Answer Key
Unit 1 Biochemistry
Answers to Unit 1 Preparation Questions
Assessing Student Readiness
(Student textbook pages 4–7)
1.Sketches should depict a central atomic nucleus
surrounded by four valence electrons (or six electrons
in total), or six neutrons and six protons within
the nucleus.
2.b
3.a
4.metal, non-metal ion
5.a.Since the atoms in a glucose molecule share valence
electrons, the bonds between these atoms must
be covalent.
b. Glucose is a stable molecule because its covalent
bonds are stable and strong.
6.a.water—molecular, atoms (both non-metals)
are joined by covalent bonds
b.carbon dioxide—molecular, atoms (both
non-metals) are joined by covalent bonds
c.glucose—molecular, atoms (two non-metals and
uncharged hydrogen) are joined by covalent bonds
d.sodium chloride—ionic, ions (a metal and a
non-metal) joined by an ionic bond
e.calcium phosphate—ionic, ions (a metal and a
polyatomic ion) joined by ionic bonds
f.methane—molecular, atoms (a non-metal and
uncharged hydrogen) joined by covalent bonds
g.oxygen—molecule or molecular element, atoms
joined by double covalent bond, but not a molecular
compound because it is formed from only one
element (two oxygen atoms)
h.ammonia—molecular, atoms (a non-metal and
uncharged hydrogen) joined by covalent bonds
7.a.chloride
b.sulphate/sulfate
c.magnesium ion
d.copper (II) ion
8.a.OHb.S2 c. Al3+
d. Fe3+
9.H2O or water
10.water → hydrogen + oxygen
11.d
2.sodium hydroxide + hydrochloric acid →
1
water + sodium chloride
13.a.Since the pH of the solution increased, the student
must have added a base to the solution.
b.One of: When the second substance was added, it
released hydroxide ions (OH-), which combined
with protons (H+) to produce water. The second
substance accepted protons (H+) from the solution.
14.a.decomposition
b.neutralization
c.decomposition
d.synthesis
e.neutralization
f.synthesis
g.decomposition
h.synthesis
i.synthesis
j.synthesis
15.pH 7
16.c
17.One of: enzymes, cell structure, membrane transport,
muscle
18.A molecule is a substance composed of two or
more atoms that are bonded together covalently. A
macromolecule is a large, complex molecule. Usually,
the subunits of a macromolecule are small molecules
bonded together.
19.An enzyme is a macromolecule that speeds up
chemical reactions.
20.c
21.e
22.b
Biology 12 Answer Key Unit 1 • MHR TR 1
23.A—prokaryotic; B—eukaryotic
•Venn diagram should show:
•Prokaryotic Cells Only—DNA is free in the
cytoplasm; usually has a cell wall; smaller; unicellular
•Prokaryotic Cells and Eukaryotic Cells—DNA;
cell membranes control passage in and out of
cell; cell membranes separate inside of cell from
extracellular environment
•Eukaryotic Cells Only—chromosomes (DNA) are
enclosed in the nucleus; plants and fungi have cell
walls, most other eukaryotes do not; membranebounded organelles, such as the Golgi apparatus;
larger; may be part of a multicellular organism
24.e
25.e
26.d
27.The cell membrane is selectively permeable in that it
allows the passage of some substances while restricting
the passage of others. Cells need to allow nutrients
and water in and allowing wastes out while preventing
harmful substances from entering.
28.Venn diagram should show:
•Diffusion Only—includes the movement of
solutes from an area of higher concentration to an
area of lower concentration; particles may move
across a membrane, within cytoplasm, or within
extracellular environment
•Both Diffusion and Osmosis—net movement of
particles is from an area of high concentration to
an area of low concentration
•Osmosis Only—refers only to the movement of
water from an area of higher water concentration
to an area of lower water concentration; involves a
semi-permeable membrane
29.Sample answer: Before osmosis, the sugar
concentration is higher on the right side of the
U-shaped tube than on the left side. During osmosis,
water moves from the left side of the selectively
permeable membrane to the right, diluting the sugar
solution on the right side. After osmosis, the solutions
on both sides of the selectively permeable membrane
have the same concentration, but there is a greater
volume on the right.
30.Water molecules will move by osmosis into the cell
because the animal cell contains a higher concentration
of solutes than its surroundings do (pure water has a
lower concentration of solutes).
31.c
2 MHR TR • Biology 12 Answer Key Unit 1
Chapter 1 The Molecules of Life
Answers to Learning Check Questions
(Student textbook page 13)
1.Elements are pure substances that cannot be broken
down into simpler components by normal means.
Atoms are the smallest units of elements that have all
of an element’s properties.
2.Carbon-12 has an atomic mass of 12 and has
6 neutrons whereas carbon-14 has 8 neutrons and
an atomic mass of 14. Carbon-12 is very stable (and
thus very common) compared to carbon-14 which is
unstable and radioactive.
3.A polar covalent bond involves the unequal sharing
of a pair of electrons between two atoms, where one
atom is more electronegative than the other. A polar
bond results in one of the atoms having a partial
negative charge (the more electronegative atom) and
the other atom having a partial positive charge. In an
ionic bond, two oppositely charged ions are attracted
to one another; one ion has a full negative charge (as
a result of its extra electron) and the other has a full
positive charge (as a result of having lost an electron).
The full charges result from one of the atoms being
so much more electronegative than the other that it is
able to fully acquire an electron from the more weakly
electronegative atom. The transfer of the electron
results in ion formation.
4.A water molecule’s intramolecular forces make it a
polar molecule. As such, its intermolecular interactions
are dominated by attraction to other polar molecules
(hydrophilic molecules since they are attracted to
water) and its inability to attract non-polar molecules
(hydrophobic molecules, since they are not attracted
to water).
5.The tendency of non-polar molecules or non-polar
portions of molecules to stay away from water. The
hydrophobic effect is very important in determining
the tertiary structure of proteins and in the structure
of many other biological molecules.
6.Sample answer: Biotechnology brings together
the study of biology and/or biochemistry with
technological advancements in many other disciplines,
which may include physics, engineering, and
chemistry. A good and often overlooked example is
the microscope, in which case advances in optics have
allowed us to observe cells in great detail, even while
they are alive. Molecular genetics is another example of
a marriage between disciplines; molecular biology and
genetics. Advances in molecular biology, such as the
ability to clone genes and sequence DNA, have allowed
us to study heritable characteristics at a level of detail
never imagined 50 years ago.
(Student textbook page 21)
7.A monomer is the basic building block of a larger
polymer; typically a long molecule that may be
composed of many hundreds or even thousands of
monomeric units joined together. The truly large
polymers belong in a class of giant molecules known
collectively as macromolecules. Examples include most
proteins, larger polysaccharides like starch, glycogen,
and cellulose, and many larger nucleic acids such as
long double-stranded DNA molecules.
8.Structurally, carbohydrates are organic molecules with
many hydroxyl groups, that often contain carbonyl
groups, and that generally have the molecular formula
(CH2O)n. Functionally, carbohydrates are used as
an energy source for cells (e.g., glucose), for energy
storage (e.g., starch and glycogen), and to provide
structural support (e.g., cellulose).
9.Because they have the same molecular formula (C6H12O6)
but have different three-dimensional structures.
10.Answers should include a Venn diagram or concept
map that shows that all three types of molecules are
carbohydrates and involved in energy use and/or
energy storage, and that disaccharides are composed
of two monosaccharides, polysaccharides are
composed of more than two monosaccharides, and
that monosaccharides are the basic building blocks
for the other two.
11.Any two of: energy storage, energy source, and
structural support.
12.All three are polysaccharides composed of many
monosaccharides linked together, they are composed
of many individual glucose monomers linked together
covalently, and are macromolecules. They differ in
that starch and cellulose are plant products, whereas
glycogen is an animal product. Another difference is
that the glucose monomers in cellulose are joined by
different glycosidic linkages than those found in starch
or glycogen. The molecules also differ in function:
cellulose is involved in structural support, but starch
and glycogen are energy storage molecules.
(Student textbook page 25)
13.Both are organic molecules and have examples that are
used for storing energy in cells. Lipids, however, have a
greater proportion of carbon and hydrogen atoms and
fewer oxygen atoms and, in general, are hydrophobic.
Carbohydrates are polar, hydrophilic molecules.
14.Lipids have many energy-rich C-H bonds in their long
hydrocarbon chains.
15.Sketches should be similar to Figure 1.12 or 1.13B
on student textbook page 22. The basic structure of
a triglyceride is a glycerol molecule with three fatty
acid chains linked to it covalently. The fatty acid
chains may be all saturated (no carbon-carbon double
bonds) or all unsaturated (one or more carboncarbon double bonds) or a mixture of saturated and
unsaturated chains. The unsaturated chain(s) may be
monounsaturated or polyunsaturated. The presence of
carbon-carbon double bonds in the fatty acid chains
makes the chains bent (or kinked) so that they do not
pack together as well. Molecules with unsaturated fatty
acid chains tend to be liquids at room temperature
whereas those with saturated fatty acyl chains (that are
quite straight and pack together well) tend to be solids
at room temperature.
16.A phospholipid molecule has a dual character in that
part of the molecule—its “head” group is polar and
thus hydrophilic, but its “tail” portion is non-polar
and thus hydrophobic. This structure is essential to its
function in that it forces the molecules to orient their
head groups toward water and their tails toward each
other when placed in an aqueous environment. This
arrangement forms the familiar phospholipid bilayer
that is the basic structure of the biological membranes
in all cells.
17.Sample answer: Cholesterol is an important component
of animal cell membranes and is the precursor used for
making many other steroids. Estrogen is an important
determinant of sexual function in females and helps
regulate the storage of fat.
18.Both are solids at room temperature. In plants such as
trees, waxes coat leaves, preventing water loss and offering
protection from insects. In animals such as ducks, a
waxy layer on feathers prevents them from getting wet,
which would add weight and make flight difficult.
(Student textbook page 28)
19.Sample answer: Given the many different ways
there are of ordering the 20 common amino acids,
proteins exhibit tremendous diversity in structure and
function. Examples include the protein fibres that lend
structural support to tendons, transport proteins such
as hemoglobin that carry oxygen in blood, enzymes
that catalyze specific biological reactions, and antibody
proteins that fight infections.
Biology 12 Answer Key Unit 1 • MHR TR 3
20.Sketches should resemble Figure 1.18 on page 25.
21.The R group gives an amino acid its distinctive shape
and properties.
22.Because proteins can be built from many different
monomers (there are about 20 common amino acids),
the many different properties of those amino acids (i.e.,
polar, non-polar, acidic, or basic, and the reactivity of
the specific functional groups found as part of their
R groups), and the many different linear sequences of
amino acids that are possible (20n, where n = length
of the protein in amino acids).
23.Primary—the linear sequence of amino acids.
Secondary—regions of repetitive structure seen in
many different proteins (e.g., alpha helices and beta
strands). Tertiary—the overall three-dimensional
shape of the protein.
Quaternary—the association of more than one
polypeptide to form an intact protein.
24.The overall three-dimensional structure of a protein
can be changed by a variety of environmental variables
and chemical or physical treatments. Anything that
could alter the intramolecular and/or intermolecular
interactions that occur in a protein may alter its
structure and thus its function as well. For example,
a change in temperature (too hot or too cold), in pH,
or ionic environment (resulting from changes in salt
concentrations). All of these changes could denature
a protein either partly or completely. Since a cell
depends on the many different functions performed by
its proteins, a change in the structure of one or more
proteins can be harmful to an organism.
(Student textbook page 36)
25.Products are a salt (an ionic compound) and water. The
reaction neutralizes both the acid and the base so that
the acid loses its acidic properties and the base loses its
basic properties.
26.When a compound is oxidized, its electrons are
donated to another molecule (which we say becomes
reduced).
27.When electrons are removed from one molecule (the
molecule being oxidized) they end up being taken by
some other molecule (the molecule being reduced)
since they are so reactive.
28.A molecule has more energy in its reduced form than
when oxidized. As covalent bonds are broken (shared
pairs of electrons are separated) energy is released.
4 MHR TR • Biology 12 Answer Key Unit 1
29.During hydrolysis, water is consumed (used as a
reactant) by the reaction. During a condensation
reaction, water is produced (released).
30.Condensation reactions—build molecules and release
a molecule of water from the H-atom and OH-group
that are removed from the combining molecules.
Hydrolysis reactions—break apart molecules and
consume a molecule of water that is used to donate an
H-atom and an OH-group to the products.
Oxidations— remove electrons from molecules,
generally by breaking apart covalent bonds and
forming a simpler, less complex product or products.
Reductions—always accompany oxidations. They add
electrons to molecules, typically building them up into
larger, more complex molecules.
(Student textbook page 39)
31.The activation energy of a reaction is the initial input
of energy needed to start the reaction. Its value is
significant because reactions with high activation
energies occur more slowly than reactions with low
activation energies. Anything that lowers the activation
energy of a reaction (such as a catalyst) will speed the
reaction. Many of the chemical reactions that occur in
cells have high activation energies and only proceed
at the rates needed for life because cells have enzymes
that catalyze those reactions.
32.When an enzyme catalyzes a biological reaction, it
needs to bind the substrate(s) at its active site to form
an enzyme-substrate complex. The enzyme destabilizes
the substrate(s) as it binds the active site, lowering the
activation energy of the reaction and allowing covalent
bonds to be broken and new bonds to form.
33.Even though many important biological reactions can
occur naturally without an enzyme, their rates are often
too slow to support the needs of the cell or organism.
Enzymes are important to biological systems because
they help speed these reactions to up the rate required
to sustain life.
34.Enzymes are unable to catalyze many different types of
reactions because the shape of an active site is usually
very substrate-specific. This specificity limits the ability
of an enzyme to perform different types of reactions on
different molecules.
