Download Chemistry Big Ideas

Enduring understanding 1.D: The origin of
living systems is explained by natural
processes.
• Essential knowledge 1.D.1 There are several hypotheses about the
natural origin of life on Earth, each with supporting scientific
evidence.
•
•
•
a. Scientific evidence supports the various models
1. Primitive Earth provided inorganic precursors from which organic molecules could
have been synthesized due to the presence of available free energy and the absence
of a significant quantity of oxygen.
2. In turn, these molecules served as monomers or building blocks for the formation
of more complex molecules, including amino acids and nucleotides. [See also 4.A.1]
3. The joining of these monomers produced polymers with the ability to replicate,
store and transfer information.
4. These complex reaction sets could have occurred in solution (organic soup
model) or as reactions on solid reactive surfaces. [See also 2.B.1]
5. The RNA World hypothesis proposes that RNA could have been the earliest
genetic material.
1.D.1 Learning Objectives
• Learning Objectives:
• LO 1.27 The student is able to describe a scientific hypothesis about
the origin of life on Earth. [See SP 1.2]
• LO 1.28 The student is able to evaluate scientific questions based
on hypotheses about the origin of life on Earth. [See SP 3.3]
• LO 1.29 The student is able to describe the reasons for revisions of
scientific hypotheses of the origin of life on Earth. [See SP 6.3]
• LO 1.30 The student is able to evaluate scientific hypotheses about
the origin of life on Earth. [See SP 6.5]
• LO 1.31 The student is able to evaluate the accuracy and legitimacy
of data to answer scientific questions about the origin of life on
Earth. [See SP 4.4]
Organic Molecules and the Origin of
Life on Earth
• Stanley Miller’s classic experiment
demonstrated the abiotic synthesis of
organic compounds
• Experiments support the idea that abiotic
synthesis of organic compounds, perhaps
near volcanoes, could have been a stage in
the origin of life
© 2011 Pearson Education, Inc.
Figure 4.2
EXPERIMENT
“Atmosphere”
CH4
Water vapor
Electrode
Condenser
Cooled “rain”
containing
organic
molecules
H2O
“sea”
Sample for chemical analysis
Cold
water
L.O. 1.28
Enduring understanding 2.A: Growth, reproduction
and maintenance of the organization of living
systems require free energy and matter.
• Essential knowledge 2.A.3: Organisms must exchange matter with
the environment to grow, reproduce and maintain organization
a. Molecules and atoms from the environment are necessary to build new molecules.
1. Carbon moves from the environment to organisms where it is used to build
carbohydrates, proteins, lipids or nucleic acids. Carbon is used in storage compounds and
cell formation in all organisms.
2. Nitrogen moves from the environment to organisms where it is used in building proteins
and nucleic acids. Phosphorus moves from the environment to organisms where it is used
in nucleic acids and certain lipids.
3. Living systems depend on properties of water that result from its polarity and hydrogen
bonding. To foster student understanding of this concept, instructors can choose an
illustrative example such as:
• Cohesion
• Adhesion
• High specific heat capacity
• Universal solvent supports reactions
• Heat of vaporization
• Heat of fusion
• Water’s thermal conductivity
• b. Surface area-to-volume ratios affect a biological
system’s ability to obtain necessary resources or
eliminate waste products.
1. As cells increase in volume, the relative surface area
decreases and demand for material resources increases;
more cellular structures are necessary to adequately
exchange materials and energy with the environment.
These limitations restrict cell size.
• Root hairs
• Cells of the alveoli
• Cells of the villi
• Microvilli
2. The surface area of the plasma membrane must be
large enough to adequately exchange materials; smaller
cells have a more favorable surface area-to-volume ratio
for exchange of materials with the environment.
• Metabolic requirements set upper limits on the
size of cells
• The surface area to volume ratio of a cell is
critical
• As the surface area increases by a factor of n2,
the volume increases by a factor of n3
• Small cells have a greater surface area relative
to volume
© 2011 Pearson Education, Inc.
