Biology Revision (January Exam) PDF

BIOLOGY JANUARY EXAM GRADE 1 1 SEMESTER 1 1. Cell structure 2.Biological molecules 3.Ezymes 4.The Cell Cycle and Mitosis FIRST TOPIC: CELL STRUCTURE 1.1 Make temporary preparations of cellular material suitable for viewing with a light microscope. How to Prepare a Temporary Slide for a Light Micro...

BIOLOGY JANUARY EXAM GRADE 1 1 SEMESTER 1 1. Cell structure 2.Biological molecules 3.Ezymes 4.The Cell Cycle and Mitosis FIRST TOPIC: CELL STRUCTURE 1.1 Make temporary preparations of cellular material suitable for viewing with a light microscope. How to Prepare a Temporary Slide for a Light Microscope: 1. Collect the Sample: Use a small, thin piece of the material (e.g onion skin or cheek cells) 2. Place on the Slide: Put the sample in the center of a clean glass slide. 3. Add a stain (Optional) Add a drop of stain (e.g, iodine for plants or methylene blue for animal cells) to make structures easier to see 4. Add a Coverslip: Gently place a coverslip over the sample at an angle to avoid air bubbles 5. Remove Extra Liquid: Blot excess stain or water with tissue paper. 6. Examine Under the Microscope: Start with low magnification, then zoom in to observe details. 1.2 Draw cells from light microscope slides and photomicrographs. Steps for Drawing Cells 1. Observe Carefully: ○ Look closely at the microscope slide or photomicrograph. ○ Identify key structures such as the cell membrane, nucleus, and cytoplasm. 2. Use Pencil Only: ○ Always draw in pencil for clear, neat lines. 3. Outline Clearly: ○ Draw the general shape of the cell (e.g., round for animal cells, rectangular for plant cells). ○ Include important visible organelles.  4. Label Accurately: ○ Add labels for structures like the nucleus, cytoplasm, cell wall (if present), and any other visible parts. 5. Keep it Simple: ○ Avoid shading or adding unnecessary details. ○ Ensure the drawing is proportionate and matches what you observe. 6. Add Magnification: ○ Write the total magnification of the microscope (e.g., x40 or x400). Tips Focus on what you can see, not what you know should be there. Keep the drawing neat and organized. Plant cell biological drawing: Bacterial cell biological drawing Animal cell drawing: 1.3 Calculate magnifcations of drawings and images and calculate actual sizes of specimens from scale drawings, photomicrographs and electron micrographs (scanning and transmission). 1.Formula to calculate Magnification: Magnification = Image size/Actual size Image Size = the measured size of the object in the drawing or photomicrograph (use a ruler) Actual size : The true size of the object (often given or calculated) 2. To Calculate Actual Size: Rearrange the formula: Actual Size = Image Size/Magnification 3. To calculate Image Size: Rearrange the formula: Image Size = Actual Size x Magnification STEPS FOR CALCULATIONS: 1. Measure Image Size: - Use a ruler to measure the image in millimeters (mm) -Convert to micrometers (um) if needed: 1mm=100um 2.Find or Calulate Magnification: - If magnification is not given, use the scale bar or formula. 3. Convert units if Necessary: - Actual sizes are often given in micrometers (um) or nanometers (nm): 1um = 100nm 4.Use the Formula: - Substitute the Values into the correct formula to find magnification, image size, or actual size. Electron Micrographs (Scanning vs. Transmission) -Scanning Electron Micrograph (SEM) : Shows 3D surface structure of the specimen. Use the same formula as above to calculate magnification or sizes. -Transmission Electron Micrograph (TEM) : Shows detailed internal structures of the specimen. Again, apply the formula to calculate magnifications or sizes. 1.4 Use an eyepiece graticule and stage micrometer scale to make measurements and use the appropriate units, millimeter (mm), micrometer (um) and nanometer (nm) 1. Understand the Tools: Eyepiece Graticule: A scale inside the eyepiece of the microscope. Stage Micrometer: A ruler with known measurements placed on the stage. 2. Calibrate the Eyepiece Graticule: Place the stage micrometer on the microscope stage. Look through the microscope and compare the stage micrometer with the eyepiece graticule to find out how much each division on the eyepiece graticule equals. ○ For example, if 10 divisions of the eyepiece graticule = 0.1 mm, then each division is 0.01 mm (or 10 micrometers, µm). 3. Measure the Object: Look at the object (e.g., a cell) and count how many divisions it covers on the eyepiece graticule. Multiply the number of divisions by the value you calculated earlier (e.g., 10 µm per division). 4. Use the Right Units: Millimeters (mm) for larger objects (1 mm = 1,000 µm). Micrometers (µm) for medium-sized objects like cells (1 µm = 1,000 nm). Nanometers (nm) for tiny structures like organelles (1 nm = 1/1,000 µm). Example: If a cell spans 15 divisions on the eyepiece graticule, and each division is 10 µm: ○ The cell size = 15×10 μm=150 μm 1.5 Identify organelles and other cell structures found in eukrayotic cells that can be seen under a light microscope, and outline their structures and functions, limited to: Cell surface membrane Nucleus Golgi apparatus Mitochondria Centrioles Chloroplasts Cell Wall Tonoplast and large permanent vacoule of plant cells Organelle Structure Function Cell Surface Membrane Thin, flexible layer made Controls entry and exit of of phosopholipids and substances. proteins. Nucleus Large, round structure Controls cell activities; with a double membrane. stores genetic matieral Contains chromatin and (DNA) nucleolus. Glogi Apparatus Stack of flattened sacs. Modifies and packages proteins into vesicles for transport Mitochondria Rod-shaped with double Site of aerobic repiration; membrane; inner produces ATP membrane folded into cristae. Centrioles Small cylindrical Involved in cell division. structures made of microtubes (in animal cells only) Chloroplasts Oval-shaped with a Site of photosynthesis in double membrane; plant cells. contains chlorophyll. Cell Wall Rigid structure made of Provides support and cellulose (plants) shape to plant cells. Large Vacoule Large central sac Stores water, nutrients surrounded by a and waste; helps tonoplast (membrane) maintain cell shape. 