CHAPTER 1. THE SCIENCE OF BIOLOGY
THE ESSENTIALS
Students need to know:
ƒ the essential components of a scientific experiment: hypothesis,
independent variable, dependent variable, confounding variables
(constants), control group(s), experimental group(s), statistical tests.
ƒ how to craft a hypothesis as an experimental prediction.
ƒ the difference between the common use of the term “theory” and the
scientific use of the term “theory”.
Key Terms
organelles
cells
tissues
organs
organ systems
population
species
biological community
ecosystem
emergent properties
deductive reasoning
inductive reasoning
hypothesis
experiment
independent variable
dependent variable
experimental control
theory
evolution
artificial selection
1
natural selection
homology (homologous)
analogous
phylogenetic tree
cell theory
DNA (deoxyribonucleic acid)
gene
genome
CHAPTER 2. THE NATURE OF MOLECULES AND THE PROPERTIES OF WATER
THE ESSENTIALS
Students need to know:
ƒ the bonds important in biological systems (covalent, hydrogen, ionic,
and van der Waals interactions) and how they form.
ƒ the comparative strength of each of the bonds: covalent (strong),
hydrogen (relatively weak), ionic (relatively weak), van der Waals
interactions (very weak).
ƒ the difference between polar and nonpolar covalent bonds.
ƒ how hydrogen bonds determine the properties of water.
ƒ the five properties of water and how each influences life on Earth.
ƒ how hydrophobic and hydrophilic portions of molecules interact with
water.
ƒ how to interpret the pH scale.
ƒ the importance of buffers in stabilizing pH (and maintaining
homeostasis) in biological systems.
Key Terms
matter
atoms
electrons
protons
neutrons
atomic number
atomic mass
element
isotopes
radioactive isotopes
half-life
neutral atoms
ions
cation
anion orbital
oxidation
reduction
energy level
valence electrons
inert
octet rule
molecule
compound
chemical bonds
ionic bonds
ionic compound
covalent bonds
double bonds
single bonds
triple bonds
structural formulas
molecular formulas
chemical reaction
reactants
products water
electronegativity
polar molecules
hydrogen bonds
cohesion
adhesion
2
surface tension
specific heat
heat of vaporization
hydration shell
hydrophobic
hydrophilic
hydrogen ion
hydroxide ion
ionization
mole
molar concentration
pH scale
buffer
Concept Map
matter
atoms
structure
elements
form
molecules &
compounds
bonds
through
electrons
neutrons
protons
isotopes
determines
negative
charge
chemical
behavior of
atom
positive
charge
chemical
reactions
weak bonds
atomic
number
can be
different
number of
neutrons
ionic or
hydrogen
bonds
often
unstable
scientific tool
harmful
strong bonds
oxidation
reduction
loss of
electron
gain of
electron
radioactive
decay
age dating
expressed as
genetic&
cellular
damage
half life
3
covalent
bonds
Multiple Choice Questions
1. An atom with a neutral charge always has
a.
b.
c.
d.
e.
an equal number of protons and neutrons.
more neutrons than protons.
an equal number of protons and electrons.
more protons than electrons.
an equal number of neutrons and electrons.
2. Covalent bonds are formed by the
a.
b.
c.
d.
e.
transfer of protons from one atom to another.
sharing of neutrons between two atoms.
movement of electrons into the nucleus of an atom.
sharing of electrons between two atoms.
joining of the nuclei of two atoms.
3. A hydrogen cation has
a.
b.
c.
d.
e.
one proton and no electrons.
two protons and one electron.
two electrons and one proton.
one proton, one neutron, and one electron.
one neutron and one electron.
4. When an atom gains an electron by transfer from another atom, the chemical reaction is called
a.
b.
c.
d.
e.
oxidation.
reduction.
covalent bonding.
ionic bonding.
radioactive decay.
5. Elements in the same column of the periodic table share the same
a.
b.
c.
d.
e.
number of protons.
atomic mass.
number of valence electrons (rule of 8).
total number of electrons (rule of 8).
number of neutrons.
6. When a crystal of sodium chloride (NaCl) is placed in water the
a.
b.
c.
d.
e.
Na+ and Cl- atoms remain paired in a tightly bound molecule.
crystal dissolves as polar H20 molecules form associations with the individual Na+ and Cl- ions.
chloride reacts with H20 molecules to form hydrochloric acid.
Na+ reacts with H20 molecules to form sodium hydroxide.
Na+ reacts with C02 molecules to form sodium bicarbonate.
4
Answers:
1. c, 2. d, 3. a, 4. b, 5. c, 6. b
Essay Questions
1. Different isotopes of an element have identical chemical properties, yet the use of isotopes plays a
critical role in many biological experiments. Explain three properties of isotopes that make them
useful to biologists.
Isotopes of an element have the same number of protons and electrons, but different
numbers of neutrons. Property 1) atoms of two different isotopes of an element have
different atomic mass. This difference in mass can be detected with sensitive measuring
devices, or portions of a biological mixture can be separated based on differences in the
mass of the isotopes. Property 2) Some isotopes are unstable; the nuclei undergo
radioactive decay. This process generates energy, which can be detected by X-ray film
and other means—so a biological molecule that contains a radioactive isotope can be
tracked within an organism by the emission of radioactivity. Property 3) This radioactive
decay occurs at a specific rate for each isotope of each element. So by measuring the
ratio of isotopes, it is possible to calculate when a biological sample was created (i.e.
when it was alive).
2. Water is essential for the existence of all forms of life on earth. If you were to design the biochemistry
of an artificial form of life that did not use water, what properties would be required for a molecule
that takes the place of water?
Properties of water—liquid, polar solvent, can be oxidized as a source of electrons for
energy storage reactions (photosynthesis), forms hydrogen bonds with itself and other
polar molecules—contributes to the 3-dimensional shape of biological molecules.
5
CHAPTER 3. THE CHEMICAL BUILDING BLOCKS OF LIFE
THE ESSENTIALS
Students need to know:
ƒ the properties of the carbon atom that make it the building block of
biological molecules.
ƒ changes to the chemical properties of hydrocarbon chains is caused by
the addition of functional groups.
ƒ the four biological macromolecules and their representative structural
formulas.
ƒ the biological functions of each of the four macromolecules.
ƒ the process of dehydration synthesis in building organic molecules and
the process of hydrolysis in breaking down organic molecules.
ƒ the four structural levels of proteins and the forces that help maintain
each level.
ƒ
the denaturing effect that heat and pH can have on protein structure.
Key Terms
hydrocarbons
functional groups
macromolecule
isomers
monomers
polymers
dehydration synthesis
hydrolysis
carbohydrates
monosaccharides
disaccharides
polysaccharides
amylose
starch
cellulose
glycogen
chitin
lipids
fats
phospholipid
triacylglycerol
saturated
unsaturated
polyunsaturated
amino acid
protein
peptide bond
polypeptide
6
primary structure
secondary structure
tertiary structure
quaternary structure
chaperone proteins
denaturation
nucleic acid
nucleotide
deoxyribonucleic acid (DNA)
ribonucleic acid (RNA)
double helix
complementary
Macromolecule
Structure
carbohydrates
(CH2O)x
Monomer or
Building Block
sugar
lipids (fats)
long hydrocarbon
chains
glycerol
fatty acids
proteins
N-C-C backbone,
amine group,
carboxyl group
amino acid
nucleic acids
phosphate group,
pentose sugar,
N base
nucleotide
Function
energy storage,
structure
energy storage,
hormones, cell
membranes
enzymes,
transport,
movement,
support,
hormones, defense
information
storage
Examples
glucose, sucrose,
cellulose,
glycogen, chitin
fats,
phospholipids,
cholesterol
hemoglobin,
insulin, digestive
enzymes, collagen
DNA, RNA
Concept Map
Carbohydrates
Function
energy
storage
Structure
structure
polymers of
sugar
monomer
monosaccharides
Formula
CH 2O
disaccharides
2 sugars
joined by
glycosidic
bonds
polysaccharides
dehydration
synthesis
polymer of
sugars joined
by glycosidic
bonds
glucose
sucrose
maltose
7
cellulose
glycogen
starch
chitin
polymer of
a-glucose
polymer of
glucose
polymer of
b-glucose
glucose chain
with N group
structure in
plants
energy
storage in
animals
energy
storage in
plants
insect
exoskeleton
Nucleic Acids
Function
Structure
information
storage &
transmission
polymers of
nucleotide
monomers
phosphatepentose
sugar
backbone
heredity
Examples
nitrogenous
base
phosphodiester
bonds
purine
adenine,
guanine
RNA
DNA
double heix
complementary
base pairing
pyrimidine
thymine,
cytosine
uracil
pairs
with
replication
mechanism
built in error
correction
mechanism
Lipids
Fats
Steroids
composed of
glycerol &
fatty acid
subunits
function
insulation &
cushions
organ
stored energy
in C-H bonds
twice energy of
carbohydrates
bonded
through
dehydration
synthesis
hormones
sex
hormones
structure
saturated
animal fats
solid at room
temperature
unsaturated
liquid at room
temperatre
8
plant oils
built from
cholesterol
precursor
Phospholipids
hydrophobic
H-C tail
hydrophillic
phosphate
head
forms bilayer
of cell
membrane
......
Vl
e.g.,digestrve
eznymes,
polymerase
enzymes,
etc.
created by
hydrophobic
&hydrophillicl'-----...1
interactions,
H bondin ,
ionic onds,
disulfide
bridges
Level of protein structure
Bonds
Primary
Covalent peptide bonds
Secondary
H bonds between adjacent R
groups
Tertiary
Hydrophobic interactions
between distant R groups,
ionic bonds, stabilized by
disulfide covalent bonds
Quaternary
Hydrophobic interactions
between nonpolar R groups
of separate subunits
Multiple Choice Questions
1. All of the following are functions of lipids in organisms EXCEPT
a.
b.
c.
d.
e.
energy storage.
membrane structure.
information storage.
chemical signaling (hormones).
enzymes.
2. All of the following are carbohydrates EXCEPT
a.
b.
c.
d.
e.
cellulose.
glucose.
adenine.
methionine.
deoxyribose.
3. All of the following are chemical properties of hydrocarbons EXCEPT
a.
b.
c.
d.
e.
composed of only carbon and hydrogen
insoluble in water
contains a large amount of stored energy
does not contain double bonds between carbon atoms
may have linear, branched or ring structures
4. The number of covalent bonds formed by an atom of carbon is determined by
a.
b.
c.
d.
e.
the number of protons in its nucleus.
the number of electron shells.
its atomic mass.
the number of electrons in its outermost electron shell.
the pH of the surrounding solution.
24
5. Cellulose is a carbohydrate that
a. contains a long chain of glucose sub-units in the alpha form.
b. contains a long chain of glucose sub-units in the beta form. c.
contains a mixture of different sugar sub-units.
d. forms fibers that are soluble in water.
e. is easily oxidized by the digestive enzymes produced by most animals.
6. Proteins are able to perform many diverse biological functions because
a. they are polymers formed from a mixture of lipids, carbohydrates, and amino acids.
b. they are polysaccharides composed of long chains of identical subunits.
c. they are polymers formed from chemically diverse sub-units which can fold into many different
shapes.
d. all proteins can function at a wide range of pH and temperature conditions.
e. they are used store genetic information and pass it on to future generations.
7. The sugar subunits in the DNA backbone are joined by what type of bonds?
a.
b.
c.
d.
e.
phosphodiester bonds
hydrogen bonds
ionic bonds
non-polar interactions
peptide bonds
8. Isomers are pairs of similar molecules that
a.
b.
c.
d.
e.
are composed of different isotopes of the same atoms.
are built with the same atoms in a different arrangement.
have a different number of double bonds.
have a different charge.
always have the same chemical and biological properties.
9. All amino acids
a.
b.
c.
d.
e.
are insoluble in water.
are polymers containing alternating sugar and amide subunits.
have carboxyl and amine functional groups.
have non-polar R groups.
contain sugar subunits.
10. Purine and pyrimidine bases
a.
b.
c.
d.
e.
cannot form hydrogen bonds due to steric hindrance.
form the sugar-phosphate backbone of the DNA molecule.
are insoluble in water.
have different types of ring structures.
are the building blocks of proteins.
Answers:
1. c, 2. c, 3. d, 4. d, 5. b, 6. c, 7. a, 8. b, 9. c, 10. d
25
Essay Questions
1. What chemical reactions cause changes in the structure of proteins when they are placed in a high or
low pH environment?
Change in charge of acidic and basic side chains leads to new attractive and repulsive
forces. Highly acidic and basic environments disrupt hydrogen bonds, because they
either add H+ or OH- ions to the solution..
2. Polymers are molecules built from long chains of repeating simple subunits. How are these molecules
capable of storing information?
Information can be stored in the sequence of different subunits in a chain, just as
information can be stored in the sequence of letters in a sentence or the sequence of 1’s
and 0’s in a computer file.
3. Describe the four different types of side chains that are attached to the core carbon atom of an amino
acid.
All amino acids have an amino group, a carboxyl group, a hydrogen atom and a variable
side chain known as the R group.
4. How does salt or vinegar preserve food?
Acid pH (vinegar) or high salt concentration denatures the enzymes (disrupts protein
structure) of microorganisms that would cause food to spoil.
5. What are some properties of DNA that make it well-suited for the storage of genetic
information?
DNA is chemically stable. Strong covalent bonds hold the sugar-phosphate backbone
together.
DNA stores information in the varying sequence of nucleotides (the genetic code).
Base pairing allows the coded sequence to be copied exactly by the synthesis of
complementary strands.
Damage to one strand can be repaired by addition of bases that match the
complementary strand.
6. How does 70% alcohol kill bacteria?
Alcohol disrupts hydrogen bonds in protein structure—alcohol molecules form new
hydrogen bonds with the amino acid side chains.
7. Describe the chemical structure of a phospholipid molecule and explain how this structure leads to the
formation of membranes when phospholipids are placed in water.
Phospholipids have long non-polar hydrocarbon chains at one end and a polar phosphate
group at the other end. This causes groups of phospholipid molecules to aggregate together
when they are placed in water as circular “fat bubbles” with their non-polar ends directed
inward and their hydrophilic polar heads facing the aqueous solution. Larger amounts of
phospholipids form a bilayer membrane with polar heads oriented out toward the water and
hydrocarbon chains oriented inward.
8. What are some advantages to an organism of storing energy in the form of fats and oils?
Fats and oils are very energy dense, so they take up less space and weigh less than an
equivalent amount of energy stored as sugars or starch. They are insoluble in water, so
they can easily be stored in one place without diffusing away or creating osmotic
imbalance.
26
CHAPTER 4. CELL STRUCTURE
THE ESSENTIALS
Students need to know:
ƒ the differences between prokaryotes and eukaryotes.
ƒ the structure and function of the organelles found in both plant and
animal cells.
ƒ the structure and function of the organelles found only in plant cells
and the structure and function of the organelles found only in animal
cells.
ƒ the origin of mitochondria and chloroplasts through endosymbiosis
ƒ the factors that limit cell size and how cell shape is related to cell
function.
ƒ the different types of cell junctions and how they affect tissue
function.
Key Terms
prokaryotes
eukaryotes
nucleoid
nucleus
nuclear envelope
nuclear pores
nucleolus
cytoplasm
cytosol
organelles
plasma membrane
cell wall
cell theory
compound microscope
transmission electron
microscope
scanning electron microscope
flagellum
organelles
chromosomes
chromatin
endomembrane system
endoplasmic reticulum (ER)
rough ER
smooth ER
Golgi apparatus
vesicles
lysosomes
peroxisome
ribosomes
mitochondria
cristae
matrix
intermembrane space
chloroplasts
grana
thylakoids
amylopast
plastids
endosymbiosis
cytoskeleton
actin filaments
27
microtubules
intermediate filaments
centrioles
centrosome
flagella
cilia
9 + 2 structure
basal body
cilia
central vacuole
primary walls
middle lamella
glycoproteins
extracellular matrix (ECM)
cytoplasmic streaming
cell junction
tight junction
anchoring junction
communicating junction
gap junction
plasmodesmata
Concept Map
Cell
structures
Functions
basic unit of
life
chromosomes,
genetic information,
replication,
RNA synthesis
nucleus
protein
synthesis
ribosomes
endomembrane
system
nuclear envelope,
rough & smooth ER,
Golgi apparatus, vesicles,
lysosomes, vacuoles
biomolecule
synthesis,
processing,
transport,
export,
storage,
digestion
energy
transformation
mitochondria,
chloroplasts
structural support,
cell movement
cytoskeleton
28
Multiple Choice Questions
1. Which of the following is true about the eukaryotic nucleus?
a.
b.
c.
d.
e.
The nucleolus is the site of most protein synthesis in the cell.
Nucleosomes are pores in the double membrane of the nuclear envelope.
Histones regulate the movement of mRNA molecules through the nuclear pores.
The nucleolus is a region where ribosomal RNA is synthesized.
Chromatin is the tightly wound complex of DNA and histone proteins that is only visible during
cell division.
2. A signal sequence is a
a. region of the chromosome that controls the expression of a gene.
b. portion of a mRNA molecule that determines if a protein will be synthesized on free ribosomes or
on the rough endoplasmic reticulum.
c. short amino acid sequence at the beginning of a protein that directs the protein to a specific
destination within the cell.
d. portion of a protein that is spliced out inside the nucleus.
e. portion of a mRNA molecule that is spliced out in the cytoplasm.
3. Which of the following are NOT found in plant cells?
a.
b.
c.
d.
e.
nucleus
endoplasmic reticulum
mitochondria
centrioles
central vacuole
4. The functions of the Golgi apparatus include the
a. detoxification of drugs.
b. packaging of proteins and lipids in secretory vesicles that are transported to other locations in the
cell.
c. synthesis of nucleotides and amino acids.
d. synthesis of ribosomes.
e. synthesis of ATP by the oxidation of glucose.
5. Mitochondria are
a.
b.
c.
d.
e.
present in animal cells and protests but not in plant cells.
sub-cellular organelles that contain their own circular DNA molecules.
organelles that contain hydrolytic enzymes that digest particles taken into the cell by endocytosis.
formed when folds of the Golgi apparatus pinch off to form vesicles.
the site of photosynthesis.
6. All of the following are true EXCEPT
a. Eukaryotic cells contain many copies of the genes encoding the ribosomal RNAs, which are
clustered together on the chromosomes.
b. Ribosomes are composed of both RNA and protein.
c. Ribosomes synthesize proteins within the nucleus by copying information from the DNA.
d. Ribosome sub-units are assembled within the nucleus in a region called the nucleolus.
e. Cells which produce large amounts of excreted proteins have large numbers of ribosomes
attached to the endoplasmic reticulum.
