SBI4U Unit 1 Biochemistry Test Review Package PDF

SBI4U Thalen Unit 1 - Biochemistry Test Review Package Objectives Lesson 1: Chemistry in Living Things Identify the six key elements (C, H, O, N, P, S) that make up biological molecules....

SBI4U Thalen Unit 1 - Biochemistry Test Review Package Objectives Lesson 1: Chemistry in Living Things Identify the six key elements (C, H, O, N, P, S) that make up biological molecules. - Carbon, hydrogen, oxygen, nitrogen, phosphorus, sulfur Describe how covalent bonds form between atoms to create essential molecules like carbohydrates, proteins, lipids, and nucleic acids. - Carbohydrates are made up of carbon hydrogen and oxygen. In carbohydrates, covalent bonds primarily occur when monosaccharides (simple sugars like glucose) link together via dehydration synthesis (removal of water). The oxygen from one hydroxyl group (–OH) and hydrogen from another hydroxyl group combine to form water, creating a glycosidic bond between two sugar molecules. - Example: In sucrose (table sugar), a glucose and fructose molecule are joined by a covalent glycosidic bond. - Proteins are made of amino acids, which contain carbon (C), hydrogen (H), oxygen (O), nitrogen (N), and sometimes sulfur (S). Covalent peptide bonds form between amino acids. This occurs when the carboxyl group (–COOH) of one amino acid reacts with the amino group (–NH₂) of another, releasing a molecule of water and forming a covalent bond that links the amino acids in a chain. This chain folds to become a protein. - Example: Proteins like enzymes or structural proteins in muscles and tissues are formed by long chains of amino acids held together by peptide bonds. - Lipids are Mainly composed of carbon (C), hydrogen (H), and oxygen (O), but with fewer oxygen atoms compared to carbohydrates. In lipids, covalent bonds form when glycerol bonds with fatty acids via ester bonds. This happens during dehydration synthesis, where hydroxyl groups from glycerol react with carboxyl groups from fatty acids, releasing water and creating a covalent ester bond. - Example: Triglycerides (fats and oils) are made when three fatty acids bond covalently to one glycerol molecule. - Nucleic acids are made of nucleotides, which include a sugar, a phosphate group, and a nitrogenous base. Covalent bonds occur in nucleic acids when nucleotides link together via phosphodiester bonds. This bond forms between the phosphate group of one nucleotide and the hydroxyl group on the sugar of another nucleotide, forming the backbone of DNA and RNA. - Example: In DNA, the sugar-phosphate backbone is held together by covalent bonds, while the nitrogenous bases pair with each other via hydrogen bonds. - In all these molecules, covalent bonds are crucial for maintaining the structure and function, allowing them to perform their biological roles effectively. Explain the concept of polarity in water molecules and how hydrogen bonds form between them. SBI4U Thalen - Polarity of water: Unequal sharing of electrons: Oxygen is more electronegative than hydrogen, meaning it has a stronger tendency to attract electrons. In a water molecule, the oxygen atom pulls the shared electrons in the covalent bonds closer to itself, creating an unequal sharing of electrons. Partial charges: As a result of this unequal sharing, the oxygen atom carries a partial negative charge (δ⁻), while each hydrogen atom carries a partial positive charge (δ⁺). This makes the water molecule polar, meaning it has a positive end and a negative end, like a magnet. - Hydrogen Bonding: Attraction between polar molecules: Because of the partial charges, the positive end of one water molecule (the hydrogen atoms) is attracted to the negative end of a neighboring water molecule (the oxygen atom). Formation of hydrogen bonds: These attractions lead to the formation of hydrogen bonds, which are weak bonds that form between the hydrogen atom of one water molecule and the oxygen atom of another. Though much weaker than covalent bonds, hydrogen bonds are crucial in determining water’s properties. Illustrate and differentiate between hydrophobic and hydrophilic molecules, showing how they interact with water. Recognize and draw the structures of important functional groups (e.g., hydroxyl, carboxyl, amine) and explain their role in giving molecules specific properties. Lesson 2: Macromolecules - Carbohydrates Define and identify monosaccharides, disaccharides, and polysaccharides through molecular structures and chemical formulas. - 1. Monosaccharides - Definition: Monosaccharides are the simplest form of carbohydrates, consisting of a single sugar unit. They are the building blocks of more complex carbohydrates. - Molecular Structure: Monosaccharides typically have a backbone of 3-7 carbon atoms, each with a hydroxyl group (–OH) attached, and either an aldehyde group (–CHO) or a ketone group (C=O) at one end. - Common Examples: - Glucose: C₆H₁₂O₆ – A hexose (6-carbon) sugar and the primary energy source in cells. - Fructose: C₆H₁₂O₆ – Another hexose sugar, structurally different from glucose, found in fruits. - Galactose: C₆H₁₂O₆ – Similar to glucose, it is commonly found in milk. - 2. Disaccharides - Definition: Disaccharides consist of two monosaccharides joined together by a covalent bond known as a glycosidic bond, which forms through a dehydration synthesis reaction (removal of water). - Molecular Structure: Disaccharides are formed by linking two monosaccharides via an oxygen bridge, with the general formula C₁₂H₂₂O₁₁ for most common disaccharides, since a molecule of water (H₂O) is lost during bonding. - Common Examples: - Sucrose (table sugar): C₁₂H₂₂O₁₁ – Formed from glucose and fructose. SBI4U Thalen - Lactose (milk