02.13 Chapter Summary
Biocast
1. Benedict’s Test for Reducing Sugars
Purpose:
- Detect the presence of reducing sugars (e.g., glucose, fructose).
Principle:
- Reducing sugars can reduce Cu²⁺ ions in Benedict’s reagent to Cu⁺, forming a brick-red precipitate of copper(I) oxide.
Procedure:
- Reagents Needed:
- Benedict’s reagent (a blue solution containing copper(II) sulfate, sodium carbonate, and sodium citrate).
- Steps:
- Add a few drops of the sample solution to a test tube.
- Add an equal volume of Benedict’s reagent.
- Heat the mixture in a boiling water bath for 5 minutes.
- Observation:
- Color change from blue to green, yellow, orange, or brick-red precipitate indicates the presence of reducing sugars.
Interpretation:
- Blue: No reducing sugars.
- Green to Brick-Red: Increasing concentrations of reducing sugars.
2. Iodine Test for Starch
Purpose:
- Identify the presence of starch.
Principle:
- Iodine interacts with the helical structure of starch (amylose) to form a blue-black complex.
Procedure:
- Reagents Needed:
- Iodine solution (iodine-potassium iodide).
- Steps:
- Add a few drops of iodine solution to the sample.
- Observation:
- A color change to blue-black indicates the presence of starch.
Interpretation:
- Yellow/Brown: No starch.
- Blue-Black: Presence of starch.
3. Emulsion Test for Lipids
Purpose:
- Detect the presence of lipids (fats and oils).
Principle:
- Lipids are insoluble in water but form a cloudy emulsion when mixed with ethanol and then water is added.
Procedure:
- Reagents Needed:
- Ethanol.
- Steps:
- Add a small amount of the sample to a test tube.
- Add ethanol and shake to dissolve any lipids.
- Carefully add water without shaking; formation of a cloudy emulsion indicates lipids.
- Observation:
- Cloudiness signifies lipid presence.
Interpretation:
- Clear Solution: No lipids.
- Cloudy Emulsion: Presence of lipids.
4. Biuret Test for Proteins
Purpose:
- Detect the presence of proteins.
Principle:
- Proteins contain peptide bonds that react with copper(II) ions in an alkaline solution to form a violet-colored complex.
Procedure:
- Reagents Needed:
- Biuret reagent (a solution of copper(II) sulfate and sodium hydroxide).
- Steps:
- Add a few drops of Biuret reagent to the sample.
- Mix gently.
- Observation:
- Formation of a violet or purple color indicates the presence of proteins.
Interpretation:
- No Color Change: No proteins.
- Violet/Purple: Presence of proteins.
5. Semi-Quantitative Benedict’s Test for Reducing Sugars
Purpose:
- Estimate the concentration of reducing sugars in a solution.
Principle:
- The rate of color change or the intensity of the precipitate correlates with sugar concentration.
Procedure:
- Standardization:
- Prepare standard solutions with known concentrations of a reducing sugar (e.g., glucose).
- Testing Unknowns:
- Perform Benedict’s test on both standard and unknown samples under identical conditions.
- Observation:
- Note the time taken for the first color change or compare the color intensity to a standard color chart.
- Estimation:
- Determine the concentration of reducing sugars in the unknown based on the comparison with standards.
Tips for Accuracy:
- Ensure all solutions are fresh.
- Compare colors against a standardized color chart for precise estimation.
- Use consistent heating times.
6. Testing for Non-Reducing Sugars
Purpose:
- Identify non-reducing sugars (e.g., sucrose).
Principle:
- Non-reducing sugars can be hydrolyzed into reducing sugars using acid before performing Benedict’s test.
Procedure:
- Reagents Needed:
- Dilute acid (e.g., dilute hydrochloric acid).
- Benedict’s reagent.
- Steps:
- Add a few drops of the sample to a test tube.
- Add dilute acid and gently heat to hydrolyze non-reducing sugars into reducing sugars.
- Neutralize the acid if necessary.
- Perform Benedict’s test on the hydrolyzed sample.