29
7. The cytoskeleton is a network of protein fibers including all of the following EXCEPT
a. actin fibers composed of two filaments of actin molecules.
b. microtubules which are hollow tubes composed of 13 filaments built from dimmers of alpha and
beta tubulin protein molecules.
c. intermediate filaments which are composed of mixed protein subunits including vimentin and
keratin.
d. dynein filaments which connect the centrioles to flagella.
e. spindle fibers are microtubules that grow out of the centrosomes to guide the movement of
chromosomes during cell division.
8. The cell theory states that
a.
b.
c.
d.
e.
all living things are composed of cells and all living cells come from other cells.
cells can form spontaneously under the proper conditions.
living cells can be made in the laboratory from non-living chemicals.
cells must be smaller than 10 microns in diameter.
all cells must belong to either the plant or animal kingdom.
Answers: 1. e, 2. c, 3. d, 4. b, 5. b, 6. c, 7. d, 8. a
30
Essay Questions
1. The surface-to-volume ratio limits the size of most cells, yet some animal nerve cells are several feet
long. Explain why a large volume is a problem for cells and how nerve cells overcome this constraint.
As a cell increases in size, volume increases much more rapidly than its surface area:
volume =(4/3)πr3 vs. surface area = 4πr2
This makes it difficult for nutrients to diffuse from the cell membrane into the center of a
large cell and for waste products to diffuse out. Nerve cells are very long and skinny, so
that no part of the cytoplasm is far from the cell membrane.
2. The nucleus is a key feature that distinguishes eukaryotic from prokaryotic cells. As a general rule,
each eukaryotic cell has a single nucleus. Discuss three exceptions to this rule (eukaryotic cells
without a nucleus or with more than one nucleus) and how these exceptions represent adaptations for
specialized cell functions.
Red blood cells and plant phloem sieve tube cells do not have a nucleus. Muscle cells and
fungal cells have many nuclei. Multiple nuclei allow control of large cells. Cells without
nuclei have simple biochemical functions and do not grow. Red blood cells have more
room for hemoglobin, sieve tubes allow rapid movement of plant nutrients.
3. Describe the movements of an excreted protein within the endomembrane system from the time
protein synthesis begins on a free ribosome until it is released outside the cell. Include an explanation
of the role of the signal sequence.
Protein synthesis begins when an mRNA reaches a free ribosome in the cytoplasm. After
the first few amino acids of the protein are joined into a polypeptide chain, this short
peptide is recognized as a signal sequence. The ribosome is then attached to the
endoplasmic reticulum at a ribosome docking site, with the signal peptide inserted
through the membrane of the ER. As the synthesis of the protein continues, it passes into
the cisternal space of the ER. When protein synthesis is complete, the protein moves in a
transport vesicle from the ER to the Golgi apparatus where it is processed. The Golgi
package the protein in a secretory vesicle, which moves to the cell membrane. The
secretory vesicle fuses with the cell membrane, releasing the protein outside the cell.
4. Explain the theory of endosymbiosis and give four facts that support the prokaryotic origin of
mitochondria and chloroplasts.
The theory of endosymbiosis states that the ancestors of modern eukaryotic cells engulfed
other smaller prokaryotic cells. These cells became mitochondria and chloroplasts.
Evidence that supports this theory includes: 1) modern cases of endosymbiosis where
blue-green algae live inside protests, 2) mitochondria have circular DNA molecules
similar to prokaryotes 3) mitochondria and chloroplasts reproduce by simple binary
fission like prokaryotes 4) mitochondria have ribosomes that are similar to prokaryote
ribosomes in size and structure.
31
CHAPTER 5. MEMBRANES
THE ESSENTIALS
Students need to know:
ƒ why cell membranes need to be selectively permeable.
ƒ the components of the cell membrane (phospholipids, proteins,
carbohydrates, and cholesterol) and the function of each.
ƒ in what direction a solute will flow when a gradient exists across a
selectively permeable membrane.
ƒ in what direction water will flow when a cell is placed in a hypotonic,
isotonic, or hypertonic solution.
ƒ the differences between simple diffusion, facilitated diffusion, and
active transport as well as cellular examples of each.
ƒ bulk transport (the differences between endocytosis and exocytosis as
well as cellular examples of each).
Key Terms
diffusion
osmosis
solvent
solutes
osmotic concentration
hyperotonic (hyperosmotic)
hypotonic (hypoosmotic)
isotonic (isoosmostic)
aquaporins
plasma membrane
phospholipids
phospholipid bilayer
selectively permeable
ion channels
fluid mosaic model
transmembrane proteins
carriers
facilitated diffusion
32
osmotic pressure
turgor pressure
endocytosis
phagocytosis
pinocytosis
exocytosis
active transport
sodium-potassium pump
Concept Map
Cell Membranes
function
structure
fluid mosaic model
phospholipid
bilayer
receptor
proteins
proteins
transmembrane
proteins
permeable barrier
passive
transport
active
transport
down
concentration
gradient
against
concentration
gradient
cholesterol
peripheral
protein
glycoprotein
transport
proteins
large molecules
simple
diffusion
facilitated
diffusion
nonpolar
molecules
polar
molecules
protein
pumps
endocytosis
phagocytosis
requires
energy
example
sodium
potassium
pump
example
aquaporin
channel for
water
Osmosis
diffusion of water
through a
semi-permeable
membrane
hyperosmotic / hypertonic
solution
hypoosmotic / hypotonic
solution
higher solute
concentration
lower water
concentration
lower solute
concentration
higher water
concentration
animal cell
shrivels
plant cell
plasmolyzes
animal cell
swells & may
burst (lyse)
turgid
plant cell
33
isoosmostic / isotonic
solution
equilibrium
"normal"
animal cell
flaccid plant
cell
exocytosis
pinocytosis
Multiple Choice Questions
1. Cell membranes are composed primarily of
a. carbohydrates and proteins.
b. phospholipids and proteins.
c. carbohydrates and phospholipids.
d. nucleic acids and phospholipids.
e. ribosomes and lipids.
2. The phospholipids of the cell membrane form
a. a bilayer that is impermeable to water and water soluble molecules.
b. a long polymer that folds into a non-polar sphere.
c. a bilayer that hides the non-polar alcohol groups inside and exposes the polar fatty acids to the
surrounding water.
d. an alpha helix.
e. a spherical fat globule which minimizes the contact of non-polar lipids with the surrounding
water.
3. All cell signal receptor proteins must
a. be embedded in the plasma membrane.
b. have a DNA binding domain.
c. have a signal molecule binding site.
d. have seven transmembrane domains.
e. act as protein kinase enzymes.
4. Which of these types of molecules can diffuse across the cell membrane without the assistance of
carrier proteins?
a.
b.
c.
d.
e.
large polar molecules such as starch
carbon dioxide and oxygen
ions such as calcium (Ca+2)
proteins
DNA
5. Receptor proteins embedded in the cell membrane
a. are polar and diffuse easily out of the cell membrane into the cytoplasm.
b. are covalently bound to the phospholipids in the cell membrane and unable to move.
c. contain one or more non-polar regions that passes through the cell membrane and polar regions
that interact with soluble molecules in the cytoplasm and outside the cell.
d. are non-polar and remain completely embedded in the lipid region of the membrane to avoid
contact with water.
e. are synthesized in the nucleus and transported to the cell membrane by mRNA.
6. Which of the following is NOT a function of proteins embedded in the cell membrane?
a.
b.
c.
d.
e.
forming pores to transport of ions and polar molecules across the membrane
active transport
protein synthesis
anchors for the cytoskeleton
hormone receptors
34
7. Which of the following requires an input of energy from the cell?
a. the increase in cell volume by osmosis when it is in a hypoosmotic medium
b. the decrease in cell volume by osmosis when it is in a hyperosmotic medium
c. the diffusion of uncharged molecules from high concentration to lower concentration through a
protein channel
d. facilitated diffusion across the plasma membrane by carrier proteins from a region of high solute
concentration to a region of lower concentration
e. active transport across the plasma membrane from a region of lower concentrations to a region of
higher concentration
8. Paracrine signaling involves
a. direct contact between membrane bound protein of one cell and receptors of another cell.
b. signal molecules that diffuse only a short distance in intracellular fluid before they are destroyed
by extracellular enzymes or removed by other cells.
c. signal molecules that move through the circulatory system and may have effects on cells in
distant organs.
d. signal molecules that travel only a short distance between the specialized signaling and target
structures of a synapse.
e. receptor mediated endocytosis.
9. A gated ion channel
a.
b.
c.
d.
e.
allows a specific type of ion or molecule to diffuse through the cell membrane very rapidly.
moves ions through the cell membrane by active transport.
allows messenger RNA to move out of the nucleus.
is usually open, but closes when is it activated by a specific signaling molecule.
acts as a membrane bound enzyme.
10. The primary function of tight junctions between cells is to
a. provide physical strength for tissues that experience stress such as muscle and skin.
b. allow for direct movement of ions and other water soluble molecules from one cell to another.
c. seal a sheet of cells into an impermeable layer, so that water and soluble molecules must move
through the cells of the sheet.
d. identify cells as ‘self’ to the immune system.
e. pride a direct connection from proteins in the extracellular matrix to cytoskeletal filaments.
Answers: 1. b, 2. a, 3. c, 4. b, 5. c, 6. c, 7. e, 8. b, 9. a, 10. c
35
Essay Questions
1. Describe the process of receptor mediated endocytosis, and explain why it is necessary for cells to
have a system to transport large molecules across the plasma membrane.
Specific regions of the cell membrane contain groups of receptors that are specific for one
or more large molecules that need to be imported into the cell. This region of the
membrane forms a clathrin lined pit. When the target molecules bind the receptor, the pit
folds into the cell to form a vesicle. The molecules bound to the receptor are now on the
inside of the vesicle. The vesicle transports the target molecules to the lysosome where
they can be digested. This system moves large molecules from the outside of the cell to
the inside of an organelle, without the need to cross two membranes.
2. How can a cell maintain internal concentrations that are much higher than the surrounding
environment of some small molecules such as Na+ ions while allowing rapid diffusion of other
molecules across the cell membrane?
The plasma membrane is selectively permeable. Some molecules are allowed to diffuse
through freely, others cannot pass, and may be pumped against the concentration
gradient by active transport
3. Describe three different adaptations that allow cells and/or organisms to live in a hypotonic
(hypoosmotic) environment.
• contractile vacuole expels excess water from the cell
• sea creatures may become isoosmotic to sea water
• plant cells prevent swelling and cell rupture by pressing against the cell wall (turgor
pressure)
• multicellular animals (and some plants) have an outer layer (skin) that is impermeable
to water, while internal cells are kept in isoosmotic fluid (blood)
4.
Explain how cell receptors are able to receive information from many different kinds of signaling
molecules (peptide and steroid hormones, neurotransmitters, amino acids), and translate these into
different actions inside the cell.
Cell surface receptor proteins have variable shapes for the region that extends outside
the cell membrane. These different shapes can form specific bonds with a wide variety of
different signaling molecules. After a receptor protein is bound by its specific ligand, the
portion of the molecule that extends inside the cell membrane is activated and transmits
its signal to the cell. The effect of this signal may to be to activate a G-protein, open a
gated membrane channel, or phosphorylate another protein.
36
CHAPTER 6. ENERGY AND METABOLISM
THE ESSENTIALS
Students need to know:
ƒ that enzymes function by lowering activation energy.
ƒ the catalytic cycle of an enzyme: how the enzyme is not permanently
altered and how a product is produced from substrate(s).
ƒ the factors that affect enzyme function: enzyme concentration,
substrate concentration, temperature, pH, salinity, competitive and
non-competitive inhibitors, activators.
ƒ how enzymes can be regulated through feedback inhibition.
ƒ the structure of the ATP (or GTP) molecule.
ƒ
how the structure of ATP enables it to transfer energy within the cell.
Key Terms
energy
kinetic energy
potential energy
thermodynamics
kilocalorie
joule
oxidation
reduction
oxidation-reduction (redox)
reactions
First Law of Thermodynamics
heat
Second Law of
Thermodynamics
entropy
free energy
endergonic
exergonic
activation energy
catalysis
enzymes
substrates
active sites
enzyme-substrate complex
inhibitor
competitive inhibitors
noncompetitive inhibitors
allosteric site
37
allosteric inhibitor
activators
cofactors
coenzyme
nicotinamide adenine
dinucleotide (NAD+)
adenosine triphosphate (ATP)
adenosine diphosphate (ADP)
metabolism
anabolism
catabolism
biochemical pathways
feedback inhibition
Concept Map
Enzymes
reduce
activation
energy
speed up
factors affect
enzyme
activity
active site
temperaure
metabolic
reactions
pH
substrate
specific
synthesis
reactions
digestion
reactions
build
molecules /
build
polymers
breakdown
molecules /
breakdown
polymers
lock and key
model
induced fit
model
cofactors
coenzymes
organic
compounds
inorganic /
metals
inhibitors
noncompetitive
inhibitors
38
competitive
inhibitors
Multiple Choice Questions
1. Chemical reactions in biological systems often involve the movement of electrons. Which of the
following is true?
a.
b.
c.
d.
e.
Oxidation reactions produce oxygen.
When it is oxidized, a molecule gains energy.
When it is oxidized, a molecule loses an electron.
A reduction reaction reduces the mass of a molecule.
When it is reduced, a molecule loses an electron.
2. The Second Law of Thermodynamics states
a.
b.
c.
d.
e.
that the Universe is constantly gaining energy.
that living things must store energy as sugar for later use.
that when one kind of energy is changed into another kind of energy, some is lost as heat.
that entropy is decreased when sunlight is absorbed by plants.
that all chemical reaction increase free energy.
3. An endergonic reaction
a.
b.
c.
d.
e.
produces products with more free energy than the reactants.
proceeds spontaneously.
reduces the entropy of the system.
release excess free energy as heat.
can be used to do work in the cell.
4. Which of the following is NOT true about enzymes?
a.
b.
c.
d.
They increase the rate of exergonic chemical reactions by lowering the activation energy.
They act as catalysts for specific chemical reactions.
They reduce the free energy produced by chemical reactions.
They bring substrates together in a position that stresses chemical bonds and makes it easier for
new bonds to form.
e. They can increase the rate of chemical reactions by more than a million fold.
5. The induced fit model refers to
a. the ability of enzymes to change their shape to fit many different substrate molecules.
b. the ability of enzymes to force other molecules to change their shape.
c. the movement of enzymes to different compartments of the cell in response to changes in the
concentration of substrate.
d. a change in enzyme shape that is caused by its binding to a substrate molecule.
e. the ability of RNA to change the shape of enzymes.
6. The rate of an enzyme-catalyzed reaction is NOT affected by
a.
b.
c.
d.
e.
Temperature.
pH.
the concentration of substrates.
the concentration of enzyme.
free energy of the reaction product.
39
7. Competitive enzyme inhibitors work by
a.
b.
c.
d.
e.
breaking down the chemical structure of the enzyme.
binding to the active site of the enzyme and blocking the substrate from reaching it.
changing the free energy of the reaction.
removing the metal ion from the active site.
changing the pH of the solution.
8. ATP stores energy
a.
b.
c.
d.
e.
by binding high energy electrons and hydrogen atoms.
in the unstable bonds between negatively charged phosphate groups.
in its internal hydrogen bonds.
by capturing energetic photons.
as a charge imbalance across the mitochondrial membrane.
9. ATP is useful to the cell because
a. the hydrolysis of the high energy phosphate bonds in ATP releases enough energy to power other
endergonic reactions.
b. it is a stable form of stored energy.
c. it requires little energy to synthesize ATP from ADP and inorganic phosphate.
d. it is regulated by allosteric inhibitors.
e. it can be imported from the surrounding environment.
10. How do very high temperatures (greater than 40°C) affect enzyme activity?
a. Higher temperatures increase enzyme activity because the rate of diffusion of substrates to the
active site is increased.
b. High temperatures denature enzymes, changing the shape of the active site and destroying
activity.
c. Higher temperatures increase enzyme activity due to increased molecular movement, making
chemical bonds easier to break and reducing activation energy.
d. High temperatures reduce enzyme activity due to the breakdown of chemical bonds in the
substrate.
e. High temperatures do not affect enzyme activity.
Answers: 1. c, 2. c, 3. a, 4. c, 5. d, 6. e, 7. b, 8. b, 9. a, 10. b
Essay Questions
1. All living things require energy in order to conduct the processes of life such as growth, reproduction,
movement, and maintaining homeostasis. Describe two different sources of external energy that can be
used by a cell, and how that energy is converted into a useable form.
Cells obtain energy from photosynthesis (capturing energy from photons) or by importing
high energy molecules from outside.
2. Many exergonic reactions, such as the oxidation of wood (fire) produce large amounts of heat, which
would destroy a cell. Other anabolic reactions require the input of large amounts of energy to
synthesize large molecules. How does a cell manage to conduct all of these reactions at a constant
temperature?
40
Cells use enzymes to reduce the activation energy of exergonic reactions. Much of the
energy of catabolic reactions is captured as ATP and NADPH, which are then used to
power other endergonic reactions in the cell. Many reactions are conducted in steps, as a
biochemical pathway, so that the input, or release of energy is not very large at each
step.
3. Why is it necessary for cells to precisely regulate chemical reactions? Why is feedback regulation an
efficient system for regulation of biochemical pathways?
Unregulated reactions could produce too much of a product and use up substrates and
energy that are needed for other functions. Feedback regulation is efficient because the
product of a pathway directly inhibits the first reaction in the pathway, so that no
unnecessary intermediates are produced and no additional mechanism is needed to sense
the level of product and turn off the pathway.
4. At a constant temperature and pH, the rate of an enzyme-catalyzed reaction is proportional to the
concentration of enzyme and substrate. Explain why, at very high concentrations of substrate, the
reaction rate becomes constant and at very high concentrations of enzyme, the reaction rate drops to
zero.
At high concentrations of substrate, all active sites on enzyme molecules are filled and the
enzyme is working at its maximum speed. Adding even more substrate will not increase
the reaction rate. At very high concentrations of enzyme, all available substrate will be
used up, and the reaction cannot continue.
41
CHAPTER 7. HOW CELLS HARVEST ENERGY
THE ESSENTIALS
Students need to know:
ƒ the fundamental differences between fermentation and cellular
respiration: the inputs, the conditions that trigger each process, and the
products they produce.
ƒ that glycolysis is a universal metabolic process in both prokaryotes
and eukaryotes.
ƒ the summary of each stage of aerobic respiration: the inputs of each,
the products from each, where each occurs, and the net ATP
production of each.
ƒ that glycolysis oxidizes glucose to produce 2 pyruvate molecules and a
net yield of 2 ATP.
ƒ how pyruvate is brought from the cytosol to the Krebs cycle (citric
acid cycle) in the mitochondria.
ƒ the value of the Krebs cycle is in the production of electron carriers,
NADH and FADH2, which will have a role in the electron transport
chain.
