of the pair is on. (Fig. 13.10) 2. When pairs of homologous chromosomes line up and the homologs separate, different combinations of maternal and paternal chromosomes result. The number of different possible combinations of paternally derived and maternally derived chromosomes in the daughter cells is 2n, where n = the number of chromosomes in the haploid state. (principle of independent assortment).

3. The creation of new combinations of alleles is called genetic recombination.

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4. Students should be able to explain how genetic variation would be affected if maternal chromosomes always lined up together on one side of the metaphase plate during meiosis I and paternal chromosomes always lined up on the other side.

D. The role of crossing over 1. During prophase of meiosis I, homologous chromosomes exchange DNA at chiasmata.

2. This crossing over “mixes up” the maternal and paternal genes on each of the homologs.

3. Crossing over introduces an incalculable amount of genetic variation, since at least one chiasma forms per homologous pair.

E. How does fertilization affect genetic variation?

1. Fusion of male and female gametes occurs randomly. a. Self-fertilization (Fig. 13.11) (1) Male and female gametes that fuse are from the same organism.

(2) Offspring are still genetically different from parents due to independent assortment and crossing over occurring prior to gamete formation. b. Outcrossing (1) Male gametes are from a different individual than female gametes.

(2) This event maximizes genetic variability.

2. The amount of possible genetic variation is calculated by multiplying the number of possible gametes produced by the female parent times the number of possible gametes produced by the male parent. (BioSkills 4)

III. What Happens When Things Go Wrong in Meiosis?

A. How do mistakes occur?

1. Down syndrome (extra copy of chromosome 21) (Fig. 13.11) 2. Nondisjunctions are chromosome separation mistakes that can occur during meiosis I or II. 3. Both members of a homologous chromosome pair may migrate to the same daughter cell. (Fig. 13.12) a. One daughter cell has two copies of the chromosome after meiosis I (n + 1). b. The other daughter cell lacks a copy of the chromosome after meiosis I (n – 1). c. Following gamete formation, if the n – 1 daughter cell is fertilized by normal sperm, the zygote is 2n – 1: monosomy. d. Following gamete formation, if the n + 1 daughter cell is fertilized by normal sperm, the zygote is 2n + 1: trisomy. e. Cells with too many or too few chromosomes are aneuploid.

4. Sister chromatids do not separate in anaphase II. B. Why do mistakes occur?

1. Random error a. There is no evidence of an inherited tendency for meiotic mistakes. Example: Down syndrome (trisomy-21) does not run in families. b. Trisomies generally involve smaller chromosomes because most trisomies of larger chromosomes are lethal before birth. c. With the exception of trisomy-21, most human trisomies involve sex chromosomes.

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2. Maternal age. Example: trisomy-21. (Fig. 13.13)

IV. Why Does Meiosis Exist?

A. Sexual reproduction occurs in only a small fraction of all organisms.

1. Bacteria, archaea, and a few animals reproduce only asexually.

2. Most algae, fungi, some land plants, and most vertebrates reproduce both asexually and sexually.

3. Sexual reproduction is the major mode of reproduction in many multicellular organisms (insects, vertebrates, and flowering plants).

B. The paradox of sex 1. John Maynard Smith (1978) proposed that natural selection should eliminate sexual reproduction from a population, based on a mathematical model. a. Asexual reproduction is more efficient because no males are required, so a population of all females can produce more offspring than one in which some individuals are male. b. If a population is composed of asexual and sexual individuals, the asexual individuals should increase in frequency and the sexual individuals should decrease.

(Fig. 13.14) c. Paradox: Stable populations of sexually reproducing organisms do exist.

C. The purifying selection hypothesis 1. Asexual reproduction passes damaged genes to all offspring.

2. In sexually reproducing species, only half of the offspring inherit the damaged allele; this means that some of the offspring do not inherit the damaged gene at all.

3. Natural selection favors the success of those offspring that did not inherit the damaged gene; this is called purifying selection.

4. Purifying selection steadily reduces the numerical advantage of asexual reproduction.

5. An example supporting this hypothesis is the observation that a sexually reproducing species of Daphnia has fewer deleterious genes than an asexually reproducing species.

D. The changing-environment hypothesis 1. Sexual reproduction provides more genetic diversity.

2. A population of genetically diverse offspring is more likely to survive changing environmental conditions than a population of genetically identical organisms.

3. Testing the changing-environment hypothesis: Moran and Lively’s studies on the roundworm, Caenorhabditis elegans. (Fig. 13.15) a. In C. elegans, individuals are either male or hermaphrodite. Hermaphrodites can either self-fertilize or outcross with males. b. Roundworms were exposed to a bacterial pathogen that killed susceptible individuals.

(1) The exposed population showed increased frequency of outcrossing, and after 32 generations the roundworms were more resistant to the pathogen, even though the bacteria had become more virulent.

(2) Unexposed roundworms maintained a lower rate of outcrossing.

(3) The same experiment with a strain of roundworms unable to outcross resulted in their extinction.

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Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. allele anaphase I anaphase II aneuploid/aneuploidy asexual reproduction autosomes bivalent centromere changing-environment hypothesis chiasma (s.), chiasmata (pl.) chiasmata/chiasma clones crossing over cytokinesis daughter cell diploid early prophase egg embryo fertilization gametes gametogenesis

gene genetic recombination haploid haploid number hexaploid homolog homologous chromosomes homologous pair karyotype kinetochore late prophase life cycle maternal chromosomes meiosis meiosis I meiosis II metaphase I metaphase II metaphase plate monosomy nondisjunction nonsister chromatids octoploid oogenesis

outcrossing paternal chromosomes ploidy polyploid primary oocytes prophase II purifying selection hypothesis reduction division self-fertilization sex chromosomes sexual reproduction sister chromatid sperm spindle apparatus synapsis synaptonemal complex telophase I telophase II tetraploid triploid trisomy trisomy-21/Down syndrome zygote

Learning Activity Modeling Meiosis Estimated duration of activity: 10–15 minutes Meiosis is a concept best demonstrated using 3-D models. You can purchase commercial chromosome sets, but one cheap and easy alternative is to use craft sticks, available in all craft stores. The sticks are similar to pipe cleaners but easier to cut and more colorful. The variety of color is useful for denoting chromosomes of maternal versus paternal origin, and you can create chromosomes of various sizes by cutting the sticks. Procedure: 1. Have students form pairs. Each pair of students will need a baggie containing 2n = 4 chromosomes. There should be two long pieces in each of two colors and two shorter pieces in each of the same two colors (to allow for DNA replication). One student can prepare the baggies for a class of 50 in less than an hour.

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2. In class, after distributing the baggies, hold up examples from one chromosome set to demonstrate what each pair of students should do. You may want to go through mitosis first because this concept is easier, and your students should already be familiar with it.

3. After working through mitosis, each student pair will replicate their chromosomes. Remind students that cell division cannot take place without this replication, followed by coiling and condensation of the chromosomes.

4. Provide your students with prompts to come up with each step of the process: a. “What occurs next?” (The homologs pair; crossing over has to be imagined, or you can declare that during this particular division, no crossing over will occur.) b. “Next?” (Chromosomes line up, still in homologous pairs, in the center area of the cell.) c. “Next?” (The homologs separate.) d. “Next?” (The progeny cells now have one copy of each type of chromosome, so they are haploid, a very difficult concept for many students.) At this stage, the alleles for each gene on the homologous chromosomes are segregated from each other. e. “Next?” (The cells are ready for meiosis II, so each replicated chromosome separates at the centromere, each sister chromatid becomes a chromosome, and each cell now has the proper amount of DNA to be ready for fertilization.) f. Also ask your students to find alternative ways to line up the chromosomes to show the principle of independent assortment.

Meiosis is difficult for students to understand because most of them have memorized the sequence without understanding its function of reducing genetic information. Instructors can hold up one or two pairs of homologous chromosomes and ask whether the cell containing them is diploid or haploid. Most students should do fine with this. Then try holding up an unreplicated chromosome pair and ask the same question. Many students will say the cell is haploid because the chromosomes are not replicated (they will not be thinking about whether the cell contains one or two sets of chromosomes). If you try holding up one replicated chromosome from each pair (as would be present at the end of meiosis I), many students will tell you that the cell is diploid. They quickly forget that the chromosomes are replicated, but since the two chromosome sets have been separated, the cell is haploid.

A useful analogy is the information in a textbook. Having two copies of the same textbook (replicated chromosomes, the sister chromatids) does not give you twice the information.

However, having two different textbooks on the same subject (homologous chromosomes) does give you additional information, explaining why cells are generally diploid.

Another step may be to ask each pair of students to draw various combinations of haploid or diploid cells, replicated and unreplicated chromosomes, or paired and unpaired homologs.

Or you may show these combinations on an overhead while asking students about the chromosomes or the cells.

An option particularly good for demonstrating crossing over is to use colored modeling clay. After rolling out dough strips of two different colors (denoting chromosomes of maternal versus paternal origin), students can form homologous pairs and then select chromatids to participate in crossing over. They can tear the dough and reassemble it to illustrate the exchange that occurs during crossing over.

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CHAPTER

14

Mendel and the Gene

Learning Objectives: Students should be able to...

• Define common genetics terms and provide examples of multiple alleles, codominance, incomplete dominance, pleiotropic genes, environmental influences on phenotype, interactions between genes, and quantitative traits.

• Explain how Mendel drew conclusions about particulate inheritance and the principle of segregation from the analysis of monohybrid crosses.

• Explain how Mendel drew conclusions about the principle of independent assortment from •

the analysis of dihybrid crosses.

Explain how chromosome movement during meiosis accounts for the principles of segre-

• gation and independent assortment.

• Describe how gene linkage affects the assortment of alleles at meiosis.

Solve genetics problems, including those involving monohybrid crosses, dihybrid crosses, different forms of dominance, and pedigree analysis.

Lecture Outline Historical overview • Mendel laid the groundwork for the chromosomal theory of inheritance by working out the rules of inheritance.

• Nineteenth-century biologists described the details of meiosis.

• In 1902, Walter Sutton and Theodor Bovari proposed the theory that meiosis causes the patterns of inheritance Mendel observed and that hereditary factors called genes are located on chromosomes.

• The chromosome theory launched the study of genetics.

I. Mendel’s Experimental System A. What questions was Mendel trying to answer?

1. Hypotheses to explain how traits are transmitted from parent to offspring: a. Blending inheritance: Hereditary determinants from two parents blend together in their offspring.

(1) Example: A white sheep mates with a black sheep, producing gray offspring. b. Inheritance of acquired characteristics: Traits present in the parents are modified through use and passed on to offspring.

(1) Example: Adult giraffes strain their necks to reach food in trees, necks become longer, and their offspring inherit the longer neck trait.

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B. The garden pea served as the first model organism in genetics.

1. Garden peas are inexpensive, easy to grow, have a relatively short generation time, and produce large numbers of seeds.

2. Model organisms are useful for studies of biological sciences because they are practical, and conclusions that are drawn from them can apply to many other species.

(BioSkills 11) 3. How did Mendel control matings? a. Under normal conditions, self-fertilization occurs when the pollen from the male parts of one plant fall onto the female parts of the same plant. (Fig. 14.1a) b. Mating can be controlled by manual pollination to ensure cross-pollination, which occurs when the pollen from one plant falls on the female parts of a different plant.

(Fig. 14.1b) 4. What traits did Mendel study? a. Peas had multiple traits that exhibited one of two phenotypes. Mendel studied seven of these variable traits in peas: seed shape, seed color, pod shape, pod color, flower color, flower and pod position, and stem length. b. Phenotypes are the observable traits of an individual. c. A summary of terms in genetics is in Table 14.1.

5. True-breeding or pure lines could be obtained.

II. Mendel’s Experiments with a Single Trait A. The monohybrid cross 1. Monohybrid crosses investigate the inheritance pattern of a single trait.

2. Evidence for the particulate inheritance hypothesis: When two true-breeding pea plants (round versus wrinkled seeds) are crossed, all of the offspring have round seeds.

(Fig. 14.2a) a. Traits did not blend together to produce intermediate phenotypes. (Fig. 14.2b) b. The genetic component for the wrinkled seed seemed to disappear.

3. Dominant and recessive traits a. Recessive traits disappeared in the F1 generation and reappeared in the F2 generation. b. The trait that was visible in the hybrids is called “dominant.” c. Dominance and recessiveness identify only which phenotype is visible in a heterozygote; they do not refer to strength or fitness.

4. A reciprocal cross: a. Question: Is the inheritance of seed shape in peas affected by whether the genetic determinant is in a male or female gamete? (Fig. 14.3) b. Hypothesis: The type of gamete does affect the inheritance of seed shape. c. Null hypothesis: The type of gamete does not affect the inheritance of seed shape. d. Experimental setup: Perform two crosses. The first cross uses pollen from a roundseeded parent and is added to the female organ of the wrinkled-seeded plant. In the reciprocal cross, pollen from a wrinkled-seeded parent is added to the female organ of a round-seeded plant. e. Prediction: Offspring phenotypes will be different in the two crosses. f. Null prediction: Offspring phenotypes will be the same in the two crosses. .

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g. Results: The offspring were the same in the two crosses (all offspring had round seeds). h. Conclusion: It makes no difference whether the genetic determinant for seed shape comes from the male gamete or the female gamete.

5. The results are general. Mendel was able to repeat them with the other six traits he investigated. (Table 14.2) a. F1 progeny showed only the dominant trait and no intermediate phenotypes. b. Reciprocal crosses produced the same results. c. The ratio of dominant to recessive traits in the F2 generation was about 3:1. d. The results were inconsistent with both the blending inheritance hypothesis and the inheritance of acquired characters hypothesis. e. Mendel proposed a new hypothesis.

B. Particulate inheritance 1. Genetic determinants (genes) act like discrete particles.

2. Genetic determinants maintain their integrity from generation to generation. a. They do not become blended. b. They do not acquire new or modified characteristics through use.

3. Genes, alleles, and genotypes a. Each individual has two versions, or alleles, of each gene. b. All the alleles present in one individual are its genotype. c. Genotype influences the expression of traits (phenotype) in the individual. d. The allele hypothesis explains the framework for dominance/recessiveness.

(1) In garden peas, the allele for round seeds is dominant and wrinkled seeds is recessive.

(2) If one allele for round seed is present, the seed would be round.

7. The principle of segregation a. Phenotype depends on which two alleles of the gene are paired in the individual.

Using the above example, RR = round seeds, Rr = round seeds, and rr = wrinkled seeds. b. The alleles of a gene segregate to different gamete cells during egg and sperm formation. c. Each gamete receives one allele of each gene (haploid state). d. Fertilization restores the diploid state; the fusion of egg and sperm brings together two alleles of each gene.

(1) Homozygous: an individual with two copies of the same allele for a particular gene (2) Heterozygous: an individual with two different alleles of a particular gene e. Punnet squares help to predict the genotypes and phenotypes of different crosses.

(1) Fig. 14.4a demonstrates what happens to alleles during crosses of two homozygous individuals (pure lines): RR and rr. The cross results in all heterozygous individuals (Rr) that all have the dominant phenotype.

(2) Fig. 14.4b demonstrates what happens to alleles during crosses of two heterozygous individuals (Rr and Rr). The cross results in 1/4 RR, 1/2 Rr, and

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1/4 rr individuals, and 3/4 have the dominant phenotype and 1/4 show the recessive phenotype.

8. Predicting offspring genotypes and phenotypes with a Punnett square a. There are five steps to using a Punnett square: (1) Write each unique gamete genotype produced by one parent in the horizontal row along the top of the column.

(2) Write each unique gamete genotype produced by the other parent in the vertical column down the left side of the diagram.

(3) Create a table under the row of gametes and to the right of the vertical column of gametes.

(4) Fill in the table with entries for the parental gamete genotypes written at the top and left side.

(5) Tally the proportions of each genotype and phenotype. b. Students should be able to state how filling in the top and side of the Punnett square is related to the principle of segregation and predict the phenotype and genotype ratios for a cross between Rr and rr peas. c. Mendel’s conclusions can be summarized as a genetic model. (Table 14.3)

III. Mendel’s Experiments with Two Traits A. The dihybrid cross 1. Mendel generated a dihybrid (RrYy) by crossing a pure line of round, yellow-seeded parents with a pure line of wrinkled, green-seeded parents (RRYY × rryy).

2. The alleles for each gene may be transmitted to gametes independently (independent assortment; Fig. 14.5a) or together (dependent assortment; Fig. 14.5b).

3. Selfing the F1 generation produced phenotypes in the ratio of 9 round yellow to 3 round green to 3 wrinkled yellow to 1 wrinkled green.

4. If the two traits are considered separately, selfing of the F1 generation produced a 3:1 ratio for round:wrinkled and for yellow:green; therefore, the data support the principle of independent assortment for two traits. (Fig. 14.5c) 5. Students should understand that an individual with the genotype AaBb produces gametes with the genotypes AB, Ab, aB, and ab. Students should also be able to predict the genotypes of the gametes produced by individuals with the genotypes AABb, PpRr, and AaPpRr. B. Using a testcross to confirm predictions 1. Mendel wanted to test the prediction that an RrYy plant produces four different types of gametes in equal proportions.

2. A test cross crosses a parent that has the dominant phenotype with a parent that is recessive for the same trait(s). (Fig. 14.6) 3. If the parent with the dominant phenotype is RrYy and the R and Y alleles assort independently: a. Then crossing RrYy with rryy predicts four types of gametes (RY, Ry, rY, and ry) from the first parent and one type of gamete (ry) from the second parent. b. It also predicts four types of offspring: RrYy, Rryy, rrYy, and rryy (in a 1:1:1:1 ratio).

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4. Test crosses can also establish the genotype of a parent with dominant phenotypes.

If the parent with dominant phenotypes is homozygous (RRYY), then all offspring will have dominant phenotypes.

5. The resulting offspring were four different phenotypes in the proportion of 1:1:1:1.

These data confirm the principle of independent assortment.

6. Mendel identified the two fundamental patterns of transmission genetics: a. Segregation of alleles into separate gametes b. Independent assortment of alleles that control different traits

IV. The Chromosome Theory of Inheritance A. Meiosis explains Mendel’s principles.

1. Paternal and maternal chromosomes can possess different alleles for genes. (Fig. 14.7) 2. These alleles segregate into different daughter cells at anaphase I of meiosis.

3. If alleles for different genes are located on different (nonhomologous) chromosomes, then they assort independently from one another at meiosis I. (Fig. 14.8) 4. The locus is the position on the chromosome that a particular gene resides.

5. The chromosome theory of inheritance was formalized by Sutton and Bavari in 1902.

B. Testing the chromosome theory 1. Thomas Hunt Morgan and his students used Drosophila melanogaster as a model organism. He needed genetic variants as well as wild-type phenotypes for his crosses.

(BioSkills 11) 2. The white-eyed mutant a. Morgan discovered the white-eyed mutant Drosophila male in the red-eyed wild-type population. (Fig. 14.9) b. Morgan mated white-eyed males to red-eyed females.

(1) All F1 flies had red eyes. c. A reciprocal cross involving red-eyed males and white-eyed females revealed a different pattern.

(1) All the F1 females had red eyes, but all the F1 males had white eyes.

3. The discovery of sex chromosomes a. Nettie Stevens discovered sex chromosomes by studying karyotypes of the beetle Tenebrio. b. Like Tenebrio, Drosophila females carry two X chromosomes, and males carry an X and a Y. c. All Drosophila sperm (gametes) have one sex chromosome; half receive the X chromosome, and half receive the Y. 4. Sex linkage and the chromosome theory (Fig. 14.9) a. Genes located on the X chromosome are described as x-linkage. b. Genes located on the Y chromosome are described as y-linkage. c. Genes located on the sex chromosome are sex-linked.

5. Punnet square explains how eye color is located on the X chromosome. (Fig. 14.10) a. There is no allele for these genes on the Y chromosome. b. Males have one X chromosome and therefore one allele. The phenotype encoded by the allele on the X chromosome is observed.

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c. Females have two X chromosomes and therefore have two copies of the allele. d. Morgan proposed, based on results from reciprocal crosses, that the gene for eye color in fruit flies is on the X chromosome; the Y chromosome does not carry an allele for this gene. (Fig. 14.11) e. Reciprocal crosses that yield different results are likely sex-linked. f. Inheritance of genes located on nonsex chromosomes is called autosomal inheritance.

6. Morgan’s data confirmed the chromosomal theory of inheritance because they indicated that chromosomes carry genes that govern certain traits.

V. Extending Mendel’s Rules A. Linkage: What happens when genes are located on the same chromosome?

1. Linkage refers to the tendency of particular alleles of different genes to be inherited together as a result of being on the same chromosome. a. Genes located on the same chromosome will not assort independently of one another except when crossing over or genetic recombination occurs. (Fig. 14.11) 2. If this is true, then linked genes should be inherited together during gamete formation, and the F1 heterozygotes should produce only two classes of gametes.

3. The role of crossing over a. Morgan tested the prediction of independent assortment involving linked genes.

(Fig. 14.12) b. Research question: Will genes undergo independent assortment if they are on the same chromosome? c. Hypothesis: Linked genes will violate the principle of independent assortment. d. Null hypothesis: Linked genes will adhere to the principle of independent assortment. e. Experiment: Cross heterozygous females that have red eyes and gray bodies with males that have white eyes and gray bodies (these genes for eye color and body color are both on the X chromosome). f. Prediction: Male offspring will have only one copy of each gene from their mother, and the two potential male offspring genotypes are XwYY and XWyY. g. Null prediction: Four male genotypes are possible: XwYY, XWyY, XwyY, and XWYY, and they will occur in equal frequencies. h. Results: Four male genotypes were observed, but they were not in equal frequency. i. Conclusion: Neither hypothesis was fully supported. Independent assortment does not apply to linked genes. j. Linked genes segregate together, except when crossing over and recombination occurs. (Fig. 14.13) k. Individuals resulting from recombination events are known as recombinants.

4. The physical distance between two loci dictates the frequency of crossing over between the loci. The smaller the distance, the less chance of crossing over.

(Quantitative Methods 14.1) 5. When two genes are far apart on a chromosome, crossing over will always separate them. These genes will appear to assort independently because their recombination frequency is 50 percent. (Fig. 14.14) .

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6. Data on the frequency of crossing over between many genes on the same chromosome can be used to create a genetic map. (BioSkill 10) (Fig. 14.15) 7. The unit of measurement for the distance between loci is the centiMorgan (cM). One cM is equal to the physical distance that produces 1 percent recombinant offspring.

8. The locations of genes can be mapped by analyzing recombination. a. In Drosophila, yellow body and white eye are 1.4cM apart, and white eye and single bristle are 19.6cM apart. Therefore, yellow body and single bristle are 21cM apart. (Fig. 14.15a) b. Mapping multiple genes can provide a genetic map. (Fig. 14.15b) B. How many alleles can a gene have?

1. Multiple allelism: more than two alleles of the same gene. Example: ABO blood group in humans; more than 500 alleles of the β-globin gene 2. Polymorphism: two or more phenotypes in a population due to multiple combinations of multiple alleles a. ABO blood groups in humans refer to the polysaccharides (or lack thereof) associated with each allele type. There are six potential genotypes for the three alleles of the I gene. (Fig. 14.16) C. Are alleles always dominant or recessive?

1. Codominance a. Heterozygotes have the phenotype associated with each individual allele. Example: ABO blood types in humans 2. Incomplete dominance a. Heterozygotes exhibit a phenotype intermediate between those of the parental generation. Example: four o’clock flowers (Fig. 14.17a) (1) When two pure lines (red and white) are mated, they produce 1/4 red, 1/2 pink, and 1/4 white (1:2:1 Genotypic and phenotypic ratios; Fig. 14.17b).

3. Molecular biology of the various dominance relationships (This discussion is not in the text, but it is helpful.) a. Genes encode for proteins, and different alleles encode for different forms of the same protein. b. Codominance results when two different forms of a membrane protein are expressed in a single cell (A and B proteins on red blood cells). c. Complete and incomplete dominance usually involve alleles that encode for enzymes.

(1) Enzyme function is dependent on amino acid sequence.

(2) Usually one allele for an enzyme makes a functional protein, and the other allele makes a dysfunctional protein.

(3) Complete dominance occurs in heterozygotes for traits in which half the amount of functional enzyme is enough to generate the full dominant phenotype.

(4) Incomplete dominance occurs in heterozygotes when half the amount of functional enzyme produces half the dominant phenotype. Example: red pigment production in four o’clock flowers D. Does each gene affect just one trait?

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E. Are all traits determined by a gene?

1. The environment affects phenotype. a. Gene-by-environment interaction: An individual’s phenotype is the combined effect of the physical environment and genotype. b. For example, the PKU phenotype can be blocked by altering the physical environment⎯in this case, diet.

2. Different genes can work together to affect one phenotype. a. The phenotype produced by an allele depends on the action of the alleles of other genes. b. For example, the comb shape in chickens depends on the interaction of the alleles of two genes. (Fig. 14.18a) c. A genetic model of gene-gene interactions demonstrates that the phenotypes associated with any one allele of one gene depends on the alleles at a second gene.

(Fig. 14.18b) F. Can Mendel’s principles explain traits that don’t fall into categories?

1. Discrete traits have no intermediate phenotypes.

2. Quantitative traits exhibit variation in individuals. Examples: height and weight a. Traits are also affected by the environment. b. Phenotypes of the trait show a continuous, bell-shaped (normal) distribution of frequencies. (BioSkills 2) (Fig. 14.19) c. Herman Nilsson-Ehle proposed that if each gene contributes a small amount to the value of a quantitative trait, then a normal distribution for a population can be observed. d. He crossed pure lines of wheat plants with dark red kernels versus white kernels.

(1) The more alleles associated with the dominant trait present, the darker the red pigmentation.

(2) The frequency of the phenotypes in the F2 generation was approximately a normal distribution. (Fig. 14.20a) (3) He proposed that three genes control kernel color in wheat, and each assorts independently. (Fig. 14.20b) (4) This type of inheritance is called polygenic inheritance.

G. Key exceptions and extensions to Mendel’s rules (Table 14.4)

VI. Applying Mendel’s Rules to Human Inheritance A. Controlled genetic crosses cannot ethically be performed to study inheritance patterns in humans.

B. Scientists use genetic family trees (pedigrees) to study inheritance patterns in humans.

C. Pedigrees record each family member’s gender and phenotype and familial relationship.

D. Identifying human alleles as recessive or dominant 1. If a trait is governed by a single gene, then a pedigree will enable one to determine whether it is inherited in a dominant or recessive pattern.

2. Patterns of inheritance: autosomal recessive traits a. Individuals with the recessive trait must be homozygous for that allele.

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b. Parents of those individuals may not exhibit the trait, but they are likely to be carriers (heterozygotes). c. Affected individuals, if mating with an unaffected individual, may have unaffected children (who are heterozygous and would be carriers). d. An example pedigree for a family in which an autosomal recessive trait (sickle-cell anemia) occurs is provided in Figure 14.21. e. Summary of patterns observed in autosomal recessive traits (1) Males and females are equally likely to be affected.

(2) Affected offspring have unaffected parents.

(3) Unaffected parents of affected offspring are carriers.

(4) Affected offspring are homozygous.

(5) If both parents are heterozygous, about 1/4 of the offspring will be affected.

3. Autosomal dominant traits a. Both homozygous dominant and heterozygous individuals express the phenotype. b. All children of homozygous dominant individuals express the phenotype. c. Affected individuals have at least one affected parent. d. An example pedigree of a family in which an autosomal dominant trait (Huntington’s disease) occurs is provided in Figure 14.21b. e. Summary of patterns observed in autosomal dominant traits: (1) Males and females are equally likely to be affected.

(2) Affected offspring have at least one affected parent.

(3) Affected offspring are heterozygous if they only have one affected parent.

(4) Unaffected offspring are homozygous recessive.

(5) If one parent is affected, approximately half of the offspring will be affected.

(6) Trait does not skip generations.

E. Identifying traits as autosomal or sex-linked 1. If a trait appears equally in males and females, it is probably autosomal.

2. If males are much more likely to express the trait, then it is probably an X-linked recessive trait. Example: hemophilia 3. An example pedigree of a family in which an X-linked recessive trait (red-green color blindness) occurs is provided in Figure 14.22a. a. If males have an X-linked recessive allele, they will express the trait. b. Females can be carriers of an X-linked recessive trait without expressing the trait. c. Summary of patterns observed in X-linked recessive traits: (1) Males are affected more than females.

(2) Traits are not transferred father to son.

(3) Affected sons are born to carrier mothers.

(4) Approximately half of the sons from a carrier mother will be affected.

(5) All daughters of affected males and unaffected, noncarrier females are carriers (heterozygous).

(6) Traits often skip a generation.

4. If an affected male has all affected daughters and no affected sons, then the trait is probably X-linked dominant.

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a. An example pedigree of a family in which an X-linked dominant trait (rickets) occurs is provided in Figure 14.22b. b. Summary of patterns observed with X-linked dominant: (1) Males and females are equally likely to be affected.

(2) All daughters of an affected father are affected, but no sons.

(3) Affected sons always have affected mothers.

(4) Approximately 1/2 of the offspring of affected mothers will be affected.

(5) Affected daughters have an affected parent (either mother or father).

(6) Trait does not skip generations.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

1:1:1:1 1:2:1 3:1 9:3:3:1 allele autosomal autosomal inheritance autosomes blending inheritance carrier centiMorgan chromosome theory of inheritance codominance cross crossing over dihybrid cross diploid discrete traits discrete traits dominance dominant F1 generation F2 generation gene genetic map genotype hemophilia

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heredity heterozygous homozygous Huntington’s disease hybrids incomplete dominance inheritance of acquired characteristics linkage linkage map (genetic map) locus Marfan syndrome mode of transmission model organism monohybrid cross multiple allelism multiple allelism mutant mutation parental generation (P1) particulate inheritance pedigree phenotype phenylalanine phenylketonuria (PKU) pleiotropic pleiotropy pollen grains

polygenic traits polymorphic principle of independent assortment principle of segregation Punnett square pure lines quantitative traits recessive recessiveness reciprocal cross recombinants recombination rickets self-fertilization/selfing sex chromosomes sex-linked inheritance sex-linked inheritance/ sex-linkage testcross trait transmission genetics wild type wild-type X chromosome X-linked inheritance Y chromosome Y-linked inheritance

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Lecture Activities Suggested Readings for Instructors before Presentation of the Chapter 14 Material In three recent articles, authors have proposed introducing basic genetics in a way that emphasizes the connection with meiosis and postpones the discussion of phenotype until later. Concerns have also been raised about dominance and recessiveness of alleles.

Allchin, D. “Mendel’s mistake?” BioQuest Notes 9(3):1–7 (1999).

Allchin, D. “Mending Mendelism.” American Biology Teacher 62(9):633–639 (2000).

Donovan, M. P. “The vocabulary of biology and the problem of semantics: ‘Dominant,’ ‘recessive,’ and the puzzling role of alleles.” Journal of College Science Teaching 26(6): 381 (1997).

Illustrating the Continuum of Dominance Relationships Estimated duration of activity: 10–15 minutes Many students come into college biology classes with the misconception that a dominant allele is the “better” or more advantageous allele. This misconception makes it difficult for these students to understand alternative inheritance patterns such as codominance, incomplete dominance, gene interaction, and so on. This activity may be helpful in illustrating the molecular basis of dominance.

Introduce this activity by reminding students that genotype governs phenotype because genes encode for proteins. Remind them that proteins function in myriad ways in cells (enzymes, structural proteins, cell-surface recognition, etc.). First use the fictitious scenarios to illustrate the point, and then move on to the real biological examples, having the students come up with the molecular mechanism that underlies the inheritance pattern.

Note: The fictitious scenarios may seem silly and oversimplified to you, but students will enjoy the break from lecture and will appreciate the visual example.

Fictitious scenarios Complete dominance: The jumping-jack protein is encoded for by the J gene. The J allele encodes for an enzyme that produces jumping-jack movement. The j allele encodes for a dysfunctional enzyme that does not produce any movement.

Have two students write J on a piece of paper and come to the front of the class. Have two other students write j on a piece of paper and come to the front of the class. Point out that each student represents a chromosome that is carrying one allele for the J gene. Create two homozygous parents by grouping like alleles. Illustrate their phenotypes by having the J students do jumping jacks and the j students remain still. Then illustrate segregation by having the alleles separate; illustrate fertilization by bringing a J and a j together. Again, illustrate phenotype by having the J student do jumping jacks while the j student remains still.

Point out that the Jj individual still exhibits the jumping-jack phenotype even though it carries only one J allele.

Incomplete dominance: The laughter protein is encoded for by the L gene. The L allele encodes for an enzyme that produces really loud laughter. The l allele encodes for an enzyme that produces no laughter at all.

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Have two students write L on a piece of paper and come to the front of the class. Have two other students write l on a piece of paper and come to the front of the class. Point out that each student represents a chromosome that is carrying one allele for the L gene. Create two homozygous parents by grouping like alleles. Illustrate their phenotypes by having the L students laugh loudly and the l students remain silent. Then illustrate segregation by having the alleles separate; illustrate fertilization by bringing an L and an l together. Again, illustrate phenotype by having the L student laugh loudly while the l student remains silent. Point out that the volume of laughing in the Ll individual was only about half that of the LL individual.

Codominance: The pen-holding protein is encoded for by a single gene. The R allele encodes for a single raised hand holding a red pen, and the B allele encodes for a single raised hand holding a blue pen.

Have two students write R on a piece of paper and come to the front of the class. Have two other students write B on a piece of paper and come to the front of the class. Point out that each student represents a chromosome that is carrying one allele for the pen-holding gene.

Create two homozygous parents by grouping like alleles. Illustrate their phenotypes by having the R students hold a red pen up in the air and the B students hold a blue pen in the air.

Then illustrate segregation by having the alleles separate; illustrate fertilization by bringing an R and a B together. Again, illustrate phenotype by having the B student hold up a blue pen while the R student holds up a red pen. Point out that the homozygous individuals could be identified by the presence of either blue or red pens. However, the heterozygote could be identified by the presence of both blue and red pens.

Real scenarios Complete dominance: Pea shape in garden pea plants is an example of complete dominance.

The W gene encodes for an enzyme that synthesizes starch granules in peas. The w gene encodes for a dysfunctional enzyme that will not synthesize the starch granules. The presence of starch granules makes the seeds round, and the absence of the starch granules results in wrinkled seeds. Heterozygotes have only half as many starch granules, yet that is enough to make the seeds round.

Incomplete dominance: Piebald (white) spotting in cats is an example of an incompletely dominant trait. Cats homozygous for the s allele do not have any white spots. Cats with the S allele are completely white. Cats that are heterozygous (Ss) are about 50 percent white.

Codominance: Sickle-cell anemia is an example of codominance in humans. Heterozygotes have one copy of the wild-type hemoglobin and one copy of the mutant hemoglobin.

Phenotypically, heterozygotes have about half the number of sickled red blood cells as homozygotes do.

Powers of Deduction Estimated duration of activity: 5 minutes (per scenario) Students often find genetics a difficult and uninteresting subject due to its complexity and the necessity of studying mainly plant examples. After discussing this chapter, you may find it helpful to use this in-class activity to capture the students’ attention and to illustrate the relevance of the information in Chapter 14. If you do not have time in class to do one or more of these exercises, you may want to consider assigning them as homework or for use in TA-led study sessions.

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A parental mystery: Mr. and Mrs. Smith have just arrived home with their new baby. As soon as they walk in the door, they get a call from the hospital maternity ward. The nurse who discharged the Smiths is calling to say she is now worried that she sent the wrong baby home with them. They return to the hospital for a battery of blood tests to clear up the confusion.

Help them interpret the results of the blood test. Could baby Emma be their daughter? Why or why not?

Hemoglobin Type (H—normal; h—mutant hemoglobin, sickle cell)

Individual

ABO Blood Type

Mr. Smith

Type A

MM

Hh

Mrs. Smith

Type B

MN

HH

Baby Emma

Type O

NN

Hh

MN Blood Type

Note: During this activity, you can point out that both blood-type systems are examples of codominance and that sickle-cell anemia is an example of an autosomal recessive trait that also exhibits codominance.

Your future children: You visit a gypsy at the county fair (for fun, of course; you don’t actually believe in fortune-telling), and she sees your future son in her crystal ball. He is a fighter pilot in the U.S. Air Force. You are laughing with some friends about the vision later, but you are curious whether your future son really might be a fighter pilot. You know that people who are color-blind cannot become Air Force pilots. Draw a pedigree of your family, and determine the probability that you will have a son who is color-blind. Of course, you will need to make some assumptions about your mate . . . just be sure to state those assumptions clearly.

Discussion Idea: What Will Your Children Look Like?

Photography booths that claim to produce pictures of what a male or female child of any given couple will look like are beginning to appear in arcades throughout the country. A couple enters the booth, a camera takes their pictures, and then a computer program analyzes and blends the morphologies of the two people to produce images of a male child and a female child they might create.

Question: In genetic terms, why will the couple be wasting their money if they buy these computer images?

Answer: In their answer, students should mention the blending hypothesis, which was rejected on the basis of Mendel’s experimental results, and the information from Chapter 13 (that each human can produce 8.4 million genetically different gametes as a result of independent assortment alone), plus the added variation introduced by crossing over (which cannot be estimated), combined with the randomness of fertilization by any given sperm (which is a possibility of 1 in 8.4 million). The enormous amount of built-in variability that exists in the process of making a new human means that siblings can look very different from one another and from their parents. The computer generates only a single image out of millions of possibilities. Furthermore, the computer image is based solely on the physical appearance and morphological features of the two potential parents; it cannot take into account the actual genotype information.

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15

DNA and the Gene: Synthesis and Repair

Learning Objectives: Students should be able to...

• Describe the secondary structure of DNA and interpret the evidence of the Hershey−Chase experiment that pointed to DNA as the hereditary material.

• Interpret the results of the Meselson−Stahl experiment to argue that DNA is replicated semiconservatively.

• Predict the sequences and their polarities of DNA and RNA molecules that would be syn• thesized using any template sequence of DNA. List the enzymes involved in leading and lagging strand replication and explain the role of • each in DNA replication.

•

Explain why one strand of DNA at the end of a chromosome cannot be replicated to its end and how telomerase solves this problem.

Describe how proofreading, mismatch repair, and nucleotide excision repair work and why they’re important.

Lecture Outline I. What Are Genes Made Of? A. The Hershey−Chase experiment 1. The T2 virus attaches to a host cell, leaves a capsule outside, and injects a material inside. (Fig. 15.1) 2. The host cell begins to produce copies of the T2 virus. a. The injected material from the virus must be the hereditary material because it triggers the production of new T2 viruses. b. The capsule does not enter the cell, so it cannot be the hereditary material. c. Analysis showed that the capsule is protein and the injected material is DNA. (Fig. 15.2) d. Conclusion: DNA, not protein, is the hereditary material.

3. Students should be able to understand the work of Hershey and Chase and predict what the investigators would have concluded if they had found both radioactive DNA and protein in the pellet.

B. The secondary structure of DNA 1. DNA’s primary structure: a long, linear polymer made up of nucleotides .

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a. Each nucleotide consists of a deoxyribose sugar, a phosphate, and a nitrogenous base. (Fig. 15.3a) b. Nucleotides are linked together with phosphodiester bonds. c. The primary structure has two components: (Fig. 15.3b) (1) A sugar-phosphate backbone (2) A series of nitrogenous bases that protrude from the backbone d. One end of the polymer has a free –OH group (3' end), and the other has a free phosphate group (5' end).

2. DNA’s secondary structure: two antiparallel strands bound together at the nitrogenous bases (Fig. 15.4) a. The two strands twist around each other to form a double helix. b. The secondary structure is stabilized by hydrogen bonds between complementary base pairs.

(1) Adenine always pairs with thymine.

(2) Guanine always pairs with cytosine.

3. Watson and Crick suggested that complementary base pairing implied that a strand of DNA could be used as a template for its own replication.

II. Testing Early Hypotheses about DNA Synthesis A. Three different models of DNA replication were proposed, inferred from the structure of DNA. 1. Semiconservative replication model a. The two original strands separate, each serving as a template for a new strand. b. Each new daughter DNA molecule would consist of one parent and one new strand.

2. Conservative replication model a. Either the bases on each original strand temporarily turn outward, or a non-DNA intermediate molecule is produced and becomes the template for synthesizing new molecules. b. The parent DNA molecule would remain unchanged; the daughter molecule would consist of two entirely new DNA strands.

3. Dispersive replication model a. The original DNA molecule is cut and unwound in short stretches, each of which is copied and then put back together. b. Each DNA strand in a new daughter DNA molecule would consist of a mixture of parent and daughter DNA. B. Meselson and Stahl distinguished between the semiconservative and conservative models.

1. Under laboratory conditions, Escherichia coli (E. coli) cells divide regularly, replicating their DNA before each round of cell division. (BioSkills 11) 2. Meselson and Stahl developed a technique to distinguish parent DNA from new DNA by tagging it with a heavy isotope of nitrogen, 15N. 3. Experiment a. E. coli cells were grown in a medium with 15N as the only nitrogen source. b. A sample of cells grown in the 15N for several generations was removed, and the

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DNA was purified (generation 0, the parental generation: all DNA has 15N).

(Fig. 15.5) c. The remaining culture was transferred to the 14N medium; a sample of cells was removed after each round of cell division and the DNA was purified (generations 1 and 2: DNA is a mix of 14N and 15N). d. The DNA samples from generations 0, 1, and 2 were subjected to density gradient centrifugation. (BioSkills 7) e. Predictions (1) If DNA replication is semiconservative, after two rounds in the 14N medium, there should be an intermediate-density band and a low-density band.

(2) If DNA replication is conservative, after two rounds in the 14N medium, there should be a high-density band and a low-density band. f. Results: The DNA from generation 2 formed two bands: one of intermediate density and one of low density. g. Conclusion: The data supported the semiconservative model of DNA replication.

III. A Model for DNA Synthesis (Table 15.1, Figs. 15.8, 15.10) A. DNA polymerase catalyzes DNA synthesis.

1. DNA polymerases add dNTPs only to the 3' end of a growing DNA strand. (Fig. 15.6) 2. DNA synthesis always proceeds in the 5' → 3' direction.

3. Students should be able to draw two lines indicating a DNA molecule, assign the

3' → 5' polarity of each strand, and then label the direction in which DNA synthesis will proceed for each strand.

B. Where does replication start?

1. Electron micrographs show that DNA replication is bidirectional. a. Replication is initiated at an origin of replication, forming a replication bubble.

(Fig. 15.7a) (1) Bacterial chromosomes have a single origin of replication. (Fig. 15.7b) (2) Eukaryotic chromosomes have multiple origins. (Fig. 15.7c) b. Replication proceeds in both directions at the same time. c. Each side of the bubble forms a Y-shaped replication fork.

C. How is the helix opened and stabilized? (Fig. 15.8) 1. DNA helicases open the replication fork by breaking hydrogen bonds between DNA strands, thus producing single strands.

2. Single-stranded binding proteins (SSBPs) stabilize single strands.

3. Topoisomerases relieve tension in the DNA molecule. a. Unraveling the two DNA strands at the replication fork creates twisting and tension farther down the molecule. b. Topoisomerases nick the DNA downstream and allow it to untwist.

D. How is the leading strand synthesized? (Fig. 15.8) 1. Primase, an RNA polymerase, builds a short RNA primer complementary to the parent strand.

2. DNA polymerase adds dNTPs to the 3' end of the primer, moving into the replication fork unzipping ahead of it. .

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a. DNA polymerase can add dNTPs only to an existing −OH group. b. The primer provides an −OH group to which DNA polymerase can add the first dNTP. 3. Behind DNA polymerase III, a doughnut-shaped structure called the sliding clamp holds the newly synthesized strand in place.

4. This process occurs continuously (without interruption) on the leading strand.

5. Students should be able to list the enzymes involved and predict the consequences if any of them are defective.

E. How is the lagging strand synthesized? (Fig. 15.10) 1. The other parent strand is antiparallel to the leading strand, so after primase adds a primer, DNA polymerase III must work away from the replication fork on this strand, called the lagging strand. (Fig. 15.9) 2. When helicase expands the replication fork, a new segment of parent strand is exposed. a. A new primer must be made to initiate replication in this section. b. This results in discontinuous replication on the lagging strand. c. The lagging strand is synthesized in short, discontinuous fragments (“Okazaki fragments” named after Reiji Ozazaki’s work using pulse-chase experiments).

3. Removing the primers and joining the fragments (Fig. 15.10) a. DNA polymerase I removes the primers at the start of each fragment. b. DNA ligase catalyzes the formation of a phosphodiester bond between the 3'–OH of one fragment and the 5'–P of the next fragment. c. Students should be able to draw what the two molecules resulting from DNA synthesis—starting from a single origin of replication—would look like if DNA ligase were defective.

4. The enzymes required for DNA synthesis are organized into the replisome.

(Fig. 15.11)

IV. Replicating the Ends of Linear Chromosomes A. The telomere is the region at the end of a linear eukaryotic chromosome.

B. Telomerase solves the end replication problem. (Fig. 15.12) 1. Replication of the leading strand can proceed to the end of the chromosome.

2. Replication of the lagging strand cannot proceed to the end of the chromosome. a. Primase can add a primer near the end of the chromosome. b. Once the primer is removed, DNA polymerase cannot replace the primer because there is no 3'−OH group onto which it can add. c. As a result, there is a bit of single-stranded DNA, a 3' overhang, on each end of the chromosome. d. This 3' overhang is degraded through a series of many replications, which results in the shortening of the chromosome.

C. Process of telomere replication (Fig. 15.13) 1. A strand of parent DNA remains unreplicated when the RNA primer is removed from the end of the lagging strand.

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2. Telomerase binds to the “overhanging” section of single-stranded parental DNA and, using its own internal template, extends the strand.

3. Telomerase extends the DNA strand by adding additional repeats.

4. Standard DNA synthesis on this template creates double-stranded DNA to prevent chromosome shortening.

D. Effect of telomere length on cell division 1. Telomerase is found primarily in cells that produce gametes. a. Cells not involved in gamete formation are somatic cells. b. Chromosomes of somatic cells gradually shorten with each mitotic division, becoming smaller with age.

2. Telomere length predicts the number of cell divisions. a. Carol Greider tested this by obtaining cells from donors aged newborn to 90 and growing these cells in culture. (BioSkills 11) b. Figure 15.4 shows the positive relationship between telomere length and number of divisions before the cells stop dividing. (BioSkills 3) 3. Telomere shortening controls the number of cell divisions, just as restoration of telomerase activity frees those limitations resulting in somatic cells dividing long past the age.

4. Telomerase is also active in cancer cells, allowing unlimited cell division.

V. Repairing Mistakes and DNA Damage A. Correcting mistakes in DNA synthesis 1. DNA replication must be extremely accurate because mistakes would accumulate during development or could be passed on to the next generation.

2. DNA polymerase moves quickly and inserts incorrect nucleotides about once every 100,000 bases. (Fig. 15.15a) 3. DNA polymerase can proofread. a. A mutant E. coli with an error rate 100 times higher than normal was found.

(1) The mutation was in the epsilon (ε) subunit of polymerase III. (2) The ε subunit of polymerase III is an exonuclease. (Fig. 15.15b) (3) Polymerase III checks nucleotides after they are added to the new strand.

(4) The ε subunit moves 3' → 5' to remove any incorrect nucleotides. b. Conclusion: Polymerase III has a proofreading function. c. Eukaryotic DNA polymerases have the same type of proofreading.

4. Proofreading by polymerase III reduces the error rate to 1 per 10 million bases, which is still not accurate enough.

5. Mismatch repair a. After DNA replication is completed, other DNA repair systems correct most remaining mismatches within several minutes. b. Mismatch repair enzymes were discovered in E. coli mutants. c. The enzyme functions like a copy editor who corrects typos.

B. Repairing damaged DNA 1. Even after replication, DNA bases can become damaged by chemicals and environmental elements. .

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2. Ultraviolet light can cause the formation of thymine dimers. (Fig. 15.16) 3. Cells fix damaged nucleotides by cutting them out and replacing them with undamaged nucleotides; the system is called nucleotide excision repair. (Fig. 15.17) a. Enzymes find damaged nucleotides by identifying irregularities in DNA’s secondary structure. b. Enzymes remove single-stranded DNA around the damage. c. DNA synthesis machinery replaces the excised nucleotides.

C. Xeroderma pigmentosum: A case study 1. Xeroderma pigmentosum (XP) is a rare, autosomal-recessive disease.

2. XP causes extreme sensitivity to sunlight; skin lesions form after brief exposure.

3. The rate of skin cancer is elevated 1,000 to 2,000 times.

4. Hypothesis: Patients with XP have a defective DNA repair system and cannot repair UV light-induced damage (Cleaver, 1968).

5. James Cleaver’s experiment: a. Cleaver exposed cell cultures from normal humans and XP patients to UV light.

(BioSkills 11) b. He supplied the cultures with radioactive thymine. c. Cells from XP patients incorporated very little radioactive thymine and had a very low survival rate after UV exposure. (Fig. 15.18) d. Cells from normal individuals incorporated large amounts of radioactive thymine and had a high survival rate. (Fig. 15.18) e. Conclusion: DNA repair is virtually nonexistent in cells from XP patients.

6. There is an increased mutation rate in XP patients because defective DNA repair can trigger mutations that lead to cancer.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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N and 14N 3' → 5' direction 35S, 32P 5' → 3' direction antiparallel strands capsid complementary base pairing conservative DNA replication density gradient centrifugation deoxyribonucleases (DNases) 146

deoxyribonucleotide triphosphates (dNTP) discontinuous replication hypothesis dispersive DNA replication DNA ligase DNA polymerase I (pol I) DNA polymerase III (pol III) DNA polymerase III epsilon subunit DNA proofreading DNA replication double helix Exonuclease

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genome helicase host lagging, or discontinuous, strand leading, or continuous strand mismatch repair nucleotid excision repair Okazaki fragments origin of replication primase primer proteases replication fork .

replisome ribonucleases (RNases) RNA polymerase RNA primers semiconservative DNA replication

single-stranded binding proteins (SSBPs) sliding clamp somatic cells T2 virus telomerase

telomere thymine dimer topoisomerase xeroderma pigmentosum (XP)

Lecture Activity Folic Acid and DNA Repair Estimated duration of activity: 5–10 minutes This lecture activity is a suggestion for engaging students while reinforcing a concept foundational to their understanding of the subject at hand.

Folic acid (folate), a vitamin B derivative, has recently received attention for its protective effects on developing fetuses. However, folic acid has also been claimed to prevent many different types of cancers as well as neurodegenerative disorders such as Alzheimer’s disease and Parkinson’s disease. Folate is involved in photorepair, a light-dependent mechanism by which thymine dimers are corrected.

At the end of the class period in which you discuss DNA repair, ask your students to research folic acid on the Internet. Ask them to come prepared to the next class period to answer the following questions. (If they need motivation to complete the assignment, write a series of multiple-choice questions or a short essay question about folic acid and its effects on the next test. This is a nice context in which to test their knowledge of DNA repair, UV light damage, and cancer.)

Questions: • What is the normal role of folate in cellular function?

• In what foods is folate normally found?

• In which disorders or diseases is a folic acid deficiency implicated?

• Why would a deficiency in folic acid increase one’s risk of having those disorders or diseases?

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CHAPTER

16

How Genes Work

Learning Objectives: Students should be able to...

• Interpret results of experiments that supported the one-gene, one-enzyme hypothesis.

• Use the central dogma of molecular biology to explain the link between genotype and phenotype.

• Use the genetic code to predict how mutations can influence the production of proteins.

• Compare and contrast different types of mutations such as point mutations versus chromosome mutations and beneficial, neutral, and deleterious mutations.

Lecture Outline I. What Do Genes Do? A. George Beadle and Edward Tatum (1941) produced an explicit hypothesis of what genes do.

1. In their experiment, they damaged a gene (null or loss of function) to study the phenotype of the resulting mutant.

B. The one-gene, one-enzyme hypothesis 1. Beadle and Tatum used Neurospora crassa, a bread mold.

2. Wild-type Neurospora grows on minimal medium because it can synthesize all the amino acids and other substances it needs.

3. They exposed Neurospora to radiation to create mutations.

4. They analyzed knock-out mutants and generated the one-gene, one-enzyme hypothesis, which asserted that each gene contains the information to make one enzyme.

C. An experimental test of the hypothesis 1. Adrian Srb and Norman Horowitz used arginine mutants in N. crassa to confirm the one-gene, one-enzyme hypothesis.

2. Their hypothesis: Arginine is synthesized in a metabolic pathway that involves ornithine and citrulline as intermediates because their structures are very similar to arginine. (Fig. 16.1) 3. They performed a genetic screen to identify mutants; they grew mutants in media containing or lacking ornithine, citrulline, or arginine. (Fig. 16.2) 4. Results: None of the mutants grew on minimal medium alone, but they identified three arginine mutants: arg 1, arg 2, and arg 3.

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5. Conclusions: a. Different Neurospora mutants had defects in different genes. b. The metabolic pathway leading to arginine synthesis could be partially deduced from the mutants. c. Ornithine is a precursor to citrulline, which is a precursor to arginine. d. The one-gene, one-enzyme hypothesis: One gene contains the information needed to code for one enzyme.

(1) The hypothesis was later revised to be the one-gene, one-polypeptide hypothesis.

6. These data confirmed what genes do; they are the instructions for making proteins.

II. The Central Dogma of Molecular Biology A. The genetic-code hypothesis 1. Francis Crick proposed that the nucleotide sequence of DNA is a code; DNA is only an information storage molecule.

2. Different combinations of bases specify each of the 20 amino acids.

3. The amino acid sequence of an enzyme is coded for by a specific stretch of DNA sequence.

4. The information in DNA can be copied for the next generation by complementary base pairing of nucleotides.

B. RNA as the intermediary between genes and proteins 1. Information does not flow from DNA directly to proteins because DNA is enclosed in the nucleus (in eukaryotes), and proteins are made on ribosomes in the cytoplasm.

2. Jacob and Monod hypothesis: RNA molecules act as intermediaries between genes and ribosomes. (Fig. 16.3) a. Prediction: Short-lived messenger RNA carries the information of DNA to the ribosomes. b. Follow-up research confirmed the mRNA hypothesis.

(1) RNA polymerase, the enzyme that synthesizes RNA, was discovered.

(2) RNA polymerase uses the information in DNA to determine the RNA sequence.

(3) RNA polymerase does not need a primer.

(4) RNA polymerase pairs uracil (not thymine) to adenine.

C. Dissecting the central dogma 1. The central dogma summarizes the flow of information in a cell.

2. The central dogma states that DNA codes for RNA, and RNA codes for protein. a. DNA is hereditary material. b. Genes are stretches of DNA that code for proteins. c. The gene sequence in DNA specifies an RNA sequence. d. The RNA sequence specifies a sequence of amino acids in the protein. e. Ultimately, genes code for proteins.

3. The roles of transcription and translation a. The process of making RNA from the DNA code is called transcription; DNA is transcribed to RNA by RNA polymerase.

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b. The process of using the RNA code to make a protein is called translation; mRNA is translated to proteins in ribosomes.

4. Linking genotypes and phenotypes a. The central dogma states that an organism’s genotype is determined by its DNA sequences, and its phenotype is a product of the proteins it produces. Example: coat color in Oldfield mice (Fig. 16.4) b. Alleles are two forms of the same gene that have slightly different sequences and thus code for slightly different proteins. (Fig. 16.4) (1) Their primary structures differ slightly.

(2) Their functions likely differ as well.

5. Exceptions to the central dogma a. Many genes code for RNA molecules that are not translated into proteins.

(1) Some RNAs are important during translation.

(2) Some RNAs regulate transcription.

(3) Some RNAs regulate which proteins are active in a cell. b. Reverse transcriptase: Some RNA viral polymerases synthesize DNA from the RNA template. Example: human immunodeficiency virus (HIV)

III. The Genetic Code A. How long is a “word” in the genetic code?

1. The genetic code is a set of rules that specify the relationship between a sequence of nucleotides in DNA or RNA and the amino acid sequence in the primary structure of a protein.

2. Gamow hypothesized a 3-base code. (Fig. 16.5) a. A 1-base code (4 different nucleotides) could specify only 4 amino acids. b. A 2-base code (4 × 4) could specify 16 amino acids. c. A 3-base code (4 × 4 × 4) could specify 64 different amino acids, more than the 20 found in living organisms. d. Prediction: The code is redundant; some amino acids are specified by more than one triplet of nucleotides.

3. Francis Crick and Sydney Brenner confirmed that codons are three bases long. a. They used a mutagen that caused additions or deletions of single nucleotides. b. A single nucleotide addition can lead to loss of gene function. c. Additions or deletions of three nucleotides restore normal function, but additions or deletions of one or two nucleotides do not. d. Conclusions: (1) The code is read in triplets of bases. Each triplet is a codon.

(2) Additions or deletions of one or two nucleotides disturb the reading frame (the groups of three nucleotides that are read as codons) and scramble the message.

(3) Additions or deletions of three nucleotides maintain the reading frame (although the sequence either gains a triplet or loses one).

B. How did researchers crack the code?

1. Marshall Nirenberg and Heinrich Matthaei developed a technique to synthesize RNAs of known sequence.

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a. They used the enzyme polynucleotide phosphorylase, which randomly catalyzes phosphodiester bond formation between ribonucleotides. b. The addition of only uracil ribonucleoside triphosphates to a reaction mixture containing the enzyme results in the formation of poly-uracil (poly-U) RNA. c. Poly-U RNA was added to an in vitro mixture containing ribosomes and amino acids.

(1) From poly-U RNA, a short protein consisting of only phenylalanine is produced.

(2) The triplet U–U–U in mRNA codes for the amino acid phenylalanine. d. Later work by others deciphered the exact amino acids specified by each codon and identified start and stop codons. (Fig. 16.6) 2. Analyzing the code a. The code is redundant: All amino acids except methionine are specified by more than one codon. b. The code is unambiguous: A given codon never codes for more than one amino acid. c. The code is nonoverlapping: One ribosome locks onto the first codon and the reading frame is established. d. The code is nearly universal: All organisms, with few exceptions, use the same codons to specify the same amino acids. e. The code is conservative: When several codons specify the same amino acid, the first two bases in the codons are almost always identical.

(1) If a transcription or translation error occurs in the third position of the codon, it is less likely to affect the amino acid in the final protein.

(2) This minimizes the phenotypic effects of small changes in DNA and increases the efficiency of the genetic code.

3. The value of knowing the code a. One can predict the amino acid sequence of a protein from the gene sequence.

(Fig. 16.7a) b. Students should be able to identify the codons in Figure 16.4 and check to make sure they are translated correctly. They should also be able to complete the exercises in Figure 16.7b, indicate the mRNA’s 5'→3' polarity, and then write the double-stranded DNA that corresponds to this mRNA. They should also be able to indicate the 5'→3' polarity of both DNA strands.

IV. What are the types and consequences of mutation?

A. What is a mutation?

1. A mutation is a permanent change in an organism’s DNA, a change in its genotype.

2. Mutations create new alleles.

B. Point mutation (Fig. 16.8) 1. DNA polymerase mistakenly inserts the wrong base.

2. Proofreading and mismatch repair fail to correct the error.

3. A change is now present in one of the bases in the DNA. 4. This change can cause a change in an amino acid in the protein and thus a change in the phenotype. (Fig. 16.4) .

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a. This change is called a missense mutation. b. A silent mutation does not change the amino acid sequence of the protein. c. A frameshift mutation results when there is a change in the reading frame from an addition or deletion of a single base pair within a protein coding region. d. A nonsense mutation occurs when a codon that specifies an amino acid is changed by a mutation to one that specifies a stop codon.

5. Impact of point mutations (Table 16.1) a. A beneficial mutation may increase fitness, that is, the ability to survive in certain environments. b. A neutral mutation has no effect on fitness. Example: silent mutations c. A deleterious mutation lowers fitness because most organisms are well adapted to their current habitat. d. Students should be able to predict the kind of mutation—missense, silent, nonsense, or frameshift—that would be created by changing the sequence 5'-ATA-3', which specifies a codon in the DNA nontemplate strand to ATC, TTA, or ACA, respectively.

C. Chromosome mutations 1. Larger-scale mutations, like polyploidy and aneuploidy, are often caused by mistakes in mitosis and meiosis.

2. Chromosome inversion: Chromosome segments can break and become detached and flipped before rejoining, which changes the sequence of genes.

