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Solution Manual For Campbell Biology, 9th Edition

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    Notes to Instructors Chapter 2 The Chemical Context of Life Chapter 3 Water and the Fitness of the Environment What is the focus of these activities? Living organisms function in the real world, so they are subject to all the laws of chemistry and physics. In addition, biological organisms and systems are variable. No two organisms are exactly alike, and no two systems are identical in form or function. As a result, our analysis of such systems tends to deal with statistical averages or probabilities. This means that it is difficult to understand biological systems without having a good basic understanding of chemistry, physics, and math (including probability and statistics). The vast majority of introductory biology students have studied inorganic chemistry in their high school and first-year college chemistry courses. Many students compartmentalize their knowledge, however. In some cases, the compartmentalization is so extreme that the students feel uncomfortable dealing with chemical formulas and ideas outside of chemistry classes. Therefore, it is generally useful to review some of the basic ideas in chemistry and, at the same time, demonstrate how they can be applied to understanding biological systems. What are the particular activities designed to do? Activity 2.1 A Quick Review of Elements and Compounds The questions in this activity are designed to help students review and understand: • atomic/molecular number, mass number, and atomic/molecular weight and how they can be used to determine the reactivity of elements; • various types of chemical bonds and how they affect the structure and energetics of molecules; and • the difference between a mole and a molar equivalent and how a knowledge of these can be used in biological applications. Activity 3.1 A Quick Review of the Properties of Water The questions in this activity are designed to help students review and understand the properties of water and how they support life. Students are asked to review these key properties: • H 2 O molecules are cohesive; they form hydrogen bonds with each other. • H 2 O molecules are adhesive; they form hydrogen bonds with polar surfaces. • Water is a liquid at normal physiological (or body) temperatures. • Water has a high specific heat. • Water has a high heat of vaporization. • Water’s greatest density occurs at 4°C. Notes to Instructors 1

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    In addition, students review pH and how it is related to both the ionization constant of pure water and the concentration of H + ions in a solution. What misconceptions or difficulties can these activities reveal? Activity 2.1 Question 1: Many students don’t understand that nutrients for plants are inorganic and most nutrients for animals (heterotrophs) are organic. Questions 2 and 3: Most students know how to balance a chemical equation. Fewer understand the relationship between molecules of a substance and moles of that substance. Similarly, most students can recite what a mole is; however, the majority have not thought about how that knowledge can be applied. Therefore, much of this first activity is devoted to making it clear that a balanced equation indicates not only the number of molecules required but also the number of moles required. It also explains why moles can be substituted for molecules in such equations. Question 4: Some students have difficulty understanding that a solution’s concentration or molarity does not change if you aliquot or subdivide the solution into smaller volumes. To test this, ask your students: “There is 10% sugar in this solution. If I pour half of it into one beaker and the other half into another beaker, what percent sugar will I have in each beaker?” More than half of the students will automatically answer 5%. Questions 5 and 6: These questions are designed to help students understand how a knowledge of balanced equations and molar equivalents can be useful in biology. Questions 7 and 8: The answers go into a little more detail than does Campbell Biology , 9th edition. Students obviously shouldn’t be asked to know the specific electronegativity of each of the elements. However, using concrete numbers may help students understand how their electronegativity is related to the type of bonds formed between elements. Activity 3.1 Most students have no difficulty stating the properties of water and the definition of pH. On the other hand, not all of them have a good understanding of how these properties are related to biological and other phenomena. Therefore, some questions ask students to relate pH values to actual concentrations of H + ions in solution and to relate the properties of water to common experiences they have had in class or in life. 2 Notes to Instructors

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    Answers Activity 2.1 A Quick Review of Elements and Compounds 1. Table 2.1 (page 32) lists the chemical elements that occur naturally in the human body. Similar percentages of these elements are found in most living organisms. Activity 2.1 3 a. In what abiotic (nonlife) chemical forms are these elements often found in nature? These elements are most commonly found as CO 2 , N 2 , and O 2 in the atmosphere and as H 2 0, PO 4 , and S compounds on Earth. b. In what chemical form(s) do animals need to obtain these elements? With the exception of oxygen and water, animals obtain the majority of these elements in the form of organic compounds. c. In what chemical form(s) do plants need to obtain these elements? Plants can obtain C as CO 2 , N as ammonia or nitrite or nitrates, phosphorus as phosphates, sulfur as sulfides or sulfates, and so on. In other words, plants obtain these elements as inorganic compounds. 2. A chemical element cannot be broken down to other forms by chemical reactions. Each element has a specific number of protons, neutrons, and electrons. a. What is the name of the following element, and how many protons, neutrons, and electrons does it have? Na 11 23 Name Sodium Number of protons 11 Number of electrons 11 Number of neutrons 12