ƒ how the electron transport chain harvests the electrons from NADH
and FADH2 to fuel chemiosmosis.
ƒ the enzymatic role of ATP synthase in the synthesis of ATP from ADP
& Pi and how a proton gradient is needed to drive this reaction.
ƒ the role of oxygen as the final electron acceptor in the electron
transport chain.
ƒ how the double membrane structure of the mitochondria enable their
function in chemiosmosis.
Key Terms
autotrophs
heterotrophs
digestion
catabolism
aerobic respiration
anaerobic respiration
fermentation
ATP synthase
substrate-level
phosphorylation
oxidative phosphorylation
glycolysis
aerobic respiration
anaerobic respiration
acetyl-CoA
Krebs cycle
electron transport chain
ATP
NADH
FADH2
42
NADH dehydrogenase
chemiosmosis
oxidation
reduction
feedback inhibition
metabolism
mitochondria
mitochondrial matrix
intermembrane space
Approach:
Cellular respiration is one of the most difficult topics in AP biology. This chapter is one in which students
(and teachers) can get lost in the details and lose track of the over-riding principles. It is often mistakenly
approached as an exercise in memorizing the steps of glycolysis, Krebs cycle, and the electron transport
chain as well as emphasizing an accounting of exact numbers of ATPs. High school students do not have to
know the biochemical details of each of these pathways. They will face that challenge in Biochemistry
class in college.
Instead, students primarily need to understand the overriding goal of the processes (transferring energy
from fuel molecules to ATP to supply energy for life processes), how that goal is achieved in different
organisms and under different conditions (anaerobic vs. aerobic), the key reactants and the products only in
terms of where they fit into the flow through the pathways, how the pathways integrate with each other,
and where they occur in the cell.
To set the stage, for this topic, three principles need to be established:
•
The point is to make ATP.
First discuss why the structure of ATP allows it to serve the function of short term energy storage.
ATP is unstable—the third phosphate group is easily donated to other molecules (repelled by the
two other negative phosphate groups) and releases enough energy to drive the endergonic
reactions of life.
• In biology, moving hydrogens moves electrons; moving electrons moves energy.
Students will see hydrogens move from fuel molecules to NAD and FAD and these molecules
will be called “electron carriers”. There is a disconnect, if they do not understand that electrons
do not move on their own around biological systems. They are carried by hydrogen atoms and
this is the way for organisms to move energy from one molecule to another.
• Oxidation and reduction reactions are coupled.
Oxidation will be seen in respiration in a number of ways—removing a hydrogen from a
molecule or cleaving off a carbon and bonding it to oxygen yielding CO2. Every time this
happens have your students look for what other molecule was reduced. It will help them track the
electron carriers (NADH and FADH2) .
With regards to the pathways, a comparative approach is beneficial—glycolysis vs. Krebs cycle vs. the
electron transport chain—and can be carried through when photosynthesis is introduced in the next
chapter. Most especially, do not get side-tracked in counting the exact number of ATP molecules that are
generated in each pathway or an exact accounting of the conversion of NADH and FADH2 to ATP. This is
focusing on unnecessary details for this level student. The critical concept in ATP accounting is that
glycolysis and the Krebs cycle produce minimal ATP and the electron transport chain produces roughly
10 times the quantity.
Glycolysis
• This is the earliest evolutionary step for harvesting energy from organic molecules. An
evolutionary perspective is useful to tie this unit to evolution and taxonomy; all organisms
perform glycolysis and the genes for glycolytic enzymes are highly conserved.
• Fuel molecules (glucose) are oxidized in small steps to produce a little bit of ATP and a little
bit of NADH.
• The initial step costs some ATP (activation energy), because glucose needs to be destabilized
to cleave it in two.
• Pyruvate becomes a branching point to fermentation or Krebs cycle, dependent on conditions
or taxonomy (kind of organism).
• Glycolysis occurs in the cytosol.
• NAD and ADP must be regenerated.
Oxidation of pyruvate
• A 3-carbon pyruvate moves to mitochondria in eukaryotes and is oxidized in small steps to a
2-carbon acetyl CoA generating CO2.
• The oxidation of pyruvate occurs in the mitochondrial
matrix.
43
Krebs cycle
• Acetyl CoA is fed into the Krebs cycle to regenerate a 6-carbon compound that is oxidized in
many small steps. Since oxidation of one molecule is accompanied by reduction of another,
each time a Krebs cycle intermediary compound is oxidized, an electron carrier (NAD or
FAD) is reduced (forming NADH or FADH2).
• Each time a carbon is cleaved off a Krebs cycle intermediary compound (6-carbon to 5carbon to 4-carbon molecules), CO2 is generated.
• A small amount of ATP is generated, but the primary value of the Krebs cycle is the
generation of electron carriers (NADH & FADH2) to be used tin the electron transport chain.
• NAD, FAD, and ADP must be regenerated.
•
The Krebs cycle occurs in the mitochondrial matrix.
Electron transport chain
• NADH and FADH2 donate their hydrogens (H+ proton and electron separate) to the electron
transport chain.
• The electrons moving through the chain drive the pumping of the protons (H+) into the
intermembrane space by transmembrane proteins. Oxygen serves a pivotal role as the final
electron acceptor pulling the electrons down the electron transport chain.
• The protons flow down their concentration gradient from the intermembrane space into the
mitochondrial matrix through ATP synthase enzyme and—like water over a water wheel that
grinds grain—they enable the work of bonding Pi to ADP to make ATP.
• NAD and FAD have been regenerated for the Krebs cycle.
• The electron transport chain occurs in the inner mitochondrial membrane.
There is a lot of new vocabulary in this unit. It is helpful to repeat familiar words alongside these new
science terms repeatedly relating them in students’ minds:
• anabolism : synthesis : building molecules : endergonic reactions
• catabolism : digestion : breaking down molecules : exergonic reactions
In addition, relating the details of respiration to everyday activities helps to anchor this topic in real life:
•
•
Where does the oxygen come from? (inhale, the other use of the term respiration)
What is oxygen used for in cellular respiration? (final electron acceptor in the electron transport
chain)
• Where does the CO2 come from in cellular respiration? (oxidation of intermediary molecules in
the Krebs cycle)
• Where does the CO2 go? (exhale, the other use of the term respiration)
• Where do the fuel molecules for respiration come from? (eating and then digestion of food)
• What is the ATP used for? (examples, an active muscle needs over 10 million ATP per second;
active transport Na+/K+ pumps in the membranes of neurons)
This chapter can be a segue into an introductory animal unit which would include nutrition, circulatory
system, gas exchange, and muscle/motor systems to show how each of these structures and processes
either supports respiration or is supported by it.
44
Concept Map
cellular
respiration
harvesting
chemical
energy
uses
anaerobic
respiration /
absence of
O2
aerobic
respiration /
presence of
O2
cytoplasm
mitochondria
enzymes
inner
ATP
synthase
reduction
reactions
oxidation
reactions
membrane
kinases
glycolysis
fermentation
ethanol in
yeast
lactic acid in
animals
Kreb's cycle
provides
produces
carbon
skeletons
electron
carriers
biosynthesis
pathways
45
NADH
transport
chain
chemiosmosis
FADH2
proton
gradient
produces
releases
energy / ATP
CO2
Multiple Choice Questions
1. What types of chemical bonds contain most of the energy in food?
a.
b.
c.
d.
e.
oxygen-hydrogen and carbon-nitrogen bonds
nitrogen-carbon and phosphorus-carbon bonds
carbon-hydrogen and carbon-oxygen bonds
sulfur-carbon and hydrogen-nitrogen bonds
hydrogen-nitrogen and hydrogen-oxygen bonds
2. The term aerobic respiration refers to
a.
b.
c.
d.
e.
the direct transformation of glucose into ATP in the absence of oxygen.
the breakdown of food molecules by reactions where oxygen is the ultimate electron acceptor.
the formation of alcohol by yeast in a closed bottle of fruit juice.
the formation of lactic acid in muscle tissue during hard work.
the synthesis of carbohydrate where oxygen is the ultimate electron donor.
3. Which of the following is NOT true about glycolysis?
a. The enzymes that carry out the reactions of glycolysis are located in the cytoplasm, not bound to
any membrane or in any organelle.
b. Glycolysis produces a total of two ATP molecules for each molecule of glucose.
c. Glycolysis occurs only in prokaryotic cells.
d. Glycolysis does not make use of the electron transport chain or ATP synthase enzyme.
e. Glycolysis is always the first stage of respiration, even when oxygen is present.
4. The initial steps of glycolysis require an input of energy. What molecules provide this energy?
a.
b.
c.
d.
e.
Phospholipids in the cell membrane transfer phosphate to glucose.
The electron transport chain provides the energy.
Phosphate is transferred from ATP to glucose.
The hydrolysis of pyruvate provides the energy.
The active transport of glucose across the cell membrane provides the energy.
5. Where are the enzymes of the Krebs cycle located?
a.
b.
c.
d.
e.
in the cytoplasm
embedded in the membranes of the endoplasmic reticulum
in the inner matrix of the mitochondrion
in the ribosome
embedded in the outer membrane of the mitochondrion
46
Questions 6-7. Use the following diagram
Cytoplasm
Glucose
A
ATP
Pyruvate
B
NADH
Intermembrane
space
CO2
AcetylCoA
NADH
E
CO2
C
ATP
FADH 2
H2 O
e-
D
ATP
NAD + and FAD
Inner mitochondrial membrane
Mitochondrion
6. In the diagram above, identify the location of the electron transport chain.
a. A
b. B
c. C
d. D
e. E
7. In the diagram above, identify the location of the glycolysis reactions.
a. A
b. B
c. C
d. D
e. E
8.
Pyruvate is the final product of glycolysis. The pyruvate dehydrogenase multi-enzyme complex
oxidizes pyruvate and creates the starting material for the Krebs cycle. The products of pyruvate
dehydrogenase are
a.
b.
c.
d.
e.
H2O and CO2.
NADPH and glyceraldehyde-3-phosphate (G3P).
NADH, CO2, and acetyl-CoA.
H2O and citrate.
citrate and ATP.
47
9. How much ATP is directly synthesized by the enzymes of the Krebs cycle from each molecule of
glucose?
a. The enzymes of the Krebs cycle produce 36 molecules of ATP from each molecule of glucose. b.
The enzymes of the Krebs cycle directly produce only 2 molecules of ATP for each molecule of
glucose.
c. The enzymes of the Krebs cycle do not directly produce any ATP, all ATP production comes
from ATP synthase driven by chemiosmosis.
d. The enzymes of the Krebs cycle produce 18 molecules of ATP from each molecule of glucose?
e. The enzymes of the Krebs cycle use up 2 molecules of ATP.
10. The function of the electron transport chain in the inner mitochondrial membrane is to
a. oxidize pyruvate to acetyl-CoA.
b. extract energy from high energy electrons and pump H+ ions out of the mitochondrial matrix.
c. transport NADH molecules out of the mitochondria so that they can be used to synthesize ATP in
the cytoplasm.
d. synthesize ATP by chemiosmosis as H+ ions are pumped into the mitochondrial matrix.
e. organize the enzymes of the Krebs cycle into a multi-enzyme complex.
Answers:
1. c, 2. b, 3. c, 4. c, 5. c, 6. d, 7. a, 8. c, 9. b, 10. b
Essay Questions
1. It is often said that the enzyme ATP synthase works like a water wheel turbine. Explain the details of
this analogy, what flows, what turns, and what work is done?
ATP synthase forms a channel in a membrane (the inner membrane of mitochondria or
chloroplasts in eukaryotes). The greater concentration of H+ ions on one side of the
membrane is like the potential energy of water above a dam. Hydrogen ions flow through
the channel like water flows over a water wheel. The energy of the turning wheel of ATP
synthase powers the synthesis of ATP from ADP and phosphate.
2. Why does the aerobic respiration of glucose into CO2 yield more energy than the anaerobic
production of alcohol or lactic acid?
More C-H bonds are broken when glucose is oxidized to CO2 than when it is transformed
into alcohol or lactic acid. The free energy of the products is lower when CO2 is produced.
48
3. Explain why the process of glycolysis is considered one of the earliest biochemical processes to
evolve in the first living cells.
All living cells are capable of glycolysis. The enzymes used in glycolysis are very similar
across all domains of life from bacteria to mammals. Glycolysis takes place in the
cytoplasm, suggesting that it predates the existence of organelles and intra-cellular
membranes.
4. Acetyl-CoA is a key molecule in both oxidative respiration and the synthesis of fat (and other large
molecules). How can the cell control its metabolism if the same molecule is used in both energy
production and biosynthesis?
The fate of acetyl-CoA is determined by the level of ATP present in the cell. At low ATP
levels, of acetyl-CoA enters the Krebs cycle to produce energy. When ATP levels are
high, the key enzyme that forms citrate from acetyl-CoA and a 4-carbon sugar is
inhibited, and acetyl-CoA is used for biosynthesis of fatty acids.
49
CHAPTER 8. PHOTOSYNTHESIS
THE ESSENTIALS
Students need to know:
ƒ how the photosystems collect light energy and convert it to chemical
energy.
ƒ how photosystem II produces ATP by chemiosmosis in the light
reactions.
ƒ how the double membrane structure of the chloroplasts enable their
function in chemiosmosis.
ƒ how photosystem I produces NADPH in the light reactions.
ƒ how the Calvin cycle uses the products of the light reactions to
synthesize sugar (G3P).
ƒ the causes of and metabolic costs of photorespiration.
ƒ the evolutionary adaptations of C4 and CAM plants that enable them to
efficiently perform the Calvin cycle in hot, dry habitats.
ƒ the commonalities and distinctions between photosynthesis in
chloroplasts and aerobic respiration in mitochondria.
Key Terms
light-dependent reactions
Calvin cycle
light-independent reactions
photosystem
photosystem II
photosystem I
carbon fixation
absorption spectrum
pigments
chlorophyll
chlorophyll a
chlorophyll b
action spectrum
carotenoid
NADPH
cyclic photophosphorylation
noncyclic
photophosphorylation
50
Calvin cycle
Rubisco
RuBP
glyceraldehyde-3-phosphate
C3 photosynthesis
photorespiration
C4 photosynthesis
crassulacean acid metabolism
(CAM)
Approach:
Once cellular respiration—with its concepts of chemiosmosis and an electron transport chain—has been
taught, students may have an easier time with the processes of photosynthesis. It is useful to remind them
along the way of the similarities and differences between the two processes. As with respiration,
photosynthesis should not become an exercise in memorizing the steps of the light reactions and the
Calvin cycle. High school students do not have to know the biochemical details of each of these pathways.
They will face that challenge in Biochemistry class in college.
Instead, students primarily need to understand the overriding goal of each process, the key reactants and
the products only in terms of where they fit into the flow through the pathways, how the pathways
integrate with each other, and where they occur in the cell. Emphasize how the light reactions use
chemiosmosis to produce ATP and NADPH (through the electron transport chain) and that these products
are then used in the Calvin cycle to convert CO2 to sugars for plant growth and in the process recycle ADP
and NADP back to the light reactions.
The light reactions can be presented from the evolutionary perspective as two separate photosystems that
evolved at separate times, but that now work in coordination as both noncyclic and cyclic
photphosphorylation. Stress why water is split (to fill the electron hole in PSII), why oxygen is produced
(merely a byproduct of the splitting of water), and that two photons of different energy are needed for PS
I/PS II photosynthesis.
It is very important not to get lost in the details and accounting of the Calvin cycle. Students only need to
know RuBP as the carbon acceptor, Rubisco as the enzyme that catalyzes this reaction which captures
(fixes) carbon, the immediate product is G3P and that the whole process is fueled by the energy of ATP
and the reducing power of NADPH.
At some point in this unit be sure to paint a picture of wonderment for students of just how incredible this
process of photosynthesis is—with the power of light and water, plants weave life out of thin air.
C4 and CAM should be seen as variations on carbon fixation, because plants had to evolve alternative
systems given the limitations of their enzymes and their need to conserve water. The key problem these
plants are trying to overcome is that Rubisco is a very inefficient enzyme in the presence of high O2. In
high O2, Rubisco bonds oxygen to RuBP rather than carbon, so the plants have to keep O2 away from
Rubsico. The key point is how carbon dioxide is grabbed out of the air—carbon fixation—and then
handed off to the Calvin cycle.
•
C4 plants separate the two steps of carbon fixation anatomically. They use two different cells to
complete the process.
• CAM plants separate the two steps of carbon fixation temporally. They do them at two different
times.
Many students may confuse photorespiration with cellular respiration, but they are two entirely different
processes. Although they both produce CO2, photorespiration is a damaging reaction because it does not
produce ATP.
This chapter can be a segue into a introductory plant unit which would include structure of plants, plant
growth, transpiration, gas exchange, and nutrient uptake to show how each of these structures and
processes either supports photosynthesis or is supported by it.
51
Concept Map
Photosynthesis
adaptations
chlorophyll
chloroplast
hot conditions
captures
C3
C4
CAM
light energy
reduces
stroma
thylakoid
photorespiration
light
reactions
caused by
Calvin cycle
high O2
concentration
reducing
power
split water
energize
electrons
reduces
NADP to
NADPH
electron
transport
chain
chemiosmosis
carbon fixation
(CO2 to RUBP)
releases
O2
produces
powers
G3P
requires
produces
used to build
proton
gradient
ATP / energy
glucose, sucrose &
other carbohydrates
52
uses
ribulose
bisphosphate
carboxylase
"Rubisco"
Multiple Choice Questions
1. What is the internal membrane system of the chloroplast that contains the photosynthetic pigments?
a.
b.
c.
d.
e.
stroma
microvilli
bundle sheath
Golgi bodies
thylakoids
2. Where does the synthesis of carbohydrates from CO2 take place?
a.
b.
c.
d.
e.
inside the thylakoid membranes
in the cytoplasm of plant cells
in the stroma of the chloroplast
in the vacuole of plant cells
in the lysosome
3. Why does the rate of chemical reactions in the Calvin cycle increase when the temperature is raised
from 25°C to 35°C, but decrease when it is raised from 35°C to 45°C?
a.
b.
c.
d.
e.
The rate of diffusion of CO2 is reduced at higher temperatures.
The enzymes of the Calvin cycle become denatured at temperatures above 35°C.
The thylakoid membranes become depolarized at temperatures above 40°C.
Water molecules cannot pass through the chloroplast membrane at temperatures above 40°C.
The pH in the chloroplast stroma increases at temperatures above 35°C.
4. Most of the dry mass of living plants (cellulose, starch, sugars, proteins) originates as
a.
b.
c.
d.
e.
hydrogen and oxygen from water.
carbon and oxygen from CO2.
nitrogen from the air.
light energy.
soil minerals.