35.Enzymes can prepare substrates for a reaction by
altering the substrate, its environment, or both.
Substrate changes can involve the alteration of bond
lengths or bond angles (stretching or bending bonds),
the addition or removal of electrons (reduction or
oxidation), and/or the addition or removal of H-ions
(protons) or functional groups. Environmental changes
may involve providing an acidic or basic environment
at the active site and/or holding substrates close together
and in the best orientation relative to one another.
36.A coenzyme is an organic molecule other than the
substrate(s) that is required for an enzyme-catalyzed
reaction. A cofactor is an inorganic molecule (e.g.,
metal ion) that is required for an enzyme catalyzed
reaction. These molecules assist enzymes in performing
the reactions they catalyze. Without them, enzymes
could not work. Many enzymes require one or more
coenzymes or cofactors for their activity.
Answers to Caption Questions
Figure 1.2 (Student textbook page 11): Based on the partial
positive and partial negative charges in the diagram, water
molecules would be predicted to interact weakly with each
other through the attraction and repulsion of those charges
(because like charges repel and unlike charges attract).
Since each water molecule has multiple partial charges,
each would be predicted to interact with at least three or
more other water molecules.
Figure 1.5 (Student textbook page 15): Methane is
a non-polar molecule. The two forms of glucose are
polar molecules.
Figure 1.7 (Student textbook page 18): Unlike the other
molecules, the triglyceride does not have a structure
composed of a series of “building blocks” (monomers)
linked one after the other to form a long chain (polymer).
Although the triglyceride has three fatty acids that are
somewhat like monomers in a polymer, they are not
linked directly to one another in a chain, but branch off
an intermediate molecule (glycerol).
Figure 1.15 (Student textbook page 23): If a phospholipid
bilayer contained many lipids with unsaturated fatty
acid chains, the bilayer would have a looser arrangement
of molecules—since the unsaturated chains would
be bent (kinked) and prevent the lipids from packing
close together.
Figure 1.25 (Student textbook page 33): If some process in
the body began to contribute H+ (hydrogen ions; protons)
to blood, the buffer system would counteract the rise in
the concentration of H+ (the increase in acidity). This
would occur because the rise in H+ concentration would
drive the equilibrium of the reaction to the left, toward the
formation of water and carbon dioxide.
Figure 1.29 (Student textbook page 37): Even though
maltose is also a disaccharide, maltose is not a substrate
for this enzyme because it has a different shape than the
normal substrate (sucrose) and so would not bind to the
enzyme’s active site in a manner that would allow the
hydrolysis to occur.
Figure 1.30 (Student textbook page 39): It would not be
active since the pH is far from optimal for that enzyme.
Answers to Section 1.1 Review Questions
(Student textbook page 17)
1.oxygen, carbon, hydrogen, and nitrogen
2.a.Radioactive material is injected into the patient.
Since cancerous tissues have a higher level of activity
than healthy tissues, the radioactive material tends
to concentrate in the cancerous areas, where it can
be seen using a PET scan.
b.By inserting the radioactive isotopes of atoms into
food, a biologist could follow the movement of these
atoms as they are digested. The biologist would hope
to learn what tissues take up the molecule, how it is
processed, and where it is either stored or released
as waste.
3.Sample answer:
atom
vary by number
of neutrons
2 or more
atoms
gain or loss of
1 or more electron
isotope
molecule
ion
unstable,
radioactive
carbonbased
type
other
(inorganic
type)
radioisotope
can be used diagnostically
when used to make
some can be
organic
molecule
4.Sample answer: a covalent bond is an intramolecular
force, and a hydrogen bond is an intermolecular
force. Intramolecular forces are stronger, since these
are the bonds that hold molecules together. If forces
between molecules were stronger than forces within
molecules, most molecules would constantly split apart
and reassemble.
Biology 12 Answer Key Unit 1 • MHR TR 5
5.Sketches should contain the detail found in Figure 1.2
on page 11 of the student textbook, with the addition
of covalent bonds identified between the oxygen atom
and each hydrogen atom.
6.a.Non-polar (same electronegativity
b.Polar (nitrogen has a higher electronegativity)
c.Polar (chlorine has a higher electronegativity)
d.Polar (oxygen has a higher electronegativity)
7.Since fluorine has a high electronegativity, it will form
a polar bond with carbon.
8.Jordan has the best answer since polarity and hydrogen
bonding do not always go together. Hydrogen bonding
is specific to molecules containing partially positive H
atoms, which are attracted to partially negative N, O, or
F atoms on other molecules.
9.A hydrogen atom will participate in hydrogen bonding
only when it is bonded with an atom that has a
significantly higher electronegativity. Since carbon
and hydrogen have comparable values, the difference
in electronegativity is not enough to make the
hydrogen atom capable of hydrogen bonding.
10.Magnets have positive and negative ends, as does a
water molecule. When magnets are brought close
together, they attract each other in a predictable manner.
Similarly, in water, the positive hydrogen atoms of one
water molecule are attracted to the negative oxygen
atoms of neighbouring water molecules. While both
magnets and water molecules have positive and negative
poles, the analogy should not be carried too far since
magnets have magnetic (north and south) poles
whereas water molecules have electrical poles.
11.The hydrophobic effect refers to the natural clumping
together of non-polar molecules, such as oil, in water.
The oil is non-polar, so it clumps together on top of
the polar water mixture.
12.a.A cation is an atom that is missing one or more
electrons (has a positive charge).
b.Sample answer: Hydrogen cations for
cellular respiration, and sodium cations for
transport mechanisms.
13.A molecular formula takes up less space while
presenting the total ratio of atoms of each element. A
structural formula illustrates the connections between
the atoms, which gives an indication of a molecule’s
function and other properties.
6 MHR TR • Biology 12 Answer Key Unit 1
14.Venn diagrams should show:
•2-D Structure Only—Appears flat, which is not the
molecule’s real shape
•2-D and 3-D Structures—Show number and types of
atoms; show which atoms are bonded to each other, 4
single lines represent its 4 single covalent bonds
•3-D Structure Only—Shows the molecule’s real
tetrahedral shape
15.These are the most abundant elements in most living
things and contribute to the high proportion of water
found in living things. Earth’s crust is not living.
Answers to Section 1.2 Review Questions
(Student textbook page 31)
1.Chains, six-sided rings, and five-sided rings.
2.a.amino acids
b.glucose, a monosaccharide
c.nucleotides
3.a.carboxyl group, amino group
b.amino, carbonyl, hydroxyl
4.Ribose, amino acid, triglyceride, cellulose—according
to their number of carbon atoms.
5.Simple sugars would be released, and blood glucose
levels would spike.
6.Fats contain long fatty acid hydrocarbon chains with
many energy-rich C-H bonds. Although carbohydrates
also contain C-H bonds, they have fewer.
7.Sample answer: C=O•••H–N; hydrogen bonding
between amino acids; intramolecular
R•••R; hydrogen bonding between R-groups; intraor intermolecular*
R+ -R; electrostatic interaction between R-groups;
intra- or intermolecular*
interior of protein
R
R
* intermolecular interactions occur between polypeptide
chains (for proteins with quaternary structure)
8.In both structures, the polar “head” is facing out, where
it can interact with water molecules, and the non-polar
“tails” are facing in, away from water molecules.
9.Eating a balanced diet allows the body to collect all
the lipids, proteins, and carbohydrates it needs from a
variety of sources. For example, fast food contains a lot
of fat and oil, but a diet containing only fast food would
have a limited variety of oils. A diet that contains fish,
nuts, legumes and other sources of natural oil will
prevent a deficiency.
10.a.Two carbonyl and two hydroxyl groups.
b.None, since the carbonyl and hydroxyl groups are
not connected to the same carbon.
11.Since hair is made of protein, it likely contains
intermolecular bonds that hold hair in its shape. Heat
might break these intermolecular bonds and denature
the proteins, resulting in straight hair.
12.Venn diagrams should show:
•Strand of DNA Only—thymine (T); sugar in the
nucleotides is deoxyribose
•DNA and RNA—composed of nucleotides;
nucleotides joined by phosphodiester bonds; found
different types of nitrogenous bases are part of the
nucleotides: adenine (A), guanine (G), cytosine (C)
•Strand of RNA Only—uracil (U), sugar in the
nucleotides is ribose
Answers to Section 1.3 Review Questions
(Student textbook page 42)
1.a.neutralization reaction
b.redox reaction
c.condensation reaction
d.hydrolysis reaction
2.Sample answers, one each of:
•neutralization reactions—
NaOH + HCl → NaCl + H2O
NH3 + HCl → NH4Cl + H2O
NaOH + CH3COOH → NaCH3COO + H2O
•redox reactions—
C3H8 + 5O2 → 3CO2 + 4H2O
•condensation reactions—
nucleotide + nucleotide → dinucleotide + H2O
monosaccharide + monosaccharide →
disaccharide + H2O
glycerol + 3 fatty acids →
fat molecule (3 condensation reactions) + 3H2O
•hydrolysis reactions—
dinucleotide + H2O → 2 nucleotides
dipeptide + H2O → 2 amino acids
fat molecule + 3H2O → glycerol + 3 fatty acids
3.Without the noncompetitive inhibitor bound to it, the
substrate can bind to the active site and be acted upon
by the enzyme.
substrate
active
site
enzyme
allosteric
site
noncompetitive
inhibitor
When a noncompetitive inhibitor binds
at the allosteric site, it induces a change
in shape that alters the active site and
shape change
makes it difficult or even impossible
for the substrate to bind normally. The
effect reduces or eliminates the enzyme activity.
4.a.formation of a polypeptide
b.obtaining glucose from glycogen
c.neutralization of stomach acid
d.photosynthesis
5.Buffers in living organism help maintain pH levels so
that enzymes and other internal systems can operate
optimally. For example, carbonic acid acts as a buffer
in human blood, along with its hydrogen carbonate ion
base. If blood becomes too basic, carbon dioxide and
water react to form carbonic acid, which dissociates
into hydrogen carbonate and acidic hydrogen ions.
6.Since the antacid increases the pH of the stomach, the
enzymes naturally occurring in the stomach will not
function at their optimal rate.
7.Diagram should include the active site and the
allosteric site, along with the substrate, enzyme,
and inhibitor. The inhibitor binds to the allosteric
site, triggering a change in enzyme conformation.
The active site of the enzyme no longer binds to the
substrate, so the enzyme cannot function.
8.Diagrams should show that both types of inhibition
prevent the substrate from binding, and hinder enzyme
activity. However, in competitive inhibition, the
inhibitor binds to the active site and in non-competitive
inhibition, the inhibitor binds to the allosteric site.
9.The active site of an enzyme binds with the substrate
to form an enzyme-substrate complex. The enzyme
lowers the activation energy for the chemical reaction,
for example, by putting stress on some of the bonds
in the substrate. With a lower activation energy, the
chemical reaction will proceed much faster.
Biology 12 Answer Key Unit 1 • MHR TR 7
10.The compound is likely a monomer, water-soluble
(not a lipid or fatty acid), and since it is acidic and has
15 to 20 atoms, it could be one of the acidic amino
acids (such as aspartic acid or glutamic acid).
11.Temperature, pH, ionic concentrations
12.a. Graphs should have temperature (from 20°C to
40°C) on the x-axis and rate of reaction (mol/s)
on the y-axis. Lines should show an increase up to
29°C, then a sharp decrease until 31°C, and then a
very slow, steady increase to 40°C.
b.The rate increases with temperature until it reaches
the optimal temperature for the enzyme (at 9 min).
Then, the activity of the enzyme decreases as it
denatures. By 11 min, the enzyme is no longer
functioning, and the reaction is proceeding on its
own. Like many non-catalyzed reactions, the rate of
reaction continues to increase with temperature.
Answers to Chapter 1 Review Questions
(Student textbook pages 49–53)
1.b
2.c
3.b
4.b
5.d
6.a
7.b
8.c
9.d
10.e
11.a
12.d
13.d
14.b
15.Carbon is found as carbon dioxide, the waste product,
as well as in the backbone of most biological molecules
(such as carbohydrates, lipids, proteins, and nucleic acids).
16.Polarity, there is an uneven electron distribution in the
water molecule.
17.Primary—chain of amino acids formed; secondary—
R groups interact and join through hydrogen bonding;
tertiary—bridges form; quaternary—two or more
tertiary subunits join.
18.Enzymes function best within a narrow temperature
range. If the temperature becomes too high, bonds
8 MHR TR • Biology 12 Answer Key Unit 1
become too weak to maintain an enzyme’s shape, so
substrate molecules will not fit (bind) and the enzyme
will not function.
19.A condensation reaction produces water and makes
a larger molecule but a hydrolysis reaction uses water
and breaks down a polymer into smaller subunits.
20.Changes the shape of the enzyme so that it cannot bind
to the substrate, so it slows or limits the reaction rate of
an enzyme without interfering with the active site.
21.A buffer ensures that factors like pH and salinity are
maintained so that the cell can continue to function in
a predictable and efficient manner.
22.Sample answer: Probably the secondary, since H-bonds
and intermolecular forces require less energy to break
than covalent bonds do.
23.It regulates the enzymatic activity by increasing or
decreasing reactions at the active site. This allows
the function of the enzyme to be turned on or off
depending on substrate levels within the cell.
24.Coenzymes are organic molecules that assist enzymes,
while cofactors are usually key elements, often metallic,
that allow an enzyme to function.
25.Any two of: extremes of higher or lower temperatures,
or changes in pH, ionic pressure, or salinity
26.Organic molecules (e.g., methane and ethanol) contain
carbon and hydrogen atoms, and inorganic molecules
(e.g., salt and water) do not.
27.Enzyme efficiency is reduced.
28.a.amino acids, condensation
b.glucose(a monosaccharide), condensation
c.nucleotides, condensation
29.Isotopes are versions of an element that differ in
neutron numbers. Radioactive isotopes are those
that are unstable and will break down, releasing high
energy radiation.