Figure 6.7
Surface area increases while
total volume remains constant
5
1
1
Total surface area
[sum of the surface areas
(height  width) of all box
sides  number of boxes]
6
150
750
Total volume
[height  width  length
 number of boxes]
1
125
125
Surface-to-volume
(S-to-V) ratio
[surface area  volume]
6
1.2
6
2.A.3 Learning Objectives
• LO 2.6 The student is able to use calculated surface area-to-volume
ratios to predict which cell(s) might eliminate wastes or procure
nutrients faster by diffusion. [See SP 2.2]
• LO 2.7 Students will be able to explain how cell size and shape
affect the overall rate of nutrient intake and the rate of waste
elimination. [See SP 6.2]
• LO 2.8 The student is able to justify the selection of data regarding
the types of molecules that an animal, plant or bacterium will take up
as necessary building blocks and excrete as waste products. [See
SP 4.1]
• LO 2.9 The student is able to represent graphically or model
quantitatively the exchange of molecules between an organism and
its environment, and the subsequent use of these molecules to build
new molecules that facilitate dynamic homeostasis, growth and
reproduction. [See SP 1.1, 1.4]
L.O. 2.6
Essential knowledge 2.B.2: Growth and dynamic homeostasis are
maintained by the constant movement of molecules across
membranes
a. Passive transport does not require the input of metabolic
energy; the net movement of molecules is from high
concentration to low concentration.
Evidence of student learning is a demonstrated understanding of
each of the following:
1. Passive transport plays a primary role in the import of
resources and the export of wastes.
2. Membrane proteins play a role in facilitated diffusion of
charged and polar molecules through a membrane.
To foster student understanding of this concept, instructors can
choose an illustrative example such as:
• Glucose transport
• Na+/K+ transport
✘✘ There is no particular membrane protein that is required for
teaching this concept.
3. External environments can be hypotonic, hypertonic or
isotonic to internal environments of cells.
Concept 7.3: Passive transport
is diffusion of a substance
across a membrane with no
energy
investment
• Diffusion is the tendency for molecules to spread
out evenly into the available space
• Although each molecule moves randomly, diffusion
of a population of molecules may be directional
• At dynamic equilibrium, as many molecules cross
the membrane in one direction as in the other
© 2011 Pearson Education, Inc.
Animation: Membrane
Selectivity
© 2011 Pearson Education, Inc.
Right-click slide / select “Play”
Animation:
Diffusion
© 2011 Pearson Education, Inc.
Right-click slide / select “Play”
Figure 7.13
Molecules of dye
Membrane (cross section)
WATER
Net diffusion
Net diffusion
Equilibrium
(a) Diffusion of one solute
Net diffusion
Net diffusion
Equilibrium
Net diffusion
Net diffusion
Equilibrium
(b) Diffusion of two solutes
• Substances diffuse down their concentration
gradient, the region along which the density of
a chemical substance increases or decreases
• No work must be done to move substances
down the concentration gradient
• The diffusion of a substance across a biological
membrane is passive transport because no
energy is expended by the cell to make it
happen
© 2011 Pearson Education, Inc.
Effects of Osmosis on Water
Balance
• Osmosis is the diffusion of water across a
selectively permeable membrane
• Water diffuses across a membrane from the
region of lower solute concentration to the
region of higher solute concentration until the
solute concentration is equal on both sides
© 2011 Pearson Education, Inc.
Figure 7.14
Lower
concentration
of solute (sugar)
Higher
concentration
of solute
Sugar
molecule
H2O
Selectively
permeable
membrane
Osmosis
Same concentration
of solute
Water Balance of Cells Without
Walls
• Tonicity is the ability of a surrounding solution
to cause a cell to gain or lose water
• Isotonic solution: Solute concentration is the
same as that inside the cell; no net water
movement across the plasma membrane
• Hypertonic solution: Solute concentration is
greater than that inside the cell; cell loses
water
• Hypotonic solution: Solute concentration is
less than that inside the cell; cell gains water
© 2011 Pearson Education, Inc.
Figure 7.15
Hypotonic
solution
(a) Animal cell
H2O
(b) Plant cell
Isotonic
solution
H2O
H2O
Lysed
Normal
H2O Cell wall H2O
H2O
Turgid (normal) Flaccid
Osmosis
Hypertonic
solution
H2O
Shriveled
H2O
Plasmolyzed
• Hypertonic or hypotonic environments create
osmotic problems for organisms
• Osmoregulation, the control of solute
concentrations and water balance, is a necessary
adaptation for life in such environments
• The protist Paramecium, which is hypertonic to its
pond water environment, has a contractile
vacuole that acts as a pump
© 2011 Pearson Education, Inc.
Figure 7.16
Contractile vacuole
Video
50 m
Water Balance of Cells with
Walls
• Cell walls help maintain water balance
• A plant cell in a hypotonic solution swells until
the wall opposes uptake; the cell is now turgid
(firm)
• If a plant cell and its surroundings are isotonic,
there is no net movement of water into the cell;
the cell becomes flaccid (limp), and the plant
may wilt
© 2011 Pearson Education, Inc.