5. The diagram below illustrates
Photon
Antenna
complex
Thylakoid
membrane
Photon
H+ + NADP+
NADPH
Fd
e-
e-
Q
e-
eH2O
pC
Proton
gradient
Plastoquinone Plastocyanin Ferredoxin
Water-splitting
enzyme
H+
1
Thylakoid
2H+
O
2
2
space
Photosystem II
b6-f complex Photosystem I NADP reductase
53
a.
b.
c.
d.
that ATP is synthesized inside the thylakoid space by chemiosmosis.
the circular movement of electrons in cyclic photophosphorylation.
the movement of CO2 in C4 photosynthesis.
the organization of the two photosystems and electron transport molecules in the thylakoid
membrane.
e. the Calvin cycle.
6. The free oxygen (O2) produced by photosynthesis comes from the
a.
b.
c.
d.
e.
breakdown of water.
breakdown of CO2.
breakdown of glucose.
synthesis of NADPH from NADP and H+.
synthesis of ATP by chemiosmosis.
7. Plant leaves are green because
a.
b.
c.
d.
e.
chlorophyll absorbs green light.
chlorophyll absorbs red and blue light and reflects green light.
plants have pigments that absorb all wavelengths of light.
chlorophyll absorbs blue light and carotenoids absorb red and orange light.
chlorophyll b emits photons of green light.
8. After a photon is absorbed by the pigments of photosynthetic photosystem II, the chlorophyll
molecule at the reaction center becomes oxidized. Then water reacts with the reaction center by
a.
b.
c.
d.
e.
donating an electron receiving an electron.
donating a hydrogen atom.
forming hydrogen bonds.
causing the hydrolysis of the chlorophyll molecule.
pumping protons into the thylakoid.
9. High levels of O2 in the chloroplast reduce the efficiency of photosynthesis. This process is called
a.
b.
c.
d.
e.
the Calvin cycle.
non-cyclic photophosporylation.
C4 photosynthesis.
crassulacean acid metabolism.
Photorespiration.
10. The photosynthetic electron transport chain pumps protons
a.
b.
c.
d.
e.
out of the chloroplast into the cytoplasm.
into the inner space within the thylakoid membrane.
into the stroma of the chloroplast.
across the plasma membrane of the plant cell.
to the antennae molecules of photosystem II.
Answers:
1. e, 2. c, 3. b, 4. b, 5. d, 6. a, 7. b, 8. a, 9. e, 10. c
54
Essay Questions
1. Mitochondria and chloroplasts both perform the essential function of providing energy to the cell.
Describe the many structural and biochemical similarities of these organelles.
Mitochondria and chloroplast both originated as free living bacteria which were
incorporated in eukaryotic cells by endosymbiosis. They both have two layers of
membranes and operate an electron transport chain across their inner membranes which
produces an excess of protons in the matrix/stroma space. This proton gradient powers
the synthesis of ATP by chemiosmosis with ATP synthase. They both have their own
circular DNA molecule which replicates independently of the host cell.
2. Why are the reactions of the Calvin cycle more accurately called “light independent” rather than
“dark reactions?”
Light drives the synthesis of ATP and NADPH by the photosystems and the electron
transport chain which takes place in the thylakoid membranes. ATP and NADPH plus
CO2 are the substrates for the synthesis of carbohydrates by the enzymes of the Calvin
cycle. These reactions can take place without the direct input of light energy if all of the
necessary substrates are present—so they are light independent. However, these
reactions do not require darkness, and in most plants they do not take place in the dark,
but are closely coupled with the “light reactions.”
3. C4 plants are able to photosynthesize more efficiently than C3 plants in hot dry conditions. Describe
the differences in structure and function (anatomy and biochemistry) between C3 and C4 plants.
C3 plants suffer from photorespiration—the O2 produced by photosynthesis reacts with
RuBP, an intermediate in the Calvin cycle producing CO2 and short circuiting the
synthesis of carbohydrates. C4 plants conduct the Calvin cycle in different cells from the
ones that perform the light dependent reactions. These cells specialized for the Calvin
cycle are called bundle sheath cells, which have cell membranes that are impermeable to
CO2 and O2. In C4 plants, is fixed into malate and transported to the bundle sheath cells,
where it is converted back to CO2 which is then converted to glucose by the Calvin cycle
enzymes.
4. Purple sulfur bacteria use hydrogen sulfide as a raw material and produce sulfur instead of CO2 as a
product of photosynthesis. Describe the parallels between the photosynthesis based on sulfur vs.
water and how this led to the hypothesis that the oxygen produced by plants comes from water.
Sulfur bacteria use H2S as the electron donor for photosystem II rather than H2O and
they produce elemental sulfur rather than oxygen as a product of photosynthesis. The
reaction looks like this:
CO2 + 2 H2S Æ (CH2O) + 2 S
Which is very similar to the formula for standard green plant photosynthesis:
CO2 + 2 H2O Æ (CH2O) + O2
In the 1930s, C.B. van Niel observed the chemistry of this sulfur bacteria photosynthesis
and hypothesized that water played a similar role to H2S in green plant photosynthesis –
so that CO2 provides both the carbon and the oxygen for carbohydrate synthesis and the
free oxygen is derived from water.
55
CHAPTER 9. CELL COMMUNICATION
THE ESSENTIALS
Students need to know:
 ƒ that cells communicate through a system of complementary-shaped molecules:
signal molecules (ligands) and protein receptors.
 ƒ the three stages of cell signaling: reception, signal transduction, and cellular
response.
 ƒ that phosphorylation is a common mechanism for activating or de- activating
protein function.
 ƒ that cell-surface receptor proteins receive signals from hydrophilic ligands which
cannot pass through the cell membrane.
 ƒ examples of the membrane receptor system such as neurotransmitters, G proteincoupled receptors, receptor tyrosine kinases (such as the response to insulin triggering
the activation of glycogen synthase).
 ƒ that intracellular receptors (within the cytosol) receive signals from hydrophobic
ligands which can easily pass through the cell membrane.
 ƒ examples of the intracellular receptor system such as steroid hormones triggering gene
expression.
 ƒ that membrane receptors can trigger secondary messenger systems within the cell
and often amplify the signal via a cascade response.
 ƒ examples of the secondary messenger systems such as the glucagon- cAMP cascade
that triggers glycogen hydrolysis and the MAP kinase cascade that advances the cell
cycle.
Key Terms
receptor protein
ligand
autocrine signaling
paracrine signaling
endocrine signaling
synaptic signaling
neurotransmitters
signal transduction
transduction
phosphorylation
protein kinase
intracellular receptors
cell surface receptors
ion channel
G protein
G protein-coupled receptors
56
second messenger
cyclic AMP/cAMP
adenylyl cyclase
receptor tyrosine kinases
cascade
MAP kinase cascade
scaffold protein
coactivators
CHAPTER 10. HOW CELLS DIVIDE
THE ESSENTIALS
Students need to know:
ƒ the structure of the eukaryotic chromosome.
ƒ the relationship between homologous chromosomes.
ƒ the structure of the replicated eukaryotic chromosome.
ƒ the biological purpose of mitosis: the production of cells with the same
genetic composition.
ƒ the stages of mitosis.
ƒ the distinctions between the process of cytokinesis in animal cells and
plant cells.
ƒ the role of kinases and cyclins in the control of the cell cycle in normal
cell division and in cancer.
ƒ the role of oncogenes, proto-oncogenes, and tumor-suppressor genes
in the development of cancer.
Key Terms
binary fission
mitosis
chromatin
nucleosome
heterochromatin
euchromatin
karyotype
haploid (n)
diploid (2n)
homologous
homologues
centromere
chromatids
cell cycle
G1
S
G2 interphase
mitosis (M)
cytokinesis (C)
G0 phase
centromere
kinetochore
condensation
75
centrioles
prophase
aster
metaphase
anaphase
telophase
cell plate
middle lamella
cyclin-dependent kinases
(Cdk) growth
factors protooncogenes
Concept Map
cell division
enables
reproduction
growth &
repair
part of
cell cycle
cyclins,
Cdk's
regulation
loss of
regulation
unicellular
multicellular
cancer
interphase
mitosis
G0
G1, S, G2
resting stage
cell growth,
organelle & enzyme
production,
DNA replication
76
prophase,
metaphase,
anaphase,
telophase
cytokinesis
cleavage
furrow
in animals
cell plate
in plants
Multiple Choice Questions
1. The picture to the right shows a cell in
a.
b.
c.
d.
e.
prophase.
metaphase.
anaphase.
telophase.
cytokinesis.
2. The DNA of a eukaryotic cell is extremely long. It is packed into the nucleus by
a.
b.
c.
d.
e.
forming a shell of tightly packed layers inside the nuclear membrane.
wrapping around the centromere like a ball of string.
disassembly into smaller pieces when the cell is not dividing.
winding around histone proteins, which then pack together to form solenoids and loops.
looping through pores in the nuclear membrane.
3. After the DNA is duplicated, how does the dividing cell ensure that the two copies of each
chromosome are partitioned so that one goes into each of the new cells?
a. The nucleus divides in two parts and one goes into each new cell. b.
Kinetochore fibers attach to sister chromatids, and pull them apart. c.
Chromosomes are randomly sorted by active transport.
d. Histone proteins attach to the telomeres and drag the sister chromatids along actin fibers to
opposite ends of the cell.
e. Crossing over ensures that portions of each chromosome are sent to the two daughter cells.
4. Eukaryotic chromosomes are composed of
a.
b.
c.
d.
e.
DNA and RNA.
only DNA.
DNA, protein, and RNA.
RNA and protein.
DNA and polysaccharides.
5. A somatic human cell is
a.
b.
c.
d.
e.
haploid.
diploid.
polyploid.
aneuploid.
a gamete.
77
6. What is the primary function of microtubules in mitosis?
a.
b.
c.
d.
e.
to regulate the replication of DNA
to constrict a band around the center of the cell until it pinches the cell into two equal halves
to guide the separation, sorting, and movement of duplicate chromosomes to the daughter cells
to divide the mitochondria and other organelles evenly between the two daughter cells
to move chromosomes through the nuclear pore complex
7. Cyclin dependent kinase enzymes are involved in all of the following EXCEPT
a.
b.
c.
d.
e.
separating replicated chromosomes during anaphase of mitosis.
regulating the cell cycle.
allowing the cell to progress from the G1 to S phase.
initiation of DNA synthesis.
phosphorylation of growth suppressing proteins such as retinoblastoma protein.
8. DNA replication takes place during
a.
b.
c.
d.
e.
prophase.
interphase.
anaphase.
telophase.
G0 phase.
9. All of the following promote the growth of cancer cells EXCEPT
a.
b.
c.
d.
e.
oncogenes.
mutations in tumor suppressor genes.
mutations that inactivate p53.
proto-oncogenes.
high activity of cyclin-dependent kinases.
10. The drug colchicine prevents the proper assembly of microtubules. What effect would this have on
cell division?
a. It would prevent DNA replication.
b. It would prevent the pinching in of the cell membrane in cytokinesis.
c. It would prevent the assembly of chromosomes on the metaphase plate and the sorting during
anaphase.
d. It would force cells to divide before the DNA was fully replicated.
e. It would prevent the reassembly of the nuclear membranes in telophase.
Answers:
1. c, 2. d, 3. b, 4. d, 5. b, 6. c, 7. a, 8. b, 9. d, 10. c
78
Essay Questions
1. Humans have 46 chromosomes, bread yeast has 16 and dogs have 78. The number of chromosomes
that a species has seems quite arbitrary, so why is it that a human generally cannot live with a single
chromosome added to or lost from the usual 46?
All mammals have a similar total number of genes, however these are divided into
different numbers of chromosomes of differing length in each species. Each human
chromosome has thousands of genes that function in a complex and tightly regulated
fashion—so the lack of one copy of these genes, or an extra copy of them, will disrupt the
function of many systems.
2. How is mitosis different in prokaryotes vs. eukaryotes?
Both prokaryotes and eukaryotes replicate DNA during interphase, but prokaryote DNA
replication begins at a single origin of replication site on the chromosome while eukaryote
DNA replicates from many origins. Prokaryotes do not have a nucleus. Prokaryotes have
a single circular chromosome. After DNA replication, the two copies of the chromosome
are attached to the cell membrane in different spots, and they end up in the two daughter
cells after cytokinesis. Prokaryotes do not use spindle fibers to guide
the chromosomes during cell division.
3. Multicellular animals have an additional layer of regulation of cell division compared to yeast and
other unicellular eukaryotes. How do animals prevent excessive cell growth that would disrupt the
organization of tissues and organs?
Multicellular animals block cell division using growth regulators. These substances are
produced in very small quantities and are required for cells to initiate cell division.
Epithelial cells also have contact inhibition—the stop growing when they touch other
cells.
4. Mitosis is sometimes described as a “dance of the chromosomes”. Explain why this is a good
metaphor.
Each chromosome has a partner—a sister chromatid. They circle around during
prophase, then all line up at metaphase. Then in anaphase, each chromosome lets go of
its partner and they march to opposite ends of the cell. In telophase and cytokinesis, the
cell splits in two, then the dance can begin again.
79
CHAPTER 11. SEXUAL REPRODUCTION AND MEIOSIS
THE ESSENTIALS
Students need to know:
ƒ the difference between asexual and sexual reproduction.
ƒ the function of meiosis and fertilization in sexual reproduction.
ƒ how chromosome number is reduced from diploid to haploid in the
process of meiosis.
ƒ the role of homologous chromosomes in meiosis.
ƒ the genetic consequences of crossing over.
ƒ the three key differences between mitosis and meiosis.
ƒ how crossing over, independent assortment, and random fertilization
increase genetic variation.
Key Terms
gametes
somatic cells
zygote
fertilization
syngamy
diploid
haploid
sexual reproduction
meiosis I
meiosis II
80
synapsis
chiasma (chiasmata)
independent assortment
asexual reproduction
parthenogenesis
Concept Map
sexual
reproduction
joining of
gametes
diploid
organisms
need to produce
fertilization
recombination
zygote
creates
variation
haploid cells
synapsis
crossing
over
creates
variation
81
gametes
independent
assortment
meiosis
meiosis I
meiosis II
separates
homologous
pairs
separates
sister
chromatids
reduction
division
4 haploid
cells
Multiple Choice Questions
1. In meiosis, all of the following are true EXCEPT
a.
b.
c.
d.
e.
the cell divides twice without a pause for growth
the number of chromosomes is reduced in half
DNA replicates twice
sister chromatids separate in the second anaphase
crossing over occurs in the first prophase
2. Meiosis produces
a.
b.
c.
d.
e.
two haploid daughter cells.
two diploid daughter cells.
four haploid daughter cells.
four diploid daughter cells.
four daughter cells with unequal number of chromosomes.
3. The segregation of genes (which produces genetic diversity) happens at what stage of meiosis?
a.
b.
c.
d.
e.
prophase I
anaphase I
cytokinesis I
metaphase II
anaphase II
4. The joining of male and female gametes is called
a.
b.
c.
d.
e.
synapsis.
cytokinesis.
parthenogenesis.
fertilization.
binary fission.
5. All of the following occur during synapsis EXCEPT
a.
b.
c.
d.
e.
DNA replication
pairing of homologous chromosomes
crossing over
formation of chiasmata
formation of the synaptonemal complex
82
6. The life cycle of ferns alternates generations as diploid and haploid cells. After meiosis, haploid cells
grow into a multicellular form. Then single cells are produced for the purpose of mating. These
mating cells are called
a.
b.
c.
d.
e.
germ-line cells.
spores.
gametes.
sporophytes.
somatic cells.
7. How is the function of spindle microtubules different in metaphase I of meiosis as compared to
mitosis?
a. In meiosis, the spindle microtubules are attached to the ends of chromosomes, rather than to the
centromeres.
b. The spindle microtubules do not attach to chromosomes in meiosis.
c. In meiosis, the spindle microtubules only attach to one of each pair of homologous chromosomes.
d. In meiosis, the spindle microtubules attach to only one side of the centromere for each pair of
sister chromatids.
e. In meiosis, the spindle microtubules are not attached to centrioles.
8. The second cell division of meiosis is similar to mitosis EXCEPT
a. The spindle microtubules bind to kinetochores and pull apart sister chromatids to opposite poles
of the dividing cell.
b. The nuclear membrane breaks down during prophase and re-forms during telophase.
c. Centrioles guide the formation of spindle microtubules.
d. After cytokinesis, the daughter cells have half the number of chromosomes as somatic cells.
e. Chromosomes consisting of sister chromatids align alone the metaphase plate.
9. Germ-line cells are
a.
b.
c.
d.
e.
haploid.
present only in females.
capable of producing all of the different specialized cell types in the body.
specialized for mitosis.
specialized for meiosis.
83
Haploid cells (n)
A
B
Haploid
form (n)
C
n
2n
Diploid
form
(2n)
E
Zygote (2n)
D
10. The diagram above shows the life cycle of the fern. At which lettered stage does meiosis occur?
a. A
b. B
c. C
d. D
e. E
Answers:
1. c, 2. c, 3. b, 4. d, 5. a, 6. c, 7. d, 8. d, 9. e, 10. a
Essay Questions
1. How does the attachment of spindle microtubules to centromeres in metaphase I of meiosis lead to the
segregation of homologous chromosomes?
In metaphase I of meiosis, the spindle microtubules bind to only one side of each
centromere. Because homologous chromosomes are aligned parallel to the metaphase
plate (the axis of the cell), the spindle microtubules from opposite ends of the cell bind
each of the homologous chromosomes. When the microtubules pull the chromosomes
apart in anaphase I, the sister chromatids remain together, but homologous pairs are
separated. Since homologous pairs may have different alleles for the same gene, the
gametes are genetically different. In mitosis, the spindle microtubules bind to
kinetochores on both sides of each centromere, so the sister chromatids are pulled apart
during anaphase. Homologous chromosomes do not line up in mitosis.
84
2. A horse has 64 chromosomes and a donkey has 62. It is possible for these animals to mate and
produce mules as offspring. Why are mules sterile?
Mules receive gametes with 32 horse chromosomes and 31 donkey chromosomes, for an
uneven total of 63. These 63 chromosomes cannot pair during meiosis, so gametes are
not formed.
3. What are the costs and benefits of sexual reproduction vs. asexual (binary fission, parthenogenesis, or
cloning).
The primary benefit of sexual reproduction is the production and maintenance of genetic
diversity by segregation of homologous chromosomes and crossing over. Sexual
reproduction allows new genes that arise as mutations in different individuals to be
recombined in various ways in subsequent generations. A diploid, sexually reproducing
population, maintains a large reserve of genetic diversity stored as recessive genes. One
obvious cost of sexual reproduction is the need for two parents of different sexes to be
present. A single individual cannot reproduce. Furthermore, there is a cost for mating
behavior—bringing together the sexes, establishing a truce between individuals for the
exchange of gametes—or the cost of spreading gametes widely across the environment
(plant pollen, sea urchin sperm).