sugar): C₁₂H₂₂O₁₁ – Formed from glucose and galactose. - Maltose (malt sugar): C₁₂H₂₂O₁₁ – Formed from two glucose molecules. - 3. Polysaccharides - Definition: Polysaccharides are large, complex carbohydrates composed of long chains of monosaccharide units. These chains can be linear or branched and can contain hundreds to thousands of monosaccharides. - Molecular Structure: Polysaccharides are formed by repeated glycosidic linkages between monosaccharide units. Their structure varies depending on the type of glycosidic bonds and branching. - Common Examples: - Starch: (C₆H₁₀O₅)ₙ – A storage form of glucose in plants, made of amylose (linear) and amylopectin (branched). - Glycogen: (C₆H₁₀O₅)ₙ – A highly branched storage form of glucose in animals, found mainly in the liver and muscles. - Cellulose: (C₆H₁₀O₅)ₙ – A structural polysaccharide in plant cell walls, with linear chains of glucose units that form strong fibers. - Each type of carbohydrate plays a unique role in biological systems, from immediate energy sources (monosaccharides) to energy storage (disaccharides and polysaccharides) and structural support (polysaccharides). Explain the role of carbohydrates in providing energy for cells, using examples like glucose and glycogen. - Carbohydrates are essential for providing energy to cells, as they are the primary and most readily available source of fuel. Through various metabolic processes, carbohydrates like glucose and glycogen are broken down to release energy in the form of ATP (adenosine triphosphate), which powers cellular activities. - 1. Glucose: Immediate Energy Source - Role in Cells: Glucose is the most common and primary energy source for cells, particularly in organisms that require quick energy. - Process of Energy Production: When glucose enters a cell, it undergoes glycolysis in the cytoplasm, where it is broken down into two molecules of pyruvate, releasing a small amount of ATP and NADH (an electron carrier). - Example: During intense exercise, muscle cells rapidly use glucose to meet their increased energy requirements. - 2. Glycogen: Stored Energy for Later Use - Role in Cells: Glycogen is a polysaccharide that serves as the main form of stored glucose in animals. It is primarily stored in the liver and muscle tissues. - Process of Energy Release: When blood glucose levels drop, glycogen is broken down into glucose molecules through a process called glycogenolysis. - In the liver: Glycogen breakdown releases glucose into the bloodstream, maintaining stable blood glucose levels, which provides energy for the whole body, including the brain and nervous system. - In muscles: Glycogen is broken down to provide glucose for muscle cells directly, which is critical during sustained physical activity. SBI4U Thalen - Example: During fasting or between meals, the liver breaks down glycogen to release glucose, supplying energy to maintain bodily functions. Demonstrate the process of dehydration synthesis to show how monosaccharides form disaccharides and polysaccharides. - Dehydration synthesis is a chemical process in which two molecules are joined together by removing a molecule of water. This process is fundamental for building larger carbohydrates, as it allows monosaccharides to link and form disaccharides and polysaccharides. - - 1. Forming Disaccharides from Monosaccharides - Process: When two monosaccharides join, one monosaccharide loses a hydroxyl group (–OH), and the other loses a hydrogen atom (H) from its hydroxyl group. These two atoms combine to form a molecule of water (H₂O). - Glycosidic Bond: The bond that forms between the two monosaccharides is called a glycosidic bond, a type of covalent bond. - Example: Glucose + Fructose → Sucrose - Glucose (C₆H₁₂O₆) and fructose (C₆H₁₂O₆) combine, releasing one molecule of water (H₂O). - The resulting disaccharide is sucrose (C₁₂H₂₂O₁₁) with a glycosidic bond connecting the two monosaccharides. - 2. Forming Polysaccharides from Multiple Monosaccharides - Process: When multiple monosaccharides link together, the same dehydration synthesis process occurs repeatedly. Each new bond formed between monosaccharides results in the removal of one water molecule. - Polysaccharide Structure: This chain can be linear or branched, depending on the type of glycosidic bonds formed. - Example: Glucose + Glucose + Glucose … → Starch or Glycogen - Starch: In plants, many glucose molecules link together in a long chain to form starch, which serves as a storage form of glucose. - Glycogen: In animals, glucose molecules form a highly branched polysaccharide known as glycogen, stored mainly in liver and muscle cells for energy. Compare the structures and functions of common polysaccharides such as starch, cellulose, and glycogen, and explain why humans can digest some but not others. - Humans can digest starch and glycogen because both have α-glycosidic bonds, which our digestive enzymes can break down into glucose. - Digestibility in Humans: Humans can digest starch because we produce the enzyme amylase, which breaks down α-1,4-glycosidic bonds in starch. This releases glucose units that can be absorbed and used as energy. - Digestibility in Humans: Humans can digest glycogen easily using enzymes that break α-1,4 and α-1,6 bonds, similar to those in starch. This makes glycogen a readily accessible energy source. - Humans cannot digest cellulose because it has β-glycosidic bonds, which we lack the enzymes to break. Instead, cellulose serves as dietary fiber, which aids in digestive health. SBI4U Thalen - In summary, the structural differences in glycosidic bonding (α vs. β) in these polysaccharides determine whether humans can digest them Describe the role of dehydration synthesis and