• In a hypertonic environment, plant cells lose
water; eventually, the membrane pulls away from
the wall, a usually lethal effect called
plasmolysis
Video
© 2011 Pearson Education, Inc.
Animation:
Osmosis
© 2011 Pearson Education, Inc.
Right-click slide / select “Play”
Short-Distance Transport of
Water Across Plasma
Membranes
• To survive, plants must balance water uptake and
loss
• Osmosis determines the net uptake or water loss
by a cell and is affected by solute concentration
and pressure
© 2011 Pearson Education, Inc.
• Water potential is a measurement that combines
the effects of solute concentration and pressure
• Water potential determines the direction of
movement of water
• Water flows from regions of higher water potential
to regions of lower water potential
• Potential refers to water’s capacity to perform work
© 2011 Pearson Education, Inc.
• Water potential is abbreviated as Ψ and measured
in a unit of pressure called the megapascal (MPa)
• Ψ = 0 MPa for pure water at sea level and at room
temperature
© 2011 Pearson Education, Inc.
How Solutes and Pressure Affect
Water Potential
• Both pressure and solute concentration affect
water potential
• This is expressed by the water potential equation:
Ψ  ΨS  ΨP
• The solute potential (ΨS) of a solution is directly
proportional to its molarity
• Solute potential is also called osmotic potential
© 2011 Pearson Education, Inc.
• Pressure potential (ΨP) is the physical pressure
on a solution
• Turgor pressure is the pressure exerted by the
plasma membrane against the cell wall, and the
cell wall against the protoplast
• The protoplast is the living part of the cell, which
also includes the plasma membrane
© 2011 Pearson Education, Inc.
Figure 36.8
Solutes have a negative
effect on  by binding
water molecules.
Positive pressure has a
positive effect on  by
pushing water.
Solutes and positive
pressure have opposing
effects on water
movement.
Negative pressure
(tension) has a negative
effect on  by pulling
water.
Pure water at equilibrium
Pure water at equilibrium
Pure water at equilibrium
Pure water at equilibrium
H2O
H2O
H2O
H2O
Adding solutes to the
right arm makes  lower
there, resulting in net
movement of water to
the right arm:
Applying positive
pressure to the right arm
makes  higher there,
resulting in net movement
of water to the left arm:
Positive
pressure
In this example, the effect
of adding solutes is
offset by positive
pressure, resulting in no
net movement of water:
Positive
pressure
Applying negative
pressure to the right arm
makes  lower there,
resulting in net movement
of water to the right arm:
Negative
pressure
Pure
water
Membrane
H2O
Solutes
Solutes
H2O
H2O
H2O
Water Movement Across Plant
Cell Membranes
• Water potential affects uptake and loss of water by
plant cells
• If a flaccid cell is placed in an environment with a
higher solute concentration, the cell will lose water
and undergo plasmolysis
• Plasmolysis occurs when the protoplast shrinks
and pulls away from the cell wall
© 2011 Pearson Education, Inc.
Figure 36.9
Initial flaccid cell:
0.4 M sucrose solution:
Plasmolyzed
cell at osmotic
equilibrium with
its surroundings
P  0
S  0.9
  0.9 MPa
P  0
S  0.9
  0.9 MPa
(a) Initial conditions: cellular   environmental 
P  0
S  0.7
  0.7 MPa
Pure water:
P  0
S  0
  0 MPa
Turgid cell
at osmotic
equilibrium with
its surroundings
P  0.7
S  0.7
  0 MPa
(b) Initial conditions: cellular   environmental 
• If a flaccid cell is placed in a solution with a lower
solute concentration, the cell will gain water and
become turgid
• Turgor loss in plants causes wilting, which can be
reversed when the plant is watered
© 2011 Pearson Education, Inc.
LO 2.12 The student is able to use representations and models to analyze
situations or solve problems qualitatively and quantitatively to investigate
whether dynamic homeostasis is maintained by the active movement of
molecules across membranes. [See SP 1.4]
Enduring understanding 3.A: Heritable
information provides for continuity of life.
• Essential knowledge 3.A.1: DNA, and in some cases RNA, is the
primary source of heritable information.
•
•
•
•
•
a. Genetic information is transmitted from one generation to the next through DNA or
RNA. 1. Genetic information is stored in and passed to subsequent generations
through DNA molecules and, in some cases, RNA molecules.