3. Chromosome translocation: Detached segments can be rejoined to a different chromosome.

4. Chromosome-level mutations can also be beneficial, neutral, or deleterious. a. They are usually deleterious; for example, they are often found in cancer cells.

(Fig. 16.9) D. The importance of mutations 1. In individuals, mutations can lead to disease or death or increased fitness.

2. In populations, mutations provide the genetic variation that makes evolution possible.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. adenine amino acid aneuploidy arg mutant arginine beneficial mutation central dogma citrulline codon 152

conservative cytosine deleterious mutation deletion duplication fitness gene expression genetic code genetic screen

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guanine histidine inversion knock-out mutant loss-of-function mutants lysine messenger RNA (mRNA) metabolic pathway methionine .

minimal medium missense mutation mutation Neurospora crassa neutral mutation non-overlapping null mutants one-gene, one-enzyme hypothesis ornithine

phenylalanine point mutation polyploidy reading frame redundant reverse transcriptase RNA polymerase silent mutation single-nucleotide addition

start codon stop codon transcription translation translocation triplet code unambiguous universal uracil

or deletion

Lecture Activities The History of Discovery Estimated duration of activity: 10–15 minutes Few students have much background in the history of biology. This chapter describes the history as well as the basic aspects of the processes by which genes work. This is an opportunity for students to discover how fast the picture of science changes—to understand that science is not a collection of facts, but many works in progress.

1. Group students into pairs and ask them to list a series of steps that occurred as research and technological advances took scientists from having no idea how heredity works up to the completed Human Genome Project, and then on to a time in the future when each gene sequence is understood and linked to some function or functions. Have them use the chapter information, but ask them also to include some steps that might logically have occurred. To help them appreciate the scientific inquiry that drives discovery, ask them to come up with a question that may have preceded each discovery. (Example: What is the molecule of heredity—DNA or protein?) Suggest that they list technical developments that made some of the steps possible.

2. After students have listed some possible steps and can visualize how new information and new techniques served to change ideas, talk as a whole group about why science will always be more than merely a set of facts.

3. Ask students to briefly describe how skepticism and questioning have helped the advancement of scientific understanding.

The Genetic Code Estimated duration of activity: 5 minutes or less While introducing the genetic code, encourage students to make their own observations about the genetic code. This may help them see for themselves that the genetic code is redundant and degenerate without being ambiguous.

When you show the codon chart, ask students to look at the chart and identify any patterns they see. It should not take them long to notice that the sets of codons encoding for a given amino acid are all identical in the first two nucleotides but different in the third. You can briefly introduce the concept of tRNA, and explain that there are more than 20 but fewer than 64 tRNAs in a cell because there is wobble in the third position of the anticodon.

Have students look at the codon chart and hypothesize how many different tRNAs are in a cell. This is also a nice opportunity to introduce the effects of mutation on the protein product. As the students are looking at the chart, ask them what would happen if the codon .

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UUU were changed to UUC (nothing; it’s a silent mutation). Ask them what would happen if the codon UUU were changed to CUU (Phe-Leu; missense mutation). You can also use examples in which the AUG start codon is mutated, resulting in no translation; or you could introduce the concept of a nonsense mutation by illustrating what would happen if a stop codon were mutated.

Even though the specifics of mutation and translation are not covered in detail until Chapter 17, introducing these concepts as you introduce the genetic code can prepare the students for what is to come in the next chapter.

Discussion Idea: Mutational Analysis You can use this activity to generate discussion in lecture and to help students understand the fundamental technique of using mutational analysis to identify genes and their functions.

An important research tool in molecular biology is to identify and determine the function of a gene by blocking its activity. Typically, researchers do this by creating mutants that lack the phenotype of interest or have an altered version of it. Then they use the mutants to map the location of the gene and determine the function of the gene product. The following scenario will help students relate a real-life problem to the way scientists search for information about genes and their function.

Scenario: You get into your car after class, and much to your dismay, it won’t start. Before calling a tow truck, you decide to call a couple of friends to help you figure out the problem.

As a class, come up with possible scenarios about what could be wrong with the car and ways that you can test them.

An important point to make with students is that researchers can’t always identify a defect exactly. For example, if spark plugs are removed from a car engine, the car won’t start. But if you didn’t see the spark plugs being removed, it’s not immediately obvious to you why the car won’t start. Careful exploration is needed to determine that the spark plugs are missing.

Once you know the plugs are missing, you can make inferences about the role of spark plugs in the normal functioning of an engine. You can then do additional testing to get more specific information on the exact role of the spark plugs.

A scientific researcher, in effect, does not “see the spark plugs being removed.” When mutations are induced in a broad way, such as by UV light or X-ray exposure, the actual site of the defect is not known. Instead, the mutant is identified solely because its phenotype for the trait of interest has been altered. This is the starting point for determining how the phenotype has been altered and for mapping the location of the mutated gene on a chromosome. Sometimes, a great deal of careful exploration is required to determine exactly how a particular gene contributes to the observed phenotype. Another complication is that the researcher must determine whether the selected mutant has only one gene defect or multiple gene defects, which would greatly complicate the interpretation of the mutant phenotype.

You can also tie this discussion to the use of model organisms in biology. A great deal of genetic information is available for the model organisms used in the study of genetics, such as Drosophila or E. coli. When a researcher introduces a mutation into a model organism, it is much easier to map the location of the mutant locus because so many other loci are already known, and the segregation of the mutant phenotype can be compared with the segregation of these other known “marker” genes. Compare this to trying to map a locus in the genome of an organism about which nothing is known. It is much easier, once a locus is identified, to look later for the same locus in an organism of interest that is not a model organism for research.

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CHAPTER

17

Transcription, RNA Processing, and Translation

Learning Objectives: Students should be able to...

• Describe the structure and function of elements needed for transciption, including the promoter, RNA polymerase core enzyme and holoenzyme, sigma factor, and template and nontemplate strands of DNA. • List major differences between transcription and RNA processing in bacteria and eukaryotes.

• Discuss major components and events in RNA processing, including introns and exons, •

splicing, the spliceosome, the 5' cap, and poly(A) tail.

Describe tRNA structure and function, including how amino acids are attached to tRNAs

• and how codons are read.

Discuss mechanisms for translation initiation, elongation, and termination.

Lecture Outline Proteins are the stuff of life.

The role of proteins • Proteins control most cell processes like chemical reactions and give cells their structure and shape.

• Cell function is dependent on protein function.

• If one or more proteins do not work correctly, the cell may become abnormal and die.

The central dogma • Proteins are synthesized in a two-step process: transcription of genes into messenger RNAs and translation of messenger mRNAs into proteins.

I. An Overview of Transcription A. Characteristics of RNA polymerase 1. This enzyme is responsible for synthesizing RNA from DNA. 2. The enzyme uses only one of the DNA strands as a template. a. This is called the template strand. b. The other strand is called the nontemplate or coding strand. c. The coding strand matches the mRNA sequence, except that it has thymine where the mRNA has uracil.

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3. RNA polymerase reads the DNA template and synthesizes a complementary RNA strand. (Fig. 17.1) 4. It builds RNA in the 5' → 3' direction by reading the DNA template.

5. It does not require a primer to begin RNA synthesis.

6. Bacteria have one RNA polymerase; eukaryotes have three distinct types: pol I, pol II, and pol III. B. Initiation: How does transcription begin in bacteria?

1. In bacteria, a protein, called sigma factor, binds to RNA polymerase before transcription begins. a. RNA polymerase + sigma = holoenzyme. b. RNA polymerase is the core enzyme. (Fig. 17.2a) 2. Bacterial promoters a. Sigma binds tightly to specific regions of DNA called promoters. b. David Pribnow analyzed the base sequence of promoters.

(1) Promoters are 40 to 50 base pairs long.

(2) They have a series of bases similar or identical to TATAAT, known as the –10 box.

(3) The –10 box is 10 base pairs upstream from where transcription will start (the +1 site).

(4) Another conserved sequence, TTGACA, occurs in the same promoters about 35 bases upstream of the start site; it is called the –35 box. c. Researchers found that sigma is required to facilitate RNA polymerase binding to DNA. d. The holoenzyme binds to DNA at specific locations called promoters. e. Promoters are sites on the DNA template where transcription begins. f. Sigma appeared to be responsible for guiding RNA polymerase to promoters.

3. Events inside the holoenzyme a. Transcription begins when sigma binds to the –35 and the –10 boxes. (Fig. 17.3) b. After sigma binds to a promoter for a bacterial gene, the DNA double helix opens. c. The template strand is threaded through a channel in RNA polymerase that leads to the active site. (Fig. 17.3) d. Ribonucleoside triphosphates (NTPs) enter another channel in RNA polymerase, move to the active site, and are incorporated into the mRNA when they are complementary to the template strand. e. Sigma is released once RNA synthesis is under way.

C. Elongation and termination 1. Elongation a. RNA polymerase moves in the 3' → 5' direction along the template DNA strand at a rate of 50 nucleotides per second. b. In the interior of the enzyme, a group of amino acids called the rudder helps steer the template and nontemplate strands through channels inside the enzyme.

(Fig. 17.3, step 3) c. It inserts a uracil whenever it encounters adenine in the template DNA.

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d. A group of amino acids called the enzyme’s zipper helps separate the new strand from the template strand. e. An enzyme’s structure is correlated with its function.

(1) Double-stranded DNA goes into and out of one groove.

(2) NTPs enter another channel.

(3) The growing RNA structure exits through the rear.

2. Termination a. Specific sequences in template DNA act as termination sites.

(1) In bacteria, the bases in the transcriptional termination sequence form complementary base pairs.

(2) This results in a hairpin structure (stem and loop), which causes the RNA strand to separate from the RNA polymerase, terminating transcription. (Fig. 17.4) b. RNA polymerase and the mRNA strand are released from the DNA template. c. Specific termination proteins facilitate the release in some bacteria.

D. Transcription in eukaryotes 1. Fundamental features of transcription are the same in bacteria and eukaryotes.

(Table 17.1) 2. Eukaryotic differences a. Eukaryotes have three polymerases: pol I, pol II, pol III. Pol II is the only polymerase that transcribes protein-coding genes.

(1) RNA polymerase I transcribes genes that code for ribosomal RNAs. (2) RNA polymerase II transcribes genes that code for proteins; thus, it synthesizes mRNAs. (3) RNA polymerase III transcribes genes that code for tRNAs and other small RNAs. b. Promoters are more diverse. Most eukaryotic promoters include a unique sequence called a TATA box. c. Eukaryotic basal transcription factors (1) Analogous to sigma proteins in prokaryotes (2) Involve many proteins (3) Not part of a holoenzyme (4) Bind to the appropriate region of DNA (5) Followed by RNA polymerase binding d. Termination involves a poly(A) signal.

II. RNA Processing in Eukaryotes: The initial product after transcription is a primary transcript that must undergo processing before becoming functional.

A. The startling discovery of split eukaryotic genes 1. In prokaryotes, when transcription of mRNA is complete, the mRNA is ready to be converted into a protein.

2. In eukaryotes, pre-mRNA is produced by transcription, and this must be processed before translation can occur.

3. Eukaryotic genes have intervening sequences called introns. a. The protein-coding region of eukaryotic genes is interrupted by stretches of noncoding DNA. .

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b. Noncoding sequences must be disposed of to make a functional mRNA. c. Eukaryotic gene organization is very different from gene organization in prokaryotes.

4. P. Sharp and colleagues detected noncoding regions in genes. a. They mixed together mRNA and DNA. b. Then they incubated the mixture under conditions that promote hybridization of the mRNA to its DNA template. c. Following hybridization, they used electron microscopy to observe the molecules in the solution. d. Result: All of the mRNA was paired to DNA, but some of the DNA looped out and was unpaired to the mRNA. (Fig. 17.5a) e. Conclusion: The template DNA had stretches of nucleotides that were not present in the mRNA, so they could not hybridize and consequently looped out away from the mRNA. A 1:1 ratio of nucleotides did not exist between the DNA and the mRNA. (Fig. 17.5b) 5. W. Gilbert called the expressed portions of genes (translated regions) exons and the intervening portions introns.

B. RNA splicing 1. Eukaryotic genes are first transcribed into a primary transcript that contains both introns and exons. (Fig. 17.6a) 2. Intron splicing (removing the extra sequences) occurs inside the nucleus.

3. Intron splicing is catalyzed by a complex of proteins and small RNAs (snRNPs, that is, small nuclear ribonucleoprotein particles, or “snurps”). (Fig. 17.6b, step 1) a. snRNPs assemble on the primary mRNA transcript, forming a spliceosome.

(Fig. 17.6b, step 2) b. The intron forms a loop that breaks when a specific adenine nucleotide in the intron RNA attacks the 5' end of the intron. c. The spliceosome mediates breakage of the intron at its 5' end, and a loop forms.

(Fig. 17.6b, step 3) d. The loop is excised, and a phosphodiester bond links the exons on either side.

(Fig. 17.6b, step 4) e. The excised intron is usually degraded to ribonucleotide monophosphates.

C. Adding caps and tails to transcripts 1. A cap of 7-methylguanylate-P-P-P is attached to the 5' end of each mRNA. (Fig. 17.7) a. It serves as a recognition signal for the translational machinery. b. It protects the transcript from degradation.

2. A tail, consisting of 100 to 250 adenines (“poly(A) tail”), is added to the 3' end of mRNA. 3. The coding sequence is flanked at the 5' and 3' ends by sequences that are not destined for translation, untranslated sequences (UTRs).

4. Caps and tails protect mRNAs from degradation by ribonucleases and increase the efficiency of translation.

5. Similarities and differences between eukaryotic and prokaryotic transcription are summarized in Table 17.1.

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III. An Introduction to Translation A. Ribosomes are the site of protein synthesis.

1. Translation is the conversion of a sequence of nucleotides in an mRNA into a sequence of amino acids in a protein.

2. Cells that synthesize lots of protein have lots of ribosomes, and vice versa.

3. Studies showed that the basic mechanism of translation was the same throughout the tree of life.

4. R. Britten and colleagues tested ribosomes as the site of protein synthesis. a. They conducted a pulse-chase experiment with 35S to label proteins and track their locations at various times during and after translation. b. Results: (1) Immediately after giving the pulse, the researchers found radioactivity in ribosomes.

(2) Two minutes later, they found radioactivity only in free proteins. c. Conclusion: Protein synthesis occurs at the ribosome, and then the protein is released from the ribosome.

B. Translation in bacteria and eukaryotes 1. Electron microscopy of Escherichia coli DNA being transcribed shows that: a. Ribosomes have attached to mRNA even before transcription is complete, while the mRNA is still linked to DNA. b. Ribosomes translate mRNA as it is being synthesized by RNA polymerase (transcription and translation occur simultaneously in bacteria). (Fig. 17.8a, b) c. Multiple ribosomes attach to each mRNA, forming a polyribosome. d. Transcription and translation are physically connected because there is no nuclear membrane.

2. In eukaryotes: a. mRNAs are released from DNA in the nucleus. b. mRNAs are then processed before leaving the nucleus. c. Then they become associated with ribosomes in the cytoplasm (transcription and translation are separate processes in eukaryotes). (Fig. 17.9) 3. Drawing a process is a good way to see how the molecules are related. (Making Models 17.1) C. How does an mRNA triplet specify an amino acid?

1. Hereditary instructions are contained in sequences of nucleotides, which are then converted into a sequence of amino acids.

2. Early hypothesis: Nucleotides of an mRNA codon chemically combine with amino acid side chains through shape or charge interactions. (Fig. 17.10a) 3. Crick hypothesis: Adapter molecules hold amino acids in place while interacting with an mRNA codon (predicted the existence of transfer RNA molecules). (Fig. 17.10b)

IV. The Structure and Function of Transfer RNA A. Requirements for translation 1. Ribosomes provide the machinery.

2. mRNAs contribute the message. .

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3. Amino acids are the building blocks.

4. ATP and GTP provide energy.

5. tRNA is important.

6. Evidence shows that amino acids are transferred from tRNAs to proteins. (Fig. 17.10) B. What do tRNAs look like?

1. Sequencing revealed that tRNAs are 75−85 nucleotides long.

2. When a tRNA is linked with an amino acid, it is known as an aminoacyl tRNA. 3. Secondary structure was inferred from the nucleotide sequence. a. Hydrogen bonding occurs between complementary bases in different parts of the tRNA molecule.

(1) Short, double-stranded regions form.

(2) The tRNA assumes a cloverleaf shape of stems and loops (stems are double-stranded; loops are single-stranded regions). (Fig.17.11a) b. All tRNAs have the sequence CCA at their 3' end; it is a binding site for an amino acid. c. One of the loops has a triplet of bases that varies in each tRNA; it is the anticodon (i.e., the nucleotides that form base pairs with an mRNA codon).

4. X-ray crystallography shows tertiary structure—the 3-D arrangement of atoms. a. The tRNA cloverleaf folds over, producing an L-shaped structure. (Fig. 17.11b) b. Each tRNA has a distinct anticodon and amino acid. c. The anticodon and amino acid attachment sites are separated by a precise distance. d. Students should be able to (1) describe where on the L-shaped structure the amino acid attaches and (2) explain the relationship between the anticodon of a tRNA and a codon in an mRNA. C. How are amino acids attached to tRNAs? 1. A tRNA that becomes covalently linked to an amino acid is called an aminoacyl tRNA. 2. The addition of amino acids to tRNAs (charging a tRNA) is catalyzed by aminoacyl tRNA synthetases. (Fig. 17.12) 3. Each of the 20 amino acids has a different aminoacyl tRNA synthetase and one or more tRNAs. 4. Amino acids are transferred from aminoacyl tRNAs to proteins synthesized in ribosomes.

D. How many tRNAs are there?

1. There are 61 different mRNA codons but only about 40 tRNAs. 2. How can an mRNA codon, for which there is no tRNA, be translated?

3. Crick’s wobble hypothesis: a. Many amino acids are specified by more than one codon. b. Codons for the same amino acid tend to differ only in the third position.

(1) Example: Codons 5'-CAA-3' and 5'-CAG-3' code for glutamine.

(2) A tRNA with the anticodon 3'-GUU-5' can pair with either 5'-CAA-3' or 5'-CAG-3'.

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(3) The third codon position is the wobble position; it often contains an inosine base that can pair with A, U, or C.

V. The Structure of Ribosomes and Their Function in Translation A. Characteristics of protein synthesis 1. The sequence of bases in an RNA message is translated into a sequence of amino acids in a polypeptide.

2. The conversion of each mRNA codon begins when the anticodon of an aminoacyl tRNA binds to the codon.

3. The conversion is complete when a peptide bond forms between the tRNA’s amino acid and the growing polypeptide.

4. Conversion occurs inside a ribosome, which is composed of two subunits. (Fig.17.13) a. There is a larger subunit and a smaller subunit. b. Both subunits are composed of multiple RNA molecules and numerous proteins. c. The small subunit holds the mRNA in place during translation. d. The large subunit is where peptide bond formation takes place. e. The large ribosomal subunit has three tRNA binding sites. (Fig. 17.13) (1) During translation, the acceptor or aminoacyl (A) site holds the tRNA with an amino acid that is about to be added to the polypeptide.

(2) The peptidyl (P) site holds the tRNA that is holding the growing polypeptide chain.

(3) The exit (E) site holds the empty tRNA that is about to be released.

5. Three-step sequence of protein synthesis: (Fig. 17.14) a. Aminoacyl tRNA diffuses into the A site; its anticodon binds to an mRNA codon. b. A peptide bond forms between the amino acid on the aminoacyl tRNA in the A site and the growing polypeptide (held by a tRNA) in the P site. c. The ribosome moves ahead, and all the tRNAs move one position down the mRNA. The tRNA in the E site exits, the tRNA in the P site moves to the E site, and the tRNA in the A site moves to the P site.

6. The polypeptide has grown by one amino acid.

7. The process occurs up to 20 times per second in bacteria and about 2 times per second in eukaryotes.

8. Protein synthesis begins at the amino end (N-terminus) of the protein and proceeds to the carboxy end (C-terminus).

B. Initiating translation 1. In bacteria, the small ribosomal subunit binds to a sequence on mRNA about six nucleotides upstream from an AUG start codon.

2. The sequence to which the small subunit binds is the Shine-Dalgarno sequence and contains all or part of 5'-AGGAGGU-3'. (Fig. 17.14, step 1) 3. The initial interaction is mediated by initiation factors. In eukaryotes, initiation factors bind to the 5' cap on mRNA. 4. An aminoacyl tRNA with N-formylmethionine (f-met) binds to the AUG codon.

(Fig. 17.14, step 2) In eukaryotes, the first amino acid is normal methionine.

5. Initiation is complete when the large ribosomal subunit binds and the tRNA-bearing f-met occupies the P site. (Fig. 17.14, step 3) .

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C. Elongation: extending the polypeptide (Fig. 17.15) 1. After the ribosome subunits join, a new aminoacyl tRNA binds to the mRNA codon in the A site of the ribosome (aminoacyl site). This reaction requires GTP. 2. A peptide bond forms between the carboxyl end of the N-formylmethionine amino acid in the P site and the amino end of the amino acid in the A site.

3. Is the ribosome an enzyme or a ribozyme? a. The active site of the ribosome is composed of rRNA, so the ribosome is a ribozyme. b. There is support for the RNA world hypothesis: Life began with RNA molecules.

4. Moving down the mRNA (Fig. 17.15, step 3) a. The tRNA in the P site moves to the E site (exit site). b. The new peptidyl tRNA moves from the A site to the P site. c. The empty tRNA is ejected from the E site. d. Translocation requires energy in the form of GTP and assistance from elongation factors. e. Translocation of the mRNA occurs, moving it through the ribosome in the 5' → 3' direction.

5. The three-part cycle of elongation (arrival of aminoacyl tRNA, peptide bond formation, translocation) repeats over and over.

6. Once elongation is under way, additional ribosomes can bind to the translational start site and begin translating.

D. Terminating translation (Fig. 17.16) 1. Three stop codons are included in the genetic code: UAA, UAG, and UGA. 2. The ribosome reaches any one of these codons on an mRNA. a. The release factor proteins enter the A site. b. The completed polypeptide is released from the tRNA in the P site. c. The ribosome separates from the mRNA. d. The two ribosomal subunits dissociate from each other, allowing subunits to attach to the start codon of another message so that translation can begin again.

E. Post-translational modifications 1. Folding a. Protein shape is critical to function. b. How does folding occur to produce the 3-D structure of an active enzyme? c. Thermodynamic hypothesis: Proteins fold into their most stable energetic state. d. Molecular chaperones facilitate protein folding.

2. Chemical modifications a. Many proteins go through the ER and Golgi apparatus and have sugar and/or lipid groups added. b. Many proteins have phosphate groups added or removed during their lifetimes. The addition or removal of this large negatively charged group changes a protein’s structure and thus changes its function.

3. The major steps of gene expression in eukaryotic cells is a multistep process.

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Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

–10 box 3' poly(A) tail –35 box 5' cap (7-methylguanylate-P-P-P) aminoacyl tRNA aminoacyl tRNA synthetase aminoacyl/A site anticodon AUG codon basal transcription factors CCA sequence coding strand core enzyme downstream elongation factors elongation phase of transcription exit/E site exons GTP holoenzyme initiation factors initiation phase of transcription inosine

introns large (50S) subunit messenger RNA (mRNA) molecular chaperones N-formylmethionine nontemplate strand nucleoside triphosphates (NTPs) peptide bond formation peptidyl/P site phosphorylation polyribosomes post-translational modification pre-mRNA primary RNA transcript promoters protein folding release factors ribosomal RNA (rRNA) ribosome ribosome binding site ribozyme RNA polymerase RNA polymerase I RNA polymerase II RNA polymerase III RNA processing RNA splicing

Shine-Dalgarno sequence/ ribosomal binding site sigma protein small (30S) subunit snRNPs spliceosomes stop codon TATA box TATAAT template strand termination phase of transcription transcription transcription termination sequence transfer RNA (tRNA) translation translation elongation translation elongation factors translation initiation translation initiation factors translation termination translocation TTGACA upstream wobble hypothesis X-ray crystallography

Lecture Activities What Makes a Firefly Glow?

Estimated duration of activity: 5–10 minutes The website http://learn.genetics.utah.edu/content/begin/dna/firefly has a really fun and interesting review of transcription and translation in the context of the question, What makes a firefly glow? If your classroom is computer equipped, this would be a nice website to go through with your students as a review.

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An In-Class Demonstration: Translation Estimated duration of activity: 15 minutes This activity is especially suited to large lecture classes, particularly if there is enough space at the front of the room to assemble sizable groups of students. (You can scale down the activity presented here for smaller classes or smaller rooms.)

In this exercise, students are turned into ribosomes, messenger RNAs, protein factors, ribozymes, aminoacyl tRNAs, and aminoacyl synthetases. You can assemble some relatively simple, inexpensive identification labels in advance (suggestions follow), and you can get a large number of students in the class directly involved in acting out the process of translation.

This activity may seem elementary to you, but students will welcome the break from lecture and appreciate the chance to visualize a complex biological process. If you don’t have time to put the activity together, assign it to a teaching assistant and have the TA direct the activity during class.

Procedure: 1. Have 24 students come to the front of the room to form an mRNA molecule. The 24 students should form a long, straight line facing the rest of the class. Start with the first student on the far left side (stage left) and proceed to the right. a. Give student 1 a card to hold up that says 5'. b. Leave student 2 unlabeled (this person is part of the mRNA 5' leader sequence). c. Give students 3, 4, 5, and 6 each a label to wear marked SD (Shine-Dalgarno sequence). d. Leave students 7 and 8 unlabeled (they are the rest of the 5' leader sequence). e. Label students 9, 10, and 11 as A, U, and G (respectively) to mark them as the AUG codon, which specifies methionine. f. Label students 12, 13, and 14 as C, C, and A (respectively) to mark them as the CCA codon, which specifies proline. g. Label students 15, 16, and 17 as G, C, and A (respectively) to mark them as the GCA codon, which specifies alanine. h. Label students 18, 19, and 20 as U, G, and A (respectively) to mark them as the UGA stop codon. i. Leave students 21, 22, and 23 unlabeled (they are the 3' trailer sequence of the mRNA). j. Give student 24 a card to hold up that says 3'.

2. Next, have nine students come to the front of the room to form the small subunit of the ribosome. These students should form a group in front of the 5' end of the mRNA molecule (between the mRNA and the audience). a. Three of the nine students will be the E site (label each with an E). Have them stand at the extreme 5' end. b. Three students will be the P site (label each with a P). Have them stand just to the right of the E site. c. Three students will be the A site (label each with an A). Have them stand just to the right of the P site.

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a. They can be assisted by a couple of students labeled IF (initiation factor). b. The small subunit students should pause at the Shine-Dalgarno sequence and then spread out across the front of the mRNA so that the three small students labeled P are positioned in front of the three mRNA students labeled AUG. 4. Off to one side of the mRNA, some students may be labeled as various transfer RNAs and others as aminoacyl synthetases. Each of them will hold a card specifying a single amino acid. a. The aminoacyl synthetase (labeled Met aa-synth) should give a card that says MET to the person labeled tRNA-met. b. The tRNA-met student, who is now holding the MET card, should go behind the mRNA molecule, directly behind the students labeled AUG. c. The tRNA-met student may be assisted in positioning by one or more IFs. 5. Now assemble a group of 11 students behind the mRNA molecule and the tRNA-met student. a. Student 1 closest to the 5' end of the mRNA should be unlabeled. b. The next three students will be the E site. c. The next three students will be the P site. d. The next student will be labeled as the 23S ribozyme. e. The next three students will be the A site. f. One student at the 3' end will be unlabeled. g. Have the students link arms in order and then line up so that the large E site students are directly aligned with the small E site students, and so forth. At this point, the ribosome is completely assembled on the mRNA, and the tRNA-met student is positioned between the small and large P-site students.

6. Next, a person labeled tRNA-pro should pick up a card labeled PRO from the appropriate aa-synth and go to the A site of the mRNA molecule.

7. The individual on the large ribosome who is labeled 23S ribozyme then forms a peptide bond between the card labeled MET and the card labeled PRO (staple the cards together or have the ribozyme string them together through previously punched holes in the cards).

8. After the peptide bond forms, the cards are handed to the person labeled tRNA-pro.

9. Translocation: All of the mRNA students link arms and each takes three steps to the left (i.e., in the direction of the 5' end). a. The tRNA students should stay with the AUG and CCA sites on the mRNA. b. Students labeled EF (elongation factor) can assist the mRNA movement. c. When translocation is complete, the tRNA-met student should be in the E site of the ribosome, and the tRNA-pro student (holding the MET-PRO peptide) should be located in the P site.

10. The tRNA-met student now leaves the E site.

11. A student labeled tRNA-ala now picks up the ALA card from the appropriate synthetase, enters the A site, and binds to the GCA codon, where the ribozyme can form a peptide bond.

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12. Repeat steps 7–10, until the UGA stop codon is positioned at the A site. a. Students labeled RF (release factor) enter the A site, instead of a tRNA. b. Release the completed peptide (of three amino acids) from the tRNA at the A site. c. Last, the large ribosome subunit students all link arms and move as a group away from the mRNA, as do the small students, to complete one round of translation.

Suggestions for labels: • Make the labels large (use 8.5- x 11-inch paper). Design them to be attached to the person’s chest using safety pins, tape, or some other rapid, easy method.

• For the cards that represent amino acids, use paper or stiffer cardboard, keeping in mind that they’ll need to be quickly and easily linked together.

• Use different colors of papers to designate different items. All the labels for students who will be part of the mRNA could be on red paper, the amino acid cards could all be orange, the small ribosome labels could be green, the large blue, and so on.

• All labels and cards may be photocopied onto colored paper from a master sheet, or printed from a computer with colored letters on white paper or black letters on colored paper.

For the procedure as outlined in steps 1–12, you’ll need the following labels and cards: a. Labels designed to go on a person’s chest 18 red labels for the mRNA students: • 4 marked SD • 3—each one marked with either A, U, or G • 3—each one marked with either C, C, or A • 3—each one marked with either G, C, or A • 3—each one marked with either U, G, or A

• 2 unmarked red labels for the leader and trailer students 9 green labels for the small students: • 3 marked E • 3 marked P • 3 marked A 11 blue labels for the large students: • 2 unmarked • 3 marked E • 3 marked P • 3 marked A 3 purple labels for the tRNA students: • 1 marked tRNA-met • 1 marked tRNA-pro • 1 marked tRNA-ala 3 white labels for the aminoacyl synthetase students: • 1 marked MET aa synth 166

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• 1 marked PRO aa synth • 1 marked ALA aa synth Optional: Labels marked IF, EF, and RF (for initiation, elongation, and release factors), if you choose to have students perform these roles. b. Cards to represent amino acids • 1 marked MET • 1 marked PRO • 1 marked ALA c. Cards for either end of the mRNA molecule • 1 marked 5' • 1 marked 3'

Discussion Ideas Because the central focus of this chapter is the link between nucleic acids and proteins, this might be a good time to consider the differences between molecules that are catalysts (proteins and RNA) and templates (DNA and RNA). Prions are unusual molecules because they seem to be able to perform both functions. How are prions different? Where do they fit in? You may want to pose these questions to the whole class, to small groups, as challenge questions for pairs of students, or as thought problems to prepare for a general discussion.

Question 1: What makes DNA and RNA good templates?

Possible answers: • DNA and RNA can act as blueprints for new synthesis from monomers.

• When DNA double strands open and beginning primers are established, synthesis proceeds in a predictable way, making a complementary copy of each existing strand.

• The start complex that includes RNA polymerase binds to the promoter region of a gene in the DNA sequence, and the RNA molecule is synthesized to be complementary to the DNA strand that follows the promoter position.

• DNA and RNA have very predictable structures under ordinary conditions.

Question 2: Why are proteins considered poor templates?

Possible answers: • Proteins cannot act as blueprints for new synthesis from monomers.

• Proteins do not have complementary pieces that can act as a template.

• Proteins have complex, three-dimensional structures that can vary drastically even under physiological conditions.

• Proteins do not have a means to regularly transmit their blueprints through generations.

Question 3: Why are proteins the best catalysts?

Possible answers: • Proteins have the ability to change three-dimensional shapes drastically, often with very small differences in their primary structure.

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• Proteins have many variable “R” groups in their amino acids to bind to other molecules in order for changes to occur.

Question 4: What do prions do?

Possible answers: • Prions are normal proteins in cells. Mutated or misfolded forms exist, and they apparently can cause normal forms to assume the mutated shape without intervention of a nucleic acid like RNA or DNA. (You could quiz students about how they could test this—treating suspensions with DNase or RNase to see whether the mutated prions still change normal ones.)

• Cases have now been documented that somehow these mutated prion forms present in food can affect normal prions in the brains of those species exposed.

• Prions seem to act as infectious agents but have no known nucleic acid.

Question 5: Are these prions like catalysts or templates?

Possible answers: • Prions do not seem to be able to act as templates to form entirely new molecules using only monomers.

• Prions change shape like other proteins that act as catalysts. The prions are misfolded compared to the normal protein.

• Prions or some information related to them survive the digestive system, absorption, and migration through the body to get to the site where they do damage. The protein that has changed shape is partially protease resistant.

• The host does not mount an immune response, which suggests that no virus is associated with the protein.

• Some explanation needs to be found for how the first change in the normal protein took place. The ability to change shape might be possible, at least due to the flexibility of shape for proteins. After all, they are good catalysts.

• Although the differently folded proteins of the infectious particles are called “mutant,” are they? Families with inherited forms of the human disease all have mutations in their normal protein gene.

• In England and France, these infectious agents have apparently crossed the species barrier from cattle and sheep to humans. This makes them even more mysterious. This transmission is different from the inherited type; presumably, some sort of agent is transmitted.

To emphasize the importance of in vitro experiments to biological research, and to focus on the differences between transcription and translation, you can present the following challenge question to the class. You can also make a connection to DNA replication, which was discussed in Chapter 15, by asking the question given in part C. Question 6: How is a cell-free transcription system different from a cell-free translation system? a. What purified components would you need to combine to achieve transcription in vitro?

Answer: Template DNA that contains one or more genes (with their promoters) Ribonucleotides—ATP, GTP, CTP, and UTP RNA polymerase enzyme

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If the gene is prokaryotic—a cell fraction that contains sigma factors If the gene is eukaryotic—a nuclear extract that contains transcription factors b. What purified components would you need to combine to achieve translation in vitro?

Answer: Ribosomes Messenger RNA Amino acids (all 20) GTP and ATP to provide energy A cell fraction that contains aminoacyl tRNAs, aminoacyl synthetases, initiation factors, elongation factors, and termination factors c. How is an in vitro DNA replication system different from transcription and translation systems? What components would be needed to achieve replication in vitro?

Answer: Template DNA Deoxyribonucleotides—dATP, dGTP, dCTP, and dTTP DNA polymerase A protein fraction from bacterial cells that contains helicases, topoisomerases, single-stranded binding proteins, and primase

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CHAPTER

18

Control of Gene Expression in Bacteria

Learning Objectives: Using the lac operon in bacteria as a model, students should be able to…

• Explain why cells need to regulate expression of their genes.

• Distinguish between transcriptional, translational, and post-translational control of gene expression.

• Distinguish positive and negative control of transcription.

• Describe the structure and function of elements of the E. coli lac operon.

• Describe global gene regulation and discuss why it’s important.

Lecture Outline I. Overview of Gene Regulation and Information Flow A. Regulation of gene expression in bacteria is critical.

1. Bacteria occupy a variety of habitats and often survive by switching among several different food sources.

2. Each type of nutrient taken up by a cell requires specific membrane transport proteins and enzymes to metabolize it.

3. Case study: intestinal bacterium Escherichia coli (E. coli) a. The bacteria utilize a wide variety of nutrients for energy and carbon. b. The available food sources vary with the diet of the host. c. E. coli competes for resources with numerous other species in the intestine. d. Survival requires the efficient utilization of available resources. e. The expression of all genes in the genome is energetically unfavorable, but expression of needed genes is vital to survival.

B. Mechanisms of regulation (Fig. 18.1) 1. Transcriptional control a. If a gene is not transcribed, no mRNA is made. b. Transcriptional control is the most energy-efficient control mechanism.

2. Translational control a. An mRNA is not translated, or the translation rate is controlled, preventing unneeded proteins from being made.

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b. Translational control is a rapid method of control that allows cells to quickly change which proteins are being produced.

3. Post-translational control a. Many proteins are manufactured in an inactive form. b. The protein is activated by chemical modification such as phosphorylation. c. Post-translational regulation is the least efficient but fastest control mechanism: Proteins are already made and can rapidly be activated.

4. Each type of control represents a trade-off between efficiency and resource use. a. The trade-off is between the speed of response and the conservation of ATP. b. Transcriptional control is slow but doesn’t use a lot of energy. c. Post-translational control is fast but uses a lot of energy.

5. All levels of control of gene expression occur in bacteria.

6. Levels of gene expression may be constant or highly variable. a. Some genes, like those involved in glycolysis, are expressed constitutively (all of the time). b. Other genes are regulated, either induced or repressed.

7. Variation in gene expression allows organisms to respond to their environment.

C. Metabolizing lactose⎯a model system (BioSkills 11) 1. E. coli can metabolize a wide variety of sugars such as glucose, lactose, and others; however, glucose is preferred.

2. The sugar must be transported into the cell with the assistance of a membrane protein.

3. The sugar is broken down in the cytoplasm by a specific set of enzymes.

4. Other enzymes are involved in using the raw material of the sugar to drive the synthesis of amino acids, vitamins, and other compounds.

5. If glucose supplies are depleted, E. coli can use other sugars such as lactose.

(Fig. 18.2) a. Lactose is a disaccharide composed of glucose and galactose. b. If lactose is available in the medium, it is transported into the cell. c. Only then will the cell produce the enzyme β-galactosidase, which breaks down lactose into glucose and galactose. d. This suggested that lactose may act as an inducer of the β-galactosidase gene.

6. Monod and Jacob investigated whether lactose utilization by the cell depends on other sugars that might be present. (Fig. 18.3) a. Cells grown on glucose only: no β-galactosidase production b. Cells grown on glucose + lactose: no β-galactosidase production c. Cells grown on lactose only: β-galactosidase production d. Conclusion: The presence of lactose without glucose stimulates the expression of the β-galactosidase gene. e. Students should be able to make a chart summarizing molecules involved in regulating lactose use in E. coli. The chart should have 2 columns, entitled “Name” and “Function,” and 7 rows: lacZ, lacY, operator, promoter, repressor, lactose, and glucose. As students read the chapter, they will fill in the “Function” column.

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II. Identifying Regulated Genes A. Replica plating to find lactose metabolism mutants 1. Researchers must generate large numbers of individuals with mutations at random sites in their genomes.

2. The mutants must be screened to find individuals with defects in the biochemical pathway of interest.

3. Jacob and Monod screened for E. coli mutants incapable of lactose utilization by using replica plating. a. Cells are plated on a “master plate” containing a solid growth medium. b. Bacterial cells are allowed to divide on the surface to produce a colony. c. Each colony was transferred to a set of other plates and observed growing under different conditions.

B. Three types of lactose metabolism mutants were yielded. (Table 18.1) 1. Some mutants cannot cleave lactose into glucose and galactose due to a lack of functional β-galactosidase. (Fig. 18.4) a. The genetic locus that codes for β-galactosidase was named lacZ. b. Cells deficient in β-galactosidase enzyme are lacZ –. c. Cells with normal β-galactosidase are lacZ+.

2. Other mutants cannot accumulate lactose in their cytoplasm due to lack of functional galactoside permease, encoded by the lacY gene.

3. A third category of mutants were constitutive for ability to use lactose, expressing lacZ and lacY regardless of whether lactose was present. a. The locus associated with constitutive lactose utilization was named lacI. (1) lacI – cells are mutants that do not need an inducer to activate lacZ and lacY. (2) lacI+ cells are wild type and require an inducer (lactose—specifically, allolactose) to activate lacZ and lacY. 4. Jacob and Monod’s discovery identified three genes involved in lactose metabolism. a. lacZ and lacY encode proteins required for metabolism and import of lactose. b. lacI encodes a regulatory function.

(1) When lactose is absent, the lacI gene or its product shuts down the expression of lacZ and lacY genes.

(2) When lactose is present, transcription of lacZ and lacY genes is induced.

5. Students should be able to describe the functions of lacZ and lacY. They should also be able to describe the effect of the lacI gene or gene product when lactose is present versus absent, and explain why these effects are logical.

6. All of these loci mapped to locations very close to one another on the bacteria’s circular chromosome, suggesting they might be transcribed together. (Fig. 18.5)

III. Negative Control of Transcription A. Transcription can be regulated in two ways.

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3. The genes involved in lactose degradation are under both positive and negative control.

4. Drawing figures like Figure 18.5 can help in understanding how transcription is controlled. (Making Models 18.1) B. A repressor encoded by lacI negatively controls lacZ and lacY expression.

1. Glucose and the lacI repressor protein prevent expression of lacZ and lacY genes.

2. Lactose (as allolactose) induces transcription of lacZ and lacY. 3. Szilard hypothesis: β-galactosidase synthesis is under negative control. (Fig. 18.6a) a. Transcription of lacZ is normally repressed, unless lactose becomes available. b. The repressor was assumed to be the lacI gene product, which binds the DNA near the promoter for lacZ and lacY. c. Lactose acts as an inducer because it interferes with the repressor. (Fig. 18.6b) d. When lactose interacts with the repressor, the repressor is unable to bind the DNA, and transcription of lacZ proceeds normally. e. Mutants that constitutively express β-galactosidase in the absence of lactose have a defective repressor. (Fig. 18.6c) 4. The idea that lacI encodes a repressor was confirmed when a functioning copy of lacI was added to constitutive mutants and lactose regulation was restored.

C. The operon model 1. Jacob and Monod made three hypotheses, based on their model that the lactose genes form an operon, a set of genes with coordinated expression transcribed as a single mRNA initiated from a single promoter. (Fig. 18.7) a. The lacZ, lacY, and lacA (codes for transacetylase, which regulates the export of sugars) genes are cotranscribed and coregulated. b. The repressor is a protein encoded by lacI. (1) It binds to DNA, preventing the transcription of lacZ, lacY, and lacA. (2) The site on the DNA to which the repressor binds is called the operator.

(3) The repressor is expressed constitutively. c. The inducer (later shown to be a derivative of lactose, allolactose) binds to the repressor, causing it to change shape.

(1) The shape change causes the repressor to fall off the DNA strand (allosteric regulation).

(2) When the repressor falls off, transcription of lacZ, lacY, and lacA proceeds.

2. Years later, experimenters have confirmed the existence of the lac operon and the role of the lac repressor in regulating its expression.

3. Students should be able to predict the effect of a mutation in the lacI gene that alters the repressor so it cannot bind to lactose, and the effect of a mutation in the operator that prevents repressor binding.

D. How does glucose regulate the lac operon?

1. If glucose is present, the lac operon is repressed, even if lactose is available.

2. Abundant glucose means E. coli does not need to catabolize lactose to produce glucose.

3. New evidence shows that the lactose permease (LacY) is inhibited by glucose. a. In the presence of glucose, lactose importation is limited.

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(1) Lactose does not accumulate in the cell.

(2) Allolactose is not formed.

(3) The repressor stays bound to the operator. b. When glucose is absent, galactoside permease is active and the operon is turned on.

4. Inducer Exclusion (Fig.18.8a) a. Glucose inhibition of the galactoside permease is an example of regulation by inducer exclusion. b. Many sugar transporters are regulated by inducer exclusion. c. This allows E. coli to preferentially use glucose.

5. Regulation of CAP (Fig. 18.8b) a. CAP binds to DNA near the promoter and is important in increasing transcription when glucose is limited. b. The new findings about inducer exclusion in the lac operon have altered our understanding of the role of the CAP activator.

E. Why has the lac operon model been so important?

1. Information gathered about the lac operon system in E. coli has been important in understanding the regulation of gene expression in other species.

2. It introduced the idea that gene expression is controlled via physical contact between regulatory molecules and specific regions of DNA. 3. The lac operon also offered an example of post-translational control. a. The repressor is expressed constitutively. b. The activity of the repressor is controlled by post-translational modification.

IV. Positive Control of Transcription

1. In positive control, an activator protein binds DNA and interacts with RNA polymerase to increase transcription. (Fig. 18.6a) 2. The arabinose (ara) operon is an example of positive control. (Fig. 18.9) 3. Students in a laboratory course discovered the mechanism of control for arabinose utilization in E. coli. a. Three genes in the ara operon encode enzymes used to metabolize arabinose. b. A nearby region encodes AraC, a regulatory protein.

4. AraC controls expression of the ara operon. a. AraC is allosterically regulated by arabinose.

(1) When arabinose is bound to AraC, two copies of the protein bind to DNA at the ara initiator region.

(2) When AraC is bound to the DNA, it interacts with RNA polymerase, increasing transcription. b. AraC is both an activator and a repressor. (Fig. 18.10) (1) When arabinose is absent, AraC remains a dimer.

(2) One subunit remains bound to the initiator, while the other binds the ara operator, blocking transcription of both the araC gene and the ara operon.

5. Students should be able to predict the effect of a mutation that removes the part of AraC that binds to RNA polymerase.

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V. Global Gene Regulation

1. Cells respond to their environments by changing gene expression. When large sets of genes are involved, global gene regulation is required.

2. Mechanisms for global gene regulation include sigma factors (Chapter 17) and grouping genes into regulons. a. Regulons are genes and operons that have the same regulatory sequences and are controlled by the same regulatory protein. b. Regulons allow a coordinated response to changing conditions in the cell’s environment.

3. Regulons can be under negative control by a repressor protein or positive control by an activator protein. a. The SOS response regulon is under negative control.

(1) Damage to DNA is the signal for removal of a repressor.

(2) Repressor removal allows expression of genes needed to repair the damage. b. The ToxR regulon of Vibrio cholerae is under positive control. (Fig. 18.11) (1) Outside the human host, genes regulated by ToxR are inactive.

(2) When V. cholerae infects a human, ToxR regulon genes are activated, leading to production of virulence factors.

(3) The resulting diarrhea acts to spread V. cholerae through the environment to new human hosts.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. allolactose allosteric regulation allosteric regulation ara operon AraC cAMP catabolite activator protein (CAP) catabolite repression constitutive constitutive mutants disaccharide galactose gene expression genetic screen

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global gene regulation glucose monosaccharide inducer lac operon lacI lacI + lactose lacY lacY + lacZ lacZ + master plate medium mutagens negative control

negative control operator operon permease positive control positive control post-translational control regulon replica plate replica plating repressor transcriptional control translational control β-galactosidase

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Lecture Activities Activity Idea: Acting Out the lac Operon Estimated duration of activity: 10–15 minutes Students often appreciate the opportunity to take a break from lecture, move around, and use their imagination while learning. Since the regulation of gene expression is a foreign concept to students (and one they have a hard time visualizing), you can use one of the following ideas to get students involved in teaching one another.

The regulation drama: Divide students into three groups. Have them stage a demonstration of the lac operon system using people and items in the classroom. Give each group a scenario: glucose and lactose present, just lactose, a little bit of glucose, and a lot of lactose. lac operon charades: Ask for student volunteers. Assign each student a role with a sign (repressor, lactose, CAP, cAMP, RNA polymerase). Designate the board as the chromosome, and draw the operator, the promoter, and the lacY and lacZ genes. Ask the actors to act out the scenario you whisper to them, and have the class figure out the scenario (glucose and lactose present, just lactose, a little bit of glucose and a lot of lactose).

Discussion Idea: Teaching with Analogies Students have a difficult time relating to and understanding gene regulation because it is so foreign to their life experiences. Therefore, using analogies may help students relate to the information. Because most of your students know how to drive a car, almost everyone can relate to automotive analogies.

Examples: • Repressors are like parking brakes; they inhibit transcription until they are released.

• The binding of inducers (like lactose) is analogous to releasing the parking brake.

However, the car will not move forward with any speed and direction until the gas pedal is pressed.

• The CAP protein is like the gas pedal that initiates movement of the car because it allows RNA polymerase binding to the promoter.

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Control of Gene Expression in Eukaryotes

Learning Objectives: Students should be able to…

• Describe how chromatin remodeling occurs and why it is important in controlling gene expression.

• Discuss how regulatory DNA sequences, transcription factors, and RNA polymerase work together to control transcription initiation.

• Describe different ways that gene expression is controlled after transcription has occurred • •

and why these mechanisms are important.

Compare and contrast how gene expression is controlled in bacteria and eukaryotes.

Explain how defects in gene regulation are linked to cancer.

Lecture Outline I. Gene Regulation in Eukaryotes—An Overview A. Gene regulation in multicellular organisms is important for differentiation and for cell communication and coordination.

B. Eukaryotes can regulate gene expression in many levels (Fig. 19.1); the first three are shared with prokaryotes, and the last three are unique to eukaryotes: 1. Transcription 2. Translation 3. Post-translational modifications to proteins 4. Chromatin remodeling: Histone proteins and DNA coiling must be altered on promoter regions of DNA before it can be transcribed.

5. RNA processing: Transcripts may be spliced using different combinations of exons.

6. mRNA stability is influenced by frequency of translation.

II. Chromatin Remodeling A. What is chromatin’s basic structure?

1. DNA is complexed with histones and other proteins to form a DNA−protein complex called chromatin. a. Electron microscopy showed that chromatin has a regular bead-like structure.

(Fig. 19.2a) b. The “beads” were called nucleosomes. (Fig. 19.2b)

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c. X-ray crystallography showed the structure of nucleosomes. (BioSkills 6) (1) A core of eight histone proteins forms each nucleosome.

(2) DNA is wrapped twice around the core of histones.

(3) Histone H1 forms a link between adjacent nucleosomes.

2. Histones bind to DNA because they are positively charged. a. The positive charge is due to many lysine and arginine amino acids. b. DNA is negatively charged due to the phosphates of the backbone.

3. Interactions between histones form higher-order structures in DNA. a. H1 proteins interact with one another and with other nucleosomes. b. The interactions produce the 30-nm DNA fiber. c. The 30-nm fibers are attached to scaffold proteins and packed into higher-order structures.

4. Summary: a. Chromosomes are made of chromatin. b. Chromatin is made of DNA and histones wrapped into nucleosomes. c. Nucleosomes are organized into 30-nm fibers. d. The 30-nm fibers are organized into higher-order structures.

5. Eukaryotic DNA has more levels of organization where each chromosome lies in its own distinct region, and this organization is important to gene expression. (Fig. 19.3) B. Evidence that chromatin structure is altered in active genes 1. Chromatin must be decondensed so that RNA polymerase can have access to genes.

2. In actively transcribed regions of the genome, chromatin is remodeled so that RNA polymerase can bind to the promoter.

3. DNA in condensed chromatin is protected from DNase. a. DNase cuts DNA; some cut at random locations. (Fig. 19.4) b. DNase cannot cut DNA efficiently if it is tightly bound to histones.

4. Experimental evidence for chromatin remodeling a. Weintraub and Groudine hypothesis: DNA of actively transcribed genes is in an open configuration.

(1) Researchers compared the structures of the chicken β-globin and ovalbumin genes.

(2) In red blood cells, the β-globin is transcribed at high levels.

(3) The ovalbumin gene is expressed in cells of the female reproductive tract but not in red blood cells.

(4) DNase I rapidly degrades DNA that is not complexed with proteins. (Fig. 19.4) (5) The reticulocyte β-globin gene is digested by DNase I, but the ovalbumin gene is not.

(6) Similar results in experiments with other genes confirm that actively transcribed DNA is less condensed. b. Histone mutants in brewer’s yeast (1) In a mutant yeast that lacks certain histones, genes that are normally inactive are transcribed at very high levels.

(2) The lack of histones prevents the normal chromatin assembly.

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c. Conclusions: (1) Histone−DNA interactions inhibit transcription.

(2) Eukaryotic genes are turned off until chromatin opens up to allow transcription.

C. How is chromatin altered?

1. DNA methylation a. DNA methyltransferase methylates C residues in DNA at specific sites. b. In mammals, the recognition sequence is CpG. c. Methylated CpG sequences are recognized by proteins that regulate chromatin condensation. d. Low levels of methylated CpG near a promoter correlate with active transcription. e. Nontranscribed genes have high levels of methylated CpG. 2. Histone modification a. Histone proteins are also covalently modified. b. Chemical groups that may be added include phosphate, methyl, acetyl, and short-chain polypeptides. c. Histone modification either promotes or decreases condensation of DNA, depending on the specific modification set. d. The histone-code hypothesis proposes that specific combinations of modifications determine DNA condensation, and therefore regulate transcription. e. Acetylation is an example. (Fig. 19.5) (1) Histone acetyltransferases (HATs) acetylate lysines in histones.

(2) Histone deacetylases (HDACs) remove those acetyl groups.

(3) Acetylation neutralizes lysine’s positive charge, reducing the histone’s ability to interact with DNA. (4) Acetylation of histones results in decondensed chromatin and increased transcription.

(5) HATs turn genes on, and HDACs turn genes off.

3. Chromatin-remodeling complexes a. Multimolecular complexes use energy from ATP to alter chromatin structure. b. Nucleosomes move along DNA or are removed completely, allowing transcription to occur.

4. Summary: a. Several processes work together to modify chromatin. b. The condensation state of chromatin regulates expression of specific genes.

D. Chromatin modifications can be inherited.

1. Different cell types in the same individual will have different patterns of DNA methylation and histone modification. a. This epigenetic information influences whether or not a gene is expressed. b. Epigenetic inheritance is the result of inheritance that is not due to DNA sequence variation. c. Epigenetic factors result in different gene expression in different cell types.

2. Epigenetic information may also be passed to the next generation. (Fig. 19.6)

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a. An example is the Hnf4a regulatory gene, associated with development of type 2 diabetes.

(1) Rats whose mothers ate a low-protein diet during pregnancy and lactation are at increased risk of developing diabetes.

(2) These diabetic rats also have lower expression of Hnf4a. b. Researchers have found that histone modifications in a regulatory region of Hnf4a correlate with expression of this gene. (Fig. 19.5) (1) Rats born to malnourished mothers had more condensed chromatin in the regulatory region than normal mice.

(2) The more condensed chromatin in the treatment group correlated with lower levels of Hnf4a expression.

III. Initiating Transcription: Regulatory Sequences and Regulatory Proteins A. Promoters are the site of RNA polymerase binding to initiate transcription.

1. Eukaryotic promoters are more complex than those of bacteria and often have 2 to 3 conserved sequences that regulate transcription.

2. A well-studied eukaryotic promoter element is the sequence TATA, the “TATA box.” a. TATA-binding protein (TBP), binds to the TATA box when it is exposed by chromatin remodeling. b. TBP binding is an early initiation step in transcription but does not guarantee that a gene will be transcribed.

B. Promoter-proximal elements are regulatory sequences near the promoter.

1. Regulatory sequences are involved in controlling gene activity.

2. Yasuji Oshima and colleagues worked with yeast mutants to understand how galactose metabolism is controlled. a. When galactose is absent, only small amounts of the enzyme required for its metabolism are produced. b. When galactose is present, transcription of the genes encoding these enzymes increases 1,000 times. c. Mutant yeast cells were discovered that did not produce any of the five enzymes. d. Hypothesis to explain these observations: (1) The five genes are regulated together, even though they are on different chromosomes.

(2) Normal cells have an activator protein that exerts positive control over these five genes.

(3) Mutant cells have disabled activator protein enzymes. e. Other researchers found that the regulatory protein binds to a short stretch of DNA upstream from the promoter for the five genes. f. Unlike bacteria, where co-regulated genes are clustered into a single operon, in eukaryotes co-regulated genes may be on different chromosomes but share a regulatory DNA sequence bound by the same regulatory protein. g. Specific regulatory sequences close to the promoter and bound by corresponding regulatory proteins are called promoter-proximal elements.

C. Enhancers are regulatory sequences far from the promoter.

1. Susumu Tonegawa and coworkers discovered enhancer sequences.

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2. Enhancers are found in all eukaryotes (but not in prokaryotes) and share certain features: (Fig. 19.7) a. Enhancers may be up to 100,000 bases away from the promoter. b. They may be located in introns. c. They may be on either the 3' or 5' ends or untranscribed sequences of the gene. d. Different types of enhancers exist. e. They may be within the gene. f. Most genes have more than one enhancer. g. Enhancers can usually bind more than one protein. h. Enhancers can work even if their normal 5' → 3' orientation is flipped. i. Enhancers can work even if they are moved to a different location in the vicinity of the gene on the same chromosome.

3. Transcriptional activators bind to enhancers, accelerating transcription initiation.

4. Silencers have the same characteristics as enhancers, but they shut down transcription when they are bound by repressors.

D. The role of transcription factors in differential gene expression 1. Activators and repressors regulate transcription and are called transcription factors.

2. Different cell types in multicellular organisms express different genes because they have different transcription factors.

3. Transcription factors are produced in response to signals.

4. In multicellular organisms, cells signal to each other during development, resulting in different tissue types.

5. In muscle cell differentiation, for example: a. A signal transduction cascade causes specific transcription factors to be produced. b. The transcription factors bind regulatory sequences. c. Muscle-specific messages are produced.

6. Differing signals in different cells lead to different tissue types.

E. How do transcription factors recognize specific DNA sequences?

1. The major and minor grooves of DNA’s double helix expose part of the A-T and G-C pairs, which are recognized by transcription factors. (Fig. 19.8a) 2. Transcription factors can interact with the DNA sequence, binding and therefore regulating gene expression. (Fig. 19.8b) 3. The specificity of interactions between DNA and transcription factors makes specialized cell responses possible.

F. A model for transcription initiation 1. In addition to regulatory transcription factors, basal transcription factors are needed.

2. General transcription factors interact with the promoter. a. General transcription factors must be present for transcription to occur. b. They do not regulate gene expression very much.

3. An example of a basal transcription factor is the TATA-binding protein (TBP), which is used in many genes.

4. Additional protein complexes, mediators, act as bridges between regulatory transcription factors, basal transcription factors, and RNA polymerase II. .

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5. Students should be able to compare and contrast the regulatory and basal transcription factors found in muscle cells versus nerve cells.

6. Model of steps in transcription initiation (Fig. 19.9) a. Transcriptional activators bind and activate chromatin-remodeling complexes and histone acetyl transferases (HATs). b. The promoter, promoter-proximal elements, and enhancers become accessible. c. Other transcription activators bind to the exposed enhancers or promoter-proximal elements, allowing mediators and activators to come together, assemble, and recruit RNA polymerase II. e. Mediators then associate with the core promotor, and RNA polymerase begins transcription.

7. Students should be able to explain why DNA forms loops near the promoter in order for transcription to begin.

8. Transcriptional activators perform two roles. a. They stimulate transcription. b. They recruit chromatin-remodeling proteins, which cannot recognize specific DNA sequences.

9. Transcriptional activators have access to DNA because chromatin is dynamic; when nucleosomes dissociate from DNA, regulatory sequences are exposed and can be bound by activators.

10. Summary: a. Transcription initiation requires many proteins: (1) Transcriptional activators bound to enhancers and promoter-proximal elements.

(2) Mediators bound to activators and basal transcription factors.

(3) Basal transcription factors bound to promoters.

(4) RNA polymerase to make the transcript. b. Eukaryotic transcription control is complex compared to that seen in prokayrotes.

IV. Post-Transcriptional Control A. Alternative splicing of mRNAs 1. Spliceosomes can connect exons together in multiple ways after intron removal, so a primary RNA transcript can yield multiple mature mRNAs. a. The mRNA transcript for tropomyosin contains 14 exons. (Fig. 19.10a) b. Different subsets of these exons are spliced together in smooth muscle cells versus in skeletal muscle cells. (Fig. 19.10b) c. Result: The tropomyosin protein is different in these two types of cells.

2. Alternative splicing is controlled by proteins that bind to mRNAs in the nucleus and interact with the spliceosome.

3. The human genome has a smaller number of genes than previously thought. a. Alternative splicing makes it possible to produce more than one protein product from a single locus. b. One gene in fruit flies gives rise to 38,000 distinct forms of mRNA. c. New definition of a gene: coding and regulatory sequences that direct the production of one or more polypeptides or RNAs

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4. Students should be able to explain why alternative splicing does not occur in bacteria and describe where it occurs in Figure 19.1.