hydrolysis reactions in the metabolism of carbohydrates - 1. Dehydration Synthesis: Building Carbohydrates - Definition: Dehydration synthesis, or condensation reaction, is the process of joining two molecules by removing a water molecule. This reaction links monosaccharides to form disaccharides and polysaccharides. - Process: In this reaction, one monosaccharide loses a hydroxyl group (–OH), while another loses a hydrogen atom (H). These atoms combine to form a molecule of water (H₂O), and the two monosaccharides are joined by a covalent glycosidic bond. - Role in Carbohydrate Metabolism: Dehydration synthesis allows cells to store glucose as polysaccharides such as glycogen (in animals) or starch (in plants). - 2. Hydrolysis: Breaking Down Carbohydrates - Definition: Hydrolysis is the process of breaking down complex molecules by adding a molecule of water. In carbohydrate metabolism, hydrolysis breaks down disaccharides and polysaccharides into simpler sugars, making them available for cellular energy. - Process: During hydrolysis, a water molecule is added to a glycosidic bond, breaking it and separating the larger carbohydrate into its smaller monosaccharide components. - Role in Carbohydrate Metabolism: Hydrolysis is crucial for releasing glucose from stored carbohydrates, providing energy for cellular processes. - Example: When energy is needed, glycogen in the liver or muscles is broken down into glucose molecules through hydrolysis, which can then be used in cellular respiration to produce ATP. Lesson 3: Macromolecules - Lipids Identify and describe the structure of different lipids (fats, phospholipids, steroids, waxes) and explain their hydrophobic properties. - 1. Fats (Triglycerides) - Structure: Fats consist of a glycerol molecule attached to three fatty acid chains. Each fatty acid is a long hydrocarbon chain with a carboxyl group (-COOH) at one end. In triglycerides, each hydroxyl group of the glycerol backbone binds to the carboxyl group of a fatty acid through dehydration synthesis, forming three ester bonds. - Saturated vs. Unsaturated: Saturated fats have no double bonds in their fatty acid chains, making them straight and solid at room temperature. Unsaturated fats contain one or more double bonds, introducing kinks that make them liquid at room temperature. - Hydrophobic Properties: The long hydrocarbon chains in triglycerides are nonpolar, making fats insoluble in water. This hydrophobic nature allows fats to store energy densely without interacting with water. - 2. Phospholipids - Structure: Phospholipids are similar to triglycerides but with one major difference: they have two fatty acid chains and one phosphate group attached to the glycerol backbone. The SBI4U Thalen phosphate group is often modified by an additional polar group, giving phospholipids both a hydrophilic (water-attracting) "head" and hydrophobic (water-repelling) "tails." - Amphipathic Nature: Phospholipids have both hydrophobic and hydrophilic parts, making them amphipathic. This structure is crucial for forming cell membranes, as the hydrophilic heads face outward toward water, while the hydrophobic tails face inward, away from water, creating a bilayer. - Hydrophobic Properties: The fatty acid tails are nonpolar, which drives them to aggregate away from water, creating a stable barrier in the membrane and contributing to the membrane's selective permeability. - 3. Steroids - Structure: Steroids have a unique structure of four fused carbon rings, with various functional groups attached to these rings. Cholesterol is a common steroid, and it serves as a precursor for other steroids like testosterone and estrogen. - Rigid and Hydrophobic: The ring structure of steroids is largely nonpolar, making them hydrophobic. Despite their unique shape compared to other lipids, steroids do not dissolve in water and are often found embedded in cell membranes or transported in the bloodstream bound to proteins. - Hydrophobic Properties: The nonpolar ring system makes steroids hydrophobic, allowing them to easily pass through cell membranes and interact with lipid environments. - 4. Waxes - Structure: Waxes are composed of long-chain fatty acids esterified to long-chain alcohols. This structure creates a highly nonpolar molecule that is solid at room temperature and extremely hydrophobic. - Water-Repellent Coating: Waxes are often found on the surfaces of plants (leaves and stems) and in animal products like earwax and beeswax, where they form protective, water-resistant layers. - Hydrophobic Properties: The long hydrocarbon chains make waxes highly water-repellent, enabling them to create protective barriers that prevent water loss in plants and protect animal skin and fur. Differentiate between saturated and unsaturated fats based on their chemical structure, and explain how their structure influences their health impacts. - 1. Saturated Fats - Chemical Structure: Saturated fats have fatty acid chains in which all carbon atoms are connected by single bonds, meaning each carbon is "saturated" with hydrogen atoms. This structure forms straight, tightly packed chains. - Physical Properties: Because of their straight structure, saturated fats can pack closely together, making them solid at room temperature. They are commonly found in animal fats, butter, and some plant oils like coconut oil - 2. Unsaturated Fats - Chemical Structure: Unsaturated fats have one or more double bonds between carbon atoms in the fatty acid chain, which causes kinks or bends. These double bonds reduce the number of hydrogen atoms attached to the carbon chain. - Monounsaturated fats have one double bond (e.g., olive oil, avocado). SBI4U Thalen - Polyunsaturated fats have two or more