2. Noneukaryotic organisms have circular chromosomes, while eukaryotic organisms
have multiple linear chromosomes, although in biology there are exceptions to this
rule.
3. Prokaryotes, viruses and eukaryotes can contain plasmids, which are small extrachromosomal, double-stranded circular DNA molecules.
4. The proof that DNA is the carrier of genetic information involved a number of
important historical experiments. These include:
i. Contributions of Watson, Crick, Wilkins, and Franklin on
the structure of DNA
ii. Avery-MacLeod-McCarty experiments
iii. Hershey-Chase experiment
b. DNA and RNA molecules have structural similarities and differences that define
function. [See also 4.A.1]
• 1. Both have three components — sugar, phosphate and a nitrogenous base —
which form nucleotide units that are connected by covalent bonds to form a linear
molecule with 3‘ and 5' ends, with the nitrogenous bases perpendicular to the sugarphosphate backbone.
• 2. The basic structural differences include:
i. DNA contains deoxyribose (RNA contains ribose).
ii. RNA contains uracil in lieu of thymine in DNA.
iii. DNA is usually double stranded, RNA is usually single stranded.
iv. The two DNA strands in double-stranded DNA are antiparallel in
directionality.
• 3. Both DNA and RNA exhibit specific nucleotide base pairing that is conserved
through evolution: adenine pairs with thymine or uracil (A-T or A-U) and
cytosine pairs with guanine (C-G).
i. Purines (G and A) have a double ring structure.
ii. Pyrimidines (C, T and U) have a single ring structure.
• 4. The sequence of the RNA bases, together with the structure of the RNA
molecule, determines RNA function.
i. mRNA carries information from the DNA to the ribosome.
ii. tRNA molecules bind specific amino acids and allow information in the mRNA
to be translated to a linear peptide sequence.
iii. rRNA molecules are functional building blocks of ribosomes.
iv. The role of RNAi includes regulation of gene expression at the level of
mRNA transcription.
•
•
•
•
•
c. Genetic information flows from a sequence of nucleotides in a gene
to a sequence of amino acids in a protein.
1. The enzyme RNA-polymerase reads the DNA molecule in the 3' to 5'
direction and synthesizes complementary mRNA molecules that determine
the order of amino acids in the polypeptide.
2. In eukaryotic cells the mRNA transcript undergoes a series of enzymeregulated modifications.
• Addition of a poly-A tail
• Addition of a GTP cap
• Excision of introns
3. Translation of the mRNA occurs in the cytoplasm on the ribosome.
4. In prokaryotic organisms, transcription is coupled to translation of the
message. Translation involves energy and many steps, including initiation,
elongation and termination.
The salient features include:
i. The mRNA interacts with the rRNA of the ribosome to initiate
translation at the (start) codon.
ii. The sequence of nucleotides on the mRNA is read in triplets called
codons.
iii. Each codon encodes a specific amino acid, which can be deduced
by using a genetic code chart. Many amino acids have more than one
codon.
3.A.1 Learning Objectives
•
•
•
•
•
•
LO 3.1 The student is able to construct scientific explanations that use the
structures and mechanisms of DNA and RNA to support the claim that DNA
and, in some cases, that RNA are the primary sources of heritable
information. [See SP 6.5]
LO 3.2 The student is able to justify the selection of data from historical
investigations that support the claim that DNA is the source of heritable
information. [See SP 4.1]
LO 3.3 The student is able to describe representations and models that
illustrate how genetic information is copied for transmission between
generations. [See SP 1.2]
LO 3.4 The student is able to describe representations and models
illustrating how genetic information is translated into polypeptides. [See SP
1.2]
LO 3.5 The student can justify the claim that humans can manipulate
heritable information by identifying at least two commonly used
technologies. [See SP 6.4]
LO 3.6 The student can predict how a change in a specific DNA or RNA
sequence can result in changes in gene expression. [See SP 6.4]
Concept 5.1: Macromolecules are
polymers, built from monomers
• A polymer is a long molecule consisting of
many similar building blocks
• These small building-block molecules are
called monomers
• Three of the four classes of life’s organic
molecules are polymers
– Carbohydrates
– Proteins
– Nucleic acids
© 2011 Pearson Education, Inc.
X
Animation: Polymers
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Figure 5.2a
(a) Dehydration reaction: synthesizing a polymer
1
2
3
Unlinked monomer
Short polymer
Dehydration removes
a water molecule,
forming a new bond.