4. If sexual reproduction has so many advantages, then why is so much research being conducted to
develop methods to clone animals?
While sexual reproduction is very useful to generate genetic diversity, it is difficult to
preserve chance combinations of genes that occur in a single individual. Animal breeders
seek the ability to produce exact copies of prize individuals. Pet lovers want an exact
genetic duplicate of a beloved pet. In medicine, cloned cells might be used to replace
damaged organs in a living person.
85
CHAPTER 12. PATTERNS OF INHERITANCE
THE ESSENTIALS
Students need to know:
ƒ Mendel’s two principles of inheritance.
ƒ the difference between a gene and an allele.
ƒ the vocabulary of genetic crosses: P, F1, F2, dominant, recessive,
homozygous, heterozygous, phenotype, and genotype.
ƒ how to properly set up a cross to solve a genetics problem, including
establishing the correct gametes and determining the proper offspring.
ƒ how to interpret a Punnett square.
ƒ how to apply the rules of probability to predict the outcomes of genetic
crosses.
ƒ how to interpret a pedigree.
ƒ
examples of recessive and dominant genetic disorders in humans.
Key Terms
character
segregation
self-fertilization
cross-fertilization
pure-breeding
dominant
recessive
first filial generation (F1)
second filial generation (F2)
hybrid
Mendelian ratio
alleles
homozygous
heterozygous
genes
locus genotype
phenotype
Punnett square
Principle of Segregation
testcross
dihybrids
pedigree
albinism
hemophilia
sickle-cell anemia
86
Principle of Independent
Assortment
polygenic inheritance
continuous variation
quantitative traits
pleiotropic
epistasis
incomplete dominance
codominance
ABO blood groups
epistasis
genetic recombination
Multiple Choice Questions
1. A number of other scientists made plant hybridization experiments before Mendel. Why is he credited
as the discoverer of the laws of heredity?
a.
b.
c.
d.
e.
Mendel kept careful quantitative records of his experiments.
Mendel published his discoveries in widely read articles.
Mendel was the first to make controlled crosses with pea plants.
Mendel discovered that genes were on chromosomes.
Mendel discovered that offspring have traits that are a blend of traits from their parents.
2. Pea plants were a good choice for Mendel’s experiments for all of the following reasons EXCEPT
a.
b.
c.
d.
e.
peas have large flowers which can easily be cross-fertilized.
there are many pure-bred strains of peas with visibly different traits.
peas have short generation times.
pea flowers do not self-fertilize.
peas are easy to grow in a small garden.
3. When Mendel crossed pure breeding white-flower peas with pure breeding purple-flower peas, the
first hybrid generation had
a.
b.
c.
d.
e.
all white flowers.
half white and half purple flowers.
3/4 white and 1/4 purple flowers.
all purple flowers.
flowers with an intermediate (light purple) color.
4. In the same white x purple experiment, Mendel self-fertilized the first generation hybrids. The second
generation hybrid plants had
a.
b.
c.
d.
e.
all white flowers.
half white and half purple flowers.
3/4 white and 1/4 purple flowers.
all purple flowers.
there were no second generation plants because the F1 hybrids were sterile.
5. After the white x purple cross, Mendel self-fertilized his hybrid pea plants for several generations.
Why did some of the hybrid plants with purple flowers not breed true—producing both purple and
white flowered offspring?
a.
b.
c.
d.
e.
The purple gene mutated in some plants.
Mendel mixed up his seeds.
The white gene was epistatic to the purple gene.
The purple allele became recessive in some plants.
Some of the purple plants were heterozygous for the purple/while gene.
6. In a test cross, what fraction of the progeny show the recessive phenotype?
a.
b.
c.
d.
e.
none
1/4
1/2
3/4
all
87
7. Unlike Mendel’s experiments, when a very tall and a very short person mate, the children are usually
intermediate in height. Why is one trait (tall or short) not dominant over the other?
a.
b.
c.
d.
e.
The gene for height in humans has incomplete dominance.
The human gene for height has multiple co-dominant alleles.
Height is not controlled by genes in humans.
Human height is controlled by multiple genes.
The gene for height is recessive.
8. The process of gene therapy for a genetic disease involves
a.
b.
c.
d.
e.
removing a defective gene from a patient.
repairing a mutation in a patient’s DNA.
injecting cells that can grow to replace damaged tissues or organs.
inserting a new copy of a gene that will allow cells to produce a needed protein.
adding new chromosomes to replace damaged or missing chromosomes in a patient’s cells.
9. T.H. Morgan studied the white eye mutation in fruit flies. He crossed a white eyed male with a
normal (red eyed) female, then crossed two of the red eyed F1 progeny. Why did he find white eyes
only in males in the F2 generation?
a.
b.
c.
d.
e.
The white eye trait is cause by a recessive allele in the X chromosome.
The white eye trait is fatal to females.
The white eye gene has incomplete dominance in females.
The white eye enzyme must be activated by male hormones.
The white eye trait is carried by mitochondria.
10. Fringed gill covers (G) are dominant over smooth gills (g) and red eyes (R) are dominant over yellow
eyes (r) in the arctic salamander. A cross is made between pure breeding fringe gilled yellow eyed
salamanders (GGrr) and a smooth gilled red eyed ones (ggRR), then the F1 hybrids are crossed with
each other. What fraction of the F2 progeny have fringed gills and yellow eyes?
a.
b.
c.
d.
e.
1/2
3/4
1/8
3/16
1/16
11. In the salamander cross in the previous question, what will happen to the number of F2 progeny with
fringed gills and yellow eyes if these two genes are tightly linked on the same chromosome?
a.
b.
c.
d.
e.
There will be fewer progeny with fringed gills and yellow eyes.
There will be more progeny with fringed gills and yellow eyes.
There will be no change in the number of progeny with fringed gills and yellow eyes.
All of the progeny will have fringed gills and yellow eyes.
None of the progeny will have fringed gills and yellow eyes.
12. Humans have two different sex chromosomes, the X and the Y. A zygote becomes a male if
a.
b.
c.
d.
e.
the sperm has an X chromosome.
the sperm has a Y chromosome.
the egg has an X chromosome.
the egg has a Y chromosome.
there is nondisjunction of the sex chromosomes.
88
Answers:
1. a, 2. d, 3. d, 4. c, 5. e, 6. c, 7. d, 8. d, 9. a, 10. d, 11. b, 12. b
Essay Questions
1. Explain how the movement of chromosomes during meiosis produces the segregation of traits
observed by Mendel in his F2 generation plants.
Homologous chromosomes line up in the center of the cell in metaphase I of meiosis, then
they are attached to spindle fibers that pull them to opposite ends of the cell. In
heterozygotes, the two different alleles are randomly chosen to go to either end of the
dividing cell. Each gamete ends up with a random mixture of chromosomes that originated
from the two different grandparents.
2. Many of the common human genetic diseases are known as “inborn errors of metabolism.” These
diseases are almost always caused by a mutation in a gene that produces an essential enzyme. Why
are these diseases usually recessive?
A genetic disease can be caused by a mutation in a gene for an essential enzyme.
However, a single copy of a mutant allele can generally be compensated if a person has
another functional copy of the gene on the homologous chromosome. Enough of the
enzyme can be made from a single copy of the gene. However, if a person has two copies
of the gene that are mutant, then the enzyme cannot be made and the disease phenotype
appears. Therefore, a single copy of the mutant allele (which makes no functional
enzyme) is recessive to the dominant healthy allele.
3. If one parent has a genetic disease (two copies of the recessive allele), and the other parent is
homozygous normal, what is the chance that each of their children will have the disease? What is the
chance that the children will carry a recessive allele for the disease? Use a Punnett square to illustrate
you answer.
All children will inherit one copy of the disease allele and one normal allele. Since the
disease allele is recessive, all children will by healthy carriers.
d
d
D
Dd
Dd
D
Dd
Dd
4. Down’s syndrome is caused by a chromosomal abnormality, trisomy of chromosome 21. How does
this occur?
Down’s syndrome is the result of nondisjunction chromosome 21 during meiosis. The
paired chromosomes do not separate during anaphase of meiosis, so one gamete gets two
copies and the other gamete gets no copies. Human embryos formed from the gamete that
has no copies of chromosome 21 do not survive beyond the early stages of pregnancy, but
embryos formed form the gamete with 2 copies can develop into a fetus with 3 copies of
chromosome 21, which is Down’s syndrome.
89
CHAPTER 13. CHROMOSOMES, MAPPING & THE MEIOSIS–INHERITANCE
CONNECTION
THE ESSENTIALS
Students need to know:
ƒ the relationship between sex chromosomes and sex determination.
ƒ how to solve a cross for a sex-linked trait.
ƒ the relationship between the chromosomal theory of inheritance, the
movement of chromosomes during meiosis, and Mendel’s principles
of inheritance.
ƒ how mutations can cause genetic diseases.
ƒ
how alteration of chromosome number can cause genetic disorders.
Key Terms
sex chromosome
X chromosome
Y chromosome
autosomal
testcross
sex-linked
pedigree
hemophilia
Barr body
maternal inheritance
crossing over
genetic recombination
genetic map
map units
recombination frequency
linked genes
marker
single-nucleotide
polymorphisms (SNPs)
sickle-cell anemia
nondisjunction
aneuploidy
monosomy
trisomy
Down’s syndrome
genomic imprinting
methylation
genetic counseling
amniocentesis
chorionic villi sampling
CHAPTER 14. DNA: THE GENETIC MATERIAL
THE ESSENTIALS
Students need to know:
ƒ the structure of DNA.
ƒ the importance of complementarity to the structure and function of
DNA.
ƒ the process of replication: the major steps, the role of key enzymes, the
semi-conservative model.
ƒ
the difference between replication, transcription, and translation.
Key Terms
genes
bacteriophages
nucleic acid
purine
pyrimidine
nucleotide
phosphodiester bond
Chargaff’s rules
complementarity
semiconservative
primer
endonucleases
107
exonucleases
leading strand
lagging strand
Okazaki fragments
replication fork
Multiple Choice Questions
1. In Griffith’s experiment with Streptococcus bacteria, what is passed from dead S bacteria to live R
bacteria that then become lethal to mice?
a.
b.
c.
d.
e.
toxic proteins
toxic polysaccharides
a virus that kills mice
genetic information that makes the r strain more virulent
radioactive sulfur incorporated in proteins
2. In the Avery, MacLeod, and McCarty experiment, what is the “transforming principle?”
a.
b.
c.
d.
e.
protein
DNA
polysaccharides
lipids
protein-digesting enzymes
3. Why are Okazaki fragments necessary for DNA replication?
a.
b.
c.
d.
e.
The DNA polymerase I enzyme cuts newly synthesized DNA into short strands.
The DNA polymerase III enzyme can only copy DNA in one direction.
The helicase enzyme can only unwind short regions of the DNA molecule.
The DNA primase enzyme can only produce short RNA templates for DNA synthesis.
Single stranded binding proteins block the progress of the DNA polymerase III enzyme.
4. In the Hershey-Chase experiment, how does radioactive phosphorus get into the bacteria infected
with T2 bacteriophage?
a.
b.
c.
d.
e.
The phage inject viral DNA that contains radioactive phosphorus into the bacteria.
The bacteria absorb radioactively labeled phosphorus directly from the growth medium.
The phage inject viral proteins that contain radioactive phosphorus into the bacteria.
The whole T2 phage enters the bacteria, carrying radioactive phosphorus in its cytoplasm.
The bacterial chromosome becomes labeled with radioactive phosphorus after DNA replication.
5. Watson and Crick used all of the following information to build their model of DNA structure,
EXCEPT:
a. DNA is a polymer of four different types of nucleotides.
b. Chargaff’s rules: in DNA the amount of adenine equals thymine and the amount of guanine
equals cytosine.
c. Rosalind Franklin’s X-ray measurements that show DNA forms a helix with a diameter of 2 nm.
d. Three base codons of DNA determine the sequence of amino acids in proteins.
e. DNA contains genetic information.
110
6. Meselson and Stahl demonstrated that after mitosis, each daughter cell contains
a. all newly synthesized DNA.
b. one strand of DNA from the original parent cell and one newly synthesized strand.
c. a DNA molecule that is a random mixture of nucleotides from the original parent and newly
synthesized nucleotides.
d. double the amount of DNA as the original parent cell.
e. half the amount of DNA as the original parent cell.
7. In bacteria, a DNA molecule is replicated by
a. copying the entire molecule in a single direction from one end to the other.
b. breaking the molecule into small fragments which are copied by DNA polymerase III, and then
reassembled.
c. partially separating the two strands at the origin of replication and copying in two directions at
once.
d. completely separating the two strands and copying each one separately.
e. DNA polymerase II binds to DNA at many sites and copies the DNA all at once.
8. What is the primary difference between DNA replication in bacteria and eukaryotes?
a. Bacteria replicate their DNA in only one direction while eukaryotes replicate in both directions at
once.
b. Eukaryotes use enzymes to copy DNA while bacteria use only RNA.
c. Eukaryotes replicate their DNA from many different origins on each chromosome while bacteria
have only a single origin of replication.
d. Bacteria copy DNA directly while eukaryotes copy DNA into an RNA intermediate, which serves
as the template for DNA synthesis.
e. Bacteria replicate their DNA in a single continuous strand while eukaryotes replicate in short
fragments that are joined together by DNA ligase.
9. Beadle and Tatum found that X-rays caused mutations in Neurospora fungus. Why were mutated
fungi unable to grow on a minimal medium that contained sugar, but no amino acids?
a.
b.
c.
d.
The X-rays destroyed all of the fungal DNA.
The X-rays blocked DNA replication.
The X-rays destroyed a protein that is needed to synthesize an amino acid.
Mutations in many different places in the DNA always blocked the function of the same enzyme
needed to synthesize all amino acids.
e. The X-rays changed the sequence of the DNA at a single spot, which eliminated an enzyme
needed to synthesize an amino acid.
10. How did sequencing the DNA of the hemoglobin gene from sickle-cell anemia patients lead to an
understanding of this disease?
a.
b.
c.
d.
e.
Sickle-cell anemia is caused by a single base change in the hemoglobin gene.
Patients with sickle-cell anemia lack hemoglobin in their blood.
Patients with sickle-cell anemia have many different mutations in their DNA.
Mutations in the hemoglobin gene are always dominant.
Sickle-cell anemia is caused by a virus that mutates DNA.
Answers:
1. d, 2. b, 3. b, 4. a, 5. d, 6. b, 7. c, 8. c, 9. e, 10. a
111
Essay Questions
1. Watson and Crick won the Nobel Prize, yet they did no experimental work and their research paper
was only one-page long. Their double helix model of DNA was just a hypothesis. The importance of
scientific discoveries are judged by their ability to explain natural phenomena. What biological
phenomena are explained by the double helix model of DNA?
The Watson-Crick model of DNA structure is the foundation for the field of molecular
biology. The fundamental concept is that DNA is a polymer with a specified sequence of
bases and two complementary strands In one stroke, it explains how DNA is replicated,
how DNA codes for protein sequences, how mutations create alleles of genes, and how
genes segregate and recombine in meiosis.
2. Why does the replication (copying) of DNA require the complicated structure of the replication fork?
The DNA molecule is a double helix with the two strands running in opposite directions
(5’-3’ for one strand and 3’-5’ for the other). A helicase enzyme is required to separate
the two strands. The DNA polymerase enzyme requires a primer that it can extend as it
adds bases complementary to the template strand, so a primase enzyme is needed. The
DNA polymerase enzyme can only synthesize DNA in one direction, so as it slides along
the molecule, one strand is copied and the other strand “lags” behind. A second DNA
polymerase copies the lagging strand in the opposite direction, in short pieces (Okazaki
fragments) as the primase makes new primers. DNA polymerase I replaces the primer
(RNA) with new DNA nucleotides. A ligase enzyme joins the Okazaki fragments into a
complete DNA strand. Additional proteins are required to help the DNA polymerase III
remain attached to the template strand and to stabilize the single stranded DNA as the
strands are separated by the helicase.
3. Chargaff found that in DNA, the amount of the amount of adenine was always equal to the amount
thymine and the amount of guanine was equal to the amount of cytosine. How does the structure of
DNA account for these measurements?
DNA is built of two complementary chains of nucleotides wound into a helix. Each
complementary pair of nucleotides in the two strands forms hydrogen bonds. The
diameter of the helix means that there is only room for bonds between one purine (with
two carbon rings) and one pyrimidine (with one carbon ring). Cytosine and guanine form
three hydrogen bonds with each other while adenine and thymine form two hydrogen
bonds. These are the only base pairs that can form, so the amounts of A equal T and the
amounts of C equal G.
112
CHAPTER 15. GENES AND HOW THEY WORK
THE ESSENTIALS
Students need to know:
ƒ the difference between replication, transcription, and translation.
ƒ the process of transcription, its machinery, and end products.
ƒ the difference between exons and introns.
ƒ how eukaryotic cells modify mRNA after transcription.
ƒ the process of translation, its machinery, and end products.
ƒ
how mutations can alter the amino acid sequence of a protein.
Key Terms
gene
one-gene/one-polypeptide
hypothesis
one-gene/one-enzyme
hypothesis
ribosomes
ribosomal RNA (rRNA)
transfer RNA (tRNA)
messenger RNA (mRNA)
Central Dogma
transcription
RNA polymerase
promoter
translation
gene expression
codons
reading frame
triplet code
transcription
template strand
coding strand
promoters
transcription bubble
TATA box
113
5’ cap
3’ poly-A tail
aminoacyl-tRNA synthetases
nonsense codons
initiation complex
initiation factors
translocation
release factors
introns
exons
spliceosome
alternative splicing
This is the core of molecular biology—the Central Dogma, as Crick coined the phrase. Understanding this
process lays the groundwork for truly understanding genetics, inheritance, evolution, and biotechnology.
Students may become confused with the terms: replication, transcription, and translation. It is helpful to
describe the rationale behind the names. Transcription is the process of copying information from one
storage system to another—such as writing out in print the words of a taped conversation or writing out
musical notes from a recorded performance. In the cell, the original “text” is double-stranded DNA; the
copy is single-stranded mRNA. Messenger RNA is simply a working copy of the cell’s DNA blueprint
for a protein. They are written in the same language—nucleotides. The cell does not use its original set of
instructions for the same reasons that a carpenter would not take the original building blueprints to the
work site—they need to remain intact and readable, so they remain safe in the company vault. Also, many
different workers need to use parts of the blueprint instructions at the same time to do all different sorts of
construction, often in different locations in the building. Similarly, transcription of mRNA allows many
different kinds of protein synthesis to occur at the same time; the DNA is not held to making only one
protein at a time.