1.6 Identify the additional Organelles and other Cell structures found in eukraytoic cells thay can be seen under an electron microscope, and outline their structures and functions, limited to: Nuclear envelope and nucleolus Rough endoplasmic reticulum Smooth endoplasmic reticulum Cristae and the presence of small circular DNA in mitochondria Ribsoomes (80s in the cytoplasm 70s in chloroplasts and mitochondria) Lysosomes Microtubules Cilia Microvilli Thylakoid membranes and the presence of small circular DNA in chloroplasts Plasmodesmata Organelle Structure Function Nuclear Envelope Double membrane with Controls entry and exit of pores. materials into the nucleus. Nucleolus Dense structure inside Produces ribosomes the nucleus Rough Endoplasmic Network of membranes Synthesizes and Reticulum (RER) studded with ribosomes transports proteins Smooth Endoplasmic Network of membranes Synthesizes and Reticulum (SER) without ribosomes transports lipids  Cristae in mitochondria Folded inner memebrane Increases surface area for of mitochondria; energy production contains small circular DNA. Small circular DNA Found inside Enables mitochondria mitochondria and and chloroplats to chloroplasts. self-replicate. Ribosomes (80s in the -80s ribosomes are Make proteins cytoplasm 70s in larger and found in the chloroplasts and cytoplasm. -70s ribosomes are mitochondria) smaller and found in mitochondria and chloroplasts. Lysosomes Small, membrane-bound Break down waste and sacs containing enzymes. old cell parts. Microtubules Tiny protein tubes. Provid support and help in cell divison Cilia Hair-like structures on Move fluids or particles the cell surface over the cell surface. Microvilli Small finger-like Increase surface area for projections on the cell absorption surface Thylakoid membranes Thylakoids are stacked Chloroplasts also have and the presence of small membranes inside small, circular DNA circular DNA in chlorplasts (from grana) chloroplasts Plasmodesmata Thylakoids are where Circular DNA allows photosynthesis occurs. chloroplasts to make some proteins 1.7 Describe and interpret photomicrographs, electron micrographs and drawings of a typical plant and animal cells. Photomicrographs: What they are: Photos taken using a light microscope. What you see: Plant cells: Rectangular shape, cell wall, nucleus, large central vacoule, and sometimes chloroplasts Animal cells: Irregular or round shape, no cell wall, no chloroplasts, nucleus visible. Electron Micrographs: What they are:High-detail images taken using a electron microscope. What you see: Common organelles: Nucleus, mitochondria, endoplasmic reticulum (ER), and ribosomes. Plant cells: Cell wall, chloroplasts with stacks (thylakouds), and a large vacoule Animal cells: Centrioles, lysosomes, no cell wall. Drawings: What they are: Simple diagrams of plant or animal cells with labeled organelles. What you see: Plant cells: Cell wall, chloroplasts, large vacoule, rectangular shape. Animal cells: No cell wall or chloroplasts, small vacuoles, round shape. Key Features to Recognize Plant cells: ○ Rigid shape due to cell wall. ○ Large vacuole in the center. ○ Chloroplasts (in photosynthetic cells). Animal cells: ○ Irregular or round shape. ○ No cell wall or chloroplasts. ○ May show centrioles or lysosomes. Organelles to Identify in Images Nucleus: Large, round or oval, controls the cell. Mitochondria: Small, oval with folded inner membrane, provides energy. Chloroplasts (plants): Disc-shaped with stacks (grana), for photosynthesis. Cell Wall (plants): Thick outer layer for support. Vacuole: Large and central in plants, small in animals. Ribosomes: Tiny dots (make proteins). ER: Network of tubes, rough ER has ribosomes, smooth ER doesn’t. 1.8 Compare the structure of typical plant and animal cells. Feature Plant Cell Animal Cell Cell Wall Present (made of Absent. cellulose) Chloroplasts Present (for Absent. photosynthesis) Vacoule Large and central Small or Absent. Shape Regular, rectangular. Irregular, round or oval. Centrioles Absent. Present. Key points Plant cells have a cell wall, chloroplasts, and a large vacoule. Animal cells do not have these features but have centrioles. 1.9 State that cells use ATP from respiration for energy-requiring processes ATP (adenosine triphosphate) is the energy source for cells. Made during respiration in mitochondria. Used for: Active transport (moving substances) Protein synthesis (making proteins) Cell division Muscle contraction (in animals) -Releases energy when broken into ADP (adenosine diphosphate) + phosphate. 