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    c. How would you determine how many grams are in a mole of any chemical element or compound? A mole of any chemical element or compound is equal to the mass number in grams of that mole or compound. For example, the mass number of Na is 23; therefore, a mole of Na has a mass of 23 g. The mass number of water is 18; therefore, a mole of water has a mass of 18 g. 4. One atom of Na can combine with one atom of Cl (chlorine) to produce one molecule of NaCl (table salt). 4 Activity 2.1 b. What information do you need to calculate or determine the following? The atomic number of an element The atomic number is equal to the number of protons (or electrons). The mass number of an element The mass number is equal to the number of protons plus the number of neutrons. The weight in daltons of one atom of an element You can estimate the weight in daltons as 1 dalton per proton or neutron.Therefore, the weight in daltons of an element is approximately equal to the number of protons plus the number of neutrons. c. What are the atomic number, mass number, and weight in daltons of the element shown in part a? 3. One mole of an element or compound contains 6.02  10 23 atoms or molecules of the element or compound. One mole of an element or compound has a mass equal to its mass number (or molecular weight) in grams. For example, 1 mole of hydrogen gas (H 2 ) contains 6.02  10 23 molecules and weighs 2 g. Atomic number Mass number Weight in daltons 11 23 23 a. What is the weight of 1 mole of pure sodium (Na)? 23 g b. How many molecules of Na are in 1 mole of Na? 6.02  10 23 a. If Cl has 17 electrons, 17 protons, and 18 neutrons, what is its mass number? 35 b. What is the mass number of NaCl? 23 + 35 = 58 c. How many grams of NaCl equal a mole of NaCl? 58 g

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    d. If you wanted to combine equal numbers of Na and Cl atoms in a flask, how much Cl would you have to add if you added 23 g of Na? (Include an explanation of the reasoning behind your answer.) 23 g of Na is equal to 1 mole of Na. A mole contains 6.02  10 23 molecules of the substance. To add an equal number of molecules of Cl, you need to add 1 mole of Cl, or 35 g. e. To make a one-molar (1 M ) solution of NaCl, you need to add 1 mol of NaCl to distilled water to make a final volume of 1 L (1,000 ml). A 1 M solution is said to have a molarity of 1. If you added 2 moles of NaCl to 1 L of distilled water, you would make a 2 M solution and its molarity would equal 2. You make up a 1 M solution of NaCl. Activity 2.1 5 5. The summary formula for photosynthesis is 6 CO 2 + 6 H 2 O → C 6 H 12 O 6 + 6 O 2 How many molecules of NaCl are in the 1 M NaCl solution? If you used 1 L of water to make the 1 M solution, you would have 6.02  10 23 molecules in the liter. How many molecules of NaCl are there per ml of the solution? To calculate the number of molecules per ml, divide 6.02  10 23 by 1,000 = 6.02  10 20 molecules/ml. f. Next, you divide this 1 M solution of NaCl into four separate flasks, putting 250 mL into each flask. How many grams of NaCl are in each flask? 58/4 = 14.5 g How many molecules of NaCl are in each flask? 6.02  10 23 divided by 4 = 1.51  10 23 How many molecules of NaCl are there per ml of distilled water? 6.02  10 20 What is the molarity of NaCl in each of the four flasks? 1 M a. How many molecules of carbon dioxide and water would a plant have to use to produce three molecules of glucose (C 6 H 12 O 6 )? For each molecule of glucose produced, 6 molecules of carbon dioxide and 6 molecules of water are consumed. Therefore, the plant would need to use 18 molecules of each. b. How many moles of carbon dioxide and water would a plant have to use to produce 2 mole of glucose? Because a mole of anything contains the same number of molecules, the plant would need to use 6 times as many moles of carbon dioxide and water, or 12 moles of each.