Translation, on the other hand, is not a simple copying process. One language—the sequence of nucleotides
that compose RNA (a language of four letters)—is changed into an entirely different language—the
sequence of amino acids (a language of twenty letters) that compose proteins. This is not unlike the process
of translating English into Chinese. Wholly different words and symbols are exchanged for one another yet
the same meaning is conveyed by both. Fortunately there are fewer semantic discrepancies in biological
translation as each codon stands for only one specific amino acid.
It makes sense that DNA replication requires a proofreading mechanism while mRNA transcription does
not. DNA replication is like printing a bound copy of a book, the process is expensive and the product
must be accurate without mistakes or changes. Transcription is a cheap photocopy process. Many copies
are made quickly and easily. If a few copies don’t turn out too well, just throw them in the trash! If the
RNA transcript is damaged and doesn’t work—no big deal, there’s another floating around. But if the
DNA copy is altered from the original, the whole existence of the cell may be compromised.
If one looks at protein synthesis as a collection of subassemblies of various molecules, it is much easier to
understand. Henry Ford did not invent the assembly line, cells did. The ribosome is the construction site
for protein synthesis. The ER is analogous to the truck or rail system that moves the product from the
assembly line to where it is needed. If you expect your students to be able to identify the A site and the P
site on the ribosome, remind them that the A site is the location where the tRNA with the single amino
acid attaches. The P site is the spot where the growing polypeptide chain is located. Obviously, the E site is
the exit.
115
When discussing the genetic code (i.e., table 15.1), point out that where there are two or more codons that
specify a single amino acid, the variation is usually in the third nucleic acid—called the “wobble”. Only
leucine and arginine are exceptions with variation in the first and third letters, the former being coded for
by CU(UCAG) and UU(AG), the latter coded by CG(UCAG) and AG(AG). Emphasize that the
universality of the genetic code is one of the best pieces of evidence of evolution from a common
ancestor.
RNA processing enables a cell to produce multiple products from one gene, exponentially expanding the
versatility of the genome.
Student Misconceptions and Common Pitfalls
•
With regards to the one-gene/one-polypeptide hypothesis it is helpful to discuss with students how
that model has been significantly modified even beyond the textbook’s discussion. This shows them
how active an area of scientific research this is and how dynamic a process science really is—not
merely hard facts on the printed page of a textbook.
The model initially evolved from one-gene/one-protein to one-gene/one-polypeptide since many
proteins do not become functional until multiple polypeptide units are united. But even that was off
the mark since alternate splicing can create different transcripts that can be translated into different
polypeptides—all from one gene.
Then it got more complicated (doesn't it always!)—this model had to again be altered when it was
realized that many genes code for RNA that is never translated to polypeptide—the many RNAs like
tRNA, rRNA, microRNA, and more.
So can we come up with a united definition of a gene? It's actually quite difficult. It has to be fairly
general, like: A gene is a region on a chromosome that is transcribed into RNA. It also includes
regulatory regions nearby.
Although this sounds rather vague, a narrower definition than that would require us to list the
exceptions. This is one of the reasons why it has been difficult to program a computer model to hunt
for genes in the genome. If you cannot define it, you cannot search for it. It requires a pre-existing
definition for a gene and—as you can see—we don't have a good enough one yet.
•
Students may find themselves in a chicken and egg conundrum when discussing enzyme regulation of
translation. Translation produces enzymes, yet enzymes—such as aminoacyl-tRNA synthetase—
catalyze translation. If students get stuck on this, reassure them that this is no more of a paradox than
cells come from cells.
•
Students may be confused by the terms intron and exon—focusing on the counterintuitive prefixes
(they think in- means something that is kept in and ex- is something that is cut out) rather than their
true derivation. It is helpful to describe the rationale behind the names: introns are intervening (or inbetween) sequences whereas exons are expressed sequences.
•
With our increased knowledge of the function of RNA comes more complexity for introductory
biology students. Students may be confused by the myriad roles of RNA. Clarify for students that
these multiple roles come from the fact that unlike DNA which only has an informational role, RNA
is able to perform such varied functions as informational (mRNA), structural (rRNA), and catalytic
(rRNA, tRNA, snRNPs, ribozymes).
•
Emphasize to students that proteins are not the only catalysts—that is a simplification from their
earlier biology classes. The discovery of ribozymes has led to a greater understanding of both
translation and the origin of life on Earth. You can introduce your students to the concept of “RNA
world” and use this new paradigm as an example of how biology is a dynamic field—our
understanding of how life works continues to evolve as we make new discoveries.
116
More about RNA World:
http://nobelprize.org/chemistry/articles/altman/
http://www.mhhe.com/biosci/genbio/life/articles/article28.mhtml
•
Students may react to the news that the genetic code is not completely universal with confusion,
wondering if this suggests that there is a flaw in the evolutionary theory that life on earth arose from a
common ancestor. Clarify for them that this is the exception that proves the rule. It is a more
parsimonious explanation that a genetic code that is 95% shared between two organisms, diverged
from a single common ancestor, rather than emerged from different origins.
Multiple Choice Questions
1. What is the process of copying DNA into mRNA called?
a.
b.
c.
d.
e.
translation
translocation
termination
transcription
transfer RNA
2. All of the following are true about eukaryotic gene expression EXCEPT:
a. most eukaryotic genes contain introns, which are spliced out of the mRNA.
b. prokaryotic mRNA molecules often contain transcripts for several different proteins, but this is
very uncommon for eukaryotic genes.
c. eukaryotic cells can begin translation of an mRNA before transcription is complete.
d. eukaryotic mRNA molecules are modified by the addition of a GTP cap at the 5’ end and a polyA tail at the 3’ end before they leave the nucleus.
e. translation of all eukaryotic proteins begins with a methionine codon.
3. Why is the DNA sequence of eukaryotic genes much longer than prokaryotic genes?
a.
b.
c.
d.
e.
Eukaryotic genes code for much larger proteins.
Eukaryotes use much larger codons to specify the amino acids in each protein.
Eukaryotic genes contain large regulatory sequences within the coding region.
There are nucleotides that function as spacers between each codon of eukaryotic genes.
The DNA sequences of eukaryotic genes have introns, which are spliced out of the mRNA.
4. The “Central Dogma” refers to what basic process in cells?
a. the semi-conservative replication of DNA by synthesis of two new strands using the two
complementary strands as templates
b. the use of feedback mechanisms whereby enzymes are inhibited by the products of the reactions
that they catalyze
c. the movement of information from DNA to RNA by transcription and then to protein by
translation
d. the base pairing of nucleotides A with T and C with G in complementary strands of DNA
e. one gene codes for one protein
117
5. What effect will adding a single base to the coding sequence of a eukaryotic gene have on the
resulting protein?
a. The entire protein sequence will be changed because an additional nucleotide will create a frame
shift.
b. A single amino acid will be changed in the resulting protein.
c. There will be no effect on the protein because DNA polymerase has an error correcting function.
d. The protein sequence encoded by one exon will be changed, but intron splicing will preserve the
rest of the protein.
e. The protein will not be synthesized because the mRNA will not be recognized by the ribosome.
6. Nirenberg discovered that a synthetic mRNA made entirely of poly-uracil (UUUUU….) produced a
poly-phenylalanine protein. What protein would be produced from a poly-G-C mRNA
(GCGCGCGC…).
a.
b.
c.
d.
This message would produce poly-histidine.
This message would produce poly-serine.
This message would produce either poly-histidine or poly-serine, depending on the reading frame.
This message would produce either poly-arginine or poly-alanine, depending on the reading
frame.
e. This message would produce poly-lysine.
7. In the translation process, how does the ribosome read information from each successive codon in
order to add the next amino acid to the growing polypeptide chain?
a. The mRNA moves through the ribosome, bringing each codon into the active site for protein
synthesis.
b. The tRNA molecules read each successive codon and transport that information to the ribosome.
c. The ribosome moves down the mRNA as it reads each codon.
d. The mRNA folds to bring each codon to the active site for protein synthesis on the ribosome.
e. Protein translation factors carry information from the mRNA to the site of protein synthesis on
the ribosome.
8. If eukaryotic genes have introns interrupting the protein coding sequence, how can the information be
read to guide amino acid synthesis?
a.
b.
c.
d.
e.
The RNA polymerase skips over introns and only transcribes exons into mRNA.
The ribosome skips over introns and only translates exons into protein.
Introns are spliced out of the DNA prior to transcription by small nuclear endonuclease proteins.
Introns are spliced out of the mRNA by ribonuclease proteins active in the cytoplasm
Introns are spliced out of the mRNA in the nucleus by small nuclear ribonuclear proteins.
118
Use the diagram below to answer questions 9 and 10.
Coding
strand
Template
strand
59
C
DNA G
39
39
A GCA T C G T A
A
A
G
A
GT
CT
T
T
C
A
AT C
39 C G
T
TA
59
T
A
G
G
C
A
U
C
G
U
CG
T
T AGCA
A
5'
B
9. In the diagram above, what is the function of the large pink structure labeled B?
a.
b.
c.
d.
e.
transcription of DNA into RNA
replication of DNA
translation of DNA into protein
translation of mRNA into protein
splicing of pre-mRNA into mature mRNA
10. The template strand of the DNA is
a.
b.
c.
d.
e.
identical to the sequence of the mRNA.
complementary to the sequence of the mRNA.
transcribed directly into protein by the ribosome.
replaced by the synthesis of new DNA during transcription.
not used by RNA polymerase.
Answers:
1. d, 2. c, 3. e, 4. c, 5. a, 6. d, 7. c, 8. e, 9. a, 10. b
119
Essay Questions
1. Why must the DNA molecule be unwound and have the two strands separated for a gene to be
expressed?
Gene expression requires that a DNA sequence be transcribed into messenger RNA by
RNA polymerase enzyme. During transcription, the template strand of DNA directs the
synthesis of mRNA by forming base pairs with the RNA nucleotides. Therefore, the DNADNA base pairs must open up in order to make room for the DNA-RNA base pairs to
form.
2. How does the RNA polymerase enzyme find where to begin transcribing a prokaryotic gene?
Every prokaryotic gene begins with a promoter sequence, located directly upstream of the
transcribed sequence on the template strand of the genomic DNA. Prokaryotic promoters
all follow a common pattern with sequences similar to TTGACA at 35 bases upstream and
TATAAT at 10 bases before the start of transcription. The sigma subunit of bacterial RNA
polymerase recognizes the TATAAT sequence and binds to the DNA. The other subunits of
RNA polymerase bind to the sigma subunit and begin to copy the DNA into RNA 10 bases
downstream from the TATAAT signal.
3. How do tRNA molecules recognize the codons of the mRNA and bring the appropriate amino acids
to the site of protein synthesis?
There are 45 different types of tRNA molecules. Each tRNA contains a 3-base anti-codon,
which forms RNA-RNA base pairs with codons of the mRNA. There is a different tRNA
for each amino acid, and some amino acids have more than one different tRNA in order
to match different codons. Some tRNAs match more than one codon due to “wobble” in
base pairing in the third position of the codon. The aminoacly-tRNA synthetase enzymes
are responsible for attaching the appropriate amino acid to each type of tRNA. There is
one tRNA synthetase for each of the 20 amino acids, so some of them recognize more
than one type of tRNA.
120
CHAPTER 16. CONTROL OF GENE EXPRESSION
THE ESSENTIALS
Students need to know:
ƒ the function of the three parts of an operon: the operator, the promoter,
the bacterial genes.
ƒ the role of the repressor protein in an operon.
ƒ the lac operon and the trp operon as examples of the control of gene
expression in bacteria.
ƒ the impact of chromatin modifications—DNA methylation and histone
acetylation—on gene expression in eukaryotes.
ƒ the role of transcription factors in control of gene expression in
eukaryotes.
ƒ
the six points at which gene expression can be regulated in eukaryotes
Key Terms
promoter
transcriptional control
post-transcriptional controls
introns
exons
operons
activators
enhancers
methylation
RNA interference
microRNAs (miRNAs)
small interfering RNAs
(siRNAs)
primary transcript
RNA splicing
snRNP (snurps)
spliceosome
translation repressor proteins
Much of the information in this chapter is beyond the scope of an AP Biology course, but there are some
core concepts that are very necessary for even the first-year students. Three main concepts need to be
conveyed in this chapter:
•
Regulation of gene expression in prokaryotes
Teach the function of operons with lac operon and trp operon as examples of inducible and
repressible systems.
• Transcription control in eukaryotes
Teach the role of transcription factors and the organization of the promoter and enhancer regions
in relation to the gene to be transcribed. Discuss the effect of chromatin structure on gene
expression, specifically the effects of packaging, methylation, and acetylation on the rate of
transcription.
• Post-transcriptional control in eukaryotes
Discuss the effects of alternative splicing, selective degradation of mRNA transcripts, and
translational controls on the rate of production of gene products. Specifically review Figure 16.20
as an overview of the different points of gene regulation in eukaryotes. Students should be able to
reproduce the concepts in this diagram—explaining the six points at which gene expression can
be regulated in eukaryotes.
Prokaryotes and multicellular eukaryotes both control gene expression, but for quite different reasons.
Bacteria must exploit the resources of a changing environment and thereby rapidly adapt to it by changing
gene expression. If they do not adapt, they die, but maintaining numerous unused enzymes is
metabolically expensive. A common pattern in prokaryotes is that gene products necessary for certain
catabolic reactions are only expressed when the substrate is present in the environment—an inducible
system. Other gene products necessary for anabolic pathways are only expressed when the cell needs to
build that particular molecule—a repressible system. Each system involves regulatory proteins that will
bind to the DNA and alter genetic expression, either by initiating expression or suppressing expression.
On the other hand, multicellular eukaryotes seek to counter environmental changes to maintain a constant
internal environment—maintaining homeostasis. To ensure this, genes must be transcribed in a specific
order over a specific time frame. Transcriptional control and post-transcriptional control are two primary
levels of gene regulation. The former is the more common method. Transcriptional gene control is achieved
by influencing the binding of RNA polymerase to the DNA helix. If a transcription factor blocks RNA
polymerase then an mRNA transcript cannot be produced and the gene is “turned off.” If a transcription
factor facilitates binding of RNA polymerase and a promoter then more mRNA transcripts are produced
and the gene is “turned on.”
Eukaryote regulation of gene expression greatly depends on the structure of the eukaryotic chromosome.
Histones affect gene transcription by physically blocking the promoter with the nucleosome they create.
Methylation, once thought to be a primary regulator in vertebrates, helps ensure that once a gene is turned
off, it stays off. Post-transcriptional control is common in eukaryotes. Researchers have found that small
RNA molecules (siRNA) seem to interfere with translation directly or the breakdown of the mRNA
before translation. The eukaryote primary mRNA transcript is a linear patchwork of coding exons and
noncoding introns. The entire sequence is made during transcription; the introns are cut out later. In many
cases, the various ways the exons can be spliced back together allows for production of different
polypeptides from just one gene. Proteins called translation factors regulate production of polypeptides
from the mRNA transcript. Translation repressor proteins can also shut down translation by preventing
the attachment of the transcript to a ribosome. Although most mRNA transcripts are very stable, some,
like those associated with regulatory proteins and growth factors, are less stable. They possess certain 3’
sequences that make them attractive to mRNA degrading enzymes. This ensures that control by these
proteins remains as transitory as it should be.
123
Multiple Choice Questions
1. What is the primary method of regulation of gene expression in both eukaryotes and prokaryotes?
a.
b.
c.
d.
e.
regulation of intron splicing by small nuclear ribonuclear particles
regulation of translation by ribosomes
regulation of transcription by DNA-binding proteins interacting with RNA polymerase
regulation of chromatin structure by acetylation of histones
regulation of DNA polymerase by microtubules
2. What feature do most types of DNA-binding proteins share?
a.
b.
c.
d.
e.
They all contain a metal atom that fits between the strands of the double helix.
They all contain short alpha-helix motifs that fin into the major groove of the DNA helix.
They all contain an RNA sub-unit that binds to specific DNA sequences.
They all contain B-sheet domains that wrap around the DNA helix.
They all form peptide dimmers that recognize palindrome (inverted repeat) sequences in the
DNA.
3. A bacterial operon contains all of the following EXCEPT
a.
b.
c.
d.
e.
a gene for a repressor protein
a promoter sequence
two or more genes for enzymes or structural proteins
a operator sequence that is bound by a DNA-binding protein
a gene for RNA polymerase
4. Which of the following affect the length of time that eukaryotic mRNA molecules remain in the
cytoplasm?
a.
b.
c.
d.
e.
alternative splicing
sequences at the 3' end of the mRNA that are recognized by RNAase enzymes
binding of translation factors to the 5' end of the mRNA
recognition and transport of the mRNA through nuclear pores
RNA editing
5. Eukaryotic transcription is regulated by all of the following EXCEPT
a.
b.
c.
d.
e.
chromatin structure (tight packing of histones)
enhancer sequences and activator proteins
restriction enzymes and ligase
basal transcription factors and transcription associated factors
coactivators and mediators
6. What is the function of the operator in the lactose operon?
a. Lactose binds directly to the operator sequence and induces transcription of the genes in the lac
operon.
b. The operator contains the DNA sequences for several enzymes that are used to absorb and digest
lactose.
c. The operator is a sequence recognized by RNA polymerase as the position on the DNA to start
transcription of the lac operon.
d. The operator is a gene that codes for a repressor protein that blocks the transcription of genes in
the lac operon.
e. The lac repressor protein binds to the operator sequence, blocking RNA polymerase from
transcribing the genes in the operon.
7. The trp operon is a perfect example of feedback regulation. How do high tryptophan levels in the cell
influence the trp operon?
a. tryptophan binds to the trp repressor protein, the repressor binds to the operator, which blocks
transcription of the genes in the operon
b. tryptophan binds to the trp inducer protein, which binds the promoter, increasing transcription of
genes in the operon
c. tryptophan binds to an RNA processing enzyme, which degrades the mRNA for tryptophan
biosynthetic enzymes
d. translation of the mRNA produced by the trp operon is blocked
e. high levels of tryptophan interfere with the binding of RNA polymerase to DNA
8. Small interfering RNAs are produced by
a.
b.
c.
d.
e.
RNA polymerase III.
restriction enzymes.
introns spliced out of primary mRNA transcripts by sNRPs.
digestion of primary mRNA transcripts by the dicer enzyme.
RNA editing.
9. The lac operon is controlled by both the lac repressor and the CAP activator protein. What
combination of factors is required for high level transcription of the lac operon?
a.
b.
c.
d.
e.
Both the lac repressor and the CAP protein are bound to the operon.
Neither the lac repressor or the CAP protein are bound to the operon.
The lac repressor is bound, but the CAP protein is not bound to the operon.
The CAP protein is bound, but the lac repressor is not bound to the operon.
The CAP protein binds to the lac repressor, preventing it from binding to the operon.