1.10 Outline key structural features of a prokaryotic cell as found in a typical bacterium, including: Unicellular Generally 1-5um diameter Peptidogylcan cell walls Circular DNA 70s ribsomes Absence of organelles surrounded by double membranes. Feature Description Unicellular Made up of a single cell that performs all life functions Size Genrally 1-5um in diameter (very small) Cell wall Made of peptidoglycan, providing structure and protection DNA Circular DNA, not enclosed in a nucleus (no membrane) Ribosomes 70s ribosomes, (smaller than eukrayotic 80s ribosomes) for protein synthesis. No Membrane-Bound Organelles No organelles like the nucleus, mitochondria, or chloroplasts. 1.11 Compare the structure of prokaryotic cell as found in a typical bacterium with the structures of typical eukaryotic cells in plants and animals. Feature Prokaryotic Cell Eukaryotic Cell (Bacterium) (Plant/Animal) Size Small, 1-5um Larger, 10-100um Cell Wall Made of peptidoglycan Plant cells: cellulos Animal cells: no cell wall DNA Circular, no nucleus Linear, enclosed in a nucleus Nucleus Absent Present (encloses DNA) Ribosomes 70s ribosomes 80s ribosomes Organelles No membrane-bound Membrane-bound organelles (e.g no organelles (e.g mitochondria) mitochondria, chloroplasts in plants) Key differences: Prokaryotic cells are smaller, simpler, and lack a nucleus and membrane-bound organelles. Eukaryotic cells are larger, more complex, with a defined nucleus and various membrane-bound organelles. 1.12 State that all viruses are non-cellular structures with a nucleic acid core (either DNA or RNA) and a capsid made of protein, and that some viruses have an outer envelope made of phospholipids. Structure of viruses: Non cellular Viruses are not made of cells Nucleic Acid Core Contain either DNA or RNA (genetic material) Capsid Surrounded by a protein coat called a capsid Envelope (in some viruses) Some viruses have an outer envelope made of phospholipids (derived from the host cell membrane). SECOND TOPIC: BIOLOGICAL MOLECULES 2.1 Describe a semi-quantitative Benedict's test on a reducing sugar solution, including standardizing the test and using the results (time to first color change or comparison to color standards) to estimate the concentration. The Benedict’s test is used to detect reducing sugars (like glucose): Procedure: 1.Prepare the Sample: Put the liquid you want to test in a test tube 2.Add Benedict’s Solution: Add an equal amount of Benedict’s reagent (blue liquid) to the test tube. 3.Heat: Place the test tube in a boiling water bath for 5 minutes. 4.Observe the Color Change: Blue: No reducing sugars. Green to Yellow: Low concentration of reducing sugars. Orange to Red: High concentration of reducing sugars. Standardizing the Test: To standardize, test known solutions with different concentrations of sugar. Note the time it takes for the first color change or compare the color to known standards. Estimating Sugar Concentration: Time to Color Change: The faster the color change, the higher the sugar concentration Color Comparison: Compare the color of the test sample to a color chart to estimate the sugar concentration This test helps helps estimate the concentration of reducing sugars based on the color change. 2.2 Describe a test to identify the presence of noon-reducing sugars, using acid hydrolysis and Benidict’s solution. To test for non-reducing sugars, such as sucrose, you first need to break them down into reducing sugars using acid hydrolysis. Here’s how you do it: Steps: 1. Boil the same with Acid: Add dilute hydrochloric acid to the sample. Heat in a boiling water bath for 5 minutes to break down the sugar into reducing sugars. 2. Neutralize the Acid: After boeing, add sodium bicarbonate (baking soda) to neutralize the acid 3. Benedict's Test: Add Benedict’s solution and heat in a boiling water bath for 5 minutes. Results: If the solution turns green, yellow, orange or red, reducing sugars are present. If it stays blue, no reducing sugars were formed, meaning there were no non-reducing sugars in the sample. SUMMARY: Non-reducing sugars must first be broken down with acid before doing the Benedict’s test. 2.3 Describe and draw the ring forms of a-glucose and b-glucose Glucose can exist in two forms, a-gluocse and β-glucose, which are both ring shapes in solution. The only difference is inn the position of the OH group on carbon 1 a-glucose: - In a-glucose, the OH group on carbon 1 is down (below the ring) B-glucose: - In β-glucose, the OH group on carbon 1 is up (above the ring) Strucure: both a-glucose and β-glucose form a 6-membered ring with 5 carbon atoms and 1 oxygen atom. 2.4 Define the terms monomer, polymer, macromolecule, monosaccharide, disaccharide and polysaccharide. Monomer: A small molecule that can join with others to from a larger molecule. Polymer: A large molecule made of many monomers joined together Macromolecule: A very large molecule, like a polymer, made of many atoms. Monosaccharide: A single sugar molecule. Example: glucose Disaccharide: Two sugar molecules joined together. Example: Sucrose Polysaccharide: Many sugar molecules joined together. Example: Starch or Cellulose 2.5 state the role of covalent bonds in joining smaller molecules together to form polymers Covalent bonds join smaller molecules (monomers) together to form polymers. Covalent bonds are strong chemical bonds where atoms share electrons When monomers join, covalent bonds are formed between them, making a long chain (polymer) 2.6 State that glucose, fructose