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Biology 12 Answer Key Unit 1 • MHR TR 1

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. OH-

b. S2-

c. Al3+

d. Fe3+

9. H2O or water

10. water → hydrogen + oxygen

11. d

12. sodium hydroxide + hydrochloric acid → 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

2 MHR TR • Biology 12 Answer Key Unit 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; membrane-bounded 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

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 carbon-carbon 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.


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 = lengthof 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.

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 QuestionsFigure 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:

radioisotope

unstable,radioactive

can be used diagnosticallywhen used to make

isotope ion

organicmolecule

atom

vary by numberof neutrons

gain or loss of1 or more electron

2 or moreatoms

carbon-basedtype

other(inorganic

type)

molecule

some can be

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.


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.

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; intra- or intermolecular*

R+ -R; electrostatic interaction between R-groups; intra- or intermolecular*

R

interior of protein

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 + H2ONH3 + HCl → NH4Cl + H2ONaOH + CH3COOH → NaCH3COO + H2O

• redox reactions— C3H8 + 5O2 → 3CO2 + 4H2O

• condensation reactions— nucleotide + nucleotide → dinucleotide + H2Omonosaccharide + monosaccharide →

disaccharide + H2O glycerol + 3 fatty acids → fat molecule (3 condensation reactions) + 3H2O• hydrolysis reactions—

dinucleotide + H2O → 2 nucleotidesdipeptide + H2O → 2 amino acidsfat 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.

substrateactive

site

allostericsite

noncompetitiveinhibitor

enzyme

When a noncompetitive inhibitor binds at the allosteric site, it induces a change in shape that alters the active site and 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.

shape change


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

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.


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.

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.

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.

52.

Conc

entr

atio

n of

Rea

ctan

t or P

rodu

ct

Time

product

reactant

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 non-catalyzed reaction rate was much less than that with the enzyme.

53.

Reac

tion

Rate

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.


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

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.

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.

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.

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.

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.

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


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.

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.

OH

OHH2O

R

H

H OH

N C C

R O

H

OHH

N CR

H HN C CC

OH

R

H

H OH

N C C


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-dissolves-like”—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


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 vesicle-mediated 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.

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 non-lipid 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.


(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

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 QuestionsFigure 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 Sketch Function

Nuclear envelope

Sketches should resemble parts of Figure 2.4 on page 60 of the student textbook.

separates nucleus from rest of cell

Nuclear pore 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

• 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 membrane-enclosed sac with enzymes → enzymes break down (synthesize) molecules → new molecules produced

7. Sample answer:

Cell Structure or Organelle 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 lipid-containing 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 inner membrane and stacks of thylakoids (grana); contains some of its own DNA

Site of photosynthesis plant

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


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.

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.

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.

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.

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.

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

12. b

13. d

14. c

15. b

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.

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.


• 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.

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.


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. Receptor-mediated 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.

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.


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

• Plant and Animal Cells—Vesicles are membrane-bound, 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.

ion pump

ions (Na+, K+, H+)

ATP

H+cytoplasm

carrier protein

ions, small and large molecules (glucose, amino acids)

ions (Na+, K+, Cl-), smaller polar molecules

extracellular �uid

channel protein

small non-polar molecules (O2)

ADP + Pi

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.


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

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 base-pairing 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; double-stranded 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.


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

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 non-polar 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.

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:

Cell Type Specialized Features Specialized Functions

Heart • arrangement of microfilaments and other proteins of the cytoskeleton into contractile units (sarcomeres)

• connections (gap junctions) between neighbouring heart cells (cardiac muscle cells)

• the contractile units allow these cells to contract and relax for the pumping action of the heart

• gap junctions allow the flow of ions between cells that act as signals to help ensure that individual cell contractions occur simultaneously

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)


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.

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.

18. A, hypertonic; B, isotonic; and C, hypotonic. 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.

answer key - manning's · pdf filebiology 12 answer key unit 1 • mhr tr 3 genetics....

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