1
2
3
Longer polymer
4
Figure 5.2b
(b) Hydrolysis: breaking down a polymer
1
2
3
Hydrolysis adds
a water molecule,
breaking a bond.
1
2
3
4
Enduring understanding 4.A: Interactions within
biological systems lead to complex properties.
• Essential knowledge 4.A.1: The subcomponents
of biological molecules and their sequence
determine the properties of that molecule.
1. In nucleic acids, biological information is encoded
in sequences of nucleotide monomers. Each nucleotide
has structural components: a five-carbon sugar
(deoxyribose or ribose), a phosphate and a nitrogen
base (adenine, thymine, guanine, cytosine or uracil).
DNA and RNA differ in function and differ slightly in
structure, and these structural differences account for
the differing functions. [See also 1.D.1, 2.A.3, 3.A.1]
Concept 5.5: Nucleic acids store,
transmit, and help express
hereditary information
• The amino acid sequence of a polypeptide is
programmed by a unit of inheritance called a
gene
• Genes are made of DNA, a nucleic acid
made of monomers called nucleotides
© 2011 Pearson Education, Inc.
The Roles of Nucleic Acids
• There are two types of nucleic acids
– Deoxyribonucleic acid (DNA)
– Ribonucleic acid (RNA)
• DNA provides directions for its own
replication
• DNA directs synthesis of messenger RNA
(mRNA) and, through mRNA, controls
protein synthesis
• Protein synthesis occurs on ribosomes
© 2011 Pearson Education, Inc.
Figure 5.25-3
DNA
1 Synthesis of
mRNA
mRNA
NUCLEUS
CYTOPLASM
mRNA
2 Movement of
mRNA into
cytoplasm
Ribosome
3 Synthesis
of protein
Polypeptide
Amino
acids
Figure 5.26
5 end
Sugar-phosphate backbone
Nitrogenous bases
Pyrimidines
5C
3C
Nucleoside
Nitrogenous
base
Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA)
Purines
5C
1C
5C
3C
Phosphate
group
3C
Sugar
(pentose)
Guanine (G)
Adenine (A)
(b) Nucleotide
Sugars
3 end
(a) Polynucleotide, or nucleic acid
Deoxyribose (in DNA)
(c) Nucleoside components
Ribose (in RNA)
The Structures of DNA and RNA
Molecules
• RNA molecules usually exist as single
polypeptide chains
• DNA molecules have two polynucleotides
spiraling around an imaginary axis, forming a
double helix
• In the DNA double helix, the two backbones
run in opposite 5→ 3 directions from each
other, an arrangement referred to as
antiparallel
• One DNA molecule includes many genes
© 2011 Pearson Education, Inc.
• The nitrogenous bases in DNA pair up and form
hydrogen bonds: adenine (A) always with
thymine (T), and guanine (G) always with
cytosine (C)
• Called complementary base pairing
• Complementary pairing can also occur between
two RNA molecules or between parts of the same
molecule
• In RNA, thymine is replaced by uracil (U) so A
and U pair
© 2011 Pearson Education, Inc.
Figure 5.27
5
3
G and C linked by 3 hydrogen bonds, A
and T by only 2
Sugar-phosphate
backbones
Hydrogen bonds
Base pair joined
by hydrogen
bonding
3
5
(a) DNA
Base pair joined
by hydrogen bonding
(b) Transfer RNA
DNA and Proteins as Tape
Measures of Evolution
• The linear sequences of nucleotides in DNA
molecules are passed from parents to offspring
• Two closely related species are more similar in
DNA than are more distantly related species
• Molecular biology can be used to assess
evolutionary kinship
© 2011 Pearson Education, Inc.
Proteins
• 2. In proteins, the specific order of amino acids
in a polypeptide (primary structure) interacts with
the environment to determine the overall shape
of the protein, which also involves secondary
tertiary and quaternary structure and, thus, its
function. The R group of an amino acid can be
categorized by chemical properties
(hydrophobic, hydrophilic and ionic), and the
interactions of these R groups determine
structure and function of that region of the
protein. [See also 1.D.1, 2.A.3, 2.B.1]
Figure 5.15-a
Enzymatic proteins
Defensive proteins
Function: Selective acceleration of chemical reactions
Example: Digestive enzymes catalyze the hydrolysis
of bonds in food molecules.
Function: Protection against disease
Example: Antibodies inactivate and help destroy
viruses and bacteria.