B. mRNA stability and RNA interference 1. The amount of protein made from an mRNA depends on the length of time the mRNA is stable in the cytoplasm. a. Casein (milk protein) mRNA is usually stable in the cytoplasm of mammary gland cells for only an hour. b. In a lactating female, the casein mRNA is stable for 30 times as long, leading to a large increase in the amount of casein protein produced. c. Shortening the poly(A) tail decreases the RNA’s stability.

2. RNA interference can control the stability of an mRNA in the cytoplasm. a. Small single-stranded pieces of RNA that are complementary to mRNA sequences are held by a protein complex. b. The regulatory RNA binds to the mRNA molecule and makes it double stranded. c. This either targets the mRNA for destruction or blocks its translation.

3. The details of RNA interference (Fig. 19.10) a. RNA polymerase makes a short piece of RNA with internal complementary sequences that bind to one another, forming a hairpin. b. The hairpin loop is clipped in the nucleus; the resulting short, double-stranded piece of RNA is exported to the cytoplasm. c. In the cytoplasm, the double-stranded RNA molecule is cut into pieces that are about 22 nucleotides long. d. The resulting pieces are taken up by RNA-induced silencing complexes (RISCs). e. The single-stranded pieces taken up by the RISCs are called microRNAs (miRNAs). f. When the miRNAs bind to complementary sequences in an mRNA, the RISC either destroys the mRNA (if the base-pair match is perfect) or inhibits translation without destroying the mRNA (if the base-pair match is not perfect). g. There are at least 500 sequences in plant and animal cells that code for miRNAs. (1) miRNAs regulate more than one mRNA. (2) miRNAs make a significant contribution to gene regulation.

(3) Research on miRNAs will determine whether they may be useful in treating disease.

4. Students should be able to describe where RNA interference occurs in Figure 19.1.

C. How is translation controlled?

1. Global regulation of translation a. A cell often benefits from shutting down translation of most mRNA. b. Protein synthesis is energy intensive, and in unfavorable conditions a cell may be able to survive its current state by cutting back translation. c. mTor is a protein kinase that plays a role in global regulation.

2. RNA interference and expression of individual RNAs a. RNA interference occurs when a tiny, single-stranded RNA held by a protein complex binds to a complementary sequence in an mRNA leading to: (1) Destruction of mRNA .

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(2) Blocking of the mRNAs translation b. RNA interference through microRNA (miRNA) (Fig. 19.11) either targets mRNAs for destruction or prevents them from being translated.

D. Post-translational control 1. Post-translational control allows the cell to respond to new conditions rapidly without waiting for transcription and translation to occur.

2. It involves greater energy expenditure by the cell than transcriptional control mechanisms because translation requires additional energy.

3. Existing proteins can be quickly activated or inactivated.

4. Mechanisms of post-translational control a. Protein activity may be changed by enzymes that add carbohydrate groups or cleave off amino acids. b. Proteins may be phosphorylated by protein kinases such as mTor. c. Targeted destruction of proteins (1) When a protein needs to be destroyed, an enzyme adds a small polypeptide, ubiquitin.

(2) Proteasomes recognize proteins with a ubiquitin tag and cut them into short segments.

V. How Does Gene Expression Compare in Bacteria and Eukaryotes? (Table 19.1) A. DNA packaging 1. Due to the structure of chromatin, in eukaryotes the default state of transcription is off.

2. Because the structure of prokaryote DNA allows RNA polymerase easy access to promoters, in prokaryotes the default state of transcription is on.

B. Complexity of transcription 1. Transcriptional control is much more complex in eukaryotes than it is in prokaryotes.

2. Many more eukaryotic proteins are involved in eukaryotic transcription.

C. Coordinated transcription 1. In bacteria, operons coordinate the expression of genes controlled by a single promoter.

2. In eukaryotes, operons are rare, but regulatory transcription factors can coordinate the expression of physically separated genes.

D. Reliance on post-transcriptional control 1. Eukaryotic genes have introns, and alternative splicing generates many more gene products than the total number of genes in a genome. Bacterial genes do not have introns, so no splicing occurs.

2. Eukaryotes rely heavily on microRNAs and regulation of mRNA stability, which are seldom seen in bacteria.

E. Multicellular eukaryotes have a need for more complex regulation.

1. Each cell type requires a unique pattern of gene expression, especially during development.

2. This may explain why the control of gene expression is so much more complex in eukaryotes.

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VI. Linking Cancer with Defects in Gene Regulation A. Uncontrolled cell growth is the disease-causing mechanism in cancer. Two types of genes may be involved: 1. Genes that stop or slow the cell cycle 2. Genes that initiate cell cycle events leading to cell growth and division B. The genetic basis of uncontrolled cell growth 1. Tumor suppressor genes encode proteins that stop or slow the cell cycle.

2. Proto-oncogenes encode proteins that trigger the progression of the cell cycle.

Mutations convert them into oncogenes.

3. Cancer results when several mutations occur within a single cell, altering both tumor suppressor genes and proto-oncogenes.

C. The p53 tumor suppressor: a case study 1. One particular tumor-suppressor gene is mutated in many human cancers.

2. p53 seems to be required when a cell is damaged. (Fig. 19.12) a. p53 levels increase dramatically in human cells exposed to UV radiation. b. A close correlation exists between DNA damage and the amount of p53 in cells. c. p53 also has a region that could bind to DNA. d. Hypothesis: p53 is a transcription factor that serves as a brake on the cell cycle making it a “guardian of the genome.”

(1) This gives the cell time to repair the damage or be targeted for destruction.

(2) If p53 is missing, damaged DNA is replicated and the cell moves closer to cancer.

3. Evidence supporting this hypothesis: a. p53 activates genes for cell cycle regulation, DNA repair, and apoptosis. b. p53 binds directly to DNA regulatory sequences of genes it controls. c. Virtually all p53 mutations associated with cancer are located in the protein’s DNA-binding site.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

.

30-nm fibers

core promotor

general transcription factors

acetylation activator alternative splicing chromatin chromatin chromatin remodeling chromatin remodeling chromatin-remodeling complexes

differential gene DNA methylation DNA methyltransferase enhancer epigenetic inheritance exon expression galactose metabolism genes gene

H1 histone histone acetyl transferases (HATs) histone code histone deacetylases (HDACs) histones intron mediator methylation

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microRNAs (miRNA) mRNA stability nucleosomes oncogenes p53 phosphorylation post-transcriptional

post-translational regulation primary RNA transcript promoter-proximal elements proteasome proto-oncogenes regulatory sequences regulatory transcription

control post-translational control post-translational modification

factors regulatory transcription factors repressors

RNA interference RNA processing RNA-induced silencing complex (RISC) silencer TATA box TATA-binding protein (TBP) transcription factors transcriptional activator tumor suppressor

Lecture Activities Motherly Advice Estimated duration of activity: 10 minutes This chapter touches on some issues that can help students understand the science behind some common motherly advice. After covering the material in this chapter, ask your students to do some additional research and be able to explain (in detail) the reasons behind the following “advice”: • Wear sunscreen with UVA and UVB protection!

• Don’t smoke!

• Tell the X-ray technician if you are pregnant.

Prokaryotes versus Eukaryotes Estimated duration of activity: 20 minutes (if the students do the work in class) This activity is tied to Chapter 18. It should help students sort out the differences between controls in prokaryotes and eukaryotes.

You can introduce this activity in the classroom (if your class is small, you could divide the class in half to form a prokaryote group and a eukaryote group). You may also want to give it as homework or use it as a review activity in a study section.

Method: Students should compare and contrast general aspects of the control of gene expression in eukaryotes (the work discussed has been done primarily in yeast and multicellular animals). Their work should include (1) possible levels of control, (2) possible molecules and sites involved in control, and (3) overall reasons to have these controls. They should consider the following examples.

Levels of control in prokaryotes: • In most cases, transcription is off or on.

• In some cases, there is a low rate of transcription when a substrate is present, but there is no accompanying positive signal.

• There is possibly a differential rate of degradation of messages or proteins.

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Levels of control in eukaryotes: • Transcription • Processing of the RNA transcript • Translation • Post-translational • Differential rates of degradation of messages or proteins Molecules and sites involved in control in prokaryotes: • There are operator sites 5' to the structural gene where control molecules can bind or be removed from binding.

• There are sites like the CAP site, which are 5' to the promoter site and change the configuration of the promoter site and make the promoter more efficient in directing synthesis of the genes under its control.

• The promoter site can be altered to allow more efficient binding of the RNA polymerase.

• Repressors can bind to operator sites to turn off RNA synthesis.

• There are molecules that can sit at sites 5' to the promoter and are able to change the promoter binding to allow for more efficient binding of RNA polymerase and therefore more efficient synthesis of the genes under control of that region.

• There is one RNA polymerase.

• Different sigma factors bind to RNA polymerase.

• Many small molecules can release repressors and change repressors to bind to operators; others can alter molecules to provide for positive control of RNA synthesis.

Molecules and sites involved in control in eukaryotes: • Enhancer and silencer sequences or sites can be 5', inside introns, or anywhere in the genome where regulatory transcription factors act.

• Promoter-proximal elements are 5' to the promoter where regulatory factors act.

• The promoter site is where a large protein complex binds, acts to change confirmation of the site (along with other factors), and then allows RNA polymerase II and other transcription factors to bind to this structure to make the core complex.

• Many regulatory transcription factors can bind to enhancers, silencers, or promoterproximal elements or be part of the core transcription complex.

• Many small molecules participate by changing the conformation of the transcription factors (phosphorylation, for example).

• There are three RNA polymerases; RNA polymerase II participates in the synthesis of mRNA, and the others synthesize other kinds of RNA. • Many different core transcription factors may or may not participate past the initiation of transcription.

Reasons to have controls: • Bacteria have limited space and time for synthesis and storage of substrates, intermediates, and products.

• Bacteria have less genomic information, so they must utilize each gene and its products efficiently. .

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• It takes energy to be prepared for changes, so bacteria often shut down unnecessary pathways.

• Bacteria are less complex, so they are able to function with less possible variety.

• Eukaryotes have more complexity and more variety compared with bacteria.

• Yeast as a single-celled organism has been studied for the basic steps.

• Multicellular organisms have the most variety because they need to have signals that • instruct cells to become specialized for various tasks.

Eukaryotic cells have more room for all of these processes taking place in various loca• tions in the cells—in many cases, only in specialized cells.

• Eukaryotic cells have much larger genomes, so they have room for more complex controls.

Eukaryotic cells have primary transcripts with intervening sequences or introns that must be removed to make the message useable. This allows differential splicing to achieve more variety with the messages than is possible in the original single transcript.

• Eukaryotic RNAs have caps and tails attached to the working messages.

• To have a faster response in critical conditions, there frequently are translational and post-translational controls.

• Eukaryotic cells have histones that with DNA form nucleosomes; to allow for transcription, these structures must be opened.

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Learning Objectives: Students should be able to…

• Explain recombinant DNA technologies and what they can achieve.

• Describe the polymerase chain reaction and provide examples of how it is used for DNA fingerprinting and other applications.

• Explain what next generation sequencing allows and why it is important to sequence DNA. • Compare and contrast the general features of prokaryotic and eukyotic genomes.

• Describe bioinformatics and explain why it is important for modern biology.

• Explain why finding genes is important and how genes can be found using linkage studies • and genome-wide association studies.

• Define lateral gene transfer and explain its role in the evolution of genomes.

• Describe how gene duplication allows the evolution of new gene functions.

• Describe how gene therapy may be used to cure a genetic disease.

Describe what function genomics, proteomics, and systems biology are and provide examples of discoveries in these research areas.

Lecture Outline Sequence of the human genome: • Published 2001 • Landmark in the history of science (Fig. 20.1) • Biologists have gone from not only understanding the molecular nature of the gene, but also knowing the molecular makeup of every gene present in our species.

• Genomics is the effort to sequence, interpret, and compare whole genomes.

• Techniques used to engineer genes are called recombinant DNA technology.

I. Recombinant DNA Technology A. Using plasmids in cloning 1. A basic factor in recombinant DNA technology is the ability to make many identical copies of the gene, DNA cloning.

2. Cloning is generally done by putting the gene of interest into a small, circular DNA molecule (plasmid). .

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a. Plasmids are physically separate from the bacterial chromosome; they are not required for normal growth and reproduction, and they replicate independently of host. b. Plasmids can carry genes for antibiotic resistance or other traits that allow them to grow in hostile environments.

3. When a plasmid is used to make copies of foreign DNA sequences, it is called a cloning vector.

B. Using restriction endonucleases and DNA ligase to cut and paste DNA 1. To cut a gene out for later insertion into a cloning vector, enzymes (restriction endonucleases) are used.

2. Restriction endonucleases cut DNA at specific base sequences called recognition sites. a. There are more than 800 restriction endonucleases known.

3. Insertion of foreign DNA into plasmids illustrated (Fig. 20.2) a. Step 1: identify a restriction site b. Step 2: add restriction endonuclease c. Step 3: sticky ends capable of hydrogen bonding d. Step 4: gene gets inserted into the plasmid e. Step 5: recombinant plasmids are inserted into a bacterial culture and allowed to replicate, making many copies (clones) of the gene of interest C. Transformation: Introducing recombinant plasmids into bacterial cells 1. If a recombinant plasmid (vector) can be inserted into a bacterial or yeast cell, the foreign DNA on the plasmid will be copied and transmitted to new cells as the host cell divides.

2. Transformation is the process of bacterial cells taking up foreign DNA and incorporating it into their genome. a. Chemical treatment or electrical shock must be applied. b. The recombinant plasmid then enters the cell. c. The cells are spread on plates so that colonies are physically isolated. d. The individual cells grow into colonies containing millions of identical cells.

D. Using reverse transcriptase to produce cDNAs 1. Reverse transcriptase is an enzyme that makes DNA from RNA. 2. The DNA it produces is called cDNA. 3. cDNA is single stranded; reverse transcriptase can synthesize a complementary strand, but usually a primer is added and DNA polymerase is used to synthesize the second strand. (BioSkills 10) E. Biotechnology in agriculture 1. Transgenic plants can be created to reduce crop losses from herbivores and pathogens.

II. The Polymerase Chain Reaction (PCR) A. Requirements of PCR 1. PCR is an in vitro DNA synthesis reaction in which a specific section of DNA is repeatedly replicated by DNA polymerase; this results in many copies of that sequence.

2. To do a PCR, you start by synthesizing short lengths of single-stranded DNA, called primers, which match sequences on either side of the gene of interest.

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3. Sequence of events in a PCR experiment: (BioSkills 10) a. Create a reaction mix with an abundant supply of the four deoxyribonucleotide triphosphates, a double-stranded piece of DNA containing the gene of interest, many copies of the two primers, and the enzyme, Taq polymerase, a heat-stable bacterial DNA polymerase used to polymerize the new DNA strands. b. Heat the reaction mixture to 95ºC to denature the double-stranded template DNA, forming single-stranded templates. c. Cool the reaction mixture to 50 to 60ºC to allow the primers to bind (annealing). d. Heat the reaction mixture to 72ºC so that Taq polymerase can synthesize the new DNA strands (extension). e. Repeat steps b–d (one PCR cycle) 20 to 30 times to obtain as many as a million copies of the gene of interest. Each cycle doubles the amount of template sequence DNA; n cycles generate 2n copies.

B. DNA Fingerprinting (DNA Profiling/DNA Typing) 1. Any technique for identifying individuals based on the unique features of their genome 2. Tandem repetitive DNA sequences fall into two major classes: a. Short tandem repeats (STRs)/microsatellites/simple sequence repeats (SSRs) b. Minisatellites/variable number tandem repeats (VNTRs) 3. DNA fingerprinting using PCR (Fig. 20.3) a. DNA fingerprinting can be used in paternity analysis, criminal cases, and in identifying remains. b. To fingerprint, investigators: (1) Obtain a DNA sample and perform PCR using primers that flank the region containing the STR. (Fig. 20.3a) (2) Once the STR is amplified, gel electrophoresis (BioSkills 10) is used to determine the number of repeats.

C. A new branch of the human family tree 1. Svante Pääbo and colleagues used PCR to obtain many copies of a region of DNA from a fossilized 50,000-year-old girl’s pinky bone found in the Denisova cave in Siberia in order to sequence it and compare it to the corresponding region in the H. sapiens genome.

2. Previous work from these researchers has been in analysis of 30,000-year-old bones from Homo neanderthalensis. a. The data supported the hypothesis that H. neanderthalensis had not interbred with H. sapiens in African populations. b. For non-African populations, 1–4 percent of each person’s genome is derived from Neanderthal ancestry.

3. When comparing the girl’s DNA to modern humans and Neanderthals, they discovered she represented a new type of human, Denisovans. a. Modern humans of southwestern Pacific, New Guinea, and Australia show traces of Denosovan DNA today.

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III. DNA Sequencing A. Finding the nucleotide sequence once the gene has been cloned 1. Knowing the sequence of a gene is useful to understand its function and to deduce the amino acid sequence and function of its product.

2. Comparing sequences is fundamental to understanding why alleles vary in function.

3. Comparing sequences of the same gene in different species allows researchers to study and infer evolutionary relationships.

B. Dideoxy sequencing (BioSkills 10) 1. The technique developed by Frederick Sanger (1977), an in vitro DNA synthesis reaction that can determine the base sequence of DNA C. “Next generation” sequencing 1. Allow for rapid sequencing of entire genomes 2. Sequence only 50–200 nucleotides, making it challenging to piece together a whole genome D. Whole-genome sequencing 1. To sequence whole genomes, researchers rely on shotgun sequencing. a. In shotgun sequencing, many copies of a genome are broken up randomly into a set of fragments of various sizes. These fragments are expected to overlap in many regions and are used as guides to put the whole genome back together. b. The process of putting the pieces together is called genome assembly (BioSkills 10) E. Bioinformatics 1. Bioinformatics is the field that fuses mathematics, computer science, and biology to manage and analyze sequence data.

2. Databases have been created to allow investigators all over the world to evaluate the similarities between newly discovered genes and genes that have been studied previously in the same or other species. a. Databases include NCBI, BLAST F. Which genomes are being sequenced, and why?

1. The first, Haemophilus influenza, which causes respiratory tract infections and earaches in children 2. Assortment of bacteria and archaea followed 3. First eukaryote, Saccharomyces cerevisiae (1996) 4. Most whole-genome sequencing represent information for evolutionary investigations, are of agricultural importance, or cause disease.

G. Which sequences are genes?

1. Once the genome is sequenced, identifying genes and important sequences is important. This is called genome annotation.

2. To annotate the genome, biologists use computer programs that scan the genome in both directions looking at each reading frame that is possible on the two strands. a. Codons consist of three bases, so three reading frames are possible on each strand.

(Fig. 20.4) b. Open reading frames of a long stretch of codons are sought for good protein coding sequences.

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c. Open reading frames in bacteria are determined by finding frames that are flanked by a stop and a start codon. Here, the computer programs also look for consensus sequences typical of promoters, ribosome binding sites, or other regulatory sites.

3. Similarities in sequence and in function between genes in different species are usually due to homology. Homology means they are similar because they are related in descent from a common ancestor.

4. Eukaryotic sequencing is more complicated because coding regions are broken up by introns. a. One strategy is to isolate mRNAs from cells, use reverse transcriptase to produce a cDNA version of a portion of each mRNA, and sequence the CDNA produce an expressed sequence tag (EST). b. Researchers use sequence-matching programs to search for matches between the EST and a sequence in genomic DNA.

VI. Insights from Genome Analysis A. The natural history of prokaryotic genomes 1. Principles of prokaryotic genomes: a. Genomes are compact, have uninterrupted coding sequences, have little space between genes, use operons, and have few regulatory sequences. In general, genome size and number of genes increase together (Fig. 20.5a); however, the same is not true in eukaryotes. (Fig. 20.5b) b. Genome size is related to living conditions (habitat vs. host). c. Most genes are not shared among species of prokaryotes. d. Genome size varies within species. e. Genomes are frequently rearranged during evolution. f. A significant portion of prokaryotic genomes have been acquired from other, distantly related species.

2. Lateral gene transfer is a major force in the evolution of prokaryotes; in eukaryotes, it is much less common. a. In eukaryotes, the capture of bacterial cells that were once predecessors of today’s mitochondria and chloroplasts is an example of lateral gene transfer. b. When lateral gene transfer occurs, it changes the tree of life. (Fig. 20.6) 3. Metagenomics or environmental sequencing is the creation of a catalog of all the genes present in a complex community of prokaryotes.

4. The natural history of eukaryotic genomes a. Challenges of sequencing eukaryotic genomes: (Fig. 20.5) (1) Genome size (2) Repeated sequences that are noncoding regions of the DNA b. Repeated DNA partially explains the immense variation in eukaryotic genome sizes; however, many of the repeated sequences are derived from transposable elements.

(1) Discovered by Barbara McClintock (2) Segments of DNA that can insert into new locations in a genome (3) Make up 45 percent of the human genome, 85 percent of the corn genome (4) Insert in the genome and therefore are transferred from mother to daughter cell .

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(5) Example of transposable element: long interspersed nuclear element (LINE) (Fig. 20.7) c. In eukaryotes, an important source of new genes is the duplication of existing genes. These genes are considered part of the same gene family due to similar structure/function and sequence.

(1) Unequal crossing over during meiosis can lead to duplication or deletion (Fig. 20.8) (2) When duplication results in an altered protein product with a beneficial new function, a new gene has been added to the genome. This process is referred to as duplication and divergence. (Fig. 20.9) (3) When duplicated genes lead to nonfunctional genes due to a stop codon in the middle of the exon, it resembles a working gene but not a functional product; it is called a pseudogene. d. The Human Genome Project took 15 years and $3 billion to assemble. Today, whole genomes can be sequenced for a low as $1,000.

(1) The composition of the human genome (Fig. 20.10) (2) Alternative splicing explains why there are so few genes in the human genome and yet great organismal diversity.

(3) Noncoding RNAs are produced when 90 percent of the genome is transcribed, making regulatory RNAs.

V. Finding and Engineering Genes: Huntington’s Disease Story A. How was the Huntington’s disease gene found?

1. Huntington’s disease is a rare but devastating disease; individuals typically become symptomatic between ages 35 and 45. a. Clumsiness, small tics, and abnormal movements are the first signs. b. Later, uncontrollable movements become more pronounced, eventually causing the individual to twist and writhe.

2. Huntington’s disease tends to run in families, so scientists hypothesized that it was hereditary and set out to identify the gene(s) involved. a. The search was led by Nancy Wexler, whose mother died of the disease.

3. Genetic markers serve as landmarks that can be used to locate other genes. a. Approach: Find a known genetic marker closely linked to the unknown gene. b. If the marker and disease gene are close together, they will tend to be inherited together. c. Genes that are close together on the same chromosome assort together during meiosis and are less likely to be separated by crossing over. d. Students should be able to explain why it’s helpful to hunt for genes using a genetic map with many genetic markers rather than only a few.

4. The types of genetic markers used to find genes have changed over time. a. Today, researchers have a large catalog of genetic markers available.

(1) Single nucleotide polymorphisms (SNPs) are sites in DNA in a population where some individuals have a different base. b. At the time of the search for the Huntington’s disease gene, restriction enzyme recognition sites were the best available markers. (Fig. 20.10) 194

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(1) The number of recognition sites for a given restriction enzyme varies among individuals because of differences in their DNA sequences.

(2) DNA from different individuals that is cut with the same restriction enzyme will have different numbers and sizes of fragments, called restriction fragment length polymorphisms (RFLPs).

5. A linkage study: Its goal is to find a genetic marker that is almost always inherited with the disease-causing allele. a. For an inherited disorder, gene hunts are more likely to be successful if large families are affected. b. For Huntington’s disease, a large affected Venezuelan family was found.

(1) In the early 1800s, a European man introduced the Huntington’s disease allele into the family.

(2) By the 1970s, more than 3,000 descendants were living in the same area.

(3) One hundred had been diagnosed with Huntington’s disease. c. To find the Huntington’s gene, researchers searched for a DNA fragment pattern that was the same in all diseased individuals but different in unaffected individuals.

(1) Several recognition sites appeared to be inherited with the Huntington’s disease allele.

(2) Statistical analysis determined which recognition site was inherited with the allele often enough that it was not due to chance.

(3) The recognition site marker was on chromosome 4.

(4) The location of the disease gene was narrowed to a region of 500,000 base pairs.

6. Pinpointing the defect a. Biologists searched for exons in the region to which the gene was mapped. b. For one gene, healthy individuals had 11 to 25 copies of a CAG codon near the 5' end, but diseased individuals had 42 or more copies. c. The Huntington’s disease locus was named IT15, and the protein product was named huntingtin.

B. How are human genes found today?

1. Genome wide association studies (GWAS): SNP alleles known and statistical and bioinformatics methods allow rapid search for association between SNP and a disease trait.

2. Researchers look for co-occurrence of a particular SNP allele and a particular trait.

When the trait and SNP occur together, the gene for the trait must lie near the known location of SNP. 3. Advantages of GWAS a. Pedigrees are only needed for one, not multiple generations because the trait can be mapped to a small region. b. Polygenetic traits can be studied. c. Alleles that contribute to a trait can be identified.

4. GWAS have found alleles that increase predisposition for: Type I and II diabetes, breast/ovarian cancer, obesity, coronary heart disease, bipolar disorder, Crohn’s disease, and rheumatoid arthritis.

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5. Huntington’s is caused by a single gene, but most GWAS find that many alleles of many different genes each contribute a small amount to disease development.

C. What are the benefits of finding a disease gene?

1. Improved understanding of the phenotype a. The molecular nature of the disease was inferred from the gene sequence.

(1) The CAG repeats code for glutamine amino acids.

(2) Long stretches of polyglutamine in proteins are known to cause protein aggregates.

(3) Brains of Huntington’s disease patients shrink due to loss of neurons and fill with aggregates of huntingtin.

(4) This explains why the disease takes so long to appear and is progressive; it takes time for huntingtin to build up, and it continues to do so over time.

(5) The allele that causes Huntington’s disease is dominant because one copy is enough to produce fatal concentrations of aggregates.

2. Transgenic organisms have alleles that have been modified by genetic engineering. b. The gene has been introduced into experimental mice.

(1) These transgenic mice show tremors, abnormal movements, loss of brain neurons, and aggression characteristic of Huntington’s disease.

(2) Animal models can be used to test potential drug treatments⎯for example, drugs that block aggregation of huntingtin.

3. Genetic testing a. Individuals who suspect they may carry the Huntington’s disease allele can now be tested. b. DNA from the individual is obtained, and the CAG repeat region is amplified by PCR. c. A positive diagnosis for Huntington’s disease is made if 40 or more repeats are found. d. Types of genetic testing done today (1) Testing for carriers of a disease so they can make decisions about marrying or having children (2) Prenatal testing of fetuses 10 to 16 weeks old using amniocentesis or chorionic villi sampling (3) Adult testing for warning of the onset of a possible disease, especially important if there is a strong environmental component D. Can gene therapy provide a cure?

1. For diseases like Huntington disease, sickle-cell anemia, and cystic fibrosis that are caused by mutation of a single gene, gene therapy would replace these defective alleles with normal ones.

2. Requirements for gene therapy: a. Defect must be in a single gene. b. The sequence of the healthy phenotype must be known. c. A method to introduce the healthy allele into the affected individual, in the correct tissue, in the correct amount, and at the correct time

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3. One method of gene therapy uses viruses as the vector to carry the engineered genes into the chromosomes of target cells. (Fig. 20.12) 4. Gene therapy has had rare successes and tragic failures, causing optimism to be little.

Renewed hope has emerged recently with improvement of vectors and success to treat blindness, a brain disorder, and a type of immune system disorder.

VI. Functional Genomics, Proteomics, and Systems Biology A. What is functional genomics?

1. Whole-genome data can be used to answer fundamental questions about how genes work together to produce a phenotype.

2. Functional genomics is the study of how all the products of a genome interact.

3. DNA microarrays represent every gene in a genome and are used to analyze changes in gene activity. (BioSkills 10) 4. Deep sequencing uses sequencing technologies to directly sequence cDNAs. B. What is proteomics?

1. Transcriptome is the complete set of genes transcribed in a particular cell.

2. Proteome is the complete set of proteins produced in a particular cell.

3. Proteomics is the large-scale study of protein function and protein–protein interactions in an organism.

4. Proteomics allows biologists to study all proteins expressed at one time.

C. What is systems biology?

1. Systems biology is based on the premise that a whole is greater than the sum of its parts. It aims to understand how interactions between individual parts of a biological system create new properties.

2. Complex properties that arise from the interaction of simpler elements are emergent properties.

3. Students should be able to explain why cell replication is an emergent property.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. alternative splicing bioinformatics biotechnology cloning vector complementary DNA (cDNA) deoxy sequencing divergence DNA cloning DNA fingerprinting DNA ligase .

DNA microarray DNA profiling DNA typing duplication emergent properties environmental sequencing expressed sequence tag (EST) functional genomics gene duplication gene family gene therapy

genetic map genetic markers genetically modified (GM) crop genome genome annotation genome-wide association study (GWAs) genomics homology Human Genome Project

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huntingtin Huntington’s disease lateral gene transfer linkage map long interspersed nuclear element (LINE) long noncoding RNA metagenomics microsatellites minisatellites nextgeneration sequencing open reading frame (ORF)

physical map plasmid polymerase chain reaction (PCR) polymorphic proteome proteomics pseudogene recombinant DNA technology restriction endonucleases short tandem repeats (STRs) shotgun sequencing simple sequence repeats (SSRs)

single nucleotide polymorphisms (SNPs) stem cells sticky ends systems bilogy transcriptome transformation transgenic transposable elements unequal crossing over variable number tandem repeats (VNTRs) vector

Lecture Activities A Debate over the Use of Genetically Modified Seed in U.S Farming Estimated duration of activity: 20 minutes This chapter has the potential to truly capture the students’ interest. You may want to devote a little extra class time to this subject to give students the opportunity to become involved and invested in it.

Divide your class into four groups. One group will be organic farmers who grow and sell corn. Their crops have recently suffered from particularly damaging infestations of the European corn borer. They believe the genetically engineered Bt-Corn that their conventionalfarming neighbors have been growing is responsible for these infestations. The second group of students will be the conventional-farming neighbors, who have enjoyed particularly good crop yields thanks to the Bt-Corn. The third group of students will be scientists from the company that makes Bt-Corn. They will be present at the debate at the request of the conventional farmers. The fourth group of students will be agricultural research scientists from a local university who are at the debate at the request of the organic farmers.

After explaining the scenario to the class and assigning the roles they will play, ask them to research their positions and come to the next class prepared to debate the other groups.

The organic farmers need to clearly state why the nearby use of Bt-Corn endangers their crops. The conventional farmers need to be able to defend their farming practices and explain the precautions they have taken to safeguard their neighbors’ crops. The representatives from the company that makes Bt-Corn need to be prepared to explain how the Bt-Corn was engineered, and why it is such a stellar product. The agricultural research scientists need to be prepared to explain how Bt-Corn has led to Bt-resistance in the corn borers.

You can mediate the debate by giving each group 3 or 4 minutes to present their arguments. To ensure that everyone participates, you may want to call on each student to make one point until all members of the group have had a chance to speak. Alternatively, you might ask each student to write a one-page position paper to use during the debate and turn in to you at the end of the activity.

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Activity Idea: Applications of Genomics Estimated duration of activity: variable (depending on how you orchestrate the activity) The material in this chapter and in surrounding chapters (Chapters 13–19) is both difficult and fascinating to students. One way to help them learn the material is to get them involved in teaching themselves and one another.

The concepts in this chapter lend themselves well to problem-based learning. You may want to consider assigning the chapter as a reading assignment and then assigning the following scenarios as a writing assignment due on the day you plan to cover this material. (You could divide the class by students’ last initials and have each student write a one- or two-page essay response to the assigned scenario.) Students may need to do some research beyond the scope of the text to complete the assignment. You can then ask for volunteers to present a summary of their response as you come to relevant subjects in lecture, or you could replace your traditional lecture with an entire class period in which you let the students teach one another.

Gene cloning, cDNA, and recombinant DNA technology: You are a marketing executive for a large biotech company that makes human insulin (“Humulin”). To date, the company has been using marketing materials containing the statement: “We were the first company to clone the human insulin gene.” However, patients are now expressing concern about buying your product due to the recent debates in the media over human cloning. You need to design a new marketing campaign that will educate these patients and reassure them that your company is not engaging in human cloning studies. In 250 words or less, write an explanation (for the layperson) of what it means to clone a gene, how cloning enables your company to produce human insulin, and how that is vastly different from human cloning.

Gene therapy: You are a genetic counselor, and one of your clients (the Smiths) has a child with SCID (severe combined immunodeficiency disease). They have read about gene therapies for this disease and would like you to explain to them what gene therapy is (in general) and how this treatment would work for their child with SCID. This couple met and befriended another couple (the Joneses), who have a child with Down syndrome. The Joneses are also interested in learning whether gene therapy would be useful in treating their child.

Explain whether or not gene therapy is a viable treatment for the child with Down syndrome.

Genetically engineered plants: Congratulations! You have just been hired as a salesperson for Monsanto, a large agricultural products company. You have a large sales quota and stand to make a lot of money in commissions if you can sell your product, Roundup-Ready Soybeans, to farmers across the country. Compose a sales pitch for these soybean seeds.

In the pitch, which you would deliver to a farmer, explain the benefits of the seeds and how they were made. Remember, these farmers are concerned about genetically modified seeds because many of them plan to sell their crops to food companies that are required to meet certain FDA regulations about using genetically modified plant products in food made for human consumption.

The Human Genome Project: The National Science Foundation has hired you as an educational consultant to work closely with the Human Genome Project. Your job is to design a high school education program to spread the word about the Human Genome Project and generate interest in genetics. Compose an outline (with paragraphs at each entry explaining the content) for a week-long (five class periods) set of lessons and/or activities for a high school general biology class. These lessons should describe the Human Genome Project, how it was accomplished (keep in mind that students may need to learn about the structure of

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DNA in order to understand how DNA is sequenced), how it will be used, and some of the controversies surrounding it.

DNA chips/microarray: You have just been named the winner on the reality show The Biotech Apprentice, and with it you have won a sales job for a biotech company that sells DNA chips. Your first sales call is at UGA with a geneticist (Deoxy Ribose, PhD), who is studying how the expression of gene X changes during rat development. Write out your sales pitch. You need to explain to him what a DNA chip is, how it is made, and how he could use them in his research.

Discussion Idea: Genetic Engineering Now is the time for discussion or debate about genetic engineering. Students have a good foundation for understanding basic molecules, cells, and genes and their expression.

Questions for discussion: • Should scientists artificially move genes among organisms?

• Should we as humans potentially change the natural relationships among organisms?

Examples: plants that can ward off insects if they were not able to do so naturally; tomatoes that are firmer so they last longer and can be shipped more easily; plants that have amino acids in them that are not there naturally; and plants that are resistant to herbicides so they can be sprayed and only the nearby weeds will die.

• Is it dangerous to eat foods that contain genes from another organism?

• Should we be spending scarce healthcare dollars to help a few people with major genetic problems when we could help hundreds or thousands with basic needs for the same amount of money?

• Cells do vary, but they have many characteristics in common. Does this mean that mixing parts of them is really not a big deal? After all, we are all related.

• Scientists have the ability to do all of these tricks and possibly discover spectacular cures.

Should they be stopped just because of some unknown future problem?

• Would it be better to give vitamin A pills to people rather than change the rice to include vitamin A in the edible part of the plant?

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Genes, Development, and Evolution

Learning Objectives: Students should be able to...

• Explain shared developmental processes.

• Compare and contrast cytoplasmic determinants and induction.

• Discuss how genomic equivalence implies that differentiation occurs through differential gene expression.

• Explain what morphogens are, what they do, how they work, and why they're important in • development.

• Discuss how genetic regulatory cascades lead to increasing specific positional information.

• Provide evidence for the conservation of chemical signals and reglatory genes used during development.

• Compare and contrast determination and differentiation.

Discuss the relationship between development and evolution.

Lecture Outline Understanding development is one of the greatest challenges in biological science. Development weaves together discoveries from genetics, biochemistry, cell biology, and evolution.

An individual develops as cells: (Table 21.1) 1. Divide 2. Signal to other cells 3. Express genes 4. Move and expand 5. Die

I. Shared Developmental Processes (Table 21.1) A. Cell division 1. For an embryo to grow and develop, new cells must be formed via cell division that is tightly regulated so that it occurs at the right time and in the right location. a. In plants and animals, most cells stop dividing when they mature. b. Some cells are maintained and grow throughout an individual’s life; these are stem cells. .

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c. Differentiation is the process by which an unspecialized cell becomes specialized. d. Location of stem cells: (1) Stem cells in plants are located in meristems.

(2) Stem cells in animals are in a variety of locations: gut, skin, and blood.

B. Cell-cell interactions 1. During development, cells communicate with other cells by sending and receiving signals. a. The signal binds to the surface of a cell, and its message is transduced inside the cell. b. Cells change their activity in response to these signals.

2. These signals usually activate transcription factors that change gene expression.

C. Cell differentiation 1. Differentiation is the process through which a cell becomes a specialized type of cell. a. This is the process of muscle cells becoming muscle cells and nerve cells becoming nerve cells.

2. There are two mechanisms for specifying a cell’s fate: a. Cytoplasmic determinants (Fig. 21.1a) are regulatory molecules that regulate the cell to a distinct developmental pathway. b. Induction (Fig. 21.1b) works through cell-cell interactions.

D. Cell movement and cell expansion 1. In addition to mitosis and apoptosis, normal animal cell development includes cell movement. a. In animal development, cells begin to move and position themselves during gastrulation to form three embryonic tissue layers (skin, gut, and other basic parts). b. During later stages of development, cells break away from their positions and migrate to new positions in the embryo to give rise to new cell types.

2. Unlike animal cells, plant cells do not move during development. a. Plant cells control the orientation of cell division. b. Controlling the orientation of cell division ultimately controls the direction of growth.

E. Programmed cell death 1. In addition to cell division, programmed cell death is important for proper development. a. During development, cell death occurs in a highly regulated process called apoptosis, or programmed cell death. b. In Caenorhabditis elegans, 131 of its roughly 1,000 cells undergo apoptosis.

(See BioSkills 11) c. When cells do not die properly during development, defects and disease result.

(Fig. 21.2)

II. Genetic Equivalence and Differential Gene Expression in Development A. Evidence that differentiated plant cells are genetically equivalent 1. In the 1950s, biologists were able to grow entire tobacco plants or carrots from a single differentiated cell taken from an adult. (BioSkills 11) 202

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2. There are countless examples of gardeners and farmers producing new plants from a small section of root or shoot. (Fig. 21.3) B. Evidence that differentiated animal cells are genetically equivalent 1. Early cloning experiments using frogs confirmed that animal cells are genetically equivalent.

2. The cloning of the first mammal, Dolly the sheep, confirmed that animal cells are genetically equivalent. (Fig. 21.4) a. Ian Wilmut and colleagues (1997) removed a diploid nucleus from a differentiated mammary gland cell and placed it in an enucleated egg.

(1) The egg was from a black-faced sheep.

(2) The nucleus was from a white-faced sheep. b. The resulting embryos were cultured and implanted into the uterus of a surrogate mother. c. After several hundred attempts, Dolly was born.

(1) Genetically, Dolly was the clone of the white-faced sheep.

(2) Dolly was fertile, grew into adulthood, and bore a lamb of her own—Bonnie. d. To date, many other mammals have been cloned (mice, cows, etc.); however, humans have not been cloned.

3. In summary, cloning plants and animals has shown that the process of cellular differentiation does not involve changes in the genetic makeup of cells but is based on differential gene expression.

C. How does differential gene expression occur?

1. Gene expression can be controlled at several different levels. a. Transcriptional control b. Alternative splicing of mRNA c. Selective destruction of mRNAs d. Speeding up or slowing down translation e. Activating or deactivating proteins after they have been translated 2. Transcriptional control is the most important level of control. a. During development, transcriptional control is the fundamental mechanism of differential gene expression. b. This type of control is exerted by regulatory transcription factors that influence chromatin remodeling and bind to promoter-proximal elements and other regulatory sites on DNA. c. To understand differentiation, it is important to understand regulatory transcription factors.

III. Regulatory Cascades Establish a Body Plan A. Signals tell cells their location.

1. Four axes determine a cell’s location: time and three body axes. (Fig. 21.5) a. Anterior−posterior axis b. Ventral−dorsal axis c. Left−right axis 2. Cells know their position in time and space by sending and receiving signals from nearby cells that activate transcription factors and affect gene transcription. .

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a. As development proceeds, certain sets of genes are activated to determine the fate of each cell.

B. Morphogens set up the major body axes.

1. Pattern formation refers to the events that determine the spatial organization of the embryo. a. During development, early signals act as master regulators that set up the three major axes in the embryo. b. These master regulators activate a network of other genes that send signals to indicate the location of cells in space. c. Cells that can detect differences in concentrations to learn their position are called morphogens.

2. Discovery of the bicoid morphogen a. Christiane Nusslein-Volhard and Eric Wieschaus (1970s) generated mutations in fruit flies and looked for body-plan defects in order to identify genes involved in pattern formation. (BioSkills 11) b. The most striking mutations they found were those that altered the segments of the embryo. c. Fruit-fly embryos are divided into head, thoracic, and abdominal segments.

(Fig. 21.6a) d. Embryos with the most dramatic mutation were missing the anterior segments, and those segments were replaced with additional posterior segments. (Fig. 21.6b) (1) Nusslein-Volhard and Wieschaus named this mutant bicoid (two-tailed).

(2) They concluded that bicoid’s gene product (a morphogen) must provide positional information, specifically anterior−posterior axis information. e. Nusslein-Volhard and Wieschaus found that the bicoid mutation was inherited in a very interesting pattern.

(1) Embryos could be homozygous for the mutation and not show the phenotype.

(2) Embryos showed the bicoid phenotype only when the mother was homozygous for the mutation.

(3) They concluded that bicoid mRNA or protein is expressed by the mother and left in the egg to be a morphogen.

3. The importance of morphogen concentration gradients a. Nusslein-Volhard and Wieschaus localized bicoid mRNA in fruit-fly embryos using in situ hybridization. (Fig. 21.7) (BioSkills 6 & 10) (1) This technique allows researchers to pinpoint where and when specific genes are expressed. (a) They made labeled copies of bicoid RNA and then incubated embryos with the labeled RNA. (b) They labeled RNA hybridized to the bicoid mRNA. (2) The technique allows researchers to visualize by fluorescence microscopy how gene expression varies among cells in space. (BioSkills 9) (3) In situ hybridization allows scientists to document differential gene expression. b. They found that bicoid mRNA is localized to the anterior end of the early embryo.

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c. Using antibodies to Bicoid protein, Nusslein-Volhard and Wieschaus studied the localization of Bicoid protein. (Fig. 21.8) (1) They found that Bicoid protein is concentrated in the anterior end but diffuses toward the posterior end.

(2) This creates a concentration gradient of Bicoid protein along the anterior− posterior axis of the embryo. d. All of these data combined indicated that Bicoid protein directs the formation of anterior structures such as the head, whereas absence of Bicoid protein leads to development of posterior segments and structures.

4. Morphogens in plant development a. In plants, the main morphogen is a hormone called auxin. b. Auxin enters cells and triggers the production of transcription factors that affect differentiation. c. Auxin is produced at the tip of the growing embryos, and concentration gradients determine whether cells will become root or shoot.

C. Regulatory genes provide increasingly specific positional information.

1. Nusslein-Volhard and Wieschaus also identified mutants that had normal anterior−posterior development but had defects in how their body segments developed later.

2. These mutants had defective segmentation genes. a. Segmentation genes are regulatory genes that act in a sequence during development. b. Their sequential expression provides progressively detailed information on where cells are in time and space.

3. What do segmentation genes do? a. Segmentation genes are expressed in a particular sequence to define body segments and their boundaries.

(1) The first genes expressed in the sequence are gap genes, whose expression is triggered by the master regulators such as Bicoid. Gap genes define the general position of segments in the anterior, middle, or posterior part of the body.

(Fig. 21.9) (2) Gap genes trigger the expression of pair-rule genes. Pair-rule genes define the boundaries of individual segments within each general region. (Fig. 21.9) (3) Pair-rule genes trigger the expression of segment polarity genes, which delineate the boundaries within individual segments. (Fig. 21.9) b. All of the segmentation genes encode transcription factors that regulate the expression of the next set of genes in the sequence.

4. Specifying segment identity a. Hox genes are turned on by the products of segmentation genes. b. Hox gene products specify each segment’s identity by activating effector genes. c. Homeotic mutants have structures in the wrong segments. (Fig. 21.10) d. Hox genes are distinguished by a sequence called the homeobox, which encodes a DNA-binding domain.

5. Expression of the segmentation genes is an example of a regulatory cascade (segmentation genes regulate each other). (Fig. 21.9)

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a. Master regulators trigger the production of other regulatory signals and transcription factors, which trigger the production of another set of signals and regulatory proteins, and so on. b. Regulatory cascades act in sequence and therefore can provide information about where cells are located in time and space.

(1) The identity and concentration gradients of signals and transcription factors vary along the three major body axes and communicate unique positional information to the cells in those areas.

(2) Each level in a regulatory cascade provides more specific information about where a cell is located.

(3) As regulatory cascades proceed over time, a cell’s fate becomes more and more finely determined.

6. Homeotic mutants that alter flower parts arise from mutations in MADS box genes.

(Fig. 21.11) D. Regulatory genes and signaling molecules are evolutionarily conserved.

1. Researchers have found clusters of Hox genes in virtually every animal studied to date.

2. The chromosomal organization of the Hox genes in all of these different species is strikingly similar.

3. Hox genes are expressed along the head-to-tail axis of mice in the same order as they are in fruit flies. (Fig. 22.12) a. These data indicate that in most animal species, Hox genes play a key role in the position of cells along the head-to-tail axis of the body. b. Biologists also hypothesize that the Hox genes are homologous, meaning that they descended from genes in a common ancestor. c. Other regulatory pathway molecules are highly conserved as well. d. Although animal bodies are very diverse, the underlying mechanisms responsible for their development are similar. e. The location of Hox genes on an organism’s chromosome correlates with the Hox gene pattern of expression in that organism’s embryos.

E. One regulator may be used many different ways.

1. During development, the same regulatory transcription factors and cell-cell signals are used in a variety of contexts. a. Regulatory proteins are not only conserved but also reused. b. Hox genes involved in the development of animal appendages are also involved in defining the anterior−posterior axis of the animal body.

2. Multicellular organisms have a toolkit of common signals, pathways, and regulatory proteins used over and over during development and can direct the development of different structures based on when and where those genes are expressed.

IV. Cells Are Determined Before They Differentiate A. Determination: the commitment of a cell to a particular developmental fate 1. Transplantation studies have shown researchers that cells gradually become determined.

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2. Before determination, cells transplanted to a new location can take on the fate appropriate for that location, but after determination cells do not change their fate.

3. Animal cells play by the rule: Once determined, (nearly) always determined. Plant cells seem to change their fate.

4. An exception in animal cells, tumor cells B. Master regulators of differentiation and development 1. Research in myoblasts by Harold Weintraub illustrated what changes inside a cell occur for it to follow a particular pathway of development. (Fig. 21.13) 2. The search for this regulatory protein was done by: a. Isolating mRNA from myoblasts b. Using reverse transcriptase to convert the mRNA to cDNAs (BioSkills 10) c. Attaching a promotor to the cDNA to ensure expression d. Introducing these modified cDNAs into muscle cells called fibroblasts e. Converting these cDNAs fibroblasts into muscle-like cells 3. Weintraub’s group called the protein produced MyoD. The MyoD gene encodes a regulatory transcription factor and binds to enhancer elements located upstream of muscle-specific genes.

4. MyoD is a master regulator. It can unleash a series of events that produce a specialized cell type, tissue, or body structure.

C. Stem cell therapy 1. Shinya Yananaka and his colleagues were able to convert differentiated adult cells into embryo-like cells by causing them to express specific regulatory transcription factors and de-differentiate back to an early embryonic cell state. These cells are called iPS cells.

2. This research holds promise for diseases caused by damaged or destroyed cells.

V. Changes in Developmental Gene Expression Drive Evolutionary Change A. Modifications to developmental pathways lead to phenotype changes in the adult organism.

1. Genetic changes in regulatory signals and pathways that alter developmental processes are the foundation of evolutionary change.

2. Evo-devo is the research field that focuses on understanding how changes in developmentally important genes have led to the evolution of new phenotypes.

3. Limb loss in snakes a. Snakes are clearly tetrapods. b. Some snakes have rudimentary pelvic and thigh bones. c. How did snakes lose the tetrapod limbs?

(1) In tetrapods, Hoxc6 and Hoxc8 are expressed together in vertebrae that form ribs, and Hoxc6 alone is expressed in the region that gives rise to forelimbs.

(Fig. 21.14a) (2) In snakes, Hoxc6 and Hoxc8 are expressed together throughout the part of the trunk where forelimbs should form. (Fig. 21.14b) (3) These data suggest that a change in the regulation of Hoxc6 led to the evolutionary loss of the forelimb.

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(4) Hindlimb loss in snakes seems to have resulted from changes in a signaling molecule encoded by the gene Sonic hedgehog.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. apoptosis/programmed cell death Bicoid protein body axes anterior posterior dorsal ventral cell proliferation cleavage plane clone cloning cytoplasmic determinants de-differentiating determination development developmental biology

differential gene expression

master regulator

differentiation embryo evo-devo gap genes gastrulation genes genetic regulatory cascade genetically equivalent homeobox homeotic mutations homologous Hox genes in situ hybridization induced pluripotent stem

meristems mitosis mitosis-promoting factor (MPF) morphogens pair-rule genes pattern formation regulatory transcription factors regulatory genes segment polarity segmentation genes segments stem cells toolkit genes zygote

cell (iPS cells) induction

Lecture Activities Understanding the Role of Genes in the Body Plan Estimated duration of activity: 10–15 minutes The purpose of this activity is to solidify students’ understanding of the sequential roles of maternal genes, gap genes, pair-rule genes, segment polarity genes, and homeotic genes in establishing the body pattern in Drosophila.

Instruct your students to draw a diagram explaining in as much detail as possible the cascade of events from maternal genes to homeotic genes as the body plan is created in Drosophila. After allowing the class about 5 minutes, select one student to draw her or his diagram at the board. You might encourage a volunteer who feels uncertain about the principles and would therefore like help from the class. Ask that student to explain vocally the rationale for each element and step of the diagram. The remaining students monitor the volunteer’s drawing, ask questions when they disagree or are confused, and offer suggestions when the volunteer gets stuck or requests help. Periodically ask questions relevant to the diagram to assess the group’s level of understanding and to keep students engaged. Also encourage all students to complete and/or modify their own diagrams as they participate in the discussion.

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The Four Fundamental Principles of Development Estimated duration of activity: 15 minutes For this activity, divide the class into groups of four. Students designate themselves as Student A, Student B, Student C, and Student D, according to the alphabetical order of their names. Then instruct them to take turns explaining one of the four fundamental principles of development as outlined in the chapter. Their explanation should take the following form: 1. Definition 2. Explanation of how the principle contributes to organismal development 3. Specific example of the principle After each student finishes her or his explanation, the other members of the group should comment, elaborate on, and/or repair misconceptions. Allow 2 or 3 minutes for each principle, and keep track of time and announce when the groups are expected to rotate to the next principle.

Assignments: Student A: Cell proliferation and programmed cell death Student B: Cell movement and expansion Student C: Cell differentiation Student D: Cell-cell interactions At the end of the activity, invite the class to share their general questions and requests for clarification. Note that this activity will be more effective if textbooks and notes are not used for prompting.

Practicing the Fundamentals of Body-Pattern Development Estimated duration of activity: 10–15 minutes For this activity, students work individually and in pairs. Provide a series of scenarios related to the fundamentals of body-pattern development. For each scenario, students are expected to predict the outcome. Present the scenarios one at a time. After each scenario, give students 1 or 2 minutes to make their prediction. They must write their prediction on a sheet of paper. Then give students a minute to share their answer with a partner and negotiate any differences. They should also explain the rationale for their predictions at this time. Give the correct prediction with an explanation. Students are given another minute of discussion with their partner to repair and clarify their understanding. Answer questions from the class at this time. The sequence is then repeated with a new scenario. Students should attempt their predictions without relying on the textbook or their notes.

Sample scenarios: 1. Suppose that when both alleles of a certain gene involved in development are inactive, several aspects of the body pattern are grossly altered in the embryo. Predict the type of function associated with the normal product of this gene.

2. Suppose a mutation in a Drosophila homeotic gene causes that gene to be activated by a gap gene to which it normally does not respond. Predict how the embryo is likely to be altered.

3. Suppose a Drosophila egg has an excessive amount of the Bicoid protein in the posterior end. Predict the phenotype of the embryo. .

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4. Suppose the following pairings of Drosophila with the indicated genotypes for bicoid are mated. Predict the phenotypes of the offspring.

• male Bb mated with female Bb • male bb mated with female Bb • male Bb mated with female bb 5. Suppose an embryo inherits two dysfunctional (inactive) alleles of a homeotic gene.

Predict the likely phenotype of the embryo.

Analyzing Primary Literature: Is Drosophila melanogaster really a good model organism for pattern formation?

Estimated duration of activity: 30 minutes (also a good out-of-class assignment) Chapter 21 contains some information supporting D. melanogaster as a model organism. The data imply that other related organisms are similar and that even less related organisms have some commonalities in their development.

Students might benefit from looking at research on some aspect of pattern formation in another organism. They can see that when investigators have studied other organisms using the same techniques, they have obtained similar results. Have your students do group presentations on primary research articles by reading primary scientific literature, interpreting it (with your help), and explaining their findings to the rest of the class.

Example 1: Abzhanov, Arhat, Aleksandar Popadic, and Thomas C. Kaufman. “Chelicerate Hox genes and the homology of arthropod segments.” Evolution and Development 1(2): 77–89 (1999).

Because insects and other arthropods develop differently from Drosophila, the purpose of this study was to see whether there is homology of segments based on the anterior boundaries of the expression of Hox genes. The investigators compared the expression patterns of UBX and ABD-A in chelicerates, myriapods, crustaceans, and insects. Results of the comparisons are that the anterior boundary has changed during the evolution of arthropods. Therefore, when we are trying to find homology in this group of organisms, it is not enough to use only Hox gene expression patterns.

The paper presents some original work on spiders and an excellent diagram showing the five organisms (four groups). Some excellent photographs of in situ hybridizations are included. The anterior border of the homeotic complex in a common ancestor was probably at the Mx2/T1 junction, but descendant lines have changed and are variable. The general patterning is similar, confirming the chapter conclusion; but certainly even these closely related organisms have important variations and cannot be directly compared.

Example 2: Patel, Nipam H. “It’s a bug’s life.” Proceedings of the National Academy of Sciences 97(9): 4442–4444 (2000).

Patel has published many papers on evolutionary comparisons of insects. This is a short piece that mentions some important points. Insecta is a diverse group with a million species, so variation should be expected. The segment polarity and homeotic classes of genes are fairly well conserved. The gap genes and pair-rule genes have more variability. More primitive insects, which might be thought of more as a common ancestor, seem to have a posterior patterning system rather than bicoid. Bicoid might be a relatively new gene, possibly a duplicate of a Hox3 gene (zen) cloned from another dipteran. The zen gene might serve two purposes in Tribolium, the flour beetle. In grasshoppers, a posterior gradient is certainly what sets up the anterior−posterior axis with no input from an anterior system. Because the more 210

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primitive insects are cellular at the time of a gradient, not a syncytium like Drosophila, a gradient will be different in whatever species and wherever it is located. The paper includes a few grasshopper pictures and a proposed insect phylogeny.

You can find many good and readable articles in Evolution and Development, Proceedings of the National Academy of Sciences, American Zoologist, Science, and Development, among other journals. The existence of variations does not diminish the use of “model organisms.”

But it is good to know there are limitations and that development can be variable even if the adult forms are similar. The developmental information can inform phylogenetic comparisons.

Discussion Ideas for Any Class Size Chapter 21 is a rich source of socially interesting science. Even nonmajors who fear taking a biology course perk up when issues of stem cell research and human cloning are mentioned.

How much more excited will science majors become about their chosen field of study if they are given the opportunity to participate in such discussions? The following sets of questions (along with the essays and box text in the chapter) can serve as springboards from which you can design and implement an in-class debate, an online debate (if you use Blackboard or some other type of classroom intranet), group presentations, or individual position papers.

Stem Cell Research Stem cell research is a scientific issue that has been front-page news in recent years. Have students answer the following questions in class (using information they have learned from the chapter), then ask them to make a list of their questions about stem cell research. Instruct each student to choose a question to research and have him or her report findings in the next class. Alternatively, you can tell students to make individual lists of questions to research in preparation for an in-class debate at your next meeting. If class time is too precious to devote to an in-class debate, set up an online debate that will take place over a week’s time. Tell students they will be graded on their participation, research effort, and ability to explain themselves in writing.

Essay review questions to stimulate discussion: • What are stem cells, and where are they found in an embryo? In an adult?

• What properties do stem cells have that make them attractive for medical applications?

• What types of diseases can stem cells be used to cure or treat? How might stem cells work for these diseases?

• What is the current source of stem cells? Why is this so controversial?

An assignment involving the subject of stem cell research will allow your students to apply the following concepts from Chapter 21: totipotency determination

positional information cell fate

differentiation

induction

Here are some references that may help you (and your students) prepare for the stem cell discussion.

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General research with stem cells: http://www.stemcellresearch.org/ Lovell-Badge, R. “The future of stem cell research.” Nature 414: 88–91 (2001). Lumelsky, N., et al. “Differentiation of embryonic stem cells to insulin-secreting structures similar to pancreatic islets.” Science 292: 1389–1394 (2001).

Humpherys, D., et al. “Epigenetic instability in ES cells and cloned mice.” Science 293: 95–97 (2001).

Anderson, D. J. “Stem cells and pattern formation in the nervous system: The possible versus the actual.” Neuron 30: 19–35 (2001).

Anderson, D. J., F. H. Gage, and I. L. Weissman. “Can stem cells cross lineage boundaries?”

Nature Medicine 7: 393–395 (2001).

Shamblott, M. J., et al. “Derivation of pluripotent stem cells from cultured human primordial germ cells.” Proceedings of the National Academy of Sciences USA 95: 13726–13731 (1998).

Wilmut, I., et al. “Viable offspring derived from fetal and adult mammalian cells.” Nature 385: 810–813 (1997).

Thomson, J. A., et al. “Embryonic stem cell lines derived from human blastocysts.” Science 282: 1145–1147 (1998).

Pardal, R., M. Clarke, and S. Morrison. ”Applying the principles of stem-cell biology to cancer.” Nature Reviews Cancer 3: 895–902 (2003).

Pluchino, S., et al. “Injection of adult neurospheres induces recovery in a chronic model of multiple sclerosis.” Nature 422: 688–694 (2003).

McKay, R., et al. “Dopamine neurons derived from embryonic stem cells function in an animal model of Parkinson’s disease.” Nature 417 (published online 20 June; doi:10.1038/nature00900) (2002).

Verfaille, C. M., et al. “Pluripotency of mesenchymal stem cells derived from adult marrow.”

Nature 417 (published online 20 June; doi:10.1038/nature00870) (2002).

Cibelli, J. B., et al. “Parthenogenetic stem cells in nonhuman primates.” Science 295: 465 (2002).

Ready, T. “Private donors breathe new life into U.S. stem cell research.” Nature Medicine 10: 320.

Research with adult stem cells: Clarke, D. L., et al. “Generalized potential of adult neural stem cells.” Science 288: 1660–1663 (2000).

Ende, N. “The Berashis cell: A review. Is it similar to the embryonic stem cell?” Journal of Medicine 31: 113–130 (2000).

Lu, S., and N. Ende. “Potential for clinical use of viable pluripotent progenitor cells in blood bank stored human umbilical cord blood.” Life Sciences 61: 1113–1123 (1997).

Zuk, P. A., et al. “Multilineage cells from human adipose tissue: Implications for cell-based therapies.” Tissue Engineering 7: 211–228 (2001).

Norton, A. “Stem cells from body fat—limitless supply.” Reuters Health, October 18 (2000).

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Beltrami, A. P., et al. “Evidence that human cardiac myocytes divide after myocardial infarction.” New England Journal of Medicine 344: 1750–1757 (2001).