- Observation:
- A positive Benedict’s test after hydrolysis indicates the presence of non-reducing sugars.
Interpretation:
- No Color Change Initially, Positive After Hydrolysis: Presence of non-reducing sugars.
- Positive Without Hydrolysis: Presence of reducing sugars (may contain both types).
7. Ring Forms of α-Glucose and β-Glucose
Structure:
- Glucose: A six-carbon monosaccharide with the molecular formula C₆H₁₂O₆.
- Ring Formation: Glucose forms a cyclic hemiacetal structure, creating a six-membered ring (pyranose form).
α-Glucose vs. β-Glucose:
- Anomeric Carbon: Carbon 1 (C1) in glucose.
- α-Glucose: –OH group on C1 is trans (below the plane) relative to the CH₂OH group on C5.
- β-Glucose: –OH group on C1 is cis (above the plane) relative to the CH₂OH group on C5.
Diagram:
- Draw the Haworth projections showing the orientation of the –OH group on the anomeric carbon for both α and β forms.
8. Key Terms Defined
- Monomer: The basic building block of a polymer (e.g., glucose for carbohydrates).
- Polymer: A large molecule composed of repeating monomer units (e.g., starch, glycogen).
- Macromolecule: A very large molecule, typically a polymer (e.g., proteins, nucleic acids).
- Monosaccharide: The simplest form of carbohydrate; single sugar unit (e.g., glucose, fructose).
- Disaccharide: A carbohydrate formed by two monosaccharides joined by a glycosidic bond (e.g., sucrose, maltose).
- Polysaccharide: A carbohydrate composed of many monosaccharide units (e.g., starch, glycogen, cellulose).
9. Role of Covalent Bonds in Polymer Formation
- Covalent Bonds: Strong chemical bonds where electrons are shared between atoms.
- Polymerization: Smaller molecules (monomers) are linked together by covalent bonds to form polymers.
- Example: In carbohydrates, glycosidic bonds (a type of covalent bond) link monosaccharides to form disaccharides and polysaccharides.
10. Reducing and Non-Reducing Sugars
- Reducing Sugars: Can donate electrons or reduce other molecules; have a free anomeric carbon.
- Examples: Glucose, Fructose, Maltose.
- Non-Reducing Sugars: Cannot donate electrons; have their anomeric carbons involved in glycosidic bonds.
- Example: Sucrose.
Test: Benedict’s solution can distinguish between reducing and non-reducing sugars.
11. Formation of Glycosidic Bonds by Condensation
- Condensation Reaction: Removal of a water molecule to form a bond between two molecules.
- Glycosidic Bond: A covalent bond joining a carbohydrate molecule to another group, which may also be a carbohydrate.
Formation Process:
- Disaccharides: Two monosaccharides join via a glycosidic bond (e.g., glucose + fructose → sucrose).
- Polysaccharides: Multiple monosaccharides linked together (e.g., glucose units in starch).
- Diagram: Show the condensation reaction forming a glycosidic bond between two glucose molecules.
12. Hydrolysis of Glycosidic Bonds and Non-Reducing Sugar Test
Hydrolysis of Glycosidic Bonds :
- Hydrolysis: The chemical breakdown of a bond through the addition of water.
- Polysaccharides/Disaccharides: Glycosidic bonds are broken down into monosaccharides.
Non-Reducing Sugar Test:
- Sucrose: Does not react with Fehling’s solution unless hydrolyzed into glucose and fructose.
- Procedure: Hydrolyze the disaccharide, then perform the reducing sugar test.
13. Molecular Structure of Polysaccharides: Starch and Glycogen
Starch:
- Amylose: Linear chains of α-glucose units connected by α-1,4-glycosidic bonds. Forms helical structures.
- Amylopectin: Branched chains with α-1,6-glycosidic bonds at branch points.
Glycogen:
- Highly branched structure with more α-1,6-glycosidic bonds than amylopectin, allowing rapid release of glucose.
Function:
- Starch: Energy storage in plants.
- Glycogen: Energy storage in animals.