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    c. Refer to the summary formula for photosynthesis. If you know the number of molecules or moles of any of the reactants used (or products produced), how would you calculate the number of molecules or moles of all of the other reactants needed and products produced? If the formula is balanced and if it is a true representation of the overall reactions that occur, then the numbers in front of each reactant and product indicate the molecular or molar equivalents required for the reactions. Note: To represent the actual reactants required and products produced, the overall formula for photosynthesis is more correctly stated as: 6 CO 2 + 12 H 2 O → C 6 H 12 O 6 + 6 O 2 + 6 H 2 O In most texts, however, this is reduced to 6 CO 2 + 6 H 2 O → C 6 H 12 O 6 + 6 O 2 6. A biologist places a plant in a closed chamber. A sensor in the chamber maintains the carbon dioxide level at the normal atmospheric concentration of 0.03%. Another sensor allows the biologist to measure the amount of oxygen produced by the plant over time. If the plant produces 0.001 mole of oxygen in an hour, how much carbon dioxide had to be added to the chamber during that hour to maintain the atmospheric concentration of 0.03%? For every mole of oxygen produced, 1 mole of carbon dioxide had to be consumed. Therefore, 0.001 mole of carbon dioxide had to be added to maintain a constant level of CO 2 in the chamber. 7. O 2 and NH 3 are both small covalent molecules found in cells. NH 3 is extremely soluble in the aqueous environment of the cell, while O 2 is relatively insoluble. What is the basis for this difference in solubility between the two molecules? In reaching your answer, draw the structures of the molecules as valence shell diagrams (as in Figure 2.12, page 38). Given these diagrams, consider the types of interactions each molecule could have with water. 6 Activity 2.1 H H H N O O Ammonia is a polar molecule much like water. The N in it is relatively negative, and the H’s are relatively positive. Polar substances tend to be more soluble in water. O 2 , on the other hand, is not polar. Ammonia (NH 3 ) Oxygen (O 2 )

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    8. Refer to pages 38–41 of Campbell Biology , 9th edition, which describe these types of chemical bonds: nonpolar and polar covalent bonds, ionic bonds, hydrogen bonds, and van der Waals interactions. The molecule diagrammed here can also be represented by the formula CH 3 COOH. Activity 2.1 7 H H C C O OH H Explain how you could determine which of the bonds between elements in this molecule are polar or nonpolar covalent bonds, ionic bonds, hydrogen bonds, and van der Waals interactions. The best way to determine the bond types is to determine each atom’s electronegativity, or its attraction for electrons. As a general rule, the more filled the outer electron shell of an atom is, the higher is its electronegativity. In addition, the fewer electron shells, the greater the electron negativity. As a result, an atom’s attraction for electrons increases as you go from left to right in the periodic table. Electronegativity values tend to decrease as you go from top to bottom of the periodic table. To determine whether bonds are ionic, polar covalent, or nonpolar covalent, you need to determine the difference in electronegativity between the atoms that make up a molecule. If the difference in electronegativity is small, the bond is likely to be nonpolar covalent. If the difference is very large, the bond is likely to be ionic. Intermediate differences produce polar covalent bonds. The following table lists specific electronegativity values for selected elements. Using specific electronegativity values, you can determine the type of bond: If the difference in electronegativity between two atoms in a compound is less than 0.5, the bond is nonpolar covalent. If the difference is between 0.5 and 1.6, the bond is polar covalent. If the difference is greater than 1.6, the bond is ionic. H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O = 3.5 F = 4.0 Na = 0.9 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl = 3.0