and maltose are reducing sugars and that sucrose is a non-reducing sugar. Glucose, fructose, and maltose are reducing sugars. -This means they can donate electrons to other molecules, which is why they change the color of Benedict’s solution during the test. Sucrose is a non-reducing sugar -It does not donate electrons in the same way, so it does not change the color of Benedicts solution unless it is first broken down into reducing sugars by hudrolysis. In summary: Reducing sugars: Glucose, fructose, maltose Non-reducing sugar: Sucrose 2.7 Describe the formation of a glysosidic bond by condensation, with reference to disaccharides including sucrose, and polysaccharides. A glycosidic bond forms when two sugar molecules join together through a condensation reaction, which releases water. Steps: 1. Condensation: Two sugar molecules (e.g glucose and fructose) come together. A water molecules is removed (one sugar gives OH, the other gives H) 2. Glycosidic Bond: The remaining oxygen links the two sugars together, forming a glycosidic bond. Example: Sucrose (table sugar) is made of glucose and fructose. The glycosidic bond is formed between them through condensation, with the release of water. In polysaccharides Starch and cellulose are made by joining many sugars (like glucose) with glycosidic bonds. In simple terms, a glycosific bond is formed when two sugars join, and water is removed in the process. 2.8 Describe the breakage of a glycosidic bond in polysaccharides and disaccharides by hydrolysis, with reference to the non-reducing sugar test. Hydrolysis is the process where a glycosidic bond where a glycosidic bond between sugar molecules is broken by adding water. Steps in Hydrolysis: 1.Water is added to the glycosidic bond btween two sugar. 2.The bond breaks, and the two sugars separate. 3.Result: You get two monosaccharides instead instead of one disaccharide or polysaccharide. Example in Disaccharides: Sucrose (table sugar) can be broken into glucose and fructose by hydrolysis (using water). Non-Reducing sugar Test: Sucrose is a non-reducing sugar and doesn’t react with Benedict’s solution After hydrolysis, it becomes glucose and fructose, which are reducing sugars and will react with Benedict’s solution. Summary: Hydrolysis breaks the bond between sugar molecules by adding water. Sucrose is a non-reducing sugar until it is hydrolyzed inot glucose and fructose. 2.9 Describe the molecular structure of the polysaccharides starch (amylose and amylopectin) and glycogen and relate their structure to their functions in living organisms: Starch: Starch is made up of two types: amylose and amylopectin 1. Amylose: Structure: A long, unbranched chain of glucose that forms a spiral shape. Function: Stores energy in plants. Its spiral shape makes it compact and not easily dissolved in water. 2. Amylopectin: Structure: A branched chain of glucose Function: Stores energy in plants and can release energy more quickly because enzymes can break it down at the branches. Glycogen: Structure: Similar to amylopectin but more branched. Function: Stores energy in animals (mainly in the liver and muscles). The branches allow for quick energy release. Summary: Amylose: Unbranched, stores energy in plants. Amylopectin: Branched, stores energy and releases it faster in plants Glycogen: Highly branched, stores and releases energy quickly in animals. 2.10 Describe the molecular structure of the polysaccharide cellulose and outline how the arrangement of cellulose molecules contributes to the function of plant cell walls. Structure of Cellulose: Cellulose is made of long straight chains of glucose molecules The glucose molecules are linked by β-1,4 glycosidic bonds. The chains are held together by hydrogen bonds, making them strong Function in Plant Cell Walls: The cellulose chains form microfibrils, which are strong fibres. These fibres make the cell wall rigid and strong, giving support to the plant cell. The strong cell wall helps the plant maintain its shape and resist damage. Summary: Cellulose is made of glucose molecules linked by strong bonds. These molecules form fibres that give the plant strength and protection. 2.11 State that triglycerides are non-polar hydrophobic molecules and describe the molecular structure of triglycerides with reference to fatty acids (saturated and unsaturated), glycerol and the formation of ester bonds. A triglyceride is made up of: 1. Glycerol: A small molecule with 3 carbon atoms 2. Fatty Acids: Long chains of carbon atoms. Saturated: No double bonds, straight chains Unsaturated: One or more double bonds, causing kinks in the chain. Ester Bonds: Triglycerides are formed when 3 fatty acids join to 1 glycerol Each fatty acid forms an ester bond with glycerol through a condensation reaction, which releases water. Summary: Triglycerides consist of glycerol and 3 fatty acids. Ester bonds form between them, making triglycerides hydrophobic and non-polar. 2.12 Relate the molecule structure of triglycerides to their functions in living organisms: Structure of Triglycerides: Glycerol + 3 fatty acids (either saturated or unsaturated) Non-polar and hydrophobic (do not dissolve in water) Functions of Triglycerides: 1. Energy Storage: - Triglycerides store a large amount of energy. When broken down, they release energy for the body to use. 2. Insulation: - In animals, triglycerides help to unsulate the body keep it warm by forming fat layers under the skin. 3. Protection: - They act as shock absorbers, protecting organs like kidneys from damage 4. Waterproofing: - Triglycerides in plants and animals help to waterproof surfaces, like the skin or leaves, preventing water loss. Summary: - Triglycerides store energy, provide insulation, offer protection, and act as waterproofing in living organisms. Their structure (glycerol + 3 fatty acids) makes them ideal for these functions. 