30.Sample answer: Almost all biological macromolecules
(except lipids) are polar, hydrophilic molecules
that dissolve readily in water, and all (except lipids)
can be commonly found in polymeric forms (e.g.,
polysaccharides, polypeptides, and polynucleotides).
One of the main functions of carbohydrates and
lipids is energy storage, whereas nucleic acids like
DNA are used in storing genetic information and
in making polypeptides (proteins) which perform a
wide variety of functions. All four types of biological
macromolecules (carbohydrates, lipids, polypeptides,
and nucleic acids) are organic molecules consisting
mainly of carbon and hydrogen atoms.
31.It would be rendered useless because this active site
would be filled.
32.It stops functioning because of pH changes in
the stomach.
33.Most mutations change the specific shape of the active
site so it no longer matches the substrate, making it
devastating to the enzyme.
34.a.Stomach acid may have hydrolysed the coating.
b.Since succinate is naturally occurring in the body,
the body has enzymes that metabolize it. This must
also occur in the small intestine, thereby dissolving
the coating and releasing the medication there.
35.Water rises in glass tubes (capilliarity) because of the
attraction of the polar molecules to the surface of
the glass. The water molecules remain together as a
column against the force of gravity as a result of the
hydrogen bonds between them. Both of these are
examples of intermolecular interactions. These two
examples of attraction are possible because water is a
polar molecule. Its polarity results from the difference
in electronegativity between oxygen and hydrogen in
a water molecule (intramolecular forces).
36.a.Enzymes perform best at a specific temperature.
A number of enzymes in the yeast work together as
a whole to metabolize the sugars. The average of all
the optimum enzyme temperatures will allow the
enzyme as a whole to produce a maximum amount
of carbon dioxide.
b.Prepare a large sample of dough and place it in
an oven at a variety of temperatures to determine
the temperature at which the activity begins
(dough rises).
c. The amount of dough, amount and quality of yeast,
the way measurements are taken, time duration for
rising, identical set-up of each trial (ideally all from
the same batch of dough), and species of yeast.
d.Let the dough rise in a beaker and measure the
change in volume. Measuring change in mass
may help estimate how much additional gas was
produced but lost to the atmosphere.
37.Since pills do not break evenly, take the mass of the
individual pieces to accurately measure the amount of
pill; measure the amount of milk/dairy, and consume
the same amount each day; set clear guidelines for
determining lactose intolerance threshold (such as
measures of stomach pain, cramps, gas, and nausea).
38.Sample answer: The active site is not attracted to the
substrate with the same affinity that is found in the
natural enzyme. Or, the artificial enzymes don’t fold
properly or aren’t modified properly, resulting in an
approximate fit.
39.a.Acid hydrolysis can be quite slow as well as leaving
an acid-sugar residue that must be neutralized to
ensure safe consumption and the integrity of the
simple sugars.
b.Enzyme-catalyzed hydrolysis, since enzymes are
specific to the reaction and can easily be denatured
once they are no longer needed.
40.There are two problems with her methodology: First,
since enzymes work quickly, it is likely the reaction
will be complete long before one hour has passed.
Second, the formation of products makes it harder for
the enzyme to determine reactants, reducing the rate
as soon as the reaction begins. The shorter the time
interval, the more accurate is the representation of
the enzyme (as long as it yields sufficient glucose to
measure accurately).
41.Since biochemists are dealing with multiple proteins,
all of them must remain intact. If one protein was
outside normal operating conditions, this could impact
the validity of the test.
42.a.Starch is not being broken down since it is too large
to leave the dialysis tubing. This is not a reflection
on enzyme activity. In this case the enzyme and
substrate do not meet.
b.True, since the alternative would allow for a positive
Benedict’s test.
c.This is irrelevant since the enzyme catalyzes
thousands of reactions per second. A single glucose
attached to an enzyme, if this was the case, would
not have an impact on the colour change.
43.Since fat is a water-insoluble macromolecule, it can be
separated through density and settling. Once a cell is
lysed, or broken open, the contents can be centrifuged
and the percent content of the fat can be calculated
either through mass or volume.
44.a.Other explanations include errors in measurement,
sub-optimal pH for enzyme activity, and naturally
lower levels of enzyme production.
b.Repeating the investigation several times, ensuring
all conditions are identical, ensuring all students
have the same time gap since their last meal to
ensure adequate amylase production, or mixing
the samples to create an average sample.
Biology 12 Answer Key Unit 1 • MHR TR 9
47.The enzyme is the hand press. It is different from a
biological enzyme since it is unable to catalyze the
reverse reaction.
48.As body temperature increases, the bacteria that are
more susceptible to temperature increases based on
their enzyme sensitivity cease to function.
49.a.hydrolysis
b.Water is consumed when a polymer is broken down
into smaller molecules by hydrolysis.
c.Flow charts should look like the original, with arrow
direction reversed.
50.In DNA, A and T are always in the same
concentrations since they are complementary bases.
This indicates that DNA is double stranded while
RNA, which does not share this pattern, must be
single stranded.
51.Time how long it takes to burn an identical mass
of each.
Concentration of Reactant or Product
46.a.Hydrogen bonds between molecules help maintain
the proper structure and function of the molecule.
For example, hydrogen bonds maintain the shape
of DNA, and the breaking and reforming of these
bonds is key to the proper functioning of the
molecule. Hydrophobic interactions are responsible
for the formation of membranes and determine the
three-dimensional shape of proteins.
b.It would change in the shape of the molecules,
which likely would affect the way they function. For
example, if the double-helix structure of DNA were
not maintained, the ways in which the molecule
performs its replication functions would be impaired
or rendered inoperative.
c.This would impair or dismantle membranes.
Life as we know it likely would not exist without
hydrophobic interactions in living systems.
52.
product
reactant
Time
For a non-catalyzed reaction the rate of reaction
will be much less. The slope of each line in the graph
would be much less, perhaps nearly flat if the noncatalyzed reaction rate was much less than that with
the enzyme.
53.
Reaction Rate
45.Probably a condensation reaction to produce a starch
or protein or nucleic acid, because whatever was
formed was likely a much larger molecule that did not
dissolve as well in water. We know that the monomers
for proteins, carbohydrates, and nucleic acids all
dissolve well in water. It’s unlikely to be a lipid, because
it wouldn’t dissolve to start with.
Concentration of Substrate
54.Answers may include foods such as pizza which
contain protein in the meat and cheese, DNA in all of
the vegetables, oil and fat in the cheese, and complex
carbohydrates in the whole wheat crust.
55.Organizers should show that DNA is the cell’s blueprint
for protein production. When proteins need to be
made, RNA is produced and serves as the template.
RNA and DNA are single and double stranded
respectively. DNA’s bases are ACGT while RNA
contains uracil (U), not thymine (T).
56.The electrons in the O-H bonds spend the majority of
their time with the oxygen atom, making it partially
negative. This in turn causes the hydrogen atoms to
become partially positive.
57.Emails should explain that at a low pH, peptide bonds
which compose enzymes are subject to hydrolysis and
therefore decomposition into their amino acids.
10 MHR TR • Biology 12 Answer Key Unit 1
58.Venn diagrams should show:
•Condensation Only—produces water, makes
a polymer
•Condensation and Hydrolysis—occur in the same
place, opposite processes
•Hydrolysis Only—uses water to break a polymer
apart, produces monomers
66.The protein is hydrophilic since it easily mixes with the
blood of the fish. If it was hydrophobic, it would not
mix and not be able to prevent freezing.
59.Diagrams should show that simple sugars are
comprised of only a few monomers, whereas complex
carbohydrates are comprised of polymers, which
provides for much greater complexity.
68.Sample answer: Hydrogen bonding occurs within
and between all four main types of biological
macromolecules and also occurs with many of their
monomers. Water also exhibits hydrogen bonding.
Since water is the solvent in which most of life’s
molecules are dissolved, and since other hydrogen
bonding is so widespread inside and outside of cells, an
understanding of hydrogen bonding is fundamentally
important while studying biochemistry.
60.Sample answer: The term oxidation refers to the loss of
electrons by a molecule (the molecule acts as a donor
of electrons). Although oxygen can act as an electron
donor many other elements can do this as well, so
oxygen is not necessary. Diagrams should show an
electron donor atom or molecule being oxidized when
it donates an electron to an electron acceptor (the atom
or molecule being reduced).
61.a.Diagrams might show the addition of two hexagons
to form two linked hexagons. Hexagons were
used to represent the six-membered rings of
glucose molecules.
b.Diagrams might show an E adding to three bars,
forming an E with three horizontal bars. The
letter E was used since there are three sites at which
condensation can occur on glycerol. Thick black
lines were used for fatty acids since these molecules
are long carbon chainsc. Diagrams might show the
addition of two balls to form two linked balls. Balls
represent the many differently shaped R-groups
amino acids (about 20) that can be used to construct
a protein.
62.An enzyme’s active site is substrate specific. Even
though both molecules are disaccharides they are
different disaccharides with different shapes. The shape
of an active site can be specific for sucrose while not
allowing maltose or other disaccharides to bind.
63.Answers should include the great majority of key terms
listed in the Chapter 1 Summary.
64.The bonds it forms may be stronger, since enzymes
are not as adept at removing fluorine as they are at
removing chlorine.
65.Glycine would be soluble in water since it contains an
H-O bond on one end and N-H bonds on the other.
In neither case are these bonds balanced by other
polar bonds.
67.Heat shock proteins tend to reduce the unfolding/
denaturing that occurs due to other proteins in the
cell. In general they provide biochemical stability by
counteracting the effects of temperature.
69.a.Bacteria would usually come from water droplets
in the air (spray, mist, sneeze) or through
direct contact.
b.Environmental conditions are optimal for growth
and the glycoprotein provides energy and cellular
building blocks to allow bacteria to grow.
c.Normally the bacteria are washed down the throat
into the stomach, where they are hydrolyzed.
70.The enzyme would have the ability to form hydrogen
bonds with the polysaccharide to hold it in place.
The active site would likely be at least the size of
two monomers since these must be stabilized as the
covalent bond is broken.
71.For molecules to have strong intermolecular bonding,
they must attract each other by a difference in
charges—partially positive and partially negative
regions that attract. These are polar molecules.
However, methane is a small non-polar molecule, so
it has a weak form of intermolecular bonding. Only
a small amount of heat energy is needed to break the
weak intermolecular bonds in methane, therefore
it has a low boiling point and can exist as a gas at
room temperature.
72.Enzymes are made of proteins and are subject to pH
sensitivity that may cause a loss of function.
73.Water molecules are polar, each having a slightly
positive end and a slightly negative end. Many
biological molecules have polar covalent bonds
involving a hydrogen atom (which is slightly positive)
and an oxygen or nitrogen atom. The polarity of the
molecules means that hydrogen bonds are common
among water molecules (the aqueous environment of
cells) and biological molecules. They help maintain
Biology 12 Answer Key Unit 1 • MHR TR 11
the structure and function of the molecules. The
molecules in canola oil are non-polar, and they do
not form hydrogen bonds. Instead of forming bonds
with biological molecules, the canola molecules would
simply clump together. As a result, life as it currently
exists could not occur in canola oil.
74.When DNA is heated the two strands separate, since
the H-bonds break.
Answers to Chapter 1 Self-Assessment Questions
(Student textbook pages 54–5)
1.d
2.a
3.e
4.b
5.e
6.d
7.b
8.d
9.c
10.b
11.Water expands since the structure is more ordered and
there are larger spaces between the molecules than in
the liquid phase.
12.I would tell my classmate that nothing happens
since the allosteric inhibitor can only bind at the
allosteric site.
13.See Figure 1.20 on page 26 of the student textbook.
The H on an amino group reacts with the OH from a
carboxylic acid, producing water and a peptide bond
(also known as an amide linkage).
14.An acid and a base react to produce water and a salt.
15.A functional group is an atom or group of atoms
attached to a molecule that gives the molecule
particular physical and chemical properties. Important
characteristics include giving a molecule a charge that
facilitates bonding with other molecules, and making it
acidic or basic.
16.a.A redox reaction is a coupled reaction involving
both an oxidation (loss of electrons by an atom
or molecule) and a reduction (the gain of those
electrons by some other atom or molecule).
Oxidations and reductions always occur together
(they are coupled to each other) and so we refer to
them with one term, redox reaction.
12 MHR TR • Biology 12 Answer Key Unit 1
b.Answers should state that an electron is lost by
sodium (oxidized) and gained by chloride (reduced),
which together form a redox reaction.
17.Chewing is part of mechanical digestion, which breaks
open the maximum number of cells so that bacteria
can access and digest the cellulose for the cow.
18.a.condensation; an H atom and an OH group are lost
from the two reactants, forming a molecule of water
and a combined larger molecule from the two reactants
b.neutralization; methylamine is a base that picks up a
proton from water and the water acts as the acid (a
proton donor); note that with the products formed
both the acid and the base have lost their acidic and
basic properties
c.condensation; it produces a water molecule
d. oxidation; each hydrogen atom is losing an electron
e.reduction; each oxygen atom is gaining two
electrons and two hydrogen ions
f.redox; hydrogen is oxidized and oxygen is reduced;
each hydrogen atom gives up one electron (a total
of four electrons) and each oxygen atom gains two
electrons (for a total of four electrons)
g.redox; C6H12O6 is losing electrons and hydrogen
ions (being oxidized) to form carbon dioxide, and
oxygen is gaining those electrons and hydrogen ions
(being reduced) to form water
h.hydrolysis; a molecule of water is consumed
in breaking apart the dimethyl ether into two
molecules of methanol
i.neutralization; an acid (acetic acid) donates a
proton to a base (methylamine) to form a salt
(an ionic compound)
j.reduction; each chlorine atom gains an electron
19.a.amino acid
b.The left side is the amino group. In the middle,
H is hydrogen, R is the R group, and C is the central
carbon atom. The right side is the carboxyl group.
c.The R group is the significant feature determining a
protein’s shape and properties.
d.Two amino acids can be joined together by a
condensation reaction to form a dipeptide, a short
polymer (protein) of just two amino acids.