double bonds (e.g., fish oil, sunflower oil). - Physical Properties: The kinks created by the double bonds prevent unsaturated fats from packing closely together, making them liquid at room temperature. They are commonly found in plant-based oils, nuts, seeds, and fish. - Illustrate how phospholipids arrange themselves in a bilayer to form cell membranes, identifying the hydrophilic heads and hydrophobic tails. Lesson 4: Macromolecules - Proteins Identify the structure of amino acids and describe how they bond together through peptide bonds to form polypeptides. - 1. Structure of an Amino Acid - Amino Group (-NH₂): This group consists of a nitrogen atom bonded to two hydrogen atoms. It is typically attached to the central carbon atom. - Carboxyl Group (-COOH): This acidic group is made up of one carbon atom double-bonded to an oxygen atom and bonded to an -OH group. The carboxyl group gives the amino acid its acidic property. - Alpha (Central) Carbon: This carbon atom is connected to both the amino group and the carboxyl group. It also bonds to a hydrogen atom and a variable R group (side chain). - R Group (Side Chain): This is the variable group that differs among amino acids, giving each one unique chemical properties. The R group can be anything from a simple hydrogen atom (as in glycine) to a more complex ring structure (as in tryptophan). - 2. Formation of Peptide Bonds - Peptide Bond Definition: A peptide bond is a covalent bond that links amino acids together to form a polypeptide. - Process: During a dehydration synthesis (condensation) reaction, the carboxyl group of one amino acid reacts with the amino group of another. This reaction releases a molecule of water (H₂O) and forms a peptide bond between the two amino acids. - Bond Formation: The bond forms between the carbon of the carboxyl group (C=O) of one amino acid and the nitrogen (N-H) of the amino group of the next amino acid. This C-N bond is known as the peptide bond. - 3. Formation of Polypeptides - Chain Structure: As more amino acids join together through peptide bonds, a long chain of amino acids, called a polypeptide, is formed. - Directionality: Polypeptides have directionality, with an N-terminus (amino end) and a C-terminus (carboxyl end). This orientation is important for protein function and folding. - Protein Folding: After the polypeptide chain is formed, it will fold into a specific three-dimensional structure based on interactions among the R groups of its amino acids. This final structure determines the protein's function. Explain the four levels of protein structure (primary, secondary, tertiary, quaternary) and how folding affects protein function. - 1. Primary Structure SBI4U Thalen - Definition: The primary structure is the unique sequence of amino acids in a polypeptide chain. This sequence is determined by the gene encoding the protein. - Bonding: The amino acids are linked by covalent peptide bonds. - Importance: The specific order of amino acids in the primary structure determines how the protein will fold and, ultimately, its function. A single change in the amino acid sequence (e.g., as in sickle cell anemia) can alter a protein’s function. - 2. Secondary Structure - Definition: The secondary structure refers to the local folding of the polypeptide chain into regular structures such as α-helices and β-pleated sheets. - Bonding: These structures are stabilized by hydrogen bonds between the backbone atoms (the oxygen of the carbonyl group and the hydrogen of the amino group) in the polypeptide chain. - Importance: The formation of α-helices and β-sheets contributes to the protein’s stability and provides a framework for further folding in the tertiary structure. - 3. Tertiary Structure - Definition: The tertiary structure is the overall three-dimensional shape of a single polypeptide chain. It results from interactions among the R groups (side chains) of the amino acids. - Bonding and Interactions: The tertiary structure is stabilized by various types of interactions, including: - Hydrogen bonds between polar side chains. - Ionic bonds between positively and negatively charged side chains. - Hydrophobic interactions where nonpolar side chains cluster away from water. - Disulfide bridges, which are covalent bonds between sulfur atoms in the R groups of cysteine amino acids. - Importance: The tertiary structure gives the protein its functional shape. For enzymes, the active site (where substrate binding occurs) is part of the tertiary structure, making it essential for catalytic activity. - 4. Quaternary Structure - Definition: The quaternary structure is the assembly of multiple polypeptide chains (subunits) into a single functional protein complex. - Bonding and Interactions: The subunits are held together by the same types of interactions seen in tertiary structure: hydrogen bonds, ionic bonds, hydrophobic interactions, and occasionally disulfide bonds. - Importance: The quaternary structure is necessary for proteins made of more than one polypeptide chain, such as hemoglobin. Proper arrangement of subunits is critical for the function of these multi-subunit proteins. - How Folding Affects Protein Function - Proper folding at each structural level is essential for the protein to achieve its functional shape. Misfolded proteins may lose their functionality or even become harmful, as in the case of prion diseases or cystic fibrosis. The unique folding pattern determines the protein’s active sites, binding sites, and overall interaction with other molecules, which are all critical for carrying out its biological role effectively. SBI4U Thalen Analyze how the structure of fibrous and globular proteins influences their role in the body, providing