Malouf, N. N., et al. “Adult-derived stem cells from the liver become myocytes in the heart in vivo.” American Journal of Pathology 158: 1929–1935 (2001).

Woodbury, D., et al. “Adult rat and human bone marrow stromal cells differentiate into neurons.” Journal of Neuroscience Research 61: 364–370 (2000).

Jackson, K. A., et al. “Regeneration of ischemic cardiac muscle and vascular endothelium by adult stem cells.” Journal of Clinical Investigation 107: 1395–1402 (2001).

Slosberg, E. D., et al. “Treatment of type 2 diabetes by adenoviral-mediated overexpression of the glucokinase regulatory protein.” Diabetes 50: 1813–1820 (2001).

Human Cloning Everyone is curious about human cloning. A discussion, debate, or writing assignment on this subject is a great way to capture students’ interest. Information about cloning in Chapter 21 will give your students some introductory information but will likely not quench their thirst or answer all of their questions about human cloning. As suggested above, you can use the following questions to get the discussion started and then ask students to pose their own questions. These questions can be the basis of an in-class or online debate or individual position papers.

Questions to get the discussion started: • What does it mean to clone an organism?

• Have you ever cloned anything? (plant cuttings) • What is the difference between cloning a gene and cloning an organism?

• Explain the technique by which the first mammal was cloned.

• How did the scientist confirm that the animal was successfully cloned?

• Why was it important for Dolly to reproduce?

• What other mammals have been cloned?

• Why are scientists interested in cloning animals?

• What are some of the positive motivations for cloning humans?

• What are some of the dangers of cloning humans?

• What are some of the technical limitations of cloning humans?

• What are some of the ethical concerns involved in cloning humans?

The following references may help you and your students prepare for this activity.

Westhusin, M. “Human cloning and human dignity: The report of the President’s Council on Bioethics.” Nature Medicine 9: 502 (2003).

Rhind, S. M., et al. “Human cloning: Can it be made safe?” Nature Reviews Genetics 4: 855–864 (2003).

Hwang, W. S., et al. “Evidence of a pluripotent human embryonic stem cell line derived from a cloned blastocyst.” Sciencexpress (published online; doi:10.1126/science.1094515) (2004).

Harris, J., and T. Takala. “A clear view of cloning.” Nature 428: 802–803 (2004). .

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CHAPTER

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Evolution by Natural Selection

Learning Objectives: Students should be able to...

• Contrast typological thinking, Lamarkian evolution, and Darwinian evolution.

• Describe the evidence that supports the theory of evolution. Define evolution, fitness, • population, and adaptation from a biological perspective. Explain Darwin’s four postulates • of natural selection and apply them to a scenario. Identify common misconceptions about • evolution and use evidence to explain why these misconceptions are innacurate.

Lecture Outline • Evolution is one of the best-supported and most important theories in the history of science.

• Evolution is one of the five attributes of life.

• Evolution has both a pattern and a process.

I. The Evolution of Evolutionary Thought A. Plato and typological thinking 1. Plato saw species as unchanging, perfect “types” created by God. (Fig. 22.1a) 2. Plato thought individual variation was an unimportant deviation from the true “type.”

B. Aristotle and the great chain of being (scale of nature) 1. Aristotle, like Plato, thought species were unchanging types.

2. Aristotle thought species could be organized into a sequence or ladder of increasing complexity, with humans at the top. (Fig. 22.1b) C. Lamarck and the idea of evolution as change through time 1. Lamarck noticed that organisms changed over time.

2. Lamarck thought animals progressed over time from “lower” to “higher” forms (like Aristotle’s ladder) via inheritance of acquired characteristics. (Fig. 22.1c) D. Darwin and Wallace and evolution by natural selection 1. Species change over time, but they do not “progress.” They are related by a common ancestry. (Fig. 22.1d) 2. A species does not have a single true “type.”

3. Individual variation is important; variation is what drives evolution.

4. This theory made predictions and was testable; that is, it was scientific.

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II. The Pattern of Evolution: Have Species Changed, and Are They Related?

A. Two predictions of Darwin’s theory: 1. Species change through time.

2. Species are related by common ancestry.

B. Evidence of change through time 1. The vastness of geologic time a. A fossil is any trace of an organism that lived in the past. (Fig. 22.2) b. The fossil record was initially organized based on the relative age of the fossils. c. The geologic time scale (1) Sedimentary rocks form layers over long times. These layers form in a chronological sequence (the geologic time scale).

(2) From the number of layers and the time it takes to deposit each one, geologists realized that Earth must be very old. d. Radiometric dating enables us to date rocks directly.

(1) Earth is 4.6 billion years old.

(2) The earliest signs of life are in rocks that are 3.4 to 3.8 billion years old.

2. Extinction changes the species present over time. a. The fossil record shows that more than 99 percent of all the species that have ever lived are now extinct. (Fig. 22.3) b. This is evidence that the species composition on Earth has changed over time.

3. Transitional features link older and younger species. a. Law of succession: Fossils found in a certain geographic region frequently resemble the species currently living in that region. This is evidence that the extinct species are related to existing species. b. Fossils with transitional features (traits intermediate between those of older and younger species) are compelling evidence that species change over time. Example: the fins-to-feet transition (Fig. 22.4) 4. Vestigial traits are evidence of change through time. a. Vestigial traits are reduced incompletely developed structures that have no function, or a reduced function, but are clearly related to functioning organs in related species. (Fig. 22.5) b. The existence of these traits challenges the theory of special creation that organisms were designed by a supernatural being and are static. c. Biologists interpret the existence of these traits as evidence that organisms change over time.

5. Current examples of change through time a. Biologists have documented hundreds of contemporary populations that are changing in response to their environment.

C. Evidence of descent from a common ancestor 1. Similar species are found in the same geographic area. a. Similar, but distinct, species are often found living close together in the same geographic area, implying that they are linked by a common ancestor. Example: Galápagos mockingbirds (Fig. 22.6a)

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b. These similar species can be diagrammed on a phylogenetic tree (a branching diagram that indicates genealogy). (Fig. 22.6b) 2. Similar species share homologies a. Homology can be recognized and studied at three levels (Table 22.1) (Making Models 22.1): (1) Genetic homology—similarities in DNA sequences (2) Developmental homology—similarities in the morphology of embryos and the fate of embryonic tissues (3) Structural homology—similarities in the structure of body parts b. The three levels of homology interact: Genetic homologies cause developmental homologies, which cause structural homologies. c. Hypotheses about homology can be tested experimentally. (Fig. 22.7) d. Homology is used extensively in contemporary biology. Examples: use of model organisms, comparative genomics (BioSkills 11) 3. Current examples of descent from a common ancestor a. Biologists have documented dozens of examples of populations undergoing speciation.

D. Evolution’s “internal consistency”⎯ the importance of independent data sets 1. When data from independent sources support a theory, this is powerful evidence that the theory is correct. (Table 22.2) a. Example: evolution of cetaceans from a terrestrial ancestor (DNA, morphology, fossil record, and other evidence all agree) (Fig. 22.8) 2. The theory of evolution by natural selection is much more consistent with the data than the pattern predicted by special creation.

E. Students should be able to determine what kind of evidence supports the hypothesis that birds evolved from dinosaurs and why the DNA sequences of chimps and humans are 96 percent similar.

III. The Process of Evolution: How Does Natural Selection Work?

A. Darwin’s inspiration 1. Darwin crossbred pigeons to observe the effects of artificial selection. (Fig. 22.9) 2. Thomas Robert Malthus’s study of human population growth inspired the idea that there is a “struggle for existence.”

B. Darwin’s four postulates, the outcome of which is evolution (Making Models 22.2): 1. Individuals vary.

2. Some trait differences are heritable.

3. More offspring are produced than can survive.

4. Individuals with traits that confer an advantage are more likely to survive and reproduce.

5. Summary: Evolution by natural selection occurs whenever heritable variation leads to differential success in survival and reproduction.

C. The biological definitions of fitness, adaptation, and selection 1. Fitness is the ability of an individual to survive and reproduce, relative to other individuals in that population.

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2. Adaptation is a heritable trait that increases the fitness of an individual in a particular environment, relative to other individuals lacking the trait.

3. Artificial selection is the process of breeders choosing characteristics in the animals or plants they breed. Natural selection is a passive process by which differential reproduction is a result of heritable variation.

IV. Evolution in Action: Recent Research on Natural Selection A. Case Study 1: How did Mycobacterium tuberculosis become resistant to antibiotics?

1. A patient history a. A patient with active tuberculosis (TB) was given the antibiotic rifampin for 40 weeks and then released when lung cultures showed no bacteria. b. Two months later, the patient had a relapse and died of rifampin-resistant TB. 2. A mutation in a bacterial gene confers resistance. a. DNA analysis showed that the patient’s TB bacteria had acquired one new mutation in the gene for RNA polymerase, the same enzyme that is the target of rifampin. b. These bacteria were likely present at low frequency before the rifampin treatment. c. When the rifampin treatment began, bacteria without the mutation were killed off, and only the bacteria with the mutation survived. (Fig. 22.10) d. Students should be able to explain why the patient relapsed and whether a family member or healthcare worker who got TB from the patient after the relapse would respond to drug therapy.

3. Testing Darwin’s postulates a. Did variation exist in the population? Yes. Research shows that populations of TB bacteria do have variation for rifampin resistance. b. Was the variation heritable? Yes. Variation in rifampin resistance (the phenotype) is due directly to genotype. c. Was there variation in survival and reproductive success? Yes. Once rifampin treatment began, only a few bacteria survived to reproduce. d. Were survival and reproduction nonrandom? Yes. Certain bacteria (those with the drug-resistant allele) were much more likely to survive and reproduce.

4. Drug resistance: A widespread problem a. Resistance (to drugs, insecticides, etc.) has evolved repeatedly in many species and is a growing public health problem. Example: Staphylococcus aureus resistant to vancomycin (Fig. 22.11) b. Students should be able to explain why the overprescription of antibiotics by doctors, or the overuse of everyday soaps and cleaners laced with antibiotics, can be a health risk.

B. Case Study 2: Why do beak sizes and shapes vary in Galápagos finches?

1. The medium ground finch population of the island Daphne Major varies in beak shape and body size, and these traits are heritable. (Fig. 22.12) 2. Selection during drought conditions a. Drought survivors had deeper beaks. (Fig. 22.13) b. Deeper beaks allowed those individuals to eat the only remaining seeds, which were tough and difficult to crack.

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c. In the next generation, the average beak size was considerably larger than it had been before the drought. (BioSkills 2) d. Summary: In only one generation, natural selection had led to a measurable change in the characteristics of the population.

3. Continued changes in the environment, continued selection, continued evolution a. In a subsequent rainy year, selection favored individuals with small, pointy beaks. b. Over 30 years of study, researchers have documented continued evolution in response to environmental events. (Fig. 25.17) 4. Which genes are under selection? a. Recent research has identified several genes that affect beak length in development, such as Bmp4. (Fig. 22.15)

V. Debunking Common Myths about Natural Selection and Adaptation A. Natural selection does not change individuals. (Table 22.3) 1. During the evolutionary process, individuals do not change, only the population changes. Examples: Galápagos finches, TB bacteria 2. Natural selection is not “Lamarkian” inheritance. a. Lamark proposed that individuals change in response to challenges posed by the environment and these changes are passed on to their offspring. b. Darwin realized that individuals do not change when selected but merely produce more or less offspring than others. c. Natural selection just sorts existing variants; it doesn’t change them.

3. Acclimatization is not adaptation. a. Individuals may change during their lifetime (acclimatization), but these changes are not passed on to offspring. (When this process occurs in a laboratory, it is called acclimation.) b. For example, if you travel to the Tibetan Plataeu, your body would acclimatize to high altitude. In contrast, populations that have lived in the Tibetan Plateau have adapted over time through genetic changes that are heritable. (Fig. 22.16) c. Students should be able to explain the difference between the evolutionary definition of adaptation and its use in everyday English.

B. Natural selection is not goal directed.

1. Mutations occur randomly; they do not occur because organisms “want” or “need” them.

2. Students should be able to explain how thorns originated in roses.

3. Evolution is not “progressive.” a. Evolution does not always result in “better” or “more advanced” organisms. b. Complex traits are often lost in evolution.

4. There is no such thing as a “higher” or “lower” organism. a. Evolution is not a “ladder”; it is more like a tree. b. Evolution simply results in organisms adapted to live in different environments; no organism is “higher” than another. (Making Models 22.3)

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C. Natural selection does not always lead to perfection.

1. Traits are not always adaptive. a. Evolution does not lead to perfect organisms. Examples: vestigial organs, silent mutations 2. Fitness trade-offs exist. a. Natural selection often results in a compromise between traits with different effects. b. Many types of trade-offs occur due to energetic constraints; resources are limited.

3. Traits are genetically constrained. a. Sometimes nonoptimal traits are propagated because they are genetically linked with an optimal trait (genetic correlation). b. One example is finch beaks. The best beak would have been a narrow deep beak, but what evolved was a wide deep beak because beak depth and beak width are genetically linked. c. Lack of genetic variation can also limit evolution.

4. Traits are historically constrained. a. Natural selection can act only on traits that existed in the ancestral population. b. The tiny ear bones in the mammalian middle ear evolved from existing jaw bones or bones supporting the jaw of the ancestors of mammals.

5. Traits are environmentally constrained. a. Natural selection often occurs in the context of a changing environment. b. In the finch example, beak shape that may be adaptive in one season may not be adaptive in the next season. c. Some environmental events are so catastrophic that organisms are wiped out regardless of which adaptations they have.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. absolute dating acclimation acclimatization adaptation allele frequencies antibiotic biological fitness cetaceans descent with modification developmental homologies evolution

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extant extinct fitness trade-off fossil fossil record genetic constraint genetic correlation genetic homologies geologic time scale great chain of being heritable traits heritable variation historical constraint

homology inheritance of acquired characters internal consistency law of succession morphology natural experiment natural selection phylogenetic tree phylogeny population population thinking radiometric dating

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relative dating resistance sedimentary rocks special creation

structural homologies transitional features tuberculosis typological thinking

vertebrates vestigial traits

Lecture Activities Practicing BioSkills: Making Concept Maps: “The Big Picture” of Evolution (see BioSkills 12 in Appendix B) Estimated duration of activity: 10–15 minutes Drawing a concept map can help students review and grasp the many new terms and concepts associated with evolution. If the class is not familiar with concept maps, you may wish to model the idea by starting a map yourself, soliciting increasing input from students as the map progresses.

If students are already familiar with concept maps, have them break into small groups and spend 5 minutes drawing their own evolution concept map. Give them a starting list of terms (see below) and ask them to add more terms on their own. Afterward, poll the class for ideas about what new terms they added. Collect the ideas into a single concept map. You can save this map and add to it in later lectures as students learn more about evolution.

Note: Refer students to “The Big Picture” concept map on evolution presented in the textbook. At this point, they will not yet know many of the terms on this map, but encourage them to look at the map as a preview of things to come. You can return to this in subsequent lectures.

Suggested starting list of terms and phrases: evolution, change in allele frequencies, evidence for common ancestry, evidence that species change, examples of natural selection, vestigial traits, transitional fossils, homology, adaptation, fitness, variation, species

Distinguishing between Lamarckian Conclusions and Natural Selection Estimated duration of activity: 5–10 minutes Many students have a tendency to conclude that a population will adapt to its environment because it “needs” to. In other words, they tend to see adaptation the way Lamarck did instead of the way Darwin did. This exercise will give your students practice in correcting those mistaken conclusions.

Provide students with the following observations, and have them explain them first in Lamarckian terms and then in terms of natural selection.

• Over time, the average length of giraffe necks has increased.

• After the Industrial Revolution, moths changed from being mostly light to being mostly dark.

• The incidence of antibiotic-resistant bacteria is on the rise. • Heterozygotes carrying the gene for sickle-cell anemia are resistant to malaria. Answer example: The Lamarckian interpretation for giraffe necks: Individual short-necked giraffes stretched their necks by reaching for food that was high in trees. They passed this new acquired trait on to their offspring, which were born with long necks. The Darwinian 220

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interpretation: The ancestral giraffe population happened to have some short-necked giraffes and some slightly longer-necked giraffes. The giraffes with slightly longer necks could reach more food (or could win more fights, etc.) and had superior survival and reproduction. Over time, the long-necked giraffes became more common in the population. Repeated over many generations, this resulted in evolution of longer and longer necks.

Identifying Evolution in Action Estimated duration of activity: 5–10 minutes Provide your students with the following list, and ask them to identify which items are most likely to result in evolution, and why. a. Insecticide is applied to a population of mosquitoes, most of which are initially killed upon exposure to the chemical, but a few survive and reproduce. b. A man spends many hours in the gym exercising and is able to increase his body weight, muscle mass, and strength, which makes him more attractive to the opposite sex. c. Some male individuals in a population of kangaroos are born with larger front paws and longer forearms. This gives them an advantage when boxing other males to compete for mates. d. Toxic levels of heavy metals are dumped as waste from coal mines. Initially, no plants will grow on the waste dump, but eventually some grasses begin to colonize the dump sites and reproduce. e. A pregnant woman spends too much time in the sun and suffers severe sunburn. The sunlight causes genetic changes to the DNA in her skin. f. A small group of beetle-eating birds is blown to an island in the ocean by a hurricane. The new environment lacks the beetles they normally eat but has various types of hard-shelled nuts. Some of the birds have larger, more pointed beaks and are able to crack the nuts open.

Answer: Items a, c, d, and f are all situations in which evolution by natural selection might occur. Items b and e are examples of Lamarckian theory; they will not lead to changes in a population because the changes that have occurred are not heritable.

Computer Simulations of Natural Selection and Darwin’s Postulates Estimated duration of activity: 20 minutes There are many computerized models of natural selection that show a simulated population changing in real time in response to selection. Students enjoy watching and experimenting with these real-time simulations, and it helps them gain a more intuitive sense of how natural selection really works. One example is Simbio’s “Darwinian Snails” simulation, which models how a population of snails with variable shell thickness will change in response to predation from a crab. If you have a computer display available in class, you can use this simulation to ask students to make predictions about what will happen if one or more of Darwin’s postulates is not true. For example, if snails vary in shell thickness but the variation is not heritable, will evolution occur? If selection is random (all snails have an equal chance of being eaten), will evolution occur?

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For additional student involvement, call students up to the front of the room to control the cursor and be “the crab.” If multiple computers are available, Darwinian Snails also works well as a game for small groups of students.

(Darwinian Snails is available for $3 to $5 (at the time of this writing) from http://simbio.com/products-college/EvoBeaker.)

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CHAPTER

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Evolutionary Processes

Learning Objectives: Students should be able to...

• Apply the Hardy-Weinberg principle to determine if natural selection is occurring on a particular gene.

• Describe the four evolutionary processes, including their effects on genetic variation and average fitness.

• Explain how nonrandom mating can affect genotype frequencies without contributing to •

evolution.

Contrast directional selection, stabilizing selection, disruptive selection, and balancing

• selection.

• Explain sexual selection and discuss why it tends to affect one sex more than the other.

• Explain genetic drift and describe the circumstances under which it is likely to cause evolution in natural populations.

• Explain gene flow and its effects on populations.

Explain the importance of mutation to evolution.

Lecture Outline A. Mechanisms of evolutionary change: • Natural selection increases the frequency of alleles that contribute to improved reproductive success.

• Genetic drift changes allele frequencies randomly.

• Gene flow changes allele frequencies when individuals join or leave a population.

• Mutation continually introduces new alleles.

I. Analyzing Change in Allele Frequencies: The Hardy-Weinberg Principle A. The gene pool concept and allele frequencies (Fig. 23.1) 1. The gene pool is all the gametes produced in one generation.

2. The symbols p and q are traditionally used to depict allele frequencies. a. p = frequency of parental A1 allele b. q = frequency of parental A2 allele 3. The frequencies must add up to 1. a. If there are just two alleles, p + q = 1.

4. Gametes are assumed to combine randomly.

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a. Genotype frequencies in the offspring can be calculated based on probabilities of gametes combining. (See BioSkills 4) 5. Deriving the Hardy-Weinberg principle (Quantitative Methods 23.1) a. Consider a population in which all gametes combine randomly. b. Prediction 1: Genotype frequencies in the offspring can be calculated from the parental allele frequencies as follows: 2 (1) The predicted frequency of the offspring A 1A1 genotype is p.

(2) The predicted frequency of the offspring A1A2 genotype is 2pq.

2 (3) The predicted frequency of the offspring A 2A2 genotype is q.

(4) The frequencies of all genotypes add up to 1, so p2 + 2pq + q2 = 1. c. Prediction 2: Offspring allele frequencies are the same as parental allele frequencies.

(1) In the absence of evolutionary forces, allele frequencies stay wherever they were originally.

(2) Allele frequencies do not tend to move toward 0.5.

(3) Dominant alleles do not tend to increase in frequency. d. Students should be able to: (1) calculate the frequencies of the three offspring genotypes if p = 0.6 and q = 0.4 in the parental gene pool; (2) use the answer to calculate the allele frequencies in the offspring gene pool; (3) determine whether evolution occurred.

6. For evolution to occur (that is, for allele frequencies to change), some outside factor must be interfering with the random combination of gametes that the model assumed at the beginning.

7. Any population that does not match the two predictions is evolving, is experiencing nonrandom mating, or both.

B. The Hardy-Weinberg model makes important assumptions.

1. For a population to be in Hardy-Weinberg equilibrium, there must be: a. Random mating (with respect to the gene in question) b. No natural selection c. No genetic drift (no random allele frequency changes) d. No gene flow via immigration or emigration e. No mutation C. How do biologists apply the Hardy-Weinberg principle to real populations?

1. The Hardy-Weinberg principle tells us what to expect if no evolution is occurring and if there is random mating.

2. Case study 1: Are MN blood-type alleles in humans in Hardy-Weinberg equilibrium? a. For every human population surveyed, genotypes at the MN locus satisfy the Hardy-Weinberg equation. (Table 23.1) b. Therefore, the M and N alleles are not currently affected by any of the four evolutionary mechanisms, and mating is random (i.e., humans do not choose their mates based on blood type).

3. Case study 2: Are HLA alleles in humans in Hardy-Weinberg equilibrium? a. Allele frequencies were calculated from observed genotype frequencies in a tribe in Arizona.

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b. The observed values did not satisfy the Hardy-Weinberg equation. There were too many heterozygotes and not enough homozygotes. (Table 23.2) c. Which of the Hardy-Weinberg assumptions was violated?

(1) Mutation, drift, and migration were thought to be negligible.

(2) Mating may not be random with respect to the HLA genotype. (i) There is evidence that people are attracted to the scent of mates whose HLA genotypes are different from their own.

(3) Heterozygous individuals may have stronger immune systems and may be healthier and more reproductively successful (heterozygote advantage).

II. Nonrandom Mating A. How does inbreeding affect allele frequencies and genotype frequencies?

1. Inbreeding is nonrandom mating between relatives.

2. Effect on allele frequencies and genotype frequencies: a. Inbreeding increases homozygosity over time. (Fig. 23.3) b. Inbreeding does not cause evolution because it doesn’t affect allele frequencies.

3. Students should be able to predict how observed genotype and allele frequencies differ from those under the Hardy-Weinberg principle when inbreeding is occurring.

B. How does inbreeding influence evolution?

1. Inbreeding depression a. Inbreeding depression is a decline in average fitness when homozygosity increases. b. There are two causes of inbreeding depression: (1) Recessive loss-of-function alleles are “exposed” in homozygotes.

(2) At many loci, homozygotes have lower fitness than heterozygotes (heterozygote advantage). c. This prediction has been verified in a wide variety of species. Example: Florida panthers (Fig. 23.4) and humans.

III. Natural Selection A. How does selection affect genetic variation?

1. Natural selection occurs in a wide variety of modes. (Table 23.3) 2. Mode 1: Directional selection a. One extreme is favored, and the average phenotype (and the other extreme) is selected against. (Fig. 23.5a) (BioSkills 2) Examples: body size and cold tolerance in cliff swallows (Fig. 23.5b) b. Directional selection changes the average value of a trait. c. It reduces genetic variation. If continued for a long time, the favored allele becomes “fixed” at 1.0, and the other allele(s) dies out. d. It is often counterbalanced by selection on a different trait (countervailing selection and fitness trade-offs). e. Students should be able to predict how genotype frequencies differ from Hardy-Weinberg equilibrium under directional selection.

3. Mode 2: Stabilizing selection a. The average phenotype is favored, and both extremes are selected against.

(Fig. 23.6a) Example: human birth weight (Fig. 23.6b) .

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b. The average value of the trait does not change over time. c. Stabilizing selection reduces genetic variation.

4. Mode 3: Disruptive selection a. Both extremes are favored, and the average phenotype is selected against.

(Fig. 23.7a) Example: beak length in black-bellied seedcrackers (Fig. 23.7b) b. Disruptive selection is relatively rare. c. It maintains genetic variation. d. It plays a role in speciation—the generation of new species.

5. Mode 4: Balancing selection a. No single allele has an advantage; instead, there is a balance among several alleles, with each having advantages in certain situations. Examples: heterozygote advantage, selection in variable environments, and frequency-dependent selection (Table 23.3) B. Sexual selection 1. Sexual selection favors individuals with heritable traits that enhance their ability to attract mates.

2. This includes selection both within the sex (intraspecific selection) and between the sexes (intraspecific selection).

3. Theory: The fundamental asymmetry of sex (Bateman-Trivers theory) a. Pattern: Sexual selection usually acts on males more strongly than females. b. Process: In most species, females invest more in their offspring than males.

(1) “Eggs are expensive, but sperm are cheap.” (Fig. 23.8) c. Predictions of the Bateman-Trivers theory (1) Females should be choosy about mates.

(2) Males should compete with one another for mates.

(3) Sexual selection should act more strongly on males than on females.

4. Female choice for “good alleles” a. Female birds often prefer colorful males, which is an indication of the male’s health and diet (and presumably the quality of his genes). b. Example: zebra finches. Nine out of 10 females prefer colorful males, and males that eat better diets are more colorful. (Fig. 23.9) c. Sometimes females choose males for their ability to help care for the young.

5. Male–male competition a. Example: northern elephant seals (Le Boeuf long-term study) (1) Males fight over the ownership of patches of beach occupied by congregations of females; large males win more battles. (Fig. 23.10a) (2) Variation in reproductive success is high in males and low in females.

(Fig. 23.10b, c) (3) Since the alleles for large body size in males have a fitness advantage, large male size has evolved in this population.

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6. Sexual dimorphism results from sexual selection. (Fig. 23.11) a. Because sexual selection tends to be much more intense in males than in females, males tend to have many more traits that function only in courtship or in male–male competition. b. Sexual selection often results in sexual dimorphism. Sexual dimorphism is any trait that differs between males and females. c. Like inbreeding, sexual selection violates the assumptions of the Hardy-Weinberg principle. d. Unlike inbreeding, sexual selection causes evolution.

7. Take-home messages a. Sexual selection is one type of natural selection. b. Collectively, these are referred to as ecological (or environmental) selection.

IV. Genetic Drift A. Genetic drift is any change in the allele frequencies in a population that is due to chance (luck, sampling error).

1. Genetic drift causes allele frequencies to drift up and down randomly over time.

B. Simulation studies of genetic drift 1. Genetic drift can be simulated by flipping a coin.

2. Computer simulations can model genetic drift over many generations. (Fig. 23.12) a. Students should be able to examine the MN blood group genotype frequencies in Table 23.1 and describe how drift could explain differences in genotype frequencies among populations.

3. Key points about genetic drift: a. Genetic drift is random with respect to fitness. b. Genetic drift is most pronounced in small populations. c. Over time, genetic drift can lead to the random loss or fixation of alleles.

C. Experimental studies of genetic drift (Fig. 23.13) 1. Bristle type in fruit flies a. 96 small populations (8 individuals each), starting with equal frequencies of forked bristles and normal bristles, were followed for 16 generations. b. Four males and females from the F1 generation were randomly selected from each of the 96 populations. c. By the 16th generation, 73% of the populations had completely lost either the forked-bristle or the normal-bristle allele.

D. What causes genetic drift in natural populations?

1. Founder effects on the green iguanas of Anguilla (Fig. 23.14a) (Making Models 23.1) a. Drift occurs when a group of individuals immigrate to a new geographic area and establish a new population (founder event). b. If the new population is small, the allele frequencies will likely be different from those in the original population (founder effect).

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2. Population bottleneck on Pingelap Atoll (Fig. 23.14b) a. A bottleneck occurs when a disease outbreak, a natural catastrophe, or another event causes a sudden reduction in population size. b. The remaining individuals likely have different allelic frequencies than the original population. c. Example: the island of the color-blind (Pingelap Atoll) (1) Only 20 people survived a typhoon and famine in 1775.

(2) The survivors happened to have a high frequency of an allele for total colorblindness (allele CNGB 3).

(3) Today, many people on the island are completely color-blind.

V. Gene Flow A. Gene flow is the movement of alleles from one population to another. (Fig. 23.15) B. Measuring gene flow between populations 1. Example: steelhead trout in Oregon (Fig. 23.16a) a. Captive bred trout reduce the fitness of wild populations due to reduced gene flow.

(Fig. 23.16b) C. Gene flow is random with respect to fitness.

1. Gene flow tends to equalize allele frequencies among populations. a. For example, human migration across continents is homogenizing allelic frequencies in human populations.

VI. Mutation A. There are three ways mutations can occur to increase genetic diversity in populations.

1. Point mutations—a change in the nucleotide sequence in a stretch of DNA. 2. Chromosome-level mutations—this often results in gene duplication.

3. Lateral (Horizontal) gene transfer—transfer of genes from one species to another.

B. Mutation as an evolutionary process 1. Mutation happens too infrequently to cause dramatic changes in allele frequencies on its own. a. Example: human mutation rates. It would take 4,000 years for a mutation to produce a change in allele frequency of 1%.

2. However, when considered across entire genomes and combined with natural selection, mutation becomes an evolutionary force.

C. Experimental studies of mutation 1. Study of mutation in Escherichia coli over 10,000 generations. (Fig. 23.17) a. Relative fitness increased dramatically over time but in a stepwise pattern. b. Each jump in fitness was caused by a novel mutation.

2. Students should be able to explain why deleterious mutations do not cause downward drops in fitness in Figure 23.17.

D. Studies of mutation in natural populations 1. Lateral gene transfer—Pea aphids acquired the ability to generate carotenoids from the genome of a fungal symbiont. (Fig. 23.18a)

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2. Gene duplication—Once genes were transferred, genes were duplicated (and diversified) to form the genome of the red pea aphid and subsequently deleted to be adaptive for the green pea aphid. (Fig. 23.18b) E. Take-home messages (Table 23.4) 1. Mutation is the ultimate source of genetic variation.

2. If mutation did not occur, evolution would eventually stop.

3. Mutation alone is usually inconsequential in changing allele frequencies at a particular gene.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. adaptation allele frequency balancing selection countervailing selection deleterious allele directional selection disruptive selection evolution

female choice fitness fitness trade-off founder effect founder event frequency-dependent

fundamental asymmetry of sex gene flow gene pool genetic bottleneck genetic drift genetic marker genetic variation genotype frequency Hardy-Weinberg principle heterozygote advantage homozygosity inbreeding inbreeding depression loss-of-function mutation male-male competition

migration mutation natural selection nonrandom mating null hypothesis parental care investment polygenic trait population population bottleneck purifying selection sampling error sexual sexual dimorphism selection speciation stabilizing selection territory

selection

Lecture Activities Illustrating the Hardy-Weinberg Principle and Evolutionary Forces (for a small class) Estimated duration of activity: 20–30 minutes This activity will help your students conceptualize the Hardy-Weinberg principle and understand how the agents of evolution described in this chapter affect allele frequencies and genotype frequencies in a population. This activity can be done in pieces during the lecture, or it can be done as a whole after you have covered all of the material in this chapter.

Take 25 white index cards and label 4 of the cards with AA, 12 with Aa, and 9 with aa, separating the letters enough that the card can be torn to separate the alleles (to simulate the formation of gametes). Also, prepare 10 colored index cards: 7 AA and 3 aa (these will be used in the gene flow example). Alternatively, you could use two bags of different-colored .

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beads, designating one color to represent allele A and the other to represent a. Give one index card to each student (pass them out randomly). You can then do any of the following exercises.

• Calculating genotype frequencies: Ask all of the homozygote dominant individuals to hold up their cards, and record that data on the board. Do the same for heterozygotes and homozygote recessives and record the data. Have the students calculate genotype frequencies for the parent population. (If you used the 25 cards described above, the genotype frequencies will be AA = 16%, Aa = 48%, and aa = 36%.)

• Calculating allele frequencies: Have everyone generate gametes by ripping their cards in half to separate the alleles. Now have everyone who has an A allele hold it up. Count them, and record the data on the board. Do the same for the a allele. Ask the students to calculate the allele frequencies for A and a. (If you used the 25 cards described above, A = 40% and a = 60%.) Ask students how they could have figured out these numbers from the genotype frequencies.

• Illustrating the Hardy-Weinberg principle: Pass a box around and ask everyone to put their “gametes” into the box. Mix up the gametes well, simulating random mating. Pass the box around again, and have everyone pick out two gametes. Collect the genotype data as before, calculate allele frequencies and genotype frequencies, and use the Hardy-Weinberg equation to determine whether this population is in equilibrium (Note: Genotype frequencies might show some “drift,” but they should be fairly close to the Hardy-Weinberg predictions. Allele frequencies will not change.)

• Illustrating natural selection and its effect on population genetics: Reestablish the original population by collecting all of the cards, mixing them up in the box, and having each student take two “gametes” from the box. Explain that the A allele codes for some trait that decreases fitness. Collect the cards from AA individuals and explain that they have died without reproducing; only individuals who have at least one a allele survive and reproduce.

Pass around the box and collect gametes from Aa and aa survivors. Mix up the gametes, pass the box around again, and have students pick out two gametes. Calculate allele and genotype frequencies and discuss how they changed from the parent population. Ask the students to predict what will happen in the next generation if only Aa and aa individuals successfully mate. (Note: The population size will get smaller in this exercise, but explain to the students that this effect is an artifact of the demonstration. In this exercise, each individual is basically reproducing only to replace itself; most organisms generate more offspring than that.)

• Illustrating genetic drift and its effect on population genetics: Reestablish the original population. Simulate a bottleneck by having a natural disaster wipe out most of the population. Randomly choose 10 individuals to “survive.” Collect genotype data from the survivors, and have the class calculate allele and genotype frequencies. Compare these values to those of the parent population, and discuss how genetic drift has affected the population’s gene pool. Have students predict how the population will change in successive generations.

• Illustrating gene flow and its effect on population genetics: Reestablish the original population. Illustrate immigration by giving some students colored index cards. Have the “immigrants” generate gametes by ripping their cards in half. Pass the box around and have everyone deposit their gametes. Then mix them up and pass the box around again.

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colored). Collect genotype data, and calculate allele and genotype frequencies. Have the students analyze how the population has changed after the immigration event. If time permits, you can have 10 individuals leave the population and repeat the exercise with the remaining individuals to illustrate emigration. You could also analyze the 10 who left as a founder population.

Wrap-up: What good is the Hardy-Weinberg principle? Many students will learn correctly how to test whether a population is in Hardy-Weinberg equilibrium but will fail to grasp why it is useful to do so. Review the results of your tests and then ask: Suppose you are a field biologist studying two different populations of, say, Grevy’s zebra. You have determined the genotypes of all individuals in both populations at a certain locus. The first zebra population is in Hardy-Weinberg equilibrium, and the second population is not. What does this tell you?

Why is it useful to know?

Answer: Testing for Hardy-Weinberg equilibrium is the best way to determine whether a certain locus is evolving. The first zebra population is not evolving at the locus tested. The second zebra population is evolving, is not mating randomly, or both. Conclude by asking students what exactly might be happening to the second population: Which evolutionary force is acting? How could they find out? Have them design follow-up experiments.

Illustrating the Hardy-Weinberg Principle and Evolutionary Forces (for a large class) The activity described above for a small class will also work in a large class. But in very large classes, use a subset of students for each exercise rather than the whole class at once. (Otherwise, it will take too much time to pass the box and tally the results.) You can move to a different set of students for each new exercise so that everybody gets to participate. Alternatively, you can make the same point by keeping a running example going on the board. The activity above can be used, but no cards will be passed out, and you will have to use probability calculations to generate the F1 generation in each example. Using the same population and showing how that population would change if exposed to each of the agents of evolution will really help students put these concepts together into a larger context.

Practicing BioSkills: “The Big Picture” of Evolution (BioSkills 15) Estimated duration of activity: 10 to 15 minutes Students will now have learned many new terms associated with evolution, and a concept map will help them tie this new knowledge together. Break the class into small groups and have them draw concept maps that connect the terms given below. Remind them that they should label the arrows so that it is as clear as possible how each term is related to the next.

(If they have already started a concept map in the preceding chapter, they can take out their earlier map and add new terms to it.)

Starting list of terms: evolution, natural selection, gene pool, genetic drift, gene flow, mutation, nonrandom mating, allele frequency, genotype frequency, Hardy-Weinberg equilibrium After students have drawn their own concept maps, encourage students to refer to “The Big Picture” concept map on evolution presented in the textbook for more ideas. Remind them that there are many different ways to draw concept maps.

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Coin-Flipping, Heterozygotes, and Genetic Drift (BioSkills 4) Estimated duration of activity: 10 to 15 minutes Some students will not be familiar with the probability concepts necessary to understand the Hardy-Weinberg principle. The equation relies on basic probability theory that can be modeled with flipping coins.

Have students form pairs. Both students in the pair should have a coin to flip. Explain that each coin flip simulates the production of a gamete with a certain allele, and each time the two students flip their two coins, they are simulating the production of an offspring with a certain genotype. Have each pair of students flip their coins at least 10 times and write down the results. If the class is small, you can have the pairs come up and write their results on the board afterward.

Now ask what happened: What was the actual frequency of the “heads” homozygote? The “tails” homozygote? The “head-tail” heterozygote? Students will likely have discovered that there are two different ways to form a heterozygote. Ask them how to mathematically describe the probability of a heterozygote. See if they can come up with the “2pq” term on their own, or even the whole Hardy-Weinberg equation.

You may wish to review the two basic rules of probability theory discussed in BioSkills 4: the Both-And rule and the Either-Or rule. (The 2pq term uses both these rules.)

You can wrap up by using this example to illustrate genetic drift. Point out that any one pair of students might have had unusual results due to sampling error. Have the students add up the overall results for a set of, say, five pairs of students, and calculate the total frequency of heterozygotes. Compare the overall frequency of heterozygotes (i.e., in 50 coin flips taken together) to the frequencies observed in the pairs of students (i.e., in just 10 coin flips). Was deviation from the expected probabilities greatest for small sample sizes or large sample sizes? Why?

Discussion Idea: Cheetahs and Population Bottlenecks Perhaps the greatest problem with preserving the global cheetah population is that these animals are now almost genetically identical to one another (due to the bottleneck effect). Ask students to discuss the following questions: • What are the problems associated with having such low genetic diversity?

• Because of this problem, many biologists think that cheetahs are doomed to go extinct in the near future. Do you agree?

• If you think cheetahs can be saved, what conservation efforts will be necessary to save them?

• Is there any way to increase genetic diversity in a local cheetah population? (For example, consider just the population of captive cheetahs in North American zoos.) How about in the global cheetah population?

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CHAPTER

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Speciation

Learning Objectives: Students should be able to...

• Compare and contrast the three most common species concepts.

• Discuss the roles that gene flow, selection, genetic drift, and mutation play in speciation.

• Contrast the mechanisms of allopatric and sympatric speciation.

• Predict what will happen when two divergent populations come back into contact.

Lecture Outline • Speciation occurs when a single ancestral group splits into two or more species.

(Fig. 24.1) • Speciation results from genetic isolation and genetic divergence.

• If gene flow ends (genetic isolation), allele frequencies in isolated populations are free to diverge via mutation, genetic drift, and selection.

• If the populations diverge enough, they become new species. (Making Models 24.1)

I. How Are Species Defined and Identified?

A. A species is an evolutionarily independent population or group of populations.

1. Biologists use different sets of criteria to identify species.

B. The biological species concept 1. Species are defined by reproductive isolation.

2. Populations are reproductively isolated if they fail to interbreed or to produce viable, fertile offspring. (Table 24.1) a. Prezygotic isolation mechanisms prevent fertilization. b. Postzygotic isolation mechanisms prevent survival or reproduction of the hybrid offspring.

3. Reproductive isolation cannot be evaluated in fossils, asexual species, or species whose ranges do not overlap.

C. The morphospecies concept 1. Species are defined by differences in size, shape, or other morphological features (polymorphic).

2. The morphospecies concept is widely applicable and can be used for fossil species and asexual species.

3. The morphospecies concept cannot identify cryptic species, and it is subjective.

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D. The phylogenetic species concept 1. A species is defined as the smallest monophyletic group on the tree of life. a. A monophyletic group consists of an ancestral population, all of its descendants, and only its descendants. (Fig 24.2a) b. A monophyletic group is identified by synapomorphies⎯homologous traits unique to that lineage. c. Example: elephants (Fig. 24.2b) 2. Advantages of the phylogenetic species concept: a. It can be applied to any populations. b. It is logical because populations are monophyletic only if they are independent of one another and isolated from gene flow.

3. The phylogenetic species concept typically recognizes many more species than any other species concept.

4. In actual practice, researchers use all three species concepts: biological, morphospecies, and phylogenetic. (Table 24.2) E. Species definitions in action: The case of the dusky seaside sparrow 1. Under the biological and morphospecies concepts, the endangered dusky seaside sparrow appeared to be a unique taxon worth saving. (Fig. 24.3a) 2. A hybridization program was started to cross duskys with another population of seaside sparrow, to save the duskys’ genes.

3. New genetic data showed that under the phylogenetic species concept, there are only two species of seaside sparrow. (Fig. 24.3b) a. Dusky seaside sparrows were found to be part of a larger group that had other surviving populations. b. In addition, the hybridization program had bred the few surviving duskys to birds from the wrong population.

II. Isolation and Divergence in Allopatry A. Speciation that begins with physical isolation is called allopatric speciation.

1. Physical isolation occurs in two ways. a. Dispersal—a population can colonize a new habitat. (Fig. 24.4a) b. Vicariance—a new physical barrier can split a widespread population into two or more isolated groups. (Fig. 24.4b) B. Allopatric speciation by dispersal 1. Example: Galápagos finches a. A population of ground finches colonized a new island. Later they were found to have larger beaks than the original population.

2. Colonization, followed by drift and selection, is thought to be responsible for speciation in many groups.

3. Students should be able to apply the model in Figure 24.4a to this example. What does the small population on the right side of the model represent? What does the red color represent? How can you modify the model to show that gene flow is still occurring?

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C. Allopatric speciation by vicariance 1. Example: trumpeters a. Many species of trumpeters within the Amazon basin are closely related. (Fig. 24.5) b. This pattern indicates that the formation of the Amazon basin via the rising Andes Mountains was a vicariance event that separated the original trumpeter species, which then caused speciation.

III. Isolation and Divergence in Sympatry A. Sympatric populations live in the same area.

1. Traditionally, biologists thought that speciation could not occur among sympatric populations because gene flow is possible. (Fig. 24.6) B. Sympatric speciation by disruptive selection 1. Populations that are not geographically isolated may become reproductively isolated by adapting to different ecological niches via disruptive selection.

2. Disruptive selection for ecological niche in flies a. Example: apple maggot and hawthorn flies (Fig. 24.7) b. Apple maggot flies occur only in North America and feed only on apple trees, but apple trees have been in North America for only 300 years. c. Apple maggot flies’ closest relatives are hawthorn flies. d. The two types of flies each prefer the scent of their own host tree, rarely land on other trees, and thus rarely meet each other. (Fig. 24.8) (1) There is very little gene flow, despite sharing the same range. e. Disruptive selection appears to be driving the two populations toward sympatric speciation. f. Students should be able to hypothesize why natural selection would favor divergence based on the observation that apple fruits drop about 3 to 4 weeks earlier in the fall than hawthorn fruits.

3. Disruptive selection for mate choice a. Sexual selection is a form of natural selection that specifically favors individuals with traits that increase their ability to obtain mates. b. Examples of disruptive selection include courtship songs in crickets, coloration in stickleback fish, courtship structure in bowerbirds, and wing pattern in butterflies. c. In Drosophila, the gene mFAS encodes for fatty acid synthase, an enzyme that catalyzes the formation of waxy hydrocarbons that make external cuticles (prevents desiccation). The waxy hydrocarbons also act as pheromones. d. Those flies that lost the ability to produce waxy hydrocarbons also were not chosen to mate with by females (prezygotic isolation), which led to speciation.

C. Sympatric speciation by polyploidization 1. In plants, a mutation known as polyploidy reduces gene flow between normal and mutant plants. a. Polyploidy is a condition of having more than two complete sets of chromosomes. b. Tetraploid (4n) individuals cannot mate with diploid (2n, normal) individuals because the resulting triploid offspring are sterile. c. Because polyploidy reduces gene flow, it can and does lead to speciation.

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2. Autopolyploidy (Fig 24.9a) a. Autopolyploidy is a doubling of the chromosome number in the offspring of one species. b. This can result in a new tetraploid species. Example: maidenhair ferns c. Students should be able to explain the origin of a tetraploid grape with extralarge fruit from a diploid population with smaller fruit.

3. Allopolyploidy (Fig. 24.9b) a. Allopolyploidy occurs when two diploid species hybridize and the hybrid offspring experiences a mutation that doubles the chromosome number so that it can produce gametes. b. Like autopolyploidy, this can result in a new, tetraploid species. Example: Tragopogon (Fig. 24.10) c. Students should be able to explain how a cross between Emmer wheat, a tetraploid, and a wild, diploid wheat gave rise to the hexaploid bread wheat grown throughout the world today.

4. Why is speciation by polyploidy so common in plants? a. In plants, gametes are produced from tissues that have undergone many rounds of mitosis, making mutations (like chromosome doubling) more likely. This increases the variation in the population. b. Many plants can self-fertilize, making it possible for a rare individual with diploid gametes to produce tetraploid offspring. c. Hybridization between plant species is common. Polyploids have a higher level of heterozygosity than diploids.

IV. What Happens When Isolated Populations Come into Contact?

A. Possible outcomes of secondary contact (Table 24.3) 1. If prezygotic isolation mechanisms do not exist, the populations may start breeding again, leading to any of the following outcomes: a. Fusion⎯if gene flow erases differences between populations b. Reinforcement⎯selection against interbreeding, leading to further divergence and evolution of prezygotic isolation mechanisms c. Founding of a stable hybrid zone d. Extinction of one population e. Creation of a new species via hybridization B. Reinforcement 1. If hybrid offspring have lower fitness than their parents, there will be strong selection against interbreeding (reinforcement).

2. Example: Closely related species of Drosophila fruit flies: Sympatric Drosophila species are seldom willing to mate with each other, but allopatric species are.

C. Hybrid zones 1. Hybrid zones are geographic areas where interbreeding occurs and hybrid offspring are common. a. This sometimes occurs when hybrids have intermediate traits that are well adapted to the environment where the two species overlap. b. Hybrid zones can be narrow or wide and long- or short-lived.

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2. Example: Townsend’s and hermit warblers. Mitochondrial DNA (mtDNA) analysis shows that the hybrid zone has been expanding to the south, apparently driving out the hermit warblers. (Fig. 24.11) 3. Hybrid zones can lead to the extinction of one of the parent species or to the formation of a new species.

D. New species through hybridization 1. If hybrids have higher fitness than either parent, a new species can arise. a. This can occur if hybrids have a unique combination of alleles that give it different characteristics than either parent.

2. Example: Helianthus sunflowers of the American West (Fig. 24.12) a. H. anomalus appears to have arisen as a hybrid. b. It inherited a unique combination of alleles that allowed it to survive in drier habitats than either parent species. c. Researchers succeeded in recreating the speciation event in the lab.

3. Students should be able to suggest one result that would have caused the researchers to reject the hybridization hypothesis in the experiment in Figure 24.12.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. allopatric speciation allopatry allopolyploidy autopolyploidy backcrossing behavioral isolation biogeography biological species concept clade colonization common ancestry cryptic species dispersal disruptive selection Endangered Species Act fusion (of populations) gametic barrier

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genetic divergence genetic isolation habitat isolation hybrid sterility hybrid viability hybrid zone hybridization lineage mechanical isolation mitochondrial DNA (mtDNA) monophyletic group morphology morphospecies concept nondisjunction phylogenetic species concept phylogenetic tree phylogeny

polyploidy postzygotic isolation prezygotic isolation reinforcement reproductive isolation secondary contact self-fertilization speciation species subspecies sympatric speciation sympatry synapomorphy temporal isolation tetraploid triploid vicariance

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Lecture Activities Practicing BioSkills: Reading Phylogenetic Trees (BioSkills 13) Estimated duration of activity: 10–20 minutes This chapter presents several opportunities to introduce students to phylogenetic trees (e.g., elephants in Figure 24.2, seaside sparrows in Figure 24.3, and trumpeters in Figure 24.5). The following exercises will help students become comfortable with phylogenetic trees. You may wish to provide a handout with the elephant tree, the trumpeter tree, and some blank space for students to draw their own trees. Refer students to BioSkills 13 in Appendix B for more information.

Tree basics: Have students label all the nodes, tips, and branches on the elephant tree; label the position of each common ancestor; and circle five monophyletic groups on the elephant tree. (Students can do this alone or in small groups, or you can ask volunteers to come to the front of the class to add elements to a tree drawn on the board.)

Branches can rotate: One of the most common errors that students make is to think that tips that happen to be drawn in close proximity are more closely related. To put it another way, students often don’t realize that branches can “rotate” without changing evolutionary relationships. They also often need assistance understanding that they need to look for the node with the common ancestor (i.e., not look at how close two tips are).

• Have students redraw the elephant tree so that forest elephants and Asian elephants are in close proximity but without changing the position of their common ancestor (i.e., rotate the bottom branch of the original tree). Ask students: Does this new tree make forest and Asian elephants more closely related? Why not?

• Using the trumpeter example, ask students to draw a phylogenetic tree based on the information given in Figure 24.5.

• Sum up by asking students: What’s the best way to determine how closely related two species are from a phylogenetic tree? (Answer: Look for the node with the common ancestor. How far back is it?)

Understanding synapomorphies: Ask students to draw their own phylogenetic trees depicting a hypothesis of relationships of species they are already familiar with (e.g., dog, wolf, cat, tiger, lemur, gorilla, human, eagle). For each monophyletic group, have them list a synapo- morphy.

(You can add a puzzling species to the list⎯hyena or sea lion, for example⎯that students will probably not know how to place. Ask them what further information they need to place the new species correctly.) When students have finished their trees, have some volunteers put some of the completed trees on the board. Sum up by listing all the synapo- morphies. (Students will likely also discover they have drawn the tree in several different ways, demonstrating there are many different correct ways of drawing a tree.)

Incorporating new information into an existing tree: Have students redraw the elephant tree to incorporate these new pieces of information: • Recent research indicates that Asian elephants from Borneo are genetically distinct from other Asian elephants.

• Genetic analysis of frozen woolly mammoth specimens has shown that woolly mammoths are more closely related to Asian elephants than they are to African elephants.

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Experimental Design: Has Speciation Occurred?

Estimated duration of activity: 10 minutes To help students integrate what they are learning about speciation with what they have learned about the scientific method, present this scenario to your students and have them design an experiment to test whether these two populations are separate species.

The scenario: The apple maggot fly (Rhagoletis pomonella) can use hawthorn berries or apples as hosts. If one population of apple maggot flies breeds and lives on hawthorn berries for many years, and another population breeds and lives on apples for many years, have they diverged into separate species?

Divide class into groups of four to six. Each group should design an experiment to determine whether speciation has taken place, using just one of the three species concepts discussed in the chapter: biological, morphospecies, or phylogenetic. (Make sure each species concept has at least one student group studying it.) Their experiments should include all the components of good scientific inquiry (e.g., hypothesis, prediction, controls, replication, etc.). Give groups 5 to 7 minutes to work, and then have them compare their ideas.

Case Studies in Speciation Estimated duration of activity: 5–10 minutes Present these short scenarios and ask students to make an initial hypothesis (with a brief explanation of which species concept they are using) as to whether speciation has occurred or may occur.

• Case study 1: Primrose polyploidy. The evening primrose, Oenothera lamarckiana, has a diploid number of 2n = 14. Another evening primrose, Oenothera gigas, has a diploid number of 2n = 28. These two plants look very similar.

• Case study 2: Drosophila. Fruit flies (D. melanogaster) from the same ancestral population are separated and raised in different temperature and humidity conditions for several years. When reunited, matings between the two populations produce sterile offspring.

However, interbreeding between populations separated but grown in identical conditions produced viable offspring.

• Case study 3: Wolbachia. Wasps infected with the bacteria Wolbachia pipiens can mate, but they do not produce viable offspring because the cytoplasm of their gametes is incompatible.

Discussion Ideas: Speciation Here are three suggestions for in-class challenge questions: • Species concepts in conflict: This chapter outlined three distinct concepts by which species can be defined and identified. Which species concept do you think is the most convincing?

Which is the least convincing? Give an example of a situation in which each concept would lead to a different conclusion regarding whether a population is a distinct species.

(For example, southern plains zebras have much less striping than northern plains zebras, but they can interbreed, with fertile offspring.) How does the idea of a subspecies complicate each of these species concepts?

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• Allopatric speciation flowchart: Draw a flowchart detailing the events of allopatric speciation. Be sure to include all of the events that will take place from beginning to end, including the mechanisms that will ensure reproductive isolation following secondary contact. Be specific and thorough. When you are finished with the drawing, choose any modern species you want and speculate about what might happen if allopatric speciation were to occur in one of its major populations.

• Reproductive isolation mechanisms: Are the mechanisms of prezygotic and postzygotic reproductive isolation mutually exclusive? Why or why not? Are these mechanisms more effective when they are combined? Which mechanisms do you think are most effective for securing reproductive isolation and therefore species integrity? Explain your answer. Give a real example of how each mechanism is used in nature. Can you think of any examples in which one or more of these mechanisms are used together in nature?

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CHAPTER

25

Phylogenies and the History of Life

Learning Objectives: Students should be able to...

• Analyze evolutionary relationships from phylogenetic trees, draw alternative trees, and build trees from small data matrices.

• Discuss the major strengths and limitations of the fossil record.

• Describe the three major eras of the Phanerozoic eon, including the major evolutionary • events and dominant taxa characteristic of each.

• Use examples to explain why adaptive radiations occur.

Compare and contrast the causes of mass extinction and background extinction.

Lecture Outline I. Tools for Studying History: Phylogenetic Trees A. Tree basics 1. A phylogeny is an evolutionary history of a group of organisms.

2. A phylogenetic tree is a graphical summary of this history.

3. The tree of life is the most universal of all phylogenetic trees depicting the evolutionary relationships among all living organisms.

4. Reading phylogenetic trees (Table 25.1) (BioSkills 13) a. Branches represent populations through time. b. The tips of the branches represent specific taxa. c. Nodes (forks) occur when an ancestral group split into multiple descendent groups. d. The root is the most ancestral branch in the tree. e. Outgroups help to root the tree by including a taxa that diverged prior to the focal group. f. A node that depicts an ancestral branch dividing into three or more descendant branches is called a polytomy. g. Branches of the tree can rotate at each node. (Making Models 25.1) h. The number of nodes is not useful in determining relationships. (Making Models 25.2) B. How do biologists estimate phylogenies?

1. Creating the data matrix

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a. Closely related species should share many characters, while more distantly related species should share fewer characters. (Fig. 25.1a) 2. Using the data matrix to estimate a tree a. The cladistic approach is based on synapomorphies (shared derived traits) that identify monophyletic groups. (Fig. 25.1b)(Table 25.2) b. Ancestral and derived traits are relative. c. Ultimately, trees with the best fit are those that exhibit the least amount of changes (parsimony). d. Cladistic analysis does not consider the length of branches, but other analyses estimate this via genetic distance or time. (Table 25.3) 3. Using the alternative hypothesis that limbs arose separately in lizards and in the ancestors of dogs and humans, students should be able to draw a revised tree (Fig. 25.1b) that shows five character changes.

C. How can biologists distinguish homology from homoplasy?

1. Homology is similarity in a trait due to common ancestry.

2. Homoplasy is similarity in a trait for some reason other than common ancestry, such as convergent evolution.

3. Are the flowers of water lilies and wild roses homologous or convergent? a. Phylogenetic evidence – water lilies and wild roses have flowers because they inherited these structures from a common ancestor that also had flowers. (Fig. 25.2) b. Structural evidence – although flower structure is diverse, most flowers are built upon the same general template. c. Genetic and developmental evidence – the genetic toolkit and developmental pattern of floral structures are similar. d. The data provide strong evidence that flowers in diverse types of plants are homologous.

4. Are streamlined bodies in dolphins and ichthyosaurs homologous or convergent? a. The streamlined bodies are convergent evolution as natural selection favored similar solutions to the problems posed by a similar way of making a living. (Fig. 25.3) D. Whale evolution: A case history 1. Data set 1: A phylogeny based on morphological traits a. Analysis seemed to indicate that whales are not closely related to artiodactyls (cows, hippos, etc.). (Fig. 25.4a) b. Whales lack a pulley-shaped astragalus ankle bone, which is the key shared derived characteristic of the artiodactyls.

2. Data set 2: A phylogeny based on DNA sequence data a. DNA analysis indicated that whales are most closely related to hippos. (Fig. 25.4b) 3. Data set 3: Transposable elements a. Analysis of SINE gene sequences confirmed that whales share four unique synapomorphies with hippos and with no other artiodactyls. (Fig. 25.4c) 4. Conclusion: Whales are closely related to hippos. a. In 2001, paleontologists have since found fossils that possess both an unusual ear bone found only in whales and a pulley shaped astragalus characteristic of the artiodactyls.

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II. Tools for Studying History: The Fossil Record A. The fossil record provides the only direct evidence about what organisms in the past looked like.

1. A fossil is a piece of physical evidence of an organism that lived in the past.

2. The fossil record is composed of all the fossils that have been found and described.

B. How do fossils form?

1. Fossils form when part or all of an organism is buried in ash, sand, mud, or some other sediment before it can decompose.

2. Types of fossils: (Table 25.4) a. Intact fossil⎯decomposition does not occur and organic remains preserved b. Compressed fossil⎯forms when sediments accumulate on top of the organism c. Cast fossil⎯organism decomposes after buried and hole is replaced with minerals d. Permineralized fossil⎯organisms decompose slowly and dissolved minerals infiltrate cells e. Trace fossil⎯sedimentation and mineralization preserve indirect evidence of an organism, like footprints, feeding marks, or feces 3. Fossilization is a very rare event. a. For example, only about 1 out of every 200 million Archeopteryx individuals was fossilized.

C. Limitations of the fossil record 1. Habitat bias: Species that live in areas with sediment deposition (beaches, mudflats, swamps) are much more likely to be fossilized.

2. Taxonomic and tissue basis: Organisms, or tissues, with hard parts are much more likely to be fossilized.

3. Temporal bias: Recent fossils are much more common than ancient fossils.

4. Abundance bias: Species that were abundant, widespread, and present on Earth for a long time are much more likely to produce fossils.

D. Life’s time line 1. Precambrian (the three earliest eons combined) (Fig. 25.5) a. Life was exclusively unicellular for most of Earth’s history. b. Oxygen was virtually absent for almost 2 billion years after the origin of life.

2. The Phanerozoic Eon is the interval from 541 mya (million years ago) to the present.

(Fig. 25.6) a. Paleozoic Era⎯origin and initial diversification of animals, land plants, and fungi, and first appearance of land animals b. Mesozoic Era⎯terrestrial environments dominated by gymnosperms and dinosaurs c. Cenozoic Era⎯terrestrial environments dominated by angiosperms and mammals 3. There have been constant changes in the climate and the locations of continents during these eras.

III. Adaptive Radiation A. Adaptive radiation is rapid diversification of a single lineage into a wide variety of adaptive forms.

1. Example: Hawaiian silverswords (Fig. 25.7) .

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B. Why do adaptive radiations occur?

1. Two general mechanisms can trigger them: new resources and new ways to exploit resources.

2. Ecological opportunity a. Adaptive radiations frequently occur when habitats are unoccupied by competitors. b. Example: Anolis lizards in the Caribbean (Fig. 25.8) (1) From different starting points, an adaptive radiation occurred on each island, eventually producing the same four ecological types of lizards. c. Students should be able to generate a hypothesis to explain why tarweeds in California and Anolis lizards on the mainland are not particularly species-rich or ecologically diverse.