R
H
O
N C C
H
OH
H
R
H
O
N C C
H
OH
H
H2O
R O
R
H
O
N C C N C C
H
OH
H
H H
20.The activation energy of a reaction is the amount of
energy required to begin a reaction. Reactions with
low activation energies proceed more quickly than
those with high activation energies. An enzyme speeds
up a reaction by lowering the activation energy of
the reaction.
21.Sample caption: The active site of an enzyme is
typically located in an indentation or pocket at the
surface of an enzyme as shown in a. The structure of
the active site is specific to its substrate. When the
substrate binds at the active site as in b., intermolecular
interactions between the substrate and enzyme induce
small changes in shape in both the substrate and the
enzyme (indicated by the arrows). The enzyme actually
binds the substrate more tightly, a phenomenon known
as induced fit.
22.See Table 1.2 on page 30 of the student textbook.
23.Sample answer: There are many advantages such
as producing many different individual product
molecules, allowing for multiple sites at which
a process can be regulated and multiple types of
regulation. This allows for the fine-tuning of a pathway
and the many products resulting from that set of
reactions. In pathways that synthesize some final
product multiple steps allow for different molecules to
contribute to that synthesis such that the process is not
entirely dependent upon a single type of molecule that
might be in limited supply. Similarly, in pathways that
break down complex molecules, perhaps for use as fuel
by the cell, multiple steps allow the energy contained
within a molecule to be released in smaller amounts.
This stepwise release of many smaller amounts of
energy is often more efficient than simply releasing
all of the energy in a molecule all at once.
24.a.In the lock-and-key model, the enzyme can be
thought of as the lock, the substrate the key, and
the active site the key hole. Just as a lock has a
specific key that fits it exactly, most enzymes have
a specific substrate that is able to bind at the active
site. Similarly, although many slightly different keys
may fit into a lock’s key hole, only a key with exactly
the right shape is correct for the lock and capable of
opening it; the lock and key have complementary
geometries. This is also similar to many enzymes
and their substrates where many different molecules
may be able to enter the active site but only one is
properly acted upon by the enzyme. In this way,
the lock-and-key model can be used to describe
enzyme specificity.
b.The lock-and-key model assumes that the system is
rigid, in other words that the lock does not affect the
shape or structure of the key and the key does not alter
the key hole or the geometry of the pins inside the
lock. In that sense it does not adequately describe the
interaction between a substrate and an enzyme, since
both are flexible and can undergo changes in shape as
the two bind together. A more accurate model of this
interaction is the “induced fit” model in which the
binding of a substrate at an active site can affect the
shape of the active site such that the enzyme actually
binds the substrate more strongly as they interact. In
addition, the shape of the substrate can be altered upon
binding to an enzyme’s active site. These changes in
shape contribute to lowering the activation energy for
the reaction.
25.Sample answer: Water is a great transporter of most
nutrients because many things will dissolve in water.
Just as people tend to make friends with people who
like the same things, water and most nutrients go
together well because they things in common. They act
a bit like tiny magnets because the electrons in their
atoms are not evenly distributed around the nucleus.
This causes the water and nutrient molecules to stick
together, helping them mix well. For example, sugar
dissolves easily in water and can be transported by it
quite readily because both have this property of being
like little magnets that attract one another. Molecules
without this property, like vegetable oil and butter, do
not dissolve well in water and are not easily transported
by water because the two are not attracted to one
another. This is why many people say “like-dissolveslike”—molecules with similar properties dissolve
in one another and are easily transported together
whereas molecules with very different properties
(like oil and water) just don’t mix well at all.
Chapter 2 The Cell and
Its Components
Answers to Learning Check Questions
(Student textbook page 63)
1.Sketches should resemble the centre of Figure 2.4 on
page 60 of the student textbook. Sample caption: The
double membrane system of the nucleus, the nuclear
envelope, surrounds the nucleoplasm (a thick, complex
solution of proteins and nucleic acids that contains the
chromosomes). Although only visible in dividing cells,
the chromosomes are composed of about an equal
mass of DNA and protein. A nucleolus is often visible
Biology 12 Answer Key Unit 1 • MHR TR 13
as a dense region containing RNA and other proteins.
Nuclear pore complexes stud the nuclear envelope and
act as gateways for the passage of materials into and
out of the nucleus.
2.These differ from one another structurally by the
presence (rough ER) or absence (smooth ER)
of ribosomes on the cytoplasmic surface of the
membrane. Both are a complex set of highly folded
membranes interconnected with one another and
with the outer membrane of the nuclear envelope.
The rough ER is involved in the production of
proteins for the endomembrane system. It not only
synthesizes proteins for use by the endomembrane
system organelles, but also synthesizes transmembrane
proteins and proteins for secretion from the cell (e.g.,
insulin, glucagon, digestive enzymes). The smooth ER
is involved in the modification of these newly made
proteins and their packaging into vesicles for transport
to a Golgi apparatus. Smooth ER also synthesizes
phospholipids, steroids, and other lipids.
3.The endomembrane system is a set of eukaryotic
membranes and organelles that are either directly
connected to one another or engage in the vesiclemediated transport of substances between one
another. The endomembrane system includes the
nuclear envelope, endoplasmic reticulum (both
smooth ER and rough ER), Golgi, lysosomes, cell
membrane, and vesicles and vacuoles of many different
types. The functions of the endomembrane system
include the synthesis, modification, and transport of
proteins, and the compartmentalization of the cell
into function-specific and often materials-specific
membrane-bound enclosures. A good example of
the compartmentalization function is the formation
of lysosomes which contain many different enzymes
for degrading biomolecules. Those enzymes are
prevented from degrading everything in the cell by
being compartmentalized and by having an acidic pH
optimum (lysosomes are acidic compartments).
4.Peroxisomes are organelles containing oxidative
enzymes for the break-down of many types of
biomolecules. These organelles are also involved in the
synthesis of bile acids and cholesterol.
5.A vacuole is a large membranous compartment. The
main difference between a vacuole and a vesicle is size;
vesicles are much smaller. There are many different
types of vacuoles (and vesicles), but many are involved
in the storage of water, ions, and other molecules.
14 MHR TR • Biology 12 Answer Key Unit 1
6.Because it sorts and packages lipids and proteins, like
a post office does. The Golgi receives material from the
ER on the cis face in vesicles and sends the packaged
material in vesicles off the trans face.
(Student textbook page 67)
7.Both mitochondria and chloroplasts possess their
own DNA molecule(s) which encode some of their
own proteins. Both are also surrounded by a double
membrane system. Both organelles are also involved in
energy transduction and redox reactions. They differ in
the form of energy that is initially converted into usable
forms for the cell—chloroplasts convert the energy of
sunlight into energy-rich organic molecules, whereas
mitochondria convert the stored energy in energy-rich
organic molecules into other molecules that can act as
usable energy. Both organelles are similar and different
in terms of their location—mitochondria are found in
both plants and animal cells, but chloroplasts are found
only in plant cells.
8.The cell wall is a rigid layer surrounding plant (but
not animal) cells, composed of proteins and/or
carbohydrates. The cell wall provides structural support
for a cell as well as a measure of protection against
mechanical injury and invasion by other organisms.
9.The cytoskeleton is a network of protein fibres
extending throughout the cytosol that provides a cell
with internal structural elements that help support
the cell and determine its shape. The cytoskeleton also
provides for the movement and subsequent placement
of organelles in the cytoplasm, the contractile activity
of muscle, and the movement of cells through their
environment as components (microtubules) of flagella
and/or cilia.
10.The three main types of protein filaments of the
cytoskeleton are microtubules, microfilaments,
and intermediate filaments. All are involved in
maintaining cell shape, in addition to other functions.
Intermediate filaments provide no motility but
do provide a structure and support including the
internal framework supporting the nucleus and sites
of anchorage for some organelles. Microtubules and
microfilaments move organelles, and are involved in
the formation and dynamic activities of the spindle
used during cell division. They are also used in cilia
and flagella for moving cells through their environment
or sweeping material across the cell surface.
Microfilaments are used during muscle contraction
and cleavage furrow formation during cell division.
11.Cilia are relatively short structures composed of an
internal shaft made of microtubules, covered with
an outer membrane. Cilia move fluid and anything
in that fluid, for example, sweeping debris along and
out of airways, or helping move an egg through the
fallopian tube.
12.Flagella are longer structures but, like cilia, are
composed of an internal shaft made of microtubules,
covered with an outer membrane. Human sperm
possesses a single flagellum used to propel the sperm in
the female genital tract in search of an egg (oocyte).
(Student textbook page 70)
13.Answers should include any two of: physically
separate the contents of the cell from its aqueous
environment; act as a selective barrier for the passage
of some molecules through its lipid bilayer; serve
as a site for cell recognition events; serve as a site
for catalysis by many membrane-bound enzymes;
transporting specific molecules into and out of the
cell using transport proteins; and serving as a site for
communication between cells through signal reception
and transduction.
14.Lipids and proteins and their associated carbohydrates.
The carbohydrate content is covalently linked to some
membrane lipids and proteins, forming glycolipids and
glycoproteins, respectively.
15.If membranes were composed only of lipids they
would be expected to disallow the passage of nonlipid substances such as polar, hydrophilic molecules.
Monosaccharides and amino acids would not be able
to cross a lipid-only membrane. It is well known,
however, that such substances can cross membranes.
This fact is part of what led to the early hypothesis
that membranes also possess proteins (many of which
were later shown to be involved in the transport of
substances across membranes).
16.According to the fluid mosaic model, a membrane
is a fluid-like (dynamic) phospholipid bilayer with a
variety of proteins. Some of the proteins partially or
completely span the lipid bilayer (integral membrane
proteins) while others are associated with either face
of the lipid bilayer (peripheral membrane proteins).
Most of the lipid and protein molecules are able to
move quite freely in the membrane, with lipids having
limited motility within their layer of the membrane
and exchanging places millions of times per second,
and proteins moving laterally as components of the
membrane and others moving about at either surface.
17.The molecules (mainly phospholipids) that make up
their basic framework are not covalently bound to one
another but are instead associated with one another
through weak intermolecular interactions that are
readily made and broken.
18.When phospholipids are mixed with water, their polar
(hydrophilic) head groups orient themselves toward
the water molecules and cluster together while their
non-polar (hydrophobic) fatty acid “tails” cluster
together because of hydrophobic interactions. This
results in a bilayer of lipids with no free edge (they
are “self-sealing”) in which the polar head groups face
toward the aqueous environment and their non-polar
tails face each other on the inside of the bilayer.
(Student textbook page 74)
19.In a cellular context, a concentration gradient is a
difference between the concentration of a substance
on either side of a membrane.
20.Diffusion involves the random movements of a
substance in space with a net movement from a region
with the higher concentration to regions with lower
concentration. Diffusion occurs because all molecules
are in motion.
21.Answers should include any three of: temperature,
pressure, molecule size, polarity, and molecular charge.
The rate of diffusion is inversely related to molecule
size. The larger a molecule is, the more difficult it is for
it to diffuse across a membrane; although small polar
molecules can cross membranes, their rates of diffusion
are generally lower than those of non-polar molecules
of the same size; in general, charged molecules and
ions cannot diffuse across a cell membrane; and
temperature and pressure both increase the rate of
diffusion since they make molecules move faster.
22.Both involve the random movement that results in
net movement of molecules from a region of higher
concentration to a region of lower concentration
and both can involve movement from one side of a
membrane to the other. Osmosis, however, refers only
to the diffusion of water molecules.
23.Isotonic, since our cells are normally not crenated
(as they would be if the environment were hypertonic)
or swollen (as they would be if the environment
were hypotonic).
24.Hypotonic, since this condition fills the central vacuole
with water, providing rigidity (turgor) to the plant,
which is required to keep it upright.
Biology 12 Answer Key Unit 1 • MHR TR 15
(Student textbook page 77)
25.Both use membrane proteins and aid in the movement
of molecules across a membrane. The processes
differ in that facilitated diffusion moves molecules
down their gradients (“with” their gradients) in a
passive manner that requires no energy input. Active
transport, on the other hand, requires a net input of
energy to move substances against their gradients.
26.Both are integral membrane proteins used to transport
molecules across a membrane. Unlike channel
proteins, carrier proteins bind the molecules they
transport across the membrane. As a result, they
undergo a change in shape during transport and move
their solutes at much lower rates than channel proteins
do. Since channel proteins do not bind the molecules
they transport, very high rates of diffusion are possible
through channel proteins.
27.ATP is a nucleotide with three phosphate groups
and is used as the main source of energy in cells. In
active transport, the hydrolysis of ATP often provides
the energy necessary to move a substance against
its gradient.
28.One component of an electrochemical gradient is a
concentration gradient. It is different because it also
includes a charge difference across the membrane
(an electrical potential component). Thus, an
electrochemical gradient involves two differences
across a membrane—a concentration difference
and an electric potential difference.
29.In primary active transport, ATP hydrolysis is used
directly by a transport protein as a source of energy for
transporting a substance across a membrane against
its concentration gradient and/or against its electric
potential gradient. In secondary active transport
the energy released from the hydrolysis of ATP is
used indirectly to power the transport of a substance
against its gradient. The ATP hydrolysis performed by
a different (primary active) transport process sets up
an electrochemical gradient that the secondary active
transporter uses to power its transport process.
