Solution Manual For Campbell Biology, 9th Edition

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Solution Manual For Campbell Biology, 9th Edition provides the perfect textbook solutions, giving you the help you need to succeed in your studies.

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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:
• H2O molecules are cohesive; they form hydrogen bonds with each other.
• H 2O 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

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

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, N2, and O2 in
the atmosphere and as
H20, PO4, 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

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 (H2) 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 1023
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

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 1023 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 CO2 + 6 H2O → C 6H12O6 + 6 O2
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 1023
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 1023 by 1,000 =
6.02 1020 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 1023
divided by 4 =
1.51 1023
How many
molecules of NaCl
are there per ml of
distilled water?
6.02 1020
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
(C6H12 O6)?
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 CO2 + 12 H2O → C 6H12O6 + 6 O2 + 6 H2O
In most texts, however, this is reduced to 6 CO 2 + 6 H2O → C 6H12O6 + 6 O2
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 CO2 in the chamber.
7. O2 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
HN 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. O2,
on the other hand, is not polar.
Ammonia (NH3)
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 3COOH.
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
(H3 O+
, often described simply as H +
) and hydroxide ions (OH). The concentration
of each of these ions in pure water is 107
. 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 log10 of the hydrogen ion
concentration.What is the pH of pure water?
The hydrogen ion concentration of pure water is 107
. The log 10 of 107 is 7.
The negative log10 of 107 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 104
, what is the pH of this solution?
The pH of a solution with a H + ion concentration of 104 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. H2O molecules are cohesive; they form hydrogen bonds with each other.
B. H2O 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 (CH2 )n
CH2OH
H
OH
OH
C O
C C
C C
O
OHH HO
OH
H H
H
HO
H
HO
H
OH
C
OH
CH3
CH3
CH3
C (CH2 )n CH3Fats, oils,
waxes,
cholesterol
Proteins:
Enzymes,
structural
proteins
Sugars, starches,
glycogen,
cellulose
Carbohydrates:
Compound
Some Simple Chemistry
Basic
components Reaction
C
C (CH2 )n
O
OHH HO
O
C
OH C (CH2 )n CH3
Product
H
Glycerol + 3 fatty acids Triglyceride or fat
Disaccharide6C hexose
dehydration
reaction
amino
group carboxyl
group
(etc.)
C
C (CH2 )n
O
OHH HO
O
R
H
C COOHH 2 N
R
H H
HOH
O
C CH 2 N
OR
H OH
C CN
H
C
OH C (CH2 )n
+ 3 H2 O
CH3
dehydration
reaction
dehydration
reaction
+ 1 H2 O
Dipeptide
+ 1 H2 O
3 H2 O
H2 O
CH2OH
H
OH
C O
C C
C C
CH2OH
C O
C C
C C
CH2OH
H HO
O
CH2OH
O
H2 O
PO4 +
+
+
+
Base
(Base = A, U, G, or C)
Ribose
RNA
DNA
Deoxyribose
Base
(Base = A, T, G, or C)
PO4
peptide
bond
hydrogen
bonds
R
H
CH2N
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 CN
H
Amino acid
Nucleic acids:
DNA,
RNA
HOCH2
H H H
1
23
4
5
HOCH2
5
H
O
CC
C C
OH OH
OH
H H H
1
23
4
H
O
CC
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
—CH2 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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Solution Manual For Campbell Biology, 9th Edition

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