2.13 Describe the molecular structure of phospholipids with reference to their hydrophilic (polar) phosphate heads and hydrophobic (non-polar) fatty acid tails. Molecular Structuar of Phospholipids: Phospholipids consist of: - A PHOSPHATE HEAD (hydrophillic - water-loving) - TWO FATTY ACID TAILS (hydrophobic - water-repelling) Key points: - The phosphate head is polar and attracts water - The fatty acid tails are non-polar and repel water. Summary: -Phospholipids have a water-loving head and water-repelling tails. This helps them form cell membranes by arranging with heads facing water and tails facing away. 2.14 Describe and draw: The general structure of an amino acid The formation and breakage of a peptide bond The general structure of an amino acid: An amino acid consists of a central carbon atom (called the alpha carbon) bonded to: - An amino group (-NH2) - A carboxyl group (-COOH) - A hydrogen atom (-H) - A variable R group (side chain) that differs for each amino acid and determines its properties. The formation and breakage of a peptide bond: FORMATION OF A PEPTIDE BOND: - A peptide bond forms when the carboxyl group (-COOH) of one amino acid reacts with the amino group (-NH2) of another amino acid. - This is a condensation reaction, releasing a molecule of water (H2O) - The bond formed between the carbon of the carboxyl group and the nitrogen of the amino group is called a peptide bond BREAKAGE OF A PEPTIDE BOND: - A peptide bond can be broken down in a hydrolysis reaction - Water is added, and the bond between the carbon and nitrogen is broken - This separates the amino acids into their original forms 2.15 Explain the meaning of the terms primary structure, secondary structure, tertiary structure and quaternary structure of proteins. PRIMARY STRUCTURE: - The sequence of amino acids in a protein chain - Held together by peptide bonds - Example: Glycine-Serine-Alanine-Tyrosine SECONDARY STRUCTURE: - The folding or coiling of the amino acid chain into specific shapes: Common shapes: - Alpha helix (spiral shape) - Beta-pleated sheet (folded sheet) -Held together by hydrogen bonds. TERTIARY STRUCTURE: - The three-dimensional shape of the protein - Formed by the bending and folding of the secondary structure. - Held together by: 1. Hydrogen bonds 2.Ionic bonds 3.Disulfide bridges -Determines the protein’s function 4. QUATERNARY STRUCTURE: - When a protein consists of two or more polypeptide chains joined together. - Example: Hemoglobin (made of 4 polypeptide chains) - Held together by the same bonds as the tertiary structure 2.16 Describe the types of interaction that hold protein molecules in shape: Hydrophobic interactions Hydrogen bonding Ionic bonding Covalent bonding including disulfide bonds 1.HYDROPHOBIC INTERACTIONS: - Happen between water-hating parts of the protein. - These parts mobe away from water, folding into the center of the protein - Help keep the protein’s shape 2. HYDROGEN BONDING: - Formed when a slightly positive hydrogen bonds with a slightly negative atom (like oxygen or nitrogen) - Found in coiled shapes (alpha-helices) or folded sheets (beta-sheets) - Help hold the protein together. 3.IONIC BONDING: - Happen between positively charged and negatively charged parts of the protein. - Help keep the proteins shape, but they can be affected by pH changes. 4.COVALENT BONDING (Disulfide Bonds): - Strong bonds formed when two sulfur atoms in the protein link together (from two cysteine amino acids) - These bonds give the protein extra strength to hold its shape. 2.17 State that globular proteins are generally soluble and have physiological roles and fibrous proteins are generally insoluble and have structural roles: 1. Globular Proteins: -Generally soluble in water -Have physiological roles, such as: 1. Enzymes (e.g amylase) 2.Hormones (e.g insulin) 3. Transport proteins (e.g hemoglobin) 2. Fibrous proteins: - Generally insoluble in water - Have structural roles, such as: 1. Collagen (in skin and bones) 2.Keratin (in hair and nails) 3.Elastin (in connective tissues) 2.18 Describe the structure of a molecule of haemoglobin as an example of globular proteins including the formation of its quaternary structure from two alpha (α) chains (α-globin), two beta (β) chains (β-globin) and a haem group. 1. Type of protein: - Hemoglobin is a globular protein - It is soluble and has a physiological role int transporting oxygen. 2.Quaternary Structure: -Hemoglobin is made up of four polypeptide chains 1.Two alpha (a) chains (a-globin) 2.Two beta (β) chains (β-globin) -These chains are held together by bonds and interactions (e.g hydrogen bonds, ionic bonds, and hydrophobic interactions) 3.Haem Group: - Each polypeptide chain contains a haem group -A haem group is a non-protein part withan iron (Fe²⁺) ion at its center. -This iron binds to oxygen molecules, allowing hemoglobin to carry oxygen in the blood. 