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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 5C 3C Nucleoside Nitrogenous base Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA) Purines 5C 1C 5C 3C Phosphate group 3C 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]