examples like keratin and hemoglobin. - 1. Fibrous Proteins - Structure: Fibrous proteins are long, elongated, and thread-like. They have extensive secondary structures, typically forming α-helices or β-sheets that coil and link to create rigid, stable fibers. This structure gives them strength and insolubility in water. - Role: The structure of fibrous proteins makes them ideal for structural and supportive functions in the body, as they are strong, stable, and resistant to stretching or deforming. - Example: Keratin - Function: Keratin is a fibrous protein found in hair, nails, and the outer layer of skin. Its helical structure and strong disulfide bonds make it very tough and resistant to wear. - Structural Suitability: The tightly coiled α-helices in keratin form strong fibers, which provide mechanical strength and protection to tissues. The disulfide bonds between cysteine residues further stabilize keratin, giving it durability in the face of physical stress. - 2. Globular Proteins - Structure: Globular proteins are more compact and spherical due to the complex folding of polypeptide chains into tertiary and sometimes quaternary structures. This folding often creates hydrophobic interiors and hydrophilic exteriors, making globular proteins generally soluble in water. - Role: The soluble, compact structure of globular proteins allows them to function in dynamic roles, such as catalysis, transport, and regulation, within cells and bodily fluids. - Example: Hemoglobin - Function: Hemoglobin is a globular protein in red blood cells responsible for transporting oxygen from the lungs to tissues and carbon dioxide back to the lungs. - Structural Suitability: Hemoglobin has a quaternary structure with four polypeptide subunits, each containing a heme group that binds oxygen. This flexible, spherical structure enables it to change shape as it binds and releases oxygen, which is essential for efficient oxygen transport and delivery. Explain how proteins can become denatured by environmental factors like heat and pH, and predict the consequences of this on protein function. - How Denaturation Occurs - Heat: - - Heat increases the kinetic energy of molecules, which can disrupt the hydrogen bonds and van der Waals forces that stabilize a protein’s secondary, tertiary, and quaternary structures. - As these bonds break, the protein unfolds, losing its functional three-dimensional structure. - pH Changes: - - Changes in pH alter the concentration of hydrogen ions (H⁺) in the environment, which can affect ionic bonds and interactions between acidic and basic side chains in the protein. - Extreme pH changes can lead to the protonation or deprotonation of amino acid side chains, destabilizing the structure and causing it to unfold. - Consequences of Denaturation on Protein Function SBI4U Thalen - Loss of Shape: Since a protein’s shape is crucial for its function, particularly in binding and catalysis, denaturation usually results in the loss of its biological activity. - Inactivation of Enzymes: Enzymes, which are proteins, depend on specific active sites for binding substrates. Denaturation disrupts these active sites, rendering the enzyme inactive and stopping the reaction it catalyzes. - Structural Protein Weakening: Structural proteins like collagen or keratin lose their supportive function when denatured, leading to weakened physical structures (e.g., weakened skin, brittle hair). - Irreversibility in Some Cases: Some proteins cannot refold to their functional shape once denatured, resulting in permanent loss of function. However, some proteins can regain their shape if normal conditions are restored, a process known as renaturation. - Examples of Functional Impact - Hemoglobin: If denatured, hemoglobin can no longer effectively bind oxygen, which can impair oxygen transport in the body. - Digestive Enzymes: Enzymes in the stomach, such as pepsin, are adapted to acidic conditions. If moved to a neutral pH, they denature, and digestion is disrupted. Solve problems involving protein structure, including identifying amino acids and their interactions within a polypeptide. Lesson 5: Macromolecules - Nucleic Acids Identify the structure of nucleotides and explain how they combine to form the backbone of DNA and RNA. - 1. Structure of a Nucleotide - Phosphate Group: A nucleotide contains one or more phosphate groups, which are negatively charged and contribute to the overall charge of the nucleic acid. - Pentose Sugar: Each nucleotide has a five-carbon sugar. In DNA, the sugar is deoxyribose, which lacks an oxygen atom on the 2' carbon; in RNA, the sugar is ribose, which has an OH group on the 2' carbon. - Nitrogenous Base: A nucleotide includes a nitrogen-containing base. There are four bases in DNA (adenine, thymine, cytosine, guanine) and four in RNA (adenine, uracil, cytosine, guanine). - 2. Formation of the DNA and RNA Backbone - Phosphodiester Bonds: Nucleotides connect to form nucleic acids via phosphodiester bonds between the phosphate group of one nucleotide and the hydroxyl group on the 3' carbon of the sugar in another nucleotide. - Backbone Structure: This bonding forms a sugar-phosphate backbone, where the alternating phosphate and sugar molecules create the structural framework of DNA and RNA. - In DNA, the backbone consists of deoxyribose-phosphate linkages. - In RNA, it consists of ribose-phosphate linkages. SBI4U Thalen - Directionality: The backbone has a 5' to 3' direction because each nucleotide has a free 5' phosphate group at one end and a free 3' hydroxyl group at the other. This directionality is critical for replication and transcription processes. - 3. Double Helix in DNA - In DNA, two complementary strands run in opposite