14. Molecular Structure of Cellulose and Its Function
Cellulose:
- Linear chains of β-glucose units linked by β-1,4-glycosidic bonds.
- Hydrogen Bonding: Forms strong hydrogen bonds between adjacent chains, creating rigid, insoluble fibers.
Function:
- Plant Cell Walls: Provides structural support and rigidity to plants.
- Diagram: Show the linear structure and hydrogen bonding in cellulose.
15. Triglycerides: Structure and Properties
Triglycerides:
- Non-Polar, Hydrophobic Molecules: Do not mix with water.
Molecular Structure:
- Glycerol: A three-carbon alcohol.
- Fatty Acids: Long hydrocarbon chains (saturated: no double bonds; unsaturated: one or more double bonds).
- Ester Bonds: Formed between the hydroxyl groups of glycerol and the carboxyl groups of fatty acids.
- Diagram: Illustrate glycerol linked to three fatty acids via ester bonds.
16. Structure-Function Relationship of Triglycerides
- Energy Storage: High-energy bonds store significant energy, making triglycerides efficient for long-term energy storage.
- Insulation and Protection: Provide thermal insulation and protect vital organs due to their hydrophobic nature and storage in adipose tissue.
17. Molecular Structure of Phospholipids
Phospholipids:
Structure:
- Hydrophilic (Polar) Heads: Contain phosphate groups, which interact with water.
- Hydrophobic (Non-Polar) Tails: Composed of fatty acid chains.
- Amphipathic Nature: Possess both hydrophilic and hydrophobic regions.
Function:
- Cell Membranes: Form bilayers with hydrophobic tails facing inward and hydrophilic heads facing outward, creating a semi-permeable membrane.
- Diagram: Depict a phospholipid molecule with distinct head and tail regions, and illustrate the bilayer structure.
18. Amino Acids and Peptide Bonds
Amino Acid Structure
- General Structure:
- Central Carbon (α-Carbon): Connected to four groups:
- Amino Group (—NH₂)
- Carboxyl Group (—COOH)
- Hydrogen Atom (—H)
- R Group (Side Chain): Varies among different amino acids, determining their properties.
- Central Carbon (α-Carbon): Connected to four groups:
- Diagram: Draw a central carbon with the four groups attached, labeling each part.
Peptide Bond Formation
Formation:
- Condensation Reaction: Carboxyl group of one amino acid reacts with amino group of another, releasing a molecule of water (H₂O).
- Peptide Bond: The bond formed is a covalent bond between the carbon of the carboxyl group and the nitrogen of the amino group (—CO—NH—).
Peptide Bond Breakage:
- Hydrolysis Reaction: Addition of water breaks the peptide bond, separating the amino acids.
19. Protein Structures
Primary Structure
- Definition: The linear sequence of amino acids in a polypeptide chain.
- Importance: Determines all subsequent levels of structure.
Secondary Structure
- Types:
- Alpha (α) Helix: Spiral structure stabilized by hydrogen bonds between backbone atoms.
- Beta (β) Pleated Sheet: Sheet-like structure formed by hydrogen bonds between adjacent polypeptide chains.
Tertiary Structure
- Definition: The three-dimensional folding of a single polypeptide chain.
- Stabilized by: Hydrophobic interactions, hydrogen bonds, ionic bonds, and disulfide bridges.
Quaternary Structure
- Definition: The arrangement of multiple polypeptide chains (subunits) into a functional protein.
- Example: Hemoglobin consists of four subunits.
20. Interactions Maintaining Protein Shape
Hydrophobic Interactions
- Description: Nonpolar side chains avoid water, clustering together inside the protein.
- Role: Stabilize the protein’s core.
Hydrogen Bonding
- Description: Attraction between hydrogen and electronegative atoms (e.g., O, N).
- Role: Stabilizes secondary and tertiary structures.
Ionic Bonding
- Description: Electrostatic attractions between positively and negatively charged side chains.
- Role: Contributes to tertiary and quaternary structures.
Covalent Bonding (Including Disulfide Bonds)
- Description: Strong bonds formed between atoms sharing electrons.