30.It is a primary active transporter that hydrolyzes one
molecule of ATP for every three sodium ions it pumps
out of a cell and every two potassium ions it pumps
in. Both ions are moved against their concentration
gradients. The pump works by undergoing a series of
shape changes powered initially by ATP hydrolysis and
then later by the binding or release of the ions and a
phosphate group that is transiently bound to the pump
during the pumping cycle. During the pumping cycle
16 MHR TR • Biology 12 Answer Key Unit 1
the E1 conformation of the protein releases K+, binds
Na+, and hydrolyzes ATP. In the E2 conformation,
the protein is transiently phosphorylated on its
cytoplasmic side, releases Na+ and binds K+.
Answers to Caption Questions
Figure 2.2 (Student textbook page 59): Chloroplasts and a
central vacuole. Chloroplasts use the energy from sunlight
to convert carbon dioxide and water into high-energy
organic molecules. The central vacuole is used to store
excess water which contributes to the turgor pressure used
by plant cells to maintain their rigidity. It also stores ions,
sugars, amino acids, and macromolecules, and has enzymes
that can break down a variety of macromolecules and
waste products.
Figure 2.14 (Student textbook page 73): The water level
in the tube rises because the tube has a solution with a
higher solute concentration than the solution in the beaker
and the two solutions are separated by a semi-permeable
membrane. The difference in solute concentrations results
in a difference in the concentration of water between the
tube (lower) and the beaker (higher). The situation results
in the net diffusion of water (osmosis) from the beaker into
the tube, causing the level of water in the tube to rise.
Figure 2.18 (Student textbook page 76): Because the
pump moves a dissimilar number of charges across the
membrane, with three positive charges being pumped
out for very two positive charges pumped in. The activity
results in a deficit of positive charge in the cell (a build-up
of negative charge).
Answers to Section 2.1 Review Questions
(Student textbook page 71)
1.Eukaryotic cells share some typical features, such as
a membrane-bound nucleus and a cell membrane
that contains the cell’s cytoplasm along with various
organelles. Although there is great diversity in form, size,
and features of all these cells, the yeast cell (a fungal cell),
plant cell, and animal cell are all eukaryotic.
2.Sample answer:
Cell Structure
or Region
Nuclear
envelope
Nuclear pore
Sketch
Function
Sketches should
resemble parts of
Figure 2.4 on page
60 of the student
textbook.
separates nucleus from
rest of cell
controls movement of
macromolecules, such
as RNA
Nucleolus
contains a select set of
genes (ribosomal DNA)
Chromatin
contain a complex of
DNA and proteins
Nucleoplasm
fills nucleus, providing
support & medium for
nuclear structures
3.Any two of: cellular organization, compartmentalization,
specialization, conducting life processes, containing
biochemical reactions, and storage
4.Venn diagram should show:
•Rough ER Only—studded with ribosomes; folds and
processes proteins made by its ribosomes
Cell Structure
or Organelle
•Rough ER and Smooth ER—part of endomembrane
system; connected to cell nucleus
•Smooth ER Only—smooth surface (has no
ribosomes); synthesizes lipids and lipid-containing
molecules; depending on the cell type, may
detoxify drugs or alcohol or produce testosterone
and estrogen
5.The endomembrane system performs a series of
related but separate functions vital to a eukaryotic cell’s
survival: the rough ER folds, processes, and packages
proteins; the smooth ER synthesizes lipids; and vesicles
transport proteins or lipids to the Golgi apparatus,
which further modifies, sorts, and packages these
molecules. Meanwhile, lysosomes (produced by the
Golgi apparatus) break down worn-out cell parts and
other materials while keeping the lysosomal enzymes
from affecting other cell components.
6.Flow charts should include: organelle is a membraneenclosed sac with enzymes → enzymes break down
(synthesize) molecules → new molecules produced
7.Sample answer:
Description
Function
Plant or Animal
Nucleus
Double-membrane-bound organelle
that contains chromatin
Stores and replicates cell’s genetic information
both
Endoplasmic
reticulum
Complex of membrane-bound tubules
and sacs; rough ER is studded with
ribosomes
Rough ER assembles membrane-proteins and proteins
for export; smooth ER synthesizes lipids and lipidcontaining molecules and detoxifies drugs and alcohol
both
Golgi apparatus Stack of curved membrane sacs
Stores, modifies, sorts, and packages proteins
both
Lysosomes
Membrane-enclosed sac of digestive
enzymes
Enzymes break down macromolecules into smaller
molecules for reuse by cell
animal (possibly
in some plants)
Peroxisomes
Membrane-enclosed sac of enzymes
called oxidases
Some enzymes break down alcohol; catalase breaks
down toxic hydrogen peroxide; in some cells, enzymes
help synthesize cholesterol and bile acids
both
Vesicles
Membrane-bound sacs
Transport and store substances
both
Vacuoles
Large, central vesicles (one per cell)
Control turgor pressure of plant cell; stores water, ions,
sugars, amino acids, and macromolecules; contains
enzymes that break down macromolecules and wastes
plant
Chloroplasts
Chorophyll-containing organelles with an Site of photosynthesis
inner membrane and stacks of thylakoids
(grana); contains some of its own DNA
Mitochondria
Organelles with smooth outer
membrane and folded inner membrane;
contains some of its own DNA
Break down high-energy organic molecules to convert
stored energy into usable energy
both
Cell wall
Rigid layer surrounding cell; composed
of cellulose and other substances
Provides protection, shape, and structural support
plant
Cytoskeleton
Internal network of protein fibres; in
some cells, forms extensions such as
cilia and flagella
Provides structure; anchors cell membrane and
organelles; provides tracks for organelles to move along;
in some cells forms cilia and flagella for movement
both
Cell membrane
Phospholipid bilayer that surrounds
cytoplasm
Separates cell’s contents from its surroundings
both
plant
Biology 12 Answer Key Unit 1 • MHR TR 17
8.Those that require the most energy (e.g., muscle cells).
9.Peroxisomes in liver cells support the functions of the
liver in breaking down toxic substances, such as alcohol,
and in helping to synthesize cholesterol and bile acids.
10.The cell membrane provides a dynamic, selective,
physical barrier between the cell’s contents and the
extracellular fluid. This barrier is essential in order to
regulate the passage of molecules and ions into and out
of the cell. Without this regulation, harmful substances
could enter freely and substances important in cellular
processes could leak out. In short, cellular processes
would fail and the cell would die.
11.Tables should include:
•Phospholipid bilayer—Each layer (or leaflet)
contains phospholipids embedded with proteins. The
hydrophobic “tails” of the phospholipids face each
other while the hydrophilic “heads” face out, toward
the cytoplasm or extracellular fluid.
•Fluid membrane—Phospholipids can move about
freely, thus can rearrange themselves to seal ruptures
in the cell membrane.
•Mosaic of proteins and other molecules—Proteins
and other molecules form a floating mosaic on or
in the membrane: peripheral proteins are bound
to the cytoplasmic or external surface; integral
proteins reach through the membrane. Glycolipids
and carbohydrate chains are attached to some
membrane proteins.
12.Sample answer: the phospholipid bilayer is like a lake,
and membrane proteins are like boats, docks, and
buoys on the lake.
13.The mitochondria (in red) are recognizable by their
round to oval shape and inner folds. The endomembrane
system (in blue) can be identified by the stacks of
sac-like structures, which may be a combination of the
endoplasmic reticulum and Golgi apparatus. The smaller
structures in blue are likely vesicles travelling between
the Golgi apparatus and other areas of the cell. The
material in green is difficult to identify, but students
may suggest that it is part of the cytoskeleton.
Answers to Section 2.2 Review Questions
(Student textbook page 81)
1.When a difference in concentration for a solute exists
on opposite sides of a membrane, the net direction of
movement for water (the solvent in which the solute is
dissolved) will be from the side of the membrane with
the lowest solute concentration toward the side with
the highest solute concentration.
18 MHR TR • Biology 12 Answer Key Unit 1
2.See top left part of Figure 2.15 on page 73 of the
student textbook.
3.Any two of:
•facilitated diffusion of ions or polar molecules
through channel proteins, or molecules such as
glucose or amino acids via carrier proteins
•active transport of solutes against their concentration
gradient, for example, via ion pumps: primary active
transport directly uses ATP; secondary active transport
uses the energy of an electrochemical gradient
•membrane-assisted transport, specifically
endocytosis, during which the cell engulfs material:
phagocytosis brings in discrete particles; pinocytosis
brings in liquid; receptor-mediated endocytosis uses
receptor proteins to bind to target molecules and
bring them in
4.The 5% NaCl solution is likely hypertonic relative
to the plant cell, thus the central vacuole will lose
water to the salt solution surrounding the cell,
and the cytoplasm will shrink, a condition known
as plasmolysis.
5.The independent variable in each case will be the solute
concentration surrounding egg membrane. A method
of observing the effects of transport through the egg
membrane would be the volume of the egg. The use of
a coloured solute may also help.
6.A channel protein is a highly specific channel through
the cell membrane that allows for the facilitated
diffusion of certain ions and polar molecules that
could not otherwise easily diffuse through the
phospholipid bilayer. Some channel proteins remain
open all the time, while others open and close in
response to signals.
7.Cholesterol helps to maintain the proper fluidity of a
membrane over a wide range of temperatures. At low
temperatures it helps to increase membrane fluidity by
preventing the close packing of phospholipids. At high
temperatures, when a membrane might become too
fluid, cholesterol helps to reduce membrane fluidity
by increasing the intermolecular forces that hold the
membrane together.
8.Venn diagrams should show:
•Endocytosis Only—In endocytosis, cell membrane
engulfs discrete particles (phagocytosis), liquids
(pinocytosis), or specific molecules bound in
clathrin-coated pits (receptor-mediated endocytosis).
In exocytosis, vesicles fuse with cell membrane and
empty contents outside cell.
•Endocytosis and Active Transport—Requires
energy from cell. Moves a substance against its
concentration gradient.
•Active Transport Only—Primary active transport
directly uses ATP; secondary active transport uses the
energy of an electrochemical gradient. Protein pumps
transport ions or molecules across cell membrane.
9.Size, polarity, and charge of substances affect their rate
of diffusion through a membrane. Smaller molecules
can generally cross a membrane with greater ease
than larger molecules. Non-polar molecules can pass
through the non-polar interior of a membrane more
readily than polar molecules. Because of the non-polar
interior of a membrane, charged molecules and ions
generally cannot diffuse unaided across a membrane
at significant rates.
Answers to Chapter 2 Review Questions
(Student textbook pages 89–93)
1.e
2.d
3.b
4.c
5.a
6.d
7.b
8.e
9.c
10.e
11.a
10.No, because harmful materials would be able to
enter the cell, nutrients and important ions would
leak out of the cell, and the cell could burst if in a
hypotonic environment or lose too much water if in
a hypertonic environment.
12.b
11.No, because all molecules are in constant motion.
Even when equally distributed across a membrane,
molecules will continue to move in and out of cells.
However, there will be no net change in concentration
on either side of the membrane.
16.Sample answer: The rough ER makes the proteins
secreted from a cell and proteins that are parts of
membranes. Free ribosomes make the proteins that
function in the cytosol.
12.Phospholipids provide a fluid bilayer that allows
some small, non-polar molecules to diffuse through
its hydrophobic interior. Ions and large molecules
need the assistance of membrane proteins (channel
proteins, carrier proteins, or pumps) to pass through
the hydrophobic interior of the membrane. In general,
non-polar amino acids coat the exterior portion of
membrane proteins (the portion that interacts with
the non-polar interior of the cell membrane). The
interior of a membrane protein (the portion that faces
the particle being transported) may be of a specific size
and shape and may have polar or charged amino acids
that interact with the particle being transported. Active
transport pumps undergo shape changes in order to
move substances across the membrane.
13.The pond alga is likely using active transport to pump
ions in, against the concentration gradient. As a result,
the concentration of ions (particularly chloride ions
and potassium ions) in the cell is much higher than in
the pond water.
13.d
14.c
15.b
17.The length of the fatty acid “tails” and the presence of
double bonds in the fatty acid tails affect membrane
fluidity. Membrane fluidity is decreased by longer tails
since they have more atoms, and thus there are more
intermolecular interactions holding the molecules
together. Membrane fluidity is increased by the
presence of double bonds since they result in “kinks”
or bends in the fatty acid tails that prevent them
from packing closely together (this allows a looser
arrangement of phospholipids in the bilayer).
18.This protects a cell. If a lysosome were to break inside a
cell and spill its digestive enzymes into the cytosol, very
little damage would be done since the enzymes don’t
work well at the neutral pH of the cytosol.
19.Any three of:
•Size—Smaller molecules diffuse more quickly
through a membrane.
•Polarity—Since the interior of a membrane is mostly
made up of the hydrophobic, non-polar fatty acid
tails of phospholipids, non-polar molecules are
able to diffuse through a membrane more readily.
However, if small enough, some polar molecules may
pass through readily while larger polar molecules
tend to be excluded.
Biology 12 Answer Key Unit 1 • MHR TR 19
•Charge—Charged molecules or ions are repelled by
the hydrophobic, non-polar interior of a membrane,
so they are almost completely unable to diffuse
through it.
•Temperature—As temperature increases, the rate
at which substances diffuse increases, since the
molecule are moving more quickly, making the
membrane more fluid.
•Pressure—Increased pressure makes molecules
vibrate faster and may also force the molecules across
a membrane.
•Concentration—As the difference in concentration of
a substance increases across a membrane, its rate of
diffusion also increases.
20.a.flagellum
b.Golgi apparatus
c.thylakoids
21.a.H
b.L
c.L
d.H
e.H
f.L
g.L
h.H
Note that e/f and g/h could be reversed as long as one
pair is moving from high (H) to low (L) and the other
pair from L to H.
22.Primary active transport directly uses energy released
by the hydrolysis of ATP. Secondary active transport
uses energy released by an electrochemical gradient.
Answers may include that the original source of the
electrochemical gradient used by a secondary active
transport mechanism is often one or more primary
active transport processes.
