2. 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  Solutions

3. Suspension & Emulsions Colloidal solution True solution Dispersed particles are aggregate of molecules. Diameter of particles > 0.1 μ. Dispersed particles are aggregate of molecules. Diameter of particles between 0.001μ and 0.1 μ. Dispersed particles are molecules or ions. Diameter of particles < 0.001μ

4. Cannot pass through any membranes. Examples: Sand in H 2 O Oil in H 2 O Pass through permeable membrane only. Examples: Starch in H 2 O Protein in H 2 O Pass through permeable and semi-permeable membrane. Examples: NaCl in H 2 O Sucrose in H 2 O

5. Cellophane sac 5ml NaCl + 5ml starch Distilled water (dist. H 2 O) 

6. Dialysis After 20 mins take 2 ml from the external solution then add AgNO 3 solution add I 2 solution 1 2

9.  Observation: In tube no. 1 white ppt. formed after addition of AgNO 3 solution. In tube no. 2 there is no change after addition of I 2 solution.

10.  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 (AgNO 3 ) this reaction occurs: AgNO 3 + NaCl NaNO 3 + AgCl white ppt.

11. 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 (I 2 ) no reaction occurs.

12. 

13. 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.  Electrical adsorption + + + + + + + + Colloidal particle negative charge positive charge

14. Filter paper stripe Methylene blue dye Light green dye Mixture of M.B. & L.G.

15. leave for 15 mins

18.  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.

19.  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.

20. 

22. I 2 solution Gelatin + Starch Starch Gelatin + I 2 solution

23. Leave it for 48 hours Leave it for 48 hours 1 2

25.  Observation: In tube no. 1 blue ring formed on the top and increases gradually downwards. In tube no. 2 there is no change.

26.  Comment: 1- Iodine solution ( I 2 ) 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.

28. FeCl 3 NaOH + ph. ph. + K 4 [Fe(CN) 6 ] + gelatin

29. Blue zone Colourless zone Leave it for 48 hours

31.  Observation: Formation of colorless zone followed by blue zone through gelatin layer.

32.  Comment: FeCl 3 Large in number and small in size & charge React with NaOH 3 NaCl Colourless zone small in number & Large in size & charge React with K 4 [Fe(CN) 6 ] Fe 4 [Fe(CN) 6 ] 3 Blue zone ionization 3 Cl - + Fe 3+

33. 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

34. 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 : 4Fe 3+ + 3 K 4 [Fe(CN) 6 ] Fe 4 [Fe(CN) 6 ] 3 Prussian blue Ferric Ferro cyanide (Prussian blue) make the blue zone

35. 

36.  Permeability It is a property of membranes posses which mean the capacity of controlling the exit and entrance of various substances.

37. Plasma membrane The plasma membrane consists of two phospholipids layers contain protein.

40. pH values 286 4 Beet discs

41. Record the time for appearing the pigment 2 4682 468 Draw the relation between pH values and time

43.  Observation: Red color formed in tubes having pH values 2,4 and 6. And yellow color formed in tube which have pH value 8.

44.  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

45. CH R NH 3 + COOH CH R NH 2 COO - in acidic medium in alkaline medium CH R NH 2 COOH Amino acid

46. 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

48. Alcohol concentrations 30%70% 50% Beet discs

49. Record the time for appearing the pigment 30% 50%70%30% 50%70% Draw the relation between alcohol concentrations and time

51.  Observation: Red color formed in all tubes 30%, 50% and 70%.

52.  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.

53. 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.

54. 

55. 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.  E nzymes

56. 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. Inhibitors are molecules that decrease enzyme activity; activators are molecules that increase enzyme activity.

57. Activity is also affected by temperature, chemical environment (e.g. pH), and the concentration of substrate.

58. IEC Classification of Enzymes Type of Reaction Catalyzed Group Name Oxidation-reduction reactions Oxidases or Dehydrogenases Transfer of functional groups Transferases Hydrolysis reactions Hydrolases Addition to double bonds or its reverse Lyases Isomerization reactions Isomerases Formation of bonds with ATP cleavage Ligases or Synthetases

61. 3 ml sucrose (substrate) 2 ml sucrase (enzyme)

62. Direct flame Reddish brown ppt. Add 5ml fehling (A:B) soln. Water bath at 37± 2° C for 15 mins

63. 

64.  Observation: Reddish brown ppt. was formed.  Comment: Invertase is a hydrolytic enzyme, at suitable condition it converts sucrose (disaccharide) into glucose and fructose (monosaccharides). C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 Invertase 37 ± 2°C Glucose fructose

65. After addition of fehling (A:B) solution to the monosaccharides, formed before, reddish brown ppt. is formed as a result of the following equation: CHO (CHOH) 4 CH 2 OH C O (CHOH) 3 CH 2 OH + CuSO 4 or COOH (CHOH) 4 CH 2 OH + Cu 2 O cuprous oxide Red brown ppt. Gluconic acid

68. 5 ml non- boiled yeast 5 ml boiled yeast Put 2 ml of glucose then add 1 ml of methylene blue dye Paraffin oil

69. Water bath at 37± 2° C for 10 mins Non-boiled yeast Boiled yeast

71.  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.

72.  Comment: In non-boiled yeast, glucose dehydrogenase enzyme is in active form. So, in suitable conditions this reaction occur : C 6 H 12 O 6 + H 2 O + M.B. C 6 H 12 O 7 + M.B. H 2 Glucose dehydrogenase 37±2°C

73. The boiled yeast contains enzyme in inactive form. So, the oxidation – reduction between M.B. and glucose not occur.

75. Potato disc Catechol atmospheric oxygen Brown colour

76.  Observation: A deep brown colour formed on the surface of potato disc.

77.  Comment: Due to the presence of oxidase enzyme which work as an oxidizing agent, it catalyze the following reaction in the presence of atmospheric oxygen:

78. + O 2 Oxidase OH 2 O O 2 + 2 H 2 O Catechol “ Phenol form” Colourless Catechol “ Quinone form” Brown colour

80. Radish disc Catechol H2O2H2O2 Brown colour

81.  Observation: A deep brown colour formed on the surface of radish disc after addition of hydrogen peroxide H 2 O 2.

82.  Comment: Radish tissues contain peroxidase enzyme which oxidize the catechol in the presence of H 2 O 2 ( as oxygen donor) as follow:

83. 2 H 2 O 2 2 H 2 O + O 2 + O 2 peroxidase OH 2 O O 2 + 2 H 2 O Catechol “ Phenol form” Colourless Catechol “ Quinone form” Brown colour

85. 12 5 ml non- boiled yeast 5 ml boiled yeast 12 H2O2H2O2

86. 

87.  Observation: In tube no. 1, effervescence occur and gaseous bubbles evolved. In tube no. 2, there is no change

88.  Comment: Any living tissue contain catalase enzyme which catalyze the breakdown of toxic H 2 O 2 into H 2 O and O 2 as follow: 2 H 2 O 2 2 H 2 O + O 2

89. This reaction did not occur in non- living tissues due to absence of catalase enzyme in active form.

90. 

91. 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).

92. 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.

93. It is a complex process occurring in plants, algae, as well as bacteria such as cyanobacteria. Photosynthetic organisms are also referred to as photoautotrophs.

94. Equation for photosynthesis 6CO 2 + 6H 2 O + Energy Energy C 6 H 12 O 6 + 6O 2 Carbon dioxide water Glucose Oxygen Chlorophyll

97. Phenol red CO 2

98. Spirogyra Leave one of them in light and the other one in dark In light In dark

100.  Observation: The red color reappeared in tube which present in light, while the yellow colour remained in tube which present in dark.

101.  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 CO 2 and water as follow:

102. CO 2 + H 2 O H 2 CO 3 In tube present in light CO 2 consumed in photosynthesis due to presence of all suitable conditions. So, the solution became neutral again and red color reappear as the following equation:

103. 6CO 2 + 2 H 2 O C 6 H 12 O 6 + 6H 2 O + 6O 2 In tube present in dark photosynthesis not occur due to the absence of light. So, CO 2 not consumed and the medium remain acidic with yellow color. Chlorophyll light

105. Duranta leaf Green region Non green region Put in alcohol and transfer it to water bath (70° C)

106. Add I 2 solution

107.  Observation: The green region of leaf became blue while non green region remain as it is.

108.  Comment: In the green region, all conditions necessary for photosynthesis are available. So, the process occurs and starch formed. 6CO 2 + 2 H 2 O C 6 H 12 O 6 + 6H 2 O + 6O 2 Chlorophyll light

109. After addition of I 2 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.

111. Water+ NaHCO 3 Elodea (hydrophyte) Light Leave 24 hour

112.  Observation: Formation of air bubbles and decreasing in water level in test tube

113.  Comment: all condition necessary for photosynthesis are available. So, photosynthesis process occurs and O 2 evolved causing formation of gaseous bubbles which lead to decrease in the level of water at the base of the test tube.

114. 6CO 2 + 2 H 2 O C 6 H 12 O 6 + 6H 2 O + 6O 2 Chlorophyll light