- Disulfide Bonds: Covalent bonds between sulfur atoms of cysteine residues.
- Role: Provide significant stability, especially in tertiary and quaternary structures.
21. Globular vs. Fibrous Proteins
Globular Proteins
- Characteristics:
- Solubility: Generally soluble in water.
- Function: Physiological roles (e.g., enzymes, hormones, transport proteins).
- Structure: Compact, spherical shapes.
Fibrous Proteins
Characteristics:
- Solubility: Generally insoluble in water.
- Function: Structural roles (e.g., collagen, keratin).
- Structure: Long, fibrous, and elongated shapes.
22. Haemoglobin: A Globular Protein
Structure of Haemoglobin
- Subunits:
- Two Alpha (α) Chains (α–globin)
- Two Beta (β) Chains (β–globin)
- Haem Groups:
- Each Subunit Contains One Haem Group
- Haem Group: Contains an iron (Fe) atom that binds oxygen.
Quaternary Structure
- Assembly: Four polypeptide chains (2 α and 2 β) form the functional protein.
- Functionality: Each subunit can bind one oxygen molecule, allowing haemoglobin to carry four oxygen molecules.
23. Structure-Function Relationship in Haemoglobin
Function of Haemoglobin
- Oxygen Transport: Binds oxygen in the lungs and releases it in tissues.
Importance of Iron
- Role of Iron in Haem Group:
- Binding Site: Iron atom binds reversibly to oxygen.
- Essential for Oxygen Transport: Without iron, haemoglobin cannot carry oxygen.
Structural Features Facilitating Function
- Quaternary Structure: Allows cooperative binding of oxygen (binding of one O₂ increases affinity for the next).
- Globular Shape: Enhances solubility and mobility in blood.
24. Collagen: A Fibrous Protein
Structure of Collagen
- Triple Helix:
- Three Polypeptide Chains: Wound together in a triple helix formation.
- Amino Acid Sequence: Repeating Glycine-Proline-X or Glycine-X-Hydroxyproline, where X is any amino acid.
Collagen Fibres
- Arrangement:
- Fibrils: Individual collagen molecules align in a staggered pattern.
- Fibres: Multiple fibrils bundle together to form strong, insoluble fibres.
Stabilizing Features
- Hydrogen Bonds: Stabilize the triple helix.
- Cross-linking: Covalent bonds between collagen molecules enhance strength.
25. Structure-Function Relationship in Collagen
Function of Collagen
- Structural Support: Provides strength and elasticity to connective tissues (e.g., skin, bones, tendons).
Structural Features Facilitating Function
- Triple Helix: Offers high tensile strength, resisting stretching.
- Fibrous Structure: Insolubility and robust fibre formation support structural integrity.
- Arrangement in Fibres: Efficient distribution of mechanical stress across tissues.
26. Hydrogen Bonding Between Water Molecules
a. Structure of Water (H₂O):
- Molecular Composition: Each water molecule consists of two hydrogen atoms covalently bonded to one oxygen atom.
- Polarity: Oxygen is more electronegative than hydrogen, creating a polar molecule with a partial negative charge (δ⁻) on the oxygen atom and partial positive charges (δ⁺) on the hydrogen atoms.
b. Formation of Hydrogen Bonds:
- Definition: A hydrogen bond is a weak electrostatic attraction between the partially positive hydrogen atom of one water molecule and the partially negative oxygen atom of another.
- Mechanism:
- Orientation: Due to the polarity, water molecules arrange themselves so that hydrogen atoms are near oxygen atoms of neighboring molecules.
- Network Formation: Each water molecule can form up to four hydrogen bonds (two through its hydrogen atoms and two through lone pairs on the oxygen atom), creating a cohesive network.
c. Significance of Hydrogen Bonds:
- Physical Properties: Contribute to high boiling and melting points relative to other similar-sized molecules.
- Cohesion and Adhesion: Enable water to exhibit strong cohesion (molecules sticking together) and adhesion (molecules sticking to other surfaces).