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    Activity 3.1 A Quick Review of the Properties of Water 1. Compounds that have the capacity to form hydrogen bonds with water are said to be hydrophilic (water loving). Those without this capacity are hydrophobic (water fearing). 8 Activity 3.1 Is the molecule on the left hydrophilic or hydrophobic? Explain your answer. This molecule is acetic acid. The  COOH group is polar, which makes this molecule hydrophilic. 2. In addition to being polar, water molecules can dissociate into hydronium ions (H 3 O + , often described simply as H + ) and hydroxide ions (OH  ). The concentration of each of these ions in pure water is 10  7 . Another way to say this is that the concentration of hydronium ions, or H + ions, is one out of every 10 million molecules. Similarly, the concentration of OH  ions is one in 10 million molecules. a. The H + ion concentration of a solution can be represented as its pH value. The pH of a solution is defined as the negative log 10 of the hydrogen ion concentration.What is the pH of pure water? The hydrogen ion concentration of pure water is 10  7 . The log 10 of 10  7 is  7. The negative log 10 of 10  7 is therefore +7. b. Refer to the diagram of the molecule of acetic acid in question 1. The COOH group can ionize to release a H + ion into solution. If you add acetic acid to water and raise the concentration of H + ions to 10  4 , what is the pH of this solution? The pH of a solution with a H + ion concentration of 10  4 is 4. 3. Life as we know it could not exist without water. All the chemical reactions of life occur in aqueous solution. Water molecules are polar and are capable of forming hydrogen bonds with other polar or charged molecules. As a result, water has the following properties: A. H 2 O molecules are cohesive; they form hydrogen bonds with each other. B. H 2 O molecules are adhesive; they form hydrogen bonds with polar surfaces. C. Water is a liquid at normal physiological (or body) temperatures. D. Water has a high specific heat. E. Water has a high heat of vaporization. F. Water’s greatest density occurs at 4°C. H H C C O OH H

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    Explain how these properties of water are related to the phenomena described in parts a–h below. More than one property may be used to explain a given phenomenon. a. During the winter, air temperatures in the northern United States can remain below 0°C for months; however, the fish and other animals living in the lakes survive. Water’s greatest density occurs at 4°C. In a lake, the 4°C water sinks below the water that is colder (or warmer). As a result, 0°C water is less dense than 4°C water. In addition, as it freezes, water takes on a crystalline structure and becomes ice. Ice has a density of about 0.92 g/cm 3 , pure water at 0°C has a density of about 0.99 g/cm 3 , and pure water at 4°C has a density of 1.0 g/cm 3 . The ice on top of a lake acts like insulation and, as a result, most deep lakes do not freeze to the bottom. b. Many substances—for example, salt (NaCl) and sucrose—dissolve quickly in water. Water is very polar. The attraction of the polar water molecules for the Na + and Cl  ions of NaCl is strong enough to allow them to dissociate and interact with water molecules (dissolve). c. When you pour water into a 25-mL graduated cylinder, a meniscus forms at the top of the water column. Water is attracted to the polar molecules that make up the glass (or plastic) cylinder. At the same time, they are attracted to each other. As a result, some of the water molecules associate with the polar molecules of the cylinder and are apparently “pulled up” the inside edge of the cylinder. d. Sweating and the evaporation of sweat from the body surface help reduce a human’s body temperature. Water has a high specific heat. The specific heat of water is 1 cal/g/°C. In other words, it takes 1 calorie of heat to change the temperature of 1 g of water 1°C. In addition, water has a high latent heat of vaporization (540 cal/g at 100°C). This can be thought of as the additional heat required to break apart polar water molecules so that they can move from the liquid to the gaseous state. As a result, evaporation (change of water from liquid to gaseous state) carries with it large amounts of heat. e. A bottle contains a liquid mixture of equal parts water and mineral oil. You shake the bottle vigorously and then set it on the table. Although the law of entropy favors maximum randomness, this mixture separates into layers of oil over water. The oil molecules are nonpolar and hydrophobic. The water molecules are polar and cohesive. As a result, the water molecules tend to interact strongly with each other and exclude the oil molecules. The oil layers on top of the water because it is less dense than water. Activity 3.1 9

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    f. Water drops that fall on a surface tend to form rounded drops or beads. Water molecules are cohesive and form hydrogen bonds with each other. As a result, a drop of water tends to bead up or become rounded. g. Water drops that fall on your car tend to bead or round up more after you polish (or wax) the car than before you polished it. The wax (or polish) is hydrophobic and therefore less polar than the surface was likely to be before you polished it. Because the adhesion between the surface and the water molecules is lower, the cohesion of the water molecules for each other appears even more dramatic. h. If you touch the edge of a paper towel to a drop of colored water, the water will move up into (or be absorbed by) the towel. The polar water molecules adhere to the cellulose in the paper towel and cohere to each other. As a result, they are drawn up into the towel. The same mechanism accounts for the movement of water molecules up capillary tubes. 10 Activity 3.1