4.Function: -The quaternary structure and haem groups allow hemoglobin to carry up to four oxygen molecules at a time. 2.19 Relate the structure of haemoglobin to its function, including the importance of iron in the haem group 1. Globular Shape: - Hemoglobin is a globular protein, making it soluble in water -This allows it to move easily through the bloodstream to transport oxygen. 2. Quaternary Structure: -Made of four polypeptide chains (two alpha chains and two beta chains) -The quaternary structure enables it to carry four oxygen molecules at a time, one on each haem group. 3. Haem Groups: -Each polypeptide chains contains a haem group with a iron (Fe²⁺) ion at the center. -The iron ion binds to oxygen, allowing hemoglobin to pick up oxygen in the lungs and release it to cells where it’s needed 4. Importance of Iron: -Without iron, hemoglobin cannot bind oxygen effectively -Iron is essential for hemoglobin’s role in oxygen transport 2.20 Describe the structure of a molecule of collagen as an example of a fibrous protein, and the arrangement of collagen molecules to form collagen fibres to their function Molecular Structure of Collagen: Collagen is made of three long chains twisted together in a spiral shape. These chains are made of amino acids like glycine, proline, and hydroxyproline, which help the structure stay strong. Arrangement into Collagen Fibres: The collagen molecules line up and overlap at the ends, forming a strong structure. Many collagen molecules group together to form thicker fibres. Function of Collagen: Collagen fibres are strong and flexible, helping tissues like skin, bones, and tendons stay strong and resist stretching. 2.21 Relate the structures of collagen molecules and collagen fibres to their function 1. Collagen Molecules: ○ Collagen is made of three twisted chains. This makes it strong and hard to stretch. 2. Collagen Fibres: ○ Many collagen molecules stick together to form thick, strong fibres. 3. Function: ○ These strong fibres help parts of the body like skin, bones, and tendons stay strong and flexible. 2.22 Explain how hydrogen bonding occurs between water molecules and relate the properties of water to its roles in living organisms, limited to solvent action, high specific heat capacity and latent heat of vaporization. Hydrogen Bonding in Water: Water molecules have a slightly negative end (oxygen) and a slightly positive end (hydrogen). These opposite charges attract each other, forming hydrogen bonds between water molecules. Properties of Water and Their Roles in Living Organisms: 1. Solvent Action: ○ Water can dissolve many substances, helping carry important things like nutrients and waste in our bodies. 2. High Specific Heat Capacity: ○ Water doesn’t change temperature easily. This helps keep the temperature in living things steady, even when the weather changes. 3. Latent Heat of Vaporization: ○ Water needs a lot of energy to turn into vapor. This helps cool organisms down when they sweat or release water vapor. THIRD TOPIC: ENZYMES 3.1 State that enzymes are globular proteins that catalyze reactions inside cells or are secreted to catalyze reactions outside cells. Enzymes are globular proteins that help speed up chemical reactions. They can work inside cells to help with things like breaking down food, or they can be released outside the cells to help with reactions, like in digestion. 3.2 Explain the mode of action of enzymes in terms of an active site, enzyme-substrate complex, lowering of activation energy and enzyme specificity, in terms of the induced-fit hypothesis. 1. Active Site: ○ Enzymes have a special area called the active site where the substance (called a substrate) fits. 2. Enzyme-Substrate Complex: ○ When the substrate enters the active site, it forms an enzyme-substrate complex. This helps the reaction happen more easily. 3. Lowering Activation Energy: ○ Enzymes lower the activation energy (the energy needed to start a reaction), making it easier for reactions to happen. 4. Enzyme Specificity (Induced-Fit Hypothesis): ○ Enzymes are specific to one substrate, meaning only certain molecules can fit into their active site. According to the induced-fit hypothesis, when the substrate enters, the enzyme changes shape slightly to fit better, like a key fitting into a lock. 3.3 Describe how to investigate the progress of enzyme-catalyzed reactions by measuring rates of formation if products using catalase and rates of disappearance of substrate using amylase. Using Catalase (measuring the formation of products): Catalase breaks down hydrogen peroxide into water and oxygen. To measure the rate of reaction, you can collect the oxygen gas produced in a syringe or using an inverted graduated cylinder. The faster the oxygen is produced, the faster the enzyme is working. Using Amylase (measuring the disappearance of substrate): Amylase breaks down starch into sugar. You can test for starch with iodine solution. Iodine turns blue-black when it reacts with starch. As amylase breaks down the starch, the blue-black color fades. By checking how long it takes for the color to disappear, you can measure how fast the starch is being broken down. 3.4 Outline the use of colorimeter for measuring the progress of enzyme-catalyzed reactions that involve color changes. 