- Surface Tension: High surface tension due to hydrogen bonding allows insects like water striders to walk on water.
Diagram:
27. Properties of Water and Its Roles in Living Organisms
a. Solvent Action
Definition:
- Water is often called the “universal solvent” because it can dissolve a wide variety of substances.
Mechanism:
- Polarity: The polar nature of water molecules allows them to surround and interact with charged ions and polar molecules, effectively separating and dispersing them in solution.
- Hydration Shells: Ions in solution are surrounded by water molecules, stabilizing them and preventing them from recombining.
Biological Significance:
- Biochemical Reactions: Facilitates enzyme-substrate interactions and metabolic pathways by dissolving reactants.
- Transport Medium: Enables the movement of nutrients, gases (like oxygen and carbon dioxide), and waste products within organisms (e.g., blood plasma).
- Cellular Environment: Maintains the proper environment for cellular processes by dissolving necessary ions and molecules.
Examples:
- Dissolution of glucose in blood plasma for energy transport.
- Transport of oxygen in blood bound to hemoglobin.
b. High Specific Heat Capacity
Definition:
- Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.
Water’s Specific Heat:
- Approximately 4.18 J/g°C, which is significantly higher than most other common substances.
Biological Significance:
- Temperature Regulation: Helps organisms maintain stable internal temperatures despite external temperature fluctuations.
- Climate Moderation: Large bodies of water (oceans, lakes) absorb and store heat, mitigating extreme temperature changes in the environment.
- Homeostasis: In humans, high water content in tissues and blood helps buffer against temperature changes, supporting metabolic functions.
Examples:
- Human bodies use water to regulate temperature through sweating.
- Coastal regions experience milder climates due to the high specific heat of ocean water.
c. Latent Heat of Vaporisation
Definition:
- The latent heat of vaporisation is the amount of heat required to convert one gram of liquid into vapor without changing its temperature.
Water’s Latent Heat:
- Approximately 2260 J/g at 100°C.
Biological Significance:
- Evaporative Cooling: Critical for temperature regulation in organisms.
- Humans: Sweating releases heat as water evaporates from the skin, cooling the body.
- Plants: Transpiration releases water vapor from leaves, cooling the plant and enabling nutrient uptake.
- Energy Transfer: The high latent heat of vaporisation allows water to carry large amounts of energy during phase changes, aiding in thermal regulation.
Examples:
- Evaporation of sweat from human skin during exercise.
- Transpiration in plants facilitating nutrient transport from roots to leaves.
Practise Questions 1
1. Describe how large biological molecules are made from smaller molecules.
Answer:
- Formation of Macromolecules:
- Large biological molecules are made by joining smaller molecules (monomers) through condensation reactions (removes water).
- Polymers (e.g., proteins, polysaccharides) are formed by repeating identical or similar monomers.
- Types of Biological Molecules:
- Polysaccharides: Built from monosaccharides (simple sugars).
- Proteins (polypeptides): Made of amino acids.
- Lipids: Formed by linking fatty acids with glycerol.
- Breaking Down Macromolecules:
- Hydrolysis reactions add water to break large molecules into smaller ones, reversing condensation.
2. Describe the structure of carbohydrates, lipids, and proteins and how their structure relates to their functions.
Answer:
- Carbohydrates:
- Monosaccharides (e.g., glucose):
- Simple sugars; quick energy sources in cells.
- Disaccharides (e.g., sucrose):
- Two monosaccharides joined by glycosidic bonds.
- Polysaccharides:
- Starch: Energy storage in plants; made of α-glucose (amylose: unbranched; amylopectin: branched).
- Glycogen: Energy storage in animals; similar to amylopectin but more branched.
- Cellulose: Structural component in plant cell walls; made of β-glucose forming strong fibers due to hydrogen bonds.
- Lipids:
- Triglycerides:
- Structure: Glycerol + 3 fatty acids joined by ester bonds.
- Hydrophobic and energy-rich; function as energy storage, insulation, and buoyancy (e.g., in marine mammals).
- Phospholipids:
- Structure: Hydrophilic phosphate head + 2 hydrophobic fatty acid tails.