Antibodies
Enzyme
Virus
Bacterium
Storage proteins
Transport proteins
Function: Storage of amino acids
Function: Transport of substances
Examples: Hemoglobin, the iron-containing protein of
vertebrate blood, transports oxygen from the lungs to
other parts of the body. Other proteins transport
molecules across cell membranes.
Examples: Casein, the protein of milk, is the major
source of amino acids for baby mammals. Plants have
storage proteins in their seeds. Ovalbumin is the
protein of egg white, used as an amino acid source
for the developing embryo.
Transport
protein
Ovalbumin
Amino acids
for embryo
Cell membrane
Figure 5.15-b
Hormonal proteins
Receptor proteins
Function: Coordination of an organism’s activities
Example: Insulin, a hormone secreted by the
pancreas, causes other tissues to take up glucose,
thus regulating blood sugar concentration
Function: Response of cell to chemical stimuli
Example: Receptors built into the membrane of a
nerve cell detect signaling molecules released by
other nerve cells.
High
blood sugar
Insulin
secreted
Normal
blood sugar
Receptor
protein
Signaling
molecules
Contractile and motor proteins
Structural proteins
Function: Movement
Examples: Motor proteins are responsible for the
undulations of cilia and flagella. Actin and myosin
proteins are responsible for the contraction of
muscles.
Function: Support
Examples: Keratin is the protein of hair, horns,
feathers, and other skin appendages. Insects and
spiders use silk fibers to make their cocoons and webs,
respectively. Collagen and elastin proteins provide a
fibrous framework in animal connective tissues.
Actin
Myosin
Collagen
Muscle tissue
100 m
Connective
tissue
60 m
Figure 5.UN01
Side chain (R group)
 carbon
Amino
group
Carboxyl
group
Figure 5.16a
Nonpolar side chains; hydrophobic
Side chain
Glycine
(Gly or G)
Methionine
(Met or M)
Alanine
(Ala or A)
Valine
(Val or V)
Phenylalanine
(Phe or F)
Leucine
(Leu or L)
Tryptophan
(Trp or W)
Isoleucine
(Ile or I)
Proline
(Pro or P)
Figure 5.16b
Polar side chains; hydrophilic
Serine
(Ser or S)
Threonine
(Thr or T)
Cysteine
(Cys or C)
Tyrosine
(Tyr or Y)
Asparagine
(Asn or N)
Glutamine
(Gln or Q)
Figure 5.16c
Electrically charged side chains; hydrophilic
Basic (positively charged)
Acidic (negatively charged)
Aspartic acid Glutamic acid
(Glu or E)
(Asp or D)
Lysine
(Lys or K)
Arginine
(Arg or R)
Histidine
(His or H)
Figure 5.17
Peptide bond
New peptide
bond forming
Side
chains
Backbone
Amino end
(N-terminus)
Peptide
bond
Carboxyl end
(C-terminus)
Figure 5.18
Groove
Groove
(a) A ribbon model
(b) A space-filling model
Figure 5.19
Antibody protein
Protein from flu virus
Figure 5.20a
Primary structure
Amino
acids
Amino end
Primary structure of transthyretin
Carboxyl end
Animation: Primary Protein Structure
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Animation: Secondary Protein Structure
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Animation: Tertiary Protein Structure
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Figure 5.20g
Quaternary structure
Transthyretin
protein
(four identical
polypeptides)
Figure 5.20h
Collagen
Figure 5.20i
Heme
Iron
 subunit
 subunit
 subunit
 subunit
Hemoglobin
Animation: Quaternary Protein Structure
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Figure 5.21
Sickle-cell hemoglobin
Normal hemoglobin
Primary
Structure
1
2
3
4
5
6
7
Secondary
and Tertiary
Structures
Quaternary
Structure
Function
Molecules do not
associate with one
another; each carries
oxygen.
Normal
hemoglobin
 subunit

Red Blood
Cell Shape

10 m


1
2
3
4
5
6
7
Exposed
hydrophobic
region
Sickle-cell
hemoglobin

 subunit

Molecules crystallize
into a fiber; capacity
to carry oxygen is
reduced.


10 m
What Determines Protein
Structure?
• In addition to primary structure, physical and
chemical conditions can affect structure
• Alterations in pH, salt concentration,
temperature, or other environmental factors
can cause a protein to unravel
• This loss of a protein’s native structure is
called denaturation
• A denatured protein is biologically inactive
© 2011 Pearson Education, Inc.