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    Notes to Instructors Chapter 4 Carbon and the Molecular Diversity of Life Chapter 5 The Structure and Function of Macromolecules What is the focus of these activities? The activities associated with Chapters 2 and 3 provided students with a review of some of the basics of inorganic chemistry. The activities associated with Chapters 4 and 5 deal with organic chemistry and biochemistry. Although most introductory biology students have had some exposure to inorganic chemistry, fewer have had courses in organic chemistry or biochemistry. Therefore, these activities are designed with these goals in mind: • Help students identify the major differences among the four main types of biological macromolecules: carbohydrates, lipids, proteins, and nucleic acids. • Recognize what functional groups are and how they can modify the general properties and functions of organic compounds. • Use their understanding from the goals above to predict how various structural modifications can affect the behavior of macromolecules. What are the particular activities designed to do? Activity 4.1/5.1 How can you identify organic macromolecules? This activity is designed to help students easily recognize carbohydrates, lipids, proteins, and nucleic acids when viewed as chemical structures. Students develop some simple rules to make it easier for them to recognize differences among the general chemical structures of carbohydrates, lipids, proteins, and nucleic acids. This type of activity is especially important for the many introductory biology students who have not yet had organic chemistry or biochemistry. Activity 4.2/5.2 What predictions can you make about the behavior of organic macromolecules if you know their structure? In this activity, students examine the general properties of organic macromolecules. In particular, they examine the properties of functional groups and how these can modify the behavior of macromolecules. This includes how modifying functional groups can affect the water solubility and the chemical reactivity of molecules. At the end of this activity, students are asked to predict how various macromolecules could react if placed in specific environments or if modified in specific ways. Answering these questions requires students to integrate and use understanding they have gained from Chapters 2–5. Notes to Instructors 11

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    What misconceptions or difficulties can these activities reveal? Activity 4.1/5.1 Question 1, Part A: This question asks for the C:H:O ratio of the various macromolecules. For carbohydrates, it is approximately 1:2:1. For many lipids, it is approximately 1:2:very few. And for proteins and nucleic acids, there is no reliable ratio of C:H:O. Many students become upset that the answer for proteins and nucleic acids is “no reliable ratio of C:H:O.” Remind them that they would not have known the C:H:O ratio was not a good predictor for these molecules unless they investigated it. Answers like this just mean we need to look for other ways of identifying these molecules. Some of these other ways include looking for specific functional groups, for example amino and carboxyl groups in amino acids and ribose versus deoxyribose and specific nucleotides to identify RNA versus DNA. Part B: After developing their own rules for identifying macromolecules, most students won’t have difficulty identifying these structures as carbohydrate, protein, lipid, or nucleic acid. It is best to present this activity in that light; that is, let students know that the purpose of the activity is to help them prove it is possible to categorize complex macromolecules using only a few simple rules. Activity 4.2/5.2 Question 1: The table asks students to look at different possible characteristics of R groups. 12 Notes to Instructors This question helps students understand, for example, that something that is polar is also hydrophilic; that is, these designations are not mutually exclusive. In addition, by learning the characteristics of key functional groups, students will have a better understanding of how modifications in macromolecular structure can lead to modifications in function. Question 4: The first three experiments ask students to examine how phospholipids (and glucose) will distribute themselves in different kinds of aqueous environments. Many students have memorized that phospholipids can form bilayers or micelles in aqueous environments, and they have the misconception that phospholipids will organize into micelles or bilayers under any circumstance. R group Basic, acidic, or neutral Polar or nonpolar Hydrophilic or hydrophobic

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    Answers Activity 4.1/5.1 How can you identify organic macromolecules? Refer to the figure (Some Simple Chemistry) on the next page when doing this activity. Part A. Answer the questions. Then use your answers to develop simple rules for identifying carbohydrates, lipids, proteins, and nucleic acids. 1. What is the approximate C:H:O ratio in each of the following types of macromolecules? Activity 4.1/5.1 13 Carbohydrates 1:2:1 Lipids 1:2:very few Proteins There is no reliable C:H:O ratio for proteins. Nucleic acids There is no reliable C:H:O ratio for nucleic acids. 2. Which of the compounds listed in question 1 can often be composed of C, H, and O alone? Carbohydrates and lipids can often be composed of C, H, and O alone. 3. Which of the compounds can be identified by looking at the C:H:O ratios alone? Only carbohydrates and some lipids can be identified using C:H:O ratios alone. 4. What other elements are commonly associated with each of these four types of macromolecules? Carbohydrates Lipids Proteins Nucleic acids Always contain P No No (except for phospholipids) No Yes Generally contain no P* Yes Yes (except for phospholipids) Yes No Always contain N No No Yes Yes Generally contain no N Yes Yes No No Frequently contain S No No Yes No Generally contain no S Yes Yes No Yes * Note: It is possible to find some exceptions in each of these categories where “Yes” is the answer to “Generally contain no ___.” For example, in reaction sequences many compounds undergo phosphorylation. However, if the natural state of the compound does not contain P (for example) the answer to “Generally contain no P” would be yes.