- Essential for cell membrane formation as they form a bilayer with heads outward and tails inward.
- Proteins:
- Primary Structure: Sequence of amino acids in a polypeptide.
- Secondary Structure: Hydrogen bonding creates α-helix or β-pleated sheets.
- Tertiary Structure: 3D shape formed by folding and bonding (e.g., hydrogen bonds, disulfide bonds).
- Quaternary Structure: Arrangement of multiple polypeptide chains.
- Types:
- Globular proteins (e.g., haemoglobin): Soluble, functional roles.
- Fibrous proteins (e.g., collagen): Insoluble, structural roles.
3. Describe and carry out biochemical tests to identify carbohydrates, lipids, and proteins.
Answer:
- Carbohydrates:
- Benedict’s Test for reducing sugars:
- Add Benedict’s reagent and heat; a color change (green to red) indicates presence.
- Non-reducing sugars test:
- Hydrolyze with acid, neutralize, then use Benedict’s test.
- Iodine Test for starch:
- Add iodine solution; a blue-black color indicates starch presence.
- Lipids:
- Emulsion Test:
- Mix sample with ethanol, then add water. A milky-white emulsion indicates lipids.
- Proteins:
- Biuret Test:
- Add Biuret reagent; a purple color change indicates proteins.
4. Explain some key properties of water that make life possible
Answer:
Solvent Properties:
- Dissolves ions and polar molecules (e.g., salts, sugars), facilitating biochemical reactions.
- Non-polar molecules (e.g., lipids) do not dissolve, aiding membrane and protein stability.
High Specific Heat Capacity:
- Absorbs and stores heat, resisting temperature changes.
- Stabilizes internal temperatures in cells and organisms and moderates climate for aquatic environments.
High Latent Heat of Vaporisation:
- Evaporation causes cooling, important in temperature regulation (e.g., sweating in animals, transpiration in plants).
Hydrogen Bonding:
- Enables liquid state at Earth’s temperatures, supports cohesive behavior, and stabilizes biological structures.
Practise Questions 2
Question 1
Describe the structure and function of carbohydrates, and explain the test for reducing sugars. (6 marks)
Mark Scheme:
- Carbohydrates are organic molecules with the general formula Cx(H₂O)y, composed of monosaccharides, disaccharides, and polysaccharides. (1 mark)
- Monosaccharides are simple sugars (e.g., glucose) that can exist in straight-chain or ring forms and are isomers (e.g., α-glucose, β-glucose). (1 mark)
- Disaccharides form when two monosaccharides are joined by a glycosidic bond via condensation reactions. (1 mark)
- Polysaccharides (e.g., starch, glycogen) consist of long chains of monosaccharides and function as energy stores or structural components. (1 mark)
- The Benedict’s test detects reducing sugars:
- Heat the sugar solution with Benedict’s reagent.