directions (antiparallel) and pair through hydrogen bonds between the nitrogenous bases, forming a double helix structure. This helical structure is stabilized by base pairing (A-T and G-C) and the sugar-phosphate backbone on the exterior. - Compare and contrast the structures and functions of DNA and RNA, highlighting key differences such as the sugar molecule and nitrogenous bases used. - Structural Differences - Sugar Molecule - DNA: Contains deoxyribose, a five-carbon sugar that lacks an oxygen atom on the 2' carbon. This makes DNA more chemically stable and less reactive. - RNA: Contains ribose, which has an -OH group on the 2' carbon, making RNA more reactive and less stable than DNA. - Nitrogenous Bases - DNA: Uses four nitrogenous bases—adenine (A), thymine (T), cytosine (C), and guanine (G). - RNA: Also has four bases but substitutes uracil (U) for thymine, pairing uracil with adenine instead. - Structure - DNA: Typically a double-stranded molecule that forms a double helix. The two strands are antiparallel and held together by hydrogen bonds between complementary base pairs (A-T and G-C). - RNA: Usually single-stranded and does not form a double helix. However, RNA can fold into complex secondary structures, forming loops and hairpins that allow it to take on various shapes. - Location in the Cell - DNA: Primarily located in the cell nucleus (in eukaryotes), with a small amount in mitochondria and chloroplasts. - RNA: Primarily found in the cytoplasm and the nucleus. Types of RNA, such as mRNA, tRNA, and rRNA, move between the nucleus and cytoplasm in eukaryotic cells. - Functional Differences - Role in Genetic Information - DNA: DNA’s primary function is to store genetic information. It serves as the hereditary material that encodes instructions for synthesizing proteins, maintaining cellular functions, and passing genetic traits to offspring. - RNA: RNA’s primary role is in protein synthesis and gene expression. It acts as a messenger and adapter molecule, helping translate DNA’s instructions into proteins. - Types and Specific Functions - DNA: There is only one main type of DNA, as its role is solely to store and protect genetic information. SBI4U Thalen - RNA: Exists in several forms, each with specialized functions: - mRNA (Messenger RNA): Carries genetic instructions from DNA to ribosomes for protein synthesis. - tRNA (Transfer RNA): Delivers amino acids to the ribosome to assemble into a protein based on the mRNA sequence. - rRNA (Ribosomal RNA): Forms part of the ribosome’s structure, catalyzing peptide bond formation during protein synthesis. Summary of Differences - FeatureDNA RNA - Sugar Deoxyribose Ribose - Nitrogenous Bases A, T, C, G A, U, C, G - Structure Double-stranded, double helix Single-stranded, varied shapes - Location Nucleus (mainly) Nucleus and cytoplasm - Function Genetic information storage Protein synthesis and gene expression - Stability More chemically stableLess stable, more reactive Demonstrate how complementary base pairing occurs in DNA (A-T, C-G) and RNA (A-U, C-G) - Demonstrate how complementary base pairing occurs in DNA (A-T, C-G) and RNA (A-U, C-G) - Adenine (A) pairs with Thymine (T): Two hydrogen bonds form between adenine and thymine, connecting them and making the pairing relatively stable. - Cytosine (C) pairs with Guanine (G): Three hydrogen bonds form between cytosine and guanine, providing a stronger connection than the A-T pairing. Solve problems involving DNA and RNA sequences, such as giving the complementary strand to a given strand of DNA. Lesson 6: Enzymes Explain how enzymes function as biological catalysts by lowering the activation energy of chemical reactions. - Enzymes are proteins that act as biological catalysts, accelerating chemical reactions in the body by lowering the activation energy—the initial energy input required for a reaction to proceed. By reducing this energy barrier, enzymes enable reactions to occur more quickly and efficiently, even at the relatively low temperatures within living organisms. - - Mechanism of Enzyme Action - Active Site and Substrate Binding: - - Each enzyme has a unique active site, a specific region where the substrate—the molecule(s) on which the enzyme acts—binds. - The shape and chemical properties of the active site are complementary to the substrate, enabling a precise "lock-and-key" or "induced fit" interaction. This binding stabilizes the substrate in a way that promotes the reaction. - Lowering Activation Energy: - SBI4U Thalen - When the substrate binds to the enzyme, the enzyme-substrate complex is formed, creating a favorable environment for the reaction. - Enzymes lower activation energy by: - Orienting substrates: Enzymes position substrates in an optimal arrangement, reducing the energy needed to initiate the reaction. - Straining substrate bonds: Enzymes can put strain on the bonds within the substrate, making them easier to break. - Providing an alternative reaction pathway: Some enzymes provide a pathway that bypasses high-energy steps, requiring less energy overall. - Stabilizing transition states: Enzymes can stabilize the high-energy transition state of a reaction, making it easier for the reaction to proceed. - Catalysis and Product Release: - - Once the reaction occurs, the enzyme releases the products, leaving the enzyme ready to bind to new substrate molecules. - Because enzymes are not consumed in the reaction, they can repeatedly catalyze the same reaction. - Impact of Lowering Activation Energy - By lowering the activation energy, enzymes increase the reaction rate significantly, allowing biochemical reactions to proceed rapidly enough to sustain life. For example, without enzymes, essential processes like