1. Colour changes in the Reaction: - Some enzymes reactions cause a color change in the solution (e.g, when a substrate is broken down into products that have a different color) 2. Using the Colorimeter: -A colorimeter shines light through the solution and measures how much how much light is absorbed by the solution -The more color there is in the solution, the more light is absorbed -By measuring how the absorption changes over time, you can track how the reaction progressing EXAMPLES: - If an enzyme reaction turns a solution from colorless to colred, the colorimeter can measure how the color intensity increases or decreases, showing how fast the reaction is happening. 3.5 Explain and describe how to investigate the effects of the following factors on the rate of enzyme-catalyzed reactions: Enzyme concentration Substrate concentration Inhibitor concentration ENZYME CONCENTRATION: What to do: Change the amount of enzyme. What happens: More enzymes usually make the reaction faster, but if there’s not enough substrate, the rate won’t increase. SUBSTRATE CONCENTRATION: What to do: Change the amount of substrate What happens: More substrate make the reaction faster, but after a point, adding more won’t help because the enzymes are already working at full speed. INHIBITOR CONCENTRATION: What to do: Add an inhibitor What happens: More inhibitors slow down the reaction because they stop the enzyme from working properly In short: More enzyme = faster reaction More substrate = faster reaction (up to a point) More inhibitor = slower reaction 3.6 Explain that the maximum rate of reaction (Vmax) is used to derive the Michaelis-Menten constant (Km), Which is used to compare the affinity of different enzymes for their substrates. 1.Vmax (Maximum Rate of Reaction): -Vmax is the fastest rate at which an enzyme can work when all the enzyme active sites are fully occupied by the substrate 2.Km (Michaelis-Menten Constant): -Km is the substrate concentration at which the enzyme work at half of its maximum rate (Vmax) -A low Km means the enzyme has a high affinity for the substrate (it binds easily), while a high Km means the enzyme has a low affinity (it doesn’t bind as easily). 3.How Km is Used: -By knowing Vmax and the substrate concentration at half of Vmax, we can calculate Km. -This helps us compare how well different enzymes bind to their substrates. Enzymes with a lower Km are more efficient at low substrate concentrations. In short, Vmax gives the maximum speed of the reaction, and Km helps us understand how tightly an enzyme binds to its substrate. A lower Km means a stronger bond. 3.7 Explain the effects of reversible inhibitors, both competitive and non-competitive, on enzyme activity. 1. Competitive Inhibitors: How they work: These inhibitors look like the substrate and compete to bind to the enzyme's active site. Effect: They slow the reaction, but adding more substrate can help the enzyme work again. 2. Non-Competitive Inhibitors: How they work: These inhibitors bind to a different part of the enzyme, changing its shape. Effect: They reduce enzyme activity, and adding more substrate won’t help because the enzyme’s shape is changed. In short: Competitive inhibitors can be outcompeted by more substrate. Non-competitive inhibitors can't be overcome by adding more substrate. 3.8 Describe the different modes of action an enzymes immobilized in alginate and the same enzyme free in solution and state the advantages of using immobilized enzymes. Immobilized Enzymes in Alginate: -The enzyme is trapped in alginate beads, and the substrate can still reach the enzyme to react. -The enzyme stays in place, so it can be reused multiple times without being lost. Free Enzymes in Solution: -The enzyme is free to move in the solution, and the substrate reacts with it. -The enzyme can be washed away after the reaction is complete. Advantages of Immobilized Enzymes: Reusability: Immobilized enzymes can be used many times. Easy Separation: It’s easy to separate the enzyme from the product after the reaction.  Increased Stability: Immobilized enzymes can handle changes in temperature and pH better. Continuous Reactions: Immobilized enzymes are useful in continuous processes, where the reaction goes on without stopping. FOURTH TOPIC: THE CELL CYCLE & MITOSIS 4.1 Describe the structure of a chromosome, limited to: DNA Histon proteins Sister chromatids Cenromere Telomeres DNA: - Chromosomes are made of long, coiled strands of DNA -DNA carries genes that contain the instructions for making proteins. HISTONE PROTEINS: -DNA wraps around histone proteins to form a compact structure. -This helps to package DNA tightly inside the cell nucleus. SISTER CHROMATIDS: -After DNA replication, a chromosome consists of two identical opies called sister chromatids. -These chromatids are joined together at the centromere. CENTROMERE: -The centromere is the region where the sister chromatids are held together. -It plays a key role during cell division by attaching to spindle fibers. TELOMERES: -Telomeres are the ends of a chromosome made of repetitive DNA -They protect the chromosome from damage and prevent it from sticking to other chromosomes. 4.2 Explain the importance of mitosis in the production of genetically identical daughter cells during: Growth of multicellular organisms Replacement of damaged or dead cells Repair of tissues by cell replacement Asexual reproduction Mitosis produces genetically identical daughter cells, which is essential for: 1.Growth of Multicellular Organisms: -Mitosis allows organisms to grow by increasing the number of cells. 