3. Morphological, physiological, or behavioral innovation as a trigger a. The evolution of key traits are a driving force behind many adaptive radiations. b. Examples: flowers for angiosperms, feathers and wings for birds C. The Cambrian explosion 1. Virtually every major group of multicellular animals appeared in the fossil record in a short time of just 50 million years.

2. Early animal fossils (Fig 25.9) a. Microfossils⎯first faunas to appear were small fossils (< 1 mm) of sponges and corals from the Ediacaran period just before the Cambrian. b. First macroscopic fossils⎯larger sponges, jellyfish, and comb jellies appeared in the Ediacaran. c. Cambrian macroscopic fossils⎯the first arthropods, mollusks, vertebrates, and other phyla; tremendous increases in size, morphological complexity, and ecological niches (Figs. 25.10) 3. What triggered the Cambrian explosion? a. Four possible causes (not mutually exclusive): (1) Higher oxygen levels made big, mobile animals possible.

(2) The evolution of predation exerted selection for hard parts, exoskeletons, and rapid movement.

(3) “New niches beget more new niches.” Animals moved off the ocean floor, which led to the development of new ecosystems in other locations.

(4) “New genes, new bodies.” The Hox genes evolved. (Fig. 25.11)

IV. Mass Extinction A. Mass versus background extinction 1. A mass extinction is the rapid extinction of at least 60% of the species on Earth within 1 million years.

2. Background extinction is the lower, average rate of extinction observed over the entire history of life when mass extinctions are not occurring.

B. How do mass extinctions differ from background extinctions?

1. Five major mass-extinction events are commonly recognized. (Fig. 25.12) 2. Background extinctions result primarily from natural selection, but mass extinctions seem to hit species randomly, rather like genetic drift.

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C. The end-Permian extinction 1. The end-Permian was the worst mass extinction of all time. It wiped out more than 90% of multicellular species.

2. We still do not understand its cause, but the clues include huge flood basalts and rapid global warming, a steep reduction in the amount of oxygen in the oceans and on land, and a dramatic drop in sea level.

D. The end-Cretaceous extinction 1. This mass extinction killed off 60% to 80% of species worldwide, including the dinosaurs.

2. Evidence for the impact hypothesis (Table 25.5) a. Evidence includes high levels of iridium worldwide, a huge crater near Mexico, and shocked quartz and microtektites found near the proposed impact site. b. Fireball of hot gas spread from impact site resulting in large soot and ash deposits in sediments dated 65 mya. c. Largest tsunami in the last 3.5 billion years disrupted ocean sediments and circulation patterns. d. Impact site contains sulfate-containing rock suggesting the triggering of extensive acid rain. e. Massive quantities of dust, ash, and soot would have blocked the sun for long periods leading to rapid global cooling and a crash in plant productivity.

4. Selectivity of the extinctions: The asteroid impact did not kill indiscriminately. a. Dinosaurs and all large-bodied marine reptiles perished. b. Mammals, crocodilians, amphibians, and turtles survived. c. An older hypothesis is that the extinction was size-selective (killing larger-bodied species), but this does not match the data. d. A newer hypothesis is that organisms capable of inactivity (hibernation, spores, etc.) survived.

5. Recovery from the extinction: Slow, taking 10 to 15 million years and eventually resulting in adaptive radiation of the mammals and birds.

E. The sixth mass extinction?

1. Many scientists propose the Earth is on the verge of a 6th mass extinction precipitated by human impacts.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. abundance bias adaptive radiation ancestral trait artiodactyl astragalus

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background extinction branches Burgess Shale faunas Cambrian explosion cast

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diversification Doushantuo microfossils ecological niche ecological opportunity Ediacaran faunas end-Cretaceous extinction end-Permian extinction fauna fossil fossil record fossilization genetic distance habitat bias homology homoplasy Hox gene

impact hypothesis iridium lineage mass extinction Mesozoic era microtektite molecular clock monophyletic group morphological innovation nodes/fork paleontologist Paleozoic era parsimony permineralized fossils phylogenetic tree phylogeny polytomy

Precambrian interval radiometric dating recovery selectivity shared derived characteristics shocked quartz SINE (short interspersed nuclear element) size selectivity synapomorphy taxon, taxa taxonomic bias temporal bias tip (terminal node) tissue bias

Lecture Activities The Cladistic Approach: A Genetic Example Estimated duration of activity: 5–10 minutes Cladistic approach: Give your students the following list of animals: human, gerbil, gorilla, dog, orangutan, chicken, and zebrafish. Have the students come up with a list of morphological characteristics that various subgroups of these animals have in common and a list of characteristics that they do not share. (For example: What traits do all the animals have in common? What traits do the mammals and the chicken have in common, but not the fish?

And so on.) When they have finished making their observations, have them look at the list of shared traits and identify shared derived characteristics that they could use to construct a phylogenetic tree using the cladistic approach. (Explain that derived characteristics are those that have been uniquely modified in just the lineage in question. Point out that not all shared characteristics are derived.) Have them construct a tree and draw it on the board.

Discussion Ideas Limitations of the Fossil Record The text lists several clear limitations of the fossil record. Have students work in pairs to make lists of the fossil record’s advantages, disadvantages, and many uses. Do the positives outweigh the negatives; that is, is the fossil record truly a useful source of information?

Explain your answer.

Causes of Mass Extinctions This chapter describes the end-Cretaceous mass extinction in detail and states that its cause was the impact of a large asteroid.

• What events do you think might have caused the other four mass extinctions?

• How could you test your ideas?

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(Further information: Definitive evidence for iridium, shocked quartz, microtektites, or impact craters has not been found for any other mass extinction, despite much searching.

However, most mass extinctions are associated with large volcanic eruptions, rapid changes in sea level, or both. See if students notice the common thread that links large volcanic eruptions, rapid changes in sea level, and meteor impacts: All three are associated with rapid global climate change.)

Global Climate Change As a Trigger for Mass Extinctions Both of the mass extinctions discussed in this chapter involved rapid global climate change.

(The other three probably did, too.) Evidence has accumulated that we are living in a period of unusually rapid global climate change today. Could we be approaching the level of global climate change that has triggered mass extinctions in the past? Solicit input from students on what evidence we would need to answer this question and how it should be gathered. Finally, ask them their personal opinions on what the best course of action should be, given that we do not know exactly what sort of global climate change caused the mass extinctions in the past.

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CHAPTER

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Bacteria and Archaea

Learning Objectives: Students should be able to...

• Explain how bacteria and archaea are the most important, diverse, and abundant organisms on the planet.

• Describe the six processes that bacteria and archaea use to produce ATP and obtain carbon building blocks.

• Describe the key roles that bacteria play in agriculture, bioremediation, and human health.

Lecture Outline • Bacteria, Archaea, and Eukarya are the three largest branches on the tree of life.

• Bacteria and archaea may look similar at first glance, but they are very different.

• Similarities: All bacteria and archaea are prokaryotic and unicellular.

• Fundamental differences: • Bacteria have cell walls made of peptidoglycan.

• Archaea have unique phospholipids in the cell membranes.

• Bacteria and archaea have different ribosome and RNA polymerase structures.

• Archaea are more closely related to Eukarya than to Bacteria.

I. Why Do Biologists Study Bacteria and Archaea?

A. Biological impact 1. Bacteria and archaea are amazingly abundant. a. A mere teaspoon of soil contains billions of microbial cells. b. A liter of seawater contains a population of microbes equivalent to that of a large human city. c. Microbes living under the ocean may make up to 10 percent of the world’s total living biomass.

2. They are found in every possible environment.

3. They are very diverse, and we are still discovering entire new phyla.

B. Some microbes thrive in extreme environments.

1. Extremophiles are bacteria that live in unusual environments. a. For example, there are bacteria that live at a pH less than 1.0, at temperatures of 0°C under the ice, and in water 5–10 times saltier than seawater. (Fig. 26.1) 2. Studying extremophiles may help us understand the origin of life, since life probably evolved in a high-temperature, anoxic environment.

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3. Astrobiologists use extremophiles as model organisms in the search for extraterrestrial life.

4. Extremophiles are useful in certain commercial and research applications. a. For example, the heat-tolerant enzyme that is necessary for PCR (the DNA-copying technique fundamental to most genetic research) is from an extremophile found in hot springs in Yellowstone National Park.

C. Medical importance 1. Some bacteria are pathogenic, meaning that they cause disease. (Table 26.1) a. Pathogenic forms come from several different lineages in the domain Bacteria. b. Pathogens tend to affect tissues at entry points into the body.

2. Koch’s postulates: Koch proposed four criteria that had to be met to prove that a specific microbe causes a certain disease (they are still used today): a. The microbe must be present in individuals suffering from the disease and absent in healthy individuals. b. The microbe must be isolated and grown in pure culture. c. Injection of the microbe (from the pure culture) into a healthy animal should cause the disease symptoms to appear. d. The microbe should be isolated again from the diseased animal and shown to be identical in size, shape, and color to the original microbe.

3. The germ theory of disease a. The germ theory is based on Koch’s postulates. b. The germ theory states that infectious diseases are caused by microbes (microscopic organisms). c. Infectious diseases are spread in three main ways: (1) from person to person (2) from bites of insects or animals (3) from ingesting contaminated food or water, or environmental exposure d. The germ theory’s immediate impact was in improving sanitation, greatly reducing mortality due to infectious disease.

4. What makes some bacterial cells pathogenic? a. Virulence, or the ability to cause disease, is a heritable trait that varies among individuals in a population. b. Current research is identifying the genes responsible for virulence in a wide variety of bacteria.

5. Some pathogenic bacteria produce resistant endospores a. Endospores are tough, thick-walled, dormant structures formed during times of environmental stress. b. Endospores contain a copy of the cell’s DNA, RNA, ribosomes, and essential enzymes. (Fig. 26.2) c. When conditions become favorable, endospores resume growth as normal cells.

6. Antibiotics are molecules that kill bacteria. a. Since their development in the late 1920s, antibiotics have been very useful in combating infectious disease.

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b. Unfortunately, many pathogens are evolving resistance to antibiotics. c. Biofilms can provide protective structures that shield bacteria from antibiotics.

D. Role in bioremediation 1. Some of the most serious pollutants are hydrophobic compounds that accumulate in sediment and in the bodies of living organisms.

2. Bioremediation strategies use bacteria to break down these compounds. a. Fertilization of contaminated sites encourages the growth of whatever existing bacteria are already on site. These bacteria often degrade the toxic compounds. b. “Seeding” adds specific bacteria that are known to use that pollutant as a food source, producing a nontoxic by-product.

II. How Do Biologists Study Bacteria and Archaea?

A. Using enrichment cultures 1. Enrichment cultures provide a specific set of living conditions (food, temperature, etc.) and can grow large populations of certain bacteria and archaea.

2. Most bacteria and archaea species were discovered when they were grown in a culture in a lab.

3. Example: discovering bacteria from the depths of Earth (Fig. 26.3) a. Samples were taken from 860 to 2,800 meters below Earth’s surface, where temperatures reach 85ºC. b. Scientists hypothesized that if anything were living down there, it would produce magnetite as a by-product of cellular respiration. c. Magnetite did appear in cultures, and microscopy confirmed the presence of previously unidentified thermophiles (bacteria that grow at only high temperatures).

B. Using metagenomics 1. Metagenomics allows biologists to name and characterize organisms that have never been seen and investigate biological processes.

2. DNA is sequenced directly from a small sample from a habitat (soil, water, etc.). The DNA sequences are then compared to known sequences to determine whether any previously undiscovered species exist in the sample. (Fig. 26.4) 3. Metagenomics revealed in one study the similarity and differences in fecal bacteria across 125 subjects. a. Most of the 1,000 species were found across all subjects, but some subjects contained uncommon species. b. The vast majority of bacterial species were from three phyla: Bacteriodetes, Firmicutes, and Actinobacteria. c. Many of the bacterial genes across the species were shared by all human subjects, suggesting the critical importance that bacteria play in human physiology.

C. Evaluating molecular phylogenies 1. Ribosomal RNA sequences have shown that the tree of life has three major lineages: Bacteria, Archaea, and Eukarya. (Fig. 26.5) a. Archaea are more closely related to Eukarya than to Bacteria.

2. Further analysis has identified several monophyletic groups within each of these domains.

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III. What Themes Occur in the Diversification of Bacteria and Archaea?

A. Genetic variation through gene transfer 1. Gene transfer occurs in three ways: a. transformation – when bacteria or viruses naturally take up DNA from the environment b. transduction – when viruses pick up DNA from one prokaryotic cell and transfer it to another cell c. conjugation – when genetic information is transferred by direct cell-to-cell contact 2. Conjugation is often followed by plasmid transfer. A plasmid is copied in one cell and transferred to another cell. (Fig. 26.6a) 3. Conjugation can also result in genetic recombination – the plasmid becomes integrated into the main bacterial chromosome (Fig. 26.6b) B. Morphological diversity 1. Size, shape, and motility (Fig. 26.7) a. Bacteria vary greatly in size. For example, more than a billion of the smallest bacterium could fit inside the largest bacterium. b. Bacteria may be filaments, spheres, rods, chains, or spirals. c. Many bacteria can swim or glide.

2. Cell wall composition (Fig. 26.8) a. Gram-positive bacteria have a cell wall with abundant peptidoglycan, which stains dark purple when exposed to a Gram stain. (Fig. 26.9a) b. Gram-negative bacteria have a cell wall with a thin layer of peptidoglycan surrounded by a phospholipid bilayer. They stain light pink. (Fig. 26.9b) c. Gram stain analysis can predict sensitivity to certain drugs.

(1) Gram-positive bacteria are often sensitive to penicillin-like drugs that disrupt peptidoglycan synthesis.

(2) Gram-negative bacteria are more likely to be affected by drugs that target bacterial ribosomes.

C. Metabolic diversity 1. Bacteria and archaea are astonishingly diverse in the ways they acquire energy to make ATP and the carbon compounds they can use as building blocks.

2. There are three ways to acquire energy to produce ATP: a. Phototrophs use light energy to energize electrons, producing ATP by photophosphorylation (light reactions of photosynthesis). b. Chemoorganotrophs oxidize organic molecules with high potential energy, such as sugars (cellular respiration, fermentation). c. Chemolithotrophs oxidize inorganic molecules with high potential energy, such as ammonia or methane (usually via cellular respiration).

3. There are two ways to acquire carbon: a. Autotrophs use carbon dioxide or methane to build their own carbon-containing compounds. b. Heterotrophs acquire carbon-containing compounds from other organisms.

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4. Overall, there are six major “feeding strategies” (the six possible combinations of three methods of acquiring energy and two methods of acquiring carbon).

(Table 26.2) a. Plants, animals, fungi, and other eukaryotes use only two strategies. b. Bacteria and archaea use all six.

5. Producing ATP via cellular respiration: variation in electron donors and acceptors a. In cellular respiration, electrons are moved from molecules with high potential energy and gradually “stepped down” to a molecule with low potential energy, using the released energy to make ATP. (Fig. 29.10) b. Eukaryotes are chemoorganotrophs that use a sugar like glucose as the electron donor and oxygen as the final electron acceptor. c. Many bacteria and archaea are chemolithotrophs that use an electron donor that is not a sugar and often an electron acceptor that is not oxygen. (Table 26.3) (1) The electron donor may be hydrogen molecules, hydrogen sulfide, ammonia, or methane.

(2) The electron acceptor may be sulfate, nitrate, carbon dioxide, or ferric ions.

6. Producing ATP via fermentation: variation in substrates a. Fermentation is a strategy for making ATP from organic molecules that requires a separate electron acceptor. b. Fermentation is less efficient than respiration in making ATP. c. Some bacteria ferment glucose to either ethanol or lactic acid. d. Other bacteria use a variety of other organic compounds as fermentable substrates.

Examples: ethanol, fatty acids, complex carbohydrates, proteins, amino acids, lactose.

7. Producing ATP via photophosphorylation: variation in electron sources and pigments a. There are three forms of photophosphorylation: (1) Light activates a pigment that transports protons across a membrane, driving the synthesis of ATP via chemiosmosis.

(2) Geothermal radiation can be used instead of light.

(3) Light can be absorbed by pigments that raise electrons to high-energy states.

The electrons are stepped down with electron transport chains, and the energy released is used to make ATP. b. The last mode of photophosphorylation uses electron transport chains that require a source of electrons.

(1) Plants and cyanobacteria use oxygenic photosynthesis. This means they use water as the source of electrons and produce oxygen as a by-product.

(2) Many bacteria use anoxygenic photosynthesis, using a molecule other than water as the source of electrons. Examples: hydrogen sulfide, ferrous iron.

(They also produce by-products other than oxygen.)

8. Obtaining building-block compounds: variation in pathways for fixing carbon a. Autotrophs can build their own carbon compounds.

(1) Plants (and cyanobacteria) fix CO2 with the enzymes of the Calvin cycle.

(2) Some bacteria also fix CO2 but use different enzymatic pathways to do so⎯that is, not the Calvin cycle.

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(3) Some bacteria use a molecule other than CO2 as the starting point. Examples: methane, carbon monoxide, methanol b. Heterotrophs obtain organic compounds from other organisms.

(1) Animals, fungi, and many bacteria and archaea use this strategy.

C. Ecological diversity and global impacts 1. Bacteria and archaea have altered the chemical composition of the oceans, the atmosphere, and terrestrial environments for billions of years.

2. The oxygen revolution a. No free O2 existed in Earth’s atmosphere for 2.3 billion years. b. Cyanobacteria were the first organisms to perform oxygenic photosynthesis, and they are responsible for the appearance of oxygen in Earth’s atmosphere.

(Fig. 26.11) (1) The fossil record shows that cyanobacteria first became numerous in oceans 2.7 to 2.5 billion years ago.

(2) Oxygen concentrations began to increase 2.3 to 2.1 billion years ago. c. Once O2 was abundant, other organisms could use it as an electron acceptor. The evolution of aerobic respiration was a crucial event in the history of life.

(Fig. 26.12) 3. Nitrogen fixation and the nitrogen cycle a. Plant growth is often limited by the availability of nitrogen, but plants cannot use molecular nitrogen (N2) from the atmosphere.

(1) Some bacteria absorb N2 from the atmosphere and fix or reduce it to NH3, ammonia, a form that plants can use.

(2) Other bacteria and archaea convert ammonia to nitrates and nitrites, resulting in a complex nitrogen cycle. (Fig. 26.13) b. Students should be able to provide a plausible explanation of what the composition of the atmosphere and what the nitrogen cycle might be like if prokaryotes did not exist.

4. Nitrate pollution a. Most farmers use synthetic fertilizers, which often contain ammonia, to add nitrogen to soils and increase crop yields. b. Bacteria convert the ammonia in fertilizer runoff into nitrates. c. This has led to worldwide pollution of aquatic ecosystems with excessive nitrates, which causes overgrowth of algae and results in anoxic “dead zones.” (Fig. 26.14)

IV. Key Lineages of Bacteria and Archaea A. Bacteria. At least 29 major phyla now known; 6 are featured here. (Table 26.4) 1. Actinobacteria a. Traits: They form rods, filaments, and sometimes mycelia. Many are heterotrophs.

Some are parasitic. b. Human and ecological impact: They cause tuberculosis and leprosy. Many antibiotics have been isolated from Streptomyces. One species is important for making Swiss cheese. Species in this group live in plant roots and fix nitrogen.

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2. Chlamydiae a. Traits: They are spherical and tiny. All are endosymbionts and live in hosts. b. Human and ecological impact: Chlamydia trachomatis infections cause blindness and urogenital tract infections in humans.

3. Cyanobacteria a. Traits: They are highly diverse in morphology. Some form colonies. All perform oxygenic photosynthesis, and some can fix nitrogen. b. Human and ecological impact: They produce much of the oxygen and nitrogen that other species need.

4. Firmicutes a. Traits: Most are rod shaped or spherical. b. Human and ecological impact: They are very common in the human gut. Some cause anthrax, botulism, tetanus, gangrene, strep throat, and other diseases. Some used to ferment milk to make yogurt or cheese.

5. Proteobacteria a. Traits: Their morphology is very diverse. They use every metabolic strategy except oxygenic photosynthesis. b. Human and ecological impact: They cause Legionnaire’s disease, cholera, food poisoning, dysentery, ulcers, diarrhea, and other diseases. Rhizobium live in the root nodules of legumes and fix nitrogen. Escherichia coli, the best-studied bacterium, is in this group.

6. Spirochaetes (spirochetes) a. Traits: They have a unique corkscrew shape, with flagella housed in a sheath. Most make ATP via fermentation. b. Human and ecological impact: They cause syphilis and Lyme disease. Many live in freshwater and marine habitats, some in anaerobic conditions.

B. Archaea. Several major phyla are known; three are featured here. (Table 26.5) 1. Crenarchaeota are perhaps the oldest lineage of archaeans. a. Traits: They have highly diverse morphology and metabolism. b. Human and ecological impacts: Some are extremophiles and may be the only life-form present in certain environments, such as extremely acid pH and extremely deep sea.

2. Euryarchaeota a. Traits: They have highly diverse morphology and metabolism. Many species produce methane (methanogens). They are found in every conceivable habitat. b. Human and ecological impacts: Ferroplasma live in piles of waste rocks near abandoned mines and produce acids that pollute nearby streams. Methanogens live in the mammalian gut.

3. Thaumarchaeota a. Traits: They are not typically extremophiles, and survive best at moderate temperatures. They are considered ammonia oxidizers. b. Human and ecological impacts: Due to their abundance, they are thought to play a major role in the Earth’s nitrogen and carbon cycles.

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Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

Actinobacteria/high-GC Gram-positives aerobic respiration anaerobic respiration anoxic anoxygenic photosynthesis antibiotic Archaea Autotroph Bacteria bioremediation Calvin cycle cell wall cellular respiration chemolithotroph chemoorganotrophs Chlamydiae Cyanobacteria dead zone electron acceptor electron donor endospores

endosymbiont

nitrogen cycle

enrichment culture Euryarchaeota extremophile fermentation Firmicutes/low-GC Gram-positives germ theory of disease Gram stain Gram-negative bacteria Gram-positive bacteria halophile heterotroph kingdom Koch’s postulates metagenomincs methanogen methanotroph microbe microbiology monophyletic group mycelium (plural: mycelia) nitrate pollution

nitrogen fixation oxygenic photosynthesis pathogen peptidoglycan photosynthesis phototroph phylum (plural: phyla) prokaryotic Proteobacteria seeding Spirochaetes Spirochetes spore sulfate reducer Thaumarchaeota Crenarchaeota thermophile tree of life unicellular universal tree

Lecture Activities Student Research Topic: Is Antibiotic Use in Livestock a Human Health Hazard?

Estimated duration of activity: 10 minutes for discussion (after students have done research on their own at home) Antibiotic resistance is a great topic to use as a learning tool because it is in the news and it integrates many of the concepts that students have been learning throughout the semester (genetics, evolution, bacterial diversity, etc.). The following mini case study can be used as a springboard for an in-class discussion or debate, an out-of-class writing assignment, a group presentation assignment, or an online debate. Of course, the students will need some time to research the issue in order to formulate an educated opinion.

Question: Thirty years ago, Salmonella infections and infections by deadly strains of E. coli were almost unknown. Now they are in the news regularly. Are these infections really more common now, and if so, why? Some have linked the rise of these infections to the

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long-term use of antibiotics in livestock, such as cows and chickens. How could this affect infections in humans? Ask students to research this on their own and report back with their findings.

You may also use this activity as an opportunity to demonstrate good ways to find reliable material (such as Google Scholar, the CDC, etc.), to explain about primary literature and secondary literature, and to warn students about the poor quality of many websites.

Student Research Topic: How Did MRSA Spread to College Campuses?

Estimated duration of activity: 10 minutes for discussion (after students have done research on their own at home) For most of the twentieth century, serious skin infections were rare in Canada and the United States because antibiotics could kill most pathogenic bacteria rapidly. That situation has changed. Many students will have heard of MRSA (methicillin-resistant Staphylococcus aureus), which can cause life-threatening skin infections and is now a persistent problem in many North American colleges. First, ask students whether they have heard of MRSA. (Athletes likely will have been told about it by their coaches. You may even have a few students who have had MRSA infections.) Then ask them to research the history of MRSA. Have them construct a time line for the following events: • Years when penicillin, methicillin, and vancomycin were introduced • Years when S. aureus first evolved resistance to penicillin, methicillin, and vancomycin • When S. aureus first moved out of hospitals and became widespread in the community (“community-associated” MRSA, or CA-MRSA) Also ask students to research how many antibiotics remain that are effective against S. aureus.

Wrap-up: Ask students why S. aureus evolved resistance so rapidly to all of our major antibiotics. (It is due to the widespread use of the antibiotics.) Have students discuss whether or not doctors should prescribe vancomycin widely, and why. Students may at this point appreciate some pointers on the proper care of skin cuts and common-sense precautions.

Game: Fact or Fiction?

Estimated duration of activity: 5 to 10 minutes Microbiology may be one of your students’ favorite subjects if they understand how dependent they are on the microbes of the world. This game is a fun way to introduce the chapter to your students, or you may use it as a platform to present the key lineages of bacteria presented in the chapter.

Read each of the following statements and have students raise their hands when they have determined whether it is fact or fiction. You may then use the bacteria mentioned to discuss a particular lineage.

1. Don’t lick the spoon of the brownie batter after the eggs have been added, because Salmonella is a normal part of the bacterial flora of chickens and is often found in eggshells.

Fact: Salmonella are members of the Proteobacteria lineage.

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2. If you have syphilis, you cannot donate blood.

Fiction: Treponema pallidum, the spirochete that causes syphilis, is killed when donated blood is refrigerated.

3. Chlamydia trachomatis infections are commonly asymptomatic, which makes them a particularly common STD. Fact: Chlamydia trachomatis is a member of the Chlamydiales lineage.

4. Streptomycin is an effective antibiotic in treating Streptomyces infections.

Fiction: Streptomyces are the source of streptomycin. Streptomyces are members of the high-GC Gram-positives lineage.

5. A sample of yogurt with active cultures will reveal masses of Lactobacillus bulgaricus cells.

Fact: Lactobacilli are members of the low-GC Gram-positives lineage.

6. You breathe oxygen thanks to cyanobacteria.

Fact: Cyanobacteria are a major lineage of bacteria.

Discussion Idea: Sexually Transmitted Disease Several types of bacteria can infect the human reproductive tract and cause disease; several other species of bacteria are non-pathogenic and live normally in these environments. The normal pH of the vagina is around 4.7. Women are most susceptible to genital infections when something changes the pH of their vagina. Using birth control pills, IUDs, and other medications can cause such a change. From this information, what can you infer about the optimal growth conditions for pathogenic and non-pathogenic bacteria that grow on genitalia?

What do they have in common, and how do they differ?

Answer: Non-pathogenic genital bacteria grow optimally in low-pH, humid environments.

Many pathogenic genital bacteria grow optimally in neutral-pH, humid environments.

Discussion Idea: Anthrax Anthrax is mainly a disease of large, herbivorous mammals such as sheep, goats, and cows.

Though treatable by antibiotics, the anthrax bacterium is particularly dangerous because of its ability to form endospores. Endospores are formed when the bacterium is exposed to air (spores are usually not found in the bodies of infected animals) and can remain viable in the soil and on surfaces for 50 years or more. Human anthrax infections can be either cutaneous (on the skin) or respiratory (always fatal). What are some ways that agricultural workers might come into contact with and contract anthrax? Discuss why this bacterium poses such a threat as a biological weapon, as opposed to other types of bacteria.

Answer: Agricultural workers may contract anthrax when they touch or inhale spores while working with infected wool or hides. Anthrax bacteria form spores quite easily, and those spores can survive for a long time on inanimate objects. Moreover, once spores germinate in a human, the infection can be fatal, so this bacterium is a particularly effective biological weapon.

Discussion Idea: Look Mom, No Cavities Dental caries, or tooth decay, are caused by contagious bacteria called Streptococcus mutans.

These bacteria feed on refined sugar and produce acid as a by-product of fermentation. The .

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acid breaks down the enamel, and protein-digesting enzymes produced by the bacteria break down other regions of the tooth. Fluoride strengthens the enamel of the tooth; therefore, fluoride treatment protects teeth against decay. Why do you think that brushing and flossing are effective preventive measures? What are some advantages and disadvantages of antibacterial toothpaste?

Answer: Brushing and flossing clear sugar and starches from teeth and gums, thus limiting the food supply for the bacteria. Although antibacterial toothpastes kill the Streptococcus mutans, they may also kill many non-pathogenic bacteria that live in our mouths and digestive tracts. In addition, the regular use of antibiotic toothpaste promotes the evolution of antibiotic-resistant bacteria in the mouth.

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CHAPTER

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Protists

Learning Objectives: Students should be able to...

• Define protists and explain why protists are considered a paraphyletic group.

• Summarize the ecological and medical importance of protists.

• Explain how endosymbiosis played a role in the origin of mitochondria and chloroplasts.

• Describe the diversity of feeding, locomotion, and reproduction seen in protists.

Lecture Outline • Protists are eukaryotes.

• Fundamental features of eukaryotes: • The nuclear envelope is a synapomorphy that defines the domain Eukarya.

• Most eukaryotic cells are large, have many organelles, and have an extensive system of structural proteins called the cytoskeleton.

• Multicellularity has evolved multiple times in eukaryotes.

• Eukaryotes reproduce either asexually via mitosis or sexually via meiosis.

• Protists are all the eukaryotes that are not fungi, green plants, or animals.

• Protists tend to live in or near water. (Fig. 27.1)

I. Why Do Biologists Study Protists?

A. Impacts on human health and welfare 1. The Irish potato famine in 1845 was caused by a protist, Phytophthora infestans. a. One million people (out of a population of 9 million) died, and millions of others emigrated to other countries.

2. Malaria a. Malaria is the world’s worst chronic health problem, affecting 200 million people worldwide each year and killing 600,000 people each year. b. It is caused by five species of the protist genus Plasmodium. c. Plasmodium cells enter the bloodstream via a mosquito bite, infects liver cells and (later) red blood cells, and destroy the cells. (Fig. 27.2) d. Plasmodium is probably the best studied of all the protists, but we still do not have effective measures to control it.

(1) Plasmodium has evolved resistance to many drugs.

(2) Efforts to develop a vaccine have not been successful.

(3) Mosquitoes have continually evolved resistance to insecticides. .

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e. Many other protists also cause human disease. (Table 27.1) 3. Harmful algal blooms a. Algal blooms occur when a unicellular aquatic species experiences rapid population growth. b. Many harmful algal blooms are caused by photosynthetic protists called dinoflagellates, which contain toxins. (Fig. 27.3) c. High levels of toxins build up in shellfish that filter the protists out of the water, and people who eat the shellfish can then get sick or die.

B. Ecological importance of protists 1. Protists are extremely abundant in their habitats. For example, there are more than 500 flagellated protists per milliliter of pond water and 60 thousand dinoflagellates per milliliter of seawater.

2. Protists play a key role in aquatic food chains as primary producers. a. Diatoms and other protists that drift in the ocean are called plankton.

3. Could protists help limit global climate change? a. Phytoplankton in the ocean take up large quantities of carbon dioxide and often carry it to the bottom of the ocean when they die. (Fig. 27.4) b. Fertilizing the ocean with iron speeds up the carbon cycle and might increase the amount of carbon taken into the marine carbon sink. c. Once at the bottom of the ocean, they may enter one of two carbon sinks: (1) Sedimentary rocks—mostly made of calcium carbonate (2) Petroleum—oil formation tends to begin with accumulation of dead bacteria, archaea, and protists

II. How Do Biologists Study Protists?

A. Microscopy: studying cell structure 1. Detailed studies of cell structure revealed that many protists have a characteristic overall form, organelles with distinctive features, or both. (BioSkills 9) a. For example, special flagella with tiny hairlike projections, known as stramenopiles, have been identified on the protist that caused the potato famine and on some types of brown algae. (Fig. 27.5) 2. Seven major groups of eukaryotes have been identified in this way, based on diagnostic morphological features (synapomorphies) that either protect or support the cell or aid in reproduction or feeding. (Table 27.2) B. Evaluating molecular phylogenies 1. DNA sequence analysis has demonstrated that the seven major eukaryotic groups are indeed monophyletic. (Fig. 27.6) (BioSkills 13) 2. Current data suggest that the first major split in the eukaryotic tree was between the Unikonta (one flagellum) and the Bikonta (two flagella).

C. Discovering new lineages via direct sequencing 1. Direct sequencing is based on collecting samples from a habitat and isolating the DNA in those samples⎯that is, without knowing which organism the DNA comes from.

2. A vast array of new protist lineages have been discovered in this way, including many protists that turn out to be as tiny as bacteria—picoplankton.

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III. What Themes Occur in the Diversification of Protists?

A. What morphological innovations evolved in protists?

1. All eukaryotes alive today have (1) either mitochondria or mitochondrial genes, (2) a nucleus and endomembrane system, and (3) a cytoskeleton.

2. Early eukaryotes were probably single-celled organisms with mitochondria, a nucleus and endomembrane system, and a cytoskeleton, but no cell wall. a. Students should be able to map the origin of the nuclear envelope and the eukaryotic flagellum in Figure 27.6.

3. Endosymbiosis and the origin of the mitochondrion a. The endosymbiosis theory (Lynn Margulis, 1970s) proposes that mitochondria evolved from an aerobic bacterium that was engulfed by an anaerobic eukaryotic cell. (Fig. 27.7) (1) This became a mutually beneficial symbiosis: The host supplied the bacterium with protection and carbon compounds, and the bacterium produced much more ATP than the host could produce on its own. b. Many lines of evidence now support this theory.

(1) Mitochondria are similar in size to α-proteobacteria.

(2) They divide independently of the host cell, and they divide by fission, as bacteria do.

(3) They have their own ribosomes and synthesize their own proteins, and the ribosomes are similar in many ways to bacterial ribosomes.

(4) They have double membranes, as would be expected if they were engulfed by another cell.

(5) They have their own chromosomes, which are circular and similar to bacterial chromosomes. c. The most conclusive evidence is that mitochondrial genes are very closely related to the genes from α-proteobacteria 4. Endosymbiosis and the origin of the chloroplasts a. All photosynthetic protists have chloroplasts. b. An extension of the endosymbiotic theory contends that the eukaryotic chloroplast originated when a protist engulfed a cyanobacterium. c. Many lines of evidence support this: (1) Chloroplasts have the same list of bacteria-like characteristics presented earlier for mitochondria.

(2) They contain circular DNA containing genes extremely similar to genes found in various species of cyanobacteria.

(3) The photosynthetic organelle of glaucophyte algae have an outer layer containing the same peptidoglycan found in cyanobacteria.

(4) There are many examples of endosymbiotic cyanobacteria living inside cells of protists or animals today. d. To explain the fact that many protists have chloroplasts containing multiple layers of membrane, researchers proposed the process of secondary endosymbiosis—an organism engulfs a photosynthetic eukaryotic cell and retains the chloroplast.

(Fig. 27.8)

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e. Figure 27.9 shows where primary and secondary endosymbiosis occurred in the phylogenetic tree of eukaryotes. f. Students should be able to assess and explain why the primary and secondary endosymbiosis events introduced in this chapter represent the most massive lateral gene transfers in the history of life, in terms of number of genes moved at once. (Chapter 20, lateral gene transfer) 5. The nuclear envelope a. The nuclear envelope likely arose as infoldings of the plasma membrane that also gave rise to the endoplasmic reticulum. (Fig. 27.10) b. Evidence: Infoldings of the plasma membrane occur in some extant bacteria, and the nuclear envelope and ER of today’s eukaryotes are continuous. c. Students should be able to explain why the above observations support the infolding hypothesis. d. The nuclear envelope allowed for the separation of transcription and translation. e. This enabled alternative splicing and gave early eukaryotes more ways to control gene expression.

6. Diversification of nuclei a. Once a nucleus evolved, it underwent diversification in the types and number of nuclei. b. Diplomonads have 2 nuclei that look identical. (Fig. 27.11a) c. Ciliates have diploid nucleus and a polyploid nucleus. (Fig. 27.11b) d. In foraminifera, red algae, and plasmodial slime molds, certain cells contain many nuclei. (Fig. 27.11c) 7. Structures for support and protection a. The diversification of protists has been associated with the evolution of innovative structures for support and protection: (1) Glasslike shells in a box-and-lid arrangement (diatoms) (Fig. 27.12a) (2) A cell wall made of cellulose plates (dinoflagellates) (Fig. 27.12b) (3) Chambers of calcium carbonate (foraminiferans) (Fig. 27.12c) (4) Coverings of tiny pebbles (other foraminiferans and some amoebae) (5) An internal support rod (parabasalids) (6) A collection of protein strips under the plasma membrane (euglenids) (Fig. 27.12d) (7) Saclike structures (alveolates) 8. Multicellularity a. Multicellular organisms contain more than one cell and have cells that are specialized for different functions and express different genes. b. The vast majority of multicellular species are eukaryotes. c. Multicellularity evolved independently in several eukaryotic lineages: green plants, fungi, animals, brown algae, slime molds, and red algae.

B. How do protists obtain food?

1. Protists have three ways of feeding: ingesting packets of food, absorption of molecules directly from the environment, and photosynthesis.

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a. Some species of protists use multiple feeding methods, and all three methods may occur within a single lineage. Example: alveolates 2. Ingestive feeding a. The large cell size of protists enabled some of them to develop a unique feeding strategy: ingesting other organisms (ingestive feeding). b. Ingestion can occur by eating live or dead organisms or scavenging loose bits of organic debris. c. Protists that feed by phagocytosis typically lack a cell wall and can move their plasma membranes around their prey using pseudopodia. (Fig. 27.13a) d. Some ingestive feeders sit and wait for prey to come by or have cilia that move the environment by them. (Fig. 27.13b) 3. Absorptive feeding a. Some absorptive-feeding protists are decomposers and feed on detritus (dead organic matter). b. Others live inside other organisms.

(1) Absorptive feeders that damage their hosts are called parasites.

4. Photosynthesis a. Organisms that produce their own food through photosynthesis are called autotrophs. b. Organisms that ingest or absorb food are called heterotrophs.

C. How do protists move?

1. Amoeboid motion is a sliding movement produced by extensions of the cell, called pseudopodia, into which the cytoplasm streams. (Fig. 27.14) 2. Other protists move using cilia or flagella. a. Flagella and cilia are identical in structure; each is composed of nine sets of doublet microtubules and two central, single microtubules. b. Flagella are long and usually single or paired (Fig. 27.15a). Cilia are short and numerous. (Fig. 27.15b) 3. Even closely related protists can use very different forms of locomotion.

D. How do protists reproduce?

1. Sexual reproduction evolved in protists and is one of the most significant evolutionary innovations of the eukaryotes. a. Asexual reproduction occurs via mitosis and cell division (protists) or fission (bacteria, archaea). Offspring are genetically identical to the parents. b. Sexual reproduction occurs via meiosis and fusion of gametes. Offspring are genetically different from the parents.

2. Sexual versus asexual reproduction a. Genetically variable offspring have a better chance of surviving in a changing environment. b. Pathogens and parasites are highly variable aspects of the environment because they evolve quickly. Thus, sexual reproduction may primarily be an adaptation to fight disease and parasites.

3. Students should be able to evaluate and explain whether it is consistent with the observation that many protists undergo meiosis only when food is scarce or population density is high. .

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4. Lifecycles⎯haploid versus diploid dominated a. Virtually every aspect of a lifecycle is variable among protists. b. The evolution of meiosis created a distinction between haploid and diploid phases in lifecycles. c. In the diploid-dominated lifecycle, meiosis occurs in the mitotically produced offspring rather than the zygote, and gametes are the only haploid cells in the lifecycle.

(Fig. 27.16a) d. The diploid phase is longer and more prominent than the haploid phase.

(Fig. 27.16a) (Making Models 27.1) 5. Lifecycles⎯alternation of generations (Fig. 27.17) a. Alternation of generations is a type of lifecycle that includes both multicellular haploid and multicellular diploid phases. b. Diploid sporophytes produce haploid spores by meiosis; haploid gametophytes produce haploid gametes by mitosis.

IV. Key Lineages of Protists (Table 27.3) • There are seven major lineages in the Eukary (six are described in this section).

• Each of the seven has at least one distinctive morphological characteristic but has also diversified greatly in other ways.

• There have been multiple independent origins of parasitism, photosynthesis, and multicellularity.

A. Amoebozoa 1. Amoebozoa lack cell walls, feed by engulfing, and move by amoeboid movement with large, lobelike pseudopodia. a. Students should be able to mark the origin of this lineage’s amoeboid form in Figure 27.6.

2. They include lobose amoebae, cellular slime molds, and plasmodial slime molds.

3. Plasmodial slime molds (Fig. 27.11) a. Traits: All form a large, multinucleated “supercell,” feed on decaying vegetation, move by cell crawling, and use sexual reproduction when food is scarce. b. Human and ecological impacts: They are important decomposers in forests. c. Students should be able to mark the origin of the supercell and the origin of this lineage’s amoeboid form in Figure. 27.6.

B. Excavata 1. Excavata have a feeding groove on one side and have lost their mitochondria. a. Students should be able to mark the loss of mitochondria in Figure 27.6.

2. They include parabasilids, diplomonads, and euglenids.

3. Parabasalids, Diplomonadids, and Euglenids a. Traits: Parabasalids have distinctive rods of cross-linked microtubules. Diplomonadids possess two nuclei, each with four flagella. Euglenoids have a system of interlocking protein structure to support the cell and some possess light sensitive eyespot. Nearly all members of Excavata reproduce only asexually.

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b. Human and ecological impacts: The parabasalid Trichomonas and the diplomoinid, Giardia intestinalis are widespread human pathogens. Euglenids are important components of freshwater food chains.

C. Plantae 1. Plantae have chloroplasts and perform photosynthesis.

2. The term Plantae is increasingly used to include glaucophyte algae (“blue-green algae”), red algae, green algae, and land plants.

3. Students should be able to mark the origin of the plant chloroplast in Figure 27.6.

4. Red algae a. Traits: They are primarily marine; their cell walls have cellulose; and they are photosynthetic, many with red pigments that allow them to live in deeper water. Most are multicellular. They are the only type of algae that lack flagella. Alternation of generations is common. b. Human and ecological impacts: They stabilize coral reefs. The nori seaweed (Porphyra) used to wrap sushi is a red alga.

D. Rhizaria 1. Rhizaria are single-celled amoebae that lack cell walls; move by amoeboid motion; and produce long, slender pseudopodia. a. Students should be able to mark the origin of this lineage’s amoeboid form in Figure 27.6. Students should also be able to explain whether amoeboid form evolved just once or independently in separate protist lineages.

2. They include foraminifera, actinopods, chlorarachniophytes, and many others.

3. Foraminiferans (Fig. 27.18) a. Traits: All are marine, have multiple nuclei per cell, produce shells stiffened by calcium carbonate, feed by engulfing, and have no locomotion; they just float in water.

They use both asexual and sexual reproduction. b. Human and ecological impacts: They are abundant in marine sediments, where they form deposits that become chalk, limestone, and marble.

E. Alveolata 1. Alveolata are unicellular and have small sacs called alveoli just under the cell membrane.

2. They include ciliates, dinoflagellates, apicomplexans, and others.

3. Students should be able to mark the origin of alveoli in Figure 27.6.

4. Ciliates, Dinoflagellates, and Apicomplexans a. Traits: Ciliate cells have two nuclei: macronucleus and micronucleus. Dinoflagellate species each has a distinct shape some maintained by cellulose plates. Some species are bioluminescent. Apicomplexans lack flagella or cilia. b. Human and ecological impacts: Ciliates are abundant marine plankton and live in the guts of grazing animals, helping them digest cellulose. Photosynthetic dinoflagellates are important primary producers in marine environments. Two important apicomplexans are Plasmodium (malaria) and Toxoplasma.

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F. Stramenopila (Heterokonta) 1. Stramenopila have a unique “hairy” flagellum at some point in their lifecycle.

2. They include water molds, diatoms, and brown algae.

3. Students should be able to mark the origin of “hairy” flagellum in Figure 27.6.

4. Water molds, Diatoms, and Brown Algae a. Traits: Some water molds are unicellular; others are multicellular with long, branching filaments called hyphae. Most diatoms are unicellular, often in a box-and-lid arrangement. Brown algae are multicellular with bodies consisting of leaflike blades, a stipe, and a holdfast. Most species of stramenopiles exhibit diploid dominant lifecycles. b. Human and ecological impacts: Water molds are important decomposers in aquatic ecosystems. Diatoms are considered the most primary producer in marine and freshwater systems. Brown algae form kelp forests in cold marine habitats.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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absorptive feeding

direct sequencing

macronucleus

alternation of generations Alveolata amoeba (plural: amoebae) amoeboid motion Amoebozoa Apicomplexa asexual reproduction Bikonta bioluminescence blade brown algae cellular slime mold Ciliata cilium (plural: cilia) conjugation cytoskeleton decomposer detritus diatom diatom diatomaceous earth Dinoflagellata dinoflagellate diplomonad

endomembrane system endosymbiosis theory engulfing euglenid Eukarya eukaryote Excavata fertilization fission flagellum flagellum (plural: flagella) food chain Foraminifera gamete gametophyte glaucophyte algae global carbon cycle global warming green algae harmful algal bloom holdfast hypha (plural: hyphae) ingestive feeding life cycle light microscopy

malaria meiosis micronucleus mitochondria mitosis multicellularity Myxogastrida nuclear envelope Opisthokonta parabasalid paraphyletic parasite petroleum (oil) picoplankton plankton Plantae plasmodial slime mold Plasmodium primary endosymbiosis primary producer protist pseudopodia red algae Rhizaria secondary endosymbiosis

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sedimentary rocks sessile sexual reproduction shell spore

sporophyte stipe Stramenopila symbiosis synapomorphy

transmission electron microscopy Unikonta water mold

Lecture Activities Mystery Protists: A Student-Directed Examination of Protist Lineages Estimated duration of activity: 10–20 minutes After presenting the general features of protists, give small groups of students a collection of data (below) describing some newly identified species of protists. Each group is responsible for classifying one protist species. The data include direct sequencing data, morphological data, and field observations of each organism. The students’ job is to go through Section 30.4 on their own and use the information to decide which group each protist belongs to. Encourage students to name their newly discovered “faux” species. After giving the groups 10–15 minutes to classify their organisms, have a spokesperson from each group present the group’s decision to the class while answering the questions below. In effect, the students will be lecturing one another on the features of each protist. You can motivate students to take notes on the other groups’ presentations by promising a test question related to the activity. If class time is limited, have each group post its conclusions on a class website or chat room.

Questions: 1. Which of the seven major eukaryotic lineages does this species belong to? Within that major lineage, which specific protist sublineage does it belong to?

2. What morphological features of this organism made you think it belonged in those taxa?

3. How and what does this organism (or its closest relatives) eat?

4. Does this organism move? What about its relatives?

5. How does this organism and its closest relatives reproduce?

6. What impact does this organism and its closest relatives have on humans or on their native ecological community?

Mystery Protist A • Direct sequence data show significant sequence homology to Cyclotella, whose shells are often harvested to be used in industrial filtering.

• Morphological analysis: It is covered by a glassy shell and contains photosynthetic pigments. Reproductive cells have two flagella.

• Field observations: It was isolated from a marine water sample and has been observed gliding via microtubules that extend from its shell.

Answer: This organism is a diatom, in the Stramenopila.

Mystery Protist B • Direct sequence data show significant homology with Giardia intestinalis, the parasite that •

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can cause severe intestinal distress in humans.

Morphological analysis: It is flagellated and has two nuclei and no cell wall.

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• Field observations: It has been observed to reproduce asexually. It was isolated from the feces of cats.

Answer: This organism is a diplomonad, in the Excavata.

Mystery Protist C • Direct sequence data show significant homology with dinoflagellates and apicomplexans.

• Morphological analysis: It has a swimming cell covered with cilia. It has two nuclei—a smaller one and a larger one.

• Field observations: It was isolated from a sample of goat feces and has been observed reproducing sexually via conjugation, trading its smaller nuclei.

Answer: This organism is a ciliate, in the Alveolata.

Mystery Protist D • Direct sequence data show high homology with Dictyostelium discoideum.

• Morphological analysis: It forms a gigantic, web-shaped cell with many nuclei, and it moves with amoeboid motion.

• Field observations: It was isolated from a forest soil sample.

Answer: This organism is a plasmodial slime mold, in the Amoebozoa.

Mystery Protist E • Direct sequence data show significant homology to Porphyra, which is harvested for sushi in eastern Asia.

• Morphological analysis: It is red, has photosynthetic pigments, and has no flagella.

• Field observations: It was isolated from an ocean-water sample near a coral reef.

Answer: This organism is a red alga, in the Plantae.

Mystery Protist F • Direct sequence data show significant homology to Pyrgo anomala, which contributes to chalk- and limestone-forming sediments.

• Morphological analysis: It has multiple nuclei. It has no cell wall but does have a shell made of calcium carbonate through which pseudopodia extend.

• Field observations: It was isolated from a marine-water sample, in which numerous individuals were found drifting passively in the water, feeding by extending pseudopodia out through openings in their shells. Numerous shells were collected from the ocean floor as well.

Answer: This organism can be classified as a foraminferan, in the Rhizaria.

Mystery Protist G • Direct sequence data show significant homology with diplomonads and parabasalids.

• Morphological analysis: It does not have a cell wall but does have a network of protein molecules just under the plasma membrane that stiffen the cell. It is photosynthetic but produces an unusual carbohydrate unlike starch.

• Field observations: It was isolated from a freshwater sample. It is easy to observe under a light microscope because it actively swims toward light.

Answer: This organism can be classified as a euglenid, in the Euglenida.

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Discussion Idea: What Good Is a Blob of Carotenoids?

Algae are photosynthetic protists that may be unicellular, colonial, or multicellular. Some algae are motile and flagellated; others are not. In addition to their chloroplasts, many unicellular algae have a separate structure that contains light-absorbing pigments, which are often carotenoids. This structure does not perform photosynthesis. It is found only in flagellated unicellular algae or flagellated colonial algae and is never present in nonmotile unicellular or multicellular algae. Based on this information, what do you suppose is the function of the structure? Why is it found only in motile unicellular or colonial algae?

Answer: This structure is called an “eyespot,” and the name gives a clue to its function.

The eyespot is sensitive to light. Photosynthetic organisms need to be bathed in light in order to carry out photosynthesis. In the water, motile unicellular or colonial algae need to position themselves so that they are in contact with light. Since nonmotile algae cannot actively position themselves at a selected location in the water, they have no need for an eyespot. The eyespot also enables algae to detect the intensity of light. If light is too intense, the motile algae can move away from it to avoid photo-oxidizing (“burning out”) their photosynthetic pigments. If the light is dim, motile algae can swim toward it to maximize their exposure.

Game Idea: What’s My Phylum?

At this point in the class, students have learned about many species of bacteria, archaea, and protists, and they may benefit from a refresher lesson to remember which organism is in which category. You can make a review exercise into a game. Call out the names of those organisms you have featured in class, or the diseases caused by them, or other key traits of those organisms. The first student, or group of students, to call out the correct answer gets a point. To encourage shy students to participate, you may have teams of students consult with one another (“Family Feud” style), with a group spokesperson then calling out the consensus answer. For an easier game, have students say only whether the organism is a bacterium or eukaryote; for an additional challenge, have students also name the specific lineage of the organism.

Sample starting list: sushi seaweed, leprosy, tuberculosis, malaria, Giardia, Salmonella, potato blight, sudden oak death, anthrax, Chlamydia, red tides.

Protist Videos Two excellent videos are available that, together, cover several important groups of protists.

Each video is approximately 25 minutes long. The videos are produced and distributed by BioMedia Associates, P.O. Box 457, Loomis, California, 95650.

• The Biology of the Ciliates has excellent footage, shot through the microscope, of Paramecium, Didinium, and Vorticella, among others. The sequence on Paramecium includes an explanation of ciliary movement and shows feeding and reproductive processes. Information about how to collect ciliates from ponds and streams is also presented.

• The Biology of Flagellates and Amoebas includes sequences on several species of Euglena as well as Volvox and Trypanosoma.

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CHAPTER

28

Green Algae and Land Plants

Learning Objectives: Students should be able to...

• Discuss the ecological importance of green algae and land plants.

• Describe the evolutionary adaptations that allowed plants to survive and reproduce on land.

• Compare and contrast the lifecycles and key traits of green algae, nonvascular plants, seedless vascular plants, gymnosperms, and angiosperms.

Lecture Outline I. Why Do Biologists Study Green Algae and Land Plants?

A. Plants provide ecosystem services.

1. Plants improve the atmosphere, surface water, and soil in ways that benefit other organisms.

2. Plants produce oxygen.

3. Plants build and hold soil. (Fig. 28.1) 4. Plants hold water and moderate climate.

5. Plants are primary producers. a. Plants are the primary food producers on Earth. b. Plants produce sugars and oils, via photosynthesis, that provide the foundation for virtually all food chains. c. Herbivores eat plants, carnivores eat herbivores, and omnivores eat many sources of food. d. Plants are the key to the carbon cycle on land.

(1) Plants take CO2 from the atmosphere and use it to make sugar.

(2) Plants fix more CO2 than they generate.

6. Students should be able to explain the potential impacts of deforestation.

B. Plants provide humans with food, fuel, fiber, building materials, and medicines.

1. Food a. Crop plants were domesticated at several different locations between 10,000 and 2,000 years ago. (Fig. 28.2a) b. Humans gradually changed the characteristics of food plants via artificial selection.

Example: kernel size in maize (Fig. 28.2b) 2. Fuel a. For thousands of years, wood was the primary source of fuel used by humans.

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b. In recent centuries, coal—formed by plants that lived hundreds of millions of years ago—has replaced wood as our primary energy source.

3. Fiber and building materials a. Cotton and other plant fibers are used in clothing and other cloth-based household items. b. Woody plants are used to make lumber and paper.

4. Medicines a. Plants contain many compounds helpful in fighting disease. (Table 28.1)

II. How Do Biologists Study Green Algae and Land Plants?

A. Analyzing morphological traits 1. Similarities between green algae and land plants a. Both have chloroplasts with chlorophyll a and b and beta-carotene. b. They have similar structure and composition of their thylakoids, cell walls, sperm, and peroxisomes. c. They both synthesize starch as a storage product. d. The ancestor of land plants was probably a multicellular green alga that lived in freshwater, similar to stoneworts. (Fig. 28.3) 2. Major morphological differences among land plants a. Nonvascular plants lack vascular tissue. (Fig. 28.4a) b. Seedless vascular plants have vascular tissue but do not make seeds. (Fig. 28.4b) c. Seed plants have vascular tissue and make seeds. (Fig. 28.4c) B. Using the fossil record (Fig. 28.5) 1. The first-known green algae lived between 700 and 725 million years ago (mya), when oxygen levels began to rise.

2. Origin of land plants a. The first-known land plants occurred 475 mya. b. These fossils resemble today’s land plants because they have sheets of waxy cuticle, spores with sporopollenin-containing walls, and spores contained within specialized structures called sporangia.

3. Silurian-Devonian explosion a. Macroscopic fossils from all major lineages of land plants are found from about 416 to 359 mya. b. Most structural adaptations to terrestrial life appeared: vascular tissue and roots. c. Plants probably colonized land with the assistance of fungi.

4. The Carboniferous period a. Extensive coal deposits formed between 359 and 299 mya. b. Most of the coal was formed from club mosses, horsetails, and ferns.

5. Diversification of gymnosperms a. Gymnosperms are prominent in the fossil record from 299 to 145 mya. b. Gymnosperms grow well in dry habitats, so plants probably colonized drier environments during this time.

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6. Diversification of angiosperms a. Fossil angiosperms appeared about 150 mya and are still dominant today.

7. Summary of the fossil record: The green algae appear first, followed by the nonvascular plants, seedless vascular plants, and seed plants. Organisms that appear late in the fossil record are much less dependent on water than groups that appear earlier.

C. Evaluating molecular phylogenies (Fig. 28.6) 1. The current molecular phylogenetic tree suggests these interpretations: a. The green algal group Zygnematophyceae is the closest living relative to land plants. b. Nonvascular plants are the most ancient living group of land plants. c. Monophyletic groups are: green plants, land plants, vascular plants, seed plants, gymnosperms, and angiosperms. d. Paraphyletic groups are: green algae, nonvascular plants, and seedless vascular plants.

2. Students should be able to explain how morphological data, the fossil record, and molecular phylogenies all support the hypothesis that land plants evolved from green algae.

III. What Themes Occur in the Diversification of Land Plants?

A. The transition to land, I: How did plants adapt to dry conditions with intense sunlight?

1. Terrestrial habitats have more sunlight and more CO2 than aquatic habitats.

2. Preventing water loss: cuticle and stomata a. The cuticle is a waxy sealant that prevents the loss of H2O but also inhibits the uptake of CO2. (Fig. 28.7a) b. Stomata are pores bounded by guard cells that allow the uptake of CO2 while controlling water loss. (Fig. 28.7b) c. Liverworts have simple pores; all other land plants today have stomata.

3. Providing protection from UV irradiation a. Early plants contained compounds that could absorb damaging UV light.

4. The importance of upright growth a. Early plants probably had a low, sprawling growth habit. b. Plants capable of vertical growth would have an advantage in competing for space and light. c. Two problems had to be overcome: transporting water up the plant from the soil and becoming rigid enough to withstand wind and gravity.

5. The origin of vascular tissue a. Fossilized plants dated to 400 mya had elongated cells containing lignin that were probably vascular tissue. (Fig. 28.8a, b) 6. Elaboration of vascular tissue: tracheids and vessels a. Tracheids have long, thin tapered ends, with lignin rings in secondary cell walls for support. Water flows through pits in secondary cell walls. Tracheids first appear in fossils dated to 380 mya. (Fig. 28.8c) b. Vessel elements are shorter and stacked end to end like a pipe. Water flows through perforations that completely lack cell-wall material. Vessels first appear in fossils dated to 250 to 270 mya. (Fig. 31.8d) c. Both tracheids and vessel elements are dead and lack cytoplasm at maturity.

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B. Mapping evolutionary changes on the phylogenetic tree (Fig. 28.9) 1. Fundamentally important adaptations to dry conditions—such as the cuticle, pores, stomata, vascular tissue, and tracheids—evolved only once.

2. Water-conducting cells evolved independently in mosses and vascular plants.

3. Vessels evolved independently in gnetophytes and angiosperms.

C. The transition to land, II: How do plants reproduce in dry conditions?

1. Desiccation-resistant sporopollenin-encased spores were one of the first innovations making the colonization of land possible.

2. Protective, complex reproductive organs a. The evolution of an elaborate gametangium protected gametes from drying and from mechanical damage. b. There are two types of gametangia: Antheridia produce sperm, and archegonia produce eggs. (Fig. 28.10) 3. Embryos nourished by parental tissues a. Land plants retain their eggs in the archegonia. Once fertilized, the zygote begins to develop on the parent plant and receives nutrients from it. b. For this reason, plants are called embryophytes.

4. Alternation of generations a. In conjugating algae, coleochaetes, and stoneworts, the multicellular form is haploid; the only diploid form is the zygote. (Fig. 28.11) b. Plants alternate between a multicellular haploid stage and a multicellular diploid stage. (Fig. 28.12) (1) The diploid sporophyte produces haploid spores by meiosis.

(2) A spore grows via mitosis to become a new gametophyte.

(3) A haploid, multicellular form called the gametophyte produces haploid gametes via mitosis.

(4) Two gametes fuse to form a diploid zygote.

(5) The zygote grows via mitosis to become the diploid multicellular form, called the sporophyte. c. Alternation of generations is a difficult topic to master—compare and contrast the lifecycle of land plants with that of green algae. (Making Models 28.1) d. Students should be able to summarize the sequence of events in a lifecycle with alternation of generations starting with a sporophyte.

5. From gametophyte-dominant to sporophyte-dominant a. In nonvascular plants, the dominant form is the haploid gametophyte. (Fig. 28.13) b. In ferns, the diploid sporophyte is larger. (Fig. 28.14) c. In gymnosperms and angiosperms, the sporophyte is dominant and the microscopic gametophyte is retained within the sporophyte. d. The transition from gametophyte-dominated lifecycles to sporophyte-dominated lifecycles is one of the most striking trends in land plant evolution. e. Students should be able to examine the photos of hornworts (a nonvascular plant) and horsetails (a seedless vascular plant) in Figure 28.15 and identify which is the gametophyte and which is the sporophyte.