23.An electrochemical gradient is a combination of two
gradients: an electric potential involving a difference
in charge across a membrane and a concentration
gradient involving a difference in the concentration
of a substance across a membrane. A concentration
gradient has only one component: a difference in
concentration of a substance across a membrane. If the
substance is charged, the concentration gradient may
produce an electric potential. The difference between
these two types of gradients is the presence or absence
of an electric potential across a membrane.
20 MHR TR • Biology 12 Answer Key Unit 1
24.Unlike channel proteins, carrier proteins bind the
solute molecules that they transport and undergo a
change in shape during the transport process. Since a
carrier must bind its solute molecules, can only bind
one or sometimes a few solute molecules at a time,
and must undergo a shape change, it has much lower
rates of transport than channel proteins do. Channel
proteins do not need to bind their solutes or undergo
a change in shape (except those that are gated) and so
can allow very rapid passage of their solutes from one
side of a membrane to the other.
25.The cell wall, chloroplasts, and the large central
vacuole are found only in plant cells. The cell wall and
central vacuole provide rigidity and structure, while
chloroplasts use light energy to convert carbon dioxide
and water into organic molecules.
26.Membrane proteins and phospholipids interact with
one another through weak intermolecular forces
that allow for their lateral movement. This fluid
characteristic of membranes is important for the
many functions of a membrane that require it to be
deformable (for example, the movement of a flagellum
or cilium, cell division, exocytosis, endocytosis, and
other vesicle formation, etc.). This characteristic also
allows a membrane to immediately repair small tears
and allows membrane proteins to undergo the required
shape changes.
27.d.During their synthesis, proteins are inserted into
the lumen of the rough ER.
e.Proteins are packaged into vesicles formed at the
rough ER.
b.Vesicles merge with the cis face of the
Golgi apparatus.
f.Proteins are modified in the Golgi apparatus.
a.Proteins are packaged into vesicles that form at the
trans face of the Golgi.
c.Vesicles fuse with the cell membrane and release
proteins by exocytosis.
28.The cell would shrink (crenate). Since the concentration
of solutes in the solution is much higher than inside the
cell, water will tend to move out of the cell.
29.Smooth ER, because rough ER is specialized for the
synthesis of proteins, whereas smooth ER can be
specialized for the synthesis of lipids.
30.Since the pump moves three sodium ions out of the
cell for every two potassium ions pumped into the cell,
the pump creates a greater concentration of positive
charge outside the cell than inside the cell, as well as a
greater concentration of sodium ions outside the cell
than inside and a greater concentration of potassium
ions inside the cell than outside. This electrochemical
gradient, or electric potential, is a source of potential
energy that can be used by the cell to do work. In
addition to the electric potential, the pump also
produces concentration gradients for both of the ions
(a higher concentration of sodium ions outside the cell
than inside and a higher concentration of potassium
ions inside the cell than outside of it). The cell can
use the resulting electrochemical gradients for doing
the work of secondary active transport. Just as plant
cells use an electrochemical gradient of protons for
importing sucrose using the H+/sucrose symporter,
animal cells can use Na+ gradients for the secondary
active import of glucose and many other substances
into cells.
31.a.Hypotonic, since under those conditions the plant
cells swell with water and turgor pressure builds.
b.Less able, since the concentration of solutes in the
external environment would be higher than normal.
This would change the environment from hypotonic
towards isotonic or even hypertonic and result
in less water inside the cell and thus less turgor
pressure to support the plant.
32.Although many polar molecules can diffuse through
lipid bilayers, the rates at which they do so may be too
low to support the activities of a cell. In such cases it
would be beneficial for a cell to have carriers for these
types of molecules to allow the molecules a greater
rate of entry or exit as needed by the cell. (When rates
of diffusion through a lipid bilayer alone are very low,
we often describe such molecules as being unable to
diffuse freely through the bilayer. In actual fact, almost
all molecules are able to diffuse through a lipid bilayer,
but many do so at rates so low that we consider the
bilayer to be impermeable to them.)
33.Mix cells with fluorescently-tagged antibody for
an integral membrane protein. → Observe by
fluorescence microscopy: the surface of the cells should
be uniformly coloured. → Photobleach a small spot on
the surface of a cell using a laser. → Observe the small
dark region created by photobleaching. → Continued
observation (at a temperature that permits the
membranes to be in the fluid state) should reveal that
the small dark photobleached area(s) disappear over
time, indicating that the other fluorescently labelled
proteins were able to move laterally into the area that
was treated with the laser.
34.The cold-tolerant plants are able to change the structure
or proportion of phospholipids in their membranes as
required at different temperatures. Chemical analysis of
the membranes should reveal that cold-tolerant plants
have a greater proportion of phospholipids with short
fatty acid tails and/or a greater proportion with double
bonds in their fatty acid tails.
35.Every 10 minutes for an hour, collect identical cells
grown in the absence of radioactive amino acids.
Examine the cells using a microscope (or by placing
them on an X ray film) for the presence of radioactive
proteins first in the rough ER and then later in the early
(cis) compartments of the Golgi apparatus, and much
later near the trans face of the Golgi apparatus.
36.Sample answer:
•carbohydrate chains for membrane glycoproteins are
added to proteins within ER or Golgi apparatus
•vesicles surround modified proteins and pinch off
from Golgi apparatus
•vesicles transport modified proteins to cell membrane
•vesicle fuses with cell membrane so that inner
surface of vesicle becomes the outer surface of the
cell membrane
•carbohydrate chains stick outward
37.Molecule A is probably transported by a channel
protein whereas molecule B is probably transported by
a carrier protein. Since channel proteins do not bind
the molecules they transport, they are able to transport
solutes at much greater rates than carrier proteins are
able to do.
38.Tables should include any three of:
•Cell recognition—Membrane glycoproteins possess
carbohydrate chains that can be recognized by
other cells.
•Transport—Transport proteins, including channel
proteins, carrier proteins, and pumps, allow for the
passage of specific substances across membranes that
would otherwise cross membranes too slowly or not
at all.
•Catalytic reactions—Many membrane proteins are
enzymes that catalyze specific chemical reactions
at membranes.
•Signal reception—Receptor proteins in cell membranes
bind to signal molecules, such as hormones.
•Signal transduction—The change in shape of receptor
proteins (when they bind signal molecules) initiates a
cellular response to the signals.
•Anchoring to cytoskeleton—Peripheral proteins and
some integral proteins help stabilize membranes and
link them to the cytoskeleton.
Biology 12 Answer Key Unit 1 • MHR TR 21
39.Phospholipids have both polar head groups and
non-polar fatty acid tails. When placed in water,
the non-polar fatty acid tails cluster together due to
hydrophobic interactions. At the same time, the polar
head groups cluster together, facing the polar water
molecules. Diagrams should look like Figure 1.15 on
page 23 of the student textbook.
40.The statement refers to two types of endocytosis, a
process in which molecules are brought into cells.
Pinocytosis is non-specific in that specific molecules
are not selected and bound by receptors. Receptormediated endocytosis involves the use of receptor
proteins that bind specific molecules and bring them
into the cell.
41.Venn diagrams should show:
•Pinocytosis Only—Imports small dissolved particles
and liquid
•Pinocytosis and Phagocytosis—Bring material into
cell by endocytosis; membrane-assisted transport;
require energy from cell
•Phagocytosis Only—Imports large particulate matter,
even whole cells
42.Any five of:
•diffusion through the lipid bilayer
•facilitated diffusion (or passive transport) by a
channel protein
•facilitated diffusion (or passive transport) by a
carrier protein
•active transport or a membrane pump
•pinocytosis
•phagocytosis
•receptor-mediated endocytosis
•endocytosis (instead of the pinocytosis, phagocytosis,
or receptor-meciated endocytosis)
•primary or secondary active transport
43.Concept maps should have a title such as Entry of Polar
Molecules Across a Lipid Bilayer. The terms diffusion,
transport protein, and membrane-assisted transport
would be appropriate for higher-order labels. Branches
from these terms should show diffusion through
a membrane contrasted with facilitated diffusion
(involving proteins), channel proteins contrasted with
carrier proteins, passive transport contrasted with
active transport, and endocytosis contrasted with
exocytosis. Additional branches could contrast primary
and secondary active transport or the binding of solute
to carrier proteins versus the flow of solute molecules
through channel proteins.
22 MHR TR • Biology 12 Answer Key Unit 1
44.Articles should make it clear that red blood cells will
crenate in a hypotonic solution and swell or burst in
a hypertonic solution. Check for a creative title and
story line and complete sentences organized in a logical
flow. The article should lead with the most important
information, or big picture, before getting into detail.
45.Venn diagrams should show:
•Channel Proteins Only—Do not bind to their
transported solute molecules; look like a tunnel
through membrane; some are always open and some
are gated; high rates of diffusion are possible
•Channel Proteins and Carrier Proteins—Allow for
the transport of ions or polar molecules; specific for
the solute molecules they transport; non-polar amino
acids usually found at exterior of protein (interact
with non-polar interior of membrane); interior of
channel or solute binding site usually composed of
polar or charged amino acids; defective proteins can
cause many diseases
•Carrier Proteins Only—Bind their transported solute
molecules; undergo a change in shape after binding
solute; not gated; provide lower rates of diffusion
46.Venn diagram should show:
•Passive Carriers Only—Move particles down their
concentration gradients; does not require expenditure
of cell’s energy
•Passive Carriers and Active Carriers—Allow for
the transport of ions or polar molecules across cell
membrane; specific to the particles they transport;
non-polar amino acids usually found at exterior
of protein (interact with non-polar interior of
membrane); interior of channel or solute binding site
usually composed of polar or charged amino acids;
bind their transported solute molecules; undergo a
change in shape after binding solute
•Primary Active Carriers Only—Move particles against
(up) their concentration gradients; requires direct
expenditure of energy, usually from ATP hydrolysis
47.Any three of:
•a change in electric potential (or electric charge)
•binding of a hormone or other molecule
•pressure
•light
48.Venn diagram should show:
•Chloroplasts Only—Inner membrane surrounds
stacks (granum) of flattened disc-like thylakoids;
thylakoid membranes contain chlorophyll; stroma
are fluid-filled space contained by inner membrane;
synthesizes energy-rich organic molecules from
carbon dioxide and water using the energy of light;
found in cytoplasm of photosynthetic cells
•Chloroplasts and Mitochondria—Surrounded
by a double membrane (outer and inner); inner
membrane surrounds a fluid-filled space; involved
in conversion of energy from one form to another;
found in cytoplasm of plant cells
•Mitochondria Only—Inner membrane is highly
folded; folds are called cristae; fluid-filled space
contained by inner membrane is the matrix; convert
energy stored in high-energy organic molecules
into useable energy for the activities and chemical
reactions required by cells; found in cytoplasm of
plant and animal cells
49.Sample answer: Polypeptides are produced by rough
ER and put into the lumen. → Polypeptides are
processed in rough ER; smooth ER synthesizes lipids.
→ Pieces of ER pinch off to enclose the protein in a
vesicle. → Vesicle goes to cis face of Golgi apparatus
and fuses with the membrane. → Protein is modified
and/or stored. → When needed, the protein is
enclosed in a vesicle formed at the trans face of
the Golgi apparatus. → The vesicle transports the
modified protein to another location in the cell or to
the cell membrane.
50.Graphic organizer such as a spider map, fishbone
diagram, or a concept map should identify key
concepts and relationships from the Chapter 2
Summary, such as the characteristics and organelles
of eukaryotic cells (2.1) and passive and active forms
of transport across the cell membrane.
51.a.Batch A
b.Batch A
c.Cholesterol helps preserve normal membrane
fluidity at temperature extremes. At high
temperatures that would otherwise increase
membrane fluidity, cholesterol helps resist the
change in fluidity by increasing the intermolecular
forces between membrane molecules. The
membranes without cholesterol would become
much more fluid and perhaps even disintegrate. At
low temperatures that would otherwise decrease
membrane fluidity, cholesterol helps the membrane
resist the change in fluidity by preventing the close
packing of membrane lipids. The membranes
without cholesterol would become much less fluid
and perhaps even become non-functional gels.
52.a.Facilitated diffusion involving a carrier protein.
b.Initially the rate of diffusion increased with increasing
concentration of the molecule. Since there can only be
a finite number of carrier proteins in a membrane for
any type of molecule, as the concentration increased,
a point was reached where all of the carrier proteins
were busy transporting the molecules and so others
had to wait their turn. It is at this point when the
steady linear increase in rate began to slow down and
plateau at high concentrations.
c.A continuous linear relationship, even at very high
concentrations of the molecule, would indicate
that the rate of entry is not limited by the number
of entry sites in the membrane. The most likely
explanation would be unaided diffusion through the
lipid bilayer or perhaps facilitated diffusion through
a non-gated channel protein.
53.C and D. Unlike the first two processes, these are
examples of primary active transport that depend
directly on the availability of useable energy in the cell.
Since mitochondria provide useable energy for cells,
their improper functioning could have a direct effect
on these types of processes.
54.Yes. Even when concentration of water is equal on
both sides of a membrane, there is still movement of
water across the membrane since all molecules are in
constant motion. There is, however, no net movement
of water across the membranes.
55.Integral proteins are embedded in the lipid bilayer (and
typically expose part of themselves at each face of a
membrane), whereas peripheral proteins are loosely
associated with one or the other leaflet of a membrane.
56.Peripheral proteins, since they are only loosely attached
to one or the other face of a membrane through weak
interactions, whereas integral proteins are embedded
in the lipid bilayer.
57.The receptor proteins must be integral since the
question states that they have a region at the exterior
side of the membrane and a region at the cytosolic side
of the membrane. Proteins that are embedded in the
membrane are integral membrane proteins. Peripheral
membrane proteins are those that are loosely attached
at either face of a membrane.