- A color change from blue to green, yellow, or brick-red indicates the presence of reducing sugars. (1 mark)
- The test is semi-quantitative, with the intensity of the color change roughly indicating the concentration of reducing sugars. (1 mark)
Question 2
Explain the differences in structure and function between starch and cellulose. (6 marks)
Mark Scheme:
Feature | Starch | Cellulose |
---|---|---|
Monomer | Composed of α-glucose. | Composed of β-glucose. |
Structure | Contains amylose (linear, helical) and amylopectin (branched). | Consists of straight, unbranched chains. |
Bonding | 1,4 glycosidic bonds; amylopectin has 1,6 branches. | 1,4 glycosidic bonds, with alternating glucose orientation. |
Hydrogen Bonds | Few hydrogen bonds within chains. | Extensive hydrogen bonding between chains for strength. |
Function | Energy storage in plants (e.g., starch granules in chloroplasts). | Structural support in plant cell walls. |
Properties | Insoluble, compact, easily hydrolyzed for energy. | Strong fibers provide high tensile strength. |
Question 3
Describe the structure of triglycerides and phospholipids, and explain their roles in living organisms. (6 marks)
Mark Scheme:
- Triglycerides consist of glycerol and three fatty acids joined by ester bonds, formed via condensation reactions. (1 mark)
- Phospholipids have a glycerol backbone, two fatty acids, and a phosphate group, forming an amphipathic molecule. (1 mark)
- The hydrophilic phosphate head interacts with water, while the hydrophobic fatty acid tails repel water. (1 mark)
- Triglycerides act as energy storage molecules, providing high energy per gram and insulation in animals. (1 mark)
- Phospholipids are crucial in forming the phospholipid bilayer of cell membranes, providing a semi-permeable barrier. (1 mark)
- Phospholipids’ amphipathic nature enables membrane fluidity and interactions with proteins for transport and signaling. (1 mark)
Question 4
Explain the Biuret test for proteins and describe the structural levels of proteins. (6 marks)
Mark Scheme:
- Biuret Test: Add Biuret reagent to the solution; a purple color indicates proteins due to the presence of peptide bonds. (1 mark)
- Primary Structure: Linear sequence of amino acids joined by peptide bonds. (1 mark)
- Secondary Structure: Folding into α-helices or β-pleated sheets stabilized by hydrogen bonds. (1 mark)
- Tertiary Structure: Three-dimensional shape due to further folding, stabilized by hydrogen bonds, ionic bonds, disulfide bonds, and hydrophobic interactions. (1 mark)
- Quaternary Structure: Assembly of two or more polypeptide chains, as seen in haemoglobin. (1 mark)
- Each structural level contributes to the protein’s function, such as active sites in enzymes or oxygen-binding in haemoglobin. (1 mark)
Question 5
Discuss the significance of water’s high specific heat capacity and its role as a solvent in biological systems. (6 marks)
Mark Scheme:
- Water has a high specific heat capacity due to hydrogen bonding, requiring significant energy to change its temperature. (1 mark)
- This property helps maintain stable temperatures in organisms and environments, essential for enzymatic reactions. (1 mark)
- As a universal solvent, water dissolves ions and polar molecules, enabling chemical reactions and transport. (1 mark)
- Dissolved substances, such as nutrients and gases, are easily transported in water (e.g., blood plasma). (1 mark)
- Water’s polarity facilitates hydrophilic interactions, while hydrophobic molecules (e.g., lipids) aggregate, stabilizing membranes. (1 mark)
- These properties make water essential for metabolic reactions, nutrient transport, and temperature regulation in living systems. (1 mark)
Question 6
Explain the differences between reducing and non-reducing sugars, and describe how you would test for each. (6 marks)
Mark Scheme:
- Reducing Sugars: Sugars that can donate electrons to other molecules, such as glucose and maltose. (1 mark)
- Test: Heat the solution with Benedict’s reagent; a color change from blue to brick-red indicates a reducing sugar. (1 mark)
- Non-Reducing Sugars: Sugars like sucrose that do not reduce Benedict’s reagent directly. (1 mark)
- Test for Non-Reducing Sugars:
- Boil with dilute acid (e.g., HCl) to hydrolyze the sugar into its monosaccharides.
- Neutralize with sodium hydroxide.
- Perform the Benedict’s test again; a positive result indicates non-reducing sugars. (2 marks)
- These tests are used to differentiate types of carbohydrates in biological samples. (1 mark)
Question 7
Describe the role of hydrogen bonding in the properties of water and biological macromolecules. (6 marks)
Mark Scheme:
- In water, hydrogen bonds between molecules provide cohesion (e.g., surface tension) and adhesion (e.g., capillary action). (1 mark)
- Hydrogen bonding gives water a high specific heat capacity, stabilizing temperature in organisms and habitats. (1 mark)
- Hydrogen bonds stabilize the secondary structure of proteins (e.g., α-helices and β-sheets). (1 mark)
- They also maintain the base-pairing in DNA (e.g., A-T and G-C). (1 mark)
- In polysaccharides, hydrogen bonds link cellulose molecules, forming strong microfibrils for structural support in plants. (1 mark)
- These bonds, while weak individually, collectively provide significant strength and stability. (1 mark)