digestion, DNA replication, and cellular respiration would occur too slowly to meet the demands of a living organism. Use the lock-and-key and induced-fit models to explain how enzymes bind to specific substrates and facilitate reactions. - 1. Lock-and-Key Model - Concept: In the lock-and-key model, the enzyme’s active site is thought to have a specific shape that fits only a particular substrate, much like a lock and key. The substrate (the "key") has a shape that exactly matches the enzyme's active site (the "lock"). - Binding Process: When the substrate encounters the enzyme, it fits precisely into the active site without any alteration in shape, forming the enzyme-substrate complex. - Reaction Facilitation: Once bound, the enzyme catalyzes the reaction by lowering the activation energy, allowing the substrate to be converted into the product(s). - Example: This model works well for enzymes with very specific substrate requirements, such as sucrase, which only binds sucrose. - Limitations: The lock-and-key model is limited in explaining how some enzymes can adjust to fit their substrates, and it doesn't account for enzymes that can work with multiple substrates or have flexible active sites. - - 2. Induced-Fit Model - Concept: In the induced-fit model, the enzyme’s active site is flexible, not rigid. The enzyme undergoes a subtle conformational change to accommodate the substrate more precisely upon binding, creating a snug fit. SBI4U Thalen - Binding Process: When the substrate approaches, the enzyme’s active site molds itself around the substrate. This dynamic "induced fit" strengthens the interaction and positions the substrate in an optimal orientation for the reaction to occur. - Reaction Facilitation: By adjusting to fit the substrate, the enzyme can better stabilize the transition state, reducing activation energy and promoting the chemical reaction. The substrate is then converted to the product(s), which are released from the enzyme. - Example: The induced-fit model is observed in enzymes like hexokinase, which changes shape to bind glucose tightly and catalyze the first step of glycolysis. - Advantages: The induced-fit model explains how enzymes can have flexibility in substrate binding and work with slightly different substrates, as well as how they increase catalytic efficiency through active-site adjustments. Analyze graphs to determine the effects of temperature, pH, enzyme concentration, and substrate concentration on enzyme activity. Differentiate between competitive and non-competitive enzyme inhibitors and explain how each affects enzyme activity. - 1. Competitive Inhibitors - Mechanism of Action: Competitive inhibitors resemble the substrate and bind directly to the enzyme’s active site, competing with the substrate for access. When the inhibitor is bound, the actual substrate cannot enter the active site, so the enzyme cannot catalyze the reaction. - Effect on Enzyme Activity: - Reversible Effect: Competitive inhibition can often be overcome by increasing the concentration of the substrate. With more substrate molecules available, they have a higher chance of outcompeting the inhibitor and binding to the active site. - Example: Malonate acts as a competitive inhibitor for the enzyme succinate dehydrogenase by mimicking the substrate succinate, blocking it from binding and undergoing the reaction. - 2. Non-Competitive Inhibitors - Mechanism of Action: Non-competitive inhibitors bind to a different site on the enzyme, known as an allosteric site, rather than the active site. This binding changes the enzyme’s shape, including the active site, making it less effective or entirely inactive in catalyzing the reaction. - Effect on Enzyme Activity: - Irreversible or Reduced Effect: Because non-competitive inhibitors do not compete with the substrate, increasing the substrate concentration does not overcome the inhibition. - Example: Cyanide acts as a non-competitive inhibitor for cytochrome c oxidase in the electron transport chain, binding to a separate site and halting cellular respiration. Explain how feedback inhibition regulates enzyme activity in metabolic pathways, using examples like the regulation of cellular respiration. - Feedback inhibition is a regulatory mechanism where the end product of a metabolic pathway inhibits an enzyme involved early in the pathway. This prevents overproduction of the product, conserving resources and maintaining cellular balance. In feedback inhibition, the final product binds to an enzyme in the pathway, typically at an allosteric site, altering the enzyme’s shape and reducing its activity. - SBI4U Thalen - How Feedback Inhibition Works in Metabolic Pathways - Pathway Structure: Metabolic pathways often involve a series of enzymes that convert substrates through several steps to form an end product. - End Product as an Inhibitor: When enough of the final product accumulates, it can inhibit an enzyme at the beginning of the pathway (often the rate-limiting enzyme), slowing down or halting the pathway. - Regulation of Production: As the product concentration decreases, inhibition is reduced, allowing the pathway to resume and produce more product as needed. Solve problems involving enzyme kinetics and inhibition, including predicting how changes in environmental conditions affect enzyme function. Lesson 7: Parts of the Cell Identify the major components of a eukaryotic cell, including the nucleus, mitochondria, endoplasmic reticulum, Golgi apparatus, lysosomes, and chloroplasts (in plant cells). Describe the functions of each organelle and how they contribute to the overall functioning of the cell. Differentiate between prokaryotic and eukaryotic cells based on their structures and