58.Since the protein was only released by destroying the
integrity of the lipid bilayer, the protein was likely an
integral protein, embedded in the bilayer rather than
loosely attached to either face of the bilayer.
59.Answers should include: Osmosis across a
semipermeable membrane allows the passage of
water but not solute molecules (urea in this example).
The left chamber has a higher concentration of urea
molecules (hypertonic) than the right chamber
(hypotonic). Water moves in both directions across the
membrane but the arrows indicate the net gain of water
into the left chamber and the net loss of water from the
right chamber.
Biology 12 Answer Key Unit 1 • MHR TR 23
60.a.an electrochemical gradient
b.Yes, since making the gradient requires an input
of energy.
c.A primary active transport process such as an
ATP-hydrolysis driven hydrogen ion pump could
pump protons from the cytosol into the interior
of a lysosome, thus lowering the lysosomal pH.
Diagrams should look like the left side of Figure 2.19
on page 77 of the student textbook.
Answers to Chapter 2 Self-Assessment
Review Questions
(Student textbook page 94–5)
1.c
2.e
3.d
4.b
5.d
6.c
7.d
8.b
9.c
10.b
11.Red blood cells (erythrocytes) lack nuclei and their
chromosomes, which is genetic information required
for reproduction.
12.Because cholesterol both increases (at low
temperatures) and decreases (at high temperatures)
membrane fluidity.
13.Paramecia need contractile vacuoles to push water out
and prevent swelling or even bursting (lysing) caused
by the natural tendency for (hypotonic) water to enter
the cell via osmosis.
14.The lysosomes are not performing their function
of breaking down lipids, possibly due to missing or
defective digestive enzymes, causing a build-up of
lipids would stored inside the lysosomes.
15.That lipids are a component of the cell membrane,
since lipid-soluble molecules would be unlikely to
pass through a hydrophilic membrane as easily.
16.Venn diagram show:
•Plant Cells Only—Central vacuole is a large vesicle
that stores water, ions, sugars, amino acids, and
macromolecules; central vacuole contains enzymes
for breaking down waste; central vacuole plays role
in turgor pressure
24 MHR TR • Biology 12 Answer Key Unit 1
•Plant and Animal Cells—Vesicles are membranebound, sac-like organelles; transport and store
substances; can fuse with cell membrane and other
organelle membranes
•Animal Cells Only—Typically contain many
small vesicles
17.Active transport requires ATP to move substances
against their concentration gradient across the cell
membrane. For ATP to release energy for active
transport, it must undergo hydrolysis, that is, it is split
into adenosine diphosphate and a phosphate group
with the addition of a water molecule.
18.The presence of a channel protein for water, since
facilitated diffusion is likely involved when molecules
cross the cell membrane at a faster rate than is possible
by simple diffusion.
19.In summer, the cell membranes must contain more
cholesterol since higher temperatures would result
in less fluidity of cell membranes (due to increased
intermolecular forces and therefore tighter packing
of cholesterol).
20.a.Ribosomes are present on the rough ER, not the
smooth ER.
b.What is the role of ribosomes on the rough ER?
c.Proteins that are part of membranes, or are for
export, are assembled on the ribosomes of the
rough ER.
21.a.
extracellular fluid
small non-polar
molecules (O2)
ions (Na+, K+, Cl-),
smaller polar molecules
channel protein
ions, small and large
molecules (glucose,
amino acids)
cytoplasm
ATP
ADP + Pi
H+
ions (Na+, K+, H+)
carrier protein
ion pump
b.Small, non-polar molecules can cross directly
through the bilayer; polar molecules and ions can
cross where the membrane proteins are located.
22.The surface area of the cell increases during exocytosis
because the vesicles that transport materials for
secretion fuse with the cell membrane, adding to its
surface area.
23.It seems that the bacteria interfere with the protein
fibres of the cytoskeleton that are responsible for
maintaining cell shape.
24.Components destined for secretion would have to be
tagged with the fluorescent labels. Calcium ions could
be injected into the cells, which could be monitored
with a microscope. The movement of the fluorescent
dye out of the cells would indicate exocytosis.
25.Muscle cells require large numbers of mitochondria
to supply sufficient energy for contraction. If the
mitochondria are not functioning and not supplying
energy to the muscles, they will become weak.
Answers to Unit 1 Review Questions
(Student textbook pages 99–103)
1.a
2.e
3.c
4.e
5.e
6.c
7.e
8.e
9.c
10.d
11.a
12.e
13.d
14.a
15.e
16.Passive transport refers to the movement of solutes
down their concentration gradients across a
membrane, without the need for energy input.
17.Channel proteins do not bind their solutes but provide
a passageway for their rapid diffusion from one side
of a membrane to another. Carrier proteins have
binding sites for their solutes and undergo a change
in shape as they move their solutes from one side of
a membrane to the other. While all channel proteins
are passive, carrier proteins may be either passive or
active depending on the specific protein. The carriers
involved in active transport use an energy input to
pump their solutes against their gradients.
18.a.Since the reaction is shown going from left to right,
the reactants are on the left and include a glycerol
molecule and three fatty acid molecules.
b.The product is a triglyceride (fat) molecule, a type of
lipid, plus three molecules of water.
c.condensation reactions
19.• Polysaccharides—monomers are monosaccharides
•Polypeptides (proteins)—monomers are amino acids
•Polynucleotides (nucleic acids such as DNA and
RNA)—monomers are nucleotides
20.• Condensation reactions—formation of a disaccharide
(and a water molecule) from two monosaccharides
•Hydrolysis reactions—formation of two
monosaccharides from a disaccharide (and a
molecule of water)
•Redox reactions—cellular respiration that uses
glucose and oxygen to produce water and carbon
dioxide (and energy); this may be separated into the
reduction and oxidation reactions
•Acid-base, or neutralization, reactions—a reaction
between an acid and a base to produce a salt and water
21.Plasmolysis occurs when a plant cell is placed into a
hypertonic environment. In such a solution, the cell
will suffer a net loss of water (mostly from the central
vacuole which is responsible for the majority of the
cell volume). Because of the fairly rigid cell wall, the
exterior shape of the cell remains relatively unchanged,
but the cell membrane will separate from the inner
aspect of the cell wall as the cytoplasm shrinks
in volume.
22.The passageway (channel) provided by a channel
protein has a roughly tubular shape. The shape and
size of the channel largely determine the shape and
size of molecules that are allowed to pass through
it. This helps to regulate the entry (and exit) of
solute molecules.
23.Both glycogen and cellulose are polysaccharides built
from glucose monomers. Glycogen has a branched,
helical structure whereas cellulose is a linear,
unbranched polymer.
24.Any two of:
•The joining of two monosaccharides (reactants) to
produce a disaccharide (product) and a molecule
of water.
•The joining of two nucleotides (reactants) to produce
a dinucleotide (product) and a molecule of water
•The addition of one or more fatty acids to a glycerol
molecule to produce a mono-, di-, or triglyceride
25.phagocytosis
26.Cell membrane, cytoplasm, and DNA within a
membrane-bound nucleus
27.A transport method in which a vacuole fuses with the
cell membrane and releases its contents outside the cell.
Biology 12 Answer Key Unit 1 • MHR TR 25
28.Proteins made by ribosomes of the rough ER are
destined to be part of the endomembrane system or
exported from the cell, whereas proteins made by free
ribosomes are destined to function in the cytosol,
nucleus, and nucleolus.
29.Any two of:
•Pinocytosis occurs when the cell takes in water and
its solutes.
•Phagocytosis occurs when the cell takes in
large particles.
•Receptor-mediated endocytosis occurs when receptor
proteins in the cell membrane bind to specific
molecules outside the cell.
30.Both intramolecular and intermolecular forces are
responsible for the 3-D shape of molecules and
thus their functions inside (and outside) cells. The
forces determine how molecules interact (i.e., which
molecules are attracted to one another and which are
repelled) and which portions of molecules are attracted
or repelled by other substances.
31.The total concentration of solutes in the cell or what
osmotic concentration of a solution would result in no
net gain or loss of water from the cell.
32.The structures in the rectangle are nuclear pore
complexes (NPCs). NPCs allow the free passage of
small molecules such as water and ions but their pore
openings can be regulated to allow for the passage of
much larger molecules such as RNAs.
33.The fact that both molecules give the same result
with the same reagent indicates that the molecules
share a common structural feature. In this case, the
common structural feature is the presence of glucose
(lactose is a disaccharide consisting of galactose and
glucose). Benedict’s solution detects the presence of
monosaccharides and some disaccharides.
34.Both fats and oils are lipids since they are composed
of C, H, and O with a high proportion of non-polar
C-H bonds that make the molecules hydrophobic.
Fat molecules are triglycerides containing saturated
fatty acid chains and are generally solids at room
temperature. Oils are triglycerides containing
unsaturated fatty acid chains and are generally liquids
at room temperature.
35.Fill the dialysis tubing bag with water, and fill a beaker
with salt. The water in the bag will leave as the differing
concentrations move toward balance.
36.Primary active transport moves the transported
substance(s) against a concentration gradient and
requires the direct input of energy, usually from the
26 MHR TR • Biology 12 Answer Key Unit 1
hydrolysis of ATP. Secondary active transport uses an
electrochemical gradient to drive the transport of one
substance against its gradient while another substance
is moved down its gradient. It is this downhill
movement of one solute that provides the energy for
the movement of the other solute against its gradient.
37.Sample answer: The alphabet, words, a sentence, and a
paragraph. If we think of the 26 letters of the alphabet
and compare those to the 20 or so common amino
acids, then words are like the primary structure of a
protein, its linear order of amino acids. The secondary
structure, the formation of regular repetitive structure
such as alpha helices and beta pleated sheets are like
words or phrases that we commonly encounter when
reading an article or story. When specific sentences
are constructed with a specific meaning, it is like
the tertiary structure of proteins—their overall shape.
Finally we can think of the quaternary structure of some
proteins, the construction of a protein from more than
one polypeptide chain, as being like the construction of
a paragraph composed of multiple sentences. It is only
when the entire paragraph is read that we understand its
full meaning, just as when the various chains of a protein
with quaternary structure come together to form a single
intact functioning protein.
38.With a malfunction that prevented transport vesicle
formation, the endomembrane system would be
severely affected. The endomembrane system depends
on the formation and movement of vesicles for the
transport of substances to specific organelles, the
export of some substances from the cell, and the
development of the endomembrane system organelles
themselves. Without the ability to form transport
vesicles, the normal functioning of a eukaryotic cell
would be greatly affected and probably result in its
eventual death.
39.Since molecular chaperones help many proteins fold
properly into their final three-dimensional shape,
a disorder that altered the function of molecular
chaperones could have drastic effects for many other
proteins in a person’s body. If a protein were unable
to fold properly, its function would be changed or
eliminated. Depending on the importance of the
protein, it could have severe consequences not only to
the function of particular proteins but on the ability of
cells and even the whole person to function properly.
40.If hydrogen bonding did not occur, double-stranded
DNA molecules could not be formed. DNA molecules
depend on hydrogen bonding between complementary
bases to form their double-stranded structure.
41.If the activity of a bacterial enzyme was shown to be
present only in bacteria, a search could be made for
molecules that inhibit its activity. Such molecules
might then be used as antibiotics. Similarly, if the
enzyme activity was present in both human cells and
bacterial cells, a search could be made for molecules
that inhibited the bacterial enzyme activity only, or
preferentially, in comparison to the human enzyme.
Those molecules might also be used as antibiotics.
42.No, biological membranes also contain other lipids
such as cholesterol in addition to a great many proteins.
Many of the phospholipids and proteins also have
carbohydrate attached to them, so a diagram would need
to be considerably more complex to properly represent
the molecules normally present in a membrane.
43.Sample answer: The transport function, since excluding
them would only allow a limited subset of solute
molecules to cross from one side of the membrane to
the other. Cell recognition, so that proteins would not
be recognized as foreign if the membrane was going
to be used in living organism like humans. Catalysis,
because including specific enzymes could help catalyse
specific reactions.
44.Hypertonic, to make the grapes shrivel (lose water)
into raisins. Problems could involve the use of a too
concentrated solution or exposure of the grapes to the
solution for too long. Both of which could produce
“dry” raisins that have lost too much water. Care would
also need to be taken in the choice of solutes used since
some could cause discolouration, poor taste, or other
problems in the final product.
45.a.channel protein; a roughly cylindrical passageway
through the protein could be used for the movement
of solutes
b.carrier protein; it appears to be binding (but not
altering) a solute that is present on both sides of
the membrane
c.receptor protein; has a specific binding site for a
solute but does not appear to be transporting or
changing the solute
d.enzymatic protein; it is acting on a solute that is
changed into some other (product) molecule
46.Sample answer: A cell in hypotonic conditions is a bit
like a swimming pool with lots of people on the deck. If
the pool is likened to a cell, then people are like water
molecules. When the park is relatively empty, many
people will be attracted to it and enter, until so many
people are in the pool that it is no longer a desirable
place to be and no more people enter the pool.
47.Diagrams should show a charged substance being
transported across a membrane, resulting in a difference
in charge on opposite sides of the membrane.
48.First, the two amino acids would bind to the enzyme at
the active site, forming the enzyme-substrate complex.
Then, binding of the substrates would involve changes
in the shape of the active site and the substrates
themselves. These shape changes would help destabilize
the bonds during the reaction. After the condensation
reaction occurs, water would be released and the
product (dipeptide) would be momentarily bound at
the active site. Finally, the product would be released
with the enzyme returning to its pre-binding shape
and the enzyme would be ready for another reaction.
Diagrams should look like Figure 1.29 on page 37 of
the student textbook, but reversed left to right, and with
the reactants shaped as balls to represent amino acids.
49.In both answer parts, evidence of research from at least
two different and credible sources should be presented
and documented.
For the first part, research will reveal efforts to find
alternative ways to produce Mo-99 are predominant,
with some references to using MRI and computerized
tomography as a completely different alternative.