Figure 5.22
tu
Normal protein
Denatured protein
Figure 5.23
Polypeptide
Correctly
folded
protein
Cap
Hollow
cylinder
Chaperonin
(fully assembled)
Steps of Chaperonin
Action:
1 An unfolded polypeptide enters the
cylinder from
one end.
2 The cap attaches, causing 3 The cap comes
the cylinder to change
off, and the
shape in such a way that
properly folded
it creates a hydrophilic
protein is
environment for the
released.
folding of the polypeptide.
• Scientists use X-ray crystallography to
determine a protein’s structure
• Another method is nuclear magnetic
resonance (NMR) spectroscopy, which does
not require protein crystallization
• Bioinformatics uses computer programs to
predict protein structure from amino acid
sequences
© 2011 Pearson Education, Inc.
Figure 5.24
EXPERIMENT
Diffracted
X-rays
X-ray
source X-ray
beam
Crystal
Digital detector
X-ray diffraction
pattern
RESULTS
RNA
DNA
RNA
polymerase II
Lipids
• 3. In general, lipids are nonpolar; however,
phospholipids exhibit structural properties,
with polar regions that interact with other
polar molecules such as water, and with
nonpolar regions where differences in
saturation determine the structure and
function of lipids. [See also 1.D.1, 2.A.3, 2.
B.1]
Concept 5.3: Lipids are a diverse
group of hydrophobic molecules
• Lipids are the one class of large biological
molecules that do not form polymers
• The unifying feature of lipids is having little or
no affinity for water
• Lipids are hydrophobic because they consist
mostly of hydrocarbons, which form nonpolar
covalent bonds
• The most biologically important lipids are fats,
phospholipids, and steroids
© 2011 Pearson Education, Inc.
Figure 5.10a
X
Fatty acid
(in this case, palmitic acid)
Glycerol
(a) One of three dehydration reactions in the synthesis of a fat
Figure 5.10b
Ester linkage
(b) Fat molecule (triacylglycerol)
X
Figure 5.11
X
(a) Saturated fat
Structural
formula of a
saturated fat
molecule
Space-filling
model of stearic
acid, a saturated
fatty acid
(b) Unsaturated fat
Structural
formula of an
unsaturated fat
molecule
Space-filling model
of oleic acid, an
unsaturated fatty
acid
Cis double bond
causes bending.
X
• A diet rich in saturated fats may contribute to
cardiovascular disease through plaque deposits
• Hydrogenation is the process of converting
unsaturated fats to saturated fats by adding
hydrogen
• Hydrogenating vegetable oils also creates
unsaturated fats with trans double bonds
• These trans fats may contribute more than
saturated fats to cardiovascular disease
© 2011 Pearson Education, Inc.
Hydrophobic tails
Hydrophilic head
Figure 5.12a
(a) Structural formula
Choline
X
Phosphate
Glycerol
Fatty acids
(b) Space-filling model
Steroids
X
• Steroids are lipids characterized by a carbon
skeleton consisting of four fused rings
• Cholesterol, an important steroid, is a
component in animal cell membranes
• Although cholesterol is essential in animals,
high levels in the blood may contribute to
cardiovascular disease
© 2011 Pearson Education, Inc.
Figure 5.14
X
Carbs
4. Carbohydrates are composed of sugar
monomers whose structures and bonding
with each other by dehydration synthesis
determine the properties and functions of
the molecules. Illustrative examples
include: cellulose versus starch.
Concept 5.2: Carbohydrates serve
as fuel and building material
• Carbohydrates include sugars and the
polymers of sugars
• The simplest carbohydrates are
monosaccharides, or single sugars
• Carbohydrate macromolecules are
polysaccharides, polymers composed of
many sugar building blocks
© 2011 Pearson Education, Inc.
Sugars
• Monosaccharides have molecular formulas
that are usually multiples of CH2O
• Glucose (C6H12O6) is the most common
monosaccharide
• Monosaccharides are classified by
– The location of the carbonyl group (as aldose
or ketose)
– The number of carbons in the carbon skeleton
© 2011 Pearson Education, Inc.