10. Enhancers differ from promoters in that they
a.
b.
c.
d.
e.
can be located far away from the transcription start site.
are not recognized by DNA-binding proteins.
are bound by repressor proteins that block transcription.
are the site of attachment of RNA polymerase.
are spliced out of the gene before transcription.
Answers:
1. c, 2. b, 3. e, 4. b, 5. c, 6. e, 7. a, 8. d, 9. d, 10. a
Essay Questions
1. Why do many DNA-binding proteins have two helix-turn-helix motifs located 3.4 nm apart? How
does this structure provide both a strong bond with DNA and the ability to recognize a specific
location on the DNA?
The 3.4 nm distance corresponds to one full turn of the DNA helix. One alpha helix of
each helix-turn-helix motif fits into the major groove of the DNA helix at two adjacent
turns. This provides twice the binding strength of a single DNA-protein interaction and
increases the number of DNA base pairs that are recognized at the protein binding site.
2. How can bacteria use DNA-binding repressor proteins in some feedback systems that respond to a
signal by inducing the expression of genes and in other systems that respond by blocking the
expression genes?
Some DNA-binding repressor proteins have their DNA-binding domain activated by
binding an effector molecule (the signal). These repressors block transcription when the
signal is present. Other DNA-binding proteins change shape to inactivate the DNAbinding domain when they are bound by an effector molecule, so the signal releases the
repression and induces transcription.
3. How do operons simplify the regulation of biochemical pathways for bacteria?
An operon includes several genes with related functions (enzymes in a pathway for
biosynthesis of a metabolite or for digestion of a food source) transcribed on a single
mRNA. This allows all of the genes to be turned on and off with a single regulatory
protein. Also, all of the genes in the pathway are transcribed and translated together so
the enzymes in the pathway are produced in equal amounts.
CHAPTER 17. BIOTECHNOLOGY
THE ESSENTIALS
Students need to know:
ƒ
the basic biotechnology tools and their application:
ƒ how restriction enzymes work and their application in other
biotechnology techniques.
ƒ
the process of gene cloning and its practical application.
ƒ
the process of gel electrophoresis and its practical application in
separating DNA fragments or protein molecules.
ƒ
the three key steps in the process of polymerase chain reaction
(PCR) and its practical application.
ƒ
the basic principles and the value of DNA sequencing
Key Terms
recombinant DNA
restriction endonucleases
(enzymes)
restriction sites
palindrome
DNA ligase
gel electrophoresis
transformation
plasmid
vector
DNA library
genomic library
reverse transcriptase
complementary DNA (cDNA)
molecular hybridization
probe
Southern blot
restriction fragment length
129
polymorphisms (RFLP)
DNA fingerprinting
DNA sequencing
dideoxynucleotide
polymerase chain reaction
(PCR)
knockout mice
gene therapy
transgenic
Approach:
This is the exciting world of biology today—full of both promise and money. This is the exciting stage at
which we are learning to manipulate the materials of heredity. There is a lot to discuss here in regards to
basic science as well as current events, politics, and ethics.
The Science
This chapter offers a great example of the dovetail between science and technology—hand-in-hand they
enable advances in each field. However, students can get lost in vocabulary and concepts in this unit if it
is presented merely as a laundry list of techniques. Establish the conceptual framework of basic
biotechnology first.
•
DNA cleavage: First, we need to cut up the genomic DNA of an organism (e.g. a human), so we
are working with small pieces.
• Recombinant DNA: Second, we insert the DNA fragments into vectors (e.g. plasmids) making
recombinant plasmids, so we have a vehicle to easily insert the DNA fragments into bacterial
cells.
• Cloning: Third, we insert the recombinant vectors into bacteria (transformation) so that we have
a biological copying mechanism to make many copies of our gene (cloning) and a biological
factory for making the protein product from that gene.
• Screening: Lastly, we need to figure out which bacterial colonies have our gene of interest.
Preliminary steps select out bacteria that didn’t take up vectors (and therefore are not resistant to
ampicillin) and also screen out those which did not receive recombinant plasmids (and therefore
can still produce blue indicator pigment by digesting X-gal sugar). Bacteria which successfully
take up recombinant vector are resistant to ampicillin and remain as white colonies. Final
screening finds which bacterial colonies contain the gene of interest by using a marker probe to
recognize (by hybridization) a short sequence of the gene of interest in the bacteria.
The polymerase chain reaction (PCR) is an indispensable technique in all modern molecular biology
laboratories. Students need to understand that PCR is a method for making many copies of a piece of
DNA without having to use the biological copying machinery of a bacterial cell. It requires two primers
that match the ends of the sequence of the target DNA fragment. PCR can be used to isolate one specific
fragment of DNA from a mixture, and its exponential nature allows a large amount of DNA to be
produced from an extremely tiny sample.
If you are not yet comfortable with biotechnology techniques, many of the biotechnology labs can be
performed through easy-to-use kits supplied by the major science supply vendors or as field trips to
biotechnology education centers. In addition, almost all of them are available as paper or computer
simulations as an alternative or as a warm up to the “wet lab.”
The Politics / Ethics
One merely needs to pick up the science section of the weekly newspaper, or a popular science magazine
to see examples of gene technology in action. As a result, it is important to discuss the implications of
such research and the necessary scientific and governmental regulations. This is one of the stronger
reasons to have some knowledge of biology, to be able to make informed decisions, and to determine if
the decisions made by those in power are indeed in the best interest of the populace. Someone will need to
make difficult decisions in the not-so-distant future. Just because science can perform certain
technological feats does not mean that it should be allowed to do so. Conversely, just because some gene
technology is potentially dangerous, does not mean that all related technology should be brought to a halt.
It’s your students who will be making the political decisions for the future of the world.
The ongoing debate over the risks and benefits of GM organisms offers an excellent opportunity to
engage students in an active scientific/political/societal issue that will surely engage their interest. Have
students do independent research on the different positions—both public opinions and scientific
evidence—as well as read articles as a class. This is a good topic for “Socratic Seminar” style of
discussion.
130
Multiple Choice Questions
1. What is the natural function of the restriction enzymes used by molecular biologists to cut DNA?
a.
b.
c.
d.
e.
Restriction enzymes are used by bacteria to replicate their own DNA.
Restriction enzymes are used by eukaryotic cells to kill bacteria.
Restriction enzymes are used by bacteria to destroy the DNA of invading viruses.
Restriction enzymes are used by eukaryotic cells to cut introns out of mRNA.
Restriction enzymes block the expression of genes that are not needed in the cell.
2. Plasmids have all of the following features EXCEPT
a.
b.
c.
d.
e.
Plasmids are small circular DNA molecules that are found in many bacteria.
A bacterial cell may have many copies of a plasmid.
Plasmids often carry genes that provide resistance to antibiotics.
A plasmids can be used to grow an entire eukaryotic chromosome in a bacterial cell.
Bacteria can grow without plasmids.
3. Phage vectors are useful to molecular biologists because
a.
b.
c.
d.
e.
they can infect both bacterial or eukaryotic cells.
they can be used to clone larger pieces of DNA than plasmids.
they can be grown in the laboratory without the need for bacterial cells as a host.
they can detect expressed portions of genomic DNA.
they can be used to insert new genes into the human genome.
4. The sticky ends created by Type II restriction enzymes consist of
a.
b.
c.
d.
e.
double stranded DNA with a free 3' OH group.
double stranded DNA with a free 5' phosphate group.
an RNA-protein complex.
DNA binding proteins.
a few bases of single stranded DNA.
5. A gene cloning experiment involves
a. transferring an entire chromosome from one organism to another.
b. using restriction enzymes to insert a piece of DNA into a vector, and growing a colony of cells
that all contain this modified vector.
c. digesting the DNA from an organism and labeling it with radioactive elements.
d. sequencing the entire genome of an organism.
e. identifying a mutation that causes a genetic disease.
6. What is the function of the enzyme DNA ligase in genetic engineering?
a.
b.
c.
d.
e.
It helps complementary bases of DNA to form hydrogen bonds.
It joins the centromeres of sister chromatids during mitosis.
It cuts DNA at specific sequences.
It copies DNA in PCR reactions.
It joins the sugar-phosphate backbone of the DNA after complementary bases have paired at
sticky ends.
131
7. A Southern blot identifies a specific piece of DNA by
a.
b.
c.
d.
e.
the binding of a radioactively labeled protein.
cutting it with a sequence specific restriction enzyme.
hybridization with a radioactively labeled single stranded DNA probe.
copying it with a heat stable DNA polymerase enzyme.
translation of the DNA into protein with isolated ribosomes.
8. The polymerase chain reaction (PCR) is useful for
a.
b.
c.
d.
e.
joining together two fragments of DNA when their sequences are not known.
making many copies of a known sequence of DNA.
cutting DNA at specific positions that have a particular sequence.
joining together sticky ends after they have been cut with a restriction enzyme.
inserting new genes into plants.
9. A genomic library contains
a.
b.
c.
d.
the DNA sequence information from all of the genes in an organism.
all of the proteins from the cells of an organism.
all of the mRNA from the cytoplasm of an organism.
all of the DNA from the chromosomes of an organism, cut into pieces and grown in set of
vectors.
e. all of the DNA sequences known to science.
10. Genetic engineering can involve all of the following EXCEPT
a. adding a single new gene to an organism that provides instructions to produce a new protein that
has never been produced by that organism before.
b. blocking the function of a gene so that an organism does not produce a specific protein.
c. providing a different copy of a gene, so that an organism can produce a functional or normal
protein when it previously had a mutant phenotype that did not produce that protein.
d. combining the complete genomes of two organisms to produce a hybrid that does not exist in
nature.
e. adding multiple genes to an organism to create a large change in its metabolism.
Answers:
1. c, 2. d, 3. b, 4. e, 5. b, 6. e, 7. c, 8. b, 9. d, 10. d
Essay Questions
1. Why is it better to make a genomic library with randomly sheared DNA, rather than with DNA cut
with a restriction enzyme?
A genomic library should contain every sequence in the genome of an organism.
Restriction enzymes cut DNA in specific places. Some areas of the genome may have
restriction sites very close together, producing very small fragments, other areas may lack
restriction sites for the enzyme, resulting in fragments too large to clone in a vector.
Also, if a restriction site cuts a gene in half, then the full sequence of that gene will not be
present in any single clone.
132
2. Why has PCR become an essential tool for police investigations? Why are the accuracy of these PCR
techniques sometimes questioned?
PCR can make many DNA copies from an extremely tiny amount of starting material.
PCR allows for the genetic analysis of very small samples of human tissue. Identification
of a person can be made with a few skin cells or a single hair. However, there is also a
great risk of contamination. A tiny amount of contaminating DNA can be amplified
together with a sample from a crime scene.
3.
Genetic engineering of plants has produced a number of commercially successful plant varieties that
are resistant to insects, viruses or fungi, tolerant of herbicides, or nutritionally improved. What is the
most significant risk associated with these engineered plants?
The risks of genetically modified foods are summarized as danger to humans due to
toxicity or allergic reactions to foreign proteins, direct harm to insects or the
environment due to the plants (toxicity to non-pest insects), and possible gene flow to
other plants (creating “super weeds”).
133
CHAPTER 18. GENOMICS
THIS CHAPTER HAS LOTS OF INFO __ KNOW THE BOLDED TERMS ONLY
Key Terms
genomics
genetic maps
physical
maps
landmarks
fluorescence in situ hybridization (FISH)
sequenced-tagged sites (STSs)
clone-by-clone sequencing
shotgun sequencing
consensus sequence
Human Genome Project
linkage disequilibrium
haplotypes
expressed sequence tag (EST)
open reading frame (ORF)
transposable elements
long interspersed elements (LINEs)
short interspersed elements (SINEs)
long terminal repeats (LTRs)
microarray
functional genomics
proteomics
transcriptome
single nucleotide polymorphisms (SNPs)
134
CHAPTER 19. CELLULAR MECHANISMS & DEVELOPMENT
THIS CHAPTER HAS LOTS OF INFO __ KNOW THE BOLDED TERMS ONLY
Key Terms
blastomeres
cleavage
animal pole
vegetal pole
blastula
gastrula
gastrulation
germ layers
neural tube
neurulation
neural crest
somites larva
metamorphosis
blastoderm discs
pupa
meristems
epidermal cells
ground tissue
vascular tissue
cotyledons
induction
135
organizers
totipotent
chimera
determination
differentiation
homeobox
Hox genes
necrosis
apoptosis
antioxidants
telomeres
CHAPTER 20. GENES WITHIN POPULATIONS
THE ESSENTIALS
Students need to know:
ƒ the difference between Darwin’s evolution by natural selection and
Lamarck’s inheritance of acquired traits,
ƒ that genetic variation in a population is necessary for evolution.
ƒ that evolution is measured as changes in allele frequencies in a
population.
ƒ the five conditions that must be met for a population to be in HardyWeinberg equilibrium.
ƒ know how to use the Hardy-Weinberg equations to calculate genotype
frequencies and allele frequencies in a population, to detect whether
that population is evolving.
ƒ the five agents of evolutionary change and an example of each.
ƒ the principle of heterozygote advantage and how sickle-cell anemia is
a key example of it.
ƒ that fitness includes both a survival advantage and enhanced
reproductive success.
Key Terms
natural selection
inheritance of acquired
characteristics
population genetics
polymorphism
Hardy-Weinberg equilibrium
allele frequencies
genotype frequencies
nonrandom mating
genetic drift
founder effect
bottleneck effect
gene flow
selection
136
artificial selection
natural selection
fitness
heterozygote advantage
disruptive selection
directional selection
stabilizing selection
•
•
•
Genetic variation exists in populations and is the raw material for evolution.
Variation in the genetic composition of living organisms is the driving force behind evolution as a
whole. Evolution does not cause mutation. Genetic variation is the result of mutations that cause
changes in the coding sequence of genes creating new alleles. Mutations are produced as a
consequence of the biochemistry of DNA replication. Variation is also created by the new
combinations of alleles of different genes created by sexual reproduction. Darwin and Wallace
proposed that evolution is the result of natural selection acting on the genetic variation that exists
in a population.
Evolutionary forces change allele frequencies and these changes can be measured.
The Hardy-Weinberg equation explains why dominant alleles do not drive out recessive alleles and
eliminate genetic variation. In large populations exhibiting random mating, the frequency of a
genotype remains constant over time. The genotypes of such populations are in equilibrium. In
contrast, evolutionary forces—natural selection, genetic drift, gene flow, sexual selection—
change allele frequencies over time. It will be necessary to review manipulation of fractions and
decimals and other basic algebraic skills so that students can work with and understand the
mathematics associated with the Hardy-Weinberg equation. Work with examples to help students
understand when and in what form to use the formulas. The AP Lab will help considerably.
Natural selection favors individuals that are more fit for their environment.
Reproductive success of an individual is defined by how long it survives, how often it mates, how
many offspring it produces per mating, and how many survive to reproduce themselves. In total,
fitness is the genetic contribution that an individual makes to the next generation. Natural
selection can be directional, disruptive, or stabilizing. Natural selection can also maintain variation
in a population through the existence of successful heterozygotes. An example of this principle is
sickle-cell anemia and its association with malaria in Central Africa. Individuals with
the heterozygous condition are more likely to survive and reproduce than either homozygote, one
which succumbs to sickle-cell anemia and the other to malaria.
137
Concept Map
evolution
defined as
p+q=1
p2 + 2pq + q2 = 1
conditions for
Hardy-Weinberg equilibrium
are met
may be
caused by
may be
caused by
in
populations
may be
caused by
which includes
individuals with
no selection,
no mutation,
random mating,
no gene flow,
large population
genetic drift
may be
due to
founder
effect
change in
allele
frequency
does not
occur wnen
may be
due to
bottleneck
natural
selection
gene flow
due to
favors
increased
fitness
migration
differential
survival &
reproductive
success
140
genetic
variation
caused by
can be
sexual
selection
differential
reproductive
success
mutation
genetic
recombination
•
The biggest obstacle is that many of the terms associated with evolution have everyday meanings that
are different or even counter to the scientific use of the term. The following concepts may be
problematic:
•
•
Theory: This term may be the greatest obstacle to acceptance of evolution in first-year students.
In everyday language, theory means a tentative explanation. You are likely to hear, “Evolution is
only a theory.” It is critical to emphasize that in science a theory is a comprehensive, wellsupported explanation for a wide rage of observations. It may be likened to a principle. The
groundwork for this understanding should be laid from the first day of class to short circuit this
misconception.
• Fitness: In everyday language, fitness means to be in good health, vigorous, and strong. Clarify
for your students that any trait that increases an organism’s reproductive success increases it
fitness. Reproductive success is defined by how long an individual survives, how often it mates,
how many offspring it produces per mating, and how many offspring survive to reproduce
themselves. In a particular case, it may actually be the smaller individuals that survive predation
better, rather than the larger, more robust ones.
• Adaptation: In everyday language, adaptation can be used to refer to an individual changing over
its lifetime to survive in its local environment. Throughout this course statements may have been
made about adaptations by individuals to survive in their environment, such as adaptations in
behavior. Students may mistakenly think that these changes accumulate over the lifetime of many
individuals and then lead to evolutionary change in a population. Emphasize that this is
Lamarckian reasoning.
• Evolution: Evolution is defined as a change in allele frequency over time. This is measurable and
testable—and therefore should not be controversial. Students mistakenly associate evolution only
with the large-scale evolutionary changes that lead to the origin of new species and give rise to
major morphological changes. Clarify for students that these are the same process just in different
timescales—small-scale, short-term changes in gene frequencies in a population from generations
to generation and the accumulated changes in gene frequencies over geologic time that give rise
to major phenotypic changes like insect wings and tetrapod limbs, and lead to the origin of new
species.
Students mistakenly frame evolution as a goal-oriented process—a progressive striving to greater
complexity that leads to the appearance of “higher” life forms. This especially arises when students
think about human evolution. There seems to be an inherent desire to see humans at the apex of the
tree of life—as the ultimate end-goal. Emphasize that evolution is not purposeful, not goal-driven,
and most definitely did not travel a predetermined path to “us.” Be careful to not mistakenly support
this misconception by referring to evolutionary changes as “advancements.” This implies that there is
a designer guiding natural selection. This pitfall reappears in the chapters on plant and animal diversity
where references may be made to “higher” or “more complex” species.
•
Unfortunately, students hold tenaciously onto the misconception that individuals evolve. The
instructor should choose his/her words carefully so as to not reinforce this misunderstanding.
Individuals are selected, and only populations evolve. Clarify that evolution—change over time—is
measuring the change in allele frequencies over time. Also clarify that when we say “individuals are
selected,” we are really meaning “individual phenotypes are selected.”