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6. Heterospory a. All nonvascular plants and most seedless vascular plants are homosporous, meaning they produce a single type of spore. (Fig. 28.16a) b. Seed plants are heterosporous, with two distinct spore-producing structures that produce two types of spores. (Fig. 28.16b) (1) Microsporangia make microspores. Microspores grow to become male gametophytes, which produce the sperm.

(2) Megasporangia make megaspores. Megaspores grow to become female gametophytes, which produce the eggs.

(3) The gametophytes of seed plants are either male or female, but never both.

7. Pollen a. A pollen grain is a tiny male gametophyte surrounded by a tough coat of sporopollenin. b. When pollen evolved, heterosporous plants no longer needed water to accomplish fertilization. Instead of swimming to the egg as a naked sperm cell, the tiny male gametophytes took to the skies.

8. Seeds a. A seed includes an embryo and a food supply surrounded by a tough coat. b. Seeds enable embryos to be dispersed to new habitats. (Fig. 28.17) c. A summary of the traits of seed plants can be seen in the lifecycle of a pine tree.

(Fig. 28.18) 9. Flowers a. Flowering plants (angiosperms) are the most successful land plants. b. Flowers contain two reproductive structures: stamens and carpels. (Fig. 28.19) (1) Stamens have anthers, which contain the microsporangia.

(2) Carpels have ovaries, containing ovules, which contain the megasporangia. (i) The evolution of the ovary was a key innovation that protects female gametophytes from insects and other predators. c. In double fertilization, a pollen grain produces two sperm cells. One sperm cell fuses with the egg to form an embryo. The other sperm cell fuses with two nuclei in the female gametophyte to form triploid endosperm.

(1) The adaptive significance of double fertilization is still not understood.

10. Pollination by insects and other animals a. Sepals and petals enclose the ovary and give flowers a wide variety of colors, shapes, and smells. b. Directed-pollination hypothesis: Flowers are an adaptation to attract specific pollinators (rather than leaving pollination to the wind), thus increasing the probability that pollination will occur. c. Evidence for this hypothesis: The characteristics of a flower (scent, flower shape, color, etc.) correlate with the characteristics of its pollinator.

(1) Examples: carrion flowers, hummingbird-pollinated flowers, bee-pollinated flowers (Fig. 28.20) d. Experimental evidence: flower color in South American petunias

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(1) Flowers with genetically altered flowers were visited differently by different pollinators. (Fig. 28.21) (2) Conclusion: Flower color is an adaptation that determines the type of pollinator visiting the flower.

11. Fruits a. A fruit is a structure derived from the ovary and encloses one or more seeds.

(Fig. 28.22a) b. A fruit is an adaptation for seed dispersal. Fruit tissues are nutritious and attractive, so animals eat the fruit and disperse the seeds via their feces. (Fig. 28.22b) 12. Summary: Once land plants could grow efficiently in dry habitats, traits associated with pollination and seed dispersal drove further diversification. (Fig. 28.23) D. The angiosperm radiation 1. The Anthophyta (angiosperms, flowering plants) represent one of the significant adaptive radiations in evolutionary history.

2. Angiosperms were traditionally divided into monocots and dicots. (Fig. 28.24) a. Monocots have one cotyledon (embryonic leaf). Examples: grasses, orchids, palms, lilies b. Dicots have two cotyledons. Examples: roses, daisies, maples 3. Genetic analysis has shown that monocots are monophyletic but dicots are not.

(Fig. 28.25) a. The term eudicot is now used for the monophyletic group that includes roses, daisies, and maples.

E. Students should be able to (1) explain why the evolution of cuticle, UV absorbing compounds and vascular tissue was important in survival and reproduction and (2) using Figure 28.6, describe where the origin of pollen, flowers, and fruits occurred.

IV. Key Lineages of Green Algae and Land Plants A. Green algae (Table 28.2) 1. Traits common to all green algae: They have double-membraned chloroplasts containing chlorophyll a and b; they are important producers in near-shore ocean and freshwater environments; and many have symbiotic relationships (lichen, freshwater protists). (Fig. 28.26) 2. Ulvophyceae (ulvophytes) a. Traits: They are marine or freshwater, most reproduce both asexually and sexually, and they often have a dormant winter stage. Many are diploid only as zygotes, although some multicellular species have alternation of generations. b. Human and ecological impacts: They are primary producers in freshwater and coastal marine environments. They include Ulva, the sea lettuce.

3. Charophyceae (stoneworts) a. Traits: Stonewarts are strictly freshwater, accumulate crusts of calcium carbonate, and have plasmodesmata. Sexual reproduction is common, with gametangia similar to those of land plants; eggs are retained on the parent; and the zygote is the only diploid stage. b. Human and ecological impacts: They form extensive beds in lakes that are food and shelter for fish, geese, and ducks. They are good indicators of water quality. .

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4. Coleochaetophyceae (coleochaetes) a. Traits: They are strictly freshwater and grow as flat sheets (with plasmodesmata) attached to aquatic plants or rocks. Asexual reproduction is common, with flagellated spores. During sexual reproduction, eggs are retained and nourished by the parent. b. Human and ecological impacts: Scientists study them for insight into how plants colonized land.

5. Zygnematophyceae (conjugating algae) a. Traits: They are freshwater and have unicellular and multicellular forms. Conjugation occurs between cells of adjacent filaments. Zygotes are the only diploid stage. b. Human and ecological impacts: Recent evidence of their close relationship to land plants.

B. Nonvascular plants (Table 28.3) 1. Traits: Most have a low, sprawling growth habit, anchored by rhizoids. All have flagellated sperm that swim to eggs, and wind-dispersed spores. Sporophytes are small and dependent on the larger gametophyte.

2. Hepaticophyta (liverworts) a. Traits: Liverworts are commonly in dense mats on damp forest floors or riverbanks.

Adaptations to land include a cuticle and pores. b. Human and ecological impacts: They contribute to soil production. Marchantia is an important model organism.

3. Bryophyta (mosses) a. Note that Bryophyta includes only the mosses, not other bryophytes. b. Traits: There is simple conducting tissue in one group, but no lignin. Mosses can often dry out and become dormant without dying. c. Human and ecological impacts: Sphagnum is one of the most abundant plants on Earth. It produces peat, which stores a large amount of carbon. Peat is widely used in gardening and as a cooking fuel.

4. Anthocerophyta (hornworts) a. Traits: Hornworts have a hornlike appearance and stomata. b. Human and ecological impacts: Some species harbor cyanobacteria that fix nitrogen.

C. Seedless vascular plants (Table 28.4) 1. This paraphyletic group forms a grade between nonvascular plants and seed plants.

2. All the seedless vascular plants have conducting tissues with lignin. The gametophyte is still physically independent, but the sporophyte is the larger phase. Sperm swim to eggs.

3. Lycophyta (lycophytes, or club mosses) a. Traits: These small plants live on the forest floor or on tropical trees. They have true roots (i.e., with vascular tissue) and unusual leaves called microphylls.

Gametophytes may live a long time, sometimes underground with symbiotic fungi. b. Human and ecological impacts: They dominated the coal-generating forests during the Carboniferous period.

4. Psilotophyta (whisk ferns)

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a. Traits: Only six distinct species remain today—all tropical. They have secondarily lost roots and leaves. Some gain nutrients from fungi in underground rhizomes, and others are epiphytes. The gametophytes are tiny and live underground. b. Human and ecological impacts: Some are popular landscaping plants.

5. Pteridophyta (ferns) a. Traits: Ferns are the most speciose of the seedless vascular plants (12,000 species) and the only ones to have large, well-developed leaves (fronds). They are homosporous. Sporangi found in clusters called sori are on the underside of the fronds. b. Human and ecological impacts: Young ferns (fiddleheads) may be eaten and many are used as ornamental plants in landscaping.

6. Equisetophyta (horsetails) a. Traits: Only 15 species remain. They prefer wet habitats and have hollow stems that allow oxygen diffusion to the roots. Asexual reproduction is common, via fragmentation or rhizomes. Rhizomes can produce two types of stems, one for photosynthesis and one for spore reproduction. The gametophytes are small and short-lived. b. Human and ecological impacts: They are rich in silica granules, which are used for scouring pots and pans (“scouring rushes”).

D. Seed plants: Gymnosperms (Table 28.5) 1. Seed plants are a monophyletic group defined by the presence of seeds and pollen grains. Seed plants are heterosporous.

2. Gymnosperms include cycads, ginkgos, the redwood group, and conifers.

3. Ginkgophyta (ginkgoes) a. Traits: Only one species is living today, one of the few deciduous gymnosperms.

Individuals are either male or female. Pollen is transported by wind. Flagellated sperm swim from the pollen grain to the egg cells. b. Human and ecological impacts: They are ornamental and used in many landscapes.

They are often planted in cities because they are tolerant of air pollution.

4. Cycadophyta (cycads) a. Traits: Cycads live in the tropics and look like palm trees. They do not make wood but are supported by stiff stems. Individuals are either male or female, but not both.

Pollen is carried by beetles, weevils, or wind. The seeds are large and brightly colored. b. Human and ecological impacts: Cycads harbor large amounts of nitrogen-fixing cyanobacteria. They are popular landscaping plants.

5. Cupressophyta (redwoods, junipers, yews) a. This was recently found to be a monophyletic group separate from the Pinophyta. b. Traits: Narrow leaves, often arranged as overlapping scales, are an adaptation to dry or cold habitats. All are pollinated by wind. All make wood. c. Human and ecological impacts: This highly rot-resistant wood is popular for home building, furniture, and similar uses. Taxol is from the Pacific yew.

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6. Pinophyta (pines, spruces, firs) a. Traits: Needlelike leaves are an adaptation to dry or cold habitats. All are pollinated by wind. The female cones take two years to mature. b. Human and ecological impacts: They dominate high-latitude and high-altitude forests, they are commercially very valuable for paper production, and their seeds are key food sources for birds and rodents.

7. Gnetophyta (gnetophytes) a. The three diverse genera are tropical vines and trees, desert-dwelling shrubs, and the unique Welwitschia. They have vessel elements in addition to tracheids. All make wood. There are clusters of sporangia (like flowers) at the ends of the stalks.

Two genera have double fertilization. b. Human and ecological impacts: The widely used antihistamine drug ephedrine was isolated from Ephedra.

E. Seed plants: Angiosperms (Table 28.6) 1. Basal angiosperms (oldest living angiosperm lineages) a. Traits: share some features with monocots (pollen with one groove) and eudicots (two seed leaves); thus, this is a paraphyletic assemblage at the base of the angiosperm lineage tree b. Human and ecological impacts: evolved prior to the vast majority of other angiosperms 2. Monocotyledons (monocots) a. Traits: embryo has a single-seed leaf and flower parts typically in multiples of three; lack secondary growth, pollen grains have a single groove, and leaf veins run parallel b. Human and ecological impacts: some of the major world crops—corn, wheat, rice, sugarcane 3. Magnoliids a. Traits: large trees, shrubs, or vines with large flowers and numerous petals and sepals; large net-veined leaves b. Human and ecological impacts: categorized as eudicots incorrectly; used for food, medicines, perfumes, and timber 4. Eudicotyledons (eudicots) a. Traits: embryos have two seed leaves; produce pollen grains with three grooves; monophyletic group sister to the monocots and magnoliids; largest group of angiosperms b. Human and ecological impacts: many species used for food, medicine, perfume, and timber; most trees in deciduous forests are eudicots

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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adaptive radiation alternation of generations antheridium (plural: antheridia) Anthocerophyta Anthophyta archegonium (plural: archegonia) artificial selection Bryophyta bryophytes carbon cycle Carboniferous carnivore carpel Charophyceae club mosses coal coleochaetes Coleochaetophyceae compound leaf cotyledon cuticle Cycadophyta cycads deciduous dicot directed-pollination hypothesis double fertilization ecosystem ecosystem services embryophyte endosperm Ephedra epiphyte Equisetophyta/ Sphenotophyta eudicot ferns fiddlehead flower flowering plants/ angiosperms frond

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fruit gametangium (plural: gametangia) gamete gametophyte gemma (plural: gemmae) ginkgoes Ginkgophyta Gnetophyta gnetophytes green algae guard cell gymnosperms Hepaticophyta herbivore heterospory homospory hornworts horsetails land plant lichen lignin liverworts Lycophyta Marchantia megasporangium (plural: megasporangia) megaspore meiosis microphyll microsporangium (plural: microsporangia) microspore mitosis monocot mosses nectar nonvascular plants omnivore ovary ovule oxygenic photosynthesis peat petal Pinophyta

plasmodesmata pollen pollen grain pollination pollinator pore primary cell wall primary producer Psilotophyta Pteridophyta Redwood group rhizoid rhizome root scouring rushes sea lettuce secondary cell wall seed seed plants seedless vascular plants sepal Silurian-Devonian explosion sorus (plural: sori) Sphagnum sporangium (plural: sporangia) spore spore sporophyte sporopollenin stamen stoma (plural: stomata) stoneworts tracheid transfer cell Ulva Ulvophyceae ulvophytes vascular tissue vessel element whisk ferns wood zygote

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Lecture Activities Plant Taxa “Jeopardy” Estimated duration of activity: 40 minutes The point of this chapter is to explain the importance of studying plants, highlight the evolutionary relationships among green algae and the various lineages of land plants, and introduce students to the diversity of plants. This “Jeopardy” game is a suggested alternative to a lecture presentation of this material. The game allows the students to learn the material and have some fun in the process.

The students will need to read the chapter and take notes on it before class. (To ensure that students are prepared, design an online quiz or hold an in-class quiz based on the game during the next class period. Alternatively, you could allow students to consult their books during the game.) As an incentive, offer prizes such as candy for the winning team or some number of bonus points on the next test. The game will take most of a 50-minute class period, but you should have a few minutes at the end of the game to highlight the major themes of the chapter.

At the beginning of class, divide students into two to four teams. If you teach a large class (more than 60 students), you may want to send half the class to a different room and have a TA run the same activity with them. Then present the following rules: 1. Each team will be given the opportunity to select a question and then answer it. If the answer is correct, that team will be awarded the appropriate number of points. If the answer is incorrect, another team will have the chance to answer. (If you have more than one team, you can draw a number out of a hat to determine the order in which they answer.)

2. There will be a 30-second time limit for team conversation before the answer must be given.

3. Each team should appoint a spokesperson. The answer given by the spokesperson will be considered the final answer.

“Jeopardy” Under the Sea (algae) 100:

The closest living relative of land plants Correct response: What is Charophyceae? (stoneworts)?

300:

The only diploid stage in an algal lifecycle Correct response: What is the zygote?

500:

You might find this lineage of algae when you visit the coast.

Correct response: What is Ulvophyceae (ulvophytes)?

Land Lovers (adaptations to land) 100:

This structure allowed sperm to fly instead of swim.

Correct response: What is pollen (or a pollen grain)?

300:

These cells are like drinking straws for a plant.

Correct response: What are tracheids or vessels?

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500:

This group has a condition analogous to pregnancy in mammals.

Correct response: What is Embryophyta (land plants)?

Can’t Live without ’Em (human uses of plants) 100:

Any organism that eats a plant Correct response: What is a herbivore?

300:

Coal and fossil fuels replaced this plant product as our nation’s main energy source.

Correct response: What is wood?

500:

This anticancer drug was originally purified from a plant.

Correct response: What is vinblastin?

Designer Plants (agriculture and artificial selection) 100:

This is a human-directed process similar to natural selection.

300:

Correct response: What is artificial selection?

Selection by humans has resulted in an increased amount of this substance in corn.

Correct response: What is oil?

500:

Artificial selection _____ genetic diversity.

Correct response: What is decreases?

What Came First? (plant phylogeny) 100:

Lycophyta or Pinophyta Correct response: What is Lycophyta?

300:

Anthocerophyta or Pteridophyta Correct response: What is Pteridophyta?

500:

Cycads or maple trees?

Correct response: What are cycads?

“Double Jeopardy” Which of These Does Not Belong? (Choose the plant that is not in the category.)

200:

Seedless vascular: moss, horsetail, fern Correct response: What is moss?

600:

Nonvascular: liverwort, club moss, hornwort Correct response: What is club moss?

1000: Dicots: lily, rose, daisy Correct response: What is lily?

The Generation Gap (lifecycles, dominant forms, etc.)

200:

A group of plants that has a dominant sporophyte with sporangia in clusters called sori Correct response: What are ferns?

600:

The term for seed plants that produce two different types of spores Correct response: What is heterosporous?

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1000: A carpet-like plant that has a dominant gametophyte that houses and nourishes the small sporophyte Correct response: What is moss?

Long-Distance Love (reproduction) 200:

After fertilization, the ovary of an angiosperm develops into this frequently edible structure.

Correct response: What is a fruit?

600:

Pollination and fertilization are separated by as much as a year’s time in these popular Christmas plants.

Correct response: What are conifers?

1000: Several seedless vascular plants send out these growths that can break off and grow independently.

Correct response: What are gemmae?

Who Am I? (characteristics) 200:

Live only in water; may be unicellular or multicellular; haploid stage dominant Correct response: What are green algae?

600:

Live in most habitats on Earth; seeds dispersed by wind and animals; coevolved with specific pollinators Correct response: What are angiosperms?

1000: Live in tropical environments; an epiphyte; no seeds, no roots; has rhizomes; used in landscaping Correct response: What are whisk ferns?

What’s in a Name? (meanings of all these crazy names) 200:

These plants have cones.

Correct response: What are conifers?

600:

These plants have leaves shaped like livers.

Correct response: What are hepaticophyta or liverworts?

1000: These angiosperms have a single “seed leaf.”

Correct response: What is a monocot?

Challenge Question: Which Came First?

Present the class with the following challenge question based on Figure 28.5 (the time line of plant fossil history). First, ask students to write their ideas on a sheet of paper, and then solicit comments from the class to put together a final answer on the board.

Question: The terrestrial plant fossil record shows only microscopic fossils for about 60 million years, beginning 476 mya. Beginning about 410 mya, the fossils are quite different; they include macroscopic fossil branches, roots, stems, and leaves from a great variety of plants.

Can you list or outline a chain of events, in order, that might have occurred during this period of plant evolution, resulting in the Silurian-Devonian explosion of new plant forms?

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Answer: 1. After the first plants established themselves on land, the land offered a huge amount of unoccupied habitat that could be exploited. Early plants were required to stay close to streams or ponds.

2. However, lots of drier habitat remained open, and those plants in which adaptations to drier conditions developed were then able to radiate into new habitats.

3. The presence of plants growing in a habitat dramatically changed that habitat, creating new conditions—shade, tree trunks on which vines and epiphytes could grow, and so forth.

Challenge Question: The Age of Gymnosperms What reason(s) can you think of for the age of gymnosperms? That is, what adaptive advantage(s) might gymnosperms have had over other plants living at that time?

Answer: The seed, which provided many advantages over spores: 1. The seed coat protects the embryo from desiccation.

2. The seed contains stored food that fuels development of the embryo during germination until it reaches sunlight and can photosynthesize.

3. Some seeds have features on the seed coat that enhance dispersal to a new location, like spines or stickers to catch on animal fur.

4. Most seeds contain inhibitors that enable them to delay germination until conditions are favorable (until water is available or winter has passed).

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CHAPTER

29

Fungi

Learning Objectives: Students should be able to...

• Describe the general morphology and ecological importance of fungi.

• Summarize the types of symbiotic relationships fungi can have with other organisms.

• Compare and contrast the lifecycles and mode of nutrition among fungi, animals, and plants.

• Identify the distinct types of reproductive structures known in fungi.

Lecture Outline • What are the fungi?

• Fungi are one of the three major lineages of multicellular eukaryotes.

• Fungi absorb carbon and energy from other organisms.

• Fungi may be decomposers or parasitic or beneficial symbiotes.

• Fungi play a key role in nutrient cycling in terrestrial ecosystems.

I. Why Do Biologists Study Fungi?

A. Fungi have important economic and ecological impacts.

1. No more than 300 species of fungi cause human disease. Examples: athlete’s foot, diaper rash, vaginitis, pneumonia, thrush 2. Soil-dwelling fungi have been the source of many important antibiotics.

3. Certain fungi destroy billions of dollars of crops each year. (Fig. 29.1) 4. Many fungi are used to make food. a. Mushrooms are a popular edible fungus. b. Yeasts are used to make bread, beer, wine, and other foods. c. Chocolate is made from the seeds of plants fermented by fungi. d. Enzymes derived from fungi improve the characteristics of several foods.

5. Fungi have destroyed significant numbers of trees in North America. Examples: chestnut blight, Dutch elm disease B. Mycorrhizal fungi provide nutrients for land plants.

1. Mycorrhizal fungi are fungi that live in close association with plant roots.

2. The fungi provide the plant with water and key nutrients, such as nitrogen and phosphorus, in exchange for sugars.

3. Without these fungi, plant growth is greatly slowed; therefore, these fungi are important to agricultural productivity. (Fig. 29.2) 284

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C. Saprophytic fungi accelerate the carbon cycle on land. (Fig. 29.3) 1. Carbon cycling during the Carboniferous period a. The coal deposits of the Carboniferous may be the result of a scarcity of saprophytic (decomposing) fungi at the time. b. The lack of fungi allowed large deposits of dead plant tissue to build up, eventually producing coal. c. Biologists hypothesize that because peatlands have an acidic pH, the fungi could not grow in them.

2. Carbon cycling at the end-Permian a. A “fungal spike” (a large increase in fungal fossils) coincides with the end-Permian mass extinction. b. The end-Permian extinction may have involved a massive die-off of land plants, resulting in a population explosion of saprophytic fungi.

3. Carbon cycling at present a. Today, saprophytic fungi speed the terrestrial carbon cycle. b. Fungi are almost the only organisms capable of digesting both cellulose and lignin, recycling carbon into glucose and carbon dioxide. c. Without fungi, carbon would be trapped in the wood for a very long time, and terrestrial environments would be far less productive.

4. Students should be able to explain how fungi help connect two major parts of the cycle: carbon fixation and carbon release.

II. How Do Biologists Study Fungi?

A. About 110,000 species of fungi have been described.

1. A Panama study found that the leaves of just two tree species housed 418 fungal species.

2. A direct sequencing study found that more than 650 species of fungi (200 previously unknown) were found in the guts of 27 species of beetles.

B. Analyzing morphological traits 1. Fungi have only two growth forms: a. Single-celled yeasts (Fig. 29.4a) b. Multicellular, filamentous mycelia (Fig. 29.4b) 2. The nature of the fungal mycelium a. Mycelia can be very long-lived and very large. b. Mycelia constantly grow in the direction of food sources and die back when food is scarce. c. Mycelia are made up of individual filaments called hyphae.

3. The nature of hyphae a. A hypha may have just one haploid nucleus or several (heterokaryotic). Most heterokaryotic hyphae are dikaryotic (two nuclei per cell, one from each parent). b. A hypha is a long, narrow filament that branches frequently. (Fig. 29.5a) c. Each hypha is subdivided by septa (walls) with pores through which cytoplasm, organelles, and other material can flow. Due to this flow, a fungal mycelium is intermediate between a multicellular organism and a large unicellular organism.

(Fig. 32.5b)

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d. Coenocytic fungi have hundreds or thousands of nuclei scattered throughout the mycelium. (Fig. 29.5c) e. Some fungi produce hyphae for prey capture and nutrient uptake. (Fig. 29.6) 4. Mycelia have a large surface area. a. The shape of mycelia gives fungi the highest surface-area-to-volume ratio of all multicellular organisms. b. The large surface area is important because fungi make their living via absorption; however, it also makes fungi susceptible to dying out. c. When conditions dry, fungal mycelium may die back partially or completely.

Spores may also be produced.

5. Reproductive structures a. Chytrids have swimming gametes and spores. (Fig. 29.7a) b. Zygomycetes produce a distinctive spore-producing structure called a zygosporangium, which develops when haploid hyphae from two individuals meet.

(Fig. 29.7b) c. Basidiomycetes (mushrooms, puffballs, etc.) form spores in small pedestal-like structures called basidia at the ends of hyphae. (Fig. 29.7c) d. Ascomycetes produce spores with specialized cells called asci. (Fig. 29.7d) e. Many fungi are quite good at asexual reproduction as well as sexual reproduction.

Some produce trillions of asexual spores called conidia. (Fig. 29.8) C. Evaluating molecular phylogenies 1. Fungi are closely related to animals. (Fig. 29.9) a. Animals and fungi both produce chitin, have similar flagella, and use glycogen to store energy.

2. What are the relationships among the major fungal groups? (Fig. 29.10) a. Microsporidians are actually fungi. b. Chytrids and zygomycetes form a paraphyletic group that is poorly resolved. c. A group called the Glomeromycota are monophyletic. d. Basidomycota and Ascomycota are each monophyletic groups, and together they form a monophyletic group. e. The sister group to fungi consists of choanoflagellates (protists) and animals.

III. What Themes Occur in the Diversification of Fungi?

A. The diversification of fungi was driven by the evolution of novel methods for absorbing nutrients from a diverse array of food sources.

B. Fungi often participate in symbioses.

1. Mutualistic relationships benefit both species, parasitic relationships benefit one species at the expense of the other, and commensal relationships benefit one species with no effect on the other.

2. Researchers used isotopes as tracers for CO2, P, and N. They showed carbon compounds transferred to the fungi from the plant and P and N transferred from the fungi to the plant. (Fig. 29.11) 3. Ectomycorrhizal fungi (EMF) (Fig. 29.12a) a. EMF are often basidiomycetes (or occasionally ascomycetes).

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b. They form a dense network of hyphae (0.1 mm thick) around plant roots, extending between root cells and into the soil. c. They are abundant in regions with warm summers and cold winters, and they are especially common on trees of temperate and boreal forests. d. EMF are the dominant nutrient-gathering organs in temperate forests.

(1) They secrete peptidases that break down proteins in decaying material, take up the amino acids, and transfer them to the host plant.

(2) They also take up and transfer phosphorus to the host plant.

(3) Experiment: Birch seedlings grown in forest soil in pots without EMF cannot acquire sufficient N and P for healthy growth.

4. Arbuscular mycorrhizal fungi (AMF; endomycorrhizal fungi) (Fig. 29.12b) a. AMF are members of Glomeromycota. b. Highly branched hyphae penetrate the cell walls of root cells and contact the plasma membrane directly. This increases the contact area for the exchange of molecules. c. AMF occur in 80 percent of land plant species, especially in grasslands and tropical forests (areas with long, warm growing seasons). d. They deliver phosphorus to plants but not nitrogen.

(1) Nitrogen is readily available in these environments because plant tissues degrade rapidly in warm climates.

(2) But phosphorus is often scarce because it leaches out of soil.

5. Endophytes a. Endophytes are fungi that live in the above ground tissues of a plant. b. Some may be beneficial. For example, endophytes in some grasses and in morning glories produce compounds that deter herbivores. c. Others may be commensals since some studies have not documented any benefit to the host plant.

6. Other symbioses a. Many plant-eating insects harbor yeast in their guts to aid in digestion or detoxification of plant compounds. b. Lichens are a mutualistic partnership between an ascomycete and either a cyanobacterium or an alga. (Fig. 29.13) c. Some ant species farm fungi inside their colonies, harvesting them for food.

(Fig. 29.14) C. What adaptations make fungi such effective decomposers?

1. Extracellular digestion a. Fungi secrete digestive enzymes outside their hyphae into their food, so digestion actually occurs extracellularly. b. The smaller, digested compounds (amino acids, sugars, etc.) can then be absorbed by the hyphae.

2. Lignin degradation a. Lignin is a heterogeneous family of strong polymers that give woody texture to plant stems. b. Basidiomycetes can degrade lignin completely to CO2, using an enzyme called lignin peroxidase. .

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c. Lignin peroxidase removes a single electron from the ring structures of lignin, creating a free radical and triggering an unpredictable series of further reactions. d. The degradation of lignin is so unpredictable that it cannot be harnessed to drive ATP production. e. Instead, fungi degrade lignin simply to get it out of the way, thereby gaining access to cellulose strands within the lignin matrix.

3. Cellulose digestion a. Once lignin peroxidase has softened the wood, long strands of cellulose can be digested with cellulases (cellulose-digesting enzymes). b. Unlike lignin peroxidase, cellulases are extremely specific in their action. For example, the fungus Trichoderma reesei has seven cellulases, each catalyzing a very specific reaction. c. Cellulose is broken down completely to glucose, which is used for ATP production.

D. Variation in reproduction 1. Spores as key reproductive cells a. Spores are the dispersal stage in the fungal lifecycle. b. Spores are produced both by asexual and sexual reproduction. c. If food becomes scarce, a mycelium responds by forming spores that can be dispersed by water, wind, or animals. d. If a spore falls on a food source, it germinates and forms a new mycelium.

2. Multiple mating types a. Many fungi have mating types (analogous to sexes) that look identical. For example, zygomycete hyphae fuse and reproduce only if they carry alleles from different mating types. b. Instead of having just two sexes, a single fungal species may have tens of thousands.

3. How does fertilization occur? a. Chytrids are the only fungi that produce gametes, and they are not different enough in size to be called sperm and egg. b. Fertilization in other fungi occurs in two distinct steps: (Fig. 29.15) (1) Plasmogamy: The cytoplasm from different cells fuses. The nuclei may remain independent; that is, the mycelium is heterokaryotic.

(2) Karyogamy: Eventually, a pair of unlike nuclei fuse to form a diploid zygote.

The fused nucleus then divides by meiosis to form spores. c. Students should be able to compare and contrast these events with the lifecycles of other eukaryotes. For example, which human cells undergo plasmogamy and karyogamy? Are any human cells similar to a heterokaryotic fungal cell?

Explain.

4. Asexual reproduction a. Many fungi reproduce asexually. (Fig. 29.15 left) b. Spore-forming structures are produced by a haploid mycelium. c. Offspring are genetic clones of the parent.

E. Four major types of lifecycles 1. Chytrid lifecycle (Fig. 29.16a) 288

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a. Chytrids are the only fungi with alternation of generations. b. Swimming gametes are produced in haploid adults by mitosis. c. Gametes fuse to form a diploid zygote that grows into a sporophyte. d. Meiosis occurs in the sporangium (a structure in the sporophyte), forming spores that disperse by swimming.

2. Zygomycete lifecycle (Fig. 29.16b) a. Hyphae from two individuals fuse (plasmogamy) to form a structure called a zygosporangium. b. Zygosporangia develop a tough coat and can survive cold, dry periods. c. Once conditions are favorable, meiosis occurs and spores are produced. d. Spores germinate into new mycelia. e. Mycelia can also reproduce asexually.

3. Basidiomycota lifecycle (Fig. 29.16c) a. Heterokaryotic mycelia form large structures including mushrooms, bracket fungi, and puffballs, all of which contain tiny pedestal structures called basidia. b. Karyogamy occurs in a basidium, followed by meiosis to produce haploid spores.

4. Ascomycota lifecycle (Fig. 29.16d) a. Short dikaryotic hyphae form saclike, spore-producing structures called asci. b. Karyogamy occurs in an ascus, followed by meiosis to produce haploid spores.

IV. Key Lineages of Fungi (Table 29.1) A. Microsporidia 1. Traits: All are single-celled and parasitic, mostly parasitizing fish and insects. They use a structure called a polar tube to enter host cells. They have an extremely small genome and no mitochondria.

2. Human and ecological impacts: Some can infect humans, although they typically cause problems only in patients with impaired immune systems. Other species affect honeybees, silkworms, and grasshoppers.

B. Chytrids 1. Chytrids are a paraphyletic group.

2. Traits: They are largely aquatic; a few that live in dry habitats have drought-tolerant spores. They are the only fungi that produce motile cells. They have an absorptive lifestyle. Many are parasitic. Many have cellulases and are decomposers, and some are mutualists in the guts of grazing animals.

3. Human and ecological impacts: They are important decomposers in freshwater habitats, and they assist grazing animals and termites in digesting cellulose. A parasitic chytrid is largely responsible for catastrophic declines in amphibian populations around the world.

C. Zygomycetes 1. Zygomycetes are a paraphyletic group.

2. Traits: They are primarily soil-dwelling saprophytes; some are parasitic and affect insects and spiders. Asexual reproduction is extremely common. Hyphae of different mating types fuse during sexual reproduction.

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3. Human and ecological impacts: The common black bread mold, Rhizopus stolonifer, is a common household pest. Some saprophytic species cause large losses to the fruit and vegetable industry.

D. Glomeromycota 1. Traits: These include all the arbuscular mycorrhizal fungi (AMF) of grasslands and tropical forests. They absorb phosphorus and nitrogen from the soil and transfer them to the plant host, in exchange for sugars. Most form spores underground. They are difficult to culture; their lifecycles not well understood.

2. Human and ecological impacts: They are extremely important for the growth and productivity of grasslands and tropical forests.

E. Basidiomycota (club fungi) 1. Traits: These important saprophytes can completely digest wood. Some are ectomycorrhizal fungi (EMF). Some are parasitic (rusts and smuts). All produce basidia during sexual reproduction. The largest subgroup forms above ground reproductive structures: mushrooms, puffballs, brackets, and so on. They have heterokaryotic mycelia.

2. Human and ecological impacts: EMF are enormously important in forestry. Mushrooms are a source of food and many produce toxins or drugs.

F. Ascomycota (sac fungi) 1. Traits: They are found in virtually every habitat. They use every possible absorptive lifestyle (EMF, endophytes, saprophytes, parasites). About half of ascomycetes grow in symbiotic associations in lichens, where they protect the alga or cyanobacterium from desiccation. They may produce spores called conidia and may produce an aboveground cup-like structure called an ascocarp (cup fungi).

2. Human and ecological impacts: They include Penicillium (source of penicillin), truffles, morels, and Saccharomyces cerevisiae (used in making beer, wine, bread). Some parasitic species can infect humans, other animals, and plants (including the devastating chestnut blight and Dutch elm disease). They dominate tundra ecosystems and are abundant in northern forests.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. arbuscular mycorrhizal fungi (AMF) ascocarp Ascomycota ascus (plural: asci) Basidiomycota basidium (plural: basidia) carbon cycle Carboniferous period 290

cellulases

ectomycorrhizal fungi (EMF)

cellulose chitin chytrid club fungi coenocytic commensal conidia cup fungi

endophyte end-Permian mass extinction extracellular digestion free radical fungal spike fungicide fungus (plural: fungi) gamete

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glomalin Glomeromycota heterokaryotic hypha (plural: hyphae) karyogamy lichen lichen-formers lignin lignin peroxidase mating type Microsporidia

microsporidian mushroom mutualistic mycelium (plural: mycelia) mycorrhizal non-lichen formers parasitic plasmogamy polar tube puffball saprophyte

septum (plural: septa) soredia soredium (plural: soredia) sporangium spore symbiosis yeast zygomycete Zygomycota zygosporangium

Lecture Activities True/False Game: There’s a FUNGUS AMONGUS! Estimated duration of activity: 5–10 minutes Except for eating mushrooms, students are probably unaware of the many human uses of fungi. This activity is designed to capture students’ interest in the study of fungi and may be used as an introduction to the chapter or as a conclusion. Present each statement and have students respond as a group to judge whether it is true or false. When there seems to be an equal mix of both responses, ask for a show of hands. Ask students to defend their answer, and then confirm or correct their response. You may also use each statement as a springboard to discuss the features of the lineage represented by that particular fungus.

Questions: 1. The processes for producing penicillin and blue cheese both use the Penicillium fungus.

True: Penicillium fungi belong to the non-lichen forming Ascomycota lineage.

2. The reproductive organs of Amanita muscaria are popular pizza toppings.

False: Amanita muscaria is a poisonous mushroom—a member of the Basidiomycota lineage.

3. The fungus Claviceps purpurea is thought to have triggered the infamous Salem witch trials.

True: Claviceps purpurea produces ergot, a poison that infects grains and is thought to have caused the hallucinations and strange behavior of the Salem witches. This fungus belongs to the Ascomycota lineage.

4. Saccharomyces cerevisiae is a common household bread mold.

False: It is the most common yeast used in bread production. S. cerevisiae is a member of the Ascomycota lineage of fungi.

5. The oils from lichens are used as a base in perfume production.

True: Lichens are members of the Ascomycota lineage of fungi.

6. Chytrids are parasites that devastate entire livestock populations.

False: Chytrids have a symbiotic relationship with cows and other herbivorous livestock. Chytrids are members of the Chytridiomycota lineage.

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7. Rhizopus stolonifer is found in many American households.

True: It causes black bread mold. R. stolonifer belongs to the Mucorales lineage of the Zygomycota phyla.

Integration questions: To encourage students to apply what they have learned about fungi, ask them the following questions about the fungal species introduced in the first part of the activity: 1. Why would a fungus make and secrete an antibiotic?

Answer: Probably to protect its food source from being eaten by bacteria 2. Describe the anatomy of a mushroom, and explain how it serves as the reproductive organ of the fungi.

Answer: See Sections 31.3 and 31.4 for the details.

3. Why would ergot, a parasite of grains like wheat, produce an animal toxin?

Possible answer: Perhaps to prevent grazing animals from eating infected grains; this ensures the survival of not only the ergot but also its food source 4. Explain, in detail, why the addition of brewer’s yeast (S. cerevisiae) makes bread rise.

Answer: Brewer’s yeast uses the carbohydrate in the bread dough as fuel for fermentation and gives off carbon dioxide as a by-product, thus making the dough rise.

5. Are lichens pure fungi? Why or why not?

Answer: No. They are a symbiosis of fungi and algae or photosynthetic bacteria.

6. What feature of members of the Mucorales lineage makes them able to have a mutualistic relationship with cattle and other herbivores?

Answer: Mucorales have cellulases that can help the herbivores break down cellulose.

7. How can black bread mold begin to grow on bread in a house that does not currently contain moldy bread?

Answer: The mold can produce spores that can be airborne or survive on surfaces until a food source becomes available. The spores may have been in the house for a while, or they may have been picked up at the grocery store.

Discussion Idea: This Is Your Brain on Fungi Fungi are the source of many psychoactive compounds, including psilocybin and LSD. As a class project, ask students (singly or in groups) to pick a certain fungus or a certain fungal hallucinogen, research it at home, and then describe their findings briefly to the rest of the class in a group discussion during the next lecture.

This is also a good opportunity to demonstrate to students how to find and screen information online. For example, you can show them one of Wikipedia’s entries (say, for Psilocybe or LSD) and then show them how to inspect and assess the citations listed at the bottom⎯and demonstrate how easy it is to alter Wikipedia entries! You may also wish to show them how to find peer-reviewed articles in Google Scholar or the database of your choice. Two good websites for this topic are: http://www.ethnopharmacology.com/fungi/ http://psychedelics.com/psilocybe/ 292

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An excellent source book is Drugs and the Brain by Solomon H. Snyder (New York: Scientific American, 1996). This book includes a section on Psilocybe and the fungus Claviceps, the chemical structures of several psychoactive compounds, the fascinating history of the discovery of LSD and its early use in psychiatry, and a section on the medical uses of the ergotamine compounds from Claviceps (used to treat migraines and to control bleeding during childbirth).

Classroom Demonstration: Fungi, Live and in Person A general discussion of fungi is best illustrated with examples of living fungi, which can be obtained or prepared relatively easily. Living fungi may be taken into the classroom and placed in several different small, open boxes that can be passed around during lecture.

• The grocery store is an excellent place to get live or preserved reproductive structures (mushrooms) of the Basidiomycota. These have particularly large caps with easily visible gills and are ideal for explaining to the class that the location of the basidia is on the gill surfaces. Students may not be aware that the common button white mushroom, the intermediate-sized crimini mushroom, and the large portobello mushroom are different stages of the same species, Agaricus bisporus (the portobello is the mature stage).

• If you have access to a gourmet foods store, you may be able to find some reproductive structures of the Ascomycota, such as morels. The surface of the morel is made up of many chambers. The ascus cells line each chamber. Morels are very expensive by the pound, but you can often purchase several of them for just a few dollars.

• If you live in an area where wild mushrooms are easy to find, you may be able to obtain a variety of fresh mushrooms that are actively releasing spores. If so, the stipe of the mushroom can be cut off, and the cap can be placed gill-side down on a white sheet of paper for several days to make a spore print. The tiny spores are often colored, and the print will show the pattern of the gills.

• Yeast (Ascomycota) is readily available in the grocery store. If placed in a sugar-water solution for a couple of hours, it will grow to high density. If a microscope is available, the yeast method of mitosis by budding is easily observed. If active yeast are mixed into dough and kept in a warm place for about two hours, their production of CO2 as a byproduct of cell respiration can be observed in the rising of the dough.

• Lichens are generally abundant in all types of habitats. Most tree branches have a variety of lichens on their surfaces, so a fallen branch may be an excellent source of these organisms. It is also easy to find rocks covered with lichens. Almost anytime lichens are collected, ascocarps (the cup-shaped sexual reproductive fruiting bodies) can be seen on their surfaces. Lichens are probably the easiest way for students to see the fruiting body that is typical of members of the Ascomycota because the ascocarps are generally easily visible with the naked eye and can be seen in excellent detail with just a hand lens.

• Alternatively, ask your students to slightly moisten a piece of bread (made without preservatives) at home and expose it to the air for a while. If the bread is then kept moist inside a bag in a warm, dark place, a good crop of fungi can usually be seen after a few days.

• A variety of fungi and molds can be ordered on agar plates from Carolina Biological Supply Company. Penicillium is the mold from which the first known antibiotic was isolated.

It grows abundantly and produces colored spores, so an agar plate of living Penicillium is good for demonstrating the mycelium morphology and the abundant spore production that most fungi exhibit. Another mold that produces a good plate of hyphae with spores of a .

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different color is Aspergillus. Some species of this genus are responsible for human lung infections. Other species of Aspergillus produce the highly toxic aflatoxin, which can be fatal at certain doses and is suspected to be a powerful carcinogen in birds and mammals, including humans.

Game Idea: What’s My Phylum?—Part II At this point in the class, students have learned about many traits of bacteria, archaea, protists, plants, and fungi. Before moving on to the animals, they may benefit from a refresher lesson to remember the traits of each group. You can expand the activity given in Chapter 30 to review all of the nonanimal lineages. Call out the names of those organisms you have featured in class (or key traits of those organisms). The first student, or group of students, to call out the correct answer gets a point. To encourage shy students to participate, you may have teams of students consult with one another (“Family Feud” style), with a group spokesperson then calling out the consensus answer. For an easier game, have students say only whether the organism is a bacterium, protist, plant, or fungus; for an additional challenge, have students also name the specific sublineage of the organism.

Sample starting list (expanding on the list suggested for the Chapter 30 activity): sushi seaweed, leprosy, moss, athlete’s foot, tuberculosis, green algae, malaria, Giardia, Salmonella, chytrid, potato blight, bryophyte, endophyte, sudden oak death, ringworm, gymnosperms, hornwort, anthrax, AMF, Chlamydia, red tides.

Trivia Game: Meet the Model Organisms (BioSkills 11) Estimated duration of activity: 15 minutes Another good way to help students review bacteria, protists, plants, and fungi is to introduce them to a few of the model organisms in each lineage. For example, ask groups of students to read (on their own) BioSkills 11 “Model Organisms.” BioSkills 11 describes Escherichia coli, Dicyostelium discoideum, Arapidopsis thaliana, Saccharomyces cerevisiae, and Drosophila melanogaster. In the next lecture, divide the class in half and tell them you are going to quiz them about these five species. When you ask a question, a single person from each team (each half of the class) should run up to the board and write the answer. The first to fully write the answer (correctly spelled) earns a point for their team. Other members of the team can verbally coach their “runner.” For the next question, each team must use a different runner. (This game design encourages all students to participate and prevents a few confident students from taking over the game. It also forces students to spell accurately, an aspect that is missing from many verbal games.)

First set of questions: Ask students to write the name of a model organism that is a plant, bacterium, and so on.

Second set of questions: Pick one of the model organisms the students named in the first round. Announce the organism to the class. Then ask the following questions: • What domain is this organism in? (You may ask follow-up questions about kingdom, phylum, etc., or about other traits of the organism: Is the organism photosynthetic? Is it considered to be a protist?)

• Name one major biological discovery that was made using this organism.

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CHAPTER

30

An Introduction to Animals

Learning Objectives: Students should be able to...

• Discuss what makes an animal an animal.

• Describe the key innovations to body plan that occurred during the early radiation of animals.

• Place the major lineages of animals correctly on a phylogenetic tree.

• Describe how animals within lineages diversified in their modes of sensing, feeding, moving, and reproducing.

Lecture Outline • The radiation of animals began with the Cambrian explosion around 550 mya.

• There are an estimated 8 to 50 million animal species alive, although only 1.4 million have been described.

I. What Is an animal?

A. Animals occur in a clade called Opisthokonta. (Fig. 30.1) B. Animals share these key traits: 1. They are multicellular eukaryotes that move under their own power.

2. Outside of sponges, they all have both nerve and muscle cells.

3. Animals are the only multicellular heterotrophs that ingest their food prior to digesting it.

4. They are the largest predators, herbivores, and detritrovores.

C. There are 30 to 35 recognized phyla. (Table 30.1)

II. What Key Innovations Occurred during the Origin of Animal Phyla?

A. There are several sources of data that suggest the same evolutionary sequence.

(Fig. 30.2) (BioSkills 13) 1. Fossils provide direct evidence of past forms but are incomplete in their coverage of the history of life.

2. Comparative morphology provides information about which embryonic, larval, or adult morphological characteristics are common and which are unique to particular lineages.

3. Comparative development provides information about patterns of gene expression and morphological change during development.

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4. Comparative genomics provides information about the relative similarity of genes or whole genomes of diverse organisms.

B. Origin of multicellularity 1. Fossil evidence—Sponges appear in the record more than 700 mya.

2. Morphological evidence—Choanocytes and sponge-feeding cells are almost identical in structure and function. (Fig. 30.3) 3. Molecular evidence—Sponges are paraphyletic, providing support of a sponge-like common ancestor to all animals. a. All possess a complex developmental tool kit of genes, suggesting a series of important genetic innovations that occur at the root of the animal tree along with multicellularity.

4. Alternative views: The Ctenophores-first hypothesis a. This hypothesis is based on the pattern of gains and losses of certain genes in the ctenophores. There is not enough evidence to reject this hypothesis.

5. Insights from the origin-of-animals debate a. The evolution of animals is more complicated than a smooth transition from simple to complex. b. Many key innovations did not arise all at once. c. Evolution did not stop within any of the lineages. (Making Models 30.1) C. Origin of embryonic tissue layers 1. All animals have tissues, which are groups of similar cells organized into structural and functional units. a. Sponges have historically been thought to lack tissues, but recent research shows that some sponges have epithelium.

2. All other animals have at least two different types of tissues that are derived from germ layers, which are tissue layers in the embryo. a. There are three germ layers: (1) Endoderm gives rise to the lining of the digestive tract.

(2) Ectoderm gives rise to the skin and nervous system.

(3) Mesoderm gives rise to the circulatory system, muscles, and internal structures such as bone and most organs. b. Diploblasts have two germ layers: ectoderm and endoderm.

(Fig. 30.4) (BioSkills 15) (1) Cnidaria and Ctenophora are traditionally considered diploblasts, but recent data indicate that some cnidarians have mesoderm. c. Triploblasts have three germ layers: ectoderm, endoderm, and mesoderm.

(1) All animals other than cnidarians, ctenophores, and sponges are triploblasts. d. All animals share homologous genes for contractile proteins. Even sponges possess the genes for actin and myosin. e. These components were elaborated upon in different lineages in different ways. The result is a functional similarity—an example of convergent evolution.

D. Origin of bilateral symmetry, cephalization, and the nervous system 1. Body symmetry is a key morphological aspect of an animal’s body plan.

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a. Ctenophores, many cnidarians, and some sponges have radial symmetry.

(Fig. 30.5a) b. Bilaterally symmetric animals have only a single plane of symmetry. (Fig. 30.5b) 2. Homology or convergent evolution? a. Cnidarians that appear to be radially symmetric can be seen to actually have bilateral symmetry upon closer inspection. (Fig. 30.6) b. Bilateral symmetry is achieved by the combination of the two genes responsible for anterior-posterior and dorsal-ventral axes formation. c. The Hox and dpp genes in Nematostella is homologous to bilaterians. (Fig. 30.7) 3. Origin of the nervous system a. Sponges have no neurons, no well-defined symmetry, and no head region. b. Cnidarians and ctenophores have a nerve net and radial symmetry but no head region. (Fig. 30.8a) (1) These animals are equally likely to encounter prey and other environmental stimuli from any direction. c. All other animals (i.e., the triploblasts) have bilateral symmetry, a head region, and a central nervous system (CNS) with ganglia. (Fig. 30.8b) (1) They tend to move in one direction and thus to encounter environmental stimuli at only one end.

(2) Evolution of bilateral symmetry is associated with cephalization: evolution of a head region that contains sensing, feeding, and information-processing structures such as eyes, mouth, and brain.

(3) This is an efficient design for directed movement, hunting, and capturing food.

E. Origin of the coelom 1. The basic bilaterian body shape is a tube within a tube. (Fig. 30.9) 2. Many triploblasts have a fluid-filled internal cavity called a coelom. a. Those that have a coelom are called coelomates. (Fig. 30.10a) b. Those that do not have a coelom are called acoelomates. (Fig. 30.10b) c. Those that retain a colem without a mesodermal lining such as nematodes are pseudocoelomates. (Fig. 30.10c) 3. Molecular data suggest that some lineages (such as flatworm ancestors) had a coelom, but then those traits were lost.

F. Origin of protostomes and deuterostomes 1. Bilaterians (triploblastic, bilaterally symmetric animals) can be split into two subgroups: protostomes and deuterostomes.

2. Protostome development a. The mouth forms first during gastrulation. b. The coelom forms via splitting of blocks of mesoderm.

3. Deuterostome development a. The anus forms first during gastrulation. b. The coelom forms via mesoderm pinching off from the gut.

4. These represent two ways of achieving the same end—a bilaterally symmetric body that contains a cavity lined with mesoderm.

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5. There are two major groups within the protostomes: a. Lophotrochozoa—include the mollusks, annelids, flatworms, and rotifers b. Ecdysozoa—include the arthropods and nematodes G. Origin of segmentation 1. Segmentation is the presence of repeated body structures. a. Vertebrates have segmented backbones. b. Invertebrates (annelids, arthropods, and perhaps mollusks) also demonstrate segmentation of differing types.

2. Is segmentation an example of convergent evolution? a. Some tool-kit genes for segmentation may have developed early in animal evolution. b. Small changes in the expression of certain tool-kit genes can result in novel numbers, shapes, and sizes of body segments and appendages.

III. What Themes Occur in the Diversification of Animals within Phyla?

A. Major lineages of animals are defined by a particular body plan, but subsequent diversification within a lineage was triggered by innovations in sensing, feeding, and movement to deal with several variables: 1. Higher oxygen levels 2. Evolution of predation 3. New niches beget more new niches 4. New genes, new bodies B. Sensory organs 1. Variation in sensory abilities a. Most animals have senses of touch, balance, smell, taste, and hearing, and at least some ability to sense light and time. (Table 30.2) b. Some animals evolved abilities to detect other stimuli, such as magnetism, electric fields, and barometric pressure. c. Animals vary greatly in the complexity and abilities of a given sensory structure.

C. Feeding 1. What animals eat: diversification of ecological roles (Table 30.3) a. Detritivores eat dead or decaying matter, herbivores eat plants or algae, carnivores eat animals, and omnivores eat both plants and animals. b. Consumers can be subclassified as predators or parasites.

(1) Predators are usually larger than their prey and kill them quickly.

(2) Parasites are usually smaller than their prey and often harvest nutrients without causing death. (i) Endoparasites live inside their hosts, and ectoparasites live outside their hosts.

2. How animals feed: four general strategies (Table 30.4) a. Animals have variation in mouthpart structures, and this correlates closely with how they eat. b. Suspension feeders (filter feeders) capture food by filtering particles out of water or air.

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(1) Examples: sponges, clams and mussels, baleen whales (2) They are common in aquatic habitats and in sessile organisms. c. Deposit feeders eat their way through a substrate on the surface.

(1) Examples: annelids (earthworms), sea cucumbers (2) All have simple mouthparts and a wormlike body shape. d. Fluid feeders suck or mop up liquids (nectar, plant sap, blood, fruit juice).

(1) Examples: butterflies, blow flies e. Mass feeders take chunks of food into their mouths.

(1) Examples: horses, snails (2) Mouthparts vary according to the type of food eaten.

D. Movement 1. Locomotion has several purposes: finding food, finding mates, escaping from predators, and dispersing to new habitats.

2. Three types of skeletal systems enable complex movements: a. Hydrostatic skeleton—derive support from flexible body wall in tension surrounding a fluid or soft tissue under compression. b. Endoskeleton—derive support from rigid structures inside the body such as bones. c. Exoskeleton—derive support from rigid structures on the outside of the body such as external armor of arthropods.

2. Limbs take on a variety of forms. (Table 30.5) a. The major types are: lobelike limbs, jointed limbs, parapodia, tube feet, and tentacles.

3. Are all animal appendages homologous? a. The great diversity of animal limbs suggested that limbs evolved independently several times. b. Recent data have challenged this view and imply that all animal limbs are homologous.

(1) The gene Dll (Distal-less) encodes for a protein that stimulates limb growth during development.

(2) The gene’s protein has been found in the developing limb buds of annelids, arthropods, echinoderms, and chordates. (Fig. 30.11) c. Conclusion: All animal appendages are homologous and are presumably derived from a simple appendage that evolved early in the history of the Bilateria.

E. Reproduction (Table 30.6) 1. Organisms live to reproduce.

2. Asexual or sexual reproduction? a. Some animal phyla, such as certain rotifers, reproduce only asexually. b. Parthenogenesis is a process by which the organism produces a diploid egg that matures into an adult without fertilization. c. Sexual reproduction is much more common in animals, as it can lead to greater genetic diversity.

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3. Where does fertilization occur? a. Internal fertilization usually involves the male inserting a sperm-transfer organ into the female, where the egg is fertilized. b. In external fertilization, the egg is fertilized outside the female’s body.

4. Where do embryos develop? a. Viviparous animals (e.g., humans and most mammals) give birth to live young that are nourished by the mother. b. Oviparous animals (e.g., chickens and crickets) deposit fertilized eggs that are nourished by yolk. c. Ovoviviparous animals (e.g., guppies and garter snakes) retain eggs inside the body, but the embryos are nourished by a yolk rather than through nutrients transferred directly from the mother. d. Most animals are oviparous.

F. Life cycles 1. Some animals undergo metamorphosis, a marked change from one developmental stage to another. a. The sexually immature form is called a larva (plural: larvae) if it looks radically different from the adult, or a juvenile if it looks like a miniature adult. b. Only adults reproduce.

2. Why does natural selection favor this indirect pathway to adulthood? (Fig. 30.12) a. Larval stages are often adapted to dispersal. b. Larval and adult stages typically feed on different foods.

3. The diversification of life cycles, as well as sensory structures, feeding strategies, ecological roles, limbs, and reproductive strategies, represent the consequences of evolutionary processes occurring over millions of years.

IV. Key Lineages of Animals: Non-Bilaterian Groups (Table 30.7) A. Porifera (sponges) 1. Habitat: Most are marine, and all are benthic.

2. Body plan a. Most are asymmetrical. b. They do not have germ layers but have specialized cell types, and a few have epithelial tissues. c. They are built around a system of tubes and pores for water currents. d. Silica or calcium carbonate spicules may provide structural support.

3. Feeding: Most are suspension feeders, creating water currents that carry food particles to feeding cells.

4. Movement: Adults are sessile, but larvae can swim via flagella.

5. Reproduction a. Sponge cells are totipotent, and asexual reproduction can occur via fragmentation. b. Most are hermaphroditic, producing both egg and sperm, but they generally cross-fertilize. c. External fertilization and oviparity are most common. A few are ovoviviparous.

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1. Habitat: All are marine, mostly planktonic.

2. Body plan: They are transparent, ciliated, gelatinous diploblasts.

3. Feeding: Ctenophores are predators that catch prey using a variety of methods (tentacles, mucus, or whole ingestion).

4. Movement: Adults move by beating cilia that are arranged in comblike plates.

5. Reproduction a. Most species are hermaphroditic and self-fertilize. b. Fertilization is usually external, but some species have internal fertilization and ovoviviparity. c. Larvae are free swimming.

C. Cnidaria (jellyfish, corals, anemones, hydroids) 1. Habitat: Most are marine.

2. Body plan a. Most are radially symmetric diploplasts; a few are triploblasts. b. Ectoderm and endoderm surround a gelatinous material called mesoglea. c. They have a gastrovascular cavity with only one opening.

3. Feeding: A unique cell type called a cnidocyte ejects a barbed spear (cnidocyst) that captures and poisons prey.

4. Movement a. Jellyfish have a life cycle consisting of a sessile polyp and a free-floating medusa. b. In polyps, the gut serves as a hydrostatic skeleton. c. Medusae swim using jet propulsion produced by pulsing of the bell.

5. Reproduction a. Polyps can reproduce asexually via budding, fission, or fragmentation. b. Sexual reproduction involves external fertilization and oviparity, with planktonic larvae.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. acoelomate adult anemones animal asymmetrical benthic bilateral symmetry Bilateria body plan brain carnivore .

central nervous system (CNS) cephalization choanocytes choanoflagellates Cnidaria cnidocyst cnidocyte coelom coelomate colony

consumer corals Ctenophora/comb jellies deposit feeder detritivore deuterostome diploblast Ecdysozoa/ecdysozoans ectoderm ectoparasite endoderm CHAPTER 30An Introduction to Animals

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endoparasite endoskeleton epithelium exoskeleton external fertilization extracellular matrix (ECM) fluid feeder ganglion (plural: ganglia) gastrulation germ layer head herbivore heterotroph hydroids hydrostatic skeleton internal fertilization invertebrates juvenile larva (plural: larvae)

limbs Lophotrochozoa/ lophotrochozoans mass feeder medusa (plural: medusae) mesoderm mesoglea metamorphosis model organism nerve net neuron nymph omnivore oviparous ovoviviparous parasite parthenogenesis phylum (plural: phyla) polyp Porifera/sponges

predator protostome pupa (plural: pupae) pupation radial symmetry radula segmentation sessile spicule suspension feeder (filter feeder) symmetry tissue triploblast tube-within-a-tube design vertebrates viviparous worm

Lecture Activities Game: Animal Taxa Jeopardy Estimated duration of activity: 40 minutes This Jeopardy game is a suggested alternative to a lecture presentation of this material. The game allows the students to learn the material and have some fun in the process.

The students will need to read the chapter and take notes on it before class. (To ensure that the students have read the chapter, design an online quiz or hold an in-class quiz based on the game during the next class period. Alternatively, you could allow students to consult their books during the game.) As an incentive, offer prizes such as candy for the winning team or some number of bonus points on the next test. The game will take most of a 50-minute class period, but you should have a few minutes at the end of the game to highlight the major themes of the chapter.

At the beginning of class, divide students into two to four teams. If you teach a large class (more than 60 students), you may want to send half the class to a different room and have a TA run the same activity with them. Present the following rules: 1. Each team will be given the opportunity to select a question and answer it. If the answer is correct, then that team will be awarded the appropriate number of points. If the answer is incorrect, then another team will have the chance to answer. (If you have more than one team, then you can draw a number out of a hat to determine the order in which they answer.)

2. There will be a 30-second time limit for team conversation before the answer must be given.

3. Each team should appoint a spokesperson. The answer given by the spokesperson will be considered the final answer.

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Animal Jeopardy Animals Are Not . . . OR Animals Do Not Have . . . (Choose the feature that is not characteristic of animals.)

100: 300: 500:

Unicellular, eukaryotic, heterotrophic Correct response: What is unicellular?

Diploid, sexual reproduction, alternation of generations Correct response: What is alternation of generations?

Muscle tissue, cell walls, cytoskeleton Correct response: What are cell walls?

How Many Tissue Layers?

100: 300: 500:

Humans Correct response: What is 3?

Sponges Correct response: What is 0?

Jellyfish Correct response: What is 2?

The Art of Symmetry 100: 300: 500:

A type of symmetry that has two or more planes Correct response: What is radial symmetry?