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    14 Activity 4.1/5.1 Lipids: H C C (CH 2 ) n CH 2 OH H OH OH C O C C C C O OH H HO OH H H H HO H HO H O H C O H CH 3 CH 3 CH 3 C (CH 2 ) n CH 3 Fats, oils, waxes, cholesterol Proteins: Enzymes, structural proteins Sugars, starches, glycogen, cellulose Carbohydrates: Compound Some Simple Chemistry Basic components Reaction C C (CH 2 ) n O OH H HO O C O H C (CH 2 ) n CH 3 Product H Glycerol + 3 fatty acids Triglyceride or fat Disaccharide 6C hexose dehydration reaction amino group carboxyl group (etc.) C C (CH 2 ) n O OH H HO O R H C COOH H 2 N R H H H OH O C C H 2 N O R H OH C C N H C O H C (CH 2 ) n + 3 H 2 O CH 3 dehydration reaction dehydration reaction + 1 H 2 O Dipeptide + 1 H 2 O 3 H 2 O H 2 O CH 2 OH H OH C O C C C C CH 2 OH C O C C C C CH 2 OH H H O O CH 2 OH O H 2 O PO 4 + + + + Base (Base = A, U, G, or C) Ribose RNA DNA Deoxyribose Base (Base = A, T, G, or C) PO 4 peptide bond hydrogen bonds R H C H 2 N P P B S P B S P B S (etc.) (etc.) P P P P P B S P B S P B B B B S S S S O C R H O OH C C N H Amino acid Nucleic acids: DNA, RNA HOCH 2 H H H 1 2 3 4 5 HOCH 2 5 H O C C C C OH OH OH H H H 1 2 3 4 H O C C C C OH OH H

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    5. Functional groups can modify the properties of organic molecules. In the following table, indicate whether each functional group is polar or nonpolar and hydrophobic or hydrophilic. Which of these functional groups are found in proteins and lipids? Activity 4.1/5.1 15 Functional group Polar or nonpolar Hydrophobic or hydrophilic Found in all proteins Found in many proteins Found in many lipids —OH Polar Hydrophilic No In some R groups In fatty acids as terminal reactive group —CH 2 Nonpolar Hydrophobic No Yes in side groups Yes —COOH Polar Hydrophilic Yes No —NH 2 Polar Hydrophilic Yes No —SH Polar Hydrophilic No No Found in cysteine No —PO 4 Polar Hydrophilic No Only if they have been phosphorylated In phospholipids 6. You want to use a radioactive tracer that will label only the protein in an RNA virus. Assume the virus is composed of only a protein coat and an RNA core. Which of the following would you use? Be sure to explain your answer. a. Radioactive P b. Radioactive N c. Radioactive S d. Radioactive C To distinguish between protein and RNA in a virus, you could use radioactively labeled S compounds. If you grew viruses on cells with radioactively labeled S compounds, the sulfhydryl groups in the virus’s protein would become labeled but the RNA would not become labeled. 7. Closely related macromolecules often have many characteristics in common. For example, they share many of the same chemical elements and functional groups. Therefore, to separate or distinguish closely related macromolecules, you need to determine how they differ and then target or label that difference. a. What makes RNA different from DNA? RNA contains ribose sugar, whereas DNA contains deoxyribose sugar. In addition, RNA contains uracil and not thymine. DNA contains thymine but not uracil. b. If you wanted to use a radioactive or fluorescent tag to label only the RNA in a cell and not the DNA, what compound(s) could you label that is/are specific for RNA? You could label either ribose or uracil.

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