organelles. Explain how the structure of the cell membrane (phospholipid bilayer) supports its function in controlling what enters and exits the cell. Use diagrams to label and describe the structure of key cell components, emphasizing their role in processes like energy production (mitochondria) and protein synthesis (ribosomes). Explain the relationship between structure and function for specialized cells in different tissues or organs (e.g., muscle cells, nerve cells). Lesson 8: Diffusion, Osmosis, and Cell Transport Define diffusion and explain how it facilitates the movement of molecules from high to low concentration across the cell membrane. Explain the concept of osmosis as the diffusion of water across a semi-permeable membrane, and describe the effects of hypotonic, hypertonic, and isotonic conditions on cells. Analyze how passive transport (diffusion and facilitated diffusion) works without energy and how specific molecules (e.g., glucose) cross the membrane via channel or carrier proteins. SBI4U Thalen Explain active transport, including the role of ATP in moving molecules against a concentration gradient (e.g., sodium-potassium pump). Describe how bulk transport processes (endocytosis and exocytosis) move large molecules into or out of cells, and differentiate between phagocytosis, pinocytosis, and receptor-mediated endocytosis. Solve problems related to predicting the movement of molecules across membranes based on concentration gradients and the type of transport involved (e.g., passive vs. active). Interpret data from experiments or diagrams showing the effects of different solutions on cells (e.g., plasmolysis, turgor pressure). Multiple Choice Practice Questions 1. What type of macromolecule is wax? a. Carbohydrate b. Lipid c. Protein d. Nucleic acid 2. Which of the following is FALSE about enzymes? a. The have an active site that binds to a specific substrate b. They work best around body temperature c. They catalyze reactions in the body d. They always require an activator to bind to them in order to work e. All of the above f. None of the above 3. Which of the following is TRUE about cell membranes? a. They become more fluid at higher temperatures b. They contain nucleic acids c. They are made primarily of phospholipids d. They become more fluid if there are no double bonds in their fatty acid chains e. All of the above f. Only A and C are correct 4. Which cellular process moves particles into the cell along their concentration gradient with the help of a hollow-tube-like protein? a. Diffusion b. Carrier proteins c. Channel proteins d. Primary active transport SBI4U Thalen e. Secondary active transport 5. In what type or reaction is a peptide bond formed a. Condensation reaction b. Hydrolysis reaction c. Redox reaction d. Double displacement reaction e. None of the above 6. What type of linkage is present between two amino acids? a. Phosphodiester linkage b. Glycosidic linkage c. Peptide linkage d. Ester linkage 7. Which of the following organelles is involved in the process of endocytosis? a. Nucleus b. Vesicle c. Vacuole d. Ribosome 8. Which of the following best describes diffusion? a. Movement of molecules against a concentration gradient. b. Movement of water molecules through a semi-permeable membrane. c. Movement of molecules from an area of low concentration to an area of high concentration. d. Movement of molecules from an area of high concentration to an area of low concentration. 9. Osmosis specifically refers to the movement of which type of molecule? a. Oxygen b. Water c. Glucose d. Sodium 10. Which statement about osmosis is correct? a. It involves the movement of solute molecules across a membrane. b. It requires energy in the form of ATP to occur. c. It occurs when water moves from an area of low solute concentration to high solute concentration. d. It involves the movement of water from a hypertonic solution to a hypotonic solution. 11. If a red blood cell is placed in a hypotonic solution, what will most likely happen to the cell? a. It will shrink. b. It will swell and possibly burst. c. It will remain the same size. d. It will lose water and become dehydrated. 12. Which of the following factors does NOT affect the rate of diffusion? a. Temperature b. Size of the molecules SBI4U Thalen c. Concentration gradient d. Volume of the solution Short Answer Practice Questions 9. Compare the different structurally defining features of the 4 types of macromolecules. Identify the monomers, type of polymer bond (i.e. glycosidic, ester, etc.) and major functional groups. 10. Compare starch and cellulose 11. What is meant by the dual nature of phospholipids? Why do they make a great cell membrane, and what increases membrane fluidity? 12. Summarize the four levels of protein structure. 13. Compare competitive vs non-competitive inhibitors. 14. What is feedback inhibition? 15. Summarize the four factors that affect enzyme activity. 16. What is the fluid mosaic model? 17. What are the four functions of proteins in the cell membrane? 18. Compare Passive and Active Transport. 19. Draw a diagram to show how an animal and a plant cell would react in the following conditions: hypertonic, hypotonic, and isotonic. 20. Compare the 3 types of endocytosis. 21. How does the Sodium-Potassium pump create an electrochemical gradient? 22. How does the Hydrogen-Sugar Pump bring sucrose into the cell against its concentration gradient? Practice Diagram Questions 1. Identify and label two functional groups on the following molecule: SBI4U Thalen 2. Circle the peptide bond in this dipeptide: 3. Circle and label the type of linkage in this disaccharide: 4. Draw a representation of what would happen if you put drops of oil in water: