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Solutions Solution = Solvent + Solute According to the size of solute particles there are three types of the solutions: 1- true solution 2- Colloidal solution 3- Suspension and emulsion solution True solution Dispersed particles are molecules or ions. Colloidal solution Suspension & Emulsions Dispersed particles are aggregate of molecules. Dispersed particles are aggregate of molecules. Diameter of Diameter of Diameter of particles < 0.001μ particles between particles > 0.1 μ. 0.001μ and 0.1 μ. Pass through permeable and semi-permeable membrane. Pass through permeable membrane only. Cannot pass through any membranes. Examples: NaCl in H2O Sucrose in H2O Examples: Starch in H2O Protein in H2O Examples: Sand in H2O Oil in H2O Cellophane sac 5ml NaCl + 5ml starch Distilled water (dist. H2O) Dialysis 1 After 20 mins take 2 ml from the external solution then 2 Observation: In tube no. 1 white ppt. formed after addition of AgNO3 solution. In tube no. 2 there is no change after addition of I2 solution. Comment: 1- Sodium chloride (NaCl) is a true solution which has small particles size so it can pass through cellophane membrane (semipermeable membrane) and after addition of silver nitrate (AgNO3) this reaction occurs: AgNO3 + NaCl NaNO3 + AgCl white ppt. 2- Starch is a colloidal solution which has large particle size than the true one so it cannot pass through cellophane membrane (semipermeable membrane), so after addition of iodine solution (I2) no reaction occurs. Electrical adsorption The particles of colloidal solutions carry electrical charges on its surface which may be positive or negative and every particles surrounded by opposite charge to form electrical double layer. Every solution its particles has the same charge. + + + Colloidal particle negative charge + + + + + positive charge Filter paper stripe Mixture of M.B. & L.G. Methylene Light green blue dye dye leave for 15 mins Observation: 1- In case of M.B., the particles of dye accumulated on the end of the filter paper stripe, while water move vertically through the stripe. 2- In case of L.G., the levels of water and dye are the same, whereas the particles of L.G. raised through filter paper stripe. Comment: Witted filter paper stripe posses -ve charge, so in case of M.B. dye, attraction force occur between the particles and the stripe in contact region as M.B. particles posses +ve charge. While in case of L.G. repulsion force occurs between the particles and the stripe as L.G. particles posses -ve charge . I2 solution Gelatin + Starch Starch Gelatin + I2 solution Leave it 1 for 48 hours Leave it for 48 hours 2 Observation: In tube no. 1 blue ring formed on the top and increases gradually downwards. In tube no. 2 there is no change. Comment: 1- Iodine solution (I2) is a true solution which have small particle size so it can pass through gelatin membrane so it can react with starch forming the blue zone: 2- Starch is a colloidal solution which have large particle size than the true one so it cannot pass through gelatin membrane and cannot react with iodine. FeCl3 NaOH + ph. ph. + K4[Fe(CN)6] + gelatin Blue zone Leave it for 48 hours Colourless zone Observation: Formation of colorless zone followed by blue zone through gelatin layer. Comment: FeCl3 ionization 3 Cl - + Large in number and small in size & charge React with NaOH Fe 3+ small in number & Large in size & charge React with K4[Fe(CN)6] 3 NaCl Fe4[Fe(CN)6]3 Colourless zone Blue zone Chloride ions are large in number and small in size so it pass through the gelatin membrane faster and react with NaOH as follow: 3 Cl- + 3 NaOH 3 NaCl Sodium chloride (NaCl) is a neutral solution so in the presence of ph.ph. as indicator it form the colourless zone while, ferric ions are small in number and large in size so it pass through the gelatin membrane slowly and react with potassium Ferro cyanide as follow : 4Fe3+ + 3 K4[Fe(CN)6] Fe4[Fe(CN)6]3 Prussian blue Ferric Ferro cyanide (Prussian blue) make the blue zone Permeability It is a property of membranes posses which mean the capacity of controlling the exit and entrance of various substances . Plasma membrane The plasma membrane consists of two phospholipids layers contain protein . Beet discs pH values 2 4 6 8 Record the time for appearing the pigment 2 4 6 8 Draw the relation between pH values and time 2 4 6 8 Observation: Red color formed in tubes having pH values 2,4 and 6. And yellow color formed in tube which have pH value 8. Comment: The plasma membrane consists of two phospholipids layers bordered by protein layer. Variation in hydrogen ion concentration (pH) affect on protein layers of membrane as protein a polypeptide which consists of amino acids. Amino acids are amphoteric , so it may react on alkaline or acidic medium as follow in acidic medium R R NH3+ CH COOH NH2 CH COOH Amino acid in alkaline medium R NH2 CH COO In both cases denaturation occurs to protein which cause plasma membrane destruction. So, plasmamembrane lose the control on exit and entrance of substances which lead to exit anthocyanin pigment easily. Anthocyanin pigment act as indicator which have red color in acidic medium and yellow color in alkaline medium Beet discs Alcohol concentrations 30% 50% 70% Record the time for appearing the pigment 30% 50% 70% Draw the relation between alcohol concentrations and time 30% 50% 70% Observation: Red color formed in all tubes 30%, 50% and 70%. Comment: The plasma membrane consists of two phospholipids layers contain protein . Alcohol is an organic solvent which effect on the phospholipid layer (fats) this cause plasma- membrane destruction. So, plasmamembrane lose the control on exit and entrance of substances (selective permeability) which lead to exit anthocyanine pigment easily in the medium. The time needed for anthocyanine appearance is reversely proportional with alcohol concentration. Enzymes Enzymes are biomolecules that catalyze (increase the rate of) chemical reactions. Almost all enzymes are proteins. In enzymatic reactions, the molecules at the beginning of the process are called substrates, and the enzyme converts them into different molecules, the products. Almost all processes in a biological cell need enzymes to occur at significant rates. Enzymes are selective for their substrates. Enzyme activity can be affected by other molecules. decrease Inhibitors enzyme are activity; molecules activators molecules that increase enzyme activity. that are Activity is also affected by temperature, chemical environment (e.g. pH), and the concentration of substrate. IEC Classification of Enzymes Group Name Type of Reaction Catalyzed Oxidases or Dehydrogenases Oxidation-reduction reactions Transfer of functional groups Transferases Hydrolases Hydrolysis reactions Lyases Addition to double bonds or its reverse Isomerases Isomerization reactions Ligases or Synthetases Formation of bonds with ATP cleavage 2 ml sucrase (enzyme) 3 ml sucrose (substrate) Add 5ml fehling (A:B) soln. Water bath at Direct flame 37± 2° C for 15 mins Reddish brown ppt. Observation: Reddish brown ppt. was formed. Comment: Invertase is a hydrolytic enzyme, at suitable condition it converts sucrose (disaccharide) into glucose and fructose (monosaccharides). C12H22O11 + H2O Invertase 37 ± 2°C C6H12O6 + C6H12O6 Glucose fructose After addition of fehling (A:B) solution to the monosaccharides, formed before, reddish brown ppt. is formed as a result of the following equation: COOH CHO (CHOH)4 or CH2OH (CHOH)4 CH2OH C O + CuSO4 (CHOH)3 CH2OH CH2OH Gluconic acid + Cu2O cuprous oxide Red brown ppt. Paraffin oil Put 2 ml of glucose then add 1 ml of methylene blue dye 5 ml non- 5 ml boiled boiled yeast yeast Water bath at 37± 2° C for 10 mins Non-boiled yeast Boiled yeast Observation: The blue colour disappeared in the tube which contain non-boiled yeast, while the blue colour remain in the tube which contain boiled yeast. Comment: In non-boiled yeast, glucose dehydrogenase enzyme is in active form. So, in suitable conditions this reaction occur : C6H12O6 + H2O + M.B. 37±2°C C6H12O7 + M.B. H2 Glucose dehydrogenase The boiled yeast contains enzyme in inactive form. So, the oxidation – reduction between M.B. and glucose not occur. Catechol atmospheric oxygen Potato disc Brown colour Observation: A deep brown colour formed on the surface of potato disc. Comment: Due to the presence of oxidase enzyme which work as an oxidizing agent, reaction it catalyze in the atmospheric oxygen: the following presence of O OH 2 + O2 OH Catechol “ Phenol form” Colourless Oxidase 2 + 2 H2O O Catechol “ Quinone form” Brown colour Catechol H 2O 2 Radish disc Brown colour Observation: A deep brown colour formed on the surface of radish disc after addition of hydrogen peroxide H2O2 . Comment: Radish tissues contain peroxidase enzyme which oxidize the catechol in the presence of H2O2 (as oxygen donor) as follow: 2 H2O2 2 H2O + O2 O OH 2 + O2 OH Catechol “ Phenol form” Colourless peroxidase 2 + 2 H2O O Catechol “ Quinone form” Brown colour 1 H 2O 2 2 5 ml non5 ml boiled boiled yeast yeast 1 2 Observation: In tube no. 1, effervescence occur and gaseous bubbles evolved. In tube no. 2, there is no change Comment: Any living tissue contain catalase enzyme which catalyze the breakdown of toxic H2O2 into H2O and O2 as follow: 2 H2O2 2 H2O + O2 This reaction did not occur in nonliving tissues due to absence of catalase enzyme in active form. Photosynthesis is a metabolic pathway that converts light energy into chemical energy. Its initial substrates are carbon dioxide and water; the energy source is sunlight (electromagnetic radiation). The end-products are oxygen and carbohydrates, such as sucrose, glucose or starch. This process is one of the most important biochemical pathways, since nearly all life on earth either directly or indirectly depends on it as a source of energy. It is a complex process occurring in plants, algae, as well as bacteria such as cyanobacteria. Photosynthetic organisms are also referred to as photoautotrophs. Equation for photosynthesis water 6CO2 + 6H2O + Energy Carbon dioxide Chlorophyll Oxygen C6H12O6 + 6O2 Glucose CO2 Phenol red In dark Leave one of them in light and the other one in dark In light Spirogyra Observation: The red color reappeared in tube which present in light, while the yellow colour remained in tube which present in dark. Comment: Phenol red has a red color in neutral medium and yellow color in acidic medium. So, in the first step, phenol red change to yellow color when the medium became acidic due to reaction between CO2 and water as follow: CO2 + H2O In tube H2CO3 present in light CO2 consumed in photosynthesis due to presence of all suitable conditions. So, the solution became neutral again and red color following equation: reappear as the 6CO2 + 2 H2O In Chlorophyll light tube C6H12O6 + 6H2O + 6O2 present in dark photosynthesis not occur due to the absence of light. So, CO2 not consumed and the medium remain acidic with yellow color. Non green region Put in alcohol and transfer it to water bath (70° C) Green region Duranta leaf Add I2 solution Observation: The green region of leaf became blue while non green region remain as it is. Comment: In the green region, all conditions necessary for photosynthesis are available . So, the process occurs and starch formed. 6CO2 + 2 H2O Chlorophyll light C6H12O6 + 6H2O + 6O2 After addition of I2 solution a blue colour was formed due to the reaction with starch. In non green region, there is no chlorophyll present, so, the photosynthesis process not occur and there is no starch formed. Light Leave 24 hour Elodea (hydrophyte) Water+ NaHCO3 Observation: Formation of air bubbles and decreasing in water level in test tube Comment: all condition photosynthesis are necessary available. for So, photosynthesis process occurs and O2 evolved causing formation of gaseous bubbles which lead to decrease in the level of water at the base of the test tube. 6CO2 + 2 H2O Chlorophyll light C6H12O6 + 6H2O + 6O2