A clustering of sense organs that often occurs in bilaterally symmetric animals Correct response: What is cephalization?

The only animal group in which larvae have bilateral symmetry but adults have radial symmetry Correct response: What are echinoderms?

Gotta Move!

100: 300: 500:

The Dll gene may have been involved in this major evolutionary innovation.

Correct response: What are appendages?

A type of skeleton that uses water instead of bones Correct response: What is a hydrostatic skeleton?

An echinoderm appendage used for moving and attacking Correct response: What are tube feet?

Food, Glorious Food!

100: 300: 500:

Baleen whales are the largest of this type of feeder.

Correct response: What is a suspension feeder?

The term for an animal that eats herbivores Correct response: What is a carnivore or a predator?

A tapeworm is an example of this type of parasite, but a leech is not.

Correct response: What is an endoparasite?

Double Jeopardy Brothers and Sisters (Choose the pair of animals that are most closely related.)

200:

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Cnidaria, Ctenophora, Porifera Correct response: What are Cnidaria and Ctenophora?

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600:

Roundworms, segmented worms, and flatworms Correct response: What are segmented worms and flatworms?

1000: Frogs, clams, and sand dollars Correct response: What are frogs and sand dollars?

There’s a Hole in My Embryo, Dear Liza, Dear Liza!

200:

The group of bilaterians in which the mesoderm develops from outpocketings of the gut Correct response: What are deuterostomes?

600: The group of bilaterians in which the mouth forms first Correct response: What are protostomes?

1000: One of the two great lineages of the protostomes Correct response: What are the Ecdysozoa or the Lophotrochozoa?

The Many Faces of Sex 200: 600:

With this type of fertilization, sex is a no-contact sport.

Correct response: What is external fertilization?

A “complete” life-changing event for insects Correct response: What is holometabolous metamorphosis?

1000: The condition in which a mother retains fertilized eggs that are nourished by yolks Correct response: What is ovoviviparous?

Who Am I? (characteristics) 200: 600:

Diploblastic, radially symmetric, aquatic Correct response: What is Cnidaria or Ctenophora?

Triploblastic, bilaterally symmetric, with a skull Correct response: What is a vertebrate?

1000: Triploblastic, coelomate, radially symmetric Correct response: What are (adult) echinoderms?

Ancient Animal Lineages 200: 600:

Diploblastic, nematocyst, polyp Correct response: What are Cnidaria?

Benthic, sessile, suspension feeders, asymmetrical Correct response: What are sponges?

1000: Lack a coelom, commonly detritivores, cilia cover body surface, triploblastic, only a few millimeters long Correct response: What are Acoelomorpha?

Class Activity: Building the Animal Tree Estimated duration of activity: 5 minutes (per lecture) Students may become overwhelmed by the many different lineages discussed in the animal diversity lectures, and they may especially need some assistance reviewing the protostome/ deuterostome and ecdysozoan/lophotrochozoan vocabulary. To help students grasp the bigger picture, have them construct a large “Animal Tree” phylogenetic diagram that they will build on with each additional lecture. (You may also have them build plant trees, fungal trees, protist trees, or others and tape them together to create one huge tree.)

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To start building the Animal Tree, bring several large sheets of poster paper (plus tape and a poster tube) to class during any (or all) of the diversity lectures. Tape the paper to a wall before class. At the end of the lecture, allow 5 minutes to have students help you draw a phylogenetic tree on the poster paper, reviewing that day’s material. You may prompt students by drawing the major lines and writing in the first initials of each of the major lineages that were covered that day. Then ask students to fill in the rest of the name of each lineage, and to name two or three key traits for that group. You may also ask for volunteers to come up and add elements to the tree, or to mark the origin of important evolutionary events. (For example, on the Animal Tree, ask for the origin of the three germ layers and bilateral symmetry.)

Encourage students to copy the tree in their own notes as you are drawing the big tree.

They will remember it better when they have drawn it themselves.

Once students have the idea, in subsequent lectures you may continue to build on the tree, either as one big group or by having students break into small groups to work on their own trees. Allowing 5 minutes at the end of each diversity lecture for tree-building will help students review the day’s material, refresh their memory of groups covered in previous days, and help them learn the evolutionary relationships. For extra incentives, you might offer extra-credit points for groups that turn in a fully completed tree at the end of the diversity lectures.

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CHAPTER

31

Protostome Animals

Learning Objectives: Students should be able to...

• Define the protostomes and explain their ecological and evolutionary importance.

• Discuss the major characteristics that differentiate the Lophotrochozoa and the Ecdysozoa.

• Describe the characteristics of the major Lophotrochozoan phyla (Platyhelminthes, Annelida, Mollusca).

• Describe the characteristics of the major Ecdysozoan phyla (Nematoda, Arthropoda)

Lecture Outline • Of the 30 known animal phyla, at least 22 are protostomes.

• Almost all animal species are protostomes. (Fig. 31.1) • Protostomes are important sources of food and materials, provide important ecosystem services, can cause significant human diseases and damage to crops, and provide two of the most important biological models (Drosophila melanogaster and Caenorhabditis elegans).

I. What Is a Protostome?

A. Protostomes are distinguished from Deuterostomes by the following: 1. During development, the initial pore forms the mouth, not the anus.

2. Isolated early embryonic cells are unable to develop into a complete embryo.

3. If a coelom forms, it forms from cavities within blocks of mesodermal cells rather than from mesodermal pockets.

B. Protostomes are a monophyletic group that can be divided into Lophotrochozoa and Ecdysozoa. (Fig. 31.2) C. The water-to-land transition 1. Phylogenetic analysis indicates that the ability to live in terrestrial environments evolved independently several times. (Fig. 31.3) 2. Several adaptations were required for animals to survive on land: a. The ability for gas exchange (1) Roundworms and earthworms are able to exchange gases across their body surface in moist environments. b. The ability to avoid dessication (1) Terrestrial arthropods and mollusks have gills or respiratory structures inside their body to minimize water loss.

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(2) Insects possess a waxy layer to minimize water loss from their body surface.

(3) Insect eggs have a thick membrane; eggs of slugs and snails have a think calcium carbonate shell. c. The ability to support their own weight D. Modular body plans 1. Multicellular animals share a common tool kit of genes that establish the animal body plan during development.

2. This tool kit can direct the development of dramatically different types and numbers of structures when expressed at different times and places.

II. What is a Lophotrochozoan?

A. The Lophotrochozoa comprise 13 phyla and include the mollusks, annelids, and flatworms. (Table 31.1) 1. The Lophotrochozoa name is derived from two distinctive morphological traits that occur in some, but not all members: a. A feeding structure called a lophophore (in three phyla) (Fig. 31.4a) b. A type of ciliated larva called a trochophore (in several marine phyla) (Fig. 31.4b) 2. Recent evidence supports the hypothesis that the spiral pattern of cleavage is a synapomorphy for this monophyletic group. (Fig. 31.5) 3. Students should be able to indicate the origin of spiral cleavage on Figure 31.2, as well as the loss of spiral cleavage in Rotifera.

B. What is a flatworm?

1. They are very flat and lack a coelom and specialized structures for gas exchange and circulation.

2. Students should be able to indicate the origin of the flattened, acoelomate body plan on Figure 31.2.

3. They are a large and diverse phylum consisting of four lineages: a. Turbellaria—free-living and prey on small protists or scavenge; digestive tract has only one opening (Fig. 31.6a) b. Cestoda—endoparasitic tapeworms that parasitize diverse vertebrates, absorbing nutrients by diffusion c. Trematoda—endo- or ectoparasitic flukes that parasitize many groups of animals (Fig. 31.6b) d. Monogenea—tiny ectoparasites that usually parasitize the gills of fish C. What is a segmented worm?

1. Most annelids have a coelom, a fully developed digestive tract with a mouth and an anus, and a segmented body. (Making Models 31.1) 2. Students should be able to indicate the origin of annelid segmentation on Figure 31.2.

3. The common ancestor of the annelids had a key synapomorphy in addition to segmentation: bristle-like extensions called chaetae that extend from parapodia. (Fig. 31.6c) 4. Annelids are traditionally divided into three groups: a. Polychaeta—diverse marine worms with many bristles

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b. Oligochaeta—earthworms that deposit feed in soils (Fig. 31.6d) c. Hirudinea—leeches; half are ectoparasites and half are predators/scavengers D. What is a mollusk?

1. The molluscan body plan: (Fig. 31.7) a. A muscular foot is used for movement. b. A visceral mass contains the internal organs. c. A mantle covers the visceral mass and may secrete a shell. d. Students should be able to indicate the origin of the molluscan body plan on Figure 31.2.

2. The successful body plan allowed the radiation of several lineages including the chitons, bivalves, gastropods, and cephalopods. (Table 31.2) 3. The foot is a muscular hydrostat a. Waves of muscular contractions sweep backward or forward, allowing individuals to crawl. (Fig. 31.8) b. The foot of cephalopods such as squid and octopuses is modified into tentacles.

4. The visceral mass separates internal organs from the hydrostatic skeleton. a. The visceral mass contains the organs and surrounding fluid. b. The coelom is reduced, and the organs occupy a hemocoel where body fluids bathe the organs directly. c. A unique feeding structure, the radula, is located at the anterior end. (Fig. 31.9) 5. The mantle has diverse functions. a. Many species secrete a calcium carbonate shell that serves a protective function against predators. b. The mantle in cephalopods is modified into a propelling siphon structure.

(Fig. 31.10)

III. What Is an Ecdysozoan?

A. While Lophotrochozoans increase their size by growing continuously, Ecdysozoans grow by molting—shedding an exoskeleton or covering rather than growing incrementally.

(Fig. 31.11) B. The Ecdysozoa comprise seven phyla. (Table 31.3) C. What is a roundworm?

1. Species in the phylum Nematoda are called roundworms or nematodes.

2. Roundworms are unsegmented worms with a pseudocoelom, a tube-within-a-tube body plan, no appendages, and an elastic cuticle that is molted during growth.

3. Lack specialized organs for respiration and circulation 4. Feed on a wide variety of foods using diverse mouthparts 5. Majority are free-living but some parasitize animals including humans: a. Pinworms infect about 40 million people in the U.S. b. Onchocerca volvulus causes an eye disease that infects about 20 million people in Africa and Latin America. c. Advanced infections of Wuchereria bancrofti causes elephantiasis. d. Trichinosis can be caused when people consume undercooked meat infected with Trichinella.

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D. What are tardigrades and velvet worms?

1. They are similar to arthropods in having a segmented body and segmented limbs.

2. Unlike arthropods, their cuticle is not hardened and their limbs are not jointed.

3. Tardigrades or “water bears” are microscopic animals that live in diverse aquatic and terrestrial habitats.

4. The onychophorans, or velvet worms, are small caterpillar-like organisms that live in moist leaf litter and prey on small invertebrates.

5. Students should be able to indicate the origin of the following traits on the ecdysozoan (red) branches of Figure 31.2: molting, cuticle, segmented body, segmented limbs.

E. What is an arthropod?

1. Arthropods have a specialized body plan. (Fig. 31.12) a. Arthropods have segmented bodies divided into sections called tagmata (singular: tagma). b. The exoskeleton is made of chitin and hardened by calcium carbonate. c. Movement occurs by contraction of muscles that pull against the exoskeleton to move jointed appendages. d. A fluid-filled hemocoel holds internal organs and circulates body fluid and functions as a hydrostatic skeleton in some larvae. e. Students should be able to indicate the origin of tagmata, exoskeleton, and jointed appendages on Figure 31.2.

2. Why has the arthropod body plan been so successful? a. As with mollusks, the modularity of arthropods is also determined by certain tool-kit genes. b. Arthropods exhibit a “serialized” body plan with diverse and functionally specialized appendages. (Fig. 31.13) 3. Origin of the wing a. There are two hypotheses for the origin of wings in insects: independent origin and gill co-option. (Fig. 31.14) b. A study using developmental genetics strongly supports the gill co-option hypothesis.

F. Arthropod diversity (Table 31.4) 1. Myriapods have long, segmented trunks a. Habitat: They are found in most terrestrial environments. b. Morphology: They have a head region and a long segmented trunk, with each segment bearing one or two pairs of legs. c. Feeding: (1) Millipedes live in rotting plant materials and are detritivores.

(2) Centipedes hunt insects and have poison-containing fangs.

2. Insects have three tagmata, unbranched appendages, and one pair of antennae (Table 31.5) a. Habitat: Insects dominate terrestrial environments. Larvae are also common in many freshwater habitats.

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b. Morphology: They have three tagmata⎯head, thorax, and abdomen⎯with wings and three pairs of legs. c. Feeding: Most insects have four sets of mouthparts that have diversified to allow them to feed in every conceivable way. d. Movement: They walk, run, swim, and fly.

3. Crustaceans have two or three tagmata, branched appendages, and two pairs of antennae. a. Include shrimp, lobster, crabs, barnacles, isopods, and copepods b. Habitat: Most are aquatic (freshwater or marine). c. Morphology: The body is divided into the cephalothorax and abdomen. Many have a carapace. They have two pairs of antennae and sophisticated compound eyes. d. Feeding (1) Barnacles and shrimp are suspension feeders that use feathery appendages to capture their prey.

(2) Crabs and lobsters are active hunters, herbivores, and scavengers that use mouthparts called mandibles to bite or chew. e. Movement: Crustaceans swim, walk, or run using highly diverse limbs.

(1) Barnacles are the only sessile crustaceans.

4. Chelicerates have two tagmata and chelicerae. a. Include spiders, ticks, mites, horseshoe crabs, daddy longlegs, and scorpions. b. Habitat: Most are terrestrial, but a few are marine. c. Morphology: The body is divided into anterior and posterior regions. The anterior region has no antennae but usually has eyes. They are named for the chelicerae, a unique pair of appendages found near the mouth. d. Feeding (1) Spiders, scorpions, and daddy longlegs are predators. Spiders often build sticky webs to trap their prey.

(2) Mites and ticks are ectoparasites that feed on the body fluids of animals.

(3) Horseshoe crabs are omnivores.

5. New phylogeny of arthropods a. The relationships of lineages within the arthropods are currently undergoing a revision. (Fig. 31.15) b. Insects and myriapods have long been considered sister groups, however, recent evidence indicates placement of the insect clade within the crustacean lineage.

G. Arthropod Metamorphosis 1. Two types of insect metamorphosis a. Hemimetabolous (incomplete) metamorphosis occurs when there is only a minor change in form from juvenile to adult. Example: aphids (Fig. 31.16a) b. Holometabolous (complete) metamorphosis occurs when there is a dramatic change in form from larva to adult. Example: mosquitoes (Fig. 31.16b) 2. What is the adaptive significance of metamorphosis? a. Holometabolous metamorphosis is 10 times more common than hemimetabolous metamorphosis.

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b. Larvae and adults from holometabolous species do not compete with each other because they use different food sources and different habitats. c. Larvae and adults can also specialize for different functions. Examples: caterpillars feed; butterflies reproduce.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text abdomen

ecdysozoans

Oligochaeta

adaptation Annelida (segmented worms) antenna (plural: antennae) Arthropoda/arthropods beak bilateral symmetry Bivalvia/bivalves carapace Cephalopoda/ cephalopods

Echiura (spoon worms) exoskeleton external fertilization foot Gastropoda/gastropods gastrulation gill head hemimetabolous hemocoel Hirudinea (leeches) holometabolous hydrostatic skeleton Insecta/insects intermediate host internal fertilization jet propulsion jointed limb lophophore Lophotrochozoa/

oligochaetes Onychophora (velvet worms) ovoviviparity parapodium (plural: parapodia) parthenogenesis Platyhelminthes (flatworms) Polychaeta polychaetes Polyplacophora (chitons) proboscis protostome radula Rotifera (rotifers) simple eye siphon Sipunculida (peanut worms) spermatophore tagma (plural: tagmata) Tardigrada (water bears) tentacle thorax Trematoda (flukes) triploblastic trochophore larva tube-within-a-tube design Turbellaria veliger visceral mass

Cestoda (tapeworms) chaeta (plural: chaetae) chelicera (plural: chelicerae) Chelicerata chelicerates chitin Clitellata coelom compound eye corona Crustacea/crustaceans cuticle definitive host detritivore direct development Ecdysozoa

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lophotrochozoans mandible mantle mantle cavity Mollusca/mollusks molting Myriapoda/myriapods nauplius Nematoda (roundworms)

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Lecture Activities Game: The Museum Curator Game Estimated duration of activity: 20–30 minutes An interesting alternative to a catalog-style lecture on the different lineages of protostomes is to involve your students in an active pursuit of this material. Although this activity requires a preparatory investment on your part, it will engage the students and lead them to investigate and apply the information in this chapter. The good news is that when you have prepared the activity once, it can be reused in future classes.

Preparation: Make color photocopies of as many protostome animals as you can find. (Use Chapter 30, Chapter 31, the Internet, books, and other materials.) Copy or print the pictures on card stock (laminating them will increase their life span for use in later semesters).

Finally, make place cards with the names of the protostome lineages from Chapter 31.

If you wish, you may also ask students to bring in specimens from the campus or grocery store (live insects in jars, seafood like lobsters, crabs, shrimp, spiders, worms, etc.). Instruct them not to bring in anything that might be dangerous or poisonous, and that any seafood should be in a glass jar or leak-proof bag. (Be sure to query the class beforehand to see whether anybody has a seafood allergy.) If the students catch insects but do not want to bring them to class alive, advise them to put the organism in an airtight jar with a cotton ball soaked in rubbing alcohol. This should euthanize the insect or worm while keeping it intact.

Extra-credit points can encourage the collection of more specimens!

Setting up the museum: Set the place cards (with the names of the protostome lineages) on tables arranged around the classroom. As students arrive in class, have them place all their specimens on a central table along with the pictures you have prepared. Organize the students into groups, assign each group to a lineage table, and give each group a random assortment of pictures and specimens. Ask the students in each group to spend a few minutes researching their lineage and learning its distinguishing morphological features, feeding habits, reproductive strategies, and habitats (in essence, they are to become the class experts on that lineage).

Then ask them to go through their photographs and specimens, select the ones that belong in their lineage, and place the others in a pile on the edge of the table. When they are done, they should split up, visit other tables, search through their “discard” pile, and collect any photos/specimens that belong on their table.

Touring the museum: When all the organisms have been properly classified, have all students sit at their respective tables. One by one, ask students at each table to present their lineage, using the pictures and specimens to illustrate that lineage’s characteristic features.

Web resources: The following website has additional information and photos of protostomes that you and your students may find useful. Click on any phylum for further information.

(The same UCMP website also has pages on plants, fungi, etc.) http://www.ucmp.berkeley.edu/phyla/metazoasy.html

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Game: Name That Protostome!

Estimated duration of activity: 5–10 minutes This activity is a time-saving and interesting alternative to cataloging all the various protostome lineages in lecture format. It requires that the students read the chapter beforehand and make a chart of the protostome lineages and their major features as laid out in Section 31.1. Have them bring that chart to class in order to participate in the game. You may wish to offer an extra-credit point for the charts, to encourage students to actually make and bring them.

At the beginning of the game, divide the class into two teams (the left and right sides of the classroom). You will read the list of characteristics, and the students must raise their hands to identify the protostome lineage you are describing. Call on the student whose hand goes up first, and award a point to that team if the answer is correct. If the answer is incorrect, give the other team an opportunity to answer. Of course, the team with the most points wins. You can decide what the prize will be (candy, bonus points on the next test, exemption from a quiz, the sheer joy of knowledge, etc.).

Protostome A: chelicerae, webs, ectoparasites, exoskeleton Answer: Arthropoda/Chelicerata Protostome B: tentacles, brains, eyes, spermatophore Answer: Mollusca/Cephalopoda Protostome C: hydrostatic skeleton, coelom, chaetae, parapodia Answer: Annelida Protostome D: no lophophore, trematode, flukes Answer: Platyhelminthes (flatworms) Protostome E: muscular ventral foot, radula, move via gliding Answer: Mollusca/Gastropoda Protostome F: carapace, mandibles, nauplius larvae Answer: Arthropoda/Crustaceans Protostome G: two hinged shells, veliger larva, feed via gills Answer: Mollusca/Bivalvia Protostome H: unsegmented, pseudocoelom, hydrostatic skeleton, some parasitic, some detritivores Answer: Nematoda (roundworms) Protostome I: biting/chewing mouthparts; many segments, each with one or two pairs of legs Answer: Arthropoda/Myriapoda Protostome J: ciliated corona Answer: Rotifera Protostome K: everywhere, segmented, six legs Answer: Arthropoda/Insecta Protostome L: eight plates on back, marine, glide on muscular foot Answer: Mollusca/Polyplacophora (chitons)

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Class Activity: Building the Animal Tree Estimated duration of activity: 5 minutes (per lecture) Drawing a large phylogenetic tree is a fun activity that helps students review the traits and relationships of the taxa that they have learned about so far. If you began an Animal Tree in Chapter 30, have students continue to build on the tree now by adding the protostome phyla and their distinctive traits. Take 5 minutes at the end of lecture for this activity. The tree may be made with the whole class contributing to one big tree that is taped to the board (you may save and add to this tree in subsequent lectures) or with small groups of students each building their own trees. This tree-building activity will help students review the day’s material, refresh their memories of taxa covered in previous days, and learn evolutionary relationships.

See the Lecture Activities for Chapter 30 for further details on building an Animal Tree as a class activity.

Game: Making Animals Out of Roots (BioSkills 15) The complex names of the animal phyla will make more sense to students if they know some of the Latin and Greek roots from which the names are derived. Read to students the English “translation” in the left column below, and ask them to try to figure out which taxon you are referring to. Refer students to BioSkills 15 for some additional Greek and Latin roots.

Tuft-wheel-animals Lophotrochozoa (Greek lopho, crest or tuft + trocho, Escaping-animals Pore-bearers Stinging-animals Comb-bearers Without-cavity-shape Flat-worms Ringed Thread-like Many-bristles Few-bristles Two-shell Many-plate-bearing Belly-foot Head-foot Jointed-foot Many-foot

wheel + zoa) Ecdysozoa (Greek ecdysis, casting off + zoa) (refers to molting) Porifera (Latin porus, pore) Cnidaria (Greek cnida, nettle) Ctenophora (Latin cteno, comb + phora) Acoelomorpha (a + coelo + morpha) (for lack of coelom) Platyhelminthes (platy + helminthes) Annelida (Latin annelus, small ring) (for segmentation) Nematoda (Greek nemat, thread) (for thread-like appearance) Polychaete (Greek poly + chaeta, mane or bristles) Oligochaete (Greek poly + chaeta, mane or bristles) Bivalve (Latin bi + valve) Polyplacophora (Latin poly + placo + phora) Gastropod (Latin gastro + pod) Cephalopod (Latin cephalo + pod) Arthropod (Latin arthro + pod) Myriapod (Latin myria + pod)

If time permits, add a brief review of the plural forms of Latin nouns. Without knowing how to make these words plural, students often become confused about whether various scientific words are singular or plural. Explain each of the three rules below and then ask students to think of examples from each category. (If they can’t, give them the singular forms in the examples below and the students will rapidly come up with the plural forms.)

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Rule 2: Latin words that end in -us are usually pluralized with -i.

Examples: focus/foci, fungus/fungi Rule 3: Latin words that end in -um are usually pluralized with -a.

Example: cilium/cilia, ovum/ova These three rules cover only some of the more common categories of Latin words, so you may wish to warn students that there will be exceptions. (Some that occur in this chapter are genus/genera, taxon/taxa, and tagma/tagmata.)

Follow up by asking students for the plurals of bacterium, chelicera, antenna, alga, and flagellum.

Discussion Idea: Why Study Fruit Flies and Worms? (BioSkills 11) BioSkills 11 introduces Drosophila melanogaster and Caenorhabditis elegans as important model organisms in genetics and developmental biology. This subject is great for initiating a discussion with your students in order to give them an appreciation for the usefulness of nonmammalian model organisms that have led to some extremely important breakthroughs in biology. Below are some sources that you may find useful in preparing for the discussion. If class time is limited, you may assign the articles to your students and have them participate in a web-based discussion or debate that is moderated by you or a TA. • The Wellcome Trust has several good articles on the use of model organisms in biology.

Go to www.wellcome.ac.uk and search for the articles “Why the Fly?”; “Why the Worm?”; and “Why Fish?”

• “When we were worms,” New Scientist, no. 2104 (Oct. 18, 1997), www.newscientist.com/article/mg15621045.100-when-we-were-worms.html

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CHAPTER

32

Deuterostome Animals

Learning Objectives: Students should be able to...

• Define the deuterostomes and explain their ecological and evolutionary importance.

• Describe the echinoderm body plan and compare the major echinoderm lineages.

• Describe the chordate body plan and compare the major chordate lineages.

• List and describe the relationships among the major lineages of vertebrates.

• Explain the importance of the following vertebrate adaptations: jaws, limbs, amniotic eggs, placentas, parental care, flight, ear bones.

• Summarize the known history of the hominid lineages, from the common ancestor shared with chimpanzees to modern humans.

Lecture Outline • All deuterostomes undergo development in a similar way.

• The gut develops from posterior to anterior; that is, the anus forms first.

• The coelom, if present, develops from outpocketings of mesoderm.

• There are four phyla of deuterostomes: (Fig. 32.1) • Echinodermata, including sea stars and sea urchins • Hemichordata, or acorn worms • Xenoturbellida, containing a single genus with two wormlike species • Chordata, including the vertebrates

I. What Is an Echinoderm?

A. The echinoderm body plan 1. Echinoderm larvae are bilateral, but adults have evolved pentaradial symmetry (five-sided symmetry). (Fig. 32.2a) 2. Echinoderms have an endoskeleton made of plates of calcium carbonate secreted inside the skin. (Fig. 32.2b) 3. Echinoderms have a system of branching fluid-filled tubes, and chambers called the water vascular system. (Fig. 32.3) a. Tube feet are elongated fluid-filled appendages of this system. b. Sections of tube feet called podia project outside the body wall to make contact with the substrate. c. The water vascular system is a sophisticated hydrostatic skeleton used for movement.

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B. Echinoderms are important consumers.

1. Sea stars are carnivores. a. For example, sea stars eat bivalves by using their podia to pry open the shells.

(Fig. 32.4a) b. Sea stars are a keystone species—their presence maintains community diversity.

(Fig. 35.4b) 2. Sea urchins are herbivores. a. They have a jawlike structure in their mouths called Aristotle’s lantern. b. They are ecologically important grazers and can overgraze kelp forests if not kept in check by predators.

3. Table 32.1 summarizes the key lineages of echinoderms.

II. What Is a Chordate?

A. Chordates contain four particular features present at some stage in their life cycle.

1. These morphological features are: a. Pharyngeal gill slits in the throat b. A bundle of nerves that runs the length of the body, called the dorsal hollow nerve cord c. A stiff but flexible notochord that runs the length of the body d. A muscular tail that extends past the anus 2. Together, these traits create a “torpedo” that can swim forward rapidly.

3. These traits may be present only briefly, but they are present in all chordates at some point in the life cycle.

4. The phylum Chordata is made up of three major lineages—the cephalochordaes, the urochordates, and the vertebrates.

B. The cephalochordates (Table 32.2) 1. The cephalochordates (lancelets or amphioxus) are small, mobile animals with a “fish-like” appearance.

2. Adults burrow into sand and suspension using pharyngeal gill slits. (Fig. 32.5a) C. The urochordates 1. The urochordates (tunicates and sea squirts) are small suspension feeders that live attached to hard substrates in the ocean. (Fig. 32.5b) 2. The mobile larvae have the diagnostic chordate traits, but the sessile adults may lose them.

3. Tunicates have an exoskeleton-like coat of polysaccharide called a tunic, a U-shaped gut, and two openings called siphons.

D. The vertebrates 1. The vertebrates include sharks, bony fishes, reptiles, and mammals. (Fig. 32.5c) a. The human spinal cord is the dorsal hollow nerve cord. b. The pharyngeal slits give rise to gill slits in fishes and are present during development in the embryos of terrestrial vertebrates. c. The notochord is present only in development, when it helps organize the body plan; for example, it helps induce the formation of vertebrae and muscles.

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III. What Is a Vertebrate?

A. Vertebrates are a monophyletic group distinguished by two traits: 1. A vertebral column 2. A cranium that encloses the brain B. The large vertebrate brain is divided into three regions: 1. The forebrain houses the sense of smell and may be elaborated into a large cerebrum.

2. The midbrain is associated with vision.

3. The hindbrain is associated with balance and vision and may include a cerebellum and medulla oblongata.

C. The most species-rich and diverse vertebrate lineages are the ray-finned fishes and the tetrapods. (Fig. 32.6)

IV. What Key Innovations Occurred during the Evolution of Vertebrates?

A. Three general themes in vertebrate evolution: (Fig. 32.7) 1. Most vertebrates are extinct.

2. Some traits evolved more than once.

3. Traits are sometimes lost.

B. Urochordates: Outgroup to vertebrates 1. According to the most recent molecular data, the urochordates are the closest living relatives of the vertebrates.

C. First vertebrates: Origin of the cranium and vertebrae 1. Fossil evidence for early vertebrates a. The first vertebrates lived about 520 mya.

(1) They had streamlined bodies, gills, a notochord, post-anal tail, and paired eyes.

(2) Fossils also suggest a cranium made of cartilage and a series of cartilaginous reinforcements of the notochord—early vertebrae. b. Students should be able to add the extinct Cambrian vertebrate, Haikouichthys, to the tree in Figure 32.7.

2. The hagfish hypothesis for early vertebrates a. Hagfish and lampreys have a notochord, gills, and a muscular post-anal tail, plus a three-part brain, paired eyes, and a cartilaginous cranium. (Table 32.3) b. Lampreys have cartilaginous reinforcements (vertebrae) on their notochord but hagfish do not. c. Recent evidence groups the lamprey and hagfish into the Cyclostomata—the ancestor to vertebrates was probably not hagfish-like in its mouthparts, but probably not lamprey-like either. d. Vertebrae could have been lost in hagfish lineages.

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2. A grade is a sequence of lineages that are paraphyletic. (Making Models 32.1) 3. Fossil evidence for the origin of the jaw a. Extensive fossil evidence has disproved a shark-like origin of jawed fishes. b. Most likely placoderm originated from the Silurian (430 mya). c. Soon after the appearance of jaw bones, teeth appeared.

4. The gill-arch hypothesis for the origin of the jaw (Fig. 32.8) a. Four lines of evidence that support this hypothesis: (1) Jaws and gill arches both consist of flattened bars of bone or cartilage that are hinged.

(2) Muscles that move gill arches and jaws are derived from the same population of cells.

(3) Jaws and gill arches are both derived from neural crest cells.

(4) Expression patterns of developmental regulatory genes are similar in jaws and gill arches. b. Modifications in the jaw structure make ray-finned fishes particularly diverse in their eating habits.

(1) In most species, the jaw protrudes to be able to nip out or bite food.

(2) Most species also have a pharyngeal jaw, consisting of modified gill arches that act as a second set of jaws at the back of the mouth. This makes food processing more efficient.

E. Origin of the bony endoskeleton 1. In some lineages from the late Silurian, cartilaginous endoskeleton became stiffened by the deposition of bone.

2. At first, bone was not part of the endoskeleton; rather, bony plates formed an exoskeleton—these later evolved into scales in some lineages.

3. As fish become more dense, the swim bladder evolved to allow fish to maintain a neutral buoyancy.

4. About 96% of all living fishes are descendants of ray-finned fishes.

5. Coelacanths and lungfishes are sometimes grouped into the lobe-finned fishes—an important evolutionary link to the tetrapods.

F. Tetrapods: Origin of the limb 1. Limbs-from-fins hypothesis a. Lungfish morphology provides clues for how vertebrates may have made the waterto-land transition. b. The fossil record includes a series of fossil species between lungfishes and tetrapods, documenting a gradual transition from a fin to a limb that could support walking. (Fig. 32.9) c. Genetic data also show that the regulatory proteins involved in fish fins and the upper parts of mammal limbs are homologous.

2. Amphibians a. The first tetrapods to live on land were amphibians. b. Most feed on land and lay eggs in the water. Most undergo metamorphosis.

(Table 32.5)

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G. Amniotes: Origins of the amniotic egg 1. Amniotic eggs allow reptiles, birds, and some mammals to lay their eggs on land.

2. These eggs have membranes surrounding a food supply, water supply, and waste repository. (Fig. 32.10) 3. The membranes provide support and extra surface area for gas exchange, allowing the production of larger, better-developed young.

4. Amniotic eggs usually have leathery shells that are not waterproof (most reptiles), but in birds the shell is hard and waterproof.

H. Mammals: Origin of lactation and fur 1. Mammals are a monophyletic group of amniotes named for their mammary glands that produce milk.

2. Mammals are endotherms, producing significant metabolic heat to maintain a high and environmentally independent body temperature.

3. Three major lineages of mammals: (Table 32.6) a. egg-laying monotremes b. pouch-bearing marsupials c. placental mammals or eutherians 4. The placenta is an organ that forms from both fetal and maternal tissue; it is rich in blood vessels and facilitates the flow of oxygen and nutrients from the mother to the embryo. (Fig. 32.11) 5. Why did viviparity and the placenta evolve? a. Offspring develop at a more constant, favorable body temperature. b. Offspring are protected. c. Offspring are portable (the mother is not tied to a nest).

H. Reptiles: Origin of scales and feathers made of keratin 1. The reptiles are a monophyletic group that represents the second major living lineage of amniotes besides mammals.

2. The four major lineages of reptiles include snakes, turtles, crocodilians, and birds.

(Table 32.7) 3. Several adaptations for life on land: a. Skin is water tight because of a layer of scales made of keratin b. Breathe air through well-developed lungs c. Lay shelled, amniote eggs d. Ectothermic reptiles regulate their body temperature very tightly 4. Feathers are specialized out-growths of the skin composed of keratin. They provide insulation, are used for display, and furnish the lift and steering required for flight.

5. Wings and flight a. Wings and flight evolved in three lineages of tetrapods: pterosaurs, bats, and birds. b. A series of fossils discovered in the early 2000s indicate that at least one lineage of dinosaurs had feathers. c. These well-preserved fossils address key questions about the evolution of birds: (1) Birds are a part of the monophyletic group called dinosaurs.

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(2) Feathers probably evolved in a series of steps, beginning with simple projections from the skin. (Fig. 32.12) (3) It is not yet clear whether flight evolved from the ground up or from the trees down. d. Once feathers evolved, further adaptations made powered, flapping flight increasingly efficient. Examples: keeled sternum, reduction in weight of skeleton, endothermy (Fig. 32.13) I. Parental care a. Parental care represents a fitness trade-off: Parents produce fewer offspring, but each offspring has a better chance of survival. b. Parental care involves supplying food, warmth, and protection. c. Many taxa exhibit some parental care (ray-finned fishes, frogs, insects), but the most extensive parental care is found in mammals and birds. (Fig. 32.14) (1) Mammals and birds often feed their young after birth (with lactation in mammals).

(2) The evolutionary success of mammals and birds has been attributed to the time and energy they invest in parental care.

V. The Primates and Hominins A. The primates 1. The lineage called Primates contains two groups: (Figs. 32.15, 32.16) a. Prosimians include lemurs, tarsiers, pottos, and lorises; most live in trees and are nocturnal. b. Anthropoids include New World and Old World monkeys, the great apes, and humans.

2. What makes a primate a primate? a. Grasping hands and feet, with nails rather than claws b. Color vision, with eyes facing forward for good depth perception c. Large brains, complex social behavior, and extensive parental care 3. What makes a great ape a great ape? a. Great apes are also called hominids. Humans are great apes. b. Great apes are larger than other primates and have long arms, short legs, and no tail. c. One lineage of great apes, the hominins, became bipedal.

B. Fossil humans 1. DNA sequence data indicate that our closest living relatives are chimpanzees and that the common ancestor of chimps and humans lived 6 to 7 mya. (Making Models 32.2) 2. More than 20 fossil hominin species have been found to date. (Table 32.8) (Fig. 32.17) 3. Gracile australopithecines (Australopithecus) (Fig. 32.18a) a. Fossils of species of these slender, small hominins have been found in Africa. b. The gracile australopithecines were bipedal.

(1) The hole in their skull, from which their spinal cord extends, points downward as it does in humans.

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4. Robust australopithecines (Paranthropus) (Fig. 32.18b) a. These hominins were stockier and heavier, with massive cheek teeth and jaws and a sagittal crest to which jaw muscles attached. b. Jaw morphology suggests that these hominins fed on nuts and seeds.

5. Early Homo (Fig. 32.18c) a. They had flatter and narrower faces, smaller jaws and teeth, and larger braincases than earlier hominins. b. They made simple tools. c. They are considered to be the first humans.

6. Recent Homo (Fig. 32.18d) a. They had even flatter faces, even smaller jaws and teeth, and even larger braincases than early Homo. b. These humans were painters and sculptors and buried their dead. c. Cro-Magnons were a specific population of recent Homo sapiens. d. Neanderthals were probably a different species, Homo neanderthalensis.

(Fig. 32.19) e. Recently a very small fossil hominid was found in Indonesia—H. floresiensis has smaller braincases and were only about 1 meter tall.

7. What can be deduced from the hominin fossil record? a. All hominins were bipedal. b. Several species of hominin coexisted at the same time and sometimes in the same place. c. Homo species have extremely large brains relative to their body size.

(1) The large brain is thought to have evolved due to selection for tool use and symbolic language.

C. The out-of-Africa hypothesis 1. Fossil evidence a. First Homo sapiens fossils appear in East Africa about 195,000 years ago. b. Between 60,000 and 30,000 years ago, H. sapiens are found throughout Europe, Asia, Africa, and Australia.

2. Molecular evidence a. Genetic comparisons of diverse human populations today match patterns in the fossil record. (Fig. 32.20) (1) Earlier groups originate in Africa (2) Later branches give rise to lineages in central Asia, Europe, East Asia, Polynesia, and the Americas. b. Genetic data support the notion that there were several waves of migrations that left Africa.

(1) These migration routes are hotly debated. (Fig. 32.21) c. Recent data suggest that H. sapiens and Neanderthals interbred in the Middle East after H. sapiens left Africa, but prior to migrating to Europe and Asia. d. Recent data also suggest that H. sapiens interbred with a population of H. erectus known as the Denisovans.

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D. Have humans stopped evolving?

1. All four processes of evolution—mutation, genetic drift, gene flow, and natural selection—are measureable human populations today.

2. Just a few variables of current topics of study: a. Humans continue to evolve with their pathogens—frequency of hemoglobin alleles vary in human populations according to the prevalence of malaria. b. The size of a newborn’s head is limited by the size of the birth canal—how will this change in response to the increase of Caesarian sections? c. A mutation that allowed humans to digest the milk sugar lactose evolved independently in multiple populations living with domesticated goats, sheep, and cows.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

Actinistia

ectothermy

Marsupiala/marsupials

Actinopterygii albumen Amniota amniotic egg Amphibia/amphibians anadromous anthropoid Asteroidea Aves bipedal bone caecilian cartilage Cephalochordata cerebellum cerebrum Chondrichthyes coelacanth cranium Crocodilia Cro-Magnon deuterostomes Dipnoi dorsal hollow nerve cord Early Homo Echinodermata/

endoskeleton endothermy Eutheria exoskeleton feathers forebrain gestation gill arches gills gnathostomes gracile australopithecines grade great ape hagfish hindbrain hominid hominin human invertebrates jaws keel lactation lampreys Lepidosauria lobe-finned fishes lungfish Mammalia mammary gland

medulla oblongata midbrain Monotremata/monotremes Myxinoidea Neanderthal neural crest cells notochord out-of-Africa hypothesis oviparous ovoviviparous parental care pentaradial symmetry Petromyzontoidea pharyngeal gill slits pharyngeal jaw placenta placental mammals/eutherians podium (plural: podia) primate Primates prosimian protrusible jaw radial symmetry Recent Homo Reptilia/reptile robust australopithecines somites swim bladder

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Teleostei Testudinia Tetrapoda tetrapods

tube feet Urochordata vertebra (plural: vertebrae) vertebrates

viviparous water vascular system

Lecture Activities Game: The Museum Curator Game (deuterostome version) Estimated duration of activity: 30–40 minutes An interesting alternative to a catalog-style lecture on the different lineages of deuterostomes is to involve your students in an active pursuit of this material. Although this activity requires a preparatory investment on your part, it will engage the students and lead them to investigate and apply the information in this chapter. The good news is that when you have prepared the activity once, it can be reused in future classes. Preparation: Make color photocopies of as many deuterostome animals as you can find. (Use Chapter 30, Chapter 32, the Internet, books, teaching specimens, and other materials.) Copy or print the pictures on card stock (laminating them will increase their life span for use in later semesters). Finally, make place cards with the names of the deuterostome lineages from Chapter 32. Setting up the museum: Set the place cards (with the names of the deuterostome lineages) on tables arranged around the classroom. Organize the students into groups, assign each group to a lineage table, and give each group a random assortment of photos and specimens. Ask the students in each group to spend a few minutes researching their lineage and learning its distinguishing morphological features, feeding habits, reproductive strategies, and habitats (in essence, they are to become the class experts on that lineage). Then ask the groups to go through their photos and specimens, select the ones that belong in their lineage, and place the others in a pile on the edge of the table. When they are done, they should split up, visit other tables, search through their “discard” pile, and collect any photos/specimens that belong on their table.

Touring the museum: When all the organisms have been properly classified, have all students sit at their respective tables. One by one, ask students at each table to present their lineage, using the pictures and specimens to illustrate that lineage’s characteristic features.

Game: Name That Deuterostome!

Estimated duration of activity: 5–10 minutes This activity is a time-saving and interesting alternative to cataloging all the various deuterostome lineages in lecture format. It requires that the students read the chapter beforehand and make a chart of the deuterostome lineages and their major features as laid out in Chapter 32. Students will be required to bring their chart to class in order to participate in the game. As an incentive offer an extra-credit point to those who bring them in.

At the beginning of the game, divide the class into two teams (the left and right sides of the classroom; it does not matter how many students are on each team). You will read the list of characteristics, and the students must raise their hands to identify the deuterostome lineage you are describing. You will identify the hand that went up first, call on that student to answer, and award a point to that team if the answer is correct. If the answer is incorrect, give the other team an opportunity to answer. Of course, the team with the most points wins. You 324

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can decide what the prize will be (candy, bonus points on the next test, exemption from a quiz, the sheer joy of knowledge, etc.).

Deuterostome A: no jaws, purely aquatic, swim by moving notochord Answer: Chordata/Myxinoidea and Petromyzontoidea Deuterostome B: tube feet, calcium carbonate teeth, use podia and spines to crawl Answer: Echinodermata/Echinoidea Deuterostome C: swim bladder, Teleostei, breathe via gills Answer: Chordata/Actinopterygii Deuterostome D: leathery beak or bill, oviparous, mammary glands Answer: Chordata/Mammalia/Monotremata Deuterostome E: cartilaginous skeleton, purely marine, most are predators Answer: Chordata/Chondrichthyes Deuterostome F: placenta, extensive parental care, well-developed teeth Answer: Chordata/Mammalia/Eutheria Deuterostome G: bony shell fused to vertebrae and ribs, no teeth, skull Answer: Chordata/Reptilia/Testudinia Deuterostome H: some species are legless; fangs, vestigial hips, and leg bones Answer: Chordata/Reptilia/Lepidosauria Deuterostome I: tube feet, internal skeleton, some eat corals Answer: Echinodermata/Asteroidia Deuterostome J: predators, sit in water with nostrils and eyes exposed, carry young in mouth Answer: Chordata/Reptilia/Crocodilia Deuterostome K: endotherms, feathers, oviparous, parental care Answer: Chordata/Reptilia/Aves Deuterostome L: “both-sides-living,” larvae are aquatic, three subgroups perform reproduction in three very different ways Answer: Chordata/Amphibia Deuterostome M: gills and lungs, muscular fins, omnivorous Answer: Chordata/Dipnoi and Actinistia Deuterostome N: extensive parental care, pouch, mammary glands, Tasmanian wolf Answer: Chordata/Mammalia/Marsupials Web resources: The following websites provide additional information and photos of deuterostomes that you and your students may find useful. http://www.ucmp.berkeley.edu/echinodermata/echinomm.html http://www.ucmp.berkeley.edu/chordata/hemichordata.html http://www.ucmp.berkeley.edu/chordata/chordata.html

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Discussion Idea: The World without Them Estimated duration of activity: 10–15 minutes Students may have heard of the TV series and book The World Without Us, which explores the idea of what would happen to the world if all humans instantly vanished. You can use this idea to get students thinking about the ecological importance of various groups. What would happen if all vertebrates on Earth suddenly vanished? All insects? All protostomes? All birds, but not mammals? All mammals, but not birds? If any of these groups suddenly disappeared, would all ecosystems on Earth collapse, or would life continue mostly unaffected?

To encourage the discussion, assign small groups of students to discuss one taxon each (insects, crustaceans, mollusks, birds, mammals, and so on, and one group for humans). Have each group spend 5 minutes discussing among themselves the ecological importance of their taxon, with the central question of whether most ecosystems on Earth could continue to function without that taxon. Have the students consider both marine and terrestrial ecosystems.

Then have each group present its conclusions to the class. Have the class vote at the end on which group is most ecologically critical. If time permits, you may organize this discussion as a formal debate of invertebrates versus vertebrates.

Class Activity: Building the Animal Tree Estimated duration of activity: 5 minutes (per lecture) Drawing a large phylogenetic tree is a fun activity that helps students review the traits and relationships of the taxa that they have learned about so far. If you began building an Animal Tree in earlier chapters, students may continue to build on the tree now by adding the deuterostome phyla and their distinctive traits. Remind students to leave room for all the chordate lineages! Take 5 minutes at the end of lecture for this activity. See the Lecture Activities for Chapter 30 for further details on building an Animal Tree as a class activity.

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CHAPTER

33

Viruses

Learning Objectives: Students should be able to...

• Describe the impact of viruses on human evolution and disease.

• Describe the six steps of a viral replicative (lytic) cycle, and contrast it with the lysogenic cycle.

• Compare and contrast the three hypotheses for the origin of viruses and explain how they • continue to evolve.

Describe how the relationship between the diversity of viruses is related to their genetic material.

Lecture Outline • Viruses are obligate intracellular parasites that are not cells or organisms. (Table 33.1) • Viruses cannot manufacture ATP, nucleic acids, or proteins on their own.

• Viruses must enter and take over a host cell to make copies of themselves.

• Viruses are not alive. They are not given scientific (genus + species) names.

• Viruses are extremely abundant. For example, in 1 liter of seawater there can be over 10 billion virions.

• Each type of virus infects a certain species or cell type, and all known organisms are parasitized by at least one kind of virus.

I. Why Do Biologists Study Viruses?

A. Viruses shape the evolution of organisms.

1. Viruses have promoted the evolution of physical barriers and the immune system.

2. Viruses have also influenced the genetic makeup of organisms via lateral gene transfer—5–8% of the human genome consists of remnants of viral genomes from past infections.

B. Recent viral epidemics in humans 1. Virtually every human organ and cell type can be infected by one or more kinds of virus. (Fig. 33.1) 2. Epidemics are widespread outbreaks of disease that infect a large number of people at one time. Examples: measles and smallpox in North American tribes 3. A pandemic is an epidemic that is worldwide: the 1918−19 Spanish flu. (Fig. 33.2)

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C. Current viral pandemics in humans: AIDS 1. In terms of the total number of people affected, the AIDS epidemic has surpassed the smallpox and measles epidemics as well as the Spanish flu pandemic. a. HIV, the human immunodeficiency virus, causes acquired immune deficiency syndrome (AIDS). b. HIV is currently the single most studied virus in the world.

2. How does HIV cause disease? a. HIV infects helper T lymphocytes and macrophages of the immune system. b. As T cell counts decline, the immune system loses its ability to fight other infections. (Fig. 36.3) c. The other infections eventually kill the host.

3. What is the scope of the AIDS pandemic? a. AIDS has already killed 36 million people worldwide, and 35 million more are thought to be infected with HIV. b. The epidemic is at its worst in sub-Saharan Africa and is accelerating in south and east Asia. c. AIDS is unlike many diseases in that it particularly attacks young adults. d. Students should be able to explain why it takes 8 to 12 years before clinical signs of AIDS appear.

II. How Do Biologists Study Viruses?

A. Two common research goals: 1. Develop vaccines to help host cells fight a viral infection.

2. Develop antiviral drugs that will stop the virus from replicating once the host is infected.

B. Viruses are tiny and require special research techniques.

1. Most viruses are 20 to 300 nm in diameter⎯much smaller than eukaryotic or bacterial cells. (Fig. 33.4) 2. Viruses can be isolated using a filter that allows the virus through but traps cells.

3. Scientists can then use the isolated virus to test virulence using Koch’s postulates.

C. Analyzing morphological traits 1. Viruses come in a wide variety of shapes. (Fig. 33.5) 2. Most viruses fall into two general categories: a. Nonenveloped viruses are simply a genome enclosed by a protein shell called a capsid. b. Enveloped viruses consist of a genome enclosed in a capsid entirely surrounded by a phospholipid bilayer called an envelope.

D. Analyzing the genetic material 1. Viruses can be categorized by the nature of their genome.

2. The RNA and DNA genomes of viruses can be either single stranded or double stranded.

3. Single stranded RNA genomes can be classified as positive sense or negative sense. a. Positive-sense RNA: Genome contains the same sequences as the mRNA required to produce viral proteins.

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b. Negative-sense RNA: Base sequences in the genome are complementary to those in viral mRNAs. c. Ambisense RNA: Genome has at least one strand that contains two regions—one is positive sense, and one is negative sense.

E. Analyzing the phases of replicative growth 1. All viruses can do replicative growth (also called the lytic cycle). (Fig. 33.6) a. Replicative growth involves six phases: entry, transcription, replication, assembly, exit, and transmission to a new host. b. Once the virus has entered the host cell, the viral genome is transcribed and new viral proteins are produced, which are then assembled to make new virions (virus particles). c. The new particles break open the host cell (usually killing it) and exit to infect other cells. d. Figure 33.7 shows the bacteriophage’s capacity for replication that results in a nonlinear growth that proceeds in a stepwise manner.

2. How do viruses enter a cell? a. Plant viruses usually enter plant cells after an insect has disrupted the cell wall. b. Viruses enter bacterial and animal cells by binding to cell surface molecules.

(1) After binding, nonenveloped viruses simply dump their genome into the cell, leaving the capsid at the cell surface.

(2) When an enveloped virus binds, its envelope and the host membrane fuse, and the entire capsid enters the cell. c. Example: What cell surface molecules does HIV use?

(1) Researchers found that people with AIDS lacked T cells with the CD4 surface protein.

(2) Experiments showed that if CD4 is blocked by an antibody, HIV cannot enter the cells. (Fig. 33.8) (3) Uncoating of the viral genome will not occur unless the virus binds to a second membrane protein called a co-receptor. This can occur both at the cell surface (Fig. 33.9a) or in the endosome. (Fig. 33.9b) (4) Drugs called fusion inhibitors can interfere with the binding process and prevent HIV from entering cells.

3. How do viruses produce proteins? a. Viruses lack translational machinery and so must use the host’s. b. Viral envelope proteins are made in the same way that the host’s own membrane proteins are made, using ribosomes bound to the ER. (Fig. 33.10a) c. Viral capsid and inner core proteins are translated on free ribosomes. (Fig. 33.10b) d. In some viruses, long viral polypeptides are cleaved into functional proteins by a viral protease.

(1) Protease inhibitors are a major class of HIV drugs that block the active site of HIV’s protease.

4. How do viruses copy their genomes? a. Most viruses contain a polymerase that copies the viral genome inside the infected host cell, using nucleotides and ATP provided by the host cell. .

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b. Viruses that have DNA genomes contain a viral DNA polymerase. c. Some RNA viruses use an enzyme called RNA replicase to copy their genome (i.e., the new copies are also made of RNA). (Fig. 33.11) (Making Models 33.1) d. Other RNA viruses transcribe their RNA genome into complementary DNA (cDNA) using reverse transcriptase. (Fig. 33.12) (1) These viruses are called retroviruses.

(2) HIV is a retrovirus.

(3) The cDNA generated by reverse transcriptase gets incorporated into the host DNA and is replicated by the host’s biosynthetic machinery.

5. How are new virions assembled? a. The capsid (made of viral protein) forms around the viral genome. b. Noncapsid proteins such as DNA polymerase are also packaged within the capsid, along with the genome. c. For enveloped viruses, envelope proteins become incorporated into the host membrane. The envelope is acquired when the virus buds out of the host cell. d. In most cases, self-assembly occurs, although the details are not yet understood.

6. How do progeny virions exit an infected cell? a. Enveloped viruses bud off from the host cell. (Fig. 33.13a) b. Nonenveloped viruses burst from the cell, lysing it in the process. (Fig. 33.13b) 7. How are virions transmitted to new hosts? a. In multicellular organisms, the new virions travel through the host, often via the bloodstream or lymphatic system.

(1) They may then be destroyed by the immune system or infect new cells. b. If the virus leaves that host entirely (e.g., through sneezing), it might then encounter a new host. c. Natural selection favors viruses that can (1) successfully replicate within their host and (2) infect new hosts. d. Students should be able to compare and contrast the replicative cycles of bacteriophage T4 and HIV. F. Analyzing how viruses coexist with host cells.

1. All viruses undergo replicative growth, but some undergo lysogeny by arresting the replicative cycle and going dormant. (Fig. 33.14) 2. This dormant state is referred to as latency in animal infections.

3. In HIV for example, blocking viral transmission from host to host is a major public health strategy. a. Treatment of venereal diseases b. Condom use c. Abstinence or monogamy

III. What Themes Occur in the Diversification of Viruses?

A. Where did viruses come from?

1. Origin in plasmids and transposable elements? a. The escaped-gene hypothesis proposes that viral genomes, as well as plasmids and transposable elements, descended from clusters of genes that escaped from bacterial or eukaryotic chromosomes long ago.

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b. The escaped genes took on a mobile, parasitic existence because they happened to contain the information needed to replicate themselves.

2. Origin in symbiotic bacteria? a. This hypothesis proposes that symbiotic bacteria living inside eukaryotic cells eventually lost the genes needed to make ATP and other macromolecules because they were able to get these things from their host. b. As a result, such bacteria could have degenerated into viruses.

3. Origin at the origin of life? a. This hypothesis proposes that viruses trace their ancestry back to the RNA-based forms of life on Earth.

4. Currently, there is no single widely accepted view—all three hypotheses may be valid.

B. Emerging viruses, emerging diseases 1. Emerging diseases are new illnesses that suddenly affect large numbers of people.

Example: Ebola virus claimed the lives of more than 11,000 people in 2014.

2. Some emerging viruses arise from genome reassortment. a. Small changes in the influenza virus are responsible for the annual need for flu shots. b. More significant changes can occur when genomic segments are randomly mixed between two strains. (Fig. 33.15) 3. Using phylogenetic trees to understand emerging viruses a. Phylogenetic analyses indicate that the two forms of HIV are derived from two different primate viruses. (Fig. 33.16) (1) HIV-1, which is more virulent, probably came from a closely related virus found in chimpanzees.

(2) HIV-2 probably came from sooty mangabeys. b. Different strains of HIV-1 probably represent different origins (e.g., HIV-1 jumped from chimps to humans several times).

4. Responding to an outbreak a. Doctors report cases to public health officials, who try to identify the pathogen and determine the mode of transmission. b. Identifying the pathogen (1) Example: SARS; in 2002, an outbreak of an unknown respiratory syndrome rocked southern China and rapidly spread to several countries.

(2) Within a few weeks, three independent groups confirmed the causative agent of SARS as a new coronavirus. c. Mode of transmission (1) Transmission of SARS was mostly limited to people who were in close contact with infected individuals.

IV. Key Lineages of Viruses A. There is no phylogeny for all viruses because viruses probably originated multiple times throughout the history of life.

B. Viruses may be grouped into seven general categories determined by the nature of their genome. (Fig. 33.17)

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1. Students should be able to associate particular types of genomes with each of the three hypotheses that have been posed to explain the origin of viruses.

C. Double-stranded DNA (dsDNA) viruses (Table 33.2) 1. Genetic material: a single molecule of linear or circular dsDNA 2. Host species: almost anything except land plants 3. Replication cycle a. Viral genes must enter the nucleus to be replicated. b. Viral genes can be replicated only during S phase, and therefore viruses are replicated only in actively dividing cells.

4. Examples: smallpox, T series and lambda bacteriophages, poxviruses, herpesviruses, adenoviruses D. Single-stranded DNA (ssDNA) viruses (Table 33.2) 1. Genetic material: a single molecule of linear or circular ssDNA 2. Host species: most infect animals; marine invertebrates hosts 3. Replication cycle a. Genome replicated via a dsDNA intermediate b. mRNA used to produce viral proteins 4. Examples: canine parvovirus type 2 E. Double-stranded RNA (dsRNA) viruses (Table 33.3) 1. Genetic material: up to 12 dsRNA molecules 2. Host species: a wide variety (fungi, plants, animals, bacteria) 3. Replication cycle: Once in the cytoplasm of the host cell, the viral dsRNA serves as a template for transcription to ssRNA molecules, which are used for translation of viral proteins.

4. Examples: bluetongue virus; some dsRNAs devastate rice, corn, and sugarcane crops F. Positive-sense single-stranded RNA ([+]ssRNA) viruses (Table 33.3) 1. Genetic material: one to three ssRNA molecules 2. Host species: mostly plants, but also bacteria, fungi, and animals 3. Replication cycle: Once in the host’s cells, the ssRNA is translated directly into a large polyprotein that is cleaved into smaller proteins.

4. Examples: Many plant diseases that turn patches of plants white, often called mottle, spotted, chlorotic, necrotic, or mosaic viruses. This group also includes the common cold, polio, and hepatitis A, C, and E. G. Negative-sense single-stranded RNA ([−]ssRNA) viruses (Table 33.3) 1. Genetic material: up to eight ssRNA molecules 2. Host species: a wide variety of plants and animals 3. Replication cycle a. Once inside the host’s cells, viral RNA polymerase transcribes the genome into mRNA. b. The viral mRNA is used to make viral proteins and is used as a template for the next generation of the ssRNA genome.

4. Examples: influenza, mumps, measles, Ebola, hantavirus, rabies

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H. Reverse-transcribing (+)ssRNA viruses (Table 33.4) 1. Genetic material: two copies of a single-stranded RNA genome 2. Host species: only vertebrates; usually birds, fish, or mammals 3. Replication cycle a. These viruses use reverse transcriptase to make a cDNA of their genome. b. Integrase splices the viral cDNA into the host genome. c. The virus may be quiescent for a time but will eventually switch to lytic growth.

4. Examples: HIV, Rous sarcoma virus, mouse mammary tumor virus, murine leukemia virus I. Reverse-transcribing dsDNA viruses (Table 33.4) 1. Genetic material: two copies of a single-stranded RNA genome 2. Host species: only vertebrates and plants 3. Replication cycle a. Genome used as a template for transcription of full length (+)ssRNA and mRNA b. Viral reverse transcriptase converts (+)ssRNA into dsDNA to replicate genome. c. mRNA used to produce viral proteins 4. Examples: Hepatitis B

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. acquired immune deficiency syndrome (AIDS) ambisense virus antibody antiviral drugs bacteriophage capsid CD4 complementary DNA (cDNA) co-receptor DNA virus double-stranded DNA (dsDNA) viruses double-stranded RNA (dsRNA) viruses emerging disease emerging virus

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envelope enveloped virus epidemic fusion fusion inhibitor helper T lymphocyte host cell human immunodeficiency virus (HIV) immune system integrase intracellular parasite Koch’s postulates latent lysogenic cycle/lysogeny lysozyme lytic cycle macrophage microscopy

negative-sense single-stranded RNA ([−]ssRNA) viruses negative-sense virus nonenveloped virus obligate pandemic plasmid positive-sense single-stranded RNA ([+]ssRNA) viruses positive-sense virus protease protease inhibitor replicative growth retrovirus reverse transcriptase RNA replicase RNA reverse-transcribing viruses/retroviruses RNA virus

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simian immunodeficiency virus (SIV) strain

symbiotic bacteria transmission transmission electron transposable element

virion virulence virus

Lecture Activities Student Research Activity: Viruses and Human Health Estimated duration of activity: 20–30 minutes if research is done outside of class; 50–60 minutes if research is done in class This activity encourages students to do some group research to learn more about how viruses can affect our health and food supply. This activity works best if done in a computer lab or a library where students have access to information resources. This would also make a good out-of-class writing assignment.