Research is ongoing; accept all supported answers.
Answers to part two will focus on production issues,
particularly related to aging facilities.
50.Drawings should look like Figure 1.18 on page 25
of the student textbook, with the addition of
“hydrophobic” on the R group label.
51.Venn diagram should show:
•RNA Only—uracil (U); ribose; single-stranded, but
double stranded regions with complementary basepairing can form
•RNA and DNA—monomers are nucleotides;
adenine (A), guanine (G), and cytosine (C); can
also form a hybrid double-stranded molecule with
a strand of RNA; phosphodiester bonds between
nucleotide monomers
•DNA Only—thymine (T); deoxyribose; doublestranded with complementary base-pairing (A pairs
with T and G pairs with C), but single-stranded
regions can form
52.Diagrams should look like the right side of Figure 1.14
on page 23 of the student textbook, with a label for the
head group and a label for the fatty acid tails.
When phospholipids are added to an aqueous
environment, they will self-assemble into a molecular
bilayer that self-seals so that no free edge is left; like
Figure 1.15 on page 23 of the student textbook.
Biology 12 Answer Key Unit 1 • MHR TR 27
53.Diagrams should look like those for question 52.
The hydrophobic effect is mainly responsible for this
phospholipid bilayer structure. Since the head groups
are polar and thus hydrophilic, they are oriented
toward the aqueous external environment and that of
the cytoplasm. The non-polar hydrophobic fatty acid
tails are oriented toward one another, which keeps
them sequestered from water.
54.Diagrams should look like Figure 1.29 on page 37 of
the student textbook, with the following labels added
from top to bottom: substrate, active site, enzyme.
If this enzyme was being affected by allosteric
inhibition we could add a separate binding site for the
allosteric inhibitor and show it binding at that site. The
allosteric inhibition would involve a change in shape of
the enzyme and its active site such that it was unable or
less able to bind its substrate. The inhibition occurs as
a result of this change in shape, brought about by the
binding of the inhibitor at the allosteric site.
55.Sketches should look like the upper left portion of
Figure 2.15 on page 73 of the student textbook with the
labels animal cell and plasma membrane added. Under
the figure, a label should say “solutes in cell = solutes
outside cell” and “no change in cell volume.”
56.Sample answer: Balls of various types and sizes could
represent vesicles and vacuoles. A stack of dinner
plates could represent the Golgi apparatus. A rumpled
bed sheet could represent the endoplasmic reticulum.
Alternatively, the balls could be used in conjunction
with string laid out in patterns to represent the
endoplasmic reticulum, Golgi, and cell membrane.
57.• maintenance of cell shape
•movement of the cell
•movement of organelles
•anchoring of organelles
•spindle formation for cell division
•cleavage furrow formation during cell division
•internal support of nucleus
•muscle contraction
58.Sketches should resemble Figures 2.20 and 2.21 on
pages 78 and 79 of the student textbook.
59.Answers should show an understanding of redox
reactions, condensation and hydrolysis reactions, and
acid-base reactions (or neutralization). A fishbone
diagram or a spider map would be good choices for
graphic organizers (a main idea web might also be
used). In a fishbone diagram the main topic along the
backbone should be related to “chemical reactions in
living systems” and the bones would be the three types
of reaction pairs. Each reaction pair should be defined
28 MHR TR • Biology 12 Answer Key Unit 1
roughly as noted in the key term definitions on text
pages 32–4 of the student textbook.
60.a.The short, acutely curved arrows indicate that a
molecule is unable to cross the membrane. The long,
slightly curved arrows indicated the diffusion of
molecules through the lipid bilayer.
b.Membranes allow some types of molecules to
diffuse quite readily, while disallowing others. Water
and small uncharged molecules are able to diffuse
through the bilayer quite readily. Charged molecules
and ions as well as macromolecules are turned away
at a lipid bilayer—they are unable to diffuse through
it at any significant rate.
c.Sample answer:
•facilitated diffusion by a channel protein and
carrier protein (one or two slides)
•primary active transport
•secondary active transport
•exocytosis and endocytosis as examples of
membrane-assisted transport (one or two slides)
61.Both primary active and/or secondary active transport
could be used to move more glucose into the cells
against its gradient. Unlike facilitated diffusion, both
of these mechanisms are able to transport a substance
against its gradient. Another, less obvious, mechanism
might involve endocytosis.
62.Sample answer: The maintenance of cell membrane
integrity illustrates the power of the hydrophobic
effect. Phospholipids in a membrane are constructed
from both polar and non-polar molecules giving
these molecules a dual personality that helps maintain
the basic framework of a membrane. The polar head
groups orient toward the aqueous environment outside
and inside the cell whereas the hydrophobic nonpolar fatty acid tails orient toward one another. These
polar and non-polar interactions are fundamentally
important to the maintenance of membrane structure.
63.Muscle cells need a tremendous amount of
usable energy for the contractions they perform.
Mitochondria provide usable energy for cells. So it
makes sense that cells with high energy requirements
(like muscle cells) have many mitochondria.
White blood cells are involved with our immune
system, and many are used to search out and
destroy invading microbes as well as dead and dying
cells. White blood cells contain many lysosomes
that can break down and recycle that material.
The main function of lysosomes is to break down
macromolecules and recycle their components for use
by the cell. So it makes sense that white blood cells
have many lysosomes.
64.a.Since olestra is so very large, it is unlikely that there
are any enzymes with active sites large enough to
bind it. If the molecule cannot be broken down
in a stepwise fashion through enzyme-catalyzed
reactions, it not surprising that it cannot provide
any food energy.
b.Olestra was invented by researchers at Proctor and
Gamble in 1968. Some of the side effects associated
with the use of olestra as a food additive include loose
stools and abdominal cramping. Olestra also adversely
affects the absorption of some vitamins. Health Canada
refused approval of olestra as a food additive mainly
because of its adverse effects on the bowel.
65.Answers should show (and document) evidence
of research from at least two different and credible
sources. Answers should define the role of enzymes in
bioremediation, noting the potential uses of the process
in various applications. Suggested problems should be
related to those applications.
66.No, enzymes do not only work at a specific temperature
and pH, but they all have optimal temperatures and pH
for their activity. The two graphs of enzyme reaction
rates with different temperatures and pH (Figures 1.30A
and B respectively, on page 39 of the student textbook)
show that while there is an optimal temperature and
pH for the enzymes, they all exhibit activity above and
below these optima. This can be seen as the rise and fall
of the reaction rate on either side of the optima.
67.Sample answers: Manufacturers will use an enteric
coating to slow the breakdown of the medication
by enzymes in the digestive system. Alternatively,
they may infer that manufacturers of time-release
medication include inhibitors to slow down the activity
of the enzyme that is delivering the medication.
Cell Type
Specialized Features
68.The treatment is likely to be unsuccessful because
lysosomal enzymes work best at an acidic pH of
about 5. Since the bloodstream is at about the same
pH as the interior of cells, near neutrality, the enzymes’
activity will be very low if they are active at all. And, if
they are active, the enzymes could digest component in
the blood or blood vessels
69.a.Since the basic mechanism of protein synthesis
is similar in all cells, it might have some effect on
protein synthesis in human cells.
b.The ribosomes of bacteria and humans have
different characteristics. So if the second antibiotic
specifically inhibits protein synthesis by bacterial
ribosomes but not human ribosomes, it is likely to
be successful.
70.The body cannot function without cholesterol since
it is an important facilitator of membrane fluidity.
Cholesterol helps maintain the proper fluidity of a
membrane at high temperatures and low temperatures.
71.Cell membrane glycoproteins with their carbohydrate
facing the exterior environment of the cell can play
important roles in autoimmune disorders. The unique
carbohydrate chains that these molecules display
at the outer surface of a cell allow many of these
glycoproteins to be recognized as normal or foreign
by other cells of the body, especially those of the
immune system. Slight alterations in the proteins that
recognize these glycoproteins could result in normal
glycoproteins being recognized as foreign. This can
result in autoimmune disorders in which normal cells
are recognized as foreign and destroyed by the body.
72.Sample answer:
Specialized Functions
Heart
• arrangement of microfilaments and other proteins of • the contractile units allow these cells to contract and relax for the
pumping action of the heart
the cytoskeleton into contractile units (sarcomeres)
• connections (gap junctions) between neighbouring • gap junctions allow the flow of ions between cells that act as signals
to help ensure that individual cell contractions occur simultaneously
heart cells (cardiac muscle cells)
Sperm
• long whip-like flagellum composed of specially
arranged microtubules and other proteins of
the cytoskeleton
• whip-like motion helps cell move through female reproductive tract
• designed to interact with egg to form the zygote
Egg
• large round cell surrounded by egg coats
• reproductive cell that accepts a sperm cell (fertilization) and its
single set of human chromosomes to produce a single cell (the
zygote) that has two sets of chromosomes (diploid)
Red blood
• basically a membranous bag of hemoglobin that
has lost all membranous organelles including the
nucleus and its chromosomes
• unique biconcave shape
• small size
• the millions of hemoglobin molecules in each cell act as carriers of
oxygen and deliver that oxygen to other cells of the body
• the shape of the cell helps maximize its surface area for the uptake
and release of oxygen
• its small size allows it to squeeze through even the tiniest of blood
vessels (capillaries)
Biology 12 Answer Key Unit 1 • MHR TR 29
73.Salt may prevent the rapid uptake of water that causes
cells to burst and lose green chlorophyll pigment into
the boiling water. There are many reports however
that indicate the addition of salt to help keep boiling
vegetables bright green has very little effect.
Answers to Unit 1 Self-Assessment Questions
(Student textbook pages 104–5)
1.d
2.e
3.a
4.d
5.a
6.e
7.b
8.a
9.b
10.b
11.Any three of: nucleus, nucleolus, ER, freely suspended
in the cytosol, mitochondria, or chloroplasts
12.The rate of diffusion should increase. Typically, ions
do not diffuse unaided through the cell membrane,
however, with more space between the phospholipids,
some diffusion could occur.
13.The benefit of this adaptation is that it prevents the
wheat cell membranes from solidifying in the cold.
Unlike saturated fatty acids, unsaturated fatty acids
have double bonds along their length, and so do not
pack close together, a feature that would help retain
membrane fluidity in the cold.
14.While one hydrogen bond is relatively weak, many
together are strong, making the DNA molecule quite
stable overall. However, since individual hydrogen
bonds break and form easily, when cell processes
require that the two DNA strands separate, the cell
can do this and then reform the hydrogen bonds.
15.Answers may include using Benedict’s solution to
test for glucose or maltose (positive results vary with
concentration—from green, if low, to yellow to orange
to red, if high); or checking the ingredients list for
carbohydrates such as maltose, glucose, or fructose,
which are sugars, but not “table sugar” (sucrose).
Either answer is acceptable.
16.Any two of: stabilizing membranes by linking them
with the cytoskeleton, reaction catalysis (as enzymes),
cell recognition, signal reception and/or transduction,
or transport across the cell membrane in combination
with an integral protein.
30 MHR TR • Biology 12 Answer Key Unit 1
17.Knowledge of the enzyme’s three-dimensional
structure could be helpful in identifying the shape of
the active site or an allosteric site. Perhaps the inhibitor,
when joined with phosphate groups, has a shape that
fits more readily into the active site or allosteric site.
8.A, hypertonic; B, isotonic; and C, hypotonic.
1
An animal cell in a hypertonic solution (A) loses water
by osmosis and so shrivels (crenates). An animal cell
in an isotonic solution (B) continues to exchange water
molecules with its surroundings, but because there is
no net movement of water in either direction, the cell’s
volume does not change. An animal cell in a hypotonic
solution (C) gains water by osmosis and may burst
(undergo lysis).
19.Enzyme activity can be regulated by the binding of
an inhibitor to the enzyme’s allosteric site. This is
called non-competitive inhibition (a form of allosteric
regulation) because the inhibitor does not compete
with a substrate for the active site, but rather causes the
conformation of the enzyme such that the active site
cannot properly interact with the substrate. Enzyme
activity can also be regulated by competitive inhibition,
in which the inhibitor competes with the substrate to
occupy the active site.
20.Answers should include any one example for each:
a.cilia, flagellum or flagella, cytoskeleton, microtubules,
microfilaments (in muscle contraction)
b.mitochondrion, chloroplast
c.lysosome, peroxisome, vesicle
d.vesicle, endomembrane system, cytoskeleton,
microtubules
21.Plant and animal cells have tremendous variety in
form, size, and structure because they are specialized
for particular tasks. Therefore, any one type of plant or
animal cell is unlikely to contain all of organelles that
biology students need to learn about.
22.Sample answer: Diffusion Across a Selectively
Permeable Membrane
At the beginning of the experiment, there is an equal
volume of water on either side of the membrane (A),
but the solution in the right half of the U-tube has
a higher concentration of solute (sugar molecules),
making the right side hypertonic to the left. In B,
the solute is unable to pass through the selectively
permeable membrane, but water can. Therefore, water
moves into the hypertonic solution, increases the
volume (and concentration) of water on the right side
while decreasing the solute concentration, and makes
the solutions on either side of the membrane isotonic
to one another.
23.When proteins denature, they unfold and lose their
normal three-dimensional shape. High temperatures
can disrupt the bonds between the R groups that
maintain the normal secondary, tertiary, and
quaternary structures of a protein.
24.Compared to carbohydrates, fats (lipids) have fewer
oxygen atoms and more carbon-hydrogen bonds. These
C-H bonds are energy rich and fats have a lot of them
in their hydrocarbon chains.
25.Answer should show an understanding of the fluid
nature of the phospholipid bilayer and mosaic
produced by the various proteins, glycoproteins,
glycolipids, and other molecules that are attached
to or inserted in the membrane.
Biology 12 Answer Key Unit 1 • MHR TR 31