2.Replacement of Damaged or Dead Cells: -Mitosis replaces old, damaged or dead cells with new, identical ones. 3.Repair of tissues: -Mitosis helps repair tissues by creating new cells to replace those lost of damaged. 4.Asexual Reproduction: -In asexual reproduction, mitosis produces offspring that are genetically identical to the parent Summary: Mitosis ensures that all new cells have the same genetic information, which is important for growth, repair, replacement, and asexual reproduction. 4.3 Outline the cell cycle including: Interphase (G1, S, and G2 phases) Mitosis Cytokinesis INTERPHASE (the longest phase, where the cell prepares for division): G1 Phase (Growth 1): - The cell grows and makes proteins needed for DNA replication S Phase (Synthesis): - DNA is copied (replicated), so each chromosome has two identical sister chromatids. G2 Phase (Growth 2): - The cell grows more and prepares organelles and energy for mitosis. MITOSIS: - The nucleus divides into two, each with a complete set of chromosomes. (Includes prophase, metaphase, anaphase, and telophase) CYTOKINESIS: - The cytoplasm divides, forming two identical daughter cells. Summary: Interphase: Cell grows and replicates DNA Mitosis: Nucleus divides Cytokinesis: Cytoplasm divides, creating two new cells 4.4 Outline the role of telemores in preventing the loss of genes from the ends of chromosomes during DNA replication ROLE OF TELOMERES: - Telomeres are repetitive DNA sequences at the ends of chromosomes - They act as a protective cap to prevent the loss of important fenes during DNA replication. WHY ARE TELOMERES IMPOTANT? 1.DNA shortening: - Each time a cell divides, the DNA at the ends of chromosomes gets shorter. -Telomeres prevent important genes from being lost 2.Protect Chromosomes: -Telomeres stop chromosomes from sticking to each other or getting damaged. SUMMARY: Telomeres protect the ends of chromosomes and prevent the loss of important genetic information during DNA replication. 4.5 Outline the role of stem cells in cell replacement and tissue repair by mitosis ROLE OF STEM CELLS IN CELL REPLACEMENT AND TISSUE REPAIR: 1.What Are Stem Cells? - Stem cells are unspecialized cells that can divide by mitosis and develop into different types of specialized cells. 2.Cell Replacement:  - Stem cells produce new cells to replace damaged or dead cells, ensuring tissues stay healthy 3. Tissue repair: - When tissues are injured, stem cells divide by mitosis to make new cells that help in healing and regeneration. Summary: Stem cells divide by mitosis to replace lost cells and repair damaged tissues, keeping the body functioning properly. 4.6 Explain how uncontrolled cell division can result in the formation of a tumor 1.What happens in Normal Cells? -Cell division is controlled to ensure cells grow, divide, and die at the right times 2. Uncontrolled Cell Divison: -If these controls fail (due to mutations or damaged DNA), cells divide continuously. -This leads to a mass of abnormal cells, called a tumor. 3. Types of tumors: -Bengin tumors: Do not spread to other parts of the body -Malignant tumors (Cancer): Spread and invade surrounding tissues. Summary: Uncontrolled cell division due to loss of regulation leads to the formation of a tumor which can be benign or malignant. 4.7 Describe the behaviour of chromosomes in plant and animal cells during mitosis and the cell cycle and the associated behaviour of the nuclear envelope, the cell surface membrane and the spindle (names of the main stages of mitosis are expected: prophase, metaphase, anaphase and telophase) 1.PROPHASE - Chromosomes chondense and become visible - Nuclear envelope breaks down - Spindle fibres begin to form 2.METAPHASE - Chromosomes line up at the equator of the cell - Spindle fibres attach to the centromeres of chromosomes 3.ANAPHASE - Sister chromatids are pulled apart to opposite poles by the spindle fibres. - Each chromatids is now a separate chromosome 4.Telophase - Chromosomes decondense and become invisible under a microscope - Nuclear envelope reforms around each set of chromosomes - Spindle fibres disappear. 4.8 Interpret photomicrographs, diagrams and microscope slides of cells in different stages of a cell cycle including the main stages of mitosis. When observing cells under a microscope or in diagrams, you can identify the stages of the cell cycle, including the stages of mitosis, based on the following features: Stages of Mitosis 1. Prophase: ○ Chromosomes are visible as thickened, condensed threads. ○ Nuclear envelope starts breaking down. ○ Spindle fibers begin forming. 2. Metaphase: ○ Chromosomes line up in the center (equator) of the cell. ○ Spindle fibers attach to the centromeres of chromosomes. 3. Anaphase: ○ Chromatids are pulled apart by spindle fibers to opposite poles of the cell. ○ The shape of the cell elongates. 4. Telophase: ○ Chromosomes gather at opposite ends of the cell. ○ Nuclear envelope reforms around each set of chromosomes. ○ Spindle fibers disappear. 5. Cytokinesis: ○ The cytoplasm divides, forming two separate cells. ○ In animal cells, the membrane pinches in. ○ In plant cells, a cell plate forms between the two new cells. Tips for Identification: Look for chromosome shapes (condensed or spread out). Check the location of chromosomes (aligned, separating, or grouped). Observe the nuclear envelope (intact or broken down). Summary: Use the visible features of chromosomes, spindle fibers, and the nuclear envelope to determine the stage of the cell cycle.