Divide students into small groups (two or three students), and have each group prepare a brief report to give to the class on any one of the following human viruses: adenoviruses

mumps virus

dengue fever virus Ebola virus hepatitis viruses herpes viruses human papillomaviruses (HPV) influenza virus measles virus molluscum contagiousum virus

parainfluenza virus parvoviruses poliovirus respiratory syncytial virus (RSV) rhinovirus rotavirus smallpox monkeypox virus (an emerging virus?)

H1N1 influenza (“swine flu”) Present the following questions to help students structure their research: 1. What type of virus is it? a. To what family of viruses does it belong? b. What other viruses are members of this same family? c. Is the virus enveloped or nonenveloped? d. What type of genome does it have? (DNA or RNA? single- or double-stranded? positive-sense, negative-sense, or ambisense?) e. What is the size of the genome? How many proteins does it encode? f. Does the virus encode its own polymerase enzyme? If so, what type of polymerase is it? g. Are any virus-encoded enzymes packed into its virus particle?

2. Where and how did the virus originate? a. Is there a particular part of the world where the virus is a serious problem? b. If so, what cultural practices and socioeconomic conditions contribute to high infection rates?

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3. What is the known history of the virus? a. From what other type of virus, or from what host, is the human virus thought to have originated? b. Where and when is this thought to have originated?

4. What part of the human body does the virus infect?

5. Is the virus lytic or lysogenic? How do new viral particles exit the host cell?

6. How virulent is the virus? a. Can it kill people? If so, what exactly causes death? b. Are some strains of the virus more virulent than others?

7. How is the virus transmitted? a. Can it spread from human to human? b. Can humans catch this virus from other animals? If so, which animals?

8. Does the virus routinely cause epidemics? Has the virus caused any pandemics?

9. What, if any, drugs have been developed to block the virus or treat its symptoms? Is a vaccine available?

10. Does the virus have a high mutation rate?

11. What is the most interesting thing you found out about this virus or this group of viruses?

Good Internet sources for online student research: www.virology.net (All The Virology on the WWW) www.scirus.com (Scirus, a comprehensive science search engine) www.cdc.gov (U.S. Centers for Disease Control and Prevention) www.scholar.google.com (Google Scholar)

Discussion Idea: Positive-Sense and Negative-Sense Viruses Estimated duration of activity: 5 minutes Students often need some assistance understanding the difference between positive-sense and negative-sense RNA viruses. Depending on your students’ backgrounds, you may first give them a brief primer or review on the structure of DNA and RNA and the rules of complementary base pairing. Then present the following scenario and questions. (Students will need access to a copy of the genetic code to figure out the peptide sequence.)

Scenario: A cell has been infected with a virus. Many viral mRNA molecules are being produced, and the cell is producing many copies of a tiny viral peptide consisting of only two amino acids.

Questions: • Suppose it is a single-stranded positive-sense RNA virus that contains the sequence AAA CCC. What will the sequence of the viral mRNA be? What will the peptide be?

Answer: The viral mRNA sequence is the same, AAA CCC. The peptide will be lysineproline.

• Suppose it is a single-stranded negative-sense RNA virus that contains the same sequence, AAA CCC. What will the sequence of the viral mRNA be? What will the peptide be?

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Answer: The viral mRNA sequence will be complementary, UUU GGG, and the peptide will be phenylalanine-glycine.

• Is it possible to know the final peptide sequence if we know the sequences for the two strands of a double-stranded virus, whether it is RNA or DNA? Answer: Not without further information. We would need to know which of the two strands is the transcribed strand.

Discussion Idea: Vaccines for Viruses If you have not already covered the animal immune system, a brief introduction about antibody-antigen recognition and immune response may be in order. After that, ask your students to devise a plan to develop a vaccine. Ask them what parts of a virus would cause the development of immunity in a host without actually causing serious disease. (Students should be able to realize that the outermost proteins of a virus are good targets for vaccine development.) Students will likely have heard about the difference between live, weakened vaccines and killed-virus vaccines; ask them to theorize which might provoke a stronger immune response, and how they could test their ideas. At this point, the students should be prepared to discuss why there are no effective vaccines for the common cold or HIV. You might complete the discussion by initiating a short informal debate on whether the seasonal flu vaccine is effective and worth receiving.

Discussion Idea: The Lysogenic-to-Lytic Switch Several viruses that infect humans are known to “lie dormant” and then manifest as disease states at some point during a person’s life. Start by asking students to think of any viral diseases that are characterized by dormancy or by lifetime recurrent infections. The first example they will likely come up with is HIV, which is described in the text. The following viruses are also known to be lysogenic in humans and then, under certain conditions, cause disease when they switch to a lytic cycle.

The chicken pox virus, varicella zoster The chicken pox virus is an enveloped dsDNA virus. Upon initial infection, it is lytic and causes chicken pox. However, in some people, it becomes lysogenic in dorsal root ganglion cells of the spinal cord. At some point later in life, usually in response to stress, the virus becomes lytic again, traveling down the dendrites of the dorsal root ganglion cells and budding out of the free nerve endings that innervate the skin. The resulting disease, called shingles, is visible as a blistering rash on the skin. Because the virus is budding out of free nerve endings that sense pain, shingles is extremely painful. It in interesting that, because dorsal root ganglion cells innervate the skin in defined patterns called dermatomes, the shingles rash shows up only on that dermatome. For more information on the chicken pox virus and shingles, visit these websites: http://www.cdc.gov/vaccines/vpd-vac/varicella/default.htm http://www.emedicine.com/neuro/topic390.htm http://www.virology-online.com/viruses/VZV.htm Warts, human papillomavirus The nonenveloped/naked dsDNA virus that causes warts is known to infect some people and not others. Although the papillomavirus does not show a true lysogenic replication cycle, it does incorporate its genome into that of the host cell. Subsequently, it induces the host cells (usually epidermal cells in humans) to replicate more rapidly than usual. As a consequence, 336

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the virus’s genome is replicated and an epidermal growth (the wart) appears. Some warts can be treated and will never return; in other cases, the virus can become latent and form a wart at some later time. For more information on human papillomaviruses, visit these websites: http://www.niaid.nih.gov/topics/genitalWarts/Pages/default.aspx http://www.virology-online.com/viruses/Papillomaviruses.htm Herpes There are several different types of herpes simplex viruses, most of which are enveloped dsDNA viruses. Once a person is infected, the herpes simplex virus is lysogenic in nerve cells. In response to immunosuppression, stress, a cold, or a fever, it becomes lytic and causes lesions in the skin of the genitalia, mouth, fingers, or eyes. For more information on herpes simplex viruses, visit these websites: http://www.virology-online.com/viruses/HSV.htm http://darwin.bio.uci.edu/~faculty/wagner/movieindex.html http://www.virology.net/Big_Virology/BVDNAherpes.html http://dermatology.about.com/od/herpessimplex/

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CHAPTER

34

Plant Form and Function

Learning Objectives: Students should be able to…

• Compare surface area/volume relationships among different shapes and explain the significance of these relationships.

• Describe examples of morphological diversity and phenotypic plasticity in roots and shoots.

• Explain the functional significance of the dermal, ground, and vascular tissue systems.

• Describe the role of apical meristems in primary growth.

• Explain the role of lateral meristems in secondary growth.

Lecture Outline I. Plant Form: Themes with Many Variations A. Plants use energy and carbon to grow and reproduce.

1. Plants need light, carbon dioxide, water, nitrogen, phosphorus, magnesium, and potassium to photosynthesize, make macromolecules, and maintain their structure.

2. The root system (consisting of a vertical taproot with lateral roots that branch horizontally) is belowground and acquires water and dissolved nutrients from the soil.

(Fig. 34.1) 3. The shoot system is aboveground and acquires carbon dioxide and light from the atmosphere. The root and shoot systems make up the plant body.

4. The vascular tissue connects the root and shoot systems by transporting water and nutrients from the roots to the shoots and by transporting glucose and other nutrients from the leaves to the rest of the plant.

B. The importance of surface area/volume relationships 1. A plant is more efficient as an absorbance-and-synthesis machine when it has a large surface area relative to its volume. (Fig. 34.2) 2. Making Models 34.1 asks students to redraw a cube and calculate surface area.

C. The root system 1. Most root systems have a main vertical taproot and numerous horizontal lateral roots.

2. Morphological diversity in root systems a. Example: herbaceous perennial prairie plants of North America (Fig. 34.3) b. Root system diversity enables many different species to live together in the same environment without directly competing for the same resources.

(1) Grasses have dense, fibrous root systems.

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(2) Prairie roses have extremely long taproots.

(3) Other plants can have thick roots that store starch.

3. Phenotypic plasticity in root systems a. Roots show a great deal of phenotypic plasticity, meaning that they are “plastic”, or changeable, depending on environmental conditions. b. For example, spruce trees that grow in waterlogged areas have shallow root systems, whereas spruce trees that grow in drier areas have deeper root systems. c. Roots actively grow into areas of soil where resources are abundant, and roots do not grow or die back in areas where resources are used up or lacking.

4. Modified roots (Table 34.1) a. Adventitious roots are lateral roots that develop from shoots. b. Anchor roots that grow from nodules on ivy help the ivy grow up a wall or tree. c. Prop roots are a type of adventitious root that helps support corn plants against wind. d. Pneumatophores are roots (e.g., on mangroves) that aid in gas exchange. e. Storage roots, such as sugar beets, store carbohydrates in their taproot for future use.

D. The shoot system 1. Features of the shoot system a. The system consists of the aboveground portions of a plant, including nodes where leaves are attached, internodes between leaves, and leaves, the photosynthetic organs that extend from the stems. b. It is a repeating series of nodes, internodes, leaves, and apical and axillary buds. c. As plants grow, the number of nodes, internodes, and leaves increases. d. Once a leaf, node, or internode forms, however, it does not increase much in size over time. e. Thus, plants grow by adding more parts rather than by increasing the size of each part.

2. Morphological diversity in shoot systems a. The way branches are added contributes to the overall shape of a shoot system and varies widely among species. b. This variation is important because it allows different species of plants to harvest light in different ways and minimizes competition, allowing plants to thrive in many different habitats. c. For example, Hawaiian silverswords are a monophyletic group of plants whose species vary in growth form, size, shape, and habitat.

(1) Some have thick, woody stems and grow as shrubs or trees.

(2) Others form sprawling cushions or mats.

(3) Several species grow as rosettes. d. The diverse silversword shoot systems are adaptations that enable the plants to obtain the resources they need for growth and reproduction in varied habitats.

3. Phenotypic plasticity in shoot systems (Fig. 34.4) a. Shoots change their growth pattern in response to variations in light availability. b. Trees growing in ample sunlight have shoot systems that are short and broad.

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c. Trees growing in areas with dense vegetation grow tall and narrow. d. A plant’s shoot system moves around to maximize its chances of capturing light.

4. Modified stems (Table 34.2) a. Cactus stems are enlarged to store water and have modified leaves called spines. b. Stolons are modified stems that run over the soil surface, producing roots and leaves at each node. Stolons function to produce new plants via asexual reproduction. c. Rhizomes are stems that grow horizontally under the ground. They function to produce new plants via asexual reproduction. d. Tubers are rhizomes that are modified for carbohydrate storage. e. Thorns are modified stems that protect the plant from predators.

E. The leaf 1. Leaves consist of an expanded blade and a petiole that arises from the lateral bud.

(Fig. 34.5a) 2. Morphological diversity in leaves a. Grass lacks petioles. b. Compound leaves are divided into a series of leaflets (Fig. 34.5b); doubly compound leaves have leaflets that are again divided into leaflets. (Fig. 34.5c) c. Cacti and evergreens have needlelike leaves to minimize water loss via transpiration. (Fig. 34.5d) d. The arrangement of leaves on a stem varies greatly among species.

(1) Some leaves alternate on a stem. (Fig. 34.6a) (2) Some leaves are paired evenly on a stem. (Fig. 34.6b) (3) Other leaves are arranged in a whorl. (Fig. 34.6c) (4) Some leaves grow in a rosette. (Fig. 34.6d) e. Each arrangement of leaves is presumed to maximize the efficiency of light capture for the plant, while minimizing transpiration water loss in drier habitats.

3. Phenotypic plasticity in leaves (Fig. 34.7) a. Leaves do not grow continuously but do exhibit phenotypic plasticity. b. Leaves that grow in the shade are large and broad, providing a large surface area to capture rare photons. c. Leaves that grow in full sunlight have a smaller surface area to minimize water loss via transpiration.

4. Modified leaves (Table 34.3) a. Onion bulbs consist of thickened leaves that store nutrients. b. The thick leaves of succulents store water. c. The tendrils of vines help the plant climb. d. Brightly colored leaves (floral mimics) of poinsettias attract pollinators. e. Leaves of carnivorous plants help to trap insect prey.

II. Plant Cells and Tissue Systems A. Features of plant cells 1. Plant cells are surrounded by a stiff cell wall made of strong cellulose fibers.

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2. Plant cells are connected by gaps in the cell wall and cell membrane called plasmodesmata. (Fig. 34.8b) 3. Plant cells have one or more storage vacuoles that contain cell sap and help the cell maintain its proper volume.

4. Photosynthetic plant cells have chloroplasts for photosynthesis.

5. A tissue is a group of cells that function as a unit. There are three primary plant tissue systems.

B. The dermal tissue system 1. Epidermal cells protect the surface. a. The epidermis (dermal tissue), the outermost single layer of cells covering the entire plant, provides protection and reduces water loss. b. Epidermal cells secrete the cuticle, a waxy compound that coats the outer surface of the plant and minimizes water loss. It also protects the plant against pathogens. c. The cuticle also reduces gas exchange, and CO2 needs to be able to enter leaves to be used in photosynthesis.

2. Stomata regulate gas exchange and water loss. a. Therefore, stomata are in the epidermis of leaves. b. Stomata are made up of specialized epidermal guard cells that regulate the opening and closing of a pore. (Fig. 34.9) c. Stomata can open to allow CO2 in and O2 out, and they close to minimize water loss.

3. Trichomes perform an array of functions. (Fig. 34.10) a. Trichomes on epidermal cells: (1) Keep the leaf surface cool by reflecting sunlight.

(2) Reduce water loss by forming a dense mat.

(3) Provide barbs or store toxic compounds that defend against herbivores.

(4) Trap and ingest insects.

C. The ground-tissue system 1. Ground tissue produces and stores valuable molecules. a. Ground tissue, which makes up the bulk of the plant, is composed of parenchyma, collenchyma, and sclerenchyma. b. Most photosynthesis and carbohydrate storage takes place in ground tissue. c. Cells in the ground tissue also make and release hormones, pigments, and defense toxins.

2. Parenchyma consists of “workhorse” cells. a. Parenchyma cells are the most abundant and versatile cells in plants. b. They are living cells with thin walls that are the primary site of photosynthesis and starch storage. (Fig. 34.11) c. They are totipotent—these cells can divide and develop into a complete plant.

(1) This trait is important in plant wound healing and asexual reproduction.

(2) If you remove part of a coleus stem, the parenchyma cells divide to form a callus, which is the beginning of a new plant. (BioSkills 11) 3. Collenchyma cells function primarily in shoot support.

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a. Collenchyma cells are living cells with unevenly thickened primary cell walls that support young stems and leaves. (Fig. 34.12) 4. Sclerenchyma: two types of specialized support cells a. Sclerenchyma cells are dead, hollow cells with thick secondary walls found throughout a mature plant. b. In addition to cellulose, the secondary cell wall of sclerenchyma cells contains lignin, a tough, rigid compound. c. Sclerenchyma cells provide support to mature stems. d. Whereas collenchyma cells can support actively growing parts of the plant because they have an expandable primary cell wall, the nonexpandable secondary cell wall of sclerenchyma cells specializes them for supporting stems and other structures after active growth has ceased. e. There are two types of sclerenchyma cells: fibers and sclereids.

(1) Fibers are extremely elongated sclerenchyma cells often used to manufacture paper, fabrics, and ropes. (Fig. 34.13a) (2) Sclereids are short sclerenchyma cells often found in the protective coats of seeds and nuts. (Fig. 34.13b) D. The vascular tissue system 1. Xylem structure a. Xylem tissue transports water and nutrients and is composed of specialized sclerenchyma cells called vessel elements and tracheids. (Fig. 34.14c) b. All vascular plants have tracheids, but only angiosperms have vessel elements. c. Tracheids are long, slender cells with tapered ends that contain pits where one tracheid contacts another. (Fig. 34.14a) (1) Pits are breaks in the secondary cell wall but not in the primary cell wall.

(2) Water moves from one tracheid to another through the pits. d. Vessel elements are shorter and wider than tracheids and have perforations where one vessel element meets another. (Fig. 34.14b) (1) Perforations are breaks in both the primary and secondary cell walls, and sometimes these cells lack cell walls at these junctions.

(2) Since vessel elements offer little resistance to water flow, they conduct water more efficiently than tracheids.

2. Phloem structure (Fig. 34.15) a. Phloem tissue transports sugars, amino acids, hormones, and toxins through specialized parenchyma cells called sieve-tube elements.

(1) Sieve-tube elements are long, thin cells that are alive at maturity.

(2) They have perforated ends called sieve plates, and they lack nuclei, chloroplasts, and other organelles. b. Companion cells are connected to the sieve-tube members via plasmodesmata.

(1) Companion cells provide the sieve-tube elements with the molecules that keep them alive.

(2) Companion cells help to load and unload carbohydrates from the sieve-tube elements.

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III. Primary Growth Extends the Plant Body A. Plants exhibit indeterminate growth; that is, they never stop growing.

1. Plant growth is the result of two processes: a. Repeated mitotic divisions in meristems b. Expansion of the volume of those cells; cell enlargement 2. Apical meristems are found at the tips of roots and shoots and are the site of primary growth.

3. Primary growth makes the plant taller and the roots longer.

4. Primary growth leads to non-woody, herbaceous stems.

B. How do apical meristems produce the primary plant body?

1. The three types of primary meristematic cells are important because they give rise to three major tissue systems that extend throughout the plant body: (Fig. 34.16) (Table 34.5) a. Protoderm gives rise to dermal tissue or epidermis. b. Ground meristem gives rise to the ground tissue system. c. Procambium gives rise to the vascular tissue system.

2. These three tissue types are distributed in the plant body according to their function.

(Fig. 34.17) C. How is the primary root system organized?

1. To access more water and nutrients, roots must continuously be growing into new regions of soil.

2. How do roots push their way through tough soil? a. The growing edge of the root is protected by the root cap. (Fig. 34.18a) b. Apical meristems in the root constantly replenish damaged cells in the root cap. c. Root cap cells can sense gravity and determine the direction of growth. d. Root cap cells also make and secrete a slimy lubricant that aids in movement through the soil.

3. Three distinct populations of cells exist behind the root cap and facilitate root growth: a. The zone of cellular division contains the apical meristem, where cells are actively dividing. b. The zone of cellular elongation contains new cells that are growing in length. c. The zone of cellular maturation contains cells that are differentiating into dermal, vascular, and ground tissue cells.

(1) Dermal cells in this zone form root hairs.

(2) Root hairs are the actual sites of water and nutrient absorption (Fig. 34.18b).

The rest of the root system provides structural support for the root hairs, conducts water and ions to the shoot, stores the products of photosynthesis, and anchors the plant in the soil.

D. How is the primary shoot system organized?

1. The top of the stem is a rounded dome of cells that make up the shoot apical meristem.

2. On both sides of this meristem, new leaves arise.

3. As in the root, the cells behind the apical meristem are enlarging and differentiating.

4. The anatomy of a eudicot stem (Fig. 34.19a) .

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a. The center of the stem, the pith, is made up of ground tissue that stores carbohydrates. b. Xylem and phloem are found in vascular bundles around the pith that run the length of the stem. c. The cortex lies between the vascular bundles and the epidermis. d. The vascular bundles form a ring around the pith.

5. In monocots, the vascular bundles are scattered throughout the pith. (Fig. 34.19b)

IV. Secondary Growth Widens Shoots and Roots A. Secondary growth increases the width of stems and roots and thus increases the amount of conducting tissue and structural support for primary growth to continue.

B. What is a cambium?

1. Lateral meristems are found in rings along the length of roots and shoots, and they are the site of secondary growth. a. Secondary growth increases the width of the stem and roots. b. Secondary growth results in the formation of wood.

2. Secondary growth results from a ring of secondary meristem or cambium that leads to an increase in girth, providing the growing plant with structural support.

3. Lateral meristems divide in a plane parallel to the long axis of the root or shoot, thus increasing the width of the structure.

4. There are two distinct types of cambium in plants that undergo secondary growth: vascular cambium and cork cambium. (Fig. 34.20) 5. Summary of the tissues and cells involved in secondary growth (Table 34.6) C. How does a cambium initiate secondary growth?

1. Within each vascular bundle, a single layer of cells between the xylem and phloem becomes meristematic—this layer of cells represents the beginning of the vascular cambium. (Fig. 34.21) 2. Likewise, a ring of cells just beneath the epidermis becomes meristematic—this represents the beginning of the cork cambium.

3. The cork cambium and vascular cambium cells divide throughout the plant’s life, producing secondary growth each year.

4. Secondary growth occurs in roots similarly to that in shoots.

D. What do vascular cambia produce?

1. The vascular cambium is a secondary meristem located between the primary xylem and primary phloem.

2. The vascular cambium (secondary meristem) produces secondary xylem and phloem, whereas the primary meristem produces primary xylem and phloem. a. Secondary phloem functions in sugar transport and forms the outer layer of tree trunks, called bark. b. Secondary xylem functions in water transport and makes up wood.

3. The vascular cambium also produces conducting parenchyma cells in rays to transport water and nutrients laterally through large stems.

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4. Students should be able to explain why a nail hammered into a tree at eye level doesn’t move up as the tree grows.

E. What do cork cambia produce?

1. The cork cambium is a secondary meristem, located near the perimeter of the stem that replaces the epidermis and forms the bark.

2. The cork cambium produces cork cells that replace the original epidermal cells.

3. The cells in the cork layer die, forming a protective outer layer.

4. Cork cells are an important element of bark, although the term bark refers to all of the vascular cambium.

5. Gas exchange can still occur between the atmosphere and living tissues inside the stem via small spongy opening called lenticels. (Fig. 34.22) F. The structure of tree trunks 1. Trees are perennial plants that live for many years.

2. Heartwood and sapwood a. As the tree trunk grows in width, the innermost xylem layers stop conducting and are infiltrated with waste products, forming gums, resins, and other mixtures. b. Heartwood is older secondary xylem that is no longer conducting water. c. Sapwood is secondary xylem toward the exterior of a stem that is transporting water. (Fig. 34.23a) 3. Annual growth rings a. In many places, the vascular cambium of trees stops growing for a certain amount of time each year; this phase is known as dormancy. b. Growth rings form because of differences in the size and cell-wall thickness of secondary xylem cells when the vascular cambium is activated after dormancy.

(1) The first cells formed after dormancy are large and have thin cell walls because water and nutrients are abundant.

(2) As the growing season ends and water and nutrients are less abundant, the cells arising from vascular cambium are smaller and have thick cell walls.

(3) Therefore, in trees that undergo this type of seasonal growth, regions of large, thin-walled cells alternate with regions of small, compact, thick-walled cells.

(4) This alternating pattern in the secondary xylem forms growth rings that can be seen in cross sections of tree trunks. (Fig. 34.23b) (5) Growth-ring size reflects the availability of water and nutrients in a given year and can be used as an indicator of past climatic changes. (Fig. 34.23c) (6) Dendrochronology is the science of dating and studying tree growth rings.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text.

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branch callus cambium cell enlargement cell sap cell wall chloroplasts collenchyma cells complex tissues cork cambium cork cells cortex cuticle dendrochronology dormancy epidermal tissue fibers fibrous roots ground tissue growth rings guard cells heartwood herb herbaceous plants indeterminate growth internodes lateral buds lateral meristem lateral roots

leaf leaf primordial lenticels lignin meristematic cells nodes palisade mesophyll parenchyma cells perennial plants petiole phloem pith plasmodesmata pore primary cell wall primary growth proplastids rays rhizomes root cap root hairs root system rosettes sapwood sclereids sclerenchyma cells secondary cell wall secondary growth secondary phloem

secondary xylem shoot apical meristem shoot system sieve-tube elements simple tissues spines spongy mesophyll stolons stomata taproot thorns tissues totipotent tracheids transpiration trichomes tubers vacuole vascular bundles vascular cambium vascular tissue vessel elements wood xylem zone of cellular division zone of cellular elongation zone of cellular maturation

Lecture Activities Show and Tell Estimated duration of activity: 5–10 minutes as you lecture Many students are visual and tactile learners and can more easily retain information they are hearing and reading if they see and touch examples. This is a great opportunity to bring examples of plants that exhibit the various types of roots, stems, and leaves presented in this chapter. Many of the specimens will be familiar to students, but many will be new. It may seem elementary to you, but for students the novelty of having something to look at and touch while listening and taking notes will enhance their retention rate. If your discussion/lecture on this chapter will last more than one class period, have the students bring in these examples to present during the second class period.

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Applying What You Know Discussion Ideas for a Class of Any Size Here are some questions that may initiate a class discussion that requires your students to process and apply what they have learned in this chapter.

• How would the morphology of a tree trunk growing in a tropical rain forest differ from that of a tree growing in a temperate climate?

Answer: In temperate climates, seasonal changes lead to periodic reactivation of the vascular cambium and the production of secondary xylem and secondary phloem from it. In such climates, the vascular cambium is activated in the spring and continues to produce new cells throughout the summer before becoming dormant in the fall and winter. The growth rings appear due to differences in cell size of the first-formed secondary xylem in the spring and the later-formed cells in the summer. Thus, a tree growing in a temperate climate would have visible growth rings. But in a tropical rain forest, where the climate is more or less constant throughout the year, the cambium is continuously producing secondary tissues and no growth rings would be produced. In cross section, the two trees would look quite different—one with distinct growth rings and one without them.

Note: Not all tropical trees exhibit continuous cambial activity; it depends on their location.

• What pathway would water follow from the soil solution to the exterior of the leaf? Trace the water’s movement through the cells, tissues, and systems.

Answer: Have either a slide or a large diagram of a root in both transverse section and longitudinal section to demonstrate this pathway: from the root hair of the epidermis, through the parenchyma cells of the cortex, the cells of the endodermis, pericycle cells, tracheids and vessels of xylem tissue, the photosynthetic parenchyma of leaf mesophyll, the stomatal pore created by guard cells, and into the atmosphere.

• What type of wood is preferable for construction: sapwood or heartwood? Why? Answer: Heartwood is preferable because it no longer conducts water and minerals and is dry. It is located at the center of the tree and is often a darker color than the sapwood. This coloration is due to the accumulation of secondary metabolites, tannins, resins, oils, gums, and so forth in the dead xylem cells. These secondary metabolites can even give the heartwood a different odor than the sapwood. Heartwood is also denser and stronger than sapwood and does not undergo the splitting that occurs in sapwood as it dries.

• How are growth rings used to predict past climatic conditions?

Answer: The width of growth rings is determined by various environmental factors, including temperature, rainfall, light, and duration of the growing season. Thus, when environmental conditions are favorable, the growth rings will be wider than during times of drought, extreme temperatures, excessive wind, and heavy predation. Examining transverse sections of the trunks of fossil trees, as well as those of living and dead trees, enables the investigator to record past climatic changes.

• How does the presence of vascular tissue influence plant size?

Answer: The presence of efficient water- and sugar-conducting tissues has several outcomes. First, plants that can transport water over long distances in specialized cells can become larger because not all cells of the organism need to be in contact with water. Second,

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the presence of lignin in the walls of xylem cells provides support and rigidity that enable plants to grow vertically. The ability to transport sugars specializes some plant parts for photosynthesis and other parts for storage of photosynthate. As a result, there can be enlarged shoot systems with expanded leaves and/or photosynthetic stems adapted for light gathering as well as extensive root systems in which photosynthetic products are stored.

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Water and Sugar Transport in Plants

Learning Objectives: Students should be able to…

• Explain how differences in water potential drives water movement in plants.

• Discuss the two components of water potential in plants.

• Explain how water potential, surface tension, cohesion, and adhesion contribute to the process of transpiration.

• Explain how active transport, osmosis, and differences in pressure contribute to the movement of sugars in phloem tissue.

Lecture Outline I. Water Potential and Water Movement A. What is water potential?

1. Transpiration occurs when two conditions are met: a. Stomata are open, usually during the day when photosynthesis is occurring. b. The air in the environment is drier than the air inside the leaves.

2. Plants replace water lost by transpiration with water that is absorbed by the roots. a. This process is passive; it does not require ATP. 3. Water moves up a plant because of differences in the potential energy of water. a. Water potential is the potential energy that water has in a particular situation compared to the potential energy of pure water at atmospheric pressure at the same temperature. b. Water always flows from areas of high water potential to areas of low water potential. c. There is a water-potential gradient between the roots and the leaves. d. Water potential is measured in units called megapascals (MPa).

B. What factors affect water potential?

1. A solution is a homogeneous mixture containing several substances. Dissolved substances are called solutes.

2. The role of solute potential a. Solute potential (ψS) is the tendency of water to move by osmosis, defined by its solute concentration relative to pure water. (Fig. 35.1a)

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b. In a cell that has a solute concentration equal to the solute concentration of the surrounding solution, the solute potentials are the same (isotonic) and no net movement of water occurs. c. For a cell in a solution that has a solute concentration less than that of the cell, the solute potentials are different and water moves into the cell.

(1) The addition of solutes lowers the solute potential.

(2) This cell is hypertonic relative to the solution in the beaker.

(3) Water moves from high solute potential to low solute potential. d. Osmosis is the process by which water moves across a semipermeable cell membrane. e. Solute potential is also known as osmotic potential.

3. The role of pressure potential a. Pressure potential (ψP) is the tendency of water to move in response to any kind of physical pressure. (Fig. 35.1b) b. In a cell, pressure potential consists of turgor pressure and wall pressure.

C. Working with water potentials 1. Water potential (ψ) is the sum of pressure potential (ψP) and solute potential (ψS). (ψ = ψP + ψS).

2. Water potential is the tendency of water to move from one location to another depending on the solute and pressure potentials at the other location.

3. Water movement in the absence of pressure (Fig. 35.2a) a. Water potentials of cells are usually negative because solute potentials of cells are always negative. b. The solute potential of pure water is 0 MPa because it contains no dissolved solutes. c. The cell solute potential is measured relative to the solute potential of pure water; thus, it is negative. d. Dissolved solutes in cells lower the water potential and reduce the probability that water will move out of the cell by osmosis.

4. Water movement in the presence of a solute potential and pressure potential (Fig 35.2b) a. The pressure potential from turgor pressure is positive inside cells; it thus increases the water potential and increases the probability that water will move out of the cell. b. Water moves from regions of high water potential to regions of low water potential. c. Cells exposed to solutions with extremely low water potential lose water and undergo plasmolysis. d. In most situations, the water potential of plant cells is lower than that of the surrounding solution, water moves into cells, and cells become turgid.

5. QUANTITATIVE: Students should be able to calculate the water potential of a cell

with ψP = 0 MPa andψS = -2 MPa. Students should indicate whether there will be an overall net movement of water into or out of the cell if that cell is placed into a solution where ψ S = -3 MPa.

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D. Water potentials in soils, plants, and the atmosphere 1. Plant tissues, root and shoot systems, and entire plants have measurable water potentials.

2. Soil and air both have water potentials.

3. Water potentials in soils a. Soil often has high water potential relative to plants. b. However, when soils are salty, (such as near coastlines), water potentials can be much lower than plants because of high solute concentrations. c. When soils are dry, water adheres tightly to soil particles, which lowers the water potential.

4. How are some plants specialized for salty or dry habitats? a. Water potential differences among soil, plants, and air establish a water-potential gradient that dictates the direction of water flow in a system. b. Plants gain water from soil and lose it to the atmosphere. c. Loss of water from plant cells results in cell shrinkage, plasmolysis, dehydration, and potential cell death.

(1) This condition can occur in irrigated fields when the water potential of soil is so low that root cells lose water to the soil rather than gaining it.

(2) Certain species of plants have adapted to live in dry or salty soil conditions where the soil’s water potential is low. (Fig. 35.3) d. Loss of water from plant tissues leads to wilting and possible plant death if not corrected. (Fig. 35.4) 5. Water potential in air a. Air often has very low water potential relative to plants. b. The pressure exerted by water vapor in the air can be low or high, depending on the conditions. c. When air is dry, there are fewer water molecules present, and the pressure they exert is low. d. When air is warm, water molecules move farther apart and exert lower pressure. e. To move up a plant, water moves down the water-potential gradient that exists between soil, its tissues, and the atmosphere. (Fig. 35.5)

II. How Does Water Move from Roots to Shoots?

A. Movement of water and solutes into the root (Fig. 35.6) 1. The epidermis is the outermost layer of cells. It protects the root and produces root hairs that increase the absorptive area of the root.

2. The cortex consists of ground tissue (parenchyma cells) that stores carbohydrates.

3. The endodermis is a ring of cells that separates the cortex from the vascular tissue.

4. The pericycle is a population of potentially meristematic cells that can produce lateral roots.

5. Vascular tissues are in the center of the root.

6. Three routes through root cortex to xylem (Fig. 35.7) a. Water moving through the symplast moves from one cell to another through the plasmodesmata connecting them.

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b. The transmembrane route is based on flow through aquaporin channels. c. Water moving through the apoplast moves between cells through the porous cell walls. Once the water reaches the endodermis, it can no longer travel through the apoplast.

7. The role of the Casparian strip a. Endodermal cells are surrounded by the waxy Casparian strip, composed of a water-repelling substance called suberin. b. The Casparian strip acts like a gasket around endodermal cells, preventing water movement through the apoplast. c. This strip forces the water to pass through the cytoplasm of endodermal cells (which act as filters) before it reaches the vascular tissue.

(1) These cells allow needed ions such as potassium to move in, but they filter out dangerous ions such as sodium.

(2) These cells prevent the movement of water out of the vascular tissue of the root, back into the soil during dry conditions. d. Students should be able to predict what would happen if a plant had a mutation that prevented synthesis of suberin and formation of the Casparian strip.

B. Water movement via root pressure 1. The endodermis is also responsible for root pressure.

2. During the night, the root accumulates ions taken up by the epidermal cells.

3. These ions are transported to the xylem, decreasing its water potential.

4. Water flows into the xylem, creating positive pressure and pushing the water up the xylem toward the leaves.

5. Since transpiration is low at night, the water being pushed into the leaves escapes through small pressure valves, resulting in guttation. (Fig. 35.8) C. Water movement via capillary action 1. When a thin glass tube is placed in a pan of water, the water creeps up the tube.

(Fig. 35.9) 2. This happens because of adhesion, cohesion, and surface tension. a. Adhesion is the attraction of unlike molecules; water forms hydrogen bonds with other polar molecules. b. Cohesion is the attraction among like molecules; water molecules bind to each other via hydrogen bonds. c. Surface tension is a pull that exists on water molecules at an air–water interface, forming a meniscus.

(1) The polar nature of water molecules leads to the formation of hydrogen bonds between water molecules.

(2) Water molecules at the air–water interface form hydrogen bonds in only one direction, pulling the surface of the water downward.

3. Only very small plants can depend on water transport via capillary action alone.

4. Larger plants depend on capillarity combined with the pulling force of transpiration to move water over long distances.

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D. The cohesion-tension theory (Fig. 35.10) 1. The role of surface tension in water transport a. As water evaporates from the leaf via transpiration, a steep water-potential gradient leads to deepening of menisci, and the inward pull (surface tension) on the remaining water molecules becomes stronger. b. This generates a pulling force on water molecules that form a continuum of cohesively bonded molecules throughout the entire plant (through the xylem, the roots, and ultimately the soil). c. This is possible because xylem forma a continuous network from roots to leaves, and all of the water molecules are bonded to each other through hydrogen bonds.

2. Creating a water-potential gradient a. In xylem, water movement is driven entirely by changes in pressure potential. b. The water in a column of xylem cells moves by bulk flow—a mass movement of molecules along a pressure gradient. c. Menisci form at the air−water interfaces of cell walls. d. If water molecules leave these menisci rapidly due to a steep water-potential gradient with the atmosphere, then fewer water molecules are available at the surface than before. e. The inward pull on the remaining water molecules at the surface becomes stronger, which increases the curve of each meniscus. f. As the menisci deepen, the surface tension increases dramatically—meaning the pull on water molecules below the surface increases.

3. The importance of secondary cell walls. a. Secondary cell walls are thicker and reinforced with tough lignin molecules that allow the vascular tissue to withstand extremely negative pressures without collapsing.

4. What evidence do biologists have for the cohesion-tension theory? a. When an actively transpiring leaf is cut, the xylem sap withdraws up into the petiole. (Fig. 35.11) b. This occurs because of the transpirational pull at the air–water interface within the spaces between spongy mesophyll cells. c. Wei, Tyree, and Stuedle designed an experiment to measure pressure changes in xylem vessels instantly and directly. (Fig. 35.12) (1) They studied the effect of two variables on xylem pressure: increased root pressure applied with a root bomb, and altered light levels at three intensities.

(2) A probe inserted directly into a xylem vessel measured pressure changes.

(3) Based on the cohesion-tension theory, increased root pressure should decrease xylem sap tension, and light should increase xylem sap tension resulting from increased transpiration.

(4) Results confirmed the theory’s predictions. (a) As pressure is added to the root system, the xylem sap tension decreases. (b) As light levels and transpiration increase, tension on the xylem sap increases.

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III. Water Absorption and Water Loss A. Expenditure of energy by plants is not required.

1. The sun provides the energy required to break the hydrogen bonds of water, causing transpiration.

2. Transpiration creates menisci at the air–water interface in leaves.

3. Hydrogen bonding of water molecules at menisci creates tension that pulls water from roots to shoots.

4. Water moves along a water-potential gradient by bulk flow.

5. Under dry conditions, stomata close to conserve water, but then the amount of carbon dioxide available for photosynthesis decreases; this is the photosynthesis-transpiration compromise.

B. Limiting water loss 1. Plants in dry environments need to cope with soils that have very low water potentials and must conserve as much water as possible.

2. Morphological traits reduce water loss in plants native to dry habitats. (Fig. 35.13) a. Thick cuticle on leaf surfaces b. Stomata on the lower epidermis c. Sunken stomata with hairs (trichomes) C. Obtaining carbon dioxide under water stress 1. CAM plants have stomata open at night to collect carbon dioxide, and they store it for use during the day. This allows them to keep stomata closed during the day.

2. C4 plants efficiently fix CO2 under hot, dry conditions, limiting the amount of time their stomata need to be open.

VI. Translocation of Sugars A. Tracing connections between sources and sinks 1. A source is a tissue where sugar enters the phloem.

2. A sink is a tissue where sugar exits the phloem.

3. Sinks and sources vary depending on the time of year. (Fig. 35.14) a. During the growing season, leaves and stems are sources; apical and lateral meristems, seeds, fruits, and storage cells in roots are sinks. b. Early in the growing season, root storage cells are sources, and developing leaves are sinks.

4. The locations of sources and sinks were demonstrated by tracking radioactively labeled sugars in sugar beets. a. Results: Sugars moved from a mature leaf, a source, to growing leaves, the sink. b. Sugars moved from leaves, the source, to roots, the sink.

5. Sources and sinks are physically related. a. Sugars from a source move to a sink on the same side of the plant. (Fig. 35.15a) b. Sources on the upper part of the stem send sugars to sinks at apical meristems; sources on the lower part send sugars to sinks in roots. (Fig. 35.15b) c. Phloem in the leaves on one side of the plant connects with phloem in the stem and roots on same side.

6. Sugars are transported rapidly.

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B. The anatomy of phloem 1. Two cell types make up phloem tissue: a. Sieve-tube elements (1) These elements are alive at maturity but usually lack nuclei and many organelles.

(2) They connect to one another end to end by sieve-like open pores called sieve plates. (Fig. 35.16) (3) They conduct phloem sap. b. Companion cells (1) Companion cells are alive at maturity and contain nuclei and numerous organelles.

(2) They function in support.

2. Phloem transports sugars. a. When bark is removed from a tree, water transport continues and sugar transport stops. b. Phloem tissue contains a high concentration of sucrose.

C. The pressure-flow hypothesis (Fig. 35.17) 1. This hypothesis for sugar movement was proposed by Ernst Münch in 1926. a. Sugars are transported from source to sink along a turgor pressure gradient in the phloem. b. The force responsible for movement is generated by turgor pressure differences at source and sink tissues.

2. Creating high pressure near sources and low pressure near sinks a. Sucrose moves from source cells to companion cells and then to sieve-tube elements. b. Phloem sap at the source has a high concentration of sucrose and low water potential. c. Adjacent xylem cells have high water potential, and water moves from the xylem into sieve-tube elements along a water-potential gradient. d. The turgor pressure at the source increases. e. Sugars move by bulk flow along the pressure gradient. f. Cells in the sink remove sucrose from the phloem sap. g. The water potential in sieve-tube elements increases as sucrose is removed until it is higher than the water potential in adjacent xylem cells. h. Water flows from sieve-tube elements into xylem along the water-potential gradient. i. The turgor pressure of phloem at sink drops. j. The outcome is a continuous loop of water flow driven by water-potential gradients between xylem and phloem and unidirectional movement of sucrose molecules driven by the pressure-potential gradient in phloem.

3. Tests of the pressure-flow hypothesis (Fig. 35.18) a. Aphids are used as a means to collect phloem for analysis. b. If aphids are severed from their stylets, sap continues to flow. c. Phloem sap is indeed under pressure, which forces it to enter the aphid. .

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D. Phloem loading 1. How are sucrose and other solutes transported across membranes? a. Sucrose is transported into phloem cells against a concentration gradient. (See Figure 35.19 for a review of membrane transport mechanisms.) b. Active transport, secondary active transport, and passive transport are all involved in moving sugars around plants.

2. How are sugars concentrated in sieve-tube elements at sources? a. Observation of pH differences between the interior and exterior of phloem cells supported the existence of the proton pump (H+-ATPase) and cotransporter.

(1) The proton pump (H+-ATPase) that transports H+ to the cell exterior creates an electrochemical gradient between the inside and outside of the cell. (Fig. 35.20) (2) The cotransporter protein uses the proton motive force as protons flow back into the cell to transport sucrose. b. Michael Sussman’s lab used the plant model Arabidopsis thaliana to isolate a mutant lacking a gene called SUC2. (BioSkills 11) (1) The SUC2 gene is known to be expressed in vascular tissue and to encode a proton-sucrose symporter.

(2) Mutant plants grew much more slowly than wild-type plants, indicating that something was wrong with the mutant. (Fig. 35.21a) (3) Chloroplasts from mutant plants accumulated huge amounts of starch, an indication that they could not efficiently transport sugar to the rest of the plant.

(Fig. 35.21b) (4) Mutant leaves were then supplied with radioactive sucrose to show they indeed retained more sucrose compared to wild-type leaves. (Fig. 35.21c) c. There is now considerable genetic and biochemical evidence to support the following model for phloem loading: (1) Proton pumps in membranes of companion cells create a strong electrochemical gradient that favors a flow of protons into companion cells.

(2) A symporter in the membranes of companion cells uses the proton gradient to bring sucrose into companion cells from sucrose cells.

(3) Once inside companion cells, sucrose moves into sieve-tube elements via plasmodesmata.

E. Phloem unloading 1. Energy is required at some point in the process.

2. The mechanism varies with different sinks in the same plant as well as between different plant species.

3. Sugar beets have two different mechanisms for unloading at different sinks. a. Growing leaves: Sugar moves passively into a sink, where energy-requiring synthesis of organic molecules occurs. (Fig. 35.22a) b. Roots: Sucrose is actively transported across the vacuolar membrane of root cells, where sucrose is stored in vacuoles against its concentration gradient. c. The membrane surrounding this organelle is called the tonoplast.

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d. It contains a protein that hydrolyzes ATP and uses the energy released to transport sucrose into the vacuole, against its concentration gradient. (Fig. 35.22b) e. The movement of sucrose into the vacuole keeps the water potential of the phloem sap near the sink low.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. active transport

isotonic

solute

adhesion apoplast bulk flow bundle-sheath cells C4 photosynthesis CAM photosynthesis capillarity Casparian strip cohesion cohesion-tension theory companion cells compromise cortex cotransporter dendograph endodermis epidermis facilitated diffusion guttation hypotonic

meniscus (plural: menisci) negative result pascal/megapascal passive transport pericycle phloem loading phloem unloading photosynthesis-transpiration plasmolysis pressure bomb pressure potential pressure-flow hypothesis proton pump root bomb root hairs root pressure rubisco sieve plates sieve-tube elements sink

solute potential/osmotic potential solution source suberin surface tension symplast symporter tonoplast translocation transpiration turgid turgor pressure vascular tissues wall pressure water potential water-potential gradient wilt xylem pressure probe xylem sap

Lecture Activities A One-Minute Paper Estimated duration of activity: Less than 5 minutes One-minute papers can be useful exercises to get students to think about what they have just heard and learned in lecture. Use these questions to get students to write a short essay about what they have learned in this chapter. Most of these questions do not have a single correct answer and involve a “big picture” view of the material in the chapter.

1. Describe an analogy for the tension-cohesion model of water transport through the xylem.

2. Which is more important to a plant: the sink or the source? Why? 3. Which root cell population is more important: the epidermal cells or the endodermal cells?

Defend your opinion. .

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In-class Challenge Questions Discussion Ideas for All Class Sizes Here are three suggestions for in-class challenge questions.

• On some mornings, water droplets can be seen on the tips of leaves of various herbaceous plants, including grasses and strawberries. This exudation of water, caused by the buildup of hydrostatic pressure in the xylem cells of the roots, is known as guttation. a. Under what conditions does guttation occur? b. Why is guttation most prominent in the early morning? c. Why doesn’t guttation occur later in the day? d. In this example, water movement is due to positive pressure, which forces water up the xylem. Why doesn’t this pressure explain the movement of water in trees?

Answers: a. Guttation occurs under cool conditions, high humidity, high soil water content, little wind, and a low rate of transpiration. b. At night, when little or no transpiration is occurring, ions are absorbed from the soil solution and the concentration of ions builds up in the xylem sap. The water potential in the roots is lowered, water is absorbed from the soil solution into the root, and hydrostatic pressure in the xylem forces water and dissolved ions up the xylem. c. Guttation does not occur during the day because transpiration is occurring. Water is lost very quickly from the leaves, and a positive pressure cannot develop in the xylem. d. When transpiration takes place, the root pressure never becomes high enough to push the water to the top of a tall tree. Some plants never develop positive hydrostatic root pressure.

• How do temperature, humidity, and wind influence the rate of transpiration from a plant?

Answer: Increasing temperature increases the rate of transpiration up to the temperature when stomata close due to factors other than temperature (that is, increasing CO2 due to increased respiration at higher temperatures). The rate of water loss from leaves is a function of the water-vapor pressure difference between the interior of the leaf and its surface—the greater the difference, the greater the water loss. Thus, decreasing humidity increases the rate of water loss, whereas increasing humidity decreases the rate of water loss. A dry wind blowing across the surface of the leaf will remove moisture from the leaf surface and increase the water-vapor pressure difference, leading to increased water loss.

• Why are crop plants that have been genetically engineered to synthesize mannitol able to survive in salty soils? Why is it important to develop plants that are capable of tolerating elevated salt concentrations in soil?

Answers: Plants that can synthesize mannitol are able to increase the cell’s solute concentration and maintain a solute potential lower than the solute potential of the soil solution.

The outcome of this elevated solute concentration is that water moves by osmosis from where the water potential is higher (the soil solution) to where the water potential is lower (the root cells of the plant). A plant that is unable to synthesize mannitol or some other

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solute would lose water and could not survive in the salty soil. Various agricultural practices, such as irrigating arid lands and over-pumping water in coastal areas, have led to an increase in soil salinity. As a result, the soil has become too salty for the cultivation of crop plants. It has become increasingly apparent that salt-tolerant crop plants must be bred or that crop plant varieties need to be modified in order to grow successfully in salty environments.

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CHAPTER

36

Plant Nutrition

Learning Objectives: Students should be able to…

• Identify the essential macro- and micronutrients that plants require to support growth.

• Explain how plants absorb useful materials and avoid toxic minerals.

• Describe the symbiotic relationships with bacteria and fungi that plants use to aid nutrient absorption.

• Identify examples of specialized methods some plants use to obtain nutrients.

Lecture Outline I. Nutritional Requirements of Plants A. Where does a plant get its mass?

1. Jean-Baptiste van Helmont researched this question through an experiment using willow trees. (Fig. 36.1) a. He planted a willow tree sapling in 90 kg of soil. After five years, the plant had gained more than 74 kg, but the soil had lost only 60 g. b. He erroneously concluded that the plant gained its mass from water.

2. Further experiments demonstrated that a plant gains some mass from carbon dioxide and some from the soil.

B. Which nutrients are essential?

1. Essential nutrients are those absolutely required for normal growth and development.

(Table 36.1) 2. Three criteria are used to define an essential nutrient: a. It is required for growth and reproduction. b. No other element can substitute. c. It is necessary for a specific structure or metabolic function.

3. Macronutrients are required in large quantities and are needed for the formation of cell macromolecules. a. Some are obtained from the atmosphere and water, and others are extracted from the soil. b. They include N, P, and K, the most common ingredients in fertilizers.

4. Micronutrients are required in small quantities and are usually enzyme cofactors.

C. What happens when key nutrients are in short supply?

1. Deficiency in an essential nutrient results in abnormal growth. (Fig. 36.2) 360

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2. A study was conducted on the effect of a copper deficiency on tomato plants grown hydroponically. (Fig. 36.3) a. One set of tomato seedlings was grown in hydroponic culture with all essential nutrients. The other set was grown with all nutrients except copper. b. Tomatoes that lacked copper had stunted roots and shoots, curled leaves, no flowers, and dark foliage.

3. Nutrient deficiency has severe effects on all plant tissues.

4. A small amount of the nutrient will cure the deficiency.

5. The effects of all essential nutrients have been studied.

II. Soil: A Dynamic Mixture of Living and Nonliving Components A. The formation and texture of soil 1. The breakdown (weathering) of rock yields particles of different size and composition.

(Fig. 36.4) 2. Decomposed organic matter is called humus.

3. Soil eventually becomes a mixture of inorganic and organic particles and living organisms. (Fig. 36.5) 4. The texture and chemical composition of soil depend on the parent rock. a. Texture affects root penetration, water holding ability, and oxygen availability. b. Chemical composition affects nutrient availability.

5. The soil composition determines what plant species are capable of growth.

B. The importance of soil conservation 1. Soil erosion occurs when soil is carried away from a site by wind or water.

2. Erosion can become a problem when an area has become devegetated. a. The Dust Bowl of the 1930s resulted from drought and poor farming practices.

(Fig. 36.6a) b. Topsoil in about 30% of all U.S. croplands is eroding faster than it is replenished.

(Fig. 39.6b) 3. Sustainable agriculture techniques can prevent or reduce unnecessary erosion: a. Planting rows of trees as windbreaks b. Minimizing tilling and plowing c. Planting crops in strips that follow the contour of hillsides C. What factors affect nutrient availability?

1. Elements are found as ions in soil.

2. The role of ionic charge and soil texture. a. Ions behave in one of two ways depending on their charge. (Fig. 36.7) b. Anions are negatively charged ions that dissolve in soil water via hydrogen bonds, except phosphate ions. c. They are absorbed by plants but can leach out of soil with rainwater. d. Cations are positively charged ions that bind to negatively charged organic matter and clay. e. They resist leaching but are difficult for plants to absorb because of the strong binding to clay.

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f. The presence of sand, clay, and organic matter affects ion availability and other soil properties, including water availability. (Table 36.2) 3. The role of soil pH a. Acidic soil usually has high concentrations of carbonic acid, phosphoric acid, and nitric acid. b. Basic or alkaline soils are common where limestone is common. c. Most plants thrive in soils with a relatively neutral pH. d. The presence of acids in soils can increase the rate at which elemental ions are released from the soil. e. Cation exchange occurs when protons or other cations bind to negative charges on soil particles and cause bound cations such as magnesium or calcium to be released.

(Fig. 36.8a) f. Cation exchange makes the cation nutrients available to nearby plant roots.

(Fig. 36.8b) g. A higher concentration of protons in the soil speeds the rate of cation exchange and makes cations more available to plants. (Fig. 36.8c)

III. Nutrient Uptake A. The root system is the site of nutrient uptake in most plants.

1. The zone of maturation, just above the root tip, is the site of absorption.

2. Epidermal cells in the zone of maturation have extensions called root hairs, which increase the surface area of roots for absorption. (Fig. 36.9) a. Root hairs contain many membrane proteins, which selectively facilitate ion passage into cells.

B. Mechanisms of nutrient uptake 1. Establishing a proton gradient a. Plants establish a proton gradient for nutrient uptake. b. Plants need to take up diffuse nutrients and concentrate them inside the cell. c. Uptake against a concentration gradient requires the expenditure of energy. d. Proton pumps (H+-ATPases) use the energy in ATP to accumulate protons outside of the cell. (Fig. 36.10a) (1) This leads to a concentration gradient that favors proton movement into the cell.

(2) This also leads to a voltage or charge difference across the membrane; the exterior is more positive.

(3) The electrical gradient produced by the proton pumps is enough to attract cations into the cell despite their concentration gradients.

2. Using a proton gradient to import cations a. Evidence of active uptake occurs in experiments with radioactive K+.

(1) When the K+ concentration outside the cell is lower than the concentration inside, uptake into the cell continues. (Fig. 36.10b) (2) Further research isolated the sequence for the gene that encodes for a protein that is responsible for the uptake of K+. b. Students should be able to explain what happens to cation uptake if (1) H+ATPases fail, and (2) if a molecule blocks a cation channel.

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3. Using a proton gradient to import anions. (Fig. 36.10c) a. Cotransporters are proteins that facilitate the diffusion of protons into the cell. b. These proteins use the energy released by the diffusion of protons to transport anions into the cell up their concentration gradients. c. Students should be able to explain what happens to anion uptake if 1) H+ATPases fail, and 2) if a molecule blocks an anion symporter.

4. Nutrient transfer via mycorrhizal fungi a. Mycorrhizae are fungi that live in association with plant roots and help to transfer nutrients. b. Ectomycorhizzal fungi (EMF) wrap around the exterior of the roots and radiate into the soil. (Fig. 36.11) c. Arbuscular mycorrhizal fungi (AMF) penetrate into root cells. d. The relationship between mycorrhizae and plant is mutually beneficial.

(1) Carbon dioxide fixed by plants is transferred to fungi as sugar.

(2) Nitrogen and phosphorus taken up by fungi are transferred to plant tissues.

(3) Fungi secrete digestive enzymes that break down dead plant material and make the nutrients available to both fungi and plants.

C. Mechanisms of ion exclusion 1. Passive exclusion a. The Casparian strip stops the movement of substances into the vascular tissue via the apoplastic pathway, forcing all substances to pass through endodermal cells to reach the vascular tissue.

(1) These cells have specific types of channels and transporters.

(2) This process limits the substances that are allowed into the vascular tissue of the roots. (Fig. 36.12) b. Root hairs also participate in passive exclusion.

(1) If root hairs lack the membrane protein required for a certain ion to enter the cell, the ion will not enter.

(2) Salt-tolerant strains of corn transport less salt into their roots, possibly because they have fewer sodium channels.

2. Active exclusion by metallothioneins and phytochelatins a. Many ions, such as copper, are toxic or dangerous to plants. b. One mechanism of neutralizing dangerous ions in plants involves metallothionein proteins and peptides called phytochelatins that bind to and sequester metal ions.

(1) Natural populations of Arabidopsis thaliana (BioSkills 11) vary in their ability to tolerate high concentrations of copper.

(2) Copper-tolerant A. thaliana individuals produce more metal-binding protein than copper-intolerant individuals.

3. Active exclusion by antiporters a. The tonoplast is the membrane that surrounds the vacuole. b. These transport proteins function as antiporters. (Fig. 36.13) (1) Antiporters send protons out of the vacuole and bring sodium in.

(2) This may allow plants to survive in salty soil.

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IV. Nitrogen Fixation A. Nitrogen fixation is the conversion of atmospheric nitrogen into compounds that plants can use.

1. All living organisms require nitrogen in large quantities for the production of amino acids and nucleic acids.

2. Molecular nitrogen, N2, is the most abundant component in the atmosphere; it is very stable and rarely participates in chemical reactions.

3. With few exceptions, living organisms are unable to directly utilize N2 from the atmosphere.

4. Only several species of bacteria can convert atmospheric N2 into ammonia (NH3), nitrites (NO–2), or nitrates (NO3).– B. The role of symbiotic bacteria 1. Nitrogen-fixing bacteria in the genus Rhizobium form mutualistic relationships with the legume family. a. Bacteria provide plants with ammonia. b. Plants provide protection in root nodules and sugars to support bacterial growth.

(Fig. 36.14) 2. Leghemoglobin binds oxygen. This keeps dissolved oxygen low so it does not affect the action of nitrogenase (and therefore nitrogen fixation).

C. How do nitrogen-fixing bacteria infect plant roots?

1. Colonization of roots by bacteria requires recognition. (Fig. 36.15) a. The bacterial cells contact the root-hair surface—they are attracted to root-hair surface by flavenoids. b. Rhizobia move through the root-hair plasma membrane called an infection thread. c. The infection thread extends into the root, invading the cortex. d. Portions of the infection thread bud off, forming membrane-bound clusters. e. The infected cortex cells divide, forming root nodules.

V. Nutritional Adaptations of Plants A. Parasitic plants 1. Parasitic plants live in close contact with their host and lower the fitness of those individuals, usually by obtaining water or nutrients.

2. Parasitic plants represent less than 1% of all plant species described thus far.

3. Dodder is a plant that can wrap itself around another plant and produce haustoria, structures that penetrate a host plant’s vascular tissue. (Fig. 36.16) 3. Most parasitic plants are photosynthetic but obtain nutrients and water from the host-plant root system, but some are heterotrophic.

B. Epiphytic plants 1. Epiphytic plants grow on the leaves or branches of trees and never make contact with soil. (Fig. 36.17a) 2. They absorb nutrients from rainwater that collects on plants upon which they grow.

3. Some have leaves that grow in rosettes that serve as tanks for holding water.

(Fig. 36.17b)

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C. Carnivorous plants 1. Carnivorous species make their own carbohydrates via photosynthesis but use carnivory as a way to supplement the nitrogen available in the environment.

2. In sundews, modified leaves develop hairs that exude a sticky substance and function like flypaper. (Fig. 36.18) 3. Venus flytrap is a photosynthetic plant that grows in bogs with low levels of nitrogen.

4. Through natural selection, Venus flytrap leaf has been modified to form a trap, showing phenotypic plasticity.

Chapter Vocabulary To emphasize the functional meanings of these terms, the list is organized by topic rather than by first mention in the chapter. It includes terms that may have been introduced in earlier chapters but are important to the current chapter as well. It also includes terms other than those highlighted in bold type in the chapter text. active exclusion active uptake antiporter arbuscular mycorrhizal fungi/AMF mutualism biomass cation exchange clay cotransporter ectomycorhizzal fungi electrochemical gradient elemental ions epiphytes essential nutrient H+-ATPases haustoria

humus hydroponic growth infection thread ions leaching leghemoglobin legumes limiting nutrients macronutrients membrane potential metallothioneins micronutrients mycorrhizae nitrogen fixation Nod factors nodules

parasitism passive exclusion passive uptake phytoremediation proton pumps rhizobia root hairs soil erosion sustainable agriculture symbiotic texture tonoplast voltage weathering zone of maturation

Lecture Activities Horticulturist for a Day Estimated duration of activity: 10–15 minutes Students may benefit from this practical activity, which requires them to link the symptoms of a plant nutritional deficiency and attempt to explain why that deficiency leads to that symptom.

Make index cards with the name of the nutrient on one side and the list of symptoms on the reverse side. Divide the class into small groups or pairs, and give each group a card. Ask students to use the information in Chapter 36 as well as the concepts they have learned about general plant (and eukaryotic cell) physiology to hypothesize why those symptoms occur when the plant is not receiving enough of that nutrient.

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