Figure 5.4
1
2
6
6
5
5
X
3
4
4
5
1
3
6
(a) Linear and ring forms
6
5
4
1
3
2
(b) Abbreviated ring structure
2
4
1
3
2
Figure 5.5
1–4
glycosidic
1 linkage 4
Glucose
Glucose
Maltose
(a) Dehydration reaction in the synthesis of maltose
1–2
glycosidic
1 linkage 2
Glucose
Fructose
(b) Dehydration reaction in the synthesis of sucrose
Sucrose
Animation: Disaccharide
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Figure 5.5
1–4
glycosidic
1 linkage 4
Glucose
Glucose
Maltose
(a) Dehydration reaction in the synthesis of maltose
1–2
glycosidic
1 linkage 2
Glucose
Fructose
(b) Dehydration reaction in the synthesis of sucrose
Sucrose
Figure 5.6
Chloroplast
Starch granules
Amylopectin
X
Amylose
(a) Starch:
1 m
a plant polysaccharide
Mitochondria
Glycogen granules
Glycogen
(b) Glycogen:
0.5 m
an animal polysaccharide
X
Animation: Polysaccharides
Right-click slide / select “Play”
© 2011 Pearson Education, Inc.
Figure 5.7b
X
1
4
(b) Starch: 1–4 linkage of  glucose monomers
1
4
(c) Cellulose: 1–4 linkage of  glucose monomers
Figure 5.8
Cellulose
microfibrils in a
plant cell wall
Cell wall
X
Microfibril
10 m
0.5 m
Cellulose
molecules
 Glucose
monomer
• Enzymes that digest starch by hydrolyzing 
linkages can’t hydrolyze  linkages in cellulose
• Cellulose in human food passes through the
digestive tract as insoluble fiber
• Some microbes use enzymes to digest
cellulose
• Many herbivores, from cows to termites, have
symbiotic relationships with these microbes
© 2011 Pearson Education, Inc.
Figure 5.9
The structure
of the chitin
monomer
Chitin forms the exoskeleton
of arthropods.
Chitin is used to make a strong and flexible
surgical thread that decomposes after the
wound or incision heals.
X
b. Directionality influences structure
and function of the polymer.
1. Nucleic acids have ends, defined by the 3' and 5'
carbons of the sugar in the nucleotide, that determine
the direction in which complementary nucleotides are
added during DNA synthesis and the direction in which
transcription occurs (from 5' to 3'). [See also 3.A.1]
2. Proteins have an amino (NH2) end and a carboxyl
(COOH) end, and consist of a linear sequence of amino
acids connected by the formation of peptide bonds by
dehydration synthesis between the amino and carboxyl
groups of adjacent monomers.
3. The nature of the bonding between carbohydrate
subunits determines their relative orientation in the
carbohydrate, which then determines the secondary
structure of the carbohydrate.
4.A.1 Learning Objectives
• LO 4.1 The student is able to explain the connection
between the sequence and the subcomponents of a
biological polymer and its properties. [See SP 7.1]
• LO 4.2 The student is able to refine representations and
models to explain how the subcomponents of a
biological polymer and their sequence determine the
properties of that polymer. [See SP 1.3]
• LO 4.3 The student is able to use models to predict and
justify that changes in the subcomponents of a biological
polymer affect the functionality of the molecule. [See SP
6.1, 6.4]
L.O. 4.1
Enduring understanding 4.B: Competition and
cooperation are important aspects of biological
systems.
• Essential knowledge 4.B.1: Interactions between
molecules affect their structure and function.
a. Change in the structure of a molecular system may result in
a change of the function of the system. [See also 3.D.3]
Learning Objective: LO 4.17 The student is able to
analyze data to identify how molecular
interactions affect structure and function. [See
SP 5.1]
Enduring understanding 4.C: Naturally occurring diversity among
and between components within biological systems affects
interactions with the environment.
• Essential knowledge 4.C.1: Variation in molecular units
provides cells with a wider range of functions.
•
a. Variations within molecular classes provide cells and organisms with a
wider range of functions. [See also 2.B.1, 3.A.1, 4.A.1, 4.A.2]
• Different types of phospholipids in cell membranes
• Different types of hemoglobin
• MHC proteins
• Chlorophylls
• Molecular diversity of antibodies in response to an antigen
b. Multiple copies of alleles or genes (gene duplication) may provide new
phenotypes. [See also 3.A.4, 3.C.1]
1. A heterozygote may be a more advantageous genotype than a
homozygote under particular conditions, since with two different alleles, the
organism has two forms of proteins that may provide functional resilience in
response to environmental stresses.
Learning Objective: LO 4.22 The student is able
to construct explanations based on evidence of
how variation in molecular units provides cells
with a wider range of functions. [See SP 6.2]