•
Students have difficulty delineating the two steps required in evolutionary change. Novel traits arise
because of random mutations and sexual recombination. A new trait that increases the fitness of
individuals in the environment will then be passed on and will increase in frequency in the population
over time. Be on the lookout for Lamarckian thinking in your students—that individuals acquire traits
during their lifetime that help them to survive and reproduce in their environment and then pass these
on to their offspring.
141
•
Even though students will voice the understanding that mutations are rare and random events, further
discussions will often reveal an underlying belief that organisms mutate “in order to adapt” to their
environment. Watch out for student statements that suggest that mutations are adaptive responses to
environmental changes and that mutations are, in some way, intentional.
•
Many students do not appreciate that populations include heritable variation and that this is an
essential condition for evolutionary change. First-year students will look at a population of an
organism and only see homogeneity (“all giraffes look alike”). They are not aware of the extent of
variation amongst individuals and may also think that all variation must be readily and visibly
apparent.
•
Students mistakenly think that evolution is an all or nothing process—survival vs. extinction.
Evolutionary success is not measured in such black and white terms; evolutionary success can mean
succeeding just a bit better—having one more offspring or one more litter over a lifetime. As this
extends with each generations, the alleles that enable this better survival and reproduction then make
up an increasing proportion of the population.
•
•
As an extension to this, students think that evolutionary change occurs only when a trait changes
in all members of a population. Clarify that evolutionary change occurs when individuals with
favorable heritable phenotypes come to make up an increasing proportion of the population. For
example, if selection favors antibiotic resistance in a population of bacteria, students may think
that all the bacteria become resistant, rather than recognizing that more bacteria become resistant
over time.
Although instructors need to emphasize the role of chance in evolution, students can mistakenly
extend that and presume a role for chance in natural selection. New alleles arise by chance mutations,
new combinations arise by the chance shuffling of genes in sexual recombination, and chance events
may change allele frequencies in small populations. However, it would be a mistake for students to
then think that natural selection works by chance. Emphasize that natural selection is not based on
chance. Natural selection favors individuals that can survive and reproduce more successfully in their
environment.
•
Students may assume that most selection is directional whereas stabilizing selection is the norm.
Organisms are generally well-adapted to their local environments and selection tends to remove
deleterious mutations—the overwhelming majority of mutations—which alter the phenotype in ways
that reduce fitness.
•
Many students get lost in the mathematics and find it hard to understand the Hardy-Weinberg
theorem. They are confused as to how and when to use each of the Hardy-Weinberg equations. The
Hardy-Weinberg theorem does not imply that allele frequencies are static; rather it clarifies the
factors that alter allele frequency. The Hardy-Weinberg polynomial equation can be used to show that
a population is not in equilibrium, and therefore undergoing evolution.
Multiple Choice Questions
1. If a population is in Hardy-Weinberg equilibrium, then
a.
b.
c.
d.
e.
it is evolving to adapt to environmental changes.
the frequency of alleles is changing with each generation.
it is NOT evolving and allele frequencies remain the same with each generation.
mutations, immigration, and selective mating are changing allele frequencies.
homozygous recessive individuals are less fit.
142
2. For a population in Hardy-Weinberg equilibrium, the frequency of the recessive allele
a.
b.
c.
d.
e.
increases with each generation until it reaches 50%.
decreases with each generation until it reaches 25%.
remains the same in every generation.
decreases due to negative selection pressure on homozygous recessive individuals.
increases due to the occurrence of new mutations.
3. If a population that contains 40% homozygous recessive individuals (blue eyes), and 60% individuals
with brown eyes (homozygous dominant and heterozygotes), what is the frequency of the dominant
allele in the population?
a.
b.
c.
d.
e.
0.6
0.4
0.32
0.64
0.8
4. What is the only factor that can change allele frequencies in populations to produce adaptive
evolutionary change?
a.
b.
c.
d.
e.
mutation
gene flow (immigration)
non-random mating
genetic drift
selection
5. How does natural selection affect the frequency of mutations?
a.
b.
c.
d.
e.
Under conditions of high selection pressure, beneficial mutations occur more frequently.
Natural selection does not affect the frequency of mutations.
All mutations increase when selection pressure is high.
When there is no selection pressure, mutations do not occur.
Mutations occur less frequently when selection pressure is high.
6. Genetic drift is increased by all of the following EXCEPT
a. small population size
b. bottleneck effects when the population size is greatly reduced, then recovers
c. founder effects when a small number of individuals are isolated and reproduce to form a new subpopulation
d. movement of individuals from one isolated population to another
e. genetic isolation of small groups within a population
7. All of the following are required for natural selection to create evolutionary change in a population
EXCEPT
a. variation in size or health of genetically identical individuals that is caused by environmental
factors
b. random genetic variation exists in a population
c. individuals that have superior traits produce more offspring than less fit individuals
d. genes that produce superior traits are passed to offspring
e. differential survival and reproduction exist (all individuals in a population do not survive to
reproduce, and some individuals produce more offspring than others)
143
8. Predators can create rapid changes in allele frequency in populations by
a.
b.
c.
d.
killing weaker mutant individuals (negative selection pressure).
killing individuals with particular traits (color, size, feeding habits).
creating smaller populations which are more likely to have genetic drift.
forcing all individuals to produce more progeny, creating more genetic diversity. e.
gene flow because heavy predation allows immigration from neighboring regions.
9. The human ABO blood group locus is co-dominant. Can the allele frequencies at this locus be in
Hardy-Weinberg equilibrium?
a.
b.
c.
d.
e.
No, the Hardy-Weinberg equilibrium only applies to dominant-recessive allele pairs.
No, blood antigens are not subject to evolution.
Yes, if the Hardy-Weinberg assumptions are met, dominance is irrelevant.
Yes, blood antigens are always in Hardy-Weinberg equilibrium.
Yes, since blood groups can not be subject to natural selection.
10. In a population of cats, the genotype frequencies for the black hair locus are 25% homozygous
dominant (BB), 71% heterozygous (Bb), and 4% homozygous recessive. This information shows that
a.
b.
c.
d.
e.
the recessive phenotype is less fit.
there is a heterozygote advantage.
the recessive allele will disappear in a few generations.
the population is not in Hardy-Weinberg equilibrium.
individuals with the homozygous dominant genotype produce more offspring.
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Answers:
1. c, 2. c, 3. c, 4. e, 5. b, 6. d, 7. a, 8. b, 9. c, 10. d
Essay Questions
1. Are natural populations ever in Hardy-Weinberg equilibrium? Compare each of the assumptions for
Hardy-Weinberg equilibrium with the actual situation for natural populations.
Assumption #1 is large population size—that may be true in many cases. Assumption #2 is
random mating. Mating is generally not random—individuals often mate within their
locality, or there may be sexual selection, competition for mates, etc. Assumption #3 is no
mutations. This is NEVER true. Mutations occur in every generation. Assumption #4 is no
immigration—this is only true for completely isolated populations (cave fish, islands,
etc). Assumption #5 is no selection. This is never true in a global sense. There is always
selection against unfit genotypes (lethal mutations, etc), sexual selection, etc. There
could be no selection if you are measuring the allele frequency of a single gene that had
no effect on fitness, or no effect on phenotype at all (a non-coding or silent mutation).
2. Immigration (gene flow) and mutation may act against the force of natural selection, maintaining
alleles that are less fit (under current conditions) in a population. Why might this be beneficial for a
species?
Conditions are always changing in natural systems—environmental conditions,
predators, availability of food, and competition with other species can change the
relative survival value of various alleles. It is valuable for a species to maintain a
background level of genetic variation, which provides the raw material for selection of
new, better adapted individuals under new conditions.
3. Evolution by natural selection is not just studied in fossils. Biologists can conduct laboratory
experiments where they impose artificial selection, and can observe the effects of natural selection in
some environments. How does the presence of the predatory pike cichlid fish effect the size and color
of guppies in isolated pools? How does this demonstrate the principles of natural selection?
The pike cichlid preys on larger and more colorful fish, selecting for a population of
small, drab colored guppies. In pools without the predator, male guppies that are larger
and more colorful attract more mates, larger females produce more eggs. Over time, this
selection causes the population of guppies becomes larger, and more colorful. These
changes were observed to occur in less than 12 years.
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CHAPTER 21. THE EVIDENCE FOR EVOLUTION
THE ESSENTIALS
Students need to know:
ƒ examples of how the fossil record provides evidence for evolution.
ƒ examples of how artificial selection provides evidence for evolution.
ƒ examples of how comparative anatomy provides evidence for
evolution.
ƒ the difference between homologous structures and analogous
structures and what each indicates about evolution.
ƒ what vestigial structures are and how they provide evidence for
evolution.
ƒ examples of how comparative embryology provides evidence for
evolution.
ƒ the principle of convergent evolution and how it supports the
mechanism of natural selection.
ƒ how variation, adaptation, heritability, reproductive success, and time
are used to explain the principle of evolution by natural selection.
ƒ how Darwin’s finches and industrial melanism illustrate these
concepts.
Key Terms
industrial melanism
artificial selection
natural selection
homologous structures
analogous structures
vestigial structures
biogeography
convergent evolution
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theory
hypothesis
Concept Map
evidence for
evolution
includes
artificial
selection
biogeography
accounts for
distribution
of species
convergent
evolution
fossil
record
evidence ofevidence of
transitional
forms
ancestral
forms
molecular
record
homology
seen in
seen in
comparative
embryology
evidence for
comparative
anatomy
universal
genetic code
reveals
similarity of
DNA &
protein
sequence
evidence for
evidence for
similarities between
neighboring species in
different habitats
similarities in appearance
between species in
similar habitat in different
regions
homologous
structures
vestigial
structures
descent from
a common
ancestor
as seen in as seen in
different
structures &
functions
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developmental
& anatomical
similarities
closely
related
species
Multiple Choice Questions
Since the subject matter of the two chapters is so intertwined, questions for both Chapter 21 and Chapter
22 have been combined and are included below.
1. Which of the following is an example of convergent evolution?
a.
b.
c.
d.
e.
similar amino acid sequences of hemoglobin in humans and chimpanzees
similar bones in the forelimbs of horses and bats
similar body shape of dolphins and sharks
different beak shapes of Galapagos finches
similar plant species on islands and the nearest continent
2. The age of a fossil is most accurately measured by
a.
b.
c.
d.
e.
determining the ratios of radioactive isotopes in the rock where fossils occur.
counting layers of sedimentary rock.
comparison of its DNA sequence with modern species.
the complexity of the organism.
measuring the amount of organic carbon in the fossilized tissue.
3. Fossils of some intermediate forms have not been found because
a. new types of organisms can appear suddenly without progressive changes determined by natural
selection.
b. fossils are very rare, fossils of only a small fraction of all species have been found.
c. all fossils are the same age.
d. species produced by punctuated equilibrium do not leave fossils.
e. the ages of many fossils are not calculated correctly.
4. The studies of changes between black and light color in populations of the peppered moth show that
a.
b.
c.
d.
e.
natural selection can quickly change allele frequencies and common phenotypes in a population.
species can always adapt to environmental changes.
predators prefer light colored moths.
dark colored moths are physiologically superior to light colored moths.
natural selection produces irreversible changes.
5. A species is defined as
a.
b.
c.
d.
e.
a population of organisms similar in size, shape, and color.
a group of organisms that live in the same habitat.
a population of organisms that are able to interbreed.
a population of organisms that have the same number of chromosomes.
a population of organisms with a common ancestor.
6. Reproductive isolation can be maintained by all of the following EXCEPT
a.
b.
c.
d.
e.
geographic separation of a population into sub-groups
genetic incompatibility that prevents fertilization of gametes
hybrid sterility
predators that selectively prey on larger individuals in a population
elaborate mating rituals
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7. The embryonic development of vertebrates provides evidence for evolution because
a. each organism passes through the entire evolutionary history of its species as it develops from a
fertilized egg to a full grown individual.
b. the more recently species have shared a common ancestor, the more similar their embryological
development.
c. a small number of mutations can convert the embryo of one species into another species.
d. the DNA sequences of embryos change as they develop.
e. unrelated organisms can develop from very similar embryos.
8. DNA sequences can be used to determine the evolutionary relationships of species because
a. organisms with similar anatomy will develop similar DNA sequences by convergent mutations.
b. mutations occur randomly in DNA at a steady rate, so the number of DNA difference is
equivalent to the time since a pair of species that shared a common ancestor.
c. DNA sequences for proteins never change, so two species that have the same protein will have
the same DNA sequence.
d. natural selection causes organisms that live in similar environmental conditions to have the same
mutations in their DNA sequences.
e. recombination moves DNA sequences to related species.
9. For some traits (such as birth weight in mammals), natural selection favors individuals that are
average and the extremes are selected against. This is know as
a.
b.
c.
d.
e.
diversifying selection.
directional selection.
adaptive radiation.
disruptive selection.
stabilizing selection.
10. Artificial selection produces rapid changes in the phenotype of organisms by
a.
b.
c.
d.
e.
changing the frequency of alleles and selecting for new combinations of traits.
stimulating the species to increase the production of new mutations.
changing the number of chromosomes.
selecting only dominant genotypes.
allowing individuals with all genetic combinations to survive and reproduce equally.
150
Answers:
1. c, 2. a, 3. b, 4. a, 5. c, 6. d, 7. b, 8. b, 9. e, 10. a
Essay Questions
1. Evolution is often misunderstood as a linear process by which an ancestor species changes into
intermediate forms, and finally reaches a modern form – such as in the diagram of whale evolution
below. How do adaptation, natural selection, and extinction bring about large changes in organisms
(macroevolution)? How does the fossil record and popular descriptions of evolution lead to the
misconception of linear change?
Fossils are rare and hard to find, so the full diversity of past species is not evident. When
an ancient ancestor and a few intermediate forms are known for a modern species, it
obscures the continuous process of speciation and extinction that creates and prunes
branches on the evolutionary tree. Museums and science educators oversimplify when
they create linear diagrams of evolutionary change from ancient to modern forms.
Modern toothed whales
2. How does the presence of vestigial structures such as hip bones in whales, eyes in blind cave fish, and
the appendix in humans provide evidence for evolution?
The existence of vestigial structures provides strong evidence that current species have
evolved from ancestors which made use of these structures. Furthermore, species with
vestigial structures share common ancestry with other species that make use of these
structures.
3. The concept of species has been very useful for biologists, but it is flawed. Describe 3 weaknesses in
the current definition of a biological species. Why is it very difficult to define species of bacteria?
The definition of a species is a reproductively isolated population – this means that
members of a species can all interbreed, but do not breed with members of any other
species. Problems include: 1) There is widespread evidence for hybridization between
different species. 2) Geographically isolated populations can interbreed, but do not. They
may have distinct phenotypes and allele frequencies. 3) Organisms that reproduce
asexually do not have interbreeding within a species. 4) Bacteria commonly exchange
genetic information between distantly related individuals.
151
CHAPTER 22. THE ORIGIN OF SPECIES
THE ESSENTIALS
Students need to know:
ƒ the difference between the biological species concept and the
ecological species concept.
ƒ how prezygotic and postzygotic isolating mechanisms maintain
reproductive isolation and examples of each.
ƒ the distinction between allopatric and sympatric speciation and the
conditions required for each to occur.
ƒ the difference between gradualism and punctuated equilibrium with
regard to the rate of evolutionary change.
Key Terms
subspecies
biological species concept
ecological species concept
reproductively isolated
reproductive isolating
mechanisms
prezygotic isolating
mechanisms
postzygotic isolating
mechanisms
pheromones
reinforcement
allopatric
sympatric
polyploidy
autopolyploidy
152
allopolyploidy adaptive
radiation character
displacement
gradualism
stasis
punctuated equilibrium
mass extinctions
CHAPTER 23. SYSTEMATICS AND THE PHYLOGENETIC REVOLUTION
THE ESSENTIALS
Students need to know:
ƒ what a phylogeny represents.
ƒ the difference between ancestral and derived characteristics.
ƒ how a cladogram is constructed from shared derived characteristics.
ƒ that a cladistics approach to taxonomy organizes groups so that all the
members are related by common ancestry.
Key Terms
systematics
phylogeny
ancestral characters
derived characters
cladistics
cladogram
clade
outgroup
principle of parsimony
molecular clock
153
monophyletic
paraphyletic
polyphyletic
CHAPTER 24. GENOME EVOLUTION
24.1
Even distantly related species may often have many genes in common. Changes in DNA codons that do not alter the
amino acid specified are termed synonymous changes.
Mouse DNA has apparently mutated twice as fast as human DNA, and insect evolution is even more rapid. Short
generation time may be responsible for these rate differences.
Plants, animals, and fungi have approximately 70% of their genes in common.
24.2
Autopolyploidy results from an error in meiosis that leads to a duplicated genome; allopolyploidy is the result of
hybridization between species (see figure 24.2).
Polyploidy has occurred numerous times in the evolution of fl owering plants, and downsizing of genomes is common.
Downsizing of a polyploid genome can be caused by unequal loss of duplicated genes (see figure 24.7 ).
Polyploidization can lead to short-term silencing of genes via methylation of cytosines in the DNA.
Transposons become highly active after polyploidization; their insertions into new positions may lead to new phenotypes.
24.3
Aneuploidy creates problems in gamete formation. It is tolerated better in plants than in animals.
Duplicated DNA is common for genes associated with growth and development, immunity, and cell-surface receptors.
Paralogues are duplicated ancestral genes; orthologues are conserved ancestral genes.
Genomes may be rearranged by moving gene locations within a chromosome or by the fusion of two chromosomes.
Conservation of synteny refers to preservation of long segments of ancestral chromosome sequences identifiable in
related species (see figure 24.10).
Some ancestral genes become inactivated as they acquire mutations and are termed pseudogenes .
Occasionally part of a gene can end up in a new spot in the genome where its function changes.
Horizontal gene transfer creates many phylogenetic questions, such as the origins of the three major domains (see figure
24.11 ).
24.4
Even when species have highly similar genes, expression of these genes may vary greatly. Posttranscriptional
differences may also contribute to species differences.
Small evolutionary changes in the FOXP2 protein and its expression may have led to human speech (see figure 24.12 ) .
24.5
Nonprotein-coding sequences are found in retrotransposon-rich regions of the genome. Noncoding DNA may contain
regulatory RNA sequences for silencing other genes.
24.6
Genome size is most often infl ated due to the presence of introns and nonprotein-coding sequences. Genome size does
not correlate with the number of genes.
As much as a 200-fold difference in genome size has been found in plants; the number of genes has a narrower range.
154
24.7
Changes in amino acid sequences of critical proteins are a likely cause of diseases, and these differences can be
identifed by genome comparison.
By comparing related organisms, researchers can focus on genes that cause diseases and devise possible treatments.
Analysis of the genomes of pathogenic organisms may provide new avenues of treatment and prevention.
24.8
Research on Arabidopsis and Medicago species may provide insight into improvements in crop production and value.
Bacterial genes that produce protective compounds may be identified and used to engineer crop plants.
155